diff --git "a/val.tsv" "b/val.tsv" new file mode 100644--- /dev/null +++ "b/val.tsv" @@ -0,0 +1,111592 @@ + guid url correct_answer annotation target_var_json answer_json condition_json question_title question_details answer answer_description answer_text full_articfle question question_description +0 a826615c-6ddd-11ea-a013-ccda262736ce https://socratic.org/questions/how-many-moles-of-sodium-carbonate-are-present-in-6-80-grams-of-sodium-carbonate 0.06 moles start physical_unit 4 5 mole mol qc_end physical_unit 4 5 9 10 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] sodium carbonate [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.06 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] sodium carbonate [=] \\pu{6.80 grams}""}]" "

How many moles of sodium carbonate are present in 6.80 grams of sodium carbonate?

" nan 0.06 moles "
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Explanation:

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The chemical formula of sodium carbonate is #""Na""_2""CO""_3""#. The molar mass of sodium carbonate is #""105.988439 g/mol""#. http://pubchem.ncbi.nlm.nih.gov/compound/sodium_carbonate

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#6.80cancel(""g Na""_2""CO""_3)xx(1""mol Na""_2""CO""_3)/(105.988439cancel(""g Na""_2""CO""_3))=""0.0642 mol Na""_2""CO""_3""#

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There are #""0.0642 mol Na""_2""CO""_3""# in #""6.80 g""#.

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Explanation:

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The chemical formula of sodium carbonate is #""Na""_2""CO""_3""#. The molar mass of sodium carbonate is #""105.988439 g/mol""#. http://pubchem.ncbi.nlm.nih.gov/compound/sodium_carbonate

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#6.80cancel(""g Na""_2""CO""_3)xx(1""mol Na""_2""CO""_3)/(105.988439cancel(""g Na""_2""CO""_3))=""0.0642 mol Na""_2""CO""_3""#

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" "
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How many moles of sodium carbonate are present in 6.80 grams of sodium carbonate?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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There are #""0.0642 mol Na""_2""CO""_3""# in #""6.80 g""#.

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Explanation:

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The chemical formula of sodium carbonate is #""Na""_2""CO""_3""#. The molar mass of sodium carbonate is #""105.988439 g/mol""#. http://pubchem.ncbi.nlm.nih.gov/compound/sodium_carbonate

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#6.80cancel(""g Na""_2""CO""_3)xx(1""mol Na""_2""CO""_3)/(105.988439cancel(""g Na""_2""CO""_3))=""0.0642 mol Na""_2""CO""_3""#

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" How many moles of sodium carbonate are present in 6.80 grams of sodium carbonate? nan +1 a827c894-6ddd-11ea-a9d8-ccda262736ce https://socratic.org/questions/what-is-the-acid-concentration-of-a-sample-of-acid-rain-with-a-ph-of-4-20 6.30 × 10^(-5) mol/L start physical_unit 6 10 acid_concentration mol/l qc_end physical_unit 6 10 15 15 ph qc_end end "[{""type"": ""physical unit"", ""value"": ""acid concentration [OF] a sample of acid rain [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""6.30 × 10^(-5) mol/L""}]" "[{""type"":""physical unit"",""value"":""PH [OF] a sample of acid rain [=] \\pu{4.20}""}]" "

What is the acid concentration of a sample of acid rain with a pH of 4.20?

" nan 6.30 × 10^(-5) mol/L "
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Explanation:

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#pH=-log_10[H^+]#

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Given that #pH# is 4.2

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#:. 4.2=-log_10[H+]#

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#10^(-4.2)=[H^+]#

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Taking antilog we get

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#[H^+]=6.3times10^(-5)# mol/L

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#6.3times10^-5#mol/L

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Explanation:

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#pH=-log_10[H^+]#

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Given that #pH# is 4.2

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#:. 4.2=-log_10[H+]#

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#10^(-4.2)=[H^+]#

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Taking antilog we get

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#[H^+]=6.3times10^(-5)# mol/L

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What is the acid concentration of a sample of acid rain with a pH of 4.20?

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+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+1 Answer +
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#6.3times10^-5#mol/L

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Explanation:

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#pH=-log_10[H^+]#

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Given that #pH# is 4.2

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#:. 4.2=-log_10[H+]#

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#10^(-4.2)=[H^+]#

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Taking antilog we get

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#[H^+]=6.3times10^(-5)# mol/L

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" What is the acid concentration of a sample of acid rain with a pH of 4.20? nan +2 a827c895-6ddd-11ea-95df-ccda262736ce https://socratic.org/questions/calculate-the-molarity-of-1-5-w-v-solution-of-pbcl2-1 5.4 × 10^(−2) mol/L start physical_unit 6 8 molarity mol/l qc_end physical_unit 6 8 4 4 mass_concentration qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] solution of PbCl2 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""5.4 × 10^(−2) mol/L""}]" "[{""type"":""physical unit"",""value"":""w/v [OF] solution of PbCl2 [=] \\pu{1.5%}""}]" "

Calculate the molarity of 1.5% w/v solution of PbCl2?

" nan 5.4 × 10^(−2) mol/L "
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Explanation:

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And thus #""molarity""# #=# #""Moles of solute""/""Volume of solution""#

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#=((1.5*g)/(278.10*g*mol^-1))/(0.100*L^-1)=5.4xx10^-2*mol*L^-1#.

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Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.

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We ASSUME that a #100*mL# volume of SOLUTION contains a #1.5*g# mass of solute..........

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Explanation:

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And thus #""molarity""# #=# #""Moles of solute""/""Volume of solution""#

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#=((1.5*g)/(278.10*g*mol^-1))/(0.100*L^-1)=5.4xx10^-2*mol*L^-1#.

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Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.

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Calculate the molarity of 1.5% w/v solution of PbCl2?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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We ASSUME that a #100*mL# volume of SOLUTION contains a #1.5*g# mass of solute..........

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Explanation:

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And thus #""molarity""# #=# #""Moles of solute""/""Volume of solution""#

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#=((1.5*g)/(278.10*g*mol^-1))/(0.100*L^-1)=5.4xx10^-2*mol*L^-1#.

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Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.

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" Calculate the molarity of 1.5% w/v solution of PbCl2? nan +3 a827c896-6ddd-11ea-8cbf-ccda262736ce https://socratic.org/questions/can-you-find-the-number-of-moles-of-argon-in-607-grams-of-argon 15.18 moles start physical_unit 8 8 mole mol qc_end physical_unit 8 8 10 11 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Argon [IN] moles""}]" "[{""type"":""physical unit"",""value"":""15.18 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] Argon [=] \\pu{607 grams}""}]" "

Can you find the number of moles of Argon in 607 grams of Argon?

" nan 15.18 moles "
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Explanation:

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The answer is 15.175 moles

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To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.

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It means that one mole of Argon weighs 40g.

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if one mole of Argon weighs 40 g , then 607 g of Argon has

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607g / 40g moles or 15.175 moles.

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15.175 moles.

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Explanation:

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The answer is 15.175 moles

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To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.

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It means that one mole of Argon weighs 40g.

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if one mole of Argon weighs 40 g , then 607 g of Argon has

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607g / 40g moles or 15.175 moles.

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Can you find the number of moles of Argon in 607 grams of Argon?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Jul 16, 2014 + +
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15.175 moles.

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Explanation:

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The answer is 15.175 moles

+

To solve this let us find out the molar mass of Argon, it is 39.948 g/mol or roughly 40 g /mol.

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It means that one mole of Argon weighs 40g.

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if one mole of Argon weighs 40 g , then 607 g of Argon has

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607g / 40g moles or 15.175 moles.

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" Can you find the number of moles of Argon in 607 grams of Argon? nan +4 a827c897-6ddd-11ea-acca-ccda262736ce https://socratic.org/questions/what-is-the-balanced-equation-for-photosynthesis-1 12H2O + 6CO2 = 6H2O + C6H12O6 + 6O2 start chemical_equation qc_end c_other Photosynthesis qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""12H2O + 6CO2 = 6H2O + C6H12O6 + 6O2""}]" "[{""type"": ""other"",""value"": ""Photosynthesis""}]" "

What is the balanced equation for photosynthesis?

" nan 12H2O + 6CO2 = 6H2O + C6H12O6 + 6O2 "
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Explanation:

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Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:

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#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#

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This reaction must occur in the presence of sunlight because light energy is required.

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The equation can also be written out in words as:

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In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.

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Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product #O_2#. In breaking apart the water energy is also released and picked up by ADP to make ATP.

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The excited #H^-# and the ATP travel to the stroma where energy from the ATP is used to combine carbon dioxide and the excited hydrogen into a molecule of glucose #C_6H_12O_6#.

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The equation is:

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#12 H_2O + 6 CO_2 -> 6 H_2O + C6H_12O_6 + 6O_2#

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Explanation:

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Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:

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#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#

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This reaction must occur in the presence of sunlight because light energy is required.

+

The equation can also be written out in words as:

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+

In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.

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+

Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product #O_2#. In breaking apart the water energy is also released and picked up by ADP to make ATP.

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The excited #H^-# and the ATP travel to the stroma where energy from the ATP is used to combine carbon dioxide and the excited hydrogen into a molecule of glucose #C_6H_12O_6#.

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" "
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What is the balanced equation for photosynthesis?

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+ + +Chemistry + + + + + +Stoichiometry + + + + + +Equation Stoichiometry + + +
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+1 Answer +
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The equation is:

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#12 H_2O + 6 CO_2 -> 6 H_2O + C6H_12O_6 + 6O_2#

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Explanation:

+
+

Photosynthesis is the combining of Carbon Dioxide and Water to make Glucose and Oxygen. The equation is:

+
+

#12 H_2O + 6 CO_2 -> 6 H_2O + C_6H_12O_6 + 6O_2#

+
+

This reaction must occur in the presence of sunlight because light energy is required.

+

The equation can also be written out in words as:

+
+

In the presence of sunlight, six moles of carbon dioxide and six moles of water react to form one molecule of glucose and six moles of oxygen.

+
+

Sunlight sends energy in packets of light called photons. These photons activate chlorophyll in the grans to break apart molecules of water. The Hydrogen atoms pick up the electrons from the energy and become energized. The Oxygen gets released as a waste product #O_2#. In breaking apart the water energy is also released and picked up by ADP to make ATP.

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The excited #H^-# and the ATP travel to the stroma where energy from the ATP is used to combine carbon dioxide and the excited hydrogen into a molecule of glucose #C_6H_12O_6#.

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Related questions
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Impact of this question
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+ 52427 views + around the world +
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+ + Creative Commons License + +
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" What is the balanced equation for photosynthesis? nan +5 a827c898-6ddd-11ea-bada-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-concentration-of-iodate-ions-in-a-saturated-solution-of 8.04 × 10^(-5) M start physical_unit 7 8 concentration mol/l qc_end chemical_equation 17 17 qc_end c_other OTHER qc_end end "[{""type"": ""physical unit"", ""value"": ""concentration [OF] iodate ions [IN] M""}]" "[{""type"":""physical unit"",""value"":""8.04 × 10^(-5) M""}]" "[{""type"": ""chemical equation"",""value"": ""Pb(IO3)2""},{""type"": ""other"",""value"": ""Ksp = 2.6 × 10^(-13)""}]" "

How do you calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, #Pb(IO_3)_2#? The #K_(sp) = 2.6 xx 10^(-13)#?

" nan 8.04 × 10^(-5) M "
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Explanation:

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The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions, #""IO""_3^(-)#, in a saturated solution of lead(II) iodate.

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The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this

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#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#

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Notice that every mole of lead(II) iodate that dissociates produces #1# mole of lead(II) cations and #color(red)(2)# moles of iodate anions in solution.

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This means that, at equilibrium, a saturated solution of lead(II) iodate will have

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#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#

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Now, the solubility product constant for this dissociation equilibrium looks like this

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#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#

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If you take #s# to be the concentration of lead(II) cations in the solution, i.e. the molar solubility of the salt, you can say that you have

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#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

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which is equivalent to

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#2.6 * 10^(-13) = 4s^3#

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Rearrange to solve for #s#

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#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#

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This means that a saturated solution of lead(II) iodate will have

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#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#

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and

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#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#

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I'll leave the answer rounded to two sig figs.

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#[""IO""_3^(-)] = 8.0 * 10^(-5)# #""M""#

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Explanation:

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The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions, #""IO""_3^(-)#, in a saturated solution of lead(II) iodate.

+

The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this

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#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#

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Notice that every mole of lead(II) iodate that dissociates produces #1# mole of lead(II) cations and #color(red)(2)# moles of iodate anions in solution.

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This means that, at equilibrium, a saturated solution of lead(II) iodate will have

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#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#

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Now, the solubility product constant for this dissociation equilibrium looks like this

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#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#

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If you take #s# to be the concentration of lead(II) cations in the solution, i.e. the molar solubility of the salt, you can say that you have

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#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

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which is equivalent to

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#2.6 * 10^(-13) = 4s^3#

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Rearrange to solve for #s#

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#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#

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This means that a saturated solution of lead(II) iodate will have

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#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#

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and

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#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#

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I'll leave the answer rounded to two sig figs.

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How do you calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, #Pb(IO_3)_2#? The #K_(sp) = 2.6 xx 10^(-13)#?

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#[""IO""_3^(-)] = 8.0 * 10^(-5)# #""M""#

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Explanation:

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The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions, #""IO""_3^(-)#, in a saturated solution of lead(II) iodate.

+

The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this

+
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#""Pb""(""IO""_ 3)_ (2(s)) rightleftharpoons ""Pb""_ ((aq))^(2+) + color(red)(2)""IO""_ (3(aq))^(-)#

+
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Notice that every mole of lead(II) iodate that dissociates produces #1# mole of lead(II) cations and #color(red)(2)# moles of iodate anions in solution.

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This means that, at equilibrium, a saturated solution of lead(II) iodate will have

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#[""IO""_3^(-)] = color(red)(2) * [""Pb""^(2+)]#

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Now, the solubility product constant for this dissociation equilibrium looks like this

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#K_(sp) = [""Pb""^(2+)] * [""IO""_3^(-)]^color(red)(2)#

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If you take #s# to be the concentration of lead(II) cations in the solution, i.e. the molar solubility of the salt, you can say that you have

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#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

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which is equivalent to

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#2.6 * 10^(-13) = 4s^3#

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Rearrange to solve for #s#

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#s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)#

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This means that a saturated solution of lead(II) iodate will have

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#[""Pb""^(2+)] = 4.02 * 10^(-5)# #""M""#

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and

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#[""IO""_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)""M"" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)""M"")))#

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I'll leave the answer rounded to two sig figs.

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Related questions
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Impact of this question
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+
+
" How do you calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, #Pb(IO_3)_2#? The #K_(sp) = 2.6 xx 10^(-13)#? nan +6 a827c899-6ddd-11ea-9c37-ccda262736ce https://socratic.org/questions/how-do-i-write-equation-for-the-combustion-of-ch4-to-give-co2-and-water CH4 + 2O2 -> CO2 + 2H2O start chemical_equation qc_end chemical_equation 9 9 qc_end chemical_equation 12 12 qc_end substance 14 14 qc_end c_other combustion qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""CH4 + 2O2 -> CO2 + 2H2O""}]" "[{""type"": ""chemical equation"",""value"": ""CH4""},{""type"": ""chemical equation"",""value"": ""CO2""},{""type"": ""substance name"",""value"": ""water""},{""type"": ""other"",""value"": ""combustion""}]" "

How do I write equation for the combustion of +CH4 to give CO2 and water?

" nan CH4 + 2O2 -> CO2 + 2H2O "
+

Explanation:

+
+

Unbalanced Equation

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + H""_2""O""#

+

Balance the hydrogen.

+

Place a coefficient of #2# in front of the #""H""_2""O""#.

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

Balance the oxygen.

+

Place a coefficient of #2# in front of the #""O""_2""#.

+

#""CH""_4"" + 2O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.

+

Number of Atoms on Left Side: #""1C; 4H; 4O""#

+

Number of Atoms on Right Side: #""1C; 4H; 4O""#

+
+
" "
+
+
+

#""CH""_4"" + 2O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+
+
+
+

Explanation:

+
+

Unbalanced Equation

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + H""_2""O""#

+

Balance the hydrogen.

+

Place a coefficient of #2# in front of the #""H""_2""O""#.

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

Balance the oxygen.

+

Place a coefficient of #2# in front of the #""O""_2""#.

+

#""CH""_4"" + 2O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.

+

Number of Atoms on Left Side: #""1C; 4H; 4O""#

+

Number of Atoms on Right Side: #""1C; 4H; 4O""#

+
+
+
" "
+

How do I write equation for the combustion of +CH4 to give CO2 and water?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 25, 2015 + +
+
+
+
+
+
+
+

#""CH""_4"" + 2O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+
+
+
+

Explanation:

+
+

Unbalanced Equation

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + H""_2""O""#

+

Balance the hydrogen.

+

Place a coefficient of #2# in front of the #""H""_2""O""#.

+

#""CH""_4"" + O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

Balance the oxygen.

+

Place a coefficient of #2# in front of the #""O""_2""#.

+

#""CH""_4"" + 2O""_2""##rarr##""CO""_2"" + 2H""_2""O""#

+

There are now equal numbers of atoms of each element on both sides of the equation, and it is now balanced.

+

Number of Atoms on Left Side: #""1C; 4H; 4O""#

+

Number of Atoms on Right Side: #""1C; 4H; 4O""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
+ 6142 views + around the world +
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+ + Creative Commons License + +
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" "How do I write equation for the combustion of +CH4 to give CO2 and water?" nan +7 a827c89a-6ddd-11ea-88b1-ccda262736ce https://socratic.org/questions/how-many-grams-are-in-3-8-m-moles-of-ch-4 0.06 grams start physical_unit 9 9 mass g qc_end physical_unit 9 9 5 7 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] CH4 [IN] grams""}]" "[{""type"":""physical unit"",""value"":""0.06 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] CH4 [=] \\pu{3.8 m moles}""}]" "

How many grams are in 3.8 m moles of #CH_4#?

" nan 0.06 grams "
+

Explanation:

+
+

In #3.80# #mmol#, there are #3.80# #xx10^(-3)##cancel(mol)# #xx# #16.0*g*cancel(mol^-1)# #=# #??# #g#.

+
+
" "
+
+
+

One mole of methane has a mass of 16.0 g.

+
+
+
+

Explanation:

+
+

In #3.80# #mmol#, there are #3.80# #xx10^(-3)##cancel(mol)# #xx# #16.0*g*cancel(mol^-1)# #=# #??# #g#.

+
+
+
" "
+

How many grams are in 3.8 m moles of #CH_4#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Dec 6, 2015 + +
+
+
+
+
+
+
+

One mole of methane has a mass of 16.0 g.

+
+
+
+

Explanation:

+
+

In #3.80# #mmol#, there are #3.80# #xx10^(-3)##cancel(mol)# #xx# #16.0*g*cancel(mol^-1)# #=# #??# #g#.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Dec 6, 2015 + +
+
+
+
+
+
+
+

#""3.8 mmol CH""_4""# has a mass of #""0.061 g""#.

+
+
+
+

Explanation:

+
+

The molar mass (mass of one mole) of #""CH""_4""##=##(1xx12.0107""g/mol"")+(4xx1.00794""g/mol"")=""16.04246 g/mol""#

+

Convert 3.8 mmol to mol.

+

#3.8cancel""mmol""xx(1""mol"")/(1000cancel""mmol"")=""0.0038 mol""#

+

Multiply moles times molar mass.

+

#0.0038cancel""mol CH""_4xx(16.04246""g CH""_4)/(1cancel""mol CH""_4)=""0.061 g CH""_4""# (rounded to two significant figures)

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 2775 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
" How many grams are in 3.8 m moles of #CH_4#? nan +8 a827ef98-6ddd-11ea-985c-ccda262736ce https://socratic.org/questions/594c04987c01495710725da1 0.50% start physical_unit 8 8 molarity_percent none qc_end physical_unit 8 8 7 7 molarity_percent qc_end physical_unit 8 8 15 16 volume qc_end physical_unit 8 8 18 19 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity percent2 [OF] solution""}]" "[{""type"":""physical unit"",""value"":""0.50%""}]" "[{""type"":""physical unit"",""value"":""Molarity percent1 [OF] solution [=] \\pu{10%}""},{""type"":""physical unit"",""value"":""Volume1 [OF] solution [=] \\pu{25.0 mL}""},{""type"":""physical unit"",""value"":""Volume2 [OF] solution [=] \\pu{500 mL}""}]" "

What is the new concentration of a #10%# solution if we dilute a sample from #""25.0 mL""# to #""500 mL""# ?

" nan 0.50% "
+

Explanation:

+
+

The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.

+
+

#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#

+
+

In your case, the volume of the solution increases from #""25.0 mL""# to #""500 mL""#, which means that the dilution factor is equal to

+
+

#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#

+
+

Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be #color(blue)(20# times lower than the concentration of the stock solution.

+

This means that the concentration of the diluted solution will be

+
+

#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#

+
+

The answer is rounded to one significant figure.

+
+
" "
+
+
+

#0.5%#

+
+
+
+

Explanation:

+
+

The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.

+
+

#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#

+
+

In your case, the volume of the solution increases from #""25.0 mL""# to #""500 mL""#, which means that the dilution factor is equal to

+
+

#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#

+
+

Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be #color(blue)(20# times lower than the concentration of the stock solution.

+

This means that the concentration of the diluted solution will be

+
+

#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#

+
+

The answer is rounded to one significant figure.

+
+
+
" "
+

What is the new concentration of a #10%# solution if we dilute a sample from #""25.0 mL""# to #""500 mL""# ?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 8, 2017 + +
+
+
+
+
+
+
+

#0.5%#

+
+
+
+

Explanation:

+
+

The idea here is that when you're performing a dilution, the concentration of the solution decreases by the same factor (known as the dilution factor) as the volume increases.

+
+

#""volume"" uarr ""by a factor""color(white)(.)color(blue)(""DF"") implies ""concentration"" darr ""by the same factor""color(white)(.)color(blue)(""DF"")#

+
+

In your case, the volume of the solution increases from #""25.0 mL""# to #""500 mL""#, which means that the dilution factor is equal to

+
+

#""DF"" = (500 color(red)(cancel(color(black)(""mL""))))/(25.0color(red)(cancel(color(black)(""mL"")))) = color(blue)(20)#

+
+

Now remember, the same dilution factor is applied to the concentration of the solution, i.e. the concentration of the diluted solution must be #color(blue)(20# times lower than the concentration of the stock solution.

+

This means that the concentration of the diluted solution will be

+
+

#""% diluted"" = (10%)/color(blue)(20) = color(darkgreen)(ul(color(black)(0.5%)))#

+
+

The answer is rounded to one significant figure.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2338 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" "What is the new concentration of a #10%# solution if we dilute a sample from #""25.0 mL""# to #""500 mL""# ?" nan +9 a827ef99-6ddd-11ea-be87-ccda262736ce https://socratic.org/questions/a-piece-of-copper-with-a-mass-of-22-6-g-initially-at-a-temperature-of-16-3-c-is- 146.55 J start physical_unit 40 42 heat_energy j qc_end physical_unit 3 3 8 9 mass qc_end physical_unit 3 3 15 16 temperature qc_end physical_unit 3 3 23 24 temperature qc_end physical_unit 3 3 32 33 specific_heat qc_end end "[{""type"": ""physical unit"", ""value"": ""needed heat [OF] this temperature change [IN] J""}]" "[{""type"":""physical unit"",""value"":""146.55 J""}]" "[{""type"":""physical unit"",""value"":""mass [OF] copper [=] \\pu{22.6 g}""},{""type"":""physical unit"",""value"":""temperature1 [OF] copper [=] \\pu{16.3 °C}""},{""type"":""physical unit"",""value"":""temperature2 [OF] copper [=] \\pu{33.1 °C}""},{""type"":""physical unit"",""value"":""specific heat [OF] copper [=] \\pu{0.386 J/(g°C)}""}]" "

A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?

" nan 146.55 J "
+

Explanation:

+
+

A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
+#Q = K M (t2-t1)# Where K is the specifc heat
+#Q= 0.386 J/g °C 22,6 g (33,1-16,3)°C #
+#Q= 146,55 J#

+
+
" "
+
+
+

146,55 J

+
+
+
+

Explanation:

+
+

A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
+#Q = K M (t2-t1)# Where K is the specifc heat
+#Q= 0.386 J/g °C 22,6 g (33,1-16,3)°C #
+#Q= 146,55 J#

+
+
+
" "
+

A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 27, 2017 + +
+
+
+
+
+
+
+

146,55 J

+
+
+
+

Explanation:

+
+

A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place?
+#Q = K M (t2-t1)# Where K is the specifc heat
+#Q= 0.386 J/g °C 22,6 g (33,1-16,3)°C #
+#Q= 146,55 J#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1642 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A piece of copper with a mass of 22.6 g initially at a temperature of 16.3 °C is heated to a temperature of 33.1 °C. Assuming the specific heat of copper is 0.386 J/(g°C), how much heat was needed for this temperature change to take place? nan +10 a827ef9a-6ddd-11ea-ad1a-ccda262736ce https://socratic.org/questions/carbon-monoxide-can-react-with-oxygen-to-form-carbon-dioxide-how-do-you-write-a- 2CO + O2 = 2CO2 start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2CO + O2 = 2CO2""}]" "[{""type"": ""other"",""value"": ""Carbon monoxide can react with oxygen to form carbon dioxide.""}]" "

Carbon monoxide can react with oxygen to form carbon dioxide. How do you write a balanced chemical reaction for this process?

" nan 2CO + O2 = 2CO2 "
+

Explanation:

+
+

#2CO+O2=2CO2#
+Balancing the equation by balancing number of atoms equal both sides.

+
+
" "
+
+
+

#2CO+O2=2CO2#

+
+
+
+

Explanation:

+
+

#2CO+O2=2CO2#
+Balancing the equation by balancing number of atoms equal both sides.

+
+
+
" "
+

Carbon monoxide can react with oxygen to form carbon dioxide. How do you write a balanced chemical reaction for this process?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 5, 2016 + +
+
+
+
+
+
+
+

#2CO+O2=2CO2#

+
+
+
+

Explanation:

+
+

#2CO+O2=2CO2#
+Balancing the equation by balancing number of atoms equal both sides.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 19881 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" Carbon monoxide can react with oxygen to form carbon dioxide. How do you write a balanced chemical reaction for this process? nan +11 a827ef9b-6ddd-11ea-9a4b-ccda262736ce https://socratic.org/questions/how-many-moles-of-nacl-are-contained-in-467-ml-of-a-244m-solution-of-nacl 0.11 moles start physical_unit 4 4 mole mol qc_end physical_unit 13 15 8 9 volume qc_end physical_unit 13 15 12 12 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] NaCl [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.11 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution of NaCl [=] \\pu{467 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] solution of NaCl [=] \\pu{0.244 M}""}]" "

How many moles of NaCl are contained in 467 mL of a .244M solution of NaCl?

" nan 0.11 moles "
+

Explanation:

+
+

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, #""NaCl""#, divided by liters of solution.

+
+

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

+
+

Simply put, dissolving one mole of a solute in one liter of solution will produce a #""1.00-M""# solution.

+

In your case, the volume of the solution is said to be equal to #""467 mL""# and the molarity of the solution to #""0.244 M""#. Since molarity tells you number of moles per liter, you can say that you will get #""0.244 moles""# of sodium chloride in one liter of this solution.

+

Since you have a little less than half of a liter, i.e. #""467 mL""#, you can expect to have fewer than half of #""0.244 moles""#.

+

More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#

+
+

Rounded to three sig figs, the answer will be

+
+

#n_(NaCl) = color(green)(""0.114 moles"")#

+
+

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a #""0.244 M""# sodium chloride solution.

+

Instead, you would say that you have a solution that is #""0.244 M""# in sodium cations, #""Na""^(+)#, and #""0.244 M""# in chloride anions, #""Cl""^(-)#.

+
+
" "
+
+
+

#""0.144 moles""#

+
+
+
+

Explanation:

+
+

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, #""NaCl""#, divided by liters of solution.

+
+

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

+
+

Simply put, dissolving one mole of a solute in one liter of solution will produce a #""1.00-M""# solution.

+

In your case, the volume of the solution is said to be equal to #""467 mL""# and the molarity of the solution to #""0.244 M""#. Since molarity tells you number of moles per liter, you can say that you will get #""0.244 moles""# of sodium chloride in one liter of this solution.

+

Since you have a little less than half of a liter, i.e. #""467 mL""#, you can expect to have fewer than half of #""0.244 moles""#.

+

More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#

+
+

Rounded to three sig figs, the answer will be

+
+

#n_(NaCl) = color(green)(""0.114 moles"")#

+
+

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a #""0.244 M""# sodium chloride solution.

+

Instead, you would say that you have a solution that is #""0.244 M""# in sodium cations, #""Na""^(+)#, and #""0.244 M""# in chloride anions, #""Cl""^(-)#.

+
+
+
" "
+

How many moles of NaCl are contained in 467 mL of a .244M solution of NaCl?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 12, 2015 + +
+
+
+
+
+
+
+

#""0.144 moles""#

+
+
+
+

Explanation:

+
+

As you know, molarity is defined as the number of moles of solute, which in your case is sodium chloride, #""NaCl""#, divided by liters of solution.

+
+

#color(blue)(""molarity"" = ""moles of solute""/""liters of solution"")#

+
+

Simply put, dissolving one mole of a solute in one liter of solution will produce a #""1.00-M""# solution.

+

In your case, the volume of the solution is said to be equal to #""467 mL""# and the molarity of the solution to #""0.244 M""#. Since molarity tells you number of moles per liter, you can say that you will get #""0.244 moles""# of sodium chloride in one liter of this solution.

+

Since you have a little less than half of a liter, i.e. #""467 mL""#, you can expect to have fewer than half of #""0.244 moles""#.

+

More specifically, the solution will contain - do not forget to convert the volume of the solution from milliliters to liters!

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n = 0.244 ""moles""/color(red)(cancel(color(black)(""L""))) * 467 * 10^(-3) color(red)(cancel(color(black)(""L""))) = ""0.1139 moles NaCl""#

+
+

Rounded to three sig figs, the answer will be

+
+

#n_(NaCl) = color(green)(""0.114 moles"")#

+
+

SIDE NOTE Because sodium chloride dissociates completely in aqueous solution, you wouldn't actually say that you have a #""0.244 M""# sodium chloride solution.

+

Instead, you would say that you have a solution that is #""0.244 M""# in sodium cations, #""Na""^(+)#, and #""0.244 M""# in chloride anions, #""Cl""^(-)#.

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+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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" How many moles of NaCl are contained in 467 mL of a .244M solution of NaCl? nan +12 a827ef9c-6ddd-11ea-a940-ccda262736ce https://socratic.org/questions/593547257c0149397db568d6 8800 g start physical_unit 4 5 mass g qc_end physical_unit 39 40 9 10 volume qc_end physical_unit 27 27 12 13 volume qc_end physical_unit 27 27 20 21 volume qc_end physical_unit 27 27 25 26 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass1 [OF] rubidium iodide [IN] g""}]" "[{""type"":""physical unit"",""value"":""8800 g""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] original solution [=] \\pu{4.6 L}""},{""type"":""physical unit"",""value"":""Volume2 [OF] solution [=] \\pu{1 L}""},{""type"":""physical unit"",""value"":""Volume3 [OF] solution [=] \\pu{13.9 L}""},{""type"":""physical unit"",""value"":""Molarity3 [OF] solution [=] \\pu{0.65 M}""}]" "

A certain amount of rubidium iodide was dissolved in #""4.6 L""#. A #""1-L""# aliquot was extracted an diluted to #""13.9 L""# to form a #""0.65 M""# solution in water. What mass of rubidium iodide was dissolved into the original solution?

" nan 8800 g "
+

Explanation:

+
+

For this kind of questions, we use the formula:

+

#(""mol"")/(""volume ""color(blue)(""L""))=""concentration "" color(blue)(""mol""xx""L""^-1)#

+

To calculate the mass of rubidium iodide (#RbI#), we need to calculate back to the original solution. We start therefore at the end of the question and work back.

+

We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:

+

#""mol""=""volume""xx""concentration""#
+#""mol""=13.9 color(red)(cancel(color(blue)(""L""))) xx0.65 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=9.035 color(blue)("" mol"")#

+

So in our 1 L solution, we had 9.035 mol #RbI#. Using the same formula again but rewritten, we can calculate the concentration of that 1 L solution:

+

#""concentration"" =(""mol"")/(""volume"")#

+

#""concentration""=(9.035 color(blue)("" mol""))/(1color(blue)(""L""))=9.035 color(blue)((""mol"")/(""L""))#

+

Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!

+

#""mol""=""volume""xx""concentration""#
+#""mol""=4.6 color(red)(cancel(color(blue)(""L""))) xx9.035 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=41.561 color(blue)("" mol"")#

+

Now we only have to convert mol #RbI# into mass units and done! We use the molar mass of #RbI# to convert. We can calculate the molar mass from adding the atom masses #u# or looking it up:

+

#""molar mass""# #RbI =212.37 color(blue)("" gram""xx""mol""^-1)#

+

#""mass ""= ""molar mass ""xx"" mol""#

+

#""mass ""=212.37 (""gram"")/color(red)(cancel(color(blue)(""mol"")))xx41.561 color(red)(cancel(color(blue)(""mol"")))=8826.31 color(blue)(""gram"")#

+

Which, when rounded to two significant figures (the lowest number given in the problem), is #color(red)(8800# #color(red)(""g"")#.

+
+
" "
+
+
+

The amount of #""RbI""# that was added is #8800# #""g""#.

+
+
+
+

Explanation:

+
+

For this kind of questions, we use the formula:

+

#(""mol"")/(""volume ""color(blue)(""L""))=""concentration "" color(blue)(""mol""xx""L""^-1)#

+

To calculate the mass of rubidium iodide (#RbI#), we need to calculate back to the original solution. We start therefore at the end of the question and work back.

+

We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:

+

#""mol""=""volume""xx""concentration""#
+#""mol""=13.9 color(red)(cancel(color(blue)(""L""))) xx0.65 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=9.035 color(blue)("" mol"")#

+

So in our 1 L solution, we had 9.035 mol #RbI#. Using the same formula again but rewritten, we can calculate the concentration of that 1 L solution:

+

#""concentration"" =(""mol"")/(""volume"")#

+

#""concentration""=(9.035 color(blue)("" mol""))/(1color(blue)(""L""))=9.035 color(blue)((""mol"")/(""L""))#

+

Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!

+

#""mol""=""volume""xx""concentration""#
+#""mol""=4.6 color(red)(cancel(color(blue)(""L""))) xx9.035 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=41.561 color(blue)("" mol"")#

+

Now we only have to convert mol #RbI# into mass units and done! We use the molar mass of #RbI# to convert. We can calculate the molar mass from adding the atom masses #u# or looking it up:

+

#""molar mass""# #RbI =212.37 color(blue)("" gram""xx""mol""^-1)#

+

#""mass ""= ""molar mass ""xx"" mol""#

+

#""mass ""=212.37 (""gram"")/color(red)(cancel(color(blue)(""mol"")))xx41.561 color(red)(cancel(color(blue)(""mol"")))=8826.31 color(blue)(""gram"")#

+

Which, when rounded to two significant figures (the lowest number given in the problem), is #color(red)(8800# #color(red)(""g"")#.

+
+
+
" "
+

A certain amount of rubidium iodide was dissolved in #""4.6 L""#. A #""1-L""# aliquot was extracted an diluted to #""13.9 L""# to form a #""0.65 M""# solution in water. What mass of rubidium iodide was dissolved into the original solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ + +
+
+ +
+ + Jun 5, 2017 + +
+
+
+
+
+
+
+

The amount of #""RbI""# that was added is #8800# #""g""#.

+
+
+
+

Explanation:

+
+

For this kind of questions, we use the formula:

+

#(""mol"")/(""volume ""color(blue)(""L""))=""concentration "" color(blue)(""mol""xx""L""^-1)#

+

To calculate the mass of rubidium iodide (#RbI#), we need to calculate back to the original solution. We start therefore at the end of the question and work back.

+

We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:

+

#""mol""=""volume""xx""concentration""#
+#""mol""=13.9 color(red)(cancel(color(blue)(""L""))) xx0.65 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=9.035 color(blue)("" mol"")#

+

So in our 1 L solution, we had 9.035 mol #RbI#. Using the same formula again but rewritten, we can calculate the concentration of that 1 L solution:

+

#""concentration"" =(""mol"")/(""volume"")#

+

#""concentration""=(9.035 color(blue)("" mol""))/(1color(blue)(""L""))=9.035 color(blue)((""mol"")/(""L""))#

+

Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!

+

#""mol""=""volume""xx""concentration""#
+#""mol""=4.6 color(red)(cancel(color(blue)(""L""))) xx9.035 (color(blue)(""mol""))/(color(red)cancel(color(blue)(""L"")))=41.561 color(blue)("" mol"")#

+

Now we only have to convert mol #RbI# into mass units and done! We use the molar mass of #RbI# to convert. We can calculate the molar mass from adding the atom masses #u# or looking it up:

+

#""molar mass""# #RbI =212.37 color(blue)("" gram""xx""mol""^-1)#

+

#""mass ""= ""molar mass ""xx"" mol""#

+

#""mass ""=212.37 (""gram"")/color(red)(cancel(color(blue)(""mol"")))xx41.561 color(red)(cancel(color(blue)(""mol"")))=8826.31 color(blue)(""gram"")#

+

Which, when rounded to two significant figures (the lowest number given in the problem), is #color(red)(8800# #color(red)(""g"")#.

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+ +
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+ +
+ + Jun 5, 2017 + +
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+

Here is another way to do this problem using the extensive property of countable numbers. I also get #""8800 g""# (two sig figs).

+

OVERVIEW:
+Think critically through the problem... What did we do? We started with some starting mols of #""RbI""# dissolved in #""4.6 L""#, and then took #""1 L""# of that, which contained a smaller chunk of the original mols.

+

That was then diluted from #""1 L""# to #""13.9 L""# to form the #""0.65 M""# solution, which contains the same number of mols as there were in the #""1 L""# solution.

+
+

Looking at the end of the question, we were given #""13.9 L""# of a #""0.65 M""# solution, which has:

+
+

#13.9 cancel""L"" xx ""0.65 mols""/cancel""L"" = ""9.035 mols RbI""# for every #""L""# of solution.

+
+

The number of mols in the #""1 L""# sample of the solution and the number of mols in the #""13.9 L""# sample are the same quantity, since just water was added to the same mols as before.

+

Mols are an extensive quantity, just like mass. If you have a greater volume of a certain concentration, you have more mols of it.

+

Therefore, in #""4.6 L""# solution (from which the #""1 L""# was taken at fixed concentration!), we have:

+
+

#""9.035 mols RbI"" xx (4.6 cancel""L"")/cancel""1 L""#

+

#=# #""41.561 mols RbI""# in that #""4.6 L""# of solution.

+
+

Therefore, you had a mass of:

+
+

#color(blue)(m_(RbI)) = 41.561 cancel""mols RbI"" xx ""212.3723 g""/cancel""1 mol RbI""#

+

#=# #""8826.41 g""#

+

#-># #color(blue)(""8800 g"")# to two sig figs

+
+

Now think back through the problem... What did we do?

+

We started with some #""41.561 mol""#s of #""RbI""# dissolved in #""4.6 L""#, and then took #""1 L""# of that, which contained #""9.035 mols RbI""# because mols are extensive.

+

That was then diluted #13.9#-fold to form the #""0.65 M""# solution, which contains the same number of mols of #""RbI""# as in the #""1 L""# sample.

+
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+
+
+
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+
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+ + Creative Commons License + +
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+
" "A certain amount of rubidium iodide was dissolved in #""4.6 L""#. A #""1-L""# aliquot was extracted an diluted to #""13.9 L""# to form a #""0.65 M""# solution in water. What mass of rubidium iodide was dissolved into the original solution?" nan +13 a827ef9d-6ddd-11ea-a90f-ccda262736ce https://socratic.org/questions/a-compound-contains-69-94-percent-iron-and-30-06-percent-oxygen-what-is-its-mole Fe2O3 start chemical_formula qc_end physical_unit 4 4 3 3 percent qc_end physical_unit 7 7 6 6 percent qc_end physical_unit 1 1 21 23 molar_mass qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Fe2O3""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] iron [=] \\pu{69.94%}""},{""type"":""physical unit"",""value"":""Percent [OF] oxygen [=] \\pu{30.06%}""},{""type"":""physical unit"",""value"":""Molar mass [OF] compound [=] \\pu{199.55 per mole}""}]" "

A compound contains 69.94 percent iron and 30.06 percent oxygen. What is its molecular formula if the molar mass of the compound is 199.55 per mole?

" nan Fe2O3 "
+

Explanation:

+
+

As with all these problems, we ASSUME, that there is a #100*g# mass of unknown compound.

+

We break this quantity down into atoms.

+

And, there is #(30.06*g)/(15.999*g*mol^-1)# #=# #1.88*mol# #O#.

+

And #(69.94*g)/(55.85*g*mol^-1)# #=# #1.25*mol# #Fe#.

+

If we divide thru by the lowest molar amount, we get an empirical formula of #FeO_(1.5)#, but because the empirical formula is the simplest whole number ratio defining constituent atoms in a species, we double this result to get WHOLE numbers, i.e. #Fe_2O_3,"" ferric oxide.""#

+

Note that an examiner would be quite justified at A level to say that #""an oxide of iron has 70% iron content""# without quoting the oxygen percentage, and expect you to realize that the balance is oxygen.

+
+
" "
+
+
+

#Fe_2O_3#

+
+
+
+

Explanation:

+
+

As with all these problems, we ASSUME, that there is a #100*g# mass of unknown compound.

+

We break this quantity down into atoms.

+

And, there is #(30.06*g)/(15.999*g*mol^-1)# #=# #1.88*mol# #O#.

+

And #(69.94*g)/(55.85*g*mol^-1)# #=# #1.25*mol# #Fe#.

+

If we divide thru by the lowest molar amount, we get an empirical formula of #FeO_(1.5)#, but because the empirical formula is the simplest whole number ratio defining constituent atoms in a species, we double this result to get WHOLE numbers, i.e. #Fe_2O_3,"" ferric oxide.""#

+

Note that an examiner would be quite justified at A level to say that #""an oxide of iron has 70% iron content""# without quoting the oxygen percentage, and expect you to realize that the balance is oxygen.

+
+
+
" "
+

A compound contains 69.94 percent iron and 30.06 percent oxygen. What is its molecular formula if the molar mass of the compound is 199.55 per mole?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 23, 2016 + +
+
+
+
+
+
+
+

#Fe_2O_3#

+
+
+
+

Explanation:

+
+

As with all these problems, we ASSUME, that there is a #100*g# mass of unknown compound.

+

We break this quantity down into atoms.

+

And, there is #(30.06*g)/(15.999*g*mol^-1)# #=# #1.88*mol# #O#.

+

And #(69.94*g)/(55.85*g*mol^-1)# #=# #1.25*mol# #Fe#.

+

If we divide thru by the lowest molar amount, we get an empirical formula of #FeO_(1.5)#, but because the empirical formula is the simplest whole number ratio defining constituent atoms in a species, we double this result to get WHOLE numbers, i.e. #Fe_2O_3,"" ferric oxide.""#

+

Note that an examiner would be quite justified at A level to say that #""an oxide of iron has 70% iron content""# without quoting the oxygen percentage, and expect you to realize that the balance is oxygen.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 14318 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" A compound contains 69.94 percent iron and 30.06 percent oxygen. What is its molecular formula if the molar mass of the compound is 199.55 per mole? nan +14 a827ef9e-6ddd-11ea-a14e-ccda262736ce https://socratic.org/questions/what-is-the-temperature-of-2-0-moles-of-a-gas-occupying-a-volume-of-5-0-l-at-2-4 75 K start physical_unit 9 9 temperature k qc_end physical_unit 9 9 5 6 mole qc_end physical_unit 9 9 14 15 volume qc_end physical_unit 9 9 17 18 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""75 K""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] gas [=] \\pu{2.0 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] gas [=] \\pu{5.0 L}""},{""type"":""physical unit"",""value"":""Pressure [OF] gas [=] \\pu{2.46 atm}""}]" "

What is the temperature of 2.0 moles of a gas occupying a volume of 5.0 L at 2.46 atm?

" nan 75 K "
+

Explanation:

+
+

This is a fairly straightforward application of the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin

+

Rearrange this equation to solve for #T#, the temperature of the gas

+
+

#PV = nRT implies T = (PV)/(nR)#

+
+

Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.

+

As it stands, moles, atm, and liters are all units used for #R#, so you're good to go.

+

Plug in your values to get

+
+

#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#

+
+

Rounded to two sig figs, the answer will be

+
+

#T = color(green)(""75 K"")#

+
+
+
" "
+
+
+

#T = ""75 K""#

+
+
+
+

Explanation:

+
+

This is a fairly straightforward application of the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin

+

Rearrange this equation to solve for #T#, the temperature of the gas

+
+

#PV = nRT implies T = (PV)/(nR)#

+
+

Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.

+

As it stands, moles, atm, and liters are all units used for #R#, so you're good to go.

+

Plug in your values to get

+
+

#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#

+
+

Rounded to two sig figs, the answer will be

+
+

#T = color(green)(""75 K"")#

+
+
+
+
" "
+

What is the temperature of 2.0 moles of a gas occupying a volume of 5.0 L at 2.46 atm?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 16, 2015 + +
+
+
+
+
+
+
+

#T = ""75 K""#

+
+
+
+

Explanation:

+
+

This is a fairly straightforward application of the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin

+

Rearrange this equation to solve for #T#, the temperature of the gas

+
+

#PV = nRT implies T = (PV)/(nR)#

+
+

Before plugging in your values, check to make sure that the units given to you match the units used in the expression of the universal gas constant.

+

As it stands, moles, atm, and liters are all units used for #R#, so you're good to go.

+

Plug in your values to get

+
+

#T = (2.46 color(red)(cancel(color(black)(""atm""))) * 5.0 color(red)(cancel(color(black)(""L""))))/(2.0 color(red)(cancel(color(black)(""moles""))) * 0.0821 (color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(color(red)(cancel(color(black)(""mol""))) * ""K"")) = ""74.91 K""#

+
+

Rounded to two sig figs, the answer will be

+
+

#T = color(green)(""75 K"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 13698 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the temperature of 2.0 moles of a gas occupying a volume of 5.0 L at 2.46 atm? nan +15 a827ef9f-6ddd-11ea-aa12-ccda262736ce https://socratic.org/questions/how-many-milliliters-of-3-00-m-hci-aq-are-required-to-react-with-8-55-g-of-zn-s 87 milliliters start physical_unit 6 6 volume ml qc_end physical_unit 6 6 4 5 molarity qc_end physical_unit 15 15 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] HCI(aq) [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""87 milliliters""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCI(aq) [=] \\pu{3.00 M}""},{""type"":""physical unit"",""value"":""Mass [OF] Zn(s) [=] \\pu{8.55 g}""}]" "

How many milliliters of 3.00 M #HCI(aq)# are required to react with 8.55 g of #Zn(s)#?

" nan 87 milliliters "
+

Explanation:

+
+

Molecular mass HCl=36.4g , Zn=65.4g

+

#2HCl + Zn -> ZnCl_2 + H_2#

+

n°mol=#""8.55g""/""65.4g""# = 0.13mol

+

In the reaction, we can see HCl doubles the n°mol of Zn, so...
+#2(0.13)= 0.26mol#

+

#M=""n°mol""/""Volume""#

+

#3=""0.26mol""/V#

+

#V=0.087L# of 3M HCl

+
+
" "
+
+
+

#V_""HCl""=0.087L#

+
+
+
+

Explanation:

+
+

Molecular mass HCl=36.4g , Zn=65.4g

+

#2HCl + Zn -> ZnCl_2 + H_2#

+

n°mol=#""8.55g""/""65.4g""# = 0.13mol

+

In the reaction, we can see HCl doubles the n°mol of Zn, so...
+#2(0.13)= 0.26mol#

+

#M=""n°mol""/""Volume""#

+

#3=""0.26mol""/V#

+

#V=0.087L# of 3M HCl

+
+
+
" "
+

How many milliliters of 3.00 M #HCI(aq)# are required to react with 8.55 g of #Zn(s)#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
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+
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+1 Answer +
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+ +
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+ +
+ + Jun 26, 2017 + +
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+
+
+
+

#V_""HCl""=0.087L#

+
+
+
+

Explanation:

+
+

Molecular mass HCl=36.4g , Zn=65.4g

+

#2HCl + Zn -> ZnCl_2 + H_2#

+

n°mol=#""8.55g""/""65.4g""# = 0.13mol

+

In the reaction, we can see HCl doubles the n°mol of Zn, so...
+#2(0.13)= 0.26mol#

+

#M=""n°mol""/""Volume""#

+

#3=""0.26mol""/V#

+

#V=0.087L# of 3M HCl

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+
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+
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+ + +
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+
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" How many milliliters of 3.00 M #HCI(aq)# are required to react with 8.55 g of #Zn(s)#? nan +16 a827efa0-6ddd-11ea-9dd2-ccda262736ce https://socratic.org/questions/how-would-you-use-the-henderson-hasselbalch-equation-to-calculate-the-ph-of-a-so 7.65 start physical_unit 13 13 ph none qc_end physical_unit 19 19 16 17 molarity qc_end physical_unit 24 24 21 22 molarity qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] solution""}]" "[{""type"":""physical unit"",""value"":""7.65""}]" "[{""type"": ""physical unit"",""value"": ""Molarity [OF] HClO [=] \\pu{0.120 M}""},{""type"": ""physical unit"",""value"": ""Molarity [OF] KClO [=] \\pu{0.185 M}""},{""type"": ""other"",""value"": ""Henderson-Hasselbalch equation""}]" "

How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO?

" nan 7.65 "
+

Explanation:

+
+

The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the #pK_a# of the weak acid.

+
+

#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#

+
+

In your case, the weak acid is hypochlorous acid, #""HClO""#. Its conjugate base, the hypochlorite anion, #""ClO""^(-)#, is delivered to the solution by one of its salts, potassium hypochlorite, #""KClO""#.

+

The acid dissociation constant, #K_a#, for hypochlorous acid is equal to #3.5 * 10^(-8)#

+

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

+

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's #pK_a#.

+

Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since

+
+

#log(1) = 0#

+
+

This tells you that if you have more conjugate base than weak acid, the log term will be greater than #1#, which will cause the pH to be higher than the #pK_a#.

+

With this in mind, plug in your values into the H-H equation to get

+
+

#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#

+

#""pH"" = 7.46 + 0.188 = color(green)(7.65)#

+
+

+ +

+
+
" "
+
+
+

#""pH"" = 7.65#

+
+
+
+

Explanation:

+
+

The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the #pK_a# of the weak acid.

+
+

#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#

+
+

In your case, the weak acid is hypochlorous acid, #""HClO""#. Its conjugate base, the hypochlorite anion, #""ClO""^(-)#, is delivered to the solution by one of its salts, potassium hypochlorite, #""KClO""#.

+

The acid dissociation constant, #K_a#, for hypochlorous acid is equal to #3.5 * 10^(-8)#

+

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

+

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's #pK_a#.

+

Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since

+
+

#log(1) = 0#

+
+

This tells you that if you have more conjugate base than weak acid, the log term will be greater than #1#, which will cause the pH to be higher than the #pK_a#.

+

With this in mind, plug in your values into the H-H equation to get

+
+

#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#

+

#""pH"" = 7.46 + 0.188 = color(green)(7.65)#

+
+

+ +

+
+
+
" "
+

How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Calculations + + +
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+1 Answer +
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+ + Nov 22, 2015 + +
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+

#""pH"" = 7.65#

+
+
+
+

Explanation:

+
+

The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the #pK_a# of the weak acid.

+
+

#color(blue)(""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""])))#

+
+

In your case, the weak acid is hypochlorous acid, #""HClO""#. Its conjugate base, the hypochlorite anion, #""ClO""^(-)#, is delivered to the solution by one of its salts, potassium hypochlorite, #""KClO""#.

+

The acid dissociation constant, #K_a#, for hypochlorous acid is equal to #3.5 * 10^(-8)#

+

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

+

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's #pK_a#.

+

Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since

+
+

#log(1) = 0#

+
+

This tells you that if you have more conjugate base than weak acid, the log term will be greater than #1#, which will cause the pH to be higher than the #pK_a#.

+

With this in mind, plug in your values into the H-H equation to get

+
+

#""pH"" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)(""M""))))/(0.120color(red)(cancel(color(black)(""M"")))))#

+

#""pH"" = 7.46 + 0.188 = color(green)(7.65)#

+
+

+ +

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+
+
+
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+ + Creative Commons License + +
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" How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO? nan +17 a82816ac-6ddd-11ea-a869-ccda262736ce https://socratic.org/questions/how-many-moles-are-in-20-0-g-he 5.00 moles start physical_unit 7 7 mole mol qc_end physical_unit 7 7 5 6 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] He [IN] moles""}]" "[{""type"":""physical unit"",""value"":""5.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] He [=] \\pu{20.0 g}""}]" "

How many moles are in 20.0 g He?

" nan 5.00 moles "
+

Explanation:

+
+

In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.

+

The molar mass of helium is #""4.0026 g/mol""# (atomic mass from the periodic table in g/mol).

+

#20.0cancel""g He""xx(1""mol He"")/(4.0026cancel""g He"")=""5.00 mol He""# rounded to three significant figures

+
+
" "
+
+
+

There are 5.00 moles He in 20.0 g He.

+
+
+
+

Explanation:

+
+

In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.

+

The molar mass of helium is #""4.0026 g/mol""# (atomic mass from the periodic table in g/mol).

+

#20.0cancel""g He""xx(1""mol He"")/(4.0026cancel""g He"")=""5.00 mol He""# rounded to three significant figures

+
+
+
" "
+

How many moles are in 20.0 g He?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+
+1 Answer +
+
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+ +
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+ +
+ + Apr 20, 2016 + +
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There are 5.00 moles He in 20.0 g He.

+
+
+
+

Explanation:

+
+

In order to determine the number of moles of a given mass of a substance, divide the given mass by its molar mass.

+

The molar mass of helium is #""4.0026 g/mol""# (atomic mass from the periodic table in g/mol).

+

#20.0cancel""g He""xx(1""mol He"")/(4.0026cancel""g He"")=""5.00 mol He""# rounded to three significant figures

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+
+
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" How many moles are in 20.0 g He? nan +18 a82816ad-6ddd-11ea-b546-ccda262736ce https://socratic.org/questions/how-would-you-write-a-dissociation-reaction-equation-for-kno-3 KNO3 -> K^+ + NO3^- start chemical_equation qc_end chemical_equation 9 9 qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""KNO3 -> K^+ + NO3^-""}]" "[{""type"": ""chemical equation"",""value"": ""KNO3""},{""type"": ""other"",""value"": ""dissociation reaction""}]" "

How would you write a dissociation reaction equation for #KNO_3#?

" nan KNO3 -> K^+ + NO3^- "
+

Explanation:

+
+

Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.

+
+
" "
+
+
+

#KNO_3 rarr K^(+) + NO_3^(-)#

+
+
+
+

Explanation:

+
+

Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.

+
+
+
" "
+

How would you write a dissociation reaction equation for #KNO_3#?

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+
+ + +Chemistry + + + + + +Solutions + + + + + +Solvation and Dissociation + + +
+
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+
+1 Answer +
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+ + Jul 9, 2016 + +
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#KNO_3 rarr K^(+) + NO_3^(-)#

+
+
+
+

Explanation:

+
+

Nitrates are generally soluble in water. In the reaction above both MASS and CHARGE are conserved, as they must be for any chemical reaction. Most of the time, alkali metal cations are along for the ride; the business end of the molecule is the nitrate anion, which can be a potent oxidant.

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+
+
+
+
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+ + +
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+
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" How would you write a dissociation reaction equation for #KNO_3#? nan +19 a82816ae-6ddd-11ea-aad6-ccda262736ce https://socratic.org/questions/if-2-112-g-of-fe-2o-3-reacts-with-0-687-g-of-al-how-much-pure-fe-will-be-produce 1.42 g start physical_unit 14 14 mass g qc_end physical_unit 4 4 1 2 mass qc_end chemical_equation 21 23 qc_end physical_unit 10 10 7 8 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] Fe [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.42 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] Fe2O3 [=] \\pu{2.112 g}""},{""type"":""chemical equation"",""value"":""Fe2O3(s) + 2Al(s) -> Al2O3(s) + 2Fe(s)""},{""type"":""physical unit"",""value"":""Mass [OF] Al [=] \\pu{0.687 g}""}]" "

If 2.112 g of #Fe_2O_3# reacts with 0.687 g of #Al#, how much pure #Fe# will be produced in the reaction #Fe_2O_3(s) + 2Al(s) -> Al_2O_3(s) + 2Fe(s)#?

" nan 1.42 g "
+

Explanation:

+
+

#""Moles of ferric oxide""=(2.112*g)/(159.69*g*mol^-1)# #=# #0.0132*mol#

+

#""Moles of aluminum""=(0.687*g)/(26.98*g*mol^-1)# #=# #0.0254*mol#

+

Aluminum is in slight deficiency, and is thus the limiting reagent. So #0.0254*molxx55.85*g*mol^-1=1.42*g#.

+

I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.

+

See here

+
+
" "
+
+
+

Approx. #1.5*g# steel will be produced.

+
+
+
+

Explanation:

+
+

#""Moles of ferric oxide""=(2.112*g)/(159.69*g*mol^-1)# #=# #0.0132*mol#

+

#""Moles of aluminum""=(0.687*g)/(26.98*g*mol^-1)# #=# #0.0254*mol#

+

Aluminum is in slight deficiency, and is thus the limiting reagent. So #0.0254*molxx55.85*g*mol^-1=1.42*g#.

+

I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.

+

See here

+
+
+
" "
+

If 2.112 g of #Fe_2O_3# reacts with 0.687 g of #Al#, how much pure #Fe# will be produced in the reaction #Fe_2O_3(s) + 2Al(s) -> Al_2O_3(s) + 2Fe(s)#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
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+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 8, 2016 + +
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+
+

Approx. #1.5*g# steel will be produced.

+
+
+
+

Explanation:

+
+

#""Moles of ferric oxide""=(2.112*g)/(159.69*g*mol^-1)# #=# #0.0132*mol#

+

#""Moles of aluminum""=(0.687*g)/(26.98*g*mol^-1)# #=# #0.0254*mol#

+

Aluminum is in slight deficiency, and is thus the limiting reagent. So #0.0254*molxx55.85*g*mol^-1=1.42*g#.

+

I should add that there is a very practical application to this reaction (which I think is still used). When setters lay train tracks, obviously they lay steel tracks over wooden or concrete sleepers. In order to join the rails between lengths they use precisely this reaction (of course they have to heat it up to get it to go!). Such welding gives a very good seal and a smooth ride.

+

See here

+
+
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+
+
+
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+
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+ + Creative Commons License + +
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" If 2.112 g of #Fe_2O_3# reacts with 0.687 g of #Al#, how much pure #Fe# will be produced in the reaction #Fe_2O_3(s) + 2Al(s) -> Al_2O_3(s) + 2Fe(s)#? nan +20 a8282440-6ddd-11ea-ba1a-ccda262736ce https://socratic.org/questions/how-heavy-will-6-14-10-25-atoms-of-gold-be 20.10 kg start physical_unit 8 8 mass kg qc_end physical_unit 6 8 3 5 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] gold [IN] kg""}]" "[{""type"":""physical unit"",""value"":""20.10 kg""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of gold [=] \\pu{6.14 ⋅ 10^25}""}]" "

How heavy will #6.14 * 10^25# atoms of gold be?

" nan 20.10 kg "
+

Explanation:

+
+

This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (#197g#/#""mol""#) to convert moles to grams.

+

I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using #F=ma# where #F# is the force is the weight in Newton, #m# is the mass, and #a# is the acceleration is gravity at Earth's surface (#9.8m#/#s^2#)

+
+
" "
+
+
+

#20.1# #kg#
+weight of #197# #N#

+
+
+
+

Explanation:

+
+

This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (#197g#/#""mol""#) to convert moles to grams.

+

I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using #F=ma# where #F# is the force is the weight in Newton, #m# is the mass, and #a# is the acceleration is gravity at Earth's surface (#9.8m#/#s^2#)

+
+
+
" "
+

How heavy will #6.14 * 10^25# atoms of gold be?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
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+
+1 Answer +
+
+
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+ + +
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+ + Dec 7, 2015 + +
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#20.1# #kg#
+weight of #197# #N#

+
+
+
+

Explanation:

+
+

This question requires you to use Avogadro's number to convert the atoms to moles, and then use the molar mass of gold (#197g#/#""mol""#) to convert moles to grams.

+

I'm assuming by ""how heavy"" you mean mass, but if you wanted to calculate weight you would need to calculate a force by using #F=ma# where #F# is the force is the weight in Newton, #m# is the mass, and #a# is the acceleration is gravity at Earth's surface (#9.8m#/#s^2#)

+
+
+
+
+
+ +
+
+
+
+
+
+
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Impact of this question
+
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" How heavy will #6.14 * 10^25# atoms of gold be? nan +21 a8282441-6ddd-11ea-9c5e-ccda262736ce https://socratic.org/questions/56bbaff77c01492cc7431178 Zn + 2HCl -> ZnCl2 + H2 start chemical_equation qc_end substance 6 6 qc_end substance 8 9 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""Zn + 2HCl -> ZnCl2 + H2""}]" "[{""type"":""substance name"",""value"":""zinc""},{""type"":""substance name"",""value"":""hydrochloric acid""}]" "

What is the chemical equation for zinc + hydrochloric acid?

" nan Zn + 2HCl -> ZnCl2 + H2 "
+

Explanation:

+
+

Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.

+
+
" "
+
+
+

#Zn+2HCl=> ZnCl_2 +H_2#

+
+
+
+

Explanation:

+
+

Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.

+
+
+
" "
+

What is the chemical equation for zinc + hydrochloric acid?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
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+ +
+
+ +
+ + Feb 10, 2016 + +
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#Zn+2HCl=> ZnCl_2 +H_2#

+
+
+
+

Explanation:

+
+

Zinc reacts with hydrochloric acid to produce zinc chloride and hidrogen.

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+
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Related questions
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+
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+
+
" What is the chemical equation for zinc + hydrochloric acid? nan +22 a8282442-6ddd-11ea-b77d-ccda262736ce https://socratic.org/questions/in-the-reaction-n-2-g-3h-2-g-2nh-3-g-n-2-10-00-m-h-2-8-00-m-and-nh-3-2-00-m-what 7.81 ⋅ 10^(−4) start physical_unit 17 18 equilibrium_constant_k none qc_end physical_unit 3 3 6 6 molarity qc_end physical_unit 7 7 7 7 molarity qc_end physical_unit 9 9 9 9 molarity qc_end chemical_equation 3 5 qc_end end "[{""type"": ""physical unit"",""value"": ""equilibrium constant K [OF] this reaction""}]" "[{""type"":""physical unit"",""value"":""7.81 ⋅ 10^(−4)""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] N2 [=] \\pu{10.00 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] H2 [=] \\pu{8.00 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] NH3 [=] \\pu{2.00 M}""},{""type"":""chemical equation"",""value"":""N2(g) + 3H2(g) -> 2NH3(g)""}]" "

In the reaction #N_2(g) + 3H_2(g) -> 2NH_3(g)#, #N_2 = 10.00 M#, #H_2 = 8.00 M#, and #NH_3 = 2.00 M#. What is the equilibrium constant #K#, for this reaction?

" nan 7.81 ⋅ 10^(−4) "
+

Explanation:

+
+

Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.

+

Notice that the concentrations of the two reactants, nitrogen gas, #""N""_2#, and hydrogen gas, #""H""_2#, are hogher than the concentration of the product, ammonia, #""NH""_3#.

+

This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant, #K_c#, to be smaller than one.

+

For your given reaction

+
+

#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#

+
+

the equilibrium constant takes the form

+
+

#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#

+
+

Plug in your values to get

+
+

#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#

+

#K_c = color(green)(7.81 * 10^(-4))#

+
+

The answer is rounded to two sig figs.

+
+
" "
+
+
+

#7.81 * 10^(-4)#

+
+
+
+

Explanation:

+
+

Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.

+

Notice that the concentrations of the two reactants, nitrogen gas, #""N""_2#, and hydrogen gas, #""H""_2#, are hogher than the concentration of the product, ammonia, #""NH""_3#.

+

This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant, #K_c#, to be smaller than one.

+

For your given reaction

+
+

#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#

+
+

the equilibrium constant takes the form

+
+

#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#

+
+

Plug in your values to get

+
+

#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#

+

#K_c = color(green)(7.81 * 10^(-4))#

+
+

The answer is rounded to two sig figs.

+
+
+
" "
+

In the reaction #N_2(g) + 3H_2(g) -> 2NH_3(g)#, #N_2 = 10.00 M#, #H_2 = 8.00 M#, and #NH_3 = 2.00 M#. What is the equilibrium constant #K#, for this reaction?

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Equilibrium Constants + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 23, 2015 + +
+
+
+
+
+
+
+

#7.81 * 10^(-4)#

+
+
+
+

Explanation:

+
+

Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.

+

Notice that the concentrations of the two reactants, nitrogen gas, #""N""_2#, and hydrogen gas, #""H""_2#, are hogher than the concentration of the product, ammonia, #""NH""_3#.

+

This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant, #K_c#, to be smaller than one.

+

For your given reaction

+
+

#""N""_text(2(g]) + color(red)(3)""H""_text(2(g]) rightleftharpoons color(blue)(2)""NH""_text(3(g])#

+
+

the equilibrium constant takes the form

+
+

#K_c = ( [""NH""_3]^color(blue)(2))/([""N""_2] * [""H""_2]^color(red)(3))#

+
+

Plug in your values to get

+
+

#K_c = (2.00)^color(blue)(2)/(10.00 * 8.00^color(red)(3)) = 4.00/(10.00 * 512)#

+

#K_c = color(green)(7.81 * 10^(-4))#

+
+

The answer is rounded to two sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 3562 views + around the world +
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+ + Creative Commons License + +
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+
" In the reaction #N_2(g) + 3H_2(g) -> 2NH_3(g)#, #N_2 = 10.00 M#, #H_2 = 8.00 M#, and #NH_3 = 2.00 M#. What is the equilibrium constant #K#, for this reaction? nan +23 a8282443-6ddd-11ea-b7ce-ccda262736ce https://socratic.org/questions/59428f6eb72cff475e6a5aef 0.26 mol/L start physical_unit 20 25 molality mol/l qc_end physical_unit 6 6 2 3 mass qc_end physical_unit 13 13 10 11 mass qc_end end "[{""type"": ""physical unit"", ""value"": ""Molality [OF] solution with respect to calcium chloride [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.26 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] CaCl2 [=] \\pu{20 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{700 g}""}]" "

A #20*g# mass of #CaCl_2# was dissolved in #700*g# of water. What is the molality of the solution with respect to #""calcium chloride""#?

" nan 0.26 mol/L "
+

Explanation:

+
+

Chemical formula of Calcium chloride= #CaCl_2#
+Molar mass of #CaCl_2# =#110.98# #gmol^-1#
+Mass of Calcium chloride (given) = #20g#

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+Now, to calculate no. of moles:
+No. of moles = #(""mass"")/""molar mass""#

+

No. of moles= #(""20g"")/""110.98g/mol""# = #0.180# moles
+#g# of solvent = # 700 g #
+Converting #g# #-># #kg#
+#kg# of solvent = #0.7kg#
+Water is solvent while #CaCl_2# is solute
+Now,

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+#Molality# = #(""0.180 moles"")/""0.7kg""#
+#Molality# = #0.25# # molal#

+
+
" "
+
+
+

#0.25# #molal#

+
+
+
+

Explanation:

+
+

Chemical formula of Calcium chloride= #CaCl_2#
+Molar mass of #CaCl_2# =#110.98# #gmol^-1#
+Mass of Calcium chloride (given) = #20g#

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+Now, to calculate no. of moles:
+No. of moles = #(""mass"")/""molar mass""#

+

No. of moles= #(""20g"")/""110.98g/mol""# = #0.180# moles
+#g# of solvent = # 700 g #
+Converting #g# #-># #kg#
+#kg# of solvent = #0.7kg#
+Water is solvent while #CaCl_2# is solute
+Now,

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+#Molality# = #(""0.180 moles"")/""0.7kg""#
+#Molality# = #0.25# # molal#

+
+
+
" "
+

A #20*g# mass of #CaCl_2# was dissolved in #700*g# of water. What is the molality of the solution with respect to #""calcium chloride""#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Jun 15, 2017 + +
+
+
+
+
+
+
+

#0.25# #molal#

+
+
+
+

Explanation:

+
+

Chemical formula of Calcium chloride= #CaCl_2#
+Molar mass of #CaCl_2# =#110.98# #gmol^-1#
+Mass of Calcium chloride (given) = #20g#

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+Now, to calculate no. of moles:
+No. of moles = #(""mass"")/""molar mass""#

+

No. of moles= #(""20g"")/""110.98g/mol""# = #0.180# moles
+#g# of solvent = # 700 g #
+Converting #g# #-># #kg#
+#kg# of solvent = #0.7kg#
+Water is solvent while #CaCl_2# is solute
+Now,

+

#Molality# = #m# = #(""moles of solute"")/""kg of solvent""#
+#Molality# = #(""0.180 moles"")/""0.7kg""#
+#Molality# = #0.25# # molal#

+
+
+
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+ +
+
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+ +
+
+ +
+ + Jun 15, 2017 + +
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+
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+
+

#""Concentration""-=0.257*""molal""#.........

+
+
+
+

Explanation:

+
+

#""Molality""# #-=# #""Moles of solute""/""Kilograms of solvent""#

+

And thus..............................................

+

#""molality""=((20*g)/(110.98*g*mol^-1))/(700*gxx10^-3*kg*g^-1)=0.257*mol*kg#.

+

Note that at (relatively!) low concentrations, the calculated #""molality""# would be almost the same as solution #""molarity""#. #""Molality""# is used because the expression is fairly independent of temperature. What is the #""molal concentration""# with respect to #""chloride ion""#?

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+
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+
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+ + Creative Commons License + +
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+
" "A #20*g# mass of #CaCl_2# was dissolved in #700*g# of water. What is the molality of the solution with respect to #""calcium chloride""#?" nan +24 a8282444-6ddd-11ea-8e0f-ccda262736ce https://socratic.org/questions/a-sample-of-nitrogen-dioxide-gas-occupies-625-cm-3-at-70-0-c-and-15-0-psi-what-i 25.85 °C start physical_unit 3 5 temperature °c qc_end physical_unit 3 5 7 8 volume qc_end physical_unit 3 5 10 11 temperature qc_end physical_unit 3 5 13 14 pressure qc_end physical_unit 3 5 25 26 volume qc_end physical_unit 3 5 13 14 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] nitrogen dioxide gas [IN] °C""}]" "[{""type"":""physical unit"",""value"":""25.85 °C""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] nitrogen dioxide gas [=] \\pu{625 cm^3}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] nitrogen dioxide gas [=] \\pu{70.0 °C}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] nitrogen dioxide gas [=] \\pu{15.0 psi}""},{""type"":""physical unit"",""value"":""Volume2 [OF] nitrogen dioxide gas [=] \\pu{545 cm^3}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] nitrogen dioxide gas [=] \\pu{15.0 psi}""}]" "

A sample of nitrogen dioxide gas occupies #625 cm^3# at 70.0 °C and 15.0 psi. What is the final Celsius temperature if the volume is #545 cm^3# at 15.0 psi?

" nan 25.85 °C "
+

Explanation:

+
+

Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.

+

The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.

+

Equation

+

#V_1/T_1=V_2/T_2#, where #V# is volume in #""cm""^3""#, and #T# is temperature in Kelvins

+

Known
+#V_1=""625 cm""^3""#
+#T_1=""70.0""^@""C""+273.15=""343.15 K""#
+#V_2=""545 cm""^3""#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#. Substitute the known values into the equation and solve.

+

#T_2=(V_2T_1)/V_1#

+

#T_2=(545""cm""^3xx343.15""K"")/(625""cm""^3)=""299 K""#

+

Final temperature in degrees Celsius

+

#""299 K""-""273.15""=""25.9""^@""C""# rounded to three significant figures

+
+
" "
+
+
+

The final temperature in degrees Celsius is #""25.9""^@""C""#.

+
+
+
+

Explanation:

+
+

Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.

+

The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.

+

Equation

+

#V_1/T_1=V_2/T_2#, where #V# is volume in #""cm""^3""#, and #T# is temperature in Kelvins

+

Known
+#V_1=""625 cm""^3""#
+#T_1=""70.0""^@""C""+273.15=""343.15 K""#
+#V_2=""545 cm""^3""#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#. Substitute the known values into the equation and solve.

+

#T_2=(V_2T_1)/V_1#

+

#T_2=(545""cm""^3xx343.15""K"")/(625""cm""^3)=""299 K""#

+

Final temperature in degrees Celsius

+

#""299 K""-""273.15""=""25.9""^@""C""# rounded to three significant figures

+
+
+
" "
+

A sample of nitrogen dioxide gas occupies #625 cm^3# at 70.0 °C and 15.0 psi. What is the final Celsius temperature if the volume is #545 cm^3# at 15.0 psi?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Measuring Gas Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 27, 2016 + +
+
+
+
+
+
+
+

The final temperature in degrees Celsius is #""25.9""^@""C""#.

+
+
+
+

Explanation:

+
+

Since the pressure and amount of gas is held constant, you can use Charles' law, the volume-temperature law, which states that at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature. This means that as the volume increases, the temperature increases, and vice versa.

+

The initial calculation will require the conversion of the Celsius temperature to Kelvins. After the calculation is complete, the new temperature will be converted from Kelvins back to degrees Celsius.

+

Equation

+

#V_1/T_1=V_2/T_2#, where #V# is volume in #""cm""^3""#, and #T# is temperature in Kelvins

+

Known
+#V_1=""625 cm""^3""#
+#T_1=""70.0""^@""C""+273.15=""343.15 K""#
+#V_2=""545 cm""^3""#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#. Substitute the known values into the equation and solve.

+

#T_2=(V_2T_1)/V_1#

+

#T_2=(545""cm""^3xx343.15""K"")/(625""cm""^3)=""299 K""#

+

Final temperature in degrees Celsius

+

#""299 K""-""273.15""=""25.9""^@""C""# rounded to three significant figures

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" A sample of nitrogen dioxide gas occupies #625 cm^3# at 70.0 °C and 15.0 psi. What is the final Celsius temperature if the volume is #545 cm^3# at 15.0 psi? nan +25 a8282445-6ddd-11ea-9875-ccda262736ce https://socratic.org/questions/what-is-the-formula-of-potassium-phosphate-in-grams-mole 212.27 grams/mole start physical_unit 5 6 molar_mass g/mol qc_end substance 5 6 qc_end end "[{""type"":""physical unit"",""value"":""Molecular [OF] potassium phosphate [IN] grams/mole""}]" "[{""type"":""physical unit"",""value"":""212.27 grams/mole""}]" "[{""type"":""substance name"",""value"":""potassium phosphate""}]" "

What is the formula of potassium phosphate in grams/mole?

" nan 212.27 grams/mole "
+

Explanation:

+
+

The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.

+

#3(39.0983) + 30.9738 + 4(15.999) = 212.2663 g mol^-1#

+
+
" "
+
+
+

The formula of potassium phosphate is #K_3PO_4# .
+Its relative formula mass is #212.2663 g mol^-1# .

+
+
+
+

Explanation:

+
+

The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.

+

#3(39.0983) + 30.9738 + 4(15.999) = 212.2663 g mol^-1#

+
+
+
" "
+

What is the formula of potassium phosphate in grams/mole?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 27, 2015 + +
+
+
+
+
+
+
+

The formula of potassium phosphate is #K_3PO_4# .
+Its relative formula mass is #212.2663 g mol^-1# .

+
+
+
+

Explanation:

+
+

The formula mass - which I believe is what you were looking for in the first place - can be found by adding up the relative molecular masses of each species in the formula unit.

+

#3(39.0983) + 30.9738 + 4(15.999) = 212.2663 g mol^-1#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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+
" What is the formula of potassium phosphate in grams/mole? nan +26 a8282446-6ddd-11ea-ba3b-ccda262736ce https://socratic.org/questions/how-do-you-balance-this-equation-fe-2-c-2o-4-3-fec-2o-4-co-2 2Fe^3+ + 3C2O4^2- -> 2Fe(C2O4) + 2CO2(g) start chemical_equation qc_end chemical_equation 6 10 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2Fe^3+ + 3C2O4^2- -> 2Fe(C2O4) + 2CO2(g)""}]" "[{""type"":""chemical equation"",""value"":""Fe2(C2O4)3 -> FeC2O4 + CO2""}]" "

How do you balance this equation: #Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2#?

" nan 2Fe^3+ + 3C2O4^2- -> 2Fe(C2O4) + 2CO2(g) "
+

Explanation:

+
+

So we approach the equation by the method of half equations:

+

#Fe(III)# is reduced to #Fe(II)#:

+

#Fe^(3+) + e^(-)rarrFe^(2+)# #(i)#

+

And #""oxalate ion""# is oxidized to #CO_2#:

+

#C_2O_4^(2-) rarr 2CO_2 + 2e^-# #(ii)#

+

Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.

+

And we add the individual half reactions together so as to remove the electrons: i.e. #2xx(i) + (ii):#

+

#2Fe^(3+) + cancel(2e^(-)) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)+ cancel(2e^-)#

+

To give (almost finally):

+

#2Fe^(3+) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)#

+

And we can add TWO #""oxalato ligands""# to EACH side:

+

#2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)#

+

Are mass and charge balanced? If they don't, what must you do?

+
+
" "
+
+
+

This is a formal redox reaction...........

+

#2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)#

+
+
+
+

Explanation:

+
+

So we approach the equation by the method of half equations:

+

#Fe(III)# is reduced to #Fe(II)#:

+

#Fe^(3+) + e^(-)rarrFe^(2+)# #(i)#

+

And #""oxalate ion""# is oxidized to #CO_2#:

+

#C_2O_4^(2-) rarr 2CO_2 + 2e^-# #(ii)#

+

Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.

+

And we add the individual half reactions together so as to remove the electrons: i.e. #2xx(i) + (ii):#

+

#2Fe^(3+) + cancel(2e^(-)) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)+ cancel(2e^-)#

+

To give (almost finally):

+

#2Fe^(3+) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)#

+

And we can add TWO #""oxalato ligands""# to EACH side:

+

#2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)#

+

Are mass and charge balanced? If they don't, what must you do?

+
+
+
" "
+

How do you balance this equation: #Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
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+ + Mar 16, 2017 + +
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+
+
+

This is a formal redox reaction...........

+

#2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)#

+
+
+
+

Explanation:

+
+

So we approach the equation by the method of half equations:

+

#Fe(III)# is reduced to #Fe(II)#:

+

#Fe^(3+) + e^(-)rarrFe^(2+)# #(i)#

+

And #""oxalate ion""# is oxidized to #CO_2#:

+

#C_2O_4^(2-) rarr 2CO_2 + 2e^-# #(ii)#

+

Both half equations conserve mass and conserve charge, as they must if they are to reflect chemical reality.

+

And we add the individual half reactions together so as to remove the electrons: i.e. #2xx(i) + (ii):#

+

#2Fe^(3+) + cancel(2e^(-)) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)+ cancel(2e^-)#

+

To give (almost finally):

+

#2Fe^(3+) +C_2O_4^(2-)rarr2Fe^(2+) +2CO_2(g)#

+

And we can add TWO #""oxalato ligands""# to EACH side:

+

#2Fe^(3+) +3C_2O_4^(2-)rarr2Fe(C_2O_4) +2CO_2(g)#

+

Are mass and charge balanced? If they don't, what must you do?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 1160 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How do you balance this equation: #Fe_2(C_2O_4)_3 -> FeC_2O_4 + CO_2#? nan +27 a8283db0-6ddd-11ea-a7a8-ccda262736ce https://socratic.org/questions/57df202cb72cff627591e2f9 8.97 L start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mass qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] fluorine gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""8.97 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] fluorine gas [=] \\pu{15.0 g}""},{""type"":""other"",""value"":""STP""}]" "

What is the volume of #""15.0 g""# of fluorine gas #(""F""_2"")# at STP?

" nan 8.97 L "
+

Explanation:

+
+

First determine the molar mass of fluorine gas #(""F""_2"")#, then determine the moles of fluorine gas. The molar mass of fluorine is its atomic mass from the periodic table in g/mol.

+

Molar Mass of #""F""_2""#
+Multiply the molar mass of #""F""# times the subscript 2 in the formula for fluorine gas, #""F""_2""#.

+

#M(""F""_2"")##=##(2 xx ""18.998 g/mol"")=""37.996 g/mol""#

+

Moles #""F""_2""#
+Divide the given mass by the molar mass.

+

#""mol F""_2""##=##(15.0cancel""g"")/(37.996cancel""g""/""mol"")=""0.39478 mol F""_2""#

+

I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.

+

Use the formula for the ideal gas law to determine the volume of the fluorine gas.

+

+

#""Current STP""##=##""0""^@""C""# or #""273.15 K""# (for gases) and #""100 kPa""#

+

#R=""8.3144598 L kPa K""^(-1) ""mol""^(-1)#
+https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962

+

Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.

+

#V=(nRT)/P#

+

#V=((0.39478cancel""mol"")xx(8.3144598 ""L"" cancel""kPa"" cancel ""K""^(-1) cancel""mol""^(-1))xx(273.15cancel""K""))/(100""kPa"")=""8.97 L""# (rounded to three significant figures)

+

If your teacher still uses #""STP""##=##""0""^@""C""# or #""273.15 K""# and #""1 atm""#, switch #""0.082057338 L atm K""^(-1) ""mol""^(-1)""# for the gas constant #R#, and #""1 atm""# for pressure #P# in the equation. The volume will be #""8.58 L""#.

+
+
" "
+
+
+

The volume of #""15.0 g F""_2""# at STP is #""8.97 L""#.

+
+
+
+

Explanation:

+
+

First determine the molar mass of fluorine gas #(""F""_2"")#, then determine the moles of fluorine gas. The molar mass of fluorine is its atomic mass from the periodic table in g/mol.

+

Molar Mass of #""F""_2""#
+Multiply the molar mass of #""F""# times the subscript 2 in the formula for fluorine gas, #""F""_2""#.

+

#M(""F""_2"")##=##(2 xx ""18.998 g/mol"")=""37.996 g/mol""#

+

Moles #""F""_2""#
+Divide the given mass by the molar mass.

+

#""mol F""_2""##=##(15.0cancel""g"")/(37.996cancel""g""/""mol"")=""0.39478 mol F""_2""#

+

I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.

+

Use the formula for the ideal gas law to determine the volume of the fluorine gas.

+

+

#""Current STP""##=##""0""^@""C""# or #""273.15 K""# (for gases) and #""100 kPa""#

+

#R=""8.3144598 L kPa K""^(-1) ""mol""^(-1)#
+https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962

+

Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.

+

#V=(nRT)/P#

+

#V=((0.39478cancel""mol"")xx(8.3144598 ""L"" cancel""kPa"" cancel ""K""^(-1) cancel""mol""^(-1))xx(273.15cancel""K""))/(100""kPa"")=""8.97 L""# (rounded to three significant figures)

+

If your teacher still uses #""STP""##=##""0""^@""C""# or #""273.15 K""# and #""1 atm""#, switch #""0.082057338 L atm K""^(-1) ""mol""^(-1)""# for the gas constant #R#, and #""1 atm""# for pressure #P# in the equation. The volume will be #""8.58 L""#.

+
+
+
" "
+

What is the volume of #""15.0 g""# of fluorine gas #(""F""_2"")# at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 25, 2016 + +
+
+
+
+
+
+
+

The volume of #""15.0 g F""_2""# at STP is #""8.97 L""#.

+
+
+
+

Explanation:

+
+

First determine the molar mass of fluorine gas #(""F""_2"")#, then determine the moles of fluorine gas. The molar mass of fluorine is its atomic mass from the periodic table in g/mol.

+

Molar Mass of #""F""_2""#
+Multiply the molar mass of #""F""# times the subscript 2 in the formula for fluorine gas, #""F""_2""#.

+

#M(""F""_2"")##=##(2 xx ""18.998 g/mol"")=""37.996 g/mol""#

+

Moles #""F""_2""#
+Divide the given mass by the molar mass.

+

#""mol F""_2""##=##(15.0cancel""g"")/(37.996cancel""g""/""mol"")=""0.39478 mol F""_2""#

+

I am keeping a couple of guard digits to reduce rounding errors. The final answer will be rounded to three significant figures.

+

Use the formula for the ideal gas law to determine the volume of the fluorine gas.

+

+

#""Current STP""##=##""0""^@""C""# or #""273.15 K""# (for gases) and #""100 kPa""#

+

#R=""8.3144598 L kPa K""^(-1) ""mol""^(-1)#
+https://en.wikipedia.org/w/index.php?title=Gas_constant&oldid=729635962

+

Rearrange the formula to isolate volume, substitute the known values into the formula and solve for volume.

+

#V=(nRT)/P#

+

#V=((0.39478cancel""mol"")xx(8.3144598 ""L"" cancel""kPa"" cancel ""K""^(-1) cancel""mol""^(-1))xx(273.15cancel""K""))/(100""kPa"")=""8.97 L""# (rounded to three significant figures)

+

If your teacher still uses #""STP""##=##""0""^@""C""# or #""273.15 K""# and #""1 atm""#, switch #""0.082057338 L atm K""^(-1) ""mol""^(-1)""# for the gas constant #R#, and #""1 atm""# for pressure #P# in the equation. The volume will be #""8.58 L""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 10165 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "What is the volume of #""15.0 g""# of fluorine gas #(""F""_2"")# at STP?" nan +28 a8283db1-6ddd-11ea-baf2-ccda262736ce https://socratic.org/questions/how-many-grams-of-potassium-sulfite-are-required-to-dissolve-in-872g-of-water-to 17.7 grams start physical_unit 4 5 mass g qc_end physical_unit 13 13 11 11 mass qc_end physical_unit 4 5 17 17 molality qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] potassium sulfite [IN] grams""}]" "[{""type"":""physical unit"",""value"":""17.7 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{872 g}""},{""type"":""physical unit"",""value"":""Molality [OF] potassium sulfite [=] \\pu{0.128 m}""}]" "

How many grams of potassium sulfite are required to dissolve in 872g of water to make a 0.128m solution?

" nan 17.7 grams "
+

Explanation:

+
+

You're looking for the mass of potassium sulfite, #""K""_2""SO""_3""#, needed to make a #""0.128 m""#, or #""0.128 molal""#, solution.

+

Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.

+

This means that in order to find a solution's molality, you need to know

+
+
    +
  • the number of moles of solute
    • +
  • the mass of the solvent expressed in kilograms
    • +
+
+

In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms

+
+

#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#

+
+

Now, a #""0.128 m""# solution contains #0.128# moles of solute per kilogram of solvent. Since your sample contains #""0.872 kg""# of solvent, it follows that it will contain

+
+

#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#

+
+

All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles

+
+

#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""17.7 g""#

+
+
+
+

Explanation:

+
+

You're looking for the mass of potassium sulfite, #""K""_2""SO""_3""#, needed to make a #""0.128 m""#, or #""0.128 molal""#, solution.

+

Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.

+

This means that in order to find a solution's molality, you need to know

+
+
    +
  • the number of moles of solute
    • +
  • the mass of the solvent expressed in kilograms
    • +
+
+

In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms

+
+

#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#

+
+

Now, a #""0.128 m""# solution contains #0.128# moles of solute per kilogram of solvent. Since your sample contains #""0.872 kg""# of solvent, it follows that it will contain

+
+

#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#

+
+

All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles

+
+

#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

How many grams of potassium sulfite are required to dissolve in 872g of water to make a 0.128m solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 1, 2016 + +
+
+
+
+
+
+
+

#""17.7 g""#

+
+
+
+

Explanation:

+
+

You're looking for the mass of potassium sulfite, #""K""_2""SO""_3""#, needed to make a #""0.128 m""#, or #""0.128 molal""#, solution.

+

Now, molality is used to express the concentration of a solution in terms of how many moles of solute it contains per kilogram of solvent.

+

This means that in order to find a solution's molality, you need to know

+
+
    +
  • the number of moles of solute
    • +
  • the mass of the solvent expressed in kilograms
    • +
+
+

In your case, you already know the mass of the solvent in grams, so the very first thing to do here is convert it to kilograms

+
+

#872 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.872 kg""#

+
+

Now, a #""0.128 m""# solution contains #0.128# moles of solute per kilogram of solvent. Since your sample contains #""0.872 kg""# of solvent, it follows that it will contain

+
+

#0.872 color(red)(cancel(color(black)(""kg solvent""))) * overbrace((""0.128 moles K""_2""SO""_3)/(1color(red)(cancel(color(black)(""kg solvent"")))))^(color(blue)(""= 0.128 m"")) = ""0.1116 moles K""_2""SO""_3#

+
+

All you have to do now is use the molar mass of potassium sulfite to figure out how many grams would contain that many moles

+
+

#0.1116 color(red)(cancel(color(black)(""moles K""_2""SO""_3))) * ""158.26 g""/(1color(red)(cancel(color(black)(""mole K""_2""SO""_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""17.7 g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6592 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many grams of potassium sulfite are required to dissolve in 872g of water to make a 0.128m solution? nan +29 a8283db2-6ddd-11ea-965c-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-30-4-nitrogen-and-69-6 NO2 start chemical_formula qc_end physical_unit 11 11 10 10 mass_percent qc_end physical_unit 14 14 13 13 mass_percent qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""NO2""}]" "[{""type"":""physical unit"",""value"":""Mass percent [OF] nitrogen [=] \\pu{30.4%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen [=] \\pu{69.6%}""}]" "

What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass?

" nan NO2 "
+

Explanation:

+
+

AS is typical in these problems, we ASSUME an #100*g# mass of unknown compound, and we divide thru by the atomic mass:

+

#""Moles of nitrogen""=(30.4*g)/(14.01*g*mol^-1)=2.17*mol.#

+

#""Moles of oxygen""=(69.6*g)/(15.999*g*mol^-1)=4.35*mol.#

+

And we divide thru by the SMALLEST molar quantity to get an empirical formula of #NO_2#...........

+
+
" "
+
+
+

#NO_2#..........

+
+
+
+

Explanation:

+
+

AS is typical in these problems, we ASSUME an #100*g# mass of unknown compound, and we divide thru by the atomic mass:

+

#""Moles of nitrogen""=(30.4*g)/(14.01*g*mol^-1)=2.17*mol.#

+

#""Moles of oxygen""=(69.6*g)/(15.999*g*mol^-1)=4.35*mol.#

+

And we divide thru by the SMALLEST molar quantity to get an empirical formula of #NO_2#...........

+
+
+
" "
+

What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 27, 2017 + +
+
+
+
+
+
+
+

#NO_2#..........

+
+
+
+

Explanation:

+
+

AS is typical in these problems, we ASSUME an #100*g# mass of unknown compound, and we divide thru by the atomic mass:

+

#""Moles of nitrogen""=(30.4*g)/(14.01*g*mol^-1)=2.17*mol.#

+

#""Moles of oxygen""=(69.6*g)/(15.999*g*mol^-1)=4.35*mol.#

+

And we divide thru by the SMALLEST molar quantity to get an empirical formula of #NO_2#...........

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 38950 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the empirical formula of a compound that contains 30.4% nitrogen and 69.6% oxygen by mass? nan +30 a8283db3-6ddd-11ea-b6cc-ccda262736ce https://socratic.org/questions/57e722fab72cff3a289a14d1 37.5% start physical_unit 5 10 mass_percent none qc_end substance 9 10 qc_end end "[{""type"": ""physical unit"",""value"": ""Percentage by mass [OF] carbon by mass in citric acid""}]" "[{""type"":""physical unit"",""value"":""37.5%""}]" "[{""type"":""substance name"",""value"":""citric acid""}]" "

What is the percentage of carbon by mass in citric acid?

" nan 37.5% "
+

Explanation:

+
+

I think you have intuitively realized the correct approach.

+

#%""element""# #=# #""mass of element in the compound""/""mass of all elements in the compound""xx100%#

+

The #""mass of all elements in the compound""# is simply equal to the sum of the atomic masses: #(6xx12.011(C)+8xx1.00794(H)+7xx15.999(O))*g*mol^-1# #=# #192.12*g*mol^-1# (the which of course is the molecular mass of the compound).

+

And thus #%C# #=# #(6xx12.011*g)/(192.12*g*mol^-1)xx100%# #=# #37.5%#.

+

And #%H# #=# #??#

+

And #%O# #=# #??#

+

Of course all the individual percentages must sum to 100%. Why?

+

Alternatively see here.

+
+
" "
+
+
+

#%C# #=# #(6xx12.011*g)/(192.12*g*mol^-1)xx100%# #=# #37.5%#.

+
+
+
+

Explanation:

+
+

I think you have intuitively realized the correct approach.

+

#%""element""# #=# #""mass of element in the compound""/""mass of all elements in the compound""xx100%#

+

The #""mass of all elements in the compound""# is simply equal to the sum of the atomic masses: #(6xx12.011(C)+8xx1.00794(H)+7xx15.999(O))*g*mol^-1# #=# #192.12*g*mol^-1# (the which of course is the molecular mass of the compound).

+

And thus #%C# #=# #(6xx12.011*g)/(192.12*g*mol^-1)xx100%# #=# #37.5%#.

+

And #%H# #=# #??#

+

And #%O# #=# #??#

+

Of course all the individual percentages must sum to 100%. Why?

+

Alternatively see here.

+
+
+
" "
+

What is the percentage of carbon by mass in citric acid?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 25, 2016 + +
+
+
+
+
+
+
+

#%C# #=# #(6xx12.011*g)/(192.12*g*mol^-1)xx100%# #=# #37.5%#.

+
+
+
+

Explanation:

+
+

I think you have intuitively realized the correct approach.

+

#%""element""# #=# #""mass of element in the compound""/""mass of all elements in the compound""xx100%#

+

The #""mass of all elements in the compound""# is simply equal to the sum of the atomic masses: #(6xx12.011(C)+8xx1.00794(H)+7xx15.999(O))*g*mol^-1# #=# #192.12*g*mol^-1# (the which of course is the molecular mass of the compound).

+

And thus #%C# #=# #(6xx12.011*g)/(192.12*g*mol^-1)xx100%# #=# #37.5%#.

+

And #%H# #=# #??#

+

And #%O# #=# #??#

+

Of course all the individual percentages must sum to 100%. Why?

+

Alternatively see here.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5011 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the percentage of carbon by mass in citric acid? nan +31 a8283db4-6ddd-11ea-934b-ccda262736ce https://socratic.org/questions/how-many-moles-are-there-in-9-00-grams-of-beryllium 1 moles start physical_unit 9 9 mole mol qc_end physical_unit 9 9 6 7 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] beryllium [IN] moles""}]" "[{""type"":""physical unit"",""value"":""1 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] beryllium [=] \\pu{9.00 grams}""}]" "

How many moles are there in 9.00 grams of beryllium?

" nan 1 moles "
+

Explanation:

+
+

Moles = mass divided by molar mass.

+

#n=m/(Mr)#

+

#=(9g)/(9g//mol)#

+

#=1 mol#

+
+
" "
+
+
+

#1# mole

+
+
+
+

Explanation:

+
+

Moles = mass divided by molar mass.

+

#n=m/(Mr)#

+

#=(9g)/(9g//mol)#

+

#=1 mol#

+
+
+
" "
+

How many moles are there in 9.00 grams of beryllium?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 20, 2015 + +
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+
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+
+
+
+

#1# mole

+
+
+
+

Explanation:

+
+

Moles = mass divided by molar mass.

+

#n=m/(Mr)#

+

#=(9g)/(9g//mol)#

+

#=1 mol#

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+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+
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+ + Creative Commons License + +
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" How many moles are there in 9.00 grams of beryllium? nan +32 a8283db5-6ddd-11ea-8405-ccda262736ce https://socratic.org/questions/how-many-grams-of-pt-are-present-in-25-0-g-of-cisplatin 16.3 grams start physical_unit 4 4 mass g qc_end physical_unit 11 11 8 9 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] Pt [IN] grams""}]" "[{""type"":""physical unit"",""value"":""16.3 grams""}]" "[{""type"": ""physical unit"",""value"": ""Mass [OF] Cisplatin [=] \\pu{25.0 g}""},{""type"": ""other"",""value"": ""Pt(NH3)2Cl2; Molar mass: 300 g mol-1""}]" "

How many grams of #""Pt""# are present in 25.0 g of Cisplatin, an anti-tumor agent?

" "
+
+

+

#Pt(NH_3)_2Cl_2#

+

Molar mass: #"" 300. g mol""^(-1)#

+

+
+
" 16.3 grams "
+

Explanation:

+
+

The idea here is that you need to use the molar mass of cisplatin, #""Pt""(""NH""_3)""Cl""_2#, to determine the compound's percent composition of platinium.

+

As you can see by looking at the compound's chemical formula, one mole of cisplatin contains

+
+
    +
  • one mole of platinum, #1 xx ""Pt""#
    • +
  • two moles of nitrogen, #2 xx ""N""#
    • +
  • six moles of hydrogen, #6 xx ""H""#
    • +
  • two moles of chlorine, #2 xx ""Cl""#
    • +
+
+

You also know that cisplatin has a molar mass of #""300. g mol""^(-1)#, which means that one mole of cisplatin has a mass of #""300 g""#.

+

Platinum has a molar mass of #""195.08 g mol""^(-1)#, which means that one mole of platinum has a mass of #""195.08 g""#.

+

Since #""300. g""# of cisplatin, i.e .one mole of cisplatin, contain #""195.08 g""# of platinum, i.e. one mole of platinum, you can say that #""100 g""# of cisplatin will contain

+
+

#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#

+
+

Since percent composition tells you the mass of a given element per #""100 g""# of compound, you can say that cisplatin is #65.03%# platinum.

+

This means that your #""25.0-g""# sample will contain

+
+

#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin

+
+

#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

+
+
+
" "
+
+
+

#""16.3 g Pt""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the molar mass of cisplatin, #""Pt""(""NH""_3)""Cl""_2#, to determine the compound's percent composition of platinium.

+

As you can see by looking at the compound's chemical formula, one mole of cisplatin contains

+
+
    +
  • one mole of platinum, #1 xx ""Pt""#
    • +
  • two moles of nitrogen, #2 xx ""N""#
    • +
  • six moles of hydrogen, #6 xx ""H""#
    • +
  • two moles of chlorine, #2 xx ""Cl""#
    • +
+
+

You also know that cisplatin has a molar mass of #""300. g mol""^(-1)#, which means that one mole of cisplatin has a mass of #""300 g""#.

+

Platinum has a molar mass of #""195.08 g mol""^(-1)#, which means that one mole of platinum has a mass of #""195.08 g""#.

+

Since #""300. g""# of cisplatin, i.e .one mole of cisplatin, contain #""195.08 g""# of platinum, i.e. one mole of platinum, you can say that #""100 g""# of cisplatin will contain

+
+

#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#

+
+

Since percent composition tells you the mass of a given element per #""100 g""# of compound, you can say that cisplatin is #65.03%# platinum.

+

This means that your #""25.0-g""# sample will contain

+
+

#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin

+
+

#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

+
+
+
+
" "
+

How many grams of #""Pt""# are present in 25.0 g of Cisplatin, an anti-tumor agent?

+
+
+

+

#Pt(NH_3)_2Cl_2#

+

Molar mass: #"" 300. g mol""^(-1)#

+

+
+
+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 3, 2016 + +
+
+
+
+
+
+
+

#""16.3 g Pt""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the molar mass of cisplatin, #""Pt""(""NH""_3)""Cl""_2#, to determine the compound's percent composition of platinium.

+

As you can see by looking at the compound's chemical formula, one mole of cisplatin contains

+
+
    +
  • one mole of platinum, #1 xx ""Pt""#
    • +
  • two moles of nitrogen, #2 xx ""N""#
    • +
  • six moles of hydrogen, #6 xx ""H""#
    • +
  • two moles of chlorine, #2 xx ""Cl""#
    • +
+
+

You also know that cisplatin has a molar mass of #""300. g mol""^(-1)#, which means that one mole of cisplatin has a mass of #""300 g""#.

+

Platinum has a molar mass of #""195.08 g mol""^(-1)#, which means that one mole of platinum has a mass of #""195.08 g""#.

+

Since #""300. g""# of cisplatin, i.e .one mole of cisplatin, contain #""195.08 g""# of platinum, i.e. one mole of platinum, you can say that #""100 g""# of cisplatin will contain

+
+

#100color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = ""65.03 g Pt""#

+
+

Since percent composition tells you the mass of a given element per #""100 g""# of compound, you can say that cisplatin is #65.03%# platinum.

+

This means that your #""25.0-g""# sample will contain

+
+

#25.0color(red)(cancel(color(black)(""g cisplatin""))) * overbrace(""65.03 g Pt""/(100color(red)(cancel(color(black)(""g cisplatin"")))))^(color(purple)(""65.03% Pt"")) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

Notice that you can get the same result without calculating the percent composition of platinum in cisplatin. All you have to do is use the mass of platinum you get in one mole of cisplatin

+
+

#25.0 color(red)(cancel(color(black)(""g cisplatin""))) * ""195.08 g Pt""/(300. color(red)(cancel(color(black)(""g cisplatin"")))) = color(green)(|bar(ul(color(white)(a/a)""16.3 g Pt""color(white)(a/a)|)))#

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+
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+
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+
" "How many grams of #""Pt""# are present in 25.0 g of Cisplatin, an anti-tumor agent?" " + + +#Pt(NH_3)_2Cl_2# +Molar mass: #"" 300. g mol""^(-1)# + + +" +33 a8283db6-6ddd-11ea-ac6c-ccda262736ce https://socratic.org/questions/58924a0b7c01490182c48fcb 6.022 × 10^23 start physical_unit 2 3 number none qc_end physical_unit 10 11 6 7 mass qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] copper atoms""}]" "[{""type"":""physical unit"",""value"":""6.022 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] copper metal [=] \\pu{63.55 g}""}]" "

How many copper atoms in a #63.55*g# mass of copper metal?

" nan 6.022 × 10^23 "
+

Explanation:

+
+

And #N_A, ""Avogadro's number""# #=# #6.022xx10^23# individual copper atoms.

+

I look on the Periodic Table, and at #Z=29#, i.e. copper, the quoted atomic mass is #63.55*g#. The quoted mass is the mass of #""Avogadro's number""# of copper atoms. And thus, given a reaction with, say, oxygen or sulfur, I can calculate an equivalent, STOICHIOMETRIC mass of sulfur or oxygen.

+

My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.

+
+
" "
+
+
+

By definition, there are #N_A, ""Avogadro's number""# of copper atoms in a #63.55*g# mass of copper wire.

+
+
+
+

Explanation:

+
+

And #N_A, ""Avogadro's number""# #=# #6.022xx10^23# individual copper atoms.

+

I look on the Periodic Table, and at #Z=29#, i.e. copper, the quoted atomic mass is #63.55*g#. The quoted mass is the mass of #""Avogadro's number""# of copper atoms. And thus, given a reaction with, say, oxygen or sulfur, I can calculate an equivalent, STOICHIOMETRIC mass of sulfur or oxygen.

+

My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.

+
+
+
" "
+

How many copper atoms in a #63.55*g# mass of copper metal?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 13, 2017 + +
+
+
+
+
+
+
+

By definition, there are #N_A, ""Avogadro's number""# of copper atoms in a #63.55*g# mass of copper wire.

+
+
+
+

Explanation:

+
+

And #N_A, ""Avogadro's number""# #=# #6.022xx10^23# individual copper atoms.

+

I look on the Periodic Table, and at #Z=29#, i.e. copper, the quoted atomic mass is #63.55*g#. The quoted mass is the mass of #""Avogadro's number""# of copper atoms. And thus, given a reaction with, say, oxygen or sulfur, I can calculate an equivalent, STOICHIOMETRIC mass of sulfur or oxygen.

+

My advice is to look at the Periodic Table, and get to know it; you will soon learn the atomic masses of the common reagents.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 12047 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many copper atoms in a #63.55*g# mass of copper metal? nan +34 a8283db7-6ddd-11ea-beb8-ccda262736ce https://socratic.org/questions/when-water-reacts-with-carbon-dioxide-in-air-or-soil-it-forms-what-kind-of-acid H2CO3 start chemical_formula qc_end substance 1 1 qc_end substance 4 9 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""H2CO3""}]" "[{""type"":""substance name"",""value"":""water""},{""type"":""substance name"",""value"":""carbon dioxide in air or soil""}]" "

When water reacts with carbon dioxide in air or soil it forms what kind of acid?

" nan H2CO3 "
+

Explanation:

+
+

#CO_2+H_2Orightleftharpoons2H^+ +CO_3^(2-)#

+

Or, to be more precise:
+#CO_2+3H_2Orightleftharpoons2H_3O^+ +CO_3^(2-)#

+

The double arrow means that there is an equilibrium between the carbon dioxide and the acid.

+
+
" "
+
+
+

It forms carbonic acid #H_2CO_3#

+
+
+
+

Explanation:

+
+

#CO_2+H_2Orightleftharpoons2H^+ +CO_3^(2-)#

+

Or, to be more precise:
+#CO_2+3H_2Orightleftharpoons2H_3O^+ +CO_3^(2-)#

+

The double arrow means that there is an equilibrium between the carbon dioxide and the acid.

+
+
+
" "
+

When water reacts with carbon dioxide in air or soil it forms what kind of acid?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Properties of Acids and Bases + + +
+
+
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+
+1 Answer +
+
+
+
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+
+ +
+
+ +
+ + Jun 9, 2016 + +
+
+
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+
+

It forms carbonic acid #H_2CO_3#

+
+
+
+

Explanation:

+
+

#CO_2+H_2Orightleftharpoons2H^+ +CO_3^(2-)#

+

Or, to be more precise:
+#CO_2+3H_2Orightleftharpoons2H_3O^+ +CO_3^(2-)#

+

The double arrow means that there is an equilibrium between the carbon dioxide and the acid.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1322 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" When water reacts with carbon dioxide in air or soil it forms what kind of acid? nan +35 a8283db8-6ddd-11ea-ad4b-ccda262736ce https://socratic.org/questions/how-many-atoms-of-oxygen-are-there-in-18-g-of-water 6.022 × 10^23 start physical_unit 2 4 number none qc_end physical_unit 11 11 8 9 mass qc_end end "[{""type"": ""physical unit"",""value"": ""number [OF] atoms of oxygen""}]" "[{""type"":""physical unit"",""value"":""6.022 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{18 g}""}]" "

How many atoms of oxygen are there in 18 g of water?

" nan 6.022 × 10^23 "
+

Explanation:

+
+

First, it's important to know that the chemical formula for water is #H_2O#. This means that one molecule of water is made of two hydrogen atoms and one oxygen atom.

+

Then, we need to find how many moles of water there are in #""18 g""#. (We need to do this, because knowing the number of moles of water will help us convert from grams to number of atoms!)

+

To find how many moles of water there are in #""18 g""#, we have to divide #""18 g""# by the number of grams in one mole of water, also known as the molar mass of water:

+

#""number of moles"" = ""mass of sample""/""molar mass""#

+

#""number of moles"" = ""18 g""/(""molar mass of H""xx2 + ""molar mass of O"")#

+

#""number of moles"" = ""18 g""/(""1.008 g/mol""xx2 + ""16.00 g/mol"") = ""18 g""/(""18.02 g/mol"") ≈ ""1 mole""#

+

In #1# mole of water, there are #6.022 xx 10^(23)# molecules—that's how the mole was defined!

+

And, since the chemical formula for water is #H_2O#, #1# mole of water corresponds to #1# mole of oxygen atoms.

+

Therefore, there will be #6.022 xx 10^(23)# oxygen atoms in #1# mole, or #""18 g""# of water.

+
+
" "
+
+
+

#6.022 xx 10^(23)# oxygen atoms.

+
+
+
+

Explanation:

+
+

First, it's important to know that the chemical formula for water is #H_2O#. This means that one molecule of water is made of two hydrogen atoms and one oxygen atom.

+

Then, we need to find how many moles of water there are in #""18 g""#. (We need to do this, because knowing the number of moles of water will help us convert from grams to number of atoms!)

+

To find how many moles of water there are in #""18 g""#, we have to divide #""18 g""# by the number of grams in one mole of water, also known as the molar mass of water:

+

#""number of moles"" = ""mass of sample""/""molar mass""#

+

#""number of moles"" = ""18 g""/(""molar mass of H""xx2 + ""molar mass of O"")#

+

#""number of moles"" = ""18 g""/(""1.008 g/mol""xx2 + ""16.00 g/mol"") = ""18 g""/(""18.02 g/mol"") ≈ ""1 mole""#

+

In #1# mole of water, there are #6.022 xx 10^(23)# molecules—that's how the mole was defined!

+

And, since the chemical formula for water is #H_2O#, #1# mole of water corresponds to #1# mole of oxygen atoms.

+

Therefore, there will be #6.022 xx 10^(23)# oxygen atoms in #1# mole, or #""18 g""# of water.

+
+
+
" "
+

How many atoms of oxygen are there in 18 g of water?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 4, 2018 + +
+
+
+
+
+
+
+

#6.022 xx 10^(23)# oxygen atoms.

+
+
+
+

Explanation:

+
+

First, it's important to know that the chemical formula for water is #H_2O#. This means that one molecule of water is made of two hydrogen atoms and one oxygen atom.

+

Then, we need to find how many moles of water there are in #""18 g""#. (We need to do this, because knowing the number of moles of water will help us convert from grams to number of atoms!)

+

To find how many moles of water there are in #""18 g""#, we have to divide #""18 g""# by the number of grams in one mole of water, also known as the molar mass of water:

+

#""number of moles"" = ""mass of sample""/""molar mass""#

+

#""number of moles"" = ""18 g""/(""molar mass of H""xx2 + ""molar mass of O"")#

+

#""number of moles"" = ""18 g""/(""1.008 g/mol""xx2 + ""16.00 g/mol"") = ""18 g""/(""18.02 g/mol"") ≈ ""1 mole""#

+

In #1# mole of water, there are #6.022 xx 10^(23)# molecules—that's how the mole was defined!

+

And, since the chemical formula for water is #H_2O#, #1# mole of water corresponds to #1# mole of oxygen atoms.

+

Therefore, there will be #6.022 xx 10^(23)# oxygen atoms in #1# mole, or #""18 g""# of water.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 73548 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many atoms of oxygen are there in 18 g of water? nan +36 a8283db9-6ddd-11ea-b484-ccda262736ce https://socratic.org/questions/a-buffer-solution-of-ph-5-15-is-to-be-prepared-from-acetic-acid-ka-1-80-x-10-5-s 2.54 start physical_unit 28 32 molar_ratio none qc_end physical_unit 1 2 6 6 ph qc_end physical_unit 12 13 16 16 ka qc_end end "[{""type"":""physical unit"",""value"":""Molar ratio [OF] acetate ion to acetic acid""}]" "[{""type"":""physical unit"",""value"":""2.54""}]" "[{""type"":""physical unit"",""value"":""PH [OF] buffer solution [=] \\pu{5.15}""},{""type"":""physical unit"",""value"":""Ka [OF] acetic acid [=] \\pu{1.80 x 10^(−5)}""}]" "

A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka = #1.80 x 10^-5#), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

" nan 2.54 "
+

Explanation:

+
+

Ethanoic acid is a weak acid which dissociates:

+

#CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^(+)#

+

The expression for #K_a# is:

+

#K_a=[[CH_3COO_((aq))^(-)][H_((aq))^(+)])/([CH_3COOH_((aq))])#

+

So the molar ratio of acetate ion to acetic acid is:

+

#([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=K_a/[[H_((aq))^+]]#

+

The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.

+

We are told that the #pH=5.15#

+

#:.-log[H_((aq))^(+)]=5.15#

+

#:.[H_((aq))^+]=7.08xx10^(-6)"" """"mol/l""#

+

#:.([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=(nCH_3COO^(-))/(nCH_3COOH)=(1.8xx10^(-5))/(7.08xx10^(-6))=2.54#

+
+
" "
+
+
+

#2.54#

+
+
+
+

Explanation:

+
+

Ethanoic acid is a weak acid which dissociates:

+

#CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^(+)#

+

The expression for #K_a# is:

+

#K_a=[[CH_3COO_((aq))^(-)][H_((aq))^(+)])/([CH_3COOH_((aq))])#

+

So the molar ratio of acetate ion to acetic acid is:

+

#([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=K_a/[[H_((aq))^+]]#

+

The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.

+

We are told that the #pH=5.15#

+

#:.-log[H_((aq))^(+)]=5.15#

+

#:.[H_((aq))^+]=7.08xx10^(-6)"" """"mol/l""#

+

#:.([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=(nCH_3COO^(-))/(nCH_3COOH)=(1.8xx10^(-5))/(7.08xx10^(-6))=2.54#

+
+
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" "
+

A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka = #1.80 x 10^-5#), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Theory + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 27, 2016 + +
+
+
+
+
+
+
+

#2.54#

+
+
+
+

Explanation:

+
+

Ethanoic acid is a weak acid which dissociates:

+

#CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^(+)#

+

The expression for #K_a# is:

+

#K_a=[[CH_3COO_((aq))^(-)][H_((aq))^(+)])/([CH_3COOH_((aq))])#

+

So the molar ratio of acetate ion to acetic acid is:

+

#([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=K_a/[[H_((aq))^+]]#

+

The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.

+

We are told that the #pH=5.15#

+

#:.-log[H_((aq))^(+)]=5.15#

+

#:.[H_((aq))^+]=7.08xx10^(-6)"" """"mol/l""#

+

#:.([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=(nCH_3COO^(-))/(nCH_3COOH)=(1.8xx10^(-5))/(7.08xx10^(-6))=2.54#

+
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+
+ +
+
+
+
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+
+
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+ + Creative Commons License + +
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" A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka = #1.80 x 10^-5#), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH? nan +37 a82864b6-6ddd-11ea-abfb-ccda262736ce https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-that-is-62-58-carbon-9-63-hydrogen-2 C6H11O2 start chemical_formula qc_end physical_unit 11 11 10 10 mass_percent qc_end physical_unit 13 13 12 12 mass_percent qc_end physical_unit 15 15 14 14 mass_percent qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""C6H11O2""}]" "[{""type"":""physical unit"",""value"":""Mass percent [OF] carbon [=] \\pu{62.58%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] hydrogen [=] \\pu{9.63%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen [=] \\pu{27.79%}""}]" "

What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen?

" nan C6H11O2 "
+

Explanation:

+
+

As is the standard with these problems, we assume #100# #g# of compound, and break the percentages up into atoms.

+

In this mass, there are #(62.58*g)/(12.011*g*mol^-1)# #C# #=# #5.21# #mol# #C#.

+

And, #(9.63*g)/(1.00794*g*mol^-1)# #H# #=# #9.55# #mol# #H#.

+

And, #(27.79*g)/(15.999*g*mol^-1)# #O# #=# #1.740# #mol# #O#.

+

We divide thru, by the smallest molar quantity (#O#), and get (almost) the empirical formula: #C_3H_5.5O#.

+

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

+

#C_6H_11O_2#

+
+
" "
+
+
+

#C_6H_11O_2# is the simplest whole number ration that defines constituent atoms in a species, and is, therefore, the empirical formula.

+
+
+
+

Explanation:

+
+

As is the standard with these problems, we assume #100# #g# of compound, and break the percentages up into atoms.

+

In this mass, there are #(62.58*g)/(12.011*g*mol^-1)# #C# #=# #5.21# #mol# #C#.

+

And, #(9.63*g)/(1.00794*g*mol^-1)# #H# #=# #9.55# #mol# #H#.

+

And, #(27.79*g)/(15.999*g*mol^-1)# #O# #=# #1.740# #mol# #O#.

+

We divide thru, by the smallest molar quantity (#O#), and get (almost) the empirical formula: #C_3H_5.5O#.

+

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

+

#C_6H_11O_2#

+
+
+
" "
+

What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 27, 2016 + +
+
+
+
+
+
+
+

#C_6H_11O_2# is the simplest whole number ration that defines constituent atoms in a species, and is, therefore, the empirical formula.

+
+
+
+

Explanation:

+
+

As is the standard with these problems, we assume #100# #g# of compound, and break the percentages up into atoms.

+

In this mass, there are #(62.58*g)/(12.011*g*mol^-1)# #C# #=# #5.21# #mol# #C#.

+

And, #(9.63*g)/(1.00794*g*mol^-1)# #H# #=# #9.55# #mol# #H#.

+

And, #(27.79*g)/(15.999*g*mol^-1)# #O# #=# #1.740# #mol# #O#.

+

We divide thru, by the smallest molar quantity (#O#), and get (almost) the empirical formula: #C_3H_5.5O#.

+

Because the empirical formula is by definition the smallest WHOLE number ratio that describes constituent atoms in a species, we DOUBLE this formula.

+

#C_6H_11O_2#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 9612 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the molecular formula of a compound that is 62.58% carbon, 9.63% hydrogen, 27.79% oxygen? nan +38 a82864b7-6ddd-11ea-88e5-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-for-fructose-given-its-percent-composition-40-00-c CH2O start chemical_formula qc_end physical_unit 12 12 11 11 mass_percent qc_end physical_unit 14 14 13 13 mass_percent qc_end physical_unit 17 17 16 16 mass_percent qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""CH2O""}]" "[{""type"":""physical unit"",""value"":""Mass percent [OF] C [=] \\pu{40.00%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] H [=] \\pu{6.72%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] O [=] \\pu{53.29%}""}]" "

What is the empirical formula for fructose given its percent composition: 40.00% #C#, 6.72% #H#, and 53.29% #O#?

" nan CH2O "
+

Explanation:

+
+

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

+

In all of these problems it is useful to assume #100# #g# of compund, and calculate the ATOMIC composition.

+

So, in #100*g# fructose, there are:

+

#(40.00*g)/(12.01*g*mol^-1)# #=# #3.33*mol*C#,

+

#(6.72*g)/(1.00794*g*mol^-1)# #=# #6.66*mol*H#,

+

#(53.29*g)/(15.999*g*mol^-1)# #=# #3.33*mol*O#,

+

If we divide thru by the smallest molar quantity, we get a formula of #CH_2O#.

+

The molecuar mass of fructose is #180.16*g*mol^-1#. How do we use these data to provide a molecular formula?

+
+
" "
+
+
+

#CH_2O#

+
+
+
+

Explanation:

+
+

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

+

In all of these problems it is useful to assume #100# #g# of compund, and calculate the ATOMIC composition.

+

So, in #100*g# fructose, there are:

+

#(40.00*g)/(12.01*g*mol^-1)# #=# #3.33*mol*C#,

+

#(6.72*g)/(1.00794*g*mol^-1)# #=# #6.66*mol*H#,

+

#(53.29*g)/(15.999*g*mol^-1)# #=# #3.33*mol*O#,

+

If we divide thru by the smallest molar quantity, we get a formula of #CH_2O#.

+

The molecuar mass of fructose is #180.16*g*mol^-1#. How do we use these data to provide a molecular formula?

+
+
+
" "
+

What is the empirical formula for fructose given its percent composition: 40.00% #C#, 6.72% #H#, and 53.29% #O#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 26, 2016 + +
+
+
+
+
+
+
+

#CH_2O#

+
+
+
+

Explanation:

+
+

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

+

In all of these problems it is useful to assume #100# #g# of compund, and calculate the ATOMIC composition.

+

So, in #100*g# fructose, there are:

+

#(40.00*g)/(12.01*g*mol^-1)# #=# #3.33*mol*C#,

+

#(6.72*g)/(1.00794*g*mol^-1)# #=# #6.66*mol*H#,

+

#(53.29*g)/(15.999*g*mol^-1)# #=# #3.33*mol*O#,

+

If we divide thru by the smallest molar quantity, we get a formula of #CH_2O#.

+

The molecuar mass of fructose is #180.16*g*mol^-1#. How do we use these data to provide a molecular formula?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 14143 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the empirical formula for fructose given its percent composition: 40.00% #C#, 6.72% #H#, and 53.29% #O#? nan +39 a82864b8-6ddd-11ea-a1a1-ccda262736ce https://socratic.org/questions/two-cubic-meters-of-a-gas-at-30-degrees-kelvin-are-heated-at-a-constant-pressure 60 degrees Kelvin start physical_unit 5 5 temperature k qc_end physical_unit 5 5 7 9 temperature qc_end physical_unit 5 5 0 2 volume qc_end physical_unit 5 5 19 19 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] gas [IN] degrees Kelvin""}]" "[{""type"":""physical unit"",""value"":""60 degrees Kelvin""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] gas [=] \\pu{30 degrees Kelvin}""},{""type"":""physical unit"",""value"":""Volume1 [OF] gas [=] \\pu{2 cubic meters}""},{""type"":""physical unit"",""value"":""Volume2 [OF] gas [=] \\pu{doubles}""}]" "

Two cubic meters of a gas at 30 degrees Kelvin are heated at a constant pressure until the volume doubles. What is the final temperature of the gas?

" nan 60 degrees Kelvin "
+

Explanation:

+
+

This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is #V_1/T_1=V_2/T_2#.

+

Given
+#V_1=""2 m""^3#
+#T_1=""30 K""#
+#V_2=""4 m""^3#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#, substitute the given values into the equation, and solve.

+

#V_1/T_1=V_2/T_2#

+

#T_2=(T_1V_2)/V_1#

+

#T_2=(30""K""xx4cancel""m""^3)/(2cancel""m""^3)=""60 K""#

+
+
" "
+
+
+

The final temperature is 60 K.

+
+
+
+

Explanation:

+
+

This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is #V_1/T_1=V_2/T_2#.

+

Given
+#V_1=""2 m""^3#
+#T_1=""30 K""#
+#V_2=""4 m""^3#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#, substitute the given values into the equation, and solve.

+

#V_1/T_1=V_2/T_2#

+

#T_2=(T_1V_2)/V_1#

+

#T_2=(30""K""xx4cancel""m""^3)/(2cancel""m""^3)=""60 K""#

+
+
+
" "
+

Two cubic meters of a gas at 30 degrees Kelvin are heated at a constant pressure until the volume doubles. What is the final temperature of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Charles' Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 25, 2016 + +
+
+
+
+
+
+
+

The final temperature is 60 K.

+
+
+
+

Explanation:

+
+

This is an example of Charles' law, which states that the volume of a given amount of gas held at constant pressure varies directly with the temperature in Kelvins. The equation to use is #V_1/T_1=V_2/T_2#.

+

Given
+#V_1=""2 m""^3#
+#T_1=""30 K""#
+#V_2=""4 m""^3#

+

Unknown
+#T_2#

+

Solution
+Rearrange the equation to isolate #T_2#, substitute the given values into the equation, and solve.

+

#V_1/T_1=V_2/T_2#

+

#T_2=(T_1V_2)/V_1#

+

#T_2=(30""K""xx4cancel""m""^3)/(2cancel""m""^3)=""60 K""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4164 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" Two cubic meters of a gas at 30 degrees Kelvin are heated at a constant pressure until the volume doubles. What is the final temperature of the gas? nan +40 a82864b9-6ddd-11ea-8c0d-ccda262736ce https://socratic.org/questions/there-are-500-ml-hcl-solutions-having-ph-3-determine-the-amount-of-naoh-solid-mr 0.02 g start physical_unit 15 16 mass g qc_end physical_unit 5 6 10 10 ph qc_end physical_unit 5 6 3 4 volume qc_end physical_unit 15 16 19 20 molar_mass qc_end physical_unit 30 31 35 35 ph qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaOH solid [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.02 g""}]" "[{""type"":""physical unit"",""value"":""PH1 [OF] HCl solution [=] \\pu{3}""},{""type"":""physical unit"",""value"":""Volume [OF] HCl solution [=] \\pu{500 mL}""},{""type"":""physical unit"",""value"":""MM [OF] NaOH solid [=] \\pu{40 g/mol}""},{""type"":""physical unit"",""value"":""PH2 [OF] the solution [=] \\pu{10}""}]" "

You have a #""500-mL""# #""HCl""# solution having #""pH"" = 3#. Determine the amount of #""NaOH""# solid #(""M""_M = ""40 g/mol"")# that must be added so that the #""pH""# of the solution is changed to #10# ?

" "
+
+

+

Volume changes due to the addition of #""NaOH""# are not considered.

+

+
+
" 0.02 g "
+

Explanation:

+
+

Notice that the solution goes from being acidic at #""pH"" = 3# to being basic at #""pH"" = 10#, so the first thing that comes to mind here is that you need to add more sodium hydroxide than the number of moles of hydrochloric acid present in the initial solution.

+

As you know, the concentration of hydronium cations is equal to

+
+

#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#

+
+

This means that the initial solution contains

+
+

#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#

+
+

Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#

+
+

Sodium hydroxide and hydrochloric acid react in a #1:1# mole ratio to produce aqueous sodium chloride and water

+
+

#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

so you know that in order to completely neutralize the acid and make the solution neutral, you need to add #0.0005# moles of sodium hydroxide.

+

So you can say that at #25^@""C""#, you will have

+
+

#""0.0005 moles NaOH added "" -> "" pH"" = 7#

+
+

Now, you want the target solution to have

+
+

#""pH"" = 10#

+
+

which implies that it must have

+
+

#""pOH"" = 14 -10 = 4#

+
+

This means that the concentration of hydroxide anions in the final solution must be equal to

+
+

#[""OH""^(-)] = 10^(-4) quad ""M""#

+
+

Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#

+
+

Therefore, you can say that if you add #0.00050# moles of sodium hydroxide to the initial solution, you will completely neutralize the acid and get the #""pH""# to #7#.

+

At that point, adding an additional #0.000050# moles of sodium hydroxide will increase the #""pH""# of the solution from #7# to #10#.

+

The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a #1:1# mole ratio--will thus be equal to

+
+

#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#

+
+

Finally, to convert the number of moles to grams, use the molar mass of the salt.

+
+

#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#

+
+

The answer is rounded to one significant figure.

+
+
" "
+
+
+

#""0.02 g NaOH""#

+
+
+
+

Explanation:

+
+

Notice that the solution goes from being acidic at #""pH"" = 3# to being basic at #""pH"" = 10#, so the first thing that comes to mind here is that you need to add more sodium hydroxide than the number of moles of hydrochloric acid present in the initial solution.

+

As you know, the concentration of hydronium cations is equal to

+
+

#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#

+
+

This means that the initial solution contains

+
+

#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#

+
+

Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#

+
+

Sodium hydroxide and hydrochloric acid react in a #1:1# mole ratio to produce aqueous sodium chloride and water

+
+

#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

so you know that in order to completely neutralize the acid and make the solution neutral, you need to add #0.0005# moles of sodium hydroxide.

+

So you can say that at #25^@""C""#, you will have

+
+

#""0.0005 moles NaOH added "" -> "" pH"" = 7#

+
+

Now, you want the target solution to have

+
+

#""pH"" = 10#

+
+

which implies that it must have

+
+

#""pOH"" = 14 -10 = 4#

+
+

This means that the concentration of hydroxide anions in the final solution must be equal to

+
+

#[""OH""^(-)] = 10^(-4) quad ""M""#

+
+

Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#

+
+

Therefore, you can say that if you add #0.00050# moles of sodium hydroxide to the initial solution, you will completely neutralize the acid and get the #""pH""# to #7#.

+

At that point, adding an additional #0.000050# moles of sodium hydroxide will increase the #""pH""# of the solution from #7# to #10#.

+

The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a #1:1# mole ratio--will thus be equal to

+
+

#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#

+
+

Finally, to convert the number of moles to grams, use the molar mass of the salt.

+
+

#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#

+
+

The answer is rounded to one significant figure.

+
+
+
" "
+

You have a #""500-mL""# #""HCl""# solution having #""pH"" = 3#. Determine the amount of #""NaOH""# solid #(""M""_M = ""40 g/mol"")# that must be added so that the #""pH""# of the solution is changed to #10# ?

+
+
+

+

Volume changes due to the addition of #""NaOH""# are not considered.

+

+
+
+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 11, 2018 + +
+
+
+
+
+
+
+

#""0.02 g NaOH""#

+
+
+
+

Explanation:

+
+

Notice that the solution goes from being acidic at #""pH"" = 3# to being basic at #""pH"" = 10#, so the first thing that comes to mind here is that you need to add more sodium hydroxide than the number of moles of hydrochloric acid present in the initial solution.

+

As you know, the concentration of hydronium cations is equal to

+
+

#color(blue)(ul(color(black)([""H""_3""O""^(+)] = 10^(-""pH""))))#

+
+

This means that the initial solution contains

+
+

#[""H""_3""O""^(+)] = 10^(-3) quad ""M""#

+
+

Now, use the volume of the solution to calculate the number of moles of hydronium cations it contains.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-3) quad ""moles H""_3""O""^(+))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.00050 moles H""_3""O""^(+)#

+
+

Sodium hydroxide and hydrochloric acid react in a #1:1# mole ratio to produce aqueous sodium chloride and water

+
+

#""HCl""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

so you know that in order to completely neutralize the acid and make the solution neutral, you need to add #0.0005# moles of sodium hydroxide.

+

So you can say that at #25^@""C""#, you will have

+
+

#""0.0005 moles NaOH added "" -> "" pH"" = 7#

+
+

Now, you want the target solution to have

+
+

#""pH"" = 10#

+
+

which implies that it must have

+
+

#""pOH"" = 14 -10 = 4#

+
+

This means that the concentration of hydroxide anions in the final solution must be equal to

+
+

#[""OH""^(-)] = 10^(-4) quad ""M""#

+
+

Since you are told to assume that the volume of the solution does not change upon the addition of the salt, you can use it to find the number of moles of hydroxide anions present in the final solution.

+
+

#500 color(red)(cancel(color(black)(""mL solution""))) * (10^(-4) quad ""moles OH""^(-))/(10^3color(red)(cancel(color(black)(""mL solution"")))) = ""0.000050 moles OH""^(-)#

+
+

Therefore, you can say that if you add #0.00050# moles of sodium hydroxide to the initial solution, you will completely neutralize the acid and get the #""pH""# to #7#.

+

At that point, adding an additional #0.000050# moles of sodium hydroxide will increase the #""pH""# of the solution from #7# to #10#.

+

The total number of moles of sodium hydroxide that you must add--remember that sodium hydroxide delivers hydroxide anions to the solution in a #1:1# mole ratio--will thus be equal to

+
+

#""0.00050 moles + 0.000050 moles = 0.00055 moles NaOH""#

+
+

Finally, to convert the number of moles to grams, use the molar mass of the salt.

+
+

#0.00055 color(red)(cancel(color(black)(""moles NaOH""))) * ""40 g""/(1color(red)(cancel(color(black)(""mole NaOH"")))) = color(darkgreen)(ul(color(black)(""0.02 g"")))#

+
+

The answer is rounded to one significant figure.

+
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+
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+
+
+
Related questions
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" "You have a #""500-mL""# #""HCl""# solution having #""pH"" = 3#. Determine the amount of #""NaOH""# solid #(""M""_M = ""40 g/mol"")# that must be added so that the #""pH""# of the solution is changed to #10# ?" " + + +Volume changes due to the addition of #""NaOH""# are not considered. + + +" +41 a82864ba-6ddd-11ea-add9-ccda262736ce https://socratic.org/questions/the-stp-volume-of-an-ideal-gas-is-2-78-m-3-what-is-the-pressure-of-the-sample-at 1.75 atm start physical_unit 15 16 pressure atm qc_end c_other STP qc_end physical_unit 4 6 8 9 volume qc_end physical_unit 15 16 18 19 volume qc_end physical_unit 15 16 21 22 temperature qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] the sample [IN] atm""}]" "[{""type"":""physical unit"",""value"":""1.75 atm""}]" "[{""type"":""other"",""value"":""STP""},{""type"":""physical unit"",""value"":""Volume1 [OF] an ideal gas [=] \\pu{2.78 m^3}""},{""type"":""physical unit"",""value"":""Volume2 [OF] the sample [=] \\pu{1.75 m^3}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the sample [=] \\pu{27 ℃}""}]" "

The STP volume of an ideal gas is #2.78# #m^3#. What is the pressure of the sample at #1.75# #m^3# and #27°C#?

" nan 1.75 atm "
+

Explanation:

+
+

The initial conditions are:
+STP means: #T_1=273K# and #P_1=1atm#
+The volume of the gas is #V_1=2.78m^3#

+

Since the volume (#V_2=1.75m^3#) and temperature (#T_2=300K#) are changing at the same time, the new pressure could be found using:

+

#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#

+

#=>P_2= (P_1V_1)/(T_1)xx(T_2)/(V_2)=(1atmxx2.78cancel(m^3))/(273cancel(K))xx(300cancel(K))/(1.75cancel(m^3))=1.75 atm#

+
+
" "
+
+
+

#P_2=1.75 atm#

+
+
+
+

Explanation:

+
+

The initial conditions are:
+STP means: #T_1=273K# and #P_1=1atm#
+The volume of the gas is #V_1=2.78m^3#

+

Since the volume (#V_2=1.75m^3#) and temperature (#T_2=300K#) are changing at the same time, the new pressure could be found using:

+

#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#

+

#=>P_2= (P_1V_1)/(T_1)xx(T_2)/(V_2)=(1atmxx2.78cancel(m^3))/(273cancel(K))xx(300cancel(K))/(1.75cancel(m^3))=1.75 atm#

+
+
+
" "
+

The STP volume of an ideal gas is #2.78# #m^3#. What is the pressure of the sample at #1.75# #m^3# and #27°C#?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Measuring Gas Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 22, 2015 + +
+
+
+
+
+
+
+

#P_2=1.75 atm#

+
+
+
+

Explanation:

+
+

The initial conditions are:
+STP means: #T_1=273K# and #P_1=1atm#
+The volume of the gas is #V_1=2.78m^3#

+

Since the volume (#V_2=1.75m^3#) and temperature (#T_2=300K#) are changing at the same time, the new pressure could be found using:

+

#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#

+

#=>P_2= (P_1V_1)/(T_1)xx(T_2)/(V_2)=(1atmxx2.78cancel(m^3))/(273cancel(K))xx(300cancel(K))/(1.75cancel(m^3))=1.75 atm#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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" The STP volume of an ideal gas is #2.78# #m^3#. What is the pressure of the sample at #1.75# #m^3# and #27°C#? nan +42 a82864bb-6ddd-11ea-b2ff-ccda262736ce https://socratic.org/questions/how-many-moles-of-co-2-form-when-58-0-g-of-butane-c-4h-10-burn-in-oxygen 3.99 moles start physical_unit 4 4 mole mol qc_end physical_unit 10 10 7 8 mass qc_end substance 14 14 qc_end chemical_equation 11 11 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] CO2 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.99 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] butane [=] \\pu{58.0 g}""},{""type"":""substance name"",""value"":""Oxygen""},{""type"":""chemical equation"",""value"":""C4H10""}]" "

How many moles of #CO_2# form when 58.0 g of butane, #C_4H_10#, burn in oxygen?

" nan 3.99 moles "
+

Explanation:

+
+

Always start with a balanced equation.

+

#""2C""_4""H""_10"" + 13O""_2""##rarr##""8CO""_2"" + 10H""_2""O""#

+

Determine the molar mass of butane.

+

Molar Mass of #""C""_4""H""_10"":#

+

#(4xx12.011""g/mol"")+(10xx1.008""g/mol"")=""58.124 g/mol""#

+

Determine the Mole Ratios for Butane and Carbon dioxide

+

#(2""mol C""_4""H""_10)/(8""mol CO""_2"")# and #(8""mol CO""_2)/(2""mol C""_4""H""_10)#

+

Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.

+

#58.0cancel(""g C""_4""H""_10)xx(1cancel(""mol C""_4""H""_10))/(58.124cancel(""g C""_4""H""_10))xx(8""mol CO""_2)/(2cancel(""mol C""_4""H""_10))=""3.99 mol CO""_2""#

+
+
" "
+
+
+

#""58.0 g butane""# in oxygen will produce #""3.99 mol carbon dioxide""#.

+
+
+
+

Explanation:

+
+

Always start with a balanced equation.

+

#""2C""_4""H""_10"" + 13O""_2""##rarr##""8CO""_2"" + 10H""_2""O""#

+

Determine the molar mass of butane.

+

Molar Mass of #""C""_4""H""_10"":#

+

#(4xx12.011""g/mol"")+(10xx1.008""g/mol"")=""58.124 g/mol""#

+

Determine the Mole Ratios for Butane and Carbon dioxide

+

#(2""mol C""_4""H""_10)/(8""mol CO""_2"")# and #(8""mol CO""_2)/(2""mol C""_4""H""_10)#

+

Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.

+

#58.0cancel(""g C""_4""H""_10)xx(1cancel(""mol C""_4""H""_10))/(58.124cancel(""g C""_4""H""_10))xx(8""mol CO""_2)/(2cancel(""mol C""_4""H""_10))=""3.99 mol CO""_2""#

+
+
+
" "
+

How many moles of #CO_2# form when 58.0 g of butane, #C_4H_10#, burn in oxygen?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 1, 2016 + +
+
+
+
+
+
+
+

#""58.0 g butane""# in oxygen will produce #""3.99 mol carbon dioxide""#.

+
+
+
+

Explanation:

+
+

Always start with a balanced equation.

+

#""2C""_4""H""_10"" + 13O""_2""##rarr##""8CO""_2"" + 10H""_2""O""#

+

Determine the molar mass of butane.

+

Molar Mass of #""C""_4""H""_10"":#

+

#(4xx12.011""g/mol"")+(10xx1.008""g/mol"")=""58.124 g/mol""#

+

Determine the Mole Ratios for Butane and Carbon dioxide

+

#(2""mol C""_4""H""_10)/(8""mol CO""_2"")# and #(8""mol CO""_2)/(2""mol C""_4""H""_10)#

+

Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.

+

#58.0cancel(""g C""_4""H""_10)xx(1cancel(""mol C""_4""H""_10))/(58.124cancel(""g C""_4""H""_10))xx(8""mol CO""_2)/(2cancel(""mol C""_4""H""_10))=""3.99 mol CO""_2""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 21411 views + around the world +
+
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+ + Creative Commons License + +
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+
" How many moles of #CO_2# form when 58.0 g of butane, #C_4H_10#, burn in oxygen? nan +43 a82864bc-6ddd-11ea-86cd-ccda262736ce https://socratic.org/questions/what-s-the-solubility-in-grams-per-liter-of-laf3-in-pure-water 1.82 ⋅ 10^(−3) grams per liter start physical_unit 8 8 solubility g/l qc_end substance 10 11 qc_end end "[{""type"":""physical unit"",""value"":""Solubility [OF] LaF3 [IN] grams per liter""}]" "[{""type"":""physical unit"",""value"":""1.82 ⋅ 10^(−3) grams per liter""}]" "[{""type"":""substance name"",""value"":""Pure water""}]" "

What's the solubility (in grams per liter) of #""LaF""_3# in pure water?

" nan 1.82 ⋅ 10^(−3) grams per liter "
+

Explanation:

+
+

In order to solve this problem, you would need the value of the solubility product constant, #K_(sp)#, for lanthanum trifluoride, #""LaF""_3#, which is usually given to you with the problem.

+

In this case, I'll pick

+
+

#K_(sp) = 2.0 * 10^(-19)#.

+
+

You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility, #s#, of lanthanum trifluoride in aqueous solution.

+

Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.

+
+

#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#

+
+

#color(purple)(""I"") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))#
+#color(purple)(""E"") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)#

+

Initially, the concentrations of the #""La""^(3+)# and #""F""^(-)# ions are equal to zero - the solid was not yet placed in water.

+

Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.

+

By definition, the solubility product constant for this equilibrium will be

+
+

#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#

+
+

This will be equivalent to

+
+

#K_(sp) = s * (color(red)(3)s)^color(red)(3)#

+

#2.0 * 10^(-19) = 27s^4#

+
+

You will thus have

+
+

#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#

+
+

Since #s# represents the molar solubility of the salt, i.e. how many moles of lanthanum trifluorice can be dissolved in a liter of water, you will have

+
+

#s = 9.3 * 10^(-6)""mol L""^(-)#

+
+

In order to express the solubility in grams per liter, #""g L""^(-1)#, use lanthanum trifluoride's molar mass

+
+

#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#

+
+
+
" "
+
+
+

#1.8 * 10^(-3) ""g L""^(-1)#

+
+
+
+

Explanation:

+
+

In order to solve this problem, you would need the value of the solubility product constant, #K_(sp)#, for lanthanum trifluoride, #""LaF""_3#, which is usually given to you with the problem.

+

In this case, I'll pick

+
+

#K_(sp) = 2.0 * 10^(-19)#.

+
+

You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility, #s#, of lanthanum trifluoride in aqueous solution.

+

Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.

+
+

#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#

+
+

#color(purple)(""I"") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))#
+#color(purple)(""E"") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)#

+

Initially, the concentrations of the #""La""^(3+)# and #""F""^(-)# ions are equal to zero - the solid was not yet placed in water.

+

Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.

+

By definition, the solubility product constant for this equilibrium will be

+
+

#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#

+
+

This will be equivalent to

+
+

#K_(sp) = s * (color(red)(3)s)^color(red)(3)#

+

#2.0 * 10^(-19) = 27s^4#

+
+

You will thus have

+
+

#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#

+
+

Since #s# represents the molar solubility of the salt, i.e. how many moles of lanthanum trifluorice can be dissolved in a liter of water, you will have

+
+

#s = 9.3 * 10^(-6)""mol L""^(-)#

+
+

In order to express the solubility in grams per liter, #""g L""^(-1)#, use lanthanum trifluoride's molar mass

+
+

#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#

+
+
+
+
" "
+

What's the solubility (in grams per liter) of #""LaF""_3# in pure water?

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Solubility Equilbria + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Dec 22, 2014 + +
+
+
+
+
+
+
+

#1.8 * 10^(-3) ""g L""^(-1)#

+
+
+
+

Explanation:

+
+

In order to solve this problem, you would need the value of the solubility product constant, #K_(sp)#, for lanthanum trifluoride, #""LaF""_3#, which is usually given to you with the problem.

+

In this case, I'll pick

+
+

#K_(sp) = 2.0 * 10^(-19)#.

+
+

You could approach this problem by using an ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart) to help you find the molar solubility, #s#, of lanthanum trifluoride in aqueous solution.

+

Since you're dealing with an insoluble ionic compound, an equilibrium will be established between the undissolved solid and the dissolved ions.

+
+

#"" "" ""LaF""_ (color(red)(3)(s)) "" ""rightleftharpoons"" "" ""La""_ ((aq))^(3+) "" ""+"" "" color(red)(3)""F""_ ((aq))^(-)#

+
+

#color(purple)(""I"") color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"") color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(3)s))#
+#color(purple)(""E"") color(white)(aaaacolor(black)(-)aaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(3)s)#

+

Initially, the concentrations of the #""La""^(3+)# and #""F""^(-)# ions are equal to zero - the solid was not yet placed in water.

+

Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here.

+

By definition, the solubility product constant for this equilibrium will be

+
+

#K_(sp) = [""La""^(3+)] * [""F""^(-)]^color(red)(3)#

+
+

This will be equivalent to

+
+

#K_(sp) = s * (color(red)(3)s)^color(red)(3)#

+

#2.0 * 10^(-19) = 27s^4#

+
+

You will thus have

+
+

#s = root(4)((2.0 * 10^(-19))/27) = 9.3 * 10^(-6)#

+
+

Since #s# represents the molar solubility of the salt, i.e. how many moles of lanthanum trifluorice can be dissolved in a liter of water, you will have

+
+

#s = 9.3 * 10^(-6)""mol L""^(-)#

+
+

In order to express the solubility in grams per liter, #""g L""^(-1)#, use lanthanum trifluoride's molar mass

+
+

#9.3 * 10^(-6) color(red)(cancel(color(black)(""mol"")))/""L"" * ""195.9 g""/(1color(red)(cancel(color(black)(""mol"")))) = color(green)(1.8 * 10^(-3) ""g L""^(-1))#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 20766 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" "What's the solubility (in grams per liter) of #""LaF""_3# in pure water?" nan +44 a82864bd-6ddd-11ea-94d6-ccda262736ce https://socratic.org/questions/at-what-temperature-is-the-concentration-of-a-saturated-solution-of-kcl-molar-ma 0 ℃ start physical_unit 8 11 temperature none qc_end physical_unit 11 11 14 15 molar_mass qc_end physical_unit 8 11 17 18 concentration qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] saturated solution of KCl""}]" "[{""type"":""physical unit"",""value"":""0 ℃""}]" "[{""type"":""physical unit"",""value"":""Molar mass [OF] KCl [=] \\pu{74.5 g}""},{""type"":""physical unit"",""value"":""Concentration [OF] saturated solution of KCl [=] \\pu{3 molal}""}]" "

At what temperature is the concentration of a saturated solution of #KCl# (molar mass 74.5 g) approximately 3 molal?

" nan 0 ℃ "
+

Explanation:

+
+

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, #""KCl""#, which looks like this

+

+

Since the solubility of potassium chloride is given per #""100 g""# of water, calculate how many moles of potassium chloride would make a #""3.5-molal""# solution in that much water.

+

Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.

+
+

#color(blue)(b = n_""solute""/m_""solvent"")#

+
+

This means that you have

+
+

#n_""solute"" = b * m_""solvent""#

+

#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#

+
+

Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles

+
+

#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#

+
+

Now take a look at the solubility graph and decide at which temperature dissolving #""26.1 g""# of potassium chloride per #""100 g""# of water will result in a saturated solution.

+

Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in #color(blue)(""blue"")#, matches the value #""26.1 g""#.

+

From the look of it, dissolving this much potassium chloride per #""100 g""# of water will produce a #""3.5-molal""# saturated solution at #0^@""C""#.

+
+
" "
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#0^@""C""#

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+
+
+

Explanation:

+
+

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, #""KCl""#, which looks like this

+

+

Since the solubility of potassium chloride is given per #""100 g""# of water, calculate how many moles of potassium chloride would make a #""3.5-molal""# solution in that much water.

+

Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.

+
+

#color(blue)(b = n_""solute""/m_""solvent"")#

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+

This means that you have

+
+

#n_""solute"" = b * m_""solvent""#

+

#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#

+
+

Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles

+
+

#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#

+
+

Now take a look at the solubility graph and decide at which temperature dissolving #""26.1 g""# of potassium chloride per #""100 g""# of water will result in a saturated solution.

+

Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in #color(blue)(""blue"")#, matches the value #""26.1 g""#.

+

From the look of it, dissolving this much potassium chloride per #""100 g""# of water will produce a #""3.5-molal""# saturated solution at #0^@""C""#.

+
+
+
" "
+

At what temperature is the concentration of a saturated solution of #KCl# (molar mass 74.5 g) approximately 3 molal?

+
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+ + +Chemistry + + + + + +Solutions + + + + + +Solubility Graphs + + +
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+1 Answer +
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+ +
+ + Feb 10, 2016 + +
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+

#0^@""C""#

+
+
+
+

Explanation:

+
+

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, #""KCl""#, which looks like this

+

+

Since the solubility of potassium chloride is given per #""100 g""# of water, calculate how many moles of potassium chloride would make a #""3.5-molal""# solution in that much water.

+

Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.

+
+

#color(blue)(b = n_""solute""/m_""solvent"")#

+
+

This means that you have

+
+

#n_""solute"" = b * m_""solvent""#

+

#n_""solute"" = ""3.5 mol"" color(red)(cancel(color(black)(""kg""^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)(""kg""))) = ""0.35 moles KCl""#

+
+

Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles

+
+

#0.35color(red)(cancel(color(black)(""moles KCl""))) * ""74.5 g""/(1color(red)(cancel(color(black)(""mole KCl"")))) = ""26.1 g""#

+
+

Now take a look at the solubility graph and decide at which temperature dissolving #""26.1 g""# of potassium chloride per #""100 g""# of water will result in a saturated solution.

+

Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in #color(blue)(""blue"")#, matches the value #""26.1 g""#.

+

From the look of it, dissolving this much potassium chloride per #""100 g""# of water will produce a #""3.5-molal""# saturated solution at #0^@""C""#.

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" At what temperature is the concentration of a saturated solution of #KCl# (molar mass 74.5 g) approximately 3 molal? nan +45 a82864be-6ddd-11ea-a414-ccda262736ce https://socratic.org/questions/how-many-molecules-of-oxygen-are-produced-by-the-decomposition-of-6-54-g-of-pota 4.82 ⋅ 10^22 start physical_unit 2 4 number none qc_end physical_unit 14 15 11 12 mass qc_end chemical_equation 19 21 qc_end end "[{""type"": ""physical unit"",""value"": ""Number [OF] molecules of oxygen""}]" "[{""type"":""physical unit"",""value"":""4.82 ⋅ 10^22""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] potassium chlorate [=] \\pu{6.54 g}""},{""type"":""chemical equation"",""value"":""2KClO3(s) -> 2KCl(s) + 3O2(g)""}]" "

How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate in the reaction #2KClO_3(s) -> 2KCl(s) + 3O_2(g)#?

" nan 4.82 ⋅ 10^22 "
+

Explanation:

+
+

We have:
+#2KClO_3(s) -> 2KCl(s) + 3O_2(g)#

+

as a balanced chemical reaction.

+

I see you want to know the amount #O_2# molecule produced when your potassium chlorate goes through decomposition.

+

Consider the graphic below:
+

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" "
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#4.819*10^22# molecules #O_2#

+
+
+
+

Explanation:

+
+

We have:
+#2KClO_3(s) -> 2KCl(s) + 3O_2(g)#

+

as a balanced chemical reaction.

+

I see you want to know the amount #O_2# molecule produced when your potassium chlorate goes through decomposition.

+

Consider the graphic below:
+

+
+
+
" "
+

How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate in the reaction #2KClO_3(s) -> 2KCl(s) + 3O_2(g)#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Dec 6, 2016 + +
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#4.819*10^22# molecules #O_2#

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+
+

Explanation:

+
+

We have:
+#2KClO_3(s) -> 2KCl(s) + 3O_2(g)#

+

as a balanced chemical reaction.

+

I see you want to know the amount #O_2# molecule produced when your potassium chlorate goes through decomposition.

+

Consider the graphic below:
+

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" How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate in the reaction #2KClO_3(s) -> 2KCl(s) + 3O_2(g)#? nan +46 a8288bc6-6ddd-11ea-bb81-ccda262736ce https://socratic.org/questions/what-pressure-would-it-take-to-compress-200-l-of-helium-gas-initially-at-1-00-at 100.00 atm start physical_unit 11 12 pressure atm qc_end physical_unit 11 12 15 16 pressure qc_end physical_unit 11 12 8 9 volume qc_end physical_unit 11 12 19 20 volume qc_end c_other constant_temperature qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] helium gas [IN] atm""}]" "[{""type"":""physical unit"",""value"":""100.00 atm""}]" "[{""type"": ""physical unit"",""value"": ""Pressure1 [OF] helium gas [=] \\pu{1.00 atm}""}, {""type"": ""physical unit"",""value"": ""Volume1 [OF] helium gas [=] \\pu{200 L}""},{""type"": ""physical unit"",""value"": ""Volume2 [OF] helium gas [=] \\pu{2.00 L}""},{""type"": ""other"",""value"": ""constant temperature""}]" "

What pressure would it take to compress 200 L of helium gas initially at 1.00 atm into a 2.00 L tank at constant temperature?

" nan 100.00 atm "
+

Explanation:

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#P_2=(P_1V_1)/V_2# #=# #(200*Lxx1.00*atm)/(2.00*L)# #=# #""A lot of atmospheres""#

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#P_1V_1=P_2V_2#, given constant temperature and given quantity of gas.

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Explanation:

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#P_2=(P_1V_1)/V_2# #=# #(200*Lxx1.00*atm)/(2.00*L)# #=# #""A lot of atmospheres""#

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" "
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What pressure would it take to compress 200 L of helium gas initially at 1.00 atm into a 2.00 L tank at constant temperature?

+
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+ + +Chemistry + + + + + +Gases + + + + + +Boyle's Law + + +
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+1 Answer +
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+ + May 6, 2016 + +
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#P_1V_1=P_2V_2#, given constant temperature and given quantity of gas.

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Explanation:

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#P_2=(P_1V_1)/V_2# #=# #(200*Lxx1.00*atm)/(2.00*L)# #=# #""A lot of atmospheres""#

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+
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" What pressure would it take to compress 200 L of helium gas initially at 1.00 atm into a 2.00 L tank at constant temperature? nan +47 a82894b4-6ddd-11ea-9d5f-ccda262736ce https://socratic.org/questions/a-hydrated-compound-has-an-analysis-of-18-29-ca-32-37-cl-and-49-34-water-what-is CaCl2 ; 6H2O start chemical_formula qc_end substance 1 2 qc_end physical_unit 8 8 7 7 mass_percent qc_end physical_unit 10 10 9 9 mass_percent qc_end physical_unit 13 13 12 12 mass_percent qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""CaCl2 ; 6H2O""}]" "[{""type"":""substance name"",""value"":""hydrated compound""},{""type"":""physical unit"",""value"":""Mass percent [OF] Ca [=] \\pu{18.29%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] Cl [=] \\pu{32.37%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] water [=] \\pu{49.34%}""}]" "

A hydrated compound has an analysis of 18.29% Ca, 32.37% Cl, and 49.34% water. What is its formula?

" nan CaCl2 ; 6H2O "
+

Explanation:

+
+

Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.

+

The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.

+

# (18.29/ 1) / ( 40/ 1) # = .45 moles Ca

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# (32.37/1) / (35.4/1) # = .91 moles Cl

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# (49.34/1) / ( 18.0/1)# = 2.74 moles # H_2O#

+

Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.

+

# .91/.45#= 2 2 Cl : 1 Ca

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#2.74/.45# = 6 # 6 H_2O # : 1 Ca.

+

So the empirical formula is # CaCl_2: 6 H_2O#

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Ca = +2 Cl = -1 so # Ca Cl_2# is balanced so

+

the empirical formula is also the molecular formula.

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+
" "
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+
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#CaCl_2;6 H_2O#

+
+
+
+

Explanation:

+
+

Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.

+

The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.

+

# (18.29/ 1) / ( 40/ 1) # = .45 moles Ca

+

# (32.37/1) / (35.4/1) # = .91 moles Cl

+

# (49.34/1) / ( 18.0/1)# = 2.74 moles # H_2O#

+

Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.

+

# .91/.45#= 2 2 Cl : 1 Ca

+

#2.74/.45# = 6 # 6 H_2O # : 1 Ca.

+

So the empirical formula is # CaCl_2: 6 H_2O#

+

Ca = +2 Cl = -1 so # Ca Cl_2# is balanced so

+

the empirical formula is also the molecular formula.

+
+
+
" "
+

A hydrated compound has an analysis of 18.29% Ca, 32.37% Cl, and 49.34% water. What is its formula?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
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+1 Answer +
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+ + Aug 26, 2016 + +
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#CaCl_2;6 H_2O#

+
+
+
+

Explanation:

+
+

Change the percentages to grams. If there was 100 grams then Ca would have 18.29 grams Chorine would have 32.37 grams and 49.34 grams of water.

+

The grams can be changed to moles by dividing the number of grams by the grams per mole that you can find from the periodic table.

+

# (18.29/ 1) / ( 40/ 1) # = .45 moles Ca

+

# (32.37/1) / (35.4/1) # = .91 moles Cl

+

# (49.34/1) / ( 18.0/1)# = 2.74 moles # H_2O#

+

Now that the moles of each element ( compound) is known it is possible to find a whole number ratio of the elements and compounds giving an empirical formula.

+

# .91/.45#= 2 2 Cl : 1 Ca

+

#2.74/.45# = 6 # 6 H_2O # : 1 Ca.

+

So the empirical formula is # CaCl_2: 6 H_2O#

+

Ca = +2 Cl = -1 so # Ca Cl_2# is balanced so

+

the empirical formula is also the molecular formula.

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+
+
+
Related questions
+ + +
+
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" A hydrated compound has an analysis of 18.29% Ca, 32.37% Cl, and 49.34% water. What is its formula? nan +48 a82894b5-6ddd-11ea-b8ce-ccda262736ce https://socratic.org/questions/58f63a0011ef6b35bb3d0d26 0.28 mol/L start physical_unit 10 11 concentration mol/l qc_end physical_unit 10 11 6 7 mass qc_end physical_unit 19 19 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] ammonium nitrate [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.28 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] ammonium nitrate [=] \\pu{18.9 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{855 mL}""}]" "

What is the concentration of an #18.9*g# mass of ammonium nitrate dissolved in an #855*mL# volume of solution?

" nan 0.28 mol/L "
+

Explanation:

+
+

#""Molarity""=((18.9*g)/(80.05*g*mol^-1))/(0.855*L)=0.276*mol*L^-1#.

+

And all we did here was to apply the definition of #""molarity""#, i.e. #""moles of solute per unit volume of solution""#.

+

Would this solution be slightly acidic or slightly basic........?

+
+
" "
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+
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#""Molarity""-=""Moles of solute""/""Volume of solution""-=0.276*mol*L^-1#.

+
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+
+

Explanation:

+
+

#""Molarity""=((18.9*g)/(80.05*g*mol^-1))/(0.855*L)=0.276*mol*L^-1#.

+

And all we did here was to apply the definition of #""molarity""#, i.e. #""moles of solute per unit volume of solution""#.

+

Would this solution be slightly acidic or slightly basic........?

+
+
+
" "
+

What is the concentration of an #18.9*g# mass of ammonium nitrate dissolved in an #855*mL# volume of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Apr 18, 2017 + +
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#""Molarity""-=""Moles of solute""/""Volume of solution""-=0.276*mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Molarity""=((18.9*g)/(80.05*g*mol^-1))/(0.855*L)=0.276*mol*L^-1#.

+

And all we did here was to apply the definition of #""molarity""#, i.e. #""moles of solute per unit volume of solution""#.

+

Would this solution be slightly acidic or slightly basic........?

+
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+ +
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+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2025 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the concentration of an #18.9*g# mass of ammonium nitrate dissolved in an #855*mL# volume of solution? nan +49 a828b354-6ddd-11ea-a11c-ccda262736ce https://socratic.org/questions/chlorine-gas-cl-2-reacts-with-potassium-bromide-kbr-to-form-potassium-chloride-a Cl2 + 2KBr -> 2KCl + Br2 start chemical_equation qc_end chemical_equation 2 2 qc_end chemical_equation 7 7 qc_end chemical_equation 14 14 qc_end substance 10 11 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""Cl2 + 2KBr -> 2KCl + Br2""}]" "[{""type"":""chemical equation"",""value"":""Cl2""},{""type"":""chemical equation"",""value"":""KBr""},{""type"":""chemical equation"",""value"":""Br2""},{""type"":""substance name"",""value"":""potassium chloride""}]" "

Chlorine gas, #Cl_2#, reacts with potassium bromide, #KBr#, to form potassium chloride and bromine, #Br_2#. How do you write the balanced equation for this single-displacement reaction?

" nan Cl2 + 2KBr -> 2KCl + Br2 "
+

Explanation:

+
+

Unbalanced Equation

+

#""Cl""_2"" + KBr""##rarr##""KCl"" + ""Br""_2""#

+

Balance the Cl

+

There are 2 Cl atoms on the left side and 1 Cl atom on the right side.

+

Add a coefficient of 2 in front of KCl on the right side.

+

#""Cl""_2"" + KBr""##rarr##""2KCl"" + ""Br""_2""#

+

There are now 2 Cl atoms on both sides.

+

Balance the Br

+

There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.

+

#""Cl""_2"" + 2KBr""##rarr##""2KCl"" + ""Br""_2""#

+

The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.

+

Note: You can substitute the words atom or atoms by mole or moles.

+
+
" "
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#""Cl""_2"" + 2KBr""##rarr##""2KCl"" + ""Br""_2""#

+
+
+
+

Explanation:

+
+

Unbalanced Equation

+

#""Cl""_2"" + KBr""##rarr##""KCl"" + ""Br""_2""#

+

Balance the Cl

+

There are 2 Cl atoms on the left side and 1 Cl atom on the right side.

+

Add a coefficient of 2 in front of KCl on the right side.

+

#""Cl""_2"" + KBr""##rarr##""2KCl"" + ""Br""_2""#

+

There are now 2 Cl atoms on both sides.

+

Balance the Br

+

There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.

+

#""Cl""_2"" + 2KBr""##rarr##""2KCl"" + ""Br""_2""#

+

The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.

+

Note: You can substitute the words atom or atoms by mole or moles.

+
+
+
" "
+

Chlorine gas, #Cl_2#, reacts with potassium bromide, #KBr#, to form potassium chloride and bromine, #Br_2#. How do you write the balanced equation for this single-displacement reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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#""Cl""_2"" + 2KBr""##rarr##""2KCl"" + ""Br""_2""#

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+
+
+

Explanation:

+
+

Unbalanced Equation

+

#""Cl""_2"" + KBr""##rarr##""KCl"" + ""Br""_2""#

+

Balance the Cl

+

There are 2 Cl atoms on the left side and 1 Cl atom on the right side.

+

Add a coefficient of 2 in front of KCl on the right side.

+

#""Cl""_2"" + KBr""##rarr##""2KCl"" + ""Br""_2""#

+

There are now 2 Cl atoms on both sides.

+

Balance the Br

+

There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr.

+

#""Cl""_2"" + 2KBr""##rarr##""2KCl"" + ""Br""_2""#

+

The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.

+

Note: You can substitute the words atom or atoms by mole or moles.

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Related questions
+ + +
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Impact of this question
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" Chlorine gas, #Cl_2#, reacts with potassium bromide, #KBr#, to form potassium chloride and bromine, #Br_2#. How do you write the balanced equation for this single-displacement reaction? nan +50 a828b355-6ddd-11ea-9c50-ccda262736ce https://socratic.org/questions/150g-of-nacl-completely-dissolve-in-1-00-kg-of-water-at-25-0-c-the-vapor-pressur 22.75 torr start physical_unit 33 34 vapor_pressure torr qc_end physical_unit 2 2 0 0 mass qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 17 18 23 24 vapor_pressure qc_end physical_unit 17 18 11 12 temperature qc_end physical_unit 33 34 11 12 temperature qc_end end "[{""type"":""physical unit"",""value"":""Vapor pressure [OF] the solution [IN] torr""}]" "[{""type"":""physical unit"",""value"":""22.75 torr""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [=] \\pu{150 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{1.00 kg}""},{""type"":""physical unit"",""value"":""Vapor pressure [OF] pure water [=] \\pu{23.8 torr}""},{""type"":""physical unit"",""value"":""Temperature [OF] pure water [=] \\pu{25.0 ℃}""},{""type"":""physical unit"",""value"":""Temperature [OF] the solution [=] \\pu{25.0 ℃}""}]" "

150g of NaCl completely dissolve in 1.00 kg of water at 25.0 C. The vapor pressure of pure water at this temperature is 23.8 torr. How would you determine the vapor pressure of the solution?

" nan 22.75 torr "
+

Explanation:

+
+

Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.

+

Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.

+

To do that, use the two compounds' respective molar masses. For sodium chloride you'll have

+
+

#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#

+
+

For water you'll have

+
+

#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#

+
+

Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.

+
+

#n_""total"" = n_""water"" + n_""NaCl""#

+

#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#

+
+

The mole faction of water will thus be

+
+

#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#

+
+

This means that the vapor pressure of the solution will be

+
+

#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.

+
+
" "
+
+
+

#""22.8 torr""#

+
+
+
+

Explanation:

+
+

Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.

+

Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.

+

To do that, use the two compounds' respective molar masses. For sodium chloride you'll have

+
+

#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#

+
+

For water you'll have

+
+

#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#

+
+

Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.

+
+

#n_""total"" = n_""water"" + n_""NaCl""#

+

#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#

+
+

The mole faction of water will thus be

+
+

#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#

+
+

This means that the vapor pressure of the solution will be

+
+

#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.

+
+
+
" "
+

150g of NaCl completely dissolve in 1.00 kg of water at 25.0 C. The vapor pressure of pure water at this temperature is 23.8 torr. How would you determine the vapor pressure of the solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 20, 2015 + +
+
+
+
+
+
+
+

#""22.8 torr""#

+
+
+
+

Explanation:

+
+

Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.

+

Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.

+

To do that, use the two compounds' respective molar masses. For sodium chloride you'll have

+
+

#150color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.44color(red)(cancel(color(black)(""g"")))) = ""2.567 moles NaCl""#

+
+

For water you'll have

+
+

#1.00color(red)(cancel(color(black)(""kg""))) * (1000color(red)(cancel(color(black)(""g""))))/(1color(red)(cancel(color(black)(""kg"")))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""55.51 moles H""_2""O""#

+
+

Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.

+
+

#n_""total"" = n_""water"" + n_""NaCl""#

+

#n_""total"" = 55.51 + 2.567 = ""58.08 moles""#

+
+

The mole faction of water will thus be

+
+

#chi_""water"" = (55.51color(red)(cancel(color(black)(""moles""))))/(58.08color(red)(cancel(color(black)(""moles"")))) = 0.956#

+
+

This means that the vapor pressure of the solution will be

+
+

#P_""sol"" = 0.956 * ""23.8 torr"" = color(green)(""22.8 torr"")#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.

+
+
+
+
+
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+
+
+
+
+
+
Related questions
+ + +
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+
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" 150g of NaCl completely dissolve in 1.00 kg of water at 25.0 C. The vapor pressure of pure water at this temperature is 23.8 torr. How would you determine the vapor pressure of the solution? nan +51 a828b356-6ddd-11ea-8593-ccda262736ce https://socratic.org/questions/if-4-04-g-of-n-combine-with-11-46-g-o-to-produce-a-compound-with-a-molar-mass-of N2O5 start chemical_formula qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 9 9 7 8 mass qc_end physical_unit 12 13 19 20 molar_mass qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""N2O5""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N [=] \\pu{4.04 g}""},{""type"":""physical unit"",""value"":""Mass [OF] O [=] \\pu{11.46 g}""},{""type"":""physical unit"",""value"":""Molar mass [OF] a compound [=] \\pu{108.0 g/mol}""}]" "

If 4.04 g of #N# combine with 11.46 g #O# to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?

" nan N2O5 "
+

Explanation:

+
+

#""Moles of nitrogen""# #=# #(4.04*g)/(14.01*g*mol^-1)# #=# #0.289*mol#

+

#""Moles of oxygen""# #=# #(11.46*g)/(15.999*g*mol^-1)# #=# #0.716*mol#

+

We divide thru by the lowest molar quantity to give an empirical formula of #NO_(2.5)# or #N_2O_5#, because the empirical formula should be integral.

+

Now the molecular formula is always a multiple of the the empirical formula:

+

So using the quoted molecular mass:

+

#108*g*mol^-1# #=# #nxx(2xx14.01+5xx15.999)*g*mol^-1#

+

Clearly, #n=1#, and the molecular formula is #N_2O_5#.

+
+
" "
+
+
+

Dinitrogen pentoxide, #N_2O_5#.

+
+
+
+

Explanation:

+
+

#""Moles of nitrogen""# #=# #(4.04*g)/(14.01*g*mol^-1)# #=# #0.289*mol#

+

#""Moles of oxygen""# #=# #(11.46*g)/(15.999*g*mol^-1)# #=# #0.716*mol#

+

We divide thru by the lowest molar quantity to give an empirical formula of #NO_(2.5)# or #N_2O_5#, because the empirical formula should be integral.

+

Now the molecular formula is always a multiple of the the empirical formula:

+

So using the quoted molecular mass:

+

#108*g*mol^-1# #=# #nxx(2xx14.01+5xx15.999)*g*mol^-1#

+

Clearly, #n=1#, and the molecular formula is #N_2O_5#.

+
+
+
" "
+

If 4.04 g of #N# combine with 11.46 g #O# to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
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+
+1 Answer +
+
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+
+
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+ + May 29, 2016 + +
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Dinitrogen pentoxide, #N_2O_5#.

+
+
+
+

Explanation:

+
+

#""Moles of nitrogen""# #=# #(4.04*g)/(14.01*g*mol^-1)# #=# #0.289*mol#

+

#""Moles of oxygen""# #=# #(11.46*g)/(15.999*g*mol^-1)# #=# #0.716*mol#

+

We divide thru by the lowest molar quantity to give an empirical formula of #NO_(2.5)# or #N_2O_5#, because the empirical formula should be integral.

+

Now the molecular formula is always a multiple of the the empirical formula:

+

So using the quoted molecular mass:

+

#108*g*mol^-1# #=# #nxx(2xx14.01+5xx15.999)*g*mol^-1#

+

Clearly, #n=1#, and the molecular formula is #N_2O_5#.

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+
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+
+
+
+
+
+
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+ + +
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+
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" If 4.04 g of #N# combine with 11.46 g #O# to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound? nan +52 a828b357-6ddd-11ea-8e86-ccda262736ce https://socratic.org/questions/what-is-the-chemical-formula-for-the-combination-of-lead-ii-and-sulfer PbS start chemical_formula qc_end substance 12 12 qc_end substance 9 10 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""PbS""}]" "[{""type"":""substance name"",""value"":""sulfer""},{""type"":""substance name"",""value"":""lead (II)""}]" "

What is the chemical formula for the combination of lead (II) and sulfer?

" nan PbS "
+

Explanation:

+
+

Lead sulfide occurs in the mineral, #""galena""#. It is as soluble as a brick.

+

The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?

+
+
" "
+
+
+

#Pb^(2+) + S^(2-) rarr PbS(s)darr#

+
+
+
+

Explanation:

+
+

Lead sulfide occurs in the mineral, #""galena""#. It is as soluble as a brick.

+

The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?

+
+
+
" "
+

What is the chemical formula for the combination of lead (II) and sulfer?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
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+
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+
+ +
+
+ +
+ + Nov 4, 2016 + +
+
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+

#Pb^(2+) + S^(2-) rarr PbS(s)darr#

+
+
+
+

Explanation:

+
+

Lead sulfide occurs in the mineral, #""galena""#. It is as soluble as a brick.

+

The given reaction illustrates conservation of mass and conservation of charge, as indeed it must if it reflects chemical reality. Why?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1978 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the chemical formula for the combination of lead (II) and sulfer? nan +53 a828e03a-6ddd-11ea-80fe-ccda262736ce https://socratic.org/questions/56dbfc0b7c01495b54f0dd18 40.31 g/mol start physical_unit 6 7 molar_mass g/mol qc_end substance 6 7 qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] magnesium oxide [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""40.31 g/mol""}]" "[{""type"":""substance name"",""value"":""magnesium oxide""}]" "

What is the molar mass of magnesium oxide?

" nan 40.31 g/mol "
+

Explanation:

+
+

How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.

+

The Periodic Table tells us that the molar mass of oxygen is #15.99# #g*mol^-1#, and the molar mass of magnesium is #24.31# #g*mol^-1#. Sum these masses to get the molar mass of magnesium oxide.

+

What do I mean by molar mass?

+
+
" "
+
+
+

#40.31# #g*mol^-1#

+
+
+
+

Explanation:

+
+

How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.

+

The Periodic Table tells us that the molar mass of oxygen is #15.99# #g*mol^-1#, and the molar mass of magnesium is #24.31# #g*mol^-1#. Sum these masses to get the molar mass of magnesium oxide.

+

What do I mean by molar mass?

+
+
+
" "
+

What is the molar mass of magnesium oxide?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 6, 2016 + +
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#40.31# #g*mol^-1#

+
+
+
+

Explanation:

+
+

How did I know this? Well, I have access to a Periodic Table, and you should have access to one as well.

+

The Periodic Table tells us that the molar mass of oxygen is #15.99# #g*mol^-1#, and the molar mass of magnesium is #24.31# #g*mol^-1#. Sum these masses to get the molar mass of magnesium oxide.

+

What do I mean by molar mass?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 17986 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the molar mass of magnesium oxide? nan +54 a828e03b-6ddd-11ea-99a8-ccda262736ce https://socratic.org/questions/if-a-reaction-has-an-actual-yield-of-5-grams-and-a-theoretical-yield-of-10-grams 50.00% start physical_unit 1 2 percent_yield none qc_end physical_unit 1 2 8 9 actual_yield qc_end physical_unit 1 2 15 16 theoretical_yield qc_end end "[{""type"":""physical unit"",""value"":""percent yield [OF] a reaction""}]" "[{""type"":""physical unit"",""value"":""50.00%""}]" "[{""type"":""physical unit"",""value"":""actual yield [OF] a reaction [=] \\pu{5 grams}""},{""type"":""physical unit"",""value"":""theoretical yield [OF] a reaction [=] \\pu{10 grams}""}]" "

If a reaction has an actual yield of 5 grams and a theoretical yield of 10 grams, what is the percent yield?

" nan 50.00% "
+

Explanation:

+
+

Let's define those three terms before we get started:

+
    +
  • +

    The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.

    +
    • +
  • +

    The actual yield is the mass of product that was obtained in an experiment.

    +
    • +
  • +

    The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.

    +
    • +
+

Now, to calculate the percent yield we use the equation below:

+

#""Percent Yield"" = (""actual yield"")/(""theoretical yield"") xx 100%#

+

We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by #100%#:

+

#""Percent Yield"" = (5 cancel""grams"")/(10 cancel""grams"")xx100% = 50%#

+

Thus, the percent yield is #50%#

+
+
" "
+
+
+

The percent yield is #50%#

+
+
+
+

Explanation:

+
+

Let's define those three terms before we get started:

+
    +
  • +

    The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.

    +
    • +
  • +

    The actual yield is the mass of product that was obtained in an experiment.

    +
    • +
  • +

    The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.

    +
    • +
+

Now, to calculate the percent yield we use the equation below:

+

#""Percent Yield"" = (""actual yield"")/(""theoretical yield"") xx 100%#

+

We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by #100%#:

+

#""Percent Yield"" = (5 cancel""grams"")/(10 cancel""grams"")xx100% = 50%#

+

Thus, the percent yield is #50%#

+
+
+
" "
+

If a reaction has an actual yield of 5 grams and a theoretical yield of 10 grams, what is the percent yield?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Aug 24, 2016 + +
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+
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+
+
+

The percent yield is #50%#

+
+
+
+

Explanation:

+
+

Let's define those three terms before we get started:

+
    +
  • +

    The percent yield of a reaction is the ratio of actual yield to the theoretical yield times one-hundred percent.

    +
    • +
  • +

    The actual yield is the mass of product that was obtained in an experiment.

    +
    • +
  • +

    The theoretical yield is the mass of product calculated from the amounts of reactants used in the experiment; the amount of product that you should get in theory.

    +
    • +
+

Now, to calculate the percent yield we use the equation below:

+

#""Percent Yield"" = (""actual yield"")/(""theoretical yield"") xx 100%#

+

We know the actual yield and the theoretical yield, so all we have to do is divide the numbers and multiply that value by #100%#:

+

#""Percent Yield"" = (5 cancel""grams"")/(10 cancel""grams"")xx100% = 50%#

+

Thus, the percent yield is #50%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6134 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" If a reaction has an actual yield of 5 grams and a theoretical yield of 10 grams, what is the percent yield? nan +55 a828e03c-6ddd-11ea-b414-ccda262736ce https://socratic.org/questions/a-certain-ionic-compound-is-found-to-contain-012-mol-of-sodium-012-mol-of-sulfur Na2S2O3 start chemical_formula qc_end physical_unit 11 11 8 9 mole qc_end physical_unit 15 15 8 9 mole qc_end physical_unit 20 20 17 18 mole qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Na2S2O3""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sodium [=] \\pu{.012 mol}""},{""type"":""physical unit"",""value"":""Mole [OF] sulfur [=] \\pu{.012 mol}""},{""type"":""physical unit"",""value"":""Mole [OF] oxygen [=] \\pu{.018 mol}""}]" "

A certain ionic compound is found to contain .012 mol of sodium, .012 mol of sulfur, and .018 mol of oxygen. What is its empirical formula?

" nan Na2S2O3 "
+

Explanation:

+
+

A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.

+

In your case, you know that this ionic compound contains

+
+
    +
  • #0.012# moles of sodium
    • +
  • #0.012# moles of sulfur
    • +
  • #0.018# moles of oxygen
    • +
+
+

To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one

+
+

#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#

+
+

Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition #1:1 : 1.5#.

+

Notice that if you multiply all the numbers by #2#, you can get rid of oxygen's fractional number and still keep the ratio #1:1:1.5#. This means that the empirical formula of the compound will be

+
+

#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#

+
+

For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate, #""Na""_2""S""_2""O""_3#.

+
+
" "
+
+
+

#""Na""_2""S""_2""O""_3#

+
+
+
+

Explanation:

+
+

A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.

+

In your case, you know that this ionic compound contains

+
+
    +
  • #0.012# moles of sodium
    • +
  • #0.012# moles of sulfur
    • +
  • #0.018# moles of oxygen
    • +
+
+

To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one

+
+

#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#

+
+

Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition #1:1 : 1.5#.

+

Notice that if you multiply all the numbers by #2#, you can get rid of oxygen's fractional number and still keep the ratio #1:1:1.5#. This means that the empirical formula of the compound will be

+
+

#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#

+
+

For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate, #""Na""_2""S""_2""O""_3#.

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" "
+

A certain ionic compound is found to contain .012 mol of sodium, .012 mol of sulfur, and .018 mol of oxygen. What is its empirical formula?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
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+1 Answer +
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+ + Nov 13, 2015 + +
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#""Na""_2""S""_2""O""_3#

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+

Explanation:

+
+

A compound's empirical formula tells you what the smallest whole number ratio between the elements that make up that compound is.

+

In your case, you know that this ionic compound contains

+
+
    +
  • #0.012# moles of sodium
    • +
  • #0.012# moles of sulfur
    • +
  • #0.018# moles of oxygen
    • +
+
+

To get the mole ratio that exists between these elements in the compound, divide these numbers by the smallest one

+
+

#""For Na: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (0.012color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (0.018color(red)(cancel(color(black)(""moles""))))/(0.012color(red)(cancel(color(black)(""moles"")))) = 1.5#

+
+

Now, since you can't have fractional numbers as mole ratios, you will need to find the smallest whole number ratio that satisfies the condition #1:1 : 1.5#.

+

Notice that if you multiply all the numbers by #2#, you can get rid of oxygen's fractional number and still keep the ratio #1:1:1.5#. This means that the empirical formula of the compound will be

+
+

#(""Na""_1""S""_1""O""_1.5)_2 implies ""Na""_2""S""_2""O""_3#

+
+

For ionic compounds, the empirical formula is always the same as the chemical formula, which is why you cn say that your unknown compound is actually sodium thiosulfate, #""Na""_2""S""_2""O""_3#.

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" A certain ionic compound is found to contain .012 mol of sodium, .012 mol of sulfur, and .018 mol of oxygen. What is its empirical formula? nan +56 a8290118-6ddd-11ea-ad11-ccda262736ce https://socratic.org/questions/the-molecular-weight-of-nacl-is-5844-grams-mole-if-you-had-a-10-molar-solution-1 0.10 moles start physical_unit 4 4 mole mol qc_end physical_unit 4 4 6 7 molecular_weight qc_end physical_unit 14 14 15 16 molarity qc_end physical_unit 14 14 12 13 mole qc_end physical_unit 25 25 22 23 mass qc_end physical_unit 14 14 27 28 volume qc_end physical_unit 14 14 40 41 volume qc_end end "[{""type"":""physical unit"",""value"":""Mole2 [OF] NaCl [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.10 moles""}]" "[{""type"":""physical unit"",""value"":""Molecular weight [OF] NaCl [=] \\pu{58.44 grams/mole}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] solution [=] \\pu{1.0 M}""},{""type"":""physical unit"",""value"":""Mole1 [OF] solution [=] \\pu{1.0 molar}""},{""type"":""physical unit"",""value"":""Mass1 [OF] salt [=] \\pu{58.44 g}""},{""type"":""physical unit"",""value"":""Volume1 [OF] solution [=] \\pu{1.0 liter}""},{""type"":""physical unit"",""value"":""Volume2 [OF] solution [=] \\pu{100 ml}""}]" "

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution?

" nan 0.10 moles "
+

Explanation:

+
+

The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.

+

+

This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.

+

In your case, you dissolve #""58.44 g""# of sodium chloride in #""1.0 L""# of water and make a #""1.0 M""# sodium chloride solution.

+

If you take

+
+

#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#

+
+

of this solution, this sample must have the same concentration as the initial solution, i.e. #""1.0 M""#.

+

Notice that the volume of the sample is

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#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#

+
+

times smaller than the volume of the initial solution, so it follows that it must contain #color(blue)(10)# times fewer moles of solute in order to have the same molarity.

+

Sodium chloride was said to have a molar mass of #""58.44 g mol""^(-1)#, which tells you that #1# mole of sodium chloride has a mass of #""58.44 g""#.

+

Since the initial solution was made by dissolving the equivalent of

+
+

#""58.44 g "" = "" 1 mole NaCl""#

+
+

in #""10 L""# of water, it follows that the #""0.1 L""# sample must contain

+
+

#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

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+

This is equivalent to saying that the #""0.1 L""# sample contains

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#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#

+
+

ALTERNATIVELY

+

You can get the same result by using the formula for molarity, which is

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#

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Rearrange this equation to solve for #n_'solute""# and plug in your values to find

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+

#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#

+

#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

+
+

Keep in mind that the volume must always be expressed in liters when working with molarity.

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" "
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#""0.1 moles NaCl""#

+
+
+
+

Explanation:

+
+

The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.

+

+

This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.

+

In your case, you dissolve #""58.44 g""# of sodium chloride in #""1.0 L""# of water and make a #""1.0 M""# sodium chloride solution.

+

If you take

+
+

#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#

+
+

of this solution, this sample must have the same concentration as the initial solution, i.e. #""1.0 M""#.

+

Notice that the volume of the sample is

+
+

#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#

+
+

times smaller than the volume of the initial solution, so it follows that it must contain #color(blue)(10)# times fewer moles of solute in order to have the same molarity.

+

Sodium chloride was said to have a molar mass of #""58.44 g mol""^(-1)#, which tells you that #1# mole of sodium chloride has a mass of #""58.44 g""#.

+

Since the initial solution was made by dissolving the equivalent of

+
+

#""58.44 g "" = "" 1 mole NaCl""#

+
+

in #""10 L""# of water, it follows that the #""0.1 L""# sample must contain

+
+

#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

+
+

This is equivalent to saying that the #""0.1 L""# sample contains

+
+

#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#

+
+

ALTERNATIVELY

+

You can get the same result by using the formula for molarity, which is

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#

+
+

Rearrange this equation to solve for #n_'solute""# and plug in your values to find

+
+

#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#

+

#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

+
+

Keep in mind that the volume must always be expressed in liters when working with molarity.

+
+
+
" "
+

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Jul 12, 2016 + +
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#""0.1 moles NaCl""#

+
+
+
+

Explanation:

+
+

The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.

+

+

This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.

+

In your case, you dissolve #""58.44 g""# of sodium chloride in #""1.0 L""# of water and make a #""1.0 M""# sodium chloride solution.

+

If you take

+
+

#100 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.1 L""#

+
+

of this solution, this sample must have the same concentration as the initial solution, i.e. #""1.0 M""#.

+

Notice that the volume of the sample is

+
+

#(1color(red)(cancel(color(black)(""L""))))/(0.1color(red)(cancel(color(black)(""L"")))) = color(blue)(10)#

+
+

times smaller than the volume of the initial solution, so it follows that it must contain #color(blue)(10)# times fewer moles of solute in order to have the same molarity.

+

Sodium chloride was said to have a molar mass of #""58.44 g mol""^(-1)#, which tells you that #1# mole of sodium chloride has a mass of #""58.44 g""#.

+

Since the initial solution was made by dissolving the equivalent of

+
+

#""58.44 g "" = "" 1 mole NaCl""#

+
+

in #""10 L""# of water, it follows that the #""0.1 L""# sample must contain

+
+

#""1 mole NaCl""/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

+
+

This is equivalent to saying that the #""0.1 L""# sample contains

+
+

#""58.44 g NaCl""/color(blue)(10) = ""5.844 g NaCl""#

+
+

ALTERNATIVELY

+

You can get the same result by using the formula for molarity, which is

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution""color(white)(a/a)|)))#

+
+

Rearrange this equation to solve for #n_'solute""# and plug in your values to find

+
+

#c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""#

+

#n_""NaCl"" = ""1.0 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters"")) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.1 moles NaCl"")color(white)(a/a)|)))#

+
+

Keep in mind that the volume must always be expressed in liters when working with molarity.

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Related questions
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" The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution? nan +57 a8290119-6ddd-11ea-b7e9-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-one-mole-of-beryllium-if-the-atomic-mass-of-beryllium-is-9-0 9.01 g start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 8 8 16 16 atomic_mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] beryllium [IN] g""}]" "[{""type"":""physical unit"",""value"":""9.01 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] beryllium [=] \\pu{1 mole}""},{""type"":""physical unit"",""value"":""Atomic mass [OF] beryllium [=] \\pu{9.01}""}]" "

What is the mass of one mole of beryllium if the atomic mass of beryllium is 9.01?

" nan 9.01 g "
+

Explanation:

+
+

A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.

+

Example 1: Water (#""H""_2""O""#) has two hydrogen atoms and one oxygen atom per molecule. The two hydrogen atoms each have an atomic mass of #1.008#, the oxygen atom has an atomic mass of #15.999#, the sum for all three atoms is #18.015#. So a mole of water molecules has a mass of #18.015"" g""#.

+

Example 2: Beryllium has an aromuc mass of #9.01#. So a mole of beryllium atoms has a mass of #9. +01"" g""#.

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" "
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#9.01"" g""#.

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+
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Explanation:

+
+

A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.

+

Example 1: Water (#""H""_2""O""#) has two hydrogen atoms and one oxygen atom per molecule. The two hydrogen atoms each have an atomic mass of #1.008#, the oxygen atom has an atomic mass of #15.999#, the sum for all three atoms is #18.015#. So a mole of water molecules has a mass of #18.015"" g""#.

+

Example 2: Beryllium has an aromuc mass of #9.01#. So a mole of beryllium atoms has a mass of #9. +01"" g""#.

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" "
+

What is the mass of one mole of beryllium if the atomic mass of beryllium is 9.01?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Jun 18, 2016 + +
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#9.01"" g""#.

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Explanation:

+
+

A mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in atomic mass units.

+

Example 1: Water (#""H""_2""O""#) has two hydrogen atoms and one oxygen atom per molecule. The two hydrogen atoms each have an atomic mass of #1.008#, the oxygen atom has an atomic mass of #15.999#, the sum for all three atoms is #18.015#. So a mole of water molecules has a mass of #18.015"" g""#.

+

Example 2: Beryllium has an aromuc mass of #9.01#. So a mole of beryllium atoms has a mass of #9. +01"" g""#.

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Related questions
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Impact of this question
+
+ 6170 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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" What is the mass of one mole of beryllium if the atomic mass of beryllium is 9.01? nan +58 a829011a-6ddd-11ea-9c46-ccda262736ce https://socratic.org/questions/what-is-the-percent-composition-of-carbon-in-acetic-acid 40.00% start physical_unit 6 9 percent_composition none qc_end substance 8 9 qc_end end "[{""type"": ""physical unit"", ""value"": ""Percent composition [OF] carbon in acetic acid""}]" "[{""type"":""physical unit"",""value"":""40.00%""}]" "[{""type"":""substance name"",""value"":""Acetic acid""}]" "

What is the percent composition of carbon in acetic acid?

" nan 40.00% "
+

Explanation:

+
+

Acetic acid: #""CH""_3""COOH""#

+

To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid #(""60.052 g/mol"")# by the mass of carbon #(""2""xx""12.011 g/mol"")# in one mole of acetic acid and multiply by #100#.

+

#(2xx12.011color(red)cancel(color(black)(""g""))/color(red)cancel(color(black)(""mol"")))/(60.052color(red)cancel(color(black)(""g""))/(color(red)cancel(color(black)(""mol""))))xx100=""40.002%""#

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" "
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The percent composition of carbon in acetic acid is #40.002%#.

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Explanation:

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+

Acetic acid: #""CH""_3""COOH""#

+

To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid #(""60.052 g/mol"")# by the mass of carbon #(""2""xx""12.011 g/mol"")# in one mole of acetic acid and multiply by #100#.

+

#(2xx12.011color(red)cancel(color(black)(""g""))/color(red)cancel(color(black)(""mol"")))/(60.052color(red)cancel(color(black)(""g""))/(color(red)cancel(color(black)(""mol""))))xx100=""40.002%""#

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" "
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What is the percent composition of carbon in acetic acid?

+
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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
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+1 Answer +
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+ + Jul 30, 2017 + +
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The percent composition of carbon in acetic acid is #40.002%#.

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Explanation:

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+

Acetic acid: #""CH""_3""COOH""#

+

To determine the percent composition of carbon in acetic acid, divide the molar mass of acetic acid #(""60.052 g/mol"")# by the mass of carbon #(""2""xx""12.011 g/mol"")# in one mole of acetic acid and multiply by #100#.

+

#(2xx12.011color(red)cancel(color(black)(""g""))/color(red)cancel(color(black)(""mol"")))/(60.052color(red)cancel(color(black)(""g""))/(color(red)cancel(color(black)(""mol""))))xx100=""40.002%""#

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" What is the percent composition of carbon in acetic acid? nan +59 a829011b-6ddd-11ea-adf7-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-selenium-dioxide SeO2 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""SeO2""}]" "[{""type"":""substance name"",""value"":""Selenium dioxide""}]" "

What is the formula for selenium dioxide?

" nan SeO2 "
+

Explanation:

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+

#SeO_2# or #""selenium (IV) oxide""#. This stuff does not smell very nice.

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" "
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#SeO_2#

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Explanation:

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#SeO_2# or #""selenium (IV) oxide""#. This stuff does not smell very nice.

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" "
+

What is the formula for selenium dioxide?

+
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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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+1 Answer +
+
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+ + Dec 12, 2016 + +
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#SeO_2#

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Explanation:

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#SeO_2# or #""selenium (IV) oxide""#. This stuff does not smell very nice.

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Related questions
+ + +
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+
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Impact of this question
+
+ 2657 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" What is the formula for selenium dioxide? nan +60 a829011c-6ddd-11ea-bc31-ccda262736ce https://socratic.org/questions/if-a-6-03-x-10-6-0-m-solution-of-a-weak-acid-is-20-ionized-what-is-the-pka-for-t 6.52 start physical_unit 19 20 pka none qc_end physical_unit 8 10 2 5 molarity qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""pKa [OF] the acid""}]" "[{""type"":""physical unit"",""value"":""6.52""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] a weak acid [=] \\pu{6.03 x 10^(-6.0) M}""},{""type"":""other"",""value"":""a weak acid is 20% ionized""}]" "

If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid?

" nan 6.52 "
+

Explanation:

+
+

If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:

+
+

#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#

+
+

#color(red)(I)"" ""1"" ""mol"" "" 0"" ""mol"" "" 0"" "" ""mol#

+

#color(red)(C)"" ""-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

#color(red)(E)"" ""1-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

Where #alpha# is the degree of dissociation of #HA#

+

If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows

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+
+

#[HA]=c(1-alpha)M#

+

#[H_3O^+]=calphaM#

+

#[A^-]=calphaM#

+
+
+
+
+

The concentration of #H_2O# is irrelevant as it acts as sovent.

+

Now the acid dissociation constant

+
+
+
+

#K_a=([H_3O^+][A^-])/[HA]^2#

+

#=((calpha)*(calpha))/(c(1-alpha))#

+

#=(calpha^2)/(1-alpha)#

+
+
+
+

Given #c=6.03xx10^-6M and alpha=20%=0.2#

+
+
+

#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#

+
+

#=(6.03xx10^-6xx0.04)/0.8#

+

#3.015xx10^-7#

+
+
+
+

The #pK_a# of the weak acid

+

#pK_a=-logK_a=-log(3.015xx10^-7)#

+
+
+
+

#=7-log3.015=6.52#

+
+
+
+
+
" "
+
+
+

#pK_a=6.52#

+
+
+
+

Explanation:

+
+

If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:

+
+

#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#

+
+

#color(red)(I)"" ""1"" ""mol"" "" 0"" ""mol"" "" 0"" "" ""mol#

+

#color(red)(C)"" ""-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

#color(red)(E)"" ""1-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

Where #alpha# is the degree of dissociation of #HA#

+

If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows

+
+
+
+
+

#[HA]=c(1-alpha)M#

+

#[H_3O^+]=calphaM#

+

#[A^-]=calphaM#

+
+
+
+
+

The concentration of #H_2O# is irrelevant as it acts as sovent.

+

Now the acid dissociation constant

+
+
+
+

#K_a=([H_3O^+][A^-])/[HA]^2#

+

#=((calpha)*(calpha))/(c(1-alpha))#

+

#=(calpha^2)/(1-alpha)#

+
+
+
+

Given #c=6.03xx10^-6M and alpha=20%=0.2#

+
+
+

#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#

+
+

#=(6.03xx10^-6xx0.04)/0.8#

+

#3.015xx10^-7#

+
+
+
+

The #pK_a# of the weak acid

+

#pK_a=-logK_a=-log(3.015xx10^-7)#

+
+
+
+

#=7-log3.015=6.52#

+
+
+
+
+
+
" "
+

If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
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+1 Answer +
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+ +
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+ +
+ + Jul 28, 2016 + +
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+
+

#pK_a=6.52#

+
+
+
+

Explanation:

+
+

If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:

+
+

#HA"" ""+H_2O""""rightleftharpoons""""H_3O^+""""+"" ""A^-#

+
+

#color(red)(I)"" ""1"" ""mol"" "" 0"" ""mol"" "" 0"" "" ""mol#

+

#color(red)(C)"" ""-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

#color(red)(E)"" ""1-alpha"" ""mol"" "" alpha"" ""mol"" "" alpha"" "" ""mol#

+

Where #alpha# is the degree of dissociation of #HA#

+

If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows

+
+
+
+
+

#[HA]=c(1-alpha)M#

+

#[H_3O^+]=calphaM#

+

#[A^-]=calphaM#

+
+
+
+
+

The concentration of #H_2O# is irrelevant as it acts as sovent.

+

Now the acid dissociation constant

+
+
+
+

#K_a=([H_3O^+][A^-])/[HA]^2#

+

#=((calpha)*(calpha))/(c(1-alpha))#

+

#=(calpha^2)/(1-alpha)#

+
+
+
+

Given #c=6.03xx10^-6M and alpha=20%=0.2#

+
+
+

#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#

+
+

#=(6.03xx10^-6xx0.04)/0.8#

+

#3.015xx10^-7#

+
+
+
+

The #pK_a# of the weak acid

+

#pK_a=-logK_a=-log(3.015xx10^-7)#

+
+
+
+

#=7-log3.015=6.52#

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+
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" If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid? nan +61 a82932b6-6ddd-11ea-ac30-ccda262736ce https://socratic.org/questions/when-the-following-equation-is-balanced-what-is-the-coefficient-for-mg-mg-s-hcl- 1 start physical_unit 11 11 coefficient none qc_end chemical_equation 12 18 qc_end end "[{""type"":""physical unit"",""value"":""Coefficient [OF] Mg""}]" "[{""type"":""physical unit"",""value"":""1""}]" "[{""type"":""chemical equation"",""value"":""Mg(s) + HCl(aq) -> MgCl2(aq) + H2(g)""}]" "

When the following equation is balanced, what is the coefficient for Mg? #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)#

" nan 1 "
+

Explanation:

+
+

So, to balance a chemical equation we make the number of each element the same on both sides.

+

We have:
+#Mg (s) + HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg= 1
+H= 1
+Cl= 1

+

RHS
+Mg= 1
+Cl=2
+H=2

+

To make Cl and H have 2 on both sides, we place a 2 in front of the HCl

+

#Mg (s) + 2HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg=1, H=2, Cl=2

+

RHS
+Mg=1, H=1, Cl=2

+

Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.

+
+
" "
+
+
+

1 is the coefficient. See below.

+
+
+
+

Explanation:

+
+

So, to balance a chemical equation we make the number of each element the same on both sides.

+

We have:
+#Mg (s) + HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg= 1
+H= 1
+Cl= 1

+

RHS
+Mg= 1
+Cl=2
+H=2

+

To make Cl and H have 2 on both sides, we place a 2 in front of the HCl

+

#Mg (s) + 2HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg=1, H=2, Cl=2

+

RHS
+Mg=1, H=1, Cl=2

+

Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.

+
+
+
" "
+

When the following equation is balanced, what is the coefficient for Mg? #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)#

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
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+
+
+1 Answer +
+
+
+
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+ +
+
+ +
+ + May 22, 2017 + +
+
+
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+
+
+

1 is the coefficient. See below.

+
+
+
+

Explanation:

+
+

So, to balance a chemical equation we make the number of each element the same on both sides.

+

We have:
+#Mg (s) + HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg= 1
+H= 1
+Cl= 1

+

RHS
+Mg= 1
+Cl=2
+H=2

+

To make Cl and H have 2 on both sides, we place a 2 in front of the HCl

+

#Mg (s) + 2HCl (aq) → MgCl_2 (aq) + H_2 (g)#

+

LHS
+Mg=1, H=2, Cl=2

+

RHS
+Mg=1, H=1, Cl=2

+

Thus the equation is balanced and the coefficient of Mg is 1, which is understood so it is not written in the final equation.

+
+
+
+
+
+ +
+
+
+
+
+
+
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" When the following equation is balanced, what is the coefficient for Mg? #Mg(s) + HCl(aq) -> MgCl_2(aq) + H_2(g)# nan +62 a82932b7-6ddd-11ea-8df5-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-zinc-bromide ZnBr2 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""ZnBr2""}]" "[{""type"":""substance name"",""value"":""Zinc bromide""}]" "

What is the formula for zinc bromide?

" nan ZnBr2 "
+

Explanation:

+
+

The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.

+

Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.

+

Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...

+

This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.

+

+ +

+

Hope this helps!

+
+
" "
+
+
+

#ZnBr_2#

+
+
+
+

Explanation:

+
+

The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.

+

Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.

+

Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...

+

This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.

+

+ +

+

Hope this helps!

+
+
+
" "
+

What is the formula for zinc bromide?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 17, 2016 + +
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+
+

#ZnBr_2#

+
+
+
+

Explanation:

+
+

The zinc ion will have a charge of +2 and the bromide ion has a charge of -1 because it is a halogen.

+

Combining the ions in a 1:2 ratio produces a compound which is electrically neutral.

+

Zinc can sometimes have other oxidation states (+1 for example) but these are pretty rare...

+

This video discusses some more examples of how to figure out a formula for a compound from its name or vice versa.

+

+ +

+

Hope this helps!

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 15185 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the formula for zinc bromide? nan +63 a82932b8-6ddd-11ea-b119-ccda262736ce https://socratic.org/questions/5967907511ef6b0e52613b2b 1.43 g start physical_unit 7 8 mass g qc_end c_other STP qc_end physical_unit 7 8 5 5 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] dioxygen gas [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.43 g""}]" "[{""type"":""other"",""value"":""STP""},{""type"":""physical unit"",""value"":""Volume [OF] dioxygen gas [=] \\pu{1 L}""}]" "

What mass is associated with #1*L# of dioxygen gas at #""STP""#?

" nan 1.43 g "
+

Explanation:

+
+

I am a bit cagy when I quote these values, because there are different standards in each different syllabus......

+

AS far as I know, #""1 mole""# of Ideal Gas occupies #22.4*L# under conditions of #""STP""#.........And of course, we know that #1*mol# of oxygen gas has a mass of #32.0*g# because we deal with the dioxygen molecule; in fact, most of the elemental gases (certainly the ones with any chemistry) are binuclear under standard conditions; this is something you need to know......

+

And thus............

+

#""Mass of dioxygen gas""# #=# #(1*Lxx32.0*g*mol^-1)/(22.4*L*mol^-1)# #~=1.5*g#.

+
+
" "
+
+
+

Well, #22.4*L# of oxygen gas at #""STP""# has a mass of #32.00*g#.

+
+
+
+

Explanation:

+
+

I am a bit cagy when I quote these values, because there are different standards in each different syllabus......

+

AS far as I know, #""1 mole""# of Ideal Gas occupies #22.4*L# under conditions of #""STP""#.........And of course, we know that #1*mol# of oxygen gas has a mass of #32.0*g# because we deal with the dioxygen molecule; in fact, most of the elemental gases (certainly the ones with any chemistry) are binuclear under standard conditions; this is something you need to know......

+

And thus............

+

#""Mass of dioxygen gas""# #=# #(1*Lxx32.0*g*mol^-1)/(22.4*L*mol^-1)# #~=1.5*g#.

+
+
+
" "
+

What mass is associated with #1*L# of dioxygen gas at #""STP""#?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
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+
+2 Answers +
+
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+
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+
+ +
+
+ +
+ + Jul 13, 2017 + +
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+

Well, #22.4*L# of oxygen gas at #""STP""# has a mass of #32.00*g#.

+
+
+
+

Explanation:

+
+

I am a bit cagy when I quote these values, because there are different standards in each different syllabus......

+

AS far as I know, #""1 mole""# of Ideal Gas occupies #22.4*L# under conditions of #""STP""#.........And of course, we know that #1*mol# of oxygen gas has a mass of #32.0*g# because we deal with the dioxygen molecule; in fact, most of the elemental gases (certainly the ones with any chemistry) are binuclear under standard conditions; this is something you need to know......

+

And thus............

+

#""Mass of dioxygen gas""# #=# #(1*Lxx32.0*g*mol^-1)/(22.4*L*mol^-1)# #~=1.5*g#.

+
+
+
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+
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+
+ +
+
+ +
+ + Jul 13, 2017 + +
+
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+
+

#1.43# #""g O""_2#

+

(or #1# #""g O""_2# with #1# significant figure)

+
+
+
+

Explanation:

+
+

We're asked to find the mass, in #""g""# of #1# #""L""# of oxygen gas at standard temperature and pressure.

+

Chemists often use this fact: One mole of an (ideal) gas at standard temperature and pressure occupies a volume of #22.4# #""L""#.

+
+

(The definition of standard temperature and pressure varies slightly depending on usage, but you'll be using this fact most of the time.)

+
+

With that being said, we can use the conversion factor

+

#(1color(white)(l)""mol"")/(22.4color(white)(l)""L"")#

+

to convert the given #1# #""L""# oxygen to moles:

+

#1cancel(""L O""_2)((1color(white)(l)""mol O""_2)/(22.4cancel(""L O""_2))) = color(red)(0.0446# #color(red)(""mol O""_2#

+

Now that we know the number of moles of #""O""_2#, we can calculate the number of grams using its molar mass (#32.00# #""g/mol""#):

+

#color(red)(0.0446)cancel(color(red)(""mol O""_2))((32.00color(white)(l)""g O""_2)/(1cancel(""mol O""_2))) = color(blue)(1.43# #color(blue)(""g O""_2#

+

Which, if you wish to round to one significant figure, is simply #colorblue)(1# #color(blue)(""g O""_2#. (Very rarely are calculations done with only one significant figure.)

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" "What mass is associated with #1*L# of dioxygen gas at #""STP""#?" nan +64 a829500c-6ddd-11ea-a1d1-ccda262736ce https://socratic.org/questions/what-is-the-maximum-amount-of-silver-in-grams-that-can-be-plated-out-of-5-0-l-of 181.80 grams start physical_unit 6 6 maximum_amount g qc_end physical_unit 19 23 22 22 mass_percent qc_end physical_unit 31 32 34 35 density qc_end physical_unit 19 20 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Maximum amount [OF] silver [IN] grams""}]" "[{""type"":""physical unit"",""value"":""181.80 grams""}]" "[{""type"":""physical unit"",""value"":""Percentage by mass [OF] AgNO3 solution containing 3.6% Ag [=] \\pu{3.6%}""},{""type"":""physical unit"",""value"":""Density [OF] the solution [=] \\pu{1.01 g/mL}""},{""type"":""physical unit"",""value"":""Volume [OF] AgNO3 solution [=] \\pu{5.0 L}""}]" "

What is the maximum amount of silver (in grams) that can be plated out of 5.0 L of an #AgNO_3# solution containing 3.6% Ag by mass? Assume that the density of the solution is 1.01 g/mL?

" nan 181.80 grams "
+

Explanation:

+
+

#""Mass of solution""# #=# #5.0*cancelLxx10^3*cancel(mL)*cancel(L^-1)xx1.01*g*cancel(mL^-1)# #=# #5050*g#

+

If silver ion concentration is #3.6%(m/m)#, then this corresponds to #5050*gxx3.6%=??*g# of silver ion.

+
+
" "
+
+
+

Under #200*g#

+
+
+
+

Explanation:

+
+

#""Mass of solution""# #=# #5.0*cancelLxx10^3*cancel(mL)*cancel(L^-1)xx1.01*g*cancel(mL^-1)# #=# #5050*g#

+

If silver ion concentration is #3.6%(m/m)#, then this corresponds to #5050*gxx3.6%=??*g# of silver ion.

+
+
+
" "
+

What is the maximum amount of silver (in grams) that can be plated out of 5.0 L of an #AgNO_3# solution containing 3.6% Ag by mass? Assume that the density of the solution is 1.01 g/mL?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solution Formation + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 24, 2016 + +
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+

Under #200*g#

+
+
+
+

Explanation:

+
+

#""Mass of solution""# #=# #5.0*cancelLxx10^3*cancel(mL)*cancel(L^-1)xx1.01*g*cancel(mL^-1)# #=# #5050*g#

+

If silver ion concentration is #3.6%(m/m)#, then this corresponds to #5050*gxx3.6%=??*g# of silver ion.

+
+
+
+
+
+ +
+
+
+
+
+
+
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" What is the maximum amount of silver (in grams) that can be plated out of 5.0 L of an #AgNO_3# solution containing 3.6% Ag by mass? Assume that the density of the solution is 1.01 g/mL? nan +65 a829500d-6ddd-11ea-83cd-ccda262736ce https://socratic.org/questions/how-many-molecules-of-carbon-dioxide-will-be-produced-if-20-molecules-of-propane 60 start physical_unit 2 5 number none qc_end physical_unit 11 13 10 10 number qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] molecules of carbon dioxide""}]" "[{""type"":""physical unit"",""value"":""60""}]" "[{""type"":""physical unit"",""value"":""Number [OF] molecules of propane [=] \\pu{20}""}]" "

How many molecules of carbon dioxide will be produced if 20 molecules of propane react?

" nan 60 "
+

Explanation:

+
+

You need a stoichiometrically balanced equation to represent the combustion:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

So are charge and mass balanced here?

+

Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.

+
+
" "
+
+
+

#""60 such molecules........""#

+
+
+
+

Explanation:

+
+

You need a stoichiometrically balanced equation to represent the combustion:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

So are charge and mass balanced here?

+

Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.

+
+
+
" "
+

How many molecules of carbon dioxide will be produced if 20 molecules of propane react?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 1, 2017 + +
+
+
+
+
+
+
+

#""60 such molecules........""#

+
+
+
+

Explanation:

+
+

You need a stoichiometrically balanced equation to represent the combustion:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

So are charge and mass balanced here?

+

Clearly, the reaction specifies 3 equiv of carbon dioxide per equiv of propane. If we start with 20 molecules of propane, these will necessarily be 60 carbon dioxide product molecules upon complete combustion.

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+
+
" How many molecules of carbon dioxide will be produced if 20 molecules of propane react? nan +66 a829500e-6ddd-11ea-91a0-ccda262736ce https://socratic.org/questions/how-much-water-must-be-evaporated-from-60-kilograms-of-a-15-salt-solution-in-ord 28.50 kilograms start physical_unit 2 5 mass kg qc_end physical_unit 12 12 11 11 mass_percent qc_end physical_unit 12 12 19 19 mass_percent qc_end physical_unit 12 13 7 8 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] water must be evaporated [IN] kilograms""}]" "[{""type"":""physical unit"",""value"":""28.50 kilograms""}]" "[{""type"":""physical unit"",""value"":""Percentage1 by mass [OF] salt [=] \\pu{15%}""},{""type"":""physical unit"",""value"":""Percentage2 by mass [OF] salt [=] \\pu{40%}""},{""type"":""physical unit"",""value"":""Mass1 [OF] salt solution [=] \\pu{60 kilograms}""}]" "

How much water must be evaporated from 60 kilograms of a 15 salt solution in order to obtain a 40% solution?

" nan 28.50 kilograms "
+

Explanation:

+
+

The initial strength of the solution is 15% m/m. [Mass/Mass]

+

That means, the salt present in the solution is = #60 * 15/100# kg = #9# kg

+

The water present = #60 - 9# kg = #51# kg

+

If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.

+

Let's assume that we have evaporated #x# kg of water.

+

According to the condition,

+

#(9)/(51-x)*100 = 40#

+

#rArr (900)/(51-x) = 40#

+

#rArr 51 - x = 900/40#

+

#rArr 51 - x = 22.5 #

+

#rArr x = 28.5#

+

That means 28.5 kg of water needs to be evaporated.

+
+
" "
+
+
+

28.5 kg

+
+
+
+

Explanation:

+
+

The initial strength of the solution is 15% m/m. [Mass/Mass]

+

That means, the salt present in the solution is = #60 * 15/100# kg = #9# kg

+

The water present = #60 - 9# kg = #51# kg

+

If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.

+

Let's assume that we have evaporated #x# kg of water.

+

According to the condition,

+

#(9)/(51-x)*100 = 40#

+

#rArr (900)/(51-x) = 40#

+

#rArr 51 - x = 900/40#

+

#rArr 51 - x = 22.5 #

+

#rArr x = 28.5#

+

That means 28.5 kg of water needs to be evaporated.

+
+
+
" "
+

How much water must be evaporated from 60 kilograms of a 15 salt solution in order to obtain a 40% solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solutions + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 30, 2017 + +
+
+
+
+
+
+
+

28.5 kg

+
+
+
+

Explanation:

+
+

The initial strength of the solution is 15% m/m. [Mass/Mass]

+

That means, the salt present in the solution is = #60 * 15/100# kg = #9# kg

+

The water present = #60 - 9# kg = #51# kg

+

If we evaporate some of the water off the solution, the salt should not change its mass, unless it decomposes or something like that.

+

Let's assume that we have evaporated #x# kg of water.

+

According to the condition,

+

#(9)/(51-x)*100 = 40#

+

#rArr (900)/(51-x) = 40#

+

#rArr 51 - x = 900/40#

+

#rArr 51 - x = 22.5 #

+

#rArr x = 28.5#

+

That means 28.5 kg of water needs to be evaporated.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2366 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How much water must be evaporated from 60 kilograms of a 15 salt solution in order to obtain a 40% solution? nan +67 a8297726-6ddd-11ea-aa2d-ccda262736ce https://socratic.org/questions/the-weight-of-a-diamond-is-given-in-carats-one-carat-is-equivalent-to-200-mg-a-p 1.00 × 10^22 start physical_unit 24 25 number none qc_end physical_unit 4 4 33 34 mass qc_end c_other OTHER qc_end c_other OTHER qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] carbon atoms""}]" "[{""type"":""physical unit"",""value"":""1.00 × 10^22""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] diamond [=] \\pu{1.00 carat}""},{""type"":""other"",""value"":""A pure diamond is made up entirely of carbon atoms.""},{""type"":""other"",""value"":""The weight of a diamond is given in carats.""},{""type"":""other"",""value"":""One carat is equivalent to 200 mg""}]" "

The weight of a diamond is given in carats. One carat is equivalent to 200 mg. A pure diamond is made up entirely of carbon atoms. How many carbon atoms make up a 1.00 carat diamond?

" nan 1.00 × 10^22 "
+

Explanation:

+
+

Convert 200 mg to g.

+

#""1000 mg=1 g""#

+

#200cancel""mg""xx(1""g"")/(1000cancel""mg"")=""0.2 g""#

+

Convert carats to grams to moles.

+

#""1.00 ct=0.2 g""#
+#1""mol C=12.0107 g C""#

+

#1.00cancel""ct C""xx(0.2cancel""g C"")/(1cancel""ct C"")xx(1""mol C"")/(12.0107cancel""g C"")=""0.02 mol C""#

+

Convert moles C to atoms C.

+

#1 ""mol C""=6.022xx10^23 ""atoms C""#

+

#0.02cancel""mol C""xx(6.022xx10^23""atoms C"")/(1cancel""mol C"")=1xx10^22"" atoms C""# (rounded to one significant figure)

+
+
" "
+
+
+

#1xx10^22""atoms C""# are in a 1.00 carat diamond.

+
+
+
+

Explanation:

+
+

Convert 200 mg to g.

+

#""1000 mg=1 g""#

+

#200cancel""mg""xx(1""g"")/(1000cancel""mg"")=""0.2 g""#

+

Convert carats to grams to moles.

+

#""1.00 ct=0.2 g""#
+#1""mol C=12.0107 g C""#

+

#1.00cancel""ct C""xx(0.2cancel""g C"")/(1cancel""ct C"")xx(1""mol C"")/(12.0107cancel""g C"")=""0.02 mol C""#

+

Convert moles C to atoms C.

+

#1 ""mol C""=6.022xx10^23 ""atoms C""#

+

#0.02cancel""mol C""xx(6.022xx10^23""atoms C"")/(1cancel""mol C"")=1xx10^22"" atoms C""# (rounded to one significant figure)

+
+
+
" "
+

The weight of a diamond is given in carats. One carat is equivalent to 200 mg. A pure diamond is made up entirely of carbon atoms. How many carbon atoms make up a 1.00 carat diamond?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 23, 2015 + +
+
+
+
+
+
+
+

#1xx10^22""atoms C""# are in a 1.00 carat diamond.

+
+
+
+

Explanation:

+
+

Convert 200 mg to g.

+

#""1000 mg=1 g""#

+

#200cancel""mg""xx(1""g"")/(1000cancel""mg"")=""0.2 g""#

+

Convert carats to grams to moles.

+

#""1.00 ct=0.2 g""#
+#1""mol C=12.0107 g C""#

+

#1.00cancel""ct C""xx(0.2cancel""g C"")/(1cancel""ct C"")xx(1""mol C"")/(12.0107cancel""g C"")=""0.02 mol C""#

+

Convert moles C to atoms C.

+

#1 ""mol C""=6.022xx10^23 ""atoms C""#

+

#0.02cancel""mol C""xx(6.022xx10^23""atoms C"")/(1cancel""mol C"")=1xx10^22"" atoms C""# (rounded to one significant figure)

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 33843 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" The weight of a diamond is given in carats. One carat is equivalent to 200 mg. A pure diamond is made up entirely of carbon atoms. How many carbon atoms make up a 1.00 carat diamond? nan +68 a8297727-6ddd-11ea-b89b-ccda262736ce https://socratic.org/questions/5919b86211ef6b7ce57ca6c8 22.68 kJ start physical_unit 10 10 energy kj qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 12 13 temperature qc_end end "[{""type"":""physical unit"",""value"":""Required energy [OF] water [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""22.68 kJ""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{72 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{75 ℃}""}]" "

How much energy is required to heat #72# #g# of water by #75^o# #C#?

" nan 22.68 kJ "
+

Explanation:

+
+

The 'specific heat of water' is #4.184# #Jg^-1K^-1#, but is often approximated as #4.2#. (Given that we have the data to only two significant digits, #4.2# is appropriate to use in this case.)

+

That means it takes #4.2# #J# to raise the temperature of one gram of water by one kelvin, which is the same size unit as one degree celsius. Because we're only talking about a change in temperature, we can use kelvin and celsius interchangeably.

+

In this case, we have a mass, #m#, of #72# #g# of water, with a specific heat, #C#, of #4.2# #Jg^-1K^-1# and we raise its temperature by #\DeltaT#, #75^o# #C#. The symbol '#\Delta#' just means 'the change in'.

+

Over all, then, the change in energy of the water, #\DeltaH#, is #22.7# #kJ#. Since this number is positive, that is an amount of energy we had to put into the water to heat it up.

+
+
" "
+
+
+

We use the equation #\DeltaH=mC\DeltaT#.

+

#\DeltaH=72 xx 4.2 xx 75 = 22,680# #J# = #22.7# #kJ#

+
+
+
+

Explanation:

+
+

The 'specific heat of water' is #4.184# #Jg^-1K^-1#, but is often approximated as #4.2#. (Given that we have the data to only two significant digits, #4.2# is appropriate to use in this case.)

+

That means it takes #4.2# #J# to raise the temperature of one gram of water by one kelvin, which is the same size unit as one degree celsius. Because we're only talking about a change in temperature, we can use kelvin and celsius interchangeably.

+

In this case, we have a mass, #m#, of #72# #g# of water, with a specific heat, #C#, of #4.2# #Jg^-1K^-1# and we raise its temperature by #\DeltaT#, #75^o# #C#. The symbol '#\Delta#' just means 'the change in'.

+

Over all, then, the change in energy of the water, #\DeltaH#, is #22.7# #kJ#. Since this number is positive, that is an amount of energy we had to put into the water to heat it up.

+
+
+
" "
+

How much energy is required to heat #72# #g# of water by #75^o# #C#?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 15, 2017 + +
+
+
+
+
+
+
+

We use the equation #\DeltaH=mC\DeltaT#.

+

#\DeltaH=72 xx 4.2 xx 75 = 22,680# #J# = #22.7# #kJ#

+
+
+
+

Explanation:

+
+

The 'specific heat of water' is #4.184# #Jg^-1K^-1#, but is often approximated as #4.2#. (Given that we have the data to only two significant digits, #4.2# is appropriate to use in this case.)

+

That means it takes #4.2# #J# to raise the temperature of one gram of water by one kelvin, which is the same size unit as one degree celsius. Because we're only talking about a change in temperature, we can use kelvin and celsius interchangeably.

+

In this case, we have a mass, #m#, of #72# #g# of water, with a specific heat, #C#, of #4.2# #Jg^-1K^-1# and we raise its temperature by #\DeltaT#, #75^o# #C#. The symbol '#\Delta#' just means 'the change in'.

+

Over all, then, the change in energy of the water, #\DeltaH#, is #22.7# #kJ#. Since this number is positive, that is an amount of energy we had to put into the water to heat it up.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1610 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How much energy is required to heat #72# #g# of water by #75^o# #C#? nan +69 a8297728-6ddd-11ea-aa88-ccda262736ce https://socratic.org/questions/how-much-heat-is-given-off-when-10-grams-of-water-are-cooled-from-50-c-to-40-c 420.00 J start physical_unit 10 10 heat_energy j qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 14 15 temperature qc_end physical_unit 10 10 17 18 temperature qc_end end "[{""type"":""physical unit"",""value"":""Heat [OF] water [IN] J""}]" "[{""type"":""physical unit"",""value"":""420.00 J""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{10 grams}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] water [=] \\pu{50 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{40 °C}""}]" "

How much heat is given off when 10 grams of water are cooled from 50°C to 40°C?

" nan 420.00 J "
+

Explanation:

+
+

Convert 10 grams to pounds:
+#10 * 0.0022 = 0.022# pound

+

Convert 50 degrees C to Fahrenheit:
+#(50 * 9 / 5) + 32 = 122# degrees Fahrenheit

+

Convert 40 degrees C to Fahrenheit:
+#(40 * 9/5) + 32 = 104# degrees Fahrenheit

+

Calculate the change in temperature:
+#122 - 104 = 18# degrees Fahrenheit

+

Heat lost in British Thermal Units (BTU):
+#0.022 * 18 = 0.396# BTU

+
+
" "
+
+
+

0.396 BTU

+
+
+
+

Explanation:

+
+

Convert 10 grams to pounds:
+#10 * 0.0022 = 0.022# pound

+

Convert 50 degrees C to Fahrenheit:
+#(50 * 9 / 5) + 32 = 122# degrees Fahrenheit

+

Convert 40 degrees C to Fahrenheit:
+#(40 * 9/5) + 32 = 104# degrees Fahrenheit

+

Calculate the change in temperature:
+#122 - 104 = 18# degrees Fahrenheit

+

Heat lost in British Thermal Units (BTU):
+#0.022 * 18 = 0.396# BTU

+
+
+
" "
+

How much heat is given off when 10 grams of water are cooled from 50°C to 40°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Oct 12, 2016 + +
+
+
+
+
+
+
+

0.396 BTU

+
+
+
+

Explanation:

+
+

Convert 10 grams to pounds:
+#10 * 0.0022 = 0.022# pound

+

Convert 50 degrees C to Fahrenheit:
+#(50 * 9 / 5) + 32 = 122# degrees Fahrenheit

+

Convert 40 degrees C to Fahrenheit:
+#(40 * 9/5) + 32 = 104# degrees Fahrenheit

+

Calculate the change in temperature:
+#122 - 104 = 18# degrees Fahrenheit

+

Heat lost in British Thermal Units (BTU):
+#0.022 * 18 = 0.396# BTU

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Oct 13, 2016 + +
+
+
+
+
+
+
+

The process will give off 420 J of heat.

+
+
+
+

Explanation:

+
+
+

The energy #q# required to heat an object is given by the formula

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))"" ""#

+
+
+

where

+

#m# is the mass
+#c# is the specific heat capacity
+#ΔT# is the change in temperature

+
+

In your problem,

+

#m = ""10 g""#
+#c = ""4.184 J·K""^""-1""""g""^""-1""#
+#ΔT = T_""f"" - T_""i"" = ""40 °C - 50 °C = -10 °C""#

+
+

∴ #q = ""10"" color(red)(cancel(color(black)(""g""))) × ""4.184 J""·color(red)(cancel(color(black)(""K""^""-1""""g""^""-1""))) × (""-10"" color(red)(cancel(color(black)(""°C"")))) = ""-420 J""#

+

The negative sign tells you that energy is being given off.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 11389 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How much heat is given off when 10 grams of water are cooled from 50°C to 40°C? nan +70 a8297729-6ddd-11ea-8ec4-ccda262736ce https://socratic.org/questions/what-is-the-mass-in-grams-of-7-20-mol-of-antimony 876.67 grams start physical_unit 10 10 mass g qc_end physical_unit 10 10 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] antimony [IN] grams""}]" "[{""type"":""physical unit"",""value"":""876.67 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] antimony [=] \\pu{7.20 mol}""}]" "

What is the mass in grams of 7.20 mol of antimony?

" nan 876.67 grams "
+

Explanation:

+
+

#Sb# has an atomic mass of #121.76# #g*mol^-1#.

+

If there are #7# #""moles""#, then #7# #cancel(""mol"")# #xx# #121.76# #g*cancel(mol^-1)##=# #??# #mol#.

+
+
" "
+
+
+

#1# #mol# of antimony has a mass of #121.76#. If there are #7.20# mol, then...........?

+
+
+
+

Explanation:

+
+

#Sb# has an atomic mass of #121.76# #g*mol^-1#.

+

If there are #7# #""moles""#, then #7# #cancel(""mol"")# #xx# #121.76# #g*cancel(mol^-1)##=# #??# #mol#.

+
+
+
" "
+

What is the mass in grams of 7.20 mol of antimony?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Feb 4, 2016 + +
+
+
+
+
+
+
+

#1# #mol# of antimony has a mass of #121.76#. If there are #7.20# mol, then...........?

+
+
+
+

Explanation:

+
+

#Sb# has an atomic mass of #121.76# #g*mol^-1#.

+

If there are #7# #""moles""#, then #7# #cancel(""mol"")# #xx# #121.76# #g*cancel(mol^-1)##=# #??# #mol#.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Feb 4, 2016 + +
+
+
+
+
+
+
+

#876.6g#

+
+
+
+

Explanation:

+
+

From the periodic table, the molar mass of Sb is #121.75g//mol#.

+

Since #n=m/(M_r)#, we can conclude that

+

#m=nxxM_r#

+

#=7.20xx121.75#

+

#=876.6g#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 7091 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the mass in grams of 7.20 mol of antimony? nan +71 a829772a-6ddd-11ea-a881-ccda262736ce https://socratic.org/questions/how-many-kilograms-of-salt-nacl-can-be-dissolved-in-2-liters-of-water-at-25-degr 0.73 kilograms start physical_unit 4 5 mass kg qc_end physical_unit 13 13 10 11 volume qc_end physical_unit 13 13 15 17 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] salt (NaCl) [IN] kilograms""}]" "[{""type"":""physical unit"",""value"":""0.73 kilograms""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{2 liters}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{25 degrees Celsius}""},{""type"":""other"",""value"":""The solubility product constant, Ksp, is 38.65.""}]" "

How many kilograms of salt (NaCl) can be dissolved in 2 liters of water at 25 degrees Celsius? The solubility product constant, Ksp, is 38.65

" nan 0.73 kilograms "
+

Explanation:

+
+

Ksp = (Na^+) x (Cl^-) = 38,65
+but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
+g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
+in 2 liters = 726 g

+
+
" "
+
+
+

726 g

+
+
+
+

Explanation:

+
+

Ksp = (Na^+) x (Cl^-) = 38,65
+but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
+g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
+in 2 liters = 726 g

+
+
+
" "
+

How many kilograms of salt (NaCl) can be dissolved in 2 liters of water at 25 degrees Celsius? The solubility product constant, Ksp, is 38.65

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Ksp + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 6, 2016 + +
+
+
+
+
+
+
+

726 g

+
+
+
+

Explanation:

+
+

Ksp = (Na^+) x (Cl^-) = 38,65
+but (Na^+) = (Cl^-) = sqrt (38,65) = 6,2 mol /L
+g/L = mol/L x g/mol = 6,2 x 58,45 = 363 g/L
+in 2 liters = 726 g

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4607 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
+
" How many kilograms of salt (NaCl) can be dissolved in 2 liters of water at 25 degrees Celsius? The solubility product constant, Ksp, is 38.65 nan +72 a829cb1a-6ddd-11ea-b8b6-ccda262736ce https://socratic.org/questions/if-it-takes-21-966-j-of-heat-energy-to-warm-750-g-of-water-what-was-the-temperat 6.97 ℃ start physical_unit 13 13 temperature °c qc_end physical_unit 13 13 3 4 heat_energy qc_end physical_unit 13 13 10 11 mass qc_end end "[{""type"":""physical unit"",""value"":""Changed temperature [OF] water [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""6.97 ℃""}]" "[{""type"":""physical unit"",""value"":""Heat energy [OF] water [=] \\pu{21966 J}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{750 g}""}]" "

If it takes 21,966 J of heat energy to warm 750 g of water, what was the temperature change?

" nan 6.97 ℃ "
+

Explanation:

+
+

Use the equation:

+

#q=mcDeltat#

+

where #q# is the heat gained or lost in Joules, #c# is specific heat capacity, #m# is mass, and #Deltat# is the temperature change, which is #(T_""final""-T_""initial"")#.

+

The specific heat of water, #(c_""H2O"")# is #4.184 ""J""/(""g""*^@""C"")#
+https://water.usgs.gov/edu/heat-capacity.html

+

Organize the Information

+

Given/Known
+#c_""H2O""=4.184 ""J""/(""g""*^@""C"")#
+#m=""750 g""#
+#q=""21966 J""#

+

Unknown: #Deltat#

+

Solution
+Rearrange the equation to isolate #Deltat#. Substitute the given/known values into the equation and solve.

+

#Deltat=q/(mxxc)#

+

#Deltat=(21966color(red)cancel(color(black)(""J"")))/(750color(red)cancel(color(black)(""g""))xx(4.184 color(red)cancel(color(black)(""J""))/(color(red)cancel(color(black)(""g""))*^@""C"")))=""7.0""^@""C""#

+
+
" "
+
+
+

The temperature change was #7.0^@""C""#.

+
+
+
+

Explanation:

+
+

Use the equation:

+

#q=mcDeltat#

+

where #q# is the heat gained or lost in Joules, #c# is specific heat capacity, #m# is mass, and #Deltat# is the temperature change, which is #(T_""final""-T_""initial"")#.

+

The specific heat of water, #(c_""H2O"")# is #4.184 ""J""/(""g""*^@""C"")#
+https://water.usgs.gov/edu/heat-capacity.html

+

Organize the Information

+

Given/Known
+#c_""H2O""=4.184 ""J""/(""g""*^@""C"")#
+#m=""750 g""#
+#q=""21966 J""#

+

Unknown: #Deltat#

+

Solution
+Rearrange the equation to isolate #Deltat#. Substitute the given/known values into the equation and solve.

+

#Deltat=q/(mxxc)#

+

#Deltat=(21966color(red)cancel(color(black)(""J"")))/(750color(red)cancel(color(black)(""g""))xx(4.184 color(red)cancel(color(black)(""J""))/(color(red)cancel(color(black)(""g""))*^@""C"")))=""7.0""^@""C""#

+
+
+
" "
+

If it takes 21,966 J of heat energy to warm 750 g of water, what was the temperature change?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 18, 2017 + +
+
+
+
+
+
+
+

The temperature change was #7.0^@""C""#.

+
+
+
+

Explanation:

+
+

Use the equation:

+

#q=mcDeltat#

+

where #q# is the heat gained or lost in Joules, #c# is specific heat capacity, #m# is mass, and #Deltat# is the temperature change, which is #(T_""final""-T_""initial"")#.

+

The specific heat of water, #(c_""H2O"")# is #4.184 ""J""/(""g""*^@""C"")#
+https://water.usgs.gov/edu/heat-capacity.html

+

Organize the Information

+

Given/Known
+#c_""H2O""=4.184 ""J""/(""g""*^@""C"")#
+#m=""750 g""#
+#q=""21966 J""#

+

Unknown: #Deltat#

+

Solution
+Rearrange the equation to isolate #Deltat#. Substitute the given/known values into the equation and solve.

+

#Deltat=q/(mxxc)#

+

#Deltat=(21966color(red)cancel(color(black)(""J"")))/(750color(red)cancel(color(black)(""g""))xx(4.184 color(red)cancel(color(black)(""J""))/(color(red)cancel(color(black)(""g""))*^@""C"")))=""7.0""^@""C""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1554 views + around the world +
+
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+ + Creative Commons License + +
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+
" If it takes 21,966 J of heat energy to warm 750 g of water, what was the temperature change? nan +73 a829eb94-6ddd-11ea-a20f-ccda262736ce https://socratic.org/questions/how-would-you-balance-the-following-chemical-equation-mercury-ii-hydroxide-phosp 3 Hg(OH)2 + 2 H3PO4 -> Hg3(PO4)2 + 6 H2O start chemical_equation qc_end chemical_equation 8 19 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""3 Hg(OH)2 + 2 H3PO4 -> Hg3(PO4)2 + 6 H2O""}]" "[{""type"":""chemical equation"",""value"":""mercury (II) hydroxide + phosphoric acid -> mercury (II) phosphate + water""}]" "

How would you balance the following chemical equation: mercury (II) hydroxide + phosphoric acid --> mercury (II) phosphate + water?

" nan 3 Hg(OH)2 + 2 H3PO4 -> Hg3(PO4)2 + 6 H2O "
+

Explanation:

+
+

Here's your unbalanced equation:

+

#""Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

Begin by fixing the obvious. We need 3 #""Hg""# on the left side and we only have 1, so add a coefficient of 3 to #""Hg(OH)""_2#

+

#""3 Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to #""H""_3""PO""_4#

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

The only thing left to the balance is the water. On the left side, we have a total of 12 #""H""# and 6 #""O""# (not counting the phosphate). This gives us exactly 6 #""H""_2""O""#, so adding a coefficient of 6 will give us our final answer:

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""6 H""_2""O""#

+

Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.

+
+
" "
+
+
+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""6 H""_2""O""#

+
+
+
+

Explanation:

+
+

Here's your unbalanced equation:

+

#""Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

Begin by fixing the obvious. We need 3 #""Hg""# on the left side and we only have 1, so add a coefficient of 3 to #""Hg(OH)""_2#

+

#""3 Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to #""H""_3""PO""_4#

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

The only thing left to the balance is the water. On the left side, we have a total of 12 #""H""# and 6 #""O""# (not counting the phosphate). This gives us exactly 6 #""H""_2""O""#, so adding a coefficient of 6 will give us our final answer:

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""6 H""_2""O""#

+

Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.

+
+
+
" "
+

How would you balance the following chemical equation: mercury (II) hydroxide + phosphoric acid --> mercury (II) phosphate + water?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 15, 2015 + +
+
+
+
+
+
+
+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""6 H""_2""O""#

+
+
+
+

Explanation:

+
+

Here's your unbalanced equation:

+

#""Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

Begin by fixing the obvious. We need 3 #""Hg""# on the left side and we only have 1, so add a coefficient of 3 to #""Hg(OH)""_2#

+

#""3 Hg(OH)""_2 + ""H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to #""H""_3""PO""_4#

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""H""_2""O""#

+

The only thing left to the balance is the water. On the left side, we have a total of 12 #""H""# and 6 #""O""# (not counting the phosphate). This gives us exactly 6 #""H""_2""O""#, so adding a coefficient of 6 will give us our final answer:

+

#""3 Hg(OH)""_2 + ""2 H""_3""PO""_4 -> ""Hg""_3(""PO""_4)_2 + ""6 H""_2""O""#

+

Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 13197 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How would you balance the following chemical equation: mercury (II) hydroxide + phosphoric acid --> mercury (II) phosphate + water? nan +74 a829eb95-6ddd-11ea-8c36-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-h-in-c9h10n2 6.85% start physical_unit 6 8 mass_percent none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Mass % [OF] H in C9H10N2""}]" "[{""type"":""physical unit"",""value"":""6.85%""}]" "[{""type"":""chemical equation"",""value"":""C9H10N2""}]" "

What is the mass % of H in C9H10N2? +

" nan 6.85% "
+

Explanation:

+
+

The molecular formula #""C""_9""H""_10""N""_2""#, indicates that in one mole of the compound, there are 9 moles of carbon atoms, 10 moles of hydrogen atoms, and 2 moles of nitrogen atoms.

+

To calculate the mass % of H atoms in #""C""_9""H""_10""N""_2""#, divide the molar mass of H atoms by the molar mass of the compound and multiply by 100.

+

Mass of #""C""_9""H""_10""N""_2"":##""146.193 g/mol""# https://www.ncbi.nlm.nih.gov/pccompound?term=C9H10N2

+

Mass of #""H""# atoms: #(10xx1.008"" g/mol"")=""10.08 g/mol""#

+

Divide and multiply by 100:

+

#(10.08color(red)cancel(color(black)(""g"")))/(146.193color(red)cancel(color(black)(""g"")))xx100=""6.895% H""#

+
+
" "
+
+
+

The mass % of #""H""# in #""C""_9""H""_10""N""_2""# is #""6.895%""#.

+
+
+
+

Explanation:

+
+

The molecular formula #""C""_9""H""_10""N""_2""#, indicates that in one mole of the compound, there are 9 moles of carbon atoms, 10 moles of hydrogen atoms, and 2 moles of nitrogen atoms.

+

To calculate the mass % of H atoms in #""C""_9""H""_10""N""_2""#, divide the molar mass of H atoms by the molar mass of the compound and multiply by 100.

+

Mass of #""C""_9""H""_10""N""_2"":##""146.193 g/mol""# https://www.ncbi.nlm.nih.gov/pccompound?term=C9H10N2

+

Mass of #""H""# atoms: #(10xx1.008"" g/mol"")=""10.08 g/mol""#

+

Divide and multiply by 100:

+

#(10.08color(red)cancel(color(black)(""g"")))/(146.193color(red)cancel(color(black)(""g"")))xx100=""6.895% H""#

+
+
+
" "
+

What is the mass % of H in C9H10N2? +

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 15, 2017 + +
+
+
+
+
+
+
+

The mass % of #""H""# in #""C""_9""H""_10""N""_2""# is #""6.895%""#.

+
+
+
+

Explanation:

+
+

The molecular formula #""C""_9""H""_10""N""_2""#, indicates that in one mole of the compound, there are 9 moles of carbon atoms, 10 moles of hydrogen atoms, and 2 moles of nitrogen atoms.

+

To calculate the mass % of H atoms in #""C""_9""H""_10""N""_2""#, divide the molar mass of H atoms by the molar mass of the compound and multiply by 100.

+

Mass of #""C""_9""H""_10""N""_2"":##""146.193 g/mol""# https://www.ncbi.nlm.nih.gov/pccompound?term=C9H10N2

+

Mass of #""H""# atoms: #(10xx1.008"" g/mol"")=""10.08 g/mol""#

+

Divide and multiply by 100:

+

#(10.08color(red)cancel(color(black)(""g"")))/(146.193color(red)cancel(color(black)(""g"")))xx100=""6.895% H""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 4936 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" "What is the mass % of H in C9H10N2? +" nan +75 a829eb96-6ddd-11ea-a2c1-ccda262736ce https://socratic.org/questions/how-many-grams-of-al-2o-3-would-form-if-12-5-moles-of-al-burned 637.25 grams start physical_unit 4 4 mass g qc_end physical_unit 11 11 8 9 mole qc_end c_other burn qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] Al2O3 [IN] grams""}]" "[{""type"":""physical unit"",""value"":""637.25 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] Al [=] \\pu{12.5 moles}""},{""type"":""other"",""value"":""burn""}]" "

How many grams of #Al_2O_3# would form if 12.5 moles of #Al# burned?

" nan 637.25 grams "
+

Explanation:

+
+

The stoichiometric equation for the combustion of aluminum is:

+

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#

+

If #12.5# #mol# of aluminum were consumed, clearly #6.25# #mol# aluminum oxide would result.

+

#""Mass of aluminum oxide""# #=# #6.25*molxx101.96*g*mol^-1# #=# #??# #g#.

+
+
" "
+
+
+

Over #600# #g# of aluminum oxide would result.

+
+
+
+

Explanation:

+
+

The stoichiometric equation for the combustion of aluminum is:

+

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#

+

If #12.5# #mol# of aluminum were consumed, clearly #6.25# #mol# aluminum oxide would result.

+

#""Mass of aluminum oxide""# #=# #6.25*molxx101.96*g*mol^-1# #=# #??# #g#.

+
+
+
" "
+

How many grams of #Al_2O_3# would form if 12.5 moles of #Al# burned?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 4, 2016 + +
+
+
+
+
+
+
+

Over #600# #g# of aluminum oxide would result.

+
+
+
+

Explanation:

+
+

The stoichiometric equation for the combustion of aluminum is:

+

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#

+

If #12.5# #mol# of aluminum were consumed, clearly #6.25# #mol# aluminum oxide would result.

+

#""Mass of aluminum oxide""# #=# #6.25*molxx101.96*g*mol^-1# #=# #??# #g#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 8540 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many grams of #Al_2O_3# would form if 12.5 moles of #Al# burned? nan +76 a829eb97-6ddd-11ea-9948-ccda262736ce https://socratic.org/questions/while-you-cook-on-your-gas-grill-3-7-grams-of-co-2-are-produced-how-many-moles-o 0.08 moles start physical_unit 17 17 mole mol qc_end physical_unit 10 10 7 8 mass qc_end chemical_equation 24 33 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] propane [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.08 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] CO2 [=] \\pu{3.7 grams}""},{""type"":""chemical equation"",""value"":""C3H8 + 5 O2 -> 3 CO2 + 4 H2O""}]" "

While you cook on your gas grill, 3.7 grams of #CO_2# are produced. How many moles of propane, #C_3H_8# are used, using the equation #C_3H_8+5O_2 -> 3CO_2+4H_2O#?

" nan 0.08 moles "
+

Explanation:

+
+

Given mass of #CO_2# = 3.7g
+Molar mass of #CO_2# = 12+2(16)= 44g/mol
+ No of moles of #CO_2# produced = #(""given mass"")/(""molar mass"")# = #3.7/44# =0.084mol

+

According to balanced equation;
+ 3 moles of #CO_2# requires = 1 mole of #C_3H_8#
+Thus, 0.084 moles of #CO_2# will require= #1/3 xx 0.084#mol of #C_3H_8#
+ = 0.028mol of #C_3H_8#
+So, 0.028mol of #C_3H_8# are used to produce 3.7g of #CO_2#

+
+
" "
+
+
+

0.084 moles of #C_3H_8#

+
+
+
+

Explanation:

+
+

Given mass of #CO_2# = 3.7g
+Molar mass of #CO_2# = 12+2(16)= 44g/mol
+ No of moles of #CO_2# produced = #(""given mass"")/(""molar mass"")# = #3.7/44# =0.084mol

+

According to balanced equation;
+ 3 moles of #CO_2# requires = 1 mole of #C_3H_8#
+Thus, 0.084 moles of #CO_2# will require= #1/3 xx 0.084#mol of #C_3H_8#
+ = 0.028mol of #C_3H_8#
+So, 0.028mol of #C_3H_8# are used to produce 3.7g of #CO_2#

+
+
+
" "
+

While you cook on your gas grill, 3.7 grams of #CO_2# are produced. How many moles of propane, #C_3H_8# are used, using the equation #C_3H_8+5O_2 -> 3CO_2+4H_2O#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 2, 2015 + +
+
+
+
+
+
+
+

0.084 moles of #C_3H_8#

+
+
+
+

Explanation:

+
+

Given mass of #CO_2# = 3.7g
+Molar mass of #CO_2# = 12+2(16)= 44g/mol
+ No of moles of #CO_2# produced = #(""given mass"")/(""molar mass"")# = #3.7/44# =0.084mol

+

According to balanced equation;
+ 3 moles of #CO_2# requires = 1 mole of #C_3H_8#
+Thus, 0.084 moles of #CO_2# will require= #1/3 xx 0.084#mol of #C_3H_8#
+ = 0.028mol of #C_3H_8#
+So, 0.028mol of #C_3H_8# are used to produce 3.7g of #CO_2#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 1430 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" While you cook on your gas grill, 3.7 grams of #CO_2# are produced. How many moles of propane, #C_3H_8# are used, using the equation #C_3H_8+5O_2 -> 3CO_2+4H_2O#? nan +77 a82a1292-6ddd-11ea-a44a-ccda262736ce https://socratic.org/questions/a-75-0-g-sample-of-a-solution-is-known-to-contain-23-8-g-of-glucose-what-is-the- 31.73% start physical_unit 21 24 mass_percent none qc_end physical_unit 5 6 1 2 mass qc_end physical_unit 14 14 11 12 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass percent [OF] glucose in this solution""}]" "[{""type"":""physical unit"",""value"":""31.73%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] a solution [=] \\pu{75.0 g}""},{""type"":""physical unit"",""value"":""Mass [OF] glucose [=] \\pu{23.8 g}""}]" "

A 75.0 g sample of a solution is known to contain 23.8 g of glucose. What is the mass percent of glucose in this solution?

" nan 31.73% "
+

Explanation:

+
+

#Mass % = ((mass.solute))/((mass.solution))x100%=(23.8g)/(75.0g)x100%#

+

#=31.73%#(w/w)

+
+
" "
+
+
+

#31.73%#(w/w)

+
+
+
+

Explanation:

+
+

#Mass % = ((mass.solute))/((mass.solution))x100%=(23.8g)/(75.0g)x100%#

+

#=31.73%#(w/w)

+
+
+
" "
+

A 75.0 g sample of a solution is known to contain 23.8 g of glucose. What is the mass percent of glucose in this solution?

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+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
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+
+
+1 Answer +
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+ +
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+ +
+ + May 27, 2017 + +
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#31.73%#(w/w)

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+

Explanation:

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+

#Mass % = ((mass.solute))/((mass.solution))x100%=(23.8g)/(75.0g)x100%#

+

#=31.73%#(w/w)

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Related questions
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Impact of this question
+
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" A 75.0 g sample of a solution is known to contain 23.8 g of glucose. What is the mass percent of glucose in this solution? nan +78 a82a1293-6ddd-11ea-ae10-ccda262736ce https://socratic.org/questions/how-many-moles-are-equal-to-2-69-10-78-atoms-of-sulfur 4.47 × 10^54 moles start physical_unit 9 9 mole mol qc_end physical_unit 7 9 6 6 number qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] sulfur [IN] moles""}]" "[{""type"":""physical unit"",""value"":""4.47 × 10^54 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of sulfur [=] \\pu{2.69⋅10^78}""}]" "

How many moles are equal to #2.69 * 10^78# atoms of sulfur?

" nan 4.47 × 10^54 moles "
+

Explanation:

+
+

How many dozens is 72? you just need to divide 72 by 12.

+

With mols it is exactly the same thing.
+1 mol is approximately # 6.022xx10^23#, so you must divide the given number by # 6.022xx10^23#:

+

#(2.69xx10^78)/ (6.022xx10^23)=4.47xx10^54#

+

Since the molar mass of Sulfur is about 32.06 g/mol you will get:

+

#4.47xx10^54xx32.06xx10^(-3)=1.43xx10^53 kg#

+

This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least #10^53 kg#

+

https://en.wikipedia.org/wiki/Universe

+
+
" "
+
+
+

#4.47xx10^54# moles

+
+
+
+

Explanation:

+
+

How many dozens is 72? you just need to divide 72 by 12.

+

With mols it is exactly the same thing.
+1 mol is approximately # 6.022xx10^23#, so you must divide the given number by # 6.022xx10^23#:

+

#(2.69xx10^78)/ (6.022xx10^23)=4.47xx10^54#

+

Since the molar mass of Sulfur is about 32.06 g/mol you will get:

+

#4.47xx10^54xx32.06xx10^(-3)=1.43xx10^53 kg#

+

This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least #10^53 kg#

+

https://en.wikipedia.org/wiki/Universe

+
+
+
" "
+

How many moles are equal to #2.69 * 10^78# atoms of sulfur?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+
+1 Answer +
+
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+ +
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+ +
+ + Mar 31, 2016 + +
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+
+

#4.47xx10^54# moles

+
+
+
+

Explanation:

+
+

How many dozens is 72? you just need to divide 72 by 12.

+

With mols it is exactly the same thing.
+1 mol is approximately # 6.022xx10^23#, so you must divide the given number by # 6.022xx10^23#:

+

#(2.69xx10^78)/ (6.022xx10^23)=4.47xx10^54#

+

Since the molar mass of Sulfur is about 32.06 g/mol you will get:

+

#4.47xx10^54xx32.06xx10^(-3)=1.43xx10^53 kg#

+

This number is extraordinarly huge. It is comparable with the mass of the whole Universe, which it's taken as being at least #10^53 kg#

+

https://en.wikipedia.org/wiki/Universe

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1164 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many moles are equal to #2.69 * 10^78# atoms of sulfur? nan +79 a82a1294-6ddd-11ea-b114-ccda262736ce https://socratic.org/questions/ten-moles-of-a-gas-are-contained-in-a-1-00-l-container-at-295-k-what-is-the-pres 242.07 atm start physical_unit 20 21 pressure atm qc_end physical_unit 3 4 0 1 mole qc_end physical_unit 11 11 9 10 volume qc_end physical_unit 4 4 13 14 temperature qc_end end "[{""type"":""physical unit"",""value"":""Pressure [OF] the gas [IN] atm""}]" "[{""type"":""physical unit"",""value"":""242.07 atm""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] a gas [=] \\pu{ten moles}""},{""type"":""physical unit"",""value"":""Volume [OF] container [=] \\pu{1.00 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] gas [=] \\pu{295 K}""}]" "

Ten moles of a gas are contained in a 1.00 L container at 295 K. What is the pressure of the gas?

" nan 242.07 atm "
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Explanation:

+
+

We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.

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+

We can use the ideal-gas equation:

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+

#ul(PV = nRT#

+
+

where

+
    +
  • +

    #P# is the pressure of the gas (what we're trying to find)

    +
    • +
  • +

    #V# is the volume occupied by the gas (given as #1.00# #""L""#

    +
    • +
  • +

    #n# is the number of moles of gas present (given as #10# #""mol""#)

    +
    • +
  • +

    #R# is the universal gas constant, equal to #0.082057(""L""·""atm"")/(""mol""·""K"")#

    +
    • +
  • +

    #T# is the absolute temperature of the gas (which must be in units of kelvin), given as #295# #""K""#

    +
    +
    • +
+

Let's rearrange the above equation to solve for the pressure, #P#:

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#P = (nRT)/V#

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+
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Plugging in known values:

+
+

#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#

+
+
+

The pressure is thus #color(red)(242color(white)(l)""atmospheres""#.

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" "
+
+
+

#P = 242# #""atm""#

+
+
+
+

Explanation:

+
+

We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.

+
+

We can use the ideal-gas equation:

+
+

#ul(PV = nRT#

+
+

where

+
    +
  • +

    #P# is the pressure of the gas (what we're trying to find)

    +
    • +
  • +

    #V# is the volume occupied by the gas (given as #1.00# #""L""#

    +
    • +
  • +

    #n# is the number of moles of gas present (given as #10# #""mol""#)

    +
    • +
  • +

    #R# is the universal gas constant, equal to #0.082057(""L""·""atm"")/(""mol""·""K"")#

    +
    • +
  • +

    #T# is the absolute temperature of the gas (which must be in units of kelvin), given as #295# #""K""#

    +
    +
    • +
+

Let's rearrange the above equation to solve for the pressure, #P#:

+
+

#P = (nRT)/V#

+
+
+

Plugging in known values:

+
+

#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#

+
+
+

The pressure is thus #color(red)(242color(white)(l)""atmospheres""#.

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+
+
" "
+

Ten moles of a gas are contained in a 1.00 L container at 295 K. What is the pressure of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
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+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 17, 2017 + +
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#P = 242# #""atm""#

+
+
+
+

Explanation:

+
+

We're asked to find the pressure of a gas, given its temperature, and volume, and number of moles.

+
+

We can use the ideal-gas equation:

+
+

#ul(PV = nRT#

+
+

where

+
    +
  • +

    #P# is the pressure of the gas (what we're trying to find)

    +
    • +
  • +

    #V# is the volume occupied by the gas (given as #1.00# #""L""#

    +
    • +
  • +

    #n# is the number of moles of gas present (given as #10# #""mol""#)

    +
    • +
  • +

    #R# is the universal gas constant, equal to #0.082057(""L""·""atm"")/(""mol""·""K"")#

    +
    • +
  • +

    #T# is the absolute temperature of the gas (which must be in units of kelvin), given as #295# #""K""#

    +
    +
    • +
+

Let's rearrange the above equation to solve for the pressure, #P#:

+
+

#P = (nRT)/V#

+
+
+

Plugging in known values:

+
+

#color(red)(P) = ((10cancel(""mol""))(0.082057(cancel(""L"")·""atm"")/(cancel(""mol"")·cancel(""K"")))(295cancel(""K"")))/(1.00cancel(""L"")) = color(red)(ulbar(|stackrel("" "")("" ""242color(white)(l)""atm"""" "")|)#

+
+
+

The pressure is thus #color(red)(242color(white)(l)""atmospheres""#.

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+
+
+
Related questions
+ + +
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Impact of this question
+
+ 3398 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" Ten moles of a gas are contained in a 1.00 L container at 295 K. What is the pressure of the gas? nan +80 a82a1295-6ddd-11ea-8661-ccda262736ce https://socratic.org/questions/58e5375711ef6b3c614e4239 12.20 start physical_unit 5 6 ph none qc_end physical_unit 14 14 9 10 molarity qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] aqueous solution""}]" "[{""type"":""physical unit"",""value"":""12.20""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [=] \\pu{0.016 mol*L^(−1)}""}]" "

What is #pH# of an aqueous solution that is #0.016*mol*L^-1# with respect to #NaOH#?

" nan 12.20 "
+

Explanation:

+
+

In aqueous solution, #pH=-log_(10)[H_3O^+]#, and likewise, #pH=-log_(10)[HO^-]#.

+

Now since we interrogate the equilibrium:

+

#2H_2Orightleftharpoons H_3O^+ + """"^(-)OH# where #K_""water""=10^(-14)#, then..........

+

#[H_3O^+][HO^-]=10^(-14)#, and we take #log_10# of both sides......

+

#log_10[H_3O^+]+ log_10[HO^-]=log_(10)10^(-14)=-14#

+

On rearrangement,

+

#14=-log_10[H_3O^+]-log_10[HO^-]=pH+pOH#.

+

And this is our defining relationship, #pH+pOH=14# (and this has to be known and used!)

+

So, we find the #pOH# of a solution that is #0.016*mol*L^-1# with respect to #NaOH#:

+

#pOH-=-log_(10)(0.016)=1.80#

+

And finally,

+

#pH=14-pOH=14-1.80=12.2#

+

For a similar problem, see this site.

+
+
" "
+
+
+

#pH=12.2#

+
+
+
+

Explanation:

+
+

In aqueous solution, #pH=-log_(10)[H_3O^+]#, and likewise, #pH=-log_(10)[HO^-]#.

+

Now since we interrogate the equilibrium:

+

#2H_2Orightleftharpoons H_3O^+ + """"^(-)OH# where #K_""water""=10^(-14)#, then..........

+

#[H_3O^+][HO^-]=10^(-14)#, and we take #log_10# of both sides......

+

#log_10[H_3O^+]+ log_10[HO^-]=log_(10)10^(-14)=-14#

+

On rearrangement,

+

#14=-log_10[H_3O^+]-log_10[HO^-]=pH+pOH#.

+

And this is our defining relationship, #pH+pOH=14# (and this has to be known and used!)

+

So, we find the #pOH# of a solution that is #0.016*mol*L^-1# with respect to #NaOH#:

+

#pOH-=-log_(10)(0.016)=1.80#

+

And finally,

+

#pH=14-pOH=14-1.80=12.2#

+

For a similar problem, see this site.

+
+
+
" "
+

What is #pH# of an aqueous solution that is #0.016*mol*L^-1# with respect to #NaOH#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
+
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+
+1 Answer +
+
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+ +
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+ +
+ + Apr 5, 2017 + +
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#pH=12.2#

+
+
+
+

Explanation:

+
+

In aqueous solution, #pH=-log_(10)[H_3O^+]#, and likewise, #pH=-log_(10)[HO^-]#.

+

Now since we interrogate the equilibrium:

+

#2H_2Orightleftharpoons H_3O^+ + """"^(-)OH# where #K_""water""=10^(-14)#, then..........

+

#[H_3O^+][HO^-]=10^(-14)#, and we take #log_10# of both sides......

+

#log_10[H_3O^+]+ log_10[HO^-]=log_(10)10^(-14)=-14#

+

On rearrangement,

+

#14=-log_10[H_3O^+]-log_10[HO^-]=pH+pOH#.

+

And this is our defining relationship, #pH+pOH=14# (and this has to be known and used!)

+

So, we find the #pOH# of a solution that is #0.016*mol*L^-1# with respect to #NaOH#:

+

#pOH-=-log_(10)(0.016)=1.80#

+

And finally,

+

#pH=14-pOH=14-1.80=12.2#

+

For a similar problem, see this site.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" What is #pH# of an aqueous solution that is #0.016*mol*L^-1# with respect to #NaOH#? nan +81 a82a1296-6ddd-11ea-bec0-ccda262736ce https://socratic.org/questions/what-molar-ratio-of-sodium-acetate-to-acetic-acid-should-be-used-to-prepare-a-bu 1.76 start physical_unit 4 8 molar_ratio none qc_end physical_unit 14 15 19 19 ph qc_end physical_unit 7 8 24 26 ka qc_end end "[{""type"":""physical unit"",""value"":""Molar ratio [OF] sodium acetate to acetic acid""}]" "[{""type"":""physical unit"",""value"":""1.76""}]" "[{""type"":""physical unit"",""value"":""PH [OF] a buffer [=] \\pu{4.5}""},{""type"":""physical unit"",""value"":""Ka [OF] acetic acid [=] \\pu{1.8 x 10^(−5)}""}]" "

What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? #K_a# acetic acid = #1.8 x 10^-5#?

" nan 1.76 "
+

Explanation:

+
+

#pH=pK_a-log_10(([""AcO""^-])/([""HOAc""]))#

+

And thus,

+

#log_10(([""AcO""^-])/([""HOAc""]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245#

+

And thus #10^0.245=([AcO^-])/([HOAc])=1.76#

+
+
" "
+
+
+

#([""AcO""^-])/([""HOAc""])=1.76#

+
+
+
+

Explanation:

+
+

#pH=pK_a-log_10(([""AcO""^-])/([""HOAc""]))#

+

And thus,

+

#log_10(([""AcO""^-])/([""HOAc""]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245#

+

And thus #10^0.245=([AcO^-])/([HOAc])=1.76#

+
+
+
" "
+

What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? #K_a# acetic acid = #1.8 x 10^-5#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Mole Ratios + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 10, 2017 + +
+
+
+
+
+
+
+

#([""AcO""^-])/([""HOAc""])=1.76#

+
+
+
+

Explanation:

+
+

#pH=pK_a-log_10(([""AcO""^-])/([""HOAc""]))#

+

And thus,

+

#log_10(([""AcO""^-])/([""HOAc""]))=pK_a-pH=-log_10(1.8xx10^-5)-4.5=4.75-4.5=0.245#

+

And thus #10^0.245=([AcO^-])/([HOAc])=1.76#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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Impact of this question
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" What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? #K_a# acetic acid = #1.8 x 10^-5#? nan +82 a82a398a-6ddd-11ea-92f7-ccda262736ce https://socratic.org/questions/a-chemist-reacted-0-05-moles-of-solid-sodium-with-water-to-form-hydroxide-soluti 1.15 g start physical_unit 7 7 mass g qc_end physical_unit 6 7 3 4 mole qc_end chemical_equation 20 26 qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.15 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] solid sodium [=] \\pu{0.05 moles}""},{""type"":""chemical equation"",""value"":""Na + H2O = NaOH + H2""},{""type"":""other"",""value"":""Solid sodium with water to form hydroxide solution""}]" "

A chemist reacted 0.05 moles of solid sodium with water to form hydroxide solution. The chemical reaction for this is: Na + H2O = NaOH + H2. What mass of sodium was reacted?

" nan 1.15 g "
+

Explanation:

+
+

Anyway you asked the mass of sodium reacted. This is simply #0.05*molxx22.99*g*mol^-1=1.15*g#.

+

Can you tell me (i) the volume of dihydrogen gas released given #298*K# temperatures, and #1*atm# pressures; and (ii) the #pH# of the resulting solution had it been diluted to #1*L#?

+

Enquiring minds want to know!

+
+
" "
+
+
+

The stoichiometrically balanced equation for this is.....

+

#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#

+
+
+
+

Explanation:

+
+

Anyway you asked the mass of sodium reacted. This is simply #0.05*molxx22.99*g*mol^-1=1.15*g#.

+

Can you tell me (i) the volume of dihydrogen gas released given #298*K# temperatures, and #1*atm# pressures; and (ii) the #pH# of the resulting solution had it been diluted to #1*L#?

+

Enquiring minds want to know!

+
+
+
" "
+

A chemist reacted 0.05 moles of solid sodium with water to form hydroxide solution. The chemical reaction for this is: Na + H2O = NaOH + H2. What mass of sodium was reacted?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 9, 2017 + +
+
+
+
+
+
+
+

The stoichiometrically balanced equation for this is.....

+

#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#

+
+
+
+

Explanation:

+
+

Anyway you asked the mass of sodium reacted. This is simply #0.05*molxx22.99*g*mol^-1=1.15*g#.

+

Can you tell me (i) the volume of dihydrogen gas released given #298*K# temperatures, and #1*atm# pressures; and (ii) the #pH# of the resulting solution had it been diluted to #1*L#?

+

Enquiring minds want to know!

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6527 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A chemist reacted 0.05 moles of solid sodium with water to form hydroxide solution. The chemical reaction for this is: Na + H2O = NaOH + H2. What mass of sodium was reacted? nan +83 a82a3df8-6ddd-11ea-b597-ccda262736ce https://socratic.org/questions/a-30-ml-volume-of-hcl-is-titrated-with-23-ml-of-0-20-m-naoh-how-would-you-calcul 0.15 M start physical_unit 22 25 molarity mol/l qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 14 14 9 10 volume qc_end physical_unit 14 14 12 13 molarity qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl in this solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.15 M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] HCl [=] \\pu{30 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{23 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [=] \\pu{0.20 M}""}]" "

A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

" nan 0.15 M "
+

Explanation:

+
+

Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of #HCl# was dissolved in a #30xx10^-3L# volume.

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Volume (#L#) #xx# concentration (#mol*L^-1#) gives an answer in #mol#. Note that #1# #mL# #=# #1xx10^-3*L#

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So #(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)# #=# #???mol*L^-1#

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Correctly! And according to the equation:

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#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(aq)#

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Explanation:

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Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of #HCl# was dissolved in a #30xx10^-3L# volume.

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Volume (#L#) #xx# concentration (#mol*L^-1#) gives an answer in #mol#. Note that #1# #mL# #=# #1xx10^-3*L#

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So #(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)# #=# #???mol*L^-1#

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A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

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+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Titration Calculations + + +
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Correctly! And according to the equation:

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#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(aq)#

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Explanation:

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Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of #HCl# was dissolved in a #30xx10^-3L# volume.

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Volume (#L#) #xx# concentration (#mol*L^-1#) gives an answer in #mol#. Note that #1# #mL# #=# #1xx10^-3*L#

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So #(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)# #=# #???mol*L^-1#

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" A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution? nan +84 a82a3df9-6ddd-11ea-aebb-ccda262736ce https://socratic.org/questions/how-much-mass-does-1-mol-of-o-2-gas-have 32.00 g start physical_unit 7 8 mass g qc_end physical_unit 7 8 4 5 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] O2 gas [IN] g""}]" "[{""type"":""physical unit"",""value"":""32.00 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] O2 gas [=] \\pu{1 mol}""}]" "

How much mass does 1 mol of #O_2# gas have?

" nan 32.00 g "
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Explanation:

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Multiply mol #""O""_2"" # by its molar mass #""31.998 g/mol""#.

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#1color(red)cancel(color(black)(""mol O""_2))xx(31.998""g O""_2)/(1color(red)cancel(color(black)(""mol O""_2 )))= ""31.998 g O""_2"" #

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#""1 mol O""_2"" # has a mass of #""31.998 g""#.

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Explanation:

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Multiply mol #""O""_2"" # by its molar mass #""31.998 g/mol""#.

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#1color(red)cancel(color(black)(""mol O""_2))xx(31.998""g O""_2)/(1color(red)cancel(color(black)(""mol O""_2 )))= ""31.998 g O""_2"" #

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How much mass does 1 mol of #O_2# gas have?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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#""1 mol O""_2"" # has a mass of #""31.998 g""#.

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Explanation:

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Multiply mol #""O""_2"" # by its molar mass #""31.998 g/mol""#.

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#1color(red)cancel(color(black)(""mol O""_2))xx(31.998""g O""_2)/(1color(red)cancel(color(black)(""mol O""_2 )))= ""31.998 g O""_2"" #

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" How much mass does 1 mol of #O_2# gas have? nan +85 a82a3dfa-6ddd-11ea-91d8-ccda262736ce https://socratic.org/questions/how-do-i-find-the-percentage-composition-of-oxygen-in-n-2o-5 74.07% start physical_unit 8 10 percent_composition none qc_end chemical_equation 10 10 qc_end end "[{""type"":""physical unit"",""value"":""Percentage composition [OF] Oxygen in N2O5""}]" "[{""type"":""physical unit"",""value"":""74.07%""}]" "[{""type"":""chemical equation"",""value"":""N2O5""}]" "

How do I find the percentage composition of Oxygen in #N_2O_5#?

" nan 74.07% "
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Explanation:

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First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of #N_2O_5 = 2(14) + 5(16) =108# g/mol.

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To find of the percent we take the element's mass in the molecule over the total molar mass. For #O# would be #((5*16)/108) * 100# or #100%-25.9% = 74.1% #

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It is #74.1%#

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Explanation:

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First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of #N_2O_5 = 2(14) + 5(16) =108# g/mol.

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To find of the percent we take the element's mass in the molecule over the total molar mass. For #O# would be #((5*16)/108) * 100# or #100%-25.9% = 74.1% #

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How do I find the percentage composition of Oxygen in #N_2O_5#?

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It is #74.1%#

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Explanation:

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First we find the molar mass of the entire molecule. So N is 14 and O is 16. Molar mass of #N_2O_5 = 2(14) + 5(16) =108# g/mol.

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To find of the percent we take the element's mass in the molecule over the total molar mass. For #O# would be #((5*16)/108) * 100# or #100%-25.9% = 74.1% #

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#N_2O_5# molecule contains 2 N atoms and 5 O atoms, So the molar mass of #N_2O_5# =2 x atomic mass of N +5xatomic mass of O = 2x14+5x16=108 g/mole
+% of N = #2*14/108*100# =25.93
+% of O =#5*16/108*100#=74.07

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" How do I find the percentage composition of Oxygen in #N_2O_5#? nan +86 a82a3dfb-6ddd-11ea-bb6a-ccda262736ce https://socratic.org/questions/what-volume-of-a-6-0-m-hcl-solution-is-required-to-make-250-0-milliliters-of-a-1 62.50 milliliters start physical_unit 6 6 volume ml qc_end physical_unit 6 6 12 13 volume qc_end physical_unit 6 6 16 17 molarity qc_end physical_unit 6 6 4 5 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] HCl solution [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""62.50 milliliters""}]" "[{""type"":""physical unit"",""value"":""Molarity2 [OF] HCl solution [=] \\pu{6.0 M}""},{""type"":""physical unit"",""value"":""Volume1 [OF] HCl solution [=] \\pu{250.0 milliliters}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] HCl solution [=] \\pu{1.5 M}""}]" "

What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#?

" nan 62.50 milliliters "
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Explanation:

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For this type of problem, we want to use the following dilution formula:
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+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

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* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

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We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

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#(C_2xxV_2)/C_1# = #V_1#

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#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

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62.5 mL of HCl is required.

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Explanation:

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For this type of problem, we want to use the following dilution formula:
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+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

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* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

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We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

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#(C_2xxV_2)/C_1# = #V_1#

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#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

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What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#?

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62.5 mL of HCl is required.

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Explanation:

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For this type of problem, we want to use the following dilution formula:
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+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

+

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

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We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

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#(C_2xxV_2)/C_1# = #V_1#

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#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

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" What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#? nan +87 a82a3dfc-6ddd-11ea-9e95-ccda262736ce https://socratic.org/questions/how-many-moles-of-nacl-are-present-in-20-0-g-of-nacl 0.34 moles start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] NaCl [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.34 moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [=] \\pu{20.0 g}""}]" "

How many moles of #NaCl# are present in 20.0 g of #NaCl#?

" nan 0.34 moles "
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Explanation:

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Molar mass of #""NaCl = 58.4 g/mol""#

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Number of moles in #""20.0 g NaCl""# is

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#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#

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#""0.342 mol""#

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Explanation:

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Molar mass of #""NaCl = 58.4 g/mol""#

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Number of moles in #""20.0 g NaCl""# is

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#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#

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How many moles of #NaCl# are present in 20.0 g of #NaCl#?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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#""0.342 mol""#

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Explanation:

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Molar mass of #""NaCl = 58.4 g/mol""#

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Number of moles in #""20.0 g NaCl""# is

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#(20.0 cancel""g"")/(58.4 cancel""g""""/mol"") = ""0.342 mol""#

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Related questions
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Impact of this question
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" How many moles of #NaCl# are present in 20.0 g of #NaCl#? nan +88 a82a608c-6ddd-11ea-a68b-ccda262736ce https://socratic.org/questions/56d9ed8f7c01497fb9da7217 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O""}]" "[{""type"":""other"",""value"":""The complete combustion of octane, C8H18(l)""}]" "

How do we represent the complete combustion of #""octane""#, #C_8H_18(l)#?

" nan 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O "
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Explanation:

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The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.

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So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?

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#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

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Explanation:

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The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.

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So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?

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How do we represent the complete combustion of #""octane""#, #C_8H_18(l)#?

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#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#

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Explanation:

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The above equation is stoichiometrically balanced: garbage out equals garbage in. If you like you can double the equation to remove the non-integral coefficient. I have never seen the need to do so, inasmuch the arithmetic is easier when you use this form with the 1 equivalent of hydrocarbon reactant.

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So 2 questions with regard to this reaction: (i) How does energy transfer in this reaction; (ii) does this represent an oxidation reduction reaction?

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We need to have the same number of moles of each substance on each side. The balanced equation is:

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#C_8H_18 + 25/2# #O_2 = 8# #CO_2 + 9# #H_2O# or #2# #C_8H_18 + 25# #O_2 = 16# #CO_2 + 18# #H_2O#

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Explanation:

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We start out with the unbalanced equation:

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#a# #C_8H_18 + b# # O_2 = c# # CO_2 + d# # H_2O#

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We need to find the values of the coefficients #a, b, c# and #d# to balance the equation.

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For the moment, let's leave #a# as 1. There are 8 moles of C on the left, so to balance we need 8 moles of C on the right, so let's make #c = 8#, because each #CO_2# contains 1.

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#C_8H_18 + b# # O_2 = 8# # CO_2 + d # #H_2O#

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There are 18 moles of H on the left so we need 18 on the right, but each #H_2O# contains 2, so we make #d = 9#.

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#C_8H_18 + b # #O_2 = 8# # CO_2 + 9 # #H_2O#

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Now we turn our attention to the right side. There are two O in each of 8 #CO_2# for a total of 16 in the carbon dioxide and one in each of 9 #H_2O# for a total of 9 in the water, so we need 25 O all together.

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On the left, each #O_2# contains two O, so one way to balance the equation is to take #25/2# of them:

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#C_8H_18 + # #25/2 O_2 = # #8# # CO_2 + 9 # #H_2O#

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If the fraction makes you uncomfortable, another way is to multiply all the coefficients by two:

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#2 # #C_8H_18 + 25# # O_2 = 16# # CO_2 + 18 # #H_2O#

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" "How do we represent the complete combustion of #""octane""#, #C_8H_18(l)#?" nan +89 a82a608d-6ddd-11ea-bf2f-ccda262736ce https://socratic.org/questions/how-do-you-balance-nh-4-3po-4-pb-no-3-4-pb-3-po-4-4-nh-4no-3 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 start chemical_equation qc_end chemical_equation 4 10 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3""}]" "[{""type"":""chemical equation"",""value"":""(NH4)3PO4 + Pb(NO3)4 -> Pb3(PO4)4 + NH4NO3""}]" "

How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 -> Pb_3(PO_4)_4 + NH_4NO_3#?

" nan 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 "
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Explanation:

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The unbalanced equation is

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#(""NH""_4)_3""PO""_4 +""Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""NH""_4""NO""_3#

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One way to balance the equation is to recognize that the polyatomic ions stay together.

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Then we can make some substitutions.

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Let #""A"" = ""NH""_4#; #""X"" = ""PO""_4#; #""Y"" = ""NO""_3#.

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Then the equation becomes

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#""A""_3""X"" + ""PbY""_4 → ""Pb""_3""X""_4 + ""AY""#

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We start with the most complicated formula, #""Pb""_3""X""_4#, and put a 1 in front of it.

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#""A""_3""X"" + ""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AYX""#

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Balance #""Pb""#

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We have fixed 3 atoms of #""Pb""# on the right, so we need 3 atoms of #""Pb""# on the left. Put a 3 in front of #""PbY""_4#.

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#""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AY""#

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Balance #""X""#

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We have fixed 4 atoms of #""X""# on the right, so we need 4 atoms of #""X""# on the left. Put a 4 in front of #""A""_3""X""#.

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Balance #""A""#

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We have fixed 12 atoms of #""A""# on the left, so we need 12 atoms of #""A""# on the right.

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#color(orange)(4)""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + color(teal)(12)""AY""#

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Every formula has a coefficient, and the equation is balanced.

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Now, we replace the original formulas in the equation.

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The balanced equation is

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#4(""NH""_4)_3""PO""_4 +""3Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""12NH""_4""NO""_3#

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#4(""NH""_4)_3""PO""_4 +""3Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""12NH""_4""NO""_3#

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Explanation:

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The unbalanced equation is

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#(""NH""_4)_3""PO""_4 +""Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""NH""_4""NO""_3#

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One way to balance the equation is to recognize that the polyatomic ions stay together.

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Then we can make some substitutions.

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Let #""A"" = ""NH""_4#; #""X"" = ""PO""_4#; #""Y"" = ""NO""_3#.

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Then the equation becomes

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#""A""_3""X"" + ""PbY""_4 → ""Pb""_3""X""_4 + ""AY""#

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We start with the most complicated formula, #""Pb""_3""X""_4#, and put a 1 in front of it.

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#""A""_3""X"" + ""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AYX""#

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Balance #""Pb""#

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We have fixed 3 atoms of #""Pb""# on the right, so we need 3 atoms of #""Pb""# on the left. Put a 3 in front of #""PbY""_4#.

+

#""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AY""#

+
+

Balance #""X""#

+

We have fixed 4 atoms of #""X""# on the right, so we need 4 atoms of #""X""# on the left. Put a 4 in front of #""A""_3""X""#.

+
+

Balance #""A""#

+

We have fixed 12 atoms of #""A""# on the left, so we need 12 atoms of #""A""# on the right.

+

#color(orange)(4)""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + color(teal)(12)""AY""#

+

Every formula has a coefficient, and the equation is balanced.

+
+

Now, we replace the original formulas in the equation.

+

The balanced equation is

+

#4(""NH""_4)_3""PO""_4 +""3Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""12NH""_4""NO""_3#

+
+
+
" "
+

How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 -> Pb_3(PO_4)_4 + NH_4NO_3#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 19, 2017 + +
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#4(""NH""_4)_3""PO""_4 +""3Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""12NH""_4""NO""_3#

+
+
+
+

Explanation:

+
+
+

The unbalanced equation is

+

#(""NH""_4)_3""PO""_4 +""Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""NH""_4""NO""_3#

+

One way to balance the equation is to recognize that the polyatomic ions stay together.

+

Then we can make some substitutions.

+

Let #""A"" = ""NH""_4#; #""X"" = ""PO""_4#; #""Y"" = ""NO""_3#.

+

Then the equation becomes

+

#""A""_3""X"" + ""PbY""_4 → ""Pb""_3""X""_4 + ""AY""#

+
+

We start with the most complicated formula, #""Pb""_3""X""_4#, and put a 1 in front of it.

+

#""A""_3""X"" + ""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AYX""#

+
+

Balance #""Pb""#

+

We have fixed 3 atoms of #""Pb""# on the right, so we need 3 atoms of #""Pb""# on the left. Put a 3 in front of #""PbY""_4#.

+

#""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + ""AY""#

+
+

Balance #""X""#

+

We have fixed 4 atoms of #""X""# on the right, so we need 4 atoms of #""X""# on the left. Put a 4 in front of #""A""_3""X""#.

+
+

Balance #""A""#

+

We have fixed 12 atoms of #""A""# on the left, so we need 12 atoms of #""A""# on the right.

+

#color(orange)(4)""A""_3""X"" + color(blue)(3)""PbY""_4 → color(red)(1)""Pb""_3""X""_4 + color(teal)(12)""AY""#

+

Every formula has a coefficient, and the equation is balanced.

+
+

Now, we replace the original formulas in the equation.

+

The balanced equation is

+

#4(""NH""_4)_3""PO""_4 +""3Pb""(""NO""_3)_4 → ""Pb""_3(""PO""_4)_4 + ""12NH""_4""NO""_3#

+
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" How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 -> Pb_3(PO_4)_4 + NH_4NO_3#? nan +90 a82a608e-6ddd-11ea-9bc4-ccda262736ce https://socratic.org/questions/arsenic-reacts-with-chlorine-to-form-a-chloride-if-1-587-g-of-arsenic-reacts-wit AsCl5 start chemical_formula qc_end c_other OTHER qc_end physical_unit 0 0 9 10 mass qc_end physical_unit 3 3 15 16 mass qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""AsCl5""}]" "[{""type"":""other"",""value"":""Arsenic reacts with chlorine to form a chloride.""},{""type"":""physical unit"",""value"":""Mass [OF] arsenic [=] \\pu{1.587 g}""},{""type"":""physical unit"",""value"":""Mass [OF] chlorine [=] \\pu{3.755 g}""}]" "

Arsenic reacts with chlorine to form a chloride. If 1.587 g of arsenic reacts with 3.755 g of chlorine what is the empirical formula of the chloride?

" nan AsCl5 "
+

Explanation:

+
+

#""Moles of arsenic""# #=# #(1.587*g)/(74.92*g*mol^-1)=0.0212*mol#.

+

#""Moles of chlorine""# #=# #(3.755*g)/(35.45*g*mol^-1)=0.106*mol#.

+

We divide thru by the lowest molar quantity to give #AsCl_5#. This is not chemical you would find in a bottle on a shelf.

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" "
+
+
+

#AsCl_5#

+
+
+
+

Explanation:

+
+

#""Moles of arsenic""# #=# #(1.587*g)/(74.92*g*mol^-1)=0.0212*mol#.

+

#""Moles of chlorine""# #=# #(3.755*g)/(35.45*g*mol^-1)=0.106*mol#.

+

We divide thru by the lowest molar quantity to give #AsCl_5#. This is not chemical you would find in a bottle on a shelf.

+
+
+
" "
+

Arsenic reacts with chlorine to form a chloride. If 1.587 g of arsenic reacts with 3.755 g of chlorine what is the empirical formula of the chloride?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
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+
+ +
+
+ +
+ + Jul 10, 2016 + +
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+
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#AsCl_5#

+
+
+
+

Explanation:

+
+

#""Moles of arsenic""# #=# #(1.587*g)/(74.92*g*mol^-1)=0.0212*mol#.

+

#""Moles of chlorine""# #=# #(3.755*g)/(35.45*g*mol^-1)=0.106*mol#.

+

We divide thru by the lowest molar quantity to give #AsCl_5#. This is not chemical you would find in a bottle on a shelf.

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+
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+
+
+
+
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+ + +
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" Arsenic reacts with chlorine to form a chloride. If 1.587 g of arsenic reacts with 3.755 g of chlorine what is the empirical formula of the chloride? nan +91 a82a608f-6ddd-11ea-884b-ccda262736ce https://socratic.org/questions/a-1-59-g-sample-of-a-metal-chloride-mcl-2-is-dissolved-in-water-and-treated-with 126.19 g/mol start physical_unit 36 36 molar_mass g/mol qc_end physical_unit 3 7 1 2 mass qc_end physical_unit 21 22 26 27 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] M [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""126.19 g/mol""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] sample of a metal chloride [=] \\pu{1.59 g}""},{""type"":""physical unit"",""value"":""Weight [OF] silver chloride [=] \\pu{3.60 g}""},{""type"":""other"",""value"":""MCl2 is dissolved in water and treated with excess aqueous silver nitrate. ""}]" "

A 1.59-g sample of a metal chloride, #MCl_2#, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.60 g. How do you calculate the molar mass of M?

" nan 126.19 g/mol "
+

Explanation:

+
+

We need a stoichiometric equation:

+

#MCl_2(aq) + 2AgNO_3(aq) rarr M(NO_3)_2(aq) + 2AgCl(s)darr#

+

#""Moles of silver chloride:""#
+#=# #(3.60*g)/(143.32*g*mol^-1)=2.51xx10^-2*mol.#

+

Given the stoichiometry of the precipitation reaction, there were #(2.51xx10^-2*mol)/2# #=# #1.26xx10^-2*mol# #MCl_2#.

+

So we have a molar quantity, and a given mass, and the #""molecular mass""# is the #""quotient""#, #""mass""/""molar quantity""#

+

#=(1.59*g)/(1.26xx10^-2*mol)# #=# #126.2*g*mol^-1#.

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+
" "
+
+
+

#""Molar mass of metal chloride""# #=# #126.2*g*mol^-1#

+
+
+
+

Explanation:

+
+

We need a stoichiometric equation:

+

#MCl_2(aq) + 2AgNO_3(aq) rarr M(NO_3)_2(aq) + 2AgCl(s)darr#

+

#""Moles of silver chloride:""#
+#=# #(3.60*g)/(143.32*g*mol^-1)=2.51xx10^-2*mol.#

+

Given the stoichiometry of the precipitation reaction, there were #(2.51xx10^-2*mol)/2# #=# #1.26xx10^-2*mol# #MCl_2#.

+

So we have a molar quantity, and a given mass, and the #""molecular mass""# is the #""quotient""#, #""mass""/""molar quantity""#

+

#=(1.59*g)/(1.26xx10^-2*mol)# #=# #126.2*g*mol^-1#.

+
+
+
" "
+

A 1.59-g sample of a metal chloride, #MCl_2#, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.60 g. How do you calculate the molar mass of M?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
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+
+
+1 Answer +
+
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+
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+ +
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+ + Dec 12, 2016 + +
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#""Molar mass of metal chloride""# #=# #126.2*g*mol^-1#

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+
+
+

Explanation:

+
+

We need a stoichiometric equation:

+

#MCl_2(aq) + 2AgNO_3(aq) rarr M(NO_3)_2(aq) + 2AgCl(s)darr#

+

#""Moles of silver chloride:""#
+#=# #(3.60*g)/(143.32*g*mol^-1)=2.51xx10^-2*mol.#

+

Given the stoichiometry of the precipitation reaction, there were #(2.51xx10^-2*mol)/2# #=# #1.26xx10^-2*mol# #MCl_2#.

+

So we have a molar quantity, and a given mass, and the #""molecular mass""# is the #""quotient""#, #""mass""/""molar quantity""#

+

#=(1.59*g)/(1.26xx10^-2*mol)# #=# #126.2*g*mol^-1#.

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+
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+
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" A 1.59-g sample of a metal chloride, #MCl_2#, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.60 g. How do you calculate the molar mass of M? nan +92 a82a6090-6ddd-11ea-b4e6-ccda262736ce https://socratic.org/questions/how-many-grams-of-hci-are-there-in-100-0-ml-of-concentrated-hcl-which-is-approxi 44.07 grams start physical_unit 4 4 mass g qc_end physical_unit 4 4 16 17 molarity qc_end physical_unit 4 12 19 19 mass_percent qc_end physical_unit 11 12 8 9 volume qc_end physical_unit 4 4 27 28 density qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] HCl [IN] grams""}]" "[{""type"":""physical unit"",""value"":""44.07 grams""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl [=] \\pu{12.1 M}""},{""type"":""physical unit"",""value"":""Percentage by weight [OF] HCl are there in 100.0 mL of concentrated HCl [=] \\pu{37%}""},{""type"":""physical unit"",""value"":""Volume [OF] concentrated HCl [=] \\pu{100.0 mL}""},{""type"":""physical unit"",""value"":""density [OF] HCl [=] \\pu{1.191 g/mL}""}]" "

How many grams of HCI are there in 100.0 mL of concentrated HCl which is approximately 12.1M and 37% HCl by weight and the density of 1.191 g/mL?

" nan 44.07 grams "
+

Explanation:

+
+

The first thing to take note of here is that the problem provides you information you don't actually need.

+

The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that #""100.0-mL""# sample volume.

+

So, you know that the solution has a density of #""1.191 g/mL""#, which means that every mililiter of this solution has a mass of #""1.191 g""#.

+

The mass of the sample you have here will thus be

+
+

#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#

+
+

Now, you know that the solution has a #37%# concentration by mass, which means that every #""100 g""# of solution will contain #""37 g""# of hydrochloric acid.

+

This means that your sample will contain

+
+

#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#

+
+

Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be

+
+

#m_""HCl"" = color(green)(""44 g HCl"")#

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" "
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#""44 g""#

+
+
+
+

Explanation:

+
+

The first thing to take note of here is that the problem provides you information you don't actually need.

+

The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that #""100.0-mL""# sample volume.

+

So, you know that the solution has a density of #""1.191 g/mL""#, which means that every mililiter of this solution has a mass of #""1.191 g""#.

+

The mass of the sample you have here will thus be

+
+

#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#

+
+

Now, you know that the solution has a #37%# concentration by mass, which means that every #""100 g""# of solution will contain #""37 g""# of hydrochloric acid.

+

This means that your sample will contain

+
+

#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#

+
+

Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be

+
+

#m_""HCl"" = color(green)(""44 g HCl"")#

+
+
+
+
" "
+

How many grams of HCI are there in 100.0 mL of concentrated HCl which is approximately 12.1M and 37% HCl by weight and the density of 1.191 g/mL?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
+
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+1 Answer +
+
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+ +
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+ + Nov 11, 2015 + +
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#""44 g""#

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+
+
+

Explanation:

+
+

The first thing to take note of here is that the problem provides you information you don't actually need.

+

The molarity of the solution is not important in determining how many grams of hydrochloric acid you have in that #""100.0-mL""# sample volume.

+

So, you know that the solution has a density of #""1.191 g/mL""#, which means that every mililiter of this solution has a mass of #""1.191 g""#.

+

The mass of the sample you have here will thus be

+
+

#100.0color(red)(cancel(color(black)(""mL""))) * ""1.191 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""119.1 g""#

+
+

Now, you know that the solution has a #37%# concentration by mass, which means that every #""100 g""# of solution will contain #""37 g""# of hydrochloric acid.

+

This means that your sample will contain

+
+

#119.1color(red)(cancel(color(black)(""g solution""))) * ""37 g HCl""/(100color(red)(cancel(color(black)(""g solution"")))) = ""44.067 g HCl""#

+
+

Rounded to two sig figs, the number of sig figs you have for the percent concentration by mass, the answer will be

+
+

#m_""HCl"" = color(green)(""44 g HCl"")#

+
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+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 61238 views + around the world +
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+ + Creative Commons License + +
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" How many grams of HCI are there in 100.0 mL of concentrated HCl which is approximately 12.1M and 37% HCl by weight and the density of 1.191 g/mL? nan +93 a82a8798-6ddd-11ea-acea-ccda262736ce https://socratic.org/questions/if-the-pressure-of-the-gas-in-a-2-31-l-balloon-is-12-atm-and-the-volume-increase 0.04 atm start physical_unit 28 32 pressure atm qc_end physical_unit 4 5 8 9 volume qc_end physical_unit 4 5 19 20 volume qc_end physical_unit 4 5 12 13 pressure qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] the air within the balloon [IN] atm""}]" "[{""type"":""physical unit"",""value"":""0.04 atm""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] the gas [=] \\pu{2.31 L}""},{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [=] \\pu{7.14 L}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] the gas [=] \\pu{0.12 atm}""}]" "

If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

" nan 0.04 atm "
+

Explanation:

+
+

#P_1V_1=P_2V_2#

+

Thus #V_2# #=# #(P_1V_1)/P_2# #=# #(2.31*Lxx0.12*atm)/(7.14*L)#.

+

The advantage of using Boyle's Law is that I can use any units I like, #""pints""#, #""bars""#, #""mm Hg""#, #""foot pounds""#, #""gallons""# so long as I am consistent.

+
+
" "
+
+
+

From Boyle's Law, #""pressure""xx""volume""=""constant""#

+

#P_2~=2/7xx0.12*atm#

+
+
+
+

Explanation:

+
+

#P_1V_1=P_2V_2#

+

Thus #V_2# #=# #(P_1V_1)/P_2# #=# #(2.31*Lxx0.12*atm)/(7.14*L)#.

+

The advantage of using Boyle's Law is that I can use any units I like, #""pints""#, #""bars""#, #""mm Hg""#, #""foot pounds""#, #""gallons""# so long as I am consistent.

+
+
+
" "
+

If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 21, 2016 + +
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+
+

From Boyle's Law, #""pressure""xx""volume""=""constant""#

+

#P_2~=2/7xx0.12*atm#

+
+
+
+

Explanation:

+
+

#P_1V_1=P_2V_2#

+

Thus #V_2# #=# #(P_1V_1)/P_2# #=# #(2.31*Lxx0.12*atm)/(7.14*L)#.

+

The advantage of using Boyle's Law is that I can use any units I like, #""pints""#, #""bars""#, #""mm Hg""#, #""foot pounds""#, #""gallons""# so long as I am consistent.

+
+
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Related questions
+ + +
+
+
+
Impact of this question
+
+ 2787 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" If the pressure of the gas in a 2.31 L balloon is .12 atm and the volume increases to 7.14 L, what will be the final pressure of the air within the balloon? nan +94 a82a8799-6ddd-11ea-8d2a-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-the-compound-that-forms-when-magnesium-bonds-with-phosph Mg3P2 start chemical_formula qc_end substance 10 10 qc_end substance 13 13 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Mg3P2""}]" "[{""type"":""substance name"",""value"":""Magnesium""},{""type"":""substance name"",""value"":""Phosphorus""}]" "

What is the formula for the compound that forms when magnesium bonds with phosphorus?

" nan Mg3P2 "
+

Explanation:

+
+

First, we know this is an ionic compound because one of the elements is a metal.

+

From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of #+2# charge. Also, phosphorus will form a #-3# ion.

+

The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.

+

So, three #Mg^(2+)# ions and two #P^(3-)# ions result in a total of +6 and -6 charges, which gives us the zero total charge.

+

A compound with three magnesium ions and two phosphorus ions would be written #Mg_3P_2#

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+
" "
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+
+

The formula would be #Mg_3P_2#. This would be an ionic compound.

+
+
+
+

Explanation:

+
+

First, we know this is an ionic compound because one of the elements is a metal.

+

From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of #+2# charge. Also, phosphorus will form a #-3# ion.

+

The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.

+

So, three #Mg^(2+)# ions and two #P^(3-)# ions result in a total of +6 and -6 charges, which gives us the zero total charge.

+

A compound with three magnesium ions and two phosphorus ions would be written #Mg_3P_2#

+
+
+
" "
+

What is the formula for the compound that forms when magnesium bonds with phosphorus?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
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+ +
+ + Jan 24, 2017 + +
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The formula would be #Mg_3P_2#. This would be an ionic compound.

+
+
+
+

Explanation:

+
+

First, we know this is an ionic compound because one of the elements is a metal.

+

From the periodic table or from the valence shells of the atoms, we can determine that magnesium forms an ion of #+2# charge. Also, phosphorus will form a #-3# ion.

+

The formula is based on combining these ions in the smallest ratio that results in a total charge of zero, because all compounds must be electrically neutral.

+

So, three #Mg^(2+)# ions and two #P^(3-)# ions result in a total of +6 and -6 charges, which gives us the zero total charge.

+

A compound with three magnesium ions and two phosphorus ions would be written #Mg_3P_2#

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" What is the formula for the compound that forms when magnesium bonds with phosphorus? nan +95 a82a879a-6ddd-11ea-8947-ccda262736ce https://socratic.org/questions/59b00c1011ef6b4c5b6ef6a7 CH2O start chemical_formula qc_end physical_unit 9 9 8 8 percent_composition qc_end physical_unit 12 12 11 11 percent_composition qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""CH2O""}]" "[{""type"":""physical unit"",""value"":""Percentage composition [OF] C [=] \\pu{40.1%}""},{""type"":""physical unit"",""value"":""Percentage composition [OF] H [=] \\pu{6.6%}""}]" "

An organic compound has a percentage composition of #40.1%# #C#, and #6.6%# #H#. What is its probable empirical formula?

" nan CH2O "
+

Explanation:

+
+

The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.

+

We assumes that we got #100*g# of some compound....

+

And thus #""moles of carbon""-=(40.1*g)/(12.011*g*mol^-1)=3.34*mol#.

+

And #""moles of hydrogen""-=(6.6*g)/(1.00794*g*mol^-1)=6.55*mol#.

+

But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....

+

And #""moles of oxygen""-=(100*g-40.1*g-6.6*g)/(15.999*g*mol^-1)=3.33*mol#.

+

And please note that here WE CAN MAKE NO OTHER ASSUMPTION. #%O# is usually UNREPORTED because there are few analytical methods for measurement of oxygen in microanalysis, and it is assumed to be the missing percentage. This is a standard practice in analysis.

+

And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....

+

..........................#CH_2O#.

+

We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.

+
+
" "
+
+
+

I makes it #CH_2O#

+
+
+
+

Explanation:

+
+

The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.

+

We assumes that we got #100*g# of some compound....

+

And thus #""moles of carbon""-=(40.1*g)/(12.011*g*mol^-1)=3.34*mol#.

+

And #""moles of hydrogen""-=(6.6*g)/(1.00794*g*mol^-1)=6.55*mol#.

+

But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....

+

And #""moles of oxygen""-=(100*g-40.1*g-6.6*g)/(15.999*g*mol^-1)=3.33*mol#.

+

And please note that here WE CAN MAKE NO OTHER ASSUMPTION. #%O# is usually UNREPORTED because there are few analytical methods for measurement of oxygen in microanalysis, and it is assumed to be the missing percentage. This is a standard practice in analysis.

+

And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....

+

..........................#CH_2O#.

+

We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.

+
+
+
" "
+

An organic compound has a percentage composition of #40.1%# #C#, and #6.6%# #H#. What is its probable empirical formula?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 6, 2017 + +
+
+
+
+
+
+
+

I makes it #CH_2O#

+
+
+
+

Explanation:

+
+

The empirical formula is the simplest while number ratio defining constituent atoms in a chemical species.

+

We assumes that we got #100*g# of some compound....

+

And thus #""moles of carbon""-=(40.1*g)/(12.011*g*mol^-1)=3.34*mol#.

+

And #""moles of hydrogen""-=(6.6*g)/(1.00794*g*mol^-1)=6.55*mol#.

+

But you have undoubtedly already noted that the quoted percentages do not add up to 100%. In this scenario IT IS ALWAYS ASSUMED that the BALANCE, the missing percentage, is DUE TO OXYGEN.....

+

And #""moles of oxygen""-=(100*g-40.1*g-6.6*g)/(15.999*g*mol^-1)=3.33*mol#.

+

And please note that here WE CAN MAKE NO OTHER ASSUMPTION. #%O# is usually UNREPORTED because there are few analytical methods for measurement of oxygen in microanalysis, and it is assumed to be the missing percentage. This is a standard practice in analysis.

+

And so we divide each molar quantity thru by the SMALLEST molar quantity to get an empirical formula of.....

+

..........................#CH_2O#.

+

We could quote a molecular formula PROVIDED that we get a measurement of molecular mass.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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Impact of this question
+
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" An organic compound has a percentage composition of #40.1%# #C#, and #6.6%# #H#. What is its probable empirical formula? nan +96 a82a879b-6ddd-11ea-a261-ccda262736ce https://socratic.org/questions/the-mass-of-a-hydrogen-atom-is-1-0-amu-and-the-mass-of-a-carbon-atom-is-12-0-amu 16.00 g/mol start physical_unit 25 25 molar_mass g/mol qc_end physical_unit 4 5 7 8 mass qc_end physical_unit 14 15 17 18 mass qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] methane [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""16.00 g/mol""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen atom [=] \\pu{1.0 amu}""},{""type"":""physical unit"",""value"":""Mass [OF] carbon atom [=] \\pu{12.0 amu}""}]" "

The mass of a hydrogen atom is 1.0 amu, and the mass of a carbon atom is 12.0 amu. What is the molar mass of methane?

" nan 16.00 g/mol "
+

Explanation:

+
+

Can you tell us the molar mass given that carbon has a molar mass of #12.0*g#, and hydrogen, #1.0*g#?

+
+
" "
+
+
+

Well, #""methane""# has a formula of #CH_4#.

+
+
+
+

Explanation:

+
+

Can you tell us the molar mass given that carbon has a molar mass of #12.0*g#, and hydrogen, #1.0*g#?

+
+
+
" "
+

The mass of a hydrogen atom is 1.0 amu, and the mass of a carbon atom is 12.0 amu. What is the molar mass of methane?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 1, 2016 + +
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+
+
+
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+
+

Well, #""methane""# has a formula of #CH_4#.

+
+
+
+

Explanation:

+
+

Can you tell us the molar mass given that carbon has a molar mass of #12.0*g#, and hydrogen, #1.0*g#?

+
+
+
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+
+ +
+
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+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+
" The mass of a hydrogen atom is 1.0 amu, and the mass of a carbon atom is 12.0 amu. What is the molar mass of methane? nan +97 a82a879c-6ddd-11ea-b475-ccda262736ce https://socratic.org/questions/to-what-temperature-will-a-50-0-g-piece-of-glass-raise-if-it-absorbs-5275-joules 231.00 ℃ start physical_unit 33 34 temperature °c qc_end physical_unit 9 9 5 6 mass qc_end physical_unit 9 9 14 15 heat_energy qc_end physical_unit 9 9 24 27 specific_heat_capacity qc_end physical_unit 9 9 36 37 initial_temperature qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] the glass [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""231.00 ℃""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] glass [=] \\pu{50.0 g}""},{""type"":""physical unit"",""value"":""Absorb heat [OF] glass [=] \\pu{5275 joules}""},{""type"":""physical unit"",""value"":""Specific heat capacity [OF] glass [=] \\pu{0.50 J/(g * ℃)}""},{""type"":""physical unit"",""value"":""Initial temperature1 [OF] glass [=] \\pu{20.0 ℃}""}]" "

To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C, if the initial temperature of the glass is 20.0°C?

" nan 231.00 ℃ "
+

Explanation:

+
+

A substance's specific heat tells you how much heat much either be added or removed from #""1 g""# of that substance in order to cause a #1^@""C""# change in temperature.

+

The change in temperature, #DeltaT#, is always calculated by subtracting the initial temperature of the sample from the final temperature of the sample.

+
+

#color(blue)(DeltaT = T_""final"" - T_""initial"")#

+
+

Now, when the substance absorbs heat, its temperature will increase, which implies that #DeltaT > 0#.

+

Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.

+

You will have to use this equation

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.

+

In your case, adding #""5275 J""# of heat to that #""50.0-g""# piece of glass will result in a temperature change of

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+
+

Plug in your values to get

+
+

#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#

+
+

So, adding that much heat to your sample will result in a #211^@""C""# increase in temperature. This means that the final temperature of the glass will be

+
+

#T_""final"" = T_""initial"" + DeltaT#

+

#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.

+
+
" "
+
+
+

#230^@""C""#

+
+
+
+

Explanation:

+
+

A substance's specific heat tells you how much heat much either be added or removed from #""1 g""# of that substance in order to cause a #1^@""C""# change in temperature.

+

The change in temperature, #DeltaT#, is always calculated by subtracting the initial temperature of the sample from the final temperature of the sample.

+
+

#color(blue)(DeltaT = T_""final"" - T_""initial"")#

+
+

Now, when the substance absorbs heat, its temperature will increase, which implies that #DeltaT > 0#.

+

Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.

+

You will have to use this equation

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.

+

In your case, adding #""5275 J""# of heat to that #""50.0-g""# piece of glass will result in a temperature change of

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+
+

Plug in your values to get

+
+

#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#

+
+

So, adding that much heat to your sample will result in a #211^@""C""# increase in temperature. This means that the final temperature of the glass will be

+
+

#T_""final"" = T_""initial"" + DeltaT#

+

#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.

+
+
+
" "
+

To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C, if the initial temperature of the glass is 20.0°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 29, 2016 + +
+
+
+
+
+
+
+

#230^@""C""#

+
+
+
+

Explanation:

+
+

A substance's specific heat tells you how much heat much either be added or removed from #""1 g""# of that substance in order to cause a #1^@""C""# change in temperature.

+

The change in temperature, #DeltaT#, is always calculated by subtracting the initial temperature of the sample from the final temperature of the sample.

+
+

#color(blue)(DeltaT = T_""final"" - T_""initial"")#

+
+

Now, when the substance absorbs heat, its temperature will increase, which implies that #DeltaT > 0#.

+

Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.

+

You will have to use this equation

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.

+

In your case, adding #""5275 J""# of heat to that #""50.0-g""# piece of glass will result in a temperature change of

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+
+

Plug in your values to get

+
+

#DeltaT = (5275 color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 0.50color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 211^@""C""#

+
+

So, adding that much heat to your sample will result in a #211^@""C""# increase in temperature. This means that the final temperature of the glass will be

+
+

#T_""final"" = T_""initial"" + DeltaT#

+

#T_""final"" = 20.0^@""C"" + 211^@""C"" = color(green)(230^@""C"")#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.

+
+
+
+
+
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+
+
+
+
+
+
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" To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C, if the initial temperature of the glass is 20.0°C? nan +98 a82ab200-6ddd-11ea-922f-ccda262736ce https://socratic.org/questions/what-is-molarity-of-a-solution-containing-10-g-of-naoh-in-500-ml-of-naoh-solutio 0.50 mol/L start physical_unit 4 5 molarity mol/l qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 15 16 12 13 volume qc_end physical_unit 10 10 21 22 molar_mass qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] a solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.50 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaOH [=] \\pu{10 g}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH solution [=] \\pu{500 mL}""},{""type"":""physical unit"",""value"":""Molar mass [OF] NaOH [=] \\pu{40 g/mol}""}]" "

What is molarity of a solution containing 10 g of NaOH in 500 mL of NaOH solution?

" "
+
+

+

Molar mass NaOH = 40 g/mol#

+

+
+
" 0.50 mol/L "
+

Explanation:

+
+

Molarity is a concentration term in which the no. of moles and the volume are related.

+

#""Molarity"" \ (c) = (""no. of moles"" \ (n))/(""volume in L"" \ (V))#

+

Here

+

#n = ""10 g""/""40 g/mol""#

+

#V = ""500 mL""= ""0.5 L""#

+

So

+

#c=n/V#

+
+

# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#

+

# = ""10 mol""/""20 L"" = ""0.5 molar""#

+
+
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" "
+
+
+

#""0.5 molar""#

+
+
+
+

Explanation:

+
+

Molarity is a concentration term in which the no. of moles and the volume are related.

+

#""Molarity"" \ (c) = (""no. of moles"" \ (n))/(""volume in L"" \ (V))#

+

Here

+

#n = ""10 g""/""40 g/mol""#

+

#V = ""500 mL""= ""0.5 L""#

+

So

+

#c=n/V#

+
+

# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#

+

# = ""10 mol""/""20 L"" = ""0.5 molar""#

+
+
+
+
" "
+

What is molarity of a solution containing 10 g of NaOH in 500 mL of NaOH solution?

+
+
+

+

Molar mass NaOH = 40 g/mol#

+

+
+
+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
+
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+ + Mar 13, 2018 + +
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#""0.5 molar""#

+
+
+
+

Explanation:

+
+

Molarity is a concentration term in which the no. of moles and the volume are related.

+

#""Molarity"" \ (c) = (""no. of moles"" \ (n))/(""volume in L"" \ (V))#

+

Here

+

#n = ""10 g""/""40 g/mol""#

+

#V = ""500 mL""= ""0.5 L""#

+

So

+

#c=n/V#

+
+

# = ""10 g""/(""40 g/mol"" \ xx \ ""0.5 L"")#

+

# = ""10 mol""/""20 L"" = ""0.5 molar""#

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" What is molarity of a solution containing 10 g of NaOH in 500 mL of NaOH solution? " + + +Molar mass NaOH = 40 g/mol# + + +" +99 a82ab201-6ddd-11ea-845d-ccda262736ce https://socratic.org/questions/magnesium-reacts-with-titanium-4-chloride-to-produce-magnesium-chloride-and-tita 2 Mg + TiCl4 -> 2 MgCl2 + Ti start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 Mg + TiCl4 -> 2 MgCl2 + Ti""}]" "[{""type"":""other"",""value"":""Magnesium reacts with titanium(4) chloride to produce magnesium chloride and titanium metal.""}]" "

Magnesium reacts with titanium(4) chloride to produce magnesium chloride and titanium metal. How do you write a balanced equation for this reaction?

" nan 2 Mg + TiCl4 -> 2 MgCl2 + Ti "
+

Explanation:

+
+

Because the magnesium ion has a +2 charge, the formula for magnesium chloride is #MgCl_2#.

+

To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of #MgCl_2#

+
+
" "
+
+
+

#2 Mg + TiCl_4 rarr 2 MgCl_2 + Ti#

+
+
+
+

Explanation:

+
+

Because the magnesium ion has a +2 charge, the formula for magnesium chloride is #MgCl_2#.

+

To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of #MgCl_2#

+
+
+
" "
+

Magnesium reacts with titanium(4) chloride to produce magnesium chloride and titanium metal. How do you write a balanced equation for this reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
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+
+2 Answers +
+
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+
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+
+ +
+ + Dec 18, 2016 + +
+
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#2 Mg + TiCl_4 rarr 2 MgCl_2 + Ti#

+
+
+
+

Explanation:

+
+

Because the magnesium ion has a +2 charge, the formula for magnesium chloride is #MgCl_2#.

+

To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2 units of #MgCl_2#

+
+
+
+
+
+ +
+
+
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+
+ +
+
+ +
+ + Dec 18, 2016 + +
+
+
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+

#TiCl_4 + 2Mg rarr Ti + 2MgCl_2#

+
+
+
+

Explanation:

+
+

We could do this by individual redox steps:

+

#TiCl_4 + 4e^(-) rarr Ti + 4Cl^-# #(i)#

+

#Mgrarr Mg^(2+) + 2e^-# #(ii)#

+

#(i) + 2xx(ii)=# #TiCl_4 +2Mgrarr Ti +2MgCl_2#

+

With all chemical reactions, 2 conditions must be absolutely satisfied: (i) mass must be balanced; and (ii) charge must be balanced. Well, is they?

+

Note that just because I can write the reaction, there is no likelihood that the reaction can be performed. Should I attempt to reduce the chloride with magnesium metal I would likely get a mess. Titanic chloride is unstable with respect to water and dioxygen.

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Impact of this question
+
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+ + Creative Commons License + +
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+
+
" Magnesium reacts with titanium(4) chloride to produce magnesium chloride and titanium metal. How do you write a balanced equation for this reaction? nan +100 a82ab202-6ddd-11ea-8582-ccda262736ce https://socratic.org/questions/bromophenol-blue-is-an-indicator-with-a-k-a-5-84xx10-5-what-is-the-of-indicator- 80.16% start physical_unit 17 21 percent none qc_end physical_unit 0 1 9 11 ka qc_end physical_unit 4 4 26 26 ph qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""% [OF] indicator in it's basic form""}]" "[{""type"":""physical unit"",""value"":""80.16%""}]" "[{""type"":""physical unit"",""value"":""Ka [OF] bromophenol blue [=] \\pu{5.84 × 10^(−5)}""},{""type"":""physical unit"",""value"":""PH [OF] indicator [=] \\pu{4.84}""},{""type"":""other"",""value"":""Given, log(5.84) = 0.7664 & antilog of 0.6064 is 4.04.""}]" "

Bromophenol blue is an indicator with a #K_a=5.84xx10^(-5)#. What is the #%# of indicator in it's basic form at a #pH=4.84#? + +Given, #log(5.84)=0.7664# & antilog of #0.6064# is #4.04#. Thank you:)

" nan 80.16% "
+

Explanation:

+
+

The idea here is that bromophenol blue acts as a weak acid in aqueous solution.

+
+

#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#

+
+

As you can see, this equilibrium is influenced by the #""pH""# of the solution, i.e. by the concentration of hydronium cations present in the solution.

+

In your case, you know that you have

+
+

#""pH"" = 4.84#

+
+

This implies that you have

+
+

#[""H""_3""O""^(+)] = 10^(-4.84)#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#

+
+

Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.

+
+

#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#

+
+

Plug in your values to find

+
+

#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#

+

#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#

+
+

Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to

+
+

#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#

+
+

This is the case because the acid dissociates in a #1:1# mole ratio to produce the conjugate base, which implies that in order to have an equilibrium concentration of #[""Bb""^(-)]# and an equilibrium concentration of #[""HBb""]#, the initial concentration of the weak acid must decrease by #[""Bb""^(-)]#.

+

This means that the percent dissociation of the weak acid will be equal to

+
+

#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#

+
+

Use equation #color(darkorange)(""(*)"")# to say that

+
+

#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#

+
+

The percent dissociation of the weak acid will thus be equal to

+
+

#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#

+

#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#

+
+

Now, you know that

+
+

#log(5.84) = 0.7664"" ""# and #"" "" 10^0.6064 = 4.04#

+
+

This means that you can write

+
+

#10^log(5.84) = 10^0.7664#

+

#5.84 = 10^0.7664#

+
+

Plug this into the equation to get

+
+

#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#

+

#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#

+
+

Finally, you will end up with

+
+

#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#

+
+

I'll leave the answer rounded to three sig figs.

+
+
" "
+
+
+

#80.2%#

+
+
+
+

Explanation:

+
+

The idea here is that bromophenol blue acts as a weak acid in aqueous solution.

+
+

#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#

+
+

As you can see, this equilibrium is influenced by the #""pH""# of the solution, i.e. by the concentration of hydronium cations present in the solution.

+

In your case, you know that you have

+
+

#""pH"" = 4.84#

+
+

This implies that you have

+
+

#[""H""_3""O""^(+)] = 10^(-4.84)#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#

+
+

Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.

+
+

#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#

+
+

Plug in your values to find

+
+

#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#

+

#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#

+
+

Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to

+
+

#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#

+
+

This is the case because the acid dissociates in a #1:1# mole ratio to produce the conjugate base, which implies that in order to have an equilibrium concentration of #[""Bb""^(-)]# and an equilibrium concentration of #[""HBb""]#, the initial concentration of the weak acid must decrease by #[""Bb""^(-)]#.

+

This means that the percent dissociation of the weak acid will be equal to

+
+

#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#

+
+

Use equation #color(darkorange)(""(*)"")# to say that

+
+

#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#

+
+

The percent dissociation of the weak acid will thus be equal to

+
+

#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#

+

#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#

+
+

Now, you know that

+
+

#log(5.84) = 0.7664"" ""# and #"" "" 10^0.6064 = 4.04#

+
+

This means that you can write

+
+

#10^log(5.84) = 10^0.7664#

+

#5.84 = 10^0.7664#

+
+

Plug this into the equation to get

+
+

#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#

+

#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#

+
+

Finally, you will end up with

+
+

#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#

+
+

I'll leave the answer rounded to three sig figs.

+
+
+
" "
+

Bromophenol blue is an indicator with a #K_a=5.84xx10^(-5)#. What is the #%# of indicator in it's basic form at a #pH=4.84#? + +Given, #log(5.84)=0.7664# & antilog of #0.6064# is #4.04#. Thank you:)

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Dynamic Equilibrium + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Oct 15, 2017 + +
+
+
+
+
+
+
+

#80.2%#

+
+
+
+

Explanation:

+
+

The idea here is that bromophenol blue acts as a weak acid in aqueous solution.

+
+

#""HBb""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""Bb""_ ((aq))^(-) + ""H""_ 3""O""_ ((aq))^(+)#

+
+

As you can see, this equilibrium is influenced by the #""pH""# of the solution, i.e. by the concentration of hydronium cations present in the solution.

+

In your case, you know that you have

+
+

#""pH"" = 4.84#

+
+

This implies that you have

+
+

#[""H""_3""O""^(+)] = 10^(-4.84)#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""Bb""^(-)] * [""H""_3""O""^(+)])/([""HBb""])#

+
+

Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.

+
+

#([""Bb""^(-)])/([""HBb""]) = K_a/([""H""_3""O""^(+)])#

+
+

Plug in your values to find

+
+

#([""Bb""^(-)])/([""HBb""]) = (5.84 * 10^(-5))/10^(-4.84)#

+

#([""Bb""^(-)])/([""HBb""]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)"" """" ""color(darkorange)(""(*)"")#

+
+

Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to

+
+

#[""HBb""]_0 = [""Bb""^(-)] + [""HBb""]#

+
+

This is the case because the acid dissociates in a #1:1# mole ratio to produce the conjugate base, which implies that in order to have an equilibrium concentration of #[""Bb""^(-)]# and an equilibrium concentration of #[""HBb""]#, the initial concentration of the weak acid must decrease by #[""Bb""^(-)]#.

+

This means that the percent dissociation of the weak acid will be equal to

+
+

#""% dissociation"" = ([""Bb""^(-)])/([""Bb""^(-)] + [""HBb""]) * 100%#

+
+

Use equation #color(darkorange)(""(*)"")# to say that

+
+

#[""Bb""^(-)] = 5.84 * 10^(-0.16) * [""HBb""]#

+
+

The percent dissociation of the weak acid will thus be equal to

+
+

#""% dissociation"" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)([""HBb""]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)([""HBb""]))) + color(red)(cancel(color(black)([""HBb""])))) * 100%#

+

#""% dissociation"" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%#

+
+

Now, you know that

+
+

#log(5.84) = 0.7664"" ""# and #"" "" 10^0.6064 = 4.04#

+
+

This means that you can write

+
+

#10^log(5.84) = 10^0.7664#

+

#5.84 = 10^0.7664#

+
+

Plug this into the equation to get

+
+

#""% dissociation"" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%#

+

#""% dissociation"" = 10^0.6064/(10^0.6064 + 1) * 100%#

+
+

Finally, you will end up with

+
+

#""% dissociation"" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))#

+
+

I'll leave the answer rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Oct 15, 2017 + +
+
+
+
+
+
+
+

#sf(80.2%)#

+
+
+
+

Explanation:

+
+

Indicators like bromophenol blue are weak acids where the undissociated (acidic) form is a different colour from the dissociated (basic) form:

+

#sf(""HIn""rightleftharpoonsH^++In^-)#

+

#sf(color(yellow)(yellow)"" ""color(blue)(blue))#

+

#sf(K_a=([H^+][In^-])/([HIn])=5.84xx10^(-5))#

+

#:.##sf(([In^-])/([HIn])=K_a/[[H^+])#

+

We know that #sf(pH=4.84)##:.##sf(-log[H^+]=4.84)# from which #sf([H^+]=1.445xx10^(-5)color(white)(x)""mol/l"")#

+

#:.##sf(([In^-])/([HIn])=(5.84xxcancel(10^(-5)))/(1.445xxcancel(10^(-5)))=4.048)#

+

To get the percentage which are dissociated we can find #sf(alpha)# the degree of dissociation, or the fraction which are dissociated from an ICE table:

+

#sf("" ""HInrightleftharpoonsH^++In^-)#

+

#sf(I"" ""1"" "" "" ""0"" "" "" ""0)#

+

#sf(C"" ""-alpha"" "" +alpha"" ""+alpha)#

+

#sf(E"" ""(1-alpha)"" ""alpha"" ""alpha)#

+

We know that #sf(alpha/((1-alpha))=4.048)#

+

#:.##sf(alpha=4.048(1-alpha))#

+

#sf(alpha=4.048-4.048alpha)#

+

#sf(5.048alpha=4.048)#

+

#sf(alpha=4.048/5.048=0.802)#

+

This means that the basic, dissociated form is 80.2 %.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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" "Bromophenol blue is an indicator with a #K_a=5.84xx10^(-5)#. What is the #%# of indicator in it's basic form at a #pH=4.84#? + +Given, #log(5.84)=0.7664# & antilog of #0.6064# is #4.04#. Thank you:)" nan +101 a82ab203-6ddd-11ea-92e7-ccda262736ce https://socratic.org/questions/i-want-to-increase-the-no3-concentration-in-a-40-liters-tank-using-ca-no3-2-4h2o 25.46 mg start physical_unit 13 13 mass mg qc_end physical_unit 18 19 21 22 concentration qc_end physical_unit 11 11 9 10 volume qc_end physical_unit 5 5 35 36 concentration qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] Ca(NO3)2.4H2O [IN] mg""}]" "[{""type"":""physical unit"",""value"":""25.46 mg""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] CaNO3 solution [=] \\pu{500 g/L}""},{""type"":""physical unit"",""value"":""Volume [OF] tank [=] \\pu{40 Liters}""},{""type"":""physical unit"",""value"":""Increased concentration [OF] NO3 [=] \\pu{2 ppm}""},{""type"":""other"",""value"":""Increase the NO3 concentration using Ca(NO3)2.4H2O""}]" "

I want to increase the NO3 concentration in a 40 Liters tank using Ca(NO3)2.4H2O. The concentration of the CaNO3 solution is 500g/L. +How much solution do I need to add increase the concentration by 2ppm? + +Thanks for helping out!

" "
+
+

+

How much calcium would be added at the same time?

+

+
+
" 25.46 mg "
+

Explanation:

+
+
+

Let's express the concentration of 2 ppm as grams per litre.

+

#""2 ppm"" = (2 × 10^""-6"" color(white)(l)""g"")/(1 color(red)(cancel(color(black)(""g water"")))) × (1000 color(red)(cancel(color(black)(""g water""))))/""1 L"" = ""0.002 g/L"" = ""2 mg/L""#

+
+

For 40 L of water, you will need

+

#40 color(red)(cancel(color(black)(""L""))) × ""2 mg""/(1 color(red)(cancel(color(black)(""L"")))) = ""80 mg"" = ""0.08 g""# of nitrate ion.

+
+

The molar mass of #""Ca""(""NO""_3)_2·""4H""_2""O""# is 236·15 g.

+

Of this, 124.01 g comes from the nitrate ions.

+

Hence, to get 0.08 g of nitrate, you will have to use

+

#0.08 color(red)(cancel(color(black)(""g NO""_3^""-""))) × (236.15 ""g Ca""(""NO""_3)_2·""4H""_2""O"")/(124.01 color(red)(cancel(color(black)(""g NO""_3^""-"")))) = ""0.15 g Ca""(""NO""_3)_2·""4H""_2""O""#

+

The required volume of concentrated salt solution is

+

#0.15 color(red)(cancel(color(black)(""g salt""))) × ""1 L salt""/(500 color(red)(cancel(color(black)(""g salt"")))) = ""0.0003 L salt"" = ""0.3 mL salt""#

+
+

There are 40.078 g of #""Ca""# in #""236.15 g Ca""(""NO""_3)_2·""4H""_2""O""#.

+

∴ The mass of calcium added is

+

#0.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O""))) × ""40.078 g Ca""/(236.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O"")))) = ""0.025 g Ca"" = ""25 mg Ca"" #.

+
+
" "
+
+
+

You must add 0.3 mL of the calcium nitrate solution. This will include 25 mg of calcium.

+
+
+
+

Explanation:

+
+
+

Let's express the concentration of 2 ppm as grams per litre.

+

#""2 ppm"" = (2 × 10^""-6"" color(white)(l)""g"")/(1 color(red)(cancel(color(black)(""g water"")))) × (1000 color(red)(cancel(color(black)(""g water""))))/""1 L"" = ""0.002 g/L"" = ""2 mg/L""#

+
+

For 40 L of water, you will need

+

#40 color(red)(cancel(color(black)(""L""))) × ""2 mg""/(1 color(red)(cancel(color(black)(""L"")))) = ""80 mg"" = ""0.08 g""# of nitrate ion.

+
+

The molar mass of #""Ca""(""NO""_3)_2·""4H""_2""O""# is 236·15 g.

+

Of this, 124.01 g comes from the nitrate ions.

+

Hence, to get 0.08 g of nitrate, you will have to use

+

#0.08 color(red)(cancel(color(black)(""g NO""_3^""-""))) × (236.15 ""g Ca""(""NO""_3)_2·""4H""_2""O"")/(124.01 color(red)(cancel(color(black)(""g NO""_3^""-"")))) = ""0.15 g Ca""(""NO""_3)_2·""4H""_2""O""#

+

The required volume of concentrated salt solution is

+

#0.15 color(red)(cancel(color(black)(""g salt""))) × ""1 L salt""/(500 color(red)(cancel(color(black)(""g salt"")))) = ""0.0003 L salt"" = ""0.3 mL salt""#

+
+

There are 40.078 g of #""Ca""# in #""236.15 g Ca""(""NO""_3)_2·""4H""_2""O""#.

+

∴ The mass of calcium added is

+

#0.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O""))) × ""40.078 g Ca""/(236.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O"")))) = ""0.025 g Ca"" = ""25 mg Ca"" #.

+
+
+
" "
+

I want to increase the NO3 concentration in a 40 Liters tank using Ca(NO3)2.4H2O. The concentration of the CaNO3 solution is 500g/L. +How much solution do I need to add increase the concentration by 2ppm? + +Thanks for helping out!

+
+
+

+

How much calcium would be added at the same time?

+

+
+
+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solving Using PPM (Parts Per Million) + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 15, 2017 + +
+
+
+
+
+
+
+

You must add 0.3 mL of the calcium nitrate solution. This will include 25 mg of calcium.

+
+
+
+

Explanation:

+
+
+

Let's express the concentration of 2 ppm as grams per litre.

+

#""2 ppm"" = (2 × 10^""-6"" color(white)(l)""g"")/(1 color(red)(cancel(color(black)(""g water"")))) × (1000 color(red)(cancel(color(black)(""g water""))))/""1 L"" = ""0.002 g/L"" = ""2 mg/L""#

+
+

For 40 L of water, you will need

+

#40 color(red)(cancel(color(black)(""L""))) × ""2 mg""/(1 color(red)(cancel(color(black)(""L"")))) = ""80 mg"" = ""0.08 g""# of nitrate ion.

+
+

The molar mass of #""Ca""(""NO""_3)_2·""4H""_2""O""# is 236·15 g.

+

Of this, 124.01 g comes from the nitrate ions.

+

Hence, to get 0.08 g of nitrate, you will have to use

+

#0.08 color(red)(cancel(color(black)(""g NO""_3^""-""))) × (236.15 ""g Ca""(""NO""_3)_2·""4H""_2""O"")/(124.01 color(red)(cancel(color(black)(""g NO""_3^""-"")))) = ""0.15 g Ca""(""NO""_3)_2·""4H""_2""O""#

+

The required volume of concentrated salt solution is

+

#0.15 color(red)(cancel(color(black)(""g salt""))) × ""1 L salt""/(500 color(red)(cancel(color(black)(""g salt"")))) = ""0.0003 L salt"" = ""0.3 mL salt""#

+
+

There are 40.078 g of #""Ca""# in #""236.15 g Ca""(""NO""_3)_2·""4H""_2""O""#.

+

∴ The mass of calcium added is

+

#0.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O""))) × ""40.078 g Ca""/(236.15 color(red)(cancel(color(black)(""g Ca""(""NO""_3)_2·""4H""_2""O"")))) = ""0.025 g Ca"" = ""25 mg Ca"" #.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" "I want to increase the NO3 concentration in a 40 Liters tank using Ca(NO3)2.4H2O. The concentration of the CaNO3 solution is 500g/L. +How much solution do I need to add increase the concentration by 2ppm? + +Thanks for helping out!" " + + +How much calcium would be added at the same time? + + +" +102 a82ab204-6ddd-11ea-9973-ccda262736ce https://socratic.org/questions/10-00mg-of-a-substance-yields-11-53mg-h2o-and-28-16mg-co2-what-is-the-empirical- C10H20O start chemical_formula qc_end physical_unit 3 4 0 1 mass qc_end physical_unit 8 8 6 7 mass qc_end physical_unit 12 12 10 11 mass qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""C10H20O""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] a substance [=] \\pu{10.00 mg}""},{""type"":""physical unit"",""value"":""Mass [OF] H2O [=] \\pu{11.53 mg}""},{""type"":""physical unit"",""value"":""Mass [OF] CO2 [=] \\pu{28.16 mg}""},{""type"":""other"",""value"":""The substance yields H2O and CO2. ""}]" "

10.00mg of a substance yields 11.53mg H2O and 28.16mg CO2. What is the empirical formula of menthol? + + +This is so confusing, help would be much appreciated!?

" nan C10H20O "
+

Explanation:

+
+

1) calculate mmoles of #CO_2# and water produced:

+

#n_""CO2"" = ""28.16 mg""/""44.01 mg/mmol"" = 0.63985 mmol#

+

these are also the mmoles of carbon in ten milligrams of substance.

+

#n_""H2O"" = ""11.53 mg""/""18.02 mg/mmol"" = 0.63985 mmol#

+

We deduce that ten milligrams of substance contain:

+

#0.63985 * 2 = 1.2797 mmol# of hydrogen (double of water mmoles)

+

We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally) #C_xH_""2x""O_y#.

+

2) calculate moles of dioxygen (#O_2#) necessary to the complete combustion, applying the law of mass conservation

+

#""products' mass"" = 28.16 mg + 11.53 mg = 39.69 mg#

+

then:

+

#""reactants' mass"" = 39.69 mg#

+

Therefore the #O_2# mass needed for combustion must be equal to

+

#m_""O2"" = 39.69 mg - 10.00 mg = 29.69 mg#

+

then we can calculate oxygen moles:

+

#n_O = ""29.69 mg""/""16.00 mg/mmol"" = 1.85563 mmol#

+

3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone

+

The number of mmoles of oxygen atoms in the products is the double of #CO_2# mmoles plus the moles of water:

+

#n_""O(products)"" = 0.63985 * 2 + 0.63985 = 1.91956 mmol# (O total in the products or reactants"")

+

The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:

+

#n_""O (substance)"" = 1.91956 - 1.85563 = 0,06393 mmol# (O combined in 10 mg substance).

+

4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.

+

We see that carbon mmoles are approximately ten times oxygen mmoles:

+

#ratio C/O = n_C/n_O = 0.63985/0.06393 = 10,00#

+

Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:

+

#C_""10""H_""20""O_1#, that is: #C_""10""H_""20""O#.

+

This is also the molecular formula of menthol .

+
+
" "
+
+
+

Empirical formula: #C_""10""H_""20""O#. It corresponds to the empirical and molecular formula of menthol.

+
+
+
+

Explanation:

+
+

1) calculate mmoles of #CO_2# and water produced:

+

#n_""CO2"" = ""28.16 mg""/""44.01 mg/mmol"" = 0.63985 mmol#

+

these are also the mmoles of carbon in ten milligrams of substance.

+

#n_""H2O"" = ""11.53 mg""/""18.02 mg/mmol"" = 0.63985 mmol#

+

We deduce that ten milligrams of substance contain:

+

#0.63985 * 2 = 1.2797 mmol# of hydrogen (double of water mmoles)

+

We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally) #C_xH_""2x""O_y#.

+

2) calculate moles of dioxygen (#O_2#) necessary to the complete combustion, applying the law of mass conservation

+

#""products' mass"" = 28.16 mg + 11.53 mg = 39.69 mg#

+

then:

+

#""reactants' mass"" = 39.69 mg#

+

Therefore the #O_2# mass needed for combustion must be equal to

+

#m_""O2"" = 39.69 mg - 10.00 mg = 29.69 mg#

+

then we can calculate oxygen moles:

+

#n_O = ""29.69 mg""/""16.00 mg/mmol"" = 1.85563 mmol#

+

3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone

+

The number of mmoles of oxygen atoms in the products is the double of #CO_2# mmoles plus the moles of water:

+

#n_""O(products)"" = 0.63985 * 2 + 0.63985 = 1.91956 mmol# (O total in the products or reactants"")

+

The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:

+

#n_""O (substance)"" = 1.91956 - 1.85563 = 0,06393 mmol# (O combined in 10 mg substance).

+

4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.

+

We see that carbon mmoles are approximately ten times oxygen mmoles:

+

#ratio C/O = n_C/n_O = 0.63985/0.06393 = 10,00#

+

Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:

+

#C_""10""H_""20""O_1#, that is: #C_""10""H_""20""O#.

+

This is also the molecular formula of menthol .

+
+
+
" "
+

10.00mg of a substance yields 11.53mg H2O and 28.16mg CO2. What is the empirical formula of menthol? + + +This is so confusing, help would be much appreciated!?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 19, 2015 + +
+
+
+
+
+
+
+

Empirical formula: #C_""10""H_""20""O#. It corresponds to the empirical and molecular formula of menthol.

+
+
+
+

Explanation:

+
+

1) calculate mmoles of #CO_2# and water produced:

+

#n_""CO2"" = ""28.16 mg""/""44.01 mg/mmol"" = 0.63985 mmol#

+

these are also the mmoles of carbon in ten milligrams of substance.

+

#n_""H2O"" = ""11.53 mg""/""18.02 mg/mmol"" = 0.63985 mmol#

+

We deduce that ten milligrams of substance contain:

+

#0.63985 * 2 = 1.2797 mmol# of hydrogen (double of water mmoles)

+

We can deduce soon that the molecule contains a number of hydrogen atoms that is the double of carbon atoms; so the empirical formula should be (provisionally) #C_xH_""2x""O_y#.

+

2) calculate moles of dioxygen (#O_2#) necessary to the complete combustion, applying the law of mass conservation

+

#""products' mass"" = 28.16 mg + 11.53 mg = 39.69 mg#

+

then:

+

#""reactants' mass"" = 39.69 mg#

+

Therefore the #O_2# mass needed for combustion must be equal to

+

#m_""O2"" = 39.69 mg - 10.00 mg = 29.69 mg#

+

then we can calculate oxygen moles:

+

#n_O = ""29.69 mg""/""16.00 mg/mmol"" = 1.85563 mmol#

+

3) calculate number of moles of oxygen in the ten milligrams of substance, applying the conservation law to oxygen alone

+

The number of mmoles of oxygen atoms in the products is the double of #CO_2# mmoles plus the moles of water:

+

#n_""O(products)"" = 0.63985 * 2 + 0.63985 = 1.91956 mmol# (O total in the products or reactants"")

+

The mmole number of oxygen atoms in the substance is obtained by subtracting the oxygen number of mmoles used to combust the substance from this total number of atoms in mmoles:

+

#n_""O (substance)"" = 1.91956 - 1.85563 = 0,06393 mmol# (O combined in 10 mg substance).

+

4) Calculate the empirical formula by taking the moles ratios and getting the minimum integer indexes of C, H and O.

+

We see that carbon mmoles are approximately ten times oxygen mmoles:

+

#ratio C/O = n_C/n_O = 0.63985/0.06393 = 10,00#

+

Therefore we conclude that putting at least one oxygen atom in the empirical formula, we must put ten times this number in the carbon index an the double in hydrogen index, so we get:

+

#C_""10""H_""20""O_1#, that is: #C_""10""H_""20""O#.

+

This is also the molecular formula of menthol .

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
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+
+ 10140 views + around the world +
+
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" "10.00mg of a substance yields 11.53mg H2O and 28.16mg CO2. What is the empirical formula of menthol? + + +This is so confusing, help would be much appreciated!?" nan +103 a82ab205-6ddd-11ea-9044-ccda262736ce https://socratic.org/questions/58114f28b72cff4778c1f503 1.00 mol*L^(−1) start physical_unit 7 10 osmolarity mol/l qc_end physical_unit 7 10 5 6 molarity qc_end end "[{""type"":""physical unit"",""value"":""Osmolarity [OF] solution of magnesium sulfate [IN] mol*L^(−1)""}]" "[{""type"":""physical unit"",""value"":""1.00 mol*L^(−1)""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] solution of magnesium sulfate [=] \\pu{0.5 mol*L^(−1)}""}]" "

What is the osmolarity of #0.5*mol*L^-1# solution of magnesium sulfate?

" nan 1.00 mol*L^(−1) "
+

Explanation:

+
+

#""Osmolarity""# is the concentration of a solution expressed as the total number of #""solute particles""# per litre.....

+

In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.

+

#MgSO_4stackrel(H_2O)rarrMg^(2+) + SO_4^(2-)#

+

Charge and mass are balanced as is absolutely required.

+

And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of #Na_2SO_4#. What is the osmolarity of #0.5*mol*L^-1# #Al(NO_3)_3#.....?

+
+
" "
+
+
+

The #""osmolarity""# of #0.5*mol*L^-1# #MgSO_4# is #1.0*mol*L^-1#....

+
+
+
+

Explanation:

+
+

#""Osmolarity""# is the concentration of a solution expressed as the total number of #""solute particles""# per litre.....

+

In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.

+

#MgSO_4stackrel(H_2O)rarrMg^(2+) + SO_4^(2-)#

+

Charge and mass are balanced as is absolutely required.

+

And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of #Na_2SO_4#. What is the osmolarity of #0.5*mol*L^-1# #Al(NO_3)_3#.....?

+
+
+
" "
+

What is the osmolarity of #0.5*mol*L^-1# solution of magnesium sulfate?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Osmolarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 11, 2017 + +
+
+
+
+
+
+
+

The #""osmolarity""# of #0.5*mol*L^-1# #MgSO_4# is #1.0*mol*L^-1#....

+
+
+
+

Explanation:

+
+

#""Osmolarity""# is the concentration of a solution expressed as the total number of #""solute particles""# per litre.....

+

In aqueous solution, magnesium sulfate speciates to give two moles of ions.......i.e.

+

#MgSO_4stackrel(H_2O)rarrMg^(2+) + SO_4^(2-)#

+

Charge and mass are balanced as is absolutely required.

+

And since magnesium sulfate gives rise to TWO EQUIVS of soluble ions, the osmolarity is equal to TWICE the molarity of #Na_2SO_4#. What is the osmolarity of #0.5*mol*L^-1# #Al(NO_3)_3#.....?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3241 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the osmolarity of #0.5*mol*L^-1# solution of magnesium sulfate? nan +104 a82ab206-6ddd-11ea-9a63-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-volume-of-oxygen-required-for-the-complete-combustion-o 0.50 dm^3 start physical_unit 7 7 volume dm^3 qc_end c_other STP qc_end c_other complete_combustion qc_end physical_unit 17 17 14 15 volume qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [IN] dm^3""}]" "[{""type"":""physical unit"",""value"":""0.50 dm^3""}]" "[{""type"":""other"",""value"":""Complete combustion""},{""type"":""physical unit"",""value"":""Volume [OF] methane [=] \\pu{0.25 dm^3}""},{""type"":""other"",""value"":""STP""}]" "

How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP?

" nan 0.50 dm^3 "
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

+
+
" "
+
+
+

The volume of oxygen required is #""0.50 dm""^3#.

+
+
+
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

+
+
+
" "
+

How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 10, 2016 + +
+
+
+
+
+
+
+

The volume of oxygen required is #""0.50 dm""^3#.

+
+
+
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 35285 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP? nan +105 a82ab207-6ddd-11ea-b48e-ccda262736ce https://socratic.org/questions/the-theoretical-yield-of-a-reaction-is-82-5-grams-but-the-reaction-actually-yiel 85.09% start physical_unit 25 26 percent_yield none qc_end physical_unit 4 5 7 8 theoretical_yield qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Percent yield [OF] this reaction""}]" "[{""type"":""physical unit"",""value"":""85.09%""}]" "[{""type"":""physical unit"",""value"":""Theoretical yield [OF] a reaction [=] \\pu{82.5 grams}""},{""type"":""other"",""value"":""Actually yields 12.3 grams less than expected.""}]" "

The theoretical yield of a reaction is 82.5 grams, but the reaction actually yields 12.3 grams less than expected. What is the percent yield for this reaction?

" nan 85.09% "
+

Explanation:

+
+

Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by #100#

+
+

#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#

+
+

So, you know that your reaction has a theoretical yield of #""82.5 g""#. This means that if all the moles of reactants that take part in the reaction end up producing moles of product according to the reaction's stoichiometric coefficients, you will get #""82.5 g""# of product.

+

In other words, the theoretical yield tells you how much product is produced for a #100%# yield.

+

Now, the reaction is said to produce #""12.3 g""# less than expected. This means that you only collected

+
+

#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#

+
+

The reaction's percent yield will thus be

+
+

#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

+
+
" "
+
+
+

#85.1%#

+
+
+
+

Explanation:

+
+

Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by #100#

+
+

#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#

+
+

So, you know that your reaction has a theoretical yield of #""82.5 g""#. This means that if all the moles of reactants that take part in the reaction end up producing moles of product according to the reaction's stoichiometric coefficients, you will get #""82.5 g""# of product.

+

In other words, the theoretical yield tells you how much product is produced for a #100%# yield.

+

Now, the reaction is said to produce #""12.3 g""# less than expected. This means that you only collected

+
+

#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#

+
+

The reaction's percent yield will thus be

+
+

#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

+
+
+
" "
+

The theoretical yield of a reaction is 82.5 grams, but the reaction actually yields 12.3 grams less than expected. What is the percent yield for this reaction?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 5, 2016 + +
+
+
+
+
+
+
+

#85.1%#

+
+
+
+

Explanation:

+
+

Percent yield is defined as the ratio between the actual yield and the theoretical yield of the reaction, multiplied by #100#

+
+

#color(blue)(|bar(ul(color(white)(a/a)""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100 color(white)(a/a)|)))#

+
+

So, you know that your reaction has a theoretical yield of #""82.5 g""#. This means that if all the moles of reactants that take part in the reaction end up producing moles of product according to the reaction's stoichiometric coefficients, you will get #""82.5 g""# of product.

+

In other words, the theoretical yield tells you how much product is produced for a #100%# yield.

+

Now, the reaction is said to produce #""12.3 g""# less than expected. This means that you only collected

+
+

#""actual yield"" = ""82.5 g"" - ""12.3 g"" = ""70.2 g""#

+
+

The reaction's percent yield will thus be

+
+

#""% yield"" = (70.2 color(red)(cancel(color(black)(""g""))))/(82.5color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""85.1%""color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 9662 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" The theoretical yield of a reaction is 82.5 grams, but the reaction actually yields 12.3 grams less than expected. What is the percent yield for this reaction? nan +106 a82ad5c1-6ddd-11ea-8643-ccda262736ce https://socratic.org/questions/given-the-equation-c-2h-6-g-o-2-g-co-2-g-h-2o-g-not-balanced-what-is-the-number- 362.36 liters start physical_unit 7 7 volume l qc_end chemical_equation 3 9 qc_end c_other STP qc_end physical_unit 3 3 24 25 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] CO2 [IN] liters""}]" "[{""type"":""physical unit"",""value"":""362.36 liters""}]" "[{""type"":""chemical equation"",""value"":""C2H6(g) + O2(g) -> CO2(g) + H2O(g)""},{""type"":""other"",""value"":""STP""},{""type"":""physical unit"",""value"":""Mass [OF] C2H6 [=] \\pu{240.0 grams}""},{""type"":""other"",""value"":""C2H6 is burned in excess oxygen gas.""}]" "

Given the equation #C_2H_6(g) + O_2(g) -> CO_2 (g) + H_2O(g)# (not balanced), what is the number of liters of #CO_2# formed at STP when 240.0 grams of #C_2H_6# is burned in excess oxygen gas?

" nan 362.36 liters "
+

Explanation:

+
+

The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version

+
+

#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#

+
+

Now, the problem provides you with a mass of ethane, #""C""_2""H""_6#, and asks for the volume of carbon dioxide, #""CO""_2#, formed by the reaction at STP, Standard Temperature and Pressure.

+

As you know, STP conditions are defined as a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Under these specific conditions, one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.

+

Use the molar mass of ethane to determine how many moles you'd get in that #""240.0-g""# sample

+
+

#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#

+
+

Now, notice that you have a #color(red)(2):color(purple)(4)# mole ratio between ethane and carbon dioxide. Since ethane reacts with excess oxygen, you can assume that all the moles will take part in the reaction.

+

This means that the reaction will produce

+
+

#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#

+
+

So, if one mole of any ideal gas occupies #""22.7 L""# at STP, this many moles will occupy

+
+

#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#

+
+

The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.

+
+
" "
+
+
+

#""362.4 L""#

+
+
+
+

Explanation:

+
+

The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version

+
+

#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#

+
+

Now, the problem provides you with a mass of ethane, #""C""_2""H""_6#, and asks for the volume of carbon dioxide, #""CO""_2#, formed by the reaction at STP, Standard Temperature and Pressure.

+

As you know, STP conditions are defined as a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Under these specific conditions, one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.

+

Use the molar mass of ethane to determine how many moles you'd get in that #""240.0-g""# sample

+
+

#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#

+
+

Now, notice that you have a #color(red)(2):color(purple)(4)# mole ratio between ethane and carbon dioxide. Since ethane reacts with excess oxygen, you can assume that all the moles will take part in the reaction.

+

This means that the reaction will produce

+
+

#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#

+
+

So, if one mole of any ideal gas occupies #""22.7 L""# at STP, this many moles will occupy

+
+

#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#

+
+

The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.

+
+
+
" "
+

Given the equation #C_2H_6(g) + O_2(g) -> CO_2 (g) + H_2O(g)# (not balanced), what is the number of liters of #CO_2# formed at STP when 240.0 grams of #C_2H_6# is burned in excess oxygen gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 21, 2016 + +
+
+
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+
+

#""362.4 L""#

+
+
+
+

Explanation:

+
+

The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version

+
+

#color(red)(2)""C""_2""H""_text(6(g]) + 7""O""_text(2(g]) -> color(purple)(4)""CO""_text(2(g]) + 6""H""_2""O""_text((g])#

+
+

Now, the problem provides you with a mass of ethane, #""C""_2""H""_6#, and asks for the volume of carbon dioxide, #""CO""_2#, formed by the reaction at STP, Standard Temperature and Pressure.

+

As you know, STP conditions are defined as a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Under these specific conditions, one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.

+

Use the molar mass of ethane to determine how many moles you'd get in that #""240.0-g""# sample

+
+

#240.0 color(red)(cancel(color(black)(""g""))) * (""1 mole C""_2""H""_6)/(30.07color(red)(cancel(color(black)(""g"")))) = ""7.9814 moles C""_2""H""_6#

+
+

Now, notice that you have a #color(red)(2):color(purple)(4)# mole ratio between ethane and carbon dioxide. Since ethane reacts with excess oxygen, you can assume that all the moles will take part in the reaction.

+

This means that the reaction will produce

+
+

#7.9814 color(red)(cancel(color(black)(""moles C""_2""H""_6))) * (color(purple)(4)"" moles CO""_2)/(color(red)(2)color(red)(cancel(color(black)(""moles C""_2""H""_6)))) = ""15.9628 moles CO""_2#

+
+

So, if one mole of any ideal gas occupies #""22.7 L""# at STP, this many moles will occupy

+
+

#15.9628color(red)(cancel(color(black)(""moles CO""_2))) * ""22.7 L""/(1color(red)(cancel(color(black)(""mole CO""_2)))) = color(green)(""362.4 L"")#

+
+

The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.

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Related questions
+ + +
+
+
+
Impact of this question
+
+ 39909 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" Given the equation #C_2H_6(g) + O_2(g) -> CO_2 (g) + H_2O(g)# (not balanced), what is the number of liters of #CO_2# formed at STP when 240.0 grams of #C_2H_6# is burned in excess oxygen gas? nan +107 a82ad5c2-6ddd-11ea-afa0-ccda262736ce https://socratic.org/questions/560452fe11ef6b477ccb79bd 187.43 pounds start physical_unit 16 17 mass pounds qc_end c_other OTHER qc_end physical_unit 16 17 19 20 mass qc_end end "[{""type"":""physical unit"",""value"":""Weight [OF] the patient [IN] pounds""}]" "[{""type"":""physical unit"",""value"":""187.43 pounds""}]" "[{""type"":""other"",""value"":""The doctor orders a medication as 5.00 mg/kg of body weight.""},{""type"":""physical unit"",""value"":""Dosage [OF] the patient [=] \\pu{425 mg}""}]" "

The doctor orders a medication as #""5.00 mg/kg""# of body weight. The dosage the nurse gives the patient is #""425 mg""#. How much does the patient weigh in pounds?

" nan 187.43 pounds "
+

Explanation:

+
+

You have a little mor econverting to do to solve this one.

+

You know that the drug dosage is set to #""5.00 mg/kg""# of body weight, and that you need to give the patient's weight in pounds.

+

To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor

+
+

#""1 kg"" ~= ""2.2046 lbs""#

+
+

This means that you have

+
+

#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#

+
+

So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be

+
+

#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#

+
+

Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be

+
+

#color(green)(""187 lbs"")#

+
+
+
" "
+
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+

#""187 lbs""#

+
+
+
+

Explanation:

+
+

You have a little mor econverting to do to solve this one.

+

You know that the drug dosage is set to #""5.00 mg/kg""# of body weight, and that you need to give the patient's weight in pounds.

+

To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor

+
+

#""1 kg"" ~= ""2.2046 lbs""#

+
+

This means that you have

+
+

#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#

+
+

So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be

+
+

#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#

+
+

Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be

+
+

#color(green)(""187 lbs"")#

+
+
+
+
" "
+

The doctor orders a medication as #""5.00 mg/kg""# of body weight. The dosage the nurse gives the patient is #""425 mg""#. How much does the patient weigh in pounds?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
+
+
+
+
+3 Answers +
+
+
+
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+
+ +
+
+ +
+ + Sep 24, 2015 + +
+
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+

#""187 lbs""#

+
+
+
+

Explanation:

+
+

You have a little mor econverting to do to solve this one.

+

You know that the drug dosage is set to #""5.00 mg/kg""# of body weight, and that you need to give the patient's weight in pounds.

+

To start, you can convert the dosage from miligrams per kilogram to miligrams per pound by using the conversion factor

+
+

#""1 kg"" ~= ""2.2046 lbs""#

+
+

This means that you have

+
+

#5.00""mg""/color(red)(cancel(color(black)(""kg""))) * (1color(red)(cancel(color(black)(""kg""))))/""2.2046 lbs"" = ""2.268 mg/lbs""#

+
+

So, your patient needs 2.268 mg of the drug per pound of body weight. This means that if you supply 425 mg, his body weight must be

+
+

#425color(red)(cancel(color(black)(""mg""))) * ""1 lbs""/(2.268color(red)(cancel(color(black)(""mg"")))) = ""187.39 lbs""#

+
+

Rounded to three sig figs, the number of sig figs you gave for the dosage and mass of drug administered, the answer will be

+
+

#color(green)(""187 lbs"")#

+
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+ +
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+ +
+ + Sep 24, 2015 + +
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The patient weighs #""187 lb""#.

+
+
+
+

Explanation:

+
+

Given/Known

+

Medication mass/body mass: #""5.00mg/kg""#

+

Dosage: #""425 mg""#

+

#""1 kg""##=##""2.2 lb""#

+

#425cancel ""mg""xx(1""kg"")/(5.00cancel ""mg"")=""85 kg""#

+

#85 cancel""kg""xx(2.2 ""lb"")/(1 cancel""kg"") =""187 lb""#

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+ +
+
+ +
+ + Jul 13, 2018 + +
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+

Just over 187 lb

+
+
+
+

Explanation:

+
+

#M = D times W# where M is the medication mass in mg, D is the dose rate in mg per kg, and W is the weight of the patient in kg.

+

Therefore #W = M/D#

+

But you want the weight of the patient in lb, so you need to build in the conversion #lb = kg times 2.205#.

+

Therefore the final equation is: #W = (M/D) times 2.205#

+

Plug in the numbers and you get 187.43 lb.

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+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 20230 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "The doctor orders a medication as #""5.00 mg/kg""# of body weight. The dosage the nurse gives the patient is #""425 mg""#. How much does the patient weigh in pounds?" nan +108 a82ad5c3-6ddd-11ea-8dff-ccda262736ce https://socratic.org/questions/the-pressure-in-a-car-tire-is-205-kpa-at-303-k-after-a-long-drive-the-pressure-i 351.78 K start physical_unit 26 30 temperature k qc_end physical_unit 4 5 10 11 temperature qc_end physical_unit 4 5 7 8 pressure qc_end physical_unit 4 5 19 20 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] the air in the tire [IN] K""}]" "[{""type"":""physical unit"",""value"":""351.78 K""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] car tire [=] \\pu{303 K}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] car tire [=] \\pu{205 kPa}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] car tire [=] \\pu{238 kPa}""}]" "

The pressure in a car tire is 205 kPa at 303 K. After a long drive, the pressure is 238 kPa. What is the temperature of the air in the tire?

" nan 351.78 K "
+

Explanation:

+
+

You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.

+

More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.

+

+

So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.

+

In your case, the pressure in the tire increased from #""205 kPa""# to #""238 kPa""#. This tells you that the temperature of the must have increased as well.

+

Mathematically, you can write this as

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""#, where

+
+

#P_1#, #T_1# - the pressure and temperature of the gas at an initial state
+#P_2#, #T_2# - the pressure and temperature of the gas at final state

+

Rearrange the equation to solve for #T_2#

+
+

#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#

+
+

Plug in your values to get

+
+

#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#

+
+

Rounded to three sig figs, the answer will be

+
+

#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#

+
+

As predicted, the temperature of the air increased.

+
+
" "
+
+
+

#""352 K""#

+
+
+
+

Explanation:

+
+

You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.

+

More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.

+

+

So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.

+

In your case, the pressure in the tire increased from #""205 kPa""# to #""238 kPa""#. This tells you that the temperature of the must have increased as well.

+

Mathematically, you can write this as

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""#, where

+
+

#P_1#, #T_1# - the pressure and temperature of the gas at an initial state
+#P_2#, #T_2# - the pressure and temperature of the gas at final state

+

Rearrange the equation to solve for #T_2#

+
+

#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#

+
+

Plug in your values to get

+
+

#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#

+
+

Rounded to three sig figs, the answer will be

+
+

#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#

+
+

As predicted, the temperature of the air increased.

+
+
+
" "
+

The pressure in a car tire is 205 kPa at 303 K. After a long drive, the pressure is 238 kPa. What is the temperature of the air in the tire?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gay Lussac's Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 8, 2016 + +
+
+
+
+
+
+
+

#""352 K""#

+
+
+
+

Explanation:

+
+

You can assume that the volume of the tire and the amount of air it contains remain constant, which means that you can focus solely on the relationship that exists between pressure and temperature under these conditions.

+

More specifically, you should know that when volume and number of moles are kept constant, temperature and pressure have a direct relationship - this is known as Gay Lussac's Law.

+

+

So, when temperature increases, pressure increases as well. Likewise, when temperature decreases, pressure decreases as well.

+

In your case, the pressure in the tire increased from #""205 kPa""# to #""238 kPa""#. This tells you that the temperature of the must have increased as well.

+

Mathematically, you can write this as

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1 = P_2/T_2color(white)(a/a)|)))"" ""#, where

+
+

#P_1#, #T_1# - the pressure and temperature of the gas at an initial state
+#P_2#, #T_2# - the pressure and temperature of the gas at final state

+

Rearrange the equation to solve for #T_2#

+
+

#P_1/T_1 = P_2/T_2 implies T_2 = P_2/P_1 * T_1#

+
+

Plug in your values to get

+
+

#T_2 = (238color(red)(cancel(color(black)(""kPa""))))/(205color(red)(cancel(color(black)(""kPa"")))) * ""303 K"" = ""351.78 K""#

+
+

Rounded to three sig figs, the answer will be

+
+

#T_2 = color(green)(|bar(ul(color(white)(a/a)""352 K""color(white)(a/a)|)))#

+
+

As predicted, the temperature of the air increased.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 13919 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" The pressure in a car tire is 205 kPa at 303 K. After a long drive, the pressure is 238 kPa. What is the temperature of the air in the tire? nan +109 a82ad5c4-6ddd-11ea-b9ad-ccda262736ce https://socratic.org/questions/after-0-600-l-of-ar-at-1-47-atm-and-221-degrees-celsius-is-mixed-with-0-200-l-of 1.55 atm start physical_unit 40 41 pressure atm qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 9 11 temperature qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 18 18 20 21 pressure qc_end physical_unit 18 18 23 25 temperature qc_end physical_unit 30 30 28 29 volume qc_end physical_unit 30 30 32 34 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Pressure [OF] the flask [IN] atm""}]" "[{""type"":""physical unit"",""value"":""1.55 atm""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] Ar [=] \\pu{0.600 L}""},{""type"":""physical unit"",""value"":""Pressure [OF] Ar [=] \\pu{1.47 atm}""},{""type"":""physical unit"",""value"":""Temperature [OF] Ar [=] \\pu{221 degrees Celsius}""},{""type"":""physical unit"",""value"":""Volume [OF] O2 [=] \\pu{0.200 L}""},{""type"":""physical unit"",""value"":""Pressure [OF] O2 [=] \\pu{441 torr}""},{""type"":""physical unit"",""value"":""Temperature [OF] O2 [=] \\pu{115 degrees Celsius}""},{""type"":""physical unit"",""value"":""Volume [OF] the flask [=] \\pu{400 mL}""},{""type"":""physical unit"",""value"":""Temperature [OF] the flask [=] \\pu{24 degrees Celsius}""}]" "

After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

" nan 1.55 atm "
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

+

#= color(blue)(1.55# #color(blue)(""atm""#

+
+
" "
+
+
+

#P = 1.55# #""atm""#

+
+
+
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

+

#= color(blue)(1.55# #color(blue)(""atm""#

+
+
+
" "
+

After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
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+ +
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+ +
+ + Jul 8, 2017 + +
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+

#P = 1.55# #""atm""#

+
+
+
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

+

#= color(blue)(1.55# #color(blue)(""atm""#

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" After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask? nan +110 a82ad5c5-6ddd-11ea-ac05-ccda262736ce https://socratic.org/questions/chemistry-mass-percent-question 18.18% start physical_unit 17 20 mass_percent none qc_end physical_unit 0 5 10 11 density qc_end physical_unit 17 20 28 29 mass qc_end physical_unit 1 1 22 23 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass% [OF] HCl in the solution""}]" "[{""type"":""physical unit"",""value"":""18.18%""}]" "[{""type"":""physical unit"",""value"":""Density [OF] a solution of HCl in water [=] \\pu{1.1 g/mL}""},{""type"":""physical unit"",""value"":""Mass [OF] HCl in the solution [=] \\pu{30.0 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{150 mL}""}]" "

A solution of HCl in water has a density of 1.1 g/mL. What is the mass% of HCl in the solution if 150 mL of the solution contains 30. g of HCl?

" nan 18.18% "
+

Explanation:

+
+

The first thing that you need to do here is to use the density of the solution to determine the mass of #""150 mL""# of this hydrochloric acid solution.

+

The density of the solution is said to be equal to #""1.1 g mL""^(-1)#, so right from the start, you know that every #""1 mL""# of solution will have a mass of #""1.1 g""#.

+

This means that your sample will have a mass of

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+

#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#

+
+

Now, in order to find the solution's percent concentration by mass, #""% m/m""#, you need to figure out the mass of hydrochloric acid present in #""100 g""# of this solution.

+

Since you know that #""165 g""# of solution contain #""30. g""# of hydrochloric acid, you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#

+
+

You can thus say that the solution has a percent concentration by mass equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#

+
+

The answer is rounded to two sig figs.

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+
" "
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#""18% m/m""#

+
+
+
+

Explanation:

+
+

The first thing that you need to do here is to use the density of the solution to determine the mass of #""150 mL""# of this hydrochloric acid solution.

+

The density of the solution is said to be equal to #""1.1 g mL""^(-1)#, so right from the start, you know that every #""1 mL""# of solution will have a mass of #""1.1 g""#.

+

This means that your sample will have a mass of

+
+

#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#

+
+

Now, in order to find the solution's percent concentration by mass, #""% m/m""#, you need to figure out the mass of hydrochloric acid present in #""100 g""# of this solution.

+

Since you know that #""165 g""# of solution contain #""30. g""# of hydrochloric acid, you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#

+
+

You can thus say that the solution has a percent concentration by mass equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#

+
+

The answer is rounded to two sig figs.

+
+
+
" "
+

A solution of HCl in water has a density of 1.1 g/mL. What is the mass% of HCl in the solution if 150 mL of the solution contains 30. g of HCl?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
+
+
+
+
+1 Answer +
+
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+ +
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+ +
+ + Sep 16, 2017 + +
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#""18% m/m""#

+
+
+
+

Explanation:

+
+

The first thing that you need to do here is to use the density of the solution to determine the mass of #""150 mL""# of this hydrochloric acid solution.

+

The density of the solution is said to be equal to #""1.1 g mL""^(-1)#, so right from the start, you know that every #""1 mL""# of solution will have a mass of #""1.1 g""#.

+

This means that your sample will have a mass of

+
+

#150 color(red)(cancel(color(black)(""mL""))) * ""1.1 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""165 g""#

+
+

Now, in order to find the solution's percent concentration by mass, #""% m/m""#, you need to figure out the mass of hydrochloric acid present in #""100 g""# of this solution.

+

Since you know that #""165 g""# of solution contain #""30. g""# of hydrochloric acid, you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""30. g HCl""/(165color(red)(cancel(color(black)(""g solution"")))) = ""18.18 g HCl""#

+
+

You can thus say that the solution has a percent concentration by mass equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m = 18% HCl"")))#

+
+

The answer is rounded to two sig figs.

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+
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" A solution of HCl in water has a density of 1.1 g/mL. What is the mass% of HCl in the solution if 150 mL of the solution contains 30. g of HCl? nan +111 a82ad5c6-6ddd-11ea-85f8-ccda262736ce https://socratic.org/questions/how-many-g-of-water-are-required-to-be-mixed-with-11-75-g-of-hgcl-in-order-to-ma 4980 g start physical_unit 4 4 mass g qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 22 22 20 21 molality qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] water [IN] g""}]" "[{""type"":""physical unit"",""value"":""4980 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] HgCl [=] \\pu{11.75 g}""},{""type"":""physical unit"",""value"":""Molality [OF] solution [=] \\pu{0.01 m}""}]" "

How many #""g""# of water are required to be mixed with #""11.75 g""# of #HgCl# in order to make a #""0.01 m""# solution?

" nan 4980 g "
+

Explanation:

+
+

#color(blue)(""We will use this equation to answer the question:"")#

+

+

We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:

+

#11.75 cancel gxx(1 mol)/(236.04cancelg)# =

+

#color (brown) (""0.0498 mol HgCl"")#

+

Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :

+

#""moles solute""/""molality"" = ""mass of solvent (kg)""#

+

#(0.0498 cancel""mol"") /(0.01 cancel""mol""/(kg)#

+

#color(brown)(""4.98kg Solvent"")#

+

Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:

+

+

Therefore,

+

#4.98cancel""kg""xx(1000g)/(1cancelkg)# #=#

+

4,980 g #H_2O#

+
+
" "
+
+
+

4,980 g #H_2O# of water is required.

+
+
+
+

Explanation:

+
+

#color(blue)(""We will use this equation to answer the question:"")#

+

+

We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:

+

#11.75 cancel gxx(1 mol)/(236.04cancelg)# =

+

#color (brown) (""0.0498 mol HgCl"")#

+

Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :

+

#""moles solute""/""molality"" = ""mass of solvent (kg)""#

+

#(0.0498 cancel""mol"") /(0.01 cancel""mol""/(kg)#

+

#color(brown)(""4.98kg Solvent"")#

+

Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:

+

+

Therefore,

+

#4.98cancel""kg""xx(1000g)/(1cancelkg)# #=#

+

4,980 g #H_2O#

+
+
+
" "
+

How many #""g""# of water are required to be mixed with #""11.75 g""# of #HgCl# in order to make a #""0.01 m""# solution?

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+
+ + +Chemistry + + + + + +Solutions + + + + + +Solution Formation + + +
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+1 Answer +
+
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+ +
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+ +
+ + Jul 13, 2016 + +
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+

4,980 g #H_2O# of water is required.

+
+
+
+

Explanation:

+
+

#color(blue)(""We will use this equation to answer the question:"")#

+

+

We know the molality (m), but not the number of moles of solute. The mass of solute can be converted into moles using the molar mass of HgCl (236.04 g/mol) as a conversion factor:

+

#11.75 cancel gxx(1 mol)/(236.04cancelg)# =

+

#color (brown) (""0.0498 mol HgCl"")#

+

Since the moles of solute and molality are known, we can rearrange the equation to solve for mass of solvent :

+

#""moles solute""/""molality"" = ""mass of solvent (kg)""#

+

#(0.0498 cancel""mol"") /(0.01 cancel""mol""/(kg)#

+

#color(brown)(""4.98kg Solvent"")#

+

Lastly, the mass of solvent has to be converted from kg to g using the following conversion factor:

+

+

Therefore,

+

#4.98cancel""kg""xx(1000g)/(1cancelkg)# #=#

+

4,980 g #H_2O#

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" "How many #""g""# of water are required to be mixed with #""11.75 g""# of #HgCl# in order to make a #""0.01 m""# solution?" nan +112 a82ad5c7-6ddd-11ea-bdfe-ccda262736ce https://socratic.org/questions/what-volume-of-concentrated-hydrochloric-acid-12-0-m-hcl-is-require-to-make-2-0- 0.50 liters start physical_unit 3 5 volume l qc_end physical_unit 3 5 6 7 molarity qc_end physical_unit 19 20 13 14 volume qc_end physical_unit 19 20 17 18 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] concentrated hydrochloric acid [IN] liters""}]" "[{""type"":""physical unit"",""value"":""0.50 liters""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] concentrated hydrochloric acid [=] \\pu{12.0 M}""},{""type"":""physical unit"",""value"":""Volume [OF] HCl solution [=] \\pu{2.0 liters}""},{""type"":""physical unit"",""value"":""Molarity [OF] HCl solution [=] \\pu{3.0 M}""}]" "

What volume of concentrated hydrochloric acid (12.0 M #HCl#) is require to make 2.0 liters of a 3.0 M #HCl# solution?

" nan 0.50 liters "
+

Explanation:

+
+

The product #C_1V_1# is #""concentration""xx""volume""#, and when we multiply the typical units we get #mol*cancel(L^-1)xxcancelL#, i.e. units of #""moles""# as required.

+

Anyway, as to your problem, #C_1V_1=C_2V_2#, because both sides have the units of moles.

+

#V_1=(C_2V_2)/C_1=(3.0*mol*L^-1xx2.0*L)/(12.0*mol*L^-1)#

+

#=1/2*L#

+

Note that we typically buy conc. #HCl# as a #32%(w/w)# solution, and this is about #10*mol*L^-1#. This question was not well proposed.

+
+
" "
+
+
+

#500*mL# of #12.0*mol*L^-1# #HCl*(aq)# are required.

+
+
+
+

Explanation:

+
+

The product #C_1V_1# is #""concentration""xx""volume""#, and when we multiply the typical units we get #mol*cancel(L^-1)xxcancelL#, i.e. units of #""moles""# as required.

+

Anyway, as to your problem, #C_1V_1=C_2V_2#, because both sides have the units of moles.

+

#V_1=(C_2V_2)/C_1=(3.0*mol*L^-1xx2.0*L)/(12.0*mol*L^-1)#

+

#=1/2*L#

+

Note that we typically buy conc. #HCl# as a #32%(w/w)# solution, and this is about #10*mol*L^-1#. This question was not well proposed.

+
+
+
" "
+

What volume of concentrated hydrochloric acid (12.0 M #HCl#) is require to make 2.0 liters of a 3.0 M #HCl# solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
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+1 Answer +
+
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+ +
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+ +
+ + Jan 17, 2017 + +
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#500*mL# of #12.0*mol*L^-1# #HCl*(aq)# are required.

+
+
+
+

Explanation:

+
+

The product #C_1V_1# is #""concentration""xx""volume""#, and when we multiply the typical units we get #mol*cancel(L^-1)xxcancelL#, i.e. units of #""moles""# as required.

+

Anyway, as to your problem, #C_1V_1=C_2V_2#, because both sides have the units of moles.

+

#V_1=(C_2V_2)/C_1=(3.0*mol*L^-1xx2.0*L)/(12.0*mol*L^-1)#

+

#=1/2*L#

+

Note that we typically buy conc. #HCl# as a #32%(w/w)# solution, and this is about #10*mol*L^-1#. This question was not well proposed.

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" What volume of concentrated hydrochloric acid (12.0 M #HCl#) is require to make 2.0 liters of a 3.0 M #HCl# solution? nan +113 a82afe86-6ddd-11ea-b5c6-ccda262736ce https://socratic.org/questions/considering-the-reaction-shown-below-how-much-thermal-energy-would-be-required-t 5.22 KJ start physical_unit 1 2 thermal_energy kj qc_end physical_unit 20 20 17 18 mass qc_end physical_unit 25 25 22 23 mass qc_end c_other OTHER qc_end chemical_equation 26 31 qc_end end "[{""type"":""physical unit"",""value"":""Thermal energy [OF] the reaction [IN] KJ""}]" "[{""type"":""physical unit"",""value"":""5.22 KJ""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen [=] \\pu{5.0 g}""},{""type"":""physical unit"",""value"":""Mass [OF] iodine [=] \\pu{25 g}""},{""type"":""other"",""value"":""deltaH(rxn) = + 53 kJ""},{""type"":""chemical equation"",""value"":""H2(g) + I2(g) -> 2 HI(g)""}]" "

Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine?

" "
+
+

+

H2(g) + I2(g) --> 2HI(g)
+deltaH(rxn) = +53 kJ

+

I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!

+

+
+
" 5.22 KJ "
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
" "
+
+
+

#""5.2 kJ""#

+
+
+
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
+
" "
+

Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine?

+
+
+

+

H2(g) + I2(g) --> 2HI(g)
+deltaH(rxn) = +53 kJ

+

I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!

+

+
+
+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 8, 2017 + +
+
+
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+
+

#""5.2 kJ""#

+
+
+
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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" Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine? " + + +H2(g) + I2(g) --> 2HI(g) +deltaH(rxn) = +53 kJ +I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you! + + +" +114 a82afe87-6ddd-11ea-a9ef-ccda262736ce https://socratic.org/questions/how-would-you-calculate-the-mass-in-grams-of-hydrogen-chloride-produced-when-4-9 15.74 grams start physical_unit 9 10 mass g qc_end physical_unit 16 17 13 14 volume qc_end c_other STP qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen chloride [IN] grams""}]" "[{""type"":""physical unit"",""value"":""15.74 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] molecular hydrogen [=] \\pu{4.9 L}""},{""type"":""other"",""value"":""STP""},{""type"":""other"",""value"":""an excess of chlorine gas""}]" "

How would you calculate the mass in grams of hydrogen chloride produced when 4.9 L of molecular hydrogen at STP reacts with an excess of chlorine gas.?

" nan 15.74 grams "
+

Explanation:

+
+

The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of #""100 kPa""# and a temperature of #0^@""C""#.

+

Moreover, at STP one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.

+

To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

For hydrogen, you would have

+
+

#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#

+
+

and for chlorine you have

+
+

#n_""chlorine"" = (P * V_""chlorine"")/(RT)#

+
+

Thus, the mole ratio between hydrogen and chlorine will be

+
+

#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#

+
+

The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.

+

So, the balanced chemical equation for this reaction is

+
+

#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#

+
+

Notice that you have a #1:color(blue)(2)# mole ratio between hydrogen gas and hydrogen chloride.

+

This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.

+

Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction

+
+

#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#

+
+

Now use the molar volume to find how many moles you'd get in this volume of gas at STP

+
+

#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#

+
+

Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles

+
+

#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#

+
+

Rounded to two sig figs, the answer will be

+
+

#m_""HCl"" = color(green)(""16 g"")#

+
+
+
" "
+
+
+

#""16 g""#

+
+
+
+

Explanation:

+
+

The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of #""100 kPa""# and a temperature of #0^@""C""#.

+

Moreover, at STP one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.

+

To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

For hydrogen, you would have

+
+

#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#

+
+

and for chlorine you have

+
+

#n_""chlorine"" = (P * V_""chlorine"")/(RT)#

+
+

Thus, the mole ratio between hydrogen and chlorine will be

+
+

#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#

+
+

The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.

+

So, the balanced chemical equation for this reaction is

+
+

#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#

+
+

Notice that you have a #1:color(blue)(2)# mole ratio between hydrogen gas and hydrogen chloride.

+

This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.

+

Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction

+
+

#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#

+
+

Now use the molar volume to find how many moles you'd get in this volume of gas at STP

+
+

#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#

+
+

Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles

+
+

#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#

+
+

Rounded to two sig figs, the answer will be

+
+

#m_""HCl"" = color(green)(""16 g"")#

+
+
+
+
" "
+

How would you calculate the mass in grams of hydrogen chloride produced when 4.9 L of molecular hydrogen at STP reacts with an excess of chlorine gas.?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Stoichiometry + + +
+
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+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Oct 23, 2015 + +
+
+
+
+
+
+
+

#""16 g""#

+
+
+
+

Explanation:

+
+

The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of #""100 kPa""# and a temperature of #0^@""C""#.

+

Moreover, at STP one mole of any ideal gas occupies exactly #""22.7 L""# - this is known as the molar volume of a gas at STP.

+

Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.

+

To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

For hydrogen, you would have

+
+

#n_""hydrogen"" = (P * V_""hydrogen"")/(RT)#

+
+

and for chlorine you have

+
+

#n_""chlorine"" = (P * V_""chlorine"")/(RT)#

+
+

Thus, the mole ratio between hydrogen and chlorine will be

+
+

#n_""hydrogen""/n_""chlorine""= (color(red)(cancel(color(black)(P))) V_""hydronge"")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_""chlorine"") = V_""hydrogen""/V_""chlorine""#

+
+

The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.

+

So, the balanced chemical equation for this reaction is

+
+

#""H""_text(2(g]) + ""Cl""_text(2(g]) -> color(blue)(2)""HCl""_text((g])#

+
+

Notice that you have a #1:color(blue)(2)# mole ratio between hydrogen gas and hydrogen chloride.

+

This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.

+

Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction

+
+

#4.9color(red)(cancel(color(black)("" L H""_2))) * (color(blue)(2)"" L HCl"")/(1color(red)(cancel(color(black)("" L H""_2)))) = ""9.8 L HCl""#

+
+

Now use the molar volume to find how many moles you'd get in this volume of gas at STP

+
+

#9.8color(red)(cancel(color(black)("" L HCl""))) * ""1 mole HCl""/(22.7color(red)(cancel(color(black)("" L HCl"")))) = ""0.4317 moles HCl""#

+
+

Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles

+
+

#0.4317color(red)(cancel(color(black)(""moles HCl""))) * ""36.461 g""/(1color(red)(cancel(color(black)(""mole HCl"")))) = ""15.74 g""#

+
+

Rounded to two sig figs, the answer will be

+
+

#m_""HCl"" = color(green)(""16 g"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
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+
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+ + Creative Commons License + +
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+
" How would you calculate the mass in grams of hydrogen chloride produced when 4.9 L of molecular hydrogen at STP reacts with an excess of chlorine gas.? nan +115 a82afe88-6ddd-11ea-a877-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-0-0235-m-hcl-solution 1.63 start physical_unit 8 9 ph none qc_end physical_unit 8 9 6 7 molarity qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] HCl solution""}]" "[{""type"":""physical unit"",""value"":""1.63""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl solution [=] \\pu{0.0235 M}""}]" "

What is the pH of a 0.0235 M #HCl# solution?

" nan 1.63 "
+

Explanation:

+
+

Concentration of HCl is #0.0235# #M#

+

#[HCl] = 0.0235# #mol#/# L#

+

as #HCl# is a strong acid, it will completely dissociate and produce #H^+# ions.

+

[ #H^+#] = #0.0235# #mol#/# L#

+

pH = #-log# [ #H^+#]

+

#pH = - log [0.0235]#

+

#pH = -(-1.628)#

+

#pH = 1.63#

+
+
" "
+
+
+

#pH = 1.63#

+
+
+
+

Explanation:

+
+

Concentration of HCl is #0.0235# #M#

+

#[HCl] = 0.0235# #mol#/# L#

+

as #HCl# is a strong acid, it will completely dissociate and produce #H^+# ions.

+

[ #H^+#] = #0.0235# #mol#/# L#

+

pH = #-log# [ #H^+#]

+

#pH = - log [0.0235]#

+

#pH = -(-1.628)#

+

#pH = 1.63#

+
+
+
" "
+

What is the pH of a 0.0235 M #HCl# solution?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + May 17, 2016 + +
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+
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+
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#pH = 1.63#

+
+
+
+

Explanation:

+
+

Concentration of HCl is #0.0235# #M#

+

#[HCl] = 0.0235# #mol#/# L#

+

as #HCl# is a strong acid, it will completely dissociate and produce #H^+# ions.

+

[ #H^+#] = #0.0235# #mol#/# L#

+

pH = #-log# [ #H^+#]

+

#pH = - log [0.0235]#

+

#pH = -(-1.628)#

+

#pH = 1.63#

+
+
+
+
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+ +
+
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+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+
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+ + Creative Commons License + +
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+
" What is the pH of a 0.0235 M #HCl# solution? nan +116 a82afe89-6ddd-11ea-a9ba-ccda262736ce https://socratic.org/questions/how-do-you-balance-na-h-2o-naoh-h-2 2 Na + 2 H2O -> 2 NaOH + 2 H2 start chemical_equation qc_end chemical_equation 4 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""2 Na + 2 H2O -> 2 NaOH + 2 H2""}]" "[{""type"":""chemical equation"",""value"":""Na + H2O -> NaOH + H2""}]" "

How do you balance #Na + H_2O -> NaOH + H_2#?

" nan 2 Na + 2 H2O -> 2 NaOH + 2 H2 "
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
" "
+
+
+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
" "
+

How do you balance #Na + H_2O -> NaOH + H_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 11, 2017 + +
+
+
+
+
+
+
+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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" How do you balance #Na + H_2O -> NaOH + H_2#? nan +117 a82afe8a-6ddd-11ea-bba8-ccda262736ce https://socratic.org/questions/if-i-have-4-00-moles-of-a-gas-at-a-pressure-of-5-6-atm-and-a-volume-of-12-liters 204.63 K start physical_unit 6 7 temperature k qc_end physical_unit 6 7 3 4 mole qc_end physical_unit 6 7 12 13 pressure qc_end physical_unit 6 7 18 19 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] a gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""204.63 K""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] a gas [=] \\pu{4.00 moles}""},{""type"":""physical unit"",""value"":""Pressure [OF] a gas [=] \\pu{5.6 atm}""},{""type"":""physical unit"",""value"":""Volume [OF] a gas [=] \\pu{12 liters}""}]" "

If I have 4.00 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

" nan 204.63 K "
+

Explanation:

+
+

We use the Ideal Gas equation, #PV=nRT#, and solve for #T#, which here has units of #""degrees Kelvin""#.

+

#T=(5.6*cancel(atm)xx12*cancel(L))/(0.0821*cancel(L)*cancel(atm)*K^-1*cancel(mol^-1)xx4.00*cancel(mol))#

+

We are left with units of #1/(K^-1)=K#, units of temperature as required.

+
+
" "
+
+
+

#T=(PV)/(nR)# #~=# #200*K#

+
+
+
+

Explanation:

+
+

We use the Ideal Gas equation, #PV=nRT#, and solve for #T#, which here has units of #""degrees Kelvin""#.

+

#T=(5.6*cancel(atm)xx12*cancel(L))/(0.0821*cancel(L)*cancel(atm)*K^-1*cancel(mol^-1)xx4.00*cancel(mol))#

+

We are left with units of #1/(K^-1)=K#, units of temperature as required.

+
+
+
" "
+

If I have 4.00 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 3, 2016 + +
+
+
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+
+
+
+

#T=(PV)/(nR)# #~=# #200*K#

+
+
+
+

Explanation:

+
+

We use the Ideal Gas equation, #PV=nRT#, and solve for #T#, which here has units of #""degrees Kelvin""#.

+

#T=(5.6*cancel(atm)xx12*cancel(L))/(0.0821*cancel(L)*cancel(atm)*K^-1*cancel(mol^-1)xx4.00*cancel(mol))#

+

We are left with units of #1/(K^-1)=K#, units of temperature as required.

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+
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" If I have 4.00 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? nan +118 a82afe8b-6ddd-11ea-8226-ccda262736ce https://socratic.org/questions/how-many-grams-of-naf-form-when-25-7g-of-hf-reacts-with-excess-sodium-silicate 32.68 grams start physical_unit 4 4 mass g qc_end physical_unit 10 10 7 8 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaF [IN] grams""}]" "[{""type"":""physical unit"",""value"":""32.68 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] HF [=] \\pu{25.7 g}""},{""type"":""other"",""value"":""HF reacts with excess sodium silicate.""}]" "

How many grams of NaF form when 25.7g of HF reacts with excess sodium silicate?

" nan 32.68 grams "
+

Explanation:

+
+

The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (#20.0*g*mol^-1)# and sodium fluoride (#42.0*g*mol^-1)#.

+

So moles of #HF# #=# #(25.7*g)/(20.0*g*mol^-1)# #=# #??# #mol#

+

Moles of #NaF# #=# #(20.0*g)/(25.7*g*mol^-1) xx42.0*g*mol^-1# #=# #??# #g#

+
+
" "
+
+
+

#2HF(aq) + Na_2SiO_3(aq) rarr 2NaFdarr + H_2SiO_3(aq)#

+
+
+
+

Explanation:

+
+

The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (#20.0*g*mol^-1)# and sodium fluoride (#42.0*g*mol^-1)#.

+

So moles of #HF# #=# #(25.7*g)/(20.0*g*mol^-1)# #=# #??# #mol#

+

Moles of #NaF# #=# #(20.0*g)/(25.7*g*mol^-1) xx42.0*g*mol^-1# #=# #??# #g#

+
+
+
" "
+

How many grams of NaF form when 25.7g of HF reacts with excess sodium silicate?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
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+ +
+
+ +
+ + Jan 12, 2016 + +
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+
+

#2HF(aq) + Na_2SiO_3(aq) rarr 2NaFdarr + H_2SiO_3(aq)#

+
+
+
+

Explanation:

+
+

The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (#20.0*g*mol^-1)# and sodium fluoride (#42.0*g*mol^-1)#.

+

So moles of #HF# #=# #(25.7*g)/(20.0*g*mol^-1)# #=# #??# #mol#

+

Moles of #NaF# #=# #(20.0*g)/(25.7*g*mol^-1) xx42.0*g*mol^-1# #=# #??# #g#

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+
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+
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+ + +
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+
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+ + Creative Commons License + +
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+
" How many grams of NaF form when 25.7g of HF reacts with excess sodium silicate? nan +119 a82afe8c-6ddd-11ea-b81d-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-a-piece-of-copper-when-3000-j-of-heat-is-applied-causing-a-4 173.37 g start physical_unit 5 8 mass g qc_end physical_unit 8 8 29 32 heat_capacity_ratio qc_end physical_unit 8 8 10 11 heat_energy qc_end physical_unit 8 8 18 19 temperature qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] a piece of copper [IN] g""}]" "[{""type"":""physical unit"",""value"":""173.37 g""}]" "[{""type"":""physical unit"",""value"":""C_{{p}} [OF] copper [=] \\pu{38452 J/(g * ℃)}""},{""type"":""physical unit"",""value"":""Applied heat [OF] copper [=] \\pu{3000 J}""},{""type"":""physical unit"",""value"":""Increased temperature [OF] copper [=] \\pu{45 ℃}""}]" "

What is the mass of a piece of copper when 3000 J of heat is applied, causing a 45 °C increase in temperature, if the #C_p# of copper is .38452 J/g °C?

" nan 173.37 g "
+

Explanation:

+
+

I would use the relationship describing heat exchanged #Q# as:
+#Q=mc_pDeltaT#
+With our data:
+#3000=m*0.38452*45#
+Rearranging:
+#m=(3000)/(0.38452*45)=173.37~~173.4g#

+
+
" "
+
+
+

I found #173.4g#

+
+
+
+

Explanation:

+
+

I would use the relationship describing heat exchanged #Q# as:
+#Q=mc_pDeltaT#
+With our data:
+#3000=m*0.38452*45#
+Rearranging:
+#m=(3000)/(0.38452*45)=173.37~~173.4g#

+
+
+
" "
+

What is the mass of a piece of copper when 3000 J of heat is applied, causing a 45 °C increase in temperature, if the #C_p# of copper is .38452 J/g °C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
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+
+
+1 Answer +
+
+
+
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+ +
+
+ +
+ + Mar 26, 2016 + +
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+

I found #173.4g#

+
+
+
+

Explanation:

+
+

I would use the relationship describing heat exchanged #Q# as:
+#Q=mc_pDeltaT#
+With our data:
+#3000=m*0.38452*45#
+Rearranging:
+#m=(3000)/(0.38452*45)=173.37~~173.4g#

+
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+
+
+
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+
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+ + Creative Commons License + +
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+
" What is the mass of a piece of copper when 3000 J of heat is applied, causing a 45 °C increase in temperature, if the #C_p# of copper is .38452 J/g °C? nan +120 a82afe8d-6ddd-11ea-ac3d-ccda262736ce https://socratic.org/questions/gas-stored-in-a-tank-at-273-k-has-a-pressure-of-388-kpa-the-safe-limit-for-the-p 580.48 K start physical_unit 0 0 temperature k qc_end physical_unit 0 0 6 7 temperature qc_end physical_unit 0 0 12 13 pressure qc_end physical_unit 0 0 21 22 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""580.48 K""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] gas [=] \\pu{273 K}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] gas [=] \\pu{388 kPa}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] gas [=] \\pu{825 kPa}""}]" "

Gas stored in a tank at 273 K has a pressure of 388 kPa. The safe limit for the pressure is 825 kPa. At what temperature will the gas reach this pressure?

" nan 580.48 K "
+

Explanation:

+
+

Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.

+

The equation for Charles' law is #P_1/T_1=P_2/T_2#.

+

#P_1=""388 kPa""#
+#T_1=""273 K""#
+#P_2=""825 kPa""#
+#T_2=""???""#

+

Rearrange the equation to isolate #T_2#, substitute the given values, and solve.

+

#T_2=(P_2T_1)/(P_1)#

+

#T_2=(825cancel""kPa""xx273""K"")/(388cancel""kPa"")=""580 K""#

+
+
" "
+
+
+

The temperature at which the gas pressure inside the tank will equal 825 kPa is 508 K.

+
+
+
+

Explanation:

+
+

Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.

+

The equation for Charles' law is #P_1/T_1=P_2/T_2#.

+

#P_1=""388 kPa""#
+#T_1=""273 K""#
+#P_2=""825 kPa""#
+#T_2=""???""#

+

Rearrange the equation to isolate #T_2#, substitute the given values, and solve.

+

#T_2=(P_2T_1)/(P_1)#

+

#T_2=(825cancel""kPa""xx273""K"")/(388cancel""kPa"")=""580 K""#

+
+
+
" "
+

Gas stored in a tank at 273 K has a pressure of 388 kPa. The safe limit for the pressure is 825 kPa. At what temperature will the gas reach this pressure?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gay Lussac's Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 24, 2015 + +
+
+
+
+
+
+
+

The temperature at which the gas pressure inside the tank will equal 825 kPa is 508 K.

+
+
+
+

Explanation:

+
+

Gay-Lussac's law states that the pressure of a gas held at constant volume is directly proportional to its temperature in Kelvins.

+

The equation for Charles' law is #P_1/T_1=P_2/T_2#.

+

#P_1=""388 kPa""#
+#T_1=""273 K""#
+#P_2=""825 kPa""#
+#T_2=""???""#

+

Rearrange the equation to isolate #T_2#, substitute the given values, and solve.

+

#T_2=(P_2T_1)/(P_1)#

+

#T_2=(825cancel""kPa""xx273""K"")/(388cancel""kPa"")=""580 K""#

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+
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+
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+ +
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+
+
+
+
+
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" Gas stored in a tank at 273 K has a pressure of 388 kPa. The safe limit for the pressure is 825 kPa. At what temperature will the gas reach this pressure? nan +121 a82afe8e-6ddd-11ea-8d07-ccda262736ce https://socratic.org/questions/it-takes-3-190-j-to-increase-the-temperature-of-a-0-400-kg-sample-of-glass-from- 8.74 J/(g · K) start physical_unit 27 30 specific_heat j/(g_·_k) qc_end physical_unit 12 14 16 17 temperature qc_end physical_unit 12 14 19 20 temperature qc_end physical_unit 12 14 10 11 mass qc_end physical_unit 12 14 2 3 energy qc_end end "[{""type"":""physical unit"",""value"":""Specific heat [OF] this type of glass [IN] J/(g · K)""}]" "[{""type"":""physical unit"",""value"":""8.74 J/(g · K)""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] sample of glass [=] \\pu{273 K}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] sample of glass [=] \\pu{308 K}""},{""type"":""physical unit"",""value"":""Mass [OF] sample of glass [=] \\pu{0.400 kg}""},{""type"":""physical unit"",""value"":""Energy took [OF] sample of glass [=] \\pu{3,190 J}""}]" "

It takes 3,190 J to increase the temperature of a 0.400 kg sample of glass from 273 K to 308 K. What is the specific heat for this type of glass?

" nan 8.74 J/(g · K) "
+

Explanation:

+
+

Specific heat is given by

+
+

#""S"" = ""Q""/""mΔT""#

+
+

Where

+
    +
  • #""Q =""# Heat added/liberated
    • +
  • #""m =""# Mass of sample
    • +
  • #""ΔT =""# Change in temperature
    • +
+

#""S"" = ""3190 J""/""400 g × (308 K – 273 K)"" = ""8.74 J/g K""#

+
+
" "
+
+
+

#""8.74 J/g K""#

+
+
+
+

Explanation:

+
+

Specific heat is given by

+
+

#""S"" = ""Q""/""mΔT""#

+
+

Where

+
    +
  • #""Q =""# Heat added/liberated
    • +
  • #""m =""# Mass of sample
    • +
  • #""ΔT =""# Change in temperature
    • +
+

#""S"" = ""3190 J""/""400 g × (308 K – 273 K)"" = ""8.74 J/g K""#

+
+
+
" "
+

It takes 3,190 J to increase the temperature of a 0.400 kg sample of glass from 273 K to 308 K. What is the specific heat for this type of glass?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 29, 2018 + +
+
+
+
+
+
+
+

#""8.74 J/g K""#

+
+
+
+

Explanation:

+
+

Specific heat is given by

+
+

#""S"" = ""Q""/""mΔT""#

+
+

Where

+
    +
  • #""Q =""# Heat added/liberated
    • +
  • #""m =""# Mass of sample
    • +
  • #""ΔT =""# Change in temperature
    • +
+

#""S"" = ""3190 J""/""400 g × (308 K – 273 K)"" = ""8.74 J/g K""#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" It takes 3,190 J to increase the temperature of a 0.400 kg sample of glass from 273 K to 308 K. What is the specific heat for this type of glass? nan +122 a82b23ee-6ddd-11ea-99c4-ccda262736ce https://socratic.org/questions/how-many-grams-of-sodium-are-in-2-5-moles 57.47 grams start physical_unit 4 4 mass g qc_end physical_unit 4 4 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium [IN] grams""}]" "[{""type"":""physical unit"",""value"":""57.47 grams""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sodium [=] \\pu{2.5 moles}""}]" "

How many grams of sodium are in 2.5 moles?

" nan 57.47 grams "
+

Explanation:

+
+

moles = mass / atomic mass

+

moles * atomic mass = mass

+

so,
+ 2.5 moles * 22.98 g / mol = 57.45 g

+
+
" "
+
+
+

57.45 g

+
+
+
+

Explanation:

+
+

moles = mass / atomic mass

+

moles * atomic mass = mass

+

so,
+ 2.5 moles * 22.98 g / mol = 57.45 g

+
+
+
" "
+

How many grams of sodium are in 2.5 moles?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Apr 5, 2016 + +
+
+
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+
+
+
+

57.45 g

+
+
+
+

Explanation:

+
+

moles = mass / atomic mass

+

moles * atomic mass = mass

+

so,
+ 2.5 moles * 22.98 g / mol = 57.45 g

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Apr 5, 2016 + +
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+
+
+

#57.4745g#

+
+
+
+

Explanation:

+
+

#1# mole of any substance weighs #=# Average atomic mass in gram.

+

Given substance is Sodium, #""Na""#; has Average atomic mass #=22.9898#.

+

Therefore, #1# mole of Sodium weighs #=22.9898g#
+#2.5# moles of Sodium weigh#=22.9898times2.5=57.4745g#

+
+
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+
+
+ +
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+
+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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+
+
" How many grams of sodium are in 2.5 moles? nan +123 a82b23ef-6ddd-11ea-9474-ccda262736ce https://socratic.org/questions/if-it-takes-25-ml-of-0-05-m-hcl-to-neutralize-345-ml-of-naoh-solution-what-is-th 3.62 × 10^(-3) M start physical_unit 14 15 concentration mol/l qc_end physical_unit 8 8 3 4 volume qc_end physical_unit 8 8 6 7 concentration qc_end physical_unit 14 15 11 12 volume qc_end c_other neutralize qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] NaOH solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""3.62 × 10^(-3) M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] HCl [=] \\pu{25 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] HCl [=] \\pu{0.05 M}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH solution [=] \\pu{345 mL}""},{""type"":""other"",""value"":""neutralize""}]" "

If it takes 25 mL of 0.05 M #HCl# to neutralize 345 mL of #NaOH# solution, what is the concentration of the #NaOH# solution?

" nan 3.62 × 10^(-3) M "
+

Explanation:

+
+

#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

+

There is 1:1 equivalence between acid and base.

+

Thus #[NaOH]# #=# #(25xx10^-3Lxx0.05*mol*L^-1)/(345xx10^-3L)#

+

#=# #??*mol*L^-1#

+
+
" "
+
+
+

#[NaOH]~=4xx10^-3*mol*L^-1#

+
+
+
+

Explanation:

+
+

#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

+

There is 1:1 equivalence between acid and base.

+

Thus #[NaOH]# #=# #(25xx10^-3Lxx0.05*mol*L^-1)/(345xx10^-3L)#

+

#=# #??*mol*L^-1#

+
+
+
" "
+

If it takes 25 mL of 0.05 M #HCl# to neutralize 345 mL of #NaOH# solution, what is the concentration of the #NaOH# solution?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 27, 2016 + +
+
+
+
+
+
+
+

#[NaOH]~=4xx10^-3*mol*L^-1#

+
+
+
+

Explanation:

+
+

#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

+

There is 1:1 equivalence between acid and base.

+

Thus #[NaOH]# #=# #(25xx10^-3Lxx0.05*mol*L^-1)/(345xx10^-3L)#

+

#=# #??*mol*L^-1#

+
+
+
+
+
+ +
+
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+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 22630 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" If it takes 25 mL of 0.05 M #HCl# to neutralize 345 mL of #NaOH# solution, what is the concentration of the #NaOH# solution? nan +124 a82b23f0-6ddd-11ea-8b7c-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-that-contains-18-8-li-16-3-c-and-64 Li2CO3 start chemical_formula qc_end physical_unit 11 11 10 10 percent_composition qc_end physical_unit 13 13 12 12 percent_composition qc_end physical_unit 16 16 15 15 percent_composition qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Li2CO3""}]" "[{""type"":""physical unit"",""value"":""Percent composition [OF] Li [=] \\pu{18.8%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] C [=] \\pu{16.3%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] O [=] \\pu{64.9%}""}]" "

What is the empirical formula for a compound that contains 18.8% Li, 16.3% C, and 64.9% O?

" nan Li2CO3 "
+

Explanation:

+
+

Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be #""Li""_2""CO""_3#. Two things to go by here

+
    +
  • the position in the periodic table for the three elements, which gives you an idea about their molar masses
    • +
  • the fact that you're dealing with an Ionic compound tells you that its empirical formula will also be its chemical formula
    • +
+

So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be #""Li""_2""CO""_3#.

+

Let's do some calculations to see if the prediction turns out to be correct.

+

Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.

+

To make calculations easier, you can pic ka #""100-g""# sample. This will allow you to convert the percent composition directly to grams. Your sample will thus contain

+
+
    +
  • #""18.8 g""# of lithium
    • +
  • #""16.3 g""# of carbon
    • +
  • #""64.9 g""# of oxygen
    • +
+
+

Next, use the molar mass of each element to determine how many moles of each you'd get in this sample

+
+

#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#

+

#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#

+

#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#

+
+

Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.

+

To get the mole ratio that exists between these three elements, divide these three values by the smallest one.

+
+

#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#

+

#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#

+
+

Since this ratio, i.e. #2:1:3#, is also the smallest whole number ratio that can exist between these three numbers, the empirical formula of the compound will indeed be

+
+

#""Li""_2""CO""_3 -># lithium carbonate

+
+
+
" "
+
+
+

#""Li""_2""CO""_3#

+
+
+
+

Explanation:

+
+

Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be #""Li""_2""CO""_3#. Two things to go by here

+
    +
  • the position in the periodic table for the three elements, which gives you an idea about their molar masses
    • +
  • the fact that you're dealing with an Ionic compound tells you that its empirical formula will also be its chemical formula
    • +
+

So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be #""Li""_2""CO""_3#.

+

Let's do some calculations to see if the prediction turns out to be correct.

+

Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.

+

To make calculations easier, you can pic ka #""100-g""# sample. This will allow you to convert the percent composition directly to grams. Your sample will thus contain

+
+
    +
  • #""18.8 g""# of lithium
    • +
  • #""16.3 g""# of carbon
    • +
  • #""64.9 g""# of oxygen
    • +
+
+

Next, use the molar mass of each element to determine how many moles of each you'd get in this sample

+
+

#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#

+

#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#

+

#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#

+
+

Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.

+

To get the mole ratio that exists between these three elements, divide these three values by the smallest one.

+
+

#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#

+

#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#

+
+

Since this ratio, i.e. #2:1:3#, is also the smallest whole number ratio that can exist between these three numbers, the empirical formula of the compound will indeed be

+
+

#""Li""_2""CO""_3 -># lithium carbonate

+
+
+
+
" "
+

What is the empirical formula for a compound that contains 18.8% Li, 16.3% C, and 64.9% O?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 3, 2016 + +
+
+
+
+
+
+
+

#""Li""_2""CO""_3#

+
+
+
+

Explanation:

+
+

Even without doing any calculation, you could predict that the empirical formula for this compound will come out to be #""Li""_2""CO""_3#. Two things to go by here

+
    +
  • the position in the periodic table for the three elements, which gives you an idea about their molar masses
    • +
  • the fact that you're dealing with an Ionic compound tells you that its empirical formula will also be its chemical formula
    • +
+

So if you're familair with ionic compounds and polyatomic ions, you can predict that the answer will be #""Li""_2""CO""_3#.

+

Let's do some calculations to see if the prediction turns out to be correct.

+

Your strategy here will be to pick a sample of this compound and use the given percent composition to find how many grams of each element you'd get in this sample.

+

To make calculations easier, you can pic ka #""100-g""# sample. This will allow you to convert the percent composition directly to grams. Your sample will thus contain

+
+
    +
  • #""18.8 g""# of lithium
    • +
  • #""16.3 g""# of carbon
    • +
  • #""64.9 g""# of oxygen
    • +
+
+

Next, use the molar mass of each element to determine how many moles of each you'd get in this sample

+
+

#""For Li: "" 18.8 color(red)(cancel(color(black)(""g""))) * ""1 mole Li""/(6.941color(red)(cancel(color(black)(""g"")))) = ""2.709 moles Li""#

+

#""For C: "" 16.3 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011color(red)(cancel(color(black)(""g"")))) = ""1.357 moles C""#

+

#""For O: "" 64.9 color(red)(cancel(color(black)(""g""))) * ""1 mole O""/(15.9994 color(red)(cancel(color(black)(""g"")))) = ""4.056 moles O""#

+
+

Now, a compound's empirical formula gives you the smallest whole number ratio that exists between the elements that make up the compound.

+

To get the mole ratio that exists between these three elements, divide these three values by the smallest one.

+
+

#""For Li: "" (2.709 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1.996 ~~ 2#

+

#""For C: "" (1.357 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For O: "" (4.056 color(red)(cancel(color(black)(""moles""))))/(1.357color(red)(cancel(color(black)(""moles"")))) = 2.989 ~~ 3#

+
+

Since this ratio, i.e. #2:1:3#, is also the smallest whole number ratio that can exist between these three numbers, the empirical formula of the compound will indeed be

+
+

#""Li""_2""CO""_3 -># lithium carbonate

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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+
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" What is the empirical formula for a compound that contains 18.8% Li, 16.3% C, and 64.9% O? nan +125 a82b23f1-6ddd-11ea-9a30-ccda262736ce https://socratic.org/questions/what-is-the-molecular-formula-of-sodium-phosphate Na3PO4 start chemical_formula qc_end substance 6 7 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Na3PO4""}]" "[{""type"":""substance name"",""value"":""Sodium phosphate""}]" "

What is the molecular formula of sodium phosphate?

" nan Na3PO4 "
+

Explanation:

+
+

This is so since sodium is in group #1# so has a valency of #1+# while phosphate is a polyatomic ion of valency #3-# and thus we need #3# sodium atoms for each #1# phosphate ion, hence, #Na_3PO_4#.
+Overall nett charge of molecule must be zero.

+

The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.

+
+
" "
+
+
+

#Na_3PO_4#

+
+
+
+

Explanation:

+
+

This is so since sodium is in group #1# so has a valency of #1+# while phosphate is a polyatomic ion of valency #3-# and thus we need #3# sodium atoms for each #1# phosphate ion, hence, #Na_3PO_4#.
+Overall nett charge of molecule must be zero.

+

The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.

+
+
+
" "
+

What is the molecular formula of sodium phosphate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 22, 2015 + +
+
+
+
+
+
+
+

#Na_3PO_4#

+
+
+
+

Explanation:

+
+

This is so since sodium is in group #1# so has a valency of #1+# while phosphate is a polyatomic ion of valency #3-# and thus we need #3# sodium atoms for each #1# phosphate ion, hence, #Na_3PO_4#.
+Overall nett charge of molecule must be zero.

+

The ions will be held together by strong ionic bonds and the molecules by strong coulombic forces of attraction.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 11699 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What is the molecular formula of sodium phosphate? nan +126 a82b23f2-6ddd-11ea-ba3e-ccda262736ce https://socratic.org/questions/for-the-reaction-2s-s-3o2-g-2-so3-s-2-0-moles-of-sulfur-are-placed-in-a-containe 85.00% start physical_unit 1 2 percent_yield none qc_end chemical_equation 3 10 qc_end physical_unit 14 14 11 12 mole qc_end physical_unit 10 10 26 27 mole qc_end end "[{""type"":""physical unit"",""value"":""Percentage yield [OF] the reaction""}]" "[{""type"":""physical unit"",""value"":""85.00%""}]" "[{""type"":""chemical equation"",""value"":""2 S(s) + 3 O2(g) -> 2 SO3(s)""},{""type"":""physical unit"",""value"":""Mole [OF] sulfur [=] \\pu{2.0 moles}""},{""type"":""physical unit"",""value"":""Mole [OF] SO3 [=] \\pu{1.7 moles}""}]" "

For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?

" nan 85.00% "
+

Explanation:

+
+

From the reaction equation coefficients, we can understand that for every #2# moles of sulfur, #2# moles of #""SO""_3# are created. So if #2.0# moles of sulfur are placed in a container, where oxygen does not limit the reaction, #2.0# moles of #""SO""_3# are expected to be formed.

+

However, #1.7# moles were collected experimentally. To show this disparity, we calculate the percentage yield, which is

+

#""% yield"" = (""experimental value""/""actual value"") * 100%#

+

So

+

#""% yield"" = (""1.7 moles""/""2.0 moles"")*100% = 85%#

+
+
" "
+
+
+

#85%#

+
+
+
+

Explanation:

+
+

From the reaction equation coefficients, we can understand that for every #2# moles of sulfur, #2# moles of #""SO""_3# are created. So if #2.0# moles of sulfur are placed in a container, where oxygen does not limit the reaction, #2.0# moles of #""SO""_3# are expected to be formed.

+

However, #1.7# moles were collected experimentally. To show this disparity, we calculate the percentage yield, which is

+

#""% yield"" = (""experimental value""/""actual value"") * 100%#

+

So

+

#""% yield"" = (""1.7 moles""/""2.0 moles"")*100% = 85%#

+
+
+
" "
+

For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Dec 14, 2017 + +
+
+
+
+
+
+
+

#85%#

+
+
+
+

Explanation:

+
+

From the reaction equation coefficients, we can understand that for every #2# moles of sulfur, #2# moles of #""SO""_3# are created. So if #2.0# moles of sulfur are placed in a container, where oxygen does not limit the reaction, #2.0# moles of #""SO""_3# are expected to be formed.

+

However, #1.7# moles were collected experimentally. To show this disparity, we calculate the percentage yield, which is

+

#""% yield"" = (""experimental value""/""actual value"") * 100%#

+

So

+

#""% yield"" = (""1.7 moles""/""2.0 moles"")*100% = 85%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5301 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield? nan +127 a82b23f3-6ddd-11ea-8984-ccda262736ce https://socratic.org/questions/56c4b5c17c0149313eba3b09 H2PO4^- start chemical_formula qc_end chemical_equation 8 8 qc_end c_other Conjugate_base qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""H2PO4^-""}]" "[{""type"":""chemical equation"",""value"":""H3PO4""},{""type"":""other"",""value"":""Conjugate base""}]" "

What is the conjugate base of #""phosphoric acid""#, #H_3PO_4#?

" nan H2PO4^- "
+

Explanation:

+
+

The conjugate acid of a base is the original species PLUS a proton, #H^+#. The conjugate base of an acid is the origianl species LESS a proton. Both MASS and CHARGE are conserved.

+

What are the conjugate bases of #HPO_4^(2-)#, and #HSO_4^-#?

+
+
" "
+
+
+

#H_2PO_4^-#

+
+
+
+

Explanation:

+
+

The conjugate acid of a base is the original species PLUS a proton, #H^+#. The conjugate base of an acid is the origianl species LESS a proton. Both MASS and CHARGE are conserved.

+

What are the conjugate bases of #HPO_4^(2-)#, and #HSO_4^-#?

+
+
+
" "
+

What is the conjugate base of #""phosphoric acid""#, #H_3PO_4#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Conjugate Acids and Conjugate Bases + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 17, 2016 + +
+
+
+
+
+
+
+

#H_2PO_4^-#

+
+
+
+

Explanation:

+
+

The conjugate acid of a base is the original species PLUS a proton, #H^+#. The conjugate base of an acid is the origianl species LESS a proton. Both MASS and CHARGE are conserved.

+

What are the conjugate bases of #HPO_4^(2-)#, and #HSO_4^-#?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 30009 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
" "What is the conjugate base of #""phosphoric acid""#, #H_3PO_4#?" nan +128 a82b23f4-6ddd-11ea-9576-ccda262736ce https://socratic.org/questions/water-h2o-is-made-via-its-formation-reaction-how-many-moles-of-water-form-when-7 1.54 x 10^1 moles start physical_unit 0 0 mole mol qc_end physical_unit 20 21 15 18 mole qc_end c_other OTHER qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] water [IN] moles""}]" "[{""type"":""physical unit"",""value"":""1.54 x 10^1 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] oxygen gas [=] \\pu{(7.69 x 10^0) moles}""},{""type"":""other"",""value"":""Answer in mol to 3 s.d. in proper scientific notation""},{""type"":""other"",""value"":""H2O, is made via its formation reaction.""}]" "

Water, H2O, is made via its formation reaction. How many moles of water form when (7.69x10^0) moles of oxygen gas react? (Answer in mol to 3 s.d. in proper scientific notation)

" "
+
+

+

yet again, i have no idea what im doing

+

+
+
" 1.54 x 10^1 moles "
+

Explanation:

+
+

Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.

+

#""2H""_2 + ""O""_2##rarr##""2H""_2""O""#

+

Multiply the given mol oxygen gas by the mol ratio #(2""mol H""_2""O"")/(1""mol O""_2"")# from the balanced equation.

+

#7.69xx10^0color(red)cancel(color(black)(""mol O""_2))xx=(2""mol H""_2""O"")/(1color(red)cancel(color(black)(""mol O""_2)))=1.54xx10^1color(white)(.)""mol H""_2""O""#

+
+
" "
+
+
+

The given amount of oxygen gas will produce #1.54xx10^1# mol water.

+
+
+
+

Explanation:

+
+

Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.

+

#""2H""_2 + ""O""_2##rarr##""2H""_2""O""#

+

Multiply the given mol oxygen gas by the mol ratio #(2""mol H""_2""O"")/(1""mol O""_2"")# from the balanced equation.

+

#7.69xx10^0color(red)cancel(color(black)(""mol O""_2))xx=(2""mol H""_2""O"")/(1color(red)cancel(color(black)(""mol O""_2)))=1.54xx10^1color(white)(.)""mol H""_2""O""#

+
+
+
" "
+

Water, H2O, is made via its formation reaction. How many moles of water form when (7.69x10^0) moles of oxygen gas react? (Answer in mol to 3 s.d. in proper scientific notation)

+
+
+

+

yet again, i have no idea what im doing

+

+
+
+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 13, 2017 + +
+
+
+
+
+
+
+

The given amount of oxygen gas will produce #1.54xx10^1# mol water.

+
+
+
+

Explanation:

+
+

Start with a balanced equation. It will be used to determine mol ratios between the product and the reactant.

+

#""2H""_2 + ""O""_2##rarr##""2H""_2""O""#

+

Multiply the given mol oxygen gas by the mol ratio #(2""mol H""_2""O"")/(1""mol O""_2"")# from the balanced equation.

+

#7.69xx10^0color(red)cancel(color(black)(""mol O""_2))xx=(2""mol H""_2""O"")/(1color(red)cancel(color(black)(""mol O""_2)))=1.54xx10^1color(white)(.)""mol H""_2""O""#

+
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+ +
+
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+
+
+
+
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+
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+
" Water, H2O, is made via its formation reaction. How many moles of water form when (7.69x10^0) moles of oxygen gas react? (Answer in mol to 3 s.d. in proper scientific notation) " + + +yet again, i have no idea what im doing + + +" +129 a82b23f5-6ddd-11ea-814b-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-250-0-milliliter-aqueous-solution-of-sodium-hydroxide- 1.55 mol/L start physical_unit 8 12 molarity mol/l qc_end physical_unit 18 18 15 16 mass qc_end physical_unit 8 12 6 7 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] aqueous solution of sodium hydroxide [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.55 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] aqueous solution of sodium hydroxide [=] \\pu{250.0 milliliter}""},{""type"":""physical unit"",""value"":""Mass [OF] solute [=] \\pu{15.5 grams}""}]" "

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute?

" nan 1.55 mol/L "
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

+
+
" "
+
+
+

#[NaOH(aq)] = 1.55# #mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

+
+
+
" "
+

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 27, 2016 + +
+
+
+
+
+
+
+

#[NaOH(aq)] = 1.55# #mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

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+ +
+
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+
+
+
+
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+
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+ + Creative Commons License + +
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+
+
" What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute? nan +130 a82b23f6-6ddd-11ea-9275-ccda262736ce https://socratic.org/questions/what-is-partial-pressure-of-oxygen-at-sea-level 0.21 atm start physical_unit 5 8 partial_pressure atm qc_end substance 5 5 qc_end end "[{""type"":""physical unit"",""value"":""Partial pressure [OF] oxygen at sea level [IN] atm""}]" "[{""type"":""physical unit"",""value"":""0.21 atm""}]" "[{""type"":""substance name"",""value"":""Oxygen""}]" "

What is partial pressure of oxygen at sea level?

" nan 0.21 atm "
+

Explanation:

+
+

This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....

+

#P_""Total""=SigmaP_""component gases""#

+

#=P_1+P_2.....+P_n#, assuming ideality.....

+

#P_""Total""=(n_1RT)/V+............(n_nRT)/V#

+

#P_""Total""=(RT)/V{n_1+n_2+....n_n}#

+

And thus #P_1/P_""Total""=((RT)/Vn_1)/((RT)/V{n_1+n_2+....n_n})#

+

#P_1/P_""Total""=(n_1)/({n_1+n_2+....n_n})=chi_(n_1)#

+

Where #chi_(n_1)=""mole fraction of the component ""#

+

And here the container is conveniently the atmosphere, which to a first approx. is #78%# dinitrogen, and #21%# dioxygen......

+

And so what is #P_(O_2)#?

+
+
" "
+
+
+

#P_(O_2)=0.21xx1*atm#

+
+
+
+

Explanation:

+
+

This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....

+

#P_""Total""=SigmaP_""component gases""#

+

#=P_1+P_2.....+P_n#, assuming ideality.....

+

#P_""Total""=(n_1RT)/V+............(n_nRT)/V#

+

#P_""Total""=(RT)/V{n_1+n_2+....n_n}#

+

And thus #P_1/P_""Total""=((RT)/Vn_1)/((RT)/V{n_1+n_2+....n_n})#

+

#P_1/P_""Total""=(n_1)/({n_1+n_2+....n_n})=chi_(n_1)#

+

Where #chi_(n_1)=""mole fraction of the component ""#

+

And here the container is conveniently the atmosphere, which to a first approx. is #78%# dinitrogen, and #21%# dioxygen......

+

And so what is #P_(O_2)#?

+
+
+
" "
+

What is partial pressure of oxygen at sea level?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
+
+
+
+
+1 Answer +
+
+
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+
+
+ +
+
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+ + Sep 4, 2017 + +
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+

#P_(O_2)=0.21xx1*atm#

+
+
+
+

Explanation:

+
+

This illustrates old Dalton's Law of Partial Pressures. In a gaseous mixture, the partial pressure exerted by a component gas is the same as if it alone occupied the container.....

+

#P_""Total""=SigmaP_""component gases""#

+

#=P_1+P_2.....+P_n#, assuming ideality.....

+

#P_""Total""=(n_1RT)/V+............(n_nRT)/V#

+

#P_""Total""=(RT)/V{n_1+n_2+....n_n}#

+

And thus #P_1/P_""Total""=((RT)/Vn_1)/((RT)/V{n_1+n_2+....n_n})#

+

#P_1/P_""Total""=(n_1)/({n_1+n_2+....n_n})=chi_(n_1)#

+

Where #chi_(n_1)=""mole fraction of the component ""#

+

And here the container is conveniently the atmosphere, which to a first approx. is #78%# dinitrogen, and #21%# dioxygen......

+

And so what is #P_(O_2)#?

+
+
+
+
+
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+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 7780 views + around the world +
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" What is partial pressure of oxygen at sea level? nan +131 a82b23f7-6ddd-11ea-b344-ccda262736ce https://socratic.org/questions/what-is-the-percentage-of-water-in-cocl-2-6h-2o 45.42% start physical_unit 5 7 percent none qc_end chemical_equation 7 7 qc_end end "[{""type"":""physical unit"",""value"":""Percentage [OF] water in CoCl2.6H2O""}]" "[{""type"":""physical unit"",""value"":""45.42%""}]" "[{""type"":""chemical equation"",""value"":""CoCl2.6H2O""}]" "

What is the percentage of water in #CoCl_2 * 6H_2O#?

" nan 45.42% "
+

Explanation:

+
+

So, #(6xx18.01*g*mol^-1)/(237.93*g*mol^-1)xx100%~=50%#

+

The hexahydrate has a beautiful purple colour. The anhydrous salt, #CoCl_2#, has an equally beautiful blue colour.

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" "
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+
+

#""%Water = "" ""Mass of water""/""Molar mass of cobaltous chloride hexahydrate""xx100%#

+
+
+
+

Explanation:

+
+

So, #(6xx18.01*g*mol^-1)/(237.93*g*mol^-1)xx100%~=50%#

+

The hexahydrate has a beautiful purple colour. The anhydrous salt, #CoCl_2#, has an equally beautiful blue colour.

+
+
+
" "
+

What is the percentage of water in #CoCl_2 * 6H_2O#?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Ionic Compounds + + +
+
+
+
+
+1 Answer +
+
+
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+ + Jun 5, 2016 + +
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#""%Water = "" ""Mass of water""/""Molar mass of cobaltous chloride hexahydrate""xx100%#

+
+
+
+

Explanation:

+
+

So, #(6xx18.01*g*mol^-1)/(237.93*g*mol^-1)xx100%~=50%#

+

The hexahydrate has a beautiful purple colour. The anhydrous salt, #CoCl_2#, has an equally beautiful blue colour.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1357 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What is the percentage of water in #CoCl_2 * 6H_2O#? nan +132 a82b23f8-6ddd-11ea-8b8b-ccda262736ce https://socratic.org/questions/589c840b11ef6b5e01fe3ac7 7.57 mg start physical_unit 12 13 mass mg qc_end physical_unit 12 13 6 9 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] acetic acid [IN] mg""}]" "[{""type"":""physical unit"",""value"":""7.57 mg""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] acetic acid [=] \\pu{1.26 × 10^(−4) mol}""}]" "

What is the mass of a #1.26xx10^-4*mol# quantity of acetic acid?

" nan 7.57 mg "
+

Explanation:

+
+

Now, the basic equation is #""moles""# #=# #""mass""/""molar mass""#....and thus #""mass""# #=# #""moles""# #xx# #""molar mass""#.

+

So here,

+

#""mass""# #=# #1.26xx10^-4*cancel(mol)xx60.05*g*cancel(mol^-1)=7.60*mg#.

+

Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.

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+
" "
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+
+

#""In acetic acid,""# #H_3""CCO""_2H#?

+
+
+
+

Explanation:

+
+

Now, the basic equation is #""moles""# #=# #""mass""/""molar mass""#....and thus #""mass""# #=# #""moles""# #xx# #""molar mass""#.

+

So here,

+

#""mass""# #=# #1.26xx10^-4*cancel(mol)xx60.05*g*cancel(mol^-1)=7.60*mg#.

+

Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.

+
+
+
" "
+

What is the mass of a #1.26xx10^-4*mol# quantity of acetic acid?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+
+2 Answers +
+
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+
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+
+ +
+ + Feb 9, 2017 + +
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#""In acetic acid,""# #H_3""CCO""_2H#?

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+
+

Explanation:

+
+

Now, the basic equation is #""moles""# #=# #""mass""/""molar mass""#....and thus #""mass""# #=# #""moles""# #xx# #""molar mass""#.

+

So here,

+

#""mass""# #=# #1.26xx10^-4*cancel(mol)xx60.05*g*cancel(mol^-1)=7.60*mg#.

+

Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.

+
+
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+
+
+ +
+
+
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+
+ +
+
+ +
+ + Apr 2, 2018 + +
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+

The mass of #""CH""_3""COOH""# is #""0.00757 g""#.

+
+
+
+

Explanation:

+
+

Multiply moles by molar mass.

+

The molar mass of acetic acid #(""CH""_3""COOH"")# is #""60.052 g/mol""#.
+https://www.ncbi.nlm.nih.gov/pccompound?term=%22acetic+acid%22

+

#1.26xx10^(-4)color(red)cancel(color(black)(""mol CH""_3""COOH""))xx(60.052""g CH""_3""COOH"")/(1color(red)cancel(color(black)(""mol CH""_3""COOH"")))=""0.00757 g CH""_3""COOH""# (rounded to three significant figures)

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+
+
+
+
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+ + +
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Impact of this question
+
+ 7252 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the mass of a #1.26xx10^-4*mol# quantity of acetic acid? nan +133 a82b23f9-6ddd-11ea-adf1-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-1-2-10-18-formula-units-of-calcium-chloride 2.21 × 10^(-4) g start physical_unit 11 12 mass g qc_end physical_unit 8 12 5 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] calcium chloride [IN] g""}]" "[{""type"":""physical unit"",""value"":""2.21 × 10^(-4) g""}]" "[{""type"":""physical unit"",""value"":""Number [OF] formula units of calcium chloride [=] \\pu{1.2 ⋅ 10^18}""}]" "

What is the mass of #1.2*10^18# formula units of calcium chloride?

" nan 2.21 × 10^(-4) g "
+

Explanation:

+
+

The thing to keep in mind about formula units is that you need #6.022 * 10^(23)# of them to have exactly #1# mole of an ionic compound #-># think Avogadro's constant here.

+

In this case, you know that #6.022 * 10^(23)# formula units of calcium chloride are needed in order to have exactly #1# mole of calcium chloride.

+

Moreover, you know that calcium chloride has a molar mass of #""110.98 g mol""^(-1)#, which means that #1# mole of calcium chloride has a mass of #""110.98 g""#.

+

So if #1# mole of calcium chloride contains #6.022 * 10^(23)# formula units and has a mass of #""110.98 g""#, you can say that #6.022 * 10^(23)# formula units of calcium chloride have a mass of #""110.98 g""#.

+

This means that your sample has a mass of

+
+

#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.

+
+
" "
+
+
+

#2.2 * 10^(-4)# #""g""#

+
+
+
+

Explanation:

+
+

The thing to keep in mind about formula units is that you need #6.022 * 10^(23)# of them to have exactly #1# mole of an ionic compound #-># think Avogadro's constant here.

+

In this case, you know that #6.022 * 10^(23)# formula units of calcium chloride are needed in order to have exactly #1# mole of calcium chloride.

+

Moreover, you know that calcium chloride has a molar mass of #""110.98 g mol""^(-1)#, which means that #1# mole of calcium chloride has a mass of #""110.98 g""#.

+

So if #1# mole of calcium chloride contains #6.022 * 10^(23)# formula units and has a mass of #""110.98 g""#, you can say that #6.022 * 10^(23)# formula units of calcium chloride have a mass of #""110.98 g""#.

+

This means that your sample has a mass of

+
+

#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.

+
+
+
" "
+

What is the mass of #1.2*10^18# formula units of calcium chloride?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 20, 2018 + +
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+

#2.2 * 10^(-4)# #""g""#

+
+
+
+

Explanation:

+
+

The thing to keep in mind about formula units is that you need #6.022 * 10^(23)# of them to have exactly #1# mole of an ionic compound #-># think Avogadro's constant here.

+

In this case, you know that #6.022 * 10^(23)# formula units of calcium chloride are needed in order to have exactly #1# mole of calcium chloride.

+

Moreover, you know that calcium chloride has a molar mass of #""110.98 g mol""^(-1)#, which means that #1# mole of calcium chloride has a mass of #""110.98 g""#.

+

So if #1# mole of calcium chloride contains #6.022 * 10^(23)# formula units and has a mass of #""110.98 g""#, you can say that #6.022 * 10^(23)# formula units of calcium chloride have a mass of #""110.98 g""#.

+

This means that your sample has a mass of

+
+

#1.2 * 10^8 color(red)(cancel(color(black)(""f. units CaCl""_2))) * ""110.98 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""f. units CaCl""_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad ""g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.

+
+
+
+
+
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+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 6148 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" What is the mass of #1.2*10^18# formula units of calcium chloride? nan +134 a82b23fa-6ddd-11ea-9f75-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-9-76x10-12-atoms-of-nitrogen 2.27 × 10^(-10) g start physical_unit 10 10 mass g qc_end physical_unit 8 10 5 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] nitrogen [IN] g""}]" "[{""type"":""physical unit"",""value"":""2.27 × 10^(-10) g""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of nitrogen [=] \\pu{9.76 × 10^12}""}]" "

What is the mass of #9.76xx10^12# atoms of nitrogen? +

" "
+
+

+
+

+
+
" 2.27 × 10^(-10) g "
+

Explanation:

+
+

To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of #9.76 * 10^(12)# atoms of nitrogen.

+

The starting point here will be the molar mass of nitrogen, which is listed as

+
+

#M_(""M N"") = ""14.00674 g mol""^(-1)#

+
+

This tells you that one mole of nitrogen has a mass of #""14.00674 g""#. Since one mole of any element contains #6.022 * 10^(23)# atoms of that element, you can say that the mass of a single atom of nitrogen will be

+
+

#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#

+
+

Now all you have to do is multiply this value by the number of atoms given to you to find

+
+

#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

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" "
+
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+

#2.27 * 10^(-10)""g""#

+
+
+
+

Explanation:

+
+

To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of #9.76 * 10^(12)# atoms of nitrogen.

+

The starting point here will be the molar mass of nitrogen, which is listed as

+
+

#M_(""M N"") = ""14.00674 g mol""^(-1)#

+
+

This tells you that one mole of nitrogen has a mass of #""14.00674 g""#. Since one mole of any element contains #6.022 * 10^(23)# atoms of that element, you can say that the mass of a single atom of nitrogen will be

+
+

#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#

+
+

Now all you have to do is multiply this value by the number of atoms given to you to find

+
+

#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

+
+
+
" "
+

What is the mass of #9.76xx10^12# atoms of nitrogen? +

+
+
+

+
+

+
+
+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 29, 2016 + +
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+
+
+

#2.27 * 10^(-10)""g""#

+
+
+
+

Explanation:

+
+

To make this problem more interesting, let's calculate the mass of a single atom of nitrogen first, then use that value as a conversion factor to determine the mass of #9.76 * 10^(12)# atoms of nitrogen.

+

The starting point here will be the molar mass of nitrogen, which is listed as

+
+

#M_(""M N"") = ""14.00674 g mol""^(-1)#

+
+

This tells you that one mole of nitrogen has a mass of #""14.00674 g""#. Since one mole of any element contains #6.022 * 10^(23)# atoms of that element, you can say that the mass of a single atom of nitrogen will be

+
+

#1 color(red)(cancel(color(black)(""atom N""))) * ""14.00674 g""/(6.022 * 10^(23)color(red)(cancel(color(black)(""atoms N"")))) = 2.326 * 10^(-23)""g""#

+
+

Now all you have to do is multiply this value by the number of atoms given to you to find

+
+

#9.76 * 10^(12) color(red)(cancel(color(black)(""atoms N""))) * (2.326 * 10^(-23)""g"")/(1color(red)(cancel(color(black)(""atom N"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.27 * 10^(-10)""g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

+ +

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+
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+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3593 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" "What is the mass of #9.76xx10^12# atoms of nitrogen? +" " + + + + + +" +135 a82b4afa-6ddd-11ea-b4b7-ccda262736ce https://socratic.org/questions/what-is-the-molality-of-benzene-in-a-solution-that-contains-6-502-g-of-benzene-i 0.51 mol/kg start physical_unit 5 8 molality mol/kg qc_end physical_unit 19 19 16 17 mass qc_end physical_unit 5 5 11 12 mass qc_end physical_unit 19 19 24 25 kf qc_end physical_unit 19 19 31 32 freezing_point_temperature qc_end end "[{""type"":""physical unit"",""value"":""Molality [OF] benzene in a solution [IN] mol/kg""}]" "[{""type"":""physical unit"",""value"":""0.51 mol/kg""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] CHCl3 [=] \\pu{162.7 g}""},{""type"":""physical unit"",""value"":""Mass [OF] benzene [=] \\pu{6.502 g}""},{""type"":""physical unit"",""value"":""Kf [OF] CHCl3 [=] \\pu{4.68 ℃⋅kg/mol}""},{""type"":""physical unit"",""value"":""Tf [OF] CHCl3 [=] \\pu{63.5 ℃}""}]" "

What is the molality of benzene in a solution that contains #""6.502 g""# of benzene in #""162.7 g""# of chloroform/#""CHCl""_3# (#K_f# for #""CHCl""_3# is #4.68^@ ""C""cdot""kg/mol""# and #T_f^""*""# of #""CHCl""_3# is #63.5^@ ""C""#)?

" nan 0.51 mol/kg "
+

Explanation:

+
+

#""6.0502 g"" * (""1 mol"")/(""78.114 g"") = ""0.083238 moles""#

+

So

+

#""molality"" = ""0.083238 moles""/(162.7 * 10^(-3) "" kg"") = ""0.51156 mol kg""^(-1)#

+
+
" "
+
+
+

Molality is defined as moles of solute per #""1 kg = 1000 g""# of solvent. Convert benzene's mass to moles, then divide by the mass of the solvent in kilograms.

+
+
+
+

Explanation:

+
+

#""6.0502 g"" * (""1 mol"")/(""78.114 g"") = ""0.083238 moles""#

+

So

+

#""molality"" = ""0.083238 moles""/(162.7 * 10^(-3) "" kg"") = ""0.51156 mol kg""^(-1)#

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" "
+

What is the molality of benzene in a solution that contains #""6.502 g""# of benzene in #""162.7 g""# of chloroform/#""CHCl""_3# (#K_f# for #""CHCl""_3# is #4.68^@ ""C""cdot""kg/mol""# and #T_f^""*""# of #""CHCl""_3# is #63.5^@ ""C""#)?

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+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
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+1 Answer +
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+ + +
+
+ +
+ + Feb 15, 2018 + +
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Molality is defined as moles of solute per #""1 kg = 1000 g""# of solvent. Convert benzene's mass to moles, then divide by the mass of the solvent in kilograms.

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+
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+

Explanation:

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+

#""6.0502 g"" * (""1 mol"")/(""78.114 g"") = ""0.083238 moles""#

+

So

+

#""molality"" = ""0.083238 moles""/(162.7 * 10^(-3) "" kg"") = ""0.51156 mol kg""^(-1)#

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+
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" "What is the molality of benzene in a solution that contains #""6.502 g""# of benzene in #""162.7 g""# of chloroform/#""CHCl""_3# (#K_f# for #""CHCl""_3# is #4.68^@ ""C""cdot""kg/mol""# and #T_f^""*""# of #""CHCl""_3# is #63.5^@ ""C""#)?" nan +136 a82b4afb-6ddd-11ea-9a56-ccda262736ce https://socratic.org/questions/a-5-82-kg-piece-of-copper-metal-is-heated-from-21-5-c-to-328-3-c-what-is-the-hea 687.45 kJ start physical_unit 23 24 heat_energy kj qc_end physical_unit 5 6 1 2 mass qc_end physical_unit 5 6 10 11 temperature qc_end physical_unit 5 6 13 14 temperature qc_end end "[{""type"":""physical unit"",""value"":""Heat absorbed [OF] the metal [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""687.45 kJ""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] copper metal [=] \\pu{5.82 kg}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] copper metal [=] \\pu{21.5 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] copper metal [=] \\pu{328.3 °C}""}]" "

A 5.82-kg piece of copper metal is heated from 21.5°C to 328.3°C. What is the heat absorbed (in kJ) by the metal?

" nan 687.45 kJ "
+

Explanation:

+
+

The heat absorbed is

+

#E=m*C*DeltaT#

+

The mass is #=5.82kg#

+

The specific heat of copper is #C=0.385kJkg^-1ºC^-1#

+

The change in temperature is

+

#DeltaT=328.3-21.5=306.8#

+

Therefore,

+

#E=5.82*0.385*306.8=687.4kJ#

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+
" "
+
+
+

The heat absorbed is #=687.4kJ#

+
+
+
+

Explanation:

+
+

The heat absorbed is

+

#E=m*C*DeltaT#

+

The mass is #=5.82kg#

+

The specific heat of copper is #C=0.385kJkg^-1ºC^-1#

+

The change in temperature is

+

#DeltaT=328.3-21.5=306.8#

+

Therefore,

+

#E=5.82*0.385*306.8=687.4kJ#

+
+
+
" "
+

A 5.82-kg piece of copper metal is heated from 21.5°C to 328.3°C. What is the heat absorbed (in kJ) by the metal?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Jul 18, 2017 + +
+
+
+
+
+
+
+

The heat absorbed is #=687.4kJ#

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+
+
+

Explanation:

+
+

The heat absorbed is

+

#E=m*C*DeltaT#

+

The mass is #=5.82kg#

+

The specific heat of copper is #C=0.385kJkg^-1ºC^-1#

+

The change in temperature is

+

#DeltaT=328.3-21.5=306.8#

+

Therefore,

+

#E=5.82*0.385*306.8=687.4kJ#

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+ + Jul 18, 2017 + +
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Here are all informations that we have :

+
    +
  • Mass of copper(m) #=5.82g#
    • +
  • Specific Heat Capacity (I will just name it SHC) of copper #=0.385 J(g°C)^-1# #rarr# simpler way to write 0.385 J/g°C
    • +
  • Change in temperature (#DeltaC#) #=328.3°C-21.5°C=306.8°C#
    • +
+

Then we start

+

Firstly the formula,

+

SHC #=(Heat)/(m*DeltaC)#

+

Then we replace the values and calculate it

+

#0.385=(Heat)/(5.82*306.8°C)#

+

#Heat=0.385*(5.82*306.8°C)#

+

Therefore, heat absorbed#=687.44676 J=687(3s.f)#

+

(I put the answer at 3 significant figures (s.f) but for further calculations use the first value obtained)

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" A 5.82-kg piece of copper metal is heated from 21.5°C to 328.3°C. What is the heat absorbed (in kJ) by the metal? nan +137 a82b4afc-6ddd-11ea-9558-ccda262736ce https://socratic.org/questions/what-quantity-of-heat-energy-must-have-been-applied-to-a-block-of-aluminum-weigh 590.67 J start physical_unit 10 13 heat_energy j qc_end physical_unit 10 13 15 16 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Heat energy [OF] a block of aluminum [IN] J""}]" "[{""type"":""physical unit"",""value"":""590.67 J""}]" "[{""type"":""physical unit"",""value"":""Weight [OF] a block of aluminum [=] \\pu{42.7 g}""},{""type"":""other"",""value"":""The temperature of the block of aluminum increased by 15.2 °C""}]" "

What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g if the temperature of the block of aluminum increased by 15.2°C?

" nan 590.67 J "
+

Explanation:

+
+

As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.

+

This means that you're going to have to track it yourself. You can find it listed here

+

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

+

as being equal to

+
+

#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#

+
+

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

In this case, you need to provide aluminium with #""0.91 J""# of heat in order to increase the temperature of #""1 g""# by #1^@""C""#.

+

You know that the block of aluminium has a mass #""42.7 g""#. This means that if you were to increase its temperature by just #1^@""C""#, you'd need

+
+

#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#

+
+

This much heat would increase the temperature of #""42.7 g""# of aluminium by #1^@""C""#. To increase its temperature by #15.2^@""C""#, you'd need #15.2# times more heat

+
+

#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#

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+

Rounded to three sig figs, the answer will be

+
+

#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

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+

Alternatively, you can use the formula

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+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

Plug in your values to get

+
+

#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#

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#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

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" "
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+

#""591 J""#

+
+
+
+

Explanation:

+
+

As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.

+

This means that you're going to have to track it yourself. You can find it listed here

+

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

+

as being equal to

+
+

#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#

+
+

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

In this case, you need to provide aluminium with #""0.91 J""# of heat in order to increase the temperature of #""1 g""# by #1^@""C""#.

+

You know that the block of aluminium has a mass #""42.7 g""#. This means that if you were to increase its temperature by just #1^@""C""#, you'd need

+
+

#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#

+
+

This much heat would increase the temperature of #""42.7 g""# of aluminium by #1^@""C""#. To increase its temperature by #15.2^@""C""#, you'd need #15.2# times more heat

+
+

#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#

+
+

Rounded to three sig figs, the answer will be

+
+

#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

+
+

Alternatively, you can use the formula

+
+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

Plug in your values to get

+
+

#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#

+

#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

+
+
+
+
" "
+

What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g if the temperature of the block of aluminum increased by 15.2°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 11, 2016 + +
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+
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+
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+
+

#""591 J""#

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+
+
+

Explanation:

+
+

As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.

+

This means that you're going to have to track it yourself. You can find it listed here

+

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

+

as being equal to

+
+

#c = ""0.91 J g""^(-1)""""^@""C""^(-1)#

+
+

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

In this case, you need to provide aluminium with #""0.91 J""# of heat in order to increase the temperature of #""1 g""# by #1^@""C""#.

+

You know that the block of aluminium has a mass #""42.7 g""#. This means that if you were to increase its temperature by just #1^@""C""#, you'd need

+
+

#42.7 color(red)(cancel(color(black)(""g""))) xx ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))""""^@""C""^(-1) = ""38.86 J"" """"^@""C""^(-1)#

+
+

This much heat would increase the temperature of #""42.7 g""# of aluminium by #1^@""C""#. To increase its temperature by #15.2^@""C""#, you'd need #15.2# times more heat

+
+

#15.2color(red)(cancel(color(black)(""""^@""C""))) xx ""38.86 J"" color(red)(cancel(color(black)(""""^@""C""^(-1)))) = ""590.67 J""#

+
+

Rounded to three sig figs, the answer will be

+
+

#""heat needed"" = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

+
+

Alternatively, you can use the formula

+
+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat added / removed
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature

+

Plug in your values to get

+
+

#q = 42.7 color(red)(cancel(color(black)(""g""))) * ""0.91 J"" color(red)(cancel(color(black)(""g""^(-1))))color(red)(cancel(color(black)(""""^@""C""^(-1)))) * 15.2color(red)(cancel(color(black)(""""^@""C"")))#

+

#q = color(green)(|bar(ul(color(white)(a/a)""591 J""color(white)(a/a)|)))#

+
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+
+ +
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+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 7007 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g if the temperature of the block of aluminum increased by 15.2°C? nan +138 a82b4afd-6ddd-11ea-a490-ccda262736ce https://socratic.org/questions/59924a04b72cff54cf340194 CaCO3 start chemical_formula qc_end physical_unit 5 6 0 1 mass qc_end physical_unit 8 9 0 1 mass qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""CaCO3""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen chloride [=] \\pu{2.5 g}""},{""type"":""physical unit"",""value"":""Mass [OF] calcium carbonate [=] \\pu{2.5 g}""},{""type"":""other"",""value"":""limiting reagent""}]" "

#2.5*g# masses each of hydrogen chloride, and calcium carbonate are reacted. Which is the limiting reagent?

" nan CaCO3 "
+

Explanation:

+
+

First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.

+

#CaCO_3 + HCl → CO_2 + CaCl_2# Basic equation, unbalanced in Cl, H and O. This is in part because it is skipping an intermediate step carbonic acid (#H_2CO_3#) formation that may then decompose into #CO_2# and water (#H_2O#).

+

#CaCO_3 + HCl → H_2CO_3 + CaCl_2# → #H_2O + CO_2 + CaCl_2# NOW we can balance it properly:

+

#CaCO_3 + 2HCl → H_2O + CO_2 + CaCl_2# Now it can be seen that it take TWO moles of #HCl# to react fully with ONE mole of #CaCO_3#. How many moles do we have?

+

#""2.5g"" (CaCO_3)/(100g/(mol)) = 0.025 ""mole""# ; #""2.5g""( HCl)/(36.5g/(mol)) = 0.068 ""mole""# Fully reacting 0.025 moles of #CaCO_3# requires 0.050 moles HCl. We have 0.068 available, so #CaCO_3# is the * LIMITING REAGENT. *

+

From our equation, see that our 0.025 mole #CaCO_3# will produce 0.025 mole each of #H_2O , CO_2 , CaCl_2# (leaving an excess of 0.018 mole of HCl in solution). Converting these into masses we obtain:
+#CO_2 = 0.025mol xx 44g/(mol) = 1.1 grams#

+

#CaCl_2 = 0.025mol xx 111g/(mol) = 2.775 grams#

+

A mass balance will confirm the correct calculations. #H_2O = 0.025mol xx 18g/(mol) = 0.45 grams# ; #HCl = 0.018mol xx 36.5g/(mol) = 0.657 grams#

+

Original Reagent mass: 2.5g + 2.5g = 5.0g
+Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g

+
+
" "
+
+
+

#CO_2 = 1.1 grams#
+#CaCl_2 = 2.78 grams#

+
+
+
+

Explanation:

+
+

First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.

+

#CaCO_3 + HCl → CO_2 + CaCl_2# Basic equation, unbalanced in Cl, H and O. This is in part because it is skipping an intermediate step carbonic acid (#H_2CO_3#) formation that may then decompose into #CO_2# and water (#H_2O#).

+

#CaCO_3 + HCl → H_2CO_3 + CaCl_2# → #H_2O + CO_2 + CaCl_2# NOW we can balance it properly:

+

#CaCO_3 + 2HCl → H_2O + CO_2 + CaCl_2# Now it can be seen that it take TWO moles of #HCl# to react fully with ONE mole of #CaCO_3#. How many moles do we have?

+

#""2.5g"" (CaCO_3)/(100g/(mol)) = 0.025 ""mole""# ; #""2.5g""( HCl)/(36.5g/(mol)) = 0.068 ""mole""# Fully reacting 0.025 moles of #CaCO_3# requires 0.050 moles HCl. We have 0.068 available, so #CaCO_3# is the * LIMITING REAGENT. *

+

From our equation, see that our 0.025 mole #CaCO_3# will produce 0.025 mole each of #H_2O , CO_2 , CaCl_2# (leaving an excess of 0.018 mole of HCl in solution). Converting these into masses we obtain:
+#CO_2 = 0.025mol xx 44g/(mol) = 1.1 grams#

+

#CaCl_2 = 0.025mol xx 111g/(mol) = 2.775 grams#

+

A mass balance will confirm the correct calculations. #H_2O = 0.025mol xx 18g/(mol) = 0.45 grams# ; #HCl = 0.018mol xx 36.5g/(mol) = 0.657 grams#

+

Original Reagent mass: 2.5g + 2.5g = 5.0g
+Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g

+
+
+
" "
+

#2.5*g# masses each of hydrogen chloride, and calcium carbonate are reacted. Which is the limiting reagent?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Limiting Reagent + + +
+
+
+
+
+3 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Aug 15, 2017 + +
+
+
+
+
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+
+

#CO_2 = 1.1 grams#
+#CaCl_2 = 2.78 grams#

+
+
+
+

Explanation:

+
+

First, you need to write out the balanced chemical reaction equation. Then you convert the given mass to moles to see which compound “limits” the reaction (the one completely used up before the other one is fully reacted). Finally, you again use the balanced equation to convert the actual moles of reagents into the respective moles of product, and then convert those moles back into masses.

+

#CaCO_3 + HCl → CO_2 + CaCl_2# Basic equation, unbalanced in Cl, H and O. This is in part because it is skipping an intermediate step carbonic acid (#H_2CO_3#) formation that may then decompose into #CO_2# and water (#H_2O#).

+

#CaCO_3 + HCl → H_2CO_3 + CaCl_2# → #H_2O + CO_2 + CaCl_2# NOW we can balance it properly:

+

#CaCO_3 + 2HCl → H_2O + CO_2 + CaCl_2# Now it can be seen that it take TWO moles of #HCl# to react fully with ONE mole of #CaCO_3#. How many moles do we have?

+

#""2.5g"" (CaCO_3)/(100g/(mol)) = 0.025 ""mole""# ; #""2.5g""( HCl)/(36.5g/(mol)) = 0.068 ""mole""# Fully reacting 0.025 moles of #CaCO_3# requires 0.050 moles HCl. We have 0.068 available, so #CaCO_3# is the * LIMITING REAGENT. *

+

From our equation, see that our 0.025 mole #CaCO_3# will produce 0.025 mole each of #H_2O , CO_2 , CaCl_2# (leaving an excess of 0.018 mole of HCl in solution). Converting these into masses we obtain:
+#CO_2 = 0.025mol xx 44g/(mol) = 1.1 grams#

+

#CaCl_2 = 0.025mol xx 111g/(mol) = 2.775 grams#

+

A mass balance will confirm the correct calculations. #H_2O = 0.025mol xx 18g/(mol) = 0.45 grams# ; #HCl = 0.018mol xx 36.5g/(mol) = 0.657 grams#

+

Original Reagent mass: 2.5g + 2.5g = 5.0g
+Final Reaction Mass: 1.1g + 2.78g + 0.45g + 0.66g = 5.0g

+
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+ + Aug 15, 2017 + +
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#""2.5 g CaCO""_3""# produces #""1.1 g CO""_2# and #""2.8 g CaCl""_2#.

+

#""2.5 g HCl""# could produce #""1.5 g CO""_2""# and #""3.8 g CaCl""_2# only if there were enough #""CaCO""_3""#.

+

#""CaCO""_3""# is the limiting reagent, also called the limiting reactant.

+
+
+
+

Explanation:

+
+

Balanced Equation

+

#""CaCO""_3 + ""2HCl""##rarr##""CO""_2 + ""CaCl""_2 + ""H""_2""O""#

+

The basic process will be:

+

#""mass reactant""##rarr##""mol reactant""##rarr##""mol product""##rarr##""mass product""#

+

First convert the masses of calcium carbonate #(""CaCO""_3)# and hydrochloric acid #(""HCl"")# to moles by dividing the given mass by the molar mass of each compound #""100.086 g/mol""# for #""CaCO""_3#, and #""36.458 g/mol""# for #""HCl""#. Since the molar mass is a fraction, #""g""/""mol""#, divide by multiplying the given mass by the inverse of the molar mass.

+

#2.5color(red)cancel(color(black)(""g CaCO""_3))xx(1""mol CaCO""_3)/(100.086color(red)cancel(color(black)(""g CaCO""_3)))=""0.0250 mol CaCO""_3""#

+

#2.5color(red)cancel(color(black)(""g HCl""))xx(1""mol HCl"")/(36.458color(red)cancel(color(black)(""g HCl"")))=""0.0686 mol HCl""#

+

#color(red)(""Calcium Carbonate: CaCO""_3""#

+

Theoretical Mass of #""CO""_2""#

+

Multiply mol #""CaCO""_3""# by the mol ratio between #""CaCO""_3""# and #""CO""_2""# from the blanced equation.

+

#0.0250color(red)cancel(color(black)(""mol CaCO""_3))xx(1""mol CO""_2)/(1color(red)cancel(color(black)(""mol CaCO""_3)))=""0.0250 mol CO""_2""#

+

Determine the mass in grams of #""CO""_2""# produced by multiplying the mol #""CO""_2# by its molar mass #(""44.009 g/mol"")#.

+

#0.0250color(red)cancel(color(black)(""mol CO""_2))xx(44.009""g CO""_2)/(1color(red)cancel(color(black)(""mol CO""_2)))=""1.1 g CO""_2""# (rounded to two sig figs due to #""2.5 g""#)

+

Theoretical Mass of #""CaCl""_2""#

+

Multiply the mol #""CaCO""_3""# by the mol ratio between #""CaCO""_3""# and #""CaCl""_2""# from the balanced equation.

+

#0.0250color(red)cancel(color(black)(""mol CaCO""_3))xx(1""mol CaCl""_2)/(1color(red)cancel(color(black)(""mol CaCO""_3)))=""0.0250 mol CaCl""_2""#

+

Determine the mass in grams of #""CaCl""_2""# by multiplying the mol #""CaCl""_2# by its molar mass #(""110.978 g/mol"")#.

+

#0.0250color(red)cancel(color(black)(""mol CaCl""_2))xx(110.978""g CaCl""_2)/(1color(red)cancel(color(black)(""mol CaCl""_2)))=""2.8 g CaCl""_2# (rounded to two sig figs due to #""2.5 g""#)

+

#color(blue)(""Hydrochloric Acid: HCl""#

+

Theoretical Mass of #""CO""_2""#

+

Multiply the mol #""HCl""# by the mol ratio between #""HCl""# and #""CO""_2""# from the balanced equation.

+

#0.0686color(red)cancel(color(black)(""mol HCl""))xx(1""mol CO""_2)/(2color(red)cancel(color(black)(""mol HCl"")))=""0.0343 mol CO""_2""#

+

Determine the mass in grams of #""CO""_2""# produced by multiplying the mol #""CO""_2# by its molar mass #(""44.009 g/mol"")#.

+

#0.0343color(red)cancel(color(black)(""mol CO""_2))xx(44.009""g CO""_2)/(1color(red)cancel(color(black)(""mol CO""_2)))=""1.5 g CO""_2""# (rounded to two sig figs due to #""2.5 g""#)

+

Theoretical Mass of #""CaCl""_2#

+

Determine the mol #""CaCl""_2#.

+

Multiply the mol #""HCl""# by the mol ratio between #""HCl""# and #""CaCl""_2""# from the balanced equation.

+

#0.0686color(red)cancel(color(black)(""mol HCl""))xx(1""mol CaCl""_2)/(2color(red)cancel(color(black)(""mol HCl"")))=""0.0343 mol CaCl""_2""#

+

Determine the mass in grams of #""CaCl""_2""# by multiplying the mol #""CaCl""_2# by its molar mass #(""110.978 g/mol"")#.

+

#0.0343color(red)cancel(color(black)(""mol CaCl""_2))xx(110.978""g CaCl""_2)/(1color(red)cancel(color(black)(""mol CaCl""_2)))=""3.8 g CaCl""_2# (rounded to two sig figs due to #""2.5 g"")#

+

Summary

+

#""2.5 g CaCO""_3""# produces #""1.1 g CO""_2# and #""2.8 g CaCl""_2#.

+

#""2.5 g HCl""# could produce #""1.5 g CO""_2""# and #""3.8 g CaCl""_2# only if there were more #""CaCO""_3""#. #""HCl""# is present in excess.

+

#""CaCO""_3""# is the limiting reagent.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Aug 15, 2017 + +
+
+
+
+
+
+
+

We need (i) a stoichiometric equation.......

+
+
+
+

Explanation:

+
+

#CaCO_3(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l) + CO_2(g)uarr#

+

And (ii) we need equivalent quantities of metal salt, and hydrogen chloride.

+

#""Moles of calcium carbonate""=(2.5*g)/(100.09*g*mol^-1)=2.49xx10^-3*mol.#

+

#""Moles of hydrogen chloride""=(2.5*g)/(36.46*g*mol^-1)=0.0686*mol.#

+

And thus the acid is in VAST stoichiometric excess, and thus calcium carbonate is the LIMITING reagent.

+

And so we gets.................

+

#2.49xx10^-3*molxx110.98*g*mol^-1=0.276*g# with respect to #""calcium chloride""#.

+

And likewise we gets...........
+#2.49xx10^-3*molxx44.01*g*mol^-1=0.110*g# with respect to #""carbon dioxide""#.

+

An extension of this question would be to ask the VOLUME gas produced under standard conditions of #1*atm# (or whatever), and #298*K#.

+
+
+
+
+
+ +
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" #2.5*g# masses each of hydrogen chloride, and calcium carbonate are reacted. Which is the limiting reagent? nan +139 a82b4afe-6ddd-11ea-b44b-ccda262736ce https://socratic.org/questions/what-is-the-molar-mass-of-p-4o-10 283.89 g/mol start physical_unit 6 6 molar_mass g/mol qc_end chemical_equation 6 6 qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] P4O10 [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""283.89 g/mol""}]" "[{""type"":""chemical equation"",""value"":""P4O10""}]" "

What is the molar mass of #P_4O_10#?

" nan 283.89 g/mol "
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
" "
+
+
+

#=283.889048# amu

+
+
+
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
+
" "
+

What is the molar mass of #P_4O_10#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 13, 2016 + +
+
+
+
+
+
+
+

#=283.889048# amu

+
+
+
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
+
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+
Impact of this question
+
+ 1300 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What is the molar mass of #P_4O_10#? nan +140 a82b4aff-6ddd-11ea-b253-ccda262736ce https://socratic.org/questions/5959a7a511ef6b614b3a9cf3 1236.23 L start physical_unit 6 7 volume l qc_end physical_unit 6 7 14 15 volume qc_end physical_unit 6 7 17 18 temperature qc_end physical_unit 6 7 20 22 pressure qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] ideal gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""1236.23 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] ideal gas [=] \\pu{2.85 L}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] ideal gas [=] \\pu{14 ℃}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] ideal gas [=] \\pu{450 mm Hg}""},{""type"":""other"",""value"":""STP""}]" "

What is the volume of an ideal gas at STP, if its volume is #""2.85 L""# at #14^@ ""C""# and #""450 mm Hg""#?

" nan 1236.23 L "
+

Explanation:

+
+

This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.

+

The equation to use is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

https://en.wikipedia.org/wiki/Gas_laws

+

Housekeeping Issues

+

Current STP

+

Standard temperature is #""0""^@""C""# or #""273.15 K""#, and standard pressure after 1982 is #10^5# #""Pa""#, usually written as #""100 kPa""# for easier use.

+

The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add #273.15# to the Celsius temperature.

+

#14^@""C""+273.15=""287.15 K""#

+

The pressure in #""mmHg""# must be converted to #""kPa""#.

+

#""1 mmHg""# #=# #""101.325 kPa""#

+

#450color(red)cancel(color(black)(""mmHg""))xx(101.325""kPa"")/(1color(red)cancel(color(black)(""mmHg"")))=""45600 kPa""#

+

I'm giving the pressure to three sig figs to reduce rounding errors.

+

Organize the data:

+

Known

+

#P_1=""45600 kPa""#

+

#V_1=""2.85 L""#

+

#T_1=""287.15 K""#

+

#P_2=""100 kPa""#

+

#T_2=""273.15 K""#

+

Unknown

+

#V_2#

+

Solution

+

Rearrange the combined gas law equation to isolate #V_2#. Insert the data and solve.

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

#V_2=(P_1V_1T_2)/(T_1P_2)#

+

#V_2=((45600color(red)cancel(color(black)(""kPa"")))xx(2.85""L"")xx(273.15color(red)cancel(color(black)(""K""))))/((287.15color(red)cancel(color(black)(""K"")))xx(100color(red)cancel(color(black)(""kPa""))))=""1200 L""# (rounded to two significant figures)

+
+
" "
+
+
+

The volume at STP is #~~""1200 L"".#

+
+
+
+

Explanation:

+
+

This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.

+

The equation to use is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

https://en.wikipedia.org/wiki/Gas_laws

+

Housekeeping Issues

+

Current STP

+

Standard temperature is #""0""^@""C""# or #""273.15 K""#, and standard pressure after 1982 is #10^5# #""Pa""#, usually written as #""100 kPa""# for easier use.

+

The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add #273.15# to the Celsius temperature.

+

#14^@""C""+273.15=""287.15 K""#

+

The pressure in #""mmHg""# must be converted to #""kPa""#.

+

#""1 mmHg""# #=# #""101.325 kPa""#

+

#450color(red)cancel(color(black)(""mmHg""))xx(101.325""kPa"")/(1color(red)cancel(color(black)(""mmHg"")))=""45600 kPa""#

+

I'm giving the pressure to three sig figs to reduce rounding errors.

+

Organize the data:

+

Known

+

#P_1=""45600 kPa""#

+

#V_1=""2.85 L""#

+

#T_1=""287.15 K""#

+

#P_2=""100 kPa""#

+

#T_2=""273.15 K""#

+

Unknown

+

#V_2#

+

Solution

+

Rearrange the combined gas law equation to isolate #V_2#. Insert the data and solve.

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

#V_2=(P_1V_1T_2)/(T_1P_2)#

+

#V_2=((45600color(red)cancel(color(black)(""kPa"")))xx(2.85""L"")xx(273.15color(red)cancel(color(black)(""K""))))/((287.15color(red)cancel(color(black)(""K"")))xx(100color(red)cancel(color(black)(""kPa""))))=""1200 L""# (rounded to two significant figures)

+
+
+
" "
+

What is the volume of an ideal gas at STP, if its volume is #""2.85 L""# at #14^@ ""C""# and #""450 mm Hg""#?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Aug 10, 2017 + +
+
+
+
+
+
+
+

The volume at STP is #~~""1200 L"".#

+
+
+
+

Explanation:

+
+

This is an example of the combined gas law, which combines Boyle's, Charles', and Gay-Lussac's laws. It shows the relationship between the pressure, volume, and temperature when the quantity of ideal gas is constant.

+

The equation to use is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

https://en.wikipedia.org/wiki/Gas_laws

+

Housekeeping Issues

+

Current STP

+

Standard temperature is #""0""^@""C""# or #""273.15 K""#, and standard pressure after 1982 is #10^5# #""Pa""#, usually written as #""100 kPa""# for easier use.

+

The Kelvin temperature scale must be used in gas problems. To convert temperature in degrees Celsius to Kelvins, add #273.15# to the Celsius temperature.

+

#14^@""C""+273.15=""287.15 K""#

+

The pressure in #""mmHg""# must be converted to #""kPa""#.

+

#""1 mmHg""# #=# #""101.325 kPa""#

+

#450color(red)cancel(color(black)(""mmHg""))xx(101.325""kPa"")/(1color(red)cancel(color(black)(""mmHg"")))=""45600 kPa""#

+

I'm giving the pressure to three sig figs to reduce rounding errors.

+

Organize the data:

+

Known

+

#P_1=""45600 kPa""#

+

#V_1=""2.85 L""#

+

#T_1=""287.15 K""#

+

#P_2=""100 kPa""#

+

#T_2=""273.15 K""#

+

Unknown

+

#V_2#

+

Solution

+

Rearrange the combined gas law equation to isolate #V_2#. Insert the data and solve.

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

#V_2=(P_1V_1T_2)/(T_1P_2)#

+

#V_2=((45600color(red)cancel(color(black)(""kPa"")))xx(2.85""L"")xx(273.15color(red)cancel(color(black)(""K""))))/((287.15color(red)cancel(color(black)(""K"")))xx(100color(red)cancel(color(black)(""kPa""))))=""1200 L""# (rounded to two significant figures)

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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+
" "What is the volume of an ideal gas at STP, if its volume is #""2.85 L""# at #14^@ ""C""# and #""450 mm Hg""#?" nan +141 a82b4b00-6ddd-11ea-8764-ccda262736ce https://socratic.org/questions/what-volume-in-milliliters-of-0-54-m-ca-oh-2-is-needed-to-completely-neutralize- 58.11 milliliters start physical_unit 7 7 volume ml qc_end physical_unit 19 20 13 14 volume qc_end physical_unit 19 20 17 18 molarity qc_end physical_unit 7 7 5 6 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] Ca(OH)2 [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""58.11 milliliters""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] HI solution [=] \\pu{241.4 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] HI solution [=] \\pu{0.26 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] Ca(OH)2 [=] \\pu{0.54 M}""}]" "

What volume, in milliliters, of 0.54 M #Ca(OH)_2# is needed to completely neutralize 241.4 mL of a .26 M #HI# solution?

" nan 58.11 milliliters "
+

Explanation:

+
+

#""Moles of HI""# #=# #241.4xx10^-3*Lxx0.26*mol*L^-1# #=# #0.063*mol*HI#

+

#0.0313*mol# #Ca(OH)_2# are required for equivalence. Now of course I could use this molar quantity to calulate the volume of calcium hydroxide required. However, there is one problem.

+

Calcium hydroxide is very sparingly soluble. I don't think you could make a #0.5*mol*L^-1# solution of this beast (I might be wrong, but I am not going to go thru the calculation). This question should have been proposed with #NaOH# solution

+
+
" "
+
+
+

#Ca(OH)_2(aq) +2HI(aq)rarr CaI_2(aq) +2H_2O(l)#

+
+
+
+

Explanation:

+
+

#""Moles of HI""# #=# #241.4xx10^-3*Lxx0.26*mol*L^-1# #=# #0.063*mol*HI#

+

#0.0313*mol# #Ca(OH)_2# are required for equivalence. Now of course I could use this molar quantity to calulate the volume of calcium hydroxide required. However, there is one problem.

+

Calcium hydroxide is very sparingly soluble. I don't think you could make a #0.5*mol*L^-1# solution of this beast (I might be wrong, but I am not going to go thru the calculation). This question should have been proposed with #NaOH# solution

+
+
+
" "
+

What volume, in milliliters, of 0.54 M #Ca(OH)_2# is needed to completely neutralize 241.4 mL of a .26 M #HI# solution?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 3, 2016 + +
+
+
+
+
+
+
+

#Ca(OH)_2(aq) +2HI(aq)rarr CaI_2(aq) +2H_2O(l)#

+
+
+
+

Explanation:

+
+

#""Moles of HI""# #=# #241.4xx10^-3*Lxx0.26*mol*L^-1# #=# #0.063*mol*HI#

+

#0.0313*mol# #Ca(OH)_2# are required for equivalence. Now of course I could use this molar quantity to calulate the volume of calcium hydroxide required. However, there is one problem.

+

Calcium hydroxide is very sparingly soluble. I don't think you could make a #0.5*mol*L^-1# solution of this beast (I might be wrong, but I am not going to go thru the calculation). This question should have been proposed with #NaOH# solution

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 9135 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What volume, in milliliters, of 0.54 M #Ca(OH)_2# is needed to completely neutralize 241.4 mL of a .26 M #HI# solution? nan +142 a82b4b01-6ddd-11ea-83a9-ccda262736ce https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-if-its-empirical-formula-is-no-2-and N3O6 start chemical_formula qc_end physical_unit 7 7 19 20 molar_mass qc_end chemical_equation 13 13 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""N3O6""}]" "[{""type"":""physical unit"",""value"":""Molar mass [OF] compound [=] \\pu{138.02 g/mole}""},{""type"":""chemical equation"",""value"":""NO2""}]" "

What is the molecular formula of a compound if its empirical formula is #NO_2# and its molar mass is 138.02 g/mole?

" nan N3O6 "
+

Explanation:

+
+

If the molar mass of 138.02 g/mol has been provided for you,
+divide it by the molar mass of #NO_2#.

+

Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
+138.02 g/mol #-:# #(1 xx 14.0 +2 xx16.0)#
+138.02 g/mol #-:# 46.0 g/mol = 3.00 (rounded to nearest whole number)

+

Since 3.00 is the whole number, multiply it by the empirical formula.

+

(#NO_2#)#xx#3.00 = #N_3O_6# is the molecular formula.

+
+
" "
+
+
+

138.02 g/mol #-:# molar mass of empirical formulae = whole number,
+then
+Molecular Formulae = Empirical Formulae #xx# whole number

+
+
+
+

Explanation:

+
+

If the molar mass of 138.02 g/mol has been provided for you,
+divide it by the molar mass of #NO_2#.

+

Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
+138.02 g/mol #-:# #(1 xx 14.0 +2 xx16.0)#
+138.02 g/mol #-:# 46.0 g/mol = 3.00 (rounded to nearest whole number)

+

Since 3.00 is the whole number, multiply it by the empirical formula.

+

(#NO_2#)#xx#3.00 = #N_3O_6# is the molecular formula.

+
+
+
" "
+

What is the molecular formula of a compound if its empirical formula is #NO_2# and its molar mass is 138.02 g/mole?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 25, 2016 + +
+
+
+
+
+
+
+

138.02 g/mol #-:# molar mass of empirical formulae = whole number,
+then
+Molecular Formulae = Empirical Formulae #xx# whole number

+
+
+
+

Explanation:

+
+

If the molar mass of 138.02 g/mol has been provided for you,
+divide it by the molar mass of #NO_2#.

+

Nitrogen has a molar mass of 14.0 g/mol, whereas oxygen is 16.0 g/mol
+138.02 g/mol #-:# #(1 xx 14.0 +2 xx16.0)#
+138.02 g/mol #-:# 46.0 g/mol = 3.00 (rounded to nearest whole number)

+

Since 3.00 is the whole number, multiply it by the empirical formula.

+

(#NO_2#)#xx#3.00 = #N_3O_6# is the molecular formula.

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
+
+
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" What is the molecular formula of a compound if its empirical formula is #NO_2# and its molar mass is 138.02 g/mole? nan +143 a82b4b02-6ddd-11ea-b35f-ccda262736ce https://socratic.org/questions/2n-2h-4-l-n-2o-4-l-3n-2-g-4h-2o-l-if-10-81-g-of-n-2h-4-is-used-what-mass-of-nitr 14.19 g start physical_unit 20 20 mass g qc_end chemical_equation 0 9 qc_end physical_unit 1 1 11 12 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] nitrogen [IN] g""}]" "[{""type"":""physical unit"",""value"":""14.19 g""}]" "[{""type"":""chemical equation"",""value"":""2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H20(l)""},{""type"":""physical unit"",""value"":""Mass [OF] N2H4 [=] \\pu{10.81 g}""}]" "

#2N_2H_4(l) + N_2O_4(l) -> 3N_2(g) + 4H_2O(l)#. If 10.81 g of #N_2H_4# is used, what mass of nitrogen is produced?

" nan 14.19 g "
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Explanation:

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+

If you have 10.81 grams of #N_2H_4# and limitless amount, in terms of #N_2O_4#, you will get

+

#(84/64)*10.81#

+

#=14.188# grams of nitrogen gas you produce.

+

This is your answer: 14.188 grams of nitrogen gan

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From 64 grams of #N_2H_4#, you get 84 grams of nitrogen gas

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+

Explanation:

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+

If you have 10.81 grams of #N_2H_4# and limitless amount, in terms of #N_2O_4#, you will get

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#(84/64)*10.81#

+

#=14.188# grams of nitrogen gas you produce.

+

This is your answer: 14.188 grams of nitrogen gan

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" "
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#2N_2H_4(l) + N_2O_4(l) -> 3N_2(g) + 4H_2O(l)#. If 10.81 g of #N_2H_4# is used, what mass of nitrogen is produced?

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+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
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+1 Answer +
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From 64 grams of #N_2H_4#, you get 84 grams of nitrogen gas

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+
+
+

Explanation:

+
+

If you have 10.81 grams of #N_2H_4# and limitless amount, in terms of #N_2O_4#, you will get

+

#(84/64)*10.81#

+

#=14.188# grams of nitrogen gas you produce.

+

This is your answer: 14.188 grams of nitrogen gan

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+
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" #2N_2H_4(l) + N_2O_4(l) -> 3N_2(g) + 4H_2O(l)#. If 10.81 g of #N_2H_4# is used, what mass of nitrogen is produced? nan +144 a82b4b03-6ddd-11ea-9914-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-0-470-m-solution-of-methylamine 12.00 start physical_unit 8 10 ph none qc_end physical_unit 8 10 6 7 molarity qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] solution of methylamine""}]" "[{""type"":""physical unit"",""value"":""12.00""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] solution of methylamine [=] \\pu{0.470 M}""}]" "

What is the pH of a 0.470 M solution of methylamine?

" nan 12.00 "
+

Explanation:

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+

We need (i) #pK_B# values for methylamine, alternatively #pK_a# values for #H_3CNH_3^+#. This site gives #pK_b=3.66#.

+

And (ii) we need a stoichiometric equation.........

+

#H_3CNH_2(aq) + H_2O(l)rightleftharpoonsH_3CNH_3^(+) +HO^-#

+

And by definition, this equilibrium is governed by the quotient.....

+

#K_b=10^(-pK_b)=10^(-3.66)=([H_3CNH_3^+][HO^-])/([H_3CNH_2])#.

+

Now if initially, #[H_3CNH_2]=0.470*mol*L^-1#, and we say the amount of dissociation is #x#, then we can write:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2])=((x)xx(x))/(0.470-x)=x^2/(0.470-x)=10^(-3.66)#.

+

This is a quadratic in #x#, which we could solve exactly, but because chemist are lazy folk, we make the approximation, that #0.366"">>""x#, and that #x^2/(0.470-x)=10^(-3.66)~=x^2/(0.470)#.

+

And thus #x_1=sqrt(10^(-3.66)xx0.470)=1.01xx10^-2#, and if we recycle this first approximation back into the equation, we gets.....

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#x_2=sqrt(10^(-3.66)xx(0.470-1.01xx10^-2))=1.00xx10^-2*mol*L^-1#

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#x_3=sqrt(10^(-3.66)xx(0.470-1.00xx10^-2))=1.00xx10^-2*mol*L^-1#

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Because the approximations have converged, we are willing to accept this value. But #x=[HO^-]# by definition; and so #pOH=-log_10(1.00xx10^-2)=+2#

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And since we know (or should know) that #pOH+pH=14#, #pH=12#.

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#pH=12#

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+
+

Explanation:

+
+

We need (i) #pK_B# values for methylamine, alternatively #pK_a# values for #H_3CNH_3^+#. This site gives #pK_b=3.66#.

+

And (ii) we need a stoichiometric equation.........

+

#H_3CNH_2(aq) + H_2O(l)rightleftharpoonsH_3CNH_3^(+) +HO^-#

+

And by definition, this equilibrium is governed by the quotient.....

+

#K_b=10^(-pK_b)=10^(-3.66)=([H_3CNH_3^+][HO^-])/([H_3CNH_2])#.

+

Now if initially, #[H_3CNH_2]=0.470*mol*L^-1#, and we say the amount of dissociation is #x#, then we can write:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2])=((x)xx(x))/(0.470-x)=x^2/(0.470-x)=10^(-3.66)#.

+

This is a quadratic in #x#, which we could solve exactly, but because chemist are lazy folk, we make the approximation, that #0.366"">>""x#, and that #x^2/(0.470-x)=10^(-3.66)~=x^2/(0.470)#.

+

And thus #x_1=sqrt(10^(-3.66)xx0.470)=1.01xx10^-2#, and if we recycle this first approximation back into the equation, we gets.....

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#x_2=sqrt(10^(-3.66)xx(0.470-1.01xx10^-2))=1.00xx10^-2*mol*L^-1#

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#x_3=sqrt(10^(-3.66)xx(0.470-1.00xx10^-2))=1.00xx10^-2*mol*L^-1#

+

Because the approximations have converged, we are willing to accept this value. But #x=[HO^-]# by definition; and so #pOH=-log_10(1.00xx10^-2)=+2#

+

And since we know (or should know) that #pOH+pH=14#, #pH=12#.

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What is the pH of a 0.470 M solution of methylamine?

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+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
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+1 Answer +
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#pH=12#

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Explanation:

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+

We need (i) #pK_B# values for methylamine, alternatively #pK_a# values for #H_3CNH_3^+#. This site gives #pK_b=3.66#.

+

And (ii) we need a stoichiometric equation.........

+

#H_3CNH_2(aq) + H_2O(l)rightleftharpoonsH_3CNH_3^(+) +HO^-#

+

And by definition, this equilibrium is governed by the quotient.....

+

#K_b=10^(-pK_b)=10^(-3.66)=([H_3CNH_3^+][HO^-])/([H_3CNH_2])#.

+

Now if initially, #[H_3CNH_2]=0.470*mol*L^-1#, and we say the amount of dissociation is #x#, then we can write:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2])=((x)xx(x))/(0.470-x)=x^2/(0.470-x)=10^(-3.66)#.

+

This is a quadratic in #x#, which we could solve exactly, but because chemist are lazy folk, we make the approximation, that #0.366"">>""x#, and that #x^2/(0.470-x)=10^(-3.66)~=x^2/(0.470)#.

+

And thus #x_1=sqrt(10^(-3.66)xx0.470)=1.01xx10^-2#, and if we recycle this first approximation back into the equation, we gets.....

+

#x_2=sqrt(10^(-3.66)xx(0.470-1.01xx10^-2))=1.00xx10^-2*mol*L^-1#

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#x_3=sqrt(10^(-3.66)xx(0.470-1.00xx10^-2))=1.00xx10^-2*mol*L^-1#

+

Because the approximations have converged, we are willing to accept this value. But #x=[HO^-]# by definition; and so #pOH=-log_10(1.00xx10^-2)=+2#

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And since we know (or should know) that #pOH+pH=14#, #pH=12#.

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" What is the pH of a 0.470 M solution of methylamine? nan +145 a82b71fe-6ddd-11ea-b572-ccda262736ce https://socratic.org/questions/what-is-the-total-pressure-of-a-container-that-has-nh-3-g-exerting-a-pressure-of 314.00 torr start physical_unit 6 7 total_pressure torr qc_end physical_unit 10 10 15 16 pressure qc_end physical_unit 17 17 22 23 pressure qc_end physical_unit 25 25 30 31 pressure qc_end end "[{""type"":""physical unit"",""value"":""Total pressure [OF] a container [IN] torr""}]" "[{""type"":""physical unit"",""value"":""314.00 torr""}]" "[{""type"":""physical unit"",""value"":""Pressure [OF] NH3(g) [=] \\pu{34 torr}""},{""type"":""physical unit"",""value"":""Pressure [OF] N2(g) [=] \\pu{225 torr}""},{""type"":""physical unit"",""value"":""Pressure [OF] H2O(g) [=] \\pu{55 torr}""}]" "

What is the total pressure of a container that has #NH_3(g)# exerting a pressure of 34 torr, #N_2(g)# exerting a pressure of 225 torr, and #H_2O(g)# exerting a pressure of 55 torr?

" nan 314.00 torr "
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Explanation:

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+

According to Dalton's law of partial pressures,

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#P_(t otal) = P_1 + P_2 + P_3 ...#

+

So the total pressure of the system is merely the partial pressures of each component added together:

+

#P_(t otal) = P_(NH_3) + P_(N_2) + P_(H_2O)#

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#= 34 t o r r + 225 t o r r + 55 t o r r = 314 t o r r#

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#314 t o r r#

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+

Explanation:

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+

According to Dalton's law of partial pressures,

+

#P_(t otal) = P_1 + P_2 + P_3 ...#

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So the total pressure of the system is merely the partial pressures of each component added together:

+

#P_(t otal) = P_(NH_3) + P_(N_2) + P_(H_2O)#

+

#= 34 t o r r + 225 t o r r + 55 t o r r = 314 t o r r#

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" "
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What is the total pressure of a container that has #NH_3(g)# exerting a pressure of 34 torr, #N_2(g)# exerting a pressure of 225 torr, and #H_2O(g)# exerting a pressure of 55 torr?

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+1 Answer +
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#314 t o r r#

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Explanation:

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+

According to Dalton's law of partial pressures,

+

#P_(t otal) = P_1 + P_2 + P_3 ...#

+

So the total pressure of the system is merely the partial pressures of each component added together:

+

#P_(t otal) = P_(NH_3) + P_(N_2) + P_(H_2O)#

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#= 34 t o r r + 225 t o r r + 55 t o r r = 314 t o r r#

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" What is the total pressure of a container that has #NH_3(g)# exerting a pressure of 34 torr, #N_2(g)# exerting a pressure of 225 torr, and #H_2O(g)# exerting a pressure of 55 torr? nan +146 a82b71ff-6ddd-11ea-9711-ccda262736ce https://socratic.org/questions/what-would-be-the-final-volume-of-a-10-4-m-v-solution-made-from-5-00-g-of-solute 48.08 mL start physical_unit 10 10 volume ml qc_end physical_unit 10 10 8 8 mass_concentration qc_end physical_unit 16 16 13 14 mass qc_end end "[{""type"":""physical unit"",""value"":""Final volume [OF] solution [IN] mL""}]" "[{""type"":""physical unit"",""value"":""48.08 mL""}]" "[{""type"":""physical unit"",""value"":""m/v [OF] solution [=] \\pu{10.4%}""},{""type"":""physical unit"",""value"":""Mass [OF] solute [=] \\pu{5.00 g}""}]" "

What would be the final volume of a 10.4% (m/v) solution made from 5.00 g of solute?

" nan 48.08 mL "
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Explanation:

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The idea here is that a solution's mass by volume percent concentration, #""m/v %""#, tells you the number of grams of solute present in exactly #""100 mL""# of the solution.

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In your case, the solution is said to have a mass by volume percent concentration of #10.4%#, which implies that every #""100 mL""# of this solution contain #""10.4 g""# of solute.

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Now, you know that your target solution must contain #""5.00 g""# of solute, so you can determine the volume of the solution by using the mass by volume percent concentration as a conversion factor.

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#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#

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+

The answer is rounded to three sig figs, the number of sig figs you have for your values.

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So, in order to make this #""10.4% m/v""# solution, you need to dissolve #""5.00 g""# of solute in enough water to get the total volume of the solution to #""48.1 mL""#.

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#""48.1 mL""#

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Explanation:

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+

The idea here is that a solution's mass by volume percent concentration, #""m/v %""#, tells you the number of grams of solute present in exactly #""100 mL""# of the solution.

+

In your case, the solution is said to have a mass by volume percent concentration of #10.4%#, which implies that every #""100 mL""# of this solution contain #""10.4 g""# of solute.

+

Now, you know that your target solution must contain #""5.00 g""# of solute, so you can determine the volume of the solution by using the mass by volume percent concentration as a conversion factor.

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+

#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#

+
+

The answer is rounded to three sig figs, the number of sig figs you have for your values.

+

So, in order to make this #""10.4% m/v""# solution, you need to dissolve #""5.00 g""# of solute in enough water to get the total volume of the solution to #""48.1 mL""#.

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What would be the final volume of a 10.4% (m/v) solution made from 5.00 g of solute?

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+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
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#""48.1 mL""#

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Explanation:

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The idea here is that a solution's mass by volume percent concentration, #""m/v %""#, tells you the number of grams of solute present in exactly #""100 mL""# of the solution.

+

In your case, the solution is said to have a mass by volume percent concentration of #10.4%#, which implies that every #""100 mL""# of this solution contain #""10.4 g""# of solute.

+

Now, you know that your target solution must contain #""5.00 g""# of solute, so you can determine the volume of the solution by using the mass by volume percent concentration as a conversion factor.

+
+

#5.00 color(red)(cancel(color(black)(""g solute""))) * ""100 mL solution""/(10.4color(red)(cancel(color(black)(""g solute"")))) = color(darkgreen)(ul(color(black)(""48.1 mL solution"")))#

+
+

The answer is rounded to three sig figs, the number of sig figs you have for your values.

+

So, in order to make this #""10.4% m/v""# solution, you need to dissolve #""5.00 g""# of solute in enough water to get the total volume of the solution to #""48.1 mL""#.

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" What would be the final volume of a 10.4% (m/v) solution made from 5.00 g of solute? nan +147 a82b85c6-6ddd-11ea-98a7-ccda262736ce https://socratic.org/questions/how-many-moles-of-aluminum-do-3-8-10-24-aluminum-atoms-represent 6.31 moles start physical_unit 4 4 mole mol qc_end physical_unit 9 10 6 8 number qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] aluminum [IN] moles""}]" "[{""type"":""physical unit"",""value"":""6.31 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] aluminum atoms [=] \\pu{3.8 ⋅ 10^24}""}]" "

How many moles of aluminum do #3.8*10^24# aluminum atoms represent?

" nan 6.31 moles "
+

Explanation:

+
+

In a mole of aluminium atoms, there exist #6.02*10^23# aluminium atoms. So here, we get:

+

#(3.8*10^24color(red)cancelcolor(black)(Al \ ""atoms""))/(6.02*10^23color(red)cancelcolor(black)(Al \ ""atoms"")""/mol"")#

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#~~6.3 \ ""mol""#

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#6.3# moles

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Explanation:

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In a mole of aluminium atoms, there exist #6.02*10^23# aluminium atoms. So here, we get:

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#(3.8*10^24color(red)cancelcolor(black)(Al \ ""atoms""))/(6.02*10^23color(red)cancelcolor(black)(Al \ ""atoms"")""/mol"")#

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#~~6.3 \ ""mol""#

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How many moles of aluminum do #3.8*10^24# aluminum atoms represent?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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#6.3# moles

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Explanation:

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In a mole of aluminium atoms, there exist #6.02*10^23# aluminium atoms. So here, we get:

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#(3.8*10^24color(red)cancelcolor(black)(Al \ ""atoms""))/(6.02*10^23color(red)cancelcolor(black)(Al \ ""atoms"")""/mol"")#

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#~~6.3 \ ""mol""#

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" How many moles of aluminum do #3.8*10^24# aluminum atoms represent? nan +148 a82b9926-6ddd-11ea-804f-ccda262736ce https://socratic.org/questions/a-carbon-compound-contains-12-8-of-carbon-2-1-of-hydrogen-and-85-1-of-bromine-th C2H4Br2 start chemical_formula qc_end physical_unit 1 1 4 4 mass_percent qc_end physical_unit 9 9 7 7 mass_percent qc_end physical_unit 13 13 11 11 mass_percent qc_end physical_unit 1 2 21 21 molecular_weight qc_end physical_unit 32 32 34 34 atomic_mass qc_end physical_unit 35 35 37 37 atomic_mass qc_end physical_unit 38 38 40 40 atomic_mass qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""C2H4Br2""}]" "[{""type"":""physical unit"",""value"":""Mass percent [OF] carbon [=] \\pu{12.8%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] hydrogen [=] \\pu{2.1%}""},{""type"":""physical unit"",""value"":""Mass percent [OF] bromine [=] \\pu{85.1%}""},{""type"":""physical unit"",""value"":""Molecular weight [OF] carbon compound [=] \\pu{187.9}""},{""type"":""physical unit"",""value"":""Atomic weight [OF] H [=] \\pu{1.008}""},{""type"":""physical unit"",""value"":""Atomic weight [OF] C [=] \\pu{12.0}""},{""type"":""physical unit"",""value"":""Atomic weight [OF] Br [=] \\pu{79.9}""}]" "

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. What is the molecular formula of the compound? +(Atomic weight: H = 1.008, C = 12.0, Br = 79.9)

" nan C2H4Br2 "
+

Explanation:

+
+

As always with these problems, it is useful to assume #100*g# of unknown compound....and thus find an empirical formula...

+

#""Moles of carbon""=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.#

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#""Moles of hydrogen""=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.#

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#""Moles of bromine""=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.#

+

And we divide thru by the LOWEST molar quantity to give an empirical formula of.....

+

#CH_2Br#

+

But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.

+

And so #187.9*g*mol^-1=nxx(12.011+2xx1.00794+79.9)*g*mol^-1#

+

And thus #n=2#, and the molecular formula is #C_2H_4Br_2#...

+

I prefer questions that quote actual microanalytical data. #C_2H_4Br_2# is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).

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#""Molecular formula""# #-=# #C_2H_4Br_2#

+
+
+
+

Explanation:

+
+

As always with these problems, it is useful to assume #100*g# of unknown compound....and thus find an empirical formula...

+

#""Moles of carbon""=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.#

+

#""Moles of hydrogen""=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.#

+

#""Moles of bromine""=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.#

+

And we divide thru by the LOWEST molar quantity to give an empirical formula of.....

+

#CH_2Br#

+

But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.

+

And so #187.9*g*mol^-1=nxx(12.011+2xx1.00794+79.9)*g*mol^-1#

+

And thus #n=2#, and the molecular formula is #C_2H_4Br_2#...

+

I prefer questions that quote actual microanalytical data. #C_2H_4Br_2# is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).

+
+
+
" "
+

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. What is the molecular formula of the compound? +(Atomic weight: H = 1.008, C = 12.0, Br = 79.9)

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
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+
+ +
+
+ +
+ + Sep 11, 2017 + +
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+

#""Molecular formula""# #-=# #C_2H_4Br_2#

+
+
+
+

Explanation:

+
+

As always with these problems, it is useful to assume #100*g# of unknown compound....and thus find an empirical formula...

+

#""Moles of carbon""=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.#

+

#""Moles of hydrogen""=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.#

+

#""Moles of bromine""=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.#

+

And we divide thru by the LOWEST molar quantity to give an empirical formula of.....

+

#CH_2Br#

+

But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.

+

And so #187.9*g*mol^-1=nxx(12.011+2xx1.00794+79.9)*g*mol^-1#

+

And thus #n=2#, and the molecular formula is #C_2H_4Br_2#...

+

I prefer questions that quote actual microanalytical data. #C_2H_4Br_2# is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 22713 views + around the world +
+
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" "A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. What is the molecular formula of the compound? +(Atomic weight: H = 1.008, C = 12.0, Br = 79.9)" nan +149 a82b9927-6ddd-11ea-a023-ccda262736ce https://socratic.org/questions/what-is-empirical-formula-for-chloroform-89-1-chlorine-0-84-hydrogen-10-06-carbo CHCl3 start chemical_formula qc_end physical_unit 7 7 6 6 molarity_percent qc_end physical_unit 9 9 8 8 molarity_percent qc_end physical_unit 11 11 10 10 molarity_percent qc_end substance 5 5 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""CHCl3""}]" "[{""type"":""physical unit"",""value"":""Percent concentration [OF] chlorine [=] \\pu{89.1%}""},{""type"":""physical unit"",""value"":""Percent concentration [OF] hydrogen [=] \\pu{0.84%}""},{""type"":""physical unit"",""value"":""Percent concentration [OF] carbon [=] \\pu{10.06%}""},{""type"":""substance name"",""value"":""chloroform""}]" "

What is empirical formula for Chloroform (89.1% Chlorine, 0.84% Hydrogen, 10.06% Carbon)?

" nan CHCl3 "
+

Explanation:

+
+

Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.

+

To make the calculations easier, you can pick a #""100.0-g""# sample of chloroform and use its percent concentration by mass to find that it contains

+
+
    +
  • #""89.1 g "" -># chlorine
    • +
  • #""0.84 g "" -># hydrogen
    • +
  • #""10.06 g "" -># carbon
    • +
+
+

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

+
+

#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#

+

#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#

+

#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#

+
+

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

+
+

#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#

+

#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#

+
+

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

+
+

#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#

+
+
+
" "
+
+
+

#""CHCl""_3#

+
+
+
+

Explanation:

+
+

Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.

+

To make the calculations easier, you can pick a #""100.0-g""# sample of chloroform and use its percent concentration by mass to find that it contains

+
+
    +
  • #""89.1 g "" -># chlorine
    • +
  • #""0.84 g "" -># hydrogen
    • +
  • #""10.06 g "" -># carbon
    • +
+
+

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

+
+

#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#

+

#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#

+

#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#

+
+

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

+
+

#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#

+

#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#

+
+

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

+
+

#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#

+
+
+
+
" "
+

What is empirical formula for Chloroform (89.1% Chlorine, 0.84% Hydrogen, 10.06% Carbon)?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 5, 2015 + +
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+
+
+
+
+
+

#""CHCl""_3#

+
+
+
+

Explanation:

+
+

Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.

+

To make the calculations easier, you can pick a #""100.0-g""# sample of chloroform and use its percent concentration by mass to find that it contains

+
+
    +
  • #""89.1 g "" -># chlorine
    • +
  • #""0.84 g "" -># hydrogen
    • +
  • #""10.06 g "" -># carbon
    • +
+
+

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

+
+

#""For Cl: "" 89.1 color(red)(cancel(color(black)(""g""))) * ""1 mole Cl""/(35.453 color(red)(cancel(color(black)(""g"")))) = ""2.513 moles C""#

+

#""For H: "" 0.84 color(red)(cancel(color(black)(""g""))) * ""1 mole H""/(1.00794color(red)(cancel(color(black)(""g"")))) = ""0.8334 moles H""#

+

#""For C: "" 10.06 color(red)(cancel(color(black)(""g""))) * ""1 mole C""/(12.011 color(red)(cancel(color(black)(""g"")))) = ""0.8376 moles C""#

+
+

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

+
+

#""For Cl: "" (2.513 color(red)(cancel(color(black)(""moles""))))/(0.8334 color(red)(cancel(color(black)(""moles"")))) = 3.0154 ~~ 3#

+

#""For H: "" (0.8334 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For C: "" (0.8376 color(red)(cancel(color(black)(""moles""))))/(0.8334color(red)(cancel(color(black)(""moles"")))) = 1.005 ~~ 1#

+
+

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

+
+

#""C""_1""H""_1""Cl""_3 implies color(green)(""CHCl""_3)#

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+
Related questions
+ + +
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Impact of this question
+
+ 37186 views + around the world +
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+ + Creative Commons License + +
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+
" What is empirical formula for Chloroform (89.1% Chlorine, 0.84% Hydrogen, 10.06% Carbon)? nan +150 a82b9928-6ddd-11ea-8fe7-ccda262736ce https://socratic.org/questions/how-do-you-balance-the-equation-for-this-reaction-c-s-so-2-g-cs-2-l-co-2-g 3 C(s) + 2 SO2(g) -> CS2(l) + 2 CO2(g) start chemical_equation qc_end chemical_equation 9 15 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""3 C(s) + 2 SO2(g) -> CS2(l) + 2 CO2(g)""}]" "[{""type"":""chemical equation"",""value"":""C(s) + SO2(g) -> CS2(l) + CO2(g)""}]" "

How do you balance the equation for this reaction: #C(s) +SO_2(g) -> CS_2(l) + CO_2(g)#?

" nan 3 C(s) + 2 SO2(g) -> CS2(l) + 2 CO2(g) "
+

Explanation:

+
+

#C(s) + SO_2 (g) → CS_2 (l) + CO_2 (g)#

+

We have:
+LHS
+C= 1
+S= 1
+O=2

+

RHS
+C=2
+S=2
+O=2

+

We balance the equation by making the number of each element the same on both sides.

+

So:
+LHS
+C= #1xx3# = 3
+S= #1xx2#= 2
+O= #2xx2#= 4

+

RHS
+C= 1 (from #CS_2#) + #(2xx1)# (from the #CO_2#) = 3
+S= 2
+O=#2xx2=4#

+

Then we rewrite the equation:

+

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

+

Hope this isn't too confusing!

+
+
" "
+
+
+

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

+
+
+
+

Explanation:

+
+

#C(s) + SO_2 (g) → CS_2 (l) + CO_2 (g)#

+

We have:
+LHS
+C= 1
+S= 1
+O=2

+

RHS
+C=2
+S=2
+O=2

+

We balance the equation by making the number of each element the same on both sides.

+

So:
+LHS
+C= #1xx3# = 3
+S= #1xx2#= 2
+O= #2xx2#= 4

+

RHS
+C= 1 (from #CS_2#) + #(2xx1)# (from the #CO_2#) = 3
+S= 2
+O=#2xx2=4#

+

Then we rewrite the equation:

+

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

+

Hope this isn't too confusing!

+
+
+
" "
+

How do you balance the equation for this reaction: #C(s) +SO_2(g) -> CS_2(l) + CO_2(g)#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+2 Answers +
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+ +
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+ +
+ + May 10, 2017 + +
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#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

+
+
+
+

Explanation:

+
+

#C(s) + SO_2 (g) → CS_2 (l) + CO_2 (g)#

+

We have:
+LHS
+C= 1
+S= 1
+O=2

+

RHS
+C=2
+S=2
+O=2

+

We balance the equation by making the number of each element the same on both sides.

+

So:
+LHS
+C= #1xx3# = 3
+S= #1xx2#= 2
+O= #2xx2#= 4

+

RHS
+C= 1 (from #CS_2#) + #(2xx1)# (from the #CO_2#) = 3
+S= 2
+O=#2xx2=4#

+

Then we rewrite the equation:

+

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

+

Hope this isn't too confusing!

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + May 10, 2017 + +
+
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+
+

#3C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

+
+
+
+

Explanation:

+
+

Given: #C(s) + SO_2(g) -> CS_2(l) + CO_2(g)#

+

According to the Law of Conservation of mass, the number of atoms of each type on the left side of the equation must be equal to the number of atoms on the right side of the equation:

+

#ul(Left"" "" Side) "" "" ul(Right"" "" Side)#
+#C - 1 "" "" C - 2#
+#S - 1 "" "" S - 2#
+#O - 2"" "" O - 2#

+

You can't change formulas, only add a different number of molecules (coefficient). Since the Carbon atoms are in two different molecules on the right, start with balancing the Sulfur:

+

#C(s) + 2SO_2(g) -> CS_2(l) + CO_2(g)#

+

#ul(Left"" "" Side) "" "" ul(Right"" "" Side)#
+#C - 1 "" "" C - 2#
+#S - 2 "" "" S - 2#
+#O - 4"" "" O - 2#

+

Add a #2# in front of the #CO_2# to balance the Oxygen:

+

#C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

+

#ul(Left"" "" Side) "" "" ul(Right"" "" Side)#
+#C - 1 "" "" C - 3#
+#S - 2 "" "" S - 2#
+#O - 4"" "" O - 4#

+

Finally, put a #3# in front of the single Carbon:

+

#3C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

+

#ul(Left"" "" Side) "" "" ul(Right"" "" Side)#
+#C - 3 "" "" C - 3#
+#S - 2 "" "" S - 2#
+#O - 4"" "" O - 4#

+

Balance!

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+
Related questions
+ + +
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Impact of this question
+
+ 1866 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How do you balance the equation for this reaction: #C(s) +SO_2(g) -> CS_2(l) + CO_2(g)#? nan +151 a82b9929-6ddd-11ea-8025-ccda262736ce https://socratic.org/questions/57dfe8bb7c01496a3b6b8573 2 Na + 2 NH3 -> 2 NaNH2 + H2 start chemical_equation qc_end substance 3 3 qc_end substance 7 8 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 Na + 2 NH3 -> 2 NaNH2 + H2""}]" "[{""type"":""substance name"",""value"":""Sodium""},{""type"":""substance name"",""value"":""Liquid ammonia""}]" "

What occurs when sodium is added to liquid ammonia?

" nan 2 Na + 2 NH3 -> 2 NaNH2 + H2 "
+

Explanation:

+
+

In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.

+

Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).

+
+
" "
+
+
+

#Na + NH_3 rarr NaNH_2 + 1/2H_2#

+
+
+
+

Explanation:

+
+

In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.

+

Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).

+
+
+
" "
+

What occurs when sodium is added to liquid ammonia?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
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+
+
+1 Answer +
+
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+ +
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+ + Sep 19, 2016 + +
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#Na + NH_3 rarr NaNH_2 + 1/2H_2#

+
+
+
+

Explanation:

+
+

In ammoniacal solution, sodium metal eventually produces stoichiometric quantities sodium amide and dihydrogen gas. When the metal is introduced to the ammonia, a deep blue colour developes that is attributed to the solvated electron that eventually reduces the hydrogen of ammonia. Sometimes electrons couple to give a bronze colour. Preparative reactions in ammonia generally use a bit of iron salts to facilitate the electron transfer.

+

Note that the given equation is precisely equivalent to the reaction of water with sodium metal (in water, the lifetime of the solvated electron is much, much shorter!).

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3329 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What occurs when sodium is added to liquid ammonia? nan +152 a82b992a-6ddd-11ea-b5f7-ccda262736ce https://socratic.org/questions/what-is-the-solubility-of-silver-nitrate-if-only-11-1-g-can-dissolve-in-5-0-g-wa 222.00 g/(100 g H2O) start physical_unit 5 6 solubility g qc_end physical_unit 5 6 18 19 temperature qc_end physical_unit 5 6 9 10 mass qc_end physical_unit 16 16 14 15 mass qc_end end "[{""type"":""physical unit"",""value"":""Solubility [OF] silver nitrate [IN] g/(100 g H2O)""}]" "[{""type"":""physical unit"",""value"":""222.00 g/(100 g H2O)""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] silver nitrate [=] \\pu{20 ℃}""},{""type"":""physical unit"",""value"":""Mass [OF] silver nitrate [=] \\pu{11.1 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{5.0 g}""}]" "

What is the solubility of silver nitrate if only 11.1 g can dissolve in 5.0 g water at 20°C?

" nan 222.00 g/(100 g H2O) "
+

Explanation:

+
+

A compound's solubility in water is usually expressed in grams per #""100 g""# of water, or, alternatively, in grams per #""100 mL""# of water.

+

In your case, you can use the information given to you to find the solubility of silver nitrate, #""AgNO""_3#, in water at #20^@""C""#, in grams per #""100 g""# of water, #""g/100 g H""_2""O""#.

+

So, you know that you can dissolve #""11.1 g""# of silver nitrate in #""5.0 g""# of water at #20^@""C""#. You can use this as a conversion factor to determine how much silver nitrate can be dissolved in #""100 g""# of water at the same temperature

+
+

#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#

+
+

This means that the solubility of the salt will be

+
+

#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve #""11.1 g""# of silver nitrate.

+
+
" "
+
+
+

#""220 g/100 g H""_2""O""#

+
+
+
+

Explanation:

+
+

A compound's solubility in water is usually expressed in grams per #""100 g""# of water, or, alternatively, in grams per #""100 mL""# of water.

+

In your case, you can use the information given to you to find the solubility of silver nitrate, #""AgNO""_3#, in water at #20^@""C""#, in grams per #""100 g""# of water, #""g/100 g H""_2""O""#.

+

So, you know that you can dissolve #""11.1 g""# of silver nitrate in #""5.0 g""# of water at #20^@""C""#. You can use this as a conversion factor to determine how much silver nitrate can be dissolved in #""100 g""# of water at the same temperature

+
+

#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#

+
+

This means that the solubility of the salt will be

+
+

#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#

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+

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve #""11.1 g""# of silver nitrate.

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+
+
" "
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What is the solubility of silver nitrate if only 11.1 g can dissolve in 5.0 g water at 20°C?

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+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Solubility Equilbria + + +
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+1 Answer +
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+ + May 27, 2016 + +
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#""220 g/100 g H""_2""O""#

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Explanation:

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A compound's solubility in water is usually expressed in grams per #""100 g""# of water, or, alternatively, in grams per #""100 mL""# of water.

+

In your case, you can use the information given to you to find the solubility of silver nitrate, #""AgNO""_3#, in water at #20^@""C""#, in grams per #""100 g""# of water, #""g/100 g H""_2""O""#.

+

So, you know that you can dissolve #""11.1 g""# of silver nitrate in #""5.0 g""# of water at #20^@""C""#. You can use this as a conversion factor to determine how much silver nitrate can be dissolved in #""100 g""# of water at the same temperature

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+

#100color(red)(cancel(color(black)(""g H""_2""O""))) * ""11.1 g AgNO""_3/(5.0color(red)(cancel(color(black)(""g H""_2""O"")))) = ""222 g""#

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+

This means that the solubility of the salt will be

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+

#""solubility AgNO""_3 = color(green)(|bar(ul(color(white)(a/a)color(black)(""220 g/100 g H""_2""O"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve #""11.1 g""# of silver nitrate.

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Related questions
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" What is the solubility of silver nitrate if only 11.1 g can dissolve in 5.0 g water at 20°C? nan +153 a82bc02e-6ddd-11ea-adc1-ccda262736ce https://socratic.org/questions/what-is-the-hydrogen-ion-concentration-in-a-urine-specimen-that-registers-a-ph-o 1.00 × 10^(-4) mol/L start physical_unit 3 4 concentration mol/l qc_end physical_unit 7 9 15 15 ph qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] hydrogen ion [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.00 × 10^(-4) mol/L""}]" "[{""type"":""physical unit"",""value"":""pH [OF] a urine specimen [=] \\pu{4}""}]" "

What is the hydrogen ion concentration in a urine specimen that registers a pH of 4 on a strip of pH paper?

" nan 1.00 × 10^(-4) mol/L "
+

Explanation:

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+

And I didn't even need a calculator.........

+

By definition, #pH=-log_10[H_3O^+]#.

+

And if #pH=4#, then...........

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#[H_3O^+]=10^-4*mol*L^-1=0.0001*mol*L^-1.#

+

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......

+

#H_3O^+ + HO^(-) rarr2H_2O#; #K_w=[H_3O^+][HO^-]=10^(-14)#. Of course, this value HAS to be measured, and here #K_w# is measured under standard conditions of #298*K# and (almost) #1*atm#.

+

Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product #axxb=10^(log_10a+log_10b)#, and it was easier to do additions than multiplications, even if we had to take antilogs to get the final product.

+

These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).

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" "
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+
+

Well, #[H_3O^+]=1xx10^-4*mol*L^-1.......#

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+
+

Explanation:

+
+

And I didn't even need a calculator.........

+

By definition, #pH=-log_10[H_3O^+]#.

+

And if #pH=4#, then...........

+

#[H_3O^+]=10^-4*mol*L^-1=0.0001*mol*L^-1.#

+

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......

+

#H_3O^+ + HO^(-) rarr2H_2O#; #K_w=[H_3O^+][HO^-]=10^(-14)#. Of course, this value HAS to be measured, and here #K_w# is measured under standard conditions of #298*K# and (almost) #1*atm#.

+

Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product #axxb=10^(log_10a+log_10b)#, and it was easier to do additions than multiplications, even if we had to take antilogs to get the final product.

+

These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).

+
+
+
" "
+

What is the hydrogen ion concentration in a urine specimen that registers a pH of 4 on a strip of pH paper?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
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+1 Answer +
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+ +
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+ +
+ + Jul 17, 2017 + +
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Well, #[H_3O^+]=1xx10^-4*mol*L^-1.......#

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+
+

Explanation:

+
+

And I didn't even need a calculator.........

+

By definition, #pH=-log_10[H_3O^+]#.

+

And if #pH=4#, then...........

+

#[H_3O^+]=10^-4*mol*L^-1=0.0001*mol*L^-1.#

+

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution.......

+

#H_3O^+ + HO^(-) rarr2H_2O#; #K_w=[H_3O^+][HO^-]=10^(-14)#. Of course, this value HAS to be measured, and here #K_w# is measured under standard conditions of #298*K# and (almost) #1*atm#.

+

Why do we use such an absurd definition? Well, back in the day, approx. 30-40 years ago BEFORE the proliferation of hand-held electronic calculators, students and engineers would routinely use log tables for lengthy calculation involving multiplication and division. because the product #axxb=10^(log_10a+log_10b)#, and it was easier to do additions than multiplications, even if we had to take antilogs to get the final product.

+

These days with a simple electronic calculator (and I bought one for a quid in a discounter's shop a couple of weeks ago) we have access to a computational power that would have astonished both Newton and Gauss (they say that Gauss in particular, a mathematical prodigy, had memorized the log tables so that he could do his calculations).

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+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5801 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the hydrogen ion concentration in a urine specimen that registers a pH of 4 on a strip of pH paper? nan +154 a82bc02f-6ddd-11ea-a094-ccda262736ce https://socratic.org/questions/57f6a1907c01490505386a87 2.70 start physical_unit 7 9 ph none qc_end physical_unit 7 9 5 6 molarity qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] solution of HCl(aq) ""}]" "[{""type"":""physical unit"",""value"":""2.70""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] solution of HCl(aq) [=] \\pu{0.002 mol*L^(−1)}""}]" "

What is #pH# of a #0.002*mol*L^-1# solution of #HCl(aq)#?

" nan 2.70 "
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Explanation:

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#[H_3O^+]# #=# #(10^-2*Lxx0.1*mol*L^-1)/(500*mLxx10^-3L*mL^-1)#

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#=# #2.0xx10^-3mol*L^-1#

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And #-log_10(2.0xx10^-3)# #=# #-(-2.70)=2.70#

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" "
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#pH=-log_10[H_3O^+]#

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Explanation:

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#[H_3O^+]# #=# #(10^-2*Lxx0.1*mol*L^-1)/(500*mLxx10^-3L*mL^-1)#

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#=# #2.0xx10^-3mol*L^-1#

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And #-log_10(2.0xx10^-3)# #=# #-(-2.70)=2.70#

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" "
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What is #pH# of a #0.002*mol*L^-1# solution of #HCl(aq)#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
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+1 Answer +
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+ + Oct 6, 2016 + +
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#pH=-log_10[H_3O^+]#

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Explanation:

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#[H_3O^+]# #=# #(10^-2*Lxx0.1*mol*L^-1)/(500*mLxx10^-3L*mL^-1)#

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#=# #2.0xx10^-3mol*L^-1#

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And #-log_10(2.0xx10^-3)# #=# #-(-2.70)=2.70#

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Related questions
+ + +
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+
Impact of this question
+
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+
" What is #pH# of a #0.002*mol*L^-1# solution of #HCl(aq)#? nan +155 a82bc030-6ddd-11ea-914e-ccda262736ce https://socratic.org/questions/58ff807bb72cff35c6f2b0cc 39.87 L start physical_unit 8 9 volume l qc_end physical_unit 8 9 6 7 mole qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] dioxygen gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""39.87 L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] dioxygen gas [=] \\pu{1.78 mol}""},{""type"":""other"",""value"":""STP""}]" "

What volume would be occupied by #1.78*mol# dioxygen gas under conditions of #""STP""#?

" nan 39.87 L "
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Explanation:

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+

The molar volume at #""STP""# varies from syllabus to syllabus. In general, #""STP""#, #""standard temperature and pressure""# specifies #1.0*atm#, and #273*K#. It is further known that #1*mol# of an Ideal Gas occupies #22.4*L# at #""STP""#.

+

And thus the #""volume""# is given by the product #""number of moles""xx""molar volume""# if we assume (reasonably) that dioxygen gas behaves ideally under the given conditions.

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#=1.78*cancel(mol)xx22.4*L*cancel(mol^-1)=??L#

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What mass does this volume represent?

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Approx. #40*L#.

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Explanation:

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The molar volume at #""STP""# varies from syllabus to syllabus. In general, #""STP""#, #""standard temperature and pressure""# specifies #1.0*atm#, and #273*K#. It is further known that #1*mol# of an Ideal Gas occupies #22.4*L# at #""STP""#.

+

And thus the #""volume""# is given by the product #""number of moles""xx""molar volume""# if we assume (reasonably) that dioxygen gas behaves ideally under the given conditions.

+

#=1.78*cancel(mol)xx22.4*L*cancel(mol^-1)=??L#

+

What mass does this volume represent?

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" "
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What volume would be occupied by #1.78*mol# dioxygen gas under conditions of #""STP""#?

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+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
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+2 Answers +
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+ + Apr 25, 2017 + +
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Approx. #40*L#.

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Explanation:

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+

The molar volume at #""STP""# varies from syllabus to syllabus. In general, #""STP""#, #""standard temperature and pressure""# specifies #1.0*atm#, and #273*K#. It is further known that #1*mol# of an Ideal Gas occupies #22.4*L# at #""STP""#.

+

And thus the #""volume""# is given by the product #""number of moles""xx""molar volume""# if we assume (reasonably) that dioxygen gas behaves ideally under the given conditions.

+

#=1.78*cancel(mol)xx22.4*L*cancel(mol^-1)=??L#

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What mass does this volume represent?

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#39.87"" L""#

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Explanation:

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At Standard Temperature and Pressure, #""STP"" (0^(@)""C"" and ""1 atm"")#, #""1 mole""# of any ideal gas occupies #22.4"" Liters of Volume""#.

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#color(white)(aaaaaaaaaaaaaaaaa)(22.4"" L"")/(1"" mol"")#

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Knowing this, we can think about this for a second without involving any math or calculations.

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If we know #1"" mole""# of a gas occupies #22.4"" Liters of volume""# at #STP#, any number of moles more than #1# would occupy more than #22.4"" Liters of volume""#. How much more volume? Well.

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#color(white)(aaaaaaaaaaaaaaaaa)(1.78 cancel(""moles""))/(1 cancel(""mole"")) = 1.78#

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This means, there is #1.78""X""# more number of moles. So, since we know all else is held constant (#""Temperature and Pressure""#), #""moles""# and #""Volume""# are directly proportional here so #""Volume""# will increase by the same magnitude #(""Avogadro's Law"")#.

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#1.78 * 22.4"" L"" = color(blue)(39.87"" L""#

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Answer: 39.87 L

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Related questions
+ + +
+
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Impact of this question
+
+ 1533 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" "What volume would be occupied by #1.78*mol# dioxygen gas under conditions of #""STP""#?" nan +156 a82bc031-6ddd-11ea-a8bb-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-2-25-moles-of-sulfuric-acid 220.68 g start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sulfuric acid [IN] g""}]" "[{""type"":""physical unit"",""value"":""220.68 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sulfuric acid [=] \\pu{2.25 moles}""}]" "

What is the mass of 2.25 moles of sulfuric acid?

" nan 220.68 g "
+

Explanation:

+
+

A useful piece of information to have when looking for the mass of #2.25# moles of sulfuric acid, #""H""_2""SO""_4#, is the mass of #1# mole of the compound.

+

Once you know the mass of #1# mole, you can use it as a conversion factor to find the mass you're looking for.

+

So, the mass of #1# mole of a compound is given by that compound's molar mass. Sulfuric acid has a molar mass of

+
+

#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#

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+

This implies that every mole of sulfuric acid has a mass of #""98.08 g""#. In your case, the mass of #2.25# moles would be equal to

+
+

#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#

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+

The answer is rounded to three sig figs.

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#""221 g""#

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Explanation:

+
+

A useful piece of information to have when looking for the mass of #2.25# moles of sulfuric acid, #""H""_2""SO""_4#, is the mass of #1# mole of the compound.

+

Once you know the mass of #1# mole, you can use it as a conversion factor to find the mass you're looking for.

+

So, the mass of #1# mole of a compound is given by that compound's molar mass. Sulfuric acid has a molar mass of

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#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#

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+

This implies that every mole of sulfuric acid has a mass of #""98.08 g""#. In your case, the mass of #2.25# moles would be equal to

+
+

#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

What is the mass of 2.25 moles of sulfuric acid?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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#""221 g""#

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Explanation:

+
+

A useful piece of information to have when looking for the mass of #2.25# moles of sulfuric acid, #""H""_2""SO""_4#, is the mass of #1# mole of the compound.

+

Once you know the mass of #1# mole, you can use it as a conversion factor to find the mass you're looking for.

+

So, the mass of #1# mole of a compound is given by that compound's molar mass. Sulfuric acid has a molar mass of

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+

#M_(""M H""_2""SO""_4) = ""98.08 g mol""^(-1)#

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This implies that every mole of sulfuric acid has a mass of #""98.08 g""#. In your case, the mass of #2.25# moles would be equal to

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#2.25 color(red)(cancel(color(black)(""moles H""_2""SO""_4))) * overbrace(""98.08 g""/(1color(red)(cancel(color(black)(""mole H""_2""SO""_4)))))^(color(blue)(""molar mass of H""_2""SO""_4)) = color(green)(|bar(ul(color(white)(a/a)color(black)(""221 g"")color(white)(a/a)|)))#

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+

The answer is rounded to three sig figs.

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Related questions
+ + +
+
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Impact of this question
+
+ 12845 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" What is the mass of 2.25 moles of sulfuric acid? nan +157 a82bc032-6ddd-11ea-9e0b-ccda262736ce https://socratic.org/questions/a-sample-of-carbon-dioxide-gas-contains-6-0-22-molecules-how-many-moles-of-carbo 0.10 moles start physical_unit 3 4 mole mol qc_end physical_unit 3 5 7 9 number qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] carbon dioxide [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.10 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] carbon dioxide gas [=] \\pu{6 ⋅ 10^22}""}]" "

A sample of carbon dioxide gas contains #6*10^22# molecules. How many moles of carbon dioxide does this represent?

" nan 0.10 moles "
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Explanation:

+
+

#""1 mol CO""_2=6.022xx10^(23) ""molecules CO""_2""#

+

Use the equality above to determine the moles of carbon dioxide as shown below.

+

#6.022xx10^22 cancel""molecules CO""_2xx(1 ""mol CO""_2)/(6.022xx10^23 cancel""molecules CO""_2)=""0.1000 mol CO""_2""# rounded to four significant figures

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+
" "
+
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+

The answer is #""0.1000 mol CO""_2""#.

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+

Explanation:

+
+

#""1 mol CO""_2=6.022xx10^(23) ""molecules CO""_2""#

+

Use the equality above to determine the moles of carbon dioxide as shown below.

+

#6.022xx10^22 cancel""molecules CO""_2xx(1 ""mol CO""_2)/(6.022xx10^23 cancel""molecules CO""_2)=""0.1000 mol CO""_2""# rounded to four significant figures

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+
+
" "
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A sample of carbon dioxide gas contains #6*10^22# molecules. How many moles of carbon dioxide does this represent?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Oct 28, 2016 + +
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The answer is #""0.1000 mol CO""_2""#.

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Explanation:

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#""1 mol CO""_2=6.022xx10^(23) ""molecules CO""_2""#

+

Use the equality above to determine the moles of carbon dioxide as shown below.

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#6.022xx10^22 cancel""molecules CO""_2xx(1 ""mol CO""_2)/(6.022xx10^23 cancel""molecules CO""_2)=""0.1000 mol CO""_2""# rounded to four significant figures

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Related questions
+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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" A sample of carbon dioxide gas contains #6*10^22# molecules. How many moles of carbon dioxide does this represent? nan +158 a82bc033-6ddd-11ea-81cc-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-ph-of-a-0-26-m-methylamine-solution 12.00 start physical_unit 10 11 ph none qc_end physical_unit 10 11 8 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] methylamine solution""}]" "[{""type"":""physical unit"",""value"":""12.00""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] methylamine solution [=] \\pu{0.26 M}""}]" "

How do you calculate the pH of a 0.26 M methylamine solution?

" nan 12.00 "
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Explanation:

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+

..........or at least you need #K_a# for methyl ammonium ion.

+

From this site, we learn that #K_b# for methylamine is #4.4xx10^-4#. I don't know whether the site is right, I will check later.

+

Now in aqueous solution, methylamine undergoes the acid base reaction:

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#H_3CNH_2(aq) + H_2O(l) rarr H_3CNH_3^+ + HO^-#

+

And we write the equilibrium equation in the usual way:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2(aq)])=4.4xx10^-4#

+

So if #x# moles of methylamine associate, we can put in some numbers:

+

#K_b=(x^2)/(0.26-x)=4.4xx10^-4#

+

As is normal, this is a quadratic in #x#, the which I can solve precisely. But since chemists are simple folk, we can make the approximation that #(0.26-x)~=0.26#. We have to justify this approximation later.

+

And so #x_1=sqrt(4.4xx10^-4xx0.26)=0.0107#.

+

#x_1#, our first approx. was indeed small compared to #0.26*mol*L^-1#, but we can recycle this approximation to see how good it was, #x_2=0.0105#, and #x_3=0.0105#. Since #x_3# has converged, this is equal to the solution we would have got by the quadratic equation.

+

So, #[HO^-]=0.0105*mol*L^-1#; #pOH=-log_10[HO^-]=#

+

#-log_10(0.0105)=1.98#

+

But #pH+pOH=14# in aqueous solution, and thus #pH=12.0#

+

Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.

+
+
" "
+
+
+

Well, for a start you need #K_b#, the base association constant of methylamine........and finally #pH=12.0#

+
+
+
+

Explanation:

+
+

..........or at least you need #K_a# for methyl ammonium ion.

+

From this site, we learn that #K_b# for methylamine is #4.4xx10^-4#. I don't know whether the site is right, I will check later.

+

Now in aqueous solution, methylamine undergoes the acid base reaction:

+

#H_3CNH_2(aq) + H_2O(l) rarr H_3CNH_3^+ + HO^-#

+

And we write the equilibrium equation in the usual way:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2(aq)])=4.4xx10^-4#

+

So if #x# moles of methylamine associate, we can put in some numbers:

+

#K_b=(x^2)/(0.26-x)=4.4xx10^-4#

+

As is normal, this is a quadratic in #x#, the which I can solve precisely. But since chemists are simple folk, we can make the approximation that #(0.26-x)~=0.26#. We have to justify this approximation later.

+

And so #x_1=sqrt(4.4xx10^-4xx0.26)=0.0107#.

+

#x_1#, our first approx. was indeed small compared to #0.26*mol*L^-1#, but we can recycle this approximation to see how good it was, #x_2=0.0105#, and #x_3=0.0105#. Since #x_3# has converged, this is equal to the solution we would have got by the quadratic equation.

+

So, #[HO^-]=0.0105*mol*L^-1#; #pOH=-log_10[HO^-]=#

+

#-log_10(0.0105)=1.98#

+

But #pH+pOH=14# in aqueous solution, and thus #pH=12.0#

+

Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.

+
+
+
" "
+

How do you calculate the pH of a 0.26 M methylamine solution?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
+
+
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+1 Answer +
+
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+ +
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+ +
+ + Jan 2, 2017 + +
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Well, for a start you need #K_b#, the base association constant of methylamine........and finally #pH=12.0#

+
+
+
+

Explanation:

+
+

..........or at least you need #K_a# for methyl ammonium ion.

+

From this site, we learn that #K_b# for methylamine is #4.4xx10^-4#. I don't know whether the site is right, I will check later.

+

Now in aqueous solution, methylamine undergoes the acid base reaction:

+

#H_3CNH_2(aq) + H_2O(l) rarr H_3CNH_3^+ + HO^-#

+

And we write the equilibrium equation in the usual way:

+

#K_b=([H_3CNH_3^+][HO^-])/([H_3CNH_2(aq)])=4.4xx10^-4#

+

So if #x# moles of methylamine associate, we can put in some numbers:

+

#K_b=(x^2)/(0.26-x)=4.4xx10^-4#

+

As is normal, this is a quadratic in #x#, the which I can solve precisely. But since chemists are simple folk, we can make the approximation that #(0.26-x)~=0.26#. We have to justify this approximation later.

+

And so #x_1=sqrt(4.4xx10^-4xx0.26)=0.0107#.

+

#x_1#, our first approx. was indeed small compared to #0.26*mol*L^-1#, but we can recycle this approximation to see how good it was, #x_2=0.0105#, and #x_3=0.0105#. Since #x_3# has converged, this is equal to the solution we would have got by the quadratic equation.

+

So, #[HO^-]=0.0105*mol*L^-1#; #pOH=-log_10[HO^-]=#

+

#-log_10(0.0105)=1.98#

+

But #pH+pOH=14# in aqueous solution, and thus #pH=12.0#

+

Now I admit this does seem like a lot of work (it was more work for me, because I had to format the equations on this editor!). But I can assure you that you can get very proficient at these sorts of problems, especially if you can work your calculator competently. Remember the approach, approximate and then justify. And then recycle the approximation. Good luck.

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+
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" How do you calculate the pH of a 0.26 M methylamine solution? nan +159 a82be7c2-6ddd-11ea-bf2b-ccda262736ce https://socratic.org/questions/calculate-the-number-of-moles-of-solute-in-497-2-l-of-0-815-m-sodium-cyanide 405.22 moles start physical_unit 6 6 mole mol qc_end physical_unit 13 14 11 12 molarity qc_end physical_unit 13 14 8 9 volume qc_end end "[{""type"":""physical unit"",""value"":""Number of moles [OF] solute [IN] moles""}]" "[{""type"":""physical unit"",""value"":""405.22 moles""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] sodium cyanide [=] \\pu{0.815 M}""},{""type"":""physical unit"",""value"":""Volume [OF] sodium cyanide [=] \\pu{497.2 L}""}]" "

Calculate the number of moles of solute in 497.2 L of 0.815 M sodium cyanide?

" nan 405.22 moles "
+

Explanation:

+
+

It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.

+

Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.

+

(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).

+
+
" "
+
+
+

405.218 moles

+
+
+
+

Explanation:

+
+

It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.

+

Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.

+

(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).

+
+
+
" "
+

Calculate the number of moles of solute in 497.2 L of 0.815 M sodium cyanide?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
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+1 Answer +
+
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+ +
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+ +
+ + Jun 30, 2017 + +
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+

405.218 moles

+
+
+
+

Explanation:

+
+

It is a 0.815 M solution, so that means that in 1 litre you have 0.815 moles of solute.

+

Therefore in 497.2 litres of the solution you have 497.2 x 0.815 = 405.218 moles of solute.

+

(By the way, the correct terminology is ""0.815 M sodium cyanide solution"" - saying ""0.815 M sodium cyanide"" is meaningless).

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+ +
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+
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+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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+
" Calculate the number of moles of solute in 497.2 L of 0.815 M sodium cyanide? nan +160 a82be7c3-6ddd-11ea-8ece-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-hcl-in-a-500-ml-sample-of-6-m-hcl 109.38 g start physical_unit 5 5 mass g qc_end physical_unit 5 5 12 13 molarity qc_end physical_unit 5 5 8 9 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] HCl [IN] g""}]" "[{""type"":""physical unit"",""value"":""109.38 g""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HCl [=] \\pu{6.0 M}""},{""type"":""physical unit"",""value"":""Volume [OF] HCl [=] \\pu{500 mL}""}]" "

What is the mass of #HCl# in a 500 mL sample of 6.0 M #HCl#?

" nan 109.38 g "
+

Explanation:

+
+

Molarity is represented by the following equation:
+

+

In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.

+

Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
+Moles of solute = (liters of solution) * (Molarity)

+

Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl

+

Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.

+

(3.0 mol)(36.46 g/mol) = 109 g HCl

+
+
" "
+
+
+

109 g of HCl

+
+
+
+

Explanation:

+
+

Molarity is represented by the following equation:
+

+

In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.

+

Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
+Moles of solute = (liters of solution) * (Molarity)

+

Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl

+

Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.

+

(3.0 mol)(36.46 g/mol) = 109 g HCl

+
+
+
" "
+

What is the mass of #HCl# in a 500 mL sample of 6.0 M #HCl#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
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+1 Answer +
+
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+ + +
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+ +
+ + Jun 5, 2016 + +
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+

109 g of HCl

+
+
+
+

Explanation:

+
+

Molarity is represented by the following equation:
+

+

In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.

+

Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
+Moles of solute = (liters of solution) * (Molarity)

+

Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl

+

Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.

+

(3.0 mol)(36.46 g/mol) = 109 g HCl

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+ + +
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+
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" What is the mass of #HCl# in a 500 mL sample of 6.0 M #HCl#? nan +161 a82c10c6-6ddd-11ea-983c-ccda262736ce https://socratic.org/questions/58f6e7b6b72cff45e0bf2c5b +4 start physical_unit 6 6 oxidation_number none qc_end chemical_equation 10 10 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] phosphorus in the salt Zr3P4O16""}]" "[{""type"":""physical unit"",""value"":""+4""}]" "[{""type"":""chemical equation"",""value"":""Zr3P4O16""}]" "

What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#?

" nan +4 "
+

Explanation:

+
+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

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Well, the parent phosphate ion is #PO_4^(3-)#...........

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Explanation:

+
+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

+
+
+
" "
+

What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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+1 Answer +
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+ + Apr 19, 2017 + +
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Well, the parent phosphate ion is #PO_4^(3-)#...........

+
+
+
+

Explanation:

+
+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

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+
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+
Related questions
+ + +
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+
Impact of this question
+
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" What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#? nan +162 a82c3768-6ddd-11ea-a578-ccda262736ce https://socratic.org/questions/it-takes-770-joules-energy-to-raise-the-temperature-of-50-0-g-of-mercury-by-110- 0.14 J/(℃ · g) start physical_unit 13 13 specific_heat j/(°c_·_g) qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 13 13 15 16 temperature qc_end physical_unit 13 13 2 3 energy qc_end end "[{""type"":""physical unit"",""value"":""Specific heat [OF] mercury [IN] J/(℃ · g)""}]" "[{""type"":""physical unit"",""value"":""0.14 J/(℃ · g)""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] mercury [=] \\pu{50.0 g}""},{""type"":""physical unit"",""value"":""Temperature raised [OF] mercury [=] \\pu{110 °C}""},{""type"":""physical unit"",""value"":""Energy [OF] mercury [=] \\pu{770 joules}""}]" "

It takes 770 joules energy to raise the temperature of 50.0 g of mercury by 110°C. What is the specific heat of mercury?

" nan 0.14 J/(℃ · g) "
+

Explanation:

+
+
+

The formula for the heat absorbed by a substance is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#

+
+
+

where

+

#q# is the quantity of heat

+

#m# is the mass of the substance

+

#c# is the specific heat capacity of the material

+

#ΔT# is the temperature change

+
+

You can rearrange the formula to calculate the specific heat capacity:

+

#c = q/(mΔT)#

+

#c = ""770 J""/""50.0 g × 110 °C"" = ""0.140 J·°C""^""-1""""g""^""-1""#

+
+
" "
+
+
+

The specific heat capacity of mercury is #""0.140 J·°C""^""-1""""g""^""-1""#.

+
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+
+

Explanation:

+
+
+

The formula for the heat absorbed by a substance is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#

+
+
+

where

+

#q# is the quantity of heat

+

#m# is the mass of the substance

+

#c# is the specific heat capacity of the material

+

#ΔT# is the temperature change

+
+

You can rearrange the formula to calculate the specific heat capacity:

+

#c = q/(mΔT)#

+

#c = ""770 J""/""50.0 g × 110 °C"" = ""0.140 J·°C""^""-1""""g""^""-1""#

+
+
+
" "
+

It takes 770 joules energy to raise the temperature of 50.0 g of mercury by 110°C. What is the specific heat of mercury?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
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+1 Answer +
+
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+ +
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+ +
+ + Apr 11, 2016 + +
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The specific heat capacity of mercury is #""0.140 J·°C""^""-1""""g""^""-1""#.

+
+
+
+

Explanation:

+
+
+

The formula for the heat absorbed by a substance is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))"" ""#

+
+
+

where

+

#q# is the quantity of heat

+

#m# is the mass of the substance

+

#c# is the specific heat capacity of the material

+

#ΔT# is the temperature change

+
+

You can rearrange the formula to calculate the specific heat capacity:

+

#c = q/(mΔT)#

+

#c = ""770 J""/""50.0 g × 110 °C"" = ""0.140 J·°C""^""-1""""g""^""-1""#

+
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" It takes 770 joules energy to raise the temperature of 50.0 g of mercury by 110°C. What is the specific heat of mercury? nan +163 a82d2512-6ddd-11ea-9d38-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-for-magnesium-hydroxide-milk-of-magnesia Mg(OH)2 start chemical_formula qc_end substance 6 10 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Mg(OH)2""}]" "[{""type"":""substance name"",""value"":""Magnesium hydroxide (milk of magnesia)""}]" "

What is the empirical formula for magnesium hydroxide (milk of magnesia)?

" nan Mg(OH)2 "
+

Explanation:

+
+

Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to #Mg^(2+)#. Since hydroxide has formulation of (#OH^-#), formula for magnesium hydroxide is #Mg(OH)_2#.

+

What are the formulae for calcium hydroxide and barium hydroxides?

+
+
" "
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+
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#Mg(OH)_2#. Look at its position on the Periodic Table.

+
+
+
+

Explanation:

+
+

Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to #Mg^(2+)#. Since hydroxide has formulation of (#OH^-#), formula for magnesium hydroxide is #Mg(OH)_2#.

+

What are the formulae for calcium hydroxide and barium hydroxides?

+
+
+
" "
+

What is the empirical formula for magnesium hydroxide (milk of magnesia)?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
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+2 Answers +
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+ + Sep 13, 2015 + +
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#Mg(OH)_2#. Look at its position on the Periodic Table.

+
+
+
+

Explanation:

+
+

Magnesium metal lies in Group II of the Periodic Table; it readily loses 2 electrons to give rise to #Mg^(2+)#. Since hydroxide has formulation of (#OH^-#), formula for magnesium hydroxide is #Mg(OH)_2#.

+

What are the formulae for calcium hydroxide and barium hydroxides?

+
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+ + Oct 26, 2015 + +
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#""Mg""(""OH"")_2#

+
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+
+

Explanation:

+
+

The thing to remember when dealing with ionic compounds is that their empirical formula is equivalent to their chemical formula .

+

In other words, if you know an ionic compound's chemical formula, you also know its empirical formula.

+

A compound's empirical formula will tell you what the smallest whole number ratio is for the atoms of the elements that make up that compound.

+

In the case of ionic compounds, you know that their chemical formula is given by the formula unit, which tells you the smallest whole number ratio that exist between the cations and anions that form said compound.

+

You know that magnesium hydroxide is composed of magnesium cations, #""Mg""^(2+)#, and hydroxide anions, #""OH""^(-)#.

+

In order to blance the positive charge of the cations, the formula unit of magnesium hydroxide will consist of two hydroxide anions.

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#""Mg""^(2+) + 2""OH""^(-) -> ""Mg""(""OH"")_2#

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This is also the compound's empirical formula.

+

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Related questions
+ + +
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+
" What is the empirical formula for magnesium hydroxide (milk of magnesia)? nan +164 a82d46e2-6ddd-11ea-8825-ccda262736ce https://socratic.org/questions/how-do-you-balance-hgo-hg-o-2 2 HgO -> 2 Hg + O2 start chemical_equation qc_end chemical_equation 4 8 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 HgO -> 2 Hg + O2""}]" "[{""type"":""chemical equation"",""value"":""HgO -> Hg + O2""}]" "

How do you balance #HgO -> Hg + O_2#?

" nan 2 HgO -> 2 Hg + O2 "
+

Explanation:

+
+

How would you remove the half coefficient on the oxygen? What does the #Delta# symbol mean here?

+
+
" "
+
+
+

#HgO + Delta rarr Hg(l) + 1/2O_2(g)uarr#

+
+
+
+

Explanation:

+
+

How would you remove the half coefficient on the oxygen? What does the #Delta# symbol mean here?

+
+
+
" "
+

How do you balance #HgO -> Hg + O_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 6, 2016 + +
+
+
+
+
+
+
+

#HgO + Delta rarr Hg(l) + 1/2O_2(g)uarr#

+
+
+
+

Explanation:

+
+

How would you remove the half coefficient on the oxygen? What does the #Delta# symbol mean here?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2516 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do you balance #HgO -> Hg + O_2#? nan +165 a82d4f5a-6ddd-11ea-936c-ccda262736ce https://socratic.org/questions/a-sample-of-argon-gas-occupies-a-volume-of-950-ml-at-25-0-c-what-volume-will-the 1029.70 mL start physical_unit 17 18 volume ml qc_end physical_unit 0 4 9 10 volume qc_end physical_unit 0 4 12 13 temperature qc_end physical_unit 17 18 21 22 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] mL""}]" "[{""type"":""physical unit"",""value"":""1029.70 mL""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of argon gas [=] \\pu{950 mL}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] A sample of argon gas [=] \\pu{25.0 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the gas [=] \\pu{50.0 °C}""},{""type"":""other"",""value"":""The pressure remains constant.""}]" "

A sample of argon gas occupies a volume of 950 mL at 25.0°C. What volume will the gas occupy at 50.0°C if the pressure remains constant?

" nan 1029.70 mL "
+

Explanation:

+
+

Formula P1 x V1 / T1 = P2 x V2 / T2

+

Fill in what you know

+

Pressure is constant so no need to put that in making the formula
+V1 / T1 = V2 / T2

+

Voulme 1= 950 mL
+Volume 2= ?
+Temperature 1 = 25 C
+Temperature 2 = 50 C

+
    +
  1. +

    Convert your temperature to Kelvin
    +C+273=K
    +Temperature 1 = 25 C + 273 = 298 K
    +Temperature 2 = 50 C + 273 = 323 K

    +
    2. +
  2. +

    Plug in to the Formula
    +950 mL/298 K = ? / 323 K

    +
    3. +
  3. +

    Rearrange the formula to make one to solve for what is missing.
    +To get 323 K out of the denominator multiply by it.
    +Making it
    +950 mL x 323 K / 298 K = ?

    +
    4. +
  4. +

    Plug it in
    +950 mL x 323 K / 298 K = 1027.9 mL

    +
    5. +
+
+
" "
+
+
+

1027.9 mL

+
+
+
+

Explanation:

+
+

Formula P1 x V1 / T1 = P2 x V2 / T2

+

Fill in what you know

+

Pressure is constant so no need to put that in making the formula
+V1 / T1 = V2 / T2

+

Voulme 1= 950 mL
+Volume 2= ?
+Temperature 1 = 25 C
+Temperature 2 = 50 C

+
    +
  1. +

    Convert your temperature to Kelvin
    +C+273=K
    +Temperature 1 = 25 C + 273 = 298 K
    +Temperature 2 = 50 C + 273 = 323 K

    +
    2. +
  2. +

    Plug in to the Formula
    +950 mL/298 K = ? / 323 K

    +
    3. +
  3. +

    Rearrange the formula to make one to solve for what is missing.
    +To get 323 K out of the denominator multiply by it.
    +Making it
    +950 mL x 323 K / 298 K = ?

    +
    4. +
  4. +

    Plug it in
    +950 mL x 323 K / 298 K = 1027.9 mL

    +
    5. +
+
+
+
" "
+

A sample of argon gas occupies a volume of 950 mL at 25.0°C. What volume will the gas occupy at 50.0°C if the pressure remains constant?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 4, 2017 + +
+
+
+
+
+
+
+

1027.9 mL

+
+
+
+

Explanation:

+
+

Formula P1 x V1 / T1 = P2 x V2 / T2

+

Fill in what you know

+

Pressure is constant so no need to put that in making the formula
+V1 / T1 = V2 / T2

+

Voulme 1= 950 mL
+Volume 2= ?
+Temperature 1 = 25 C
+Temperature 2 = 50 C

+
    +
  1. +

    Convert your temperature to Kelvin
    +C+273=K
    +Temperature 1 = 25 C + 273 = 298 K
    +Temperature 2 = 50 C + 273 = 323 K

    +
    2. +
  2. +

    Plug in to the Formula
    +950 mL/298 K = ? / 323 K

    +
    3. +
  3. +

    Rearrange the formula to make one to solve for what is missing.
    +To get 323 K out of the denominator multiply by it.
    +Making it
    +950 mL x 323 K / 298 K = ?

    +
    4. +
  4. +

    Plug it in
    +950 mL x 323 K / 298 K = 1027.9 mL

    +
    5. +
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 23040 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" A sample of argon gas occupies a volume of 950 mL at 25.0°C. What volume will the gas occupy at 50.0°C if the pressure remains constant? nan +166 a82d4f5b-6ddd-11ea-85d4-ccda262736ce https://socratic.org/questions/if-the-quantity-of-metal-in-a-metallic-oxide-is-60-what-is-its-equivalent-weight 12.00 g start physical_unit 4 4 equivalent_weight g qc_end physical_unit 4 8 10 10 quantity qc_end end "[{""type"":""physical unit"",""value"":""Equivalent weight [OF] metal [IN] g""}]" "[{""type"":""physical unit"",""value"":""12.00 g""}]" "[{""type"":""physical unit"",""value"":""Quantity [OF] metal in a metallic oxide [=] \\pu{60%}""}]" "

If the quantity of metal in a metallic oxide is 60%, what is its equivalent weight?

" nan 12.00 g "
+

Explanation:

+
+

The idea here is that you're looking for the mass of this unknown metal that can combine with #""8 g""# of oxygen, which is what the equivalent mass of the metal tells you in this context.

+

To make the calculations easier, pick a #""100-g""# sample of this metal oxide. According to its percent composition, this sample will contain

+
+
    +
  • #60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""#
    • +
  • #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
    • +
+
+

So, you know that #""60 g""# of this unknown metal combine with #""40 g""# of oxygen, which means that #""8 g""# of oxygen will combine with

+
+

#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#

+
+

Therefore, you can say that the equivalent mass of the metal is equal to

+
+

#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#

+
+
+
" "
+
+
+

#""12 g""#

+
+
+
+

Explanation:

+
+

The idea here is that you're looking for the mass of this unknown metal that can combine with #""8 g""# of oxygen, which is what the equivalent mass of the metal tells you in this context.

+

To make the calculations easier, pick a #""100-g""# sample of this metal oxide. According to its percent composition, this sample will contain

+
+
    +
  • #60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""#
    • +
  • #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
    • +
+
+

So, you know that #""60 g""# of this unknown metal combine with #""40 g""# of oxygen, which means that #""8 g""# of oxygen will combine with

+
+

#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#

+
+

Therefore, you can say that the equivalent mass of the metal is equal to

+
+

#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#

+
+
+
+
" "
+

If the quantity of metal in a metallic oxide is 60%, what is its equivalent weight?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 13, 2017 + +
+
+
+
+
+
+
+

#""12 g""#

+
+
+
+

Explanation:

+
+

The idea here is that you're looking for the mass of this unknown metal that can combine with #""8 g""# of oxygen, which is what the equivalent mass of the metal tells you in this context.

+

To make the calculations easier, pick a #""100-g""# sample of this metal oxide. According to its percent composition, this sample will contain

+
+
    +
  • #60%color(white)(.)""metal "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 60 g metal""#
    • +
  • #40%color(white)(.)""oxygen "" stackrel(color(white)(color(blue)(""100 g sample"")aaa))(rarr) "" 40 g oxygen""#
    • +
+
+

So, you know that #""60 g""# of this unknown metal combine with #""40 g""# of oxygen, which means that #""8 g""# of oxygen will combine with

+
+

#8 color(red)(cancel(color(black)(""g oxygen""))) * ""60 g metal""/(40color(red)(cancel(color(black)(""g oxygen"")))) = ""12 g metal""#

+
+

Therefore, you can say that the equivalent mass of the metal is equal to

+
+

#color(darkgreen)(ul(color(black)(""equivalent mass of metal in 60% metal oxide = 12 g"")))#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4929 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" If the quantity of metal in a metallic oxide is 60%, what is its equivalent weight? nan +167 a82d4f5c-6ddd-11ea-b886-ccda262736ce https://socratic.org/questions/how-many-water-molecules-are-in-a-block-of-ice-containing-2-50-mol-of-water 1.51 × 10^24 start physical_unit 2 3 number none qc_end physical_unit 2 2 11 12 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] water molecules""}]" "[{""type"":""physical unit"",""value"":""1.51 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] water [=] \\pu{2.50 mol}""}]" "

How many water molecules are in a block of ice containing 2.50 mol of water?

" nan 1.51 × 10^24 "
+

Explanation:

+
+

This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.

+

In your case, the sample is said to contain #2.50# moles of water. Now, a mole is simply a very, very large collection of molecules. In order to have #1# mole of a molecular compound you need your sample to contain #6.022 * 10^(23)# molecules of that compound.

+
+

#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number

+
+

This is what Avogadro's number is all about -- the number of molecules needed to form #1# mole of a molecular compound.

+

So, now that you know how many molecules are needed to have #1# mole of water, use Avogadro's number as a conversion factor to calculate the number of molecules present in your sample

+
+

#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#1.51 * 10^(24)#

+
+
+
+

Explanation:

+
+

This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.

+

In your case, the sample is said to contain #2.50# moles of water. Now, a mole is simply a very, very large collection of molecules. In order to have #1# mole of a molecular compound you need your sample to contain #6.022 * 10^(23)# molecules of that compound.

+
+

#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number

+
+

This is what Avogadro's number is all about -- the number of molecules needed to form #1# mole of a molecular compound.

+

So, now that you know how many molecules are needed to have #1# mole of water, use Avogadro's number as a conversion factor to calculate the number of molecules present in your sample

+
+

#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

How many water molecules are in a block of ice containing 2.50 mol of water?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 10, 2016 + +
+
+
+
+
+
+
+

#1.51 * 10^(24)#

+
+
+
+

Explanation:

+
+

This is a pretty straightforward example of how to sue Avogadro's number to figure out the number of molecules present in a given sample.

+

In your case, the sample is said to contain #2.50# moles of water. Now, a mole is simply a very, very large collection of molecules. In order to have #1# mole of a molecular compound you need your sample to contain #6.022 * 10^(23)# molecules of that compound.

+
+

#color(blue)(|bar(ul(color(white)(a/a)""1 mole"" = 6.022 * 10^(23)""molecules"" color(white)(a/a)|))) -># Avogadro's number

+
+

This is what Avogadro's number is all about -- the number of molecules needed to form #1# mole of a molecular compound.

+

So, now that you know how many molecules are needed to have #1# mole of water, use Avogadro's number as a conversion factor to calculate the number of molecules present in your sample

+
+

#2.50 color(red)(cancel(color(black)(""moles H""_2""O""))) * (6.022 * 10^(23)""molec. H""_2""O"")/(1color(red)(cancel(color(black)(""mole H""_2""O"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.51 * 10^(24)""molec. H""_2""O"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 35440 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many water molecules are in a block of ice containing 2.50 mol of water? nan +168 a82d4f5d-6ddd-11ea-aa4d-ccda262736ce https://socratic.org/questions/how-many-molecules-of-glucose-are-in-a-xenopus-oocyte-if-the-total-volume-is-10- 6.02 × 10^15 start physical_unit 2 4 number none qc_end physical_unit 4 4 15 16 volume qc_end physical_unit 4 4 23 24 concentration qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] molecules of glucose""}]" "[{""type"":""physical unit"",""value"":""6.02 × 10^15""}]" "[{""type"":""physical unit"",""value"":""Total volume [OF] glucose [=] \\pu{10 ul}""},{""type"":""physical unit"",""value"":""Concentration [OF] glucose [=] \\pu{10 mM}""}]" "

How many molecules of glucose are in a xenopus oocyte if the total volume is 10 ul and the concentration of glucose is 10 mM?

" nan 6.02 × 10^15 "
+

Explanation:

+
+
+

The formula for molarity is:

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#

+
+
+

or

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#

+
+
+

In your problem,

+

#M = ""10 mmol/L"" = 1.0 × 10^""-3""color(white)(l) ""mol·L""^""-1""#
+#V = ""10 µL"" = 1.0 × 10^""-5""color(white)(l) ""L""#

+

∴ #n = MV = (1.0 × 10^""-3""color(white)(l) ""mol"")/(1 color(red)(cancel(color(black)(""L"")))) × 1.0 ×10^""-5"" color(red)(cancel(color(black)(""L""))) = 1.0 × 10^""-8""color(white)(l) ""mol""#

+

Now, we can use Avogadro's number to calculate the number of molecules.

+

#""Molecules"" = 1.0 × 10^""-8"" color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""molecules"")/(1 color(red)(cancel(color(black)(""mol"")))) = 6.0 × 10^15color(white)(l) ""molecules""#

+
+
" "
+
+
+

There are #6.0 × 10^15color(white)(l) ""molecules""# of glucose in the oocyte.

+
+
+
+

Explanation:

+
+
+

The formula for molarity is:

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#

+
+
+

or

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#

+
+
+

In your problem,

+

#M = ""10 mmol/L"" = 1.0 × 10^""-3""color(white)(l) ""mol·L""^""-1""#
+#V = ""10 µL"" = 1.0 × 10^""-5""color(white)(l) ""L""#

+

∴ #n = MV = (1.0 × 10^""-3""color(white)(l) ""mol"")/(1 color(red)(cancel(color(black)(""L"")))) × 1.0 ×10^""-5"" color(red)(cancel(color(black)(""L""))) = 1.0 × 10^""-8""color(white)(l) ""mol""#

+

Now, we can use Avogadro's number to calculate the number of molecules.

+

#""Molecules"" = 1.0 × 10^""-8"" color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""molecules"")/(1 color(red)(cancel(color(black)(""mol"")))) = 6.0 × 10^15color(white)(l) ""molecules""#

+
+
+
" "
+

How many molecules of glucose are in a xenopus oocyte if the total volume is 10 ul and the concentration of glucose is 10 mM?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 21, 2016 + +
+
+
+
+
+
+
+

There are #6.0 × 10^15color(white)(l) ""molecules""# of glucose in the oocyte.

+
+
+
+

Explanation:

+
+
+

The formula for molarity is:

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) ""Molarity"" = ""moles""/""litres""color(white)(a/a)|)))"" ""#

+
+
+

or

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) M = n/Vcolor(white)(a/a)|)))"" ""#

+
+
+

In your problem,

+

#M = ""10 mmol/L"" = 1.0 × 10^""-3""color(white)(l) ""mol·L""^""-1""#
+#V = ""10 µL"" = 1.0 × 10^""-5""color(white)(l) ""L""#

+

∴ #n = MV = (1.0 × 10^""-3""color(white)(l) ""mol"")/(1 color(red)(cancel(color(black)(""L"")))) × 1.0 ×10^""-5"" color(red)(cancel(color(black)(""L""))) = 1.0 × 10^""-8""color(white)(l) ""mol""#

+

Now, we can use Avogadro's number to calculate the number of molecules.

+

#""Molecules"" = 1.0 × 10^""-8"" color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""molecules"")/(1 color(red)(cancel(color(black)(""mol"")))) = 6.0 × 10^15color(white)(l) ""molecules""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1464 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many molecules of glucose are in a xenopus oocyte if the total volume is 10 ul and the concentration of glucose is 10 mM? nan +169 a82d4f5e-6ddd-11ea-850f-ccda262736ce https://socratic.org/questions/assume-that-you-have-2-59-mol-of-aluminum-how-many-atoms-of-aluminum-do-you-have 1.56 × 10^24 start physical_unit 10 12 number none qc_end physical_unit 7 7 4 5 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] atoms of aluminum""}]" "[{""type"":""physical unit"",""value"":""1.56 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] aluminum [=] \\pu{2.59 mol}""}]" "

Assume that you have 2.59 mol of aluminum. How many atoms of aluminum do you have?

" nan 1.56 × 10^24 "
+

Explanation:

+
+

#N_A# #=# #6.022xx10^23#. What are the masses of #1# #mol# of #Al#, and #2.59# #""mol""# aluminum?

+
+
" "
+
+
+

#2.59# #""mols""# of aluminum ATOMS corresponds to #2.59xxN_A#, where #N_A# is Avogadro's number.

+
+
+
+

Explanation:

+
+

#N_A# #=# #6.022xx10^23#. What are the masses of #1# #mol# of #Al#, and #2.59# #""mol""# aluminum?

+
+
+
" "
+

Assume that you have 2.59 mol of aluminum. How many atoms of aluminum do you have?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 7, 2016 + +
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+
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+
+

#2.59# #""mols""# of aluminum ATOMS corresponds to #2.59xxN_A#, where #N_A# is Avogadro's number.

+
+
+
+

Explanation:

+
+

#N_A# #=# #6.022xx10^23#. What are the masses of #1# #mol# of #Al#, and #2.59# #""mol""# aluminum?

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+
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" Assume that you have 2.59 mol of aluminum. How many atoms of aluminum do you have? nan +170 a82d6ed2-6ddd-11ea-9069-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-3-moles-of-barium 411.99 g start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] barium [IN] g""}]" "[{""type"":""physical unit"",""value"":""411.99 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] barium [=] \\pu{3 moles}""}]" "

What is the mass of 3 moles of barium?

" nan 411.99 g "
+

Explanation:

+
+

So if there is a quantity of #3# moles, there is a mass of #3*molxx137.33*g*mol^-1=??g#.

+
+
" "
+
+
+

Barium metal has a molar mass of #137.33*g*mol^-1#.

+
+
+
+

Explanation:

+
+

So if there is a quantity of #3# moles, there is a mass of #3*molxx137.33*g*mol^-1=??g#.

+
+
+
" "
+

What is the mass of 3 moles of barium?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 5, 2016 + +
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+
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+
+
+
+

Barium metal has a molar mass of #137.33*g*mol^-1#.

+
+
+
+

Explanation:

+
+

So if there is a quantity of #3# moles, there is a mass of #3*molxx137.33*g*mol^-1=??g#.

+
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+
+
+ +
+
+
+
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+
+
Related questions
+ + +
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+
Impact of this question
+
+ 3527 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the mass of 3 moles of barium? nan +171 a82d6ed3-6ddd-11ea-ad6f-ccda262736ce https://socratic.org/questions/if-you-dilute-a-solution-that-has-2-95-liters-of-a-8-m-down-to-1-425-l-what-is-t 16.56 M start physical_unit 3 4 molarity mol/l qc_end physical_unit 3 4 11 12 molarity qc_end physical_unit 3 4 7 8 volume qc_end physical_unit 3 4 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity2 [OF] a solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""16.56 M""}]" "[{""type"":""physical unit"",""value"":""Molarity1 [OF] a solution [=] \\pu{8 M}""},{""type"":""physical unit"",""value"":""Volume1 [OF] a solution [=] \\pu{2.95 liters}""},{""type"":""physical unit"",""value"":""Volume2 [OF] a solution [=] \\pu{1.425 L}""}]" "

If you dilute a solution that has 2.95 liters of a 8 M down to 1.425 L, what is the new molarity?

" nan 16.56 M "
+

Explanation:

+
+

The product #""concentration""xx""volume""# has units of #""moles""#.........and given a dilution of a solute, of course the amount of solute will remain constant.

+

In this problem, #V_2# is apparently SMALLER than #V_1#; you have CONCENTRATED the solution by (somehow) removing the solvent. With water this would be a hard thing to do. Would you review this question?

+
+
" "
+
+
+

The old equation is #C_1V_1=C_2V_2#

+

But I think you have worded the question improperly......

+
+
+
+

Explanation:

+
+

The product #""concentration""xx""volume""# has units of #""moles""#.........and given a dilution of a solute, of course the amount of solute will remain constant.

+

In this problem, #V_2# is apparently SMALLER than #V_1#; you have CONCENTRATED the solution by (somehow) removing the solvent. With water this would be a hard thing to do. Would you review this question?

+
+
+
" "
+

If you dilute a solution that has 2.95 liters of a 8 M down to 1.425 L, what is the new molarity?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 5, 2017 + +
+
+
+
+
+
+
+

The old equation is #C_1V_1=C_2V_2#

+

But I think you have worded the question improperly......

+
+
+
+

Explanation:

+
+

The product #""concentration""xx""volume""# has units of #""moles""#.........and given a dilution of a solute, of course the amount of solute will remain constant.

+

In this problem, #V_2# is apparently SMALLER than #V_1#; you have CONCENTRATED the solution by (somehow) removing the solvent. With water this would be a hard thing to do. Would you review this question?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1241 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" If you dilute a solution that has 2.95 liters of a 8 M down to 1.425 L, what is the new molarity? nan +172 a82d6ed4-6ddd-11ea-94f9-ccda262736ce https://socratic.org/questions/what-volume-does-22-4-g-of-chlorine-gas-occupy-at-stp 7.08 L start physical_unit 6 7 volume l qc_end physical_unit 6 7 3 4 mass qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] chlorine gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""7.08 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] chlorine gas [=] \\pu{22.4 g}""},{""type"":""other"",""value"":""STP""}]" "

What volume does 22.4 g of chlorine gas occupy at STP?

" nan 7.08 L "
+

Explanation:

+
+

Because we are at STP, we must use the ideal gas law equation
+#PxxV = nxxRxxT#.

+
    +
  • P represents pressure (must have units of atm)
    • +
  • V represents volume (must have units of liters)
    • +
  • n represents the number of moles
    • +
  • R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
    • +
  • T represents the temperature, which must be in Kelvins.
    • +
+

Next, list your known and unknown variables. Our only unknown is the volume of #Cl_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#
+The only issue is the mass of #Cl_2(g)#, we need to convert it into moles of #Cl_2(g)# in order to use the formula.
+#22.4cancel""g"" xx (1molCl_2)/(70.90cancel""g"")# = 0.316 mol #Cl_2#

+

Now all we have to do is rearrange the equation and solve for V like so:
+#V = (nxxRxxT)/P#
+#V = (0.316molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
+#V = 7.08 L#

+
+
" "
+
+
+

#7.08 L#

+
+
+
+

Explanation:

+
+

Because we are at STP, we must use the ideal gas law equation
+#PxxV = nxxRxxT#.

+
    +
  • P represents pressure (must have units of atm)
    • +
  • V represents volume (must have units of liters)
    • +
  • n represents the number of moles
    • +
  • R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
    • +
  • T represents the temperature, which must be in Kelvins.
    • +
+

Next, list your known and unknown variables. Our only unknown is the volume of #Cl_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#
+The only issue is the mass of #Cl_2(g)#, we need to convert it into moles of #Cl_2(g)# in order to use the formula.
+#22.4cancel""g"" xx (1molCl_2)/(70.90cancel""g"")# = 0.316 mol #Cl_2#

+

Now all we have to do is rearrange the equation and solve for V like so:
+#V = (nxxRxxT)/P#
+#V = (0.316molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
+#V = 7.08 L#

+
+
+
" "
+

What volume does 22.4 g of chlorine gas occupy at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 15, 2016 + +
+
+
+
+
+
+
+

#7.08 L#

+
+
+
+

Explanation:

+
+

Because we are at STP, we must use the ideal gas law equation
+#PxxV = nxxRxxT#.

+
    +
  • P represents pressure (must have units of atm)
    • +
  • V represents volume (must have units of liters)
    • +
  • n represents the number of moles
    • +
  • R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
    • +
  • T represents the temperature, which must be in Kelvins.
    • +
+

Next, list your known and unknown variables. Our only unknown is the volume of #Cl_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#
+The only issue is the mass of #Cl_2(g)#, we need to convert it into moles of #Cl_2(g)# in order to use the formula.
+#22.4cancel""g"" xx (1molCl_2)/(70.90cancel""g"")# = 0.316 mol #Cl_2#

+

Now all we have to do is rearrange the equation and solve for V like so:
+#V = (nxxRxxT)/P#
+#V = (0.316molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
+#V = 7.08 L#

+
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+ +
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+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 14700 views + around the world +
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+ + Creative Commons License + +
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+
" What volume does 22.4 g of chlorine gas occupy at STP? nan +173 a82d6ed5-6ddd-11ea-b4f8-ccda262736ce https://socratic.org/questions/what-is-the-formula-of-tin-ll-chromate SnCrO4 start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""SnCrO4""}]" "[{""type"":""substance name"",""value"":""tin (ll) chromate""}]" "

What is the formula of tin (ll) chromate?

" nan SnCrO4 "
+

Explanation:

+
+

And thus #""stannous chromate""# is composed of #Sn^(2+)# and #CrO_4^(2-)# ions in equal quantities. There must be some specialist use......

+
+
" "
+
+
+

#SnCrO_4...........#

+
+
+
+

Explanation:

+
+

And thus #""stannous chromate""# is composed of #Sn^(2+)# and #CrO_4^(2-)# ions in equal quantities. There must be some specialist use......

+
+
+
" "
+

What is the formula of tin (ll) chromate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 30, 2017 + +
+
+
+
+
+
+
+

#SnCrO_4...........#

+
+
+
+

Explanation:

+
+

And thus #""stannous chromate""# is composed of #Sn^(2+)# and #CrO_4^(2-)# ions in equal quantities. There must be some specialist use......

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4807 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What is the formula of tin (ll) chromate? nan +174 a82d6ed6-6ddd-11ea-b7ba-ccda262736ce https://socratic.org/questions/a-buffer-solution-is-prepared-by-mixing-1-0-mole-of-ha-and-1-0-mole-of-naa-into- 0.01 start physical_unit 39 40 ph none qc_end c_other OTHER qc_end physical_unit 32 32 27 28 volume qc_end physical_unit 32 32 30 31 molarity qc_end physical_unit 39 40 36 37 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Change in pH [OF] the buffer""}]" "[{""type"":""physical unit"",""value"":""0.01""}]" "[{""type"":""other"",""value"":""A buffer solution is prepared by mixing 1.0 mole of HA and 1.0 mole of NaA into 1.0 L distilled water.""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{25.0 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [=] \\pu{0.20 M}""},{""type"":""physical unit"",""value"":""Volume [OF] the buffer [=] \\pu{500 ml}""},{""type"":""other"",""value"":""Given Ka(HA) is 1.75 × 10^(-5) M.""}]" "

A buffer solution is prepared by mixing 1.0 mole of HA and 1.0 mole of NaA into 1.0 L distilled water. Calculate the change in pH when 25.0 mL of 0.20 M NaOH is added into 500 mL of the buffer? Given Ka(HA) is 1.75×10^(-5) M

" nan 0.01 "
+

Explanation:

+
+

I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.

+

So, your solution is said to contain #""HA""#, a weak acid, and #""NaA""#, the salt of its conjugate base, the #""A""^(-)# anion.

+

You know that this buffer is prepared by dissolving #""1 mole""# of each chemical species in #""1.0 L""# of distilled water. Use this information to find the molarity of the weak acid and of the conjugate base

+
+

#color(blue)(c = n/V)#

+

#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#

+
+

Keep in mind that the salt dissociates in a #1:1# mole ratio to form sodium cations, #""Na""^(+)#, and the conjugate base of the acid, #""A""^(-)#.

+

Now, you take #""500 mL""# of this buffer solution. Calculate how many moles of weak acid and conjugate base will be present in this sample

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#

+
+

To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!

+
+

#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#

+
+

#color(purple)(""I"")"" "" "" ""1.0"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""0"" "" "" "" "" "" "" "" "" "" ""1.0#
+#color(purple)(""C"")"" ""(-x)"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""(+x)"" "" "" "" "" "" ""(+x)#
+#color(purple)(""E"")"" ""1.0-x"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""x"" "" "" "" "" "" "" "" ""1.0+x#

+

By definition, the acid dissociation constant, #K_a#, will be equal to

+
+

#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#

+
+

Because the value of #K_a# is so small, you can use the approximation

+
+

#1.0 +- x ~~ 1.0#

+
+

This will get you

+
+

#1.75 * 10^(-5) = (x * 1)/(1) = x#

+
+

Here #x# represents the equilibrium concentration of the hydronium ions, so

+
+

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

+

#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#

+
+

SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's #pK_a# when equal concentrations of weak acid and conjugate base are present.

+

Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.

+

This means that you can expect the pH of the solution to increase a little after the addition of the strong base.

+

The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base

+
+

#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#

+
+

Notice that the hydroxide anions react in a #1:1# mole ratio with the weak acid and produce conjugate base in the same #1:1# mole ratio.

+

Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#

+
+

This means that after the neutralization reaction takes place, you will be left with

+
+

#n_(OH^(-)) = ""0 moles"" -># completely consumed

+

#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#

+

#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#

+
+

The new volume of the buffer will be

+
+

#V_""total"" = V_""initial"" + V_""base""#

+

#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#

+
+

Now calculate the new molarities of the weak acid and conjugate base

+
+

#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#

+

#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#

+
+

Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be #""1.0 M""#.

+

Once again, you will have

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

In this case, this will be equal to

+
+

#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#

+
+

Using the same approximation, you can say that

+
+

#1.75 * 10^(-5) = x * 0.9619/0.9429#

+
+

Therefore,

+
+

#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#

+
+

The pH of the solution will now be

+
+

#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#

+
+

The change in pH can be calculated as

+
+

#Delta_""pH"" = ""pH""_1 - ""pH""_0#

+

#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#

+
+
+
" "
+
+
+

Here's what I got.

+
+
+
+

Explanation:

+
+

I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.

+

So, your solution is said to contain #""HA""#, a weak acid, and #""NaA""#, the salt of its conjugate base, the #""A""^(-)# anion.

+

You know that this buffer is prepared by dissolving #""1 mole""# of each chemical species in #""1.0 L""# of distilled water. Use this information to find the molarity of the weak acid and of the conjugate base

+
+

#color(blue)(c = n/V)#

+

#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#

+
+

Keep in mind that the salt dissociates in a #1:1# mole ratio to form sodium cations, #""Na""^(+)#, and the conjugate base of the acid, #""A""^(-)#.

+

Now, you take #""500 mL""# of this buffer solution. Calculate how many moles of weak acid and conjugate base will be present in this sample

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#

+
+

To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!

+
+

#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#

+
+

#color(purple)(""I"")"" "" "" ""1.0"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""0"" "" "" "" "" "" "" "" "" "" ""1.0#
+#color(purple)(""C"")"" ""(-x)"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""(+x)"" "" "" "" "" "" ""(+x)#
+#color(purple)(""E"")"" ""1.0-x"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""x"" "" "" "" "" "" "" "" ""1.0+x#

+

By definition, the acid dissociation constant, #K_a#, will be equal to

+
+

#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#

+
+

Because the value of #K_a# is so small, you can use the approximation

+
+

#1.0 +- x ~~ 1.0#

+
+

This will get you

+
+

#1.75 * 10^(-5) = (x * 1)/(1) = x#

+
+

Here #x# represents the equilibrium concentration of the hydronium ions, so

+
+

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

+

#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#

+
+

SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's #pK_a# when equal concentrations of weak acid and conjugate base are present.

+

Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.

+

This means that you can expect the pH of the solution to increase a little after the addition of the strong base.

+

The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base

+
+

#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#

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+

Notice that the hydroxide anions react in a #1:1# mole ratio with the weak acid and produce conjugate base in the same #1:1# mole ratio.

+

Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer

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+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#

+
+

This means that after the neutralization reaction takes place, you will be left with

+
+

#n_(OH^(-)) = ""0 moles"" -># completely consumed

+

#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#

+

#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#

+
+

The new volume of the buffer will be

+
+

#V_""total"" = V_""initial"" + V_""base""#

+

#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#

+
+

Now calculate the new molarities of the weak acid and conjugate base

+
+

#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#

+

#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#

+
+

Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be #""1.0 M""#.

+

Once again, you will have

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

In this case, this will be equal to

+
+

#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#

+
+

Using the same approximation, you can say that

+
+

#1.75 * 10^(-5) = x * 0.9619/0.9429#

+
+

Therefore,

+
+

#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#

+
+

The pH of the solution will now be

+
+

#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#

+
+

The change in pH can be calculated as

+
+

#Delta_""pH"" = ""pH""_1 - ""pH""_0#

+

#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#

+
+
+
+
" "
+

A buffer solution is prepared by mixing 1.0 mole of HA and 1.0 mole of NaA into 1.0 L distilled water. Calculate the change in pH when 25.0 mL of 0.20 M NaOH is added into 500 mL of the buffer? Given Ka(HA) is 1.75×10^(-5) M

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+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Calculations + + +
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+2 Answers +
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+ + Jan 24, 2016 + +
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+

Here's what I got.

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Explanation:

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+

I'll show you hot to solve this one without using the Henderson - Hasselbalch equation. This method is a bit long, but I think that it's a great help to understanding the general idea behind how buffers work.

+

So, your solution is said to contain #""HA""#, a weak acid, and #""NaA""#, the salt of its conjugate base, the #""A""^(-)# anion.

+

You know that this buffer is prepared by dissolving #""1 mole""# of each chemical species in #""1.0 L""# of distilled water. Use this information to find the molarity of the weak acid and of the conjugate base

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+

#color(blue)(c = n/V)#

+

#[""HA""] = [""A""^(-)] = ""1.0 moles""/""1.0 L"" = ""1.0 M""#

+
+

Keep in mind that the salt dissociates in a #1:1# mole ratio to form sodium cations, #""Na""^(+)#, and the conjugate base of the acid, #""A""^(-)#.

+

Now, you take #""500 mL""# of this buffer solution. Calculate how many moles of weak acid and conjugate base will be present in this sample

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(HA) = n_(A^(-)) = ""1.0 M"" * 500 * 10^(-3)""L"" = ""0.50 moles""#

+
+

To find the pH of this buffer solution, you need to use an ICE table - remember to use the molarities of the weak acid and conjugate base!

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+

#"" """"HA""_text((aq]) + ""H""_2""O""_text((l]) "" ""rightleftharpoons"" "" ""H""_3""O""_text((aq])^(+) "" ""+"" "" ""A""_text((aq])^(-)#

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+

#color(purple)(""I"")"" "" "" ""1.0"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""0"" "" "" "" "" "" "" "" "" "" ""1.0#
+#color(purple)(""C"")"" ""(-x)"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""(+x)"" "" "" "" "" "" ""(+x)#
+#color(purple)(""E"")"" ""1.0-x"" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""x"" "" "" "" "" "" "" "" ""1.0+x#

+

By definition, the acid dissociation constant, #K_a#, will be equal to

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+

#K_a = ( [""H""_3""O""^(+)] * [""A""^(-)])/([""HA""]) = (x * (1+x))/(1-x) = 1.75 * 10^(-5)#

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+

Because the value of #K_a# is so small, you can use the approximation

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+

#1.0 +- x ~~ 1.0#

+
+

This will get you

+
+

#1.75 * 10^(-5) = (x * 1)/(1) = x#

+
+

Here #x# represents the equilibrium concentration of the hydronium ions, so

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+

#color(blue)(""pH"" = - log([""H""_3""O""^(+)]))#

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#""pH""_0 = - log(1.75 * 10^(-5)) = 4.757#

+
+

SIDE NOTE This is why the pH of a buffer solution is said to be equal to the acid's #pK_a# when equal concentrations of weak acid and conjugate base are present.

+

Now, sodium hydroxide is a strong base, which means that it will neutralize the weak acid to produce water and the conjugate base of the acid.

+

This means that you can expect the pH of the solution to increase a little after the addition of the strong base.

+

The balanced chemical equation for this reaction looks like this - I'll use the hydroxide anion to represent the strong base

+
+

#""HA""_text((aq]) + ""OH""_text((aq])^(-) -> ""A""_text((aq])^(-) + ""H""_2""O""_text((l])#

+
+

Notice that the hydroxide anions react in a #1:1# mole ratio with the weak acid and produce conjugate base in the same #1:1# mole ratio.

+

Use the molarity and volume of the sodium hydroxide solution to find how many moles of strong base are added to the buffer

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = ""0.20 M"" * 25.0 * 10^(-3)""L"" = ""0.0050 moles OH""^(-)#

+
+

This means that after the neutralization reaction takes place, you will be left with

+
+

#n_(OH^(-)) = ""0 moles"" -># completely consumed

+

#n_(HA) = ""0.50 moles"" - ""0.0050 moles"" = ""0.495 moles HA""#

+

#n_(A^(-)) = ""0.50 moles"" + ""0.0050 moles"" = ""0.505 moles A""^(-)#

+
+

The new volume of the buffer will be

+
+

#V_""total"" = V_""initial"" + V_""base""#

+

#V_""total"" = ""500 mL"" + ""25.0 mL"" = ""525 mL""#

+
+

Now calculate the new molarities of the weak acid and conjugate base

+
+

#[""HA""] = ""0.495 moles""/(525 * 10^(-3)""L"") = ""0.9429 M""#

+

#[""A""^(-)] = ""0.505 moles""/(525 * 10^(-3)""L"") = ""0.9619 M""#

+
+

Now it's back to the ICE table. To keep the answer at a reasonable length, I won't add it again. This time, the initial concentrations of the weak acid and conjugate base will no longer be #""1.0 M""#.

+

Once again, you will have

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

In this case, this will be equal to

+
+

#K_a = (x * (0.9619 + x))/(0.9429 + x) = 1.75 * 10^(-5)#

+
+

Using the same approximation, you can say that

+
+

#1.75 * 10^(-5) = x * 0.9619/0.9429#

+
+

Therefore,

+
+

#x = (1.75 * 0.9429)/0.9619 * 10^(-5) = 1.7154 * 10^(-5)#

+
+

The pH of the solution will now be

+
+

#""pH""_1 = - log(1.7154 * 10^(-5)) = 4.766#

+
+

The change in pH can be calculated as

+
+

#Delta_""pH"" = ""pH""_1 - ""pH""_0#

+

#Delta_""pH"" = 4.766 - 4.757 = color(green)(0.009)#

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+ +
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+ +
+ + Jan 24, 2016 + +
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#Δ""pH"" = +0.01#

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+
+

Explanation:

+
+
+

The strategy to follow is

+
    +
  1. Write the chemical equation for the buffer.
    2. +
  2. Calculate the #""pH""# of the original buffer.
    3. +
  3. Calculate the moles of base added.
    4. +
  4. Calculate the new moles of #""HA""# and #""A""^-#
    5. +
  5. Calculate the #""pH""# of the new solution.
    6. +
  6. Calculate the change in #""pH""#.
    7. +
+
+

1. Chemical Equation

+

#""HA"" + ""H""_2""O"" ⇌ H_3""O""^(+)+ ""A""^(-)#; #K_""a"" = 1.75 × 10^-5#

+
+

2. #""pH""# of Buffer

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#color(white)(XXXXXXl)""HA"" + ""H""_2""O"" ⇌ H_3""O""^+ + ""A""^-#; #K_""a"" = 1.75 × 10^-5#
+#""E/mol·L""^-1: 1.0 color(white)(XXXXXXXXXXll)1.0#

+

#[""HA""] = [""A""^-] = ""1.0 mol""/""1.0 L"" = ""1.0 mol/L""#

+

We can use the Henderson-Hasselbalch equation to calculate the #""pH""#.

+

#""pH"" = ""p""K_""a"" + log((""[A""^(-)""]"")/""[HA]"") = -log(1.75× 10^-5) + log(1.0/1.0) = 4.757 + 0 = 4.757#

+
+

3. Moles of base added

+

#""Moles of OH""^(-) = 0.0250 color(red)(cancel(color(black)(""L""))) × ""0.20 mol""/(1 color(red)(cancel(color(black)(""L"")))) = ""0.0050 mol""#

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+

4. New moles of #""HA""# and #""A""^-#

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#color(white)(XXXXXX)""HA"" + ""H""_2""O"" ⇌ H_3""O""^+ + ""A""^-#
+#""I/mol:""color(white)(XXl) 0.50color(white)(XXXXXXXXXX) 0.50#
+#""C/mol:""color(white)(l) -0.0050color(white)(XXXXXXXl) +0.0050#
+#""E/mol:""color(white)(XX)0.495color(white)(XXXXXXXXXl) 0.505#

+

You are using only half of the original buffer, so you are starting with 0.50 mol each of #""HA""# and #""A""^-#.

+

The added #""OH""^-# will decrease the amount of #""HA""# and increase the amount of #""A""^-# by 0.0050 mol.

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+

5. #""pH""# of new solution

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Since #""HA""# and #""A""^-# are in the same solution, the ratio of their concentrations is the same as the ratio of the moles.

+

#""pH"" = ""p""K_""a"" + log((""[A""^(-)""]"")/""[HA]"") = 4.757 +log(0.505/0.495) = 4.757 + 0.009 = 4.766#

+

6. Change in pH

+

#""ΔpH = 4.766 – 4.757 = 0.009""#

+

We usually express #""pH""# to only two decimal places, so

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#""ΔpH = 0.01""#

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" A buffer solution is prepared by mixing 1.0 mole of HA and 1.0 mole of NaA into 1.0 L distilled water. Calculate the change in pH when 25.0 mL of 0.20 M NaOH is added into 500 mL of the buffer? Given Ka(HA) is 1.75×10^(-5) M nan +175 a82d6ed7-6ddd-11ea-a519-ccda262736ce https://socratic.org/questions/how-many-grams-of-nacl-in-50-ml-of-0-75-m-nacl-solution 2.19 grams start physical_unit 4 4 mass g qc_end physical_unit 11 12 6 7 volume qc_end physical_unit 11 12 9 10 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [IN] grams""}]" "[{""type"":""physical unit"",""value"":""2.19 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] NaCl solution [=] \\pu{50 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaCl solution [=] \\pu{0.75 M}""}]" "

How many grams of #NaCl# in 50 mL of 0.75 M #NaCl# solution?

" nan 2.19 grams "
+

Explanation:

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+

The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is #M = (mol)/L#

+

For this #NaCl# solution we have a Molarity of 0.75 in 50 mL of solution.

+

Therefore

+

M = 0.75 M
+V = 50 mL or 0.050 L
+moles = #x#

+

#0.75M = (x mol)/(0.050 L)#

+

#0.050 L (0.75 M) = x mol#

+

#0.0375 mol = x#

+

Use the molar mass of #NaCl# to convert the moles to grams.

+

#0.0375 cancel(mol NaCl) x (58.44 g NaCl)/(1 cancel(mol NaCl)) = 2.19 grams NaCl#

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+
" "
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+
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#2.19 grams NaCl#

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+
+
+

Explanation:

+
+

The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is #M = (mol)/L#

+

For this #NaCl# solution we have a Molarity of 0.75 in 50 mL of solution.

+

Therefore

+

M = 0.75 M
+V = 50 mL or 0.050 L
+moles = #x#

+

#0.75M = (x mol)/(0.050 L)#

+

#0.050 L (0.75 M) = x mol#

+

#0.0375 mol = x#

+

Use the molar mass of #NaCl# to convert the moles to grams.

+

#0.0375 cancel(mol NaCl) x (58.44 g NaCl)/(1 cancel(mol NaCl)) = 2.19 grams NaCl#

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+
+
" "
+

How many grams of #NaCl# in 50 mL of 0.75 M #NaCl# solution?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ +
+ + Mar 10, 2016 + +
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#2.19 grams NaCl#

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+
+
+

Explanation:

+
+

The MOLARITY of a solution is determined by the ratio of moles of solute, (the material being dissolved), compared to the liters of solution. the equation is #M = (mol)/L#

+

For this #NaCl# solution we have a Molarity of 0.75 in 50 mL of solution.

+

Therefore

+

M = 0.75 M
+V = 50 mL or 0.050 L
+moles = #x#

+

#0.75M = (x mol)/(0.050 L)#

+

#0.050 L (0.75 M) = x mol#

+

#0.0375 mol = x#

+

Use the molar mass of #NaCl# to convert the moles to grams.

+

#0.0375 cancel(mol NaCl) x (58.44 g NaCl)/(1 cancel(mol NaCl)) = 2.19 grams NaCl#

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" How many grams of #NaCl# in 50 mL of 0.75 M #NaCl# solution? nan +176 a82d6ed8-6ddd-11ea-a3b1-ccda262736ce https://socratic.org/questions/in-the-reaction-cac-2-s-2h-2o-l-c-2h-2-g-ca-oh-2-aq-if-23-g-of-cac-2-are-consume 12.93 g start physical_unit 6 6 mass g qc_end chemical_equation 3 10 qc_end physical_unit 3 3 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] H2O [IN] g""}]" "[{""type"":""physical unit"",""value"":""12.93 g""}]" "[{""type"":""chemical equation"",""value"":""CaC2(s) + 2 H2O(l) -> C2H2(g) + Ca(OH)2(aq)""},{""type"":""physical unit"",""value"":""Mass [OF] CaC2 [=] \\pu{23 g}""}]" "

In the reaction #CaC_2(s) + 2H_2O(l) -> C_2H_2(g) + Ca(OH)_2(aq)#, if 23 g of #CaC_2# are consumed in this reaction, how much #H_2O# is needed?

" nan 12.93 g "
+

Explanation:

+
+

Moles of calcium carbide: #(23*g)/(64.10*g*mol^-1)# #=# #0.359# #mol#.

+

Given the equation, #""2 equiv""# of water are required:

+

#0.359xx2*mol# water, #=# #0.718# #mol# water.

+

#""Equivalent mass""# #=# #0.718*molxx18.01*g*mol^-1# #=# #""approx. 12.9 g""# water.

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" "
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Approx. #13# #mL# of water.

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+
+

Explanation:

+
+

Moles of calcium carbide: #(23*g)/(64.10*g*mol^-1)# #=# #0.359# #mol#.

+

Given the equation, #""2 equiv""# of water are required:

+

#0.359xx2*mol# water, #=# #0.718# #mol# water.

+

#""Equivalent mass""# #=# #0.718*molxx18.01*g*mol^-1# #=# #""approx. 12.9 g""# water.

+
+
+
" "
+

In the reaction #CaC_2(s) + 2H_2O(l) -> C_2H_2(g) + Ca(OH)_2(aq)#, if 23 g of #CaC_2# are consumed in this reaction, how much #H_2O# is needed?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
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+2 Answers +
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+ +
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+ +
+ + Feb 26, 2016 + +
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Approx. #13# #mL# of water.

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+

Explanation:

+
+

Moles of calcium carbide: #(23*g)/(64.10*g*mol^-1)# #=# #0.359# #mol#.

+

Given the equation, #""2 equiv""# of water are required:

+

#0.359xx2*mol# water, #=# #0.718# #mol# water.

+

#""Equivalent mass""# #=# #0.718*molxx18.01*g*mol^-1# #=# #""approx. 12.9 g""# water.

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+ + Feb 26, 2016 + +
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You will need #""13 g H""_2""O""#.

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+
+

Explanation:

+
+

Balanced Equation
+#""CaC""_2(""s"")+""2H""_2""O(l)""##rarr##""C""_2""H""_2(""g"")+""Ca(OH)""_2(""aq"")""#

+

We will make the following conversions to answer this question.

+

#""mass CaC""_2##rarr##""mol CaC""_2##rarr##""mol H""_2""O""##rarr##""mass H""_2""O""#

+

In order to do this, we need to determine the molar mass of each compound. We can calculate it or look it up on a reputable source. I like to use The PubChem Project.

+

Molar Masses

+

#""CaCl""_2"":# #""64.0994 g/mol""#
+https://pubchem.ncbi.nlm.nih.gov/compound/6352#section=Top

+

#""H""_2""O"":# #""18.01528 g/mol""#
+https://pubchem.ncbi.nlm.nih.gov/compound/962#section=Top

+

We also need the mole ratio between the two compounds.

+

From the balanced equation we can see that the mole ratio between #""CaC""_2""# and #""H""_2""O""# is #""1 mol CaC""_2: ""2 mol H""_2""O""#.

+

Solution

+

Convert #""mass CaC""_2##rarr##""mol CaC""_2# by dividing the given mass by its molar mass.

+

#23cancel""g CaC""_2xx(1""mol CaC""_2)/(64.0994cancel""g CaCl""_2)=""0.358817711242 mol CaC""_2""# (I will round to two significant figures at the end. Meanwhile I'm keeping the calculator results until the end.

+

Convert #""mol CaC""_2""##rarr##""mol H""_2""O""# by multiplying the mol #""CaC""_2""# by the mole ratio with water in the numerator.

+

#0.358817711242cancel""mol CaC""_2xx(2""mol H""_2""O"")/(1cancel""mol CaC""_2)=""0.717635422484 mol H""_2""O""#

+

Convert #""mol H""_2""O""##rarr##""mass H""_2""O""# by multiplying the mol #""H""_2""O""# by its molar mass.

+

#0.717635422484cancel""mol H""_2""O""xx(18.01528""g H""_2""O"")/(1cancel""mol H""_2""O"")=13""g H""_2""O""#

+

We can put all of the steps together like this:

+

#23cancel""g CaC""_2xx(1""mol CaC""_2)/(64.0994cancel""g CaC""_2)xx(2""mol H""_2""O"")/(1cancel""mol CaC""_2)xx(18.01528""g H""_2""O"")/(1cancel""mol H""_2""O"")=""13 g H""_2""O""#

+

This way you can do the calculations all at once and round at the end, which reduces rounding errors.

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+
+
" In the reaction #CaC_2(s) + 2H_2O(l) -> C_2H_2(g) + Ca(OH)_2(aq)#, if 23 g of #CaC_2# are consumed in this reaction, how much #H_2O# is needed? nan +177 a82d6ed9-6ddd-11ea-865d-ccda262736ce https://socratic.org/questions/an-avogadro-s-constant-amount-of-any-element-is-equivalent-to-what 6.022 × 10^23 start physical_unit 5 6 avogadro_constant none qc_end c_other Avogadro_constant qc_end end "[{""type"":""physical unit"",""value"":""Avogadro's constant [OF] any element""}]" "[{""type"":""physical unit"",""value"":""6.022 × 10^23""}]" "[{""type"":""other"",""value"":""Avogadro's constant""}]" "

An Avogadro's constant amount of any element is equivalent to what?

" nan 6.022 × 10^23 "
+

Explanation:

+
+

Any substance containing as many particles as there are in 1 mole of carbon-12.
+1 mole of Carbon-12 contains #6.022xx10^23# particles.

+

Therefore, Avogadro's constant contains #6.022xx10^23# unit particles of whatever element is asked for.

+
+
" "
+
+
+

Avogadro's constant contains #6.022xx10^23# unit particles of whatever element is asked for.

+
+
+
+

Explanation:

+
+

Any substance containing as many particles as there are in 1 mole of carbon-12.
+1 mole of Carbon-12 contains #6.022xx10^23# particles.

+

Therefore, Avogadro's constant contains #6.022xx10^23# unit particles of whatever element is asked for.

+
+
+
" "
+

An Avogadro's constant amount of any element is equivalent to what?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Nov 12, 2016 + +
+
+
+
+
+
+
+

Avogadro's constant contains #6.022xx10^23# unit particles of whatever element is asked for.

+
+
+
+

Explanation:

+
+

Any substance containing as many particles as there are in 1 mole of carbon-12.
+1 mole of Carbon-12 contains #6.022xx10^23# particles.

+

Therefore, Avogadro's constant contains #6.022xx10^23# unit particles of whatever element is asked for.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 2684 views + around the world +
+
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+ + Creative Commons License + +
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" An Avogadro's constant amount of any element is equivalent to what? nan +178 a82d6eda-6ddd-11ea-b5ad-ccda262736ce https://socratic.org/questions/how-do-you-write-and-balance-the-chemical-equation-that-relates-to-this-word-equ ZnCO3(s) ->[\Delta] ZnO(s) + CO2(g) start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""ZnCO3(s) ->[\\Delta] ZnO(s) + CO2(g)""}]" "[{""type"":""other"",""value"":""Solid zinc carbonate is heated and broken down into solid zinc oxide and carbon dioxide gas.""}]" "

How do you write and balance the chemical equation that relates to this word equation: Solid zinc carbonate is heated and broken down into solid zinc oxide and carbon dioxide gas?

" nan ZnCO3(s) ->[\Delta] ZnO(s) + CO2(g) "
+

Explanation:

+
+

What does the #Delta# symbol mean? Is the equation balanced with respect to mass and charge?

+
+
" "
+
+
+

#ZnCO_3(s) + Delta rarr ZnO(s) + CO_2(g)uarr#

+
+
+
+

Explanation:

+
+

What does the #Delta# symbol mean? Is the equation balanced with respect to mass and charge?

+
+
+
" "
+

How do you write and balance the chemical equation that relates to this word equation: Solid zinc carbonate is heated and broken down into solid zinc oxide and carbon dioxide gas?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 2, 2017 + +
+
+
+
+
+
+
+

#ZnCO_3(s) + Delta rarr ZnO(s) + CO_2(g)uarr#

+
+
+
+

Explanation:

+
+

What does the #Delta# symbol mean? Is the equation balanced with respect to mass and charge?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5720 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do you write and balance the chemical equation that relates to this word equation: Solid zinc carbonate is heated and broken down into solid zinc oxide and carbon dioxide gas? nan +179 a82d6edb-6ddd-11ea-a356-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-25-92-percent-nitrogen-1 N2O5 start chemical_formula qc_end physical_unit 11 11 10 10 mass_percent qc_end physical_unit 14 14 13 13 mass_percent qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""N2O5""}]" "[{""type"":""physical unit"",""value"":""Mass percent [OF] nitrogen [=] \\pu{25.92 percent}""},{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen [=] \\pu{74.07 percent}""}]" "

What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

" nan N2O5 "
+

Explanation:

+
+

As with all these problems we assume a #100*g# mass of compound. And then we divide the individual elemental masses by the ATOMIC masses of each constituent.

+

#""Moles of nitrogen""# #=# #(25.92*g)/(14.01*g*mol)=1.85*mol#.

+

#""Moles of oxygen""# #=# #(74.07*g)/(15.999*g*mol)=4.62*mol#.

+

And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.

+

#N:(1.85*mol)/(1.85*mol)=1; O:(4.62*mol)/(1.85*mol)=2.50#

+

Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:

+

#N_2O_5#, #""dinitrogen pentoxide""#. Of course, this is an empirical formula; we would need a molecular weight determination before we declared that #N_2O_5# was the mollykule.

+
+
" "
+
+
+

#N_2O_5#

+
+
+
+

Explanation:

+
+

As with all these problems we assume a #100*g# mass of compound. And then we divide the individual elemental masses by the ATOMIC masses of each constituent.

+

#""Moles of nitrogen""# #=# #(25.92*g)/(14.01*g*mol)=1.85*mol#.

+

#""Moles of oxygen""# #=# #(74.07*g)/(15.999*g*mol)=4.62*mol#.

+

And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.

+

#N:(1.85*mol)/(1.85*mol)=1; O:(4.62*mol)/(1.85*mol)=2.50#

+

Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:

+

#N_2O_5#, #""dinitrogen pentoxide""#. Of course, this is an empirical formula; we would need a molecular weight determination before we declared that #N_2O_5# was the mollykule.

+
+
+
" "
+

What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 19, 2016 + +
+
+
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+
+
+
+

#N_2O_5#

+
+
+
+

Explanation:

+
+

As with all these problems we assume a #100*g# mass of compound. And then we divide the individual elemental masses by the ATOMIC masses of each constituent.

+

#""Moles of nitrogen""# #=# #(25.92*g)/(14.01*g*mol)=1.85*mol#.

+

#""Moles of oxygen""# #=# #(74.07*g)/(15.999*g*mol)=4.62*mol#.

+

And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.

+

#N:(1.85*mol)/(1.85*mol)=1; O:(4.62*mol)/(1.85*mol)=2.50#

+

Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:

+

#N_2O_5#, #""dinitrogen pentoxide""#. Of course, this is an empirical formula; we would need a molecular weight determination before we declared that #N_2O_5# was the mollykule.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 11808 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen? nan +180 a82d6edc-6ddd-11ea-8648-ccda262736ce https://socratic.org/questions/how-do-you-write-the-balanced-neutralization-equation-for-the-reaction-between-a C2H4O2 + NaOH -> C2H3O2Na + H2O start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""C2H4O2 + NaOH -> C2H3O2Na + H2O""}]" "[{""type"":""other"",""value"":""The reaction between acetic acid and sodium hydroxide""}]" "

How do you write the balanced neutralization equation for the reaction between acetic acid and sodium hydroxide?

" nan C2H4O2 + NaOH -> C2H3O2Na + H2O "
+

Explanation:

+
+

acetic acid (ethanoic acid): #CH_3COOH#

+

#CH_3COOH = C_2H_4O_2#

+

sodium hydroxide: #NaOH#

+

neutralisation reactions always produce a salt and water.

+

in this case, the salt is sodium ethanoate
+(#CH_3COONa, C_2H_3O_2Na#)

+

equation:

+

#C_2H_4O_2 + NaOH = C_2H_3O_2Na + H_2O#

+

carbon: #2# on the left, #2# on the right
+sodium: #1# on the left, #1# on the right
+hydrogen: #5# on the left, #5# on the right
+oxygen: #3# on the left, #3# on the right.

+

this means that the equation is balanced, with #1:1:1:1# ratio.

+
+
" "
+
+
+

#C_2H_4O_2 + NaOH = C_2H_3O_2Na + H_2O#

+
+
+
+

Explanation:

+
+

acetic acid (ethanoic acid): #CH_3COOH#

+

#CH_3COOH = C_2H_4O_2#

+

sodium hydroxide: #NaOH#

+

neutralisation reactions always produce a salt and water.

+

in this case, the salt is sodium ethanoate
+(#CH_3COONa, C_2H_3O_2Na#)

+

equation:

+

#C_2H_4O_2 + NaOH = C_2H_3O_2Na + H_2O#

+

carbon: #2# on the left, #2# on the right
+sodium: #1# on the left, #1# on the right
+hydrogen: #5# on the left, #5# on the right
+oxygen: #3# on the left, #3# on the right.

+

this means that the equation is balanced, with #1:1:1:1# ratio.

+
+
+
" "
+

How do you write the balanced neutralization equation for the reaction between acetic acid and sodium hydroxide?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 19, 2018 + +
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+
+
+
+
+
+

#C_2H_4O_2 + NaOH = C_2H_3O_2Na + H_2O#

+
+
+
+

Explanation:

+
+

acetic acid (ethanoic acid): #CH_3COOH#

+

#CH_3COOH = C_2H_4O_2#

+

sodium hydroxide: #NaOH#

+

neutralisation reactions always produce a salt and water.

+

in this case, the salt is sodium ethanoate
+(#CH_3COONa, C_2H_3O_2Na#)

+

equation:

+

#C_2H_4O_2 + NaOH = C_2H_3O_2Na + H_2O#

+

carbon: #2# on the left, #2# on the right
+sodium: #1# on the left, #1# on the right
+hydrogen: #5# on the left, #5# on the right
+oxygen: #3# on the left, #3# on the right.

+

this means that the equation is balanced, with #1:1:1:1# ratio.

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" How do you write the balanced neutralization equation for the reaction between acetic acid and sodium hydroxide? nan +181 a82d94e8-6ddd-11ea-800c-ccda262736ce https://socratic.org/questions/a-flask-contains-oxygen-carbon-dioxide-and-nitrogen-gases-the-partial-pressures- 25.70 atm start physical_unit 29 30 total_pressure atm qc_end physical_unit 3 3 15 16 partial_pressure qc_end physical_unit 4 5 17 18 partial_pressure qc_end physical_unit 7 7 20 21 partial_pressure qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Total pressure [OF] the flask [IN] atm""}]" "[{""type"":""physical unit"",""value"":""25.70 atm""}]" "[{""type"":""physical unit"",""value"":""Partial pressure [OF] oxygen [=] \\pu{11.4 atm}""},{""type"":""physical unit"",""value"":""Partial pressure [OF] carbon dioxide [=] \\pu{5.6 atm}""},{""type"":""physical unit"",""value"":""Partial pressure [OF] nitrogen [=] \\pu{8.7 atm}""},{""type"":""other"",""value"":""A flask contains oxygen, carbon dioxide, and nitrogen gases.""}]" "

A flask contains oxygen, carbon dioxide, and nitrogen gases. The partial pressures of each are 11.4 atm, 5.6 atm, and 8.7 atm respectively. What is the total pressure in the flask in atm?

" nan 25.70 atm "
+

Explanation:

+
+

The total pressure is the sum of the pressures of the three gases in the flask.

+

#""pressure""_""total"" =""11.4 atm + 5.6 atm + 8.7 atm"" = ""25.7 atm""#

+
+
" "
+
+
+

The total pressure is 25.7 atm.

+
+
+
+

Explanation:

+
+

The total pressure is the sum of the pressures of the three gases in the flask.

+

#""pressure""_""total"" =""11.4 atm + 5.6 atm + 8.7 atm"" = ""25.7 atm""#

+
+
+
" "
+

A flask contains oxygen, carbon dioxide, and nitrogen gases. The partial pressures of each are 11.4 atm, 5.6 atm, and 8.7 atm respectively. What is the total pressure in the flask in atm?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 26, 2016 + +
+
+
+
+
+
+
+

The total pressure is 25.7 atm.

+
+
+
+

Explanation:

+
+

The total pressure is the sum of the pressures of the three gases in the flask.

+

#""pressure""_""total"" =""11.4 atm + 5.6 atm + 8.7 atm"" = ""25.7 atm""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" A flask contains oxygen, carbon dioxide, and nitrogen gases. The partial pressures of each are 11.4 atm, 5.6 atm, and 8.7 atm respectively. What is the total pressure in the flask in atm? nan +182 a82d94e9-6ddd-11ea-b0e8-ccda262736ce https://socratic.org/questions/if-82-5-ml-of-0-723-m-h-2so-4-titrates-completely-with-34-0-ml-of-al-oh-3-soluti 1.17 M start physical_unit 13 14 molarity mol/l qc_end c_other OTHER qc_end physical_unit 13 14 10 11 volume qc_end physical_unit 14 14 1 2 volume qc_end physical_unit 6 6 4 5 molarity qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] Al(OH)3 solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.17 M""}]" "[{""type"":""other"",""value"":""Titrates completely.""},{""type"":""physical unit"",""value"":""Volume [OF] Al(OH)3 solution [=] \\pu{34.0 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{82.5 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] H2SO4 [=] \\pu{0.723 M}""}]" "

If 82.5 mL of 0.723 M #H_2SO_4# titrates completely with 34.0 mL of #Al(OH)_3# solution, what is the molarity of the #Al(OH)_3# solution?

" nan 1.17 M "
+

Explanation:

+
+

First, let's write the equation for this titration reaction:

+

#3H_2SO_4 (aq) + 2Al(OH)_3 (aq) rarr Al_2(SO_4)_3 (aq) + 6H_2O (l)#

+

The solubility of #Al(OH)_3# is relatively low, but to simplify this we'll still use #(aq)# in the equation.

+

For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:

+

#molH_2SO_4 = (0.0825L)(0.723 (mol)/L) = 0.0596 mol H_2SO_4#

+

Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration, #Al(OH)_3#:

+

#0.0596 mol H_2SO_4((2 mol Al(OH)_3)/(3 mol H_2SO_4)) = 0.0397 mol Al(OH)_3#

+

Finally, we'll use this value and its known volume to find its molar concentration:

+

#M = (0.0397 mol Al(OH)_3)/(0.0340 L) = 1.17 (mol)/L#

+
+
" "
+
+
+

#1.17 (mol)/L#

+
+
+
+

Explanation:

+
+

First, let's write the equation for this titration reaction:

+

#3H_2SO_4 (aq) + 2Al(OH)_3 (aq) rarr Al_2(SO_4)_3 (aq) + 6H_2O (l)#

+

The solubility of #Al(OH)_3# is relatively low, but to simplify this we'll still use #(aq)# in the equation.

+

For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:

+

#molH_2SO_4 = (0.0825L)(0.723 (mol)/L) = 0.0596 mol H_2SO_4#

+

Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration, #Al(OH)_3#:

+

#0.0596 mol H_2SO_4((2 mol Al(OH)_3)/(3 mol H_2SO_4)) = 0.0397 mol Al(OH)_3#

+

Finally, we'll use this value and its known volume to find its molar concentration:

+

#M = (0.0397 mol Al(OH)_3)/(0.0340 L) = 1.17 (mol)/L#

+
+
+
" "
+

If 82.5 mL of 0.723 M #H_2SO_4# titrates completely with 34.0 mL of #Al(OH)_3# solution, what is the molarity of the #Al(OH)_3# solution?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Titration Calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 16, 2017 + +
+
+
+
+
+
+
+

#1.17 (mol)/L#

+
+
+
+

Explanation:

+
+

First, let's write the equation for this titration reaction:

+

#3H_2SO_4 (aq) + 2Al(OH)_3 (aq) rarr Al_2(SO_4)_3 (aq) + 6H_2O (l)#

+

The solubility of #Al(OH)_3# is relatively low, but to simplify this we'll still use #(aq)# in the equation.

+

For any titration problem like this, where you try and find the concentration of one substance when you know the other concentration and volumes of both, the next step is to convert the standard solution (the one with known volume and concentration) to moles:

+

#molH_2SO_4 = (0.0825L)(0.723 (mol)/L) = 0.0596 mol H_2SO_4#

+

Now we'll use the stoichiometrically equivalent values to find the moles of the substance with unknown concentration, #Al(OH)_3#:

+

#0.0596 mol H_2SO_4((2 mol Al(OH)_3)/(3 mol H_2SO_4)) = 0.0397 mol Al(OH)_3#

+

Finally, we'll use this value and its known volume to find its molar concentration:

+

#M = (0.0397 mol Al(OH)_3)/(0.0340 L) = 1.17 (mol)/L#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1285 views + around the world +
+
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+ + Creative Commons License + +
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+
" If 82.5 mL of 0.723 M #H_2SO_4# titrates completely with 34.0 mL of #Al(OH)_3# solution, what is the molarity of the #Al(OH)_3# solution? nan +183 a82d94ea-6ddd-11ea-89a0-ccda262736ce https://socratic.org/questions/how-would-you-balance-na2b4o7-h2so4-h2o-h3bo3-na2so4 Na2B4O7 + H2SO4 + 5 H2O -> 4 H3BO3 + Na2SO4 start chemical_equation qc_end chemical_equation 4 12 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""Na2B4O7 + H2SO4 + 5 H2O -> 4 H3BO3 + Na2SO4""}]" "[{""type"":""chemical equation"",""value"":""Na2B4O7 + H2SO4 + H2O -> H3BO3 + Na2SO4""}]" "

How would you balance Na2B4O7 + H2SO4 + H2O --> H3BO3 + Na2SO4? +

" nan Na2B4O7 + H2SO4 + 5 H2O -> 4 H3BO3 + Na2SO4 "
+

Explanation:

+
+

You first balance the metals, namely Sodium (#Na#), but in this case it is already balanced, 2 on the left of the arrow and 2 on the right.

+

Next you do the non-metals, starting with anything but Oxygen (#O#) and Hydrogen (#H#), which in this case is Sulphur but since its part of a polyatomic ion, being Suphate (#SO_4#) you balance it as a whole, in this case they are both balanced though.

+

Next is Boron (#B#), on the left there is 4 and on the right there is one, so you add a 4 before the #H_3BO_3# to balance it.

+

The you balance the Hydrogen (#H#) atoms, on the left there is 4 and on the right there is 12 after putting a 4 in front of (#H_3BO_3#).
+So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).

+

Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.

+

Remember that the coefficients should be the lowest value possible, and must not be fractions.

+

Hope I helped :)

+
+
" "
+
+
+

#Na_2B_4O_7# + #H_2SO_4# + #5H_2O# #rarr# #4H_3BO_3# + #Na_2SO_4#

+
+
+
+

Explanation:

+
+

You first balance the metals, namely Sodium (#Na#), but in this case it is already balanced, 2 on the left of the arrow and 2 on the right.

+

Next you do the non-metals, starting with anything but Oxygen (#O#) and Hydrogen (#H#), which in this case is Sulphur but since its part of a polyatomic ion, being Suphate (#SO_4#) you balance it as a whole, in this case they are both balanced though.

+

Next is Boron (#B#), on the left there is 4 and on the right there is one, so you add a 4 before the #H_3BO_3# to balance it.

+

The you balance the Hydrogen (#H#) atoms, on the left there is 4 and on the right there is 12 after putting a 4 in front of (#H_3BO_3#).
+So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).

+

Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.

+

Remember that the coefficients should be the lowest value possible, and must not be fractions.

+

Hope I helped :)

+
+
+
" "
+

How would you balance Na2B4O7 + H2SO4 + H2O --> H3BO3 + Na2SO4? +

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
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+2 Answers +
+
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+ +
+
+ +
+ + Oct 26, 2015 + +
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#Na_2B_4O_7# + #H_2SO_4# + #5H_2O# #rarr# #4H_3BO_3# + #Na_2SO_4#

+
+
+
+

Explanation:

+
+

You first balance the metals, namely Sodium (#Na#), but in this case it is already balanced, 2 on the left of the arrow and 2 on the right.

+

Next you do the non-metals, starting with anything but Oxygen (#O#) and Hydrogen (#H#), which in this case is Sulphur but since its part of a polyatomic ion, being Suphate (#SO_4#) you balance it as a whole, in this case they are both balanced though.

+

Next is Boron (#B#), on the left there is 4 and on the right there is one, so you add a 4 before the #H_3BO_3# to balance it.

+

The you balance the Hydrogen (#H#) atoms, on the left there is 4 and on the right there is 12 after putting a 4 in front of (#H_3BO_3#).
+So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).

+

Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.

+

Remember that the coefficients should be the lowest value possible, and must not be fractions.

+

Hope I helped :)

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Oct 26, 2015 + +
+
+
+
+
+
+
+

#Na_2B_4O_7# + #H_2SO_4# +#5H_2O# = #4H_3BO_3# + #Na_2SO_4#

+
+
+
+

Explanation:

+
+

It does not matter how long the equation is. You need to always start with tallying the number of atoms (based on the subscripts).

+

#Na_2B_4O_7# + #H_2SO_4# +#H_2O# = #H_3BO_3# + #Na_2SO_4#

+

Left side:

+

Na = 2
+B = 4
+O = 7 + 4 + 1 (DO NOT ADD IT UP YET)
+H = 2 + 2 (DO NOT ADD IT UP YET)
+S= 1

+

Right side:

+

Na = 2
+B = 1
+O = 3 + 4 (DO NOT ADD IT UP YET)
+H = 3
+S= 1

+

Start with the element easiest to balance, in this case boron (B). Since B is part of the substance #H_3BO_3#, whatever number you choose to multiply with B should also be applied to H and O atoms, respectively.

+

Left side:

+

Na = 2
+B = 4
+O = 7 + 4 + 1
+H = 2 + 2
+S= 1

+

Right side:

+

Na = 2
+B = 1 x 4 = 4
+O = (3 x 4) + 4
+H = (3 x 4) = 12
+S= 1

+

#Na_2B_4O_7# + #H_2SO_4# +#H_2O# = #color (red) (4)H_3BO_3# + #Na_2SO_4#

+

Now, the number of H on your right side is 12. There are two H atoms on your left side that you can choose to multiply by 5 (since both values are 2) to balance. Choose the least complicated substance or the one with least number of other atoms attached to it. In this case, it is #H_2O# (2 atoms attached to H from #H_2O# vs. 6 atoms attached to H from #H_2SO_4#).

+

Left side:

+

Na = 2
+B = 4
+O = 7 + 4 + (1 x 5)
+H = 2 + (2 x 5) = 12
+S= 1

+

Right side:

+

Na = 2
+B = 1 x 4 = 4
+O = (3 x 4) + 4
+H = (3 x 4) = 12
+S= 1

+

#Na_2B_4O_7# + #H_2SO_4# +#color (red) (5)H_2O# = #color (red) (4)H_3BO_3# + #Na_2SO_4#

+

Now, look at your tally sheet and see it everything is balanced out.

+

Left side:

+

Na = 2
+B = 4
+O = 7 + 4 + (1 x 5) = 16
+H = 2 + (2 x 5) = 12
+S= 1

+

Right side:

+

Na = 2
+B = 1 x 4 = 4
+O = (3 x 4) + 4 = 16
+H = (3 x 4) = 12
+S= 1

+

The equation is now balanced.

+

#Na_2B_4O_7# + #H_2SO_4# +#color (red) (5)H_2O# = #color (red) (4)H_3BO_3# + #Na_2SO_4#

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" "How would you balance Na2B4O7 + H2SO4 + H2O --> H3BO3 + Na2SO4? +" nan +184 a82d94eb-6ddd-11ea-9bf0-ccda262736ce https://socratic.org/questions/what-is-the-molality-of-a-solution-of-phosphoric-acid-h-3po-4-that-contains-24-5 2.50 mol/kg start physical_unit 10 10 molality mol/kg qc_end physical_unit 26 26 23 24 mass qc_end physical_unit 10 10 13 14 mass qc_end physical_unit 10 10 20 21 molar_mass qc_end end "[{""type"":""physical unit"",""value"":""Molality [OF] H3PO4 [IN] mol/kg""}]" "[{""type"":""physical unit"",""value"":""2.50 mol/kg""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2O [=] \\pu{100 g}""},{""type"":""physical unit"",""value"":""Mass [OF] H3PO4 [=] \\pu{24.5 g}""},{""type"":""physical unit"",""value"":""Molar mass [OF] H3PO4 [=] \\pu{98.0 g/mol}""}]" "

What is the molality of a solution of phosphoric acid, #H_3PO_4# that contains 24.5 g of phosphoric acid (molar mass 98.0 g) in 100 g of #H_2O#?

" nan 2.50 mol/kg "
+

Explanation:

+
+
+

The formula for molality #b# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#

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+
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#""Moles of H""_3""PO""_4 = 24.5 color(red)(cancel(color(black)(""g H""_3""PO""_4))) × (""1 mol H""_3""PO""_4)/(98.0 color(red)(cancel(color(black)(""g H""_3""PO""_4)))) = ""0.250 mol H""_3""PO""_4#

+
+

#""Mass of solvent"" = ""100 g"" = ""0.100 kg""#

+
+

∴ #b = ""0.250 mol""/""0.100 kg"" = ""2.50 mol/kg""#

+
+
" "
+
+
+

The molality is 2.50 mol/kg.

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+
+
+

Explanation:

+
+
+

The formula for molality #b# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#

+
+
+

#""Moles of H""_3""PO""_4 = 24.5 color(red)(cancel(color(black)(""g H""_3""PO""_4))) × (""1 mol H""_3""PO""_4)/(98.0 color(red)(cancel(color(black)(""g H""_3""PO""_4)))) = ""0.250 mol H""_3""PO""_4#

+
+

#""Mass of solvent"" = ""100 g"" = ""0.100 kg""#

+
+

∴ #b = ""0.250 mol""/""0.100 kg"" = ""2.50 mol/kg""#

+
+
+
" "
+

What is the molality of a solution of phosphoric acid, #H_3PO_4# that contains 24.5 g of phosphoric acid (molar mass 98.0 g) in 100 g of #H_2O#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
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+1 Answer +
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+ +
+ + Jul 6, 2016 + +
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+

The molality is 2.50 mol/kg.

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+
+

Explanation:

+
+
+

The formula for molality #b# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) b = ""moles of solute""/""kilograms of solvent""color(white)(a/a)|)))"" ""#

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+
+

#""Moles of H""_3""PO""_4 = 24.5 color(red)(cancel(color(black)(""g H""_3""PO""_4))) × (""1 mol H""_3""PO""_4)/(98.0 color(red)(cancel(color(black)(""g H""_3""PO""_4)))) = ""0.250 mol H""_3""PO""_4#

+
+

#""Mass of solvent"" = ""100 g"" = ""0.100 kg""#

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+

∴ #b = ""0.250 mol""/""0.100 kg"" = ""2.50 mol/kg""#

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+
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+
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+ + Creative Commons License + +
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" What is the molality of a solution of phosphoric acid, #H_3PO_4# that contains 24.5 g of phosphoric acid (molar mass 98.0 g) in 100 g of #H_2O#? nan +185 a82d94ec-6ddd-11ea-823f-ccda262736ce https://socratic.org/questions/how-many-moles-of-potassium-nitrate-are-produced-when-two-miles-of-potassium-pho 6.00 moles start physical_unit 4 5 mole mol qc_end physical_unit 12 13 9 10 mole qc_end physical_unit 19 20 9 10 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] potassium nitrate [IN] moles""}]" "[{""type"":""physical unit"",""value"":""6.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] potassium phosphate [=] \\pu{2 moles}""},{""type"":""physical unit"",""value"":""Mole [OF] aluminum nitrate [=] \\pu{2 moles}""}]" "

How many moles of potassium nitrate are produced when two miles of potassium phosphate react with two moles of aluminum nitrate?

" nan 6.00 moles "
+

Explanation:

+
+

First write balanced chemical equation for this ion-exchange reaction :

+

#K_3PO_4+Al(NO_3)_3->3KNO_3+AlPO_4#.

+

This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.

+

So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.

+

Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.

+
+
" "
+
+
+

#6# moles

+
+
+
+

Explanation:

+
+

First write balanced chemical equation for this ion-exchange reaction :

+

#K_3PO_4+Al(NO_3)_3->3KNO_3+AlPO_4#.

+

This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.

+

So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.

+

Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.

+
+
+
" "
+

How many moles of potassium nitrate are produced when two miles of potassium phosphate react with two moles of aluminum nitrate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
+
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+ +
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+ +
+ + Feb 13, 2016 + +
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#6# moles

+
+
+
+

Explanation:

+
+

First write balanced chemical equation for this ion-exchange reaction :

+

#K_3PO_4+Al(NO_3)_3->3KNO_3+AlPO_4#.

+

This balanced chemical equation represents the mole ratio in which the chemicals combine and form products.

+

So clearly 1 mole of each reactant combines to form 3 moles potassium nitrate and 1 mole aluminium phosphate.

+

Hence, by ratio and proportion, 2 moles of each reactant will produce 6 moles potassium nitrate and 2 moles aluminium phosphate.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many moles of potassium nitrate are produced when two miles of potassium phosphate react with two moles of aluminum nitrate? nan +186 a82d94ed-6ddd-11ea-aba6-ccda262736ce https://socratic.org/questions/how-many-moles-of-ba-oh-2-are-present-in-225-ml-of-0-800-m-ba-oh-2 0.18 moles start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 volume qc_end physical_unit 4 4 11 12 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Ba(OH)2 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.18 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] Ba(OH)2 [=] \\pu{225 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] Ba(OH)2 [=] \\pu{0.800 M}""}]" "

How many moles of #Ba(OH)_2# are present in 225 mL of 0.800 M #Ba(OH)_2#?

" nan 0.18 moles "
+

Explanation:

+
+

Molarity is represented by the equation below:

+

+

We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.

+

225 mL can be converted into liters by using the conversion factor 1000mL = 1L.

+

When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.

+

Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

+

Moles of solute = #""liters of solution"" xx Molarity #

+

Moles of solute = #0.225Lxx0.800M#

+

#0.180mol#

+
+
" "
+
+
+

0.180 moles of #Ba(OH)_2# are present.

+
+
+
+

Explanation:

+
+

Molarity is represented by the equation below:

+

+

We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.

+

225 mL can be converted into liters by using the conversion factor 1000mL = 1L.

+

When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.

+

Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

+

Moles of solute = #""liters of solution"" xx Molarity #

+

Moles of solute = #0.225Lxx0.800M#

+

#0.180mol#

+
+
+
" "
+

How many moles of #Ba(OH)_2# are present in 225 mL of 0.800 M #Ba(OH)_2#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Mole Ratios + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Jul 7, 2016 + +
+
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+
+
+
+

0.180 moles of #Ba(OH)_2# are present.

+
+
+
+

Explanation:

+
+

Molarity is represented by the equation below:

+

+

We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.

+

225 mL can be converted into liters by using the conversion factor 1000mL = 1L.

+

When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.

+

Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

+

Moles of solute = #""liters of solution"" xx Molarity #

+

Moles of solute = #0.225Lxx0.800M#

+

#0.180mol#

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+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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+
" How many moles of #Ba(OH)_2# are present in 225 mL of 0.800 M #Ba(OH)_2#? nan +187 a82dbcba-6ddd-11ea-aa2e-ccda262736ce https://socratic.org/questions/if-water-is-added-to-50-ml-of-a-0-04m-solution-so-that-it-fills-a-200-ml-beaker- 0.01 M start physical_unit 11 11 concentration mol/l qc_end physical_unit 11 11 5 6 volume qc_end physical_unit 11 11 9 10 concentration qc_end physical_unit 1 3 5 6 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration2 [OF] solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.01 M""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] solution [=] \\pu{50 ml}""},{""type"":""physical unit"",""value"":""Concentration1 [OF] solution [=] \\pu{0.04 M}""},{""type"":""physical unit"",""value"":""Volume2 [OF] water is added [=] \\pu{50 ml}""}]" "

If water is added to 50 ml of a 0.04M solution so that it fills a 200 mL beaker, what is the final concentration?

" nan 0.01 M "
+

Explanation:

+
+

#""Concentration""# #=# #""Moles of solute""/""Volume of solution in litres""#

+

#=# #(50xx10^-3*Lxx0.04*mol*L^-1)/(200xx10^-3L)#

+

#=# #(50xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(200xxcancel(10^-3L))#

+

#=# #(1xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(4xxcancel(10^-3L))#

+

#=# #0.01*mol*L^-1#

+
+
" "
+
+
+

Final concentration #=# #0.01*mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Concentration""# #=# #""Moles of solute""/""Volume of solution in litres""#

+

#=# #(50xx10^-3*Lxx0.04*mol*L^-1)/(200xx10^-3L)#

+

#=# #(50xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(200xxcancel(10^-3L))#

+

#=# #(1xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(4xxcancel(10^-3L))#

+

#=# #0.01*mol*L^-1#

+
+
+
" "
+

If water is added to 50 ml of a 0.04M solution so that it fills a 200 mL beaker, what is the final concentration?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Measuring Concentration + + +
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+1 Answer +
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+ +
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+ + Sep 28, 2016 + +
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Final concentration #=# #0.01*mol*L^-1#.

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+
+

Explanation:

+
+

#""Concentration""# #=# #""Moles of solute""/""Volume of solution in litres""#

+

#=# #(50xx10^-3*Lxx0.04*mol*L^-1)/(200xx10^-3L)#

+

#=# #(50xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(200xxcancel(10^-3L))#

+

#=# #(1xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(4xxcancel(10^-3L))#

+

#=# #0.01*mol*L^-1#

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+
Related questions
+ + +
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Impact of this question
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" If water is added to 50 ml of a 0.04M solution so that it fills a 200 mL beaker, what is the final concentration? nan +188 a82dbcbb-6ddd-11ea-ab94-ccda262736ce https://socratic.org/questions/if-solution-contains-0-85-mol-of-oh-how-many-moles-of-h-would-be-required-to-rea 0.85 mol start physical_unit 11 11 mole mol qc_end physical_unit 6 6 3 4 mole qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] H^+ [IN] mol""}]" "[{""type"":""physical unit"",""value"":""0.85 mol""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] OH [=] \\pu{0.85 mol}""},{""type"":""other"",""value"":""reach the equivalence point in a titration""}]" "

If solution contains 0.85 mol of OH, how many moles of H+ would be required to reach the equivalence point in a titration?

" nan 0.85 mol "
+

Explanation:

+
+

The balanced equation clearly specifies a 1:1 equivalence.

+

Alternatively, we could write:

+

#HO^(-) + H_3O^(+) rarr 2H_2O#

+

And the fashion seems to favour this representation.

+

The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at #298# #K#, can be written as #[H_3O^+][HO^-]=10^(-14)#. And if there are #0.85*mol# #HO^-#, there must be a corresponding molar quantity of #H^+#.

+
+
" "
+
+
+

#H^+ + HO^(-) rarr H_2O#.

+

#0.85*mol# #HO^-# are required.

+
+
+
+

Explanation:

+
+

The balanced equation clearly specifies a 1:1 equivalence.

+

Alternatively, we could write:

+

#HO^(-) + H_3O^(+) rarr 2H_2O#

+

And the fashion seems to favour this representation.

+

The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at #298# #K#, can be written as #[H_3O^+][HO^-]=10^(-14)#. And if there are #0.85*mol# #HO^-#, there must be a corresponding molar quantity of #H^+#.

+
+
+
" "
+

If solution contains 0.85 mol of OH, how many moles of H+ would be required to reach the equivalence point in a titration?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Titration Calculations + + +
+
+
+
+
+1 Answer +
+
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+
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+
+ +
+
+ +
+ + Sep 18, 2016 + +
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#H^+ + HO^(-) rarr H_2O#.

+

#0.85*mol# #HO^-# are required.

+
+
+
+

Explanation:

+
+

The balanced equation clearly specifies a 1:1 equivalence.

+

Alternatively, we could write:

+

#HO^(-) + H_3O^(+) rarr 2H_2O#

+

And the fashion seems to favour this representation.

+

The given equations are equivalent they represent the acid/base equilibrium that occurs in water, which at #298# #K#, can be written as #[H_3O^+][HO^-]=10^(-14)#. And if there are #0.85*mol# #HO^-#, there must be a corresponding molar quantity of #H^+#.

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" If solution contains 0.85 mol of OH, how many moles of H+ would be required to reach the equivalence point in a titration? nan +189 a82dbcbc-6ddd-11ea-8e1a-ccda262736ce https://socratic.org/questions/what-concentration-of-formic-acid-will-result-in-a-solution-with-ph-1-90-the-val 0.89 M start physical_unit 3 4 concentration mol/l qc_end physical_unit 3 4 13 13 ph qc_end physical_unit 3 4 22 24 ka qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] formic acid [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.89 M""}]" "[{""type"":""physical unit"",""value"":""PH [OF] formic acid [=] \\pu{1.90}""},{""type"":""physical unit"",""value"":""Ka [OF] formic acid [=] \\pu{1.8 × 10^(−4)}""}]" "

What concentration of formic acid will result in a solution with #""pH"" = 1.90#? +The value of #K_a# for formic acid is #1.8 * 10^(−4)#

" nan 0.89 M "
+

Explanation:

+
+

The idea here is that the #""pH""# of the solution will give you the equilibrium concentration of hydronium cations in this formic acid solution.

+

Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use #""HA""# to denote the formic acid and #""A""^(-)# to denote the formate anions.

+
+

#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#

+
+

As you know, the #""pH""# of the solution is given by

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

This implies that the concentration of hydronium cations is equal to

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#

+
+

Now, notice that in order for the ionization of the weak acid to produce #1# mole of hydronium cations and #1# mole of conjugate base, #1# mole of formic acid must ionize.

+

If you take #[""HA""]# to be the equilibrium concentration of formic acid, you can say that the initial concentration of the acid is equal to

+
+

#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#

+
+

This is equivalent to saying that in order to get an equilibrium concentration of #[""H""_3""O""^(+)]# in the solution, the initial concentration of the acid must decrease by #[""H""_3""O""^(+)]#.

+
+

#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

Since you know that, at equilibrium, you have

+
+

#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#

+
+

you can rewrite the expression you have for the acid dissociation constant as

+
+

#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#

+
+

which is, of course, equivalent to

+
+

#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#

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+

Rearrange to solve for the initial concentration of the formic acid

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+

#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#

+

#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#

+
+

Finally, plug in your values to get

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+

#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#

+
+

The answer is rounded to two sig figs, the number of decimal places you have for the #""pH""# of the solution.

+
+
" "
+
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#""0.89 M""#

+
+
+
+

Explanation:

+
+

The idea here is that the #""pH""# of the solution will give you the equilibrium concentration of hydronium cations in this formic acid solution.

+

Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use #""HA""# to denote the formic acid and #""A""^(-)# to denote the formate anions.

+
+

#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#

+
+

As you know, the #""pH""# of the solution is given by

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

This implies that the concentration of hydronium cations is equal to

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#

+
+

Now, notice that in order for the ionization of the weak acid to produce #1# mole of hydronium cations and #1# mole of conjugate base, #1# mole of formic acid must ionize.

+

If you take #[""HA""]# to be the equilibrium concentration of formic acid, you can say that the initial concentration of the acid is equal to

+
+

#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#

+
+

This is equivalent to saying that in order to get an equilibrium concentration of #[""H""_3""O""^(+)]# in the solution, the initial concentration of the acid must decrease by #[""H""_3""O""^(+)]#.

+
+

#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

Since you know that, at equilibrium, you have

+
+

#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#

+
+

you can rewrite the expression you have for the acid dissociation constant as

+
+

#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#

+
+

which is, of course, equivalent to

+
+

#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#

+
+

Rearrange to solve for the initial concentration of the formic acid

+
+

#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#

+

#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#

+
+

Finally, plug in your values to get

+
+

#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#

+
+

The answer is rounded to two sig figs, the number of decimal places you have for the #""pH""# of the solution.

+
+
+
" "
+

What concentration of formic acid will result in a solution with #""pH"" = 1.90#? +The value of #K_a# for formic acid is #1.8 * 10^(−4)#

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
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+2 Answers +
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+ + Oct 20, 2017 + +
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#""0.89 M""#

+
+
+
+

Explanation:

+
+

The idea here is that the #""pH""# of the solution will give you the equilibrium concentration of hydronium cations in this formic acid solution.

+

Formic acid is a weak acid, which means that it only partially ionizes to produce formate anions, the conjugate base of formic acid, and hydronium cations. For the sake of simplicity, I'll use #""HA""# to denote the formic acid and #""A""^(-)# to denote the formate anions.

+
+

#""HA""_ ((aq)) + ""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""A""_ ((aq))^(-)#

+
+

As you know, the #""pH""# of the solution is given by

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

This implies that the concentration of hydronium cations is equal to

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"")color(white)(.)""M""#

+
+

Now, notice that in order for the ionization of the weak acid to produce #1# mole of hydronium cations and #1# mole of conjugate base, #1# mole of formic acid must ionize.

+

If you take #[""HA""]# to be the equilibrium concentration of formic acid, you can say that the initial concentration of the acid is equal to

+
+

#[""HA""]_ 0 = [""HA""] + [""H""_ 3""O""^(+)]#

+
+

This is equivalent to saying that in order to get an equilibrium concentration of #[""H""_3""O""^(+)]# in the solution, the initial concentration of the acid must decrease by #[""H""_3""O""^(+)]#.

+
+

#[""HA""] = [""HA""]_ 0 - [""H""_ 3""O""^(+)]#

+
+

Now, by definition, the acid dissociation constant is equal to

+
+

#K_a = ([""H""_3""O""^(+)] * [""A""^(-)])/([""HA""])#

+
+

Since you know that, at equilibrium, you have

+
+

#[""H""_3""O""^(+)] = [""A""^(-)] = 10^(-""pH"")#

+
+

you can rewrite the expression you have for the acid dissociation constant as

+
+

#K_a = ([""H""_3""O""^(+)]^2)/([""HA""]_0 - [""H""_3""O""^(+)])#

+
+

which is, of course, equivalent to

+
+

#K_a = (10^(-""pH""))^2/([""HA""]_0 - 10^(-""pH""))#

+
+

Rearrange to solve for the initial concentration of the formic acid

+
+

#[""HA""]_0 * K_a = 10^(-2""pH"") + 10^(-""pH"") * K_a#

+

#[""HA""]_0 = 10^(-2""pH"")/K_a + 10^(-""pH"")#

+
+

Finally, plug in your values to get

+
+

#[""HA""]_ 0 = (10^(-2 * 1.90))/(1.8 * 10^(-4)) + 10^(-1.90) = color(darkgreen)(ul(color(black)(""0.89 M"")))#

+
+

The answer is rounded to two sig figs, the number of decimal places you have for the #""pH""# of the solution.

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+ + Oct 20, 2017 + +
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The concentration is 0.89 mol/L.

+
+
+
+

Explanation:

+
+
+

Step 1. Calculate #[""H""_3""O""^""+""]#

+

#[""H""_3""O""^""+""] = 10^""-pH""color(white)(l) ""mol/L"" = 10^""-1.9""color(white)(l)""mol/L"" = ""0.0126 mol/L""#

+
+

Step 2. Calculate the concentration of formic acid

+

We can use an ICE table to organize our calculations.

+

Let #c# be the initial concentration of the acid.

+

#color(white)(mmmmmmll)""HA"" + ""H""_2""O"" ⇌ ""H""_3""O""^""+"" + ""A""^""-""#
+#""I/mol·L""^""-1"":color(white)(mml)c color(white)(mmmmmmm)0color(white)(mmm)0#
+#""C/mol·L""^""-1"":color(white)(mll)""-""xcolor(white)(mmmmmml)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(m)c-xcolor(white)(mmmmmm)xcolor(white)(mmm)x#

+
+

In this problem, #x = ""0.015 85""#.

+

So, our ICE table becomes

+

#color(white)(mmmmmmmmll)""HA"" + ""H""_2""O"" ⇌ ""H""_3""O""^""+"" color(white)(ll)+color(white)(mll) ""A""^""-""#
+#""I/mol·L""^""-1"":color(white)(mmmm)c color(white)(mmmmmmml)0color(white)(mmmmml)0#
+#""C/mol·L""^""-1"":color(white)(llm)""-0.0126""color(white)(mmmml)""+0.0126""color(white)(mm)""+0.0126""#
+#""E/mol·L""^""-1"":color(white)(m)c-""0.0126""color(white)(mmmml)""0.0126""color(white)(mmm)""0.0126""#

+
+

#K_text(a) = ([""H""_3""O""^""+""][""A""^""-""])/([""HA""]) = (""0.0126 × 0.0126"")/(c-""0.0126"") = 1.8 × 10^""-4""#

+

#1.585 × 10^""-4"" = 1.8 × 10^""-4""c - 2.268 × 10^""-6""#

+

#158.5 = 180c +2.268#

+

#x =(158.5+2.268)/180 = 0.89#

+

#[""HA""] = ""0.89 mol/L""#

+
+

Check:

+

#K_text(a) = ([""H""_3""O""^""+""][""A""^""-""])/([""HA""]) = (""0.0126 × 0.0126"")/(0.89-0.0126) = 1.8 × 10^""-4""#

+

It checks!

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" "What concentration of formic acid will result in a solution with #""pH"" = 1.90#? +The value of #K_a# for formic acid is #1.8 * 10^(−4)#" nan +190 a82e0c2c-6ddd-11ea-b5cc-ccda262736ce https://socratic.org/questions/598502117c0149746998f49b 0.29 mol*L^(−1) start physical_unit 27 27 concentration mol/l qc_end physical_unit 7 7 1 2 volume qc_end physical_unit 7 7 5 6 concentration qc_end physical_unit 18 18 16 17 concentration qc_end physical_unit 18 18 12 13 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] [Na^+] [IN] mol*L^(−1)""}]" "[{""type"":""physical unit"",""value"":""0.29 mol*L^(−1)""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] Na2SO4 [=] \\pu{18 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] Na2SO4 [=] \\pu{0.200 mol*L^(−1)}""},{""type"":""physical unit"",""value"":""Concentration [OF] NaCl [=] \\pu{0.150 mol*L^(−1)}""},{""type"":""physical unit"",""value"":""Volume [OF] NaCl [=] \\pu{15 mL}""}]" "

An #18*mL# volume of #0.200*mol*L^-1# #Na_2SO_4# is mixed with a #15*mL# volume of #0.150*mol*L^-1# #NaCl#. What is the new concentration with respect to #[Na^+]#?

" nan 0.29 mol*L^(−1) "
+

Explanation:

+
+

We assume that the volumes are additive, and we work out the molar concentration of the new solution.

+

And thus.......#[Na^+]#.........

+

#=(18.0xx10^-3*Lxx0.20*(mol)/L+15xx10^-3*Lxx0.150*(mol)/L)/((15.0+18.0)xx10^-3L)#

+

#=0.177*mol*L^-1#............

+

What are #[Cl^-]# and #[SO_4^(2-)]#....?

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" "
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We use the relationship #""concentration""=""number of moles""/""volume of solution""#, and get......

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+
+
+

Explanation:

+
+

We assume that the volumes are additive, and we work out the molar concentration of the new solution.

+

And thus.......#[Na^+]#.........

+

#=(18.0xx10^-3*Lxx0.20*(mol)/L+15xx10^-3*Lxx0.150*(mol)/L)/((15.0+18.0)xx10^-3L)#

+

#=0.177*mol*L^-1#............

+

What are #[Cl^-]# and #[SO_4^(2-)]#....?

+
+
+
" "
+

An #18*mL# volume of #0.200*mol*L^-1# #Na_2SO_4# is mixed with a #15*mL# volume of #0.150*mol*L^-1# #NaCl#. What is the new concentration with respect to #[Na^+]#?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+2 Answers +
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+ + Aug 5, 2017 + +
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We use the relationship #""concentration""=""number of moles""/""volume of solution""#, and get......

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+
+
+

Explanation:

+
+

We assume that the volumes are additive, and we work out the molar concentration of the new solution.

+

And thus.......#[Na^+]#.........

+

#=(18.0xx10^-3*Lxx0.20*(mol)/L+15xx10^-3*Lxx0.150*(mol)/L)/((15.0+18.0)xx10^-3L)#

+

#=0.177*mol*L^-1#............

+

What are #[Cl^-]# and #[SO_4^(2-)]#....?

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#[""Na""^""+""] = ""0.286 mol/L""#

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+
+
+

Explanation:

+
+
+

Step 1. Calculate the total moles of #""Na""^""+""#

+

(a) Moles of #""Na""^""+""# in #""Na""_2""SO""_4#

+

#""Moles of Na""^""+"" = 0.0180 color(red)(cancel(color(black)(""L Na""_2""SO""_4))) × (0.200 color(red)(cancel(color(black)(""mol Na""_2""SO""_4))))/(1 color(red)(cancel(color(black)(""L Na""_2""SO""_4)))) × ""2 mol Na""^""+""/(1 color(red)(cancel(color(black)(""mol Na""_2""SO""_4)))) = ""0.007 20 mol Na""^""+""#

+
+

(b) Moles of #""Na""^""+""# in #""NaCl""#

+

#""Moles of Na""^""+"" = 0.0150 color(red)(cancel(color(black)(""L NaCl""))) × (0.150 color(red)(cancel(color(black)(""mol NaCl""))))/(1 color(red)(cancel(color(black)(""L NaCl"")))) × ""1 mol Na""^""+""/(1 color(red)(cancel(color(black)(""mol NaCl"")))) = ""0.002 25 mol Na""^""+""#

+
+

(c) Total moles of #""Na""^""+""#

+

#""Total moles"" = ""0.007 20 mol + 0.002 25 mol"" = ""0.009 45 mol""#

+
+

Step 2. Calculate the total volume of the solution

+

#""Total volume"" = ""18.0 mL + 15.0 mL"" = ""33.0 mL"" = ""0.0330 L""#

+
+

Step 3. Calculate the total concentration of #""Na""^""+""#

+

#[""Na""^""+""] = ""0.009 45 mol""/""0.0330 L"" = ""0.286 mol/L""#

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" An #18*mL# volume of #0.200*mol*L^-1# #Na_2SO_4# is mixed with a #15*mL# volume of #0.150*mol*L^-1# #NaCl#. What is the new concentration with respect to #[Na^+]#? nan +191 a82e0c2d-6ddd-11ea-878f-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-caffeine C4H5N2O start chemical_formula qc_end substance 6 6 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""C4H5N2O""}]" "[{""type"":""substance name"",""value"":""caffeine""}]" "

What is the empirical formula of caffeine?

" nan C4H5N2O "
+

Explanation:

+
+

To find the empirical formula for caffeine we begin with the Molecular (true) Formula

+

#C_8H_10N_4O_2#

+

We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.

+

#C_4H_5N_2O#

+

This is the Empirical Formula.

+

I hope this was beneficial.
+SMARTERTEACHER

+
+
" "
+
+
+

#""C""_4""H""_5""N""_2""O""#

+
+
+
+

Explanation:

+
+

To find the empirical formula for caffeine we begin with the Molecular (true) Formula

+

#C_8H_10N_4O_2#

+

We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.

+

#C_4H_5N_2O#

+

This is the Empirical Formula.

+

I hope this was beneficial.
+SMARTERTEACHER

+
+
+
" "
+

What is the empirical formula of caffeine?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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+1 Answer +
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+ + May 15, 2014 + +
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#""C""_4""H""_5""N""_2""O""#

+
+
+
+

Explanation:

+
+

To find the empirical formula for caffeine we begin with the Molecular (true) Formula

+

#C_8H_10N_4O_2#

+

We can then reduce the Molecular Formula to the Empirical (simple) Formula by dividing each of the subscripts by the greatest common factor. In this case we divide by 2.

+

#C_4H_5N_2O#

+

This is the Empirical Formula.

+

I hope this was beneficial.
+SMARTERTEACHER

+
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+
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+
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+
+
Related questions
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+ + Creative Commons License + +
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+
" What is the empirical formula of caffeine? nan +192 a82e0c2e-6ddd-11ea-899c-ccda262736ce https://socratic.org/questions/what-is-the-conjugate-base-of-hno-2 NO2^- start chemical_formula qc_end chemical_equation 6 6 qc_end c_other Conjugate_base qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""NO2^-""}]" "[{""type"":""chemical equation"",""value"":""HNO2""},{""type"":""other"",""value"":""Conjugate base""}]" "

What is the conjugate base of #HNO_2#?

" nan NO2^- "
+

Explanation:

+
+

The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (#H^+#) from it (every Brønsted-Lowry acid has a conjugate base, and vice versa). To exemplify this in a chemical reaction, let's have nitrous acid react with water:

+
+

#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#

+
+

Here, the Brønsted-Lowry acid, #HNO_2#, has donated a proton to #H_2O# to form #NO_2^-# and the hydronium ion, #H_3O^+#. This is the forward reaction; in the reverse reaction, #NO_2^-# is now the Brønsted-Lowry base (conjugate of #HNO_2#) because it accepts a proton from hydronium (conjugate acid of #H_2O#) to form nitrous acid again.

+
+
" "
+
+
+

#NO_2^-#

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+
+
+

Explanation:

+
+

The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (#H^+#) from it (every Brønsted-Lowry acid has a conjugate base, and vice versa). To exemplify this in a chemical reaction, let's have nitrous acid react with water:

+
+

#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#

+
+

Here, the Brønsted-Lowry acid, #HNO_2#, has donated a proton to #H_2O# to form #NO_2^-# and the hydronium ion, #H_3O^+#. This is the forward reaction; in the reverse reaction, #NO_2^-# is now the Brønsted-Lowry base (conjugate of #HNO_2#) because it accepts a proton from hydronium (conjugate acid of #H_2O#) to form nitrous acid again.

+
+
+
" "
+

What is the conjugate base of #HNO_2#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Conjugate Acids and Conjugate Bases + + +
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+2 Answers +
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+ + +
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+ +
+ + May 13, 2017 + +
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#NO_2^-#

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+
+
+

Explanation:

+
+

The conjugate base of any Brønsted-Lowry acid can be found by removing a proton (#H^+#) from it (every Brønsted-Lowry acid has a conjugate base, and vice versa). To exemplify this in a chemical reaction, let's have nitrous acid react with water:

+
+

#HNO_2 (aq) + H_2O (l) rightleftharpoons NO_2^(-) (aq) + H_3O^(+) (aq)#

+
+

Here, the Brønsted-Lowry acid, #HNO_2#, has donated a proton to #H_2O# to form #NO_2^-# and the hydronium ion, #H_3O^+#. This is the forward reaction; in the reverse reaction, #NO_2^-# is now the Brønsted-Lowry base (conjugate of #HNO_2#) because it accepts a proton from hydronium (conjugate acid of #H_2O#) to form nitrous acid again.

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+ +
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+ +
+ + May 13, 2017 + +
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+
+

#HNO_2 + HOH => H_3O^+ + NO_2^-#
+#NO_2^-# is the conjugate base of #HNO_2#

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+
+
+

Explanation:

+
+

The term 'conjugate base' in the chemical community is typically used in association with the term 'conjugate acid' and comes from the Bronsted-Lowry theory of Acids and Bases. The textbook definition is based upon the proton transfer relationship between an acidic substance that can donate a hydrogen ion (proton) to an alkaline substance that can accept the hydrogen ion forming a stronger bond with the accepting substance. The Bronsted-Acid is strictly defined as a proton donor and the Bronsted-Base a proton acceptior. These definitions imply the need for a chemical reaction in order to show 'donator' substance and 'acceptor' substance. The application origin is found in the 'Bronsted-Lowry' theory of acids and bases.

+

The following can be classified as a molecular form Bronsted-Lowry Acid-Base Rxn.

+

HCl(aq) + NaOH(aq) => NaCl(aq) + HOH(l)

+

In aqueous media, the the proton from the Hydrochloric Acid molecule is transferred to the Hydroxide ion of the Sodium Hydroxide molecule producing Sodium Chloride and molecular water. The essential driving force of the reaction is the tendency for the transferring Hydrogen to form a stronger, more stable bond with the Hydroxide ion and thus a weaker acid.

+

The products, once formed, are now themselves acids and bases but are referred to as 'Conjugate Acid' and 'Conjugate Base'. The water ( H-OH ) which now has the transferred Hydrogen is now a proton donor, or 'Conjugate Acid' of the reactant base (#OH^-#).The Sodium ion (#Na^+#) is a spectator ion in the reactive system and is nonreactive and functions as the counter ion to the anions in the reaction process. The 'Conjugate Base' is the Chloride anion #(Cl^-)# which has the potential of functioning as a proton acceptor. However neither the #H-OH or Cl^-# have strong enough electrostatic properties to interact in a proton transfer process of transferring the Hydrogen back to the Hydroxide ion. Nevertheless, the product species are generally referred to conjugate species of the Bronsted-Lowry proton transfer process.

+

Now, it should be noted that anions of negative charge are generally, and collectively referred to in the chemical community linguistically as 'Conjugate Bases' if the species are negative and as a 'Conjugate Acids' if the species are positive. The association is referred to as a 'Conjugate Acid-Base Pair'.

+

Some Examples:

+

The 'Conjugate Acid-Base Pair' for ...
+#HNO_2# => Acid = #HNO_2# and Conj Base = #NO_2^-#
+#HOBr# => Acid = #HOBr# and Conj Base #OBr^-#
+#HCO_3^-# => Acid = #HCO_3^-# and Conj Base #CO_3^(2-)#

+

The difference between a Conjugate Base and Acid of the conjugate base is the presence of 1 Hydrogen. That is, to any anion, adding 1 Hydrogen, gives the acid of that conjugate base.

+

Generally: #Anion^-# + #H^+# => #H-Anion#
+#Anion^-# => Conjugate Base of the acid #H-Anion#

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" What is the conjugate base of #HNO_2#? nan +193 a82e328c-6ddd-11ea-8932-ccda262736ce https://socratic.org/questions/how-many-kilojoules-are-released-when-8-2-g-of-water-condenses-at-100-c-and-cool 188.32 kilojoules start physical_unit 9 9 heat_energy kj qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 9 9 12 13 temperature qc_end physical_unit 9 9 17 18 temperature qc_end end "[{""type"":""physical unit"",""value"":""Heat released [OF] water [IN] kilojoules""}]" "[{""type"":""physical unit"",""value"":""188.32 kilojoules""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{8.2 g}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] water [=] \\pu{100 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{15 °C}""}]" "

How many kilojoules are released when 8.2 g of water condenses at 100°C and cools to 15°C?

" nan 188.32 kilojoules "
+

Explanation:

+
+

There are two steps to this thermochemistry process

+

+

First we are going to calculate the condensation process at #100^oC#.
+Then calculate the cooling of the liquid from #100^oC to 15^oC#

+

Step 1 #Q = mC_p#

+

#Q = 8.2g (-2261 J/g) = -185,402 J#

+

Step 2 #Q = m(T_f-T_i)C_p#

+

#Q = 8.2g(15^o-100^oC)4.18J/(g^oC)= -2,913.46 J#

+

#-185,402 + (-2,913.46) = -188315.46 J#

+

#Q = 1.9 x10^5 J#

+

#Q = 1.9 x10^2 kJ#

+

+ +

+
+
" "
+
+
+

#-185,402 + (-2,913.46) = -188315.46 J#

+

#Q = 1.9 x10^5 J#

+

#Q = 1.9 x10^2 kJ#

+
+
+
+

Explanation:

+
+

There are two steps to this thermochemistry process

+

+

First we are going to calculate the condensation process at #100^oC#.
+Then calculate the cooling of the liquid from #100^oC to 15^oC#

+

Step 1 #Q = mC_p#

+

#Q = 8.2g (-2261 J/g) = -185,402 J#

+

Step 2 #Q = m(T_f-T_i)C_p#

+

#Q = 8.2g(15^o-100^oC)4.18J/(g^oC)= -2,913.46 J#

+

#-185,402 + (-2,913.46) = -188315.46 J#

+

#Q = 1.9 x10^5 J#

+

#Q = 1.9 x10^2 kJ#

+

+ +

+
+
+
" "
+

How many kilojoules are released when 8.2 g of water condenses at 100°C and cools to 15°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
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+1 Answer +
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+ + Jun 2, 2016 + +
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#-185,402 + (-2,913.46) = -188315.46 J#

+

#Q = 1.9 x10^5 J#

+

#Q = 1.9 x10^2 kJ#

+
+
+
+

Explanation:

+
+

There are two steps to this thermochemistry process

+

+

First we are going to calculate the condensation process at #100^oC#.
+Then calculate the cooling of the liquid from #100^oC to 15^oC#

+

Step 1 #Q = mC_p#

+

#Q = 8.2g (-2261 J/g) = -185,402 J#

+

Step 2 #Q = m(T_f-T_i)C_p#

+

#Q = 8.2g(15^o-100^oC)4.18J/(g^oC)= -2,913.46 J#

+

#-185,402 + (-2,913.46) = -188315.46 J#

+

#Q = 1.9 x10^5 J#

+

#Q = 1.9 x10^2 kJ#

+

+ +

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+ +
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+
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+
Related questions
+ + +
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Impact of this question
+
+ 13913 views + around the world +
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+ + Creative Commons License + +
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+
" How many kilojoules are released when 8.2 g of water condenses at 100°C and cools to 15°C? nan +194 a82e328d-6ddd-11ea-a1a3-ccda262736ce https://socratic.org/questions/at-25-c-the-molar-solubility-of-silver-phosphate-is-1-8-10-5-mol-l-1-how-do-you- 2.86 × 10^(-2) M start physical_unit 20 21 equilibrium_constant_k mol/l qc_end physical_unit 7 8 1 2 temperature qc_end physical_unit 7 8 10 13 molar_solubility qc_end end "[{""type"":""physical unit"",""value"":""Ksp [OF] this salt [IN] M""}]" "[{""type"":""physical unit"",""value"":""2.86 × 10^(-2) M""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] silver phosphate [=] \\pu{25 °C}""},{""type"":""physical unit"",""value"":""Molar solubility [OF] silver phosphate [=] \\pu{1.8 × 10^(−5) mol*L^(-1)}""}]" "

At 25 °C, the molar solubility of silver phosphate is #1.8 × 10^-5# mol L-1. How do you calculate Ksp for this salt? + +

" nan 2.86 × 10^(-2) M "
+

Explanation:

+
+

Let's start by writing the chemical reaction for the dissociation of silver phosphate:

+

#Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq)#

+

Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.

+

#Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq)#

+

Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.

+

#1.8xx10^(-5) M = (3X)^(3) xx (X)# (always multiply when finding the molar solubility).

+

Now take #3^(3)# which is 27 and divide the Ksp by 27, so you can get all of the X's by themselves.

+

#(1.8xx10^(-5) M)/27 # = #6.67xx10^(-7)M#

+

Now you're left with #X^(3) xx X#, so multiply the X's to get #X^(4)#. Take the fourth root of the #6.67xx10^(-7)M# to obtain the value of x.

+

(#6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M#

+

The value of x that we just obtained is our molar solubility.

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+
" "
+
+
+

#2.9xx10^(-2) M#

+
+
+
+

Explanation:

+
+

Let's start by writing the chemical reaction for the dissociation of silver phosphate:

+

#Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq)#

+

Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.

+

#Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq)#

+

Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.

+

#1.8xx10^(-5) M = (3X)^(3) xx (X)# (always multiply when finding the molar solubility).

+

Now take #3^(3)# which is 27 and divide the Ksp by 27, so you can get all of the X's by themselves.

+

#(1.8xx10^(-5) M)/27 # = #6.67xx10^(-7)M#

+

Now you're left with #X^(3) xx X#, so multiply the X's to get #X^(4)#. Take the fourth root of the #6.67xx10^(-7)M# to obtain the value of x.

+

(#6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M#

+

The value of x that we just obtained is our molar solubility.

+
+
+
" "
+

At 25 °C, the molar solubility of silver phosphate is #1.8 × 10^-5# mol L-1. How do you calculate Ksp for this salt? + +

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Ksp + + +
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+1 Answer +
+
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+ + Jun 10, 2016 + +
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+

#2.9xx10^(-2) M#

+
+
+
+

Explanation:

+
+

Let's start by writing the chemical reaction for the dissociation of silver phosphate:

+

#Ag_3PO_4(s) rightleftharpoons 3Ag^(+)(aq) +PO_4^(-3)(aq)#

+

Now, set the Ksp value equal to the products (you don't care about the reactants because it's a solid). Ag is raised to the 3rd power because the coefficient is 3.

+

#Ksp=1.8xx10^(-5) M = [Ag^(+)]^(3) (aq) +[PO_4^(3-)] (aq)#

+

Replace each reactant with the letter x because that's what you're trying to find. Since the coefficient in front of silver is 3, a 3 must be placed in front of the x and it must be raised to the 3rd power.

+

#1.8xx10^(-5) M = (3X)^(3) xx (X)# (always multiply when finding the molar solubility).

+

Now take #3^(3)# which is 27 and divide the Ksp by 27, so you can get all of the X's by themselves.

+

#(1.8xx10^(-5) M)/27 # = #6.67xx10^(-7)M#

+

Now you're left with #X^(3) xx X#, so multiply the X's to get #X^(4)#. Take the fourth root of the #6.67xx10^(-7)M# to obtain the value of x.

+

(#6.67xx10^(-7)M)^(1/4) = 2.9xx10^(-2) M#

+

The value of x that we just obtained is our molar solubility.

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+
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" "At 25 °C, the molar solubility of silver phosphate is #1.8 × 10^-5# mol L-1. How do you calculate Ksp for this salt? + +" nan +195 a82e328e-6ddd-11ea-83ce-ccda262736ce https://socratic.org/questions/294-g-of-potassium-dichromate-contains-52-g-of-chromium-and-39-g-of-potassium-wh 69.05% start physical_unit 21 24 mass_percent none qc_end physical_unit 3 4 0 1 mass qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 3 3 11 12 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen in the compound""}]" "[{""type"":""physical unit"",""value"":""69.05%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] potassium dichromate [=] \\pu{294 g}""},{""type"":""physical unit"",""value"":""Mass [OF] chromium [=] \\pu{52 g}""},{""type"":""physical unit"",""value"":""Mass [OF] potassium [=] \\pu{39 g}""}]" "

294 g of potassium dichromate contains 52 g of chromium and 39 g of potassium. What is the mass percent of oxygen in the compound?

" nan 69.05% "
+

Explanation:

+
+

You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.

+

#""%Oxygen by mass""# #=# #((294-52-39)*g)/(294*g)xx100%# #=# #??%#

+

I make that about #66%#, but don't trust my arithmetic!

+
+
" "
+
+
+

Well, clearly it is #""Mass of oxygen""/""Mass of potassium dichromate""# #xx# #100%#.

+
+
+
+

Explanation:

+
+

You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.

+

#""%Oxygen by mass""# #=# #((294-52-39)*g)/(294*g)xx100%# #=# #??%#

+

I make that about #66%#, but don't trust my arithmetic!

+
+
+
" "
+

294 g of potassium dichromate contains 52 g of chromium and 39 g of potassium. What is the mass percent of oxygen in the compound?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
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+1 Answer +
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+ + May 21, 2016 + +
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Well, clearly it is #""Mass of oxygen""/""Mass of potassium dichromate""# #xx# #100%#.

+
+
+
+

Explanation:

+
+

You have already down the hard yards: the mass of oxygen is the mass of the sample LESS the mass of the metals.

+

#""%Oxygen by mass""# #=# #((294-52-39)*g)/(294*g)xx100%# #=# #??%#

+

I make that about #66%#, but don't trust my arithmetic!

+
+
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+ +
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+
+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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" 294 g of potassium dichromate contains 52 g of chromium and 39 g of potassium. What is the mass percent of oxygen in the compound? nan +196 a82e328f-6ddd-11ea-bdc7-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-aluminum-nitrite Al(NO2)3 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""Al(NO2)3""}]" "[{""type"":""substance name"",""value"":""Aluminum nitrite""}]" "

What is the formula for aluminum nitrite?

" nan Al(NO2)3 "
+

Explanation:

+
+

Aluminum nitrite consists of the aluminum cation #""Al""^(3+)""# and the polyatomic nitrite anion #""NO""_2^(-)""#. Since an ionic compound must be neutral, the number of each ion must result in an overall charge of zero.

+

Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is #""Al(NO""_2)_3""#.

+
+
" "
+
+
+

The formula for aluminum nitrite is #""Al(NO""_2)_3""#.

+
+
+
+

Explanation:

+
+

Aluminum nitrite consists of the aluminum cation #""Al""^(3+)""# and the polyatomic nitrite anion #""NO""_2^(-)""#. Since an ionic compound must be neutral, the number of each ion must result in an overall charge of zero.

+

Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is #""Al(NO""_2)_3""#.

+
+
+
" "
+

What is the formula for aluminum nitrite?

+
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+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
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+1 Answer +
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+ + Apr 18, 2016 + +
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The formula for aluminum nitrite is #""Al(NO""_2)_3""#.

+
+
+
+

Explanation:

+
+

Aluminum nitrite consists of the aluminum cation #""Al""^(3+)""# and the polyatomic nitrite anion #""NO""_2^(-)""#. Since an ionic compound must be neutral, the number of each ion must result in an overall charge of zero.

+

Since the aluminum ion has a charge of 3+, and the nitrite ion has a charge of 1-, there need to be three nitrite ions for every aluminum ion. Therefore, the chemical formula for aluminum nitrite is #""Al(NO""_2)_3""#.

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+
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" What is the formula for aluminum nitrite? nan +197 a82e3290-6ddd-11ea-93b2-ccda262736ce https://socratic.org/questions/calculate-the-volume-of-ammonia-gas-produced-at-stp-when-140g-of-nitrogen-gas-re 224.00 L start physical_unit 4 5 volume l qc_end c_other STP qc_end physical_unit 13 14 10 11 mass qc_end physical_unit 20 21 17 18 mass qc_end physical_unit 23 23 23 24 atomic_mass qc_end physical_unit 24 24 24 25 atomic_mass qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] ammonia gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""224.00 L""}]" "[{""type"":""other"",""value"":""STP""},{""type"":""physical unit"",""value"":""Mass [OF] nitrogen gas [=] \\pu{140 g}""},{""type"":""physical unit"",""value"":""Mass [OF] hydrogen gas [=] \\pu{30 g}""},{""type"":""physical unit"",""value"":""Atomic mass [OF] N [=] \\pu{14 u}""},{""type"":""physical unit"",""value"":""Atomic mass [OF] H [=] \\pu{1 u}""}]" "

Calculate the volume of ammonia gas produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas.(atomic mass: N=14u,H=1u) how to do this?

" nan 224.00 L "
+

Explanation:

+
+

Simply, the reaction is #N_2+3H_2=2NH_3#
+#140g# #""N""_2# means #140/28=5# mol;
+#30g# #H_2# means #30/2=15#mol
+From 1 mol of #N_2# 2 mol #NH_3# is prepared in perfect presence of #H_2#
+Hence, #10#mol of #NH_3# will be produced.
+We know,1mol of any ideal gas at STP (#0^@ ""C""# and #""1 atm""#) contains 22.4L of the ideal gas.

+

Produced ammonia is #224"" L""#.

+
+
" "
+
+
+

Ans:#color(red)((10×22.4L=224""L"")#

+
+
+
+

Explanation:

+
+

Simply, the reaction is #N_2+3H_2=2NH_3#
+#140g# #""N""_2# means #140/28=5# mol;
+#30g# #H_2# means #30/2=15#mol
+From 1 mol of #N_2# 2 mol #NH_3# is prepared in perfect presence of #H_2#
+Hence, #10#mol of #NH_3# will be produced.
+We know,1mol of any ideal gas at STP (#0^@ ""C""# and #""1 atm""#) contains 22.4L of the ideal gas.

+

Produced ammonia is #224"" L""#.

+
+
+
" "
+

Calculate the volume of ammonia gas produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas.(atomic mass: N=14u,H=1u) how to do this?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Equation Stoichiometry + + +
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+1 Answer +
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+ + Aug 18, 2017 + +
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Ans:#color(red)((10×22.4L=224""L"")#

+
+
+
+

Explanation:

+
+

Simply, the reaction is #N_2+3H_2=2NH_3#
+#140g# #""N""_2# means #140/28=5# mol;
+#30g# #H_2# means #30/2=15#mol
+From 1 mol of #N_2# 2 mol #NH_3# is prepared in perfect presence of #H_2#
+Hence, #10#mol of #NH_3# will be produced.
+We know,1mol of any ideal gas at STP (#0^@ ""C""# and #""1 atm""#) contains 22.4L of the ideal gas.

+

Produced ammonia is #224"" L""#.

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" Calculate the volume of ammonia gas produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas.(atomic mass: N=14u,H=1u) how to do this? nan +198 a82e3291-6ddd-11ea-a441-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-magnesium-sulfate-heptahydrate MgSO4.7H2O start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] magnesium sulfate heptahydrate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""MgSO4.7H2O""}]" "[{""type"":""substance name"",""value"":""Magnesium sulfate heptahydrate""}]" "

What is the formula for magnesium sulfate heptahydrate?

" nan MgSO4.7H2O "
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

+
+
" "
+
+
+

#MgSO_4*7H_2O#

+
+
+
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

+
+
+
" "
+

What is the formula for magnesium sulfate heptahydrate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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+1 Answer +
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+ + Mar 12, 2018 + +
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+

#MgSO_4*7H_2O#

+
+
+
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

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+
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" What is the formula for magnesium sulfate heptahydrate? nan +199 a82e3292-6ddd-11ea-b9c3-ccda262736ce https://socratic.org/questions/mg-oh-2-has-a-solubility-of-9-0x10-5-m-what-is-the-value-of-ksp-for-mg-oh-2 2.90 × 10^(-12) start physical_unit 0 0 equilibrium_constant_k none qc_end physical_unit 0 0 5 8 solubility qc_end end "[{""type"":""physical unit"",""value"":""Ksp [OF] Mg(OH)2""}]" "[{""type"":""physical unit"",""value"":""2.90 × 10^(-12)""}]" "[{""type"":""physical unit"",""value"":""Solubility [OF] Mg(OH)2 [=] \\pu{9.0 x 10^(-5) M}""}]" "

Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?

" "
+
+

+

Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?

+

+
+
" 2.90 × 10^(-12) "
+

Explanation:

+
+
+

The chemical equation for the equilibrium is

+

#color(white)(mmmmmm)""Mg(OH)""_2""(s)"" ⇌ ""Mg""^(2+)""(aq)"" + 2""OH""^""-""""(aq)""#
+#""E:/mol·L""^""-1"" color(white)(mmmmmmmmllm)x color(white)(mmmmmll)2x#

+
+

#K_""sp"" = [""Mg""^(2+)][""OH""^""-""]^2 = x(2x)^2 = 4x^3#

+

#x = 9.0 × 10^""-5""#

+

∴ #K_""sp"" = = 4(9.0 × 10^""-5"")^3 = 2.9 × 10^""-12""#

+
+
" "
+
+
+

#K_""sp"" = 2.9 × 10^""-12""#

+
+
+
+

Explanation:

+
+
+

The chemical equation for the equilibrium is

+

#color(white)(mmmmmm)""Mg(OH)""_2""(s)"" ⇌ ""Mg""^(2+)""(aq)"" + 2""OH""^""-""""(aq)""#
+#""E:/mol·L""^""-1"" color(white)(mmmmmmmmllm)x color(white)(mmmmmll)2x#

+
+

#K_""sp"" = [""Mg""^(2+)][""OH""^""-""]^2 = x(2x)^2 = 4x^3#

+

#x = 9.0 × 10^""-5""#

+

∴ #K_""sp"" = = 4(9.0 × 10^""-5"")^3 = 2.9 × 10^""-12""#

+
+
+
" "
+

Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?

+
+
+

+

Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2?

+

+
+
+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Solubility Equilbria + + +
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+1 Answer +
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+ + Jun 2, 2016 + +
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#K_""sp"" = 2.9 × 10^""-12""#

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+
+
+

Explanation:

+
+
+

The chemical equation for the equilibrium is

+

#color(white)(mmmmmm)""Mg(OH)""_2""(s)"" ⇌ ""Mg""^(2+)""(aq)"" + 2""OH""^""-""""(aq)""#
+#""E:/mol·L""^""-1"" color(white)(mmmmmmmmllm)x color(white)(mmmmmll)2x#

+
+

#K_""sp"" = [""Mg""^(2+)][""OH""^""-""]^2 = x(2x)^2 = 4x^3#

+

#x = 9.0 × 10^""-5""#

+

∴ #K_""sp"" = = 4(9.0 × 10^""-5"")^3 = 2.9 × 10^""-12""#

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+
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+ + +
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+
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+ + Creative Commons License + +
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+
" Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2? " + + +Mg(OH)2 has a solubility of 9.0x10^-5 M. What is the value of Ksp for Mg(OH)2? + + +" +200 a82e5858-6ddd-11ea-b489-ccda262736ce https://socratic.org/questions/581f69017c014944234fd2b9 0.17 moles start physical_unit 18 18 mole mol qc_end substance 3 3 qc_end substance 5 6 qc_end physical_unit 0 1 8 9 temperature qc_end physical_unit 0 1 11 13 pressure qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 23 23 33 35 vapor_pressure qc_end physical_unit 23 23 8 9 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] hydrogen [IN] mole""}]" "[{""type"":""physical unit"",""value"":""0.17 moles""}]" "[{""type"":""substance name"",""value"":""magnesium""},{""type"":""substance name"",""value"":""hydrochloric acid""},{""type"":""physical unit"",""value"":""Temperature [OF] the reaction [=] \\pu{30 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] the reaction [=] \\pu{817.36 mm Hg}""},{""type"":""physical unit"",""value"":""Volume [OF] hydrogen [=] \\pu{4.03 L}""},{""type"":""physical unit"",""value"":""vapour pressure [OF] water [=] \\pu{33.4 mm Hg}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{30 ℃}""},{""type"":""other"",""value"":""hydrogen that was collected over water.""}]" "

The reaction of magnesium with hydrochloric acid at 30 °C and 817.36 mmHg produced 4.03 L of hydrogen that was collected over water. The vapour pressure of water at 30 °C is 33.4 mmHg. How many moles of hydrogen were formed?

" nan 0.17 moles "
+

Explanation:

+
+

The Reaction:

+

#Mg(s)+2HCl(aq)\rightarrowMgCl_2(aq)+H_2(g)#

+
+

Data given:

+

Pressure #P=""817.36 mmHg""#
+Volume #V=""4.03 L""#
+Temperature #T=""303 K""#

+
+

Work :

+

#P_""total"" = P_""H₂"" + P_""H₂O""#

+

At 303 K, #P_""H₂O"" = ""33.4 mmHg""#

+

∴ #P_""H₂"" = P_""total"" - P_""H₂O"" = ""817.36 mmHg"" - ""33.4 mmHg"" = ""783.96 mmHg"" = ""1.0315 atm""#

+

Apply the Ideal Gas Law #PV=nRT# to get #n = \frac(PV)(RT)#

+

The constant #R = ""0.0821 L·atm·K""^""-1""""mol""^""-1""#

+

#\frac(""1.0315 atm""\times""4.03 L"")(""0.0821 L·atm·K""^""-1""""mol""^""-1""\times""303 K"") = 1.67\times10^-1color(white)(l)""mol""#

+
+
" "
+
+
+

#1.67\times10^-1color(white)(l)""mol""#

+
+
+
+

Explanation:

+
+

The Reaction:

+

#Mg(s)+2HCl(aq)\rightarrowMgCl_2(aq)+H_2(g)#

+
+

Data given:

+

Pressure #P=""817.36 mmHg""#
+Volume #V=""4.03 L""#
+Temperature #T=""303 K""#

+
+

Work :

+

#P_""total"" = P_""H₂"" + P_""H₂O""#

+

At 303 K, #P_""H₂O"" = ""33.4 mmHg""#

+

∴ #P_""H₂"" = P_""total"" - P_""H₂O"" = ""817.36 mmHg"" - ""33.4 mmHg"" = ""783.96 mmHg"" = ""1.0315 atm""#

+

Apply the Ideal Gas Law #PV=nRT# to get #n = \frac(PV)(RT)#

+

The constant #R = ""0.0821 L·atm·K""^""-1""""mol""^""-1""#

+

#\frac(""1.0315 atm""\times""4.03 L"")(""0.0821 L·atm·K""^""-1""""mol""^""-1""\times""303 K"") = 1.67\times10^-1color(white)(l)""mol""#

+
+
+
" "
+

The reaction of magnesium with hydrochloric acid at 30 °C and 817.36 mmHg produced 4.03 L of hydrogen that was collected over water. The vapour pressure of water at 30 °C is 33.4 mmHg. How many moles of hydrogen were formed?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
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+
+
+1 Answer +
+
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+ + +
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+ +
+ + Nov 12, 2016 + +
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#1.67\times10^-1color(white)(l)""mol""#

+
+
+
+

Explanation:

+
+

The Reaction:

+

#Mg(s)+2HCl(aq)\rightarrowMgCl_2(aq)+H_2(g)#

+
+

Data given:

+

Pressure #P=""817.36 mmHg""#
+Volume #V=""4.03 L""#
+Temperature #T=""303 K""#

+
+

Work :

+

#P_""total"" = P_""H₂"" + P_""H₂O""#

+

At 303 K, #P_""H₂O"" = ""33.4 mmHg""#

+

∴ #P_""H₂"" = P_""total"" - P_""H₂O"" = ""817.36 mmHg"" - ""33.4 mmHg"" = ""783.96 mmHg"" = ""1.0315 atm""#

+

Apply the Ideal Gas Law #PV=nRT# to get #n = \frac(PV)(RT)#

+

The constant #R = ""0.0821 L·atm·K""^""-1""""mol""^""-1""#

+

#\frac(""1.0315 atm""\times""4.03 L"")(""0.0821 L·atm·K""^""-1""""mol""^""-1""\times""303 K"") = 1.67\times10^-1color(white)(l)""mol""#

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+
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" The reaction of magnesium with hydrochloric acid at 30 °C and 817.36 mmHg produced 4.03 L of hydrogen that was collected over water. The vapour pressure of water at 30 °C is 33.4 mmHg. How many moles of hydrogen were formed? nan +201 a82e5f7e-6ddd-11ea-878c-ccda262736ce https://socratic.org/questions/5704349611ef6b1a120c3424 0.42 Osmol/L start physical_unit 5 6 osmolarity osmol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 6 6 14 15 volume qc_end end "[{""type"":""physical unit"",""value"":""osmolarity [OF] a solution [IN] Osmol/L""}]" "[{""type"":""physical unit"",""value"":""0.42 Osmol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] urea [=] \\pu{25 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{0.4163 L}""}]" "

What is the osmolarity of a solution that contains #""25 g""# of urea in #""0.4163 L""# of solution?

" nan 0.42 Osmol/L "
+

Explanation:

+
+

A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.

+

A particle that contributes to a solution's osmotic pressure is called an osmole.

+

Now, urea, #""CO""(""NH""_2)_2#, is a non-electrolyte, which means that it does not dissociate into ions when dissolved in water. Simply put, every molecule of urea dissolved in solution will not dissociate to produce additional particles of solute.

+

This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.

+

Use the molar mass of urea to calculate how many moles you have in that sample

+
+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#

+
+

Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to

+
+

#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.

+
+
" "
+
+
+

#""0.42 Osmol L""^(-1)#

+
+
+
+

Explanation:

+
+

A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.

+

A particle that contributes to a solution's osmotic pressure is called an osmole.

+

Now, urea, #""CO""(""NH""_2)_2#, is a non-electrolyte, which means that it does not dissociate into ions when dissolved in water. Simply put, every molecule of urea dissolved in solution will not dissociate to produce additional particles of solute.

+

This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.

+

Use the molar mass of urea to calculate how many moles you have in that sample

+
+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#

+
+

Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to

+
+

#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.

+
+
+
" "
+

What is the osmolarity of a solution that contains #""25 g""# of urea in #""0.4163 L""# of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Osmolarity + + +
+
+
+
+
+1 Answer +
+
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+ +
+
+ +
+ + Jul 15, 2016 + +
+
+
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+
+

#""0.42 Osmol L""^(-1)#

+
+
+
+

Explanation:

+
+

A solution's osmolarity basically tells you the number of moles of particles that contribute to the solution's osmotic pressure present in one liter of solution.

+

A particle that contributes to a solution's osmotic pressure is called an osmole.

+

Now, urea, #""CO""(""NH""_2)_2#, is a non-electrolyte, which means that it does not dissociate into ions when dissolved in water. Simply put, every molecule of urea dissolved in solution will not dissociate to produce additional particles of solute.

+

This implies that every mole of urea dissolved in solution will produce one osmole of particles of solute.

+

Use the molar mass of urea to calculate how many moles you have in that sample

+
+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole urea""/(60.055color(red)(cancel(color(black)(""g"")))) = ""0.4163 moles urea""#

+
+

Assuming that the volume of the solutionThis means that the solution's osmolarity will be equal to

+
+

#0.4163 color(white)(a)color(red)(cancel(color(black)(""moles"")))/""1 L"" * ""1 Osmol""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.42 Osmol L""^(-1))color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the volume of the solution.

+
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+
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+
+
+
Related questions
+ + +
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Impact of this question
+
+ 5586 views + around the world +
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+ + Creative Commons License + +
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+
" "What is the osmolarity of a solution that contains #""25 g""# of urea in #""0.4163 L""# of solution?" nan +202 a82e5f7f-6ddd-11ea-a069-ccda262736ce https://socratic.org/questions/3-6-g-of-oxalic-acid-h-2c-2o-4-2h-2o-are-dissolved-and-the-volume-is-made-upto-1 0.38 N start physical_unit 21 25 normality n qc_end physical_unit 5 5 0 1 mass qc_end physical_unit 21 22 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Normality [OF] the solution for redox titration [IN] N""}]" "[{""type"":""physical unit"",""value"":""0.38 N""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2C2O4.2H2O [=] \\pu{3.6 g}""},{""type"":""physical unit"",""value"":""Volume [OF] the solution [=] \\pu{150 mL}""}]" "

3.6 #g# of oxalic acid (#H_2C_2O_4 * 2H_2O#) are dissolved and the volume is made upto 150 #ml#. Calculate the normality of the solution for redox titration?

" nan 0.38 N "
+

Explanation:

+
+

The purpose of using oxalic acid dihydrate, #H_2C_2O_4 * 2H_2O#, in a redox titration is that the oxalate ion, #C_2O_4^(2-)#, will act as a reducing agent and reduce, for example, potassium permanganate, #KMnO_4#, to the #Mn^(2+)# ion.

+

In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.

+

These electrons will help you determine the equivalents needed to calculate the solution's normality.

+

Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this

+

#6H_((aq))^(+) + 5C_2O_(4(aq))^(2-) + 2MnO_(4(aq))^(-) -> 2Mn_((aq))^(2+) + 10CO_(2(g)) + 8H_2O_((l))#

+

I will show you how to balance the oxidation half-reaction, in which the oxalate ion, #C_2O_4^(2-)#, is oxidized to carbon dioxide.

+

This half-reaction will tell you how many electrons are being lost by the reducing agent.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> stackrel(color(blue)(+4))(C)O_2#

+

The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.

+

However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2#

+

This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2 + 2e^(-)#

+

This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have

+

#""1 mole ""H_2C_2O_4 * 2H_2O = ""2 moles equivalents""#

+

Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.

+

Oxalic acid dihydrate contains

+

#(90.03cancel(""g/mol""))/(126.07cancel(""g/mol"")) * 100 = ""71.4% oxalic acid""#

+

by mass. This means that your 3.6-g sample of dihydrate actually contains

+

#3.6cancel(""g dihydrate"") * ""71.4 g oxalic acid""/(100cancel(""g dihydrate"")) = ""2.57 g""# #C_2H_2O_4#

+

The number of moles of oxalic acid will be

+

#2.57cancel(""g"") * ""1 mole""/(90.03 cancel(""g"")) = ""0.0285 moles""#

+

This means that you get

+

#0.0285cancel(""moles"") * ""2 moles of eq.""/(1cancel(""mole"")) = ""0.057 moles eq.""#

+

The solution's normality will thus be

+

#N = ""no. of moles eq.""/V#

+

#N = ""0.057 moles eq.""/(150 * 10^(-3)""L"") = color(green)(""0.38 N"")#

+
+
" "
+
+
+

Normality: 0.38 N.

+
+
+
+

Explanation:

+
+

The purpose of using oxalic acid dihydrate, #H_2C_2O_4 * 2H_2O#, in a redox titration is that the oxalate ion, #C_2O_4^(2-)#, will act as a reducing agent and reduce, for example, potassium permanganate, #KMnO_4#, to the #Mn^(2+)# ion.

+

In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.

+

These electrons will help you determine the equivalents needed to calculate the solution's normality.

+

Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this

+

#6H_((aq))^(+) + 5C_2O_(4(aq))^(2-) + 2MnO_(4(aq))^(-) -> 2Mn_((aq))^(2+) + 10CO_(2(g)) + 8H_2O_((l))#

+

I will show you how to balance the oxidation half-reaction, in which the oxalate ion, #C_2O_4^(2-)#, is oxidized to carbon dioxide.

+

This half-reaction will tell you how many electrons are being lost by the reducing agent.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> stackrel(color(blue)(+4))(C)O_2#

+

The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.

+

However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2#

+

This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2 + 2e^(-)#

+

This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have

+

#""1 mole ""H_2C_2O_4 * 2H_2O = ""2 moles equivalents""#

+

Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.

+

Oxalic acid dihydrate contains

+

#(90.03cancel(""g/mol""))/(126.07cancel(""g/mol"")) * 100 = ""71.4% oxalic acid""#

+

by mass. This means that your 3.6-g sample of dihydrate actually contains

+

#3.6cancel(""g dihydrate"") * ""71.4 g oxalic acid""/(100cancel(""g dihydrate"")) = ""2.57 g""# #C_2H_2O_4#

+

The number of moles of oxalic acid will be

+

#2.57cancel(""g"") * ""1 mole""/(90.03 cancel(""g"")) = ""0.0285 moles""#

+

This means that you get

+

#0.0285cancel(""moles"") * ""2 moles of eq.""/(1cancel(""mole"")) = ""0.057 moles eq.""#

+

The solution's normality will thus be

+

#N = ""no. of moles eq.""/V#

+

#N = ""0.057 moles eq.""/(150 * 10^(-3)""L"") = color(green)(""0.38 N"")#

+
+
+
" "
+

3.6 #g# of oxalic acid (#H_2C_2O_4 * 2H_2O#) are dissolved and the volume is made upto 150 #ml#. Calculate the normality of the solution for redox titration?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Titration Calculations + + +
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+1 Answer +
+
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+ + Jul 12, 2015 + +
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+

Normality: 0.38 N.

+
+
+
+

Explanation:

+
+

The purpose of using oxalic acid dihydrate, #H_2C_2O_4 * 2H_2O#, in a redox titration is that the oxalate ion, #C_2O_4^(2-)#, will act as a reducing agent and reduce, for example, potassium permanganate, #KMnO_4#, to the #Mn^(2+)# ion.

+

In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.

+

These electrons will help you determine the equivalents needed to calculate the solution's normality.

+

Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this

+

#6H_((aq))^(+) + 5C_2O_(4(aq))^(2-) + 2MnO_(4(aq))^(-) -> 2Mn_((aq))^(2+) + 10CO_(2(g)) + 8H_2O_((l))#

+

I will show you how to balance the oxidation half-reaction, in which the oxalate ion, #C_2O_4^(2-)#, is oxidized to carbon dioxide.

+

This half-reaction will tell you how many electrons are being lost by the reducing agent.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> stackrel(color(blue)(+4))(C)O_2#

+

The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.

+

However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2#

+

This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.

+

#stackrel(color(blue)(+3))(C_2) O_4""""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2 + 2e^(-)#

+

This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have

+

#""1 mole ""H_2C_2O_4 * 2H_2O = ""2 moles equivalents""#

+

Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.

+

Oxalic acid dihydrate contains

+

#(90.03cancel(""g/mol""))/(126.07cancel(""g/mol"")) * 100 = ""71.4% oxalic acid""#

+

by mass. This means that your 3.6-g sample of dihydrate actually contains

+

#3.6cancel(""g dihydrate"") * ""71.4 g oxalic acid""/(100cancel(""g dihydrate"")) = ""2.57 g""# #C_2H_2O_4#

+

The number of moles of oxalic acid will be

+

#2.57cancel(""g"") * ""1 mole""/(90.03 cancel(""g"")) = ""0.0285 moles""#

+

This means that you get

+

#0.0285cancel(""moles"") * ""2 moles of eq.""/(1cancel(""mole"")) = ""0.057 moles eq.""#

+

The solution's normality will thus be

+

#N = ""no. of moles eq.""/V#

+

#N = ""0.057 moles eq.""/(150 * 10^(-3)""L"") = color(green)(""0.38 N"")#

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" 3.6 #g# of oxalic acid (#H_2C_2O_4 * 2H_2O#) are dissolved and the volume is made upto 150 #ml#. Calculate the normality of the solution for redox titration? nan +203 a82e8536-6ddd-11ea-83f8-ccda262736ce https://socratic.org/questions/how-do-you-balance-the-equation-al-hcl-alcl-3-h2 2 Al + 6 HCl -> 2 AlCl3 + 3 H2 start chemical_equation qc_end chemical_equation 6 12 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 Al + 6 HCl -> 2 AlCl3 + 3 H2""}]" "[{""type"":""chemical equation"",""value"":""Al + HCl -> AlCl3 + H2""}]" "

How do you balance the equation #Al + HCl -> AlCl_3 + H2#?

" nan 2 Al + 6 HCl -> 2 AlCl3 + 3 H2 "
+

Explanation:

+
+

This is NOT the stoichiometric equation:

+

#Al(s) + 3HCl(aq) rarr AlCl_3(aq) + 2H_2(g)#

+

How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients, #1/2# or #3/2#!

+
+
" "
+
+
+

You balance it stoichiometrically.

+
+
+
+

Explanation:

+
+

This is NOT the stoichiometric equation:

+

#Al(s) + 3HCl(aq) rarr AlCl_3(aq) + 2H_2(g)#

+

How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients, #1/2# or #3/2#!

+
+
+
" "
+

How do you balance the equation #Al + HCl -> AlCl_3 + H2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+2 Answers +
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+ + Nov 30, 2015 + +
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You balance it stoichiometrically.

+
+
+
+

Explanation:

+
+

This is NOT the stoichiometric equation:

+

#Al(s) + 3HCl(aq) rarr AlCl_3(aq) + 2H_2(g)#

+

How would you modify it so that the left had side is equivalent to the right hand side? You can certainly use half-coeffcients, #1/2# or #3/2#!

+
+
+
+
+
+ +
+
+
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+ +
+
+ +
+ + Nov 30, 2015 + +
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+

#""2Al(s)"" + ""6HCl(aq)""##rarr##""2AlCl""_3(""aq"") + ""3H""_2(""g"")""#

+
+
+
+

Explanation:

+
+

#""Al(s)"" + ""HCl(aq)""##rarr##""AlCl""_3(""aq"") + ""H""_2(""g"")""# is unbalanced.

+

Balance Cl.
+There are 3 Cl atoms on the product side and 1 on the reactant side. Add a coefficient of 3 in front of #""HCl""#.

+

#""Al(s)"" + ""3HCl(aq)""##rarr##""AlCl""_3(""aq"") + ""H""_2(""g"")""#

+

There are now 3 Cl atoms on both sides.

+

Balance H.
+There are 3 H atoms on the reactant side and 2 on the product side. 3 and 2 are multiples of 6. So change the coefficient in front of HCl from 3 to 6, and add a coefficient of 3 in front of the #""H""_2""#.

+

#""Al(s)"" + ""6HCl(aq)""##rarr##""AlCl""_3(""aq"") + ""3H""_2(""g"")""#

+

There are now 6 H atoms on both sides.

+

Go back to the Cl. There are now 6 Cl on the reactant side and 3 on the product side. Add a coefficient of 2 in front of #""AlCl""_3""#.

+

#""Al(s)"" + ""6HCl(aq)""##rarr##""2AlCl""_3(""aq"") + ""3H""_2(""g"")""#

+

There are now 6 Cl atoms on both sides.

+

Balance the Al
+There is 1 Al on the reactant side and 2 Al on the product side. Add a coefficient in front of Al on the reactant side.

+

#""2Al(s)"" + ""6HCl(aq)""##rarr##""2AlCl""_3(""aq"") + ""3H""_2(""g"")""#

+

There are now 2 Al on both sides and the equation is balanced.

+

Reactants: #""2 Al atoms"",##""6 H atoms"",##""6 Cl atoms""#
+Products: #""2 Al atoms"",##""6 H atoms"",##""6 Cl atoms""#

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+
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" How do you balance the equation #Al + HCl -> AlCl_3 + H2#? nan +204 a82e8537-6ddd-11ea-9623-ccda262736ce https://socratic.org/questions/58ce9b21b72cff427819c3a2 26.47 g start physical_unit 9 9 mass g qc_end physical_unit 6 6 3 4 mass qc_end c_other combustion qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] water [IN] g""}]" "[{""type"":""physical unit"",""value"":""26.47 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] methanol [=] \\pu{23.6 g}""},{""type"":""other"",""value"":""combust""}]" "

If we combust #23.6*g# of methanol, how much water will be produced?

" nan 26.47 g "
+

Explanation:

+
+

And we need a stoichiometric equation that represents the complete combustion of methanol:

+

#H_3COH(l) + 3/2O_2(g) rarr CO_2(g) + 2H_2O(l)#

+

Is it balanced? It looks like it to me.

+

And so we combust #23.6*g# of methanol, which represents a molar quantity of:

+

#(23.6*g)/(32.04*g*mol^-1)=0.737*mol#

+

Given the stoichiometry, at most, we can get #2xx0.737*mol=1.47*mol# water. And this represents a mass of #1.47*molxx18.01*g*mol^-1=26.5*g#, which was a book option. Claro?

+

If you are unsatisfied, I am willing to try again.

+
+
" "
+
+
+

I get #26.5*g# of water from the combustion.......

+
+
+
+

Explanation:

+
+

And we need a stoichiometric equation that represents the complete combustion of methanol:

+

#H_3COH(l) + 3/2O_2(g) rarr CO_2(g) + 2H_2O(l)#

+

Is it balanced? It looks like it to me.

+

And so we combust #23.6*g# of methanol, which represents a molar quantity of:

+

#(23.6*g)/(32.04*g*mol^-1)=0.737*mol#

+

Given the stoichiometry, at most, we can get #2xx0.737*mol=1.47*mol# water. And this represents a mass of #1.47*molxx18.01*g*mol^-1=26.5*g#, which was a book option. Claro?

+

If you are unsatisfied, I am willing to try again.

+
+
+
" "
+

If we combust #23.6*g# of methanol, how much water will be produced?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
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+1 Answer +
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+ + Mar 22, 2017 + +
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+

I get #26.5*g# of water from the combustion.......

+
+
+
+

Explanation:

+
+

And we need a stoichiometric equation that represents the complete combustion of methanol:

+

#H_3COH(l) + 3/2O_2(g) rarr CO_2(g) + 2H_2O(l)#

+

Is it balanced? It looks like it to me.

+

And so we combust #23.6*g# of methanol, which represents a molar quantity of:

+

#(23.6*g)/(32.04*g*mol^-1)=0.737*mol#

+

Given the stoichiometry, at most, we can get #2xx0.737*mol=1.47*mol# water. And this represents a mass of #1.47*molxx18.01*g*mol^-1=26.5*g#, which was a book option. Claro?

+

If you are unsatisfied, I am willing to try again.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 17061 views + around the world +
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+ + Creative Commons License + +
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+
" If we combust #23.6*g# of methanol, how much water will be produced? nan +205 a82e8538-6ddd-11ea-b365-ccda262736ce https://socratic.org/questions/when-56-grams-of-nacl-are-dissolved-in-enough-water-to-create-a-125-gram-solutio 44.80% start physical_unit 4 4 mass_percent none qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 15 15 13 14 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Concentration in percent by mass [OF] NaCl""}]" "[{""type"":""physical unit"",""value"":""44.80%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [=] \\pu{56 grams}""},{""type"":""physical unit"",""value"":""Mass [OF] solution [=] \\pu{125 grams}""},{""type"":""other"",""value"":""NaCl are dissolved in enough water to create a solution""}]" "

When 56 grams of #NaCl# are dissolved in enough water to create a 125 gram solution, what is the solution's concentration, expressed in percent by mass?

" nan 44.80% "
+

Explanation:

+
+

And thus #""Mass %, sodium chloride solution""# #=# #(56*g)/(125*g)xx100%# #=# #??%#

+
+
" "
+
+
+

#""Mass %""# #=# #""Mass of solute""/""Mass of solution""xx100%#

+
+
+
+

Explanation:

+
+

And thus #""Mass %, sodium chloride solution""# #=# #(56*g)/(125*g)xx100%# #=# #??%#

+
+
+
" "
+

When 56 grams of #NaCl# are dissolved in enough water to create a 125 gram solution, what is the solution's concentration, expressed in percent by mass?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
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+1 Answer +
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+ + Oct 6, 2016 + +
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#""Mass %""# #=# #""Mass of solute""/""Mass of solution""xx100%#

+
+
+
+

Explanation:

+
+

And thus #""Mass %, sodium chloride solution""# #=# #(56*g)/(125*g)xx100%# #=# #??%#

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+ +
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+
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+
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Impact of this question
+
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+ + Creative Commons License + +
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+
" When 56 grams of #NaCl# are dissolved in enough water to create a 125 gram solution, what is the solution's concentration, expressed in percent by mass? nan +206 a82f4c9e-6ddd-11ea-a394-ccda262736ce https://socratic.org/questions/a-mixture-of-two-gases-has-a-total-pressure-of-6-7-atm-if-one-gas-has-a-partial- 2.60 atm start physical_unit 28 30 partial_pressure atm qc_end physical_unit 0 4 10 11 total_pressure qc_end physical_unit 13 14 20 21 partial_pressure qc_end end "[{""type"":""physical unit"",""value"":""Partial pressure [OF] the other gas [IN] atm""}]" "[{""type"":""physical unit"",""value"":""2.60 atm""}]" "[{""type"":""physical unit"",""value"":""Total pressure [OF] a mixture of two gases [=] \\pu{6.7 atm}""},{""type"":""physical unit"",""value"":""Partial pressure [OF] one gas [=] \\pu{4.1 atm}""}]" "

A mixture of two gases has a total pressure of 6.7 atm. If one gas has a partial pressure of 4.1 atm, what is the partial pressure of the other gas?

" nan 2.60 atm "
+

Explanation:

+
+

Before we begin let me introduce Dalton's Law of Partial Pressures equation:
+

+

Where #P_T# is the total pressure of all gases in the mixture and #P_1#, #P_2#, etc. are the partial pressures of each gas.

+

Based on what you've given me, we know the total pressure, #P_T#, and one of the partial pressures (I'll just say #P_1#). We want to find #P_2#, so all we have to do is rearrange to equation to obtain the value of the second pressure:

+

#P_2 = P_T - P_1#

+

#P_2 = 6.7 atm - 4.1 atm#

+

Therefore, #P_2# = 2.6 atm

+
+
" "
+
+
+

The partial pressure of the other gas is #color(brown)(2.6 atm.#

+
+
+
+

Explanation:

+
+

Before we begin let me introduce Dalton's Law of Partial Pressures equation:
+

+

Where #P_T# is the total pressure of all gases in the mixture and #P_1#, #P_2#, etc. are the partial pressures of each gas.

+

Based on what you've given me, we know the total pressure, #P_T#, and one of the partial pressures (I'll just say #P_1#). We want to find #P_2#, so all we have to do is rearrange to equation to obtain the value of the second pressure:

+

#P_2 = P_T - P_1#

+

#P_2 = 6.7 atm - 4.1 atm#

+

Therefore, #P_2# = 2.6 atm

+
+
+
" "
+

A mixture of two gases has a total pressure of 6.7 atm. If one gas has a partial pressure of 4.1 atm, what is the partial pressure of the other gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
+
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+
+
+1 Answer +
+
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+ +
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+ +
+ + Jul 14, 2016 + +
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+

The partial pressure of the other gas is #color(brown)(2.6 atm.#

+
+
+
+

Explanation:

+
+

Before we begin let me introduce Dalton's Law of Partial Pressures equation:
+

+

Where #P_T# is the total pressure of all gases in the mixture and #P_1#, #P_2#, etc. are the partial pressures of each gas.

+

Based on what you've given me, we know the total pressure, #P_T#, and one of the partial pressures (I'll just say #P_1#). We want to find #P_2#, so all we have to do is rearrange to equation to obtain the value of the second pressure:

+

#P_2 = P_T - P_1#

+

#P_2 = 6.7 atm - 4.1 atm#

+

Therefore, #P_2# = 2.6 atm

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+
+
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+
+
" A mixture of two gases has a total pressure of 6.7 atm. If one gas has a partial pressure of 4.1 atm, what is the partial pressure of the other gas? nan +207 a82f6fe2-6ddd-11ea-81c5-ccda262736ce https://socratic.org/questions/what-is-the-balanced-equation-for-the-incomplete-combustion-of-nonane-c-9h-2o C9H20(l) + 27/2 O2(g) -> 8 CO2(g) + CO(g) + 10 H2O(l); C9H20(l) + 13 O2(g) -> 8 CO2(g) + C(s) + 10 H2O(l); C9H20(l) + 25/2 O2(g) -> 7 CO2(g) + CO(g) + C(s) + 10 H2O(l) start chemical_equation qc_end chemical_equation 10 10 qc_end c_other Incomplete_combustion qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""C9H20(l) + 27/2 O2(g) -> 8 CO2(g) + CO(g) + 10 H2O(l); C9H20(l) + 13 O2(g) -> 8 CO2(g) + C(s) + 10 H2O(l); C9H20(l) + 25/2 O2(g) -> 7 CO2(g) + CO(g) + C(s) + 10 H2O(l)""}]" "[{""type"":""chemical equation"",""value"":""C9H2O""},{""type"":""other"",""value"":""Incomplete combustion""}]" "

What is the balanced equation for the incomplete combustion of nonane #C_9H_2O#?

" nan C9H20(l) + 27/2 O2(g) -> 8 CO2(g) + CO(g) + 10 H2O(l); C9H20(l) + 13 O2(g) -> 8 CO2(g) + C(s) + 10 H2O(l); C9H20(l) + 25/2 O2(g) -> 7 CO2(g) + CO(g) + C(s) + 10 H2O(l) "
+

Explanation:

+
+

I think this is the primary component of #""kerosene""#.

+

#""Complete combustion:""#

+

#C_9H_20(l) + 14O_2(g) rarr 9CO_2(g) + 10H_2O(l)#

+

#""Incomplete combustion (i):""#

+

#C_9H_20(l) + 27/2O_2(g) rarr 8CO_2(g) + CO(g) + 10H_2O(l)#

+

#""Incomplete combustion (ii):""#

+

#C_9H_20(l) + 13O_2(g) rarr 8CO_2(g) + C(s) + 10H_2O(l)#

+

#""Incomplete combustion (iii):""#

+

#C_9H_20(l) + 25/2O_2(g) rarr 7CO_2(g) + CO(g) + C(s) + 10H_2O(l)#

+

All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.

+
+
" "
+
+
+

Well, for a start, #""nonane""# #-=# #C_9H_20#..........

+
+
+
+

Explanation:

+
+

I think this is the primary component of #""kerosene""#.

+

#""Complete combustion:""#

+

#C_9H_20(l) + 14O_2(g) rarr 9CO_2(g) + 10H_2O(l)#

+

#""Incomplete combustion (i):""#

+

#C_9H_20(l) + 27/2O_2(g) rarr 8CO_2(g) + CO(g) + 10H_2O(l)#

+

#""Incomplete combustion (ii):""#

+

#C_9H_20(l) + 13O_2(g) rarr 8CO_2(g) + C(s) + 10H_2O(l)#

+

#""Incomplete combustion (iii):""#

+

#C_9H_20(l) + 25/2O_2(g) rarr 7CO_2(g) + CO(g) + C(s) + 10H_2O(l)#

+

All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.

+
+
+
" "
+

What is the balanced equation for the incomplete combustion of nonane #C_9H_2O#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 11, 2017 + +
+
+
+
+
+
+
+

Well, for a start, #""nonane""# #-=# #C_9H_20#..........

+
+
+
+

Explanation:

+
+

I think this is the primary component of #""kerosene""#.

+

#""Complete combustion:""#

+

#C_9H_20(l) + 14O_2(g) rarr 9CO_2(g) + 10H_2O(l)#

+

#""Incomplete combustion (i):""#

+

#C_9H_20(l) + 27/2O_2(g) rarr 8CO_2(g) + CO(g) + 10H_2O(l)#

+

#""Incomplete combustion (ii):""#

+

#C_9H_20(l) + 13O_2(g) rarr 8CO_2(g) + C(s) + 10H_2O(l)#

+

#""Incomplete combustion (iii):""#

+

#C_9H_20(l) + 25/2O_2(g) rarr 7CO_2(g) + CO(g) + C(s) + 10H_2O(l)#

+

All I have done here is made sure that garbage out equals garbage in (have I?). I have assumed that SOME of the hydrocarbon reactant is incompletely combusted. What do I mean by this? The actual degree of combustion would have to be measured by some means of analysis. We could certainly represent it.

+
+
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+
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+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 27508 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the balanced equation for the incomplete combustion of nonane #C_9H_2O#? nan +208 a82f6fe3-6ddd-11ea-acd9-ccda262736ce https://socratic.org/questions/how-do-i-find-the-mass-of-copper-ii-hydroxide-which-is-formed-when-excess-copper 2.20 g start physical_unit 7 8 mass g qc_end physical_unit 26 29 19 20 volume qc_end physical_unit 26 29 23 25 molarity qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] copper(II) hydroxide [IN] g""}]" "[{""type"":""physical unit"",""value"":""2.20 g""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution of sodium hydroxide [=] \\pu{100 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] solution of sodium hydroxide [=] \\pu{0.450 mol L^(−1)}""},{""type"":""other"",""value"":""Excess copper(II) sulfate is added to the solution of sodium hydroxide""}]" "

How do I find the mass of copper(II) hydroxide which is formed when excess copper(II) sulfate is added to 100mL of a 0.450 mol #L^-1# solution of sodium hydroxide?

" nan 2.20 g "
+

Explanation:

+
+

An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide, #""Cu""(""OH"")_2#, to get the mass formed by the reaction.

+

So, your sodium hydroxide solution contains

+
+

#C = n/V implies n = C * V#

+

#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#

+
+

The net ionic equation for this double replacement reaction looks like this

+
+

#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#

+
+

So, #color(red)(2)# moles of hydroxide ions will react with one mole of copper(II) ions and form one mole of copper(II) hydroxide.

+

This means that the reaction will produce

+
+

#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#

+
+

To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass

+
+

#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(Cu(OH)_2) = color(green)(""2.20 g"")#

+
+
+
" "
+
+
+

#""2.20 g""#

+
+
+
+

Explanation:

+
+

An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide, #""Cu""(""OH"")_2#, to get the mass formed by the reaction.

+

So, your sodium hydroxide solution contains

+
+

#C = n/V implies n = C * V#

+

#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#

+
+

The net ionic equation for this double replacement reaction looks like this

+
+

#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#

+
+

So, #color(red)(2)# moles of hydroxide ions will react with one mole of copper(II) ions and form one mole of copper(II) hydroxide.

+

This means that the reaction will produce

+
+

#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#

+
+

To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass

+
+

#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(Cu(OH)_2) = color(green)(""2.20 g"")#

+
+
+
+
" "
+

How do I find the mass of copper(II) hydroxide which is formed when excess copper(II) sulfate is added to 100mL of a 0.450 mol #L^-1# solution of sodium hydroxide?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Oct 3, 2015 + +
+
+
+
+
+
+
+

#""2.20 g""#

+
+
+
+

Explanation:

+
+

An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide, #""Cu""(""OH"")_2#, to get the mass formed by the reaction.

+

So, your sodium hydroxide solution contains

+
+

#C = n/V implies n = C * V#

+

#n_(OH^(-)) = ""0.450 M"" * 100 * 10^(-3)""L"" = ""0.0450 moles OH""""""^(-)#

+
+

The net ionic equation for this double replacement reaction looks like this

+
+

#""Cu""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-) -> ""Cu""(""OH"")_text(2(s]) darr#

+
+

So, #color(red)(2)# moles of hydroxide ions will react with one mole of copper(II) ions and form one mole of copper(II) hydroxide.

+

This means that the reaction will produce

+
+

#0.0450color(red)(cancel(color(black)(""moles OH""""""^(-)))) * (""1 mole Cu""(""OH"")_2)/(color(red)(2)color(red)(cancel(color(black)(""moles OH""""""^(-))))) = ""0.0225 moles Cu""(""OH"")_2#

+
+

To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass

+
+

#0.0225color(red)(cancel(color(black)(""moles OH""""""^(-)))) * ""97.56 g""/(1color(red)(cancel(color(black)(""mole OH""""""^(-))))) = ""2.195 g Cu""(""OH"")_2#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(Cu(OH)_2) = color(green)(""2.20 g"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 9308 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do I find the mass of copper(II) hydroxide which is formed when excess copper(II) sulfate is added to 100mL of a 0.450 mol #L^-1# solution of sodium hydroxide? nan +209 a82f6fe4-6ddd-11ea-b2f1-ccda262736ce https://socratic.org/questions/1-00x10-2-ml-of-a-12-4-m-hcl-is-added-to-water-to-bring-the-total-volume-of-the- 1.51 M start physical_unit 30 32 concentration mol/l qc_end physical_unit 8 9 0 3 volume qc_end physical_unit 8 9 6 7 concentration qc_end physical_unit 30 32 23 24 volume qc_end substance 13 13 qc_end end "[{""type"":""physical unit"",""value"":""Concentration2 [OF] this new solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.51 M""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] HCl solution [=] \\pu{1.00 × 10^2 mL}""},{""type"":""physical unit"",""value"":""concentration1 [OF] HCl solution [=] \\pu{12.4 M}""},{""type"":""physical unit"",""value"":""Volume2 [OF] this new solution [=] \\pu{0.820 L}""},{""type"":""substance name"",""value"":""water""}]" "

#1.00 *10^2# #""mL""# of a #""12.4-M""# #""HCl""# solution is added to water to bring the total volume of the solution to #""0.820 L""#. What is the concentration of this new solution?

" nan 1.51 M "
+

Explanation:

+
+

You know that you're diluting #1.00 * 10^2 quad ""mL""# of a #""12-4M""# hydrochloric acid solution by adding it to enough water to get the total volume to

+
+

#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#

+
+

Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor, #""DF""#.

+
+

#""DF"" = V_""diluted""/V_""stock""#

+
+

Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

+
+

#""DF"" = c_""stock""/c_""diluted""#

+
+

In your case, the dilution factor is equal to

+
+

#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#

+
+

You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was #color(blue)(8.20)# times higher than the concentration of the diluted solution, and so

+
+

#c_""diluted"" = c_""stock""/color(blue)(8.20)#

+

#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#

+
+

The answer is rounded to three sig figs.

+

So, if you start with #1.00 * 10^3 quad ""mL""# of a #""12.4-M""# hydrochloric acid solution and add it to enough water to get its total volume of #""0.820 L""#, you will end up with #""0.820 L""# of a #""1.51-M""# hydrochloric acid solution.

+

Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!

+
+
" "
+
+
+

#""1.51 M""#

+
+
+
+

Explanation:

+
+

You know that you're diluting #1.00 * 10^2 quad ""mL""# of a #""12-4M""# hydrochloric acid solution by adding it to enough water to get the total volume to

+
+

#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#

+
+

Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor, #""DF""#.

+
+

#""DF"" = V_""diluted""/V_""stock""#

+
+

Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

+
+

#""DF"" = c_""stock""/c_""diluted""#

+
+

In your case, the dilution factor is equal to

+
+

#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#

+
+

You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was #color(blue)(8.20)# times higher than the concentration of the diluted solution, and so

+
+

#c_""diluted"" = c_""stock""/color(blue)(8.20)#

+

#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#

+
+

The answer is rounded to three sig figs.

+

So, if you start with #1.00 * 10^3 quad ""mL""# of a #""12.4-M""# hydrochloric acid solution and add it to enough water to get its total volume of #""0.820 L""#, you will end up with #""0.820 L""# of a #""1.51-M""# hydrochloric acid solution.

+

Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!

+
+
+
" "
+

#1.00 *10^2# #""mL""# of a #""12.4-M""# #""HCl""# solution is added to water to bring the total volume of the solution to #""0.820 L""#. What is the concentration of this new solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 19, 2018 + +
+
+
+
+
+
+
+

#""1.51 M""#

+
+
+
+

Explanation:

+
+

You know that you're diluting #1.00 * 10^2 quad ""mL""# of a #""12-4M""# hydrochloric acid solution by adding it to enough water to get the total volume to

+
+

#0.820 color(red)(cancel(color(black)(""L""))) * (10^3 quad ""mL"")/(1color(red)(cancel(color(black)(""L"")))) = ""820. mL""#

+
+

Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor, #""DF""#.

+
+

#""DF"" = V_""diluted""/V_""stock""#

+
+

Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

+
+

#""DF"" = c_""stock""/c_""diluted""#

+
+

In your case, the dilution factor is equal to

+
+

#""DF"" = (820. color(red)(cancel(color(black)(""mL""))))/(1.00 * 10^2color(red)(cancel(color(black)(""L"")))) = color(blue)(8.20)#

+
+

You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was #color(blue)(8.20)# times higher than the concentration of the diluted solution, and so

+
+

#c_""diluted"" = c_""stock""/color(blue)(8.20)#

+

#c_""diluted"" = ""12.4 M""/color(blue)(8.20) = color(darkgreen)(ul(color(black)(""1.51 M"")))#

+
+

The answer is rounded to three sig figs.

+

So, if you start with #1.00 * 10^3 quad ""mL""# of a #""12.4-M""# hydrochloric acid solution and add it to enough water to get its total volume of #""0.820 L""#, you will end up with #""0.820 L""# of a #""1.51-M""# hydrochloric acid solution.

+

Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3354 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "#1.00 *10^2# #""mL""# of a #""12.4-M""# #""HCl""# solution is added to water to bring the total volume of the solution to #""0.820 L""#. What is the concentration of this new solution?" nan +210 a82f96b8-6ddd-11ea-92a9-ccda262736ce https://socratic.org/questions/a-sample-of-gas-occupies-a-volume-of-70-9-ml-as-it-expands-it-does-118-9-j-of-wo 1.21 L start physical_unit 35 36 volume l qc_end physical_unit 0 3 8 9 volume qc_end physical_unit 35 36 15 16 energy qc_end physical_unit 35 36 27 28 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""1.21 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of gas [=] \\pu{70.9 mL}""},{""type"":""physical unit"",""value"":""Energy [OF] the gas [=] \\pu{118.9 J}""},{""type"":""physical unit"",""value"":""Constant pressure [OF] the gas [=] \\pu{783 torr}""}]" "

A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

" nan 1.21 L "
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
" "
+
+
+

#V_2=1.07L#

+
+
+
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
+
" "
+

A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Oct 1, 2016 + +
+
+
+
+
+
+
+

#V_2=1.07L#

+
+
+
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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" A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas? nan +211 a82fa486-6ddd-11ea-af04-ccda262736ce https://socratic.org/questions/the-concentration-of-a-solution-of-ammonia-nh3-is-1-5-m-v-what-is-the-molar-conc 0.08 mol/L start physical_unit 25 26 molarity mol/l qc_end physical_unit 7 7 9 10 concentration qc_end physical_unit 31 31 28 29 volume qc_end physical_unit 4 4 22 23 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molar concentration [OF] this solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.08 mol/L""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] NH3 [=] \\pu{1.5% m/v}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{250 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{25.0 mL}""},{""type"":""other"",""value"":""Diluting this solution with water.""}]" "

The concentration of a solution of ammonia, #""NH""_3# is #""1.5% m/v""#. What is +the molar concentration of a solution produced by diluting #""25.0 mL""# of this +solution with #""250 mL""# of water?

" nan 0.08 mol/L "
+

Explanation:

+
+

Start by calculating the number of moles of ammonia present in the #""25.0-mL""# sample of #""1.5% m/v""# ammonia solution.

+

As you know, a #""1.5% m/v""# solution contains #""1.5 g""# of solute, which in your case is ammonia, for every #""100 mL""# of solution.

+

This means that your sample will contain

+
+

#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#

+
+

Use the molar mass of ammonia to convert this to moles

+
+

#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#

+
+

Now, you're diluting this sample by adding #""250 mL""# of water. The total volume of the diluted solution will be

+
+

#""25.0 mL + 250 mL = 275 mL""#

+
+

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the #""25.0-mL""# sample.

+

As you know, the molarity of a solution tells you the number of moles of solute present for every #""1 L"" = 10^3# #""mL""# of solution.

+

In your case, the number of moles of ammonia present in #10^3# #""mL""# of this diluted solution will be equal to

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#

+
+

So, if you have #0.080# moles of ammonia in #10^3color(white)(.)""mL"" = ""1 L""# of the diluted solution, you can say that the molarity of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#

+
+

The answer is rounded to two sig figs.

+
+
" "
+
+
+

#""0.080 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

Start by calculating the number of moles of ammonia present in the #""25.0-mL""# sample of #""1.5% m/v""# ammonia solution.

+

As you know, a #""1.5% m/v""# solution contains #""1.5 g""# of solute, which in your case is ammonia, for every #""100 mL""# of solution.

+

This means that your sample will contain

+
+

#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#

+
+

Use the molar mass of ammonia to convert this to moles

+
+

#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#

+
+

Now, you're diluting this sample by adding #""250 mL""# of water. The total volume of the diluted solution will be

+
+

#""25.0 mL + 250 mL = 275 mL""#

+
+

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the #""25.0-mL""# sample.

+

As you know, the molarity of a solution tells you the number of moles of solute present for every #""1 L"" = 10^3# #""mL""# of solution.

+

In your case, the number of moles of ammonia present in #10^3# #""mL""# of this diluted solution will be equal to

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#

+
+

So, if you have #0.080# moles of ammonia in #10^3color(white)(.)""mL"" = ""1 L""# of the diluted solution, you can say that the molarity of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#

+
+

The answer is rounded to two sig figs.

+
+
+
" "
+

The concentration of a solution of ammonia, #""NH""_3# is #""1.5% m/v""#. What is +the molar concentration of a solution produced by diluting #""25.0 mL""# of this +solution with #""250 mL""# of water?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 27, 2017 + +
+
+
+
+
+
+
+

#""0.080 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

Start by calculating the number of moles of ammonia present in the #""25.0-mL""# sample of #""1.5% m/v""# ammonia solution.

+

As you know, a #""1.5% m/v""# solution contains #""1.5 g""# of solute, which in your case is ammonia, for every #""100 mL""# of solution.

+

This means that your sample will contain

+
+

#25.0 color(red)(cancel(color(black)(""mL solution""))) * ""1.5 g NH""_3/(100color(red)(cancel(color(black)(""mL solution"")))) = ""0.375 g NH""_3#

+
+

Use the molar mass of ammonia to convert this to moles

+
+

#0.375 color(red)(cancel(color(black)(""g""))) * ""1 mole NH""_3/(17.031color(red)(cancel(color(black)(""g"")))) = ""0.02202 moles NH""_3#

+
+

Now, you're diluting this sample by adding #""250 mL""# of water. The total volume of the diluted solution will be

+
+

#""25.0 mL + 250 mL = 275 mL""#

+
+

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the #""25.0-mL""# sample.

+

As you know, the molarity of a solution tells you the number of moles of solute present for every #""1 L"" = 10^3# #""mL""# of solution.

+

In your case, the number of moles of ammonia present in #10^3# #""mL""# of this diluted solution will be equal to

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * overbrace(""0.02202 moles H""_3/(275 color(red)(cancel(color(black)(""mL solution"")))))^(color(blue)(""the known composition of the diluted solution"")) = ""0.080 moles NH""_3#

+
+

So, if you have #0.080# moles of ammonia in #10^3color(white)(.)""mL"" = ""1 L""# of the diluted solution, you can say that the molarity of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molarity = 0.080 mol L""^(-1))))#

+
+

The answer is rounded to two sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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+
" "The concentration of a solution of ammonia, #""NH""_3# is #""1.5% m/v""#. What is +the molar concentration of a solution produced by diluting #""25.0 mL""# of this +solution with #""250 mL""# of water?" nan +212 a82fa487-6ddd-11ea-b9c1-ccda262736ce https://socratic.org/questions/how-many-ethanol-molecules-are-present-in-0-35-moles 2.11 × 10^23 start physical_unit 2 3 number none qc_end physical_unit 2 2 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] ethanol molecules""}]" "[{""type"":""physical unit"",""value"":""2.11 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] ethanol [=] \\pu{0.35 moles}""}]" "

How many ethanol molecules are present in 0.35 moles?

" nan 2.11 × 10^23 "
+

Explanation:

+
+

So, I have 0.35 moles of ethanol:

+

thus I have #0.35xxN_A# ethanol molecules .

+

Now #N_A = 6.022xx10^23#. And if I have 1 mole of ethanol I have a mass of 46.07 g. What is the mass of 0.35 moles of ethanol?

+
+
" "
+
+
+

If there are 0.35 moles of something, then there is the number #0.35xxN_A# of that something, where #N_A# #=# Avogadro's number.

+
+
+
+

Explanation:

+
+

So, I have 0.35 moles of ethanol:

+

thus I have #0.35xxN_A# ethanol molecules .

+

Now #N_A = 6.022xx10^23#. And if I have 1 mole of ethanol I have a mass of 46.07 g. What is the mass of 0.35 moles of ethanol?

+
+
+
" "
+

How many ethanol molecules are present in 0.35 moles?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 31, 2015 + +
+
+
+
+
+
+
+

If there are 0.35 moles of something, then there is the number #0.35xxN_A# of that something, where #N_A# #=# Avogadro's number.

+
+
+
+

Explanation:

+
+

So, I have 0.35 moles of ethanol:

+

thus I have #0.35xxN_A# ethanol molecules .

+

Now #N_A = 6.022xx10^23#. And if I have 1 mole of ethanol I have a mass of 46.07 g. What is the mass of 0.35 moles of ethanol?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many ethanol molecules are present in 0.35 moles? nan +213 a82fa488-6ddd-11ea-a229-ccda262736ce https://socratic.org/questions/in-standardization-8-77-ml-of-naoh-neutralized-1-522-g-of-khp-given-the-molar-ma 0.85 mol/L start physical_unit 26 27 molarity mol/l qc_end physical_unit 26 27 2 3 volume qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 18 19 molar_mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.85 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] NaOH solution [=] \\pu{8.77 mL}""},{""type"":""physical unit"",""value"":""Mass [OF] KHP [=] \\pu{1.522 g}""},{""type"":""physical unit"",""value"":""Molar mass [OF] KHP [=] \\pu{204.22 g/mol}""},{""type"":""other"",""value"":""In standardization""}]" "

In standardization, 8.77 mL of #NaOH# neutralized 1.522 g of #KHP#. Given the molar mass of #KHP# is 204.22 g/mol, what is the molarity of the #NaOH# solution?

" nan 0.85 mol/L "
+

Explanation:

+
+

#KHP#, #""potassium hydrogen phthalate""#, #1,2-C_6H_4CO_2HCO_2^(-)K^(+)#, is the potassium salt of the diacid, #""pthalic acid""#, #1,2-C_6H_4(CO_2H)_2#.

+

+

It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

+

#1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""""^(-)OH rarr ""1,2-""C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O#

+

And thus moles of #NaOH# #-=# #""KHP""#, and thus............

+

#[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1#.

+
+
" "
+
+
+

#""Molarity""=0.8500*mol*L^-1#

+
+
+
+

Explanation:

+
+

#KHP#, #""potassium hydrogen phthalate""#, #1,2-C_6H_4CO_2HCO_2^(-)K^(+)#, is the potassium salt of the diacid, #""pthalic acid""#, #1,2-C_6H_4(CO_2H)_2#.

+

+

It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

+

#1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""""^(-)OH rarr ""1,2-""C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O#

+

And thus moles of #NaOH# #-=# #""KHP""#, and thus............

+

#[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1#.

+
+
+
" "
+

In standardization, 8.77 mL of #NaOH# neutralized 1.522 g of #KHP#. Given the molar mass of #KHP# is 204.22 g/mol, what is the molarity of the #NaOH# solution?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 13, 2017 + +
+
+
+
+
+
+
+

#""Molarity""=0.8500*mol*L^-1#

+
+
+
+

Explanation:

+
+

#KHP#, #""potassium hydrogen phthalate""#, #1,2-C_6H_4CO_2HCO_2^(-)K^(+)#, is the potassium salt of the diacid, #""pthalic acid""#, #1,2-C_6H_4(CO_2H)_2#.

+

+

It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

+

#1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""""^(-)OH rarr ""1,2-""C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O#

+

And thus moles of #NaOH# #-=# #""KHP""#, and thus............

+

#[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6480 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" In standardization, 8.77 mL of #NaOH# neutralized 1.522 g of #KHP#. Given the molar mass of #KHP# is 204.22 g/mol, what is the molarity of the #NaOH# solution? nan +214 a82fa489-6ddd-11ea-8da6-ccda262736ce https://socratic.org/questions/how-many-grams-of-silver-chloride-can-be-produced-if-you-start-with-4-62-grams-o 6.36 grams start physical_unit 4 5 mass g qc_end chemical_equation 18 26 qc_end physical_unit 16 17 13 14 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] silver chloride [IN] grams""}]" "[{""type"":""physical unit"",""value"":""6.36 grams""}]" "[{""type"":""chemical equation"",""value"":""2 AgNO3(aq) + BaCl2(aq) -> 2 AgCl2(s) + Ba(NO3)2(aq)""},{""type"":""physical unit"",""value"":""Mass [OF] barium chloride [=] \\pu{4.62 g}""}]" "

How many grams of silver chloride can be produced if you start with #""4.62 g""# of barium chloride? + +

" "
+
+

+

#2""AgNO""_3(aq) + ""BaCl""_2(aq) -> 2""AgCl""(s) + ""Ba""(""NO""_3)_2(aq)#

+

+
+
" 6.36 grams "
+

Explanation:

+
+

The balanced chemical equation that describes this double replacement reaction

+
+

#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#

+
+

tells you that when #1# mole of barium chloride is consumed by the reaction, #2# moles of silver chloride are produced. Since barium chloride and silver chloride have a #1:2# mole ratio, you can expect the number of moles of barium chloride consumed by the reaction and the number of moles of silver chloride produced by the reaction to be in a #1:2# ratio.

+

In other words, you have

+
+

#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#

+
+

Now, in order to be able to use this #1:2# mole ratio, you need to convert the mass of barium chloride to moles. To do that, use the molar mass of barium chloride.

+
+

#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#

+
+

You can now say that the reaction will produce

+
+

#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#

+
+

Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.

+
+

#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""6.36 g AgCl""#

+
+
+
+

Explanation:

+
+

The balanced chemical equation that describes this double replacement reaction

+
+

#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#

+
+

tells you that when #1# mole of barium chloride is consumed by the reaction, #2# moles of silver chloride are produced. Since barium chloride and silver chloride have a #1:2# mole ratio, you can expect the number of moles of barium chloride consumed by the reaction and the number of moles of silver chloride produced by the reaction to be in a #1:2# ratio.

+

In other words, you have

+
+

#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#

+
+

Now, in order to be able to use this #1:2# mole ratio, you need to convert the mass of barium chloride to moles. To do that, use the molar mass of barium chloride.

+
+

#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#

+
+

You can now say that the reaction will produce

+
+

#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#

+
+

Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.

+
+

#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

How many grams of silver chloride can be produced if you start with #""4.62 g""# of barium chloride? + +

+
+
+

+

#2""AgNO""_3(aq) + ""BaCl""_2(aq) -> 2""AgCl""(s) + ""Ba""(""NO""_3)_2(aq)#

+

+
+
+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 29, 2018 + +
+
+
+
+
+
+
+

#""6.36 g AgCl""#

+
+
+
+

Explanation:

+
+

The balanced chemical equation that describes this double replacement reaction

+
+

#2""AgNO""_ (3(aq)) + ""BaCl""_ (2(aq)) -> 2""AgCl""_ ((s)) darr + ""Ba""(""NO""_ 3)_ (2(aq))#

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+

tells you that when #1# mole of barium chloride is consumed by the reaction, #2# moles of silver chloride are produced. Since barium chloride and silver chloride have a #1:2# mole ratio, you can expect the number of moles of barium chloride consumed by the reaction and the number of moles of silver chloride produced by the reaction to be in a #1:2# ratio.

+

In other words, you have

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#(""moles of BaCl""_2 quad ""consumed"")/(""moles of AgCl produced"") = 1/2#

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Now, in order to be able to use this #1:2# mole ratio, you need to convert the mass of barium chloride to moles. To do that, use the molar mass of barium chloride.

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#4.62 color(red)(cancel(color(black)(""g""))) * ""1 mole BaCl""_2/(208.23 color(red)(cancel(color(black)(""g"")))) = ""0.02219 moles BaCl""_2#

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+

You can now say that the reaction will produce

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+

#0.02219 color(red)(cancel(color(black)(""moles BaCl""_2))) * ""2 moles AgCl""/(1color(red)(cancel(color(black)(""mole BaCl""_2)))) = ""0.04438 moles AgCl""#

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+

Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.

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+

#0.04438 color(red)(cancel(color(black)(""moles AgCl""))) * ""143.32 g""/(1color(red)(cancel(color(black)(""mole AgCl"")))) = color(darkgreen)(ul(color(black)(""6.36 g"")))#

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+

The answer is rounded to three sig figs.

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" "How many grams of silver chloride can be produced if you start with #""4.62 g""# of barium chloride? + +" " + + +#2""AgNO""_3(aq) + ""BaCl""_2(aq) -> 2""AgCl""(s) + ""Ba""(""NO""_3)_2(aq)# + + +" +215 a82fa48a-6ddd-11ea-a0a5-ccda262736ce https://socratic.org/questions/591b2b677c01497a33a1bd63 2 CO + 3 H2 -> 2 CH2OH start chemical_equation qc_end chemical_equation 12 12 qc_end chemical_equation 16 16 qc_end chemical_equation 19 19 qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 CO + 3 H2 -> 2 CH2OH""}]" "[{""type"":""chemical equation"",""value"":""CO""},{""type"":""chemical equation"",""value"":""H2""},{""type"":""chemical equation"",""value"":""CH2OH""}]" "

What is the balanced chemical equation for the reaction between carbon monoxide, #CO#, and hydrogen gas, #H_2#, to make #CH_2OH#?

" nan 2 CO + 3 H2 -> 2 CH2OH "
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Explanation:

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+

The unbalanced equation is:

+

#CO + H_2 -> CH_2OH#

+

Since there is already one mole of carbon #(C)# and one mole of oxygen #(O)# on each side of the equation, we only need to balance hydrogen #(H)#. There are 2 moles on the left and 3 on the right. We can multiply the number of moles of #H_2# by #3/2#:

+

#CO + 3/2H_2 -> CH_2OH#

+

This is a correct, balanced chemical equation.

+

If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:

+

#2CO + 3H_2 -> 2CH_2OH#

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" "
+
+
+

The balanced equation is:

+

#CO + 3/2H_2 -> CH_2OH#

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+
+
+

Explanation:

+
+

The unbalanced equation is:

+

#CO + H_2 -> CH_2OH#

+

Since there is already one mole of carbon #(C)# and one mole of oxygen #(O)# on each side of the equation, we only need to balance hydrogen #(H)#. There are 2 moles on the left and 3 on the right. We can multiply the number of moles of #H_2# by #3/2#:

+

#CO + 3/2H_2 -> CH_2OH#

+

This is a correct, balanced chemical equation.

+

If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:

+

#2CO + 3H_2 -> 2CH_2OH#

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+
" "
+

What is the balanced chemical equation for the reaction between carbon monoxide, #CO#, and hydrogen gas, #H_2#, to make #CH_2OH#?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
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+1 Answer +
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+ + May 21, 2017 + +
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The balanced equation is:

+

#CO + 3/2H_2 -> CH_2OH#

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+
+
+

Explanation:

+
+

The unbalanced equation is:

+

#CO + H_2 -> CH_2OH#

+

Since there is already one mole of carbon #(C)# and one mole of oxygen #(O)# on each side of the equation, we only need to balance hydrogen #(H)#. There are 2 moles on the left and 3 on the right. We can multiply the number of moles of #H_2# by #3/2#:

+

#CO + 3/2H_2 -> CH_2OH#

+

This is a correct, balanced chemical equation.

+

If fractional coefficients make you uncomfortable, though, it would be possible to multiply all coefficients by 2:

+

#2CO + 3H_2 -> 2CH_2OH#

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" What is the balanced chemical equation for the reaction between carbon monoxide, #CO#, and hydrogen gas, #H_2#, to make #CH_2OH#? nan +216 a82fbda2-6ddd-11ea-9363-ccda262736ce https://socratic.org/questions/a-rolaids-tablet-contains-calcium-carbonate-for-neutralizing-stomach-acid-if-a-r 674.87 milligrams start physical_unit 26 32 mass mg qc_end c_other OTHER qc_end physical_unit 20 21 15 16 volume qc_end physical_unit 20 21 18 19 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] calcium carbon are in a Rolaids tablet [IN] milligrams""}]" "[{""type"":""physical unit"",""value"":""674.87 milligrams""}]" "[{""type"":""other"",""value"":""A Rolaids tablet contains calcium carbonate for neutralizing stomach acid.""},{""type"":""physical unit"",""value"":""Volume [OF] hydrochloric acid [=] \\pu{24.65 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] hydrochloric acid [=] \\pu{0.547 M}""}]" "

A Rolaids tablet contains calcium carbonate for neutralizing stomach acid. If a Rolaids tablet neutralizes 24.65 mL of 0.547 M hydrochloric acid, how many milligrams of calcium carbon are in a Rolaids tablet?

" nan 674.87 milligrams "
+

Explanation:

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+

Your starting point here will be the balanced chemical equation for this neutralization reaction.

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Calcium carbonate, #""CaCO""_3#, will react with hydrochloric acid, #""HCl""#, to produce aqueous calcium chloride, #""CaCl""_2#, water, and carbon dioxide, #""CO""_2#

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#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#

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+

Notice that it takes #color(red)(2)# moles of hydrochloric acid to neutralize #1# mole of calcium carbonate. This means that the reaction will always consume twice as many moles of hydrochloric acid than of calcium carbonate.

+

The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained

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+

#color(blue)(c = n/V implies n = c * V)#

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#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#

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+

Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table

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#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#

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Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass

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#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#

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+

Expressed in milligrams, the answer will be

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#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#

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+

The answer is rounded to three sig figs.

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" "
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#""645 mg""#

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+

Explanation:

+
+

Your starting point here will be the balanced chemical equation for this neutralization reaction.

+

Calcium carbonate, #""CaCO""_3#, will react with hydrochloric acid, #""HCl""#, to produce aqueous calcium chloride, #""CaCl""_2#, water, and carbon dioxide, #""CO""_2#

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#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#

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+

Notice that it takes #color(red)(2)# moles of hydrochloric acid to neutralize #1# mole of calcium carbonate. This means that the reaction will always consume twice as many moles of hydrochloric acid than of calcium carbonate.

+

The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#

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+

Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table

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+

#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#

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+

Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass

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+

#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#

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+

Expressed in milligrams, the answer will be

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#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#

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+

The answer is rounded to three sig figs.

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+
" "
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A Rolaids tablet contains calcium carbonate for neutralizing stomach acid. If a Rolaids tablet neutralizes 24.65 mL of 0.547 M hydrochloric acid, how many milligrams of calcium carbon are in a Rolaids tablet?

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+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
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+1 Answer +
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+ + Jan 12, 2016 + +
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#""645 mg""#

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+

Explanation:

+
+

Your starting point here will be the balanced chemical equation for this neutralization reaction.

+

Calcium carbonate, #""CaCO""_3#, will react with hydrochloric acid, #""HCl""#, to produce aqueous calcium chloride, #""CaCl""_2#, water, and carbon dioxide, #""CO""_2#

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+

#""CaCO""_ (3(s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l)) + ""CO""_ (2(g)) uarr#

+
+

Notice that it takes #color(red)(2)# moles of hydrochloric acid to neutralize #1# mole of calcium carbonate. This means that the reaction will always consume twice as many moles of hydrochloric acid than of calcium carbonate.

+

The problem provides you with the molarity and volume of the hydrochloric acid solution. This means that you can use the definition of molarity to find how many moles of acid it contained

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(HCl) = ""0.547 M"" * 24.65 * 10^(-3)""L"" = ""0.013484 moles HCl""#

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+

Use the aforementioned mole ratio to find the number of moles of calcium carbonate that must have been present in the Rolaids table

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+

#0.013484 color(red)(cancel(color(black)(""moles HCl""))) * (""1 mole CaCO""_3)/(color(red)(2)color(red)(cancel(color(black)(""moles HCl"")))) = ""0.0067420 moles CaCO""_3#

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+

Finally, to find the mass of calcium carbonate that contains this many moles, use the compound's molar mass

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+

#0.0067420 color(red)(cancel(color(black)(""moles CaCO""_3))) * ""100.1 g""/(1color(red)(cancel(color(black)(""mole CaCO""_3)))) = ""0.645 g""#

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+

Expressed in milligrams, the answer will be

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#0.645 color(red)(cancel(color(black)(""g""))) * (10^3""mg"")/(1color(red)(cancel(color(black)(""g"")))) = color(green)(""645 mg"")#

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+

The answer is rounded to three sig figs.

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" A Rolaids tablet contains calcium carbonate for neutralizing stomach acid. If a Rolaids tablet neutralizes 24.65 mL of 0.547 M hydrochloric acid, how many milligrams of calcium carbon are in a Rolaids tablet? nan +217 a82fbda3-6ddd-11ea-8ea4-ccda262736ce https://socratic.org/questions/how-many-n-atoms-are-there-in-one-mole-of-n-2 1.20 × 10^24 start physical_unit 2 3 number none qc_end physical_unit 10 10 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] N atoms""}]" "[{""type"":""physical unit"",""value"":""1.20 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] N2 [=] \\pu{one mole}""}]" "

How many #N# atoms are there in one mole of #N_2#?

" nan 1.20 × 10^24 "
+

Explanation:

+
+

You know that one molecule of nitrogen gas, #""N""_color(blue)(2)#, contains two atoms of nitrogen, #color(blue)(2) xx ""N""#.

+

Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have #6.022 * 10^(23)# particles #-># this is known as Avogadro's constant and acts as the definition of the mole.

+

So, in one mole of nitrogen gas you have #6.022 * 10^(23)# molecules of nitrogen gas, #""N""_2#. But since each individual molecule consists of #2# atoms of nitrogen, the number of moles of nitrogen atoms will be twice that of nitrogen gas molecules.

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+

#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #

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# = 1.2044 * 10^(24)""taoms of N""#

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Alternatively, you can express this as #color(blue)(2) xx N_""A""#, where #N_""A""# is Avogadro's constant..

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" "
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#2 xx N_""A""#

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Explanation:

+
+

You know that one molecule of nitrogen gas, #""N""_color(blue)(2)#, contains two atoms of nitrogen, #color(blue)(2) xx ""N""#.

+

Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have #6.022 * 10^(23)# particles #-># this is known as Avogadro's constant and acts as the definition of the mole.

+

So, in one mole of nitrogen gas you have #6.022 * 10^(23)# molecules of nitrogen gas, #""N""_2#. But since each individual molecule consists of #2# atoms of nitrogen, the number of moles of nitrogen atoms will be twice that of nitrogen gas molecules.

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#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #

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# = 1.2044 * 10^(24)""taoms of N""#

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Alternatively, you can express this as #color(blue)(2) xx N_""A""#, where #N_""A""# is Avogadro's constant..

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" "
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How many #N# atoms are there in one mole of #N_2#?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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#2 xx N_""A""#

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Explanation:

+
+

You know that one molecule of nitrogen gas, #""N""_color(blue)(2)#, contains two atoms of nitrogen, #color(blue)(2) xx ""N""#.

+

Now, a mole is simply a very, very large collection of particles. In order to have one mole of things, let's say particles, you need to have #6.022 * 10^(23)# particles #-># this is known as Avogadro's constant and acts as the definition of the mole.

+

So, in one mole of nitrogen gas you have #6.022 * 10^(23)# molecules of nitrogen gas, #""N""_2#. But since each individual molecule consists of #2# atoms of nitrogen, the number of moles of nitrogen atoms will be twice that of nitrogen gas molecules.

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#6.022 * 10^(23)color(red)(cancel(color(black)(""molecules N""_2))) * (color(blue)(2)color(white)(a)""atoms of N"")/(1color(red)(cancel(color(black)(""molecule N""_2)))) #

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# = 1.2044 * 10^(24)""taoms of N""#

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Alternatively, you can express this as #color(blue)(2) xx N_""A""#, where #N_""A""# is Avogadro's constant..

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" How many #N# atoms are there in one mole of #N_2#? nan +218 a82fbda4-6ddd-11ea-abab-ccda262736ce https://socratic.org/questions/59bc1ccb11ef6b55b9bd7338 1.01 mol/L start physical_unit 26 26 molarity mol/l qc_end physical_unit 5 6 1 2 mass qc_end physical_unit 13 14 9 10 mass qc_end physical_unit 22 23 19 20 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] Na^+ [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.01 mol/L""}]" "[{""type"":""physical unit"",""value"":""mass [OF] sodium chloride [=] \\pu{5 g}""},{""type"":""physical unit"",""value"":""mass [OF] sodium sulfate [=] \\pu{8.2 g}""},{""type"":""physical unit"",""value"":""Volume [OF] aqueous solution [=] \\pu{200 mL}""}]" "

A #5*g# mass of sodium chloride, and an #8.2*g# mass of sodium sulfate are dissolved in a #200*mL# of aqueous solution. What is #[Na^+]#?

" nan 1.01 mol/L "
+

Explanation:

+
+

#""Moles of NaCl""=(5.0*g)/(58.44*g*mol^-1)=0.0856*mol#.

+

#""Moles of""# #Na_2SO_4=(8.2*g)/(142.04*g*mol^-1)=0.0577*mol#.

+

And thus #[Na^+]=(0.0856*mol+2xx0.0577*mol)/(200*mLxx10^-3*L*mL^-1)#

+

#=??*mol*L^-1#.

+

Why did I double the molar quantity with respect to #Na_2SO_4#. Did I make a mistake?

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" "
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Well I make #[Na^+]=1.00*mol*L^-1#

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Explanation:

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+

#""Moles of NaCl""=(5.0*g)/(58.44*g*mol^-1)=0.0856*mol#.

+

#""Moles of""# #Na_2SO_4=(8.2*g)/(142.04*g*mol^-1)=0.0577*mol#.

+

And thus #[Na^+]=(0.0856*mol+2xx0.0577*mol)/(200*mLxx10^-3*L*mL^-1)#

+

#=??*mol*L^-1#.

+

Why did I double the molar quantity with respect to #Na_2SO_4#. Did I make a mistake?

+
+
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" "
+

A #5*g# mass of sodium chloride, and an #8.2*g# mass of sodium sulfate are dissolved in a #200*mL# of aqueous solution. What is #[Na^+]#?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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Well I make #[Na^+]=1.00*mol*L^-1#

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Explanation:

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+

#""Moles of NaCl""=(5.0*g)/(58.44*g*mol^-1)=0.0856*mol#.

+

#""Moles of""# #Na_2SO_4=(8.2*g)/(142.04*g*mol^-1)=0.0577*mol#.

+

And thus #[Na^+]=(0.0856*mol+2xx0.0577*mol)/(200*mLxx10^-3*L*mL^-1)#

+

#=??*mol*L^-1#.

+

Why did I double the molar quantity with respect to #Na_2SO_4#. Did I make a mistake?

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Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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+
" A #5*g# mass of sodium chloride, and an #8.2*g# mass of sodium sulfate are dissolved in a #200*mL# of aqueous solution. What is #[Na^+]#? nan +219 a82fbda5-6ddd-11ea-a919-ccda262736ce https://socratic.org/questions/58f9047311ef6b1669a29ed7 12.03 start physical_unit 11 11 ph none qc_end physical_unit 11 11 6 7 molarity qc_end physical_unit 11 11 14 16 kb qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] CH3NH2""}]" "[{""type"":""physical unit"",""value"":""12.03""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] CH3NH2 [=] \\pu{0.26 mol/L}""},{""type"":""physical unit"",""value"":""Kb [OF] CH3NH2 [=] \\pu{4.4 × 10^(-4)}""}]" "

What is the pH of a 0.26 mol/L solution of methylamine (#""CH""_3""NH""_2, K_text(b) = 4.4 × 10^""-4""#)?

" nan 12.03 "
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Explanation:

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+
+

We can use an ICE table to calculate the concentrations of the ions in solution.

+

The chemical equation is

+

#""CH""_3""NH""_2 + ""H""_2""O"" ⇌ ""CH""_3""NH""_3^""+"" + ""OH""^""-""; K_text(b) = 4.4 × 10^""-4""#

+

Let's rewrite this as

+

#color(white)(mmmmmmmmm)""B + H""_2""O"" ⇌ ""BH""^""+"" + ""OH""^""-""#
+#""I/mol·L""^""-1"":color(white)(mmm)0.26color(white)(mmmmmm)0color(white)(mmm)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmmmmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(ml)0.26-xcolor(white)(mmmmml)xcolor(white)(mmll)x#

+

#K_text(b) = ([""BH""^""+""][""OH""^""-""])/([""B""]) = (x × x)/(0.26-x) = x^2/(0.26-x) = 4.4 × 10^""-4""#

+
+

Check for negligibility:

+

#0.26/(4.4 × 10^""-4"") = 5900 ≫ 400#.

+

∴ #x ≪ 0.26#

+
+

Then

+

#x^2/0.26 = 4.4 × 10^""-4""#

+

#x^2 = 0.26 × 4.4 × 10^""-4"" = 1.14 × 10^""-4""#

+

#x = 1.07 × 10^""-2""#

+

#[""OH""^""-""] = x color(white)(l)""mol/L"" = 1.07 × 10^""-2""color(white)(l)""mol/L""#

+
+

#""pOH"" = ""-log""[""OH""^""-""] = ""-log""(1.07 × 10^""-2"") = 1.97#

+

#""pH"" = ""14.00 - pH"" = ""14.00 - 1.97"" = 12.03#

+
+
" "
+
+
+

pH = 12.03

+
+
+
+

Explanation:

+
+
+

We can use an ICE table to calculate the concentrations of the ions in solution.

+

The chemical equation is

+

#""CH""_3""NH""_2 + ""H""_2""O"" ⇌ ""CH""_3""NH""_3^""+"" + ""OH""^""-""; K_text(b) = 4.4 × 10^""-4""#

+

Let's rewrite this as

+

#color(white)(mmmmmmmmm)""B + H""_2""O"" ⇌ ""BH""^""+"" + ""OH""^""-""#
+#""I/mol·L""^""-1"":color(white)(mmm)0.26color(white)(mmmmmm)0color(white)(mmm)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmmmmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(ml)0.26-xcolor(white)(mmmmml)xcolor(white)(mmll)x#

+

#K_text(b) = ([""BH""^""+""][""OH""^""-""])/([""B""]) = (x × x)/(0.26-x) = x^2/(0.26-x) = 4.4 × 10^""-4""#

+
+

Check for negligibility:

+

#0.26/(4.4 × 10^""-4"") = 5900 ≫ 400#.

+

∴ #x ≪ 0.26#

+
+

Then

+

#x^2/0.26 = 4.4 × 10^""-4""#

+

#x^2 = 0.26 × 4.4 × 10^""-4"" = 1.14 × 10^""-4""#

+

#x = 1.07 × 10^""-2""#

+

#[""OH""^""-""] = x color(white)(l)""mol/L"" = 1.07 × 10^""-2""color(white)(l)""mol/L""#

+
+

#""pOH"" = ""-log""[""OH""^""-""] = ""-log""(1.07 × 10^""-2"") = 1.97#

+

#""pH"" = ""14.00 - pH"" = ""14.00 - 1.97"" = 12.03#

+
+
+
" "
+

What is the pH of a 0.26 mol/L solution of methylamine (#""CH""_3""NH""_2, K_text(b) = 4.4 × 10^""-4""#)?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 21, 2017 + +
+
+
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+
+
+
+

pH = 12.03

+
+
+
+

Explanation:

+
+
+

We can use an ICE table to calculate the concentrations of the ions in solution.

+

The chemical equation is

+

#""CH""_3""NH""_2 + ""H""_2""O"" ⇌ ""CH""_3""NH""_3^""+"" + ""OH""^""-""; K_text(b) = 4.4 × 10^""-4""#

+

Let's rewrite this as

+

#color(white)(mmmmmmmmm)""B + H""_2""O"" ⇌ ""BH""^""+"" + ""OH""^""-""#
+#""I/mol·L""^""-1"":color(white)(mmm)0.26color(white)(mmmmmm)0color(white)(mmm)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmmmmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(ml)0.26-xcolor(white)(mmmmml)xcolor(white)(mmll)x#

+

#K_text(b) = ([""BH""^""+""][""OH""^""-""])/([""B""]) = (x × x)/(0.26-x) = x^2/(0.26-x) = 4.4 × 10^""-4""#

+
+

Check for negligibility:

+

#0.26/(4.4 × 10^""-4"") = 5900 ≫ 400#.

+

∴ #x ≪ 0.26#

+
+

Then

+

#x^2/0.26 = 4.4 × 10^""-4""#

+

#x^2 = 0.26 × 4.4 × 10^""-4"" = 1.14 × 10^""-4""#

+

#x = 1.07 × 10^""-2""#

+

#[""OH""^""-""] = x color(white)(l)""mol/L"" = 1.07 × 10^""-2""color(white)(l)""mol/L""#

+
+

#""pOH"" = ""-log""[""OH""^""-""] = ""-log""(1.07 × 10^""-2"") = 1.97#

+

#""pH"" = ""14.00 - pH"" = ""14.00 - 1.97"" = 12.03#

+
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+
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" "What is the pH of a 0.26 mol/L solution of methylamine (#""CH""_3""NH""_2, K_text(b) = 4.4 × 10^""-4""#)?" nan +220 a82fbda6-6ddd-11ea-beba-ccda262736ce https://socratic.org/questions/5724db2011ef6b2c2e68414f 18.02 g/mol start physical_unit 6 6 molar_mass g/mol qc_end substance 6 6 qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] water [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""18.02 g/mol""}]" "[{""type"":""substance name"",""value"":""water""}]" "

What is the molar mass of water?

" nan 18.02 g/mol "
+

Explanation:

+
+

A #""water molecule""# contains #3# #""atoms""#: #2xxH#, and #1xxO#. That is, those 3 atoms, when appropriately chemically bound, constitute a water molecule.

+

We know that #18.01*g# of water contains #N_A# water molecules, where #N_A=""Avogadro's number""=6.022xx10^23*mol^-1#, and we get the mass by adding the atomic masses appropriately, i.e. #(2xx1.00794+15.999)*g*mol^-1#.

+

How many atoms in an #18*g# mass of water?

+
+
" "
+
+
+

A water molecule contains #3# #""atoms""#. These #3# #""atoms""# constitute #1# #""water molecule""#.

+
+
+
+

Explanation:

+
+

A #""water molecule""# contains #3# #""atoms""#: #2xxH#, and #1xxO#. That is, those 3 atoms, when appropriately chemically bound, constitute a water molecule.

+

We know that #18.01*g# of water contains #N_A# water molecules, where #N_A=""Avogadro's number""=6.022xx10^23*mol^-1#, and we get the mass by adding the atomic masses appropriately, i.e. #(2xx1.00794+15.999)*g*mol^-1#.

+

How many atoms in an #18*g# mass of water?

+
+
+
" "
+

What is the molar mass of water?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 18, 2017 + +
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+
+
+
+

A water molecule contains #3# #""atoms""#. These #3# #""atoms""# constitute #1# #""water molecule""#.

+
+
+
+

Explanation:

+
+

A #""water molecule""# contains #3# #""atoms""#: #2xxH#, and #1xxO#. That is, those 3 atoms, when appropriately chemically bound, constitute a water molecule.

+

We know that #18.01*g# of water contains #N_A# water molecules, where #N_A=""Avogadro's number""=6.022xx10^23*mol^-1#, and we get the mass by adding the atomic masses appropriately, i.e. #(2xx1.00794+15.999)*g*mol^-1#.

+

How many atoms in an #18*g# mass of water?

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+
" What is the molar mass of water? nan +221 a82fbda7-6ddd-11ea-be04-ccda262736ce https://socratic.org/questions/an-aqueous-solution-has-a-mass-of-490-grams-containing-8-5-x-10-3-gram-of-calciu 17.35 ppm start physical_unit 22 26 concentration ppm qc_end physical_unit 0 2 7 8 mass qc_end physical_unit 15 16 10 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] calcium ions in this solution [IN] ppm""}]" "[{""type"":""physical unit"",""value"":""17.35 ppm""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] an aqueous solution [=] \\pu{490 grams}""},{""type"":""physical unit"",""value"":""Mass [OF] calcium ions [=] \\pu{8.5 × 10^(-3) gram}""}]" "

An aqueous solution has a mass of #490# grams containing #8.5 xx 10^""-3""# gram of calcium ions. What is the concentration of calcium ions in this solution?

" nan 17.35 ppm "
+

Explanation:

+
+

Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.

+

A solution's ppm concentration tells you the number of grams of solute present for every

+
+

#10^6 = 1,000,000#

+
+

grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to

+
+

#""mass solution"" = ""490 g"" + ""0.0085 g"" #

+

#""mass solution = 490.0085 g""#

+
+

to be equal to the mass of water, the solvent.

+
+

#""490.0085 g "" ~~ "" 490 g""#

+
+

You can thus say that, in your case, a #""1 ppm""# solution will contain #""1 g""# of calcium cations for every #10^6# #""g""# of water.

+

Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in #10^6# #""g""# of solution.

+
+

#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#

+
+

Since this represents the number of grams of calcium cations present in #10^6# #""g""# of solution, you can say that the solution's ppm concentration is

+
+

#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
" "
+
+
+

#""17 ppm""#

+
+
+
+

Explanation:

+
+

Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.

+

A solution's ppm concentration tells you the number of grams of solute present for every

+
+

#10^6 = 1,000,000#

+
+

grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to

+
+

#""mass solution"" = ""490 g"" + ""0.0085 g"" #

+

#""mass solution = 490.0085 g""#

+
+

to be equal to the mass of water, the solvent.

+
+

#""490.0085 g "" ~~ "" 490 g""#

+
+

You can thus say that, in your case, a #""1 ppm""# solution will contain #""1 g""# of calcium cations for every #10^6# #""g""# of water.

+

Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in #10^6# #""g""# of solution.

+
+

#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#

+
+

Since this represents the number of grams of calcium cations present in #10^6# #""g""# of solution, you can say that the solution's ppm concentration is

+
+

#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
+
" "
+

An aqueous solution has a mass of #490# grams containing #8.5 xx 10^""-3""# gram of calcium ions. What is the concentration of calcium ions in this solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solving Using PPM (Parts Per Million) + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 7, 2017 + +
+
+
+
+
+
+
+

#""17 ppm""#

+
+
+
+

Explanation:

+
+

Judging by the values you have, the most helpful concentration that can be calculated here is that in parts per million, ppm.

+

A solution's ppm concentration tells you the number of grams of solute present for every

+
+

#10^6 = 1,000,000#

+
+

grams of solution. Now, notice that you can approximate the mass of the solution, which is equal to

+
+

#""mass solution"" = ""490 g"" + ""0.0085 g"" #

+

#""mass solution = 490.0085 g""#

+
+

to be equal to the mass of water, the solvent.

+
+

#""490.0085 g "" ~~ "" 490 g""#

+
+

You can thus say that, in your case, a #""1 ppm""# solution will contain #""1 g""# of calcium cations for every #10^6# #""g""# of water.

+

Use the known composition of the solution as a conversion factor to calculate the number of grams of calcium cations present in #10^6# #""g""# of solution.

+
+

#10^6 color(red)(cancel(color(black)(""g solution""))) * (8.5 * 10^(-3)color(white)(.)""g Ca""^(2+))/(490color(red)(cancel(color(black)(""g solution"")))) = ""17.35 g Ca""^(2+)#

+
+

Since this represents the number of grams of calcium cations present in #10^6# #""g""# of solution, you can say that the solution's ppm concentration is

+
+

#color(darkgreen)(ul(color(black)(""concentration"" = ""17 ppm Ca""^(2+))))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+ + Creative Commons License + +
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+
" "An aqueous solution has a mass of #490# grams containing #8.5 xx 10^""-3""# gram of calcium ions. What is the concentration of calcium ions in this solution?" nan +222 a82ff414-6ddd-11ea-b5de-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-2-moles-of-a-compound-dissolved-in-4-l-of-water 0.50 M start physical_unit 9 9 molarity mol/l qc_end physical_unit 8 9 5 6 mole qc_end physical_unit 15 15 12 13 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] compound [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.50 M""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] a compound [=] \\pu{2 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{4 L}""},{""type"":""other"",""value"":""a compound dissolved in 4 L of water.""}]" "

What is the molarity of 2 moles of a compound dissolved in 4 L of water?

" nan 0.50 M "
+

Explanation:

+
+

Molarity is the ratio of moles of solute to liters of solution.

+

#M = ""moles solute""/""litres solution""#

+

#""2 mol""/""4 L"" = ""0.5 mol/L""#

+

The symbol for molarity is mol/L.

+
+
" "
+
+
+

M = 0.5 mol/L

+
+
+
+

Explanation:

+
+

Molarity is the ratio of moles of solute to liters of solution.

+

#M = ""moles solute""/""litres solution""#

+

#""2 mol""/""4 L"" = ""0.5 mol/L""#

+

The symbol for molarity is mol/L.

+
+
+
" "
+

What is the molarity of 2 moles of a compound dissolved in 4 L of water?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Jun 21, 2017 + +
+
+
+
+
+
+
+

M = 0.5 mol/L

+
+
+
+

Explanation:

+
+

Molarity is the ratio of moles of solute to liters of solution.

+

#M = ""moles solute""/""litres solution""#

+

#""2 mol""/""4 L"" = ""0.5 mol/L""#

+

The symbol for molarity is mol/L.

+
+
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+
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+ +
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+
+
+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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+
+
" What is the molarity of 2 moles of a compound dissolved in 4 L of water? nan +223 a82ff415-6ddd-11ea-8875-ccda262736ce https://socratic.org/questions/58ddcd467c01490435f39c24 26.46 g start physical_unit 12 13 mass g qc_end physical_unit 5 5 1 2 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] carbon dioxide [IN] g""}]" "[{""type"":""physical unit"",""value"":""26.46 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] propane [=] \\pu{8.8 g}""},{""type"":""other"",""value"":""Completely combusted""}]" "

An #8.8*g# mass of propane is completely combusted. What mass of carbon dioxide will result?

" nan 26.46 g "
+

Explanation:

+
+

WE have the stoichiometric equation:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.

+

And then we simply access the molar quantity of propane:

+

#=(8.8*g)/(44.10*g*mol^-1)=0.200*mol#

+

And since, per the stoichiometric equation, if #0.200*mol# #""propane""# is completely combusted, #3xx0.200*molxx44.10*g*mol^-1=??*g# #CO_2(g)# will result.

+

At normal pressure and temperature, #1*atm#, and #298*K#, what volume will the evolved gas occupy?

+
+
" "
+
+
+

Approx. #26*g#..........................

+
+
+
+

Explanation:

+
+

WE have the stoichiometric equation:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.

+

And then we simply access the molar quantity of propane:

+

#=(8.8*g)/(44.10*g*mol^-1)=0.200*mol#

+

And since, per the stoichiometric equation, if #0.200*mol# #""propane""# is completely combusted, #3xx0.200*molxx44.10*g*mol^-1=??*g# #CO_2(g)# will result.

+

At normal pressure and temperature, #1*atm#, and #298*K#, what volume will the evolved gas occupy?

+
+
+
" "
+

An #8.8*g# mass of propane is completely combusted. What mass of carbon dioxide will result?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
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+
+
+2 Answers +
+
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+
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+ +
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+ +
+ + Mar 31, 2017 + +
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+

Approx. #26*g#..........................

+
+
+
+

Explanation:

+
+

WE have the stoichiometric equation:

+

#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#

+

Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.

+

And then we simply access the molar quantity of propane:

+

#=(8.8*g)/(44.10*g*mol^-1)=0.200*mol#

+

And since, per the stoichiometric equation, if #0.200*mol# #""propane""# is completely combusted, #3xx0.200*molxx44.10*g*mol^-1=??*g# #CO_2(g)# will result.

+

At normal pressure and temperature, #1*atm#, and #298*K#, what volume will the evolved gas occupy?

+
+
+
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+ +
+
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+
+ +
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+ +
+ + Mar 31, 2017 + +
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+
+

24 litres at RTP (298K) or 22.4 litres at STP (273K)

+
+
+
+

Explanation:

+
+

1) The balanced equation gives us the mole ratio propane:oxygen which is 1:5 - this means that every mole of propane will require 5 moles of oxygen for complete combustion.

+

2) However we don't have one mole of propane, we have 8.8g which, using moles =mass/molar mass, equals 8.8/44 = 0.2 moles. (RMM of propane = 12x3 + 1x8 = 44, therefore molar mass = 44g/mol).

+

3) Applying the mole ratio, it follows that 0.2 moles of propane will react with 0.2x5 = 1 mole of oxygen.

+

4) One mole of any gas (assuming ideal behaviour, which is a perfectly reasonable assumption, especially with a gas like oxygen with small and non-polar molecules) occupies 24 litres at RTP and 22.4 at STP. Presto!

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Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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+
+
" An #8.8*g# mass of propane is completely combusted. What mass of carbon dioxide will result? nan +224 a82ff416-6ddd-11ea-b7a9-ccda262736ce https://socratic.org/questions/a-certain-beverage-contains-7-700-ethanol-c-2h-6o-by-mass-what-volume-of-this-be 1909.04 milliliters start physical_unit 12 13 volume ml qc_end physical_unit 6 6 17 18 mole qc_end physical_unit 6 6 4 4 percent_composition qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] this beverage [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""1909.04 milliliters""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] C2H6O [=] \\pu{3.150 moles}""},{""type"":""physical unit"",""value"":""percent composition [OF] C2H6O [=] \\pu{7.700%}""}]" "

A certain beverage contains 7.700% ethanol #(C_2H_6O)# by mass. What volume of this beverage in milliliters contains 3.150 moles of ethanol?

" nan 1909.04 milliliters "
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use ethanol's molar mass to determine how many grams would contain the given number of moles

    +
    • +
  • +

    use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol

    +
    • +
  • +

    do a quick search for the density of an ethanol solution that has that percent concentration of ethanol

    +
    • +
+

So, ethanol's molar mass is equal to #""46.07 g/mol""#, which means that every mole of ethanol will have a mass of #""46.07 g""#. In your case, the given number of moles will have a mass of

+
+

#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#

+
+

As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by #100#

+
+

#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#

+
+

Now, a #""7.700% w/w""# ethanol solution will contain #""7.700 g""# of ethanol for every #""100 mL""# of solution. This means that you'd get #""145.12 g""# of ethanol in

+
+

#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#

+
+

Now, to get the volume of this solution, you need to use its density, which you can find here

+

http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/

+

The density of a #""7.700% w/w""# ethanol solution is about #""0.98725 g/mL""# at room temperature. This means that you have

+
+

#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#

+
+

The answer is rounded to four sig figs.

+
+
" "
+
+
+

#""1909 mL""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use ethanol's molar mass to determine how many grams would contain the given number of moles

    +
    • +
  • +

    use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol

    +
    • +
  • +

    do a quick search for the density of an ethanol solution that has that percent concentration of ethanol

    +
    • +
+

So, ethanol's molar mass is equal to #""46.07 g/mol""#, which means that every mole of ethanol will have a mass of #""46.07 g""#. In your case, the given number of moles will have a mass of

+
+

#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#

+
+

As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by #100#

+
+

#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#

+
+

Now, a #""7.700% w/w""# ethanol solution will contain #""7.700 g""# of ethanol for every #""100 mL""# of solution. This means that you'd get #""145.12 g""# of ethanol in

+
+

#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#

+
+

Now, to get the volume of this solution, you need to use its density, which you can find here

+

http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/

+

The density of a #""7.700% w/w""# ethanol solution is about #""0.98725 g/mL""# at room temperature. This means that you have

+
+

#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#

+
+

The answer is rounded to four sig figs.

+
+
+
" "
+

A certain beverage contains 7.700% ethanol #(C_2H_6O)# by mass. What volume of this beverage in milliliters contains 3.150 moles of ethanol?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 10, 2015 + +
+
+
+
+
+
+
+

#""1909 mL""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use ethanol's molar mass to determine how many grams would contain the given number of moles

    +
    • +
  • +

    use the given percent concentration by mass to find the volume of solution that would contain that many grams of ethanol

    +
    • +
  • +

    do a quick search for the density of an ethanol solution that has that percent concentration of ethanol

    +
    • +
+

So, ethanol's molar mass is equal to #""46.07 g/mol""#, which means that every mole of ethanol will have a mass of #""46.07 g""#. In your case, the given number of moles will have a mass of

+
+

#3.150 color(red)(cancel(color(black)(""moles C""_2""H""_6""O""))) * ""46.07 g""/(1color(red)(cancel(color(black)(""mole C""_2""H""_6""O"")))) = ""145.12 g""#

+
+

As you know, a solution's percent concentration by mass is defined as the ratio between the mass of the solute and the total mass of the solution, multiplied by #100#

+
+

#color(blue)(""% w/w"" = ""mass of solute""/""mass of solution"" xx 100)#

+
+

Now, a #""7.700% w/w""# ethanol solution will contain #""7.700 g""# of ethanol for every #""100 mL""# of solution. This means that you'd get #""145.12 g""# of ethanol in

+
+

#145.12 color(red)(cancel(color(black)(""g C""_2""H""_6""O""))) * ""100 g solution""/(7.700color(red)(cancel(color(black)(""g C""_2""H""_6""O"")))) = ""1884.7 g solution""#

+
+

Now, to get the volume of this solution, you need to use its density, which you can find here

+

http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/

+

The density of a #""7.700% w/w""# ethanol solution is about #""0.98725 g/mL""# at room temperature. This means that you have

+
+

#1884.7 color(red)(cancel(color(black)(""g""))) * "" 1mL""/(0.98725 color(red)(cancel(color(black)(""g"")))) = color(green)(""1909 mL"")#

+
+

The answer is rounded to four sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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Impact of this question
+
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" A certain beverage contains 7.700% ethanol #(C_2H_6O)# by mass. What volume of this beverage in milliliters contains 3.150 moles of ethanol? nan +225 a8300c36-6ddd-11ea-946d-ccda262736ce https://socratic.org/questions/ethane-c-2h-6-reacts-with-molecular-oxygen-to-produce-carbon-dioxide-and-water-h 2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O""}]" "[{""type"":""other"",""value"":""C2H6 reacts with molecular oxygen to produce carbon dioxide and water.""}]" "

Ethane, #C_2H_6#, reacts with molecular oxygen to produce carbon dioxide and water. How do you write the balanced chemical equation for this reaction?

" nan 2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O "
+

Explanation:

+
+

This is a combustion reaction of an alkane with oxygen. It is exothermic (#DeltaH<0#), and can be balanced in normal methods for balancing chemical equations ie ensuring that the total number of atoms of each element is the same on both sides of the equation.

+
+
" "
+
+
+

#2C_2H_6+7O_2->4CO_2+6H_2O#

+
+
+
+

Explanation:

+
+

This is a combustion reaction of an alkane with oxygen. It is exothermic (#DeltaH<0#), and can be balanced in normal methods for balancing chemical equations ie ensuring that the total number of atoms of each element is the same on both sides of the equation.

+
+
+
" "
+

Ethane, #C_2H_6#, reacts with molecular oxygen to produce carbon dioxide and water. How do you write the balanced chemical equation for this reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 9, 2016 + +
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+
+
+
+
+
+

#2C_2H_6+7O_2->4CO_2+6H_2O#

+
+
+
+

Explanation:

+
+

This is a combustion reaction of an alkane with oxygen. It is exothermic (#DeltaH<0#), and can be balanced in normal methods for balancing chemical equations ie ensuring that the total number of atoms of each element is the same on both sides of the equation.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 36711 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" Ethane, #C_2H_6#, reacts with molecular oxygen to produce carbon dioxide and water. How do you write the balanced chemical equation for this reaction? nan +226 a8300c37-6ddd-11ea-b990-ccda262736ce https://socratic.org/questions/a-143-1-g-sample-of-a-compound-contains-53-4-g-of-carbon-6-9-g-of-hydrogen-43-g- 27.81% start physical_unit 31 34 mass_percent none qc_end physical_unit 3 6 1 2 mass qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 15 15 12 13 mass qc_end physical_unit 19 19 16 17 mass qc_end substance 24 24 qc_end end "[{""type"":""physical unit"",""value"":""Mass percent [OF] oxygen in the compound""}]" "[{""type"":""physical unit"",""value"":""27.81%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] sample of a compound [=] \\pu{143.1 g}""},{""type"":""physical unit"",""value"":""Mass [OF] carbon [=] \\pu{53.4 g}""},{""type"":""physical unit"",""value"":""Mass [OF] hydrogen [=] \\pu{6.9 g}""},{""type"":""physical unit"",""value"":""Mass [OF] nitrogen [=] \\pu{43 g}""},{""type"":""substance name"",""value"":""Oxygen""}]" "

A 143.1 g sample of a compound contains 53.4 g of carbon, 6.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. What is the mass percent of oxygen in the compound?

" nan 27.81% "
+

Explanation:

+
+

Let me start by writing a simple equation

+

#143.1=53.4+6.9+43+x#

+

x stands for amount of oxygen in grams

+

When you solve this equation, #x=143.1-(53.4+6.9+43)#

+

#x=143.1-103.3#

+

# x=39.8 g#

+

Now in percent form

+

Mass percent of oxygen #100*39.8/143.1#

+

#=27.8%#

+
+
" "
+
+
+

27.8% oxygen

+
+
+
+

Explanation:

+
+

Let me start by writing a simple equation

+

#143.1=53.4+6.9+43+x#

+

x stands for amount of oxygen in grams

+

When you solve this equation, #x=143.1-(53.4+6.9+43)#

+

#x=143.1-103.3#

+

# x=39.8 g#

+

Now in percent form

+

Mass percent of oxygen #100*39.8/143.1#

+

#=27.8%#

+
+
+
" "
+

A 143.1 g sample of a compound contains 53.4 g of carbon, 6.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. What is the mass percent of oxygen in the compound?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 24, 2016 + +
+
+
+
+
+
+
+

27.8% oxygen

+
+
+
+

Explanation:

+
+

Let me start by writing a simple equation

+

#143.1=53.4+6.9+43+x#

+

x stands for amount of oxygen in grams

+

When you solve this equation, #x=143.1-(53.4+6.9+43)#

+

#x=143.1-103.3#

+

# x=39.8 g#

+

Now in percent form

+

Mass percent of oxygen #100*39.8/143.1#

+

#=27.8%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3931 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A 143.1 g sample of a compound contains 53.4 g of carbon, 6.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. What is the mass percent of oxygen in the compound? nan +227 a8300c38-6ddd-11ea-847e-ccda262736ce https://socratic.org/questions/the-temperature-of-15-ml-of-a-strong-acid-increases-by-2-degree-centigrade-when- 2.00 ℃ start physical_unit 8 8 temperature °c qc_end physical_unit 6 8 4 5 volume qc_end physical_unit 6 8 30 31 volume qc_end physical_unit 18 19 4 5 volume qc_end physical_unit 18 19 30 31 volume qc_end physical_unit 8 8 11 12 temperature qc_end end "[{""type"":""physical unit"",""value"":""Increased temperature2 [OF] solution [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""2.00 ℃""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] strong acid solution [=] \\pu{15 mL}""},{""type"":""physical unit"",""value"":""Volume2 [OF] strong acid solution [=] \\pu{5 mL}""},{""type"":""physical unit"",""value"":""Volume1 [OF] strong base [=] \\pu{15 mL}""},{""type"":""physical unit"",""value"":""Volume2 [OF] strong base [=] \\pu{5 mL}""},{""type"":""physical unit"",""value"":""Increased temperature1 [OF] solution [=] \\pu{2 ℃}""}]" "

The temperature of a #""15-mL""# strong acid solution increases by #2^@""C""# when #""15 mL""# of a strong base is added. By how much should the temperature increase if #""5 mL""# of each are mixed?

" nan 2.00 ℃ "
+

Explanation:

+
+

QUICK ANSWER

+

The temperature of the solution will increase by #2^@""C""# regardless of how much strong acid and strong base you're mixing because the enthalpy change of neutralization, #DeltaH_""neut""#, is constant at constant pressure.

+

Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.

+
+

#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#

+
+

The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with

+
+

#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

+
+

#color(white)(a/a)#

+

DETAILED EXPLANATION

+

The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have

+
+

#""heat given off for 5 mL "" < "" heat given off for 15 mL""#

+
+

because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.

+

However, you will also have

+
+

#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#

+
+

so look for these two factors to cancel each other out and result in the same #""2-""""""^@""C""# increase in the temperature of the resulting solution.

+

A generic thermochemical equation that can describe this neutralization reaction looks like this

+
+

#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#

+
+

This tells you that when #1# mole of a strong acid reacts with #1# mole of a strong base to produce #1# mole of water, the reaction gives off #x# #""kJ""# of heat.

+

So, assuming that the strong acid solution and the strong base solution have concentrations equal to #c# #""M""#, you can say that you have

+
+

#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and #"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#

+
+

This means that when you mix #""15 mL""# of each reactant, the reaction will give off

+
+

#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 15 * x)# #""J""# of heat.

+

Similarly, when you mix #""5 mL""# of each reactant, assuming, of course, that their concentrations are still equal to #c #""M""#, the reaction will give off

+
+

#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 5 * x)# #""J""# of heat.

+

So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.

+

Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.

+

More specifically, you know that

+
+

#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#

+
+

Here

+
+
    +
  • #q_""absorbed""# is the heat absorbed by the solution
    • +
  • #m_""sol""# is the mass of the solution
    • +
  • #c_""sol""# is the specific heat of the solution
    • +
  • #DeltaT# is the change in temperature
    • +
+
+

When you mix #""15 mL""# of each reactant, the total volume of the solution will be

+
+

#V_""total"" = ""15 mL"" + ""15 mL""#

+

#V_""total"" = ""30 mL""#

+
+

If you take #rho# #""g mL""^(-1)# to be the density of the solution, you can say that the total mass of the solution will be

+
+

#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#

+
+

This means that you have

+
+

#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#

+
+

which gets you

+
+

#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#

+

#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

+
+

Similarly, you can say that when you mix #""5 mL""# of each reactant, you have

+
+

#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#

+
+

This time, you will have

+
+

#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#

+
+

which gets you

+
+

#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#

+

#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

+
+

As you can see, you have

+
+

#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

+
+
+
" "
+
+
+

The temperature of the solution will increase by #2^@""C""#.

+
+
+
+

Explanation:

+
+

QUICK ANSWER

+

The temperature of the solution will increase by #2^@""C""# regardless of how much strong acid and strong base you're mixing because the enthalpy change of neutralization, #DeltaH_""neut""#, is constant at constant pressure.

+

Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.

+
+

#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#

+
+

The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with

+
+

#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

+
+

#color(white)(a/a)#

+

DETAILED EXPLANATION

+

The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have

+
+

#""heat given off for 5 mL "" < "" heat given off for 15 mL""#

+
+

because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.

+

However, you will also have

+
+

#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#

+
+

so look for these two factors to cancel each other out and result in the same #""2-""""""^@""C""# increase in the temperature of the resulting solution.

+

A generic thermochemical equation that can describe this neutralization reaction looks like this

+
+

#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#

+
+

This tells you that when #1# mole of a strong acid reacts with #1# mole of a strong base to produce #1# mole of water, the reaction gives off #x# #""kJ""# of heat.

+

So, assuming that the strong acid solution and the strong base solution have concentrations equal to #c# #""M""#, you can say that you have

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+

#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and #"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#

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+

This means that when you mix #""15 mL""# of each reactant, the reaction will give off

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+

#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 15 * x)# #""J""# of heat.

+

Similarly, when you mix #""5 mL""# of each reactant, assuming, of course, that their concentrations are still equal to #c #""M""#, the reaction will give off

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+

#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 5 * x)# #""J""# of heat.

+

So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.

+

Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.

+

More specifically, you know that

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+

#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#

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+

Here

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+
    +
  • #q_""absorbed""# is the heat absorbed by the solution
    • +
  • #m_""sol""# is the mass of the solution
    • +
  • #c_""sol""# is the specific heat of the solution
    • +
  • #DeltaT# is the change in temperature
    • +
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When you mix #""15 mL""# of each reactant, the total volume of the solution will be

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+

#V_""total"" = ""15 mL"" + ""15 mL""#

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#V_""total"" = ""30 mL""#

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+

If you take #rho# #""g mL""^(-1)# to be the density of the solution, you can say that the total mass of the solution will be

+
+

#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#

+
+

This means that you have

+
+

#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#

+
+

which gets you

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+

#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#

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#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

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+

Similarly, you can say that when you mix #""5 mL""# of each reactant, you have

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+

#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#

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+

This time, you will have

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+

#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#

+
+

which gets you

+
+

#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#

+

#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

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+

As you can see, you have

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+

#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

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" "
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The temperature of a #""15-mL""# strong acid solution increases by #2^@""C""# when #""15 mL""# of a strong base is added. By how much should the temperature increase if #""5 mL""# of each are mixed?

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+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
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+1 Answer +
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The temperature of the solution will increase by #2^@""C""#.

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Explanation:

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+

QUICK ANSWER

+

The temperature of the solution will increase by #2^@""C""# regardless of how much strong acid and strong base you're mixing because the enthalpy change of neutralization, #DeltaH_""neut""#, is constant at constant pressure.

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Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.

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#""more acid + base"" => {(""heat given off "" uarr color(blue)("" by a factor f"")), (""total volume of the solution "" uarr color(blue)("" by the same factor f"")) :}#

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+

The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with

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#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

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#color(white)(a/a)#

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DETAILED EXPLANATION

+

The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have

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#""heat given off for 5 mL "" < "" heat given off for 15 mL""#

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+

because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.

+

However, you will also have

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#""volume of solution for 5 mL "" < "" volume of solution for 15 mL""#

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+

so look for these two factors to cancel each other out and result in the same #""2-""""""^@""C""# increase in the temperature of the resulting solution.

+

A generic thermochemical equation that can describe this neutralization reaction looks like this

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+

#""HA""_ ((aq)) + ""BOH""_ ((aq)) -> ""H""_ 2""O""_ ((l)) + ""BA""_ ((aq)),"" ""DeltaH_""neut"" = +xcolor(white)(.)""kJ mol""^(-1)#

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+

This tells you that when #1# mole of a strong acid reacts with #1# mole of a strong base to produce #1# mole of water, the reaction gives off #x# #""kJ""# of heat.

+

So, assuming that the strong acid solution and the strong base solution have concentrations equal to #c# #""M""#, you can say that you have

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+

#((c * 15)/10^3) color(white)(.)""moles HA"" "" ""# and #"" "" ((c * 15)/10^3)color(white)(.)""moles BOH""#

+
+

This means that when you mix #""15 mL""# of each reactant, the reaction will give off

+
+

#((c * 15)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ((c * 15 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 15 * x)# #""J""# of heat.

+

Similarly, when you mix #""5 mL""# of each reactant, assuming, of course, that their concentrations are still equal to #c #""M""#, the reaction will give off

+
+

#((3 * 5)/10^3) color(red)(cancel(color(black)(""moles HA""))) * (x color(white)(.)""kJ"")/(1color(red)(cancel(color(black)(""mole HA"")))) = ( (c * 5 * x)/10^3)color(white)(.)""kJ""#

+
+

of heat, the equivalent of #(c * 5 * x)# #""J""# of heat.

+

So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.

+

Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.

+

More specifically, you know that

+
+

#color(blue)(ul(color(black)(q_""absorbed"" = m_""sol"" * c_""sol"" * DeltaT)))#

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+

Here

+
+
    +
  • #q_""absorbed""# is the heat absorbed by the solution
    • +
  • #m_""sol""# is the mass of the solution
    • +
  • #c_""sol""# is the specific heat of the solution
    • +
  • #DeltaT# is the change in temperature
    • +
+
+

When you mix #""15 mL""# of each reactant, the total volume of the solution will be

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+

#V_""total"" = ""15 mL"" + ""15 mL""#

+

#V_""total"" = ""30 mL""#

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+

If you take #rho# #""g mL""^(-1)# to be the density of the solution, you can say that the total mass of the solution will be

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+

#30 color(red)(cancel(color(black)(""mL solution""))) * (rho color(white)(.)""g"")/(1color(red)(cancel(color(black)(""mL solution"")))) = (30 * rho)color(white)(.)""g""#

+
+

This means that you have

+
+

#(c * 15 * x) color(red)(cancel(color(black)(""J""))) = (30 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""15 mL""#

+
+

which gets you

+
+

#DeltaT_""15 mL"" = ((c * 15 * x)/(30 * rho * c_""sol"")) """"^@""C""#

+

#color(blue)(ul(color(black)(DeltaT_""15 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

+
+

Similarly, you can say that when you mix #""5 mL""# of each reactant, you have

+
+

#V_""total"" = ""5 mL"" + ""5 mL"" = ""10 mL""#

+
+

This time, you will have

+
+

#(c * 5 * x) color(red)(cancel(color(black)(""J""))) = (10 * rho) color(red)(cancel(color(black)(""g""))) * c_""sol"" color(red)(cancel(color(black)(""J""))) color(red)(cancel(color(black)(""g"")))^(-1) """"^@""C""^(-1) * DeltaT_""5 mL""#

+
+

which gets you

+
+

#DeltaT_""5 mL"" = ((c * 5 * x)/(10 * rho * c_""sol""))""""^@""C""#

+

#color(blue)(ul(color(black)(DeltaT_""5 mL"" = (1/2 * (c * x)/(rho * c_""sol""))""""^@""C"")))#

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+

As you can see, you have

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#color(darkgreen)(ul(color(black)(DeltaT_""15 mL"" = DeltaT_""5 mL"" = 2^@""C"")))#

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" "The temperature of a #""15-mL""# strong acid solution increases by #2^@""C""# when #""15 mL""# of a strong base is added. By how much should the temperature increase if #""5 mL""# of each are mixed? " nan +228 a83032f0-6ddd-11ea-9d1f-ccda262736ce https://socratic.org/questions/583e29bb7c014963f9c8b7bd 8.79 start physical_unit 22 24 ph none qc_end physical_unit 5 6 3 4 molarity qc_end physical_unit 5 6 0 1 volume qc_end physical_unit 15 16 13 14 molarity qc_end physical_unit 15 16 10 11 volume qc_end end "[{""type"":""physical unit"",""value"":""PH [OF] the resulting solution""}]" "[{""type"":""physical unit"",""value"":""8.79""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] sodium hydroxide [=] \\pu{0.1 M}""},{""type"":""physical unit"",""value"":""Volume [OF] sodium hydroxide [=] \\pu{200 ml}""},{""type"":""physical unit"",""value"":""Molarity [OF] ethanoic acid [=] \\pu{0.2 M}""},{""type"":""physical unit"",""value"":""Volume [OF] ethanoic acid [=] \\pu{100 ml}""}]" "

200 ml of 0.1 M sodium hydroxide is mixed with 100 ml of 0.2 M ethanoic acid. What is the pH of the resulting solution?

" nan 8.79 "
+

Explanation:

+
+

The alkali neutralises the acid:

+

#sf(CH_3COOH+NaOHrarrCH_3COO^(-)Na^(+)+H_2O)#

+

The moles reacting (n) of each species is given by:

+

#sf(n_(OH^-)=cxxv=0.1xx200/1000=0.02)#

+

#sf(n_(CH_3COOH)=cxxv=0.2xx100/1000=0.02)#

+

This shows that they are added in the same mole ratio as defined by the equation.

+

We can, therefore, say that the no. moles of #sf(CH_3COO^(-))# formed = 0.02.

+

A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:

+

#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#

+

From an ICE table we can use this expression which applies to a weak base:

+

#sf(pOH=1/2(pK_b-logb)#

+

Where #sf(b)# is the concentration of the base which, in this case, is #sf(CH_3COO^(-))#.

+

The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L

+

#sf([CH_3COO^(-)]=n/v=0.02/0.3=0.0666color(white)(x)""mol/l"")#

+

You need to look up the #sf(pK_a)# value for ethanoic acid which = 4.76.

+

To get #sf(pK_b)# you use:

+

#sf(pK_(a)+ pK_(b)=14)#

+

#:.##sf(pK_b=14-pK_a=14-4.76=9.24)#

+

Putting in the numbers:

+

#sf(pOH=1/2[9.24-log(0.0666)]#

+

#sf(pOH=1/2[9.24-(-1.176)]=5.208)#

+

#sf(pH+pOH=14)#

+

#:.##sf(pH=14-pOH=14-5.208=8.8)#

+

As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.

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" "
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+
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#sf(pH=8.8)#

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+
+
+

Explanation:

+
+

The alkali neutralises the acid:

+

#sf(CH_3COOH+NaOHrarrCH_3COO^(-)Na^(+)+H_2O)#

+

The moles reacting (n) of each species is given by:

+

#sf(n_(OH^-)=cxxv=0.1xx200/1000=0.02)#

+

#sf(n_(CH_3COOH)=cxxv=0.2xx100/1000=0.02)#

+

This shows that they are added in the same mole ratio as defined by the equation.

+

We can, therefore, say that the no. moles of #sf(CH_3COO^(-))# formed = 0.02.

+

A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:

+

#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#

+

From an ICE table we can use this expression which applies to a weak base:

+

#sf(pOH=1/2(pK_b-logb)#

+

Where #sf(b)# is the concentration of the base which, in this case, is #sf(CH_3COO^(-))#.

+

The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L

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#sf([CH_3COO^(-)]=n/v=0.02/0.3=0.0666color(white)(x)""mol/l"")#

+

You need to look up the #sf(pK_a)# value for ethanoic acid which = 4.76.

+

To get #sf(pK_b)# you use:

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#sf(pK_(a)+ pK_(b)=14)#

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#:.##sf(pK_b=14-pK_a=14-4.76=9.24)#

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Putting in the numbers:

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#sf(pOH=1/2[9.24-log(0.0666)]#

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#sf(pOH=1/2[9.24-(-1.176)]=5.208)#

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#sf(pH+pOH=14)#

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#:.##sf(pH=14-pOH=14-5.208=8.8)#

+

As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.

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+
" "
+

200 ml of 0.1 M sodium hydroxide is mixed with 100 ml of 0.2 M ethanoic acid. What is the pH of the resulting solution?

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+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Calculations + + +
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+2 Answers +
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+ +
+ + Dec 2, 2016 + +
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#sf(pH=8.8)#

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+
+
+

Explanation:

+
+

The alkali neutralises the acid:

+

#sf(CH_3COOH+NaOHrarrCH_3COO^(-)Na^(+)+H_2O)#

+

The moles reacting (n) of each species is given by:

+

#sf(n_(OH^-)=cxxv=0.1xx200/1000=0.02)#

+

#sf(n_(CH_3COOH)=cxxv=0.2xx100/1000=0.02)#

+

This shows that they are added in the same mole ratio as defined by the equation.

+

We can, therefore, say that the no. moles of #sf(CH_3COO^(-))# formed = 0.02.

+

A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:

+

#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#

+

From an ICE table we can use this expression which applies to a weak base:

+

#sf(pOH=1/2(pK_b-logb)#

+

Where #sf(b)# is the concentration of the base which, in this case, is #sf(CH_3COO^(-))#.

+

The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L

+

#sf([CH_3COO^(-)]=n/v=0.02/0.3=0.0666color(white)(x)""mol/l"")#

+

You need to look up the #sf(pK_a)# value for ethanoic acid which = 4.76.

+

To get #sf(pK_b)# you use:

+

#sf(pK_(a)+ pK_(b)=14)#

+

#:.##sf(pK_b=14-pK_a=14-4.76=9.24)#

+

Putting in the numbers:

+

#sf(pOH=1/2[9.24-log(0.0666)]#

+

#sf(pOH=1/2[9.24-(-1.176)]=5.208)#

+

#sf(pH+pOH=14)#

+

#:.##sf(pH=14-pOH=14-5.208=8.8)#

+

As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.

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+ + Dec 2, 2016 + +
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#""pH"" = 8.8#

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+

Explanation:

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+
+

This is really a two-part question:

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    +
  1. What is in the mixture after the reaction is complete?
    2. +
  2. What is the #""pH""# of the mixture?
    3. +
+
+

What's in the mixture?

+

I like to use an ICE table to keep track of the stoichiometry calculations.

+

#color(white)(mmmmll)""HAc"" +color(white)(l) ""OH"" → ""H""_2""O"" + ""Ac""^""-""#; #K_text(a) = 1.76 × 10^""-5""#
+#""I/mol:""color(white)(ml)0.020color(white)(ml)0.020color(white)(mmmmmll)0#
+#""Cmol:""color(white)(m)""-0.020""color(white)(m)""-0.020""color(white)(mmmm)""+0.020""#
+#""E/mol:""color(white)(mm)0color(white)(mmml)0color(white)(mmmmmll)0.020#

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#""Initial moles HAc"" = 0.100 color(red)(cancel(color(black)(""L""))) × ""0.2 mol""/(1 color(red)(cancel(color(black)(""L"")))) = ""0.020 mol""#

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#""Initial moles OH""^""-""= 0.200 color(red)(cancel(color(black)(""L""))) × ""0.1 mol""/(1 color(red)(cancel(color(black)(""L"")))) = ""0.020 mol""#

+

We have 300 mL of a solution that contains 0.020 mol of #""Ac""^""-""#.

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+

2. What's the #""pH""# of the solution?

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#""HAc""# is a weak acid, and #""Ac""^""-""# is its conjugate base.

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The solution will be basic.

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#K_""b"" = K_""w""/K_""a"" = (1.00 × 10^""-14"")/(1.76 × 10^""-5"") = 5.68 × 10^""-10""#

+

Now, we set up another ICE table (with molarities) to calculate the #""pH""# at equilibrium.

+

#[""Ac""^""-""] = ""0.020 mol""/""0.300 L"" = ""0.067 mol/L""#

+

#color(white)(XXXXXX)""Ac""^""-"" + ""H""_2""O"" ⇌ ""HAc"" + ""HO""^""-""#
+#""I/mol:""color(white)(Xm) 0.067color(white)(XXXXXXll) 0color(white)(mmll)0#
+#""C/mol:""color(white)(mml) ""-""xcolor(white)(XXXXXXlll) ""+""xcolor(white)(mll)""+""x#
+#""E/mol:""color(white)(Xl)""0.067-""xcolor(white)(XXXXXlll) xcolor(white)(mmll)x#

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#K_b = ([""HAc""][""HO""^""-""])/([""Ac""^""-""]) = x^2/(0.067-x) = 5.68 × 10^""-10""#

+

Check if #x ≪ 0.067#.

+

#0.067/(2.10 × 10^""-10"") = 3.2 × 10^8 > 400#; ∴ we can ignore #x#.

+

#x^2/0.067 = 5.68 × 10^""-10""#

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#x^2 = 0.067 × 5.68 × 10^""-10"" = 3.8 × 10^""-11""#

+

#x = 6.2 × 10^""-6""#

+

#[""HO""^""-""] = 6.2 × 10^""-6"" color(white)(l)""mol/L""#

+

#""pOH"" = ""-log""(6.2 × 10^""-6"") = 5.21#

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#""pH"" = ""p""K_""w"" - ""pOH"" = ""14.00 - 5.21"" = 8.8#

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" 200 ml of 0.1 M sodium hydroxide is mixed with 100 ml of 0.2 M ethanoic acid. What is the pH of the resulting solution? nan +229 a830354c-6ddd-11ea-bd43-ccda262736ce https://socratic.org/questions/how-many-moles-of-gas-are-in-a-volume-of-63-3-l-at-stp 2.82 moles start physical_unit 4 4 mole mol qc_end c_other STP qc_end physical_unit 4 4 10 11 volume qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] gas [IN] moles""}]" "[{""type"":""physical unit"",""value"":""2.82 moles""}]" "[{""type"":""other"",""value"":""STP""},{""type"":""physical unit"",""value"":""Volume [OF] gas [=] \\pu{63.3 L}""}]" "

How many moles of gas are in a volume of 63.3 L at STP?

" nan 2.82 moles "
+

Explanation:

+
+

At STP of 273.15 K #(""0""^@""C"")# and a pressure of #10^5# Pa, usually written as 100 kPa, the molar volume of an ideal gas is 22.710980 L/mol. (This STP has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) since 1990.) http://goldbook.iupac.org/S05910.html

+

In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

#""63.3 L""/""27.710980 mol/L""=""2.33 moles""# (rounded to three significant figures)

+

Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.

+

#""63.3 L""/""22.414 mol/L""=""2.82 moles""# (rounded to three significant figures)

+
+
" "
+
+
+

At STP of 273.15 K and 100 kPa, 63.3 L of an ideal gas would contain 2.33 moles.

+

At STP of 273.15 K and 1 atm, 63.3 L of an ideal gas would contain 2.82 moles.

+
+
+
+

Explanation:

+
+

At STP of 273.15 K #(""0""^@""C"")# and a pressure of #10^5# Pa, usually written as 100 kPa, the molar volume of an ideal gas is 22.710980 L/mol. (This STP has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) since 1990.) http://goldbook.iupac.org/S05910.html

+

In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

#""63.3 L""/""27.710980 mol/L""=""2.33 moles""# (rounded to three significant figures)

+

Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.

+

#""63.3 L""/""22.414 mol/L""=""2.82 moles""# (rounded to three significant figures)

+
+
+
" "
+

How many moles of gas are in a volume of 63.3 L at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 31, 2016 + +
+
+
+
+
+
+
+

At STP of 273.15 K and 100 kPa, 63.3 L of an ideal gas would contain 2.33 moles.

+

At STP of 273.15 K and 1 atm, 63.3 L of an ideal gas would contain 2.82 moles.

+
+
+
+

Explanation:

+
+

At STP of 273.15 K #(""0""^@""C"")# and a pressure of #10^5# Pa, usually written as 100 kPa, the molar volume of an ideal gas is 22.710980 L/mol. (This STP has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) since 1990.) http://goldbook.iupac.org/S05910.html

+

In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

#""63.3 L""/""27.710980 mol/L""=""2.33 moles""# (rounded to three significant figures)

+

Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
+https://en.m.wikipedia.org/wiki/Molar_volume

+

Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.

+

#""63.3 L""/""22.414 mol/L""=""2.82 moles""# (rounded to three significant figures)

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 4363 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" How many moles of gas are in a volume of 63.3 L at STP? nan +230 a830354d-6ddd-11ea-a744-ccda262736ce https://socratic.org/questions/58e9aaa611ef6b6b5697db47 O2 start chemical_formula qc_end physical_unit 8 8 6 7 mass qc_end physical_unit 17 18 6 7 mass qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""chemical formula""}]" "[{""type"":""chemical equation"",""value"":""O2""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] pentane [=] \\pu{12.5 lb}""},{""type"":""physical unit"",""value"":""Mass [OF] dioxygen gas [=] \\pu{12.5 lb}""},{""type"":""other"",""value"":""Limiting reagent""}]" "

What is the limiting reagent when #12.5*lb# pentane are reacted with a #12.5*lb# mass of dioxygen gas?

" nan O2 "
+

Explanation:

+
+

And (ii) we need equivalent quantities of dioxygen and pentane. We know that #""1 lb""# #-=# #454*g#:

+

#""Moles of pentane""=(12.5*lbxx454*g*lb^-1)/(72.15*g*mol^-1)=78.7*mol#

+

#""Moles of dioxygen""=(12.5*lbxx454*g*lb^-1)/(32.0*g*mol^-1)=177.3*mol#

+

Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus #O_2# is the limiting reagent. Complete combustion requires #8xx78.7*mol# dioxygen gas. What is this as a mass?

+

By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....

+
+
" "
+
+
+

We need (i) a stoichiometric equation:

+

#C_5H_12 + 8O_2 rarr 5CO_2(g) + 6H_2O#

+
+
+
+

Explanation:

+
+

And (ii) we need equivalent quantities of dioxygen and pentane. We know that #""1 lb""# #-=# #454*g#:

+

#""Moles of pentane""=(12.5*lbxx454*g*lb^-1)/(72.15*g*mol^-1)=78.7*mol#

+

#""Moles of dioxygen""=(12.5*lbxx454*g*lb^-1)/(32.0*g*mol^-1)=177.3*mol#

+

Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus #O_2# is the limiting reagent. Complete combustion requires #8xx78.7*mol# dioxygen gas. What is this as a mass?

+

By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....

+
+
+
" "
+

What is the limiting reagent when #12.5*lb# pentane are reacted with a #12.5*lb# mass of dioxygen gas?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Limiting Reagent + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Apr 9, 2017 + +
+
+
+
+
+
+
+

We need (i) a stoichiometric equation:

+

#C_5H_12 + 8O_2 rarr 5CO_2(g) + 6H_2O#

+
+
+
+

Explanation:

+
+

And (ii) we need equivalent quantities of dioxygen and pentane. We know that #""1 lb""# #-=# #454*g#:

+

#""Moles of pentane""=(12.5*lbxx454*g*lb^-1)/(72.15*g*mol^-1)=78.7*mol#

+

#""Moles of dioxygen""=(12.5*lbxx454*g*lb^-1)/(32.0*g*mol^-1)=177.3*mol#

+

Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus #O_2# is the limiting reagent. Complete combustion requires #8xx78.7*mol# dioxygen gas. What is this as a mass?

+

By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Apr 9, 2017 + +
+
+
+
+
+
+
+

The limiting reactant is oxygen.

+
+
+
+

Explanation:

+
+
+

Here's another way to identify the limiting reactant.

+

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

+
+

1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

+

#M_r:color(white)(mmmmm) 72.15color(white)(mm)32.00#
+#color(white)(mmmmmmm)""C""_5""H""_12 +color(white)(ll) ""8O""_2 → ""5CO""_2 + ""6H""_2""O""#
+#""Amt/lb-mol:""color(white)(ll)0.1733color(white)(ml)0.3906#
+#""Divide by:""color(white)(mmml)1color(white)(mmmm)8#
+#""Moles rxn:""color(white)(mll)0.1733color(white)(ml)""0.048 83""#

+
+

Note: We do not have to stick with gram-moles. We can use pound-moles because the numbers will still be in the same ratio.

+

#""Moles of C""_5""H""_12 = 12.5 color(red)(cancel(color(black)(""lb C""_5""H""_12))) × (""1 lb-mol C""_5""H""_12)/(72.15 color(red)(cancel(color(black)(""lb C""_5""H""_12)))) = ""0.1733 lb-mol C""_5""H""_12#

+

#""Moles of O""_2 = 12.5 color(red)(cancel(color(black)(""lb O""_2))) × (""1 lb-mol O""_2)/(32.00 color(red)(cancel(color(black)(""lb O""_2)))) = ""0.3906 lb-mol O""_2#

+
+

2. Identify the limiting reactant

+

An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:

+

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

+

I did that for you in the table above.

+

#""O""_2# is the limiting reactant because it gives the fewest moles of reaction.

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+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+
" What is the limiting reagent when #12.5*lb# pentane are reacted with a #12.5*lb# mass of dioxygen gas? nan +231 a830354e-6ddd-11ea-9fe0-ccda262736ce https://socratic.org/questions/write-the-reaction-which-has-a-heat-of-reaction-equal-to-heat-of-formation-for-h-1 1/2 H2(g) + 1/2 Cl2(g) -> HCl(g) start chemical_equation qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""balanced_equation""}]" "[{""type"":""chemical equation"",""value"":""1/2 H2(g) + 1/2 Cl2(g) -> HCl(g)""}]" "[{""type"":""other"",""value"":""A heat of reaction equal to heat of formation for HCl(g)""}]" "

Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

" "
+
+

+

Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

+

+
+
" 1/2 H2(g) + 1/2 Cl2(g) -> HCl(g) "
+

Explanation:

+
+

A compound's standard heat of formation, or standard enthalpy of formation, #DeltaH_f^@#, represents the change in enthalpy that accompanies the formation of one mole of that compound from its constituent elements in their standard state.

+

In your case, the chemical reaction that has a change in enthalpy equal to #DeltaH_f^@# will describe the formation of one mole of hydrogen chloride, #""HCl""#, from its constituent elements in their standard state.

+

Now, one molecule of hydrogen chloride contains one atom of hydrogen, #""H""#, and one atom of chlorine, #""Cl""#.

+

The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.

+

+

This means that you can write

+
+

#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#

+
+

This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.

+

In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by #2#. This will get you

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#

+
+

This equation will have

+
+

#DeltaH_""rxn""^@ = DeltaH_f^@#

+
+

because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.

+
+
" "
+
+
+

#1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g))#

+
+
+
+

Explanation:

+
+

A compound's standard heat of formation, or standard enthalpy of formation, #DeltaH_f^@#, represents the change in enthalpy that accompanies the formation of one mole of that compound from its constituent elements in their standard state.

+

In your case, the chemical reaction that has a change in enthalpy equal to #DeltaH_f^@# will describe the formation of one mole of hydrogen chloride, #""HCl""#, from its constituent elements in their standard state.

+

Now, one molecule of hydrogen chloride contains one atom of hydrogen, #""H""#, and one atom of chlorine, #""Cl""#.

+

The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.

+

+

This means that you can write

+
+

#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#

+
+

This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.

+

In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by #2#. This will get you

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#

+
+

This equation will have

+
+

#DeltaH_""rxn""^@ = DeltaH_f^@#

+
+

because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.

+
+
+
" "
+

Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

+
+
+

+

Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ?

+

+
+
+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 4, 2016 + +
+
+
+
+
+
+
+

#1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g))#

+
+
+
+

Explanation:

+
+

A compound's standard heat of formation, or standard enthalpy of formation, #DeltaH_f^@#, represents the change in enthalpy that accompanies the formation of one mole of that compound from its constituent elements in their standard state.

+

In your case, the chemical reaction that has a change in enthalpy equal to #DeltaH_f^@# will describe the formation of one mole of hydrogen chloride, #""HCl""#, from its constituent elements in their standard state.

+

Now, one molecule of hydrogen chloride contains one atom of hydrogen, #""H""#, and one atom of chlorine, #""Cl""#.

+

The key now is to remember that hydrogen and chlorine exist as diatomic elements in their standard state.

+

+

This means that you can write

+
+

#""H""_ (2(g)) + ""Cl""_ (2(g)) -> 2""HCl""_((g))#

+
+

This represents one form of the balanced chemical equation that describes the synthesis of hydrogen chloride.

+

In order for the enthalpy change of reaction to be equal to that of the standard enthalpy change of formation, you must divide everything by #2#. This will get you

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)(1/2""H""_ (2(g)) + 1/2""Cl""_ (2(g)) -> ""HCl""_((g)))color(white)(a/a)|)))#

+
+

This equation will have

+
+

#DeltaH_""rxn""^@ = DeltaH_f^@#

+
+

because it describes the formation of one mole of hydrogen chloride from its constituent elements in their standard state.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 36026 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ? " + + +Write the reaction which has a heat of reaction equal to heat of formation for HCl(g) ? + + +" +232 a830354f-6ddd-11ea-b085-ccda262736ce https://socratic.org/questions/how-many-molecules-are-there-in-2-80-kg-of-hydrazine-n-2h-2 5.26 × 10^25 start physical_unit 10 10 number none qc_end physical_unit 10 10 6 7 mass qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] N2H2""}]" "[{""type"":""physical unit"",""value"":""5.26 × 10^25""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N2H2 [=] \\pu{2.80 kg}""}]" "

How many molecules are there in 2.80 kg of hydrazine, #N_2H_2#?

" nan 5.26 × 10^25 "
+

Explanation:

+
+

#""Moles of hydrazine""# #=# #""Mass of hydrazine""/""Molar mass of hydrazine""#

+

#=(2.8xx10^3*g)/(32.05*g*mol^-1)~=87*mol.#

+

And if there are #87*mol#, there are #87*molxxN_A#, where #N_A=""the avocado number""=6.022xx10^23*mol^-1#,

+

And #87*molxxN_A=5.26xx10^25# individual hydrazine molecules.

+

#N_2H_2# is #""diazene""#, which can exist in two isomeric forms. Why?

+
+
" "
+
+
+

#""Hydrazine is""# #N_2H_4#

+
+
+
+

Explanation:

+
+

#""Moles of hydrazine""# #=# #""Mass of hydrazine""/""Molar mass of hydrazine""#

+

#=(2.8xx10^3*g)/(32.05*g*mol^-1)~=87*mol.#

+

And if there are #87*mol#, there are #87*molxxN_A#, where #N_A=""the avocado number""=6.022xx10^23*mol^-1#,

+

And #87*molxxN_A=5.26xx10^25# individual hydrazine molecules.

+

#N_2H_2# is #""diazene""#, which can exist in two isomeric forms. Why?

+
+
+
" "
+

How many molecules are there in 2.80 kg of hydrazine, #N_2H_2#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 4, 2017 + +
+
+
+
+
+
+
+

#""Hydrazine is""# #N_2H_4#

+
+
+
+

Explanation:

+
+

#""Moles of hydrazine""# #=# #""Mass of hydrazine""/""Molar mass of hydrazine""#

+

#=(2.8xx10^3*g)/(32.05*g*mol^-1)~=87*mol.#

+

And if there are #87*mol#, there are #87*molxxN_A#, where #N_A=""the avocado number""=6.022xx10^23*mol^-1#,

+

And #87*molxxN_A=5.26xx10^25# individual hydrazine molecules.

+

#N_2H_2# is #""diazene""#, which can exist in two isomeric forms. Why?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5285 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many molecules are there in 2.80 kg of hydrazine, #N_2H_2#? nan +233 a8303550-6ddd-11ea-a0ff-ccda262736ce https://socratic.org/questions/a-60-0-ml-sample-of-co-2-gas-is-collected-over-water-at-70-0-and-101-3-kpa-what- 33.53 mL start physical_unit 22 24 volume ml qc_end physical_unit 3 6 1 2 volume qc_end physical_unit 10 10 12 13 temperature qc_end physical_unit 10 10 15 16 pressure qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] the dry gas [IN] mL""}]" "[{""type"":""physical unit"",""value"":""33.53 mL""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] sample of CO2 gas [=] \\pu{60.0 mL}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{70.0 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] water [=] \\pu{101.3 kPa}""},{""type"":""other"",""value"":""STP""}]" "

A 60.0 mL sample of #CO_2# gas is collected over water at 70.0° and 101.3 kPa. What is the volume of the dry gas at STP?

" nan 33.53 mL "
+

Explanation:

+
+

The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.

+

+

In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide, #""CO""_2#, and molecules of water, #""H""_2""O""#.

+

This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.

+
+

#color(blue)(P_""total"" = sum_i P_i)"" ""#, where

+
+

#P_""total""# - the total pressure of the mixture
+#P_i# - the partial pressure of gas #i#

+

In this case, the vapor pressure of water at #70.0^@""C""# is equal to #""31.09 kPa""#

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

This means that you have

+
+

#P_""total"" = P_(CO_2) + P_(H_2O)#

+

#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#

+
+

Now, STP conditions imply a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Since you're looking to find the volume of the gas at STP, and keeping in mind that the number of moles of gas remains constant, you can use the combined gas law equation

+
+

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""#, where

+
+

#P_1#, #V_1#, #T_1# - the pressure, volume, and temperature of the gas at an initial state
+#P_2#, #V_2#, #T_2# - the pressure, volume, and temperature of the gas at a final state

+

Plug in your values and solve for #V_2# - do not forget to convert the temperature from degrees Celsius to Kelvin!

+
+

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

+

#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#

+

#V_2 = color(green)(""33.5 mL"")#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""33.5 mL""#

+
+
+
+

Explanation:

+
+

The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.

+

+

In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide, #""CO""_2#, and molecules of water, #""H""_2""O""#.

+

This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.

+
+

#color(blue)(P_""total"" = sum_i P_i)"" ""#, where

+
+

#P_""total""# - the total pressure of the mixture
+#P_i# - the partial pressure of gas #i#

+

In this case, the vapor pressure of water at #70.0^@""C""# is equal to #""31.09 kPa""#

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

This means that you have

+
+

#P_""total"" = P_(CO_2) + P_(H_2O)#

+

#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#

+
+

Now, STP conditions imply a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Since you're looking to find the volume of the gas at STP, and keeping in mind that the number of moles of gas remains constant, you can use the combined gas law equation

+
+

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""#, where

+
+

#P_1#, #V_1#, #T_1# - the pressure, volume, and temperature of the gas at an initial state
+#P_2#, #V_2#, #T_2# - the pressure, volume, and temperature of the gas at a final state

+

Plug in your values and solve for #V_2# - do not forget to convert the temperature from degrees Celsius to Kelvin!

+
+

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

+

#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#

+

#V_2 = color(green)(""33.5 mL"")#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

A 60.0 mL sample of #CO_2# gas is collected over water at 70.0° and 101.3 kPa. What is the volume of the dry gas at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Combined Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 23, 2015 + +
+
+
+
+
+
+
+

#""33.5 mL""#

+
+
+
+

Explanation:

+
+

The important thing to remember about gases that are being collected over water is that the total pressure measured includes the vapor pressure of water at that given temperature.

+

+

In this case, the collection tube will hold a gaseous mixture that contains molecules of carbon dioxide, #""CO""_2#, and molecules of water, #""H""_2""O""#.

+

This means that you can use Dalton's Law of Partial Pressures to determine the partial pressure of carbon dioxide in this mixture.

+
+

#color(blue)(P_""total"" = sum_i P_i)"" ""#, where

+
+

#P_""total""# - the total pressure of the mixture
+#P_i# - the partial pressure of gas #i#

+

In this case, the vapor pressure of water at #70.0^@""C""# is equal to #""31.09 kPa""#

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

This means that you have

+
+

#P_""total"" = P_(CO_2) + P_(H_2O)#

+

#P_(CO_2) = ""101.3 kPa"" - ""31.09 kPa"" = ""70.21 kPa""#

+
+

Now, STP conditions imply a temperature of #0^@""C""# and a pressure of #""100 kPa""#. Since you're looking to find the volume of the gas at STP, and keeping in mind that the number of moles of gas remains constant, you can use the combined gas law equation

+
+

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)"" ""#, where

+
+

#P_1#, #V_1#, #T_1# - the pressure, volume, and temperature of the gas at an initial state
+#P_2#, #V_2#, #T_2# - the pressure, volume, and temperature of the gas at a final state

+

Plug in your values and solve for #V_2# - do not forget to convert the temperature from degrees Celsius to Kelvin!

+
+

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

+

#V_2 = (70.21 color(red)(cancel(color(black)(""kPa""))))/(100.0color(red)(cancel(color(black)(""kPa"")))) * ( (273.15 + 0)color(red)(cancel(color(black)(""K""))))/((273.15 + 70.0)color(red)(cancel(color(black)(""K"")))) * ""60.0 mL""#

+

#V_2 = color(green)(""33.5 mL"")#

+
+

The answer is rounded to three sig figs.

+
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+
+
+
+
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" A 60.0 mL sample of #CO_2# gas is collected over water at 70.0° and 101.3 kPa. What is the volume of the dry gas at STP? nan +234 a8305a12-6ddd-11ea-8d78-ccda262736ce https://socratic.org/questions/58a5b0a011ef6b26a86c6b0a 20.00 g/eq start physical_unit 21 22 equivalent_weight g/eq qc_end physical_unit 7 8 1 2 volume qc_end physical_unit 7 8 5 6 molarity qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Equivalent weight [OF] the metal [IN] g/eq""}]" "[{""type"":""physical unit"",""value"":""20.00 g/eq""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] hydrochloric acid [=] \\pu{100 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] hydrochloric acid [=] \\pu{2.0 mol*L^(−1)}""},{""type"":""other"",""value"":""Hydrochloric acid titrated a sample of metal carbonate.""}]" "

A #100*mL# volume of #2.0*mol*L^-1# hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

" nan 20.00 g/eq "
+

Explanation:

+
+

First we need a chemical equation:

+

#MCO_3(s) + 2HCl(aq) rarr MCl_2(aq) + H_2O(l) + CO_2(g)uarr#

+

#""Moles of HCl""=0.100*Lxx2*mol*L^-1=0.200*mol#

+

Given the equation, the #""metal carbonate""# represents #0.100*mol#, and thus, given the equation, the equivalent weight of the #""metal carbonate""# is #100.0*g*mol^-1#. This equivalent weight is CONSISTENT with a formula of #CaCO_3#.

+
+
" "
+
+
+

The equivalent weight of the metal is #100.0*g*mol^-1#

+
+
+
+

Explanation:

+
+

First we need a chemical equation:

+

#MCO_3(s) + 2HCl(aq) rarr MCl_2(aq) + H_2O(l) + CO_2(g)uarr#

+

#""Moles of HCl""=0.100*Lxx2*mol*L^-1=0.200*mol#

+

Given the equation, the #""metal carbonate""# represents #0.100*mol#, and thus, given the equation, the equivalent weight of the #""metal carbonate""# is #100.0*g*mol^-1#. This equivalent weight is CONSISTENT with a formula of #CaCO_3#.

+
+
+
" "
+

A #100*mL# volume of #2.0*mol*L^-1# hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+3 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Feb 16, 2017 + +
+
+
+
+
+
+
+

The equivalent weight of the metal is #100.0*g*mol^-1#

+
+
+
+

Explanation:

+
+

First we need a chemical equation:

+

#MCO_3(s) + 2HCl(aq) rarr MCl_2(aq) + H_2O(l) + CO_2(g)uarr#

+

#""Moles of HCl""=0.100*Lxx2*mol*L^-1=0.200*mol#

+

Given the equation, the #""metal carbonate""# represents #0.100*mol#, and thus, given the equation, the equivalent weight of the #""metal carbonate""# is #100.0*g*mol^-1#. This equivalent weight is CONSISTENT with a formula of #CaCO_3#.

+
+
+
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+ +
+
+
+
+
+ +
+
+ +
+ + Oct 21, 2017 + +
+
+
+
+
+
+
+

Equivalent weight of metal is 20 g/eq

+
+
+
+

Explanation:

+
+

For HCl
+Molarity = Normality

+

Gram equivalents of HCl = Normality × Volume in litres
+#= 2 N × 100 × 10^-3 L#
+#= 0.2# eq

+

For complete neutralisation
+Gram equivalents of HCl = Gram equivalents of metal carbonate

+

Equivalent weight of metal carbonate #= text(Weight of metal carbonate) / text(Gram equivalents of metal carbonate)#

+

#= (10 g) / (0.2 text(eq)) = 50# g/eq

+

Equivalent weight of carbonate #(CO_3^(2-)) = (12 + (3 × 16))/2 = 30# g/eq

+

Equivalent weight of metal carbonate = Equivalent weight of metal + Equivalent weight of carbonate

+

50 g/eq = x + 30 g/eq
+x = 20 g/eq

+

∴ Equivalent weight of metal is 20 g/eq

+
+
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+ +
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+
+ +
+
+ +
+ + Oct 21, 2017 + +
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+

The number of moles of #HCL# reacted with Carbonate salt is

+

#=""molarity of acid"" xx "" volume of acid in L""#

+

#=2 mol""/""Lxx 0.1L=0.2mol#

+

So the mass of #HCl# reacted #= 0.2xx36.5g#, where molar mass of #HCl =36.5g""/""mol#

+

Now we know from the law of equivalent proportion that the ratio of masses of two reacting substances for neutralization is equal to the ratio of their equivalent masses.

+

Hence we can say

+

#""Equivalent mass of metal carboate"" /(""Equivalent mas of ""HCl)=""Reacting mass of metal carboate"" /(""Equivalent mas of ""HCl)#

+

#=>(""Equivalent mass of metal (x=?)+ equivalent mass of ""CO_3^(2-))/(""Equivalent mas of ""HCl)=(0.2xx36.5)/36.5#

+

#=>(x + 30""g/equiv"")/(36.5""g/equiv"")=(10g)/(0.2xx36.5g)#

+

#=>color(red)(x=20""g/equiv"")#

+

Please note
+Equivalent mass of #CO_3^(2-)= ("" formula mass of "" +CO_3^(2-))/""valency ""#

+

#=(12+3xx16)/2=30""g/equiv""#

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+
+
+
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+
" A #100*mL# volume of #2.0*mol*L^-1# hydrochloric acid titrated a sample of metal carbonate. What is the equivalent weight of the metal? nan +235 a8305a13-6ddd-11ea-8278-ccda262736ce https://socratic.org/questions/a-solution-of-sodium-chloride-in-water-has-a-vapor-pressure-of-19-6-torr-at-25-d 17.30% start physical_unit 24 29 mole_fraction none qc_end physical_unit 3 6 12 13 vapor_pressure qc_end physical_unit 3 6 15 17 temperature qc_end end "[{""type"":""physical unit"",""value"":""mole fraction [OF] NaCl solute particles in this solution""}]" "[{""type"":""physical unit"",""value"":""17.30%""}]" "[{""type"":""physical unit"",""value"":""vapor pressure [OF] sodium chloride in water [=] \\pu{19.6 torr}""},{""type"":""physical unit"",""value"":""Temperature [OF] sodium chloride in water [=] \\pu{25 degrees Celsius}""}]" "

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25 degrees Celsius. What is the mole fraction of NaCl solute particles in this solution?

" nan 17.30% "
+

Explanation:

+
+

The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at #25^@""C""#.

+

You can use an online calculator to find that the vapor pressure of pure water at #25^@""C""# is equal to about #""23.7 torr""#.

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

So, plug in your values into the above equation and solve for #chi_""water""#

+
+

#chi_""water"" = P_""sol""/P_""water""^@#

+

#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#

+
+

Since the solution only contains water and sodium chloride, you can say that

+
+

#chi_""water"" + chi_""NaCl"" = 1#

+
+

This means that the mole fraction of sodium chloride is

+
+

#chi_""NaCl"" = 1 - chi_""water""#

+

#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#

+
+
+
" "
+
+
+

#0.173#

+
+
+
+

Explanation:

+
+

The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at #25^@""C""#.

+

You can use an online calculator to find that the vapor pressure of pure water at #25^@""C""# is equal to about #""23.7 torr""#.

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

So, plug in your values into the above equation and solve for #chi_""water""#

+
+

#chi_""water"" = P_""sol""/P_""water""^@#

+

#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#

+
+

Since the solution only contains water and sodium chloride, you can say that

+
+

#chi_""water"" + chi_""NaCl"" = 1#

+
+

This means that the mole fraction of sodium chloride is

+
+

#chi_""NaCl"" = 1 - chi_""water""#

+

#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#

+
+
+
+
" "
+

A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25 degrees Celsius. What is the mole fraction of NaCl solute particles in this solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 21, 2015 + +
+
+
+
+
+
+
+

#0.173#

+
+
+
+

Explanation:

+
+

The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.

+
+

#color(blue)(P_""sol"" = chi_""solvent"" * P_""solvent""^@)"" ""#, where

+
+

#P_""sol""# - the vapor pressure of the solution
+#chi_""solvent""# - the mole fraction of the solvent
+#P_""solvent""^@# - the vapor pressure of the pure solvent

+

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at #25^@""C""#.

+

You can use an online calculator to find that the vapor pressure of pure water at #25^@""C""# is equal to about #""23.7 torr""#.

+

http://www.endmemo.com/chem/vaporpressurewater.php

+

So, plug in your values into the above equation and solve for #chi_""water""#

+
+

#chi_""water"" = P_""sol""/P_""water""^@#

+

#chi_""water"" = (19.6color(red)(cancel(color(black)(""torr""))))/(23.7color(red)(cancel(color(black)(""torr"")))) = 0.827#

+
+

Since the solution only contains water and sodium chloride, you can say that

+
+

#chi_""water"" + chi_""NaCl"" = 1#

+
+

This means that the mole fraction of sodium chloride is

+
+

#chi_""NaCl"" = 1 - chi_""water""#

+

#chi_""NaCl"" = 1 - 0.827 = color(green)(0.173)#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
+ 16812 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25 degrees Celsius. What is the mole fraction of NaCl solute particles in this solution? nan +236 a830812e-6ddd-11ea-8740-ccda262736ce https://socratic.org/questions/how-many-grams-are-in-12-4-10-15-atoms-of-neon 4.16 × 10^(-7) grams start physical_unit 10 10 mass g qc_end physical_unit 8 10 5 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] neon [IN] grams""}]" "[{""type"":""physical unit"",""value"":""4.16 × 10^(-7) grams""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of neon [=] \\pu{12.4 ⋅ 10^15}""}]" "

How many grams are in #12.4*10^15# atoms of neon?

" nan 4.16 × 10^(-7) grams "
+

Explanation:

+
+

And thus the mass of #12.4xx10^15# #""neon atoms""# #=#

+

#(12.4xx10^15)/(6.022xx10^(23)*mol^-1)xx20.18*g*mol^-1# #=# #??g#

+
+
" "
+
+
+

Well I know that #6.022xx10^(23)# atoms of neon have a mass of #20.18*g#. How do I know this?

+
+
+
+

Explanation:

+
+

And thus the mass of #12.4xx10^15# #""neon atoms""# #=#

+

#(12.4xx10^15)/(6.022xx10^(23)*mol^-1)xx20.18*g*mol^-1# #=# #??g#

+
+
+
" "
+

How many grams are in #12.4*10^15# atoms of neon?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 27, 2016 + +
+
+
+
+
+
+
+

Well I know that #6.022xx10^(23)# atoms of neon have a mass of #20.18*g#. How do I know this?

+
+
+
+

Explanation:

+
+

And thus the mass of #12.4xx10^15# #""neon atoms""# #=#

+

#(12.4xx10^15)/(6.022xx10^(23)*mol^-1)xx20.18*g*mol^-1# #=# #??g#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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+
" How many grams are in #12.4*10^15# atoms of neon? nan +237 a830aeba-6ddd-11ea-abe4-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-17-g-of-nh-3-in-50-l-of-solutio 1.00 mol/L start physical_unit 5 6 molarity mol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 6 6 14 15 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] a solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.00 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NH3 [=] \\pu{17 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{0.50 L}""}]" "

What is the molarity of a solution that contains 17 g of #NH_3# in .50 L of solution?

" nan 1.00 mol/L "
+

Explanation:

+
+

Molarity (M) = number of moles of solute (n) / volume of solution ( L)

+

number of moles of solute , n = mass of solute (g) / molar mass ( g/mol)

+

n = (17 g / 34 g/mol)

+

n = 1/2 = 0.5 mol

+

M = 0.5 mol / 0.50 L

+

M = 1 mol / L

+
+
" "
+
+
+

1 molar

+
+
+
+

Explanation:

+
+

Molarity (M) = number of moles of solute (n) / volume of solution ( L)

+

number of moles of solute , n = mass of solute (g) / molar mass ( g/mol)

+

n = (17 g / 34 g/mol)

+

n = 1/2 = 0.5 mol

+

M = 0.5 mol / 0.50 L

+

M = 1 mol / L

+
+
+
" "
+

What is the molarity of a solution that contains 17 g of #NH_3# in .50 L of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 17, 2016 + +
+
+
+
+
+
+
+

1 molar

+
+
+
+

Explanation:

+
+

Molarity (M) = number of moles of solute (n) / volume of solution ( L)

+

number of moles of solute , n = mass of solute (g) / molar mass ( g/mol)

+

n = (17 g / 34 g/mol)

+

n = 1/2 = 0.5 mol

+

M = 0.5 mol / 0.50 L

+

M = 1 mol / L

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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+
+
" What is the molarity of a solution that contains 17 g of #NH_3# in .50 L of solution? nan +238 a830aebb-6ddd-11ea-920a-ccda262736ce https://socratic.org/questions/a-student-found-that-53-2-ml-of-a-0-232m-solution-of-naoh-was-required-to-titrat 0.49 M start physical_unit 35 38 molarity mol/l qc_end physical_unit 10 12 4 5 volume qc_end physical_unit 10 12 8 9 molarity qc_end physical_unit 20 23 17 18 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] the acetic acid solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.49 M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution of NaOH [=] \\pu{53.2 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] solution of NaOH [=] \\pu{0.232 M}""},{""type"":""physical unit"",""value"":""Volume [OF] an acetic acid solution [=] \\pu{25.0 mL}""}]" "

A student found that 53.2 mL of a 0.232M solution of NaOH was required to titrate 25.0 mL of an acetic acid solution of unknown molarity to the endpoint. What is the molarity of the acetic acid solution?

" nan 0.49 M "
+

Explanation:

+
+

The key to this problem is the balanced chemical equation for this neutralization reaction.

+

Acetic acid, #""CH""_3""COOH""#, a weak acid, will react with the hydroxide anions, #""OH""^(-)#, delivered to the reaction by the sodium hydroxide solution, to form acetate anions, #""CH""_3""COO""^(-)#, and water, according to the balanced chemical equation

+
+

#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#

+
+

As you can see, the acid and the base react in a #1:1# mole ratio. This tells that in order to get to the equivalence point, i.e. have a complete neutralization, you must mix equal numbers of moles of each reactant.

+

The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#

+
+

This means that the acid solution must have contained #0.01234# moles of acetic acid.

+

The molarity of the acetic acid solution was

+
+

#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

It's worth noting that the pH of the solution at equivalence point will not be equal to #7#, like you have when a strong acid and a strong base neutralize each other completely.

+

Instead, the pH of the solution at equivalence point will be greater than #7#. That happens because the acetate anions, #""CH""_3""COO""^(-)#, will react with water to reform some of the acetic acid and produce hydroxide anions, which in turn will make the solution basic.

+

+
+
" "
+
+
+

#""0.494 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

The key to this problem is the balanced chemical equation for this neutralization reaction.

+

Acetic acid, #""CH""_3""COOH""#, a weak acid, will react with the hydroxide anions, #""OH""^(-)#, delivered to the reaction by the sodium hydroxide solution, to form acetate anions, #""CH""_3""COO""^(-)#, and water, according to the balanced chemical equation

+
+

#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#

+
+

As you can see, the acid and the base react in a #1:1# mole ratio. This tells that in order to get to the equivalence point, i.e. have a complete neutralization, you must mix equal numbers of moles of each reactant.

+

The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#

+
+

This means that the acid solution must have contained #0.01234# moles of acetic acid.

+

The molarity of the acetic acid solution was

+
+

#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

It's worth noting that the pH of the solution at equivalence point will not be equal to #7#, like you have when a strong acid and a strong base neutralize each other completely.

+

Instead, the pH of the solution at equivalence point will be greater than #7#. That happens because the acetate anions, #""CH""_3""COO""^(-)#, will react with water to reform some of the acetic acid and produce hydroxide anions, which in turn will make the solution basic.

+

+
+
+
" "
+

A student found that 53.2 mL of a 0.232M solution of NaOH was required to titrate 25.0 mL of an acetic acid solution of unknown molarity to the endpoint. What is the molarity of the acetic acid solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 12, 2016 + +
+
+
+
+
+
+
+

#""0.494 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

The key to this problem is the balanced chemical equation for this neutralization reaction.

+

Acetic acid, #""CH""_3""COOH""#, a weak acid, will react with the hydroxide anions, #""OH""^(-)#, delivered to the reaction by the sodium hydroxide solution, to form acetate anions, #""CH""_3""COO""^(-)#, and water, according to the balanced chemical equation

+
+

#""CH""_3""COOH""_text((aq]) + ""OH""_text((aq])^(-) -> ""CH""_3""COO""_text((aq])^(-) + ""H""_2""O""_text((l])#

+
+

As you can see, the acid and the base react in a #1:1# mole ratio. This tells that in order to get to the equivalence point, i.e. have a complete neutralization, you must mix equal numbers of moles of each reactant.

+

The problem provides you with the molarity and volume of the sodium hydroxide solution, which means that you can use the definition of molarity to find how many moles of hydroxide anions were needed to completely neutralize the acid.

+
+

#color(blue)(|bar(ul(color(white)(a/a)c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution""color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#n_(OH^(-)) = ""0.232 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * 53.2 * 10^(-3)color(red)(cancel(color(black)(""L""))) = ""0.01234 moles OH""^(-)#

+
+

This means that the acid solution must have contained #0.01234# moles of acetic acid.

+

The molarity of the acetic acid solution was

+
+

#[""CH""_3""COOH""] = ""0.01234 moles""/(25.0 * 10^(-3)""L"") = color(green)(|bar(ul(color(white)(a/a)""0.494 mol L""^(-1)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+

It's worth noting that the pH of the solution at equivalence point will not be equal to #7#, like you have when a strong acid and a strong base neutralize each other completely.

+

Instead, the pH of the solution at equivalence point will be greater than #7#. That happens because the acetate anions, #""CH""_3""COO""^(-)#, will react with water to reform some of the acetic acid and produce hydroxide anions, which in turn will make the solution basic.

+

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+
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" A student found that 53.2 mL of a 0.232M solution of NaOH was required to titrate 25.0 mL of an acetic acid solution of unknown molarity to the endpoint. What is the molarity of the acetic acid solution? nan +239 a830aebc-6ddd-11ea-8950-ccda262736ce https://socratic.org/questions/how-many-molecules-of-ammonia-are-formed-in-the-equation-n-2-3h-2-2nh-3 2 start physical_unit 2 4 number none qc_end chemical_equation 10 16 qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] molecules of ammonia""}]" "[{""type"":""physical unit"",""value"":""2""}]" "[{""type"":""chemical equation"",""value"":""N2 + 3 H2 -> 2 NH3""}]" "

How many molecules of ammonia are formed in the equation: #N_2 + 3H_2 → 2NH_3#?

" nan 2 "
+

Explanation:

+
+

The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.

+
+
" "
+
+
+

Well clearly stoichiometry dictates that 2 equiv of ammonia are formed in the stoichiometric reaction. But there is a catch......

+
+
+
+

Explanation:

+
+

The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.

+
+
+
" "
+

How many molecules of ammonia are formed in the equation: #N_2 + 3H_2 → 2NH_3#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 30, 2016 + +
+
+
+
+
+
+
+

Well clearly stoichiometry dictates that 2 equiv of ammonia are formed in the stoichiometric reaction. But there is a catch......

+
+
+
+

Explanation:

+
+

The formation of ammonia is probably one of the best studied equilibrium reactions. This is an equilibrium reaction that requires high pressures, and high temperatures to provide acceptable rates. The one thing this reaction has got going for it is that the ammonia product is condensable (perhaps on a cold finger), whereas the reactants are essentially non-condensable.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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" How many molecules of ammonia are formed in the equation: #N_2 + 3H_2 → 2NH_3#? nan +240 a830aebd-6ddd-11ea-8442-ccda262736ce https://socratic.org/questions/5703114a7c0149124535e5aa 2 Na3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)2 + 6 NaNO3 start chemical_equation qc_end physical_unit 7 7 11 12 mass qc_end physical_unit 9 9 17 18 mass qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the equation""}]" "[{""type"":""chemical equation"",""value"":""2 Na3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)2 + 6 NaNO3""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] Na3PO4 [=] \\pu{17.3 g}""},{""type"":""physical unit"",""value"":""Mass [OF] Pb(NO3)2 [=] \\pu{12.4 g}""}]" "

How would we write the equation between #Na_3PO_4# and #Pb(NO_3)_2#, given #17.3*g# of the former, and #12.4*g# of the latter?

" nan 2 Na3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)2 + 6 NaNO3 "
+

Explanation:

+
+

#""Moles of lead nitrate""# #=# #(12.4*cancelg)/(331.2*cancel(g)*mol^-1)# #=# #0.0374 *mol#.

+

#""Moles of trisodium phosphate""# #=# #(17.3*g)/(163.94*g*mol^-1)# #=# #0.106*mol#.

+

There is a slight excess of lead nitrate: #3xx0.0374*mol"" lead nitrate""=0.112*mol#. See if you can work the excess in grams.

+

Note that if we did the actual experiment, we would probably get #PbHPO_4(s)#... i.e. #""lead(II)biphosphate...""#

+
+
" "
+
+
+

#2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) rarr Pb_3(PO_4)_2(s)darr + 6NaNO_3(aq)#

+
+
+
+

Explanation:

+
+

#""Moles of lead nitrate""# #=# #(12.4*cancelg)/(331.2*cancel(g)*mol^-1)# #=# #0.0374 *mol#.

+

#""Moles of trisodium phosphate""# #=# #(17.3*g)/(163.94*g*mol^-1)# #=# #0.106*mol#.

+

There is a slight excess of lead nitrate: #3xx0.0374*mol"" lead nitrate""=0.112*mol#. See if you can work the excess in grams.

+

Note that if we did the actual experiment, we would probably get #PbHPO_4(s)#... i.e. #""lead(II)biphosphate...""#

+
+
+
" "
+

How would we write the equation between #Na_3PO_4# and #Pb(NO_3)_2#, given #17.3*g# of the former, and #12.4*g# of the latter?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
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+
+1 Answer +
+
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+
+
+
+ +
+
+ +
+ + Apr 5, 2016 + +
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+
+
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#2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) rarr Pb_3(PO_4)_2(s)darr + 6NaNO_3(aq)#

+
+
+
+

Explanation:

+
+

#""Moles of lead nitrate""# #=# #(12.4*cancelg)/(331.2*cancel(g)*mol^-1)# #=# #0.0374 *mol#.

+

#""Moles of trisodium phosphate""# #=# #(17.3*g)/(163.94*g*mol^-1)# #=# #0.106*mol#.

+

There is a slight excess of lead nitrate: #3xx0.0374*mol"" lead nitrate""=0.112*mol#. See if you can work the excess in grams.

+

Note that if we did the actual experiment, we would probably get #PbHPO_4(s)#... i.e. #""lead(II)biphosphate...""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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+
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+ + Creative Commons License + +
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+
" How would we write the equation between #Na_3PO_4# and #Pb(NO_3)_2#, given #17.3*g# of the former, and #12.4*g# of the latter? nan +241 a830cf24-6ddd-11ea-ad2c-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-53-73-fe-and-46-27-of- Fe2S3 start chemical_formula qc_end physical_unit 11 11 10 10 percent qc_end physical_unit 15 15 13 13 percent qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""Fe2S3""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] Fe [=] \\pu{53.73%}""},{""type"":""physical unit"",""value"":""Percent [OF] S [=] \\pu{46.27%}""}]" "

What is the empirical formula of a compound that contains 53.73% Fe and 46.27% of S?

" nan Fe2S3 "
+

Explanation:

+
+

Your strategy here will be to

+
+
    +
  • pick a sample of this ionic compound
    • +
  • use the molar masses of iron and of sulfur to find how many moles of each it contains
    • +
  • find the smallest whole number ratio that exists between iron and sulfur in the compound
    • +
+
+

To make the calculations easier, pick a #""100-g""# sample of this unknown compound. This sample with contain, as given by the known percent composition

+
+
    +
  • #""53.73 g Fe""#
    • +
  • #""46.27 g S""#
    • +
+
+

Convert these masses to moles

+
+

#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#

+

#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#

+
+

Next, divide both values by the smallest one

+
+

#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#

+
+

Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by #2#. You will have

+
+

#""For Fe: "" 2 xx 1 = 2#

+

#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#

+
+

The empirical formula of the unknown compound will thus be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#

+
+
+
" "
+
+
+

#""Fe""_2""S""_3#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
+
    +
  • pick a sample of this ionic compound
    • +
  • use the molar masses of iron and of sulfur to find how many moles of each it contains
    • +
  • find the smallest whole number ratio that exists between iron and sulfur in the compound
    • +
+
+

To make the calculations easier, pick a #""100-g""# sample of this unknown compound. This sample with contain, as given by the known percent composition

+
+
    +
  • #""53.73 g Fe""#
    • +
  • #""46.27 g S""#
    • +
+
+

Convert these masses to moles

+
+

#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#

+

#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#

+
+

Next, divide both values by the smallest one

+
+

#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#

+
+

Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by #2#. You will have

+
+

#""For Fe: "" 2 xx 1 = 2#

+

#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#

+
+

The empirical formula of the unknown compound will thus be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#

+
+
+
+
" "
+

What is the empirical formula of a compound that contains 53.73% Fe and 46.27% of S?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
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+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Sep 5, 2016 + +
+
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+

#""Fe""_2""S""_3#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
+
    +
  • pick a sample of this ionic compound
    • +
  • use the molar masses of iron and of sulfur to find how many moles of each it contains
    • +
  • find the smallest whole number ratio that exists between iron and sulfur in the compound
    • +
+
+

To make the calculations easier, pick a #""100-g""# sample of this unknown compound. This sample with contain, as given by the known percent composition

+
+
    +
  • #""53.73 g Fe""#
    • +
  • #""46.27 g S""#
    • +
+
+

Convert these masses to moles

+
+

#""For Fe: "" 53.73 color(red)(cancel(color(black)(""g""))) * ""1 mole Fe""/(55.845color(red)(cancel(color(black)(""g"")))) = 0.962#

+

#""For S: "" 46.27 color(red)(cancel(color(black)(""g""))) * ""1 mole S""/(32.066color(red)(cancel(color(black)(""g"")))) = 1.44#

+
+

Next, divide both values by the smallest one

+
+

#""For Fe: "" (0.95 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1#

+

#""For S: "" (1.44 color(red)(cancel(color(black)(""moles""))))/(0.95color(red)(cancel(color(black)(""moles"")))) = 1.52#

+
+

Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by #2#. You will have

+
+

#""For Fe: "" 2 xx 1 = 2#

+

#""For S: "" 2 xx 1.52 = 3.04 ~~ 3#

+
+

The empirical formula of the unknown compound will thus be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)(""Fe""_2""S""_3)color(white)(a/a)|)))#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 15603 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the empirical formula of a compound that contains 53.73% Fe and 46.27% of S? nan +242 a830cf25-6ddd-11ea-b981-ccda262736ce https://socratic.org/questions/the-concentration-of-a-water-solution-of-nacl-is-2-48-m-and-it-contains-806-g-of 116.82 g start physical_unit 7 7 mass g qc_end physical_unit 3 7 9 10 concentration qc_end physical_unit 4 4 14 15 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [IN] g""}]" "[{""type"":""physical unit"",""value"":""116.82 g""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] a water solution of NaCl [=] \\pu{2.48 M}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{806 g}""}]" "

The concentration of a water solution of #NaCl# is 2.48 m, and it contains 806 g of water. How much #NaCl# is in the solution?

" nan 116.82 g "
+

Explanation:

+
+

This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.

+

As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!

+
+

#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#

+
+

So, a #""1-molal""# solution will contain #""1 mole""# of solute for every #""1 kg""# of solvent.

+

Now, your solution is said to have a molality of #""2.48 molal""#. This means that you would get #2.48# moles of sodium chloride for every #""1 kg""# of solvent.

+

However, you are told that your solution contains less than #""1 kg""# of water, your solvent. More specifically, it contains

+
+

#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#

+
+

of water. This means that you can expect it to contain fewer than #2.48# moles. Indeed, you can plug in your values to find

+
+

#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#

+
+

Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams

+
+

#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(NaCl) = color(green)(""117 g"")#

+
+
+
" "
+
+
+

#""117 g""#

+
+
+
+

Explanation:

+
+

This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.

+

As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!

+
+

#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#

+
+

So, a #""1-molal""# solution will contain #""1 mole""# of solute for every #""1 kg""# of solvent.

+

Now, your solution is said to have a molality of #""2.48 molal""#. This means that you would get #2.48# moles of sodium chloride for every #""1 kg""# of solvent.

+

However, you are told that your solution contains less than #""1 kg""# of water, your solvent. More specifically, it contains

+
+

#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#

+
+

of water. This means that you can expect it to contain fewer than #2.48# moles. Indeed, you can plug in your values to find

+
+

#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#

+
+

Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams

+
+

#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(NaCl) = color(green)(""117 g"")#

+
+
+
+
" "
+

The concentration of a water solution of #NaCl# is 2.48 m, and it contains 806 g of water. How much #NaCl# is in the solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 21, 2016 + +
+
+
+
+
+
+
+

#""117 g""#

+
+
+
+

Explanation:

+
+

This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.

+

As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!

+
+

#color(blue)(""molality"" = ""moles of solute""/""kilograms of solvent"")#

+
+

So, a #""1-molal""# solution will contain #""1 mole""# of solute for every #""1 kg""# of solvent.

+

Now, your solution is said to have a molality of #""2.48 molal""#. This means that you would get #2.48# moles of sodium chloride for every #""1 kg""# of solvent.

+

However, you are told that your solution contains less than #""1 kg""# of water, your solvent. More specifically, it contains

+
+

#806 color(red)(cancel(color(black)(""g""))) * "" 1 kg""/(10^3color(red)(cancel(color(black)(""g"")))) = ""0.806 kg""#

+
+

of water. This means that you can expect it to contain fewer than #2.48# moles. Indeed, you can plug in your values to find

+
+

#0.806 color(red)(cancel(color(black)(""kg water""))) * overbrace(""2.48 moles NaCl""/(1color(red)(cancel(color(black)(""kg water"")))))^(color(purple)(""the given molality"")) = ""1.9989 moles NaCl""#

+
+

Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams

+
+

#1.9989 color(red)(cancel(color(black)(""moles NaCl""))) * ""58.443 g""/(1color(red)(cancel(color(black)(""mole NaCl"")))) = ""116.82 g""#

+
+

Rounded to three sig figs, the answer will be

+
+

#m_(NaCl) = color(green)(""117 g"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 9959 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" The concentration of a water solution of #NaCl# is 2.48 m, and it contains 806 g of water. How much #NaCl# is in the solution? nan +243 a830cf26-6ddd-11ea-96cd-ccda262736ce https://socratic.org/questions/56391ba6581e2a728f785d13 1.42 g start physical_unit 15 15 maximum_amount g qc_end physical_unit 3 3 1 2 mass qc_end physical_unit 8 8 6 7 mass qc_end end "[{""type"":""physical unit"",""value"":""Maximum amount [OF] Fe [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.42 g""}]" "[{""type"":""physical unit"",""value"":""Amount [OF] Fe2O3 [=] \\pu{2.112 g}""},{""type"":""physical unit"",""value"":""Amount [OF] Al [=] \\pu{0.687 g}""}]" "

If #""2.112 g Fe""_2""O""_3""# reacts with #""0.687 g Al""#, what is the maximum amount of #""Fe""# that can be produced?

" nan 1.42 g "
+

Explanation:

+
+

#""Fe""_2""O""_3(""s"")"" + 2Al(s)""##rarr##""Al""_2""O""_3(""s"")"" + 2Fe(s)""#

+

We need to find the limiting reactant (also called limiting reagent).

+

Iron(III) Oxide

+
    +
  1. +

    First divide the given mass of #""Fe""_2""O""_3""# by its molar mass of #""159.687 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Fe""_2""O""_3""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""2.112 g Fe""_2""O""_3""#.

    +
    5. +
+

#2.112cancel""g Fe""_2""O""_3xx(1cancel""mol Fe""_2""O""_3)/(159.687cancel""g Fe""_2""O""_3)xx(2cancel""mol Fe"")/(1cancel""mol Fe""_2""O""_3)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.4772 g Fe""#

+

Aluminum

+
    +
  1. +

    First divide the given mass of #""Al""# by its molar mass of #""26.9815 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Al""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""0.687 g Al""#.

    +
    5. +
+

#0.687cancel""g Al""xx(1cancel""mol Al"")/(26.9815cancel""g Al"")xx(2cancel""mol Fe"")/(2cancel""mol Al"")xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.422 g Fe""#

+

Aluminum is the limiting reactant. The maximum amount of pure #""Fe""# that can be produced by this reaction is #""1.422 g""#.

+
+
" "
+
+
+

Aluminum is the limiting reactant. The maximum amount of pure #""Fe""# that can be produced by this reaction is #""1.422 g""#.

+
+
+
+

Explanation:

+
+

#""Fe""_2""O""_3(""s"")"" + 2Al(s)""##rarr##""Al""_2""O""_3(""s"")"" + 2Fe(s)""#

+

We need to find the limiting reactant (also called limiting reagent).

+

Iron(III) Oxide

+
    +
  1. +

    First divide the given mass of #""Fe""_2""O""_3""# by its molar mass of #""159.687 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Fe""_2""O""_3""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""2.112 g Fe""_2""O""_3""#.

    +
    5. +
+

#2.112cancel""g Fe""_2""O""_3xx(1cancel""mol Fe""_2""O""_3)/(159.687cancel""g Fe""_2""O""_3)xx(2cancel""mol Fe"")/(1cancel""mol Fe""_2""O""_3)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.4772 g Fe""#

+

Aluminum

+
    +
  1. +

    First divide the given mass of #""Al""# by its molar mass of #""26.9815 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Al""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""0.687 g Al""#.

    +
    5. +
+

#0.687cancel""g Al""xx(1cancel""mol Al"")/(26.9815cancel""g Al"")xx(2cancel""mol Fe"")/(2cancel""mol Al"")xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.422 g Fe""#

+

Aluminum is the limiting reactant. The maximum amount of pure #""Fe""# that can be produced by this reaction is #""1.422 g""#.

+
+
+
" "
+

If #""2.112 g Fe""_2""O""_3""# reacts with #""0.687 g Al""#, what is the maximum amount of #""Fe""# that can be produced?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 3, 2015 + +
+
+
+
+
+
+
+

Aluminum is the limiting reactant. The maximum amount of pure #""Fe""# that can be produced by this reaction is #""1.422 g""#.

+
+
+
+

Explanation:

+
+

#""Fe""_2""O""_3(""s"")"" + 2Al(s)""##rarr##""Al""_2""O""_3(""s"")"" + 2Fe(s)""#

+

We need to find the limiting reactant (also called limiting reagent).

+

Iron(III) Oxide

+
    +
  1. +

    First divide the given mass of #""Fe""_2""O""_3""# by its molar mass of #""159.687 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Fe""_2""O""_3""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""2.112 g Fe""_2""O""_3""#.

    +
    5. +
+

#2.112cancel""g Fe""_2""O""_3xx(1cancel""mol Fe""_2""O""_3)/(159.687cancel""g Fe""_2""O""_3)xx(2cancel""mol Fe"")/(1cancel""mol Fe""_2""O""_3)xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.4772 g Fe""#

+

Aluminum

+
    +
  1. +

    First divide the given mass of #""Al""# by its molar mass of #""26.9815 g/mol""#.

    +
    2. +
  2. +

    Next multiply times the mole ratio of #""Fe""# and #""Al""# from the balanced equation.

    +
    3. +
  3. +

    Then multiply times the molar mass of #""Fe""#, which is #""55.845 g/mol""#.

    +
    4. +
  4. +

    This will give you the amount of pure #""Fe""# that can be produced from #""0.687 g Al""#.

    +
    5. +
+

#0.687cancel""g Al""xx(1cancel""mol Al"")/(26.9815cancel""g Al"")xx(2cancel""mol Fe"")/(2cancel""mol Al"")xx(55.845""g Fe"")/(1cancel""mol Fe"")=""1.422 g Fe""#

+

Aluminum is the limiting reactant. The maximum amount of pure #""Fe""# that can be produced by this reaction is #""1.422 g""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2427 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "If #""2.112 g Fe""_2""O""_3""# reacts with #""0.687 g Al""#, what is the maximum amount of #""Fe""# that can be produced?" nan +244 a830f628-6ddd-11ea-bfd2-ccda262736ce https://socratic.org/questions/a-gold-ring-weighing-20-g-at-25-c-is-dropped-into-a-beaker-containing-15-g-of-bo 0.13 J/(g * ℃) start physical_unit 1 1 specific_heat_capacity j/(°c_·_g) qc_end physical_unit 0 2 4 5 mass qc_end physical_unit 0 2 7 8 temperature qc_end physical_unit 18 19 15 16 mass qc_end physical_unit 19 19 27 28 temperature qc_end physical_unit 19 19 37 40 heat_capacity qc_end end "[{""type"":""physical unit"",""value"":""Specific heat capacity [OF] gold [IN] J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""0.13 J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""Weight [OF] a gold ring [=] \\pu{20 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] a gold ring [=] \\pu{25 ℃}""},{""type"":""physical unit"",""value"":""Weight [OF] boiling water [=] \\pu{15 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] water [=] \\pu{97 ℃}""},{""type"":""physical unit"",""value"":""Heat capacity [OF] water [=] \\pu{4.184 J/(g * ℃)}""}]" "

A gold ring weighing 20 g at 25 °C is dropped into a beaker containing 15 g of boiling water. The final temperature of the water is 97 °C. +Recall that the heat capacity of water is 4.184 J/g °C. + +What is the specific heat capacity of gold?

" nan 0.13 J/(g * ℃) "
+

Explanation:

+
+

The first thing to mention here is that water's specific heat at around #100^@""C""# is not actually equal to #4.181 ""J""/(""g"" """"^@""C"")#, but with approximately #4.218""J""/(""g"" """"^@""C"")#.

+

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

+

However, since this is the value given to you in the problem, you must use it.

+

So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.

+

Mathematically, this is written as

+
+

#color(blue)(-q_""water"" = q_""metal"")#

+
+

Here the negative sign is used because #q_""water""# will be negative, since it represents heat lost by the water.

+

The equation that establishes a relationship between heat lost/gained and temperature change looks like this

+
+

#color(blue)(q = - m * c * DeltaT)"" ""#, where

+
+

#q# - heat lost/gained;
+#m# - the mass of the sample;
+#c# - the specific heat of the material
+#DeltaT# - the change in temperature, defined as the final temperatue minus the initial temperature

+

In your case, you would have

+
+

#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#

+
+

Rearrange this equation to solve for #c_""metal""#

+
+

#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#

+
+

Plug in your values to get

+
+

#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#

+

#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#

+
+

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring

+
+

#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#

+
+

+ +

+
+
" "
+
+
+

#0.13""J""/(""g"" """"^@""C"")#

+
+
+
+

Explanation:

+
+

The first thing to mention here is that water's specific heat at around #100^@""C""# is not actually equal to #4.181 ""J""/(""g"" """"^@""C"")#, but with approximately #4.218""J""/(""g"" """"^@""C"")#.

+

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

+

However, since this is the value given to you in the problem, you must use it.

+

So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.

+

Mathematically, this is written as

+
+

#color(blue)(-q_""water"" = q_""metal"")#

+
+

Here the negative sign is used because #q_""water""# will be negative, since it represents heat lost by the water.

+

The equation that establishes a relationship between heat lost/gained and temperature change looks like this

+
+

#color(blue)(q = - m * c * DeltaT)"" ""#, where

+
+

#q# - heat lost/gained;
+#m# - the mass of the sample;
+#c# - the specific heat of the material
+#DeltaT# - the change in temperature, defined as the final temperatue minus the initial temperature

+

In your case, you would have

+
+

#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#

+
+

Rearrange this equation to solve for #c_""metal""#

+
+

#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#

+
+

Plug in your values to get

+
+

#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#

+

#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#

+
+

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring

+
+

#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#

+
+

+ +

+
+
+
" "
+

A gold ring weighing 20 g at 25 °C is dropped into a beaker containing 15 g of boiling water. The final temperature of the water is 97 °C. +Recall that the heat capacity of water is 4.184 J/g °C. + +What is the specific heat capacity of gold?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Oct 29, 2015 + +
+
+
+
+
+
+
+

#0.13""J""/(""g"" """"^@""C"")#

+
+
+
+

Explanation:

+
+

The first thing to mention here is that water's specific heat at around #100^@""C""# is not actually equal to #4.181 ""J""/(""g"" """"^@""C"")#, but with approximately #4.218""J""/(""g"" """"^@""C"")#.

+

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

+

However, since this is the value given to you in the problem, you must use it.

+

So, the idea here is that the heat gained by the metal will be equal to the heat lost by the water. The final temperature of the water will thus be equal to the final temperature of the ring.

+

Mathematically, this is written as

+
+

#color(blue)(-q_""water"" = q_""metal"")#

+
+

Here the negative sign is used because #q_""water""# will be negative, since it represents heat lost by the water.

+

The equation that establishes a relationship between heat lost/gained and temperature change looks like this

+
+

#color(blue)(q = - m * c * DeltaT)"" ""#, where

+
+

#q# - heat lost/gained;
+#m# - the mass of the sample;
+#c# - the specific heat of the material
+#DeltaT# - the change in temperature, defined as the final temperatue minus the initial temperature

+

In your case, you would have

+
+

#- overbrace(m_""water"" * c_""water"" * DeltaT_""water"")^(color(blue)(=q_""water"")) = overbrace(m_""metal"" * c_""metal"" * DeltaT_""metal"")^(color(red)(=q_""metal""))#

+
+

Rearrange this equation to solve for #c_""metal""#

+
+

#c_""metal"" = -m_""water""/m_""metal"" * (DeltaT_""water"")/(DeltaT_""metal"") * c_""water""#

+
+

Plug in your values to get

+
+

#c_""metal"" = (15color(red)(cancel(color(black)(""g""))))/(20color(red)(cancel(color(black)(""g"")))) * ((97 - 100)color(red)(cancel(color(black)(""""^@""C""))))/((97 - 25)color(red)(cancel(color(black)(""""^@""C"")))) * 4.184""J""/(""g"" """"^@""C"")#

+

#c_""metal"" = 0.13075""J""/(""g"" """"^@""C"")#

+
+

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring

+
+

#c_""metal"" = color(green)(0.13""J""/(""g"" """"^@""C""))#

+
+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3312 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "A gold ring weighing 20 g at 25 °C is dropped into a beaker containing 15 g of boiling water. The final temperature of the water is 97 °C. +Recall that the heat capacity of water is 4.184 J/g °C. + +What is the specific heat capacity of gold?" nan +245 a830f629-6ddd-11ea-94c5-ccda262736ce https://socratic.org/questions/20-0-g-of-calcium-carbonate-is-heated-to-produce-carbon-dioxide-according-to-the 17.62 L start physical_unit 44 46 volume l qc_end physical_unit 3 4 0 1 mass qc_end chemical_equation 15 19 qc_end physical_unit 20 21 29 31 temperature qc_end physical_unit 20 21 37 38 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] carbon dioxide gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""17.62 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] calcium carbonate [=] \\pu{20.0 g}""},{""type"":""chemical equation"",""value"":""CaCO3(s) -> CaO(s) + CO2(g)""},{""type"":""physical unit"",""value"":""Temperature [OF] the reaction [=] \\pu{800.0 degrees C}""},{""type"":""physical unit"",""value"":""Constant pressure [OF] the reaction [=] \\pu{1.0 atm}""}]" "

20.0 g of calcium carbonate is heated to produce carbon dioxide according to +the equation +CaCO3(s) → CaO(s) + CO2(g)?

" "
+
+

+

20.0 g of calcium carbonate is heated to produce carbon dioxide according to
+the equation
+CaCO3(s) → CaO(s) + CO2(g)
+If the reaction is carried out at a temperature of 800.0 degrees C and a constant pressure of
+1.0 atm is maintained, what volume of carbon dioxide gas is recovered?

+

+
+
" 17.62 L "
+

Explanation:

+
+

There are #(20.0*g)/(100.0*g*mol^-1)# #=# #0.20*mol*CaCO_3#

+

Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.

+

At #1# #atm# pressure, and #1073*K#, this gas will have a volume of:

+

#V=(nRT)/P# #=# #(0.20*molxx0.0821*L*atm*K^-1*mol^-1xx1073*K)/(1*atm)#

+

#=# #??L#

+
+
" "
+
+
+

#CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr#

+
+
+
+

Explanation:

+
+

There are #(20.0*g)/(100.0*g*mol^-1)# #=# #0.20*mol*CaCO_3#

+

Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.

+

At #1# #atm# pressure, and #1073*K#, this gas will have a volume of:

+

#V=(nRT)/P# #=# #(0.20*molxx0.0821*L*atm*K^-1*mol^-1xx1073*K)/(1*atm)#

+

#=# #??L#

+
+
+
" "
+

20.0 g of calcium carbonate is heated to produce carbon dioxide according to +the equation +CaCO3(s) → CaO(s) + CO2(g)?

+
+
+

+

20.0 g of calcium carbonate is heated to produce carbon dioxide according to
+the equation
+CaCO3(s) → CaO(s) + CO2(g)
+If the reaction is carried out at a temperature of 800.0 degrees C and a constant pressure of
+1.0 atm is maintained, what volume of carbon dioxide gas is recovered?

+

+
+
+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 22, 2016 + +
+
+
+
+
+
+
+

#CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr#

+
+
+
+

Explanation:

+
+

There are #(20.0*g)/(100.0*g*mol^-1)# #=# #0.20*mol*CaCO_3#

+

Given the stoichiometry, an equivalent molar quantity of carbon dioxide gas will be evolved.

+

At #1# #atm# pressure, and #1073*K#, this gas will have a volume of:

+

#V=(nRT)/P# #=# #(0.20*molxx0.0821*L*atm*K^-1*mol^-1xx1073*K)/(1*atm)#

+

#=# #??L#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 7595 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" "20.0 g of calcium carbonate is heated to produce carbon dioxide according to +the equation +CaCO3(s) → CaO(s) + CO2(g)?" " + + +20.0 g of calcium carbonate is heated to produce carbon dioxide according to +the equation +CaCO3(s) → CaO(s) + CO2(g) +If the reaction is carried out at a temperature of 800.0 degrees C and a constant pressure of +1.0 atm is maintained, what volume of carbon dioxide gas is recovered? + + +" +246 a830f62a-6ddd-11ea-9200-ccda262736ce https://socratic.org/questions/oxalic-acid-h-2c-2o-4-occurs-as-the-potassium-or-calcium-salt-in-many-plants-inc 0.57 mol/L start physical_unit 2 2 molarity mol/l qc_end c_other OTHER qc_end physical_unit 2 2 24 25 mole qc_end physical_unit 30 31 33 34 density qc_end end "[{""type"":""physical unit"",""value"":""Molar concentration [OF] H2C2O4 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.57 mol/L""}]" "[{""type"":""other"",""value"":""H2C2O4, occurs as the potassium or calcium salt in many plants""},{""type"":""physical unit"",""value"":""Mole [OF] H2C2O4 [=] \\pu{0.585 molal}""},{""type"":""physical unit"",""value"":""Density [OF] the solution [=] \\pu{1.022 g/mL}""}]" "

Oxalic acid, #H_2C_2O_4#, occurs as the potassium or calcium salt in many plants, including rhubarb and spinach. An aqueous solution of oxalic acid is 0.585 molal #H_2C_2O_4#. The density of the solution is 1.022 g/mL. What is the molar concentration?

" nan 0.57 mol/L "
+

Explanation:

+
+

The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.

+

To make the calculations easier, let's pick a sample that contains exactly #""1 kg""# of water, the solvent. By definition, this sample will contain #0.585# moles of oxalic acid because its molality is equal to #""0.585 molal""#, or #""0.585 mol kg""^(-1)#.

+

Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.

+
+

#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#

+
+

This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to #""1 kg"" = 10^3 quad ""g""#, will be

+
+

#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#

+
+

Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly #""1 L"" = 10^3 quad ""mL""# of this solution.

+

Use the density of the solution to find the total volume of the sample

+
+

#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#

+
+

Since you know that this sample contains #0.585# moles of oxalic acid, you can say that #10^3 quad ""mL""# of this solution will contain

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#

+
+

This means that the solution has a molarity of

+
+

#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#

+
+

The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.

+
+
" "
+
+
+

#""0.568 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.

+

To make the calculations easier, let's pick a sample that contains exactly #""1 kg""# of water, the solvent. By definition, this sample will contain #0.585# moles of oxalic acid because its molality is equal to #""0.585 molal""#, or #""0.585 mol kg""^(-1)#.

+

Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.

+
+

#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#

+
+

This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to #""1 kg"" = 10^3 quad ""g""#, will be

+
+

#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#

+
+

Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly #""1 L"" = 10^3 quad ""mL""# of this solution.

+

Use the density of the solution to find the total volume of the sample

+
+

#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#

+
+

Since you know that this sample contains #0.585# moles of oxalic acid, you can say that #10^3 quad ""mL""# of this solution will contain

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#

+
+

This means that the solution has a molarity of

+
+

#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#

+
+

The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.

+
+
+
" "
+

Oxalic acid, #H_2C_2O_4#, occurs as the potassium or calcium salt in many plants, including rhubarb and spinach. An aqueous solution of oxalic acid is 0.585 molal #H_2C_2O_4#. The density of the solution is 1.022 g/mL. What is the molar concentration?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 15, 2018 + +
+
+
+
+
+
+
+

#""0.568 mol L""^(-1)#

+
+
+
+

Explanation:

+
+

The first thing that you need to do here is to pick a sample of this solution and use its molality to determine how many moles of solute it contains.

+

To make the calculations easier, let's pick a sample that contains exactly #""1 kg""# of water, the solvent. By definition, this sample will contain #0.585# moles of oxalic acid because its molality is equal to #""0.585 molal""#, or #""0.585 mol kg""^(-1)#.

+

Next, use the molar mass of oxalic acid to calculate how many grams of solute are present in this sample.

+
+

#0.585 color(red)(cancel(color(black)(""moles H""_2""C""_2""O""_4))) * ""90.03 g""/(1color(red)(cancel(color(black)(""mole H""_2""C""_2""O""_4)))) = ""52.68 g""#

+
+

This means that the total mass of the sample, which includes the mass of the solute and the mass of the solvent, which is equal to #""1 kg"" = 10^3 quad ""g""#, will be

+
+

#10^3 quad ""g"" + ""52.68 g"" = ""1052.68 g""#

+
+

Now, in order to find the molarity of the solution, you need to find the number of moles of solute present in exactly #""1 L"" = 10^3 quad ""mL""# of this solution.

+

Use the density of the solution to find the total volume of the sample

+
+

#1052.68 color(red)(cancel(color(black)(""g""))) * ""1 mL""/(1.022color(red)(cancel(color(black)(""g"")))) = ""1030.0 mL""#

+
+

Since you know that this sample contains #0.585# moles of oxalic acid, you can say that #10^3 quad ""mL""# of this solution will contain

+
+

#10^3 color(red)(cancel(color(black)(""mL solution""))) * (""0.585 moles H""_2""C""_2""O""_4)/(1030.0color(red)(cancel(color(black)(""mL solution"")))) = ""0.568 moles H""_2""C""_2""O""_4#

+
+

This means that the solution has a molarity of

+
+

#color(darkgreen)(ul(color(black)(""molarity"" = ""0.568 mol L""^(-1))))#

+
+

The answer is rounded to three sig figs, the number of sig figs you have for the molality of the solution.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6310 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" Oxalic acid, #H_2C_2O_4#, occurs as the potassium or calcium salt in many plants, including rhubarb and spinach. An aqueous solution of oxalic acid is 0.585 molal #H_2C_2O_4#. The density of the solution is 1.022 g/mL. What is the molar concentration? nan +247 a830f62b-6ddd-11ea-b998-ccda262736ce https://socratic.org/questions/a-sample-of-nitrogen-gas-has-a-volume-of-15ml-at-a-pressure-of-0-50-atm-what-is- 0.25 atm start physical_unit 3 4 pressure atm qc_end physical_unit 3 4 15 16 pressure qc_end physical_unit 3 4 9 10 volume qc_end physical_unit 3 4 27 28 volume qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] nitrogen gas [IN] atm""}]" "[{""type"":""physical unit"",""value"":""0.25 atm""}]" "[{""type"":""physical unit"",""value"":""Pressure1 [OF] nitrogen gas [=] \\pu{0.50 atm}""},{""type"":""physical unit"",""value"":""Volume1 [OF] nitrogen gas [=] \\pu{15 mL}""},{""type"":""physical unit"",""value"":""Volume2 [OF] nitrogen gas [=] \\pu{30.0 mL}""}]" "

A sample of nitrogen gas has a volume of 15mL at a pressure of 0.50 atm. What is the pressure if the volume is changed to 30.0mL?

" nan 0.25 atm "
+

Explanation:

+
+

The answer can be determined by using Boyle's Law:
+

+

Let's identify our known and unknown variables.

+

#color(brown)(""Knowns:""#
+#P_1#= 0.50 atm
+#V_1#= 15 mL
+#V_2#= 30.0 mL

+

#color(blue)(""Unknowns:""#
+#P_2#

+

Rearrange the equation to solve for the final pressure by dividing both sides by #V_2# to get #P_2# by itself like this:

+

#P_2=(P_1xxV_1)/V_2#

+

Plug in your given values to obtain the final pressure:

+

#P_2=(0.50\atm xx 15\ cancel""mL"")/(30.0\cancel""mL"")# = #0.25 atm#

+
+
" "
+
+
+

That sample of gas has a new pressure of #0.25 atm#

+
+
+
+

Explanation:

+
+

The answer can be determined by using Boyle's Law:
+

+

Let's identify our known and unknown variables.

+

#color(brown)(""Knowns:""#
+#P_1#= 0.50 atm
+#V_1#= 15 mL
+#V_2#= 30.0 mL

+

#color(blue)(""Unknowns:""#
+#P_2#

+

Rearrange the equation to solve for the final pressure by dividing both sides by #V_2# to get #P_2# by itself like this:

+

#P_2=(P_1xxV_1)/V_2#

+

Plug in your given values to obtain the final pressure:

+

#P_2=(0.50\atm xx 15\ cancel""mL"")/(30.0\cancel""mL"")# = #0.25 atm#

+
+
+
" "
+

A sample of nitrogen gas has a volume of 15mL at a pressure of 0.50 atm. What is the pressure if the volume is changed to 30.0mL?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 12, 2016 + +
+
+
+
+
+
+
+

That sample of gas has a new pressure of #0.25 atm#

+
+
+
+

Explanation:

+
+

The answer can be determined by using Boyle's Law:
+

+

Let's identify our known and unknown variables.

+

#color(brown)(""Knowns:""#
+#P_1#= 0.50 atm
+#V_1#= 15 mL
+#V_2#= 30.0 mL

+

#color(blue)(""Unknowns:""#
+#P_2#

+

Rearrange the equation to solve for the final pressure by dividing both sides by #V_2# to get #P_2# by itself like this:

+

#P_2=(P_1xxV_1)/V_2#

+

Plug in your given values to obtain the final pressure:

+

#P_2=(0.50\atm xx 15\ cancel""mL"")/(30.0\cancel""mL"")# = #0.25 atm#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 11658 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" A sample of nitrogen gas has a volume of 15mL at a pressure of 0.50 atm. What is the pressure if the volume is changed to 30.0mL? nan +248 a8311da4-6ddd-11ea-b37d-ccda262736ce https://socratic.org/questions/58d36eb611ef6b503b7158fa 1.52 start physical_unit 28 30 ph none qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 7 8 concentration qc_end physical_unit 18 18 14 15 volume qc_end physical_unit 18 18 20 21 concentration qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] the final solution""}]" "[{""type"":""physical unit"",""value"":""1.52""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] HBr(aq) [=] \\pu{25 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] HBr(aq) [=] \\pu{0.050 mol/L}""},{""type"":""physical unit"",""value"":""Volume [OF] KOH(aq) [=] \\pu{10 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] KOH(aq) [=] \\pu{0.020 mol/L}""}]" "

What is the #pH# of the final solution...?

" "
+
+

+

A #25*mL# volume of #HBr(aq)# at #0.050*mol*L^-1# concentration is mixed with a #10*mL# volume of #KOH(aq)# at #0.020*mol*L^-1# concentration. What is the #pH# of the final solution?

+

+
+
" 1.52 "
+

Explanation:

+
+

And thus we need to find the amount of substance of both #""hydrogen bromide""#, #""hydrobromic acid""#, and #""potassium hydroxide""#.

+

We use the relationship, #""Concentration""=""Moles of solute""/""Volume of solution""#, OR

+

#""Concentration""xx""Volume""=""Moles of solute""#

+

#""Moles of HBr""=25xx10^-3cancelLxx0.050*mol*cancel(L^-1)=1.25xx10^-3*mol.#

+

#""Moles of KOH""=10xx10^-3cancelLxx0.020*mol*cancel(L^-1)=0.200xx10^-3*mol.#

+

Note that I converted the #mL# volume to #L# by using the relationship: #1*mL-=1xx10^-3*L#

+

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are #((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)# #HBr# remaining.

+

So #[HBr]=((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)#

+

#=(1.05*molxx10^-3)/(35xx10^-3L)=0.030*mol*L^-1# with respect to #HBr#.

+

#pH=-log_10[H_3O^+]=-log_10(0.030)=1.52#

+
+
" "
+
+
+

A stoichiometric equation is required:

+

#HBr(aq) + KOH(aq) rarr NaBr(aq) + H_2O(l)#

+

We get (finally) #pH=1.52#

+
+
+
+

Explanation:

+
+

And thus we need to find the amount of substance of both #""hydrogen bromide""#, #""hydrobromic acid""#, and #""potassium hydroxide""#.

+

We use the relationship, #""Concentration""=""Moles of solute""/""Volume of solution""#, OR

+

#""Concentration""xx""Volume""=""Moles of solute""#

+

#""Moles of HBr""=25xx10^-3cancelLxx0.050*mol*cancel(L^-1)=1.25xx10^-3*mol.#

+

#""Moles of KOH""=10xx10^-3cancelLxx0.020*mol*cancel(L^-1)=0.200xx10^-3*mol.#

+

Note that I converted the #mL# volume to #L# by using the relationship: #1*mL-=1xx10^-3*L#

+

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are #((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)# #HBr# remaining.

+

So #[HBr]=((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)#

+

#=(1.05*molxx10^-3)/(35xx10^-3L)=0.030*mol*L^-1# with respect to #HBr#.

+

#pH=-log_10[H_3O^+]=-log_10(0.030)=1.52#

+
+
+
" "
+

What is the #pH# of the final solution...?

+
+
+

+

A #25*mL# volume of #HBr(aq)# at #0.050*mol*L^-1# concentration is mixed with a #10*mL# volume of #KOH(aq)# at #0.020*mol*L^-1# concentration. What is the #pH# of the final solution?

+

+
+
+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 23, 2017 + +
+
+
+
+
+
+
+

A stoichiometric equation is required:

+

#HBr(aq) + KOH(aq) rarr NaBr(aq) + H_2O(l)#

+

We get (finally) #pH=1.52#

+
+
+
+

Explanation:

+
+

And thus we need to find the amount of substance of both #""hydrogen bromide""#, #""hydrobromic acid""#, and #""potassium hydroxide""#.

+

We use the relationship, #""Concentration""=""Moles of solute""/""Volume of solution""#, OR

+

#""Concentration""xx""Volume""=""Moles of solute""#

+

#""Moles of HBr""=25xx10^-3cancelLxx0.050*mol*cancel(L^-1)=1.25xx10^-3*mol.#

+

#""Moles of KOH""=10xx10^-3cancelLxx0.020*mol*cancel(L^-1)=0.200xx10^-3*mol.#

+

Note that I converted the #mL# volume to #L# by using the relationship: #1*mL-=1xx10^-3*L#

+

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are #((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)# #HBr# remaining.

+

So #[HBr]=((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)#

+

#=(1.05*molxx10^-3)/(35xx10^-3L)=0.030*mol*L^-1# with respect to #HBr#.

+

#pH=-log_10[H_3O^+]=-log_10(0.030)=1.52#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5193 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the #pH# of the final solution...? " + + +A #25*mL# volume of #HBr(aq)# at #0.050*mol*L^-1# concentration is mixed with a #10*mL# volume of #KOH(aq)# at #0.020*mol*L^-1# concentration. What is the #pH# of the final solution? + + +" +249 a8313b6c-6ddd-11ea-a9d0-ccda262736ce https://socratic.org/questions/the-molecular-mass-of-octane-is-14-22-g-mol-what-is-the-mass-of-22-05-mol-of-oct 2518.55 g start physical_unit 4 4 mass g qc_end physical_unit 4 4 6 7 molecular_weight qc_end physical_unit 4 4 13 14 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] octane [IN] g""}]" "[{""type"":""physical unit"",""value"":""2518.55 g""}]" "[{""type"":""physical unit"",""value"":""Molecular mass [OF] octane [=] \\pu{114.22 g/mol}""},{""type"":""physical unit"",""value"":""Mole [OF] octane [=] \\pu{22.05 mol}""}]" "

The molecular mass of octane is 14.22 g/mol. What is the mass of 22.05 mol of octane?

" nan 2518.55 g "
+

Explanation:

+
+

So if you have #22.05*mol# of this juice, there are #114.23*g*cancel(mol^-1)xx22.05*cancel(mol)=2519*g#, probably enuff to propel my motor for 30 km.

+
+
" "
+
+
+

The molecular mass of octane is #114.23* g*mol^-1#. So........

+
+
+
+

Explanation:

+
+

So if you have #22.05*mol# of this juice, there are #114.23*g*cancel(mol^-1)xx22.05*cancel(mol)=2519*g#, probably enuff to propel my motor for 30 km.

+
+
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The molecular mass of octane is 14.22 g/mol. What is the mass of 22.05 mol of octane?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+1 Answer +
+
+
+
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+ +
+
+ +
+ + Aug 15, 2016 + +
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+

The molecular mass of octane is #114.23* g*mol^-1#. So........

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+
+
+

Explanation:

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+

So if you have #22.05*mol# of this juice, there are #114.23*g*cancel(mol^-1)xx22.05*cancel(mol)=2519*g#, probably enuff to propel my motor for 30 km.

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" The molecular mass of octane is 14.22 g/mol. What is the mass of 22.05 mol of octane? nan +250 a8314446-6ddd-11ea-9e10-ccda262736ce https://socratic.org/questions/a-sample-of-gas-is-collected-over-water-at-a-temperature-of-35-0-c-when-the-baro 699.80 mmHg start physical_unit 28 30 partial_pressure mmhg qc_end physical_unit 1 3 12 13 temperature qc_end physical_unit 1 3 20 21 barometric_pressure qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Partial pressure [OF] the dry gas [IN] mmHg""}]" "[{""type"":""physical unit"",""value"":""699.80 mmHg""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] sample of gas [=] \\pu{35.0 ℃}""},{""type"":""physical unit"",""value"":""Barometric pressure [OF] sample of gas [=] \\pu{742.0 torr}""},{""type"":""other"",""value"":""A sample of gas is collected over water.""}]" "

A sample of gas is collected over water at a temperature of 35.0°C when the barometric pressure reading is 742.0 torr. What is the partial pressure of the dry gas?

" nan 699.80 mmHg "
+

Explanation:

+
+

#P_(""collected"")# #=# #P_""gas"" + P_""SVP""#, #P_""SVP""# is given above, and is called the saturated vapour pressure. This should have been quoted with the question.

+

Thus #P_""gas""~=700*mm*Hg#.

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+
" "
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+
+

This site tells me that vapour pressure of water at #35.0# #""""^@C# #=# #42.2*mm*Hg#.

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+
+
+

Explanation:

+
+

#P_(""collected"")# #=# #P_""gas"" + P_""SVP""#, #P_""SVP""# is given above, and is called the saturated vapour pressure. This should have been quoted with the question.

+

Thus #P_""gas""~=700*mm*Hg#.

+
+
+
" "
+

A sample of gas is collected over water at a temperature of 35.0°C when the barometric pressure reading is 742.0 torr. What is the partial pressure of the dry gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
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+
+
+1 Answer +
+
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+ +
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+ +
+ + May 9, 2016 + +
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+

This site tells me that vapour pressure of water at #35.0# #""""^@C# #=# #42.2*mm*Hg#.

+
+
+
+

Explanation:

+
+

#P_(""collected"")# #=# #P_""gas"" + P_""SVP""#, #P_""SVP""# is given above, and is called the saturated vapour pressure. This should have been quoted with the question.

+

Thus #P_""gas""~=700*mm*Hg#.

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+
+
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+ + Creative Commons License + +
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" A sample of gas is collected over water at a temperature of 35.0°C when the barometric pressure reading is 742.0 torr. What is the partial pressure of the dry gas? nan +251 a8314447-6ddd-11ea-b87e-ccda262736ce https://socratic.org/questions/3-45-g-of-an-unknown-gas-at-55-c-and-1-10-atm-is-stored-in-a-2-55-l-flask-what-i 1.35 g/L start physical_unit 24 25 density g/l qc_end physical_unit 3 5 0 1 mass qc_end physical_unit 3 5 7 8 temperature qc_end physical_unit 3 5 10 11 pressure qc_end physical_unit 18 18 16 17 volume qc_end end "[{""type"":""physical unit"",""value"":""Density [OF] the gas [IN] g/L""}]" "[{""type"":""physical unit"",""value"":""1.35 g/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] an unknown gas [=] \\pu{3.45 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] an unknown gas [=] \\pu{55 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] an unknown gas [=] \\pu{1.10 atm}""},{""type"":""physical unit"",""value"":""Volume [OF] flask [=] \\pu{2.55 L}""}]" "

3.45 g of an unknown gas at 55°C and 1.10 atm is stored in a 2.55-L flask. What is the density of the gas?

" nan 1.35 g/L "
+

Explanation:

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+

1) Use PV = nRT to determine moles of gas present in gas:

+

(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)

+

n = 0.104 mol

+

2) Get molar mass of gas using calculated moles:

+

0.104 mol x M g/mol = 3.45g

+

M = 3.45 g / 0.104 mol = 33.17 g/mol

+

Density = Molar Mass x P / RT

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= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K

+

= 1.35 g /L

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" "
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+

1.35 g/L

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+
+

Explanation:

+
+

1) Use PV = nRT to determine moles of gas present in gas:

+

(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)

+

n = 0.104 mol

+

2) Get molar mass of gas using calculated moles:

+

0.104 mol x M g/mol = 3.45g

+

M = 3.45 g / 0.104 mol = 33.17 g/mol

+

Density = Molar Mass x P / RT

+

= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K

+

= 1.35 g /L

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+
+
" "
+

3.45 g of an unknown gas at 55°C and 1.10 atm is stored in a 2.55-L flask. What is the density of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+
+1 Answer +
+
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+
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+ +
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+ +
+ + Apr 11, 2016 + +
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+
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+

1.35 g/L

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+
+
+

Explanation:

+
+

1) Use PV = nRT to determine moles of gas present in gas:

+

(1.10 atm) (2.55 L) = (n) (0.08206) (328 K)

+

n = 0.104 mol

+

2) Get molar mass of gas using calculated moles:

+

0.104 mol x M g/mol = 3.45g

+

M = 3.45 g / 0.104 mol = 33.17 g/mol

+

Density = Molar Mass x P / RT

+

= 33.17 g/mol x 1.10 atm / 0.0821 L mol / atm .K x 328 K

+

= 1.35 g /L

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+ +
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+
+
Related questions
+ + +
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Impact of this question
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+ + Creative Commons License + +
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" 3.45 g of an unknown gas at 55°C and 1.10 atm is stored in a 2.55-L flask. What is the density of the gas? nan +252 a8316b64-6ddd-11ea-b246-ccda262736ce https://socratic.org/questions/what-is-the-total-pressure-of-the-gases-in-the-flask-at-this-point 724.57 Pa start physical_unit 6 10 total_pressure pa qc_end physical_unit 9 10 33 34 temperature qc_end chemical_equation 39 43 qc_end physical_unit 24 24 90 91 deltag^0 qc_end physical_unit 41 41 94 95 deltag^0 qc_end physical_unit 9 10 17 18 volume qc_end physical_unit 24 24 21 22 mole qc_end end "[{""type"":""physical unit"",""value"":""Total pressure [OF] the gases in the flask [IN] Pa""}]" "[{""type"":""physical unit"",""value"":""724.57 Pa""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] the flask [=] \\pu{227 ℃}""},{""type"":""chemical equation"",""value"":""PCl5(g) <=> PCl3(g) + Cl2(g)""},{""type"":""physical unit"",""value"":""DeltaG^0 [OF] PCl5 [=] \\pu{-305.0 J/mol}""},{""type"":""physical unit"",""value"":""DeltaG^0 [OF] PCl3 [=] \\pu{-267.8 J/mol}""},{""type"":""physical unit"",""value"":""Volume [OF] the flask [=] \\pu{1.30 L}""},{""type"":""physical unit"",""value"":""Mole [OF] PCl5(g) [=] \\pu{0.120 mol}""}]" "

What is the total pressure of the gases in the flask at this point?

" "
+
+

+

Introduced into a 1.30 −L flask is 0.120 mol of #PCl_5(g)#; the flask is held at a temperature of 227 °C until equilibrium is established.
+#PCl_5(g)⇌PCl_3(g)+Cl_2(g)#

+

What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.]

+

From what I know, I need to apply #\DeltaG°#, but I don't know whether I need that for #PCl_5# or #PCl_3#.
+Here's the information I got from the book...
+for #PCl_5#: -305.0
+for #PCl_3#: -267.8

+

+
+
" 724.57 Pa "
+

Explanation:

+
+
+

Yes, you need #ΔG^@#, but at 500 K, not 298 K.

+
+

Calculate #ΔG^@# at 500 K

+

You can calculate it from the values tabulated at 298 K.

+

#color(white)(mmmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#Δ_text(f)H^@""/kJ·mol""^""-1"": color(white)(l)""-398.9""color(white)(m)""-306.4""color(white)(mm)0#
+#S^@""/J·K""^""-1""""mol""^""-1"":color(white)(ml)353color(white)(mm)311.7color(white)(mll)223#

+

#Δ_text(r)H^@ = sumΔ_text(f)H^@(""products"") - sumΔ_text(f)H^@(""reactants"") = ""(-306.4 + 398.9) kJ/mol"" = ""92.5 kJ/mol""#

+

#Δ_text(r)S^@ = sumS^@(""products"") - sumS^@(""reactants"") = (311.7 + 223 - 353) color(white)(l)""kJ/mol"" = ""181.7 kJ/mol""#

+
+

#ΔG^@ = ΔH^@ -TΔS^@#

+

∴ At 227 °C (500 K),

+

#ΔG^@ = ""92 500 J·mol""^""-1"" -500 ""K"" × ""181.7 J""·""K""^""-1""""mol""^""-1"" = ""(92 500 - 90 850) J·mol""^""-1"" = ""1650 J·mol""^""-1""#

+
+

Calculate #K#

+

#ΔG^@ = ""-""RTlnK#

+

#lnK =(-ΔG^@)/(RT)= (""-1650 J·mol""^""-1"")/(8.314 ""J·K""^""-1""""mol""^""-1"" ×500 ""K"") = ""-0.397""#

+

#K = e^""-0.397"" = 0.672#

+
+

Calculate equilibrium concentrations

+

Finally, we can set up an ICE table to calculate the equilibrium concentrations.

+

#color(white)(mmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#""I/mol·L""^""-1"": color(white)(mm)0.0923color(white)(mmm)0color(white)(mmll)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(m)0.0923 -xcolor(white)(mll)xcolor(white)(mmll)x#

+

#[""PCl""_5]_0 = ""0.120 mol""/""1.30 L"" = ""0.0923 mol/L""#

+

#K_text(c) = ([""PCl""_3][""Cl""_2])/([""PCl""_5]) = x^2/(0.0923-x)= 0.672#

+
+

Test for negligibility:

+

#0.0923/0.672 = 0.14 < 400#. ∴ #x# is not negligible. We must solve a quadratic equation.

+
+

#x^2 = 0.672(0.0923-x) = 0.0620 - 0.672x#

+

#x^2 + 0.672x - 0.0620 = 0#

+

#x = 0.0820#

+
+

Calculate total pressure

+

#""Total concentration"" = (0.0923 - x + x + x"") mol·L""^""-1"" = ""(0.0923 + 0.0820) mol/L"" = ""0.1743 mol/L""#

+

#p = (nRT)/V = cRT = (0.1743 color(red)(cancel(color(black)(""mol·L""^""-1""))) × ""0.083 14 bar""·color(red)(cancel(color(black)(""L·K""^""-1""""mol""^""-1""))) × 500 color(red)(cancel(color(black)(""K"")))) = ""7.25 bar""#

+
+
" "
+
+
+

Warning! Long Answer. #p_text(tot) = ""7.25 bar""#

+
+
+
+

Explanation:

+
+
+

Yes, you need #ΔG^@#, but at 500 K, not 298 K.

+
+

Calculate #ΔG^@# at 500 K

+

You can calculate it from the values tabulated at 298 K.

+

#color(white)(mmmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#Δ_text(f)H^@""/kJ·mol""^""-1"": color(white)(l)""-398.9""color(white)(m)""-306.4""color(white)(mm)0#
+#S^@""/J·K""^""-1""""mol""^""-1"":color(white)(ml)353color(white)(mm)311.7color(white)(mll)223#

+

#Δ_text(r)H^@ = sumΔ_text(f)H^@(""products"") - sumΔ_text(f)H^@(""reactants"") = ""(-306.4 + 398.9) kJ/mol"" = ""92.5 kJ/mol""#

+

#Δ_text(r)S^@ = sumS^@(""products"") - sumS^@(""reactants"") = (311.7 + 223 - 353) color(white)(l)""kJ/mol"" = ""181.7 kJ/mol""#

+
+

#ΔG^@ = ΔH^@ -TΔS^@#

+

∴ At 227 °C (500 K),

+

#ΔG^@ = ""92 500 J·mol""^""-1"" -500 ""K"" × ""181.7 J""·""K""^""-1""""mol""^""-1"" = ""(92 500 - 90 850) J·mol""^""-1"" = ""1650 J·mol""^""-1""#

+
+

Calculate #K#

+

#ΔG^@ = ""-""RTlnK#

+

#lnK =(-ΔG^@)/(RT)= (""-1650 J·mol""^""-1"")/(8.314 ""J·K""^""-1""""mol""^""-1"" ×500 ""K"") = ""-0.397""#

+

#K = e^""-0.397"" = 0.672#

+
+

Calculate equilibrium concentrations

+

Finally, we can set up an ICE table to calculate the equilibrium concentrations.

+

#color(white)(mmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#""I/mol·L""^""-1"": color(white)(mm)0.0923color(white)(mmm)0color(white)(mmll)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(m)0.0923 -xcolor(white)(mll)xcolor(white)(mmll)x#

+

#[""PCl""_5]_0 = ""0.120 mol""/""1.30 L"" = ""0.0923 mol/L""#

+

#K_text(c) = ([""PCl""_3][""Cl""_2])/([""PCl""_5]) = x^2/(0.0923-x)= 0.672#

+
+

Test for negligibility:

+

#0.0923/0.672 = 0.14 < 400#. ∴ #x# is not negligible. We must solve a quadratic equation.

+
+

#x^2 = 0.672(0.0923-x) = 0.0620 - 0.672x#

+

#x^2 + 0.672x - 0.0620 = 0#

+

#x = 0.0820#

+
+

Calculate total pressure

+

#""Total concentration"" = (0.0923 - x + x + x"") mol·L""^""-1"" = ""(0.0923 + 0.0820) mol/L"" = ""0.1743 mol/L""#

+

#p = (nRT)/V = cRT = (0.1743 color(red)(cancel(color(black)(""mol·L""^""-1""))) × ""0.083 14 bar""·color(red)(cancel(color(black)(""L·K""^""-1""""mol""^""-1""))) × 500 color(red)(cancel(color(black)(""K"")))) = ""7.25 bar""#

+
+
+
" "
+

What is the total pressure of the gases in the flask at this point?

+
+
+

+

Introduced into a 1.30 −L flask is 0.120 mol of #PCl_5(g)#; the flask is held at a temperature of 227 °C until equilibrium is established.
+#PCl_5(g)⇌PCl_3(g)+Cl_2(g)#

+

What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.]

+

From what I know, I need to apply #\DeltaG°#, but I don't know whether I need that for #PCl_5# or #PCl_3#.
+Here's the information I got from the book...
+for #PCl_5#: -305.0
+for #PCl_3#: -267.8

+

+
+
+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Spontaneous and Non-Spontaneous Processes + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 24, 2017 + +
+
+
+
+
+
+
+

Warning! Long Answer. #p_text(tot) = ""7.25 bar""#

+
+
+
+

Explanation:

+
+
+

Yes, you need #ΔG^@#, but at 500 K, not 298 K.

+
+

Calculate #ΔG^@# at 500 K

+

You can calculate it from the values tabulated at 298 K.

+

#color(white)(mmmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#Δ_text(f)H^@""/kJ·mol""^""-1"": color(white)(l)""-398.9""color(white)(m)""-306.4""color(white)(mm)0#
+#S^@""/J·K""^""-1""""mol""^""-1"":color(white)(ml)353color(white)(mm)311.7color(white)(mll)223#

+

#Δ_text(r)H^@ = sumΔ_text(f)H^@(""products"") - sumΔ_text(f)H^@(""reactants"") = ""(-306.4 + 398.9) kJ/mol"" = ""92.5 kJ/mol""#

+

#Δ_text(r)S^@ = sumS^@(""products"") - sumS^@(""reactants"") = (311.7 + 223 - 353) color(white)(l)""kJ/mol"" = ""181.7 kJ/mol""#

+
+

#ΔG^@ = ΔH^@ -TΔS^@#

+

∴ At 227 °C (500 K),

+

#ΔG^@ = ""92 500 J·mol""^""-1"" -500 ""K"" × ""181.7 J""·""K""^""-1""""mol""^""-1"" = ""(92 500 - 90 850) J·mol""^""-1"" = ""1650 J·mol""^""-1""#

+
+

Calculate #K#

+

#ΔG^@ = ""-""RTlnK#

+

#lnK =(-ΔG^@)/(RT)= (""-1650 J·mol""^""-1"")/(8.314 ""J·K""^""-1""""mol""^""-1"" ×500 ""K"") = ""-0.397""#

+

#K = e^""-0.397"" = 0.672#

+
+

Calculate equilibrium concentrations

+

Finally, we can set up an ICE table to calculate the equilibrium concentrations.

+

#color(white)(mmmmmmmm)""PCl""_5 ⇌ ""PCl""_3 + ""Cl""_2#
+#""I/mol·L""^""-1"": color(white)(mm)0.0923color(white)(mmm)0color(white)(mmll)0#
+#""C/mol·L""^""-1"":color(white)(mmm)""-""xcolor(white)(mmm)""+""xcolor(white)(mm)""+""x#
+#""E/mol·L""^""-1"":color(white)(m)0.0923 -xcolor(white)(mll)xcolor(white)(mmll)x#

+

#[""PCl""_5]_0 = ""0.120 mol""/""1.30 L"" = ""0.0923 mol/L""#

+

#K_text(c) = ([""PCl""_3][""Cl""_2])/([""PCl""_5]) = x^2/(0.0923-x)= 0.672#

+
+

Test for negligibility:

+

#0.0923/0.672 = 0.14 < 400#. ∴ #x# is not negligible. We must solve a quadratic equation.

+
+

#x^2 = 0.672(0.0923-x) = 0.0620 - 0.672x#

+

#x^2 + 0.672x - 0.0620 = 0#

+

#x = 0.0820#

+
+

Calculate total pressure

+

#""Total concentration"" = (0.0923 - x + x + x"") mol·L""^""-1"" = ""(0.0923 + 0.0820) mol/L"" = ""0.1743 mol/L""#

+

#p = (nRT)/V = cRT = (0.1743 color(red)(cancel(color(black)(""mol·L""^""-1""))) × ""0.083 14 bar""·color(red)(cancel(color(black)(""L·K""^""-1""""mol""^""-1""))) × 500 color(red)(cancel(color(black)(""K"")))) = ""7.25 bar""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5459 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" What is the total pressure of the gases in the flask at this point? " + + +Introduced into a 1.30 −L flask is 0.120 mol of #PCl_5(g)#; the flask is held at a temperature of 227 °C until equilibrium is established. +#PCl_5(g)⇌PCl_3(g)+Cl_2(g)# +What is the total pressure of the gases in the flask at this point? [Hint: Use data from Appendix D in the textbook and appropriate relationships from this chapter.] +From what I know, I need to apply #\DeltaG°#, but I don't know whether I need that for #PCl_5# or #PCl_3#. +Here's the information I got from the book... +for #PCl_5#: -305.0 +for #PCl_3#: -267.8 + + +" +253 a8316b65-6ddd-11ea-9357-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-01-mole-of-silicon-atoms 0.28 g start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] silicon atoms [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.28 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] silicon atoms [=] \\pu{0.01 mole}""}]" "

What is the mass of .01 mole of silicon atoms?

" nan 0.28 g "
+

Explanation:

+
+

#m(Si)=nM=0.01mol*28.0gmol^-1=0.280g#

+
+
" "
+
+
+

0.28g

+
+
+
+

Explanation:

+
+

#m(Si)=nM=0.01mol*28.0gmol^-1=0.280g#

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" "
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What is the mass of .01 mole of silicon atoms?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Jul 6, 2016 + +
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0.28g

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Explanation:

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#m(Si)=nM=0.01mol*28.0gmol^-1=0.280g#

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Impact of this question
+
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+ + Creative Commons License + +
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" What is the mass of .01 mole of silicon atoms? nan +254 a8316b66-6ddd-11ea-ad8e-ccda262736ce https://socratic.org/questions/if-a-0-050g-sample-of-the-boron-hydride-burns-completely-in-o-2-what-will-be-the 0.03 atm start physical_unit 19 21 pressure atm qc_end physical_unit 4 8 2 3 mass qc_end c_other OTHER qc_end physical_unit 26 26 24 25 volume qc_end physical_unit 26 26 28 29 temperature qc_end end "[{""type"":""physical unit"",""value"":""Pressure [OF] the gaseous water [IN] atm""}]" "[{""type"":""physical unit"",""value"":""0.03 atm""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] sample of the boron hydride [=] \\pu{0.050 g}""},{""type"":""other"",""value"":""Burns completely in O2.""},{""type"":""physical unit"",""value"":""Volume [OF] flask [=] \\pu{4.25 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] flask [=] \\pu{30.0 ℃}""}]" "

If a 0.050g sample of the boron hydride burns completely in #O_2#, what will be the pressure of the gaseous water In a 4.25-L flask at 30.0°C?

" nan 0.03 atm "
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Explanation:

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You are presumed to have #(0.050*g)/(27.67*g*mol^-1) = ?? mol# diborane.

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By the stoichiometry of the reaction listed, for each mole of diborane that combusts, #3*mol# of water are produced.

+

Assuming ideality, #P# #=# #(nRT)/V# #=# #3xx((0.050*g)/(27.67*g*mol^-1))(xx0.0821*L*atm*K^-1mol^-1xx303K)xx(1/(4.25*L))# #=# #?? atm#.

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Why did I use #303*K# instead of the given temperature in Centigrade? Would it make a difference if I used the formula #BH_3# instead of that of diborane?

+

Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.

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" "
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I assume that you mean diborane, #B_2H_6#.
+#B_2H_6(g) + 3O_2 rarr B_2O_3(s) + 3H_2O(g)uarr#

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+
+

Explanation:

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+

You are presumed to have #(0.050*g)/(27.67*g*mol^-1) = ?? mol# diborane.

+

By the stoichiometry of the reaction listed, for each mole of diborane that combusts, #3*mol# of water are produced.

+

Assuming ideality, #P# #=# #(nRT)/V# #=# #3xx((0.050*g)/(27.67*g*mol^-1))(xx0.0821*L*atm*K^-1mol^-1xx303K)xx(1/(4.25*L))# #=# #?? atm#.

+

Why did I use #303*K# instead of the given temperature in Centigrade? Would it make a difference if I used the formula #BH_3# instead of that of diborane?

+

Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.

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+
" "
+

If a 0.050g sample of the boron hydride burns completely in #O_2#, what will be the pressure of the gaseous water In a 4.25-L flask at 30.0°C?

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+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+1 Answer +
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+ + Dec 10, 2015 + +
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I assume that you mean diborane, #B_2H_6#.
+#B_2H_6(g) + 3O_2 rarr B_2O_3(s) + 3H_2O(g)uarr#

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+
+
+

Explanation:

+
+

You are presumed to have #(0.050*g)/(27.67*g*mol^-1) = ?? mol# diborane.

+

By the stoichiometry of the reaction listed, for each mole of diborane that combusts, #3*mol# of water are produced.

+

Assuming ideality, #P# #=# #(nRT)/V# #=# #3xx((0.050*g)/(27.67*g*mol^-1))(xx0.0821*L*atm*K^-1mol^-1xx303K)xx(1/(4.25*L))# #=# #?? atm#.

+

Why did I use #303*K# instead of the given temperature in Centigrade? Would it make a difference if I used the formula #BH_3# instead of that of diborane?

+

Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.

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" If a 0.050g sample of the boron hydride burns completely in #O_2#, what will be the pressure of the gaseous water In a 4.25-L flask at 30.0°C? nan +255 a8319254-6ddd-11ea-81d8-ccda262736ce https://socratic.org/questions/what-is-a-balanced-chemical-equation-for-the-synthesis-of-sodium-bromide-from-so 2 Na + Br2 -> 2 NaBr start chemical_equation qc_end substance 10 11 qc_end substance 10 10 qc_end substance 15 15 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the synthesis""}]" "[{""type"":""chemical equation"",""value"":""2 Na + Br2 -> 2 NaBr""}]" "[{""type"":""substance name"",""value"":""sodium bromide""},{""type"":""substance name"",""value"":""sodium""},{""type"":""substance name"",""value"":""bromine""}]" "

What is a balanced chemical equation for the synthesis of sodium bromide from sodium and bromine?

" nan 2 Na + Br2 -> 2 NaBr "
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Explanation:

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Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of #Br_2#.

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#2Na + Br_2 rarr 2 NaBr#

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Explanation:

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Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of #Br_2#.

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" "
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What is a balanced chemical equation for the synthesis of sodium bromide from sodium and bromine?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+2 Answers +
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+ + Mar 31, 2017 + +
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#2Na + Br_2 rarr 2 NaBr#

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Explanation:

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Because bromine exists in the form of a diatomic molecule, the equation must be based on one molecule of #Br_2#.

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#Na(s) + 1/2Br_2(l) rarr NaBr(s)#

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Explanation:

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This is not a reaction I would care to do. Bromine is one of the most corrosive substances you use in a lab, and demands a great deal of respect.

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Is this a redox equation? How do you know?

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" What is a balanced chemical equation for the synthesis of sodium bromide from sodium and bromine? nan +256 a8319255-6ddd-11ea-acf8-ccda262736ce https://socratic.org/questions/what-is-the-mass-in-grams-of-2-10-10-21-atoms-of-copper-cu 0.22 g start physical_unit 13 13 mass g qc_end physical_unit 10 12 7 9 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] Cu [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.22 g""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of copper [=] \\pu{2.10 × 10^21}""}]" "

What is the mass in grams of #2.10 * 10^21# atoms of copper, #Cu#?

" nan 0.22 g "
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Explanation:

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+

So we know that #""Avogadro's number""(N_A)# of individual copper atoms have a mass of #63.55# #g#.

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Thus #(2.10xx10^21""copper atoms""xx63.55*g*mol^-1)/(6.022140857(74)×10^23 ""copper atoms "" mol^-1)# #=# #??g#

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There are #N_A# individual copper atoms in a moles of copper metal, and this has a mass of #63.55# #g#.

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Explanation:

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+

So we know that #""Avogadro's number""(N_A)# of individual copper atoms have a mass of #63.55# #g#.

+

Thus #(2.10xx10^21""copper atoms""xx63.55*g*mol^-1)/(6.022140857(74)×10^23 ""copper atoms "" mol^-1)# #=# #??g#

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" "
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What is the mass in grams of #2.10 * 10^21# atoms of copper, #Cu#?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Apr 2, 2016 + +
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There are #N_A# individual copper atoms in a moles of copper metal, and this has a mass of #63.55# #g#.

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Explanation:

+
+

So we know that #""Avogadro's number""(N_A)# of individual copper atoms have a mass of #63.55# #g#.

+

Thus #(2.10xx10^21""copper atoms""xx63.55*g*mol^-1)/(6.022140857(74)×10^23 ""copper atoms "" mol^-1)# #=# #??g#

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Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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+
" What is the mass in grams of #2.10 * 10^21# atoms of copper, #Cu#? nan +257 a8319256-6ddd-11ea-a587-ccda262736ce https://socratic.org/questions/for-the-reaction-2x-3y-4z-rarr-5w-initially-if-1-mole-of-x-3-mole-of-y-and-4-mol 50.00% start physical_unit 45 46 percent_yield none qc_end chemical_equation 3 13 qc_end physical_unit 4 4 16 17 mole qc_end physical_unit 7 7 20 21 mole qc_end physical_unit 10 10 25 26 mole qc_end physical_unit 13 13 32 33 mole qc_end end "[{""type"":""physical unit"",""value"":""Yield [OF] this reaction""}]" "[{""type"":""physical unit"",""value"":""50.00%""}]" "[{""type"":""chemical equation"",""value"":""2 x + 3 y + 4 z -> 5 w""},{""type"":""physical unit"",""value"":""Mole [OF] x [=] \\pu{1 mole}""},{""type"":""physical unit"",""value"":""Mole [OF] y [=] \\pu{3 moles}""},{""type"":""physical unit"",""value"":""Mole [OF] z [=] \\pu{4 moles}""},{""type"":""physical unit"",""value"":""Mole [OF] w [=] \\pu{1.25 moles}""}]" "

For the reaction,#2x#+#3y#+#4z## rarr## 5w# + +Initially if 1 mole of #x#,3 mole of #y# and 4 mole of #z# is taken and 1.25 mole of #w# is obtained then what is the % yield of this reaction? +

" nan 50.00% "
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Explanation:

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+

This is a limiting reactant problem.

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We know that we will need a balanced equation and moles of each reactant.

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+

1. Gather all the information in one place with the experimental number of moles below the formulas.

+

#color(white)(mmmmmm)2x + 3y + 4z = 5w#
+#""Amt/mol:""color(white)(ml)1color(white)(mm)3""color(white)(mm)4#
+#""Divide by:""color(white)(ml)2color(white)(mm)3color(white)(mm)4#
+#""Moles rxn:""color(white)(m)0.5color(white)(ml)1color(white)(mm)1#

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+

2. Identify the limiting reactant

+

An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:

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You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

+

I did that for you in the table above.

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#x# is the limiting reactant because it gives the fewest moles of reaction.

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+

3. Calculate the theoretical yield of #w#

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#""Theoretical yield"" = 1 color(red)(cancel(color(black)(""mol""color(white)(l) x))) × (""5 mol "" w)/(2 color(red)(cancel(color(black)(""mol ""x)))) = ""2.5 mol""color(white)(l)w#

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+

4. Calculate the percent yield

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#""Percent yield"" = ""Actual yield""/""Theoretical yield"" × 100 % = (1.25 color(red)(cancel(color(black)(""mol""))))/(2.5 color(red)(cancel(color(black)(""mol"")))) × 100 % = 50 %#

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" "
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The percent yield is 50 %.

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+
+

Explanation:

+
+
+

This is a limiting reactant problem.

+

We know that we will need a balanced equation and moles of each reactant.

+
+

1. Gather all the information in one place with the experimental number of moles below the formulas.

+

#color(white)(mmmmmm)2x + 3y + 4z = 5w#
+#""Amt/mol:""color(white)(ml)1color(white)(mm)3""color(white)(mm)4#
+#""Divide by:""color(white)(ml)2color(white)(mm)3color(white)(mm)4#
+#""Moles rxn:""color(white)(m)0.5color(white)(ml)1color(white)(mm)1#

+
+

2. Identify the limiting reactant

+

An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:

+

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

+

I did that for you in the table above.

+

#x# is the limiting reactant because it gives the fewest moles of reaction.

+
+

3. Calculate the theoretical yield of #w#

+

#""Theoretical yield"" = 1 color(red)(cancel(color(black)(""mol""color(white)(l) x))) × (""5 mol "" w)/(2 color(red)(cancel(color(black)(""mol ""x)))) = ""2.5 mol""color(white)(l)w#

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+

4. Calculate the percent yield

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#""Percent yield"" = ""Actual yield""/""Theoretical yield"" × 100 % = (1.25 color(red)(cancel(color(black)(""mol""))))/(2.5 color(red)(cancel(color(black)(""mol"")))) × 100 % = 50 %#

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" "
+

For the reaction,#2x#+#3y#+#4z## rarr## 5w# + +Initially if 1 mole of #x#,3 mole of #y# and 4 mole of #z# is taken and 1.25 mole of #w# is obtained then what is the % yield of this reaction? +

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+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
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+1 Answer +
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+ + Nov 15, 2016 + +
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The percent yield is 50 %.

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+

Explanation:

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+

This is a limiting reactant problem.

+

We know that we will need a balanced equation and moles of each reactant.

+
+

1. Gather all the information in one place with the experimental number of moles below the formulas.

+

#color(white)(mmmmmm)2x + 3y + 4z = 5w#
+#""Amt/mol:""color(white)(ml)1color(white)(mm)3""color(white)(mm)4#
+#""Divide by:""color(white)(ml)2color(white)(mm)3color(white)(mm)4#
+#""Moles rxn:""color(white)(m)0.5color(white)(ml)1color(white)(mm)1#

+
+

2. Identify the limiting reactant

+

An easy way to identify the limiting reactant is to calculate the ""moles of reaction"" each will give:

+

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

+

I did that for you in the table above.

+

#x# is the limiting reactant because it gives the fewest moles of reaction.

+
+

3. Calculate the theoretical yield of #w#

+

#""Theoretical yield"" = 1 color(red)(cancel(color(black)(""mol""color(white)(l) x))) × (""5 mol "" w)/(2 color(red)(cancel(color(black)(""mol ""x)))) = ""2.5 mol""color(white)(l)w#

+
+

4. Calculate the percent yield

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#""Percent yield"" = ""Actual yield""/""Theoretical yield"" × 100 % = (1.25 color(red)(cancel(color(black)(""mol""))))/(2.5 color(red)(cancel(color(black)(""mol"")))) × 100 % = 50 %#

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Related questions
+ + +
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Impact of this question
+
+ 11022 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" "For the reaction,#2x#+#3y#+#4z## rarr## 5w# + +Initially if 1 mole of #x#,3 mole of #y# and 4 mole of #z# is taken and 1.25 mole of #w# is obtained then what is the % yield of this reaction? +" nan +258 a8319257-6ddd-11ea-9c7e-ccda262736ce https://socratic.org/questions/how-do-you-write-calcium-nitrogen-calcium-nitride 3 Ca + N2 -> Ca3N2z start chemical_equation qc_end substance 4 4 qc_end substance 6 6 qc_end substance 8 9 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""3 Ca + N2 -> Ca3N2z""}]" "[{""type"":""substance name"",""value"":""Calcium""},{""type"":""substance name"",""value"":""Nitrogen""},{""type"":""substance name"",""value"":""Calcium nitride""}]" "

How do you write ""calcium + nitrogen -> calcium nitride""?

" nan 3 Ca + N2 -> Ca3N2z "
+

Explanation:

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+

Calcium is a metal, so its formula will simply be #""Ca""# .

+

Nitrogen is a diatomic molecular compound, making it #""N""_2# .

+

Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or #""Ca""^(2+)#. The nitride ion is a 3- ion, or #""N""^(3-)""#. As such, to balance the charges we require 3x calcium ions for every 2x nitride ions, or #""Ca""_3""N""_2#.

+

#:. Ca_((s)) + N_(2(g)) -> Ca_3N_(2(s))#

+

Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have #1# calcium, and on the right we have #3#. We can balance this simply by increasing the number of calciums on the left to #3#, as demonstrated in the answer above.

+

I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.

+

It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.

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" "
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#3Ca_((s)) + N_(2(g)) -> Ca_3N_(2(s))#

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Explanation:

+
+

Calcium is a metal, so its formula will simply be #""Ca""# .

+

Nitrogen is a diatomic molecular compound, making it #""N""_2# .

+

Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or #""Ca""^(2+)#. The nitride ion is a 3- ion, or #""N""^(3-)""#. As such, to balance the charges we require 3x calcium ions for every 2x nitride ions, or #""Ca""_3""N""_2#.

+

#:. Ca_((s)) + N_(2(g)) -> Ca_3N_(2(s))#

+

Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have #1# calcium, and on the right we have #3#. We can balance this simply by increasing the number of calciums on the left to #3#, as demonstrated in the answer above.

+

I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.

+

It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.

+
+
+
" "
+

How do you write ""calcium + nitrogen -> calcium nitride""?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
+
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+1 Answer +
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+ + Dec 28, 2015 + +
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#3Ca_((s)) + N_(2(g)) -> Ca_3N_(2(s))#

+
+
+
+

Explanation:

+
+

Calcium is a metal, so its formula will simply be #""Ca""# .

+

Nitrogen is a diatomic molecular compound, making it #""N""_2# .

+

Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. The calcium ion is a 2+ ion, or #""Ca""^(2+)#. The nitride ion is a 3- ion, or #""N""^(3-)""#. As such, to balance the charges we require 3x calcium ions for every 2x nitride ions, or #""Ca""_3""N""_2#.

+

#:. Ca_((s)) + N_(2(g)) -> Ca_3N_(2(s))#

+

Balancing the equation, then, you will notice that the number of nitrogens on each side balances However, on the left side of the equation we have #1# calcium, and on the right we have #3#. We can balance this simply by increasing the number of calciums on the left to #3#, as demonstrated in the answer above.

+

I have also provided the states of each reactant and the product at room temperature, though without further information on reaction conditions these cannot be confirmed as correct and so I recommend leaving them out unless there is further advice given.

+

It may seem unnatural to refer to these species as ""calciums"" and the like, but it is important to be aware that we are not always talking in terms of atoms: in this case, we may also be referring to the ions on the right hand side of the equation.

+
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Related questions
+ + +
+
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+
Impact of this question
+
+ 43777 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
" "How do you write ""calcium + nitrogen -> calcium nitride""?" nan +259 a8319258-6ddd-11ea-a8da-ccda262736ce https://socratic.org/questions/what-can-you-say-about-the-proportion-of-hydrogen-ions-and-hydroxide-ions-in-a-s 10^10 start physical_unit 8 15 ratio none qc_end physical_unit 15 15 21 21 ph qc_end end "[{""type"":""physical unit"",""value"":""Proportion [OF] hydrogen ions and hydroxide ions in a solution""}]" "[{""type"":""physical unit"",""value"":""10^10""}]" "[{""type"":""physical unit"",""value"":""pH [OF] solution [=] \\pu{2}""}]" "

What can you say about the proportion of hydrogen ions and hydroxide ions in a solution that has a pH of 2?

" nan 10^10 "
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Explanation:

+
+

In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions, #""H""_3""O""^(+)#, and hydroxide ions, #""OH""^(-)#.

+

Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.

+
+

#""pH"" = - log( [""H""_3""O""^(+)])#

+
+

Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions

+
+

#""pOH"" = - log( [""OH""^(-)])#

+
+

For aqueous solutions, you can say that

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Here #14# is actually equal to #- log(K_W)#, #K_W# being the ion product constant for water's self-ionization reaction.

+

So, take a look at your solution. You know that its pH is equal to #2#. Automatically, you can say that its pOH will be

+
+

#""pOH"" = 14 - ""pH""#

+

#""pOH"" = 14 - 2 = 12#

+
+

Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.

+

Use the log definitions of the pH and pOH to get

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#

+
+

and

+
+

#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#

+
+

This means that a solution that has a pH equal to #2# will have #10^10# more hydronium ions that hydroxide ions, since

+
+

#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#

+
+
+
" "
+
+
+

You can say that it contains #10^(10)# more hydronium ions that hydroxide ions.

+
+
+
+

Explanation:

+
+

In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions, #""H""_3""O""^(+)#, and hydroxide ions, #""OH""^(-)#.

+

Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.

+
+

#""pH"" = - log( [""H""_3""O""^(+)])#

+
+

Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions

+
+

#""pOH"" = - log( [""OH""^(-)])#

+
+

For aqueous solutions, you can say that

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Here #14# is actually equal to #- log(K_W)#, #K_W# being the ion product constant for water's self-ionization reaction.

+

So, take a look at your solution. You know that its pH is equal to #2#. Automatically, you can say that its pOH will be

+
+

#""pOH"" = 14 - ""pH""#

+

#""pOH"" = 14 - 2 = 12#

+
+

Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.

+

Use the log definitions of the pH and pOH to get

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#

+
+

and

+
+

#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#

+
+

This means that a solution that has a pH equal to #2# will have #10^10# more hydronium ions that hydroxide ions, since

+
+

#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#

+
+
+
+
" "
+

What can you say about the proportion of hydrogen ions and hydroxide ions in a solution that has a pH of 2?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 7, 2015 + +
+
+
+
+
+
+
+

You can say that it contains #10^(10)# more hydronium ions that hydroxide ions.

+
+
+
+

Explanation:

+
+

In order for an aqueous solution to be neutral, you need it to contain equal concentrations of hydronium ions, #""H""_3""O""^(+)#, and hydroxide ions, #""OH""^(-)#.

+

Now, a solution's pH is calculated by taking the negative common logarithm (this is simply a base 10 log) of the concentration of hydronium ions.

+
+

#""pH"" = - log( [""H""_3""O""^(+)])#

+
+

Likewise, a solution's pOH is calculated by taking the negative common logarithm of the concentration of hydroxide ions

+
+

#""pOH"" = - log( [""OH""^(-)])#

+
+

For aqueous solutions, you can say that

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Here #14# is actually equal to #- log(K_W)#, #K_W# being the ion product constant for water's self-ionization reaction.

+

So, take a look at your solution. You know that its pH is equal to #2#. Automatically, you can say that its pOH will be

+
+

#""pOH"" = 14 - ""pH""#

+

#""pOH"" = 14 - 2 = 12#

+
+

Now, a lower pH is equivalent to a higher concentration of hydronium ions, and implicitly a lower concentration of hydroxide ions.

+

Use the log definitions of the pH and pOH to get

+
+

#[""H""_3""O""^(+)] = 10^(-""pH"") = 10^(-2)""M""#

+
+

and

+
+

#[""OH""^(-)] = 10^(-""pOH"") = 10^(-12)""M""#

+
+

This means that a solution that has a pH equal to #2# will have #10^10# more hydronium ions that hydroxide ions, since

+
+

#([""H""_3""O""^(+)])/([""OH""^(-)]) = (10^(-2)color(red)(cancel(color(black)(""M""))))/(10^(-12)color(red)(cancel(color(black)(""M"")))) = 10^10#

+
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+
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+
+
+
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+
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" What can you say about the proportion of hydrogen ions and hydroxide ions in a solution that has a pH of 2? nan +260 a8319259-6ddd-11ea-aca6-ccda262736ce https://socratic.org/questions/what-are-the-oxidation-states-of-the-atoms-in-a-diatomic-gas 0 start physical_unit 6 11 oxidation_state none qc_end substance 10 11 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation states [OF] the atoms in a diatomic gas""}]" "[{""type"":""physical unit"",""value"":""0""}]" "[{""type"":""substance name"",""value"":""Diatomic gas""}]" "

What are the oxidation states of the atoms in a diatomic gas?

" nan 0 "
+

Explanation:

+
+

The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.

+

When we do this for #H_2#, or #X_2#, or #N-=N#, or #O=O#, we get sharing of electrons, and a #0# oxidation state. Of course, if it is a diatomic gas such as #H-Cl#, we break the bond and the electronic charge is distributed to the most electronegative atom, here the halide, i.e. #H^(I+)# and #Cl^(-I)#. With an interhalogen, say #Br-Cl#, we get #Br^(I+)# and #Cl^(-I)#

+
+
" "
+
+
+

If it is elemental gas, the oxidation number is #0#.

+
+
+
+

Explanation:

+
+

The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.

+

When we do this for #H_2#, or #X_2#, or #N-=N#, or #O=O#, we get sharing of electrons, and a #0# oxidation state. Of course, if it is a diatomic gas such as #H-Cl#, we break the bond and the electronic charge is distributed to the most electronegative atom, here the halide, i.e. #H^(I+)# and #Cl^(-I)#. With an interhalogen, say #Br-Cl#, we get #Br^(I+)# and #Cl^(-I)#

+
+
+
" "
+

What are the oxidation states of the atoms in a diatomic gas?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 9, 2016 + +
+
+
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+
+
+
+

If it is elemental gas, the oxidation number is #0#.

+
+
+
+

Explanation:

+
+

The oxidation number of an atom in a chemical bond, is the charge left on the atom of interest when all the bonding pairs of electrons are broken, with the charge devolving to the most electronegative atom.

+

When we do this for #H_2#, or #X_2#, or #N-=N#, or #O=O#, we get sharing of electrons, and a #0# oxidation state. Of course, if it is a diatomic gas such as #H-Cl#, we break the bond and the electronic charge is distributed to the most electronegative atom, here the halide, i.e. #H^(I+)# and #Cl^(-I)#. With an interhalogen, say #Br-Cl#, we get #Br^(I+)# and #Cl^(-I)#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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" What are the oxidation states of the atoms in a diatomic gas? nan +261 a831b97a-6ddd-11ea-aecb-ccda262736ce https://socratic.org/questions/if-the-initial-of-an-ideal-gas-at-2-250-atm-is-62-00-c-what-final-temperature-wo 260.57 K start physical_unit 5 6 temperature k qc_end physical_unit 5 6 11 12 temperature qc_end physical_unit 5 6 8 9 pressure qc_end physical_unit 5 6 24 25 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] ideal gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""260.57 K""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] ideal gas [=] \\pu{62.00 ℃}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] ideal gas [=] \\pu{2.250 atm}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] ideal gas [=] \\pu{1.750 atm}""}]" "

If the initial of an ideal gas at 2.250 atm is 62.00°c, what final temperature would cause the pressure to be reduced to 1.750 atm?

" nan 260.57 K "
+

Explanation:

+
+

From Gay Lussac’s Law

+
+

#color(white)(...)""P ∝ T""#

+

#""P""_1/""T""_1 = ""P""_2/""T""_2#

+
+

#""T""_2 = ""P""_2 × ""T""_1/""P""_1 #

+

#color(white)(""T""_2) = 1.750 cancel""atm"" × ""(62.00 + 273) K""/(2.250 cancel""atm"")#

+

#color(white)(""T""_2) = 1.750 × ""335 K""/2.250#

+

#color(white)(""T""_2) = 260.5\ ""K""#

+
+
" "
+
+
+

#""260.5 K""#

+
+
+
+

Explanation:

+
+

From Gay Lussac’s Law

+
+

#color(white)(...)""P ∝ T""#

+

#""P""_1/""T""_1 = ""P""_2/""T""_2#

+
+

#""T""_2 = ""P""_2 × ""T""_1/""P""_1 #

+

#color(white)(""T""_2) = 1.750 cancel""atm"" × ""(62.00 + 273) K""/(2.250 cancel""atm"")#

+

#color(white)(""T""_2) = 1.750 × ""335 K""/2.250#

+

#color(white)(""T""_2) = 260.5\ ""K""#

+
+
+
" "
+

If the initial of an ideal gas at 2.250 atm is 62.00°c, what final temperature would cause the pressure to be reduced to 1.750 atm?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gay Lussac's Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 2, 2018 + +
+
+
+
+
+
+
+

#""260.5 K""#

+
+
+
+

Explanation:

+
+

From Gay Lussac’s Law

+
+

#color(white)(...)""P ∝ T""#

+

#""P""_1/""T""_1 = ""P""_2/""T""_2#

+
+

#""T""_2 = ""P""_2 × ""T""_1/""P""_1 #

+

#color(white)(""T""_2) = 1.750 cancel""atm"" × ""(62.00 + 273) K""/(2.250 cancel""atm"")#

+

#color(white)(""T""_2) = 1.750 × ""335 K""/2.250#

+

#color(white)(""T""_2) = 260.5\ ""K""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
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" If the initial of an ideal gas at 2.250 atm is 62.00°c, what final temperature would cause the pressure to be reduced to 1.750 atm? nan +262 a831e346-6ddd-11ea-b729-ccda262736ce https://socratic.org/questions/5996891311ef6b0566add301 45.43 L start physical_unit 3 4 volume l qc_end physical_unit 16 17 12 13 mass qc_end c_other STP qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] carbon dioxide [IN] L""}]" "[{""type"":""physical unit"",""value"":""45.43 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] calcium carbonate [=] \\pu{200 g}""},{""type"":""other"",""value"":""STP""},{""type"":""other"",""value"":""Fierce heating.""}]" "

What volume of carbon dioxide would evolve from fierce heating of a #200*g# mass of #""calcium carbonate""# on heating under standard conditions of #""STP""#?

" nan 45.43 L "
+

Explanation:

+
+

We interrogate the decomposition reaction....

+

#CaCO_3(s) + Delta rarrCaO(s) + CO_2(g)uarr#

+

And given complete decomposition we have a molar quantity of #(200*g)/(100.09*g*mol^-1)=2.0*mol# WITH RESPECT TO CALCIUM CARBONATE, AND CARBON DIOXIDE. And as you know this molar quantity also represents a (large!) number with respect to gaseous molecules.

+

And the volume expressed by this quantity is simply given by the old Ideal Gas equation..#V=(nRT)/P#

+

#-=(2.0*molxx0.0821*(L*atm)/(K*mol)xx273.15*K)/((100*kPa)/(101.3*kPa*atm^-1))#

+

#=??*L#

+

And the number of molecules is simply #2*molxx6.022xx10^23*mol^-1=??#

+
+
" "
+
+
+

Well, #""STP""# specifies a temperature of #273.15*K# and an absolute pressure of exactly #100 *kPa#. I gets under #50*L#.

+
+
+
+

Explanation:

+
+

We interrogate the decomposition reaction....

+

#CaCO_3(s) + Delta rarrCaO(s) + CO_2(g)uarr#

+

And given complete decomposition we have a molar quantity of #(200*g)/(100.09*g*mol^-1)=2.0*mol# WITH RESPECT TO CALCIUM CARBONATE, AND CARBON DIOXIDE. And as you know this molar quantity also represents a (large!) number with respect to gaseous molecules.

+

And the volume expressed by this quantity is simply given by the old Ideal Gas equation..#V=(nRT)/P#

+

#-=(2.0*molxx0.0821*(L*atm)/(K*mol)xx273.15*K)/((100*kPa)/(101.3*kPa*atm^-1))#

+

#=??*L#

+

And the number of molecules is simply #2*molxx6.022xx10^23*mol^-1=??#

+
+
+
" "
+

What volume of carbon dioxide would evolve from fierce heating of a #200*g# mass of #""calcium carbonate""# on heating under standard conditions of #""STP""#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Equation Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 18, 2017 + +
+
+
+
+
+
+
+

Well, #""STP""# specifies a temperature of #273.15*K# and an absolute pressure of exactly #100 *kPa#. I gets under #50*L#.

+
+
+
+

Explanation:

+
+

We interrogate the decomposition reaction....

+

#CaCO_3(s) + Delta rarrCaO(s) + CO_2(g)uarr#

+

And given complete decomposition we have a molar quantity of #(200*g)/(100.09*g*mol^-1)=2.0*mol# WITH RESPECT TO CALCIUM CARBONATE, AND CARBON DIOXIDE. And as you know this molar quantity also represents a (large!) number with respect to gaseous molecules.

+

And the volume expressed by this quantity is simply given by the old Ideal Gas equation..#V=(nRT)/P#

+

#-=(2.0*molxx0.0821*(L*atm)/(K*mol)xx273.15*K)/((100*kPa)/(101.3*kPa*atm^-1))#

+

#=??*L#

+

And the number of molecules is simply #2*molxx6.022xx10^23*mol^-1=??#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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" "What volume of carbon dioxide would evolve from fierce heating of a #200*g# mass of #""calcium carbonate""# on heating under standard conditions of #""STP""#? " nan +263 a83207f0-6ddd-11ea-a50c-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-magnesium-sulfate-heptahydrate MgSO4.7H2O start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] magnesium sulfate heptahydrate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""MgSO4.7H2O""}]" "[{""type"":""substance name"",""value"":""Magnesium sulfate heptahydrate""}]" "

What is the formula for magnesium sulfate heptahydrate?

" nan MgSO4.7H2O "
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

+
+
" "
+
+
+

#MgSO_4*7H_2O#

+
+
+
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

+
+
+
" "
+

What is the formula for magnesium sulfate heptahydrate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 12, 2018 + +
+
+
+
+
+
+
+

#MgSO_4*7H_2O#

+
+
+
+

Explanation:

+
+

Magnesium is #Mg#, which carries a +2 charge
+Sulphate is #SO_4#, which carries a -2 charge.

+

Hence, magnesium sulphate is #MgSO_4#.

+

""Hepta-"" refers to ""seven"", ""hydrate"" refers to ""water"", so we have to add #7H_2O#.

+

Formula: #MgSO_4*7H_2O#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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" What is the formula for magnesium sulfate heptahydrate? nan +264 a83207f1-6ddd-11ea-bfd5-ccda262736ce https://socratic.org/questions/564404f7581e2a61ea824d0e 70.00 milliequivalents start physical_unit 4 5 mass milliequivalents qc_end physical_unit 20 20 15 16 molarity qc_end physical_unit 28 28 22 23 molarity qc_end physical_unit 12 12 8 9 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium cations [IN] milliequivalents""}]" "[{""type"":""physical unit"",""value"":""70.00 milliequivalents""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] Cl− [=] \\pu{40 mEq/L}""},{""type"":""physical unit"",""value"":""Molarity [OF] HPO4^2− [=] \\pu{15 mEq/L}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{per liter}""}]" "

How many milliequivalents of sodium cations are present per liter of a solution that contains #""40. mEq/L""# of chloride anions, #""Cl""^(-)#, and #""15 mEq/L""# of hydrogen phosphate anions, #""HPO""_4^(2-)# ?

" nan 70.00 milliequivalents "
+

Explanation:

+
+

The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.

+

The two anions present in solution are

+
    +
  • the chloride anion, #""Cl""^(-)#, which has a concentration of #""40. mEq/L""#
    • +
  • the hydrogen phosphate anion, #""HPO""_4^(2-)#, which has a concentration of #""15 mEq/L""#
    • +
+

As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.

+

In this case, every liter of solution will contain #""40. mEq""# of chloride anions and #""15 mEq""# of hydrogen phosphate anions, which means that the total number of milliequivalents for the anions will be

+
+

#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#

+
+

You know that a solution is electrically neutral, so

+
+

#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#

+
+

Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.

+

This means that the solution will contain #""55 mEq""# of sodium for every liter of solution, which is equivalent to a concentration of

+
+

#[""Na""^(+)] = color(green)(""55 mEq/L"")#

+
+
+
" "
+
+
+

#""55 mEq/L""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.

+

The two anions present in solution are

+
    +
  • the chloride anion, #""Cl""^(-)#, which has a concentration of #""40. mEq/L""#
    • +
  • the hydrogen phosphate anion, #""HPO""_4^(2-)#, which has a concentration of #""15 mEq/L""#
    • +
+

As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.

+

In this case, every liter of solution will contain #""40. mEq""# of chloride anions and #""15 mEq""# of hydrogen phosphate anions, which means that the total number of milliequivalents for the anions will be

+
+

#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#

+
+

You know that a solution is electrically neutral, so

+
+

#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#

+
+

Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.

+

This means that the solution will contain #""55 mEq""# of sodium for every liter of solution, which is equivalent to a concentration of

+
+

#[""Na""^(+)] = color(green)(""55 mEq/L"")#

+
+
+
+
" "
+

How many milliequivalents of sodium cations are present per liter of a solution that contains #""40. mEq/L""# of chloride anions, #""Cl""^(-)#, and #""15 mEq/L""# of hydrogen phosphate anions, #""HPO""_4^(2-)# ?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 5, 2015 + +
+
+
+
+
+
+
+

#""55 mEq/L""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.

+

The two anions present in solution are

+
    +
  • the chloride anion, #""Cl""^(-)#, which has a concentration of #""40. mEq/L""#
    • +
  • the hydrogen phosphate anion, #""HPO""_4^(2-)#, which has a concentration of #""15 mEq/L""#
    • +
+

As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.

+

In this case, every liter of solution will contain #""40. mEq""# of chloride anions and #""15 mEq""# of hydrogen phosphate anions, which means that the total number of milliequivalents for the anions will be

+
+

#""no. of mEq anions"" = ""40. mEq"" + ""15 mEq"" = ""55 mEq""#

+
+

You know that a solution is electrically neutral, so

+
+

#color(blue)(""no. of mEq anions "" = "" no. of mEq cations"")#

+
+

Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.

+

This means that the solution will contain #""55 mEq""# of sodium for every liter of solution, which is equivalent to a concentration of

+
+

#[""Na""^(+)] = color(green)(""55 mEq/L"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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" "How many milliequivalents of sodium cations are present per liter of a solution that contains #""40. mEq/L""# of chloride anions, #""Cl""^(-)#, and #""15 mEq/L""# of hydrogen phosphate anions, #""HPO""_4^(2-)# ?" nan +265 a8322eba-6ddd-11ea-8a5a-ccda262736ce https://socratic.org/questions/gases-are-sold-in-large-cylinders-for-laboratory-use-what-pressure-in-atmosphere 47.30 atm start physical_unit 22 22 pressure atm qc_end physical_unit 22 22 17 18 mass qc_end physical_unit 22 22 26 27 temperature qc_end physical_unit 32 32 30 31 volume qc_end end "[{""type"":""physical unit"",""value"":""Pressure exerted [OF] O2 [IN] atm""}]" "[{""type"":""physical unit"",""value"":""47.30 atm""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] O2 [=] \\pu{2500 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] O2 [=] \\pu{22 ℃}""},{""type"":""physical unit"",""value"":""Volume [OF] cylinder [=] \\pu{40.0 L}""}]" "

Gases are sold in large cylinders for laboratory use. What pressure in atmospheres will be exerted by 2500 g of oxygen gas (O2) when stored at 22°C in a 40.0 L cylinder?

" nan 47.30 atm "
+

Explanation:

+
+

Assuming near ideal gas behavior where #n = (PV)/(RT)#, with a constant volume, we use the given values with the appropriate gas constant value R to find the pressure.
+Rearranging the equation we have #P = (n*R*T)/V #
+V = 40.0 L
+R = 0.0820575 L-atm/K-mol
+T = 22'C = 295'K
+n = (2500g/32.0 g/mol) = 78.125 moles of #O_2#.

+

#P = (78.125*0.0820575*295)/40.0#

+

P = 47.3atm

+
+
" "
+
+
+

47.3 atm

+
+
+
+

Explanation:

+
+

Assuming near ideal gas behavior where #n = (PV)/(RT)#, with a constant volume, we use the given values with the appropriate gas constant value R to find the pressure.
+Rearranging the equation we have #P = (n*R*T)/V #
+V = 40.0 L
+R = 0.0820575 L-atm/K-mol
+T = 22'C = 295'K
+n = (2500g/32.0 g/mol) = 78.125 moles of #O_2#.

+

#P = (78.125*0.0820575*295)/40.0#

+

P = 47.3atm

+
+
+
" "
+

Gases are sold in large cylinders for laboratory use. What pressure in atmospheres will be exerted by 2500 g of oxygen gas (O2) when stored at 22°C in a 40.0 L cylinder?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Measuring Gas Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 9, 2016 + +
+
+
+
+
+
+
+

47.3 atm

+
+
+
+

Explanation:

+
+

Assuming near ideal gas behavior where #n = (PV)/(RT)#, with a constant volume, we use the given values with the appropriate gas constant value R to find the pressure.
+Rearranging the equation we have #P = (n*R*T)/V #
+V = 40.0 L
+R = 0.0820575 L-atm/K-mol
+T = 22'C = 295'K
+n = (2500g/32.0 g/mol) = 78.125 moles of #O_2#.

+

#P = (78.125*0.0820575*295)/40.0#

+

P = 47.3atm

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 8318 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" Gases are sold in large cylinders for laboratory use. What pressure in atmospheres will be exerted by 2500 g of oxygen gas (O2) when stored at 22°C in a 40.0 L cylinder? nan +266 a832391e-6ddd-11ea-85b3-ccda262736ce https://socratic.org/questions/menthol-the-substance-we-can-smell-in-mentholated-cough-drops-is-composed-of-c-h C10H20O start chemical_formula qc_end c_other OTHER qc_end physical_unit 0 0 18 19 mass qc_end physical_unit 29 30 26 27 mass qc_end physical_unit 35 35 32 33 mass qc_end c_other Combusted qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] menthol [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""C10H20O""}]" "[{""type"":""other"",""value"":""Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O.""},{""type"":""physical unit"",""value"":""mass [OF] menthol [=] \\pu{0.1005 g}""},{""type"":""physical unit"",""value"":""mass [OF] carbon dioxide [=] \\pu{0.2829 g}""},{""type"":""physical unit"",""value"":""mass [OF] water [=] \\pu{0.1159 g}""},{""type"":""other"",""value"":""Combusted""}]" "

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of carbon dioxide and 0.1159 g of water. What is the empirical formula for menthol?

" nan C10H20O "
+

Explanation:

+
+
+

We can calculate the masses of #""C""# and #""H""# from the masses of their oxides (#""CO""_2# and #""H""_2""O""#).

+

#""Mass of C"" = 0.2829 color(red)(cancel(color(black)(""g CO""_2))) × ""12.01 g C""/(44.01 color(red)(cancel(color(black)(""g CO""_2)))) = ""0.077 20 g C""#

+

#""Mass of H"" = 0.1159 color(red)(cancel(color(black)(""g H""_2""O""))) × ""2.016 g H""/(18.02 color(red)(cancel(color(black)(""g H""_2""O"")))) = ""0.012 97 g H""#

+

#""Mass of O"" = ""Mass of menthol - mass of C - mass of O"" = ""0.1005 g - 0.077 20 g - 0.012 97 g"" = ""0.01 033 g""#

+
+

Now, we must convert these masses to moles and find their ratios.

+

From here on, I like to summarize the calculations in a table.

+

#""Element""color(white)(Xll) ""Mass/g""color(white)(Xmll) ""Moles""color(white)(mml) ""Ratio"" color(white)(m)""Integers""#
+#stackrel(—————————————————-——)(color(white)(ll)""C"" color(white)(XXXml)""0.077 20"" color(white)(mll)""0.006 428"" +color(white)(Xlll)9.956color(white)(Xm)10#
+#color(white)(ll)""H"" color(white)(XXXXl)""0.012 97"" color(white)(mll)""0.012 87"" color(white)(mlll)19.93 color(white)(XXll)20#
+#color(white)(ll)""O"" color(white)(mmmml)""0.010 33""color(white)(mll)""0.000 6456""color(white)(ml)1color(white)(mmmmll)1#

+

The empirical formula is #""C""_10""H""_20""O""#.

+
+
" "
+
+
+

The empirical formula is #""C""_10""H""_20""O""#.

+
+
+
+

Explanation:

+
+
+

We can calculate the masses of #""C""# and #""H""# from the masses of their oxides (#""CO""_2# and #""H""_2""O""#).

+

#""Mass of C"" = 0.2829 color(red)(cancel(color(black)(""g CO""_2))) × ""12.01 g C""/(44.01 color(red)(cancel(color(black)(""g CO""_2)))) = ""0.077 20 g C""#

+

#""Mass of H"" = 0.1159 color(red)(cancel(color(black)(""g H""_2""O""))) × ""2.016 g H""/(18.02 color(red)(cancel(color(black)(""g H""_2""O"")))) = ""0.012 97 g H""#

+

#""Mass of O"" = ""Mass of menthol - mass of C - mass of O"" = ""0.1005 g - 0.077 20 g - 0.012 97 g"" = ""0.01 033 g""#

+
+

Now, we must convert these masses to moles and find their ratios.

+

From here on, I like to summarize the calculations in a table.

+

#""Element""color(white)(Xll) ""Mass/g""color(white)(Xmll) ""Moles""color(white)(mml) ""Ratio"" color(white)(m)""Integers""#
+#stackrel(—————————————————-——)(color(white)(ll)""C"" color(white)(XXXml)""0.077 20"" color(white)(mll)""0.006 428"" +color(white)(Xlll)9.956color(white)(Xm)10#
+#color(white)(ll)""H"" color(white)(XXXXl)""0.012 97"" color(white)(mll)""0.012 87"" color(white)(mlll)19.93 color(white)(XXll)20#
+#color(white)(ll)""O"" color(white)(mmmml)""0.010 33""color(white)(mll)""0.000 6456""color(white)(ml)1color(white)(mmmmll)1#

+

The empirical formula is #""C""_10""H""_20""O""#.

+
+
+
" "
+

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of carbon dioxide and 0.1159 g of water. What is the empirical formula for menthol?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 31, 2016 + +
+
+
+
+
+
+
+

The empirical formula is #""C""_10""H""_20""O""#.

+
+
+
+

Explanation:

+
+
+

We can calculate the masses of #""C""# and #""H""# from the masses of their oxides (#""CO""_2# and #""H""_2""O""#).

+

#""Mass of C"" = 0.2829 color(red)(cancel(color(black)(""g CO""_2))) × ""12.01 g C""/(44.01 color(red)(cancel(color(black)(""g CO""_2)))) = ""0.077 20 g C""#

+

#""Mass of H"" = 0.1159 color(red)(cancel(color(black)(""g H""_2""O""))) × ""2.016 g H""/(18.02 color(red)(cancel(color(black)(""g H""_2""O"")))) = ""0.012 97 g H""#

+

#""Mass of O"" = ""Mass of menthol - mass of C - mass of O"" = ""0.1005 g - 0.077 20 g - 0.012 97 g"" = ""0.01 033 g""#

+
+

Now, we must convert these masses to moles and find their ratios.

+

From here on, I like to summarize the calculations in a table.

+

#""Element""color(white)(Xll) ""Mass/g""color(white)(Xmll) ""Moles""color(white)(mml) ""Ratio"" color(white)(m)""Integers""#
+#stackrel(—————————————————-——)(color(white)(ll)""C"" color(white)(XXXml)""0.077 20"" color(white)(mll)""0.006 428"" +color(white)(Xlll)9.956color(white)(Xm)10#
+#color(white)(ll)""H"" color(white)(XXXXl)""0.012 97"" color(white)(mll)""0.012 87"" color(white)(mlll)19.93 color(white)(XXll)20#
+#color(white)(ll)""O"" color(white)(mmmml)""0.010 33""color(white)(mll)""0.000 6456""color(white)(ml)1color(white)(mmmmll)1#

+

The empirical formula is #""C""_10""H""_20""O""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 27073 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of carbon dioxide and 0.1159 g of water. What is the empirical formula for menthol? nan +267 a83255ee-6ddd-11ea-a42a-ccda262736ce https://socratic.org/questions/57ad66ab7c01497d4ab17bbb 7.02 start physical_unit 6 6 ph none qc_end physical_unit 10 11 8 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] solution""}]" "[{""type"":""physical unit"",""value"":""7.02""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] sodium hydroxide [=] \\pu{10^(−8) M}""}]" "

What is the pH of a solution of #10^(-8)# M sodium hydroxide ?

" nan 7.02 "
+

Explanation:

+
+

Right from the start, you should be able to predict that the pH of the solution will be higher than #7#. This is the case because the solution contains a strong base.

+

You could look at the concentration of the base and say that the pH will only be slightly higher than #7#, but it must come out to be higher than #7#.

+

The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to

+
+

#K_W = 10^(-14) -># water's ionization constant

+
+

In pure water, water undergoes self-ionization to form hydronium cations, #""H""_3""O""^(+)# and hydroxide anions, #""OH""^(-)#, as described by the following equilibrium reaction

+
+

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

By definition, the ionization constant is equal to

+
+

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#

+
+

If you take #x# to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

+
+

#K_w = x * x = x^2#

+
+

This gets you

+
+

#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#

+
+

So, the self-ionization of water produces

+
+

#[""H""_3""O""^(+)] = 10^(-7)""M ""# and #"" "" [""OH""^(-)] = 10^(-7)""M""#

+
+

at room temperature. Now, your solution contains sodium hydroxide, #""NaOH""#, a strong base that dissociates completely to produce hydroxide anions in a #1:1# mole ratio.

+

Therefore, you're adding

+
+

#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#

+
+

to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.

+

If you take #y# to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

+
+

#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)#

+

This time, the ionization constant will be equal to

+
+

#K_W = y * (10^(-8) + y)#

+

#K_w = y^2 + 10^(-8) * y#

+
+

This will get you

+
+

#y^2 + 10^(-8) * y - 10^(-14) = 0#

+
+

This quadratic equation will produce two solutions, one positive and one negative. Since #y# represents concentration, pick the positive one

+
+

#y = 9.51 * 10^(-8)#

+
+

This means that the equilibrium concentration of hydronium cations will be

+
+

#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#

+
+

The pH of the solution is given by

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

Plug in your value to find

+
+

#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#

+
+

As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.

+
+
" "
+
+
+

#""pH"" = 7.02#

+
+
+
+

Explanation:

+
+

Right from the start, you should be able to predict that the pH of the solution will be higher than #7#. This is the case because the solution contains a strong base.

+

You could look at the concentration of the base and say that the pH will only be slightly higher than #7#, but it must come out to be higher than #7#.

+

The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to

+
+

#K_W = 10^(-14) -># water's ionization constant

+
+

In pure water, water undergoes self-ionization to form hydronium cations, #""H""_3""O""^(+)# and hydroxide anions, #""OH""^(-)#, as described by the following equilibrium reaction

+
+

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

By definition, the ionization constant is equal to

+
+

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#

+
+

If you take #x# to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

+
+

#K_w = x * x = x^2#

+
+

This gets you

+
+

#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#

+
+

So, the self-ionization of water produces

+
+

#[""H""_3""O""^(+)] = 10^(-7)""M ""# and #"" "" [""OH""^(-)] = 10^(-7)""M""#

+
+

at room temperature. Now, your solution contains sodium hydroxide, #""NaOH""#, a strong base that dissociates completely to produce hydroxide anions in a #1:1# mole ratio.

+

Therefore, you're adding

+
+

#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#

+
+

to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.

+

If you take #y# to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

+
+

#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)#

+

This time, the ionization constant will be equal to

+
+

#K_W = y * (10^(-8) + y)#

+

#K_w = y^2 + 10^(-8) * y#

+
+

This will get you

+
+

#y^2 + 10^(-8) * y - 10^(-14) = 0#

+
+

This quadratic equation will produce two solutions, one positive and one negative. Since #y# represents concentration, pick the positive one

+
+

#y = 9.51 * 10^(-8)#

+
+

This means that the equilibrium concentration of hydronium cations will be

+
+

#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#

+
+

The pH of the solution is given by

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

Plug in your value to find

+
+

#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#

+
+

As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.

+
+
+
" "
+

What is the pH of a solution of #10^(-8)# M sodium hydroxide ?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Aug 12, 2016 + +
+
+
+
+
+
+
+

#""pH"" = 7.02#

+
+
+
+

Explanation:

+
+

Right from the start, you should be able to predict that the pH of the solution will be higher than #7#. This is the case because the solution contains a strong base.

+

You could look at the concentration of the base and say that the pH will only be slightly higher than #7#, but it must come out to be higher than #7#.

+

The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to

+
+

#K_W = 10^(-14) -># water's ionization constant

+
+

In pure water, water undergoes self-ionization to form hydronium cations, #""H""_3""O""^(+)# and hydroxide anions, #""OH""^(-)#, as described by the following equilibrium reaction

+
+

#2""H""_ 2""O""_ ((l)) rightleftharpoons ""H""_ 3""O""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

By definition, the ionization constant is equal to

+
+

#K_W = [""H""_3""O""^(+)] * [""OH""^(-)]#

+
+

If you take #x# to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

+
+

#K_w = x * x = x^2#

+
+

This gets you

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#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#

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So, the self-ionization of water produces

+
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#[""H""_3""O""^(+)] = 10^(-7)""M ""# and #"" "" [""OH""^(-)] = 10^(-7)""M""#

+
+

at room temperature. Now, your solution contains sodium hydroxide, #""NaOH""#, a strong base that dissociates completely to produce hydroxide anions in a #1:1# mole ratio.

+

Therefore, you're adding

+
+

#[""OH""^(-)] = [""NaOH""] = 10^(-8)""M""#

+
+

to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.

+

If you take #y# to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

+
+

#"" ""2""H""_ 2""O""_ ((l)) rightleftharpoons"" "" ""H""_ 3""O""_ ((aq))^(+) "" ""+"" "" "" """"OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)#

+

This time, the ionization constant will be equal to

+
+

#K_W = y * (10^(-8) + y)#

+

#K_w = y^2 + 10^(-8) * y#

+
+

This will get you

+
+

#y^2 + 10^(-8) * y - 10^(-14) = 0#

+
+

This quadratic equation will produce two solutions, one positive and one negative. Since #y# represents concentration, pick the positive one

+
+

#y = 9.51 * 10^(-8)#

+
+

This means that the equilibrium concentration of hydronium cations will be

+
+

#[""H""_3""O""^(+)] = 9.51 * 10^(-8)""M""#

+
+

The pH of the solution is given by

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

Plug in your value to find

+
+

#""pH"" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#

+
+

As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.

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+ + Aug 14, 2016 + +
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#sf(pH=7.02)#

+
+
+
+

Explanation:

+
+

This is a tiny concentration for NaOH so I would expect the pH to be close to 7 or slightly greater.

+

Because the concentration is so small, we must take into account the ions that are produced from the auto - ionisation of water:

+

#sf(H_2OrightleftharpoonsH^++OH^-)#

+

For which #sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)""mol""^2.""l""^(-2))# at #sf(25^@C)#.

+

This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are #sf(10^(-7)color(white)(x)""mol/l"")# respectively.

+

To get the total hydroxide concentration you might think you simply add #sf(10^(-7))# and #sf(10^(-8))#. This is not the case as these will not be equilibrium concentrations.

+

We need to apply Le Chatelier's Principle.

+

If we do a thought experiment we can imagine some pure water to which a tiny amount of sodium hydroxide is added such that the concentration of NaOH is #sf(10^-8color(white)(x)""mol/l""#.

+

We have now disturbed a system at equilibrium by adding extra #sf(OH^-)# ions. The system will act to oppose that change by reducing the number of #sf(OH^-)# ions and shifting to the left.

+

The reaction quotient #sf(Q)# is given by #sf([H^+][OH^-])#. This is now greater than #sf(K_w)# so the system responds by shifting the position of equilibrium to the left such that #sf(Q=K_w)#.

+

We can set up an ICE table using equilibrium concentrations to show this:

+

#"" ""sf(H_2O"" ""rightleftharpoons"" ""H^+"" ""+"" ""OH^-) +#

+

#sf(color(red)(I)color(white)(xxxxxxxxx)"" ""10^-7"" ""(10^(-7)+10^(-8))#

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#sf(color(red)(C)color(white)(xxxxxxx)"" ""-xcolor(white)(xxx)"" ""-x)#

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#sf(color(red)(E)color(white)(xxxxxxxx)"" ""(10^(-7)-x)"" ""(1.1xx10^(-7)-x))#

+

This gives us:

+

#sf((10^-7-x)(1.1xx10^(-7)-x)=10^(-14))#

+

If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for #sf(x)#. I won't go into that here but it gives, ignoring the -ve root:

+

#sf(x=0.49xx10^(-8))#

+

We can now get the equilibrium concentration of #sf([H^+])#:

+

#sf([H^+]=10^(-7)-(0.49xx10^(-8))=9.51xx10^(-8)color(white)(x)""mol/l"")#

+

#sf(pH=-log[H^+]=-log[9.51xx10^(-8)]=color(red)(7.02))#

+

This is an example of ""The Common Ion Effect"", hydroxide being the common ion in question.

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" What is the pH of a solution of #10^(-8)# M sodium hydroxide ? nan +268 a83255ef-6ddd-11ea-84aa-ccda262736ce https://socratic.org/questions/how-would-you-find-the-molar-enthalpy-of-vaporization-for-a-substance-given-3-21 8.85 kJ/mol start physical_unit 10 11 molar_enthalpy_of_vaporization kj/mol qc_end physical_unit 16 17 13 14 mole qc_end physical_unit 16 17 19 20 heat_energy qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Molar enthalpy of vaporization [OF] a substance [IN] kJ/mol""}]" "[{""type"":""physical unit"",""value"":""8.85 kJ/mol""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] the substance [=] \\pu{3.21 mol}""},{""type"":""physical unit"",""value"":""Energy absorbed [OF] the substance [=] \\pu{28.4 kJ}""},{""type"":""other"",""value"":""the substance changes from a liquid to a gas""}]" "

How would you find the molar enthalpy of vaporization for a substance, given 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when the substance changes from a liquid to a gas?

" nan 8.85 kJ/mol "
+

Explanation:

+
+

The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid #-># vapor phase change.

+

Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.

+

Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.

+

So, you know that you have #3.21# moles of a substance and that #""28.4 kJ""# of energy must be absorbed in order for the phase change to take place.

+

Mathematically, this can be expressed as

+
+

#color(blue)(q = n * DeltaH_""vap"") "" ""#, where

+
+

#q# - the amount of heat absorbed
+#n# - the number of moles of the substance
+#DeltaH_""vap""# - the molar enthalpy change of vaporization

+

Plug your values into this equation and rearrange to solve for #DeltaH_""vap""#

+
+

#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#

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#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#

+
+

Rounded to thre sig figs, the answer will be

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+

#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#

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" "
+
+
+

#""8.85 kJ/mol""#

+
+
+
+

Explanation:

+
+

The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid #-># vapor phase change.

+

Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.

+

Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.

+

So, you know that you have #3.21# moles of a substance and that #""28.4 kJ""# of energy must be absorbed in order for the phase change to take place.

+

Mathematically, this can be expressed as

+
+

#color(blue)(q = n * DeltaH_""vap"") "" ""#, where

+
+

#q# - the amount of heat absorbed
+#n# - the number of moles of the substance
+#DeltaH_""vap""# - the molar enthalpy change of vaporization

+

Plug your values into this equation and rearrange to solve for #DeltaH_""vap""#

+
+

#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#

+

#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#

+
+

Rounded to thre sig figs, the answer will be

+
+

#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#

+
+
+
+
" "
+

How would you find the molar enthalpy of vaporization for a substance, given 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when the substance changes from a liquid to a gas?

+
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+ + +Chemistry + + + + + +Thermochemistry + + + + + +Thermochemistry of Phase Changes + + +
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+1 Answer +
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+ + Nov 2, 2015 + +
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#""8.85 kJ/mol""#

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+
+
+

Explanation:

+
+

The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid #-># vapor phase change.

+

Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.

+

Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.

+

So, you know that you have #3.21# moles of a substance and that #""28.4 kJ""# of energy must be absorbed in order for the phase change to take place.

+

Mathematically, this can be expressed as

+
+

#color(blue)(q = n * DeltaH_""vap"") "" ""#, where

+
+

#q# - the amount of heat absorbed
+#n# - the number of moles of the substance
+#DeltaH_""vap""# - the molar enthalpy change of vaporization

+

Plug your values into this equation and rearrange to solve for #DeltaH_""vap""#

+
+

#q = n * DeltaH_""vap"" implies DeltaH_""vap"" = q/n#

+

#DeltaH_""vap"" = ""28.4 kJ""/""3.21 moles"" = ""8.8474 kJ/mol""#

+
+

Rounded to thre sig figs, the answer will be

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#DeltaH_""vap"" = color(green)(""8.85 kJ/mol"")#

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" How would you find the molar enthalpy of vaporization for a substance, given 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when the substance changes from a liquid to a gas? nan +269 a83255f0-6ddd-11ea-b3da-ccda262736ce https://socratic.org/questions/59b9f5e411ef6b3142c8e5ad C2H5O start chemical_formula qc_end c_other OTHER qc_end physical_unit 7 7 7 7 composition_by_mass qc_end physical_unit 8 8 8 8 composition_by_mass qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] organic sample [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""C2H5O""}]" "[{""type"":""other"",""value"":""An organic sample has C, H, and the balance was oxygen.""},{""type"":""physical unit"",""value"":""Composition by mass [OF] C [=] \\pu{53.31%}""},{""type"":""physical unit"",""value"":""Composition by mass [OF] H [=] \\pu{11.18%}""}]" "

An organic sample has composition by mass, #C:53.31%;H:11.18%;#, and the balance was oxygen. What is its empirical formula?

" nan C2H5O "
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Explanation:

+
+

We assume a #100*g# mass of compound....

+

And thus there are #(53.31*g)/(12.011*g*mol^-1)=4.44*mol# with respect to #C#....

+

And #(11.18*g)/(1.00794*g*mol^-1)=11.09*mol# with respect to #H#....

+

And #(35.51*g)/(16.00*g*mol^-1)=2.22*mol# with respect to #O#....

+

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of #C_2H_5O#.

+

This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is #C_4H_10O_2#. The question has not been drawn from actual data, and the values have been interpolated.

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" "
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#C_2H_5O#.......Are you sure you quoted the elemental compositon properly?

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Explanation:

+
+

We assume a #100*g# mass of compound....

+

And thus there are #(53.31*g)/(12.011*g*mol^-1)=4.44*mol# with respect to #C#....

+

And #(11.18*g)/(1.00794*g*mol^-1)=11.09*mol# with respect to #H#....

+

And #(35.51*g)/(16.00*g*mol^-1)=2.22*mol# with respect to #O#....

+

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of #C_2H_5O#.

+

This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is #C_4H_10O_2#. The question has not been drawn from actual data, and the values have been interpolated.

+
+
+
" "
+

An organic sample has composition by mass, #C:53.31%;H:11.18%;#, and the balance was oxygen. What is its empirical formula?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
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+1 Answer +
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#C_2H_5O#.......Are you sure you quoted the elemental compositon properly?

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+
+
+

Explanation:

+
+

We assume a #100*g# mass of compound....

+

And thus there are #(53.31*g)/(12.011*g*mol^-1)=4.44*mol# with respect to #C#....

+

And #(11.18*g)/(1.00794*g*mol^-1)=11.09*mol# with respect to #H#....

+

And #(35.51*g)/(16.00*g*mol^-1)=2.22*mol# with respect to #O#....

+

We divide thru by the lowest molar quantity, that of oxygen to get a trial empirical formula.....of #C_2H_5O#.

+

This organic formula is a bit whack, in that I think it contains a peroxo linkage (or it's a diether or diol); it must do if its MOLECULAR formula is #C_4H_10O_2#. The question has not been drawn from actual data, and the values have been interpolated.

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" An organic sample has composition by mass, #C:53.31%;H:11.18%;#, and the balance was oxygen. What is its empirical formula? nan +270 a8327d4a-6ddd-11ea-a9ac-ccda262736ce https://socratic.org/questions/what-would-be-the-formula-for-diphosphorus-pentoxide P2O5 start chemical_formula qc_end substance 6 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] diphosphorus pentoxide [IN] default""}]" "[{""type"":""chemical equation"",""value"":""P2O5""}]" "[{""type"":""substance name"",""value"":""Diphosphorus pentoxide""}]" "

What would be the formula for diphosphorus pentoxide?

" nan P2O5 "
+

Explanation:

+
+

#P_2O_5# is of course an empirical formula. The actual oxide is #P_4O_10#, with an adamantane structure.

+
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" "
+
+
+

#P_2O_5# as written, but this is an empirical formula.

+
+
+
+

Explanation:

+
+

#P_2O_5# is of course an empirical formula. The actual oxide is #P_4O_10#, with an adamantane structure.

+
+
+
" "
+

What would be the formula for diphosphorus pentoxide?

+
+
+ + +Chemistry + + + + + +Covalent Bonds + + + + + +Covalent Formulas and Nomenclature + + +
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+2 Answers +
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+ + May 11, 2016 + +
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#P_2O_5# as written, but this is an empirical formula.

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Explanation:

+
+

#P_2O_5# is of course an empirical formula. The actual oxide is #P_4O_10#, with an adamantane structure.

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+ + May 11, 2016 + +
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Here's my thought process:

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  • From name: phosphorus #-># #""P""#
    • +
  • ""di"" #-># #2#
    • +
  • From name: oxide #-># #""O""#
    • +
  • Anion stem: ""ox""
    • +
  • ""pent"" (instead of ""penta"", since ""ox"" starts with a vowel) #-># #5#
    • +
+

#-># #""P""_2""O""_5# (empirical formula)

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" What would be the formula for diphosphorus pentoxide? nan +271 a83280a6-6ddd-11ea-b76d-ccda262736ce https://socratic.org/questions/what-is-the-concentration-of-hcl-if-its-ph-is-3 1.00 × 10^(-3) M start physical_unit 5 5 concentration mol/l qc_end physical_unit 5 5 10 10 ph qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] HCl [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.00 × 10^(-3) M""}]" "[{""type"":""physical unit"",""value"":""pH [OF] HCl [=] \\pu{3}""}]" "

What is the concentration of HCl if its pH is 3?

" nan 1.00 × 10^(-3) M "
+

Explanation:

+
+

To find the concentration of something that dissociates into #H^+# and something else from the pH, we use the formula:

+

#10^(-pH)#

+

So this becomes:

+

#10^(-3) "" M""# #quad#(M stands for molar)

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" "
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#10^(-3) "" M""#

+
+
+
+

Explanation:

+
+

To find the concentration of something that dissociates into #H^+# and something else from the pH, we use the formula:

+

#10^(-pH)#

+

So this becomes:

+

#10^(-3) "" M""# #quad#(M stands for molar)

+
+
+
" "
+

What is the concentration of HCl if its pH is 3?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+2 Answers +
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#10^(-3) "" M""#

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Explanation:

+
+

To find the concentration of something that dissociates into #H^+# and something else from the pH, we use the formula:

+

#10^(-pH)#

+

So this becomes:

+

#10^(-3) "" M""# #quad#(M stands for molar)

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#10^-3 \ ""M""=0.001 \ ""M""#

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+
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Explanation:

+
+

The #""pH""# of an aqueous solution is given by the equation,

+

#""pH""=-log[H^+]#

+
+
    +
  • #[H^+]# is the hydrogen ion concentration in terms of molarity
    • +
+
+

The dissociation equation for hydrochloric acid is:

+

#HCl(aq)->H^+(aq)+Cl^(-)(aq)#

+

So, one mole of hydrochloric acid contains one mole of hydrogen ions.

+

Therefore, the hydrogen ion concentration will be,

+

#3=-log[H^+]#

+

#[H^+]=10^-3 \ ""M""#

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#=0.001 \ ""M""#

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" What is the concentration of HCl if its pH is 3? nan +272 a83280a7-6ddd-11ea-a0e9-ccda262736ce https://socratic.org/questions/lets-say-we-are-given-this-balanced-equation-cao-2hcl-cacl-2-h-2o-60-4g-of-cao-i 104.92 g start physical_unit 13 13 mass g qc_end chemical_equation 8 15 qc_end physical_unit 8 8 16 17 mass qc_end physical_unit 11 11 23 24 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] CaCl2 [IN] g""}]" "[{""type"":""physical unit"",""value"":""104.92 g""}]" "[{""type"":""chemical equation"",""value"":""CaO + 2 HCl -> CaCl2 + H2O""},{""type"":""physical unit"",""value"":""Mass [OF] CaO [=] \\pu{60.4 g}""},{""type"":""physical unit"",""value"":""Mass [OF] HCl [=] \\pu{69.0 g}""}]" "

Lets say we are given this balanced equation: #CaO + 2HCl -> CaCl_2 + H_2O#. 60.4g of #CaO# is reacted with 69.0g of #HCl#. How much #CaCl_2# is produced? + +

" nan 104.92 g "
+

Explanation:

+
+

Start by writing out the balanced chemical equation that describes your reaction

+
+

#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#

+
+

You need #color(red)(2)# moles of hydrochloric acid for every mole of calcium oxide that takes part in the reaction.

+

You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds

+
+

#M_(""M CaO"") = ""56.08 g mol""^(-1)#

+

#M_(""M HCl"") = ""36.46 g mol""^(-1)#

+
+

A #1:color(red)(2)# mole ratio will thus be equivalent to

+
+

#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio

+
+

So, you need #""72.92 g""# of hydrochloric acid for every #""56.08 g""# of calcium oxide that take part in the reaction.

+

You know that you have a sample of #""60.4 g""# of calcium oxide available. This many grams of calcium oxide would require

+
+

#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#

+
+

Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react

+
+

#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#

+
+

you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.

+

Use the molar mass of calcium chloride to convert the #color(red)(2):1# mole ratio that exists between hydrochloric acid and calcium chloride into a gram ratio

+
+

#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#

+
+

You will have

+
+

#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio

+
+

Since all the hydrochloric acid will react, you can say that the reaction will produce

+
+

#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""105 g CaCl""_2#

+
+
+
+

Explanation:

+
+

Start by writing out the balanced chemical equation that describes your reaction

+
+

#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#

+
+

You need #color(red)(2)# moles of hydrochloric acid for every mole of calcium oxide that takes part in the reaction.

+

You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds

+
+

#M_(""M CaO"") = ""56.08 g mol""^(-1)#

+

#M_(""M HCl"") = ""36.46 g mol""^(-1)#

+
+

A #1:color(red)(2)# mole ratio will thus be equivalent to

+
+

#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio

+
+

So, you need #""72.92 g""# of hydrochloric acid for every #""56.08 g""# of calcium oxide that take part in the reaction.

+

You know that you have a sample of #""60.4 g""# of calcium oxide available. This many grams of calcium oxide would require

+
+

#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#

+
+

Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react

+
+

#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#

+
+

you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.

+

Use the molar mass of calcium chloride to convert the #color(red)(2):1# mole ratio that exists between hydrochloric acid and calcium chloride into a gram ratio

+
+

#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#

+
+

You will have

+
+

#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio

+
+

Since all the hydrochloric acid will react, you can say that the reaction will produce

+
+

#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

Lets say we are given this balanced equation: #CaO + 2HCl -> CaCl_2 + H_2O#. 60.4g of #CaO# is reacted with 69.0g of #HCl#. How much #CaCl_2# is produced? + +

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Limiting Reagent + + +
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+1 Answer +
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+ + Jun 30, 2016 + +
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#""105 g CaCl""_2#

+
+
+
+

Explanation:

+
+

Start by writing out the balanced chemical equation that describes your reaction

+
+

#""CaO""_ ((s)) + color(red)(2)""HCl""_ ((aq)) -> ""CaCl""_ (2(aq)) + ""H""_ 2""O""_ ((l))#

+
+

You need #color(red)(2)# moles of hydrochloric acid for every mole of calcium oxide that takes part in the reaction.

+

You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds

+
+

#M_(""M CaO"") = ""56.08 g mol""^(-1)#

+

#M_(""M HCl"") = ""36.46 g mol""^(-1)#

+
+

A #1:color(red)(2)# mole ratio will thus be equivalent to

+
+

#(56.08 color(red)(cancel(color(black)(""g mol""^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)(""g mol""^(-1))))) = 56.08/72.92 -># gram ratio

+
+

So, you need #""72.92 g""# of hydrochloric acid for every #""56.08 g""# of calcium oxide that take part in the reaction.

+

You know that you have a sample of #""60.4 g""# of calcium oxide available. This many grams of calcium oxide would require

+
+

#60.4 color(red)(cancel(color(black)(""g CaO""))) * ""72.92 g HCl""/(56.08color(red)(cancel(color(black)(""g CaO"")))) = ""78.54 g HCl""#

+
+

Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react

+
+

#overbrace(""69.0 g"")^(color(blue)(""what you have"")) < overbrace(""78.54 g"")^(color(darkgreen)(""what you need""))#

+
+

you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.

+

Use the molar mass of calcium chloride to convert the #color(red)(2):1# mole ratio that exists between hydrochloric acid and calcium chloride into a gram ratio

+
+

#M_(""M CaCl""_2) = ""111.0 g mol""^(-1)#

+
+

You will have

+
+

#(color(red)(2) * 36.46 color(red)(cancel(color(black)(""g mol""^(-1)))))/(111.0color(red)(cancel(color(black)(""g mol""^(-1))))) = 72.92/111.0 -># gram ratio

+
+

Since all the hydrochloric acid will react, you can say that the reaction will produce

+
+

#69.0 color(red)(cancel(color(black)(""g HCl""))) * ""111.0 g CaCl""_2/(72.92color(red)(cancel(color(black)(""g HCl"")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(""105 g CaCl""_2)color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

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+
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" "Lets say we are given this balanced equation: #CaO + 2HCl -> CaCl_2 + H_2O#. 60.4g of #CaO# is reacted with 69.0g of #HCl#. How much #CaCl_2# is produced? + +" nan +273 a83280a8-6ddd-11ea-882a-ccda262736ce https://socratic.org/questions/0-040mol-of-nh4-2ni-so4-2-6h2o-is-dissolved-in-water-to-give-200cm3-of-aqueous-s 0.20 mol*dm^(–3) start physical_unit 22 23 concentration mol/dm^3 qc_end physical_unit 3 3 0 1 mole qc_end physical_unit 13 14 10 11 volume qc_end substance 7 7 qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] ammonium ions [IN] mol*dm^(–3)""}]" "[{""type"":""physical unit"",""value"":""0.20 mol*dm^(–3)""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] (NH4)2Ni(SO4)2.6H2O [=] \\pu{0.040 mol}""},{""type"":""physical unit"",""value"":""Volume [OF] aqueous solution [=] \\pu{200 cm^3}""},{""type"":""substance name"",""value"":""Water""}]" "

0.040mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200cm3 of aqueous solution. What is the concentration, in mol dm–3, of ammonium ions?

" nan 0.20 mol*dm^(–3) "
+

Explanation:

+
+

#""Concentration with respect to the nickel salt""#

+

#=(0.040*mol)/(0.200*dm^3)=0.20*mol*dm^-3#

+

But given the formulation of the salt, #(NH_4)_2[Ni +(SO_4)_2]*6H_2O#, there are 2 equiv of ammonium per equiv of salt. And thus #[NH_4^+]=0.40*mol*dm^-3#

+
+
" "
+
+
+

#""Concentration""=""Moles of solute""/""Volume of solution""......=0.40*mol*dm^-3.#

+
+
+
+

Explanation:

+
+

#""Concentration with respect to the nickel salt""#

+

#=(0.040*mol)/(0.200*dm^3)=0.20*mol*dm^-3#

+

But given the formulation of the salt, #(NH_4)_2[Ni +(SO_4)_2]*6H_2O#, there are 2 equiv of ammonium per equiv of salt. And thus #[NH_4^+]=0.40*mol*dm^-3#

+
+
+
" "
+

0.040mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200cm3 of aqueous solution. What is the concentration, in mol dm–3, of ammonium ions?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + May 11, 2017 + +
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#""Concentration""=""Moles of solute""/""Volume of solution""......=0.40*mol*dm^-3.#

+
+
+
+

Explanation:

+
+

#""Concentration with respect to the nickel salt""#

+

#=(0.040*mol)/(0.200*dm^3)=0.20*mol*dm^-3#

+

But given the formulation of the salt, #(NH_4)_2[Ni +(SO_4)_2]*6H_2O#, there are 2 equiv of ammonium per equiv of salt. And thus #[NH_4^+]=0.40*mol*dm^-3#

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" 0.040mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200cm3 of aqueous solution. What is the concentration, in mol dm–3, of ammonium ions? nan +274 a832a3f0-6ddd-11ea-b282-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-volume-of-oxygen-required-for-the-complete-combustion-o 0.50 dm^3 start physical_unit 7 7 volume dm^3 qc_end c_other OTHER qc_end physical_unit 17 17 14 15 volume qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [IN] dm^3""}]" "[{""type"":""physical unit"",""value"":""0.50 dm^3""}]" "[{""type"":""other"",""value"":""Complete combustion""},{""type"":""physical unit"",""value"":""Volume [OF] methane [=] \\pu{0.25 dm^3}""},{""type"":""other"",""value"":""STP""}]" "

How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP?

" nan 0.50 dm^3 "
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

+
+
" "
+
+
+

The volume of oxygen required is #""0.50 dm""^3#.

+
+
+
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

+
+
+
" "
+

How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
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+1 Answer +
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+ +
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+ + Sep 10, 2016 + +
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+

The volume of oxygen required is #""0.50 dm""^3#.

+
+
+
+

Explanation:

+
+
+

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

+

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

+
+

The balanced equation for the combustion is

+
+
+

#color(white)(l)""CH""_4 + ""2O""_2 → ""CO""_2 + ""2H""_2""O""#
+#""1 dm""^3color(white)(ll)""2 dm""^3#

+
+
+

According to Gay-Lussac, #""1 dm""^3color(white)(l) ""of CH""_4# requires #""2 dm""^3color(white)(l) ""of O""_2#.

+
+

∴ #""Volume of O""_2 = 0.25 color(red)(cancel(color(black)(""dm""^3 color(white)(l)""CH""_4))) × (""2 dm""^3color(white)(l) ""O""_2)/(1 color(red)(cancel(color(black)(""dm""^3color(white)(l) ""CH""_4)))) = ""0.50 dm""^3color(white)(l)""O""_2#

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" How do you calculate the volume of oxygen required for the complete combustion of 0.25 #dm^3# of methane at STP? nan +275 a832a3f1-6ddd-11ea-9953-ccda262736ce https://socratic.org/questions/how-would-you-complete-this-reaction-naoh-al-h-2 2 NaOH + 2 Al + 2 H2O -> 3 H2 + 2 NaAlO2 start chemical_equation qc_end chemical_equation 6 12 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this reaction""}]" "[{""type"":""chemical equation"",""value"":""2 NaOH + 2 Al + 2 H2O -> 3 H2 + 2 NaAlO2""}]" "[{""type"":""chemical equation"",""value"":""NaOH + Al -> H2 + ?""}]" "

How would you complete this reaction? #NaOH + Al -> H_2 + ?#

" nan 2 NaOH + 2 Al + 2 H2O -> 3 H2 + 2 NaAlO2 "
+

Explanation:

+
+

Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
+2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2

+
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" "
+
+
+

2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2

+
+
+
+

Explanation:

+
+

Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
+2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2

+
+
+
" "
+

How would you complete this reaction? #NaOH + Al -> H_2 + ?#

+
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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
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+1 Answer +
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+ + Dec 5, 2016 + +
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2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2

+
+
+
+

Explanation:

+
+

Since water is required for the complete reaction or if NaOH is taken in liquid [Lye form], H2O already exists. Hence reaction is as under :-
+2NaOH + 2Al + 2H2O = 2NaAlO2 [ Sodium aluminate] + 3 H2

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" How would you complete this reaction? #NaOH + Al -> H_2 + ?# nan +276 a832a3f2-6ddd-11ea-88cd-ccda262736ce https://socratic.org/questions/a-25-0-ml-sample-of-0-105-m-hcl-was-titrated-with-315-ml-of-naoh-what-is-the-con 8.33 × 10^(-3) M start physical_unit 14 14 concentration mol/l qc_end physical_unit 7 7 1 2 volume qc_end physical_unit 7 7 5 6 concentration qc_end physical_unit 14 14 11 12 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] NaOH [IN] M""}]" "[{""type"":""physical unit"",""value"":""8.33 × 10^(-3) M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] HCl [=] \\pu{25.0 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] HCl [=] \\pu{0.105 M}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{315 mL}""}]" "

A 25.0 mL sample of 0.105 M #HCl# was titrated with 315 mL of #NaOH#. What is the concentration of the #NaOH#?

" nan 8.33 × 10^(-3) M "
+

Explanation:

+
+

We're asked to find the molar concentration of the #""NaCl""# solution given some titration data.

+

Let's first write the chemical equation for this reaction:

+

#""NaOH""(aq) + ""HCl"" (aq) rarr ""NaCl"" (aq) + ""H""_2""O"" (l)#

+

Using the molarity equation, we can find the number of moles of #""HCl""# that reacted:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#""mol solute"" = (""molarity"")(""L soln"")#

+

#""mol HCl"" = (0.105""mol""/(cancel(""L"")))(0.0250cancel(""L"")) = 0.00263# #""mol HCl""#

+

(volume converted to liters)

+

Now, using the coefficients of the chemical reaction, we can determine the number of moles of #""NaOH""# that reacted:

+

#0.00263cancel(""mol HCl"")((1color(white)(l)""mol NaOH"")/(1cancel(""mol HCl""))) = 0.00263# #""mol NaOH""#

+

Lastly, we'll use the molarity equation (using given volume of #""NaOH soln""#) again to determine the molarity of the sodium hydroxide solution:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#M_ ""NaOH"" = (0.00263color(white)(l)""mol"")/(0.315color(white)(l)""L"") = color(red)(0.00833M#

+

(volume converted to liters)

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+
" "
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#0.00833M#

+
+
+
+

Explanation:

+
+

We're asked to find the molar concentration of the #""NaCl""# solution given some titration data.

+

Let's first write the chemical equation for this reaction:

+

#""NaOH""(aq) + ""HCl"" (aq) rarr ""NaCl"" (aq) + ""H""_2""O"" (l)#

+

Using the molarity equation, we can find the number of moles of #""HCl""# that reacted:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#""mol solute"" = (""molarity"")(""L soln"")#

+

#""mol HCl"" = (0.105""mol""/(cancel(""L"")))(0.0250cancel(""L"")) = 0.00263# #""mol HCl""#

+

(volume converted to liters)

+

Now, using the coefficients of the chemical reaction, we can determine the number of moles of #""NaOH""# that reacted:

+

#0.00263cancel(""mol HCl"")((1color(white)(l)""mol NaOH"")/(1cancel(""mol HCl""))) = 0.00263# #""mol NaOH""#

+

Lastly, we'll use the molarity equation (using given volume of #""NaOH soln""#) again to determine the molarity of the sodium hydroxide solution:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#M_ ""NaOH"" = (0.00263color(white)(l)""mol"")/(0.315color(white)(l)""L"") = color(red)(0.00833M#

+

(volume converted to liters)

+
+
+
" "
+

A 25.0 mL sample of 0.105 M #HCl# was titrated with 315 mL of #NaOH#. What is the concentration of the #NaOH#?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Titration Calculations + + +
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+1 Answer +
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+
+

#0.00833M#

+
+
+
+

Explanation:

+
+

We're asked to find the molar concentration of the #""NaCl""# solution given some titration data.

+

Let's first write the chemical equation for this reaction:

+

#""NaOH""(aq) + ""HCl"" (aq) rarr ""NaCl"" (aq) + ""H""_2""O"" (l)#

+

Using the molarity equation, we can find the number of moles of #""HCl""# that reacted:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#""mol solute"" = (""molarity"")(""L soln"")#

+

#""mol HCl"" = (0.105""mol""/(cancel(""L"")))(0.0250cancel(""L"")) = 0.00263# #""mol HCl""#

+

(volume converted to liters)

+

Now, using the coefficients of the chemical reaction, we can determine the number of moles of #""NaOH""# that reacted:

+

#0.00263cancel(""mol HCl"")((1color(white)(l)""mol NaOH"")/(1cancel(""mol HCl""))) = 0.00263# #""mol NaOH""#

+

Lastly, we'll use the molarity equation (using given volume of #""NaOH soln""#) again to determine the molarity of the sodium hydroxide solution:

+

#""molarity"" = ""mol solute""/""L soln""#

+

#M_ ""NaOH"" = (0.00263color(white)(l)""mol"")/(0.315color(white)(l)""L"") = color(red)(0.00833M#

+

(volume converted to liters)

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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" A 25.0 mL sample of 0.105 M #HCl# was titrated with 315 mL of #NaOH#. What is the concentration of the #NaOH#? nan +277 a832a3f3-6ddd-11ea-93b8-ccda262736ce https://socratic.org/questions/how-many-moles-of-a-gas-will-occupy-3-62-liters-at-stp 82.21 moles start physical_unit 4 5 mole mol qc_end physical_unit 4 5 8 9 volume qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] a gas [IN] moles""}]" "[{""type"":""physical unit"",""value"":""82.21 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] a gas [=] \\pu{3.62 liters}""},{""type"":""other"",""value"":""STP""}]" "

How many moles of a gas will occupy 3.62 liters at STP?

" nan 82.21 moles "
+

Explanation:

+
+

We first need to define #""STP""#. The most current definition of #""STP""# for gases is #""273.15 K""# and #1xx10^5 ""Pa""#.
+http://goldbook.iupac.org/S05910.html

+

Most people use #""100 kPa""# to avoid having to work with scientific notation or dealing with all those zeroes. #1xx10^5 ""Pa""##=##100""kPa""#

+

At #100""kPa""#, the molar volume of a gas is #""22.710981 mol/L""#.
+http://jfi.uchicago.edu/~ntt/formula/node197.html
+https://en.m.wikipedia.org/wiki/Molar_volume

+

To determine the moles of a gas, divide the given volume by the molar volume.

+

#3.62cancel""L""xx(22.710981""mol"")/(1cancel""L"")=""82.2 L""#

+
+
" "
+
+
+

At #""STP""#, #""3.62 L""# of an ideal gas would contain #""82.2 moles""#.

+
+
+
+

Explanation:

+
+

We first need to define #""STP""#. The most current definition of #""STP""# for gases is #""273.15 K""# and #1xx10^5 ""Pa""#.
+http://goldbook.iupac.org/S05910.html

+

Most people use #""100 kPa""# to avoid having to work with scientific notation or dealing with all those zeroes. #1xx10^5 ""Pa""##=##100""kPa""#

+

At #100""kPa""#, the molar volume of a gas is #""22.710981 mol/L""#.
+http://jfi.uchicago.edu/~ntt/formula/node197.html
+https://en.m.wikipedia.org/wiki/Molar_volume

+

To determine the moles of a gas, divide the given volume by the molar volume.

+

#3.62cancel""L""xx(22.710981""mol"")/(1cancel""L"")=""82.2 L""#

+
+
+
" "
+

How many moles of a gas will occupy 3.62 liters at STP?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 8, 2016 + +
+
+
+
+
+
+
+

At #""STP""#, #""3.62 L""# of an ideal gas would contain #""82.2 moles""#.

+
+
+
+

Explanation:

+
+

We first need to define #""STP""#. The most current definition of #""STP""# for gases is #""273.15 K""# and #1xx10^5 ""Pa""#.
+http://goldbook.iupac.org/S05910.html

+

Most people use #""100 kPa""# to avoid having to work with scientific notation or dealing with all those zeroes. #1xx10^5 ""Pa""##=##100""kPa""#

+

At #100""kPa""#, the molar volume of a gas is #""22.710981 mol/L""#.
+http://jfi.uchicago.edu/~ntt/formula/node197.html
+https://en.m.wikipedia.org/wiki/Molar_volume

+

To determine the moles of a gas, divide the given volume by the molar volume.

+

#3.62cancel""L""xx(22.710981""mol"")/(1cancel""L"")=""82.2 L""#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" How many moles of a gas will occupy 3.62 liters at STP? nan +278 a832cb52-6ddd-11ea-8e3b-ccda262736ce https://socratic.org/questions/5919e39111ef6b3efa82056d 9.107 × 10^(−34) kilograms start physical_unit 1 3 mass kg qc_end physical_unit 1 3 8 11 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass2 [OF] a single electron [IN] kilograms""}]" "[{""type"":""physical unit"",""value"":""9.107 × 10^(−34) kilograms""}]" "[{""type"":""physical unit"",""value"":""Mass1 [OF] a single electron [=] \\pu{9.107 × 10^(−31) g}""}]" "

If a single electron has a mass of #9.107xx10^-31*g#, what is the mass of an electron in kilograms?

" nan 9.107 × 10^(−34) kilograms "
+

Explanation:

+
+

Now #1*g-=10^-3*kg#

+

And so #9.107xx10^-28*cancelgxx10^-3*kg*cancel(g^-1)=#

+

#=9.107xx10^-31*kg#

+

All I have done is to use a #10^3# factor to replace the prefix #""kilo""#.

+
+
" "
+
+
+

#""Mass of electron""=9.107xx10^-31*kg..........#

+
+
+
+

Explanation:

+
+

Now #1*g-=10^-3*kg#

+

And so #9.107xx10^-28*cancelgxx10^-3*kg*cancel(g^-1)=#

+

#=9.107xx10^-31*kg#

+

All I have done is to use a #10^3# factor to replace the prefix #""kilo""#.

+
+
+
" "
+

If a single electron has a mass of #9.107xx10^-31*g#, what is the mass of an electron in kilograms?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 15, 2017 + +
+
+
+
+
+
+
+

#""Mass of electron""=9.107xx10^-31*kg..........#

+
+
+
+

Explanation:

+
+

Now #1*g-=10^-3*kg#

+

And so #9.107xx10^-28*cancelgxx10^-3*kg*cancel(g^-1)=#

+

#=9.107xx10^-31*kg#

+

All I have done is to use a #10^3# factor to replace the prefix #""kilo""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1055 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
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+
+
" If a single electron has a mass of #9.107xx10^-31*g#, what is the mass of an electron in kilograms? nan +279 a832cb53-6ddd-11ea-b9ab-ccda262736ce https://socratic.org/questions/calculate-how-many-moles-of-nh3-form-when-3-55-moles-of-n2h4-completely-react-ac 4.73 moles start physical_unit 5 5 mole mol qc_end chemical_equation 18 27 qc_end physical_unit 11 11 8 9 mole qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] NH3 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""4.73 moles""}]" "[{""type"":""chemical equation"",""value"":""3 N2H4 (l) -> 4 NH3 (g) + N2 (g)""},{""type"":""physical unit"",""value"":""Mole [OF] N2H4 [=] \\pu{3.55 moles}""},{""type"":""other"",""value"":""Completely react.""}]" "

Calculate how many moles of NH3 form when 3.55 moles of N2H4 completely react according to the equation: 3N2H4 (l) ---> 4NH3 (g) + N2 (g) ?

" nan 4.73 moles "
+

Explanation:

+
+

What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!

+

The steps to solving mole-mole problems like this are

+
    +
  • +

    write the balanced chemical equation for the reaction (this is given)

    +
    • +
  • +

    divide the number of moles of the given known substance (#3.55#) by that substance's coefficient in the chemical equation (#3#)

    +
    • +
  • +

    multiply that number by the coefficient of the substance you're trying to find (#4#)

    +
    • +
+

Using simple dimensional analysis, it looks like this:

+

#3.55cancel(""mol N""_2""H""_4)(larr ""given amount"")((4""mol NH""_3(""coefficient""))/(3cancel(""mol N""_2""H""_4)(""coefficient"")))#

+

#= color(red)(4.73# #color(red)(""mol NH""_3#

+

rounded to #3# significant figures, the amount given in the problem.

+

Thus, if the reaction goes to completion, #3.55# moles of #""N""_2""H""_4# will yield #color(red)(4.73# moles of #""NH""_3#.

+
+
" "
+
+
+

#4.73# #""mol NH""_3#

+
+
+
+

Explanation:

+
+

What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!

+

The steps to solving mole-mole problems like this are

+
    +
  • +

    write the balanced chemical equation for the reaction (this is given)

    +
    • +
  • +

    divide the number of moles of the given known substance (#3.55#) by that substance's coefficient in the chemical equation (#3#)

    +
    • +
  • +

    multiply that number by the coefficient of the substance you're trying to find (#4#)

    +
    • +
+

Using simple dimensional analysis, it looks like this:

+

#3.55cancel(""mol N""_2""H""_4)(larr ""given amount"")((4""mol NH""_3(""coefficient""))/(3cancel(""mol N""_2""H""_4)(""coefficient"")))#

+

#= color(red)(4.73# #color(red)(""mol NH""_3#

+

rounded to #3# significant figures, the amount given in the problem.

+

Thus, if the reaction goes to completion, #3.55# moles of #""N""_2""H""_4# will yield #color(red)(4.73# moles of #""NH""_3#.

+
+
+
" "
+

Calculate how many moles of NH3 form when 3.55 moles of N2H4 completely react according to the equation: 3N2H4 (l) ---> 4NH3 (g) + N2 (g) ?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 15, 2017 + +
+
+
+
+
+
+
+

#4.73# #""mol NH""_3#

+
+
+
+

Explanation:

+
+

What we're doing here is calculating basic mole-mole relationships, something that you'll be doing quite a bit!

+

The steps to solving mole-mole problems like this are

+
    +
  • +

    write the balanced chemical equation for the reaction (this is given)

    +
    • +
  • +

    divide the number of moles of the given known substance (#3.55#) by that substance's coefficient in the chemical equation (#3#)

    +
    • +
  • +

    multiply that number by the coefficient of the substance you're trying to find (#4#)

    +
    • +
+

Using simple dimensional analysis, it looks like this:

+

#3.55cancel(""mol N""_2""H""_4)(larr ""given amount"")((4""mol NH""_3(""coefficient""))/(3cancel(""mol N""_2""H""_4)(""coefficient"")))#

+

#= color(red)(4.73# #color(red)(""mol NH""_3#

+

rounded to #3# significant figures, the amount given in the problem.

+

Thus, if the reaction goes to completion, #3.55# moles of #""N""_2""H""_4# will yield #color(red)(4.73# moles of #""NH""_3#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 37589 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" Calculate how many moles of NH3 form when 3.55 moles of N2H4 completely react according to the equation: 3N2H4 (l) ---> 4NH3 (g) + N2 (g) ? nan +280 a832cb54-6ddd-11ea-bb36-ccda262736ce https://socratic.org/questions/how-do-i-calculate-molality-to-percent-by-mass 25.74% start physical_unit 12 12 mass_percent none qc_end physical_unit 12 13 10 11 molality qc_end end "[{""type"":""physical unit"",""value"":""Percent by mass [OF] RbNO3(aq)""}]" "[{""type"":""physical unit"",""value"":""25.74%""}]" "[{""type"":""physical unit"",""value"":""Molality [OF] RbNO3(aq) solution [=] \\pu{2.35 mol/kg}""}]" "

How do I calculate molality to percent by mass?

" "
+
+

+

Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass.

+

+
+
" 25.74% "
+

Explanation:

+
+

Your goal here is to figure out the number of grams of solute present for every #""100 g""# of the solution, i.e. the solution's percent concentration by mass, #""% m/m""#.

+

Now, you know that the solution has a molality equal to #""2.35 mol kg""^(-1)#. This tells you that this solution contains #2.35# moles of rubidium nitrate, the solute, for every #""1 kg""# of water, the solvent.

+

To make the calculations easier, pick a sample of this solution that contains exactly #""1 kg"" = 10^3 quad ""g""# of water. As we've said, this sample will also contain #2.35# moles of rubidium nitrate.

+

Use the molar mass of the solute to convert the number of moles to grams

+
+

#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#

+
+

This means that the total mass of the sample is equal to

+
+

#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#

+
+

So, you know that you have #""346.56 g""# of rubidium nitrate in #""1346.56 g""# of the solution, so you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#

+
+

This means that the solution's percent concentration by mass is equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))#

+
+

This tells you that you get #""25.7 g""# of rubidium nitrate for every #""100 g""# of the solution.

+
+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#25.7%#

+
+
+
+

Explanation:

+
+

Your goal here is to figure out the number of grams of solute present for every #""100 g""# of the solution, i.e. the solution's percent concentration by mass, #""% m/m""#.

+

Now, you know that the solution has a molality equal to #""2.35 mol kg""^(-1)#. This tells you that this solution contains #2.35# moles of rubidium nitrate, the solute, for every #""1 kg""# of water, the solvent.

+

To make the calculations easier, pick a sample of this solution that contains exactly #""1 kg"" = 10^3 quad ""g""# of water. As we've said, this sample will also contain #2.35# moles of rubidium nitrate.

+

Use the molar mass of the solute to convert the number of moles to grams

+
+

#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#

+
+

This means that the total mass of the sample is equal to

+
+

#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#

+
+

So, you know that you have #""346.56 g""# of rubidium nitrate in #""1346.56 g""# of the solution, so you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#

+
+

This means that the solution's percent concentration by mass is equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))#

+
+

This tells you that you get #""25.7 g""# of rubidium nitrate for every #""100 g""# of the solution.

+
+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

How do I calculate molality to percent by mass?

+
+
+

+

Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass.

+

+
+
+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 21, 2018 + +
+
+
+
+
+
+
+

#25.7%#

+
+
+
+

Explanation:

+
+

Your goal here is to figure out the number of grams of solute present for every #""100 g""# of the solution, i.e. the solution's percent concentration by mass, #""% m/m""#.

+

Now, you know that the solution has a molality equal to #""2.35 mol kg""^(-1)#. This tells you that this solution contains #2.35# moles of rubidium nitrate, the solute, for every #""1 kg""# of water, the solvent.

+

To make the calculations easier, pick a sample of this solution that contains exactly #""1 kg"" = 10^3 quad ""g""# of water. As we've said, this sample will also contain #2.35# moles of rubidium nitrate.

+

Use the molar mass of the solute to convert the number of moles to grams

+
+

#2.35 color(red)(cancel(color(black)(""moles RbNO""_3))) * ""147.473 g""/(1color(red)(cancel(color(black)(""mole RbNO""_3)))) = ""346.56 g""#

+
+

This means that the total mass of the sample is equal to

+
+

#""346.56 g "" + quad 10^3 quad ""g"" = ""1346.56 g""#

+
+

So, you know that you have #""346.56 g""# of rubidium nitrate in #""1346.56 g""# of the solution, so you can say that #""100 g""# of this solution will contain

+
+

#100 color(red)(cancel(color(black)(""g solution""))) * ""346.56 g RbNO""_3/(1346.56 color(red)(cancel(color(black)(""g solution"")))) = ""25.7 g RbNO""_3#

+
+

This means that the solution's percent concentration by mass is equal to

+
+

#color(darkgreen)(ul(color(black)(""% m/m"" = ""25.7% RbNO""_3)))#

+
+

This tells you that you get #""25.7 g""# of rubidium nitrate for every #""100 g""# of the solution.

+
+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 46124 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do I calculate molality to percent by mass? " + + +Convert 2.35 mol/kg RbNO3 (aq) solution to percent by mass. + + +" +281 a832f1fe-6ddd-11ea-87f8-ccda262736ce https://socratic.org/questions/588a715a7c01496394c345fd 2400.00 J start physical_unit 13 13 heat_energy j qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 13 13 25 28 specific_heat qc_end physical_unit 13 13 15 16 temperature qc_end end "[{""type"":""physical unit"",""value"":""Heat needed [OF] foam [IN] J""}]" "[{""type"":""physical unit"",""value"":""2400.00 J""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] foam [=] \\pu{1 kg}""},{""type"":""physical unit"",""value"":""Specific heat [OF] foam [=] \\pu{1200 J/(kg * K)}""},{""type"":""physical unit"",""value"":""Temperature increased [OF] foam [=] \\pu{2 K}""}]" "

How much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""# knowing that the specific heat of foam is #""1200 J kg""^(-1)""K""^(-1)# ?

" nan 2400.00 J "
+

Explanation:

+
+

The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually #""1 g""#, by #1^@""C""# or by #""1 K""#.

+

In your case, the specific heat of foam is given as #""1200 J kg""^(-1)""K""^(-1)#, which means that one unit of mass is #""1 kg""#. You can thus say that in order to increase the temperature of #""1 kg""# of foam by #""1 K""#, you must provide it with #""1200 J""# of heat.

+

Now, you must figure out how much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""#.

+

In this case, you know that you need #""1200 J""# to increase its temperature by #""1 K""#, so you can say that another #""1200 J""# will increase its temperature by #""1 K""# again.

+

In other words, you will need

+
+

#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#

+
+

I'll leave the answer rounded to two sig figs.

+
+
" "
+
+
+

#""2400 J""#

+
+
+
+

Explanation:

+
+

The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually #""1 g""#, by #1^@""C""# or by #""1 K""#.

+

In your case, the specific heat of foam is given as #""1200 J kg""^(-1)""K""^(-1)#, which means that one unit of mass is #""1 kg""#. You can thus say that in order to increase the temperature of #""1 kg""# of foam by #""1 K""#, you must provide it with #""1200 J""# of heat.

+

Now, you must figure out how much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""#.

+

In this case, you know that you need #""1200 J""# to increase its temperature by #""1 K""#, so you can say that another #""1200 J""# will increase its temperature by #""1 K""# again.

+

In other words, you will need

+
+

#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#

+
+

I'll leave the answer rounded to two sig figs.

+
+
+
" "
+

How much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""# knowing that the specific heat of foam is #""1200 J kg""^(-1)""K""^(-1)# ?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 27, 2017 + +
+
+
+
+
+
+
+

#""2400 J""#

+
+
+
+

Explanation:

+
+

The specific heat of a substance tells you how much heat is needed in order to increase the temperature of one unit of mass of that substance, usually #""1 g""#, by #1^@""C""# or by #""1 K""#.

+

In your case, the specific heat of foam is given as #""1200 J kg""^(-1)""K""^(-1)#, which means that one unit of mass is #""1 kg""#. You can thus say that in order to increase the temperature of #""1 kg""# of foam by #""1 K""#, you must provide it with #""1200 J""# of heat.

+

Now, you must figure out how much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""#.

+

In this case, you know that you need #""1200 J""# to increase its temperature by #""1 K""#, so you can say that another #""1200 J""# will increase its temperature by #""1 K""# again.

+

In other words, you will need

+
+

#2 color(red)(cancel(color(black)(""K""))) * overbrace(""1200 J""/(1color(red)(cancel(color(black)(""K"")))))^(color(blue)(""needed for 1 kg of foam"")) = color(darkgreen)(ul(color(black)(""2400 J"")))#

+
+

I'll leave the answer rounded to two sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
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+
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+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
+
+
+
" "How much heat is needed to increase the temperature of #""1 kg""# of foam by #""2 K""# knowing that the specific heat of foam is #""1200 J kg""^(-1)""K""^(-1)# ?" nan +282 a832f1ff-6ddd-11ea-9cf2-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-3-25-moles-of-nano-3-in-250-ml- 13.00 mol/L start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 6 6 14 15 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] NaNO3 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""13.00 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] NaNO3 [=] \\pu{3.25 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{250 mL}""}]" "

What is the molarity of a solution that contains 3.25 moles of #NaNO_3# in 250 mL of solution?

" nan 13.00 mol/L "
+

Explanation:

+
+

#""Molarity"" = ""Moles of solute (moles)""/""Volume of solution (L)""#.

+

For your problem, #""3.25 moles sodium nitrate""/(250xx10^-3*L)#

+

#=# #??*mol*L^-1#

+

The general aqueous solubility of nitrate species allows such a high concentration to be achieved.

+
+
" "
+
+
+

#> ""10 moles per litre""#

+
+
+
+

Explanation:

+
+

#""Molarity"" = ""Moles of solute (moles)""/""Volume of solution (L)""#.

+

For your problem, #""3.25 moles sodium nitrate""/(250xx10^-3*L)#

+

#=# #??*mol*L^-1#

+

The general aqueous solubility of nitrate species allows such a high concentration to be achieved.

+
+
+
" "
+

What is the molarity of a solution that contains 3.25 moles of #NaNO_3# in 250 mL of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 25, 2016 + +
+
+
+
+
+
+
+

#> ""10 moles per litre""#

+
+
+
+

Explanation:

+
+

#""Molarity"" = ""Moles of solute (moles)""/""Volume of solution (L)""#.

+

For your problem, #""3.25 moles sodium nitrate""/(250xx10^-3*L)#

+

#=# #??*mol*L^-1#

+

The general aqueous solubility of nitrate species allows such a high concentration to be achieved.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
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+
+ 8568 views + around the world +
+
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+ + Creative Commons License + +
+
+
+
+
+
+
" What is the molarity of a solution that contains 3.25 moles of #NaNO_3# in 250 mL of solution? nan +283 a832f200-6ddd-11ea-b0ff-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-0-080-moles-nahco-3-in-1250-ml-of-solution 0.064 mol/L start physical_unit 7 7 molarity mol/l qc_end physical_unit 7 7 5 6 mole qc_end physical_unit 12 12 9 10 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] NaHCO3 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.064 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] NaHCO3 [=] \\pu{0.080 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{1250 mL}""}]" "

What is the molarity of 0.080 moles #NaHCO_3# in 1250 mL of solution?

" nan 0.064 mol/L "
+

Explanation:

+
+

#""Molarity""# #=# #""Amount of substance (moles)""/""Volume of solution (L)""#

+

#=# #(0.080*mol)/(1.250*L)# #=# #0.064""*mol*L^-1# with respect to sodium bicarbonate.

+
+
" "
+
+
+

#""Molarity = 0.064""*mol*L^-1#

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Amount of substance (moles)""/""Volume of solution (L)""#

+

#=# #(0.080*mol)/(1.250*L)# #=# #0.064""*mol*L^-1# with respect to sodium bicarbonate.

+
+
+
" "
+

What is the molarity of 0.080 moles #NaHCO_3# in 1250 mL of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 10, 2016 + +
+
+
+
+
+
+
+

#""Molarity = 0.064""*mol*L^-1#

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Amount of substance (moles)""/""Volume of solution (L)""#

+

#=# #(0.080*mol)/(1.250*L)# #=# #0.064""*mol*L^-1# with respect to sodium bicarbonate.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1380 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the molarity of 0.080 moles #NaHCO_3# in 1250 mL of solution? nan +284 a832f201-6ddd-11ea-a1a5-ccda262736ce https://socratic.org/questions/how-many-oxygen-atoms-are-represented-by-mgso-4-7h-2o 11 start physical_unit 2 3 number none qc_end chemical_equation 7 7 qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] oxygen atoms""}]" "[{""type"":""physical unit"",""value"":""11""}]" "[{""type"":""chemical equation"",""value"":""MgSO4.7H2O""}]" "

How many oxygen atoms are represented by #MgSO_4 * 7H_2O#?

" nan 11 "
+

Explanation:

+
+

In #""MgSO""_4#, the number of oxygen atoms is represented by the subscript directly next to the #""O""#.

+

In #""7H""_2""O""#, the coefficient #7# can be multiplied throughout each element in the molecule.

+

#""MgS""color(red)(""O""_4)rarr""4 O atoms""#

+

#color(red)(""7"")""H""_2color(red)(""O"")rarr""7 O atoms""#

+

There is a total of #4+7=11# oxygen atoms.

+
+
" "
+
+
+

#11#

+
+
+
+

Explanation:

+
+

In #""MgSO""_4#, the number of oxygen atoms is represented by the subscript directly next to the #""O""#.

+

In #""7H""_2""O""#, the coefficient #7# can be multiplied throughout each element in the molecule.

+

#""MgS""color(red)(""O""_4)rarr""4 O atoms""#

+

#color(red)(""7"")""H""_2color(red)(""O"")rarr""7 O atoms""#

+

There is a total of #4+7=11# oxygen atoms.

+
+
+
" "
+

How many oxygen atoms are represented by #MgSO_4 * 7H_2O#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 20, 2015 + +
+
+
+
+
+
+
+

#11#

+
+
+
+

Explanation:

+
+

In #""MgSO""_4#, the number of oxygen atoms is represented by the subscript directly next to the #""O""#.

+

In #""7H""_2""O""#, the coefficient #7# can be multiplied throughout each element in the molecule.

+

#""MgS""color(red)(""O""_4)rarr""4 O atoms""#

+

#color(red)(""7"")""H""_2color(red)(""O"")rarr""7 O atoms""#

+

There is a total of #4+7=11# oxygen atoms.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5909 views + around the world +
+
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+ + Creative Commons License + +
+
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+
+
+
+
" How many oxygen atoms are represented by #MgSO_4 * 7H_2O#? nan +285 a832f202-6ddd-11ea-a248-ccda262736ce https://socratic.org/questions/58dd5f6411ef6b3a98181106 CrCl3 start chemical_formula qc_end physical_unit 6 7 5 5 mass_percent qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] chromium chloride salt [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""CrCl3""}]" "[{""type"":""physical unit"",""value"":""Percentage by mass [OF] metal content [=] \\pu{32.8%}""}]" "

A chromium chloride salt contains #32.8%# metal content by mass. What is its empirical formula?

" nan CrCl3 "
+

Explanation:

+
+

In all problems of this sort, it is convenient to assume a #100*g# mass of compound, and then work out the individual molar quantities of each element:

+

#""Moles of chromium""=(32.8*g)/(52.0*g*mol^-1)=0.631*mol#

+

#""Moles of chlorine""=(67.2*g)/(35.45*g*mol^-1)=1.90*mol#

+

We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of #CrCl_3#.

+
+
" "
+
+
+

We assume a #100*g# mass of compound, and get an empirical formula of #CrCl_3#.

+
+
+
+

Explanation:

+
+

In all problems of this sort, it is convenient to assume a #100*g# mass of compound, and then work out the individual molar quantities of each element:

+

#""Moles of chromium""=(32.8*g)/(52.0*g*mol^-1)=0.631*mol#

+

#""Moles of chlorine""=(67.2*g)/(35.45*g*mol^-1)=1.90*mol#

+

We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of #CrCl_3#.

+
+
+
" "
+

A chromium chloride salt contains #32.8%# metal content by mass. What is its empirical formula?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 30, 2017 + +
+
+
+
+
+
+
+

We assume a #100*g# mass of compound, and get an empirical formula of #CrCl_3#.

+
+
+
+

Explanation:

+
+

In all problems of this sort, it is convenient to assume a #100*g# mass of compound, and then work out the individual molar quantities of each element:

+

#""Moles of chromium""=(32.8*g)/(52.0*g*mol^-1)=0.631*mol#

+

#""Moles of chlorine""=(67.2*g)/(35.45*g*mol^-1)=1.90*mol#

+

We divide thru by the SMALLEST molar quantity, that of the metal, to get an empirical formula of #CrCl_3#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2562 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A chromium chloride salt contains #32.8%# metal content by mass. What is its empirical formula? nan +286 a832f203-6ddd-11ea-b528-ccda262736ce https://socratic.org/questions/how-many-atoms-are-in-1-0-mole-of-o-2 6.02 × 10^23 start physical_unit 2 2 number none qc_end physical_unit 8 8 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] atoms""}]" "[{""type"":""physical unit"",""value"":""6.02 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] O2 [=] \\pu{1.0 mole}""}]" "

How many atoms are in 1.0 mole of #O_2#?

" nan 6.02 × 10^23 "
+

Explanation:

+
+

#1# #""mole""# of stuff specifies #""Avogadro's number, ""N_A,# #6.022xx10^23# individual items of stuff.

+

Here I have #N_A xx O_2# molecules. Clearly, there are #2xxN_A# oxygen atoms. What is the mass of this number of oxygen atoms?

+
+
" "
+
+
+

This question is the same as if your were asked, #""how many gloves are there in 1 dozen pairs of gloves?""#
+Can you answer that?

+
+
+
+

Explanation:

+
+

#1# #""mole""# of stuff specifies #""Avogadro's number, ""N_A,# #6.022xx10^23# individual items of stuff.

+

Here I have #N_A xx O_2# molecules. Clearly, there are #2xxN_A# oxygen atoms. What is the mass of this number of oxygen atoms?

+
+
+
" "
+

How many atoms are in 1.0 mole of #O_2#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 5, 2016 + +
+
+
+
+
+
+
+

This question is the same as if your were asked, #""how many gloves are there in 1 dozen pairs of gloves?""#
+Can you answer that?

+
+
+
+

Explanation:

+
+

#1# #""mole""# of stuff specifies #""Avogadro's number, ""N_A,# #6.022xx10^23# individual items of stuff.

+

Here I have #N_A xx O_2# molecules. Clearly, there are #2xxN_A# oxygen atoms. What is the mass of this number of oxygen atoms?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1204 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many atoms are in 1.0 mole of #O_2#? nan +287 a832f204-6ddd-11ea-b673-ccda262736ce https://socratic.org/questions/an-aqueous-solution-containing-1-00-g-of-bovine-insulin-a-protein-not-ionized-pe 6.00 × 10^3 g/mol start physical_unit 7 8 molar_mass g/mol qc_end physical_unit 1 2 13 14 volume qc_end physical_unit 7 8 4 5 mass qc_end physical_unit 1 2 20 22 osmotic_pressure qc_end physical_unit 1 2 24 25 temperature qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] bovine insulin [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""6.00 × 10^3 g/mol""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] aqueous solution [=] \\pu{per liter}""},{""type"":""physical unit"",""value"":""mass [OF] bovine insulin [=] \\pu{1.00 g}""},{""type"":""physical unit"",""value"":""Osmotic pressure [OF] aqueous solution [=] \\pu{3.1 mm Hg}""},{""type"":""physical unit"",""value"":""Temperature [OF] aqueous solution [=] \\pu{25 ℃}""}]" "

An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. How do you calculate the molar mass of bovine insulin?

" nan 6.00 × 10^3 g/mol "
+

Explanation:

+
+
+

The formula for osmotic pressure #Pi# is

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#

+
+
+

where

+

#c =# the molar concentration of the solute
+#R =# the universal gas constant
+#T =# the Kelvin temperature of the solution

+

Since #c = ""moles""/""litres"" = n/V#, we can write

+
+
+

#Pi = (nRT)/V#

+
+
+

Since #n =""mass""/""molar mass""= m/M#, we can write

+
+
+

#Pi = (nRT)/(MV)#

+
+
+

We can solve this equation for the molar mass and get

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#

+
+
+

In this problem

+

#m = ""1.00 g""#
+#R = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
+#T = ""25 °C = 298.15 K""#
+#Pi = 3.1 color(red)(cancel(color(black)(""mmHg""))) × ""1 atm""/(760 color(red)(cancel(color(black)(""mmHg"")))) = ""0.004 08 atm""#
+#V = ""1 L""#

+

∴ #M = (""1.00 g"" × ""0.082 06"" color(red)(cancel(color(black)(""L·atm·K""^""-1"")))""mol""^""-1"" × 298.15 color(red)(cancel(color(black)(""K""))))/(""0.00 408""color(red)(cancel(color(black)(""atm""))) × 1 color(red)(cancel(color(black)(""L"")))) = 6.0 × 10^3color(white)(l)""g/mol""#

+
+
" "
+
+
+

The molar mass is #6.0 ×10^3color(white)(l)""g/mol""#.

+
+
+
+

Explanation:

+
+
+

The formula for osmotic pressure #Pi# is

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#

+
+
+

where

+

#c =# the molar concentration of the solute
+#R =# the universal gas constant
+#T =# the Kelvin temperature of the solution

+

Since #c = ""moles""/""litres"" = n/V#, we can write

+
+
+

#Pi = (nRT)/V#

+
+
+

Since #n =""mass""/""molar mass""= m/M#, we can write

+
+
+

#Pi = (nRT)/(MV)#

+
+
+

We can solve this equation for the molar mass and get

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#

+
+
+

In this problem

+

#m = ""1.00 g""#
+#R = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
+#T = ""25 °C = 298.15 K""#
+#Pi = 3.1 color(red)(cancel(color(black)(""mmHg""))) × ""1 atm""/(760 color(red)(cancel(color(black)(""mmHg"")))) = ""0.004 08 atm""#
+#V = ""1 L""#

+

∴ #M = (""1.00 g"" × ""0.082 06"" color(red)(cancel(color(black)(""L·atm·K""^""-1"")))""mol""^""-1"" × 298.15 color(red)(cancel(color(black)(""K""))))/(""0.00 408""color(red)(cancel(color(black)(""atm""))) × 1 color(red)(cancel(color(black)(""L"")))) = 6.0 × 10^3color(white)(l)""g/mol""#

+
+
+
" "
+

An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. How do you calculate the molar mass of bovine insulin?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Osmolarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 31, 2018 + +
+
+
+
+
+
+
+

The molar mass is #6.0 ×10^3color(white)(l)""g/mol""#.

+
+
+
+

Explanation:

+
+
+

The formula for osmotic pressure #Pi# is

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))"" ""#

+
+
+

where

+

#c =# the molar concentration of the solute
+#R =# the universal gas constant
+#T =# the Kelvin temperature of the solution

+

Since #c = ""moles""/""litres"" = n/V#, we can write

+
+
+

#Pi = (nRT)/V#

+
+
+

Since #n =""mass""/""molar mass""= m/M#, we can write

+
+
+

#Pi = (nRT)/(MV)#

+
+
+

We can solve this equation for the molar mass and get

+
+
+

#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))"" ""#

+
+
+

In this problem

+

#m = ""1.00 g""#
+#R = ""0.082 06 L·atm·K""^""-1""""mol""^""-1""#
+#T = ""25 °C = 298.15 K""#
+#Pi = 3.1 color(red)(cancel(color(black)(""mmHg""))) × ""1 atm""/(760 color(red)(cancel(color(black)(""mmHg"")))) = ""0.004 08 atm""#
+#V = ""1 L""#

+

∴ #M = (""1.00 g"" × ""0.082 06"" color(red)(cancel(color(black)(""L·atm·K""^""-1"")))""mol""^""-1"" × 298.15 color(red)(cancel(color(black)(""K""))))/(""0.00 408""color(red)(cancel(color(black)(""atm""))) × 1 color(red)(cancel(color(black)(""L"")))) = 6.0 × 10^3color(white)(l)""g/mol""#

+
+
+
+
+
+ +
+
+
+
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" An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. How do you calculate the molar mass of bovine insulin? nan +288 a832f205-6ddd-11ea-b057-ccda262736ce https://socratic.org/questions/a-certain-mass-of-water-was-heated-with-41-840-joules-raising-its-temperature-fr 1540.67 grams start physical_unit 24 25 mass g qc_end physical_unit 24 25 14 15 temperature qc_end physical_unit 24 25 17 18 temperature qc_end physical_unit 24 25 8 9 heat_energy qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] the water [IN] grams""}]" "[{""type"":""physical unit"",""value"":""1540.67 grams""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] the water [=] \\pu{22.0 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the water [=] \\pu{28.5 ℃}""},{""type"":""physical unit"",""value"":""Added heat [OF] the water [=] \\pu{41,840 Joules}""}]" "

A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5°C. What is the mass of the water, in grams?

" nan 1540.67 grams "
+

Explanation:

+
+

Water specific heat is 4.178 J/g-°K.
+The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)

+

41840J = 4.178 J/g-°K * Xg * deltaT

+

Xg = 41840J / (4.178 * 6.5) = 1541g water

+
+
" "
+
+
+

1541g

+
+
+
+

Explanation:

+
+

Water specific heat is 4.178 J/g-°K.
+The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)

+

41840J = 4.178 J/g-°K * Xg * deltaT

+

Xg = 41840J / (4.178 * 6.5) = 1541g water

+
+
+
" "
+

A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5°C. What is the mass of the water, in grams?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 31, 2016 + +
+
+
+
+
+
+
+

1541g

+
+
+
+

Explanation:

+
+

Water specific heat is 4.178 J/g-°K.
+The temperature change is 28.5 – 22.0 = 6.5 (because it is a difference, °C or °K doesn't matter)

+

41840J = 4.178 J/g-°K * Xg * deltaT

+

Xg = 41840J / (4.178 * 6.5) = 1541g water

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5°C. What is the mass of the water, in grams? nan +289 a8331b88-6ddd-11ea-9016-ccda262736ce https://socratic.org/questions/what-is-the-concentration-if-358-g-of-nano-3-is-present-in-555-ml-of-water 7.59 × 10^(-3) mol/L start physical_unit 8 8 concentration mol/l qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 15 15 12 13 volume qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] NaNO3 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""7.59 × 10^(-3) mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaNO3 [=] \\pu{0.358 g}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{555 mL}""}]" "

What is the concentration if #.358# #g# of #NaNO_3# is present in #555# #mL# of water?

" nan 7.59 × 10^(-3) mol/L "
+

Explanation:

+
+

Concentration here is simply (amount of substance)/(volume of solution).

+

So #(0.358*cancel(g))/(85.00*cancel(g)*mol^-1)# #xx# #1/(555xx10^(-3)*L)# #=# #??*mol*L^-1#

+
+
" "
+
+
+

#7.59xx10^(-3) mol*L^-1# in sodium nitrate

+
+
+
+

Explanation:

+
+

Concentration here is simply (amount of substance)/(volume of solution).

+

So #(0.358*cancel(g))/(85.00*cancel(g)*mol^-1)# #xx# #1/(555xx10^(-3)*L)# #=# #??*mol*L^-1#

+
+
+
" "
+

What is the concentration if #.358# #g# of #NaNO_3# is present in #555# #mL# of water?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 8, 2015 + +
+
+
+
+
+
+
+

#7.59xx10^(-3) mol*L^-1# in sodium nitrate

+
+
+
+

Explanation:

+
+

Concentration here is simply (amount of substance)/(volume of solution).

+

So #(0.358*cancel(g))/(85.00*cancel(g)*mol^-1)# #xx# #1/(555xx10^(-3)*L)# #=# #??*mol*L^-1#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1328 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What is the concentration if #.358# #g# of #NaNO_3# is present in #555# #mL# of water? nan +290 a8331b89-6ddd-11ea-a70c-ccda262736ce https://socratic.org/questions/how-many-grams-of-mgcl-2-would-be-required-to-produce-a-4-5-m-solution-with-a-vo 856.90 grams start physical_unit 4 4 mass g qc_end physical_unit 13 13 18 19 volume qc_end physical_unit 4 4 11 12 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] MgCl2 [IN] grams""}]" "[{""type"":""physical unit"",""value"":""856.90 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{2.0 L}""},{""type"":""physical unit"",""value"":""Molarity [OF] MgCl2 [=] \\pu{4.5 M}""}]" "

How many grams of #MgCl_2# would be required to produce a 4.5 M solution with a volume of 2.0 L?

" nan 856.90 grams "
+

Explanation:

+
+

For starters, we know that a #""1-M""# solution contains #1# mole of solute for every #""1.0 L""# of solution.

+

This implies that a #""4.5-M""# magnesium chloride solution will contain #4.5# moles of magnesium chloride, the solute, for every #""1.0 L""# of solution.

+

So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in #""2.0 L""# of solution to #4.5# moles in #'1.0 L""# of solution.

+
+

#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#

+
+

Use cross multiplication to get

+
+

#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#

+
+

This will be equal to

+
+

#color(red)(?) = ""9.0 moles MgCl""_2#

+
+

So, you know that you can get a #""4.5-M""# magnesium chloride solution by dissolving #9.0# moles of magnesium chloride in enough water to have a total volume of #""2.0 L""# of solution.

+

To convert this to grams, use the compound's molar mass

+
+

#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
" "
+
+
+

#""860 g MgCl""_2#

+
+
+
+

Explanation:

+
+

For starters, we know that a #""1-M""# solution contains #1# mole of solute for every #""1.0 L""# of solution.

+

This implies that a #""4.5-M""# magnesium chloride solution will contain #4.5# moles of magnesium chloride, the solute, for every #""1.0 L""# of solution.

+

So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in #""2.0 L""# of solution to #4.5# moles in #'1.0 L""# of solution.

+
+

#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#

+
+

Use cross multiplication to get

+
+

#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#

+
+

This will be equal to

+
+

#color(red)(?) = ""9.0 moles MgCl""_2#

+
+

So, you know that you can get a #""4.5-M""# magnesium chloride solution by dissolving #9.0# moles of magnesium chloride in enough water to have a total volume of #""2.0 L""# of solution.

+

To convert this to grams, use the compound's molar mass

+
+

#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
+
" "
+

How many grams of #MgCl_2# would be required to produce a 4.5 M solution with a volume of 2.0 L?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 7, 2017 + +
+
+
+
+
+
+
+

#""860 g MgCl""_2#

+
+
+
+

Explanation:

+
+

For starters, we know that a #""1-M""# solution contains #1# mole of solute for every #""1.0 L""# of solution.

+

This implies that a #""4.5-M""# magnesium chloride solution will contain #4.5# moles of magnesium chloride, the solute, for every #""1.0 L""# of solution.

+

So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in #""2.0 L""# of solution to #4.5# moles in #'1.0 L""# of solution.

+
+

#(color(red)(?)color(white)(.)""moles MgCl""_2)/""2.0 L solution"" = overbrace(""4.5 moles MgCl""_2/(""1.0 L solution""))^(color(blue)(""= 4.5 M solution""))#

+
+

Use cross multiplication to get

+
+

#color(red)(?) = (2.0 color(red)(cancel(color(black)(""L solution""))))/(1.0color(red)(cancel(color(black)(""L solution"")))) * ""4.5 moles MgCl""_2#

+
+

This will be equal to

+
+

#color(red)(?) = ""9.0 moles MgCl""_2#

+
+

So, you know that you can get a #""4.5-M""# magnesium chloride solution by dissolving #9.0# moles of magnesium chloride in enough water to have a total volume of #""2.0 L""# of solution.

+

To convert this to grams, use the compound's molar mass

+
+

#9.0 color(red)(cancel(color(black)(""moles MgCl""_2))) * ""95.211 g""/(1color(red)(cancel(color(black)(""mole MgCl""_2)))) = color(darkgreen)(ul(color(black)(""860 g"")))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for your values.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 48966 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many grams of #MgCl_2# would be required to produce a 4.5 M solution with a volume of 2.0 L? nan +291 a8331b8a-6ddd-11ea-83ca-ccda262736ce https://socratic.org/questions/what-is-the-initial-pressure-of-a-gas-having-an-initial-temperature-of-90-5k-an- 9.32 atm start physical_unit 6 7 pressure atm qc_end physical_unit 7 7 13 14 temperature qc_end physical_unit 7 7 19 20 volume qc_end physical_unit 7 7 24 25 pressure qc_end physical_unit 7 7 30 31 temperature qc_end physical_unit 7 7 37 38 volume qc_end end "[{""type"":""physical unit"",""value"":""Pressure1 [OF] a gas [IN] atm""}]" "[{""type"":""physical unit"",""value"":""9.32 atm""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] gas [=] \\pu{90.5 K}""},{""type"":""physical unit"",""value"":""Volume1 [OF] gas [=] \\pu{40.3 L}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] gas [=] \\pu{0.83 atm}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] gas [=] \\pu{0.54 K}""},{""type"":""physical unit"",""value"":""Volume2 [OF] gas [=] \\pu{2.7 L}""}]" "

What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

" nan 9.32 atm "
+

Explanation:

+
+

Use the combined gas law. #(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Given
+#V_1=""40.3 L""#
+#T_1=""90.5 K""#
+#P_2=""0.83 atm""#
+#V_2=""2.7 L""#
+#T_2=""0.54 K""#

+

Unknown
+#P_1#

+

Equation
+#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Solution
+Rearrange the equation to isolate #P_1# and solve.

+

#P_1=(P_2V_2T_1)/(T_2V_1)#

+

#P_1=((0.83""atm"" xx2.7cancel""L""xx90.5cancel""K""))/((0.54cancel""K""xx40.3cancel""L""))=""9.3 atm""#

+
+
" "
+
+
+

The initial volume is #""9.3 atm""#.

+
+
+
+

Explanation:

+
+

Use the combined gas law. #(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Given
+#V_1=""40.3 L""#
+#T_1=""90.5 K""#
+#P_2=""0.83 atm""#
+#V_2=""2.7 L""#
+#T_2=""0.54 K""#

+

Unknown
+#P_1#

+

Equation
+#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Solution
+Rearrange the equation to isolate #P_1# and solve.

+

#P_1=(P_2V_2T_1)/(T_2V_1)#

+

#P_1=((0.83""atm"" xx2.7cancel""L""xx90.5cancel""K""))/((0.54cancel""K""xx40.3cancel""L""))=""9.3 atm""#

+
+
+
" "
+

What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Combined Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 5, 2015 + +
+
+
+
+
+
+
+

The initial volume is #""9.3 atm""#.

+
+
+
+

Explanation:

+
+

Use the combined gas law. #(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Given
+#V_1=""40.3 L""#
+#T_1=""90.5 K""#
+#P_2=""0.83 atm""#
+#V_2=""2.7 L""#
+#T_2=""0.54 K""#

+

Unknown
+#P_1#

+

Equation
+#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Solution
+Rearrange the equation to isolate #P_1# and solve.

+

#P_1=(P_2V_2T_1)/(T_2V_1)#

+

#P_1=((0.83""atm"" xx2.7cancel""L""xx90.5cancel""K""))/((0.54cancel""K""xx40.3cancel""L""))=""9.3 atm""#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
Impact of this question
+
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+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
" What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L? nan +292 a8331b8b-6ddd-11ea-8e0a-ccda262736ce https://socratic.org/questions/how-do-you-balance-this-equation-mno-2-al-mn-al-2o-3 3 MnO2 + 4 Al -> 3 Mn + 2 Al2O3 start chemical_equation qc_end chemical_equation 6 12 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this equation""}]" "[{""type"":""chemical equation"",""value"":""3 MnO2 + 4 Al -> 3 Mn + 2 Al2O3""}]" "[{""type"":""chemical equation"",""value"":""MnO2 + Al -> Mn + Al2O3""}]" "

How do you balance this equation? #MnO_2 + Al -> Mn + Al_2O_3#

" nan 3 MnO2 + 4 Al -> 3 Mn + 2 Al2O3 "
+

Explanation:

+
+

Since we have a multiple of #3# oxygens on the right hand side we require a multiplier of #3# on the left hand side.

+

If we have #3MnO_2# on the left hand side, then that is a total of #3# manganeses and #6# oxygens, requiring #3Mn+2Al_2O_3# on the right hand side.

+

Then we need #4# aluminiums on the left hand side.

+

Hence the balanced form of the equation is:

+
+

#3MnO_2+4Al -> 3Mn+2Al_2O_3#

+
+
+
" "
+
+
+

#3MnO_2+4Al -> 3Mn+2Al_2O_3#

+
+
+
+

Explanation:

+
+

Since we have a multiple of #3# oxygens on the right hand side we require a multiplier of #3# on the left hand side.

+

If we have #3MnO_2# on the left hand side, then that is a total of #3# manganeses and #6# oxygens, requiring #3Mn+2Al_2O_3# on the right hand side.

+

Then we need #4# aluminiums on the left hand side.

+

Hence the balanced form of the equation is:

+
+

#3MnO_2+4Al -> 3Mn+2Al_2O_3#

+
+
+
+
" "
+

How do you balance this equation? #MnO_2 + Al -> Mn + Al_2O_3#

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 7, 2016 + +
+
+
+
+
+
+
+

#3MnO_2+4Al -> 3Mn+2Al_2O_3#

+
+
+
+

Explanation:

+
+

Since we have a multiple of #3# oxygens on the right hand side we require a multiplier of #3# on the left hand side.

+

If we have #3MnO_2# on the left hand side, then that is a total of #3# manganeses and #6# oxygens, requiring #3Mn+2Al_2O_3# on the right hand side.

+

Then we need #4# aluminiums on the left hand side.

+

Hence the balanced form of the equation is:

+
+

#3MnO_2+4Al -> 3Mn+2Al_2O_3#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1098 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
+
" How do you balance this equation? #MnO_2 + Al -> Mn + Al_2O_3# nan +293 a833401e-6ddd-11ea-ac9c-ccda262736ce https://socratic.org/questions/a-gas-sample-containing-only-so2-pf3-and-co-has-the-following-mass-percentage-29 150.00 torr start physical_unit 8 8 partial_pressure torr qc_end c_other OTHER qc_end physical_unit 5 5 14 15 mass_percent qc_end physical_unit 6 6 17 18 mass_percent qc_end physical_unit 2 2 38 39 total_pressure qc_end end "[{""type"":""physical unit"",""value"":""Partial pressure [OF] CO [IN] torr""}]" "[{""type"":""physical unit"",""value"":""150.00 torr""}]" "[{""type"":""other"",""value"":""A gas sample contains only SO2, PF3, and CO.""},{""type"":""physical unit"",""value"":""Mass percentage [OF] SO2 [=] \\pu{29.1 %}""},{""type"":""physical unit"",""value"":""Mass percentage [OF] PF3 [=] \\pu{61.8 %}""},{""type"":""physical unit"",""value"":""Total pressure [OF] sample [=] \\pu{684 torr}""}]" "

A gas sample containing only SO2, PF3, and CO has the following mass percentage: 29.1 % SO2, 61.8 % PF3. How would you calculate the partial pressure in torr, of CO if the total pressure of sample is 684 torr?

" nan 150.00 torr "
+

Explanation:

+
+

The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.

+

This means that in order to find the partial pressure of carbon monoxide, you need to find

+
+
    +
  • the number of moles of carbon dioxide found in the mixture
    • +
  • the total number of moles present in the mixture
    • +
+
+

So, you know that the percent composition of the mixture is #29.1%# #""SO""_2# and #""61.8%# #""PF""_3#. Since the mixture only contains three gases, you can say that the percent composition of carbon monoxide will be equal to

+
+

#100% - (29.1% + 61.8%) = 9.1%#

+
+

Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a #""100.0-g""# sample of this mixture.

+

Using the percentages given, you can say that this sample will contain

+
+
    +
  • #""29.1 g SO""_2#
    • +
  • #""61.8 g PF""_3#
    • +
  • #""9.1 g CO""#
    • +
+
+

Use the respective molar masses of the three gases to find their number of moles

+
+

#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#

+

#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#

+

#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#

+
+

The total number of moles present in the mixture will be

+
+

#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#

+
+

Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture

+
+

#chi_(CO) = n_(CO)/n_""total""#

+

#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#

+
+

The partial pressure of carbon monoxide will thus be

+
+

#P_(CO) = chi_(CO) xx P_""total""#

+

#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#P_(CO) = ""150. torr""#

+
+
+
+

Explanation:

+
+

The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.

+

This means that in order to find the partial pressure of carbon monoxide, you need to find

+
+
    +
  • the number of moles of carbon dioxide found in the mixture
    • +
  • the total number of moles present in the mixture
    • +
+
+

So, you know that the percent composition of the mixture is #29.1%# #""SO""_2# and #""61.8%# #""PF""_3#. Since the mixture only contains three gases, you can say that the percent composition of carbon monoxide will be equal to

+
+

#100% - (29.1% + 61.8%) = 9.1%#

+
+

Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a #""100.0-g""# sample of this mixture.

+

Using the percentages given, you can say that this sample will contain

+
+
    +
  • #""29.1 g SO""_2#
    • +
  • #""61.8 g PF""_3#
    • +
  • #""9.1 g CO""#
    • +
+
+

Use the respective molar masses of the three gases to find their number of moles

+
+

#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#

+

#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#

+

#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#

+
+

The total number of moles present in the mixture will be

+
+

#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#

+
+

Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture

+
+

#chi_(CO) = n_(CO)/n_""total""#

+

#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#

+
+

The partial pressure of carbon monoxide will thus be

+
+

#P_(CO) = chi_(CO) xx P_""total""#

+

#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

A gas sample containing only SO2, PF3, and CO has the following mass percentage: 29.1 % SO2, 61.8 % PF3. How would you calculate the partial pressure in torr, of CO if the total pressure of sample is 684 torr?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 1, 2015 + +
+
+
+
+
+
+
+

#P_(CO) = ""150. torr""#

+
+
+
+

Explanation:

+
+

The idea here is that the partial pressure exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction.

+

This means that in order to find the partial pressure of carbon monoxide, you need to find

+
+
    +
  • the number of moles of carbon dioxide found in the mixture
    • +
  • the total number of moles present in the mixture
    • +
+
+

So, you know that the percent composition of the mixture is #29.1%# #""SO""_2# and #""61.8%# #""PF""_3#. Since the mixture only contains three gases, you can say that the percent composition of carbon monoxide will be equal to

+
+

#100% - (29.1% + 61.8%) = 9.1%#

+
+

Your strategy now will be to pick a sample of this mixture and determine the mass of each gas it contains. To make the calculations easier, let's say that you have a #""100.0-g""# sample of this mixture.

+

Using the percentages given, you can say that this sample will contain

+
+
    +
  • #""29.1 g SO""_2#
    • +
  • #""61.8 g PF""_3#
    • +
  • #""9.1 g CO""#
    • +
+
+

Use the respective molar masses of the three gases to find their number of moles

+
+

#""For SO""_2:"" "" 29.1 color(red)(cancel(color(black)(""g""))) * ""1 mole SO""_2/(64.06color(red)(cancel(color(black)(""g"")))) = ""0.4543 moles SO""_2#

+

#""For PF""_3: "" "" 61.8 color(red)(cancel(color(black)(""g""))) * ""1 mole PF""_3/(87.97color(red)(cancel(color(black)(""g"")))) = ""0.7025 moles PF""_3#

+

#""For CO"": "" "" 9.1color(red)(cancel(color(black)(""g""))) * ""1 mole CO""/(28.01color(red)(cancel(color(black)(""g"")))) = ""0.3249 moles CO""#

+
+

The total number of moles present in the mixture will be

+
+

#n_""total"" = 0.4543 + 0.7025 + 0.3249 = ""1.4817 moles""#

+
+

Now, the mole fraction of carbon monoxide will be equal to the number of moles of carbon monoxide divided by the total number of moles present in the mixture

+
+

#chi_(CO) = n_(CO)/n_""total""#

+

#chi_(CO) = ( 0.3249 color(red)(cancel(color(black)(""moles""))))/(1.4817 color(red)(cancel(color(black)(""moles"")))) = 0.2193#

+
+

The partial pressure of carbon monoxide will thus be

+
+

#P_(CO) = chi_(CO) xx P_""total""#

+

#P_(CO) = 0.2193 * ""684 torr"" = color(green)(""150. torr"")#

+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
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" A gas sample containing only SO2, PF3, and CO has the following mass percentage: 29.1 % SO2, 61.8 % PF3. How would you calculate the partial pressure in torr, of CO if the total pressure of sample is 684 torr? nan +294 a833401f-6ddd-11ea-9ba5-ccda262736ce https://socratic.org/questions/if-the-theoretical-yield-of-a-reaction-is-145-g-and-the-actual-yield-is-0-104-g- 71.72% start physical_unit 5 6 percent_yield none qc_end physical_unit 5 6 8 9 theoretical_yield qc_end physical_unit 5 6 15 16 actual_yield qc_end end "[{""type"":""physical unit"",""value"":""Percent yield [OF] a reaction""}]" "[{""type"":""physical unit"",""value"":""71.72%""}]" "[{""type"":""physical unit"",""value"":""Theoretical yield [OF] a reaction [=] \\pu{0.145 g}""},{""type"":""physical unit"",""value"":""Actual yield [OF] a reaction [=] \\pu{0.104 g}""}]" "

If the theoretical yield of a reaction is .145 g and the actual yield is 0.104 g, what is the percent yield?

" nan 71.72% "
+

Explanation:

+
+

A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical #""100 g""# of product.

+

Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by #100#.

+
+

#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#

+
+

In your case, the reaction is said to have a theoretical yield of #""0.145 g""#. This means that if all the moles of reactants that take part in the reaction are converted to moles of product, the reaction will produce #""0.145 g""# of product.

+

The actual yield of the reaction is said to #""0.104 g""#, which means that the reaction fails to reach the theoretical mass of product formed, i.e. it doesn't have a #100%# yield.

+

The percent yield of the reaction will be

+
+

#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs

+
+

This means that for every #""100 g""# of product that can theoretically be produced by the reaction, you only get #""71.7 g""# of product.

+

+ +

+
+
" "
+
+
+

#71.7%#

+
+
+
+

Explanation:

+
+

A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical #""100 g""# of product.

+

Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by #100#.

+
+

#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#

+
+

In your case, the reaction is said to have a theoretical yield of #""0.145 g""#. This means that if all the moles of reactants that take part in the reaction are converted to moles of product, the reaction will produce #""0.145 g""# of product.

+

The actual yield of the reaction is said to #""0.104 g""#, which means that the reaction fails to reach the theoretical mass of product formed, i.e. it doesn't have a #100%# yield.

+

The percent yield of the reaction will be

+
+

#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs

+
+

This means that for every #""100 g""# of product that can theoretically be produced by the reaction, you only get #""71.7 g""# of product.

+

+ +

+
+
+
" "
+

If the theoretical yield of a reaction is .145 g and the actual yield is 0.104 g, what is the percent yield?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 4, 2016 + +
+
+
+
+
+
+
+

#71.7%#

+
+
+
+

Explanation:

+
+

A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical #""100 g""# of product.

+

Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by #100#.

+
+

#color(blue)(|bar(ul(color(blue)(""% yield"" = ""what you actually get""/""what you should theoretically get"" xx 100color(white)(a/a)|)))#

+
+

In your case, the reaction is said to have a theoretical yield of #""0.145 g""#. This means that if all the moles of reactants that take part in the reaction are converted to moles of product, the reaction will produce #""0.145 g""# of product.

+

The actual yield of the reaction is said to #""0.104 g""#, which means that the reaction fails to reach the theoretical mass of product formed, i.e. it doesn't have a #100%# yield.

+

The percent yield of the reaction will be

+
+

#""% yield"" = (0.104 color(red)(cancel(color(black)(""g""))))/(0.145color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""71.7%color(white)(a/a)|))) -># rounded to three sig figs

+
+

This means that for every #""100 g""# of product that can theoretically be produced by the reaction, you only get #""71.7 g""# of product.

+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+
" If the theoretical yield of a reaction is .145 g and the actual yield is 0.104 g, what is the percent yield? nan +295 a8334020-6ddd-11ea-a35d-ccda262736ce https://socratic.org/questions/howe-would-you-calculate-the-number-of-moles-of-cl-atoms-in-3-61x10-24-formula-u 5.99 moles start physical_unit 9 10 mole mol qc_end physical_unit 15 19 12 14 number qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Cl atoms [IN] moles""}]" "[{""type"":""physical unit"",""value"":""5.99 moles""}]" "[{""type"":""physical unit"",""value"":""Number [OF] formula units of magnesium chloride [=] \\pu{3.61 × 10^24}""}]" "

Howe would you calculate the number of moles of Cl atoms in #3.61x10^24# formula units of magnesium chloride, #MgCl_2#?

" nan 5.99 moles "
+

Explanation:

+
+

Each formula unit of magnesium chloride, #""MgCl""_2""#, contains two chlorine, #""Cl""#, atoms.

+

Multiply the number of formula units by the conversion factor

+

#(2color(white)(.)""atoms Cl"")/(1color(white)(.)""formula unit MgCl""_2#.

+

#3.61xx10^24color(red)cancel(color(black)(""formula units MgCl""_2))xx(2color(white)(.)""atoms Cl"")/(1color(red)cancel(color(black)(""formula units MgCl""_2)))=7.22xx10^24""Cl atoms""#

+
+
" "
+
+
+

Use Avogadro’s number to calculate the number of moles of #MgCl_2#. The number of moles of Cl is twice as much, because the ratio of Cl in #MgCl_2# to #MgCl_2# is 2:1.

+

#(3.61*10^24)/(6.022*10^23) = 5.99#
+Therefore, there are 12 moles of Cl.

+
+
+
" "
+

Howe would you calculate the number of moles of Cl atoms in #3.61x10^24# formula units of magnesium chloride, #MgCl_2#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Mar 26, 2017 + +
+
+
+
+
+
+
+

Use Avogadro’s number to calculate the number of moles of #MgCl_2#. The number of moles of Cl is twice as much, because the ratio of Cl in #MgCl_2# to #MgCl_2# is 2:1.

+

#(3.61*10^24)/(6.022*10^23) = 5.99#
+Therefore, there are 12 moles of Cl.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Mar 26, 2017 + +
+
+
+
+
+
+
+

There are #7.22xx10^24# atoms #""Cl""# in #3.61xx10^24# formula units of #""MgCl""_2""#.

+
+
+
+

Explanation:

+
+

Each formula unit of magnesium chloride, #""MgCl""_2""#, contains two chlorine, #""Cl""#, atoms.

+

Multiply the number of formula units by the conversion factor

+

#(2color(white)(.)""atoms Cl"")/(1color(white)(.)""formula unit MgCl""_2#.

+

#3.61xx10^24color(red)cancel(color(black)(""formula units MgCl""_2))xx(2color(white)(.)""atoms Cl"")/(1color(red)cancel(color(black)(""formula units MgCl""_2)))=7.22xx10^24""Cl atoms""#

+
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+ +
+
+
+
+
+
+
Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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+
" Howe would you calculate the number of moles of Cl atoms in #3.61x10^24# formula units of magnesium chloride, #MgCl_2#? nan +296 a8336724-6ddd-11ea-82b1-ccda262736ce https://socratic.org/questions/the-temperature-of-50-0-g-of-water-was-raised-to-50-0-c-by-the-addition-of-1-0-k 45.22 °C start physical_unit 27 28 temperature °c qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 6 6 10 11 temperature qc_end physical_unit 6 6 16 17 heat_energy qc_end end "[{""type"":""physical unit"",""value"":""Temperature1 [OF] the water [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""45.22 °C""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{50.0 g}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{50.0 ℃}""},{""type"":""physical unit"",""value"":""Added heat energy [OF] water [=] \\pu{1.0 kJ}""}]" "

The temperature of 50.0 g of water was raised to 50.0°C by the addition of 1.0 kJ of heat energy. What was the initial temperature of the water?

" nan 45.22 °C "
+

Explanation:

+
+

The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.

+

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - heat absorbed
+#m# - the mass of the sample
+#c# - the specific heat of water, equal to #4.18""J""/(""g"" """"^@""C"")#
+#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

+

Plug in your values and solve for #DeltaT# - do not forget to convert the added heat from kilojoules to Joules

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+

#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#

+
+

So, if the temperature of the water changed by #4.785^@""C""#, and its final temperature is now #50.0^@""C""#, it follows that its initial temperature was

+
+

#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#

+

#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#

+
+
+
" "
+
+
+

#45.2^@""C""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.

+

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - heat absorbed
+#m# - the mass of the sample
+#c# - the specific heat of water, equal to #4.18""J""/(""g"" """"^@""C"")#
+#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

+

Plug in your values and solve for #DeltaT# - do not forget to convert the added heat from kilojoules to Joules

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+

#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#

+
+

So, if the temperature of the water changed by #4.785^@""C""#, and its final temperature is now #50.0^@""C""#, it follows that its initial temperature was

+
+

#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#

+

#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#

+
+
+
+
" "
+

The temperature of 50.0 g of water was raised to 50.0°C by the addition of 1.0 kJ of heat energy. What was the initial temperature of the water?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 25, 2015 + +
+
+
+
+
+
+
+

#45.2^@""C""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.

+

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - heat absorbed
+#m# - the mass of the sample
+#c# - the specific heat of water, equal to #4.18""J""/(""g"" """"^@""C"")#
+#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

+

Plug in your values and solve for #DeltaT# - do not forget to convert the added heat from kilojoules to Joules

+
+

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

+

#DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)(""J""))))/(50.0color(red)(cancel(color(black)(""g""))) * 4.18color(red)(cancel(color(black)(""J"")))/(color(red)(cancel(color(black)(""g""))) """"^@""C"")) = 4.785^@""C""#

+
+

So, if the temperature of the water changed by #4.785^@""C""#, and its final temperature is now #50.0^@""C""#, it follows that its initial temperature was

+
+

#DeltaT = T_""f"" - T_""i"" implies T_""i"" = T_""f"" - DeltaT#

+

#T_""i"" = 50.0^@""C"" - 4.785^@""C"" = color(green)(45.2^@""C"")#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
+
" The temperature of 50.0 g of water was raised to 50.0°C by the addition of 1.0 kJ of heat energy. What was the initial temperature of the water? nan +297 a8337378-6ddd-11ea-8fc9-ccda262736ce https://socratic.org/questions/a-10-0-ml-sample-of-a-monoprotic-acid-is-titrated-with-45-5-ml-of-0-200-m-naoh-w 0.91 M start physical_unit 22 23 concentration mol/l qc_end physical_unit 5 7 1 2 volume qc_end physical_unit 16 16 11 12 volume qc_end physical_unit 16 16 14 15 concentration qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] the acid [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.91 M""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] a monoprotic acid [=] \\pu{10.0 mL}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{45.5 mL}""},{""type"":""physical unit"",""value"":""Concentration [OF] NaOH [=] \\pu{0.200 M}""}]" "

A 10.0 mL sample of a monoprotic acid is titrated with 45.5 mL of 0.200 M #NaOH#. What is the concentration of the acid?

" nan 0.91 M "
+

Explanation:

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A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide, #""NaOH""#, is added to the solution.

+

If you take #""HA""# to be the general formula of a monoprotic acid, you can say that the balanced chemical equation that describes this neutralization reaction looks like this

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+

#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#

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+

Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.

+

As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.

+

Now, let's assume that the monoprotic acid solution has a molarity equal to #""0.200 M""#. In this case, the titration would require #""45.5 mL""# of acid solution, since solutions of equal volumes and equal molarities contain the same number of moles of solute.

+

However, you know that the acid solution has a volume of #""10.0 mL""#. This volume is

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#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#

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+

smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is #color(blue)(4.55)# times more concentrated than the sodium hydroxide solution.

+

You thus have

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+

#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#

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+

Therefore, you can say for a fact that #""10.0 mL""# of #""0.910 M""# monoprotic acid solution contains the same number of moles of solute as #""45.5 mL""# of #""0.200 M""# sodium hydroxide solution.

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#""0.910 M""#

+
+
+
+

Explanation:

+
+

A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide, #""NaOH""#, is added to the solution.

+

If you take #""HA""# to be the general formula of a monoprotic acid, you can say that the balanced chemical equation that describes this neutralization reaction looks like this

+
+

#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.

+

As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.

+

Now, let's assume that the monoprotic acid solution has a molarity equal to #""0.200 M""#. In this case, the titration would require #""45.5 mL""# of acid solution, since solutions of equal volumes and equal molarities contain the same number of moles of solute.

+

However, you know that the acid solution has a volume of #""10.0 mL""#. This volume is

+
+

#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#

+
+

smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is #color(blue)(4.55)# times more concentrated than the sodium hydroxide solution.

+

You thus have

+
+

#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#

+
+

Therefore, you can say for a fact that #""10.0 mL""# of #""0.910 M""# monoprotic acid solution contains the same number of moles of solute as #""45.5 mL""# of #""0.200 M""# sodium hydroxide solution.

+
+
+
" "
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A 10.0 mL sample of a monoprotic acid is titrated with 45.5 mL of 0.200 M #NaOH#. What is the concentration of the acid?

+
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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Aug 6, 2016 + +
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#""0.910 M""#

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+
+

Explanation:

+
+

A monoprotic acid can contribute one proton to the neutralization reaction that takes place when sodium hydroxide, #""NaOH""#, is added to the solution.

+

If you take #""HA""# to be the general formula of a monoprotic acid, you can say that the balanced chemical equation that describes this neutralization reaction looks like this

+
+

#""HA""_ ((aq)) + ""NaOH""_ ((aq)) -> ""NaA""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

Iin order to have a complete neutralization, you need to add equal numbers of acid and of base.

+

As you know, molarity is defined as the number of moles of solute present in one liter of solution. You already know the molarity of the sodium hydroxide solution.

+

Now, let's assume that the monoprotic acid solution has a molarity equal to #""0.200 M""#. In this case, the titration would require #""45.5 mL""# of acid solution, since solutions of equal volumes and equal molarities contain the same number of moles of solute.

+

However, you know that the acid solution has a volume of #""10.0 mL""#. This volume is

+
+

#(45.5 color(red)(cancel(color(black)(""mL""))))/(10.0 color(red)(cancel(color(black)(""mL"")))) = color(blue)(4.55)#

+
+

smaller than what you'd need if the acid solution had the same molarity as the sodium hydroxide solution. This can only mean that the acid solution is #color(blue)(4.55)# times more concentrated than the sodium hydroxide solution.

+

You thus have

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+

#c_(""HA"") = color(blue)(4.55) * ""0.200 M"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.910 M"")color(white)(a/a)|)))#

+
+

Therefore, you can say for a fact that #""10.0 mL""# of #""0.910 M""# monoprotic acid solution contains the same number of moles of solute as #""45.5 mL""# of #""0.200 M""# sodium hydroxide solution.

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" A 10.0 mL sample of a monoprotic acid is titrated with 45.5 mL of 0.200 M #NaOH#. What is the concentration of the acid? nan +298 a8337379-6ddd-11ea-aaec-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-50-grams-of-naoh-dissolved-in-2 5.00 mol/L start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 6 6 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""5.00 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaOH [=] \\pu{50 grams}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{250 ml}""}]" "

What is the molarity of a solution that contains 50 grams of NaOH dissolved in 250 ml solution?

" nan 5.00 mol/L "
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Explanation:

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And thus #""Molarity""=((50*g)/(40.0*g*mol^-1))/(250*mLxx10^-3*L*mL^-1)#

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So #[NaOH(aq)]=5.0*mol*L^-1#.

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#""Molarity""# #=""Moles of solute""/""Volume of solution""=5.0*mol*L^-1#

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Explanation:

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And thus #""Molarity""=((50*g)/(40.0*g*mol^-1))/(250*mLxx10^-3*L*mL^-1)#

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So #[NaOH(aq)]=5.0*mol*L^-1#.

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" "
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What is the molarity of a solution that contains 50 grams of NaOH dissolved in 250 ml solution?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Mar 2, 2017 + +
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#""Molarity""# #=""Moles of solute""/""Volume of solution""=5.0*mol*L^-1#

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Explanation:

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And thus #""Molarity""=((50*g)/(40.0*g*mol^-1))/(250*mLxx10^-3*L*mL^-1)#

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So #[NaOH(aq)]=5.0*mol*L^-1#.

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" What is the molarity of a solution that contains 50 grams of NaOH dissolved in 250 ml solution? nan +299 a833737a-6ddd-11ea-a934-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-iron-iii-sulfide Fe2S3 start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] iron (III) sulfide [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Fe2S3""}]" "[{""type"":""substance name"",""value"":""iron (III) sulfide""}]" "

What is the formula for iron (III) sulfide?

" nan Fe2S3 "
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Explanation:

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+

The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.

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In an ionic molecule the formula is determined by balancing the charges on the ions.

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For this case the roman numeral (III) tells us the iron has a charge of #+3# or #Fe^(+3)#

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The the element sulfur has a charge of #-2# or #S^(-2)#

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To balance these ion charges it takes two #+3# iron ions to balance three #-2# sulfide ions.

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The molecule formula is #Fe_2S_3#

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The molecule formula is #Fe_2S_3#

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Explanation:

+
+

The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.

+

In an ionic molecule the formula is determined by balancing the charges on the ions.

+

For this case the roman numeral (III) tells us the iron has a charge of #+3# or #Fe^(+3)#

+

The the element sulfur has a charge of #-2# or #S^(-2)#

+

To balance these ion charges it takes two #+3# iron ions to balance three #-2# sulfide ions.

+

The molecule formula is #Fe_2S_3#

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" "
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What is the formula for iron (III) sulfide?

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+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
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+1 Answer +
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+ + Jul 9, 2016 + +
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The molecule formula is #Fe_2S_3#

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+
+
+

Explanation:

+
+

The molecule iron (III) sulfide is an ionic molecule between the metal Iron and the non-metal sulfur.

+

In an ionic molecule the formula is determined by balancing the charges on the ions.

+

For this case the roman numeral (III) tells us the iron has a charge of #+3# or #Fe^(+3)#

+

The the element sulfur has a charge of #-2# or #S^(-2)#

+

To balance these ion charges it takes two #+3# iron ions to balance three #-2# sulfide ions.

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The molecule formula is #Fe_2S_3#

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Related questions
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+ + Creative Commons License + +
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" What is the formula for iron (III) sulfide? nan +300 a833737b-6ddd-11ea-8b7a-ccda262736ce https://socratic.org/questions/how-would-you-balance-the-following-equation-ba-h2o-ba-oh-2-h2 Ba + 2 H2O -> Ba(OH)2 + H2 start chemical_equation qc_end chemical_equation 7 13 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the following equation""}]" "[{""type"":""chemical equation"",""value"":""Ba + 2 H2O -> Ba(OH)2 + H2""}]" "[{""type"":""chemical equation"",""value"":""Ba + H2O -> Ba(OH)2 + H2""}]" "

How would you balance the following equation: +Ba + H2O --> Ba(OH)2 + H2?

" nan Ba + 2 H2O -> Ba(OH)2 + H2 "
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Explanation:

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As for any chemical reaction, it is balanced stoichiometrically. What does ""#""stoichiometric""#"" mean. It means that for every reactant particle, there is a corresponding product particle. Is that true for the given equation? How would I represent the corresponding reaction between potassium and water?

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#Ba(s) + 2H_2O(l) rarr Ba(OH)_2(aq) + H_2(g)uarr#

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Explanation:

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+

As for any chemical reaction, it is balanced stoichiometrically. What does ""#""stoichiometric""#"" mean. It means that for every reactant particle, there is a corresponding product particle. Is that true for the given equation? How would I represent the corresponding reaction between potassium and water?

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+
+
" "
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How would you balance the following equation: +Ba + H2O --> Ba(OH)2 + H2?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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+ + Jan 26, 2016 + +
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#Ba(s) + 2H_2O(l) rarr Ba(OH)_2(aq) + H_2(g)uarr#

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Explanation:

+
+

As for any chemical reaction, it is balanced stoichiometrically. What does ""#""stoichiometric""#"" mean. It means that for every reactant particle, there is a corresponding product particle. Is that true for the given equation? How would I represent the corresponding reaction between potassium and water?

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Related questions
+ + +
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+
+
Impact of this question
+
+ 5599 views + around the world +
+
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+ + Creative Commons License + +
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" "How would you balance the following equation: +Ba + H2O --> Ba(OH)2 + H2?" nan +301 a8338e54-6ddd-11ea-821c-ccda262736ce https://socratic.org/questions/what-is-the-hydrate-formula-for-something-that-is-0243-mol-bai-2-and-098-mol-h-2 BaI2.4H2O start chemical_formula qc_end physical_unit 11 11 9 10 mole qc_end physical_unit 15 15 13 14 mole qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the hydrate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""BaI2.4H2O""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] BaI2 [=] \\pu{0.0243 mol}""},{""type"":""physical unit"",""value"":""Mole [OF] H2O [=] \\pu{0.098 mol}""}]" "

What is the hydrate formula for something that is .0243 mol #BaI_2# and .098 mol #H_2O#?

" nan BaI2.4H2O "
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Explanation:

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The idea here is that you need to use the number of moles of anhydrous barium iodide, #""BaI""_2#, and of water, #""H""_2""O""#, to find the mole ratio that exists between the anhydrous salt and its water of hydration in the hydrate.

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So, divide both values by the smallest one to find

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#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#

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#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#

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+

This means that one formula unit of this hydrate will contain #1# mole of barium iodide and #4# moles of water. Its chemical formula will thus be

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#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate

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#""BaI""_2 * 4""H""_2""O""#

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Explanation:

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The idea here is that you need to use the number of moles of anhydrous barium iodide, #""BaI""_2#, and of water, #""H""_2""O""#, to find the mole ratio that exists between the anhydrous salt and its water of hydration in the hydrate.

+

So, divide both values by the smallest one to find

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#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#

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#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#

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+

This means that one formula unit of this hydrate will contain #1# mole of barium iodide and #4# moles of water. Its chemical formula will thus be

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#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate

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" "
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What is the hydrate formula for something that is .0243 mol #BaI_2# and .098 mol #H_2O#?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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#""BaI""_2 * 4""H""_2""O""#

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Explanation:

+
+

The idea here is that you need to use the number of moles of anhydrous barium iodide, #""BaI""_2#, and of water, #""H""_2""O""#, to find the mole ratio that exists between the anhydrous salt and its water of hydration in the hydrate.

+

So, divide both values by the smallest one to find

+
+

#""For BaI""_2: color(white)(a)(0.0243 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 1#

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#""For H""_2""O: "" (0.098 color(red)(cancel(color(black)(""moles""))))/(0.0243color(red)(cancel(color(black)(""moles"")))) = 4.033 ~~ 4#

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+

This means that one formula unit of this hydrate will contain #1# mole of barium iodide and #4# moles of water. Its chemical formula will thus be

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#color(green)(|bar(ul(color(white)(a/a)color(black)(""BaI""_2 * 4""H""_2""O"")color(white)(a/a)|))) -># barium iodide tetrahydrate

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" What is the hydrate formula for something that is .0243 mol #BaI_2# and .098 mol #H_2O#? nan +302 a8338e55-6ddd-11ea-bbbc-ccda262736ce https://socratic.org/questions/the-degree-of-dissociation-of-ca-no3-2-in-a-dilute-aqueous-solution-containing-1 747.94 mmHg start physical_unit 44 45 vapor_pressure mmhg qc_end physical_unit 16 16 12 13 mass qc_end physical_unit 21 21 18 19 mass qc_end physical_unit 7 10 23 24 temperature qc_end physical_unit 5 10 26 26 degree_of_dissociation qc_end physical_unit 21 21 34 35 vapor_pressure qc_end end "[{""type"":""physical unit"",""value"":""Vapour pressure [OF] the solution [IN] mmHg""}]" "[{""type"":""physical unit"",""value"":""747.94 mmHg""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] salt [=] \\pu{14 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{200 g}""},{""type"":""physical unit"",""value"":""Temperature [OF] a dilute aqueous solution [=] \\pu{100 ℃}""},{""type"":""physical unit"",""value"":""Degree of dissociation [OF] Ca(NO3)2 in a dilute aqueous solution [=] \\pu{70%}""},{""type"":""physical unit"",""value"":""Vapour pressure [OF] water [=] \\pu{760 mmHg}""}]" "

The degree of dissociation of #""Ca""(""NO""_3)_2# in a dilute aqueous solution containing 14g of the salt per 200 g of water at #100^@""C""# is 70% .If the vapour pressure of water is 760 mmHg , what will be the vapour pressure of the solution ?

" nan 747.94 mmHg "
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Explanation:

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The idea here is that you need to find the number of moles of particles of solute present in your solution.

+

Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.

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Notice that a formula unit of calcium nitrate, #""Ca""(""NO""_3)_2#, contains

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    +
  • one mole of calcium cations, #1 xx ""Ca""^(2+)#
    • +
  • two moles of nitrate anions, #2 xx ""NO""_3^(-)#
    • +
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This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions

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#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#

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+

In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.

+

Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution

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+

#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#

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+

At #100%# dissociation, this many moles of calcium nitrate would produce

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#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#

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+

However, you know that the salt has a #70%# degree of dissociation in water at #100^@""C""#. This means that the actual number of moles of particles of solute present in solution will be

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+

#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#

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Use water's molar mass to determine how many moles of solvent you have present in the solution

+
+

#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#

+
+

Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent, #chi_""solvent""#, and on the vapor pressure of the pure solvent, #P_""solvent""^@#

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#

+
+

The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution

+
+

#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#

+
+

The vapor pressure of the solution will thus be

+
+

#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.

+
+
" "
+
+
+

#""745 mmHg""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to find the number of moles of particles of solute present in your solution.

+

Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.

+

Notice that a formula unit of calcium nitrate, #""Ca""(""NO""_3)_2#, contains

+
+
    +
  • one mole of calcium cations, #1 xx ""Ca""^(2+)#
    • +
  • two moles of nitrate anions, #2 xx ""NO""_3^(-)#
    • +
+
+

This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions

+
+

#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#

+
+

In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.

+

Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution

+
+

#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#

+
+

At #100%# dissociation, this many moles of calcium nitrate would produce

+
+

#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#

+
+

However, you know that the salt has a #70%# degree of dissociation in water at #100^@""C""#. This means that the actual number of moles of particles of solute present in solution will be

+
+

#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#

+
+

Use water's molar mass to determine how many moles of solvent you have present in the solution

+
+

#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#

+
+

Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent, #chi_""solvent""#, and on the vapor pressure of the pure solvent, #P_""solvent""^@#

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#

+
+

The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution

+
+

#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#

+
+

The vapor pressure of the solution will thus be

+
+

#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.

+
+
+
" "
+

The degree of dissociation of #""Ca""(""NO""_3)_2# in a dilute aqueous solution containing 14g of the salt per 200 g of water at #100^@""C""# is 70% .If the vapour pressure of water is 760 mmHg , what will be the vapour pressure of the solution ?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Partial Pressure + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 16, 2016 + +
+
+
+
+
+
+
+

#""745 mmHg""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to find the number of moles of particles of solute present in your solution.

+

Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.

+

Notice that a formula unit of calcium nitrate, #""Ca""(""NO""_3)_2#, contains

+
+
    +
  • one mole of calcium cations, #1 xx ""Ca""^(2+)#
    • +
  • two moles of nitrate anions, #2 xx ""NO""_3^(-)#
    • +
+
+

This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions

+
+

#""Ca""(""NO""_ 3)_ (2(aq)) -> ""Ca""_ ((aq))^(2+) + 2""NO""_(3(aq))^(-)#

+
+

In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.

+

Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution

+
+

#14 color(red)(cancel(color(black)(""g""))) * (""1 mole Ca""(""NO""_3)_2)/(164.09color(red)(cancel(color(black)(""g"")))) = ""0.0853 moles Ca""(""NO""_3)_2#

+
+

At #100%# dissociation, this many moles of calcium nitrate would produce

+
+

#0.0853color(red)(cancel(color(black)(""moles Ca""(""NO""_3)_2))) * ""3 moles ions""/(1color(red)(cancel(color(black)(""mole Ca""(""NO""_3)_2)))) = ""0.256 moles ions""#

+
+

However, you know that the salt has a #70%# degree of dissociation in water at #100^@""C""#. This means that the actual number of moles of particles of solute present in solution will be

+
+

#0.256 color(red)(cancel(color(black)(""moles ions""))) * overbrace(""70 moles ions""/(100color(red)(cancel(color(black)(""moles ions"")))))^(color(purple)(""= 70% dissociation"")) = ""0.179 moles ions""#

+
+

Use water's molar mass to determine how many moles of solvent you have present in the solution

+
+

#200color(red)(cancel(color(black)(""g""))) * (""1 mole H""_2""O"")/(18.015color(red)(cancel(color(black)(""g"")))) = ""11.1 moles H""_2""O""#

+
+

Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent, #chi_""solvent""#, and on the vapor pressure of the pure solvent, #P_""solvent""^@#

+
+

#color(blue)(|bar(ul(color(white)(a/a)P_""solution"" = chi_""solvent"" xx P_""solvent""^@color(white)(a/a)|)))#

+
+

The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution

+
+

#chi_""water"" = (11.1 color(red)(cancel(color(black)(""moles""))))/((11.1 + 0.179color(red)(cancel(color(black)(""moles""))))) = 0.984#

+
+

The vapor pressure of the solution will thus be

+
+

#P_""solution"" = 0.984 xx ""760 mmHg"" = color(green)(|bar(ul(color(white)(a/a)""745 mmHg""color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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+
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+
" "The degree of dissociation of #""Ca""(""NO""_3)_2# in a dilute aqueous solution containing 14g of the salt per 200 g of water at #100^@""C""# is 70% .If the vapour pressure of water is 760 mmHg , what will be the vapour pressure of the solution ? " nan +303 a833b57a-6ddd-11ea-b978-ccda262736ce https://socratic.org/questions/when-the-formula-equation-fe-3o-4-al-al-2o-3-fe-is-correctly-balanced-what-is-th 9 start physical_unit 10 10 coefficient none qc_end chemical_equation 4 10 qc_end end "[{""type"":""physical unit"",""value"":""Coefficient [OF] Fe""}]" "[{""type"":""physical unit"",""value"":""9""}]" "[{""type"":""chemical equation"",""value"":""Fe3O4 + Al -> Al2O3 + Fe""}]" "

When the formula equation #Fe_3O_4 + Al -> Al_2O_3 + Fe# is correctly balanced what is the coefficient of #Fe#?

" nan 9 "
+

Explanation:

+
+

And so the oxides are reduced to iron...

+

#FeO+2H^+ +2e^(-)rarrFe + H_2O# #(i)#

+

#Fe_2O_3 +6H^+ +6e^(-)rarr2Fe+3H_2O# #(ii)#

+

And aluminim metal is oxidized to alumina...

+

#2Al +3H_2OrarrAl_2O_3+6H^+ +6e^-# #(iii)#

+

And we add #3xx(i)+3xx(ii)+4xx(iii)#

+

#3Fe_2O_3 +24H^+ +18e^(-)+3FeO+6H^+ +6e^(-)+8Al +12H_2Orarr3Fe + 3H_2O+6Fe+9H_2O+4Al_2O_3+24H^+ +24e^-#

+

And now we cancel common reagents...

+

#underbrace(3Fe_2O_3 +3FeO)_(3Fe_3O_4) +8Al rarr6Fe +4Al_2O_3+Delta#

+

And finally....

+

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

+

And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....

+
+
" "
+
+
+

Well...#Fe_3O_4# is a mixed valence oxide...#Fe_3O_4-=FeO*Fe_2O_3#

+

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

+
+
+
+

Explanation:

+
+

And so the oxides are reduced to iron...

+

#FeO+2H^+ +2e^(-)rarrFe + H_2O# #(i)#

+

#Fe_2O_3 +6H^+ +6e^(-)rarr2Fe+3H_2O# #(ii)#

+

And aluminim metal is oxidized to alumina...

+

#2Al +3H_2OrarrAl_2O_3+6H^+ +6e^-# #(iii)#

+

And we add #3xx(i)+3xx(ii)+4xx(iii)#

+

#3Fe_2O_3 +24H^+ +18e^(-)+3FeO+6H^+ +6e^(-)+8Al +12H_2Orarr3Fe + 3H_2O+6Fe+9H_2O+4Al_2O_3+24H^+ +24e^-#

+

And now we cancel common reagents...

+

#underbrace(3Fe_2O_3 +3FeO)_(3Fe_3O_4) +8Al rarr6Fe +4Al_2O_3+Delta#

+

And finally....

+

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

+

And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....

+
+
+
" "
+

When the formula equation #Fe_3O_4 + Al -> Al_2O_3 + Fe# is correctly balanced what is the coefficient of #Fe#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 27, 2018 + +
+
+
+
+
+
+
+

Well...#Fe_3O_4# is a mixed valence oxide...#Fe_3O_4-=FeO*Fe_2O_3#

+

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

+
+
+
+

Explanation:

+
+

And so the oxides are reduced to iron...

+

#FeO+2H^+ +2e^(-)rarrFe + H_2O# #(i)#

+

#Fe_2O_3 +6H^+ +6e^(-)rarr2Fe+3H_2O# #(ii)#

+

And aluminim metal is oxidized to alumina...

+

#2Al +3H_2OrarrAl_2O_3+6H^+ +6e^-# #(iii)#

+

And we add #3xx(i)+3xx(ii)+4xx(iii)#

+

#3Fe_2O_3 +24H^+ +18e^(-)+3FeO+6H^+ +6e^(-)+8Al +12H_2Orarr3Fe + 3H_2O+6Fe+9H_2O+4Al_2O_3+24H^+ +24e^-#

+

And now we cancel common reagents...

+

#underbrace(3Fe_2O_3 +3FeO)_(3Fe_3O_4) +8Al rarr6Fe +4Al_2O_3+Delta#

+

And finally....

+

#3Fe_3O_4 +8Al rarr9Fe +4Al_2O_3#

+

And this looks balanced to me... but please check my 'rithmetic; all care taken but no responsibility admitted. And this is an important reaction practically. Where there is a join in the track of a railway line, they fill it with the iron oxide and alumina, and have at it with a blow torch. The aluminum oxidizes to alumina (and is expelled from the join), and the molten iron welds the track together....

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 8323 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" When the formula equation #Fe_3O_4 + Al -> Al_2O_3 + Fe# is correctly balanced what is the coefficient of #Fe#? nan +304 a833b57b-6ddd-11ea-8d4c-ccda262736ce https://socratic.org/questions/how-do-you-balance-naclo-3-nacl-o-2 2 NaClO3 -> 2 NaCl + 3 O2 start chemical_equation qc_end chemical_equation 4 8 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""2 NaClO3 -> 2 NaCl + 3 O2""}]" "[{""type"":""chemical equation"",""value"":""NaClO3 -> NaCl + O2""}]" "

How do you balance #NaClO_3 -> NaCl + O_2#?

" nan 2 NaClO3 -> 2 NaCl + 3 O2 "
+

Explanation:

+
+

To balance the reaction, if it is happening at the solid state (decomposition):

+

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#

+

However, to balance the reaction, if it is happening in aqueous medium:

+

#NaClO_3(aq)->NaCl(aq)+O_2(g)#

+

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from #(-2)# in #NaClO_3# to #(0)# in #O_2#.

+

The oxidation numbers for #Na# and #Cl# are not changing and they are #(+1)# and #(-1)# respectively.

+

To balance this reaction then:
+#NaClO_3(aq)+2H^(+)(aq)+2e^(-)->NaCl(aq)+O_2(g)+H_2O(l)#

+

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent #NaClO_3# such as for example #Na_2SO_3#.

+

The oxidation half equation will then be:

+

#SO_3^(2-)(aq)+H_2O(l)->SO_4^(2-)(aq)+2H^(+)(aq)+2e^(-)#

+

The redox reaction will thus be:

+

Oxidation: #SO_3^(2-)(aq)+cancel(color(green)(H_2O(l)))->SO_4^(2-)(aq) +cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))#

+

Reduction: #NaClO_3(aq)+cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))->NaCl(aq)+O_2(g)+cancel(color(green)(H_2O(l)))#

+

RedOx:
+#NaClO_3(aq)+SO_3^(2-)(aq)->NaCl(aq)+O_2(g)+SO_4^(2-)(aq)#
+or
+#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

+

Here is a video that explains how to balance a redox reaction in acidic medium:
+Balancing Redox Reactions | Acidic Medium.
+ +

+

And here is another one on how to balance a redox reaction in basic medium:
+Balancing Redox Reactions | Basic Medium.
+ +

+
+
" "
+
+
+

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#
+or
+#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

+
+
+
+

Explanation:

+
+

To balance the reaction, if it is happening at the solid state (decomposition):

+

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#

+

However, to balance the reaction, if it is happening in aqueous medium:

+

#NaClO_3(aq)->NaCl(aq)+O_2(g)#

+

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from #(-2)# in #NaClO_3# to #(0)# in #O_2#.

+

The oxidation numbers for #Na# and #Cl# are not changing and they are #(+1)# and #(-1)# respectively.

+

To balance this reaction then:
+#NaClO_3(aq)+2H^(+)(aq)+2e^(-)->NaCl(aq)+O_2(g)+H_2O(l)#

+

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent #NaClO_3# such as for example #Na_2SO_3#.

+

The oxidation half equation will then be:

+

#SO_3^(2-)(aq)+H_2O(l)->SO_4^(2-)(aq)+2H^(+)(aq)+2e^(-)#

+

The redox reaction will thus be:

+

Oxidation: #SO_3^(2-)(aq)+cancel(color(green)(H_2O(l)))->SO_4^(2-)(aq) +cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))#

+

Reduction: #NaClO_3(aq)+cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))->NaCl(aq)+O_2(g)+cancel(color(green)(H_2O(l)))#

+

RedOx:
+#NaClO_3(aq)+SO_3^(2-)(aq)->NaCl(aq)+O_2(g)+SO_4^(2-)(aq)#
+or
+#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

+

Here is a video that explains how to balance a redox reaction in acidic medium:
+Balancing Redox Reactions | Acidic Medium.
+ +

+

And here is another one on how to balance a redox reaction in basic medium:
+Balancing Redox Reactions | Basic Medium.
+ +

+
+
+
" "
+

How do you balance #NaClO_3 -> NaCl + O_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 16, 2016 + +
+
+
+
+
+
+
+

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#
+or
+#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

+
+
+
+

Explanation:

+
+

To balance the reaction, if it is happening at the solid state (decomposition):

+

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#

+

However, to balance the reaction, if it is happening in aqueous medium:

+

#NaClO_3(aq)->NaCl(aq)+O_2(g)#

+

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from #(-2)# in #NaClO_3# to #(0)# in #O_2#.

+

The oxidation numbers for #Na# and #Cl# are not changing and they are #(+1)# and #(-1)# respectively.

+

To balance this reaction then:
+#NaClO_3(aq)+2H^(+)(aq)+2e^(-)->NaCl(aq)+O_2(g)+H_2O(l)#

+

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent #NaClO_3# such as for example #Na_2SO_3#.

+

The oxidation half equation will then be:

+

#SO_3^(2-)(aq)+H_2O(l)->SO_4^(2-)(aq)+2H^(+)(aq)+2e^(-)#

+

The redox reaction will thus be:

+

Oxidation: #SO_3^(2-)(aq)+cancel(color(green)(H_2O(l)))->SO_4^(2-)(aq) +cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))#

+

Reduction: #NaClO_3(aq)+cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))->NaCl(aq)+O_2(g)+cancel(color(green)(H_2O(l)))#

+

RedOx:
+#NaClO_3(aq)+SO_3^(2-)(aq)->NaCl(aq)+O_2(g)+SO_4^(2-)(aq)#
+or
+#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

+

Here is a video that explains how to balance a redox reaction in acidic medium:
+Balancing Redox Reactions | Acidic Medium.
+ +

+

And here is another one on how to balance a redox reaction in basic medium:
+Balancing Redox Reactions | Basic Medium.
+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 5089 views + around the world +
+
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+ + Creative Commons License + +
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+
" How do you balance #NaClO_3 -> NaCl + O_2#? nan +305 a833caa8-6ddd-11ea-9dc6-ccda262736ce https://socratic.org/questions/a-balloon-full-of-air-has-a-volume-of-2-75-l-at-a-temperature-of-18-c-what-is-th 3.01 L start physical_unit 19 20 volume l qc_end physical_unit 19 20 9 10 volume qc_end physical_unit 19 20 15 16 temperature qc_end physical_unit 19 20 23 24 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the balloon [IN] L""}]" "[{""type"":""physical unit"",""value"":""3.01 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] the balloon [=] \\pu{2.75 L}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] the balloon [=] \\pu{18 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the balloon [=] \\pu{45 °C}""},{""type"":""other"",""value"":""A balloon full of air""}]" "

A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume at 45#""^@#C?

" nan 3.01 L "
+

Explanation:

+
+

Start off with what you are given.

+

#V_1#: 2.75L
+#T_1#: #18^@#C

+

#V_2#: ?
+#T_2#: #45^@#C

+

If you know your gas laws, you have to utilise a certain gas law called Charles' Law:

+

#V_1/T_1#=#V_2/T_2#

+

#V_1# is the initial volume, #T_1# is initial temperature, #V_2# is final volume, #T_2# is final temperature.

+

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

+

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

+

#2.75/291#=#V_2/318#

+

By cross multiplying, we have...

+

291#V_2#= 874.5

+

Dividing both sides by 291 to isolate #V_2#, we get...

+

#V_2#= 3.005...

+

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

+

Therefore, #V_2#= 3.01 L

+
+
" "
+
+
+

3.01 L

+
+
+
+

Explanation:

+
+

Start off with what you are given.

+

#V_1#: 2.75L
+#T_1#: #18^@#C

+

#V_2#: ?
+#T_2#: #45^@#C

+

If you know your gas laws, you have to utilise a certain gas law called Charles' Law:

+

#V_1/T_1#=#V_2/T_2#

+

#V_1# is the initial volume, #T_1# is initial temperature, #V_2# is final volume, #T_2# is final temperature.

+

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

+

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

+

#2.75/291#=#V_2/318#

+

By cross multiplying, we have...

+

291#V_2#= 874.5

+

Dividing both sides by 291 to isolate #V_2#, we get...

+

#V_2#= 3.005...

+

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

+

Therefore, #V_2#= 3.01 L

+
+
+
" "
+

A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume at 45#""^@#C?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 3, 2017 + +
+
+
+
+
+
+
+

3.01 L

+
+
+
+

Explanation:

+
+

Start off with what you are given.

+

#V_1#: 2.75L
+#T_1#: #18^@#C

+

#V_2#: ?
+#T_2#: #45^@#C

+

If you know your gas laws, you have to utilise a certain gas law called Charles' Law:

+

#V_1/T_1#=#V_2/T_2#

+

#V_1# is the initial volume, #T_1# is initial temperature, #V_2# is final volume, #T_2# is final temperature.

+

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

+

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

+

#2.75/291#=#V_2/318#

+

By cross multiplying, we have...

+

291#V_2#= 874.5

+

Dividing both sides by 291 to isolate #V_2#, we get...

+

#V_2#= 3.005...

+

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

+

Therefore, #V_2#= 3.01 L

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 71708 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
" "A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume at 45#""^@#C?" nan +306 a833dc9e-6ddd-11ea-ace5-ccda262736ce https://socratic.org/questions/a-sample-of-o-2-gas-is-stored-at-30-0-c-and-755-torr-if-the-volume-was-125ml-how 0.16 g start physical_unit 22 23 mass g qc_end physical_unit 3 4 8 9 temperature qc_end physical_unit 3 4 11 12 pressure qc_end physical_unit 3 4 17 18 volume qc_end end "[{""type"":""physical unit"",""value"":""Weight [OF] the oxygen [IN] g""}]" "[{""type"":""physical unit"",""value"":""0.16 g""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] O2 gas [=] \\pu{30.0 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] O2 gas [=] \\pu{755 torr}""},{""type"":""physical unit"",""value"":""Volume [OF] O2 gas [=] \\pu{125 mL}""}]" "

A sample of #O_2# gas is stored at 30.0 °C and 755 torr. If the volume was 125mL, how much did the oxygen weigh?

" nan 0.16 g "
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample

    +
    • +
  • +

    use oxygen gas' molar mass to find how many grams would contain that many moles

    +
    • +
+

The ideal gas law equation looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821 (""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas

+

Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.

+

More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.

+

So, rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

This means that you have

+
+

#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#

+
+

As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of #""31.9988 g/mol""#, which means that one mole of oxygen as will have a mass of #""31.9988 g""#.

+

This means that you sample will have a mass of

+
+

#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#

+
+

+ +

+
+
" "
+
+
+

#""0.160 g""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample

    +
    • +
  • +

    use oxygen gas' molar mass to find how many grams would contain that many moles

    +
    • +
+

The ideal gas law equation looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821 (""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas

+

Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.

+

More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.

+

So, rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

This means that you have

+
+

#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#

+
+

As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of #""31.9988 g/mol""#, which means that one mole of oxygen as will have a mass of #""31.9988 g""#.

+

This means that you sample will have a mass of

+
+

#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#

+
+

+ +

+
+
+
" "
+

A sample of #O_2# gas is stored at 30.0 °C and 755 torr. If the volume was 125mL, how much did the oxygen weigh?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 17, 2015 + +
+
+
+
+
+
+
+

#""0.160 g""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to

+
    +
  • +

    use the ideal gas law equation to determine how many moles of oxygen gas you have in that sample

    +
    • +
  • +

    use oxygen gas' molar mass to find how many grams would contain that many moles

    +
    • +
+

The ideal gas law equation looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas present in the sample
+#R# - the universal gas constant, usually given as #0.0821 (""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas

+

Now, before plugging in your values, you need to make sure that the units given to you match those used in the expression of the universal gas constant.

+

More specifically, you need to have a pressure in atm, a temperature in Kelvin, and a volume in liters, so don't forget to keep track of the unit conversions needed to get you to these units.

+

So, rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

This means that you have

+
+

#n = (755/760color(red)(cancel(color(black)(""atm""))) * 125 * 10^(-3)color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 30.0)color(red)(cancel(color(black)(""K"")))) = ""0.004989 moles O""_2#

+
+

As you know, a substance's molar mass tells you what the mass of one mole of that substance is. In this case, oxygen has a molar mass of #""31.9988 g/mol""#, which means that one mole of oxygen as will have a mass of #""31.9988 g""#.

+

This means that you sample will have a mass of

+
+

#0.004989 color(red)(cancel(color(black)(""moles""))) * ""31.9988 g""/(1color(red)(cancel(color(black)(""mole"")))) = color(green)(""0.160 g O""_2)#

+
+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 8137 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" A sample of #O_2# gas is stored at 30.0 °C and 755 torr. If the volume was 125mL, how much did the oxygen weigh? nan +307 a833dc9f-6ddd-11ea-a98b-ccda262736ce https://socratic.org/questions/what-mass-of-water-will-change-its-temperature-by-3-c-when-525-j-of-heat-is-adde 36.46 g start physical_unit 3 3 mass g qc_end physical_unit 3 3 9 10 temperature qc_end physical_unit 3 3 12 13 heat_energy qc_end physical_unit 3 3 26 29 specific_heat qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] water [IN] g""}]" "[{""type"":""physical unit"",""value"":""36.46 g""}]" "[{""type"":""physical unit"",""value"":""Temperature changed [OF] water [=] \\pu{3 ℃}""},{""type"":""physical unit"",""value"":""Heat added [OF] water [=] \\pu{525 J}""},{""type"":""physical unit"",""value"":""Specific heat [OF] water [=] \\pu{4.8 J/(g * ℃)}""}]" "

What mass of water will change its temperature by 3 C when 525 J of heat is added to it? The specific heat of water is 4-.8 J/gC

" nan 36.46 g "
+

Explanation:

+
+

To obtain the mass of water, let's use the equation below:
+

+

Based on the information you've provided, we know the following variables:

+

#DeltaT# = #3^oC#
+#Q# = 525 J
+#C# = #4.8J/(gxx^oC)#

+

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by #C# and #Delta T# to get m by itself like this:

+

#m = Q/(DeltaTxxC)#

+

Now, we just plug in the known values:

+

#m = (525cancelJ)/(3^cancel""oC""xx4.8cancelJ/(gxx^(cancel""oC"")))#

+

#m = 36.5 g #

+
+
" "
+
+
+

#m= 36.5g#

+
+
+
+

Explanation:

+
+

To obtain the mass of water, let's use the equation below:
+

+

Based on the information you've provided, we know the following variables:

+

#DeltaT# = #3^oC#
+#Q# = 525 J
+#C# = #4.8J/(gxx^oC)#

+

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by #C# and #Delta T# to get m by itself like this:

+

#m = Q/(DeltaTxxC)#

+

Now, we just plug in the known values:

+

#m = (525cancelJ)/(3^cancel""oC""xx4.8cancelJ/(gxx^(cancel""oC"")))#

+

#m = 36.5 g #

+
+
+
" "
+

What mass of water will change its temperature by 3 C when 525 J of heat is added to it? The specific heat of water is 4-.8 J/gC

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 18, 2016 + +
+
+
+
+
+
+
+

#m= 36.5g#

+
+
+
+

Explanation:

+
+

To obtain the mass of water, let's use the equation below:
+

+

Based on the information you've provided, we know the following variables:

+

#DeltaT# = #3^oC#
+#Q# = 525 J
+#C# = #4.8J/(gxx^oC)#

+

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by #C# and #Delta T# to get m by itself like this:

+

#m = Q/(DeltaTxxC)#

+

Now, we just plug in the known values:

+

#m = (525cancelJ)/(3^cancel""oC""xx4.8cancelJ/(gxx^(cancel""oC"")))#

+

#m = 36.5 g #

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 25092 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What mass of water will change its temperature by 3 C when 525 J of heat is added to it? The specific heat of water is 4-.8 J/gC nan +308 a8340376-6ddd-11ea-8429-ccda262736ce https://socratic.org/questions/if-you-combine-360-0-ml-of-water-at-25-00-c-and-120-0-ml-of-water-at-95-00-c-wha 42.50 ℃ start physical_unit 24 25 temperature °c qc_end physical_unit 6 6 3 4 volume qc_end physical_unit 6 6 8 9 temperature qc_end physical_unit 6 6 16 17 temperature qc_end physical_unit 6 6 11 12 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature3 [OF] the mixture [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""42.50 ℃""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] water [=] \\pu{360.0 mL}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] water [=] \\pu{25.00 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{95.00 ℃}""},{""type"":""physical unit"",""value"":""Volume2 [OF] water [=] \\pu{120.0 mL}""}]" "

If you combine 360.0 mL of water at 25.00°C and 120.0 mL of water at 95.00°C, what is the final temperature of the mixture?

" nan 42.50 ℃ "
+

Explanation:

+
+

The answer to this problem depends on whether or not you should approximate the density of water to be equal to #""1.0 g mL""^(-1)#.

+

Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value #""1.0 g mL""^(-1)# is only an approximation.

+

The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.

+
+

#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#

+
+

The minus sign is used here because heat lost carries a negative sign.

+

Your go-to equation here will be

+
+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat gained / lost
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

+

Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture, #T_f#.

+

So, the change in temperature for the two samples will be

+
+

#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#

+

#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#

+
+

If you take the density to be equal to #""1.0 g mL""^(-1)#, then the two volumes are equivalent to

+
+

#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and #"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#

+
+

Use equation #color(orange)(""(*)"")# top write

+
+

#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#

+
+

This will get you

+
+

#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#

+

#480.0 * T_f = 20400^@""C""#

+

#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#

+
+

The answer is rounded to four sig figs.

+

SIDE NOTE If you use the actual densities of water at #25.00^@""C""# and #95.00^@""C""#, you will end up with a different answer

+
+

#T_f = 38.76^@""C""#

+
+

As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here

+

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

+
+
" "
+
+
+

#42.50^@""C""#

+
+
+
+

Explanation:

+
+

The answer to this problem depends on whether or not you should approximate the density of water to be equal to #""1.0 g mL""^(-1)#.

+

Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value #""1.0 g mL""^(-1)# is only an approximation.

+

The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.

+
+

#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#

+
+

The minus sign is used here because heat lost carries a negative sign.

+

Your go-to equation here will be

+
+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat gained / lost
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

+

Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture, #T_f#.

+

So, the change in temperature for the two samples will be

+
+

#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#

+

#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#

+
+

If you take the density to be equal to #""1.0 g mL""^(-1)#, then the two volumes are equivalent to

+
+

#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and #"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#

+
+

Use equation #color(orange)(""(*)"")# top write

+
+

#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#

+
+

This will get you

+
+

#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#

+

#480.0 * T_f = 20400^@""C""#

+

#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#

+
+

The answer is rounded to four sig figs.

+

SIDE NOTE If you use the actual densities of water at #25.00^@""C""# and #95.00^@""C""#, you will end up with a different answer

+
+

#T_f = 38.76^@""C""#

+
+

As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here

+

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

+
+
+
" "
+

If you combine 360.0 mL of water at 25.00°C and 120.0 mL of water at 95.00°C, what is the final temperature of the mixture?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 12, 2016 + +
+
+
+
+
+
+
+

#42.50^@""C""#

+
+
+
+

Explanation:

+
+

The answer to this problem depends on whether or not you should approximate the density of water to be equal to #""1.0 g mL""^(-1)#.

+

Since no information about density was provide, I assume that this is what you must do. However, it's important to note that water's density varies with temperature, and that the value #""1.0 g mL""^(-1)# is only an approximation.

+

The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.

+
+

#color(blue)(-q_""lost"" = q_""absorbed"")"" "" "" ""color(orange)(""(*)"")#

+
+

The minus sign is used here because heat lost carries a negative sign.

+

Your go-to equation here will be

+
+

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))"" ""#, where

+
+

#q# - the amount of heat gained / lost
+#m# - the mass of the sample
+#c# - the specific heat of the substance
+#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

+

Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture, #T_f#.

+

So, the change in temperature for the two samples will be

+
+

#""For the hot sample: "" DeltaT_""hot"" = T_f - 95.00^@""C""#

+

#""For the warm sample: "" DeltaT_""warm"" = T_f - 25.00^@""C""#

+
+

If you take the density to be equal to #""1.0 g mL""^(-1)#, then the two volumes are equivalent to

+
+

#360.0 color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""360.0 g"" "" ""# and #"" ""120.0color(red)(cancel(color(black)(""mL""))) * ""1.0 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""120.0 g""#

+
+

Use equation #color(orange)(""(*)"")# top write

+
+

#overbrace(-120.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 95.00^@""C""))^(color(purple)(""heat lost by the hot sample"")) = overbrace(360.0color(red)(cancel(color(black)(""g""))) * color(red)(cancel(color(black)(c_""water""))) * (T_f - 25.00)^@""C"")^(color(blue)(""heat gained by warm sample""))#

+
+

This will get you

+
+

#-120.0 * T_f + 11400^@""C"" = 360.0 * T_f - 9000^@""C""#

+

#480.0 * T_f = 20400^@""C""#

+

#T_f = (20400^@""C"")/480.0 = color(green)(|bar(ul(color(white)(a/a)42.50^@""C""color(white)(a/a)|)))#

+
+

The answer is rounded to four sig figs.

+

SIDE NOTE If you use the actual densities of water at #25.00^@""C""# and #95.00^@""C""#, you will end up with a different answer

+
+

#T_f = 38.76^@""C""#

+
+

As practice, you should try redoing the calculations using the actual densities of water at those two temperatures. You can find info on that here

+

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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+
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" If you combine 360.0 mL of water at 25.00°C and 120.0 mL of water at 95.00°C, what is the final temperature of the mixture? nan +309 a8340377-6ddd-11ea-b4f4-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-in-which-10-g-of-agno-3-is-dissolved-in-500-m 0.12 mol/L start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 6 6 16 17 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] AgNO3 [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.12 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] AgNO3 [=] \\pu{10.0 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{500 mL}""}]" "

What is the molarity of a solution in which 10.0g of #AgNO_3# is dissolved in 500. mL of solution?

" nan 0.12 mol/L "
+

Explanation:

+
+

Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.

+
+

#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#

+
+

This means that in order to find a solution's molarity you must know two things

+
+
    +
  • the number of moles of solute present
    • +
  • the volume of the solution
    • +
+
+

In order to find the number of moles of silver nitrate, #""AgNO""_3#, present in that #""10.0-g""# sample, use the compound's molar mass

+
+

#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#

+
+

Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor

+
+

#""1 L"" = 10^3""mL""#

+
+

Plug in your values to get

+
+

#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#

+

#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""0.118 M""#

+
+
+
+

Explanation:

+
+

Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.

+
+

#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#

+
+

This means that in order to find a solution's molarity you must know two things

+
+
    +
  • the number of moles of solute present
    • +
  • the volume of the solution
    • +
+
+

In order to find the number of moles of silver nitrate, #""AgNO""_3#, present in that #""10.0-g""# sample, use the compound's molar mass

+
+

#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#

+
+

Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor

+
+

#""1 L"" = 10^3""mL""#

+
+

Plug in your values to get

+
+

#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#

+

#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

What is the molarity of a solution in which 10.0g of #AgNO_3# is dissolved in 500. mL of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
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+ +
+
+ +
+ + Mar 5, 2016 + +
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+
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+

#""0.118 M""#

+
+
+
+

Explanation:

+
+

Molarity is simply a measure of the concentration of a solution in terms of how many moles of solute are present per liter of solution.

+
+

#color(blue)(|bar(ul(""molarity"" = ""moles of solute""/""liters of solution""))|)#

+
+

This means that in order to find a solution's molarity you must know two things

+
+
    +
  • the number of moles of solute present
    • +
  • the volume of the solution
    • +
+
+

In order to find the number of moles of silver nitrate, #""AgNO""_3#, present in that #""10.0-g""# sample, use the compound's molar mass

+
+

#10.0 color(red)(cancel(color(black)(""g""))) * ""1 mole AgNO""_3/(169.87color(red)(cancel(color(black)(""g"")))) = ""0.05887 moles AgNO""_3#

+
+

Now, do not forget that molarity is expressed per liter of solution. This means that you're going to have to convert the volume from milliliters to liters by using the conversion factor

+
+

#""1 L"" = 10^3""mL""#

+
+

Plug in your values to get

+
+

#color(blue)( |bar(ul(c = n_""solute""/V_""solution""))|)#

+

#c = ""0.05887 moles""/(500. * 10^(-3)""L"") = color(green)(|bar(ul(""0.118 M""))|)#

+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+ + Creative Commons License + +
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" What is the molarity of a solution in which 10.0g of #AgNO_3# is dissolved in 500. mL of solution? nan +310 a8340378-6ddd-11ea-a580-ccda262736ce https://socratic.org/questions/563489f311ef6b4aa976b4fd +6 start physical_unit 6 10 oxidation_number none qc_end chemical_equation 11 11 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] the metal in manganate ion""}]" "[{""type"":""physical unit"",""value"":""+6""}]" "[{""type"":""chemical equation"",""value"":""MnO4^2-""}]" "

What is the oxidation number of the metal in #""manganate ion""#, #MnO_4^(2-)#?

" nan +6 "
+

Explanation:

+
+

The oxdiation number of manganese in the anion #MnO_4^(2-)# is clearly #VI+#, utilizing standard oxidation numbers of #-II# for oxygen. What is the oxidation state of the metal in permanganate ion, #MnO_4^-#?

+
+
" "
+
+
+

This is the manganate ion, where manganese has a #VI+# oxidation state.

+
+
+
+

Explanation:

+
+

The oxdiation number of manganese in the anion #MnO_4^(2-)# is clearly #VI+#, utilizing standard oxidation numbers of #-II# for oxygen. What is the oxidation state of the metal in permanganate ion, #MnO_4^-#?

+
+
+
" "
+

What is the oxidation number of the metal in #""manganate ion""#, #MnO_4^(2-)#?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Oct 31, 2015 + +
+
+
+
+
+
+
+

This is the manganate ion, where manganese has a #VI+# oxidation state.

+
+
+
+

Explanation:

+
+

The oxdiation number of manganese in the anion #MnO_4^(2-)# is clearly #VI+#, utilizing standard oxidation numbers of #-II# for oxygen. What is the oxidation state of the metal in permanganate ion, #MnO_4^-#?

+
+
+
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+
+ +
+
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+ +
+
+ +
+ + Oct 31, 2015 + +
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+

#Mn^""+6""#

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+
+
+

Explanation:

+
+

The first thing you need to realize is there are two bonds in this formula: the ionic bond between #NH_4^+# and the #MnO_4^""2-""# and the covalent bond in the #MnO_4^""2-""# ion. You can only do this by knowing at first look which one is the cation and which is the anion. As the cations are only a handful, it would be good if you can memorize them.

+

With this kind of substances, it would be helpful if you separate the ions first since it would break down your equation into a more simpler one.

+

Hence,
+#(NH_4)_2MnO_4# #rarr# #2NH_4^+# + #MnO_4^""2-""#

+

Now that you separated the ions, you can safely ignore the ammonium ion for the time being, as it would not affect your calculations for the oxidation state of #Mn# atom.

+

Thus,

+

#MnO_4^""2-""# = -2 (oxidation state based on the charge)

+

#O# atoms almost always have a charge of -2 (with one possible exception in peroxide ions).

+

Thus we have the equation:

+

#color (red) x# + [(4) (-2)] = - 2 where #color (red) x# is the oxidation state on #Mn# atom

+

#color (red) x# + (-8) = - 2

+

#color (red) x# = - 2 + (+8)

+

#color (red) x# = +6

+

Therefore, the oxidation state of #Mn# atom is +6.

+

Still in doubt? Let's try solving #x#, this time, including the #NH_4^+# ion.

+

#(NH_4)_2MnO_4# = 0 since the substance is neutrally charge

+

[(+1) (2)] + #color (red) x# + [(4) (-2)] = 0

+

(+2) + #color (red) x# + (-8) = 0

+

#color (red) x# + (-6) = 0

+

#color (red) x# = +6

+

Same answer. :)

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+
+
+
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+
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+
" "What is the oxidation number of the metal in #""manganate ion""#, #MnO_4^(2-)#?" nan +311 a8340379-6ddd-11ea-945e-ccda262736ce https://socratic.org/questions/when-150-cm-3-of-water-freezes-162-cm-3-of-ice-is-formed-how-do-you-calculate-th 8.00% start physical_unit 21 21 percent none qc_end c_other OTHER qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 9 9 6 7 volume qc_end end "[{""type"":""physical unit"",""value"":""Percentage increase [OF] volume""}]" "[{""type"":""physical unit"",""value"":""8.00%""}]" "[{""type"":""other"",""value"":""When water freezes, ice is formed.""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{150 cm^3}""},{""type"":""physical unit"",""value"":""Volume [OF] ice [=] \\pu{162 cm^3}""}]" "

When 150 #cm^3# of water freezes, 162 #cm^3# of ice is formed. How do you calculate the percentage increase in the volume?

" nan 8.00% "
+

Explanation:

+
+

I would use fractions for #%# and volumes in the form:

+

#""total amount""/""increase""#

+

The volume increases of #162-150=12cm^3# so with fractions:

+

#(100%)/(x%)=(150cm^3)/(12cm^3)#

+

the percentual increase will be:

+

#x%=100%*(12cancel(cm^3))/(150cancel(cm^3))=8%#

+
+
" "
+
+
+

I got #8%#

+
+
+
+

Explanation:

+
+

I would use fractions for #%# and volumes in the form:

+

#""total amount""/""increase""#

+

The volume increases of #162-150=12cm^3# so with fractions:

+

#(100%)/(x%)=(150cm^3)/(12cm^3)#

+

the percentual increase will be:

+

#x%=100%*(12cancel(cm^3))/(150cancel(cm^3))=8%#

+
+
+
" "
+

When 150 #cm^3# of water freezes, 162 #cm^3# of ice is formed. How do you calculate the percentage increase in the volume?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 15, 2017 + +
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+
+
+
+
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+

I got #8%#

+
+
+
+

Explanation:

+
+

I would use fractions for #%# and volumes in the form:

+

#""total amount""/""increase""#

+

The volume increases of #162-150=12cm^3# so with fractions:

+

#(100%)/(x%)=(150cm^3)/(12cm^3)#

+

the percentual increase will be:

+

#x%=100%*(12cancel(cm^3))/(150cancel(cm^3))=8%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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+
" When 150 #cm^3# of water freezes, 162 #cm^3# of ice is formed. How do you calculate the percentage increase in the volume? nan +312 a8342e74-6ddd-11ea-bd9d-ccda262736ce https://socratic.org/questions/how-many-grams-of-potassium-chloride-are-produced-if-25-g-of-potassium-chlorate- 15.21 grams start physical_unit 4 5 mass g qc_end physical_unit 12 13 9 10 mass qc_end chemical_equation 18 24 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] potassium chloride [IN] grams""}]" "[{""type"":""physical unit"",""value"":""15.21 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] potassium chlorate [=] \\pu{25 g}""},{""type"":""chemical equation"",""value"":""2 KClO3 -> 2 KCl + 3O2""}]" "

How many grams of potassium chloride are produced if 25 g of potassium chlorate decompose in the reaction #2KClO_3 -> 2KCl + 3O_2#?

" nan 15.21 grams "
+

Explanation:

+
+

The general pattern for solving these types of problems is

+
    +
  1. Find ""moles"" of the known substance
    2. +
  2. Write down the mole ratio using the coefficients from the balanced equation
    3. +
  3. rearrange to solve for the moles ""unknown""
    4. +
  4. Convert moles ""unknown"" to required units (i.e. mass, volume, concentration)
    5. +
+

So to solve the problem:

+
    +
  1. +

    Molecular mass of #KClO_3# = 39.10 + 35.45 + 3(16.00)
    + = 122.55

    +
    2. +
  2. +

    Moles of #KClO_3# = #25/122.55# = 0.204 mol

    +
    3. +
  3. +

    mol ratio: #mol KCl // mol KClO_3# = #2/2#
    +so
    + mol of KCl = #2/2 xx mol KClO_3#

    +

    mol of KCl = #2/2 xx 0.204#

    +

    mol of KCl = 0.204 mol

    +
    4. +
+

4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
+mass KCl = #0.204 xx 74.55# = 15.208 g

+

Therefore 15.2 g of KCl would be formed.

+
+
" "
+
+
+

15.2 g of KCl would be produced

+
+
+
+

Explanation:

+
+

The general pattern for solving these types of problems is

+
    +
  1. Find ""moles"" of the known substance
    2. +
  2. Write down the mole ratio using the coefficients from the balanced equation
    3. +
  3. rearrange to solve for the moles ""unknown""
    4. +
  4. Convert moles ""unknown"" to required units (i.e. mass, volume, concentration)
    5. +
+

So to solve the problem:

+
    +
  1. +

    Molecular mass of #KClO_3# = 39.10 + 35.45 + 3(16.00)
    + = 122.55

    +
    2. +
  2. +

    Moles of #KClO_3# = #25/122.55# = 0.204 mol

    +
    3. +
  3. +

    mol ratio: #mol KCl // mol KClO_3# = #2/2#
    +so
    + mol of KCl = #2/2 xx mol KClO_3#

    +

    mol of KCl = #2/2 xx 0.204#

    +

    mol of KCl = 0.204 mol

    +
    4. +
+

4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
+mass KCl = #0.204 xx 74.55# = 15.208 g

+

Therefore 15.2 g of KCl would be formed.

+
+
+
" "
+

How many grams of potassium chloride are produced if 25 g of potassium chlorate decompose in the reaction #2KClO_3 -> 2KCl + 3O_2#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 6, 2016 + +
+
+
+
+
+
+
+

15.2 g of KCl would be produced

+
+
+
+

Explanation:

+
+

The general pattern for solving these types of problems is

+
    +
  1. Find ""moles"" of the known substance
    2. +
  2. Write down the mole ratio using the coefficients from the balanced equation
    3. +
  3. rearrange to solve for the moles ""unknown""
    4. +
  4. Convert moles ""unknown"" to required units (i.e. mass, volume, concentration)
    5. +
+

So to solve the problem:

+
    +
  1. +

    Molecular mass of #KClO_3# = 39.10 + 35.45 + 3(16.00)
    + = 122.55

    +
    2. +
  2. +

    Moles of #KClO_3# = #25/122.55# = 0.204 mol

    +
    3. +
  3. +

    mol ratio: #mol KCl // mol KClO_3# = #2/2#
    +so
    + mol of KCl = #2/2 xx mol KClO_3#

    +

    mol of KCl = #2/2 xx 0.204#

    +

    mol of KCl = 0.204 mol

    +
    4. +
+

4.. molecular mass of KCl = 39.10 + 35.45 = 74.55
+mass KCl = #0.204 xx 74.55# = 15.208 g

+

Therefore 15.2 g of KCl would be formed.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 18350 views + around the world +
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" How many grams of potassium chloride are produced if 25 g of potassium chlorate decompose in the reaction #2KClO_3 -> 2KCl + 3O_2#? nan +313 a8342e75-6ddd-11ea-8ef1-ccda262736ce https://socratic.org/questions/you-dissolve-0-74-g-of-potassium-chloride-in-500-ml-of-water-what-is-the-molarit 0.02 mol/L start physical_unit 17 18 molarity mol/l qc_end physical_unit 5 6 2 3 mass qc_end physical_unit 11 11 8 9 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] the solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.02 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] potassium chloride [=] \\pu{0.74 g}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{500 mL}""}]" "

You dissolve 0.74 g of potassium chloride in 500 mL of water. What is the molarity of the solution?

" nan 0.02 mol/L "
+

Explanation:

+
+

#""Molarity""=((0.74*cancelg)/(74.55*cancelg*mol^-1))/(500*cancel(mL)xx10^-3*L*cancel(mL^-1))=??*mol*L^-1#.

+

I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e. #mol*L^-1#, and we got one!

+

Remember #1/(mol^-1)=1/(1/(mol))=mol# as required.............

+
+
" "
+
+
+

#""Molarity""=""Moles of solute""/""Volume of solution (L)""=0.02*mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Molarity""=((0.74*cancelg)/(74.55*cancelg*mol^-1))/(500*cancel(mL)xx10^-3*L*cancel(mL^-1))=??*mol*L^-1#.

+

I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e. #mol*L^-1#, and we got one!

+

Remember #1/(mol^-1)=1/(1/(mol))=mol# as required.............

+
+
+
" "
+

You dissolve 0.74 g of potassium chloride in 500 mL of water. What is the molarity of the solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 23, 2017 + +
+
+
+
+
+
+
+

#""Molarity""=""Moles of solute""/""Volume of solution (L)""=0.02*mol*L^-1#.

+
+
+
+

Explanation:

+
+

#""Molarity""=((0.74*cancelg)/(74.55*cancelg*mol^-1))/(500*cancel(mL)xx10^-3*L*cancel(mL^-1))=??*mol*L^-1#.

+

I include the units, even tho it is a bit a pfaff, to ensure that I have done the right order of operation. It is all too easy to divide when we should have multiplied, and vice versa. We want an answer with units of concentration, i.e. #mol*L^-1#, and we got one!

+

Remember #1/(mol^-1)=1/(1/(mol))=mol# as required.............

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 5353 views + around the world +
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+ + Creative Commons License + +
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" You dissolve 0.74 g of potassium chloride in 500 mL of water. What is the molarity of the solution? nan +314 a8342e76-6ddd-11ea-85e6-ccda262736ce https://socratic.org/questions/how-many-grams-of-ammonium-sulfate-are-needed-to-make-a-0-25-l-solution-at-a-con 198.21 grams start physical_unit 4 5 mass g qc_end physical_unit 13 13 11 12 volume qc_end physical_unit 4 5 18 19 concentration qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] ammonium sulfate [IN] grams""}]" "[{""type"":""physical unit"",""value"":""198.21 grams""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{0.25 L}""},{""type"":""physical unit"",""value"":""Concentration [OF] ammonium sulfate [=] \\pu{6 M}""}]" "

How many grams of ammonium sulfate are needed to make a 0.25 L solution at a concentration of 6 M?

" nan 198.21 grams "
+

Explanation:

+
+

Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate, #(""NH""_4)_2""SO""_4#, it must contain.

+

Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.

+

So, a solution's molarity tells you how many moles of solute you get per liter of solution.

+

In this case, the solution must have a molarity of #""6 mol L""^(-1)#, which is equivalent to saying that it must contain #6# moles of ammonium sulfate per liter of solution.

+

Use this value as a conversion factor to find the number of moles that you'd get in #""0.25 L""# of solution

+
+

#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#

+
+

Now use ammonium sulfate's molar mass, which is listed as being equal to #""132.14 g mol""^(-1)#, to find how many grams you'd need in order to get #1.5# moles

+
+

#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#

+
+

Rounded to one significant figure, the answer will be

+
+

#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#

+
+
+
" "
+
+
+

#""200 g""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate, #(""NH""_4)_2""SO""_4#, it must contain.

+

Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.

+

So, a solution's molarity tells you how many moles of solute you get per liter of solution.

+

In this case, the solution must have a molarity of #""6 mol L""^(-1)#, which is equivalent to saying that it must contain #6# moles of ammonium sulfate per liter of solution.

+

Use this value as a conversion factor to find the number of moles that you'd get in #""0.25 L""# of solution

+
+

#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#

+
+

Now use ammonium sulfate's molar mass, which is listed as being equal to #""132.14 g mol""^(-1)#, to find how many grams you'd need in order to get #1.5# moles

+
+

#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#

+
+

Rounded to one significant figure, the answer will be

+
+

#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#

+
+
+
+
" "
+

How many grams of ammonium sulfate are needed to make a 0.25 L solution at a concentration of 6 M?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 9, 2016 + +
+
+
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+
+
+
+

#""200 g""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use the volume and molarity of the solution to determine how many moles of ammonium sulfate, #(""NH""_4)_2""SO""_4#, it must contain.

+

Once you know the number of moles of solute needed to make that solution, use ammonium sulfate's molar mass to determine how many grams would contain that many moles.

+

So, a solution's molarity tells you how many moles of solute you get per liter of solution.

+

In this case, the solution must have a molarity of #""6 mol L""^(-1)#, which is equivalent to saying that it must contain #6# moles of ammonium sulfate per liter of solution.

+

Use this value as a conversion factor to find the number of moles that you'd get in #""0.25 L""# of solution

+
+

#0.25 color(red)(cancel(color(black)(""L solution""))) * (""6 moles""color(white)(a) (""NH""_4)_2""SO""_4)/(1color(red)(cancel(color(black)(""L solution"")))) = ""1.5 moles"" color(white)(a)(""NH""_4)_2""SO""_4#

+
+

Now use ammonium sulfate's molar mass, which is listed as being equal to #""132.14 g mol""^(-1)#, to find how many grams you'd need in order to get #1.5# moles

+
+

#1.5color(red)(cancel(color(black)(""moles""color(white)(a)(""NH""_4)_2""SO""_4))) * ""132.14 g""/(1color(red)(cancel(color(black)(""mole""color(white)(a)(""NH""_4)_2""SO""_4)))) = ""198.21 g""#

+
+

Rounded to one significant figure, the answer will be

+
+

#""mass of ammonium sulfate"" = color(green)(|bar(ul(color(white)(a/a)""200 g""color(white)(a/a)|)))#

+
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+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
+ 48595 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many grams of ammonium sulfate are needed to make a 0.25 L solution at a concentration of 6 M? nan +315 a8342e77-6ddd-11ea-bdd8-ccda262736ce https://socratic.org/questions/for-the-reaction-2na-s-cl-2-g-2nacl-s-how-many-grams-of-nacl-could-be-produced-f 67.90 grams start physical_unit 9 9 mass g qc_end c_other STP qc_end chemical_equation 3 9 qc_end physical_unit 4 4 19 20 mass qc_end physical_unit 6 6 24 25 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [IN] grams""}]" "[{""type"":""physical unit"",""value"":""67.90 grams""}]" "[{""type"":""other"",""value"":""STP""},{""type"":""chemical equation"",""value"":""2 Na(s) + Cl2(g) -> 2 NaCl(s)""},{""type"":""physical unit"",""value"":""Mass [OF] Na [=] \\pu{103.0 g}""},{""type"":""physical unit"",""value"":""Volume [OF] Cl2 [=] \\pu{13.0 L}""}]" "

For the reaction #2Na(s) + Cl_2(g) -> 2NaCl(s)#, how many grams of #NaCl# could be produced from 103.0 g of Na and 13.0 L of #Cl_2# (at STP)?

" nan 67.90 grams "
+

Explanation:

+
+

You have to search the limiting reagent between Na and #Cl_2#
+You have # n= ""mass""/""molecular mass"" = (103,0 g)/(23,0 (g/(mol))) = 4,47 mol # of Na.
+As 1 mole occupies 22,4 L at STP you have # n= (13,0L)/(22,4(L/(mol))) = 0,580# mol of #Cl_2#
+The limiting reagent is #Cl_2#.
+since from the balanced reaction you have 2 mol of NaCl from one of #Cl_2#, making the proportion from 0,580 mol of #Cl_2# you get 1,16 (mol) of NaCl that are
+# 1,16 (mol) 58,5 g/(mol) = 67,9 g# of NaCl

+
+
" "
+
+
+

67,9 g

+
+
+
+

Explanation:

+
+

You have to search the limiting reagent between Na and #Cl_2#
+You have # n= ""mass""/""molecular mass"" = (103,0 g)/(23,0 (g/(mol))) = 4,47 mol # of Na.
+As 1 mole occupies 22,4 L at STP you have # n= (13,0L)/(22,4(L/(mol))) = 0,580# mol of #Cl_2#
+The limiting reagent is #Cl_2#.
+since from the balanced reaction you have 2 mol of NaCl from one of #Cl_2#, making the proportion from 0,580 mol of #Cl_2# you get 1,16 (mol) of NaCl that are
+# 1,16 (mol) 58,5 g/(mol) = 67,9 g# of NaCl

+
+
+
" "
+

For the reaction #2Na(s) + Cl_2(g) -> 2NaCl(s)#, how many grams of #NaCl# could be produced from 103.0 g of Na and 13.0 L of #Cl_2# (at STP)?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Jan 28, 2017 + +
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+
+
+
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+
+

67,9 g

+
+
+
+

Explanation:

+
+

You have to search the limiting reagent between Na and #Cl_2#
+You have # n= ""mass""/""molecular mass"" = (103,0 g)/(23,0 (g/(mol))) = 4,47 mol # of Na.
+As 1 mole occupies 22,4 L at STP you have # n= (13,0L)/(22,4(L/(mol))) = 0,580# mol of #Cl_2#
+The limiting reagent is #Cl_2#.
+since from the balanced reaction you have 2 mol of NaCl from one of #Cl_2#, making the proportion from 0,580 mol of #Cl_2# you get 1,16 (mol) of NaCl that are
+# 1,16 (mol) 58,5 g/(mol) = 67,9 g# of NaCl

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+
Related questions
+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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+
" For the reaction #2Na(s) + Cl_2(g) -> 2NaCl(s)#, how many grams of #NaCl# could be produced from 103.0 g of Na and 13.0 L of #Cl_2# (at STP)? nan +316 a8342e78-6ddd-11ea-9d04-ccda262736ce https://socratic.org/questions/what-is-the-specific-heat-of-ice-at-0-degrees-celsius 2.03 J/(℃ * g) start physical_unit 6 6 specific_heat j/(°c_·_g) qc_end physical_unit 6 6 8 10 temperature qc_end end "[{""type"":""physical unit"",""value"":""Specific heat [OF] ice [IN] J/(℃ * g)""}]" "[{""type"":""physical unit"",""value"":""2.03 J/(℃ * g)""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] ice [=] \\pu{0 degrees Celsius}""}]" "

What is the specific heat of ice at 0 degrees Celsius?

" nan 2.03 J/(℃ * g) "
+

Explanation:

+
+

The specific heat capacity of ice is equal to #2.03J/(""""^@C.g)#.

+

Note that this is not the same as the ones of water (#4.18J/(""""^@C.g)#) or steam (#2.01J/(""""^@C.g)#).

+

Here is a video that explains this topic in much more details:
+Thermochemistry | Enthalpy and Calorimetry.
+ +

+
+
" "
+
+
+

#2.03J/(""""^@C.g)#

+
+
+
+

Explanation:

+
+

The specific heat capacity of ice is equal to #2.03J/(""""^@C.g)#.

+

Note that this is not the same as the ones of water (#4.18J/(""""^@C.g)#) or steam (#2.01J/(""""^@C.g)#).

+

Here is a video that explains this topic in much more details:
+Thermochemistry | Enthalpy and Calorimetry.
+ +

+
+
+
" "
+

What is the specific heat of ice at 0 degrees Celsius?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 8, 2016 + +
+
+
+
+
+
+
+

#2.03J/(""""^@C.g)#

+
+
+
+

Explanation:

+
+

The specific heat capacity of ice is equal to #2.03J/(""""^@C.g)#.

+

Note that this is not the same as the ones of water (#4.18J/(""""^@C.g)#) or steam (#2.01J/(""""^@C.g)#).

+

Here is a video that explains this topic in much more details:
+Thermochemistry | Enthalpy and Calorimetry.
+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 13363 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the specific heat of ice at 0 degrees Celsius? nan +317 a83451b4-6ddd-11ea-a2ed-ccda262736ce https://socratic.org/questions/how-would-you-determine-the-percent-yield-for-the-reaction-between-15-8g-of-nh3- 78.19% start physical_unit 8 9 percent_yield none qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 23 24 20 21 mass qc_end c_other OTHER qc_end substance 26 26 qc_end end "[{""type"":""physical unit"",""value"":""Percent yield [OF] the reaction""}]" "[{""type"":""physical unit"",""value"":""78.19%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NH3 [=] \\pu{15.8 g}""},{""type"":""physical unit"",""value"":""Mass [OF] NO gas [=] \\pu{21.8 g}""},{""type"":""other"",""value"":""Excess oxygen.""},{""type"":""substance name"",""value"":""Water""}]" "

How would you determine the percent yield for the reaction between 15.8g of NH3 and excess oxygen to produce 21.8g of NO gas and water?

" nan 78.19% "
+

Explanation:

+
+

Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to #NO_2#. Can you write this redox reaction?

+

As to your question, we start with #(15.8*g)/(17.0*g*mol^-1) = 0.929# #mol# of ammonia. We get #(21.8*g)/(30*g*mol^-1)# #=# #0.727# #mol#. Because it's 1 ammonia to 1 nitrous oxide, the yield is simply, #(0.727*mol)/(0.929*mol) xx 100% = ??#

+
+
" "
+
+
+

Rxn: #4NH_3(aq) + 5O_2(g) rarr 4NO(g) + 6H_2O(l)#

+
+
+
+

Explanation:

+
+

Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to #NO_2#. Can you write this redox reaction?

+

As to your question, we start with #(15.8*g)/(17.0*g*mol^-1) = 0.929# #mol# of ammonia. We get #(21.8*g)/(30*g*mol^-1)# #=# #0.727# #mol#. Because it's 1 ammonia to 1 nitrous oxide, the yield is simply, #(0.727*mol)/(0.929*mol) xx 100% = ??#

+
+
+
" "
+

How would you determine the percent yield for the reaction between 15.8g of NH3 and excess oxygen to produce 21.8g of NO gas and water?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
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+
+
+1 Answer +
+
+
+
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+ +
+
+ +
+ + Nov 23, 2015 + +
+
+
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+

Rxn: #4NH_3(aq) + 5O_2(g) rarr 4NO(g) + 6H_2O(l)#

+
+
+
+

Explanation:

+
+

Is the reaction above a redox reaction? What is oxidized, and what is reduced? What are the oxidation states? Nitrous oxide is capable of being further oxidized to #NO_2#. Can you write this redox reaction?

+

As to your question, we start with #(15.8*g)/(17.0*g*mol^-1) = 0.929# #mol# of ammonia. We get #(21.8*g)/(30*g*mol^-1)# #=# #0.727# #mol#. Because it's 1 ammonia to 1 nitrous oxide, the yield is simply, #(0.727*mol)/(0.929*mol) xx 100% = ??#

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+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 14725 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
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+
+
" How would you determine the percent yield for the reaction between 15.8g of NH3 and excess oxygen to produce 21.8g of NO gas and water? nan +318 a83451b5-6ddd-11ea-81d2-ccda262736ce https://socratic.org/questions/how-many-moles-of-sodium-hydroxide-are-needed-to-neutralize-1-mole-of-phosphoric 3.00 moles start physical_unit 4 5 mole mol qc_end physical_unit 13 14 10 11 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] sodium hydroxide [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] phosphoric acid [=] \\pu{1 mole}""}]" "

How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid?

" nan 3.00 moles "
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

+
+
" "
+
+
+

You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.

+
+
+
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

+
+
+
" "
+

How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + May 19, 2014 + +
+
+
+
+
+
+
+

You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.

+
+
+
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 39909 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid? nan +319 a83451b6-6ddd-11ea-a853-ccda262736ce https://socratic.org/questions/considering-the-reaction-shown-below-how-much-thermal-energy-would-be-required-t 5.22 kJ start physical_unit 1 2 thermal_energy kj qc_end physical_unit 20 20 17 18 mass qc_end physical_unit 25 25 22 23 mass qc_end physical_unit 1 2 35 36 deltah(rxn) qc_end chemical_equation 26 31 qc_end end "[{""type"":""physical unit"",""value"":""Thermal energy [OF] the reaction [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""5.22 kJ""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] hydrogen [=] \\pu{5.0 g}""},{""type"":""physical unit"",""value"":""Mass [OF] iodine [=] \\pu{25 g}""},{""type"":""physical unit"",""value"":""DeltaH(rxn) [OF] the reaction [=] \\pu{53 kJ}""},{""type"":""chemical equation"",""value"":""H2(g) + I2(g) -> 2 HI(g)""}]" "

Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine?

" "
+
+

+

H2(g) + I2(g) --> 2HI(g)
+deltaH(rxn) = +53 kJ

+

I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!

+

+
+
" 5.22 kJ "
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
" "
+
+
+

#""5.2 kJ""#

+
+
+
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
+
" "
+

Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine?

+
+
+

+

H2(g) + I2(g) --> 2HI(g)
+deltaH(rxn) = +53 kJ

+

I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you!

+

+
+
+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 8, 2017 + +
+
+
+
+
+
+
+

#""5.2 kJ""#

+
+
+
+

Explanation:

+
+

The thermochemical equation given to you tells you how much heat is needed in order to produce #2# moles of hydrogen iodide.

+

In other words, you know that in order to produce #2# moles of hydrogen iodide, you need #1# mole of hydrogen gas, #1# mole of iodine, and #""53 kJ""# of heat.

+
+

#DeltaH_""rxn"" = + ""53 kJ""#

+
+

The plus sign tells you that this reaction taken is #""53 kJ""# of heat, i.e. the reaction is endothermic.

+
+

#color(blue)(ul(color(black)(""1 mole I""_2 color(white)(.)""and 1 mole H""_2 -> ""53 kJ of heat consumed"")))#

+
+

So, convert the samples of hydrogen gas and iodine to moles by using the molar masses of the two reactants.

+
+

#5.0 color(red)(cancel(color(black)(""g""))) * ""1 mole H""_2/(2.016color(red)(cancel(color(black)(""g"")))) = ""2.48 moles H""_2#

+

#25 color(red)(cancel(color(black)(""g""))) * ""1 mole I""_2/(253.81color(red)(cancel(color(black)(""g"")))) = ""0.0985 moles I""_2#

+
+

The reaction consumes hydrogen gas and iodine in a #1:1# mole ratio, so you can say that iodine will act as a limiting reagent here because you have fewer moles of iodine than of hydrogen gas.

+

Therefore, the reaction will consume #0.0985# moles of iodine and of hydrogen gas and produce.

+

This means that the reaction will require

+
+

#0.0985 color(red)(cancel(color(black)(""moles I""_2))) * ""53 kJ""/(1color(red)(cancel(color(black)(""mole I""_2)))) = color(darkgreen)(ul(color(black)(""5.2 kJ"")))#

+
+

of heat. The answer is rounded to two sig figs. You can thus say that you have

+
+

#DeltaH_ (""rxn for 0.0985 moles I""_2 color(white)(.)""and H""_2) = + ""5.2 kJ""#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4566 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" Considering the reaction shown below, how much thermal energy would be required to run the reaction with 5.0 g of hydrogen and 25 g of iodine? " + + +H2(g) + I2(g) --> 2HI(g) +deltaH(rxn) = +53 kJ +I happen to know the answer is 5.2 kJ, but I'm not sure how to get to it. I'm studying for finals and need help. Thank you! + + +" +320 a83451b7-6ddd-11ea-a761-ccda262736ce https://socratic.org/questions/how-many-n-2o-4-molecules-are-contained-in-76-3-g-n-2o-4 4.99 × 10^23 start physical_unit 2 3 number none qc_end physical_unit 2 2 7 8 mass qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] N2O4 molecules""}]" "[{""type"":""physical unit"",""value"":""4.99 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N2O4 [=] \\pu{76.3 g}""}]" "

How many #N_2O_4# molecules are contained in 76.3 g#N_2O_4#?

" nan 4.99 × 10^23 "
+

Explanation:

+
+

#""Moles of dinitrogen tetroxide""# #=# #(76.3*g)/(92.011*g*mol^-1)#. This gives an answer in #""moles""#. Now it is a fact that in #1# #""mole""# of stuff there are #6.022xx10^23# individual items of that stuff:

+

#""Number of ""N_2O_4"" molecules""# #=# #6.022xx10^23mol^-1xx(76.3*g)/(92.011*g*mol^-1)# #=# #??"" molecules.""#

+
+
" "
+
+
+

We work out (i) the number of moles of #N_2O_4#, and (ii) convert this to molecules.

+
+
+
+

Explanation:

+
+

#""Moles of dinitrogen tetroxide""# #=# #(76.3*g)/(92.011*g*mol^-1)#. This gives an answer in #""moles""#. Now it is a fact that in #1# #""mole""# of stuff there are #6.022xx10^23# individual items of that stuff:

+

#""Number of ""N_2O_4"" molecules""# #=# #6.022xx10^23mol^-1xx(76.3*g)/(92.011*g*mol^-1)# #=# #??"" molecules.""#

+
+
+
" "
+

How many #N_2O_4# molecules are contained in 76.3 g#N_2O_4#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 10, 2016 + +
+
+
+
+
+
+
+

We work out (i) the number of moles of #N_2O_4#, and (ii) convert this to molecules.

+
+
+
+

Explanation:

+
+

#""Moles of dinitrogen tetroxide""# #=# #(76.3*g)/(92.011*g*mol^-1)#. This gives an answer in #""moles""#. Now it is a fact that in #1# #""mole""# of stuff there are #6.022xx10^23# individual items of that stuff:

+

#""Number of ""N_2O_4"" molecules""# #=# #6.022xx10^23mol^-1xx(76.3*g)/(92.011*g*mol^-1)# #=# #??"" molecules.""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4035 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many #N_2O_4# molecules are contained in 76.3 g#N_2O_4#? nan +321 a83451b8-6ddd-11ea-8c45-ccda262736ce https://socratic.org/questions/at-what-temperature-will-14-0-g-of-h-2-occupy-a-volume-of-15-5l-at-a-pressure-of 54.77 K start physical_unit 7 7 temperature k qc_end physical_unit 7 7 4 5 mass qc_end physical_unit 7 7 18 19 pressure qc_end physical_unit 7 7 12 13 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature [OF] H2 [IN] K""}]" "[{""type"":""physical unit"",""value"":""54.77 K""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2 [=] \\pu{14.0 g}""},{""type"":""physical unit"",""value"":""Pressure [OF] H2 [=] \\pu{204 kPa}""},{""type"":""physical unit"",""value"":""Volume [OF] H2 [=] \\pu{15.5 L}""}]" "

At what temperature will 14.0 g of #H_2# occupy a volume of 15.5L at a pressure of 204 kPa?

" nan 54.77 K "
+

Explanation:

+
+

Of course, we need to find the appropriate gas constant.

+

This site gives me #R= 8.314*L*kPa* K^-1* mol^-1#.

+

Given this, we plug in the numbers:

+

#T=(204*kPaxx15.5*L)/((14.0*g)/(2.016*g*mol^-1)xx8.314*L*kPa* K^-1* mol^-1)#

+

We should get an answer in #1/K^-1=1/(1/K)=K# as required for an #""absolute temperature""#.

+

The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.

+
+
" "
+
+
+

#T=(PV)/(nR)~=50*K#

+
+
+
+

Explanation:

+
+

Of course, we need to find the appropriate gas constant.

+

This site gives me #R= 8.314*L*kPa* K^-1* mol^-1#.

+

Given this, we plug in the numbers:

+

#T=(204*kPaxx15.5*L)/((14.0*g)/(2.016*g*mol^-1)xx8.314*L*kPa* K^-1* mol^-1)#

+

We should get an answer in #1/K^-1=1/(1/K)=K# as required for an #""absolute temperature""#.

+

The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.

+
+
+
" "
+

At what temperature will 14.0 g of #H_2# occupy a volume of 15.5L at a pressure of 204 kPa?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 19, 2017 + +
+
+
+
+
+
+
+

#T=(PV)/(nR)~=50*K#

+
+
+
+

Explanation:

+
+

Of course, we need to find the appropriate gas constant.

+

This site gives me #R= 8.314*L*kPa* K^-1* mol^-1#.

+

Given this, we plug in the numbers:

+

#T=(204*kPaxx15.5*L)/((14.0*g)/(2.016*g*mol^-1)xx8.314*L*kPa* K^-1* mol^-1)#

+

We should get an answer in #1/K^-1=1/(1/K)=K# as required for an #""absolute temperature""#.

+

The temperature is so low because there is a large molar quantity of dihydrogen. At this temperature we would anticipate some deviation from ideality. We are in no position to assess this.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1079 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" At what temperature will 14.0 g of #H_2# occupy a volume of 15.5L at a pressure of 204 kPa? nan +322 a83451b9-6ddd-11ea-88f6-ccda262736ce https://socratic.org/questions/how-much-heat-is-required-to-raise-the-temperature-of-a-50-0-g-block-of-iron-by- 224.50 J start physical_unit 13 15 heat_energy j qc_end physical_unit 13 15 11 12 mass qc_end physical_unit 13 15 17 18 temperature qc_end physical_unit 15 15 24 27 specific_heat qc_end end "[{""type"":""physical unit"",""value"":""Heat required [OF] block of iron [IN] J""}]" "[{""type"":""physical unit"",""value"":""224.50 J""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] block of iron [=] \\pu{50.0 g}""},{""type"":""physical unit"",""value"":""Temperature raised [OF] block of iron [=] \\pu{10.0 ℃}""},{""type"":""physical unit"",""value"":""Specific heat [OF] iron [=] \\pu{0.449 J/(g * ℃)}""}]" "

How much heat is required to raise the temperature of a 50.0 g block of iron by +#10.0^@""C""# ? Specific heat of iron: #""0.449 J/g""""""^@""C""#

" nan 224.50 J "
+

Explanation:

+
+

The key to this problem is the specific heat of iron, which is said to be equal to

+
+

#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#

+
+

As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of #""1 g""# by #1^@""C""#.

+

Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!

+

So, how much heat would be needed to increase the temperature of a #""50.0 g""# block of iron by #1^@""C""# ?

+
+

#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#

+
+

This means that in order to increase the temperature of #""50.0 g""# of iron by #1^@""C""# you need to provide it with #""22.45 J""# of heat.

+

You can thus say that in order to increase the temperature of the block by #10.0^@""C""#, you must provide ten times as much heat

+
+

#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
" "
+
+
+

#""225 J""#

+
+
+
+

Explanation:

+
+

The key to this problem is the specific heat of iron, which is said to be equal to

+
+

#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#

+
+

As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of #""1 g""# by #1^@""C""#.

+

Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!

+

So, how much heat would be needed to increase the temperature of a #""50.0 g""# block of iron by #1^@""C""# ?

+
+

#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#

+
+

This means that in order to increase the temperature of #""50.0 g""# of iron by #1^@""C""# you need to provide it with #""22.45 J""# of heat.

+

You can thus say that in order to increase the temperature of the block by #10.0^@""C""#, you must provide ten times as much heat

+
+

#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
" "
+

How much heat is required to raise the temperature of a 50.0 g block of iron by +#10.0^@""C""# ? Specific heat of iron: #""0.449 J/g""""""^@""C""#

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 15, 2016 + +
+
+
+
+
+
+
+

#""225 J""#

+
+
+
+

Explanation:

+
+

The key to this problem is the specific heat of iron, which is said to be equal to

+
+

#c_""iron"" = ""0.449 J g""^(-1)""""^@""C""^(-1)#

+
+

As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of #""1 g""# by #1^@""C""#.

+

Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!

+

So, how much heat would be needed to increase the temperature of a #""50.0 g""# block of iron by #1^@""C""# ?

+
+

#50.0 color(red)(cancel(color(black)(""g""))) * overbrace(""0.449 J"" /(1color(red)(cancel(color(black)(""g""))) """"^@""C""))^(color(blue)(""the specific heat of iron"")) = ""22.45 J"" """"^@""C""^(-1)#

+
+

This means that in order to increase the temperature of #""50.0 g""# of iron by #1^@""C""# you need to provide it with #""22.45 J""# of heat.

+

You can thus say that in order to increase the temperature of the block by #10.0^@""C""#, you must provide ten times as much heat

+
+

#10.0 color(red)(cancel(color(black)(""""^@""C""))) * overbrace(""22.45 J""/(1color(red)(cancel(color(black)(""""^@""C"")))))^(color(blue)(""for 50.0 g of iron"")) = color(green)(bar(ul(|color(white)(a/a)color(black)(""225 J"")color(white)(a/a)|)))#

+
+

The answer is rounded to three sig figs.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5049 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "How much heat is required to raise the temperature of a 50.0 g block of iron by +#10.0^@""C""# ? Specific heat of iron: #""0.449 J/g""""""^@""C""#" nan +323 a8347b42-6ddd-11ea-ab3c-ccda262736ce https://socratic.org/questions/find-the-number-of-co2-that-exert-a-pressure-of-785torrs-at-a-volume-of-32-5l-an 1.34 moles start physical_unit 4 4 number mol qc_end physical_unit 4 4 10 11 pressure qc_end physical_unit 4 4 16 17 volume qc_end physical_unit 4 4 21 22 temperature qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] CO2 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""1.34 moles""}]" "[{""type"":""physical unit"",""value"":""Pressure exerted [OF] CO2 [=] \\pu{785 torr}""},{""type"":""physical unit"",""value"":""Volume [OF] CO2 [=] \\pu{32.5 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] CO2 [=] \\pu{32 ℃}""}]" "

Find the number of #""CO""_2# that exert a pressure of #""785 torr""# at a volume of #""32.5 L""# and a temperature #32^@""C""#?

" "
+
+

+

find the number of co2 that exert a pressure of 785torrs at a volume of 32.5L and a temperature 32c ?

+

+
+
" 1.34 moles "
+

Explanation:

+
+

I'm assuming that the question wants you to determine the number of moles of carbon dioxide, #""CO""_2#, that would occupy that volume under those specific conditions for pressure and temperature.

+

Your tool of choice here will be the ideal gas law equation

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, equal to #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the absolute temperature of the gas

+

Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of #R# do not match those given to you.

+

This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of #R#. The conversion factors you'll need are

+
+

#""1 atm "" = "" 760 torr""#

+

#t [""K""] = 273.15 + t [""""^@""C""]#

+
+

Rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

Plug in your values to get

+
+

#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#

+
+

Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

+
+

#n = color(green)(""1.3 moles CO""_2)#

+
+
+
" "
+
+
+

#""1.3 moles CO""_2#

+
+
+
+

Explanation:

+
+

I'm assuming that the question wants you to determine the number of moles of carbon dioxide, #""CO""_2#, that would occupy that volume under those specific conditions for pressure and temperature.

+

Your tool of choice here will be the ideal gas law equation

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, equal to #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the absolute temperature of the gas

+

Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of #R# do not match those given to you.

+

This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of #R#. The conversion factors you'll need are

+
+

#""1 atm "" = "" 760 torr""#

+

#t [""K""] = 273.15 + t [""""^@""C""]#

+
+

Rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

Plug in your values to get

+
+

#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#

+
+

Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

+
+

#n = color(green)(""1.3 moles CO""_2)#

+
+
+
+
" "
+

Find the number of #""CO""_2# that exert a pressure of #""785 torr""# at a volume of #""32.5 L""# and a temperature #32^@""C""#?

+
+
+

+

find the number of co2 that exert a pressure of 785torrs at a volume of 32.5L and a temperature 32c ?

+

+
+
+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 25, 2016 + +
+
+
+
+
+
+
+

#""1.3 moles CO""_2#

+
+
+
+

Explanation:

+
+

I'm assuming that the question wants you to determine the number of moles of carbon dioxide, #""CO""_2#, that would occupy that volume under those specific conditions for pressure and temperature.

+

Your tool of choice here will be the ideal gas law equation

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, equal to #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the absolute temperature of the gas

+

Now, the problem provides you with everything that you need in order to solve for the number of moles of gas. However, notice that some of the units used in the expression of #R# do not match those given to you.

+

This means that you're going to have to do a couple of units conversions to get your units to match those used in expression of #R#. The conversion factors you'll need are

+
+

#""1 atm "" = "" 760 torr""#

+

#t [""K""] = 273.15 + t [""""^@""C""]#

+
+

Rearrange the ideal gas law equation to solve for #n#

+
+

#PV = nRT implies n = (PV)/(RT)#

+
+

Plug in your values to get

+
+

#n = (785/760color(red)(cancel(color(black)(""atm""))) * 32.5color(red)(cancel(color(black)(""L""))))/(0.0821(color(red)(cancel(color(black)(""atm""))) * color(red)(cancel(color(black)(""L""))))/(""mol"" * color(red)(cancel(color(black)(""K"")))) * (273.15 + 32)color(red)(cancel(color(black)(""K"")))) = ""1.3399 moles""#

+
+

Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

+
+

#n = color(green)(""1.3 moles CO""_2)#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3426 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "Find the number of #""CO""_2# that exert a pressure of #""785 torr""# at a volume of #""32.5 L""# and a temperature #32^@""C""#?" " + + +find the number of co2 that exert a pressure of 785torrs at a volume of 32.5L and a temperature 32c ? + + +" +324 a8347b43-6ddd-11ea-aa70-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-has-93-moles-of-naoh-in-1-5-liters-of-so 0.62 mol/L start physical_unit 5 6 molarity mol/l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 6 6 14 15 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] a solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.62 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] NaOH [=] \\pu{0.93 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{1.5 liters}""}]" "

What is the molarity of a solution that has .93 moles of #NaOH# in 1.5 liters of solution?

" nan 0.62 mol/L "
+

Explanation:

+
+

The equation relating moles and volume is #""M""=""n""/""V""#, where #""M""# is molarity, #""n""# is moles, and #""V""# is volume in liters.

+

Given values are #""0.93 mol NaOH""#, and #""1.5 L solution""#.

+

Substitute the given values into the equation.

+

#""M""=(0.93""mol NaOH"")/(1.5""L solution"")=""0.62 mol/L""#

+
+
" "
+
+
+

The molarity of the solution is #""0.62 mol/L""#.

+
+
+
+

Explanation:

+
+

The equation relating moles and volume is #""M""=""n""/""V""#, where #""M""# is molarity, #""n""# is moles, and #""V""# is volume in liters.

+

Given values are #""0.93 mol NaOH""#, and #""1.5 L solution""#.

+

Substitute the given values into the equation.

+

#""M""=(0.93""mol NaOH"")/(1.5""L solution"")=""0.62 mol/L""#

+
+
+
" "
+

What is the molarity of a solution that has .93 moles of #NaOH# in 1.5 liters of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 22, 2016 + +
+
+
+
+
+
+
+

The molarity of the solution is #""0.62 mol/L""#.

+
+
+
+

Explanation:

+
+

The equation relating moles and volume is #""M""=""n""/""V""#, where #""M""# is molarity, #""n""# is moles, and #""V""# is volume in liters.

+

Given values are #""0.93 mol NaOH""#, and #""1.5 L solution""#.

+

Substitute the given values into the equation.

+

#""M""=(0.93""mol NaOH"")/(1.5""L solution"")=""0.62 mol/L""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1175 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the molarity of a solution that has .93 moles of #NaOH# in 1.5 liters of solution? nan +325 a8347b44-6ddd-11ea-b055-ccda262736ce https://socratic.org/questions/the-decomposition-of-potassium-chlorate-kclo-3-is-used-as-a-source-of-oxygen-in- 1.61 kg start physical_unit 3 4 mass kg qc_end physical_unit 12 12 24 25 mole qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] potassium chlorate [IN] kg""}]" "[{""type"":""physical unit"",""value"":""1.61 kg""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] oxygen [=] \\pu{19.7 mol}""},{""type"":""other"",""value"":""The decomposition of potassium KClO3 is used as a source of oxygen in the laboratory.""}]" "

The decomposition of potassium chlorate (#KClO_3#) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

" nan 1.61 kg "
+

Explanation:

+
+

Normally, this rxn must be calalyzed with a little #Mn(IV)# salt, but we address the question with this stoichiometry.

+

We spit out a #19.7*mol# quantity of dioxygen gas. Given the stoichiometry there were #2/3xx19.7*mol# #""potassium chlorate""#.

+

This represents a mass of #2/3xx19.7*molxx122.55*g*mol^-1# of #""potassium chlorate""# #~~# #1.6*kg#.

+

This is not a reaction that you do at home.

+
+
" "
+
+
+

#KClO_3(s) + Delta rarr KCl(s) + 3/2O_2(g)#

+
+
+
+

Explanation:

+
+

Normally, this rxn must be calalyzed with a little #Mn(IV)# salt, but we address the question with this stoichiometry.

+

We spit out a #19.7*mol# quantity of dioxygen gas. Given the stoichiometry there were #2/3xx19.7*mol# #""potassium chlorate""#.

+

This represents a mass of #2/3xx19.7*molxx122.55*g*mol^-1# of #""potassium chlorate""# #~~# #1.6*kg#.

+

This is not a reaction that you do at home.

+
+
+
" "
+

The decomposition of potassium chlorate (#KClO_3#) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Decomposition Reactions + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 25, 2017 + +
+
+
+
+
+
+
+

#KClO_3(s) + Delta rarr KCl(s) + 3/2O_2(g)#

+
+
+
+

Explanation:

+
+

Normally, this rxn must be calalyzed with a little #Mn(IV)# salt, but we address the question with this stoichiometry.

+

We spit out a #19.7*mol# quantity of dioxygen gas. Given the stoichiometry there were #2/3xx19.7*mol# #""potassium chlorate""#.

+

This represents a mass of #2/3xx19.7*molxx122.55*g*mol^-1# of #""potassium chlorate""# #~~# #1.6*kg#.

+

This is not a reaction that you do at home.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 7546 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" The decomposition of potassium chlorate (#KClO_3#) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen? nan +326 a8347b45-6ddd-11ea-aba5-ccda262736ce https://socratic.org/questions/calculate-the-mass-present-in-0-75-mole-of-sodium-chloride 43.80 g start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] sodium chloride [IN] g""}]" "[{""type"":""physical unit"",""value"":""43.80 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sodium chloride [=] \\pu{0.75 mole}""}]" "

Calculate the mass present in 0.75 mole of sodium chloride?

" nan 43.80 g "
+

Explanation:

+
+

#""no. of moles"" xx ""molar mass"" = ""grams (or mass)""#

+

The molar mass of sodium chloride is #""58.4 g/mol""#, so

+

#""0.75 mol"" xx ""58.4 g/mol"" = ""44 g""#

+
+
" "
+
+
+

#""44 g""#

+
+
+
+

Explanation:

+
+

#""no. of moles"" xx ""molar mass"" = ""grams (or mass)""#

+

The molar mass of sodium chloride is #""58.4 g/mol""#, so

+

#""0.75 mol"" xx ""58.4 g/mol"" = ""44 g""#

+
+
+
" "
+

Calculate the mass present in 0.75 mole of sodium chloride?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Mar 20, 2018 + +
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+

#""44 g""#

+
+
+
+

Explanation:

+
+

#""no. of moles"" xx ""molar mass"" = ""grams (or mass)""#

+

The molar mass of sodium chloride is #""58.4 g/mol""#, so

+

#""0.75 mol"" xx ""58.4 g/mol"" = ""44 g""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6386 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" Calculate the mass present in 0.75 mole of sodium chloride? nan +327 a8347b46-6ddd-11ea-88ef-ccda262736ce https://socratic.org/questions/5806731d11ef6b0f490563a6 2.12 × 10^23 start physical_unit 2 3 number none qc_end physical_unit 12 13 8 9 mass qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] phosphorus atoms""}]" "[{""type"":""physical unit"",""value"":""2.12 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] phosphorus pentoxide [=] \\pu{25.0 g}""}]" "

How many phosphorus atoms are present in a #25.0*g# mass of #""phosphorus pentoxide""#?

" nan 2.12 × 10^23 "
+

Explanation:

+
+

Number of moles of #P_2O_5# #=# #(25.0*g)/(141.95*g*mol^-1)# #=# #0.176*mol#, with respect to #P_2O_5#.

+

There are thus #2xx0.176*mol# of #""phosphorus""#.

+

#P_2O_5# is an empirical formula. The actual molecule is #P_4O_10#, which is an extremely hygroscopic white powder. Do the different formulations make a difference to the number of phosphorus atoms present?

+
+
" "
+
+
+

#""0.352 moles of phosphorus atoms are present in this quantity.""#

+
+
+
+

Explanation:

+
+

Number of moles of #P_2O_5# #=# #(25.0*g)/(141.95*g*mol^-1)# #=# #0.176*mol#, with respect to #P_2O_5#.

+

There are thus #2xx0.176*mol# of #""phosphorus""#.

+

#P_2O_5# is an empirical formula. The actual molecule is #P_4O_10#, which is an extremely hygroscopic white powder. Do the different formulations make a difference to the number of phosphorus atoms present?

+
+
+
" "
+

How many phosphorus atoms are present in a #25.0*g# mass of #""phosphorus pentoxide""#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Oct 18, 2016 + +
+
+
+
+
+
+
+

#""0.352 moles of phosphorus atoms are present in this quantity.""#

+
+
+
+

Explanation:

+
+

Number of moles of #P_2O_5# #=# #(25.0*g)/(141.95*g*mol^-1)# #=# #0.176*mol#, with respect to #P_2O_5#.

+

There are thus #2xx0.176*mol# of #""phosphorus""#.

+

#P_2O_5# is an empirical formula. The actual molecule is #P_4O_10#, which is an extremely hygroscopic white powder. Do the different formulations make a difference to the number of phosphorus atoms present?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4665 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "How many phosphorus atoms are present in a #25.0*g# mass of #""phosphorus pentoxide""#?" nan +328 a8347b47-6ddd-11ea-a8de-ccda262736ce https://socratic.org/questions/what-is-the-mass-percent-of-hydrogen-in-ammonium-phosphate-nh-4-3po-4 8.05% start physical_unit 6 9 mass_percent none qc_end chemical_equation 10 10 qc_end end "[{""type"":""physical unit"",""value"":""Mass percent [OF] hydrogen in ammonium phosphate""}]" "[{""type"":""physical unit"",""value"":""8.05%""}]" "[{""type"":""chemical equation"",""value"":""(NH4)3PO4""}]" "

What is the mass percent of hydrogen in ammonium phosphate, #(NH_4)_3PO_4#?

" nan 8.05% "
+

Explanation:

+
+

#(3xx1.00794*g*mol^-1)/(149.09*g*mol^-1)xx100%# #""""approx# #2%#

+
+
" "
+
+
+

#""Mass of hydrogen""/""Mass of ammonium phosphate""# #xx100%=??%#

+
+
+
+

Explanation:

+
+

#(3xx1.00794*g*mol^-1)/(149.09*g*mol^-1)xx100%# #""""approx# #2%#

+
+
+
" "
+

What is the mass percent of hydrogen in ammonium phosphate, #(NH_4)_3PO_4#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 1, 2016 + +
+
+
+
+
+
+
+

#""Mass of hydrogen""/""Mass of ammonium phosphate""# #xx100%=??%#

+
+
+
+

Explanation:

+
+

#(3xx1.00794*g*mol^-1)/(149.09*g*mol^-1)xx100%# #""""approx# #2%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5793 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the mass percent of hydrogen in ammonium phosphate, #(NH_4)_3PO_4#? nan +329 a8347b48-6ddd-11ea-923a-ccda262736ce https://socratic.org/questions/20ml-of-orange-juice-citric-acid-weighing-20-31g-was-neutralized-by-20-4ml-of-na 0.58% start physical_unit 38 43 mass_percent none qc_end chemical_equation 23 30 qc_end physical_unit 3 6 0 1 volume qc_end physical_unit 3 4 8 9 mass qc_end physical_unit 18 18 13 14 volume qc_end physical_unit 18 18 15 16 molarity qc_end end "[{""type"":""physical unit"",""value"":""Percentage by mass [OF] citric acid in the orange juice""}]" "[{""type"":""physical unit"",""value"":""0.58%""}]" "[{""type"":""chemical equation"",""value"":""C6H8O7 + 3 NaOH -> Na3C6H5O7 + H2O""},{""type"":""physical unit"",""value"":""Volume [OF] orange juice (citric acid) [=] \\pu{20 mL}""},{""type"":""physical unit"",""value"":""Weight [OF] orange juice [=] \\pu{20.31 g}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{20.4 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [=] \\pu{0.0900 M}""}]" "

20mL of orange juice (citric acid), weighing 20.31g, was neutralized by 20.4mL, 0.0900M of NaOH. Given the following equation: C6H8O7 + 3NaOH --> Na3C6H5O7 + H2O. What is the percentage by mass of citric acid in the orange juice? +

" nan 0.58% "
+

Explanation:

+
+

The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.

+

So, start with the balanced chemical equation for this neutralization reaction

+
+

#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#

+
+

Notice that your reaction needs #color(red)(3)# moles of sodium hydroxide for every mole of citric acid. This means that you can find the number of moles of citric acid by using the molarity of the sodium hydroxide solution and this moel ratio.

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#

+
+

This means that the orange juice contained

+
+

#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#

+
+

Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles

+
+

#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#

+
+

Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by #100#.

+
+

#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#

+
+

In your case, the percent concentration by mass of the orange juice will be

+
+

#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#

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" "
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+

#""0.579%""#

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+
+

Explanation:

+
+

The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.

+

So, start with the balanced chemical equation for this neutralization reaction

+
+

#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#

+
+

Notice that your reaction needs #color(red)(3)# moles of sodium hydroxide for every mole of citric acid. This means that you can find the number of moles of citric acid by using the molarity of the sodium hydroxide solution and this moel ratio.

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#

+
+

This means that the orange juice contained

+
+

#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#

+
+

Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles

+
+

#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#

+
+

Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by #100#.

+
+

#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#

+
+

In your case, the percent concentration by mass of the orange juice will be

+
+

#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#

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+
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" "
+

20mL of orange juice (citric acid), weighing 20.31g, was neutralized by 20.4mL, 0.0900M of NaOH. Given the following equation: C6H8O7 + 3NaOH --> Na3C6H5O7 + H2O. What is the percentage by mass of citric acid in the orange juice? +

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
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+1 Answer +
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+ + Nov 1, 2015 + +
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#""0.579%""#

+
+
+
+

Explanation:

+
+

The idea here is that you need to find the moles of citric acid by using the molarity of the sodium hydroxide solution, then use the mass of the orange juice to find the percent concentration by mass of citric acid.

+

So, start with the balanced chemical equation for this neutralization reaction

+
+

#""C""_5""H""_8""O""_text(7(aq]) + color(red)(3)""NaOH""_text((aq]) -> ""Na""_3""C""_6""H""_5""O""_text(7(aq]) + ""H""_2""O""_text((l])#

+
+

Notice that your reaction needs #color(red)(3)# moles of sodium hydroxide for every mole of citric acid. This means that you can find the number of moles of citric acid by using the molarity of the sodium hydroxide solution and this moel ratio.

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_""NaOH"" = ""0.090 M"" * 20.4 * 10^(-3)""L"" = ""0.001836 moles NaOH""#

+
+

This means that the orange juice contained

+
+

#0.001836color(red)(cancel(color(black)(""moles NaOH""))) * ""1 mole citric acid""/(color(red)(3)color(red)(cancel(color(black)(""moles NaOH"")))) = ""0.0006120 moles C""_3""H""_8""O""_7#

+
+

Now that you know how many moles of citric acid you had in the orange juice, use its molar mass to determine how many moles of citric acid would contain this many moles

+
+

#0.000612color(red)(cancel(color(black)(""moles""))) * ""192.124 g""/(1color(red)(cancel(color(black)(""mole"")))) = ""0.11758 g C""_3""H""_8""O""_7#

+
+

Now, a solution's percent concentration by mass is defined as the ratio between the mass of the solute, which in your case is citric acid, and the total mass of the solution, multiplied by #100#.

+
+

#color(blue)(""%m/m"" = ""mass of solute""/""mass of soluteion"" xx 100)#

+
+

In your case, the percent concentration by mass of the orange juice will be

+
+

#""%m/m"" = (0.11758color(red)(cancel(color(black)(""g""))))/(20.31color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(""0.579% C""_3""H""_8""O""_7)#

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" "20mL of orange juice (citric acid), weighing 20.31g, was neutralized by 20.4mL, 0.0900M of NaOH. Given the following equation: C6H8O7 + 3NaOH --> Na3C6H5O7 + H2O. What is the percentage by mass of citric acid in the orange juice? +" nan +330 a8347b49-6ddd-11ea-a902-ccda262736ce https://socratic.org/questions/when-the-redox-equation-cr-3-aq-3mn-s-mn-2-aq-cr-s-is-completely-balanced-what-w 2 start physical_unit 4 4 coefficient none qc_end chemical_equation 4 11 qc_end end "[{""type"":""physical unit"",""value"":""Coefficient [OF] Cr^3+(aq)""}]" "[{""type"":""physical unit"",""value"":""2""}]" "[{""type"":""chemical equation"",""value"":""Cr^3+(aq) + 3 Mn(s) -> Mn^2+(aq) + Cr(s)""}]" "

When the redox equation #Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)# is completely balanced, what will the coefficient of #Cr^(3+)(aq)# be?

" nan 2 "
+

Explanation:

+
+

Manganese is oxidized:

+

#stackrel0Mn(s)rarrMn^(2+) + 2e^(-)# #(i)#

+

Chromium is reduced:

+

#Cr^(3+) + 3e^(-)rarrstackrel0Cr(s)# #(ii)#

+

And so we take #3xx(i) + 2xx(ii)# to eliminate the electrons:

+

#3stackrel0Mn(s)+2Cr^(3+) rarr3Mn^(2+) + 2stackrel(0)Cr#

+

Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.

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" "
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#2...........#

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Explanation:

+
+

Manganese is oxidized:

+

#stackrel0Mn(s)rarrMn^(2+) + 2e^(-)# #(i)#

+

Chromium is reduced:

+

#Cr^(3+) + 3e^(-)rarrstackrel0Cr(s)# #(ii)#

+

And so we take #3xx(i) + 2xx(ii)# to eliminate the electrons:

+

#3stackrel0Mn(s)+2Cr^(3+) rarr3Mn^(2+) + 2stackrel(0)Cr#

+

Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.

+
+
+
" "
+

When the redox equation #Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)# is completely balanced, what will the coefficient of #Cr^(3+)(aq)# be?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Redox Reactions + + +
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+1 Answer +
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#2...........#

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Explanation:

+
+

Manganese is oxidized:

+

#stackrel0Mn(s)rarrMn^(2+) + 2e^(-)# #(i)#

+

Chromium is reduced:

+

#Cr^(3+) + 3e^(-)rarrstackrel0Cr(s)# #(ii)#

+

And so we take #3xx(i) + 2xx(ii)# to eliminate the electrons:

+

#3stackrel0Mn(s)+2Cr^(3+) rarr3Mn^(2+) + 2stackrel(0)Cr#

+

Charge is balanced, and mass is balanced, and so this is a reasonable representation of reality. Whether the reaction actually works, I don't know, and cannot be bothered to look up redox tables.

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" When the redox equation #Cr^(3+)(aq) + 3Mn(s) -> Mn^(2+)(aq) + Cr(s)# is completely balanced, what will the coefficient of #Cr^(3+)(aq)# be? nan +331 a8349fae-6ddd-11ea-8f06-ccda262736ce https://socratic.org/questions/weight-of-50-atoms-of-an-element-is-3000-a-m-u-calculate-the-number-of-atoms-in- 2.41 × 10^24 start physical_unit 14 19 number none qc_end physical_unit 5 6 8 9 mass qc_end physical_unit 3 6 2 2 number qc_end physical_unit 5 6 16 17 mass qc_end end "[{""type"":""physical unit"",""value"":""Number2 [OF] atoms in 240 grams of element""}]" "[{""type"":""physical unit"",""value"":""2.41 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Weight1 [OF] an element [=] \\pu{3000 amu}""},{""type"":""physical unit"",""value"":""Number1 [OF] atoms of an element [=] \\pu{50}""},{""type"":""physical unit"",""value"":""Weight2 [OF] an element [=] \\pu{240 grams}""}]" "

Weight of 50 atoms of an element is 3000 a.m.u. Calculate the number of atoms in 240 grams of element?

" nan 2.41 × 10^24 "
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Explanation:

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#""Mass of 1 atom"" = 1 color(red)(cancel(color(black)(""atom""))) × ""3000 u""/(50 color(red)(cancel(color(black)(""atoms"")))) = ""60.0 u""#

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#""Mass of 1 mol of atoms = 60.0 g""#

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#""Number of moles"" = 240 color(red)(cancel(color(black)(""g""))) × ""1 mol""/(60.0 color(red)(cancel(color(black)(""g"")))) = ""4.00 mol""#

+

#""Number of atoms"" = 4.00 color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""atoms"")/(1 color(red)(cancel(color(black)(""mol"")))) = 2.41 × 10^24color(white)(l) ""atoms""#

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There are #2.41 × 10^24 color(white)(l)""atoms""#.

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Explanation:

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#""Mass of 1 atom"" = 1 color(red)(cancel(color(black)(""atom""))) × ""3000 u""/(50 color(red)(cancel(color(black)(""atoms"")))) = ""60.0 u""#

+

#""Mass of 1 mol of atoms = 60.0 g""#

+

#""Number of moles"" = 240 color(red)(cancel(color(black)(""g""))) × ""1 mol""/(60.0 color(red)(cancel(color(black)(""g"")))) = ""4.00 mol""#

+

#""Number of atoms"" = 4.00 color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""atoms"")/(1 color(red)(cancel(color(black)(""mol"")))) = 2.41 × 10^24color(white)(l) ""atoms""#

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" "
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Weight of 50 atoms of an element is 3000 a.m.u. Calculate the number of atoms in 240 grams of element?

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+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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There are #2.41 × 10^24 color(white)(l)""atoms""#.

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Explanation:

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#""Mass of 1 atom"" = 1 color(red)(cancel(color(black)(""atom""))) × ""3000 u""/(50 color(red)(cancel(color(black)(""atoms"")))) = ""60.0 u""#

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#""Mass of 1 mol of atoms = 60.0 g""#

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#""Number of moles"" = 240 color(red)(cancel(color(black)(""g""))) × ""1 mol""/(60.0 color(red)(cancel(color(black)(""g"")))) = ""4.00 mol""#

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#""Number of atoms"" = 4.00 color(red)(cancel(color(black)(""mol""))) × (6.022 × 10^23color(white)(l) ""atoms"")/(1 color(red)(cancel(color(black)(""mol"")))) = 2.41 × 10^24color(white)(l) ""atoms""#

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" Weight of 50 atoms of an element is 3000 a.m.u. Calculate the number of atoms in 240 grams of element? nan +332 a8349faf-6ddd-11ea-90a0-ccda262736ce https://socratic.org/questions/58f6e7b6b72cff45e0bf2c5b +4 start physical_unit 6 10 oxidation_number none qc_end chemical_equation 10 10 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] phosphorus in the salt Zr3P4O16""}]" "[{""type"":""physical unit"",""value"":""+4""}]" "[{""type"":""chemical equation"",""value"":""Zr3P4O16""}]" "

What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#?

" nan +4 "
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Explanation:

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+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

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" "
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Well, the parent phosphate ion is #PO_4^(3-)#...........

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Explanation:

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+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

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" "
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What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#?

+
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+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
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+1 Answer +
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Well, the parent phosphate ion is #PO_4^(3-)#...........

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Explanation:

+
+

........And so the metal oxidation state would be #stackrel(+IV)M#.

+

And thus its perhalide formula would be #MX_4#, i.e. #M^(4+) + 4xxX^-#.

+

You might get a #""zirconium (IV) phosphate""#, you can certainly get #""zirconium tetrahalide""#.

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" What is the oxidation number of phosphorus in the salt #Zr_3P_4O_16#? nan +333 a8349fb0-6ddd-11ea-9d97-ccda262736ce https://socratic.org/questions/if-i-have-21-moles-of-gas-held-at-a-pressure-of-3800-torr-and-a-temperature-of-6 310.19 L start physical_unit 25 26 volume l qc_end physical_unit 6 6 18 19 temperature qc_end physical_unit 6 6 3 4 mole qc_end physical_unit 6 6 12 13 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] the gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""310.19 L""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] gas [=] \\pu{627 ℃}""},{""type"":""physical unit"",""value"":""Mole [OF] gas [=] \\pu{21 moles}""},{""type"":""physical unit"",""value"":""Pressure [OF] gas [=] \\pu{3800 torr}""}]" "

If I have 21 moles of gas held at a pressure of 3800 torr and a temperature of 627°C what is the volume of the gas?

" nan 310.19 L "
+

Explanation:

+
+

Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
+

+

Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding #273# to the temperature given in degrees Celsius:

+

+

Therefore, #627^(o)C + 273 = 900K#

+

The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
+
+Thus, #3800cancel""torr""xx(1atm)/(760cancel""torr"")# = 5atm

+

R has the same value no matter what chemical species you are dealing with.

+

Now we know P,T, n, and R.

+

All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:

+

#(cancelPV)/cancelP = (nRT)/P#

+

#V=(nRT)/P#

+

Finally, plug in your known values like so:

+

#V = (21cancel""mol""xx0.08206Lcancel""atm""/cancel""molK""xx900cancelK)/(5cancel""atm"")#

+

#V = 310 L#

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" "
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The gas has a volume of #310L#

+
+
+
+

Explanation:

+
+

Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
+

+

Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding #273# to the temperature given in degrees Celsius:

+

+

Therefore, #627^(o)C + 273 = 900K#

+

The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
+
+Thus, #3800cancel""torr""xx(1atm)/(760cancel""torr"")# = 5atm

+

R has the same value no matter what chemical species you are dealing with.

+

Now we know P,T, n, and R.

+

All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:

+

#(cancelPV)/cancelP = (nRT)/P#

+

#V=(nRT)/P#

+

Finally, plug in your known values like so:

+

#V = (21cancel""mol""xx0.08206Lcancel""atm""/cancel""molK""xx900cancelK)/(5cancel""atm"")#

+

#V = 310 L#

+
+
+
" "
+

If I have 21 moles of gas held at a pressure of 3800 torr and a temperature of 627°C what is the volume of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
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+1 Answer +
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+ + Aug 12, 2016 + +
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The gas has a volume of #310L#

+
+
+
+

Explanation:

+
+

Because we are given the pressure, number of moles, and temperature, we will have to use the ideal gas law equation:
+

+

Based on the units that are associated with each variable, the number of moles have good units. The temperature has to be converted to Kelvins; we can do this by adding #273# to the temperature given in degrees Celsius:

+

+

Therefore, #627^(o)C + 273 = 900K#

+

The next issue is that pressure has units of torr instead of atmospheres; we can use the following relationship to give the correct units:
+
+Thus, #3800cancel""torr""xx(1atm)/(760cancel""torr"")# = 5atm

+

R has the same value no matter what chemical species you are dealing with.

+

Now we know P,T, n, and R.

+

All we have to do is rearrange the equation to solve for V. We can do this by dividing by the pressure on both sides of the equation:

+

#(cancelPV)/cancelP = (nRT)/P#

+

#V=(nRT)/P#

+

Finally, plug in your known values like so:

+

#V = (21cancel""mol""xx0.08206Lcancel""atm""/cancel""molK""xx900cancelK)/(5cancel""atm"")#

+

#V = 310 L#

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" If I have 21 moles of gas held at a pressure of 3800 torr and a temperature of 627°C what is the volume of the gas? nan +334 a8349fb1-6ddd-11ea-acbf-ccda262736ce https://socratic.org/questions/57f89f7411ef6b54be557b99 55.52 mol/L start physical_unit 5 5 concentration mol/l qc_end substance 5 5 qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] water [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""55.52 mol/L""}]" "[{""type"":""substance name"",""value"":""water""}]" "

What is the concentration of water?

" nan 55.52 mol/L "
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#.

+

For #1*L# of water, we have:

+

#(1000*g)/(18.01*g*mol^-1)xx1/(1*L)# #=# #??*mol*L^-1#.

+

Water is an exceptionally concentrated solvent.

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" "
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Around about #55*mol*L^-1#

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+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#.

+

For #1*L# of water, we have:

+

#(1000*g)/(18.01*g*mol^-1)xx1/(1*L)# #=# #??*mol*L^-1#.

+

Water is an exceptionally concentrated solvent.

+
+
+
" "
+

What is the concentration of water?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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Around about #55*mol*L^-1#

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Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#.

+

For #1*L# of water, we have:

+

#(1000*g)/(18.01*g*mol^-1)xx1/(1*L)# #=# #??*mol*L^-1#.

+

Water is an exceptionally concentrated solvent.

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+ + Creative Commons License + +
+
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+
+
" What is the concentration of water? nan +335 a8349fb2-6ddd-11ea-a79c-ccda262736ce https://socratic.org/questions/if-you-combine-250-0-ml-of-water-at-25-c-and-120-0-ml-of-water-at-95-c-what-is-t 47.07 ℃ start physical_unit 24 25 temperature °c qc_end physical_unit 6 6 3 4 volume qc_end physical_unit 6 6 8 9 temperature qc_end physical_unit 6 6 11 12 volume qc_end physical_unit 6 6 16 17 temperature qc_end end "[{""type"":""physical unit"",""value"":""Temperature3 [OF] the mixture [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""47.07 ℃""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] water [=] \\pu{250.0 mL}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] water [=] \\pu{25 ℃}""},{""type"":""physical unit"",""value"":""Volume2 [OF] water [=] \\pu{120.0 mL}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{95 ℃}""}]" "

If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture?

" nan 47.07 ℃ "
+

Explanation:

+
+

We can use a heat equilibrium equation to find this value.

+

#DeltaH = m(T_f-T_i)C_p#

+

#DeltaH =# Change in Heat
+#m =# Mass
+#T=# Tempurature
+#C_p=# Specific Heat

+

#DeltaH# cold #H_2O# = -#DeltaH# hot #H_2O#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

Since the density of water is #1 g/(mL)# the volume of water is also equal to the mass of the water in grams.

+

Cold #H_2O#

+

#m = 250.0g#
+#T_f= ?#
+#T_i = 25^oC#
+#C_p = 4.18 J/(g^oC)#

+

Cold #H_2O#

+

#m = 120.0g#
+#T_f= ?#
+#T_i = 95^oC#
+#C_p = 4.18 J/(g^oC)#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

#250.0g(T_f-25^oC)4.18 J/(g^oC) = -[120.0g(T_f-95^oC)4.18 J/(g^oC)]#

+

#1,045T_^oC - 26,125J/(g^oC) = -501.6T_f^oC + 47,652J/(g^oC)#

+

# (cancel(1,546.6)T_f^oC)/(cancel(1,546.6)) = (73,777)/(1,546.6)#

+

#T_f^oC = 47.07^oC#

+

+ +

+
+
" "
+
+
+

#T_f^oC = 47.07^oC#

+
+
+
+

Explanation:

+
+

We can use a heat equilibrium equation to find this value.

+

#DeltaH = m(T_f-T_i)C_p#

+

#DeltaH =# Change in Heat
+#m =# Mass
+#T=# Tempurature
+#C_p=# Specific Heat

+

#DeltaH# cold #H_2O# = -#DeltaH# hot #H_2O#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

Since the density of water is #1 g/(mL)# the volume of water is also equal to the mass of the water in grams.

+

Cold #H_2O#

+

#m = 250.0g#
+#T_f= ?#
+#T_i = 25^oC#
+#C_p = 4.18 J/(g^oC)#

+

Cold #H_2O#

+

#m = 120.0g#
+#T_f= ?#
+#T_i = 95^oC#
+#C_p = 4.18 J/(g^oC)#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

#250.0g(T_f-25^oC)4.18 J/(g^oC) = -[120.0g(T_f-95^oC)4.18 J/(g^oC)]#

+

#1,045T_^oC - 26,125J/(g^oC) = -501.6T_f^oC + 47,652J/(g^oC)#

+

# (cancel(1,546.6)T_f^oC)/(cancel(1,546.6)) = (73,777)/(1,546.6)#

+

#T_f^oC = 47.07^oC#

+

+ +

+
+
+
" "
+

If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 4, 2016 + +
+
+
+
+
+
+
+

#T_f^oC = 47.07^oC#

+
+
+
+

Explanation:

+
+

We can use a heat equilibrium equation to find this value.

+

#DeltaH = m(T_f-T_i)C_p#

+

#DeltaH =# Change in Heat
+#m =# Mass
+#T=# Tempurature
+#C_p=# Specific Heat

+

#DeltaH# cold #H_2O# = -#DeltaH# hot #H_2O#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

Since the density of water is #1 g/(mL)# the volume of water is also equal to the mass of the water in grams.

+

Cold #H_2O#

+

#m = 250.0g#
+#T_f= ?#
+#T_i = 25^oC#
+#C_p = 4.18 J/(g^oC)#

+

Cold #H_2O#

+

#m = 120.0g#
+#T_f= ?#
+#T_i = 95^oC#
+#C_p = 4.18 J/(g^oC)#

+

#m(T_f-T_i)C_p = -[m(T_f-T_i)C_p]#

+

#250.0g(T_f-25^oC)4.18 J/(g^oC) = -[120.0g(T_f-95^oC)4.18 J/(g^oC)]#

+

#1,045T_^oC - 26,125J/(g^oC) = -501.6T_f^oC + 47,652J/(g^oC)#

+

# (cancel(1,546.6)T_f^oC)/(cancel(1,546.6)) = (73,777)/(1,546.6)#

+

#T_f^oC = 47.07^oC#

+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
+ 16518 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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" If you combine 250.0 mL of water at 25°C and 120.0 mL of water at 95°C, what is the final temperature of the mixture? nan +336 a8349fb3-6ddd-11ea-807c-ccda262736ce https://socratic.org/questions/an-unknown-compound-is-analyzed-and-found-to-be-composed-of-14-79-nitrogen-50-68 ZnN2O6 start chemical_formula qc_end physical_unit 12 12 11 11 percent_composition qc_end physical_unit 14 14 13 13 percent_composition qc_end physical_unit 17 17 16 16 percent_composition qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""ZnN2O6""}]" "[{""type"":""physical unit"",""value"":""Percent composition [OF] nitrogen [=] \\pu{14.79%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] oxygen [=] \\pu{50.68%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] zinc [=] \\pu{34.53%}""}]" "

An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

" nan ZnN2O6 "
+

Explanation:

+
+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

+

Your values will look like this :

+
    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

+

#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

+

#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

+

#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

+

#1.051 / 0.528 = 1.99#

+

#3.168 / 0.528 = 6#

+

#0.528 / 0.528 = 1#

+

To find the smallest whole number ratio, multiply each value by #2#.

+

#1.99 * 2 ~~ 4 #

+

#6 * 2 = 12#

+

#1 * 2 = 2#

+

Your formula is now:

+

#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

+

#""ZnN""_2""O""_6#

+

(Remember that if the subscript of an element is #1#, then no subscript is necessary).

+
+
" "
+
+
+

The empirical formula is #""ZnN""_2""O""_6#.

+
+
+
+

Explanation:

+
+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

+

Your values will look like this :

+
    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

+

#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

+

#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

+

#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

+

#1.051 / 0.528 = 1.99#

+

#3.168 / 0.528 = 6#

+

#0.528 / 0.528 = 1#

+

To find the smallest whole number ratio, multiply each value by #2#.

+

#1.99 * 2 ~~ 4 #

+

#6 * 2 = 12#

+

#1 * 2 = 2#

+

Your formula is now:

+

#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

+

#""ZnN""_2""O""_6#

+

(Remember that if the subscript of an element is #1#, then no subscript is necessary).

+
+
+
" "
+

An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ + +
+
+ +
+ + Jun 18, 2018 + +
+
+
+
+
+
+
+

The empirical formula is #""ZnN""_2""O""_6#.

+
+
+
+

Explanation:

+
+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

+

Your values will look like this :

+
    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

+

#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

+

#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

+

#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

+

#1.051 / 0.528 = 1.99#

+

#3.168 / 0.528 = 6#

+

#0.528 / 0.528 = 1#

+

To find the smallest whole number ratio, multiply each value by #2#.

+

#1.99 * 2 ~~ 4 #

+

#6 * 2 = 12#

+

#1 * 2 = 2#

+

Your formula is now:

+

#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

+

#""ZnN""_2""O""_6#

+

(Remember that if the subscript of an element is #1#, then no subscript is necessary).

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Jun 18, 2018 + +
+
+
+
+
+
+
+

#ZnN_2O_6#

+
+
+
+

Explanation:

+
+

As always we ASSUME a #100*g# mass of compound, and we assess the EMPIRICAL formula by dividing the elemental masses thru by the ATOMIC masses of each element...

+

#""Moles of zinc""-=(34.53*g)/(65.4*g*mol^-1)=0.528*mol#

+

#""Moles of nitrogen""-=(14.79*g)/(14.01*g*mol^-1)=1.056*mol#

+

#""Moles of oxygen""-=(50.68*g)/(15.999*g*mol^-1)=3.168*mol#

+

And so we divide thru by the SMALLEST molar quantity to get a trial empirical formula of...

+

#Zn_((0.528*mol)/(0.528*mol))N_((1.056*mol)/(0.528*mol))O_((3.168*mol)/(0.528*mol))-=ZnN_2O_6#...(and I hope you can see those quotients....because I had to magnify them...)

+

And of course this is zinc nitrate...#Zn(NO_3)_2#. Note that normally you would NEVER be quoted percentage oxygen. You would be quoted percentage metal, percentage nitrogen, and then you would be expected to get percentage oxygen by difference. Oxygen is a difficult element to analyze...

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" An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound? nan +337 a8349fb4-6ddd-11ea-88fd-ccda262736ce https://socratic.org/questions/how-many-moles-of-o-2-are-required-to-generate-18-moles-of-h-2o-in-the-reaction- 25.00 moles start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mole qc_end chemical_equation 16 26 qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] O2 [IN] moles""}]" "[{""type"":""physical unit"",""value"":""25.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] H2O [=] \\pu{18 moles}""},{""type"":""chemical equation"",""value"":""2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O""}]" "

How many moles of #O_2# are required to generate 18 moles of #H_2O# in the reaction #2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O#?

" nan 25.00 moles "
+

Explanation:

+
+

#18mol H_2 O x ((25 molO_2)/ (18molH_2 O)) = 25 mol O_2#

+

Moles of water cancel, leaving you with moles of Oxygen.
+Factor out 18/18=1.
+25 x 1 = 25

+
+
" "
+
+
+

You will need 25 moles, but this is not based on the equation alone.

+
+
+
+

Explanation:

+
+

#18mol H_2 O x ((25 molO_2)/ (18molH_2 O)) = 25 mol O_2#

+

Moles of water cancel, leaving you with moles of Oxygen.
+Factor out 18/18=1.
+25 x 1 = 25

+
+
+
" "
+

How many moles of #O_2# are required to generate 18 moles of #H_2O# in the reaction #2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Mole Ratios + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 12, 2016 + +
+
+
+
+
+
+
+

You will need 25 moles, but this is not based on the equation alone.

+
+
+
+

Explanation:

+
+

#18mol H_2 O x ((25 molO_2)/ (18molH_2 O)) = 25 mol O_2#

+

Moles of water cancel, leaving you with moles of Oxygen.
+Factor out 18/18=1.
+25 x 1 = 25

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 1826 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" How many moles of #O_2# are required to generate 18 moles of #H_2O# in the reaction #2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O#? nan +338 a8349fb5-6ddd-11ea-b974-ccda262736ce https://socratic.org/questions/how-many-moles-of-hydrogen-will-be-needed-to-react-with-2-moles-of-nitrogen-in-t 6.00 moles start physical_unit 4 4 mole mol qc_end chemical_equation 18 22 qc_end physical_unit 14 14 11 12 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] hydrogen [IN] moles""}]" "[{""type"":""physical unit"",""value"":""6.00 moles""}]" "[{""type"":""chemical equation"",""value"":""N2 + H2 -> NH3""},{""type"":""physical unit"",""value"":""Mole [OF] nitrogen [=] \\pu{2 moles}""}]" "

How many moles of hydrogen will be needed to react with 2 moles of nitrogen in the reaction #N_2 + H_2 -> NH_3#?

" nan 6.00 moles "
+

Explanation:

+
+

The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.

+

Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of #4# moles of #""NITROGEN""# atoms.

+

Thus, if #2# moles of dinitrogen gas are specified, #6# moles of dihydrogen gas are required.

+
+
" "
+
+
+

The balanced equation for dinitrogen reduction is:
+#N_2(g) + 3H_2(g) rarr 2NH_3(g)#.

+

#2# #""moles""# #N_2# requires #6# moles #H_2#.

+
+
+
+

Explanation:

+
+

The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.

+

Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of #4# moles of #""NITROGEN""# atoms.

+

Thus, if #2# moles of dinitrogen gas are specified, #6# moles of dihydrogen gas are required.

+
+
+
" "
+

How many moles of hydrogen will be needed to react with 2 moles of nitrogen in the reaction #N_2 + H_2 -> NH_3#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Mole Ratios + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 28, 2016 + +
+
+
+
+
+
+
+

The balanced equation for dinitrogen reduction is:
+#N_2(g) + 3H_2(g) rarr 2NH_3(g)#.

+

#2# #""moles""# #N_2# requires #6# moles #H_2#.

+
+
+
+

Explanation:

+
+

The equation above represents probably the most important inorganic reaction on the planet in that it allows the formation of ammonia and nitrates for nitrogenous fertilizer.

+

Your question specified 2 moles of nitrogen. I would be quite justified in assuming that the questioner meant 2 moles of dinitrogen gas, i.e. the equivalent of #4# moles of #""NITROGEN""# atoms.

+

Thus, if #2# moles of dinitrogen gas are specified, #6# moles of dihydrogen gas are required.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 19827 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" How many moles of hydrogen will be needed to react with 2 moles of nitrogen in the reaction #N_2 + H_2 -> NH_3#? nan +339 a834c69e-6ddd-11ea-87ca-ccda262736ce https://socratic.org/questions/what-volume-of-7-91m-solution-of-nitric-acid-hno3-is-just-sufficient-to-react-wi 1.68 × 10^(-2) L start physical_unit 5 8 volume l qc_end physical_unit 9 9 3 4 molarity qc_end physical_unit 21 21 16 17 mass qc_end chemical_equation 22 34 qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] solution of nitric acid [IN] L""}]" "[{""type"":""physical unit"",""value"":""1.68 × 10^(-2) L""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] HNO3 [=] \\pu{7.91 M}""},{""type"":""physical unit"",""value"":""Mass [OF] PbO2 [=] \\pu{15.9 g}""},{""type"":""chemical equation"",""value"":""2 PbO2 + 4 HNO3 -> 2 PbNO3 + 2 H2O + O2""}]" "

What volume of 7.91M solution of nitric acid, HNO3 is just sufficient to react with 15.9 g of lead dioxide, PbO2 ? +2PbO2 + 4HNO3 ---> 2PbNO3 + 2H20 + O2

" nan 1.68 × 10^(-2) L "
+

Explanation:

+
+

First, convert the grams of lead dioxide into moles.

+

#mols PbO_2 = (15.9 g)/(239.198 g/(mol)) = 0.066472128 mols#

+

Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:

+

#mols HNO_3 = 0.132944255 mols#

+

Because #M = (mols)/L#, you can rearrange the formula to get:

+

#L = (mols)/M#

+

#""liters nitric acid"" = (0.132944255 mols)/(7.91M)#

+

#""liters nitric acid"" = 0.016807112L#

+

Seeing that you only have 3 sig figs, the answer would be:

+

#1.68 * 10^(-2)L# of a 7.91M solution of nitric acid

+
+
" "
+
+
+

#1.68 * 10^(-2)L#

+
+
+
+

Explanation:

+
+

First, convert the grams of lead dioxide into moles.

+

#mols PbO_2 = (15.9 g)/(239.198 g/(mol)) = 0.066472128 mols#

+

Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:

+

#mols HNO_3 = 0.132944255 mols#

+

Because #M = (mols)/L#, you can rearrange the formula to get:

+

#L = (mols)/M#

+

#""liters nitric acid"" = (0.132944255 mols)/(7.91M)#

+

#""liters nitric acid"" = 0.016807112L#

+

Seeing that you only have 3 sig figs, the answer would be:

+

#1.68 * 10^(-2)L# of a 7.91M solution of nitric acid

+
+
+
" "
+

What volume of 7.91M solution of nitric acid, HNO3 is just sufficient to react with 15.9 g of lead dioxide, PbO2 ? +2PbO2 + 4HNO3 ---> 2PbNO3 + 2H20 + O2

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+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Equation Stoichiometry + + +
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+1 Answer +
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+ + Apr 26, 2017 + +
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#1.68 * 10^(-2)L#

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+
+

Explanation:

+
+

First, convert the grams of lead dioxide into moles.

+

#mols PbO_2 = (15.9 g)/(239.198 g/(mol)) = 0.066472128 mols#

+

Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:

+

#mols HNO_3 = 0.132944255 mols#

+

Because #M = (mols)/L#, you can rearrange the formula to get:

+

#L = (mols)/M#

+

#""liters nitric acid"" = (0.132944255 mols)/(7.91M)#

+

#""liters nitric acid"" = 0.016807112L#

+

Seeing that you only have 3 sig figs, the answer would be:

+

#1.68 * 10^(-2)L# of a 7.91M solution of nitric acid

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" "What volume of 7.91M solution of nitric acid, HNO3 is just sufficient to react with 15.9 g of lead dioxide, PbO2 ? +2PbO2 + 4HNO3 ---> 2PbNO3 + 2H20 + O2" nan +340 a834c69f-6ddd-11ea-a88c-ccda262736ce https://socratic.org/questions/how-many-moles-of-potassium-chloride-kcl-are-needed-to-make-100-0-ml-of-a-2-0-m- 0.20 moles start physical_unit 6 6 mole mol qc_end physical_unit 17 18 11 12 volume qc_end physical_unit 17 18 15 16 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] KCl [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.20 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] KCl solution [=] \\pu{100.0 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] KCl solution [=] \\pu{2.0 M}""}]" "

How many moles of potassium chloride, KCl, are needed to make 100.0 mL of a 2.0 M KCl solution?

" nan 0.20 moles "
+

Explanation:

+
+

The thing to remember about a solution's molarity is that you can express it as a fraction that has #""1 L""# of solution as the denominator.

+

In your case, a #""2.0-M""# potassium chloride solution contains #2.0# moles of potassium chloride, the solute, for every #""1 L""# of solution, which means that you can write it as

+
+

#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#

+
+

Now, you should know that

+
+

#""1 L"" = 10^3# #""mL""#

+
+

This means that you can rewrite the molarity of the solution as

+
+

#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

So, you need to figure out how many moles of potassium chloride must be dissolved in water to make #""100.0 mL""# of #""2.0 M""# solution.

+

In other words, you must find the number of moles that when dissolved in #""100.0 mL""# are equivalent to #2.0# moles dissolved in #10^3# #""mL""# of solution.

+
+

#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

Solve this equation to find

+
+

#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#

+

#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs

+
+

Therefore, you can say that if you dissolve #0.20# moles of potassium chloride in enough water to make #""100.0 mL""# of solution, you will have a #""2.0-M""# solution.

+
+
" "
+
+
+

#""0.20 moles KCl""#

+
+
+
+

Explanation:

+
+

The thing to remember about a solution's molarity is that you can express it as a fraction that has #""1 L""# of solution as the denominator.

+

In your case, a #""2.0-M""# potassium chloride solution contains #2.0# moles of potassium chloride, the solute, for every #""1 L""# of solution, which means that you can write it as

+
+

#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#

+
+

Now, you should know that

+
+

#""1 L"" = 10^3# #""mL""#

+
+

This means that you can rewrite the molarity of the solution as

+
+

#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

So, you need to figure out how many moles of potassium chloride must be dissolved in water to make #""100.0 mL""# of #""2.0 M""# solution.

+

In other words, you must find the number of moles that when dissolved in #""100.0 mL""# are equivalent to #2.0# moles dissolved in #10^3# #""mL""# of solution.

+
+

#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

Solve this equation to find

+
+

#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#

+

#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs

+
+

Therefore, you can say that if you dissolve #0.20# moles of potassium chloride in enough water to make #""100.0 mL""# of solution, you will have a #""2.0-M""# solution.

+
+
+
" "
+

How many moles of potassium chloride, KCl, are needed to make 100.0 mL of a 2.0 M KCl solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Jun 10, 2017 + +
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#""0.20 moles KCl""#

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+
+
+

Explanation:

+
+

The thing to remember about a solution's molarity is that you can express it as a fraction that has #""1 L""# of solution as the denominator.

+

In your case, a #""2.0-M""# potassium chloride solution contains #2.0# moles of potassium chloride, the solute, for every #""1 L""# of solution, which means that you can write it as

+
+

#""2.0 M"" = ""2.0 moles KCl""/""1 L solution""#

+
+

Now, you should know that

+
+

#""1 L"" = 10^3# #""mL""#

+
+

This means that you can rewrite the molarity of the solution as

+
+

#""2.0 M"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

So, you need to figure out how many moles of potassium chloride must be dissolved in water to make #""100.0 mL""# of #""2.0 M""# solution.

+

In other words, you must find the number of moles that when dissolved in #""100.0 mL""# are equivalent to #2.0# moles dissolved in #10^3# #""mL""# of solution.

+
+

#(color(blue)(?)color(white)(.)""moles KCl"")/""100.0 mL solution"" = ""2.0 moles KCl""/(10^3color(white)(.)""mL solution"")#

+
+

Solve this equation to find

+
+

#color(blue)(?) = (100.0 color(red)(cancel(color(black)(""mL solution""))))/(10^3color(red)(cancel(color(black)(""mL solution"")))) * ""2.0 moles""#

+

#color(blue)(?) = ""0.20 moles"" -># rounded to two sig figs

+
+

Therefore, you can say that if you dissolve #0.20# moles of potassium chloride in enough water to make #""100.0 mL""# of solution, you will have a #""2.0-M""# solution.

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" How many moles of potassium chloride, KCl, are needed to make 100.0 mL of a 2.0 M KCl solution? nan +341 a834c6a0-6ddd-11ea-bae4-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-containing-26-57-g-potassium-35-36- KCrO3 start chemical_formula qc_end physical_unit 11 11 9 10 mass qc_end physical_unit 14 14 12 13 mass qc_end physical_unit 18 18 16 17 mass qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""KCrO3""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] potassium [=] \\pu{26.57 g}""},{""type"":""physical unit"",""value"":""Mass [OF] chromium [=] \\pu{35.36 g}""},{""type"":""physical unit"",""value"":""Mass [OF] oxygen [=] \\pu{38.07 g}""}]" "

What is the empirical formula for a compound containing 26.57 g potassium, 35.36 g chromium, and 38.07 g oxygen?

" nan KCrO3 "
+

Explanation:

+
+

Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

+

#K -> (26.57g)/(39.098gmol^-1) = 0.68molK#
+#Cr -> (35.36g)/(51.996gmol^-1) = 0.68molCr#
+#O -> (38.07g)/(15.999gmol^-1) = 2.34molO#

+

Now divide each number of moles by the lowest one, in this case #0.68#.

+

#(0.68molK)/(0.68) = 1molK#
+#(0.68molCr)/(0.68) = 1molCr#
+#(2.34molO)/(0.68) = 3.44molO#

+

Round to the nearest whole number

+

#K_1Cr_1O_3 = KCrO_3#

+
+
" "
+
+
+

#KCrO_3#

+
+
+
+

Explanation:

+
+

Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

+

#K -> (26.57g)/(39.098gmol^-1) = 0.68molK#
+#Cr -> (35.36g)/(51.996gmol^-1) = 0.68molCr#
+#O -> (38.07g)/(15.999gmol^-1) = 2.34molO#

+

Now divide each number of moles by the lowest one, in this case #0.68#.

+

#(0.68molK)/(0.68) = 1molK#
+#(0.68molCr)/(0.68) = 1molCr#
+#(2.34molO)/(0.68) = 3.44molO#

+

Round to the nearest whole number

+

#K_1Cr_1O_3 = KCrO_3#

+
+
+
" "
+

What is the empirical formula for a compound containing 26.57 g potassium, 35.36 g chromium, and 38.07 g oxygen?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
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+2 Answers +
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+ + Apr 4, 2016 + +
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#KCrO_3#

+
+
+
+

Explanation:

+
+

Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

+

#K -> (26.57g)/(39.098gmol^-1) = 0.68molK#
+#Cr -> (35.36g)/(51.996gmol^-1) = 0.68molCr#
+#O -> (38.07g)/(15.999gmol^-1) = 2.34molO#

+

Now divide each number of moles by the lowest one, in this case #0.68#.

+

#(0.68molK)/(0.68) = 1molK#
+#(0.68molCr)/(0.68) = 1molCr#
+#(2.34molO)/(0.68) = 3.44molO#

+

Round to the nearest whole number

+

#K_1Cr_1O_3 = KCrO_3#

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#""Empirical formula""# #-=# #K_2Cr_2O_7#

+
+
+
+

Explanation:

+
+

We got #100 *g# of stuff, and we calculate the number of atoms element by element:

+

#K# #=# #(26.57*g)/(39.1*g*mol^-1)# #=# #0.680*mol# #K#

+

#Cr# #=# #(35.36*g)/(52.00*g*mol^-1)# #=# #0.680*mol# #Cr#

+

#O# #=# #(38.07*g)/(16.0*g*mol^-1)# #=# #2.38*mol# #O#

+

Note that NORMALLY you would NEVER be given the percentage oxygen. Why not? Because there are very few ways to measure the proportion of this gas. At a 1st year undergrad level this question would have proposed an oxide of chromium that contained #26.57%# potassium, and #35.36%# chromium, and expected the student to twig that the missing percentage was due to the oxygen.

+

When we divide thru by the lowest number of moles (#0.68*mol#) we get an #""empirical formula""# of....

+

#K_((0.680*mol)/(0.680*mol))Cr_((0.680*mol)/(0.680*mol))O_((2.38*mol)/(0.680*mol))#, i.e. #KCrO_(3.5)#.

+

But by definition, the empirical formula is the simplest WHOLE number ratio defining constituent atoms in a species, so we must double this provisional formula to give:

+

#K_2Cr_2O_7#, #""i.e. potassium dichromate""#.

+

And this is a crystalline, orange powder, that is widely used in oxidations of organic materials....

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+
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+ + Creative Commons License + +
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" What is the empirical formula for a compound containing 26.57 g potassium, 35.36 g chromium, and 38.07 g oxygen? nan +342 a834c6a1-6ddd-11ea-9179-ccda262736ce https://socratic.org/questions/given-the-equation-2h-2o-2h-2-o-2-how-many-moles-of-h-o-would-be-required-to-pro 5.00 moles start physical_unit 4 4 mole mol qc_end chemical_equation 3 9 qc_end physical_unit 9 9 20 21 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] H2O [IN] moles""}]" "[{""type"":""physical unit"",""value"":""5.00 moles""}]" "[{""type"":""chemical equation"",""value"":""2 H2O -> 2 H2 + O2""},{""type"":""physical unit"",""value"":""Mole [OF] O2 [=] \\pu{2.5 moles}""}]" "

Given the equation #2H_2O ->2H_2 + O_2#, how many moles of #H_2O# would be required to produce 2.5 moles of #O_2#?

" nan 5.00 moles "
+

Explanation:

+
+

Since the chemical equation,

+
+

#color(red)2H_2O->2H_2+color(blue)1O_2#

+
+

requires a specific number of moles of the reactant, then a specific number of moles of the products are created.

+

With this knowledge, you can create a mole ratio.

+

The general format of a mole ratio is as follows:

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#

+
+

In your case, you're looking for the moles of #H_2O# required to make #2.5# moles of #O_2#.

+

Thus, your mole ratio would be:

+
+

#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#

+

#color(darkorange)(rArr)#where #x# represents the moles of #H_2O# required

+
+

From this point on, your goal is to solve for #x# to find the moles of #H_2O# required to produce #2.5# moles of #O_2#.

+
+

#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#

+

#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#

+
+
+
" "
+
+
+

#5# moles of #H_2O#

+
+
+
+

Explanation:

+
+

Since the chemical equation,

+
+

#color(red)2H_2O->2H_2+color(blue)1O_2#

+
+

requires a specific number of moles of the reactant, then a specific number of moles of the products are created.

+

With this knowledge, you can create a mole ratio.

+

The general format of a mole ratio is as follows:

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#

+
+

In your case, you're looking for the moles of #H_2O# required to make #2.5# moles of #O_2#.

+

Thus, your mole ratio would be:

+
+

#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#

+

#color(darkorange)(rArr)#where #x# represents the moles of #H_2O# required

+
+

From this point on, your goal is to solve for #x# to find the moles of #H_2O# required to produce #2.5# moles of #O_2#.

+
+

#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#

+

#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#

+
+
+
+
" "
+

Given the equation #2H_2O ->2H_2 + O_2#, how many moles of #H_2O# would be required to produce 2.5 moles of #O_2#?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Mole Ratios + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 16, 2016 + +
+
+
+
+
+
+
+

#5# moles of #H_2O#

+
+
+
+

Explanation:

+
+

Since the chemical equation,

+
+

#color(red)2H_2O->2H_2+color(blue)1O_2#

+
+

requires a specific number of moles of the reactant, then a specific number of moles of the products are created.

+

With this knowledge, you can create a mole ratio.

+

The general format of a mole ratio is as follows:

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)((""required based on balanced equation"")/(""product based on balanced equation"")=(""required"")/(""product""))color(white)(a/a)|)))#

+
+

In your case, you're looking for the moles of #H_2O# required to make #2.5# moles of #O_2#.

+

Thus, your mole ratio would be:

+
+

#(color(red)2color(white)(i)molcolor(white)(i)H_2O)/(color(blue)1color(white)(i)molcolor(white)(i)O_2)=x/(2.5color(white)(i)molcolor(white)(i)O_2)#

+

#color(darkorange)(rArr)#where #x# represents the moles of #H_2O# required

+
+

From this point on, your goal is to solve for #x# to find the moles of #H_2O# required to produce #2.5# moles of #O_2#.

+
+

#x=2.5color(purple)cancelcolor(black)(molcolor(white)(i)O_2)xx(2color(white)(i)molcolor(white)(i)H_2O)/(1color(purple)cancelcolor(black)(molcolor(white)(i)O_2))#

+

#x=color(green)(|bar(ul(color(white)(a/a)5color(white)(i)molcolor(white)(i)H_2Ocolor(white)(a/a)|)))#

+
+
+
+
+
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+ +
+
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+
+
+
+
Related questions
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Impact of this question
+
+ 2614 views + around the world +
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+ + Creative Commons License + +
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" Given the equation #2H_2O ->2H_2 + O_2#, how many moles of #H_2O# would be required to produce 2.5 moles of #O_2#? nan +343 a834c6a2-6ddd-11ea-b8d9-ccda262736ce https://socratic.org/questions/given-the-equation-pb-so4-2-4-lino3-pb-no3-4-2-li2so4-how-many-grams-of-lithium- 313.55 grams start physical_unit 16 17 mass g qc_end chemical_equation 3 11 qc_end physical_unit 26 27 23 24 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] lithium nitrate [IN] grams""}]" "[{""type"":""physical unit"",""value"":""313.55 grams""}]" "[{""type"":""chemical equation"",""value"":""Pb(SO4)2 + 4 LiNO3 -> Pb(NO3)4 + 2 Li2SO4""},{""type"":""physical unit"",""value"":""Mass [OF] lithium sulfate [=] \\pu{250 grams}""},{""type"":""other"",""value"":""Adequate amount of lead (IV) sulfate.""}]" "

Given the equation: +Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 Li2SO4, +how many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? + +Thanks lots.?

" nan 313.55 grams "
+

Explanation:

+
+

Balanced Equation
+#""Pb(SO""_4)_2(""aq"") + ""4LiNO""_3(""aq"")""##rarr##""Pb(NO""_3)_4(""aq"")"" + 2Li""_2""(SO""_4)(""aq"")""#

+

This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.

+

This reaction will not actually occur. All you will have is a mixture of #""Pb""^(4+)""# ions, #""SO""_4""""^(2-)""# ions, #""Li""^(+)""# ions, and #""NO""_3""""^(-)""# ions.

+

However, I will work it out so you can see how a problem of this type is to be done.

+

Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
+ You can calculate it or find it in a resource.
+#""LiNO""_3"":##""68.9459 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/Lithium_nitrate
+#""Li""_2""SO""_4"":##""109.9446 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/66320

+

This type of problem follows the pattern: #""mass of product""##rarr##""moles of product""##rarr##""moles of reactant""##rarr##""mass of reactant""#.

+

Step 2: Convert mass of #""Li""_2""SO""_4""# to moles #""LiSO""_4""#.
+Divide the mass of lithium sulfate by its molar mass.

+

#250cancel""g LiSO""_4xx(1""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)=""2.274 mol LiSO""_4""#

+

I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.

+

Step 3: Convert moles of #""Li""_2""SO""_4""# to moles of #""LiNO""_3""#.
+Multiply moles of #""Li""_2""SO""_4""# times the mole ratio between #""Li""_2""SO""_4""# and #""LiNO""_3""# from the balanced equation.

+

#2.274cancel""mol Li""_2""SO""_4xx(4""mol LiNO""_3)/(2cancel""mol LiSO""_4)=""4.548 mol LiNO""_3""#

+

Step 4: Convert moles #""LiNO""_3""# to mass #""LiNO""_3""#.
+Multiply the moles #""LiNO""_3""# times its molar mass.

+

#4.548cancel""mol LiNO""_3xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+

This problem can be worked out all at once as follows.

+

#250cancel""g LiSO""_4xx(1cancel""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)xx(4cancel""mol LiNO""_3)/(2cancel""mol LiSO""_4)xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+
+
" "
+
+
+

If this reaction were to occur, #""310 g LiNO""_3""# would be needed to make #""250 g Li""_2""SO""_4""#

+
+
+
+

Explanation:

+
+

Balanced Equation
+#""Pb(SO""_4)_2(""aq"") + ""4LiNO""_3(""aq"")""##rarr##""Pb(NO""_3)_4(""aq"")"" + 2Li""_2""(SO""_4)(""aq"")""#

+

This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.

+

This reaction will not actually occur. All you will have is a mixture of #""Pb""^(4+)""# ions, #""SO""_4""""^(2-)""# ions, #""Li""^(+)""# ions, and #""NO""_3""""^(-)""# ions.

+

However, I will work it out so you can see how a problem of this type is to be done.

+

Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
+ You can calculate it or find it in a resource.
+#""LiNO""_3"":##""68.9459 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/Lithium_nitrate
+#""Li""_2""SO""_4"":##""109.9446 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/66320

+

This type of problem follows the pattern: #""mass of product""##rarr##""moles of product""##rarr##""moles of reactant""##rarr##""mass of reactant""#.

+

Step 2: Convert mass of #""Li""_2""SO""_4""# to moles #""LiSO""_4""#.
+Divide the mass of lithium sulfate by its molar mass.

+

#250cancel""g LiSO""_4xx(1""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)=""2.274 mol LiSO""_4""#

+

I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.

+

Step 3: Convert moles of #""Li""_2""SO""_4""# to moles of #""LiNO""_3""#.
+Multiply moles of #""Li""_2""SO""_4""# times the mole ratio between #""Li""_2""SO""_4""# and #""LiNO""_3""# from the balanced equation.

+

#2.274cancel""mol Li""_2""SO""_4xx(4""mol LiNO""_3)/(2cancel""mol LiSO""_4)=""4.548 mol LiNO""_3""#

+

Step 4: Convert moles #""LiNO""_3""# to mass #""LiNO""_3""#.
+Multiply the moles #""LiNO""_3""# times its molar mass.

+

#4.548cancel""mol LiNO""_3xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+

This problem can be worked out all at once as follows.

+

#250cancel""g LiSO""_4xx(1cancel""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)xx(4cancel""mol LiNO""_3)/(2cancel""mol LiSO""_4)xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+
+
+
" "
+

Given the equation: +Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 Li2SO4, +how many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? + +Thanks lots.?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 14, 2015 + +
+
+
+
+
+
+
+

If this reaction were to occur, #""310 g LiNO""_3""# would be needed to make #""250 g Li""_2""SO""_4""#

+
+
+
+

Explanation:

+
+

Balanced Equation
+#""Pb(SO""_4)_2(""aq"") + ""4LiNO""_3(""aq"")""##rarr##""Pb(NO""_3)_4(""aq"")"" + 2Li""_2""(SO""_4)(""aq"")""#

+

This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of the products is a solid precipitate, an insoluble gas, or water. In this case, none of those is a product. All of the compounds are aqueous.

+

This reaction will not actually occur. All you will have is a mixture of #""Pb""^(4+)""# ions, #""SO""_4""""^(2-)""# ions, #""Li""^(+)""# ions, and #""NO""_3""""^(-)""# ions.

+

However, I will work it out so you can see how a problem of this type is to be done.

+

Step 1: You need the molar masses of lithium nitrate and lithium sulfate.
+ You can calculate it or find it in a resource.
+#""LiNO""_3"":##""68.9459 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/Lithium_nitrate
+#""Li""_2""SO""_4"":##""109.9446 g/mol""# http://pubchem.ncbi.nlm.nih.gov/compound/66320

+

This type of problem follows the pattern: #""mass of product""##rarr##""moles of product""##rarr##""moles of reactant""##rarr##""mass of reactant""#.

+

Step 2: Convert mass of #""Li""_2""SO""_4""# to moles #""LiSO""_4""#.
+Divide the mass of lithium sulfate by its molar mass.

+

#250cancel""g LiSO""_4xx(1""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)=""2.274 mol LiSO""_4""#

+

I am including a couple of guard units to reduce rounding errors. I will round the final answer to two significant figures.

+

Step 3: Convert moles of #""Li""_2""SO""_4""# to moles of #""LiNO""_3""#.
+Multiply moles of #""Li""_2""SO""_4""# times the mole ratio between #""Li""_2""SO""_4""# and #""LiNO""_3""# from the balanced equation.

+

#2.274cancel""mol Li""_2""SO""_4xx(4""mol LiNO""_3)/(2cancel""mol LiSO""_4)=""4.548 mol LiNO""_3""#

+

Step 4: Convert moles #""LiNO""_3""# to mass #""LiNO""_3""#.
+Multiply the moles #""LiNO""_3""# times its molar mass.

+

#4.548cancel""mol LiNO""_3xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+

This problem can be worked out all at once as follows.

+

#250cancel""g LiSO""_4xx(1cancel""mol LiSO""_4)/(109.9446cancel""g LiSO""_4)xx(4cancel""mol LiNO""_3)/(2cancel""mol LiSO""_4)xx(68.9459""g LiNO""_3)/(1cancel""mol LiNO""_3)=""310 g LiNO""_3""# (rounded to two significant figures)

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
+ 51073 views + around the world +
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+ + Creative Commons License + +
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+
" "Given the equation: +Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 Li2SO4, +how many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction? + +Thanks lots.?" nan +344 a834c6a3-6ddd-11ea-9295-ccda262736ce https://socratic.org/questions/what-is-the-amount-of-energy-required-to-raise-the-temperature-of-1-g-of-water-1 4.18 J start physical_unit 15 15 heat_energy j qc_end physical_unit 15 15 16 17 temperature qc_end physical_unit 15 15 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Amount of energy [OF] water [IN] J""}]" "[{""type"":""physical unit"",""value"":""4.18 J""}]" "[{""type"":""physical unit"",""value"":""Temperature raised [OF] water [=] \\pu{1 ℃}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{1 g}""}]" "

What is the amount of energy required to raise the temperature of 1 g of water 1°C?

" nan 4.18 J "
+

Explanation:

+
+

Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.

+
+
" "
+
+
+

The specific heat of water, #C_s#, is #4.184J/(g*°K)#.

+
+
+
+

Explanation:

+
+

Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.

+
+
+
" "
+

What is the amount of energy required to raise the temperature of 1 g of water 1°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 7, 2017 + +
+
+
+
+
+
+
+

The specific heat of water, #C_s#, is #4.184J/(g*°K)#.

+
+
+
+

Explanation:

+
+

Specific heat is the heat capacity per gram of the substance of interest. Heat capacity is the amount of heat energy a body can cold.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 4914 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" What is the amount of energy required to raise the temperature of 1 g of water 1°C? nan +345 a834c6a4-6ddd-11ea-9697-ccda262736ce https://socratic.org/questions/how-do-you-write-the-formula-for-barium-chloride BaCl2 start chemical_formula qc_end substance 7 8 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] barium chloride [IN] default""}]" "[{""type"":""chemical equation"",""value"":""BaCl2""}]" "[{""type"":""substance name"",""value"":""Barium chloride""}]" "

How do you write the formula for barium chloride?

" nan BaCl2 "
+

Explanation:

+
+

Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.

+

So we have Barium as #Ba^(+2)#

+

Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.

+

So we have Chloride as #Cl^-#

+

We're not finished though! The charges do not balance!

+

To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as #Cl_2#.

+

Now they balance; you have one barium #Ba^(+2)# and two chloride #Cl^-# #Cl^-#

+

The final formula is written as #BaCl_2#.

+

Hope that helps!

+
+
" "
+
+
+

#BaCl_2#

+
+
+
+

Explanation:

+
+

Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.

+

So we have Barium as #Ba^(+2)#

+

Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.

+

So we have Chloride as #Cl^-#

+

We're not finished though! The charges do not balance!

+

To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as #Cl_2#.

+

Now they balance; you have one barium #Ba^(+2)# and two chloride #Cl^-# #Cl^-#

+

The final formula is written as #BaCl_2#.

+

Hope that helps!

+
+
+
" "
+

How do you write the formula for barium chloride?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 22, 2016 + +
+
+
+
+
+
+
+

#BaCl_2#

+
+
+
+

Explanation:

+
+

Barium is listed under Group 2A of the Periodic Table. Looking at the column, Barium is Ba. The elements listed in Group 2A have a charge of +2.

+

So we have Barium as #Ba^(+2)#

+

Now let's move to Chloride. Don't bother looking at the table, because it is not listed. Chloride is an ion of Chlorine, which is from Group 7A and we know anything in that column has a charge of -1. So Chlorine has gained an electron and becomes Chloride. Note that this is an anion and has a negative charge. You can see how easy it was to change the name, you just change the suffix to -ide.

+

So we have Chloride as #Cl^-#

+

We're not finished though! The charges do not balance!

+

To balance it we simply add another Chloride so that you have two negatives balancing the two positives. You'll write it as #Cl_2#.

+

Now they balance; you have one barium #Ba^(+2)# and two chloride #Cl^-# #Cl^-#

+

The final formula is written as #BaCl_2#.

+

Hope that helps!

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
+
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+
+
" How do you write the formula for barium chloride? nan +346 a834edb4-6ddd-11ea-b3eb-ccda262736ce https://socratic.org/questions/576099ca11ef6b03ca019437 2.88 mol/L start physical_unit 6 6 molarity mol/l qc_end physical_unit 13 14 9 10 mass qc_end physical_unit 25 25 22 23 volume qc_end end "[{""type"":""physical unit"",""value"":""Molar concentration [OF] HCl [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""2.88 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] the acid [=] \\pu{10.5 g}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{100 mL}""}]" "

What is the molar concentration of #HCl# if a #10.5*g# mass of the acid is made up to a volume of #100*mL# in water?

" nan 2.88 mol/L "
+

Explanation:

+
+

We can work out the concentration of the hydrochloric acid:

+

#(10.500*g)/(36.46*g*mol^-1)xx1/(0.100*L)# #=# #??mol*L^-1#.

+

#""Ammonium hydroxide""# is unknown. There is ammonia in water. This question makes no chemical sense.

+
+
" "
+
+
+

There is no such beast as ammonium hydroxide. There is #NH_3*H_2O#, at various concentrations.

+
+
+
+

Explanation:

+
+

We can work out the concentration of the hydrochloric acid:

+

#(10.500*g)/(36.46*g*mol^-1)xx1/(0.100*L)# #=# #??mol*L^-1#.

+

#""Ammonium hydroxide""# is unknown. There is ammonia in water. This question makes no chemical sense.

+
+
+
" "
+

What is the molar concentration of #HCl# if a #10.5*g# mass of the acid is made up to a volume of #100*mL# in water?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Limiting Reagent + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 3, 2016 + +
+
+
+
+
+
+
+

There is no such beast as ammonium hydroxide. There is #NH_3*H_2O#, at various concentrations.

+
+
+
+

Explanation:

+
+

We can work out the concentration of the hydrochloric acid:

+

#(10.500*g)/(36.46*g*mol^-1)xx1/(0.100*L)# #=# #??mol*L^-1#.

+

#""Ammonium hydroxide""# is unknown. There is ammonia in water. This question makes no chemical sense.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1300 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the molar concentration of #HCl# if a #10.5*g# mass of the acid is made up to a volume of #100*mL# in water? nan +347 a834edb5-6ddd-11ea-b1a9-ccda262736ce https://socratic.org/questions/styrene-has-the-empirical-formula-ch-it-has-a-molar-mass-of-104-g-what-is-the-mu 8 start physical_unit 20 22 multiplier none qc_end physical_unit 0 0 12 13 molar_mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Multiplier [OF] the molecular formula""}]" "[{""type"":""physical unit"",""value"":""8""}]" "[{""type"":""physical unit"",""value"":""Molar mass [OF] Styrene [=] \\pu{104 g}""},{""type"":""other"",""value"":""Styrene has the empirical formula CH.""}]" "

Styrene has the empirical formula #CH#. It has a molar mass of 104 g. What is the multiplier to get the molecular formula?

" nan 8 "
+

Explanation:

+
+

Given what I have said:

+

#(""Empirical formula"")xxn=""Molecular formula""#

+

#(CH)xxn=104*g*mol^-1#

+

#(12.01+1.01)*g*mol^-1xxn=104*g*mol^-1#

+

Clearly, #n=8#, and the molecular formula is #C_8H_8#.

+

Of course this is reasonable inasmuch that we know that the styrene monomer is #C_6H_5CH=CH_2#, i.e. vinyl benzene.

+
+
" "
+
+
+

The molecular formula is always a whole number multiple of the empirical formula. Of course, the multiple might be #1#.

+
+
+
+

Explanation:

+
+

Given what I have said:

+

#(""Empirical formula"")xxn=""Molecular formula""#

+

#(CH)xxn=104*g*mol^-1#

+

#(12.01+1.01)*g*mol^-1xxn=104*g*mol^-1#

+

Clearly, #n=8#, and the molecular formula is #C_8H_8#.

+

Of course this is reasonable inasmuch that we know that the styrene monomer is #C_6H_5CH=CH_2#, i.e. vinyl benzene.

+
+
+
" "
+

Styrene has the empirical formula #CH#. It has a molar mass of 104 g. What is the multiplier to get the molecular formula?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 6, 2016 + +
+
+
+
+
+
+
+

The molecular formula is always a whole number multiple of the empirical formula. Of course, the multiple might be #1#.

+
+
+
+

Explanation:

+
+

Given what I have said:

+

#(""Empirical formula"")xxn=""Molecular formula""#

+

#(CH)xxn=104*g*mol^-1#

+

#(12.01+1.01)*g*mol^-1xxn=104*g*mol^-1#

+

Clearly, #n=8#, and the molecular formula is #C_8H_8#.

+

Of course this is reasonable inasmuch that we know that the styrene monomer is #C_6H_5CH=CH_2#, i.e. vinyl benzene.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 27813 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" Styrene has the empirical formula #CH#. It has a molar mass of 104 g. What is the multiplier to get the molecular formula? nan +348 a834edb6-6ddd-11ea-8399-ccda262736ce https://socratic.org/questions/if-250-0-ml-of-a-0-96-m-solution-of-acetic-acid-c-2h-4o-2-are-diluted-to-800-0-m 0.30 M start physical_unit 24 26 molarity mol/l qc_end physical_unit 7 7 1 2 volume qc_end physical_unit 11 11 5 6 molarity qc_end physical_unit 7 7 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity2 [OF] the final solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""0.30 M""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] solution [=] \\pu{250.0 mL}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] C2H4O2 [=] \\pu{0.96 M}""},{""type"":""physical unit"",""value"":""Volume2 [OF] solution [=] \\pu{800.0 mL}""}]" "

If 250.0 mL of a 0.96 M solution of acetic acid (#C_2H_4O_2#) are diluted to 800.0 mL, what will be the approximate molarity of the final solution?

" nan 0.30 M "
+

Explanation:

+
+

One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.

+

This will then allow you to find the molarity of the diluted solution.

+

So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#

+
+

Here you have

+

#V_""final""# - the volume of the diluted solution
+#V_""initial""# - the volume of the starting solution

+

In your case, you have

+
+

#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#

+
+

This means that the initial solution was #3.2# times more concentrated than the diluted solution. You thus have

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#

+
+

which gets you

+
+

#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.

+
+
" "
+
+
+

#""0.30 M""#

+
+
+
+

Explanation:

+
+

One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.

+

This will then allow you to find the molarity of the diluted solution.

+

So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#

+
+

Here you have

+

#V_""final""# - the volume of the diluted solution
+#V_""initial""# - the volume of the starting solution

+

In your case, you have

+
+

#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#

+
+

This means that the initial solution was #3.2# times more concentrated than the diluted solution. You thus have

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#

+
+

which gets you

+
+

#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.

+
+
+
" "
+

If 250.0 mL of a 0.96 M solution of acetic acid (#C_2H_4O_2#) are diluted to 800.0 mL, what will be the approximate molarity of the final solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 29, 2016 + +
+
+
+
+
+
+
+

#""0.30 M""#

+
+
+
+

Explanation:

+
+

One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the dilution factor.

+

This will then allow you to find the molarity of the diluted solution.

+

So, the dilution factor, which tells you how concentrated the starting solution is compared with the dilution solution, is calculated like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""D.F."" = V_""final""/V_""initial""color(white)(a/a)|)))#

+
+

Here you have

+

#V_""final""# - the volume of the diluted solution
+#V_""initial""# - the volume of the starting solution

+

In your case, you have

+
+

#""D.F."" = (800.0 color(red)(cancel(color(black)(""mL""))))/(250.0color(red)(cancel(color(black)(""mL"")))) = 3.2#

+
+

This means that the initial solution was #3.2# times more concentrated than the diluted solution. You thus have

+
+

#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_""concentrated"" = ""D.F."" xx c_""diluted"")color(white)(a/a)|)))#

+
+

which gets you

+
+

#c_""diluted"" = ""0.96 M""/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)(""0.30 M"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the concentrated solution.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 16399 views + around the world +
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+ + Creative Commons License + +
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" If 250.0 mL of a 0.96 M solution of acetic acid (#C_2H_4O_2#) are diluted to 800.0 mL, what will be the approximate molarity of the final solution? nan +349 a834edb7-6ddd-11ea-a1c7-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-c2h4o2 CH2O start chemical_formula qc_end chemical_equation 6 6 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] C2H4O2 [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""CH2O""}]" "[{""type"":""chemical equation"",""value"":""C2H4O2""}]" "

What is the empirical formula of C2H4O2?

" nan CH2O "
+

Explanation:

+
+

The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.

+

In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.

+

So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.

+

In your case, the molecular formula is given to be #""C""_2""H""_4""O""_2#. What that tells you is that the molecule contains

+
+
    +
  • two atoms of carbon
    • +
  • four atoms of hydrogen
    • +
  • two atoms of oxygen
    • +
+
+

But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?

+

The minimum number of atoms an element can contriubte to a building block is #1#, since you can't have half an atom or one atom and a half as whole numbers.

+

With this in mind, notice that if you divide all these values by #2#, you will get

+
+
    +
  • one atom of carbon
    • +
  • two atoms of hydrogen
    • +
  • one atom of oxygen
    • +
+
+

This tells you that the building block for this particular molecule is #""CH""_2""O""#. This will thus be the compound's empirical formula.

+

In other words, if you take two such building blocks, #""CH""_2""O""#. you can build the molecule #""C""_2""H""_4""O""_2#.

+

+ +

+
+
" "
+
+
+

#""CH""_2""O""#

+
+
+
+

Explanation:

+
+

The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.

+

In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.

+

So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.

+

In your case, the molecular formula is given to be #""C""_2""H""_4""O""_2#. What that tells you is that the molecule contains

+
+
    +
  • two atoms of carbon
    • +
  • four atoms of hydrogen
    • +
  • two atoms of oxygen
    • +
+
+

But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?

+

The minimum number of atoms an element can contriubte to a building block is #1#, since you can't have half an atom or one atom and a half as whole numbers.

+

With this in mind, notice that if you divide all these values by #2#, you will get

+
+
    +
  • one atom of carbon
    • +
  • two atoms of hydrogen
    • +
  • one atom of oxygen
    • +
+
+

This tells you that the building block for this particular molecule is #""CH""_2""O""#. This will thus be the compound's empirical formula.

+

In other words, if you take two such building blocks, #""CH""_2""O""#. you can build the molecule #""C""_2""H""_4""O""_2#.

+

+ +

+
+
+
" "
+

What is the empirical formula of C2H4O2?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Oct 22, 2015 + +
+
+
+
+
+
+
+

#""CH""_2""O""#

+
+
+
+

Explanation:

+
+

The empirical formula tells you what the smallest integer ratio between the atoms that make up a substance is.

+

In essence, the empirical formula can be thought of as the building block of a substance. The molecular formula will thus tell you how many building blocks are needed to form that substance.

+

So, for a given molecule, the empirical formula will tell you what is the minimum number of atoms of each element needed to form the building block for the molecular formula.

+

In your case, the molecular formula is given to be #""C""_2""H""_4""O""_2#. What that tells you is that the molecule contains

+
+
    +
  • two atoms of carbon
    • +
  • four atoms of hydrogen
    • +
  • two atoms of oxygen
    • +
+
+

But what is the minimum number of atoms of each element that you need in order to be able to start building the molecule?

+

The minimum number of atoms an element can contriubte to a building block is #1#, since you can't have half an atom or one atom and a half as whole numbers.

+

With this in mind, notice that if you divide all these values by #2#, you will get

+
+
    +
  • one atom of carbon
    • +
  • two atoms of hydrogen
    • +
  • one atom of oxygen
    • +
+
+

This tells you that the building block for this particular molecule is #""CH""_2""O""#. This will thus be the compound's empirical formula.

+

In other words, if you take two such building blocks, #""CH""_2""O""#. you can build the molecule #""C""_2""H""_4""O""_2#.

+

+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 32814 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the empirical formula of C2H4O2? nan +350 a834edb8-6ddd-11ea-ad63-ccda262736ce https://socratic.org/questions/how-many-mg-atoms-are-in-3-24-moles-of-mg-2-68-x-1024-atoms-of-cu-equal-how-many 1.95 × 10^24 start physical_unit 2 3 number none qc_end physical_unit 2 2 6 7 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] Mg atoms""}]" "[{""type"":""physical unit"",""value"":""1.95 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] Mg [=] \\pu{3.24 moles}""}]" "

How many Mg atoms are in 3.24 moles of Mg?

" nan 1.95 × 10^24 "
+

Explanation:

+
+

We must use Avogadro's number: #6.022*10^23#

+

We set up the problem like this:

+

#3.24 "" moles Mg""*# #(6.022*10^23 "" Mg atoms"")/(1 "" moles Mg"")#

+

#""moles Mg""# cancel

+

#3.24 cancel("" moles Mg"")*# #(6.022*10^23 "" Mg atoms"")/(1 cancel("" moles Mg"")#

+

Then we just compute

+

#3.24 *##(6.022*10^23 "" Mg atoms"")/1#

+

#=1.95*10^24 "" Mg atoms""#

+
+
" "
+
+
+

#1.95*10^24 "" Mg atoms""#

+
+
+
+

Explanation:

+
+

We must use Avogadro's number: #6.022*10^23#

+

We set up the problem like this:

+

#3.24 "" moles Mg""*# #(6.022*10^23 "" Mg atoms"")/(1 "" moles Mg"")#

+

#""moles Mg""# cancel

+

#3.24 cancel("" moles Mg"")*# #(6.022*10^23 "" Mg atoms"")/(1 cancel("" moles Mg"")#

+

Then we just compute

+

#3.24 *##(6.022*10^23 "" Mg atoms"")/1#

+

#=1.95*10^24 "" Mg atoms""#

+
+
+
" "
+

How many Mg atoms are in 3.24 moles of Mg?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 6, 2017 + +
+
+
+
+
+
+
+

#1.95*10^24 "" Mg atoms""#

+
+
+
+

Explanation:

+
+

We must use Avogadro's number: #6.022*10^23#

+

We set up the problem like this:

+

#3.24 "" moles Mg""*# #(6.022*10^23 "" Mg atoms"")/(1 "" moles Mg"")#

+

#""moles Mg""# cancel

+

#3.24 cancel("" moles Mg"")*# #(6.022*10^23 "" Mg atoms"")/(1 cancel("" moles Mg"")#

+

Then we just compute

+

#3.24 *##(6.022*10^23 "" Mg atoms"")/1#

+

#=1.95*10^24 "" Mg atoms""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 12628 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many Mg atoms are in 3.24 moles of Mg? nan +351 a834edb9-6ddd-11ea-ab9a-ccda262736ce https://socratic.org/questions/58c507757c01494b4c240650 +3 start physical_unit 6 8 charge none qc_end chemical_equation 15 15 qc_end end "[{""type"":""physical unit"",""value"":""Charge [OF] the metal ion""}]" "[{""type"":""physical unit"",""value"":""+3""}]" "[{""type"":""chemical equation"",""value"":""[M(SO4)(NH3)5]^+Cl^−""}]" "

What is the likely charge on the metal ion with a complex whose formula is #[M(SO_4)(NH_3)_5]^+Cl^(-)#?

" nan +3 "
+

Explanation:

+
+

#[M(SO_4)(NH_3)_5]^(+)Cl^(-)rarr[M(SO_4)(NH_3)_5]^+ + Cl^-#

+

And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......

+

#[M(SO_4)(NH_3)_5]^+ rarr[M(SO_4)]^(+) +5NH_3#

+

to leave #[M(SO_4)]^(+)#; but sulfato as a ligand is formulated as #SO_4^(2-):#

+

#[M(SO_4)]^(+) rarr SO_4^(2-) +M^(3+)#

+

So #M# whatever it was (and it is likely to be a chromium or a cobalt salt) was #M^(3+)#. And thus the oxidation state of the metal centre was #+III#.

+

Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.

+

Capisce?

+

Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The #X# symbol is reserved for the halides, and a chemist would have formulated the question of the basis of #[M(SO_4)(NH_3)_5]^+#.

+
+
" "
+
+
+

Well because the complex as written has a single chloride counterion, we can write..........

+

#[M(SO_4)(NH_3)_5]^(+)Cl^(-)rarr[M(SO_4)(NH_3)_5]^+ + Cl^-#

+
+
+
+

Explanation:

+
+

#[M(SO_4)(NH_3)_5]^(+)Cl^(-)rarr[M(SO_4)(NH_3)_5]^+ + Cl^-#

+

And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......

+

#[M(SO_4)(NH_3)_5]^+ rarr[M(SO_4)]^(+) +5NH_3#

+

to leave #[M(SO_4)]^(+)#; but sulfato as a ligand is formulated as #SO_4^(2-):#

+

#[M(SO_4)]^(+) rarr SO_4^(2-) +M^(3+)#

+

So #M# whatever it was (and it is likely to be a chromium or a cobalt salt) was #M^(3+)#. And thus the oxidation state of the metal centre was #+III#.

+

Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.

+

Capisce?

+

Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The #X# symbol is reserved for the halides, and a chemist would have formulated the question of the basis of #[M(SO_4)(NH_3)_5]^+#.

+
+
+
" "
+

What is the likely charge on the metal ion with a complex whose formula is #[M(SO_4)(NH_3)_5]^+Cl^(-)#?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 12, 2017 + +
+
+
+
+
+
+
+

Well because the complex as written has a single chloride counterion, we can write..........

+

#[M(SO_4)(NH_3)_5]^(+)Cl^(-)rarr[M(SO_4)(NH_3)_5]^+ + Cl^-#

+
+
+
+

Explanation:

+
+

#[M(SO_4)(NH_3)_5]^(+)Cl^(-)rarr[M(SO_4)(NH_3)_5]^+ + Cl^-#

+

And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......

+

#[M(SO_4)(NH_3)_5]^+ rarr[M(SO_4)]^(+) +5NH_3#

+

to leave #[M(SO_4)]^(+)#; but sulfato as a ligand is formulated as #SO_4^(2-):#

+

#[M(SO_4)]^(+) rarr SO_4^(2-) +M^(3+)#

+

So #M# whatever it was (and it is likely to be a chromium or a cobalt salt) was #M^(3+)#. And thus the oxidation state of the metal centre was #+III#.

+

Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.

+

Capisce?

+

Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The #X# symbol is reserved for the halides, and a chemist would have formulated the question of the basis of #[M(SO_4)(NH_3)_5]^+#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1299 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the likely charge on the metal ion with a complex whose formula is #[M(SO_4)(NH_3)_5]^+Cl^(-)#? nan +352 a834edba-6ddd-11ea-bb06-ccda262736ce https://socratic.org/questions/2-00-grams-of-a-straight-chained-hydrocarbon-contains-1-68grams-of-carbon-the-re C4H10 start chemical_formula qc_end physical_unit 3 5 0 1 mass qc_end physical_unit 10 10 7 8 mass qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] this hydrocarbon [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""C4H10""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] a straight-chained hydrocarbon [=] \\pu{2.00 grams}""},{""type"":""physical unit"",""value"":""Mass [OF] carbon [=] \\pu{1.68 grams}""},{""type"":""other"",""value"":""A straight-chained hydrocarbon contains carbon, the rest hydrogen.""}]" "

2.00 grams of a straight-chained hydrocarbon contains 1.68grams of carbon, the rest hydrogen. +1. Calculate the empirical formula of this hydrocarbon.?

" nan C4H10 "
+

Explanation:

+
+
+

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

+

We must convert the masses of #""C""# and #""H""# to moles and then find the ratio.

+
+

Step 1. Calculate the mass of #""H""#

+

#""Mass of H"" = ""mass of hydrocarbon - mass of C"" = ""2.00 g - 1.68 g = 0.32 g""#

+
+

Step 2. Calculate the moles of #""C""# and #""H""#

+

#""Moles of C"" = 1.68 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01 color(red)(cancel(color(black)(""g C"")))) = ""0.1399 mol C""#

+

#""Moles of H"" = 0.32 color(red)(cancel(color(black)(""g H""))) × ""1 mol H""/(1.008 color(red)(cancel(color(black)(""g H"")))) = ""0.317 mol H""#

+
+

From this point on, I like to summarize the calculations in a table.

+

#bbul(""Element""color(white)(m) ""Mass/g""color(white)(m) ""Moles""color(white)(m) ""Ratio""color(white)(m)×4color(white)(m)""Integers"")#
+#color(white)(mm)""C"" color(white)(XXXm)1.68 color(white)(Xmll)0.1399 color(white)(Xm)1color(white)(mmml)4color(white)(mmmml)4#
+#color(white)(mm)""H"" color(white)(XXXm)0.32 color(white)(mmll)0.317 color(white)(Xml)2.27color(white)(mm)9.08 color(white)(mmm)9#

+

The molar ratio is #""C:H = 4:9""#.

+

The empirical formula is #""C""_4""H""_9#.

+
+

Here is a video that illustrates how to determine an empirical formula.

+

+ +

+
+
" "
+
+
+

The empirical formula is #""C""_4""H""_9#.

+
+
+
+

Explanation:

+
+
+

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

+

We must convert the masses of #""C""# and #""H""# to moles and then find the ratio.

+
+

Step 1. Calculate the mass of #""H""#

+

#""Mass of H"" = ""mass of hydrocarbon - mass of C"" = ""2.00 g - 1.68 g = 0.32 g""#

+
+

Step 2. Calculate the moles of #""C""# and #""H""#

+

#""Moles of C"" = 1.68 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01 color(red)(cancel(color(black)(""g C"")))) = ""0.1399 mol C""#

+

#""Moles of H"" = 0.32 color(red)(cancel(color(black)(""g H""))) × ""1 mol H""/(1.008 color(red)(cancel(color(black)(""g H"")))) = ""0.317 mol H""#

+
+

From this point on, I like to summarize the calculations in a table.

+

#bbul(""Element""color(white)(m) ""Mass/g""color(white)(m) ""Moles""color(white)(m) ""Ratio""color(white)(m)×4color(white)(m)""Integers"")#
+#color(white)(mm)""C"" color(white)(XXXm)1.68 color(white)(Xmll)0.1399 color(white)(Xm)1color(white)(mmml)4color(white)(mmmml)4#
+#color(white)(mm)""H"" color(white)(XXXm)0.32 color(white)(mmll)0.317 color(white)(Xml)2.27color(white)(mm)9.08 color(white)(mmm)9#

+

The molar ratio is #""C:H = 4:9""#.

+

The empirical formula is #""C""_4""H""_9#.

+
+

Here is a video that illustrates how to determine an empirical formula.

+

+ +

+
+
+
" "
+

2.00 grams of a straight-chained hydrocarbon contains 1.68grams of carbon, the rest hydrogen. +1. Calculate the empirical formula of this hydrocarbon.?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 27, 2018 + +
+
+
+
+
+
+
+

The empirical formula is #""C""_4""H""_9#.

+
+
+
+

Explanation:

+
+
+

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

+

We must convert the masses of #""C""# and #""H""# to moles and then find the ratio.

+
+

Step 1. Calculate the mass of #""H""#

+

#""Mass of H"" = ""mass of hydrocarbon - mass of C"" = ""2.00 g - 1.68 g = 0.32 g""#

+
+

Step 2. Calculate the moles of #""C""# and #""H""#

+

#""Moles of C"" = 1.68 color(red)(cancel(color(black)(""g C""))) × ""1 mol C""/(12.01 color(red)(cancel(color(black)(""g C"")))) = ""0.1399 mol C""#

+

#""Moles of H"" = 0.32 color(red)(cancel(color(black)(""g H""))) × ""1 mol H""/(1.008 color(red)(cancel(color(black)(""g H"")))) = ""0.317 mol H""#

+
+

From this point on, I like to summarize the calculations in a table.

+

#bbul(""Element""color(white)(m) ""Mass/g""color(white)(m) ""Moles""color(white)(m) ""Ratio""color(white)(m)×4color(white)(m)""Integers"")#
+#color(white)(mm)""C"" color(white)(XXXm)1.68 color(white)(Xmll)0.1399 color(white)(Xm)1color(white)(mmml)4color(white)(mmmml)4#
+#color(white)(mm)""H"" color(white)(XXXm)0.32 color(white)(mmll)0.317 color(white)(Xml)2.27color(white)(mm)9.08 color(white)(mmm)9#

+

The molar ratio is #""C:H = 4:9""#.

+

The empirical formula is #""C""_4""H""_9#.

+
+

Here is a video that illustrates how to determine an empirical formula.

+

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" "2.00 grams of a straight-chained hydrocarbon contains 1.68grams of carbon, the rest hydrogen. +1. Calculate the empirical formula of this hydrocarbon.?" nan +353 a834edbb-6ddd-11ea-902c-ccda262736ce https://socratic.org/questions/solubility-of-mg-oh-2-is-1-6-x-10-4-mol-l-at-298-k-what-is-its-solubility-produc 1.64 × 10^(-11) start physical_unit 2 2 solubility_product none qc_end physical_unit 2 2 4 7 solubility qc_end physical_unit 2 2 9 10 temperature qc_end end "[{""type"":""physical unit"",""value"":""Solubility product [OF] Mg(OH)2""}]" "[{""type"":""physical unit"",""value"":""1.64 × 10^(-11)""}]" "[{""type"":""physical unit"",""value"":""Solubility [OF] Mg(OH)2 [=] \\pu{1.6 x 10^(−4) mol/L}""},{""type"":""physical unit"",""value"":""Temperature [OF] Mg(OH)2 [=] \\pu{298 K}""}]" "

Solubility of #Mg(OH)_2# is #1.6# x #10^-4# #""mol/L""# at #298# #K#. What is its solubility product?

" nan 1.64 × 10^(-11) "
+

Explanation:

+
+

Magnesium hydroxide, #""Mg""(""OH"")_2#, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

+

Magnesium hydroxide dissociates only partially to form magnesium cations, #""Mg""^(2+)#, and hydroxide anions, #""OH""^(-)#

+
+

#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#

+
+

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

+

In your case, a molar solubility of

+
+

#s = 1.6 * 10^(-4)#

+
+

means that you can only dissolve #1.6 * 10^(-4)# moles of magnesium in a liter of water at that temperature.

+

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces #1# mole of magnesium cations and #color(red)(2)# moles of hydroxide anions.

+

This tells you that if you successfully dissolve #1.6 * 10^(-4)# moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

+
+

#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#

+
+

and

+
+

#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#

+
+

Since we're working with one liter of solution, you can cay that

+
+

#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#

+

#[""OH""^(-)] = 3.2 * 10^(-4)""M""#

+
+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

Plug in these values to get

+
+

#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#

+

#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs

+
+

The listed value for magnesium hydroxide's solubility product is #1.6 * 10^(-11)#, so this is an excellent result.

+

http://www.wiredchemist.com/chemistry/data/solubility-product-constants

+
+
" "
+
+
+

#1.6 * 10^(-11)#

+
+
+
+

Explanation:

+
+

Magnesium hydroxide, #""Mg""(""OH"")_2#, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

+

Magnesium hydroxide dissociates only partially to form magnesium cations, #""Mg""^(2+)#, and hydroxide anions, #""OH""^(-)#

+
+

#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#

+
+

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

+

In your case, a molar solubility of

+
+

#s = 1.6 * 10^(-4)#

+
+

means that you can only dissolve #1.6 * 10^(-4)# moles of magnesium in a liter of water at that temperature.

+

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces #1# mole of magnesium cations and #color(red)(2)# moles of hydroxide anions.

+

This tells you that if you successfully dissolve #1.6 * 10^(-4)# moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

+
+

#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#

+
+

and

+
+

#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#

+
+

Since we're working with one liter of solution, you can cay that

+
+

#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#

+

#[""OH""^(-)] = 3.2 * 10^(-4)""M""#

+
+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

Plug in these values to get

+
+

#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#

+

#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs

+
+

The listed value for magnesium hydroxide's solubility product is #1.6 * 10^(-11)#, so this is an excellent result.

+

http://www.wiredchemist.com/chemistry/data/solubility-product-constants

+
+
+
" "
+

Solubility of #Mg(OH)_2# is #1.6# x #10^-4# #""mol/L""# at #298# #K#. What is its solubility product?

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Ksp + + +
+
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+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 31, 2016 + +
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+
+
+

#1.6 * 10^(-11)#

+
+
+
+

Explanation:

+
+

Magnesium hydroxide, #""Mg""(""OH"")_2#, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.

+

Magnesium hydroxide dissociates only partially to form magnesium cations, #""Mg""^(2+)#, and hydroxide anions, #""OH""^(-)#

+
+

#""Mg""(""OH"")_text(2(s]) rightleftharpoons ""Mg""_text((aq])^(2+) + color(red)(2)""OH""_text((aq])^(-)#

+
+

For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.

+

In your case, a molar solubility of

+
+

#s = 1.6 * 10^(-4)#

+
+

means that you can only dissolve #1.6 * 10^(-4)# moles of magnesium in a liter of water at that temperature.

+

Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces #1# mole of magnesium cations and #color(red)(2)# moles of hydroxide anions.

+

This tells you that if you successfully dissolve #1.6 * 10^(-4)# moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be

+
+

#n_(Mg^(2+)) = 1 xx 1.6 * 10^(-4) = 1.6 * 10^(-4)""moles Mg""^(2+)#

+
+

and

+
+

#n_(OH^(-)) = color(red)(2) xx 1.6 * 10^(-4) = 3.2 * 10^(-4)""moles""#

+
+

Since we're working with one liter of solution, you can cay that

+
+

#[""Mg""^(2+)] = 1.6 * 10^(-4)""M""#

+

#[""OH""^(-)] = 3.2 * 10^(-4)""M""#

+
+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

Plug in these values to get

+
+

#K_(sp) = 1.6 * 10^(-4) * (3.2 * 10^(-4))^color(red)(2)#

+

#K_(sp) = color(green)(1.6 * 10^(-11)) -># rounded to two sig figs

+
+

The listed value for magnesium hydroxide's solubility product is #1.6 * 10^(-11)#, so this is an excellent result.

+

http://www.wiredchemist.com/chemistry/data/solubility-product-constants

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" "Solubility of #Mg(OH)_2# is #1.6# x #10^-4# #""mol/L""# at #298# #K#. What is its solubility product?" nan +354 a834edbc-6ddd-11ea-90e2-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-solution-produced-by-dissolving-4-moles-of-sugar-in-2- 2.00 mol/L start physical_unit 5 6 molarity mol/l qc_end physical_unit 13 13 10 11 mole qc_end physical_unit 18 18 15 16 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] a solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""2.00 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] sugar [=] \\pu{4 moles}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{2 L}""}]" "

What is the molarity of a solution produced by dissolving 4 moles of sugar in 2 L of water?

" nan 2.00 mol/L "
+

Explanation:

+
+

#M = (n_2)/(V_1)=# (number of moles of the solute{mol} )#/#(volume of the solvent {L})

+

#M = (n_2)/(V_1) = (4)/(2)= 2 M = 2 molL^-1#

+
+
" "
+
+
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#M = (n_2)/(V_1) = (4)/(2)= 2 M = 2 molL^-1#

+
+
+
+

Explanation:

+
+

#M = (n_2)/(V_1)=# (number of moles of the solute{mol} )#/#(volume of the solvent {L})

+

#M = (n_2)/(V_1) = (4)/(2)= 2 M = 2 molL^-1#

+
+
+
" "
+

What is the molarity of a solution produced by dissolving 4 moles of sugar in 2 L of water?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Mar 3, 2016 + +
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#M = (n_2)/(V_1) = (4)/(2)= 2 M = 2 molL^-1#

+
+
+
+

Explanation:

+
+

#M = (n_2)/(V_1)=# (number of moles of the solute{mol} )#/#(volume of the solvent {L})

+

#M = (n_2)/(V_1) = (4)/(2)= 2 M = 2 molL^-1#

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" What is the molarity of a solution produced by dissolving 4 moles of sugar in 2 L of water? nan +355 a83514be-6ddd-11ea-827f-ccda262736ce https://socratic.org/questions/what-volume-of-a-6-0-m-hcl-solution-is-required-to-make-250-0-milliliters-of-a-1 62.50 milliliters start physical_unit 6 7 volume ml qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 6 7 12 13 volume qc_end physical_unit 6 7 16 17 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] HCl solution [IN] milliliters""}]" "[{""type"":""physical unit"",""value"":""62.50 milliliters""}]" "[{""type"":""physical unit"",""value"":""Molarity2 [OF] HCl solution [=] \\pu{6.0 M}""},{""type"":""physical unit"",""value"":""Volume1 [OF] HCl solution [=] \\pu{250.0 milliliters}""},{""type"":""physical unit"",""value"":""Molarity1 [OF] HCl solution [=] \\pu{1.5 M}""}]" "

What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#?

" nan 62.50 milliliters "
+

Explanation:

+
+

For this type of problem, we want to use the following dilution formula:
+
+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

+

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

+

We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

+

#(C_2xxV_2)/C_1# = #V_1#

+

#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

+
+
" "
+
+
+

62.5 mL of HCl is required.

+
+
+
+

Explanation:

+
+

For this type of problem, we want to use the following dilution formula:
+
+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

+

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

+

We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

+

#(C_2xxV_2)/C_1# = #V_1#

+

#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

+
+
+
" "
+

What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Dilution Calculations + + +
+
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+
+
+1 Answer +
+
+
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+
+
+ +
+
+ +
+ + Jun 12, 2016 + +
+
+
+
+
+
+
+

62.5 mL of HCl is required.

+
+
+
+

Explanation:

+
+

For this type of problem, we want to use the following dilution formula:
+
+#C_1# is the initial concentration
+#V_1# is the initial volume
+#C_2# is the final concentration
+#V_2# is the final volume

+

* TIP. * Whenever you are diluting a solution (going from a highly concentrated substance to a less concentrated substance by increasing the volume of the solvent) you always use this equation.

+

We know the initial concentration, final volume, and final concentration. All we have to do rearrange the equation to solve for the initial volume:

+

#(C_2xxV_2)/C_1# = #V_1#

+

#(1.5 cancel""M"" xx250.0mL)/(6.0cancel""M"") # = 62.5 mL

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+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 62536 views + around the world +
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+ + Creative Commons License + +
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+
" What volume of a 6.0 M #HCl# solution is required to make 250.0 milliliters of a 1.5 M #HCl#? nan +356 a83514bf-6ddd-11ea-a02e-ccda262736ce https://socratic.org/questions/591dcf3f11ef6b6cb5fd49b2 2 C6H14 + 19 O2 -> 12 CO2 + 14 H2O start chemical_equation qc_end substance 6 6 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the complete combustion""}]" "[{""type"":""chemical equation"",""value"":""2 C6H14 + 19 O2 -> 12 CO2 + 14 H2O""}]" "[{""type"":""substance name"",""value"":""Hexanes""}]" "

How is the complete combustion of hexanes represented in a chemical reaction?

" nan 2 C6H14 + 19 O2 -> 12 CO2 + 14 H2O "
+

Explanation:

+
+

Is charge balanced? Tick. Is mass balanced? Tick.

+

The #Delta# symbol represents the fact that this is an exothermic reaction, and the #""heat""# evolved may be utilized to do #""work""#.

+

Most such combustion reactions, certainly in the #""internal combustion engine""#, exhibit incomplete reaction, and both #CO(g)# and particulate #C(s)# are oxidation products.

+
+
" "
+
+
+

#C_6H_14(l) + 19/2O_2(g) rarr 6CO_2(g) +7H_2O(l) +Delta#

+
+
+
+

Explanation:

+
+

Is charge balanced? Tick. Is mass balanced? Tick.

+

The #Delta# symbol represents the fact that this is an exothermic reaction, and the #""heat""# evolved may be utilized to do #""work""#.

+

Most such combustion reactions, certainly in the #""internal combustion engine""#, exhibit incomplete reaction, and both #CO(g)# and particulate #C(s)# are oxidation products.

+
+
+
" "
+

How is the complete combustion of hexanes represented in a chemical reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
+
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+
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+2 Answers +
+
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+ +
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+ +
+ + May 18, 2017 + +
+
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+
+

#C_6H_14(l) + 19/2O_2(g) rarr 6CO_2(g) +7H_2O(l) +Delta#

+
+
+
+

Explanation:

+
+

Is charge balanced? Tick. Is mass balanced? Tick.

+

The #Delta# symbol represents the fact that this is an exothermic reaction, and the #""heat""# evolved may be utilized to do #""work""#.

+

Most such combustion reactions, certainly in the #""internal combustion engine""#, exhibit incomplete reaction, and both #CO(g)# and particulate #C(s)# are oxidation products.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + May 18, 2017 + +
+
+
+
+
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+
+

#2C_6H_14 (l) + 19O_2 (g) rarr 12CO_2 (g)+ 14 H_2O (g)#

+
+
+
+

Explanation:

+
+

Write the (first unbalanced) chemical equation:

+

#C_6H_14 (l) + O_2 (g) rarr CO_2 (g)+ H_2O (g)#

+

Balanced:

+

#2C_6H_14 (l) + 19O_2 (g) rarr 12CO_2 (g)+ 14 H_2O (g)#

+
+
+
+
+
+ +
+
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+
+
Related questions
+ + +
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Impact of this question
+
+ 3611 views + around the world +
+
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+
" How is the complete combustion of hexanes represented in a chemical reaction? nan +357 a83514c0-6ddd-11ea-9901-ccda262736ce https://socratic.org/questions/what-is-the-mass-in-grams-of-one-molecule-of-ethnic-acid-ch-3cooh 9.96 × 10^(-23) grams start physical_unit 12 12 mass g qc_end physical_unit 8 11 7 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] CH3COOH [IN] grams""}]" "[{""type"":""physical unit"",""value"":""9.96 × 10^(-23) grams""}]" "[{""type"":""physical unit"",""value"":""Number [OF] molecule of ethnic acid [=] \\pu{one}""}]" "

What is the mass in grams of one molecule of ethnic acid #CH_3COOH#?

" nan 9.96 × 10^(-23) grams "
+

Explanation:

+
+

1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass #M_r=12+3+12+16+16+1=60g//mol#.

+

But 1 mole contains an Avagadro number of molecules, ie #N_A=6.023xx10^23# molecules.

+

Therefore 1 molecule of this weak acid has a mass of #60/(6.023xx10^23)=9.96181xx10^(-23)g#

+
+
" "
+
+
+

#9.96181xx10^(-23)g#

+
+
+
+

Explanation:

+
+

1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass #M_r=12+3+12+16+16+1=60g//mol#.

+

But 1 mole contains an Avagadro number of molecules, ie #N_A=6.023xx10^23# molecules.

+

Therefore 1 molecule of this weak acid has a mass of #60/(6.023xx10^23)=9.96181xx10^(-23)g#

+
+
+
" "
+

What is the mass in grams of one molecule of ethnic acid #CH_3COOH#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
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+ +
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+ +
+ + Dec 16, 2015 + +
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#9.96181xx10^(-23)g#

+
+
+
+

Explanation:

+
+

1 mole of the organic carboxylic acid ethanoic acid (acetic acid) has a mass equal to its molar mass #M_r=12+3+12+16+16+1=60g//mol#.

+

But 1 mole contains an Avagadro number of molecules, ie #N_A=6.023xx10^23# molecules.

+

Therefore 1 molecule of this weak acid has a mass of #60/(6.023xx10^23)=9.96181xx10^(-23)g#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 14521 views + around the world +
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+ + Creative Commons License + +
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+
+
" What is the mass in grams of one molecule of ethnic acid #CH_3COOH#? nan +358 a83514c1-6ddd-11ea-ba2a-ccda262736ce https://socratic.org/questions/what-is-the-balanced-equation-for-the-chemical-reaction-mg3n2-2h2o-mg-oh-2-nh3 Mg3N2 + 6 H2O -> 3 Mg(OH)2 + 2 NH3 start chemical_equation qc_end chemical_equation 9 16 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the chemical reaction""}]" "[{""type"":""chemical equation"",""value"":""Mg3N2 + 6 H2O -> 3 Mg(OH)2 + 2 NH3""}]" "[{""type"":""chemical equation"",""value"":""Mg3N2 + 2 H2O -> Mg(OH)2 + NH3""}]" "

What is the balanced equation for the chemical reaction #Mg_3N_2+2H_2O -> Mg(OH)_2+NH_3#?

" nan Mg3N2 + 6 H2O -> 3 Mg(OH)2 + 2 NH3 "
+

Explanation:

+
+

Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.

+

For example, we have #""Mg""_3""N""_2# on the left side and on the right side on the original equation we see magnesium as #""Mg""(""OH"")_2#.

+

On the left side, there are 3 moles of #""Mg""#. Now making sure we have exactly 3 moles of #""Mg""# on the right side, we add a #3# in front of the compound #""Mg""(""OH"")_2# to make it #3""Mg""(""OH"")_2#. Do same for all.

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#""Mg""_ 3""N""_2(s) + 6""H""_2 ""O""(l) -> 3""Mg""(""OH"")_2(aq) + 2""NH""_3(aq)#

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+
+
+

Explanation:

+
+

Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.

+

For example, we have #""Mg""_3""N""_2# on the left side and on the right side on the original equation we see magnesium as #""Mg""(""OH"")_2#.

+

On the left side, there are 3 moles of #""Mg""#. Now making sure we have exactly 3 moles of #""Mg""# on the right side, we add a #3# in front of the compound #""Mg""(""OH"")_2# to make it #3""Mg""(""OH"")_2#. Do same for all.

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+
+
" "
+

What is the balanced equation for the chemical reaction #Mg_3N_2+2H_2O -> Mg(OH)_2+NH_3#?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+2 Answers +
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#""Mg""_ 3""N""_2(s) + 6""H""_2 ""O""(l) -> 3""Mg""(""OH"")_2(aq) + 2""NH""_3(aq)#

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+
+

Explanation:

+
+

Now, the first thing is to monitor each element as an individual and count the number of moles present in each on the left side and equate it to exactly the same number of moles to the same element on the right side.

+

For example, we have #""Mg""_3""N""_2# on the left side and on the right side on the original equation we see magnesium as #""Mg""(""OH"")_2#.

+

On the left side, there are 3 moles of #""Mg""#. Now making sure we have exactly 3 moles of #""Mg""# on the right side, we add a #3# in front of the compound #""Mg""(""OH"")_2# to make it #3""Mg""(""OH"")_2#. Do same for all.

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+ + Mar 3, 2018 + +
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#Mg_3N_2(s) + 6H_2O(l)rarr3Mg(OH)_2(aq) + 2NH_3(aq)#

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+
+

Explanation:

+
+

And the question you ask yourself...is mass balanced, and is charge balanced...and so you count up the atoms on each side.... There are three metals atoms, 2 nitrogens, 12 hydrogens, and 6 oxygens ON BOTH SIDES of the equation....so the given reaction at least conforms to the principle of conservation of mass. It also conforms to experiment....

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" What is the balanced equation for the chemical reaction #Mg_3N_2+2H_2O -> Mg(OH)_2+NH_3#? nan +359 a83514c2-6ddd-11ea-ba34-ccda262736ce https://socratic.org/questions/how-do-you-balance-the-following-reaction-c-6h-6-o-2-co-2-h-2o 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O start chemical_equation qc_end chemical_equation 7 17 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] reaction""}]" "[{""type"":""chemical equation"",""value"":""2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O""}]" "[{""type"":""chemical equation"",""value"":""? C6H6 + ? O2 -> ? CO2 + ? H2O""}]" "

How do you balance the following reaction: #?C_6H_6 + ?O_2 -> ?CO_2 + ?H_2O#

" nan 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O "
+

Explanation:

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+

The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.

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Start with a ""2"" in front of the benzene #(C_6H_6)# and you should reach this result

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#2C_6H_6 + 15O_2 rarr12 CO_2+6H_2O#

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Explanation:

+
+

The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.

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" "
+

How do you balance the following reaction: #?C_6H_6 + ?O_2 -> ?CO_2 + ?H_2O#

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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+ + Jan 12, 2017 + +
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Start with a ""2"" in front of the benzene #(C_6H_6)# and you should reach this result

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#2C_6H_6 + 15O_2 rarr12 CO_2+6H_2O#

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+
+

Explanation:

+
+

The problem that can arise in trying to balance a combustion equation is the fact that water has only one O atom, but oxygen gas supplies them in pairs. The ""trick"" to getting the job done is to make certain the coefficient of water is an even number, even if this means you don't start with the usual technique of placing a ""1"" in front of a compound.

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+
Related questions
+ + +
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Impact of this question
+
+ 1607 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" How do you balance the following reaction: #?C_6H_6 + ?O_2 -> ?CO_2 + ?H_2O# nan +360 a83514c3-6ddd-11ea-b370-ccda262736ce https://socratic.org/questions/molten-iron-is-extremely-hot-averaging-about-1-500-c-the-specific-heat-of-iron-i-1 8.68 × 10^5 J start physical_unit 0 1 heat_energy j qc_end physical_unit 0 1 7 8 temperature qc_end physical_unit 1 1 15 18 specific_heat qc_end physical_unit 0 1 36 37 temperature qc_end physical_unit 0 1 28 29 mass qc_end end "[{""type"":""physical unit"",""value"":""Heat released [OF] molten iron [IN] J""}]" "[{""type"":""physical unit"",""value"":""8.68 × 10^5 J""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] molten iron [=] \\pu{1,500 ℃}""},{""type"":""physical unit"",""value"":""Specific heat [OF] iron [=] \\pu{0.46 J/(g * ℃)}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] molten iron [=] \\pu{25 ℃}""},{""type"":""physical unit"",""value"":""Mass [OF] molten iron [=] \\pu{1 kg}""}]" "

Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)?

" nan 8.68 × 10^5 J "
+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

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900 Kj

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+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

+
+
+
" "
+

Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
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+3 Answers +
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+ + Jun 24, 2017 + +
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900 Kj

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+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

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I got #""1020 kJ""# were RELEASED into the atmosphere, ignoring phase changes between the #alpha#, #delta#, and #gamma# phases and just looking at the temperature changes.

+

You can get more context here:
+https://en.wikipedia.org/wiki/Iron#Phase_diagram_and_allotropes

+

and you can examine the specific heat capacity variations more closely here:
+http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Mask=2&Type=JANAFS&Plot=on#JANAFS

+

On another note, this #""1020 kJ""# is quite a bit higher than what one would normally expect to get, #""655.5 kJ""#, due to taking into account the huge variation in heat capacity across #1475^@ ""C""#.

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If you simply assume a #C_P# of #""0.46 J/g""cdot""K""# throughout, you would get #""655.5 kJ""# instead (#656# to three sig figs).

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+

There is a HUGE assumption here that iron's specific heat capacity doesn't change from #25^@ ""C""# to #1500^@ ""C""#, which is clearly not true. Here is the phase diagram of iron:

+

+

Since all these phases at #""1 bar""# are solids, we are safe in assuming there is no major enthalpy of solid-solid phase transitions to worry about.

+

However, the specific heat capacity #C_P# at constant pressure changes drastically as we transition through #alpha#, #gamma#, and #delta# phases:

+

[

+

The wonky curve is the #alpha# and #delta# phase, and the linear curve is the #gamma# phase. Here's how I would treat this:

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  • #alpha#-phase, from #""298.15 K""# up to #""700 K""# (#426.85^@ ""C""#), using an average of #C_P ~~ ""29.656 J/mol""cdot""K""# (at #~~ ""500 K""#), or #""0.531 J/g""cdot""K""#.
    • +
  • #alpha#-phase, from #""700 K""# to #""935 K""# (#661.85^@ ""C""#) using an average of #C_P ~~ ""40.149 J/mol""cdot""K""# (at #~~ ""816 K""#), or #""0.719 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""935 K""# to #""1042 K""# (#768.85^@ ""C""#) using an average of #C_P ~~ ""59.442 J/mol""cdot""K""# (at #~~ ""1010 K""#), or #""1.064 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""1042 K""# to #""1100 K""# (#826.85^@ ""C""#) using an average of #C_P ~~ ""65.743 J/mol""cdot""K""# (at #~~ ""1068 K""#), or #""1.177 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""1100 K""# to #""1183.15 K""# (#910^@ ""C""#, the #alpha->gamma# transition temperature) using an average of #C_P ~~ ""43.029 J/mol""cdot""K""# (at #~~ ""1150 K""#), or #""0.770 J/g""cdot""K""#
    • +
  • #gamma#-phase, from #""1183.15 K""# to #""1667.15 K""# (#1394^@ ""C""#, the #gamma->delta# transition temperature) using an average of #C_P ~~ ""35.856 J/mol""cdot""K""# (at #~~ ""1420 K""#), or #""0.642 J/g""cdot""K""#
    • +
  • #delta#-phase, from #""1667.15 K""# to #""1773.15 K""# (#1500^@ ""C""#!), using an average of #C_P ~~ ""41.764 J/mol""cdot""K""# (at #~~ ""1722 K""#), or #""0.748 J/g""cdot""K""#.
    • +
+

Aren't you glad we aren't doing phase changes? :-)

+

So, we would have the heat of cooling as the negative of the heat of heating:

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+

#q_""cool"" = -(q_1 + . . . + q_7)#

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#= -m(C_(P1)DeltaT_(0->1) + . . . + C_(P7)DeltaT_(6->7))#

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+

I'll leave the units out, but you know that they are #""J/g""cdot""K""# for #C_P# and #""K""# for #T#. The mass is in #""g""#.

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+

#= -1000 cdot [0.531(700 - 298.15) + 0.719(935 - 700) + 1.064(1042 - 935) + 1.177(1100 - 1042) + 0.770(1183.15 - 1100) + 0.642(1667.15 - 1183.15) + 0.748(1773.15 - 1667.15)]#

+
+

Each phase then approximately contributes:

+
+

#= overbrace(-""628487 J"")^(alpha"" phase"") + overbrace(-""310728 J"")^(gamma"" phase"") + overbrace(-""79288 J"")^(delta"" phase"")#

+

#~~# #-1.020 xx 10^(6)# #""J""#,

+
+

or about #color(blue)(-""1020 kJ"")#, to three sig figs.

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Thermal history of cooling cast iron

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Explanation:

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Thermal history of cooling cast iron from #1500^oC# to #25^oC# ...
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" Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)? nan +361 a83514c4-6ddd-11ea-a46d-ccda262736ce https://socratic.org/questions/how-do-you-write-the-balanced-equation-for-the-combustion-of-propane C3H8 + 5 O2 -> 3 CO2 + 4 H2O start chemical_equation qc_end substance 11 11 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the combustion""}]" "[{""type"":""chemical equation"",""value"":""C3H8 + 5 O2 -> 3 CO2 + 4 H2O""}]" "[{""type"":""substance name"",""value"":""Propane""}]" "

How do you write the balanced equation for the combustion of propane?

" nan C3H8 + 5 O2 -> 3 CO2 + 4 H2O "
+

Explanation:

+
+

Propane has a chemical formula of #C_3H_8#

+

Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.

+

#C_3H_8 +5O_2 ->3CO_2 + 4H_2O#

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#C_3H_8 +5O_2 ->3CO_2 + 4H_2O#

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+
+

Explanation:

+
+

Propane has a chemical formula of #C_3H_8#

+

Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.

+

#C_3H_8 +5O_2 ->3CO_2 + 4H_2O#

+
+
+
" "
+

How do you write the balanced equation for the combustion of propane?

+
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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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+ + Apr 28, 2016 + +
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#C_3H_8 +5O_2 ->3CO_2 + 4H_2O#

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+
+

Explanation:

+
+

Propane has a chemical formula of #C_3H_8#

+

Combustion is when a hydro-carbon like propane containing carbon and hydrogen is burned in oxygen, releasing carbon dioxide and water.

+

#C_3H_8 +5O_2 ->3CO_2 + 4H_2O#

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" How do you write the balanced equation for the combustion of propane? nan +362 a83514c5-6ddd-11ea-888f-ccda262736ce https://socratic.org/questions/what-concentration-of-so3-2-is-in-equilibrium-with-ag2so3-s-and-1-80-10-3-m-ag-t 4.63 × 10^(–9) M start physical_unit 3 3 concentration mol/l qc_end c_other OTHER qc_end physical_unit 14 14 10 13 concentration qc_end physical_unit 8 8 21 23 equilibrium_constant_k qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] SO3^2- [IN] M""}]" "[{""type"":""physical unit"",""value"":""4.63 × 10^(–9) M""}]" "[{""type"":""other"",""value"":""SO3^2- is in equilibrium with Ag2SO3(s) and Ag+.""},{""type"":""physical unit"",""value"":""Concentration [OF] Ag+ [=] \\pu{1.80 × 10^(−3) M}""},{""type"":""physical unit"",""value"":""Ksp [OF] Ag2SO3 [=] \\pu{1.50 × 10^(–14)}""}]" "

What concentration of #""SO""_3^(2–)# is in equilibrium with #""Ag""_2""SO""_3(s)# and #1.80 * 10^-3# #""M""# #""Ag""^(+)# ? The #K_(sp)# of #""Ag""_2""SO""_3# is #1.50 * 10^–14#

" "
+
+

+

One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations.

+

+
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" 4.63 × 10^(–9) M "
+

Explanation:

+
+

The expression of the solubility product constant for silver sulfite looks like this

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

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This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.

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#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce #color(red)(2)# moles of silver(I) cations means that the equilibrium concentration of the silver(I) cations must be raised to the power of #color(red)(2)#.

+

Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.

+

So something like

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#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
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is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.

+

In your case, you have--I'll do the calculations with the added units for the solubility product constant!

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#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#

+
+

So you can say that the equilibrium concentration of the sulfite anions will be equal to

+
+

#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#

+
+

The answer is rounded to three sig figs.

+

So, a quick recap

+
+

#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Here #color(red)(2)# is a coefficient!

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

Here #color(red)(2)# must be an exponent, so don't use it as a coefficient!

+
+
+
+
" "
+
+
+

#4.63 * 10^(-9)# #""M""#

+
+
+
+

Explanation:

+
+

The expression of the solubility product constant for silver sulfite looks like this

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.

+
+

#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce #color(red)(2)# moles of silver(I) cations means that the equilibrium concentration of the silver(I) cations must be raised to the power of #color(red)(2)#.

+

Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.

+

So something like

+
+

#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.

+

In your case, you have--I'll do the calculations with the added units for the solubility product constant!

+
+

#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#

+
+

So you can say that the equilibrium concentration of the sulfite anions will be equal to

+
+

#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#

+
+

The answer is rounded to three sig figs.

+

So, a quick recap

+
+

#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Here #color(red)(2)# is a coefficient!

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

Here #color(red)(2)# must be an exponent, so don't use it as a coefficient!

+
+
+
+
+
" "
+

What concentration of #""SO""_3^(2–)# is in equilibrium with #""Ag""_2""SO""_3(s)# and #1.80 * 10^-3# #""M""# #""Ag""^(+)# ? The #K_(sp)# of #""Ag""_2""SO""_3# is #1.50 * 10^–14#

+
+
+

+

One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations.

+

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+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Dynamic Equilibrium + + +
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+1 Answer +
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+ + Feb 20, 2018 + +
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#4.63 * 10^(-9)# #""M""#

+
+
+
+

Explanation:

+
+

The expression of the solubility product constant for silver sulfite looks like this

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.

+
+

#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce #color(red)(2)# moles of silver(I) cations means that the equilibrium concentration of the silver(I) cations must be raised to the power of #color(red)(2)#.

+

Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.

+

So something like

+
+

#K_(sp) = [color(red)(2)""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.

+

In your case, you have--I'll do the calculations with the added units for the solubility product constant!

+
+

#1.50 * 10^(-14) quad ""M""^3 = (1.80 * 10^(-3))^color(red)(2) quad ""M""^color(red)(2) * [""SO""_3^(2-)]#

+
+

So you can say that the equilibrium concentration of the sulfite anions will be equal to

+
+

#[S""O""_3^(2-)] = (1.50 * 10^(-14) quad ""M""^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)(""M""^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad ""M"")))#

+
+

The answer is rounded to three sig figs.

+

So, a quick recap

+
+

#""Ag""_ 2""SO""_ (3(s)) rightleftharpoons color(red)(2)""Ag""_ ((aq))^(+) + ""SO""_ (3(aq))^(2-)#

+
+

Here #color(red)(2)# is a coefficient!

+
+

#K_(sp) = [""Ag""^(+)]^color(red)(2) * [""SO""_3^(2-)]#

+
+

Here #color(red)(2)# must be an exponent, so don't use it as a coefficient!

+
+
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+
+
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+ +
+
+
+
+
+
+
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" "What concentration of #""SO""_3^(2–)# is in equilibrium with #""Ag""_2""SO""_3(s)# and #1.80 * 10^-3# #""M""# #""Ag""^(+)# ? The #K_(sp)# of #""Ag""_2""SO""_3# is #1.50 * 10^–14#" " + + +One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations. + + +" +363 a83514c6-6ddd-11ea-b86f-ccda262736ce https://socratic.org/questions/how-can-i-write-the-formula-for-aluminum-chloride AlCl3 start chemical_formula qc_end substance 7 8 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] aluminum chloride [IN] default""}]" "[{""type"":""chemical equation"",""value"":""AlCl3""}]" "[{""type"":""substance name"",""value"":""Aluminum chloride""}]" "

How can I write the formula for aluminum chloride?

" nan AlCl3 "
+

Explanation:

+
+

We know that aluminum has a valency of #+3# because it has three valence electron which it can lose easily. Similarly, chlorine has a valency of #-1# because it can gain an electron.

+

So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.

+
+

#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#

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#""AlCl""_3#

+
+
+
+

Explanation:

+
+

We know that aluminum has a valency of #+3# because it has three valence electron which it can lose easily. Similarly, chlorine has a valency of #-1# because it can gain an electron.

+

So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.

+
+

#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#

+
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" "
+

How can I write the formula for aluminum chloride?

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+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
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+1 Answer +
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#""AlCl""_3#

+
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+
+

Explanation:

+
+

We know that aluminum has a valency of #+3# because it has three valence electron which it can lose easily. Similarly, chlorine has a valency of #-1# because it can gain an electron.

+

So in writing the formula, there is a cross of the two valencies because three chlorine atoms will have to react with one aluminium atom to gain one electron each.

+
+

#[""Al""]^(3+) + 3[""Cl""]^(-) -> ""AlCl""_3#

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" How can I write the formula for aluminum chloride? nan +364 a83514c7-6ddd-11ea-a777-ccda262736ce https://socratic.org/questions/how-many-electrons-are-lost-or-gained-in-forming-each-as-3 3 start physical_unit 2 2 number none qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Number lost or gained [OF] electrons""}]" "[{""type"":""physical unit"",""value"":""3""}]" "[{""type"":""other"",""value"":""each As^(−3)""}]" "

How many electrons are lost or gained in forming each #As^(-3)#?

" nan 3 "
+

Explanation:

+
+

Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.

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" "
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3 electrons are gained.

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Explanation:

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+

Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.

+
+
+
" "
+

How many electrons are lost or gained in forming each #As^(-3)#?

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+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Ionic Compounds + + +
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+1 Answer +
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3 electrons are gained.

+
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Explanation:

+
+

Oxidation states with ""n+"" mean there are n number of electrons lost or removed from the total number of electrons of the ground state element . ""n-"" on the other hand means there are n number of electrons gained or added to the total number of electrons of the ground state element.

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+
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" How many electrons are lost or gained in forming each #As^(-3)#? nan +365 a83514c8-6ddd-11ea-bacf-ccda262736ce https://socratic.org/questions/how-do-you-balance-na-h-2o-naoh-h-2 2 Na + 2 H2O -> 2 NaOH + 2 H2 start chemical_equation qc_end chemical_equation 4 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""2 Na + 2 H2O -> 2 NaOH + 2 H2""}]" "[{""type"":""chemical equation"",""value"":""Na + H2O -> NaOH + H2""}]" "

How do you balance #Na + H_2O -> NaOH + H_2#?

" nan 2 Na + 2 H2O -> 2 NaOH + 2 H2 "
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
" "
+
+
+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
" "
+

How do you balance #Na + H_2O -> NaOH + H_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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#2Na + 2H_2O -> 2NaOH + 2H_2#

+
+
+
+

Explanation:

+
+

First, let's rewrite #H_2O# as #HOH#. It may seem weird, but when water reacts, it actually forms hydrogen and hydroxide ions. So, the ""new"" equation is:

+

#Na + HOH -> NaOH + H_2#

+

Now we can see how many of each element/polyatomic ion we have on each side of the equation.

+

Left:
+Na - 1
+H - 1
+OH - 1

+

Right:
+Na - 1
+OH -1
+H -2

+

Looking at what we have, there is an unequal amount of hydrogen on the left side and the right side. We have only 1 on the left and 2 on the right. Let's double everything on the left. Now we get:

+

#2Na + 2HOH -> NaOH + H_2#

+

Since we have 2 of everything on the left, we should make sure that the right side is accounted for. As mentioned in our original count, the #H_2# is accounted for. The Na and OH still need accounted for. We have two of each on the left side, but only one of each on the right side. This means that we need to form one more NaOH. When we do that, we get:

+

#2Na + 2HOH -> 2NaOH + 2H_2#

+

Changing back #HOH# to #H_2O#, we get:

+

#2Na + 2H_2O -> 2NaOH + 2H_2#

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" How do you balance #Na + H_2O -> NaOH + H_2#? nan +366 a8353bd0-6ddd-11ea-bfab-ccda262736ce https://socratic.org/questions/598a066311ef6b77a930881c 409.05 K start physical_unit 4 4 temperature k qc_end physical_unit 4 4 6 7 temperature qc_end physical_unit 4 4 0 1 volume qc_end physical_unit 4 4 9 10 pressure qc_end physical_unit 4 4 24 25 pressure qc_end physical_unit 4 4 17 18 volume qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] gas [IN] K""}]" "[{""type"":""physical unit"",""value"":""409.05 K""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] gas [=] \\pu{303 K}""},{""type"":""physical unit"",""value"":""Volume1 [OF] gas [=] \\pu{2.50 L}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] gas [=] \\pu{100 kPa}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] gas [=] \\pu{90.0 kPa}""},{""type"":""physical unit"",""value"":""Volume2 [OF] gas [=] \\pu{3.75 L}""}]" "

#""2.50 L""# of a gas at #""303 K""# and #""100 kPa""# are heated. The new volume is #""3.75 L""# and the new pressure is #""90.0 kPa""#. What is the new temperature?

" nan 409.05 K "
+

Explanation:

+
+

This question is an example of the combined gas law. The formula is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Organize the data:

+

Known

+

#P_1=""100 kPa""#

+

#V_1=""2.50 L""#

+

#T_1=""303 K""#

+

#P_2=""90.0 kPa""#

+

#V_2=""3.75 L""#

+

Unknown

+

#T_2#

+

Solution

+

Rearrange the formula to isolate #T_2#. Insert your data and solve.

+

#T_2=(P_2V_2T_1)/(P_1V_1)#

+

#T_2=((90.0color(red)cancel(color(black)(""kPa""))xx3.75color(red)cancel(color(black)(""L""))xx303""K""))/((100color(red)cancel(color(black)(""kPa""))xx2.50color(red)cancel(color(black)(""L""))))=""409K""# rounded to three significant figures

+

(#""100 kPa""# is an exact number so it has an infinite number of significant figures.)

+
+
" "
+
+
+

#""409 K""# to three significant figures.

+
+
+
+

Explanation:

+
+

This question is an example of the combined gas law. The formula is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Organize the data:

+

Known

+

#P_1=""100 kPa""#

+

#V_1=""2.50 L""#

+

#T_1=""303 K""#

+

#P_2=""90.0 kPa""#

+

#V_2=""3.75 L""#

+

Unknown

+

#T_2#

+

Solution

+

Rearrange the formula to isolate #T_2#. Insert your data and solve.

+

#T_2=(P_2V_2T_1)/(P_1V_1)#

+

#T_2=((90.0color(red)cancel(color(black)(""kPa""))xx3.75color(red)cancel(color(black)(""L""))xx303""K""))/((100color(red)cancel(color(black)(""kPa""))xx2.50color(red)cancel(color(black)(""L""))))=""409K""# rounded to three significant figures

+

(#""100 kPa""# is an exact number so it has an infinite number of significant figures.)

+
+
+
" "
+

#""2.50 L""# of a gas at #""303 K""# and #""100 kPa""# are heated. The new volume is #""3.75 L""# and the new pressure is #""90.0 kPa""#. What is the new temperature?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Combined Gas Law + + +
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+1 Answer +
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+ + Aug 9, 2017 + +
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#""409 K""# to three significant figures.

+
+
+
+

Explanation:

+
+

This question is an example of the combined gas law. The formula is:

+

#(P_1V_1)/T_1=(P_2V_2)/T_2#

+

Organize the data:

+

Known

+

#P_1=""100 kPa""#

+

#V_1=""2.50 L""#

+

#T_1=""303 K""#

+

#P_2=""90.0 kPa""#

+

#V_2=""3.75 L""#

+

Unknown

+

#T_2#

+

Solution

+

Rearrange the formula to isolate #T_2#. Insert your data and solve.

+

#T_2=(P_2V_2T_1)/(P_1V_1)#

+

#T_2=((90.0color(red)cancel(color(black)(""kPa""))xx3.75color(red)cancel(color(black)(""L""))xx303""K""))/((100color(red)cancel(color(black)(""kPa""))xx2.50color(red)cancel(color(black)(""L""))))=""409K""# rounded to three significant figures

+

(#""100 kPa""# is an exact number so it has an infinite number of significant figures.)

+
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" "#""2.50 L""# of a gas at #""303 K""# and #""100 kPa""# are heated. The new volume is #""3.75 L""# and the new pressure is #""90.0 kPa""#. What is the new temperature? " nan +367 a8353bd1-6ddd-11ea-b70b-ccda262736ce https://socratic.org/questions/an-ideal-gas-in-a-sealed-container-has-an-initial-volume-of-2-35-l-at-constant-p 117.77 ℃ start physical_unit 1 2 temperature °c qc_end physical_unit 1 2 22 23 temperature qc_end physical_unit 1 2 12 13 volume qc_end physical_unit 1 2 29 30 volume qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Temperature1 [OF] ideal gas [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""117.77 ℃""}]" "[{""type"":""physical unit"",""value"":""Temperature2 [OF] ideal gas [=] \\pu{18 ℃}""},{""type"":""physical unit"",""value"":""Volume1 [OF] ideal gas [=] \\pu{2.35 L}""},{""type"":""physical unit"",""value"":""Volume2 [OF] ideal gas [=] \\pu{1.75 L}""},{""type"":""other"",""value"":""Constant pressure.""}]" "

An ideal gas in a sealed container has an initial volume of 2.35 L. At constant pressure, it is cooled to to 18°C where its final volume is 1.75 L. What was the initial temperature?

" nan 117.77 ℃ "
+

Explanation:

+
+

We use Charles' Law:
+#V_1/T_1=V_2/T_2#

+

Plug in the values:
+#(2.35L)/T_1=(1.75L)/(291K)#

+

And don't forget to convert degrees Celsius to Kelvin! Just add 273.

+

Solve the algebra:
+#2.35L# x #291K# = #683.85 L*K#

+

#683.85 L*K# x #1/(1.75L)# = #391K#

+

You can convert it back to degrees Celsius by subtracting 273.

+
+
" "
+
+
+

391 K or 118 degrees Celsius

+
+
+
+

Explanation:

+
+

We use Charles' Law:
+#V_1/T_1=V_2/T_2#

+

Plug in the values:
+#(2.35L)/T_1=(1.75L)/(291K)#

+

And don't forget to convert degrees Celsius to Kelvin! Just add 273.

+

Solve the algebra:
+#2.35L# x #291K# = #683.85 L*K#

+

#683.85 L*K# x #1/(1.75L)# = #391K#

+

You can convert it back to degrees Celsius by subtracting 273.

+
+
+
" "
+

An ideal gas in a sealed container has an initial volume of 2.35 L. At constant pressure, it is cooled to to 18°C where its final volume is 1.75 L. What was the initial temperature?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Charles' Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 3, 2018 + +
+
+
+
+
+
+
+

391 K or 118 degrees Celsius

+
+
+
+

Explanation:

+
+

We use Charles' Law:
+#V_1/T_1=V_2/T_2#

+

Plug in the values:
+#(2.35L)/T_1=(1.75L)/(291K)#

+

And don't forget to convert degrees Celsius to Kelvin! Just add 273.

+

Solve the algebra:
+#2.35L# x #291K# = #683.85 L*K#

+

#683.85 L*K# x #1/(1.75L)# = #391K#

+

You can convert it back to degrees Celsius by subtracting 273.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 18095 views + around the world +
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+ + Creative Commons License + +
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" An ideal gas in a sealed container has an initial volume of 2.35 L. At constant pressure, it is cooled to to 18��C where its final volume is 1.75 L. What was the initial temperature? nan +368 a8353bd2-6ddd-11ea-914b-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-calcium-sulfate-dihydrate CaSO4.2H2O start chemical_formula qc_end substance 5 7 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] calcium sulfate dihydrate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""CaSO4.2H2O""}]" "[{""type"":""substance name"",""value"":""Calcium sulfate dihydrate""}]" "

What is the formula for calcium sulfate dihydrate?

" nan CaSO4.2H2O "
+

Explanation:

+
+

Calcium sulfate dihydrate is an ionic substance composed of #Ca^(2+)# and #SO_4^(2-)#. The associated water molecules are known as waters of crystallization, which are tacked on to the substance when it crystallizes (from water). They could be removed on prolonged drying.

+
+
" "
+
+
+

#CaSO_4*2H_2O#

+
+
+
+

Explanation:

+
+

Calcium sulfate dihydrate is an ionic substance composed of #Ca^(2+)# and #SO_4^(2-)#. The associated water molecules are known as waters of crystallization, which are tacked on to the substance when it crystallizes (from water). They could be removed on prolonged drying.

+
+
+
" "
+

What is the formula for calcium sulfate dihydrate?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 30, 2015 + +
+
+
+
+
+
+
+

#CaSO_4*2H_2O#

+
+
+
+

Explanation:

+
+

Calcium sulfate dihydrate is an ionic substance composed of #Ca^(2+)# and #SO_4^(2-)#. The associated water molecules are known as waters of crystallization, which are tacked on to the substance when it crystallizes (from water). They could be removed on prolonged drying.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3673 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the formula for calcium sulfate dihydrate? nan +369 a8353bd3-6ddd-11ea-916b-ccda262736ce https://socratic.org/questions/594df9bab72cff693af235a4 6.01 L start physical_unit 7 8 volume l qc_end c_other OTHER qc_end physical_unit 21 21 18 19 volume qc_end physical_unit 21 21 23 24 concentration qc_end c_other NTP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] dihydrogen gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""6.01 L""}]" "[{""type"":""other"",""value"":""EXCESS zinc metal.""},{""type"":""physical unit"",""value"":""Volume [OF] H2SO4 [=] \\pu{500 cm^3}""},{""type"":""physical unit"",""value"":""Concentration [OF] H2SO4 [=] \\pu{0.50 mol/L}""},{""type"":""other"",""value"":""NTP""}]" "

Given EXCESS zinc metal, what volume of dihydrogen gas at #""NTP""# will be evolved from the action of #500*cm^3# of #H_2SO_4# at #0.50*mol*L^-1# concentration?

" nan 6.01 L "
+

Explanation:

+
+

And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........

+

As far as I know #""NTP""# specifies #1*atm# pressure and #293.15*K#, and the molar volume of an Ideal Gas at #NTP# is #24.05*L#

+

#""Moles of sulfuric acid""=500*cm^3xx10^-3*L*cm^-3xx0.5*mol*L^-1=0.25*mol#

+

And thus #0.25*mol# dihydrogen gas will be evolved.

+

And so volume at #""NTP""=0.25*molxx24.05*L*mol^-1~=6*L#

+
+
" "
+
+
+

We interrogate the redox reaction......

+

#Zn(s) + H_2SO_4(aq) rarr ZnSO_4(aq) + H_2(g)uarr#

+
+
+
+

Explanation:

+
+

And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........

+

As far as I know #""NTP""# specifies #1*atm# pressure and #293.15*K#, and the molar volume of an Ideal Gas at #NTP# is #24.05*L#

+

#""Moles of sulfuric acid""=500*cm^3xx10^-3*L*cm^-3xx0.5*mol*L^-1=0.25*mol#

+

And thus #0.25*mol# dihydrogen gas will be evolved.

+

And so volume at #""NTP""=0.25*molxx24.05*L*mol^-1~=6*L#

+
+
+
" "
+

Given EXCESS zinc metal, what volume of dihydrogen gas at #""NTP""# will be evolved from the action of #500*cm^3# of #H_2SO_4# at #0.50*mol*L^-1# concentration?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 24, 2017 + +
+
+
+
+
+
+
+

We interrogate the redox reaction......

+

#Zn(s) + H_2SO_4(aq) rarr ZnSO_4(aq) + H_2(g)uarr#

+
+
+
+

Explanation:

+
+

And thus ONE equiv of dihydrogen results from each equiv of sulfuric acid (and also from each equiv of zinc metal).........

+

As far as I know #""NTP""# specifies #1*atm# pressure and #293.15*K#, and the molar volume of an Ideal Gas at #NTP# is #24.05*L#

+

#""Moles of sulfuric acid""=500*cm^3xx10^-3*L*cm^-3xx0.5*mol*L^-1=0.25*mol#

+

And thus #0.25*mol# dihydrogen gas will be evolved.

+

And so volume at #""NTP""=0.25*molxx24.05*L*mol^-1~=6*L#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 1318 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
" "Given EXCESS zinc metal, what volume of dihydrogen gas at #""NTP""# will be evolved from the action of #500*cm^3# of #H_2SO_4# at #0.50*mol*L^-1# concentration?" nan +370 a8353bd4-6ddd-11ea-b116-ccda262736ce https://socratic.org/questions/how-many-grams-of-iron-are-in-79-2-g-of-iron-ill-oxide 55.40 grams start physical_unit 4 4 mass g qc_end physical_unit 10 12 7 8 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] iron [IN] grams""}]" "[{""type"":""physical unit"",""value"":""55.40 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] iron (Ill) oxide [=] \\pu{79.2 g}""}]" "

How many grams of iron are in 79.2 g of iron (Ill) oxide?

" nan 55.40 grams "
+

Explanation:

+
+

We calculate (i) a molar quantity of iron oxide:

+

#(79.2*g)/(159.69*g*mol^-1)=0.496*mol#.

+

And then (ii) convert this molar quantity into a mass of iron:

+

#""Mass of iron""=2xx0.496*molxx55.85*g*mol^-1=55.4*g#.

+

Why did I multiply the molar quantity by 2?

+
+
" "
+
+
+

#""Mass of iron ""~=# #""55""*""g""#

+
+
+
+

Explanation:

+
+

We calculate (i) a molar quantity of iron oxide:

+

#(79.2*g)/(159.69*g*mol^-1)=0.496*mol#.

+

And then (ii) convert this molar quantity into a mass of iron:

+

#""Mass of iron""=2xx0.496*molxx55.85*g*mol^-1=55.4*g#.

+

Why did I multiply the molar quantity by 2?

+
+
+
" "
+

How many grams of iron are in 79.2 g of iron (Ill) oxide?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 27, 2017 + +
+
+
+
+
+
+
+

#""Mass of iron ""~=# #""55""*""g""#

+
+
+
+

Explanation:

+
+

We calculate (i) a molar quantity of iron oxide:

+

#(79.2*g)/(159.69*g*mol^-1)=0.496*mol#.

+

And then (ii) convert this molar quantity into a mass of iron:

+

#""Mass of iron""=2xx0.496*molxx55.85*g*mol^-1=55.4*g#.

+

Why did I multiply the molar quantity by 2?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4859 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many grams of iron are in 79.2 g of iron (Ill) oxide? nan +371 a8353bd5-6ddd-11ea-a69d-ccda262736ce https://socratic.org/questions/5932b7b5b72cff624f76d732 3.00 moles start physical_unit 4 5 mole mol qc_end physical_unit 16 16 10 11 volume qc_end physical_unit 16 16 14 15 molarity qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] potassium chloride [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.00 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] KCl(aq) [=] \\pu{3 L}""},{""type"":""physical unit"",""value"":""Molarity [OF] KCl(aq) [=] \\pu{1 mol/L}""}]" "

How many moles of potassium chloride are present in a #3*L# volume of #1*mol*L^-1# #KCl(aq)#?

" nan 3.00 moles "
+

Explanation:

+
+

#""Molarity""=""Moles of solute""/""Volume of solution""#.........

+

And thus for #""moles of solute""#, we take the product,

+

#""Molarity""xx""Volume of solution""#

+

#=3*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#

+

#=3.0*mol#

+

Given that #""Avogadro's number""-=6.022xx10^23*mol^-1#, how many ATOMS are in this quantity?

+
+
" "
+
+
+

Well, I gets a #3*""mole""# quantity of #KCl#.......

+
+
+
+

Explanation:

+
+

#""Molarity""=""Moles of solute""/""Volume of solution""#.........

+

And thus for #""moles of solute""#, we take the product,

+

#""Molarity""xx""Volume of solution""#

+

#=3*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#

+

#=3.0*mol#

+

Given that #""Avogadro's number""-=6.022xx10^23*mol^-1#, how many ATOMS are in this quantity?

+
+
+
" "
+

How many moles of potassium chloride are present in a #3*L# volume of #1*mol*L^-1# #KCl(aq)#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 3, 2017 + +
+
+
+
+
+
+
+

Well, I gets a #3*""mole""# quantity of #KCl#.......

+
+
+
+

Explanation:

+
+

#""Molarity""=""Moles of solute""/""Volume of solution""#.........

+

And thus for #""moles of solute""#, we take the product,

+

#""Molarity""xx""Volume of solution""#

+

#=3*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#

+

#=3.0*mol#

+

Given that #""Avogadro's number""-=6.022xx10^23*mol^-1#, how many ATOMS are in this quantity?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1813 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many moles of potassium chloride are present in a #3*L# volume of #1*mol*L^-1# #KCl(aq)#? nan +372 a8353bd6-6ddd-11ea-8894-ccda262736ce https://socratic.org/questions/how-much-heat-do-you-need-to-raise-the-temperature-of-150-g-of-ice-from-30-c-to- 4.73 × 10^3 J start physical_unit 14 14 heat_energy j qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 14 14 19 20 temperature qc_end physical_unit 14 14 11 12 mass qc_end end "[{""type"":""physical unit"",""value"":""Heat needed [OF] ice [IN] J""}]" "[{""type"":""physical unit"",""value"":""4.73 × 10^3 J""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] ice [=] \\pu{-30 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] ice [=] \\pu{-15 ℃}""},{""type"":""physical unit"",""value"":""Mass [OF] ice [=] \\pu{150 g}""}]" "

How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°C?

" nan 4.73 × 10^3 J "
+

Explanation:

+
+

Required Heat to raise the given temperature by #Delta theta# is given by:
+#H = m*c*Delta theta#, where, m is the mass, c is the specific heat capacity of the material and #Delta theta# is change in temperature. Using for ice c = 0.5 Cal/g/°C,
+We get,
+#H=150*0.5*(-15-(-30)#
+#H=75*15=1125 cal#

+
+
" "
+
+
+

1125 calories

+
+
+
+

Explanation:

+
+

Required Heat to raise the given temperature by #Delta theta# is given by:
+#H = m*c*Delta theta#, where, m is the mass, c is the specific heat capacity of the material and #Delta theta# is change in temperature. Using for ice c = 0.5 Cal/g/°C,
+We get,
+#H=150*0.5*(-15-(-30)#
+#H=75*15=1125 cal#

+
+
+
" "
+

How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°C?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 30, 2016 + +
+
+
+
+
+
+
+

1125 calories

+
+
+
+

Explanation:

+
+

Required Heat to raise the given temperature by #Delta theta# is given by:
+#H = m*c*Delta theta#, where, m is the mass, c is the specific heat capacity of the material and #Delta theta# is change in temperature. Using for ice c = 0.5 Cal/g/°C,
+We get,
+#H=150*0.5*(-15-(-30)#
+#H=75*15=1125 cal#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5455 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°C? nan +373 a8353bd7-6ddd-11ea-ba69-ccda262736ce https://socratic.org/questions/calculate-the-mass-in-kilograms-of-2-6-10-23-molecules-of-so2 2.77 × 10^(-2) kilograms start physical_unit 11 11 mass kg qc_end physical_unit 9 11 6 8 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] SO2 [IN] kilograms""}]" "[{""type"":""physical unit"",""value"":""2.77 × 10^(-2) kilograms""}]" "[{""type"":""physical unit"",""value"":""Number [OF] molecules of SO2 [=] \\pu{2.6 × 10^23}""}]" "

Calculate the mass in kilograms of #2.6xx10^23# molecules of #""SO""_2# ?

" nan 2.77 × 10^(-2) kilograms "
+

Explanation:

+
+

We were given #2.6 xx 10^23# molecules of #""SO""_2#. That means that

+

#N = 2.6 xx 10^23 \ ""molecules""#

+

Use

+

#n (""number of moles"") = N/(N_A (""Avogadro's constant""))#

+

to get

+

#n = (2.6 xx 10^23\ ""molecules"") / (6.02 xx 10^23\ ""molecules/mol"") = ""0.432 moles""#

+

Now use

+

#m = n xx M (""molar mass"")#

+

to get

+

#m = ""0.432 mol"" xx ""64.066 g/mol"" = ""27.677 g""#

+

So

+

#m (""in kilograms"") = ""27.677 g"" / (10^3 \ ""g/kg"") = ""0.028 kg""#

+
+
" "
+
+
+

#""0.028 kg""#

+
+
+
+

Explanation:

+
+

We were given #2.6 xx 10^23# molecules of #""SO""_2#. That means that

+

#N = 2.6 xx 10^23 \ ""molecules""#

+

Use

+

#n (""number of moles"") = N/(N_A (""Avogadro's constant""))#

+

to get

+

#n = (2.6 xx 10^23\ ""molecules"") / (6.02 xx 10^23\ ""molecules/mol"") = ""0.432 moles""#

+

Now use

+

#m = n xx M (""molar mass"")#

+

to get

+

#m = ""0.432 mol"" xx ""64.066 g/mol"" = ""27.677 g""#

+

So

+

#m (""in kilograms"") = ""27.677 g"" / (10^3 \ ""g/kg"") = ""0.028 kg""#

+
+
+
" "
+

Calculate the mass in kilograms of #2.6xx10^23# molecules of #""SO""_2# ?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Jun 10, 2018 + +
+
+
+
+
+
+
+

#""0.028 kg""#

+
+
+
+

Explanation:

+
+

We were given #2.6 xx 10^23# molecules of #""SO""_2#. That means that

+

#N = 2.6 xx 10^23 \ ""molecules""#

+

Use

+

#n (""number of moles"") = N/(N_A (""Avogadro's constant""))#

+

to get

+

#n = (2.6 xx 10^23\ ""molecules"") / (6.02 xx 10^23\ ""molecules/mol"") = ""0.432 moles""#

+

Now use

+

#m = n xx M (""molar mass"")#

+

to get

+

#m = ""0.432 mol"" xx ""64.066 g/mol"" = ""27.677 g""#

+

So

+

#m (""in kilograms"") = ""27.677 g"" / (10^3 \ ""g/kg"") = ""0.028 kg""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+
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+ + Creative Commons License + +
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+
+
" "Calculate the mass in kilograms of #2.6xx10^23# molecules of #""SO""_2# ?" nan +374 a83562e4-6ddd-11ea-ba28-ccda262736ce https://socratic.org/questions/589f3cc611ef6b70005b972d 52.00 kcal/mol start physical_unit 15 15 deltah^{0}_{f} kcal/mol qc_end chemical_equation 1 6 qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""DeltaH^{0}_{f} [OF] H(g) [IN] kcal/mol""}]" "[{""type"":""physical unit"",""value"":""52.00 kcal/mol""}]" "[{""type"":""chemical equation"",""value"":""H2 + [\\Delta] -> 2 H""},{""type"":""other"",""value"":""DeltaH = 104 kcal/mol""}]" "

Given #H_2+Delta rarr 2dotH# #;DeltaH=104*kcal*mol^-1#, what is #DeltaH_f^@# for #dotH(g)#?

" nan 52.00 kcal/mol "
+

Explanation:

+
+

The #""enthalpy of atomization""# is the energy required to produce one mole of ATOMS from an element in its standard state under standard conditions.......

+

We were given data for the reaction........

+

#H_2(g) +Deltararr2dotH; DeltaH^@=104*kcals*mol^-1 #

+

Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e. #""atoms""#, from one mole of gaseous #""dihydrogen MOLECULES""# (which of course is the standard state of #H_2#) , given the prior definition,

+

#DeltaH_""atomization""^@(""Hydrogen"")=52*kcal*mol^-1#.

+

Agreed?

+
+
" "
+
+
+

#DeltaH_""atomization""^@(""Hydrogen"")=52*kcal*mol^-1#.

+
+
+
+

Explanation:

+
+

The #""enthalpy of atomization""# is the energy required to produce one mole of ATOMS from an element in its standard state under standard conditions.......

+

We were given data for the reaction........

+

#H_2(g) +Deltararr2dotH; DeltaH^@=104*kcals*mol^-1 #

+

Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e. #""atoms""#, from one mole of gaseous #""dihydrogen MOLECULES""# (which of course is the standard state of #H_2#) , given the prior definition,

+

#DeltaH_""atomization""^@(""Hydrogen"")=52*kcal*mol^-1#.

+

Agreed?

+
+
+
" "
+

Given #H_2+Delta rarr 2dotH# #;DeltaH=104*kcal*mol^-1#, what is #DeltaH_f^@# for #dotH(g)#?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 5, 2017 + +
+
+
+
+
+
+
+

#DeltaH_""atomization""^@(""Hydrogen"")=52*kcal*mol^-1#.

+
+
+
+

Explanation:

+
+

The #""enthalpy of atomization""# is the energy required to produce one mole of ATOMS from an element in its standard state under standard conditions.......

+

We were given data for the reaction........

+

#H_2(g) +Deltararr2dotH; DeltaH^@=104*kcals*mol^-1 #

+

Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e. #""atoms""#, from one mole of gaseous #""dihydrogen MOLECULES""# (which of course is the standard state of #H_2#) , given the prior definition,

+

#DeltaH_""atomization""^@(""Hydrogen"")=52*kcal*mol^-1#.

+

Agreed?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
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+ + Creative Commons License + +
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+
" Given #H_2+Delta rarr 2dotH# #;DeltaH=104*kcal*mol^-1#, what is #DeltaH_f^@# for #dotH(g)#? nan +375 a83562e5-6ddd-11ea-b210-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-2-94-times-10-24-molecules-of-decane-c-10h-22 694.72 g start physical_unit 11 11 mass g qc_end physical_unit 8 11 5 7 number qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] C10H22 [IN] g""}]" "[{""type"":""physical unit"",""value"":""694.72 g""}]" "[{""type"":""physical unit"",""value"":""Number [OF] molecules of decane C10H22 [=] \\pu{2.94 × 10^24}""}]" "

What is the mass of #2.94 times 10^24# molecules of decane #C_10H_22#?

" nan 694.72 g "
+

Explanation:

+
+

#6.022xx10^23# individual items of stuff specifies a #""mole of stuff""#. The #""mole""# relates the (sub) micro world of atoms and molecules to the macro world of grams and litres.

+

And thus...............................................

+

#""mass""=(2.94xx10^24*""molecules"")/(6.022xx10^23*""molecules""*""mol""^-1)xx142.3*g*""mol""^-1~=700*g#

+
+
" "
+
+
+

Well, #6.022xx10^23# individual decane molecules have a mass of #142.3*g#

+
+
+
+

Explanation:

+
+

#6.022xx10^23# individual items of stuff specifies a #""mole of stuff""#. The #""mole""# relates the (sub) micro world of atoms and molecules to the macro world of grams and litres.

+

And thus...............................................

+

#""mass""=(2.94xx10^24*""molecules"")/(6.022xx10^23*""molecules""*""mol""^-1)xx142.3*g*""mol""^-1~=700*g#

+
+
+
" "
+

What is the mass of #2.94 times 10^24# molecules of decane #C_10H_22#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 5, 2017 + +
+
+
+
+
+
+
+

Well, #6.022xx10^23# individual decane molecules have a mass of #142.3*g#

+
+
+
+

Explanation:

+
+

#6.022xx10^23# individual items of stuff specifies a #""mole of stuff""#. The #""mole""# relates the (sub) micro world of atoms and molecules to the macro world of grams and litres.

+

And thus...............................................

+

#""mass""=(2.94xx10^24*""molecules"")/(6.022xx10^23*""molecules""*""mol""^-1)xx142.3*g*""mol""^-1~=700*g#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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Impact of this question
+
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+ + Creative Commons License + +
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+
" What is the mass of #2.94 times 10^24# molecules of decane #C_10H_22#? nan +376 a83562e6-6ddd-11ea-895d-ccda262736ce https://socratic.org/questions/potassium-chlorate-kclo-3-decomposes-to-form-potassium-chloride-kcl-and-oxygen-g 2 KClO3 -> 2 KCl + 3 O2 start chemical_equation qc_end chemical_equation 2 2 qc_end chemical_equation 8 8 qc_end substance 10 11 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this decomposition reaction""}]" "[{""type"":""chemical equation"",""value"":""2 KClO3 -> 2 KCl + 3 O2""}]" "[{""type"":""chemical equation"",""value"":""KClO3""},{""type"":""chemical equation"",""value"":""KCl""},{""type"":""substance name"",""value"":""Oxygen gas""}]" "

Potassium chlorate, #KClO_3#, decomposes to form potassium chloride, #KCl# and oxygen gas. How do you write a balanced equation for this decomposition reaction?

" nan 2 KClO3 -> 2 KCl + 3 O2 "
+

Explanation:

+
+

#""KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""O""_2(""g"")""#

+

Notice that the number of atoms of #""K""# and #""Cl""# are the same on both sides, but the numbers of #""O""# atoms are not. There are 3 #""O""# atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6 #""O""# atoms on both sides.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""3O""_2(""g"")""#

+

Now the #""K"" and ""Cl""# atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of #""KCl""#.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""2KCl(s)"" + ""3O""_2(""g"")""#

+

The equation is now balanced with 2 #""K""# atoms, 2 #""Cl""# atoms, and 6 #""O""# atoms on both sides.

+
+
" "
+
+
+

#""2KClO""_3(""s"")"" + heat""##rarr##""2KCl(s)"" + ""3O""_2(""g"")""#

+
+
+
+

Explanation:

+
+

#""KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""O""_2(""g"")""#

+

Notice that the number of atoms of #""K""# and #""Cl""# are the same on both sides, but the numbers of #""O""# atoms are not. There are 3 #""O""# atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6 #""O""# atoms on both sides.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""3O""_2(""g"")""#

+

Now the #""K"" and ""Cl""# atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of #""KCl""#.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""2KCl(s)"" + ""3O""_2(""g"")""#

+

The equation is now balanced with 2 #""K""# atoms, 2 #""Cl""# atoms, and 6 #""O""# atoms on both sides.

+
+
+
" "
+

Potassium chlorate, #KClO_3#, decomposes to form potassium chloride, #KCl# and oxygen gas. How do you write a balanced equation for this decomposition reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 8, 2015 + +
+
+
+
+
+
+
+

#""2KClO""_3(""s"")"" + heat""##rarr##""2KCl(s)"" + ""3O""_2(""g"")""#

+
+
+
+

Explanation:

+
+

#""KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""O""_2(""g"")""#

+

Notice that the number of atoms of #""K""# and #""Cl""# are the same on both sides, but the numbers of #""O""# atoms are not. There are 3 #""O""# atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6 #""O""# atoms on both sides.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""KCl(s)"" + ""3O""_2(""g"")""#

+

Now the #""K"" and ""Cl""# atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of #""KCl""#.

+

#""2KClO""_3(""s"")"" + heat""##rarr##""2KCl(s)"" + ""3O""_2(""g"")""#

+

The equation is now balanced with 2 #""K""# atoms, 2 #""Cl""# atoms, and 6 #""O""# atoms on both sides.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 142546 views + around the world +
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+ + Creative Commons License + +
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" Potassium chlorate, #KClO_3#, decomposes to form potassium chloride, #KCl# and oxygen gas. How do you write a balanced equation for this decomposition reaction? nan +377 a83562e7-6ddd-11ea-b34f-ccda262736ce https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-of-uranium-and-fluorine-that-is-comp UF6 start chemical_formula qc_end physical_unit 9 9 16 16 percent qc_end physical_unit 11 11 19 19 percent qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""UF6""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] uranium [=] \\pu{67.6%}""},{""type"":""physical unit"",""value"":""Percent [OF] fluorine [=] \\pu{32.4%}""}]" "

What is the empirical formula of a compound of uranium and fluorine that is composed of 67.6% uranium and 32.4% fluorine?

" nan UF6 "
+

Explanation:

+
+

The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.

+

Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.

+

Masses
+#""U"":##67.6%rArr""67.6 g""#
+#""F"":##32.4%rArr""32.4 g""#

+

Molar Masses
+#""U"":##""238.02891 g/mol""#
+#""F"":##""18.99840316 g/mol""#

+

Moles
+Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).

+

#""U"":##67.6cancel""g U""xx(1""mol U"")/(238.02891cancel""g U"")=""0.284 mol U""#
+#""F"":##32.4cancel""g F""xx(1""mol F"")/(18.99840316cancel""g F"")=""1.71 mol F""#

+

Mole Ratios
+Divide the number of moles of each element by the lowest number of moles.

+

#""U"":##(0.284cancel""mol"")/(0.284cancel""mol"")=1.00#

+

#""F"":##(1.71cancel""mol"")/(0.284cancel""mol"")=6.02~~6#

+

The empirical formula for this compound is #""UF""_6""#.

+
+
" "
+
+
+

The empirical formula for the compound containing uranium and fluorine is #""UF""_6""#.

+
+
+
+

Explanation:

+
+

The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.

+

Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.

+

Masses
+#""U"":##67.6%rArr""67.6 g""#
+#""F"":##32.4%rArr""32.4 g""#

+

Molar Masses
+#""U"":##""238.02891 g/mol""#
+#""F"":##""18.99840316 g/mol""#

+

Moles
+Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).

+

#""U"":##67.6cancel""g U""xx(1""mol U"")/(238.02891cancel""g U"")=""0.284 mol U""#
+#""F"":##32.4cancel""g F""xx(1""mol F"")/(18.99840316cancel""g F"")=""1.71 mol F""#

+

Mole Ratios
+Divide the number of moles of each element by the lowest number of moles.

+

#""U"":##(0.284cancel""mol"")/(0.284cancel""mol"")=1.00#

+

#""F"":##(1.71cancel""mol"")/(0.284cancel""mol"")=6.02~~6#

+

The empirical formula for this compound is #""UF""_6""#.

+
+
+
" "
+

What is the empirical formula of a compound of uranium and fluorine that is composed of 67.6% uranium and 32.4% fluorine?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 16, 2015 + +
+
+
+
+
+
+
+

The empirical formula for the compound containing uranium and fluorine is #""UF""_6""#.

+
+
+
+

Explanation:

+
+

The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.

+

Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.

+

Masses
+#""U"":##67.6%rArr""67.6 g""#
+#""F"":##32.4%rArr""32.4 g""#

+

Molar Masses
+#""U"":##""238.02891 g/mol""#
+#""F"":##""18.99840316 g/mol""#

+

Moles
+Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).

+

#""U"":##67.6cancel""g U""xx(1""mol U"")/(238.02891cancel""g U"")=""0.284 mol U""#
+#""F"":##32.4cancel""g F""xx(1""mol F"")/(18.99840316cancel""g F"")=""1.71 mol F""#

+

Mole Ratios
+Divide the number of moles of each element by the lowest number of moles.

+

#""U"":##(0.284cancel""mol"")/(0.284cancel""mol"")=1.00#

+

#""F"":##(1.71cancel""mol"")/(0.284cancel""mol"")=6.02~~6#

+

The empirical formula for this compound is #""UF""_6""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 19367 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the empirical formula of a compound of uranium and fluorine that is composed of 67.6% uranium and 32.4% fluorine? nan +378 a83562e8-6ddd-11ea-b109-ccda262736ce https://socratic.org/questions/a-sample-of-oxygen-gas-has-a-volume-of-150-0-ml-at-a-pressure-of-0-947-atm-what- 142.05 mL start physical_unit 22 23 volume ml qc_end physical_unit 0 4 9 10 volume qc_end physical_unit 0 4 15 16 pressure qc_end physical_unit 22 23 29 30 pressure qc_end c_other constant_temperature qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] mL""}]" "[{""type"":""physical unit"",""value"":""142.05 mL""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of oxygen gas [=] \\pu{150.0 mL}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] A sample of oxygen gas [=] \\pu{0.947 atm}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] the gas [=] \\pu{1.000 atm}""},{""type"":""other"",""value"":""ConstantTemperature""}]" "

A sample of oxygen gas has a volume of 150.0 mL at a pressure of 0.947 atm. What will the volume of the gas be at a pressure of 1.000 atm if the temperature remains constant?

" nan 142.05 mL "
+

Explanation:

+
+

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

+

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

+

In which from the definition, the equation is derived as;

+

#Pquadpropquad 1/V# or #P=k/V# or #PV=k#

+

#P=# Pressure of gas
+#V=# Volume of gas
+#k=# Constant

+

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

+

#P_iV_i=P_fV_f#

+

From the information given in this question;

+

#P_i=# Initial pressure of gas #=0.947atm#
+#P_f=# Final pressure of gas #=1.000atm#
+#V_i=# Initial volume of gas #=150mL#
+#V_f=# Final volume of gas #=?mL#

+

Calculating #V_f#;

+

#(0.947atm)(150mL)=(1.000atm)V_f#

+

#V_f=(142.05cancel(atm)mL)/(1.000cancel(atm))#

+

#V_f=142.05mL#

+
+
" "
+
+
+

#142.05mL#

+
+
+
+

Explanation:

+
+

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

+

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

+

In which from the definition, the equation is derived as;

+

#Pquadpropquad 1/V# or #P=k/V# or #PV=k#

+

#P=# Pressure of gas
+#V=# Volume of gas
+#k=# Constant

+

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

+

#P_iV_i=P_fV_f#

+

From the information given in this question;

+

#P_i=# Initial pressure of gas #=0.947atm#
+#P_f=# Final pressure of gas #=1.000atm#
+#V_i=# Initial volume of gas #=150mL#
+#V_f=# Final volume of gas #=?mL#

+

Calculating #V_f#;

+

#(0.947atm)(150mL)=(1.000atm)V_f#

+

#V_f=(142.05cancel(atm)mL)/(1.000cancel(atm))#

+

#V_f=142.05mL#

+
+
+
" "
+

A sample of oxygen gas has a volume of 150.0 mL at a pressure of 0.947 atm. What will the volume of the gas be at a pressure of 1.000 atm if the temperature remains constant?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Boyle's Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 6, 2016 + +
+
+
+
+
+
+
+

#142.05mL#

+
+
+
+

Explanation:

+
+

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

+

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

+

In which from the definition, the equation is derived as;

+

#Pquadpropquad 1/V# or #P=k/V# or #PV=k#

+

#P=# Pressure of gas
+#V=# Volume of gas
+#k=# Constant

+

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

+

#P_iV_i=P_fV_f#

+

From the information given in this question;

+

#P_i=# Initial pressure of gas #=0.947atm#
+#P_f=# Final pressure of gas #=1.000atm#
+#V_i=# Initial volume of gas #=150mL#
+#V_f=# Final volume of gas #=?mL#

+

Calculating #V_f#;

+

#(0.947atm)(150mL)=(1.000atm)V_f#

+

#V_f=(142.05cancel(atm)mL)/(1.000cancel(atm))#

+

#V_f=142.05mL#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 32841 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A sample of oxygen gas has a volume of 150.0 mL at a pressure of 0.947 atm. What will the volume of the gas be at a pressure of 1.000 atm if the temperature remains constant? nan +379 a83562e9-6ddd-11ea-95bf-ccda262736ce https://socratic.org/questions/a-sample-of-argon-has-a-volume-of-205-cm-3-when-it-s-temperatures-is-44-degrees- 228.93 cm^3 start physical_unit 31 32 volume cm^3 qc_end physical_unit 0 3 8 9 volume qc_end physical_unit 0 3 14 16 temperature qc_end physical_unit 0 3 21 24 pressure qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the argon [IN] cm^3""}]" "[{""type"":""physical unit"",""value"":""228.93 cm^3""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of argon [=] \\pu{205 cm^3}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] A sample of argon [=] \\pu{-44 degrees C}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] A sample of argon [=] \\pu{712 mm of Hg}""},{""type"":""other"",""value"":""STP""}]" "

A sample of argon has a volume of 205 #cm^3# when it's temperatures is -44 degrees C. and its pressure is 712 mm of Hg. What would be the volume of the argon at STP?

" nan 228.93 cm^3 "
+

Explanation:

+
+

#T# is always expressed as #""temperatura assoluta""#, which sets absolute zero at #0*K-=-273.15# #""""^@C#. And depending on your syllabus, #STP# specifies a temperature of #273.15*K#, and a pressure of #1*atm#. And we know that #1*atm# will support a column of mercury that is #760*mm# high.........

+

We want #V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#.....

+

And already we see from the quotient we ARE going to get an answer with #""VOLUME units........""# as required.

+

#V_2=(712*mm*Hgxx205*cm^3xx273.15*K)/(760*mm*Hgxx229.15*K)#

+

#~=230*cm^3#...........

+
+
" "
+
+
+

We use the combined gas law, which states for a GIVEN quantity of gas..........#(P_1V_1)/T_1=(P_2V_2)/T_2#; and we eventually get #V_2~=230*cm^3#

+
+
+
+

Explanation:

+
+

#T# is always expressed as #""temperatura assoluta""#, which sets absolute zero at #0*K-=-273.15# #""""^@C#. And depending on your syllabus, #STP# specifies a temperature of #273.15*K#, and a pressure of #1*atm#. And we know that #1*atm# will support a column of mercury that is #760*mm# high.........

+

We want #V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#.....

+

And already we see from the quotient we ARE going to get an answer with #""VOLUME units........""# as required.

+

#V_2=(712*mm*Hgxx205*cm^3xx273.15*K)/(760*mm*Hgxx229.15*K)#

+

#~=230*cm^3#...........

+
+
+
" "
+

A sample of argon has a volume of 205 #cm^3# when it's temperatures is -44 degrees C. and its pressure is 712 mm of Hg. What would be the volume of the argon at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 3, 2017 + +
+
+
+
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+
+
+

We use the combined gas law, which states for a GIVEN quantity of gas..........#(P_1V_1)/T_1=(P_2V_2)/T_2#; and we eventually get #V_2~=230*cm^3#

+
+
+
+

Explanation:

+
+

#T# is always expressed as #""temperatura assoluta""#, which sets absolute zero at #0*K-=-273.15# #""""^@C#. And depending on your syllabus, #STP# specifies a temperature of #273.15*K#, and a pressure of #1*atm#. And we know that #1*atm# will support a column of mercury that is #760*mm# high.........

+

We want #V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#.....

+

And already we see from the quotient we ARE going to get an answer with #""VOLUME units........""# as required.

+

#V_2=(712*mm*Hgxx205*cm^3xx273.15*K)/(760*mm*Hgxx229.15*K)#

+

#~=230*cm^3#...........

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+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3489 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" A sample of argon has a volume of 205 #cm^3# when it's temperatures is -44 degrees C. and its pressure is 712 mm of Hg. What would be the volume of the argon at STP? nan +380 a83562ea-6ddd-11ea-bf36-ccda262736ce https://socratic.org/questions/you-need-to-produce-a-buffer-solution-that-has-ph-5-06-you-already-have-a-soluti 20.90 millimoles start physical_unit 27 27 mole mmol qc_end physical_unit 4 6 10 10 ph qc_end physical_unit 21 22 18 19 mole qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] acetate [IN] millimoles""}]" "[{""type"":""physical unit"",""value"":""20.90 millimoles""}]" "[{""type"":""physical unit"",""value"":""pH [OF] a buffer solution [=] \\pu{5.06}""},{""type"":""physical unit"",""value"":""Mole [OF] acetic acid [=] \\pu{10 mmols}""},{""type"":""other"",""value"":""Acetate is the conjugate base of acetic acid.""}]" "

You need to produce a buffer solution that has pH 5.06. You already have a solution that contains 10 mmols of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

" "
+
+

+

The pKa of acetic acid is 4.74.

+

+
+
" 20.90 millimoles "
+

Explanation:

+
+

Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of #5.06#.

+

The Henderson - Hasselbalch equation looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#

+
+

Your buffer solution contains acetic acid, #""CH""_3""COOH""#, a weak acid, and the acetate anion, #""CH""_3""COO""^(-)#, its conjugate base. Moreover, you know that the #pK_a# of acetic acid is equal to #4.74#.

+

This means that you have

+
+

#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.

+

Notice that the pH of the target solution is higher than the #pK_a# of acetic acid

+
+

#5.06 > pK_a#

+
+

This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.

+

Rearrange the H - H equation to isolate the log term on one side

+
+

#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

This will be equivalent to

+
+

#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

which will give you

+
+

#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#

+
+

As predicted, the concentration of the conjugate base must be higher than that of the weak acid

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#

+
+

Now, the volume of the solution is the same for both chemical species, let's say #V#, which means that you'll have

+
+

#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#

+
+

Since the solution is said to contain #10# mmoles of acetic acid, it follows that you must add

+
+

#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs.

+
+
" "
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+
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#""21 mmoles""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of #5.06#.

+

The Henderson - Hasselbalch equation looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#

+
+

Your buffer solution contains acetic acid, #""CH""_3""COOH""#, a weak acid, and the acetate anion, #""CH""_3""COO""^(-)#, its conjugate base. Moreover, you know that the #pK_a# of acetic acid is equal to #4.74#.

+

This means that you have

+
+

#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.

+

Notice that the pH of the target solution is higher than the #pK_a# of acetic acid

+
+

#5.06 > pK_a#

+
+

This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.

+

Rearrange the H - H equation to isolate the log term on one side

+
+

#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

This will be equivalent to

+
+

#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

which will give you

+
+

#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#

+
+

As predicted, the concentration of the conjugate base must be higher than that of the weak acid

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#

+
+

Now, the volume of the solution is the same for both chemical species, let's say #V#, which means that you'll have

+
+

#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#

+
+

Since the solution is said to contain #10# mmoles of acetic acid, it follows that you must add

+
+

#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs.

+
+
+
" "
+

You need to produce a buffer solution that has pH 5.06. You already have a solution that contains 10 mmols of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

+
+
+

+

The pKa of acetic acid is 4.74.

+

+
+
+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Calculations + + +
+
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+
+1 Answer +
+
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+ +
+
+ +
+ + Apr 17, 2016 + +
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#""21 mmoles""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use the Henderson - Hasselbalch equation to find the ratio that must exist between the concentrations of the weak acid and of the conjugate in order to have a solution that has a pH of #5.06#.

+

The Henderson - Hasselbalch equation looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = pK_a + log( ([""conjugate base""])/([""weak acid""]))color(white)(a/a)|)))#

+
+

Your buffer solution contains acetic acid, #""CH""_3""COOH""#, a weak acid, and the acetate anion, #""CH""_3""COO""^(-)#, its conjugate base. Moreover, you know that the #pK_a# of acetic acid is equal to #4.74#.

+

This means that you have

+
+

#5.06 = 4.74 + log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

Now, before doing any calculations, try to predict how many mmoles of acetate anions you'd need relative to the number of mmoles of acetic acid.

+

Notice that the pH of the target solution is higher than the #pK_a# of acetic acid

+
+

#5.06 > pK_a#

+
+

This tells you that the target solution must contain more conjugate base than weak acid. Implicitly, this solution must contain more moles of acetate anions than of acetic acid.

+

Rearrange the H - H equation to isolate the log term on one side

+
+

#0.32 = log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

This will be equivalent to

+
+

#10^0.32 = 10^log( ([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]))#

+
+

which will give you

+
+

#([""CH""_3""COO""^(-)])/([""CH""_3""COOH""]) = 10^0.32 = 2.09#

+
+

As predicted, the concentration of the conjugate base must be higher than that of the weak acid

+
+

#color(green)(|bar(ul(color(white)(a/a)color(black)([""CH""_3""COO""^(-)] = 2.09 xx [""CH""_3""COOH""])color(white)(a/a)|)))#

+
+

Now, the volume of the solution is the same for both chemical species, let's say #V#, which means that you'll have

+
+

#n_(CH_3COO^(-))/color(red)(cancel(color(black)(V))) = 2.09 xx n_(CH_3COOH)/color(red)(cancel(color(black)(V)))#

+
+

Since the solution is said to contain #10# mmoles of acetic acid, it follows that you must add

+
+

#n_(CH_3COO^(-)) = 2.09 xx ""10 mmoles"" = color(green)(|bar(ul(color(white)(a/a)""21 mmoles CH""_3""COO""^(-)color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs.

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+
+
Related questions
+ + +
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Impact of this question
+
+ 6233 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" You need to produce a buffer solution that has pH 5.06. You already have a solution that contains 10 mmols of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? " + + +The pKa of acetic acid is 4.74. + + +" +381 a83562eb-6ddd-11ea-abf8-ccda262736ce https://socratic.org/questions/5a58c6f87c014928e6f18f7b MPO4 start chemical_formula qc_end c_other OTHER qc_end substance 3 4 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] salt [IN] default""}]" "[{""type"":""chemical equation"",""value"":""MPO4""}]" "[{""type"":""other"",""value"":""A metal whose oxide is M2O3.""},{""type"":""substance name"",""value"":""Phosphoric acid""}]" "

What salt would phosphoric acid form with a metal whose oxide is #M_2O_3#?

" nan MPO4 "
+

Explanation:

+
+

Based on #""M""_2""O""_3#, #""M""# has a valency of #3#. The phosphate group in phosphoric acid #(""H""_3""PO""_4)# has a charge of #3-#.

+

So for each #""PO""_4^(3-)# we need one #""M""^(3+)# to give a neutral salt. i.e #""MPO""_4#.

+
+
" "
+
+
+

#""MPO""_4#

+
+
+
+

Explanation:

+
+

Based on #""M""_2""O""_3#, #""M""# has a valency of #3#. The phosphate group in phosphoric acid #(""H""_3""PO""_4)# has a charge of #3-#.

+

So for each #""PO""_4^(3-)# we need one #""M""^(3+)# to give a neutral salt. i.e #""MPO""_4#.

+
+
+
" "
+

What salt would phosphoric acid form with a metal whose oxide is #M_2O_3#?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Ionic Compounds + + +
+
+
+
+
+2 Answers +
+
+
+
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+
+ + +
+
+ +
+ + Jan 13, 2018 + +
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#""MPO""_4#

+
+
+
+

Explanation:

+
+

Based on #""M""_2""O""_3#, #""M""# has a valency of #3#. The phosphate group in phosphoric acid #(""H""_3""PO""_4)# has a charge of #3-#.

+

So for each #""PO""_4^(3-)# we need one #""M""^(3+)# to give a neutral salt. i.e #""MPO""_4#.

+
+
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+ +
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+ +
+ + Jan 13, 2018 + +
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+

Well, clearly we gots an #M^(3+)# cation.....to give #M^(3+)PO_4^(3-)#...

+
+
+
+

Explanation:

+
+

We gots #M_2O_3#. The typical oxidation state of oxygen in an oxide (the which we clearly got) is #-II#...electrical neutrality demands that we have #M^(3+)#...i.e. #2xx(+3)+3xx(-2)=0#..as required for a neutral salt. And so its phosphate is formulated as #M^(3+)PO_4^(3-)#

+

On the other hand (if you are an undergrad), phosphoric acid is ONLY a diacid in water...tritation with sodium hydroxide yields a stoichiometric endpoint at #Na_2^(+)HPO_4^(2-)#...

+

And so we might have a biphosphate species of the form #M_2(HPO_4)_3#...but given the boundary conditions of the problem, clearly the answer is the former....

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+ +
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+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 2033 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" What salt would phosphoric acid form with a metal whose oxide is #M_2O_3#? nan +382 a83589f6-6ddd-11ea-a6d4-ccda262736ce https://socratic.org/questions/calculate-the-molar-mass-in-g-mol-of-diacetyl-given-that-in-the-gas-phase-100-de-1 85.83 g/mol start physical_unit 7 8 molar_mass g/mol qc_end physical_unit 7 8 15 17 temperature qc_end physical_unit 24 26 19 20 pressure qc_end physical_unit 24 26 22 23 mass qc_end physical_unit 24 26 31 32 volume qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] diacetyl (butanedione) [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""85.83 g/mol""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] diacetyl (butanedione) [=] \\pu{100 degrees Celsius}""},{""type"":""physical unit"",""value"":""Pressure [OF] sample of diacetyl [=] \\pu{747 torr}""},{""type"":""physical unit"",""value"":""Mass [OF] sample of diacetyl [=] \\pu{0.3060 g}""},{""type"":""physical unit"",""value"":""Volume [OF] sample of diacetyl [=] \\pu{0.111 L}""}]" "

Calculate the molar mass in #""g""/""mol""#of diacetyl (butanedione) given that in the gas phase #100# degrees Celsius and #747# torr, a #0.3060# g sample of diacetyl occupies a volume of #0.111L#?

" nan 85.83 g/mol "
+

Explanation:

+
+

You can use the ideal gas law to answer this question. The equation is:

+

#PV=nRT#,

+

where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins. Add #273.15# to the Celsius temperature to get the temperature in Kelvins: #100^@""C"" + 273.15=""373 K""#.

+

We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.

+

Known

+

#m=""0.3060 g""#

+

#P=""747 torr""#

+

#V=""0.111 L""#

+

#R=62.364color(white)(.)""L torr K""^(-1) ""mol""^(-1)#
+https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law

+

#T=""373 K""#

+

Unknown

+

#n=?#

+

#""molar mass = ? g/mol""#

+

Solve for #n#.
+Rearrange the equation to isolate #n#. Insert the known data and solve.

+

#n=(PV)/(RT)#

+

#n=(747color(red)cancel(color(black)(""torr""))xx0.111color(red)cancel(color(black)(""L"")))/((62.364color(white)(.)color(red)cancel(color(black)(""L""))color(red)cancel(color(black)(""torr"")) color(red)cancel(color(black)(""K"")))^(-1) ""mol""^(-1)xx373color(red)cancel(color(black)(""K"")))=""0.003565 mol""#

+

Now that you have moles, divide the mass of diacetyl given in the question by the moles.

+

#""Molar mass diacetyl"" = (0.3060""g diacetyl"")/(0.003565""mol diacetyl"")=""85.83 g/mol diacetyl""# rounded to four significant figures

+

The molecular formula for diacetyl is #(""CH""_3""CO)""_2# or #""C""_4""H""_6""O""_2""# and its known molar mass to three sig figs is #""86.09 g/mol""#. https://www.ncbi.nlm.nih.gov/pccompound?term=%22diacetyl%22

+

#""Percent error"" = abs((""known value""-""experimental value"")/(""accepted value""))xx100#

+

#""Percent error""=abs((86.09-85.83)/(86.09))xx100=""0.3020%""#

+
+
" "
+
+
+

The molar mass of diacetyl, #""C""_4""H""_6""O""_2""#, as calculated based on the question parameters is #""85.83 g/mol""#. Its actual molar mass is #""86.09 g/mol""#.

+
+
+
+

Explanation:

+
+

You can use the ideal gas law to answer this question. The equation is:

+

#PV=nRT#,

+

where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins. Add #273.15# to the Celsius temperature to get the temperature in Kelvins: #100^@""C"" + 273.15=""373 K""#.

+

We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.

+

Known

+

#m=""0.3060 g""#

+

#P=""747 torr""#

+

#V=""0.111 L""#

+

#R=62.364color(white)(.)""L torr K""^(-1) ""mol""^(-1)#
+https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law

+

#T=""373 K""#

+

Unknown

+

#n=?#

+

#""molar mass = ? g/mol""#

+

Solve for #n#.
+Rearrange the equation to isolate #n#. Insert the known data and solve.

+

#n=(PV)/(RT)#

+

#n=(747color(red)cancel(color(black)(""torr""))xx0.111color(red)cancel(color(black)(""L"")))/((62.364color(white)(.)color(red)cancel(color(black)(""L""))color(red)cancel(color(black)(""torr"")) color(red)cancel(color(black)(""K"")))^(-1) ""mol""^(-1)xx373color(red)cancel(color(black)(""K"")))=""0.003565 mol""#

+

Now that you have moles, divide the mass of diacetyl given in the question by the moles.

+

#""Molar mass diacetyl"" = (0.3060""g diacetyl"")/(0.003565""mol diacetyl"")=""85.83 g/mol diacetyl""# rounded to four significant figures

+

The molecular formula for diacetyl is #(""CH""_3""CO)""_2# or #""C""_4""H""_6""O""_2""# and its known molar mass to three sig figs is #""86.09 g/mol""#. https://www.ncbi.nlm.nih.gov/pccompound?term=%22diacetyl%22

+

#""Percent error"" = abs((""known value""-""experimental value"")/(""accepted value""))xx100#

+

#""Percent error""=abs((86.09-85.83)/(86.09))xx100=""0.3020%""#

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+
" "
+

Calculate the molar mass in #""g""/""mol""#of diacetyl (butanedione) given that in the gas phase #100# degrees Celsius and #747# torr, a #0.3060# g sample of diacetyl occupies a volume of #0.111L#?

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+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+2 Answers +
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+ + May 26, 2017 + +
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The molar mass of diacetyl, #""C""_4""H""_6""O""_2""#, as calculated based on the question parameters is #""85.83 g/mol""#. Its actual molar mass is #""86.09 g/mol""#.

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+
+
+

Explanation:

+
+

You can use the ideal gas law to answer this question. The equation is:

+

#PV=nRT#,

+

where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins. Add #273.15# to the Celsius temperature to get the temperature in Kelvins: #100^@""C"" + 273.15=""373 K""#.

+

We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.

+

Known

+

#m=""0.3060 g""#

+

#P=""747 torr""#

+

#V=""0.111 L""#

+

#R=62.364color(white)(.)""L torr K""^(-1) ""mol""^(-1)#
+https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law

+

#T=""373 K""#

+

Unknown

+

#n=?#

+

#""molar mass = ? g/mol""#

+

Solve for #n#.
+Rearrange the equation to isolate #n#. Insert the known data and solve.

+

#n=(PV)/(RT)#

+

#n=(747color(red)cancel(color(black)(""torr""))xx0.111color(red)cancel(color(black)(""L"")))/((62.364color(white)(.)color(red)cancel(color(black)(""L""))color(red)cancel(color(black)(""torr"")) color(red)cancel(color(black)(""K"")))^(-1) ""mol""^(-1)xx373color(red)cancel(color(black)(""K"")))=""0.003565 mol""#

+

Now that you have moles, divide the mass of diacetyl given in the question by the moles.

+

#""Molar mass diacetyl"" = (0.3060""g diacetyl"")/(0.003565""mol diacetyl"")=""85.83 g/mol diacetyl""# rounded to four significant figures

+

The molecular formula for diacetyl is #(""CH""_3""CO)""_2# or #""C""_4""H""_6""O""_2""# and its known molar mass to three sig figs is #""86.09 g/mol""#. https://www.ncbi.nlm.nih.gov/pccompound?term=%22diacetyl%22

+

#""Percent error"" = abs((""known value""-""experimental value"")/(""accepted value""))xx100#

+

#""Percent error""=abs((86.09-85.83)/(86.09))xx100=""0.3020%""#

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+ + May 26, 2017 + +
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#85.9 ""g""/""mol""#

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+

Explanation:

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+

What we can do here is calculate the density of the diacetyl, and use that to directly calculate the molar mass. We will use the equation

+

#M = (dRT)/P#

+

where #M# is the molar mass of the substance,
+#d# is its density, in #""g""/""L""#,
+#R# is the universal gas constant, equal to #0.08206 (""L""-""atm"")/(""mol""-""K"")#,
+#T# is the absolute temperature (in #""K""#), and
+#P# is the pressure of the gas (in #""atm""#)

+

Where is this equation derived from? Read the steps below if you would like to know, otherwise, skip to the next step.

+
+

Well, let's recall our ideal-gas equation, and rearrange it to solve for units similar to that of density, #""mol""/""L""#, which is #n/V#:

+

#PV = nRT#

+

#P = (nRT)/V#

+

#P/(RT) = n/V#

+

Now, let's multiply both sides of the equation by #M#, the molar mass with units #""g""/""mol""#:

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#(PM)/(RT) = (nM)/V#

+

If we list the right side of the equation in terms of units, we have

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#cancel(""mol"")/""L"" xx ""g""/cancel(""mol"") = ""g""/""L""#

+

Which is the units for density. Thus, the value #(nM)/V# is the density of the gas, and if we plug this back into our equation:

+

#(PM)/(RT) = (nM)/V = d#

+

Thus, #d = (MP)/(RT)#, and rearranging to solve for the molar mass yields our original equation, #M = (dRT)/P#.

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+

The density of the diacetyl is

+

#d = (0.3060""g"")/(0.111 ""L"") = 2.76 ""g""/""L""#

+

The temperature, in Kelvin, is

+

#100^oC + 273 = 373 ""K""#

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and the pressure, in atmospheres, is

+

#747 cancel(""torr"")((1 ""atm"")/(760 cancel(""torr""))) = 0.983 ""atm""#

+

Finally, plugging in all our known variables, we have

+

#M = ((2.76 ""g""/cancel(""L""))(0.08206 (cancel(""L"")-cancel(""atm""))/(""mol""-cancel(""K"")))(373 cancel(""K"")))/(0.983 cancel(""atm"")) = color(red)(85.9 ""g""/""mol""#

+

To check this, the molecular formula of diacetyl is #(""CH""_3""CO"")_2#, and from this the molar mass is found to be #86.1 ""g""/""mol""#, which measures up well with our result.

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" "Calculate the molar mass in #""g""/""mol""#of diacetyl (butanedione) given that in the gas phase #100# degrees Celsius and #747# torr, a #0.3060# g sample of diacetyl occupies a volume of #0.111L#?" nan +383 a83591c1-6ddd-11ea-994e-ccda262736ce https://socratic.org/questions/how-many-moles-of-methanol-must-react-with-excess-oxygen-to-produce-5-0-l-of-car 0.22 moles start physical_unit 4 4 mole mol qc_end physical_unit 15 16 12 13 volume qc_end c_other OTHER qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] methanol [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.22 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] carbon dioxide [=] \\pu{5.0 L}""},{""type"":""other"",""value"":""Excess oxygen.""},{""type"":""other"",""value"":""STP""}]" "

How many moles of methanol must react with excess oxygen to produce 5.0 L of carbon dioxide at STP?

" nan 0.22 moles "
+

Explanation:

+
+

The equation for the combustion of methanol is:

+

#2CH_3OH(l) + 3O_2(g) → 2 CO_2(g) + 4 H_2O(g)#

+

Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is #(5/22.4) = 0.223# mol.

+

Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.

+

Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.

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+
" "
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+
+

0.223 moles

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+
+

Explanation:

+
+

The equation for the combustion of methanol is:

+

#2CH_3OH(l) + 3O_2(g) → 2 CO_2(g) + 4 H_2O(g)#

+

Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is #(5/22.4) = 0.223# mol.

+

Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.

+

Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.

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+
+
" "
+

How many moles of methanol must react with excess oxygen to produce 5.0 L of carbon dioxide at STP?

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+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
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+1 Answer +
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+ + Jul 25, 2017 + +
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0.223 moles

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+
+

Explanation:

+
+

The equation for the combustion of methanol is:

+

#2CH_3OH(l) + 3O_2(g) → 2 CO_2(g) + 4 H_2O(g)#

+

Firstly you need to work out how many moles of carbon dioxide are in 5 litres at STP. If you assume that it is an idea gas (it isn't, but it's close enough for this sort of purpose) then 1 mole at STP occupies 22.4 litres. Therefore, at STP 5 litres is #(5/22.4) = 0.223# mol.

+

Looking at the equation, it tells you that 2 moles of carbon dioxide are produced by complete combustion of 2 moles of methanol. Therefore, if you have 0.223 moles of carbon dioxide, you would need 0.223 moles of methanol.

+

Molar mass of methanol is 32.04 g/mol so you would need 0.223 x 32.04 = 7.14 g of methanol.

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" How many moles of methanol must react with excess oxygen to produce 5.0 L of carbon dioxide at STP? nan +384 a835b11c-6ddd-11ea-bf4c-ccda262736ce https://socratic.org/questions/if-given-0-09-moles-of-sodium-sulfate-in-12-ml-of-solution-what-is-the-concentra 7.50 mol/L start physical_unit 5 6 concentration mol/l qc_end physical_unit 11 11 8 9 volume qc_end physical_unit 5 6 2 3 mole qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] sodium sulfate [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""7.50 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{12 mL}""},{""type"":""physical unit"",""value"":""Mole [OF] sodium sulfate [=] \\pu{0.090 moles}""}]" "

If given 0.090 moles of sodium sulfate in 12 mL of solution, what is the concentration?

" nan 7.50 mol/L "
+

Explanation:

+
+

In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate, #""Na""_2""SO""_4#, you get in one liter of solution.

+

That is how molarity was defined -- the number of moles of solute in one liter of solution.

+

So, you know that you have #0.090# moles of solute in #""12 mL""# of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in

+
+

#""1 L"" = 10^3""mL""#

+
+

of solution. Convert the volume from milliliters to liters first

+
+

#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#

+
+

If #""0.012 L""# of solution contain #0.090# moles of solute, it follows that #""1 L""# will contain

+
+

#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#

+
+

So, if #""1 L""# of solution contains #7.5# moles of solute, it follows that the solution's molarity is

+
+

#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs.

+
+
" "
+
+
+

#""7.5 M""#

+
+
+
+

Explanation:

+
+

In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate, #""Na""_2""SO""_4#, you get in one liter of solution.

+

That is how molarity was defined -- the number of moles of solute in one liter of solution.

+

So, you know that you have #0.090# moles of solute in #""12 mL""# of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in

+
+

#""1 L"" = 10^3""mL""#

+
+

of solution. Convert the volume from milliliters to liters first

+
+

#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#

+
+

If #""0.012 L""# of solution contain #0.090# moles of solute, it follows that #""1 L""# will contain

+
+

#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#

+
+

So, if #""1 L""# of solution contains #7.5# moles of solute, it follows that the solution's molarity is

+
+

#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#

+
+

The answer is rounded to two sig figs.

+
+
+
" "
+

If given 0.090 moles of sodium sulfate in 12 mL of solution, what is the concentration?

+
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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Jul 7, 2016 + +
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#""7.5 M""#

+
+
+
+

Explanation:

+
+

In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate, #""Na""_2""SO""_4#, you get in one liter of solution.

+

That is how molarity was defined -- the number of moles of solute in one liter of solution.

+

So, you know that you have #0.090# moles of solute in #""12 mL""# of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in

+
+

#""1 L"" = 10^3""mL""#

+
+

of solution. Convert the volume from milliliters to liters first

+
+

#12 color(red)(cancel(color(black)(""mL""))) * ""1 L""/(10^3color(red)(cancel(color(black)(""mL"")))) = ""0.012 L""#

+
+

If #""0.012 L""# of solution contain #0.090# moles of solute, it follows that #""1 L""# will contain

+
+

#1 color(red)(cancel(color(black)(""L solution""))) * (""0.09 moles Na""_2""SO""_4)/(0.012color(red)(cancel(color(black)(""L solution"")))) = ""7.5 moles Na""_2""SO""_4#

+
+

So, if #""1 L""# of solution contains #7.5# moles of solute, it follows that the solution's molarity is

+
+

#""molarity"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""7.5 mol L""^(-1) = ""7.5 molar"" = ""7.5 M"")color(white)(a/a)|)))#

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+

The answer is rounded to two sig figs.

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" If given 0.090 moles of sodium sulfate in 12 mL of solution, what is the concentration? nan +385 a835b11d-6ddd-11ea-bbb3-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-beer-in-which-the-hydrogen-ion-concentration-is-5-6-times-10-7 6.25 start physical_unit 5 5 ph none qc_end physical_unit 9 10 13 16 concentration qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] beer""}]" "[{""type"":""physical unit"",""value"":""6.25""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] hydrogen ion [=] \\pu{5.6 × 10^(−7) M}""}]" "

What is the pH of beer in which the hydrogen ion concentration is #5.6 \times 10^ -7# M?

" nan 6.25 "
+

Explanation:

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#pH = -log[H^+]#

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#pH = -log (5.6 × 10^-7) = 7 - log 5.6 = 6.25#

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#pH = 6.25#

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Explanation:

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#pH = -log[H^+]#

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+
+

#pH = -log (5.6 × 10^-7) = 7 - log 5.6 = 6.25#

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" "
+

What is the pH of beer in which the hydrogen ion concentration is #5.6 \times 10^ -7# M?

+
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+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+1 Answer +
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+ + Nov 28, 2017 + +
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#pH = 6.25#

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Explanation:

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#pH = -log[H^+]#

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#pH = -log (5.6 × 10^-7) = 7 - log 5.6 = 6.25#

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+
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" What is the pH of beer in which the hydrogen ion concentration is #5.6 \times 10^ -7# M? nan +386 a835b11e-6ddd-11ea-a140-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-hydroarsenic-acid H3AsO4 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] Hydroarsenic acid [IN] default""}]" "[{""type"":""chemical equation"",""value"":""H3AsO4""}]" "[{""type"":""substance name"",""value"":""Hydroarsenic acid""}]" "

What is the formula for Hydroarsenic acid?

" nan H3AsO4 "
+

Explanation:

+
+

The is the arsenical analogue of phosphoric acid, #H_3PO_4#. #""Arsenate""#, and #""phosphate""# salts behave very similarly; i.e. in general their salts are as soluble as bricks.

+
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" "
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+
+

Perhaps #H_3AsO_4#...........

+
+
+
+

Explanation:

+
+

The is the arsenical analogue of phosphoric acid, #H_3PO_4#. #""Arsenate""#, and #""phosphate""# salts behave very similarly; i.e. in general their salts are as soluble as bricks.

+
+
+
" "
+

What is the formula for Hydroarsenic acid?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 11, 2017 + +
+
+
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+

Perhaps #H_3AsO_4#...........

+
+
+
+

Explanation:

+
+

The is the arsenical analogue of phosphoric acid, #H_3PO_4#. #""Arsenate""#, and #""phosphate""# salts behave very similarly; i.e. in general their salts are as soluble as bricks.

+
+
+
+
+
+ +
+
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+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 7134 views + around the world +
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+ + Creative Commons License + +
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+
" What is the formula for Hydroarsenic acid? nan +387 a835b11f-6ddd-11ea-931e-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-calcium-bisulfate Ca(HSO4)2 start chemical_formula qc_end substance 5 6 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] calcium bisulfate [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Ca(HSO4)2""}]" "[{""type"":""substance name"",""value"":""calcium bisulfate""}]" "

What is the formula for calcium bisulfate?

" nan Ca(HSO4)2 "
+

Explanation:

+
+

The salt is composed of #Ca^(2+)# and #HSO_4^-# ions, hence the #1:2# composition.

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+
" "
+
+
+

#Ca(HSO_4)_2#

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Explanation:

+
+

The salt is composed of #Ca^(2+)# and #HSO_4^-# ions, hence the #1:2# composition.

+
+
+
" "
+

What is the formula for calcium bisulfate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
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+ +
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+ +
+ + Dec 29, 2016 + +
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+

#Ca(HSO_4)_2#

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Explanation:

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+

The salt is composed of #Ca^(2+)# and #HSO_4^-# ions, hence the #1:2# composition.

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+
+
" What is the formula for calcium bisulfate? nan +388 a835b120-6ddd-11ea-b8f0-ccda262736ce https://socratic.org/questions/calculate-the-molality-of-a-solution-made-by-dissolving-115-grams-of-nano3-in-50 2.71 mol/kg start physical_unit 4 5 molality mol/kg qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 17 17 14 15 volume qc_end physical_unit 12 12 24 25 molar_mass qc_end physical_unit 17 17 32 33 density qc_end end "[{""type"":""physical unit"",""value"":""Molality [OF] a solution [IN] mol/kg""}]" "[{""type"":""physical unit"",""value"":""2.71 mol/kg""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaNO3 [=] \\pu{115 g}""},{""type"":""physical unit"",""value"":""Volume [OF] water [=] \\pu{500 mL}""},{""type"":""physical unit"",""value"":""Molar mass [OF] NaNO3 [=] \\pu{85.0 g/mol}""},{""type"":""physical unit"",""value"":""Density [OF] water [=] \\pu{1.00 g/mL}""}]" "

Calculate the molality of a solution made by dissolving #""115 g""# of #""NaNO""_3# in #""500. mL""# of water? The molar mass of #""NaNO""_3# is #""85.0 g/mol""# and the density of water is #""1.00 g/mL""#.

" nan 2.71 mol/kg "
+

Explanation:

+
+

As you know, the molality of a solution tells you the number of moles of solute present for every #""1 kg""# of the solvent.

+

This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.

+
+

#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#

+
+

Next, use the molar mass of the solute to determine how many moles are present in the sample.

+
+

#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#

+
+

So, you know that this solution will contain #1.353# moles of sodium nitrate, the solute, for #""500. g""# of water, the solvent.

+

In order to find the molality of the solution, you must figure out how many moles of solute would be present for #""1 kg"" = 10^3# #""g""# of water.

+
+

#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#

+
+

You can thus say that the molality of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#

+
+

The answer is rounded to three sig figs.

+

So, you know that this solution will contain #2.71# moles of solute for every #""1 kg""# of solvent.

+
+
" "
+
+
+

#""2.71 mol kg""^(-1)#

+
+
+
+

Explanation:

+
+

As you know, the molality of a solution tells you the number of moles of solute present for every #""1 kg""# of the solvent.

+

This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.

+
+

#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#

+
+

Next, use the molar mass of the solute to determine how many moles are present in the sample.

+
+

#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#

+
+

So, you know that this solution will contain #1.353# moles of sodium nitrate, the solute, for #""500. g""# of water, the solvent.

+

In order to find the molality of the solution, you must figure out how many moles of solute would be present for #""1 kg"" = 10^3# #""g""# of water.

+
+

#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#

+
+

You can thus say that the molality of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#

+
+

The answer is rounded to three sig figs.

+

So, you know that this solution will contain #2.71# moles of solute for every #""1 kg""# of solvent.

+
+
+
" "
+

Calculate the molality of a solution made by dissolving #""115 g""# of #""NaNO""_3# in #""500. mL""# of water? The molar mass of #""NaNO""_3# is #""85.0 g/mol""# and the density of water is #""1.00 g/mL""#.

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
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+
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+1 Answer +
+
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+ +
+ + Dec 20, 2017 + +
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#""2.71 mol kg""^(-1)#

+
+
+
+

Explanation:

+
+

As you know, the molality of a solution tells you the number of moles of solute present for every #""1 kg""# of the solvent.

+

This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.

+
+

#500. color(red)(cancel(color(black)(""mL""))) * ""1.00 g""/(1color(red)(cancel(color(black)(""mL"")))) = ""500. g""#

+
+

Next, use the molar mass of the solute to determine how many moles are present in the sample.

+
+

#115 color(red)(cancel(color(black)(""g""))) * ""1 mole NanO""_3/(85.0color(red)(cancel(color(black)(""g"")))) = ""1.353 moles NaNO""_3#

+
+

So, you know that this solution will contain #1.353# moles of sodium nitrate, the solute, for #""500. g""# of water, the solvent.

+

In order to find the molality of the solution, you must figure out how many moles of solute would be present for #""1 kg"" = 10^3# #""g""# of water.

+
+

#10^3 color(red)(cancel(color(black)(""g water""))) * ""1.353 moles NaNO""_3/(500. color(red)(cancel(color(black)(""g water"")))) = ""2.706 moles NaNO""_3#

+
+

You can thus say that the molality of the solution is equal to

+
+

#color(darkgreen)(ul(color(black)(""molality""))) = ""2.706 mol kg""^(-1) ~~ color(darkgreen)(ul(color(black)(""2.71 mol kg""^(-1)))#

+
+

The answer is rounded to three sig figs.

+

So, you know that this solution will contain #2.71# moles of solute for every #""1 kg""# of solvent.

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Related questions
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" "Calculate the molality of a solution made by dissolving #""115 g""# of #""NaNO""_3# in #""500. mL""# of water? The molar mass of #""NaNO""_3# is #""85.0 g/mol""# and the density of water is #""1.00 g/mL""#." nan +389 a835b121-6ddd-11ea-892f-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-250-0-milliliter-aqueous-solution-of-sodium-hydroxide- 1.55 mol/L start physical_unit 8 12 molarity mol/l qc_end physical_unit 8 12 6 7 volume qc_end physical_unit 18 18 15 16 mass qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] aqueous solution of sodium hydroxide [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.55 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] aqueous solution of sodium hydroxide [=] \\pu{250.0 milliliter}""},{""type"":""physical unit"",""value"":""Mass [OF] solute [=] \\pu{15.5 grams}""}]" "

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute?

" nan 1.55 mol/L "
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

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#[NaOH(aq)] = 1.55# #mol*L^-1#.

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Explanation:

+
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#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

+
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+
" "
+

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute?

+
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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Feb 27, 2016 + +
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#[NaOH(aq)] = 1.55# #mol*L^-1#.

+
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+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute (moles)""/""Volume of solution (Litres)""# #=# #((15.5*g)/(40.0*g*mol^-1))/(0.2500*L)# #=# #??# #mol*L^-1#

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+
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+ + Creative Commons License + +
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+
" What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute? nan +390 a835d81c-6ddd-11ea-ab8e-ccda262736ce https://socratic.org/questions/what-would-be-the-ph-of-a-0-503-m-solution-of-ammonia-nh3-at-room-temperature-th 11.47 start physical_unit 9 12 ph none qc_end physical_unit 12 12 7 8 molarity qc_end c_other OTHER qc_end physical_unit 11 11 21 23 kb qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] solution of ammonia (NH3)""}]" "[{""type"":""physical unit"",""value"":""11.47""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] NH3 [=] \\pu{0.503 M}""},{""type"":""other"",""value"":""Room temperature.""},{""type"":""physical unit"",""value"":""Kb [OF] ammonia [=] \\pu{1.78 x 10^(−5)}""}]" "

What would be the pH of a 0.503 M solution of ammonia (NH3) at room temperature? The #K_b# of ammonia is #1.78 x 10^-5#

" nan 11.47 "
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Explanation:

+
+

We interrogate the equilibrium...............

+

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^-#

+

#K_""eq""=([NH_4^+][HO^-])/([NH_3(aq)])=1.78xx10^-5#

+

Now we know that #[NH_3(aq)]=0.503*mol*L^-1# initially, and call the amount of association it undergoes #x#.

+

So at equilibrium, #[NH_4^+]=[HO^-]=x#; and #[NH_3(aq)]=(0.503-x)*mol*L^-1#.

+

#K_""eq""=1.78xx10^-5=x^2/(0.503-x)#

+

This is quadratic in #x#, which we could solve exactly, however, we make the approximation that #(0.503-x)~=0.503*mol*L^-1#.

+

Thus #x_1=sqrt(1.78xx10^-5xx0.503)=0.00299*mol*L^-1#

+

Now that we an approximation for #x#, we can we substitute this value back into the equation.

+

And thus..........................................................

+

#x_2=sqrt(1.78xx10^-5xx(0.503-0.00299))=0.00298*mol*L^-1#

+

Since the values have converged, we can take this value as the true value.

+

And so #[HO^-]=[NH_4^+]=2.98xx10^-3*mol*L^-1#, and thus #pOH=-log_(10)[HO^-]=-log_10(2.98xx10^-3)=2.53#.

+

But we need #pOH#, however, we know (how?) that #pH+pOH=14#.

+

And so #pH=14-2.53=??#

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" "
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#pH~=11.5#

+
+
+
+

Explanation:

+
+

We interrogate the equilibrium...............

+

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^-#

+

#K_""eq""=([NH_4^+][HO^-])/([NH_3(aq)])=1.78xx10^-5#

+

Now we know that #[NH_3(aq)]=0.503*mol*L^-1# initially, and call the amount of association it undergoes #x#.

+

So at equilibrium, #[NH_4^+]=[HO^-]=x#; and #[NH_3(aq)]=(0.503-x)*mol*L^-1#.

+

#K_""eq""=1.78xx10^-5=x^2/(0.503-x)#

+

This is quadratic in #x#, which we could solve exactly, however, we make the approximation that #(0.503-x)~=0.503*mol*L^-1#.

+

Thus #x_1=sqrt(1.78xx10^-5xx0.503)=0.00299*mol*L^-1#

+

Now that we an approximation for #x#, we can we substitute this value back into the equation.

+

And thus..........................................................

+

#x_2=sqrt(1.78xx10^-5xx(0.503-0.00299))=0.00298*mol*L^-1#

+

Since the values have converged, we can take this value as the true value.

+

And so #[HO^-]=[NH_4^+]=2.98xx10^-3*mol*L^-1#, and thus #pOH=-log_(10)[HO^-]=-log_10(2.98xx10^-3)=2.53#.

+

But we need #pOH#, however, we know (how?) that #pH+pOH=14#.

+

And so #pH=14-2.53=??#

+
+
+
" "
+

What would be the pH of a 0.503 M solution of ammonia (NH3) at room temperature? The #K_b# of ammonia is #1.78 x 10^-5#

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+1 Answer +
+
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+ +
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+ +
+ + Apr 15, 2017 + +
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#pH~=11.5#

+
+
+
+

Explanation:

+
+

We interrogate the equilibrium...............

+

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^-#

+

#K_""eq""=([NH_4^+][HO^-])/([NH_3(aq)])=1.78xx10^-5#

+

Now we know that #[NH_3(aq)]=0.503*mol*L^-1# initially, and call the amount of association it undergoes #x#.

+

So at equilibrium, #[NH_4^+]=[HO^-]=x#; and #[NH_3(aq)]=(0.503-x)*mol*L^-1#.

+

#K_""eq""=1.78xx10^-5=x^2/(0.503-x)#

+

This is quadratic in #x#, which we could solve exactly, however, we make the approximation that #(0.503-x)~=0.503*mol*L^-1#.

+

Thus #x_1=sqrt(1.78xx10^-5xx0.503)=0.00299*mol*L^-1#

+

Now that we an approximation for #x#, we can we substitute this value back into the equation.

+

And thus..........................................................

+

#x_2=sqrt(1.78xx10^-5xx(0.503-0.00299))=0.00298*mol*L^-1#

+

Since the values have converged, we can take this value as the true value.

+

And so #[HO^-]=[NH_4^+]=2.98xx10^-3*mol*L^-1#, and thus #pOH=-log_(10)[HO^-]=-log_10(2.98xx10^-3)=2.53#.

+

But we need #pOH#, however, we know (how?) that #pH+pOH=14#.

+

And so #pH=14-2.53=??#

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+ + +
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+
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+ + Creative Commons License + +
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" What would be the pH of a 0.503 M solution of ammonia (NH3) at room temperature? The #K_b# of ammonia is #1.78 x 10^-5# nan +391 a835d81d-6ddd-11ea-bbe7-ccda262736ce https://socratic.org/questions/how-do-you-write-the-balanced-half-equation-for-the-oxidation-of-methanal-formal CH2O + O2 -> CO2 + H2O start chemical_equation qc_end chemical_equation 14 14 qc_end chemical_equation 16 16 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] half equation""}]" "[{""type"":""chemical equation"",""value"":""CH2O + O2 -> CO2 + H2O""}]" "[{""type"":""chemical equation"",""value"":""CH2O""},{""type"":""chemical equation"",""value"":""CO2""}]" "

How do you write the balanced half equation for the oxidation of methanal (formaldehyde), #CH_2O# to #CO_2#?

" nan CH2O + O2 -> CO2 + H2O "
+

Explanation:

+
+

Carbon is oxidized from #C^0# to #C^(IV)#.

+

Where do the electrons go? To oxygen:

+

#O_2 + 4e^(-) rarr 2O^(2-)# #ii.#

+

We add (i) and (ii) to get:

+

#H_2C=O + H_2O + O_2 + cancel(4e^(-))rarr 2O^(2-) + CO_2 + 4H^+ + cancel(4e^-)#

+

Now on the right side we have #2O^(2-) + 4H^(+)-=2H_2O#

+

Thus:

+

#H_2C=O + cancel(H_2O) + O_2 rarr CO_2 + cancel(2)H_2O #

+

To give (finally!):

+

#H_2C=O + O_2 rarr CO_2 + H_2O#

+

As the balanced redox equation.

+

Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons

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#H_2C=O + H_2O rarr CO_2 + 4H^+ + 4e^-# #i.#

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Explanation:

+
+

Carbon is oxidized from #C^0# to #C^(IV)#.

+

Where do the electrons go? To oxygen:

+

#O_2 + 4e^(-) rarr 2O^(2-)# #ii.#

+

We add (i) and (ii) to get:

+

#H_2C=O + H_2O + O_2 + cancel(4e^(-))rarr 2O^(2-) + CO_2 + 4H^+ + cancel(4e^-)#

+

Now on the right side we have #2O^(2-) + 4H^(+)-=2H_2O#

+

Thus:

+

#H_2C=O + cancel(H_2O) + O_2 rarr CO_2 + cancel(2)H_2O #

+

To give (finally!):

+

#H_2C=O + O_2 rarr CO_2 + H_2O#

+

As the balanced redox equation.

+

Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons

+
+
+
" "
+

How do you write the balanced half equation for the oxidation of methanal (formaldehyde), #CH_2O# to #CO_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Redox Reactions + + +
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+1 Answer +
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+ + Sep 4, 2016 + +
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#H_2C=O + H_2O rarr CO_2 + 4H^+ + 4e^-# #i.#

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+
+
+

Explanation:

+
+

Carbon is oxidized from #C^0# to #C^(IV)#.

+

Where do the electrons go? To oxygen:

+

#O_2 + 4e^(-) rarr 2O^(2-)# #ii.#

+

We add (i) and (ii) to get:

+

#H_2C=O + H_2O + O_2 + cancel(4e^(-))rarr 2O^(2-) + CO_2 + 4H^+ + cancel(4e^-)#

+

Now on the right side we have #2O^(2-) + 4H^(+)-=2H_2O#

+

Thus:

+

#H_2C=O + cancel(H_2O) + O_2 rarr CO_2 + cancel(2)H_2O #

+

To give (finally!):

+

#H_2C=O + O_2 rarr CO_2 + H_2O#

+

As the balanced redox equation.

+

Carbon has been oxidized, and dioxygen has been reduced. Mass and charge are balanced as required. This is a lot of work for a simple oxidation reaction. And of course you would never do this formally. You would (i) balance the carbons, (ii) then the hydrogens, and (iii) finally the oxygens, all directly, without all that tedious mucking about with electrons

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Related questions
+ + +
+
+
+
Impact of this question
+
+ 4210 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" How do you write the balanced half equation for the oxidation of methanal (formaldehyde), #CH_2O# to #CO_2#? nan +392 a835d81e-6ddd-11ea-bb62-ccda262736ce https://socratic.org/questions/590b4d7d7c01491a80acb7ce 2 Li + 2 H2O -> 2 LiOH + H2 start chemical_equation qc_end substance 7 8 qc_end substance 10 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the reaction""}]" "[{""type"":""chemical equation"",""value"":""2 Li + 2 H2O -> 2 LiOH + H2""}]" "[{""type"":""substance name"",""value"":""Lithium metal""},{""type"":""substance name"",""value"":""Water""}]" "

How do we represent the reaction of lithium metal with water?

" nan 2 Li + 2 H2O -> 2 LiOH + H2 "
+

Explanation:

+
+

Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.

+

Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......

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" "
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#Li(s) + H_2O(l) rarr LiOH(aq) + 1/2H_2(g)uarr#

+
+
+
+

Explanation:

+
+

Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.

+

Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......

+
+
+
" "
+

How do we represent the reaction of lithium metal with water?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
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+1 Answer +
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+ + May 4, 2017 + +
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#Li(s) + H_2O(l) rarr LiOH(aq) + 1/2H_2(g)uarr#

+
+
+
+

Explanation:

+
+

Is the given equation balanced with respect to mass and charge? If the answer is NON, it cannot be accepted as a representation of chemical reality.

+

Lithium is an active metal, and a strong enuff reductant to reduce the water molecule to give dihydrogen gas and the metal hydroxide......

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+
+
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" How do we represent the reaction of lithium metal with water? nan +393 a835d81f-6ddd-11ea-803c-ccda262736ce https://socratic.org/questions/when-0-560-g-of-na-s-reacts-with-excess-f-2-g-to-form-naf-s-13-8-kj-of-heat-is-e -566.74 kJ/mol start physical_unit 27 27 standard_enthalpy kj/mol qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 27 27 12 13 heat_energy qc_end c_other OTHER qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Standard enthalpy [OF] formation [IN] kJ/mol""}]" "[{""type"":""physical unit"",""value"":""-566.74 kJ/mol""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] Na(s) [=] \\pu{0.560 g}""},{""type"":""physical unit"",""value"":""Heat evolved [OF] formation [=] \\pu{13.8 kJ}""},{""type"":""other"",""value"":""Na(s) reacts with excess F2(g) to form NaF(s).""},{""type"":""other"",""value"":""Standard-slate conditions.""}]" "

When 0.560 g of #Na#(s) reacts with excess #F_2#(g) to form #NaF#(s), 13.8 kJ of heat is evolved at standard-slate conditions. What is the standard enthalpy of formation?

" nan -566.74 kJ/mol "
+

Explanation:

+
+

Your starting point here will be the balanced chemical equation for this synthesis reaction

+
+

#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#

+
+

Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this

+
+

#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#

+
+

Now, notice that you have a #1:1# mole ratio between sodium and sodium fluoride. This means that in order for the reaction to produce one mole of sodium fluoride, one mole of sodium must take part in the reaction.

+

You know that when #""0.560 g""# of sodium react, the reaction gives off #""13.8 kJ""# of heat. Use sodium's molar mass to determine how many moles of sodium would give off this much heat

+
+

#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#

+
+

So, if #0.02435# moles of sodium are needed to give off #""13.8 kJ""# of heat, how much heat will be given off when one mole of sodium will react?

+
+

#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#

+
+

Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be

+
+

#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#

+
+

The answer is rounded to three sig figs.

+

The listed value for sodium fluoride's standard enthalpy of formation is #-""569 kJ/mol""#, so this is a very good result.

+

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf

+
+
" "
+
+
+

#DeltaH_""f""^@ = -""567 kJ/mol""#

+
+
+
+

Explanation:

+
+

Your starting point here will be the balanced chemical equation for this synthesis reaction

+
+

#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#

+
+

Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this

+
+

#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#

+
+

Now, notice that you have a #1:1# mole ratio between sodium and sodium fluoride. This means that in order for the reaction to produce one mole of sodium fluoride, one mole of sodium must take part in the reaction.

+

You know that when #""0.560 g""# of sodium react, the reaction gives off #""13.8 kJ""# of heat. Use sodium's molar mass to determine how many moles of sodium would give off this much heat

+
+

#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#

+
+

So, if #0.02435# moles of sodium are needed to give off #""13.8 kJ""# of heat, how much heat will be given off when one mole of sodium will react?

+
+

#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#

+
+

Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be

+
+

#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#

+
+

The answer is rounded to three sig figs.

+

The listed value for sodium fluoride's standard enthalpy of formation is #-""569 kJ/mol""#, so this is a very good result.

+

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf

+
+
+
" "
+

When 0.560 g of #Na#(s) reacts with excess #F_2#(g) to form #NaF#(s), 13.8 kJ of heat is evolved at standard-slate conditions. What is the standard enthalpy of formation?

+
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+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
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+1 Answer +
+
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+
+ +
+ + Dec 18, 2015 + +
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#DeltaH_""f""^@ = -""567 kJ/mol""#

+
+
+
+

Explanation:

+
+

Your starting point here will be the balanced chemical equation for this synthesis reaction

+
+

#2""Na""_text((s]) + ""F""_text(2(g]) -> 2""NaF""_text((s])#

+
+

Now, the standard enthalpy of formation is always given for the formation of one mole of a substance. In this case, the chemical equation that describes the formation of one mole of sodium fluoride looks like this

+
+

#""Na""_text((s]) + 1/2""F""_text(2(g]) -> ""NaF""_text((s])#

+
+

Now, notice that you have a #1:1# mole ratio between sodium and sodium fluoride. This means that in order for the reaction to produce one mole of sodium fluoride, one mole of sodium must take part in the reaction.

+

You know that when #""0.560 g""# of sodium react, the reaction gives off #""13.8 kJ""# of heat. Use sodium's molar mass to determine how many moles of sodium would give off this much heat

+
+

#0.560 color(red)(cancel(color(black)(""g""))) * ""1 mole Na""/(23.0color(red)(cancel(color(black)(""g"")))) = ""0.02435 moles Na""#

+
+

So, if #0.02435# moles of sodium are needed to give off #""13.8 kJ""# of heat, how much heat will be given off when one mole of sodium will react?

+
+

#1 color(red)(cancel(color(black)(""mole Na""))) * ""13.8 kJ""/(0.02435color(red)(cancel(color(black)(""moles Na"")))) = ""566.7 kJ""#

+
+

Now, when heat is given off, the standard enthalpy change of formation carries a negative sign. This means that the enthalpy change of formation for sodium fluoride will be

+
+

#DeltaH_""f""^@ = - color(green)(""567 kJ/mol"")#

+
+

The answer is rounded to three sig figs.

+

The listed value for sodium fluoride's standard enthalpy of formation is #-""569 kJ/mol""#, so this is a very good result.

+

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf

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Related questions
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" When 0.560 g of #Na#(s) reacts with excess #F_2#(g) to form #NaF#(s), 13.8 kJ of heat is evolved at standard-slate conditions. What is the standard enthalpy of formation? nan +394 a835d820-6ddd-11ea-87b1-ccda262736ce https://socratic.org/questions/5710bd0e11ef6b5f0ab02dcc 18 start physical_unit 2 7 number none qc_end chemical_equation 7 7 qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] electrons in the nitrite ion, NO2^-""}]" "[{""type"":""physical unit"",""value"":""18""}]" "[{""type"":""chemical equation"",""value"":""NO2^-""}]" "

How many electrons in the nitrite ion, #NO_2^(-)#?

" nan 18 "
+

Explanation:

+
+

No. of valence electrons: #=2xx6+5+1=18#.

+

So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.

+

Of course there is a resonance isomer: #""""^(-)O-N(=O)#.

+

The neutral nitrogen atom has a formal lone pair in each mesomer.

+

How would you describe the structure of this ion based on VSEPR?

+
+
" "
+
+
+

#(O=)N-O^-#

+
+
+
+

Explanation:

+
+

No. of valence electrons: #=2xx6+5+1=18#.

+

So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.

+

Of course there is a resonance isomer: #""""^(-)O-N(=O)#.

+

The neutral nitrogen atom has a formal lone pair in each mesomer.

+

How would you describe the structure of this ion based on VSEPR?

+
+
+
" "
+

How many electrons in the nitrite ion, #NO_2^(-)#?

+
+
+ + +Chemistry + + + + + +Covalent Bonds + + + + + +Resonance + + +
+
+
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+1 Answer +
+
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+ + Apr 30, 2016 + +
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#(O=)N-O^-#

+
+
+
+

Explanation:

+
+

No. of valence electrons: #=2xx6+5+1=18#.

+

So around each atom (from left to right above), there are 6, and 5, and 7 valence electrons. Because there are 7 valence electrons around the rightmost oxygen (i.e. 9 electrons in total including the inner shell electrons), this oxygen has a formal negative charge.

+

Of course there is a resonance isomer: #""""^(-)O-N(=O)#.

+

The neutral nitrogen atom has a formal lone pair in each mesomer.

+

How would you describe the structure of this ion based on VSEPR?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
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Impact of this question
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+ + Creative Commons License + +
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" How many electrons in the nitrite ion, #NO_2^(-)#? nan +395 a835d821-6ddd-11ea-aaba-ccda262736ce https://socratic.org/questions/what-is-the-number-of-moles-of-gas-in-20-0-l-of-oxygen-at-stp 0.89 moles start physical_unit 5 7 number mol qc_end physical_unit 12 12 9 10 volume qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] moles of gas [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.89 moles""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [=] \\pu{20.0 L}""},{""type"":""other"",""value"":""STP""}]" "

What is the number of moles of gas in 20.0 L of oxygen at STP?

" nan 0.89 moles "
+

Explanation:

+
+

# 20.0/22.4# = .89285 rounding this off to three significant numbers

+

gives 0.893 moles.

+
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" "
+
+
+

An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are .893 moles of Oxygen in 20.0 liters.

+
+
+
+

Explanation:

+
+

# 20.0/22.4# = .89285 rounding this off to three significant numbers

+

gives 0.893 moles.

+
+
+
" "
+

What is the number of moles of gas in 20.0 L of oxygen at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
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+1 Answer +
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+ + Oct 13, 2016 + +
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An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are .893 moles of Oxygen in 20.0 liters.

+
+
+
+

Explanation:

+
+

# 20.0/22.4# = .89285 rounding this off to three significant numbers

+

gives 0.893 moles.

+
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+
Related questions
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Impact of this question
+
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+ + Creative Commons License + +
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" What is the number of moles of gas in 20.0 L of oxygen at STP? nan +396 a835d822-6ddd-11ea-9081-ccda262736ce https://socratic.org/questions/59aee5d011ef6b4c5b6c7357 Cr2O7^2- + 6 Fe^2+ + 14 H^+ -> 2 Cr^3+ + 6 Fe^3+ + 7 H2O start chemical_equation qc_end substance 8 9 qc_end substance 11 12 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the redox reaction""}]" "[{""type"":""chemical equation"",""value"":""Cr2O7^2- + 6 Fe^2+ + 14 H^+ -> 2 Cr^3+ + 6 Fe^3+ + 7 H2O""}]" "[{""type"":""substance name"",""value"":""Dichromate ion""},{""type"":""substance name"",""value"":""Ferrous ion""}]" "

How do we represent the redox reaction between dichromate ion and ferrous ion stoichiometrically?

" nan Cr2O7^2- + 6 Fe^2+ + 14 H^+ -> 2 Cr^3+ + 6 Fe^3+ + 7 H2O "
+

Explanation:

+
+

#""Reduction half-equation""#

+

#Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr2Cr^(3+)+7H_2O(l)#

+

#""Oxidation half-equation""#

+

#Fe^(2+) rarrFe^(3+)+e^(-) #

+

And so we takes one of the former and six of the latter......

+

#Cr_2O_7^(2-) +6Fe^(2+) + 14H^(+) rarr2Cr^(3+)+6Fe^(3+)+7H_2O(l)#

+
+
" "
+
+
+

#Cr_2O_7^(2-) +6Fe^(2+) + 14H^(+) rarr2Cr^(3+)+6Fe^(3+)+7H_2O(l)#

+
+
+
+

Explanation:

+
+

#""Reduction half-equation""#

+

#Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr2Cr^(3+)+7H_2O(l)#

+

#""Oxidation half-equation""#

+

#Fe^(2+) rarrFe^(3+)+e^(-) #

+

And so we takes one of the former and six of the latter......

+

#Cr_2O_7^(2-) +6Fe^(2+) + 14H^(+) rarr2Cr^(3+)+6Fe^(3+)+7H_2O(l)#

+
+
+
" "
+

How do we represent the redox reaction between dichromate ion and ferrous ion stoichiometrically?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation and Reduction Reactions + + +
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+1 Answer +
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+ + Sep 5, 2017 + +
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#Cr_2O_7^(2-) +6Fe^(2+) + 14H^(+) rarr2Cr^(3+)+6Fe^(3+)+7H_2O(l)#

+
+
+
+

Explanation:

+
+

#""Reduction half-equation""#

+

#Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr2Cr^(3+)+7H_2O(l)#

+

#""Oxidation half-equation""#

+

#Fe^(2+) rarrFe^(3+)+e^(-) #

+

And so we takes one of the former and six of the latter......

+

#Cr_2O_7^(2-) +6Fe^(2+) + 14H^(+) rarr2Cr^(3+)+6Fe^(3+)+7H_2O(l)#

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Related questions
+ + +
+
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+
Impact of this question
+
+ 2780 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" How do we represent the redox reaction between dichromate ion and ferrous ion stoichiometrically? nan +397 a835d823-6ddd-11ea-838f-ccda262736ce https://socratic.org/questions/when-3-grams-of-nacl-are-added-to-75-25-grams-of-water-what-is-the-change-in-the 0.42 ℃ start physical_unit 11 11 freezing_point_temperature °c qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 11 11 8 9 mass qc_end end "[{""type"":""physical unit"",""value"":""Freezing point Changed [OF] water [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""0.42 ℃""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [=] \\pu{3 grams}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{75.25 grams}""}]" "

When 3 grams of #""NaCl""# are added to 75.25 grams of water, what is the change in the water's freezing point?

" nan 0.42 ℃ "
+

Explanation:

+
+

The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride, #""NaCl""#, in #""75.25 g""# of pure water.

+

As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.

+

The equation that allows you to calculate freezing-point depression looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#

+
+

Here

+

#DeltaT_f# - the freezing-point depression;
+#i# - the van't Hoff factor
+#K_f# - the cryoscopic constant of the solvent;
+#b# - the molality of the solution

+

Water's cryoscopic constant is listed as

+
+

#K_f = 1.86^@""C kg mol""^(-1)#

+
+

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

+

Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)#

+
+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

Notice that every mole of sodium chloride that is dissolved in aqueous solution produces #2# moles of ions.

+

This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to

+
+

#i = 2 -># one mole dissolved, two moles of ions produced

+
+

The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in #""1 kg""# of solvent.

+

Use sodium chloride's molar mass to convert the grams of solute to moles

+
+

#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#

+
+

Convert the mass of water from grams to kilograms

+
+

#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#

+
+

The molality of the solution will be equal to

+
+

#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#

+
+

Plug in your values to calculate the value of #DeltaT_f#

+
+

#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#

+

#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.

+

So, this value tells you that the freezing point of the solution will be #0.42^@""C""# lower than the freezing point of pure water.

+
+
" "
+
+
+

#0.42^@""C""#

+
+
+
+

Explanation:

+
+

The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride, #""NaCl""#, in #""75.25 g""# of pure water.

+

As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.

+

The equation that allows you to calculate freezing-point depression looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#

+
+

Here

+

#DeltaT_f# - the freezing-point depression;
+#i# - the van't Hoff factor
+#K_f# - the cryoscopic constant of the solvent;
+#b# - the molality of the solution

+

Water's cryoscopic constant is listed as

+
+

#K_f = 1.86^@""C kg mol""^(-1)#

+
+

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

+

Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)#

+
+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

Notice that every mole of sodium chloride that is dissolved in aqueous solution produces #2# moles of ions.

+

This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to

+
+

#i = 2 -># one mole dissolved, two moles of ions produced

+
+

The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in #""1 kg""# of solvent.

+

Use sodium chloride's molar mass to convert the grams of solute to moles

+
+

#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#

+
+

Convert the mass of water from grams to kilograms

+
+

#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#

+
+

The molality of the solution will be equal to

+
+

#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#

+
+

Plug in your values to calculate the value of #DeltaT_f#

+
+

#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#

+

#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.

+

So, this value tells you that the freezing point of the solution will be #0.42^@""C""# lower than the freezing point of pure water.

+
+
+
" "
+

When 3 grams of #""NaCl""# are added to 75.25 grams of water, what is the change in the water's freezing point?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 29, 2016 + +
+
+
+
+
+
+
+

#0.42^@""C""#

+
+
+
+

Explanation:

+
+

The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride, #""NaCl""#, in #""75.25 g""# of pure water.

+

As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.

+

The equation that allows you to calculate freezing-point depression looks like this

+
+

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#

+
+

Here

+

#DeltaT_f# - the freezing-point depression;
+#i# - the van't Hoff factor
+#K_f# - the cryoscopic constant of the solvent;
+#b# - the molality of the solution

+

Water's cryoscopic constant is listed as

+
+

#K_f = 1.86^@""C kg mol""^(-1)#

+
+

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

+

Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)#

+
+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

Notice that every mole of sodium chloride that is dissolved in aqueous solution produces #2# moles of ions.

+

This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to

+
+

#i = 2 -># one mole dissolved, two moles of ions produced

+
+

The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in #""1 kg""# of solvent.

+

Use sodium chloride's molar mass to convert the grams of solute to moles

+
+

#3 color(red)(cancel(color(black)(""g""))) * ""1 mole NaCl""/(58.443color(red)(cancel(color(black)(""g"")))) = ""0.08555 moles NaCl""#

+
+

Convert the mass of water from grams to kilograms

+
+

#75.25 color(red)(cancel(color(black)(""g""))) * ""1 kg""/(10^3color(red)(cancel(color(black)(""kg"")))) = ""0.7525 kg""#

+
+

The molality of the solution will be equal to

+
+

#b = ""0.08555 moles""/""0.7525 kg"" = ""0.1137 mol kg""^(-1)#

+
+

Plug in your values to calculate the value of #DeltaT_f#

+
+

#DeltaT_f = 2 * 1.86^@""C"" color(red)(cancel(color(black)(""kg""))) color(red)(cancel(color(black)(""mol""^(-1)))) * 0.1137color(red)(cancel(color(black)(""mol""))) color(red)(cancel(color(black)(""kg"")))#

+

#DeltaT_f = color(green)(|bar(ul(color(white)(a/a)color(black)(0.42^@""C"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.

+

So, this value tells you that the freezing point of the solution will be #0.42^@""C""# lower than the freezing point of pure water.

+
+
+
+
+
+ +
+
+
+
+
+
+
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" "When 3 grams of #""NaCl""# are added to 75.25 grams of water, what is the change in the water's freezing point?" nan +398 a835f8dc-6ddd-11ea-9049-ccda262736ce https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-that-has-a-molecular-mass-of-54-and--1 C4H6 start chemical_formula qc_end physical_unit 6 7 14 14 molecular_weight qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a compound [IN] molecular""}]" "[{""type"":""chemical equation"",""value"":""C4H6""}]" "[{""type"":""physical unit"",""value"":""Molecular mass [OF] a compound [=] \\pu{54}""},{""type"":""other"",""value"":""The empirical formula of the compound is C2H3.""}]" "

What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

" nan C4H6 "
+

Explanation:

+
+

You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.

+

#color(blue)(""Empirical Mass of C""_2""H""_3""#
+Multiply the subscript of each element by its atomic mass and add.

+

#""C""_2""H""_3:##(2xx12.011""u C"") + (3xx1.008 ""u H"")=""27.046 u C""_2""H""_3""#

+

#color(blue)(""Determine Molecular Formula""#
+Divide the molecular mass by the empirical mass.

+

#54/(27.046)=2.0# rounded to two sig figs.

+

Multiply the subscripts in the empirical formula by 2 to get the molecular formula.

+

#""""C_(2xx2)""""H_(3xx2)""=C""_4""H""_6""#

+

The molecular formula is #""C""_4""H""_6""#.

+
+
" "
+
+
+

The molecular formula is #""C""_4""H""_6""#.

+

Refer to the explanation for the process.

+
+
+
+

Explanation:

+
+

You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.

+

#color(blue)(""Empirical Mass of C""_2""H""_3""#
+Multiply the subscript of each element by its atomic mass and add.

+

#""C""_2""H""_3:##(2xx12.011""u C"") + (3xx1.008 ""u H"")=""27.046 u C""_2""H""_3""#

+

#color(blue)(""Determine Molecular Formula""#
+Divide the molecular mass by the empirical mass.

+

#54/(27.046)=2.0# rounded to two sig figs.

+

Multiply the subscripts in the empirical formula by 2 to get the molecular formula.

+

#""""C_(2xx2)""""H_(3xx2)""=C""_4""H""_6""#

+

The molecular formula is #""C""_4""H""_6""#.

+
+
+
" "
+

What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + May 23, 2017 + +
+
+
+
+
+
+
+

The molecular formula is #""C""_4""H""_6""#.

+

Refer to the explanation for the process.

+
+
+
+

Explanation:

+
+

You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.

+

#color(blue)(""Empirical Mass of C""_2""H""_3""#
+Multiply the subscript of each element by its atomic mass and add.

+

#""C""_2""H""_3:##(2xx12.011""u C"") + (3xx1.008 ""u H"")=""27.046 u C""_2""H""_3""#

+

#color(blue)(""Determine Molecular Formula""#
+Divide the molecular mass by the empirical mass.

+

#54/(27.046)=2.0# rounded to two sig figs.

+

Multiply the subscripts in the empirical formula by 2 to get the molecular formula.

+

#""""C_(2xx2)""""H_(3xx2)""=C""_4""H""_6""#

+

The molecular formula is #""C""_4""H""_6""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
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+
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+ + Creative Commons License + +
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" What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3? nan +399 a835f8dd-6ddd-11ea-9eae-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-number-of-mol-of-a-gas-that-is-present-in-a-5-67-l-cont 0.17 mol start physical_unit 9 10 mole mol qc_end physical_unit 18 18 16 17 volume qc_end physical_unit 9 10 20 21 temperature qc_end physical_unit 23 24 29 30 pressure qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] a gas [IN] mol""}]" "[{""type"":""physical unit"",""value"":""0.17 mol""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] container [=] \\pu{5.67 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] a gas [=] \\pu{44.7 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] the gas [=] \\pu{614.0 mmHg}""}]" "

How do you calculate the number of mol of a gas that is present in a 5.67 L container at 44.7 C, if the gas exerts a pressure of 614.0 mmHg.?

" nan 0.17 mol "
+

Explanation:

+
+

#n=(PV)/(RT)# #=# #(((614*mm*Hg)/(760*mm*Hg*atm^-1)xx5.67*L))/(0.0821*L*atm*K^-1*mol^-1xx317.9*K)#

+

#~=0.17*mol#

+

This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant, #R#. The most convenient unit of pressure is still the #""atmosphere""#, and this is a unit that is very intuitive. Given this, most chemists would use #mm*Hg#, knowing that #760*mm*Hg-=1*atm#, or rather that #1# #atm# of pressure will support a column of mercury that is #760*mm# high. And thus conversion of #mm*Hg# to #""atmospheres""# makes the choice of gas constant easy.

+
+
" "
+
+
+

#n=(PV)/(RT)# #~=0.17*mol#

+
+
+
+

Explanation:

+
+

#n=(PV)/(RT)# #=# #(((614*mm*Hg)/(760*mm*Hg*atm^-1)xx5.67*L))/(0.0821*L*atm*K^-1*mol^-1xx317.9*K)#

+

#~=0.17*mol#

+

This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant, #R#. The most convenient unit of pressure is still the #""atmosphere""#, and this is a unit that is very intuitive. Given this, most chemists would use #mm*Hg#, knowing that #760*mm*Hg-=1*atm#, or rather that #1# #atm# of pressure will support a column of mercury that is #760*mm# high. And thus conversion of #mm*Hg# to #""atmospheres""# makes the choice of gas constant easy.

+
+
+
" "
+

How do you calculate the number of mol of a gas that is present in a 5.67 L container at 44.7 C, if the gas exerts a pressure of 614.0 mmHg.?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 19, 2016 + +
+
+
+
+
+
+
+

#n=(PV)/(RT)# #~=0.17*mol#

+
+
+
+

Explanation:

+
+

#n=(PV)/(RT)# #=# #(((614*mm*Hg)/(760*mm*Hg*atm^-1)xx5.67*L))/(0.0821*L*atm*K^-1*mol^-1xx317.9*K)#

+

#~=0.17*mol#

+

This simply an application of the Ideal Gas Equation. Units of pressure are always a problem, because these units dictate the choice of gas constant, #R#. The most convenient unit of pressure is still the #""atmosphere""#, and this is a unit that is very intuitive. Given this, most chemists would use #mm*Hg#, knowing that #760*mm*Hg-=1*atm#, or rather that #1# #atm# of pressure will support a column of mercury that is #760*mm# high. And thus conversion of #mm*Hg# to #""atmospheres""# makes the choice of gas constant easy.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 1209 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" How do you calculate the number of mol of a gas that is present in a 5.67 L container at 44.7 C, if the gas exerts a pressure of 614.0 mmHg.? nan +400 a835f8de-6ddd-11ea-a0d2-ccda262736ce https://socratic.org/questions/what-is-the-balanced-equation-of-the-reaction-between-gaseous-propane-and-and-ox C3H8 + 5 O2 -> 3 CO2 + 4 H2O start chemical_equation qc_end substance 9 10 qc_end substance 12 13 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the reaction""}]" "[{""type"":""chemical equation"",""value"":""C3H8 + 5 O2 -> 3 CO2 + 4 H2O""}]" "[{""type"":""substance name"",""value"":""Gaseous propane""},{""type"":""substance name"",""value"":""Oxygen gas""}]" "

What is the balanced equation of the reaction between gaseous propane and and oxygen gas?

" nan C3H8 + 5 O2 -> 3 CO2 + 4 H2O "
+

Explanation:

+
+

#""(i) balance the carbon as carbon dioxide...""#

+

#""(ii) balance the hydrogens as water...""#

+

#""(iii) ...and THEN balance the oxygens......""#

+

And so we completely COMBUST propane, #C_3H_8#..

+

#""(i)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g))_""carbons balanced"" #

+

#""(ii)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g)+4H_2O(l))_""carbons and hydrogens balanced"" #

+

#""(iii)""# #underbrace(C_3H_8(g) + 5O_2(g) rarr 3CO_2(g)+4H_2O(l))_""the entire equation balanced"" #

+

Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...

+

#underbrace(C_4H_10(g) + 13/2O_2(g))_""266 g"" rarr underbrace(4CO_2(g) + 5H_2O(l))_""266 g""#

+

...OR....

+

#underbrace(2C_4H_10(g) + 13O_2(g))_""532 g"" rarr underbrace(8CO_2(g) + 10H_2O(l))_""532 g""#

+

Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.

+
+
" "
+
+
+

The typical rigmarole is to....

+
+
+
+

Explanation:

+
+

#""(i) balance the carbon as carbon dioxide...""#

+

#""(ii) balance the hydrogens as water...""#

+

#""(iii) ...and THEN balance the oxygens......""#

+

And so we completely COMBUST propane, #C_3H_8#..

+

#""(i)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g))_""carbons balanced"" #

+

#""(ii)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g)+4H_2O(l))_""carbons and hydrogens balanced"" #

+

#""(iii)""# #underbrace(C_3H_8(g) + 5O_2(g) rarr 3CO_2(g)+4H_2O(l))_""the entire equation balanced"" #

+

Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...

+

#underbrace(C_4H_10(g) + 13/2O_2(g))_""266 g"" rarr underbrace(4CO_2(g) + 5H_2O(l))_""266 g""#

+

...OR....

+

#underbrace(2C_4H_10(g) + 13O_2(g))_""532 g"" rarr underbrace(8CO_2(g) + 10H_2O(l))_""532 g""#

+

Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.

+
+
+
" "
+

What is the balanced equation of the reaction between gaseous propane and and oxygen gas?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Jun 20, 2018 + +
+
+
+
+
+
+
+

The typical rigmarole is to....

+
+
+
+

Explanation:

+
+

#""(i) balance the carbon as carbon dioxide...""#

+

#""(ii) balance the hydrogens as water...""#

+

#""(iii) ...and THEN balance the oxygens......""#

+

And so we completely COMBUST propane, #C_3H_8#..

+

#""(i)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g))_""carbons balanced"" #

+

#""(ii)""# #underbrace(C_3H_8(g) + O_2(g) rarr 3CO_2(g)+4H_2O(l))_""carbons and hydrogens balanced"" #

+

#""(iii)""# #underbrace(C_3H_8(g) + 5O_2(g) rarr 3CO_2(g)+4H_2O(l))_""the entire equation balanced"" #

+

Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...

+

#underbrace(C_4H_10(g) + 13/2O_2(g))_""266 g"" rarr underbrace(4CO_2(g) + 5H_2O(l))_""266 g""#

+

...OR....

+

#underbrace(2C_4H_10(g) + 13O_2(g))_""532 g"" rarr underbrace(8CO_2(g) + 10H_2O(l))_""532 g""#

+

Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Jun 20, 2018 + +
+
+
+
+
+
+
+

#C_3H_8 + 5O_2 → 3CO_2 +4 H_2O#

+
+
+
+

Explanation:

+
+

The chemical equation for this reaction is #C_3H_8 + O_2 → CO_2 + H_2O#, producing heat. To balance the equation, start by balancing #C#,

+

#C_3H_8 + O_2 → 3CO_2 + H_2O#

+

then #H#,

+

#C_3H_8 + O_2 → 3CO_2 +4 H_2O#

+

and finally #O#. Since the product side contains #(3)(2)+4(1)=10# moles of oxygen, reactant #O_2# must be multiplied by 5 to have 10 moles as well.

+

The balanced chemical equation would be: #C_3H_8 + 5O_2 → 3CO_2 +4 H_2O#.

+

HOWEVER,

+

if too much or too little of #O_2# is present, the equations become

+

#2C_3H_8+9O_2→4CO_2+2CO+8H_2O#, producing carbon monoxide (soot)

+

or

+

#C_3H_8+2O_2→3C+4H_2O# producing carbon, respectively.

+

Source: https://en.wikipedia.org/wiki/Propane#Properties_and_reactions

+
+
+
+
+
+ +
+
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+ + Creative Commons License + +
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" What is the balanced equation of the reaction between gaseous propane and and oxygen gas? nan +401 a835f8df-6ddd-11ea-af50-ccda262736ce https://socratic.org/questions/what-is-the-molar-mass-of-p-4o-10 283.89 g/mol start physical_unit 6 6 molar_mass g/mol qc_end chemical_equation 6 6 qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] P4O10 [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""283.89 g/mol""}]" "[{""type"":""chemical equation"",""value"":""P4O10""}]" "

What is the molar mass of #P_4O_10#?

" nan 283.89 g/mol "
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
" "
+
+
+

#=283.889048# amu

+
+
+
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
+
" "
+

What is the molar mass of #P_4O_10#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 13, 2016 + +
+
+
+
+
+
+
+

#=283.889048# amu

+
+
+
+

Explanation:

+
+

For such mass related problems we use average mass of elements or compounds. A verge mass of element is dependent on its isotopes and respective fraction found naturally.
+In the given question we need to know average mass of Phosphorus and Oxygen.
+average mass of Phosphorus P = 30.973762 amu
+average mass of Oxygen O = 15.9994 amu
+Molar mass of #""P""_4""O""_10=4xx30.973762+10xx15.9994#
+#=283.889048# amu

+
+
+
+
+
+ +
+
+
+
+
+
+
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+ + +
+
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+
Impact of this question
+
+ 1301 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the molar mass of #P_4O_10#? nan +402 a835f8e0-6ddd-11ea-a325-ccda262736ce https://socratic.org/questions/if-the-initial-temperature-of-an-ideal-gas-at-2-250-atm-is-62-00-o-c-what-final- -49.72 ℃ start physical_unit 6 7 temperature °c qc_end physical_unit 6 7 12 13 temperature qc_end physical_unit 6 7 9 10 pressure qc_end physical_unit 6 7 25 26 pressure qc_end end "[{""type"":""physical unit"",""value"":""Temperature2 [OF] ideal gas [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""-49.72 ℃""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] ideal gas [=] \\pu{62.00 ℃}""},{""type"":""physical unit"",""value"":""Pressure1 [OF] ideal gas [=] \\pu{2.250 atm}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] ideal gas [=] \\pu{1.500 atm}""}]" "

If the initial temperature of an ideal gas at 2.250 atm is 62.00 #""^o#C, what final temperature would cause the pressure to be reduced to 1.500 atm?

" nan -49.72 ℃ "
+

Explanation:

+
+

So, we convert the temperatures to the absolute scale, and solve the quotient:

+

#T_2=(T_1xxP_2)/P_1=(335.15*Kxx1.500*atm)/(2.250*atm)=223*K#

+

You should convert this temperature back to the #""Celsius scale""#. It is intuitively sound that a decrease in temperature should result in a decrease in temperature. Why?

+
+
" "
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+
+

With a constant amount of gas, #P_1/T_1=P_2/T_2#, according to Charles' Law.

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+
+
+

Explanation:

+
+

So, we convert the temperatures to the absolute scale, and solve the quotient:

+

#T_2=(T_1xxP_2)/P_1=(335.15*Kxx1.500*atm)/(2.250*atm)=223*K#

+

You should convert this temperature back to the #""Celsius scale""#. It is intuitively sound that a decrease in temperature should result in a decrease in temperature. Why?

+
+
+
" "
+

If the initial temperature of an ideal gas at 2.250 atm is 62.00 #""^o#C, what final temperature would cause the pressure to be reduced to 1.500 atm?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+1 Answer +
+
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+ + Nov 3, 2016 + +
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+

With a constant amount of gas, #P_1/T_1=P_2/T_2#, according to Charles' Law.

+
+
+
+

Explanation:

+
+

So, we convert the temperatures to the absolute scale, and solve the quotient:

+

#T_2=(T_1xxP_2)/P_1=(335.15*Kxx1.500*atm)/(2.250*atm)=223*K#

+

You should convert this temperature back to the #""Celsius scale""#. It is intuitively sound that a decrease in temperature should result in a decrease in temperature. Why?

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+
+
+
+
+
+
Related questions
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" "If the initial temperature of an ideal gas at 2.250 atm is 62.00 #""^o#C, what final temperature would cause the pressure to be reduced to 1.500 atm?" nan +403 a835f8e1-6ddd-11ea-aac2-ccda262736ce https://socratic.org/questions/57b0bb8b7c0149322d167cc2 11.90 g start physical_unit 5 5 mass g qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 5 5 16 17 mass qc_end physical_unit 8 9 22 22 percent_yield qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] ammonia [IN] g""}]" "[{""type"":""physical unit"",""value"":""11.90 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] nitrogen [=] \\pu{14 g}""},{""type"":""physical unit"",""value"":""Mass [OF] ammonia [=] \\pu{12 g}""},{""type"":""physical unit"",""value"":""Yield [OF] the reaction [=] \\pu{70%}""}]" "

What is the mass of ammonia formed if the reaction between 14 g of nitrogen and 12 g of ammonia gives a 70 % yield?

" nan 11.90 g "
+

Explanation:

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+
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We must first identify the limiting reactant, and then we calculate the theoretical yield.

+

We start with the balanced equation.

+

#color(white)(mmmmmmml)"" N""_2 +color(white)(l) ""3H""_2 → color(white)(ll)""2NH""_3#

+

#""MM/g·mol""^""-1"": "" ""28.01color(white)(ll) 2.016color(white)(mml) 17.03#

+
+

(a) Identify the limiting reactant

+

We calculate the amount of #""NH""_3# that can form from each reactant.

+
+

Calculate the moles of #""N""_2#.

+

#""Moles of N""_2 = 14 color(red)(cancel(color(black)(""g N""_2))) × ""1 mol N""_2/(28.01 color(red)(cancel(color(black)(""g N""_2)))) = ""0.500 mol N""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""N""_2#

+

#0.500color(red)(cancel(color(black)(""mol N""_2))) × ""2 mol NH""_3/(1 color(red)(cancel(color(black)(""mol N""_2)))) = ""1.00 mol NH""_3#

+
+

Calculate the moles of #""H""_2#

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#""Moles of H""_2 = 12 color(red)(cancel(color(black)(""g H""_2))) × ""1 mol H""_2/(2.016 color(red)(cancel(color(black)(""g H""_2)))) = ""5.95 mol H""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""H""_2#

+

#5.95 color(red)(cancel(color(black)(""mol H""_2))) × ""2 mol NH""_3/(3 color(red)(cancel(color(black)(""mol H""_2)))) = ""3.97 mol NH""_3#

+

The limiting reactant is #""N""_2#, because it produces fewer moles of #""NH""_3#.

+
+

(b) Calculate the theoretical yield of #""NH""_3#.

+

#""Theoretical yield"" = 1.00 color(red)(cancel(color(black)(""mol NH""_3))) × ""17.01 g NH""_3/(1 color(red)(cancel(color(black)(""mol NH""_3)))) = ""17.0 g NH""_3#

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+

(c) Calculate the actual yield of #""NH""_3#

+

If % yield = 75 %, then

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#""Actual yield"" = 17.0 color(red)(cancel(color(black)(""g theoretical""))) × ""75 g actual""/(100 color(red)(cancel(color(black)(""g theoretical"")))) = ""13 g actual""#

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" "
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The mass of ammonia will be 13 g.

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+
+

Explanation:

+
+
+

We must first identify the limiting reactant, and then we calculate the theoretical yield.

+

We start with the balanced equation.

+

#color(white)(mmmmmmml)"" N""_2 +color(white)(l) ""3H""_2 → color(white)(ll)""2NH""_3#

+

#""MM/g·mol""^""-1"": "" ""28.01color(white)(ll) 2.016color(white)(mml) 17.03#

+
+

(a) Identify the limiting reactant

+

We calculate the amount of #""NH""_3# that can form from each reactant.

+
+

Calculate the moles of #""N""_2#.

+

#""Moles of N""_2 = 14 color(red)(cancel(color(black)(""g N""_2))) × ""1 mol N""_2/(28.01 color(red)(cancel(color(black)(""g N""_2)))) = ""0.500 mol N""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""N""_2#

+

#0.500color(red)(cancel(color(black)(""mol N""_2))) × ""2 mol NH""_3/(1 color(red)(cancel(color(black)(""mol N""_2)))) = ""1.00 mol NH""_3#

+
+

Calculate the moles of #""H""_2#

+

#""Moles of H""_2 = 12 color(red)(cancel(color(black)(""g H""_2))) × ""1 mol H""_2/(2.016 color(red)(cancel(color(black)(""g H""_2)))) = ""5.95 mol H""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""H""_2#

+

#5.95 color(red)(cancel(color(black)(""mol H""_2))) × ""2 mol NH""_3/(3 color(red)(cancel(color(black)(""mol H""_2)))) = ""3.97 mol NH""_3#

+

The limiting reactant is #""N""_2#, because it produces fewer moles of #""NH""_3#.

+
+

(b) Calculate the theoretical yield of #""NH""_3#.

+

#""Theoretical yield"" = 1.00 color(red)(cancel(color(black)(""mol NH""_3))) × ""17.01 g NH""_3/(1 color(red)(cancel(color(black)(""mol NH""_3)))) = ""17.0 g NH""_3#

+
+

(c) Calculate the actual yield of #""NH""_3#

+

If % yield = 75 %, then

+

#""Actual yield"" = 17.0 color(red)(cancel(color(black)(""g theoretical""))) × ""75 g actual""/(100 color(red)(cancel(color(black)(""g theoretical"")))) = ""13 g actual""#

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" "
+

What is the mass of ammonia formed if the reaction between 14 g of nitrogen and 12 g of ammonia gives a 70 % yield?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Percent Yield + + +
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+1 Answer +
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+ + Aug 14, 2016 + +
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The mass of ammonia will be 13 g.

+
+
+
+

Explanation:

+
+
+

We must first identify the limiting reactant, and then we calculate the theoretical yield.

+

We start with the balanced equation.

+

#color(white)(mmmmmmml)"" N""_2 +color(white)(l) ""3H""_2 → color(white)(ll)""2NH""_3#

+

#""MM/g·mol""^""-1"": "" ""28.01color(white)(ll) 2.016color(white)(mml) 17.03#

+
+

(a) Identify the limiting reactant

+

We calculate the amount of #""NH""_3# that can form from each reactant.

+
+

Calculate the moles of #""N""_2#.

+

#""Moles of N""_2 = 14 color(red)(cancel(color(black)(""g N""_2))) × ""1 mol N""_2/(28.01 color(red)(cancel(color(black)(""g N""_2)))) = ""0.500 mol N""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""N""_2#

+

#0.500color(red)(cancel(color(black)(""mol N""_2))) × ""2 mol NH""_3/(1 color(red)(cancel(color(black)(""mol N""_2)))) = ""1.00 mol NH""_3#

+
+

Calculate the moles of #""H""_2#

+

#""Moles of H""_2 = 12 color(red)(cancel(color(black)(""g H""_2))) × ""1 mol H""_2/(2.016 color(red)(cancel(color(black)(""g H""_2)))) = ""5.95 mol H""_2#

+
+

Calculate moles of #""NH""_3# formed from the #""H""_2#

+

#5.95 color(red)(cancel(color(black)(""mol H""_2))) × ""2 mol NH""_3/(3 color(red)(cancel(color(black)(""mol H""_2)))) = ""3.97 mol NH""_3#

+

The limiting reactant is #""N""_2#, because it produces fewer moles of #""NH""_3#.

+
+

(b) Calculate the theoretical yield of #""NH""_3#.

+

#""Theoretical yield"" = 1.00 color(red)(cancel(color(black)(""mol NH""_3))) × ""17.01 g NH""_3/(1 color(red)(cancel(color(black)(""mol NH""_3)))) = ""17.0 g NH""_3#

+
+

(c) Calculate the actual yield of #""NH""_3#

+

If % yield = 75 %, then

+

#""Actual yield"" = 17.0 color(red)(cancel(color(black)(""g theoretical""))) × ""75 g actual""/(100 color(red)(cancel(color(black)(""g theoretical"")))) = ""13 g actual""#

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" What is the mass of ammonia formed if the reaction between 14 g of nitrogen and 12 g of ammonia gives a 70 % yield? nan +404 a835f8e2-6ddd-11ea-9984-ccda262736ce https://socratic.org/questions/in-a-certain-acidic-solution-at-25-degrees-celsius-h-is-100-times-greater-than-o 1.00 × 10^6 M start physical_unit 21 24 value mol/l qc_end physical_unit 1 4 6 8 temperature qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Value [OF] [OH^-] for the solution [IN] M""}]" "[{""type"":""physical unit"",""value"":""1.00 × 10^6 M""}]" "[{""type"":""physical unit"",""value"":""Temperature [OF] a certain acidic solution [=] \\pu{25 degrees Celsius}""},{""type"":""other"",""value"":""[H^+] is 100 times greater than [OH^-].""}]" "

In a certain acidic solution at 25 degrees Celsius, [H+] is 100 times greater than [OH-]. What is the value for [OH-] for the solution ?

" nan 1.00 × 10^6 M "
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Explanation:

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+

We have #[""H""^+]=100[""OH""^-]# at #25^@sf(C)#. We also know that, at #25^@sfC#, #""K""_""w""=1xx10^14# and that #""K""_""w""-=[""H""^+][""OH""^-]#

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#[""H""^+][""OH""^-]=100[""OH""^-]^2=1xx10^14# #""M""#

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#[""OH""^-]^2=1xx10^12# #""M""#

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#[""OH""^-]=1xx10^6# #""M""#

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#[""OH""^-]=1xx10^6# #""M""#

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Explanation:

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We have #[""H""^+]=100[""OH""^-]# at #25^@sf(C)#. We also know that, at #25^@sfC#, #""K""_""w""=1xx10^14# and that #""K""_""w""-=[""H""^+][""OH""^-]#

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#[""H""^+][""OH""^-]=100[""OH""^-]^2=1xx10^14# #""M""#

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#[""OH""^-]^2=1xx10^12# #""M""#

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#[""OH""^-]=1xx10^6# #""M""#

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" "
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In a certain acidic solution at 25 degrees Celsius, [H+] is 100 times greater than [OH-]. What is the value for [OH-] for the solution ?

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+ + +Chemistry + + + + + +Solutions + + + + + +Solutions + + +
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+1 Answer +
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#[""OH""^-]=1xx10^6# #""M""#

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Explanation:

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We have #[""H""^+]=100[""OH""^-]# at #25^@sf(C)#. We also know that, at #25^@sfC#, #""K""_""w""=1xx10^14# and that #""K""_""w""-=[""H""^+][""OH""^-]#

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#[""H""^+][""OH""^-]=100[""OH""^-]^2=1xx10^14# #""M""#

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#[""OH""^-]^2=1xx10^12# #""M""#

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#[""OH""^-]=1xx10^6# #""M""#

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Related questions
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Impact of this question
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" In a certain acidic solution at 25 degrees Celsius, [H+] is 100 times greater than [OH-]. What is the value for [OH-] for the solution ? nan +405 a835f8e3-6ddd-11ea-a81c-ccda262736ce https://socratic.org/questions/59b5446fb72cff2a246920c4 1.99 g start physical_unit 3 3 mass g qc_end physical_unit 16 16 12 13 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] BF3 [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.99 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{0.53 g}""},{""type"":""other"",""value"":""BF3 contains the same number of molecules as water.""}]" "

What mass of #BF_3# contains the same number of molecules as a #0.53*g# mass of water?

" nan 1.99 g "
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Explanation:

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+

#""Moles of water""-=(0.53*g)/(18.01*g*mol^-1)=0.0294*mol#...

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Now the mole is simply a number, i.e. #1*mol# specifies #6.022xx10^23# individual molecules of whatever stuff you got.

+

ANd so we multiply this molar quantity by the molar mass of #BF_3#....

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#-= 0.0294*molxx67.82*g*mol^-1=2.0*g#.........

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" "
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And so you want the MASS of #BF_3# that corresponds to the same number of molecules in a #0.53*g# mass of water?

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+
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Explanation:

+
+

#""Moles of water""-=(0.53*g)/(18.01*g*mol^-1)=0.0294*mol#...

+

Now the mole is simply a number, i.e. #1*mol# specifies #6.022xx10^23# individual molecules of whatever stuff you got.

+

ANd so we multiply this molar quantity by the molar mass of #BF_3#....

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#-= 0.0294*molxx67.82*g*mol^-1=2.0*g#.........

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" "
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What mass of #BF_3# contains the same number of molecules as a #0.53*g# mass of water?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Sep 10, 2017 + +
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And so you want the MASS of #BF_3# that corresponds to the same number of molecules in a #0.53*g# mass of water?

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Explanation:

+
+

#""Moles of water""-=(0.53*g)/(18.01*g*mol^-1)=0.0294*mol#...

+

Now the mole is simply a number, i.e. #1*mol# specifies #6.022xx10^23# individual molecules of whatever stuff you got.

+

ANd so we multiply this molar quantity by the molar mass of #BF_3#....

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#-= 0.0294*molxx67.82*g*mol^-1=2.0*g#.........

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Impact of this question
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" What mass of #BF_3# contains the same number of molecules as a #0.53*g# mass of water? nan +406 a8362222-6ddd-11ea-a9c8-ccda262736ce https://socratic.org/questions/how-would-you-determine-the-ph-of-0-10-m-nh-3-solution-nh-3-is-a-weak-base-with- 11.12 start physical_unit 9 10 ph none qc_end physical_unit 9 10 7 8 molarity qc_end physical_unit 9 9 21 23 kb qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] NH3 solution""}]" "[{""type"":""physical unit"",""value"":""11.12""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] NH3 solution [=] \\pu{0.10 M}""},{""type"":""physical unit"",""value"":""Kb [OF] NH3 [=] \\pu{1.8 × 10^(−5)}""},{""type"":""other"",""value"":""NH3 is a weak base.""}]" "

How would you determine the pH of 0.10 M #NH_3# solution? #NH_3# is a weak base with a #K_b# equal to #1.8 x 10^-5#.

" nan 11.12 "
+

Explanation:

+
+

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#

+

And so #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

+

We put in some numbers, and we assume that #x*mol*L^-1# of ammonia ASSOCIATES, and thus at equilibrium, we know that #x=[NH_4^+]=[HO^-]#, and #[NH_3]=(0.10-x)*mol*L^-1#.

+

And we plug in the numbers to our equilibrium expression..

+

#K_b=1.8xx10^-5=x^2/(0.10-x)#

+

And this is a quadratic in #x#, which we could solve EXACTLY if we were so-minded, but because chemists are workshy, we make the approximation that #0.10-x~=0.10#. Note that we must justify this approx. later.

+

And thus #x_1=sqrt(1.8xx10^-5xx0.10)=1.34xx10^-3*mol*L^-1#.

+

And now that we have an approximation for #x#, we may use #x_1# in our expression to get a second approximation........

+

#x_2=1.33xx10^-3*mol*L^-1#

+

#x_3=1.33xx10^-3*mol*L^-1#

+

This method is as exact as if we used the quadratic equation, and so.....

+

#x_3=[HO^-]=1.33xx10^-3*mol*L^-1#; #pOH=-log_10([HO^-])=2.88#; and #pH=14-pOH=11.1.#

+

We know that in aqueous solution, #14=pOH+pH#, and so if you knows #[HO^-]# youze also knows #[H_3O^+]#.

+
+
" "
+
+
+

Well, we interrogate the equilibrium.......and gets #pH=11.1.#

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+
+
+

Explanation:

+
+

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#

+

And so #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

+

We put in some numbers, and we assume that #x*mol*L^-1# of ammonia ASSOCIATES, and thus at equilibrium, we know that #x=[NH_4^+]=[HO^-]#, and #[NH_3]=(0.10-x)*mol*L^-1#.

+

And we plug in the numbers to our equilibrium expression..

+

#K_b=1.8xx10^-5=x^2/(0.10-x)#

+

And this is a quadratic in #x#, which we could solve EXACTLY if we were so-minded, but because chemists are workshy, we make the approximation that #0.10-x~=0.10#. Note that we must justify this approx. later.

+

And thus #x_1=sqrt(1.8xx10^-5xx0.10)=1.34xx10^-3*mol*L^-1#.

+

And now that we have an approximation for #x#, we may use #x_1# in our expression to get a second approximation........

+

#x_2=1.33xx10^-3*mol*L^-1#

+

#x_3=1.33xx10^-3*mol*L^-1#

+

This method is as exact as if we used the quadratic equation, and so.....

+

#x_3=[HO^-]=1.33xx10^-3*mol*L^-1#; #pOH=-log_10([HO^-])=2.88#; and #pH=14-pOH=11.1.#

+

We know that in aqueous solution, #14=pOH+pH#, and so if you knows #[HO^-]# youze also knows #[H_3O^+]#.

+
+
+
" "
+

How would you determine the pH of 0.10 M #NH_3# solution? #NH_3# is a weak base with a #K_b# equal to #1.8 x 10^-5#.

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+1 Answer +
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+ + Jul 26, 2017 + +
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Well, we interrogate the equilibrium.......and gets #pH=11.1.#

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Explanation:

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#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#

+

And so #K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#

+

We put in some numbers, and we assume that #x*mol*L^-1# of ammonia ASSOCIATES, and thus at equilibrium, we know that #x=[NH_4^+]=[HO^-]#, and #[NH_3]=(0.10-x)*mol*L^-1#.

+

And we plug in the numbers to our equilibrium expression..

+

#K_b=1.8xx10^-5=x^2/(0.10-x)#

+

And this is a quadratic in #x#, which we could solve EXACTLY if we were so-minded, but because chemists are workshy, we make the approximation that #0.10-x~=0.10#. Note that we must justify this approx. later.

+

And thus #x_1=sqrt(1.8xx10^-5xx0.10)=1.34xx10^-3*mol*L^-1#.

+

And now that we have an approximation for #x#, we may use #x_1# in our expression to get a second approximation........

+

#x_2=1.33xx10^-3*mol*L^-1#

+

#x_3=1.33xx10^-3*mol*L^-1#

+

This method is as exact as if we used the quadratic equation, and so.....

+

#x_3=[HO^-]=1.33xx10^-3*mol*L^-1#; #pOH=-log_10([HO^-])=2.88#; and #pH=14-pOH=11.1.#

+

We know that in aqueous solution, #14=pOH+pH#, and so if you knows #[HO^-]# youze also knows #[H_3O^+]#.

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" How would you determine the pH of 0.10 M #NH_3# solution? #NH_3# is a weak base with a #K_b# equal to #1.8 x 10^-5#. nan +407 a8362223-6ddd-11ea-b8ba-ccda262736ce https://socratic.org/questions/59958e5ab72cff3ca5c480a4 1.25 mol/L start physical_unit 7 8 molarity mol/l qc_end physical_unit 7 7 10 11 mass qc_end physical_unit 18 18 15 16 mass qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.25 mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaOH [=] \\pu{20 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{400 g}""}]" "

What is the molarity of an aqueous NaOH solution containing 20 g of NaOH and 400 g of water?

" nan 1.25 mol/L "
+
+

Well, molarity is temperature-dependent, so I will assume #25^@ ""C""# and #""1 atm""#...

+

and I got #~~# #""1 M""#, because you have only allowed yourself one significant figure...

+
+

And at these conditions, #rho_(H_2O) = ""0.9970749 g/mL""#, so that

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+

#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#

+

#=# #""401.17 mL""#

+
+

And so, the molarity is given by:

+
+

#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)

+

#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#

+
+

We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of #""2.13 g/cm""^3# of #""NaOH""# at #25^@ ""C""# to find out its volume contribution.

+
+

#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#

+

#=# #""0.009390 L""#

+
+

And so, the solvent volume will rise by about #""9.39 mL""# (about a #2.3%# increase) upon addition of solute. The molarity should then be...

+
+

#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#

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#=# #ulcolor(blue)(""1.218 M"")#

+

(Had you assumed #V_(""soln"") ~~ V_""solvent""#, you would have gotten about #""1.25 M""#.)

+
+

But you have only allowed yourself one measly significant figure, so I guess you only have a #""1 M""# solution... I guess it'll be inaccurate!

+
+
" "
+
+
+

Well, molarity is temperature-dependent, so I will assume #25^@ ""C""# and #""1 atm""#...

+

and I got #~~# #""1 M""#, because you have only allowed yourself one significant figure...

+
+

And at these conditions, #rho_(H_2O) = ""0.9970749 g/mL""#, so that

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+

#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#

+

#=# #""401.17 mL""#

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+

And so, the molarity is given by:

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+

#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)

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#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#

+
+

We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of #""2.13 g/cm""^3# of #""NaOH""# at #25^@ ""C""# to find out its volume contribution.

+
+

#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#

+

#=# #""0.009390 L""#

+
+

And so, the solvent volume will rise by about #""9.39 mL""# (about a #2.3%# increase) upon addition of solute. The molarity should then be...

+
+

#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#

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#=# #ulcolor(blue)(""1.218 M"")#

+

(Had you assumed #V_(""soln"") ~~ V_""solvent""#, you would have gotten about #""1.25 M""#.)

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+

But you have only allowed yourself one measly significant figure, so I guess you only have a #""1 M""# solution... I guess it'll be inaccurate!

+
+
+
" "
+

What is the molarity of an aqueous NaOH solution containing 20 g of NaOH and 400 g of water?

+
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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+2 Answers +
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+ + Aug 17, 2017 + +
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Well, molarity is temperature-dependent, so I will assume #25^@ ""C""# and #""1 atm""#...

+

and I got #~~# #""1 M""#, because you have only allowed yourself one significant figure...

+
+

And at these conditions, #rho_(H_2O) = ""0.9970749 g/mL""#, so that

+
+

#400 cancel(""g H""_2""O"") xx ""1 mL""/(0.9970749 cancel""g"")#

+

#=# #""401.17 mL""#

+
+

And so, the molarity is given by:

+
+

#""M"" = ""mols solute""/""L solution""# (and NOT solvent!)

+

#= [""20 g NaOH"" xx (""1 mol NaOH"")/(""(22.989 + 15.999 + 1.0079 g) NaOH"")]/(401.17 xx 10^(-3) ""L solvent"" + V_""solute"")#

+
+

We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of #""2.13 g/cm""^3# of #""NaOH""# at #25^@ ""C""# to find out its volume contribution.

+
+

#20 cancel""g NaOH"" xx (cancel""1 mL"")/(2.13 cancel""g"") xx ""1 L""/(1000 cancel""mL"")#

+

#=# #""0.009390 L""#

+
+

And so, the solvent volume will rise by about #""9.39 mL""# (about a #2.3%# increase) upon addition of solute. The molarity should then be...

+
+

#color(blue)([""NaOH""]) = [""0.5001 mols NaOH""]/(401.17 xx 10^(-3) ""L solvent"" + ""0.009390 L NaOH"")#

+

#=# #ulcolor(blue)(""1.218 M"")#

+

(Had you assumed #V_(""soln"") ~~ V_""solvent""#, you would have gotten about #""1.25 M""#.)

+
+

But you have only allowed yourself one measly significant figure, so I guess you only have a #""1 M""# solution... I guess it'll be inaccurate!

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+ + Aug 17, 2017 + +
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1.25M

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Explanation:

+
+

The other posted solution is detailed and accurate, but possibly ""over kill"" for this venue. While 20 technically does have only one significant figure without a designated decimal point, I will also assume it to be 20. for calculations.

+

Also assuming STP for a general chemistry question of this sort, I calculate the moles of NaOH as 0.5 and use the (estimated) density of water at 1g/mL to get 400mL of solvent.

+

0.5M/400mL = 1.25M solution.

+

Right on, or certainly within any of the error probabilities, or assumptions of the more ""rigorous"" answer. You see, chemistry doesn't have to be intimidating.

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" What is the molarity of an aqueous NaOH solution containing 20 g of NaOH and 400 g of water? nan +408 a8362224-6ddd-11ea-a8c0-ccda262736ce https://socratic.org/questions/what-volume-of-1-6-m-potassium-hydroxide-would-be-needed-in-order-to-neutralize- 156.25 mL start physical_unit 5 6 volume ml qc_end physical_unit 19 20 14 15 volume qc_end physical_unit 5 6 3 4 molarity qc_end physical_unit 19 20 17 18 molarity qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] potassium hydroxide [IN] mL""}]" "[{""type"":""physical unit"",""value"":""156.25 mL""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] hydrochloric acid [=] \\pu{100.0 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] potassium hydroxide [=] \\pu{1.6 M}""},{""type"":""physical unit"",""value"":""Molarity [OF] hydrochloric acid [=] \\pu{2.50 M}""}]" "

What volume of 1.6 M potassium hydroxide would be needed in order to neutralize 100.0 mL of 2.50 M hydrochloric acid?

" nan 156.25 mL "
+

Explanation:

+
+

The first thing to do here is write a balanced chemical equation that describes this neutralization reaction

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+

#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

Potassium hydroxide, #""KOH""#, and hydrochloric acid, #""HCl""#, react in a #1:1# mole ratio to produce aqueous potassium chloride, #""KCl""#, and water.

+

This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.

+

Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

+
+

You will have

+
+

#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#

+

#n_""HCl"" = ""0.250 moles HCl""#

+
+

This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

+
+

You will have

+
+

#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#

+
+

I'll leave the answer rounded to three sig figs and express it in milliliters

+
+

#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#

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" "
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+
+

#""156 mL""#

+
+
+
+

Explanation:

+
+

The first thing to do here is write a balanced chemical equation that describes this neutralization reaction

+
+

#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

Potassium hydroxide, #""KOH""#, and hydrochloric acid, #""HCl""#, react in a #1:1# mole ratio to produce aqueous potassium chloride, #""KCl""#, and water.

+

This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.

+

Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

+
+

You will have

+
+

#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#

+

#n_""HCl"" = ""0.250 moles HCl""#

+
+

This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

+
+

You will have

+
+

#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#

+
+

I'll leave the answer rounded to three sig figs and express it in milliliters

+
+

#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#

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" "
+

What volume of 1.6 M potassium hydroxide would be needed in order to neutralize 100.0 mL of 2.50 M hydrochloric acid?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
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+1 Answer +
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+ + Jun 7, 2016 + +
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#""156 mL""#

+
+
+
+

Explanation:

+
+

The first thing to do here is write a balanced chemical equation that describes this neutralization reaction

+
+

#""KOH""_ ((aq)) + ""HCl""_ ((aq)) -> ""KCl""_ ((aq)) + ""H""_ 2""O""_ ((l))#

+
+

Potassium hydroxide, #""KOH""#, and hydrochloric acid, #""HCl""#, react in a #1:1# mole ratio to produce aqueous potassium chloride, #""KCl""#, and water.

+

This tells you that in order to have a complete neutralization, you need to mix equal numbers of moles of each reactant.

+

Now, the problem provides you with the molarity and volume of the hydrochloric acid solution, which you can use to determine how many moles of acid were needed for the reaction

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies n_""solute"" = c * V_""solution"")color(white)(a/a)|)))#

+
+

You will have

+
+

#n_""HCl"" = ""2.50 mol"" color(red)(cancel(color(black)(""L""^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)(""L""))))^(color(blue)(""volume in liters""))#

+

#n_""HCl"" = ""0.250 moles HCl""#

+
+

This is exactly how many moles of potassium hydroxide you must add to the reaction. All you have to do now is use the molarity of the potassium hydroxide solution to figure out what volume would contain that many moles

+
+

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_""solute""/V_""solution"" implies V_""solution"" = n_""solute""/c)color(white)(a/a)|)))#

+
+

You will have

+
+

#V_""KOH"" = (0.250color(red)(cancel(color(black)(""moles""))))/(1.6color(red)(cancel(color(black)(""mol"")))""L""^(-1)) = ""0.1563 mL""#

+
+

I'll leave the answer rounded to three sig figs and express it in milliliters

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+

#""volume of KOH solution"" = color(green)(|bar(ul(color(white)(a/a)color(black)(""156 mL"")color(white)(a/a)|)))#

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" What volume of 1.6 M potassium hydroxide would be needed in order to neutralize 100.0 mL of 2.50 M hydrochloric acid? nan +409 a8362225-6ddd-11ea-b5aa-ccda262736ce https://socratic.org/questions/56b388c511ef6b6a00afe5e4 1.20 × 10^(−2) mol/L start physical_unit 5 9 concentration mol/l qc_end physical_unit 20 21 13 16 concentration qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] chloride ions in a solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""1.20 × 10^(−2) mol/L""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] calcium chloride [=] \\pu{6 × 10^(−3) mol/L}""}]" "

What is the concentration of chloride ions in a solution whose concentration is #6xx10^-3*mol*L^-1# with respect to #""calcium chloride""#?

" nan 1.20 × 10^(−2) mol/L "
+

Explanation:

+
+

You have quoted a solution of #CaCl_2# that is #6xx10^-3# #mol*L^-1#, and I assume this to be this concentration with respect to the compound itself (i.e. #[CaCl_2]# #=# #6.0xx10^(-3)# #mol*L^-1# this is typically understood!).

+

Because this a #CaCl_2# solution, for each #Ca^(2+)#, there are #2# equiv #Cl^-# ion. Chloride ion concentration is effectively doubled to give #1.2xx10^-2# #mol*L^-1#.

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" "
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#1.2xx10^-2# #mol*L^-1# in chloride ion.

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+
+

Explanation:

+
+

You have quoted a solution of #CaCl_2# that is #6xx10^-3# #mol*L^-1#, and I assume this to be this concentration with respect to the compound itself (i.e. #[CaCl_2]# #=# #6.0xx10^(-3)# #mol*L^-1# this is typically understood!).

+

Because this a #CaCl_2# solution, for each #Ca^(2+)#, there are #2# equiv #Cl^-# ion. Chloride ion concentration is effectively doubled to give #1.2xx10^-2# #mol*L^-1#.

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" "
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What is the concentration of chloride ions in a solution whose concentration is #6xx10^-3*mol*L^-1# with respect to #""calcium chloride""#?

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+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
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+1 Answer +
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+ + Feb 4, 2016 + +
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#1.2xx10^-2# #mol*L^-1# in chloride ion.

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+

Explanation:

+
+

You have quoted a solution of #CaCl_2# that is #6xx10^-3# #mol*L^-1#, and I assume this to be this concentration with respect to the compound itself (i.e. #[CaCl_2]# #=# #6.0xx10^(-3)# #mol*L^-1# this is typically understood!).

+

Because this a #CaCl_2# solution, for each #Ca^(2+)#, there are #2# equiv #Cl^-# ion. Chloride ion concentration is effectively doubled to give #1.2xx10^-2# #mol*L^-1#.

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Related questions
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Impact of this question
+
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" "What is the concentration of chloride ions in a solution whose concentration is #6xx10^-3*mol*L^-1# with respect to #""calcium chloride""#?" nan +410 a8364746-6ddd-11ea-853a-ccda262736ce https://socratic.org/questions/how-many-moles-of-solid-aluminum-are-needed-to-react-with-nitric-acid-in-order-t 3.00 moles start physical_unit 4 5 mole mol qc_end substance 11 12 qc_end physical_unit 20 21 17 18 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] solid aluminum [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.00 moles""}]" "[{""type"":""substance name"",""value"":""Nitric acid""},{""type"":""physical unit"",""value"":""Mole [OF] hydrogen gas [=] \\pu{5.0 moles}""}]" "

How many moles of solid aluminum are needed to react with nitric acid in order to produce 5.0 moles of hydrogen gas?

" nan 3.00 moles "
+

Explanation:

+
+

We need a stoichiometric equation that represents the oxidation of aluminum metal.....

+

#Al(s) + 3HNO_3(aq) rarr Al(NO_3)_3(aq) + 3/2H_2(g)uarr#

+

And thus each mole of metal generates #3/2# moles dihydrogen gas......And so if we want a #5*mol# quantity of #H_2# we take the quotient......

+

#(5*mol)/(3/2*mol)=10/3*mol# #""aluminum metal""#. Given the molar mass of aluminum is #27.0*g#, what quantity of metal are needed for the given reaction?

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" "
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A bit over #3*mol# aluminum are required...........

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+
+

Explanation:

+
+

We need a stoichiometric equation that represents the oxidation of aluminum metal.....

+

#Al(s) + 3HNO_3(aq) rarr Al(NO_3)_3(aq) + 3/2H_2(g)uarr#

+

And thus each mole of metal generates #3/2# moles dihydrogen gas......And so if we want a #5*mol# quantity of #H_2# we take the quotient......

+

#(5*mol)/(3/2*mol)=10/3*mol# #""aluminum metal""#. Given the molar mass of aluminum is #27.0*g#, what quantity of metal are needed for the given reaction?

+
+
+
" "
+

How many moles of solid aluminum are needed to react with nitric acid in order to produce 5.0 moles of hydrogen gas?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Equation Stoichiometry + + +
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+1 Answer +
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+ + Aug 2, 2017 + +
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A bit over #3*mol# aluminum are required...........

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+
+

Explanation:

+
+

We need a stoichiometric equation that represents the oxidation of aluminum metal.....

+

#Al(s) + 3HNO_3(aq) rarr Al(NO_3)_3(aq) + 3/2H_2(g)uarr#

+

And thus each mole of metal generates #3/2# moles dihydrogen gas......And so if we want a #5*mol# quantity of #H_2# we take the quotient......

+

#(5*mol)/(3/2*mol)=10/3*mol# #""aluminum metal""#. Given the molar mass of aluminum is #27.0*g#, what quantity of metal are needed for the given reaction?

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" How many moles of solid aluminum are needed to react with nitric acid in order to produce 5.0 moles of hydrogen gas? nan +411 a8364747-6ddd-11ea-b620-ccda262736ce https://socratic.org/questions/a-sample-of-seawater-contains-0-000245-g-of-sodium-chloride-per-ml-of-solution-h 1.23 × 10^(-2) g start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mass qc_end physical_unit 0 3 10 11 volume qc_end physical_unit 24 25 21 22 volume qc_end end "[{""type"":""physical unit"",""value"":""Mass2 [OF] sodium chloride [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.23 × 10^(-2) g""}]" "[{""type"":""physical unit"",""value"":""Mass1 [OF] sodium chloride [=] \\pu{0.000245 g}""},{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of seawater [=] \\pu{per mL}""},{""type"":""physical unit"",""value"":""Volume2 [OF] this solution [=] \\pu{50.0 mL}""}]" "

A sample of seawater contains 0.000245 g of sodium chloride per mL of solution. How much sodium chloride is contained in 50.0 mL of this solution?

" nan 1.23 × 10^(-2) g "
+

Explanation:

+
+

Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
+#(g/(mL)) xx mL) = g#

+

#0.000245g xx 50.0 = 0.01225g#

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" "
+
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+

#0.01225g#

+
+
+
+

Explanation:

+
+

Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
+#(g/(mL)) xx mL) = g#

+

#0.000245g xx 50.0 = 0.01225g#

+
+
+
" "
+

A sample of seawater contains 0.000245 g of sodium chloride per mL of solution. How much sodium chloride is contained in 50.0 mL of this solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solutions + + +
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+2 Answers +
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+ +
+ + Oct 1, 2017 + +
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+

#0.01225g#

+
+
+
+

Explanation:

+
+

Using Dimensional Analysis we can set up the correct equation to find the desired result. If we want mass (grams) from a concentration and volume it is this:
+#(g/(mL)) xx mL) = g#

+

#0.000245g xx 50.0 = 0.01225g#

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+ + Oct 1, 2017 + +
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#""50 mL""# of the given #""NaCl""# solution contains #""0.0123 g""# of #""NaCl""#.

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+

Explanation:

+
+

This is basically a density problem, where the density is known:

+

#""0.000245 g/mL NaCl""# solution.

+

In this question, the mass of #""NaCl""# in #""50 mL""# of solution is unknown.

+

#""density""=""mass""/""volume""#

+

To find mass when density and volume are known, we can rearrange the density formula to isolate mass.

+

#""mass""=""density""xx""volume""#

+

Plug in the known values and solve for mass.

+

#50.0color(red)cancel(color(black)(""mL NaCl sol'n""))xx(0.000245""g NaCl"")/(1color(red)cancel(color(black)(""mL NaCl sol'n"")))=""0.0123 g NaCl""=1.23xx10^(-2) ""g NaCl""# (rounded to three significant figures)

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" A sample of seawater contains 0.000245 g of sodium chloride per mL of solution. How much sodium chloride is contained in 50.0 mL of this solution? nan +412 a8366e66-6ddd-11ea-817e-ccda262736ce https://socratic.org/questions/after-0-600-l-of-ar-at-1-47-atm-and-221-degrees-celsius-is-mixed-with-0-200-l-of 1.55 atm start physical_unit 40 41 pressure atm qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 9 11 temperature qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 18 18 20 21 pressure qc_end physical_unit 18 18 23 25 temperature qc_end physical_unit 40 41 28 29 volume qc_end physical_unit 40 41 32 34 temperature qc_end end "[{""type"":""physical unit"",""value"":""Pressure [OF] the flask [IN] atm""}]" "[{""type"":""physical unit"",""value"":""1.55 atm""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] Ar [=] \\pu{0.600 L}""},{""type"":""physical unit"",""value"":""Pressure [OF] Ar [=] \\pu{1.47 atm}""},{""type"":""physical unit"",""value"":""Temperature [OF] Ar [=] \\pu{221 degrees Celsius}""},{""type"":""physical unit"",""value"":""Volume [OF] O2 [=] \\pu{0.200 L}""},{""type"":""physical unit"",""value"":""Pressure [OF] O2 [=] \\pu{441 torr}""},{""type"":""physical unit"",""value"":""Temperature [OF] O2 [=] \\pu{115 degrees Celsius}""},{""type"":""physical unit"",""value"":""Volume [OF] the flask [=] \\pu{400 mL}""},{""type"":""physical unit"",""value"":""Temperature [OF] the flask [=] \\pu{24 degrees Celsius}""}]" "

After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

" nan 1.55 atm "
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

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#= color(blue)(1.55# #color(blue)(""atm""#

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" "
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#P = 1.55# #""atm""#

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+
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

+

#= color(blue)(1.55# #color(blue)(""atm""#

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+
" "
+

After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

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+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+1 Answer +
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+ + Jul 8, 2017 + +
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#P = 1.55# #""atm""#

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+
+
+

Explanation:

+
+

Here's how I go about doing this.

+

What we can do is use the ideal gas equation for both #""Ar""# and #""O""_2# to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (#400# #""mL""#) and temperature (#24^""o""""C""#).

+

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

+

#n_ ""Ar"" = ((1.47cancel(""atm""))(0.600cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(494cancel(""K""))) = 0.0218# #""mol Ar""#

+

#n_ (""O""_2) = ((0.580cancel(""atm""))(0.200cancel(""L"")))/((0.082057(cancel(""L"")•cancel(""atm""))/(""mol""•cancel(""K"")))(388cancel(""K""))) = 0.00365# #""mol O""_2#

+

#n_""total"" = 0.0218# #""mol Ar""# #+ 0.00365# #""mol O""_2# #= 0.0254# #""mol""#

+

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

+

#P = (nRT)/V = ((0.0254cancel(""mol""))(0.082057(cancel(""L"")•""atm"")/(cancel(""mol"")•cancel(""K"")))(297cancel(""K"")))/(0.400cancel(""L""))#

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#= color(blue)(1.55# #color(blue)(""atm""#

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" After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask? nan +413 a8366e67-6ddd-11ea-86e6-ccda262736ce https://socratic.org/questions/how-do-you-balance-nh-3-l-o-2-g-no-g-h-2o-l 4 NH3(l) + 5 O2(g) -> 4 NO(g) + 6 H2O(l) start chemical_equation qc_end chemical_equation 4 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""4 NH3(l) + 5 O2(g) -> 4 NO(g) + 6 H2O(l)""}]" "[{""type"":""chemical equation"",""value"":""NH3(l) + O2(g) -> NO(g) + H2O(l)""}]" "

How do you balance #NH_3(l) + O_2(g) -> NO(g) + H_2O(l)#?

" nan 4 NH3(l) + 5 O2(g) -> 4 NO(g) + 6 H2O(l) "
+

Explanation:

+
+

Given unbalanced equation is

+

#""NH""_3(l) + ""O""_2(g) -> ""NO""(g) + ""H""_2""O""(l)#
+As a thumb rule #""H"" and ""O""# are left for the last. As we see that other than these two atoms there is only one atom, #""N""#. That atom is already balanced.

+

We start with #""H"" and ""O""#. We observe that out of these two, hydrogen has more number of atoms in the equation. So we take it first.

+

Let the balanced equation be: (notice that number of N atoms have been kept balanced)
+#x""NH""_3(l) + y""O""_2(g) -> x""NO""(g) + b""H""_2""O""(l)#

+

To balance hydrogen atoms we multiply ammonia molecule with 2 (set #x=2#) on reactants side and water molecule by 3 (set #b=3#) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set #x=2#.
+The equation looks like

+

#""2NH""_3(l) + y""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#

+

For the remaining unbalanced element:
+Total number of oxygen atoms on products side becomes #2+3=5#. Which gives us number of oxygen molecules as #y=5/2# on the reactants side.

+

#""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#
+ We know that #y# needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

+

#2xx(""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l))#
+#""4NH""_3(l) + 5""O""_2(g) -> 4""NO""(g) + 6""H""_2""O""(l)#

+
+
" "
+
+
+

#""4NH""_3(l) + 5""O""_2(g) -> 4""NO""(g) + 6""H""_2""O""(l)#

+
+
+
+

Explanation:

+
+

Given unbalanced equation is

+

#""NH""_3(l) + ""O""_2(g) -> ""NO""(g) + ""H""_2""O""(l)#
+As a thumb rule #""H"" and ""O""# are left for the last. As we see that other than these two atoms there is only one atom, #""N""#. That atom is already balanced.

+

We start with #""H"" and ""O""#. We observe that out of these two, hydrogen has more number of atoms in the equation. So we take it first.

+

Let the balanced equation be: (notice that number of N atoms have been kept balanced)
+#x""NH""_3(l) + y""O""_2(g) -> x""NO""(g) + b""H""_2""O""(l)#

+

To balance hydrogen atoms we multiply ammonia molecule with 2 (set #x=2#) on reactants side and water molecule by 3 (set #b=3#) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set #x=2#.
+The equation looks like

+

#""2NH""_3(l) + y""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#

+

For the remaining unbalanced element:
+Total number of oxygen atoms on products side becomes #2+3=5#. Which gives us number of oxygen molecules as #y=5/2# on the reactants side.

+

#""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#
+ We know that #y# needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

+

#2xx(""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l))#
+#""4NH""_3(l) + 5""O""_2(g) -> 4""NO""(g) + 6""H""_2""O""(l)#

+
+
+
" "
+

How do you balance #NH_3(l) + O_2(g) -> NO(g) + H_2O(l)#?

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+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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#""4NH""_3(l) + 5""O""_2(g) -> 4""NO""(g) + 6""H""_2""O""(l)#

+
+
+
+

Explanation:

+
+

Given unbalanced equation is

+

#""NH""_3(l) + ""O""_2(g) -> ""NO""(g) + ""H""_2""O""(l)#
+As a thumb rule #""H"" and ""O""# are left for the last. As we see that other than these two atoms there is only one atom, #""N""#. That atom is already balanced.

+

We start with #""H"" and ""O""#. We observe that out of these two, hydrogen has more number of atoms in the equation. So we take it first.

+

Let the balanced equation be: (notice that number of N atoms have been kept balanced)
+#x""NH""_3(l) + y""O""_2(g) -> x""NO""(g) + b""H""_2""O""(l)#

+

To balance hydrogen atoms we multiply ammonia molecule with 2 (set #x=2#) on reactants side and water molecule by 3 (set #b=3#) on the product side. Simultaneously multiplying nitric oxide molecule on the product side with 2 We have already set #x=2#.
+The equation looks like

+

#""2NH""_3(l) + y""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#

+

For the remaining unbalanced element:
+Total number of oxygen atoms on products side becomes #2+3=5#. Which gives us number of oxygen molecules as #y=5/2# on the reactants side.

+

#""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l)#
+ We know that #y# needs to be whole number. Therefore multiply the equation with 2 to obtain balanced equation.

+

#2xx(""2NH""_3(l) + 5/2""O""_2(g) -> 2""NO""(g) + 3""H""_2""O""(l))#
+#""4NH""_3(l) + 5""O""_2(g) -> 4""NO""(g) + 6""H""_2""O""(l)#

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+
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" How do you balance #NH_3(l) + O_2(g) -> NO(g) + H_2O(l)#? nan +414 a8366e68-6ddd-11ea-a8a2-ccda262736ce https://socratic.org/questions/how-many-c-atoms-are-there-in-0-502-mole-of-c 3.02 × 10^23 start physical_unit 2 3 number none qc_end physical_unit 2 2 7 8 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] C atoms""}]" "[{""type"":""physical unit"",""value"":""3.02 × 10^23""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] C [=] \\pu{0.502 mole}""}]" "

How many #C# atoms are there in 0.502 mole of #C#?

" nan 3.02 × 10^23 "
+

Explanation:

+
+

The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to #6.0221409xx 10^23#, so there are approximately #6.0221409xx 10^23# atoms in a single mole of a substance.

+

To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:

+

Number of #C# atoms #= 5.02 xx 6.0221409xx 10^23 = 3.0231147xx10^24#

+
+
" "
+
+
+

#3.0231147xx10^24#

+
+
+
+

Explanation:

+
+

The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to #6.0221409xx 10^23#, so there are approximately #6.0221409xx 10^23# atoms in a single mole of a substance.

+

To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:

+

Number of #C# atoms #= 5.02 xx 6.0221409xx 10^23 = 3.0231147xx10^24#

+
+
+
" "
+

How many #C# atoms are there in 0.502 mole of #C#?

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+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
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+1 Answer +
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+ + Nov 9, 2015 + +
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#3.0231147xx10^24#

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+
+
+

Explanation:

+
+

The number of atoms in a single mole of any substance is defined by Avogadro's Number, a constant which is equal to #6.0221409xx 10^23#, so there are approximately #6.0221409xx 10^23# atoms in a single mole of a substance.

+

To find the number of atoms in a given amount of substance, simply multiply the amount in moles of the substance by Avogadro's Number:

+

Number of #C# atoms #= 5.02 xx 6.0221409xx 10^23 = 3.0231147xx10^24#

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+
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+ + +
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+
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+ + Creative Commons License + +
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" How many #C# atoms are there in 0.502 mole of #C#? nan +415 a8366e69-6ddd-11ea-9578-ccda262736ce https://socratic.org/questions/how-would-you-balance-h2-o2-h2o-2 2 H2 + O2 -> 2 H2O start chemical_equation qc_end chemical_equation 4 8 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""2 H2 + O2 -> 2 H2O""}]" "[{""type"":""chemical equation"",""value"":""H2 + O2 -> H2O""}]" "

How would you balance: H2 + O2 --> H2O? +

" nan 2 H2 + O2 -> 2 H2O "
+

Explanation:

+
+

#H_2+O_2rarrH_2O#

+

Notice the imbalance of atoms. On the left side, the reactant side, there are #2# hydrogen atoms and #2# oxygen atoms. On the right side, the product side, there are #2# hydrogen atoms and only #1# oxygen atom.

+

An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change #H_2O# to #3H_2O# but we can't go from #H_2O# to #H_3O#.

+

In order to deal with the original imbalance, the #2# oxygens on the left and #1# on the right, we can change the coefficient of the #H_2O# molecule from #1# to #2#.

+

#H_2+O_2rarr2H_2O#

+

Now we have the same amount of oxygens on each side, #2#, but we have an unequal amount of hydrogen—there are #2# on the left and #4# on the right. We should achieve the least common multiple of the existing numbers, which is #4#. We can make the #H_2# on the left have a coefficient of #2#, giving us the #4# hydrogens that we need on that side of the equation, to match the #4# on the right.

+

#2H_2+O_2rarr2H_2O#

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+
" "
+
+
+

#2H_2+O_2rarr2H_2O#

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+
+
+

Explanation:

+
+

#H_2+O_2rarrH_2O#

+

Notice the imbalance of atoms. On the left side, the reactant side, there are #2# hydrogen atoms and #2# oxygen atoms. On the right side, the product side, there are #2# hydrogen atoms and only #1# oxygen atom.

+

An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change #H_2O# to #3H_2O# but we can't go from #H_2O# to #H_3O#.

+

In order to deal with the original imbalance, the #2# oxygens on the left and #1# on the right, we can change the coefficient of the #H_2O# molecule from #1# to #2#.

+

#H_2+O_2rarr2H_2O#

+

Now we have the same amount of oxygens on each side, #2#, but we have an unequal amount of hydrogen—there are #2# on the left and #4# on the right. We should achieve the least common multiple of the existing numbers, which is #4#. We can make the #H_2# on the left have a coefficient of #2#, giving us the #4# hydrogens that we need on that side of the equation, to match the #4# on the right.

+

#2H_2+O_2rarr2H_2O#

+
+
+
" "
+

How would you balance: H2 + O2 --> H2O? +

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 6, 2015 + +
+
+
+
+
+
+
+

#2H_2+O_2rarr2H_2O#

+
+
+
+

Explanation:

+
+

#H_2+O_2rarrH_2O#

+

Notice the imbalance of atoms. On the left side, the reactant side, there are #2# hydrogen atoms and #2# oxygen atoms. On the right side, the product side, there are #2# hydrogen atoms and only #1# oxygen atom.

+

An important thing to remember when balancing equations is that the molecules themselves may not be changed—only their coefficients can. For example, we can change #H_2O# to #3H_2O# but we can't go from #H_2O# to #H_3O#.

+

In order to deal with the original imbalance, the #2# oxygens on the left and #1# on the right, we can change the coefficient of the #H_2O# molecule from #1# to #2#.

+

#H_2+O_2rarr2H_2O#

+

Now we have the same amount of oxygens on each side, #2#, but we have an unequal amount of hydrogen—there are #2# on the left and #4# on the right. We should achieve the least common multiple of the existing numbers, which is #4#. We can make the #H_2# on the left have a coefficient of #2#, giving us the #4# hydrogens that we need on that side of the equation, to match the #4# on the right.

+

#2H_2+O_2rarr2H_2O#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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" "How would you balance: H2 + O2 --> H2O? +" nan +416 a8366e6a-6ddd-11ea-bdad-ccda262736ce https://socratic.org/questions/a-metallic-element-x-forms-a-carbonate-with-the-formula-x-co-3-2-what-would-the- XF4 start chemical_formula qc_end c_other OTHER qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] fluoride compound of element X [IN] default""}]" "[{""type"":""chemical equation"",""value"":""XF4""}]" "[{""type"":""other"",""value"":""A metallic element X forms a carbonate with the formula X(CO3)2.""}]" "

A metallic element #X# forms a carbonate with the formula #X(CO_3)_2#. What would the corresponding formula be of the fluoride compound of element #X#?

" nan XF4 "
+

Explanation:

+
+

The starting species is #X(CO_3)_2#. And thus, separating the ions, we get #X^(4+)# and #2xxCO_3^(2-)#.

+

Fluorine is an excellent oxidant and gives #F^-# ions upon reduction, so to preserve electrical neutrality we get #XF_4#.

+
+
" "
+
+
+

#XF_4#

+
+
+
+

Explanation:

+
+

The starting species is #X(CO_3)_2#. And thus, separating the ions, we get #X^(4+)# and #2xxCO_3^(2-)#.

+

Fluorine is an excellent oxidant and gives #F^-# ions upon reduction, so to preserve electrical neutrality we get #XF_4#.

+
+
+
" "
+

A metallic element #X# forms a carbonate with the formula #X(CO_3)_2#. What would the corresponding formula be of the fluoride compound of element #X#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Determining Formula + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 5, 2017 + +
+
+
+
+
+
+
+

#XF_4#

+
+
+
+

Explanation:

+
+

The starting species is #X(CO_3)_2#. And thus, separating the ions, we get #X^(4+)# and #2xxCO_3^(2-)#.

+

Fluorine is an excellent oxidant and gives #F^-# ions upon reduction, so to preserve electrical neutrality we get #XF_4#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3370 views + around the world +
+
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+ +
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+ + Creative Commons License + +
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+
+
" A metallic element #X# forms a carbonate with the formula #X(CO_3)_2#. What would the corresponding formula be of the fluoride compound of element #X#? nan +417 a836955a-6ddd-11ea-8872-ccda262736ce https://socratic.org/questions/in-a-0-70-m-solution-of-benzoic-acid-what-percentage-of-the-molecules-are-ionize 87.29% start physical_unit 12 14 percent none qc_end physical_unit 4 7 2 3 molarity qc_end chemical_equation 49 53 qc_end c_other OTHER qc_end physical_unit 57 58 60 60 pka qc_end end "[{""type"":""physical unit"",""value"":""Percentage [OF] molecules are ionized""}]" "[{""type"":""physical unit"",""value"":""87.29%""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] solution of benzoic acid [=] \\pu{0.70 M}""},{""type"":""chemical equation"",""value"":""C6H5COOH <=> C6H5COO− + H+""},{""type"":""other"",""value"":""Benzoic acid is a weak acid that has antimicrobial properties.""},{""type"":""physical unit"",""value"":""pKa [OF] this reaction [=] \\pu{4.2}""}]" "

In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized (CHEMISTRY 1B problem)?

" "
+
+

+

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:
+#C_6H_5COOH⇌C_6H_5COO^−+H^+#
+The #pKa# of this reaction is #4.2#. In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized?
+Express the percentage numerically using two significant figures.

+
+
    +
  • Work
    +
    +It's incomplete, but I fulfilled the formula, and got (below)...
    • +
  • Answer result
    +
    • +
+

+
+
" 87.29% "
+

Explanation:

+
+

#8.5\times10^(-2)# M NaOCl

+

#K_a(HClO)=2.9\times10^(-8)#
+#K_b(ClO^-)=(1.0\times10^(-14))/(2.9\times10^(-8))#

+
+
" "
+
+
+

#pH=10.23#

+

( explanation under construction by question owner )

+
+
+
+

Explanation:

+
+

#8.5\times10^(-2)# M NaOCl

+

#K_a(HClO)=2.9\times10^(-8)#
+#K_b(ClO^-)=(1.0\times10^(-14))/(2.9\times10^(-8))#

+
+
+
" "
+

In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized (CHEMISTRY 1B problem)?

+
+
+

+

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:
+#C_6H_5COOH⇌C_6H_5COO^−+H^+#
+The #pKa# of this reaction is #4.2#. In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized?
+Express the percentage numerically using two significant figures.

+
+
    +
  • Work
    +
    +It's incomplete, but I fulfilled the formula, and got (below)...
    • +
  • Answer result
    +
    • +
+

+
+
+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 14, 2017 + +
+
+
+
+
+
+
+

#pH=10.23#

+

( explanation under construction by question owner )

+
+
+
+

Explanation:

+
+

#8.5\times10^(-2)# M NaOCl

+

#K_a(HClO)=2.9\times10^(-8)#
+#K_b(ClO^-)=(1.0\times10^(-14))/(2.9\times10^(-8))#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 8589 views + around the world +
+
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+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
+
" In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized (CHEMISTRY 1B problem)? " + + +Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water: +#C_6H_5COOH⇌C_6H_5COO^−+H^+# +The #pKa# of this reaction is #4.2#. In a #0.70 M# solution of benzoic acid, what percentage of the molecules are ionized? +Express the percentage numerically using two significant figures. + + +Work + +It's incomplete, but I fulfilled the formula, and got (below)... +Answer result + + + + +" +418 a836955b-6ddd-11ea-81d9-ccda262736ce https://socratic.org/questions/how-much-heat-in-calories-is-given-off-when-1-25-grams-of-silver-is-cooled-from- 1.43 calories start physical_unit 12 12 heat_energy cal qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 16 17 temperature qc_end physical_unit 12 12 19 20 temperature qc_end physical_unit 12 12 27 30 specific_heat qc_end end "[{""type"":""physical unit"",""value"":""Heat given off [OF] silver [IN] calories""}]" "[{""type"":""physical unit"",""value"":""1.43 calories""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] silver [=] \\pu{1.25 grams}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] silver [=] \\pu{100.0 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] silver [=] \\pu{80.0 ℃}""},{""type"":""physical unit"",""value"":""Specific heat [OF] silver [=] \\pu{0.057 cal/(g * ℃)}""}]" "

How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 C to 80.0 C? (The specific heat of silver is 0.057 cal/g C)?

" nan 1.43 calories "
+

Explanation:

+
+

The heat is

+

#Q=m*s*DeltaT#

+

mass, #m=1.25g#

+

specific heat, #s=0.057calg^-1ºC^-1#

+

#Delta T=100-80=20ºC#

+

Heat liberated is #=1.25*0.057*20=1.425cal#

+
+
" "
+
+
+

The heat is #=1.425cal#

+
+
+
+

Explanation:

+
+

The heat is

+

#Q=m*s*DeltaT#

+

mass, #m=1.25g#

+

specific heat, #s=0.057calg^-1ºC^-1#

+

#Delta T=100-80=20ºC#

+

Heat liberated is #=1.25*0.057*20=1.425cal#

+
+
+
" "
+

How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 C to 80.0 C? (The specific heat of silver is 0.057 cal/g C)?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 14, 2017 + +
+
+
+
+
+
+
+

The heat is #=1.425cal#

+
+
+
+

Explanation:

+
+

The heat is

+

#Q=m*s*DeltaT#

+

mass, #m=1.25g#

+

specific heat, #s=0.057calg^-1ºC^-1#

+

#Delta T=100-80=20ºC#

+

Heat liberated is #=1.25*0.057*20=1.425cal#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2555 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 C to 80.0 C? (The specific heat of silver is 0.057 cal/g C)? nan +419 a836955c-6ddd-11ea-b6af-ccda262736ce https://socratic.org/questions/what-volume-in-liters-does-2-895-moles-of-oxygen-occupy-at-stp 64.89 liters start physical_unit 8 8 volume l qc_end physical_unit 8 8 5 6 mole qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] oxygen [IN] liters""}]" "[{""type"":""physical unit"",""value"":""64.89 liters""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] oxygen [=] \\pu{2.895 moles}""},{""type"":""other"",""value"":""STP""}]" "

What volume (in liters) does 2.895 moles of oxygen occupy at STP?

" nan 64.89 liters "
+

Explanation:

+
+

Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:

+

+

I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.

+

List your known and unknown variables. Our only unknown is the volume of #O_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm.

+

Now we have to rearrange the equation to solve for V:

+

# (nxxRxxT)/P#

+

#V = (2.895cancel""mol""xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#

+

#V = 64.9 L#

+
+
" "
+
+
+

At STP, 2.895 moles of #O_2# occupies a volume of 64.9 L.

+
+
+
+

Explanation:

+
+

Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:

+

+

I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.

+

List your known and unknown variables. Our only unknown is the volume of #O_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm.

+

Now we have to rearrange the equation to solve for V:

+

# (nxxRxxT)/P#

+

#V = (2.895cancel""mol""xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#

+

#V = 64.9 L#

+
+
+
" "
+

What volume (in liters) does 2.895 moles of oxygen occupy at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 2, 2016 + +
+
+
+
+
+
+
+

At STP, 2.895 moles of #O_2# occupies a volume of 64.9 L.

+
+
+
+

Explanation:

+
+

Since we are at STP and are given only one set of conditions, we have to use the ideal gas law equation:

+

+

I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.

+

List your known and unknown variables. Our only unknown is the volume of #O_2(g)#. Our known variables are P,n,R, and T.

+

At STP, the temperature is 273K and the pressure is 1 atm.

+

Now we have to rearrange the equation to solve for V:

+

# (nxxRxxT)/P#

+

#V = (2.895cancel""mol""xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#

+

#V = 64.9 L#

+
+
+
+
+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" What volume (in liters) does 2.895 moles of oxygen occupy at STP? nan +420 a836955d-6ddd-11ea-b696-ccda262736ce https://socratic.org/questions/what-is-the-molarity-of-a-2-0-l-sodium-hydroxide-solution-containing-10-0-grams- 0.13 mol/L start physical_unit 8 10 molarity mol/l qc_end physical_unit 8 10 6 7 volume qc_end physical_unit 15 15 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] sodium hydroxide solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.13 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] sodium hydroxide solution [=] \\pu{2.0 L}""},{""type"":""physical unit"",""value"":""Mass [OF] solute [=] \\pu{10.0 grams}""}]" "

What is the molarity of a 2.0 L sodium hydroxide solution containing 10.0 grams of solute?

" nan 0.13 mol/L "
+

Explanation:

+
+

We take the quotient, #((10.0*g)/(40.00*g*mol^-1))/(2.0*L)=0.125*mol*L^-1#.

+

Note the dimensional consistency of the answer....#1/(mol^-1)=1/(1/(mol))=mol#.

+

What are the concentrations with respect to #Na^+# and #HO^-# ions?

+

What is the pH of this solution?

+
+
" "
+
+
+

#""Molarity""=0.125*mol*L^-1#

+
+
+
+

Explanation:

+
+

We take the quotient, #((10.0*g)/(40.00*g*mol^-1))/(2.0*L)=0.125*mol*L^-1#.

+

Note the dimensional consistency of the answer....#1/(mol^-1)=1/(1/(mol))=mol#.

+

What are the concentrations with respect to #Na^+# and #HO^-# ions?

+

What is the pH of this solution?

+
+
+
" "
+

What is the molarity of a 2.0 L sodium hydroxide solution containing 10.0 grams of solute?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 11, 2017 + +
+
+
+
+
+
+
+

#""Molarity""=0.125*mol*L^-1#

+
+
+
+

Explanation:

+
+

We take the quotient, #((10.0*g)/(40.00*g*mol^-1))/(2.0*L)=0.125*mol*L^-1#.

+

Note the dimensional consistency of the answer....#1/(mol^-1)=1/(1/(mol))=mol#.

+

What are the concentrations with respect to #Na^+# and #HO^-# ions?

+

What is the pH of this solution?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 4036 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" What is the molarity of a 2.0 L sodium hydroxide solution containing 10.0 grams of solute? nan +421 a836955e-6ddd-11ea-be32-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-solution-with-h-2-3-x-10-6 5.64 start physical_unit 5 6 ph none qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] a solution""}]" "[{""type"":""physical unit"",""value"":""5.64""}]" "[{""type"":""other"",""value"":""[H+] = 2.3 × 10^(−6)""}]" "

What is the pH of a solution with [H+]= #2.3 x 10^-6#?

" nan 5.64 "
+

Explanation:

+
+

#pH=-log_10[H_3O^+]#

+

#=-log_(10){2.3xx10^-6}#

+

#=-{-5.64}#

+

#=5.64#

+

If you are not confident with the logarithmic function please voice your concern. See here for the relationship between #pH# and #pOH# in aqueous solution.

+
+
" "
+
+
+

#pH=5.64#

+
+
+
+

Explanation:

+
+

#pH=-log_10[H_3O^+]#

+

#=-log_(10){2.3xx10^-6}#

+

#=-{-5.64}#

+

#=5.64#

+

If you are not confident with the logarithmic function please voice your concern. See here for the relationship between #pH# and #pOH# in aqueous solution.

+
+
+
" "
+

What is the pH of a solution with [H+]= #2.3 x 10^-6#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 8, 2016 + +
+
+
+
+
+
+
+

#pH=5.64#

+
+
+
+

Explanation:

+
+

#pH=-log_10[H_3O^+]#

+

#=-log_(10){2.3xx10^-6}#

+

#=-{-5.64}#

+

#=5.64#

+

If you are not confident with the logarithmic function please voice your concern. See here for the relationship between #pH# and #pOH# in aqueous solution.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 10435 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
+
" What is the pH of a solution with [H+]= #2.3 x 10^-6#? nan +422 a836955f-6ddd-11ea-868e-ccda262736ce https://socratic.org/questions/how-is-the-henderson-hasselbalch-equation-used-to-calculate-the-ratio-of-h-2co-3 1 : 10 start physical_unit 11 15 ratio none qc_end physical_unit 15 15 20 20 ph qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Ratio [OF] H2CO3 to HCO3^- in blood""}]" "[{""type"":""physical unit"",""value"":""1 : 10""}]" "[{""type"":""physical unit"",""value"":""pH [OF] blood [=] \\pu{7.40}""},{""type"":""other"",""value"":""Henderson-Hasselbalch equation""}]" "

How is the Henderson-Hasselbalch equation used to calculate the ratio of #H_2CO_3# to #HCO_3^-# in blood having a pH of 7.40?

" nan 1 : 10 "
+

Explanation:

+
+

#pH = pKa + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#Ka_1(H_2CO_3) = 4.5x10^(-7)#
+=> #pKa = -logK_a# = #-log(4.5x10^(-7))# =#6.4#

+

Given #pH = 7.4#

+

Substitute into HH Equation and solve for #([[HCO_3^-]]/[[H_2CO_3]])#

+

#7.4 = 6.4 + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#log_10([[HCO_3^-]]/[[H_2CO_3]]) = 7.4 - 6.4 = 1.0#

+

#([[HCO_3^-]]/[[H_2CO_3]]) = 10^1.0 = 10#

+

=> #([HCO_3^-]:[H_2CO_3]) = 10:1#

+
+
" "
+
+
+

#([HCO_3^-]:[H_2CO_3]) = 10:1#

+
+
+
+

Explanation:

+
+

#pH = pKa + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#Ka_1(H_2CO_3) = 4.5x10^(-7)#
+=> #pKa = -logK_a# = #-log(4.5x10^(-7))# =#6.4#

+

Given #pH = 7.4#

+

Substitute into HH Equation and solve for #([[HCO_3^-]]/[[H_2CO_3]])#

+

#7.4 = 6.4 + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#log_10([[HCO_3^-]]/[[H_2CO_3]]) = 7.4 - 6.4 = 1.0#

+

#([[HCO_3^-]]/[[H_2CO_3]]) = 10^1.0 = 10#

+

=> #([HCO_3^-]:[H_2CO_3]) = 10:1#

+
+
+
" "
+

How is the Henderson-Hasselbalch equation used to calculate the ratio of #H_2CO_3# to #HCO_3^-# in blood having a pH of 7.40?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Acids and Bases + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 29, 2017 + +
+
+
+
+
+
+
+

#([HCO_3^-]:[H_2CO_3]) = 10:1#

+
+
+
+

Explanation:

+
+

#pH = pKa + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#Ka_1(H_2CO_3) = 4.5x10^(-7)#
+=> #pKa = -logK_a# = #-log(4.5x10^(-7))# =#6.4#

+

Given #pH = 7.4#

+

Substitute into HH Equation and solve for #([[HCO_3^-]]/[[H_2CO_3]])#

+

#7.4 = 6.4 + log_10([[HCO_3^-]]/[[H_2CO_3]])#

+

#log_10([[HCO_3^-]]/[[H_2CO_3]]) = 7.4 - 6.4 = 1.0#

+

#([[HCO_3^-]]/[[H_2CO_3]]) = 10^1.0 = 10#

+

=> #([HCO_3^-]:[H_2CO_3]) = 10:1#

+
+
+
+
+
+ +
+
+
+
+
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" How is the Henderson-Hasselbalch equation used to calculate the ratio of #H_2CO_3# to #HCO_3^-# in blood having a pH of 7.40? nan +423 a8369560-6ddd-11ea-923f-ccda262736ce https://socratic.org/questions/59bae82db72cff7c58c4df24 Co2O3 start chemical_formula qc_end physical_unit 17 17 13 14 mass qc_end physical_unit 24 24 20 21 mass qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] a oxide of cobalt [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""Co2O3""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] metal [=] \\pu{1.216 g}""},{""type"":""physical unit"",""value"":""Mass [OF] oxygen [=] \\pu{0.495 g}""}]" "

What is the empirical formula of a oxide of cobalt that contains a #1.216*g# mass of metal, and a #0.495*g# mass of oxygen?

" nan Co2O3 "
+

Explanation:

+
+

We interrogate the molar quantities of metal and oxygen.....

+

#""Moles of cobalt""=(1.216*g)/(58.9*g*mol^-1)=0.0207*mol.#

+

#""Moles of oxygen""=(0.495*g)/(16.0*g*mol^-1)=0.0309*mol.#

+

We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....

+

#Co_((0.0207*mol)/(0.0207*mol))O_((0.0309*mol)/(0.0207*mol))=CoO_(1.49)#....

+

But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....

+

#Co_2O_3#.....

+

This is not a good question inasmuch as #Co_2O_3# is poorly characterized and possibly unknown. There are #CoO#, and #Co_3O_4#, a mixed cobalt oxide of #CoO#, and #Co_2O_3#....The person who set the question was not an inorganic chemist.

+
+
" "
+
+
+

#Co_2O_3#......#""cobaltic oxide""#

+
+
+
+

Explanation:

+
+

We interrogate the molar quantities of metal and oxygen.....

+

#""Moles of cobalt""=(1.216*g)/(58.9*g*mol^-1)=0.0207*mol.#

+

#""Moles of oxygen""=(0.495*g)/(16.0*g*mol^-1)=0.0309*mol.#

+

We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....

+

#Co_((0.0207*mol)/(0.0207*mol))O_((0.0309*mol)/(0.0207*mol))=CoO_(1.49)#....

+

But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....

+

#Co_2O_3#.....

+

This is not a good question inasmuch as #Co_2O_3# is poorly characterized and possibly unknown. There are #CoO#, and #Co_3O_4#, a mixed cobalt oxide of #CoO#, and #Co_2O_3#....The person who set the question was not an inorganic chemist.

+
+
+
" "
+

What is the empirical formula of a oxide of cobalt that contains a #1.216*g# mass of metal, and a #0.495*g# mass of oxygen?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
+
+
+
+
+2 Answers +
+
+
+
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+
+ +
+
+ +
+ + Sep 18, 2017 + +
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+
+

#Co_2O_3#......#""cobaltic oxide""#

+
+
+
+

Explanation:

+
+

We interrogate the molar quantities of metal and oxygen.....

+

#""Moles of cobalt""=(1.216*g)/(58.9*g*mol^-1)=0.0207*mol.#

+

#""Moles of oxygen""=(0.495*g)/(16.0*g*mol^-1)=0.0309*mol.#

+

We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....

+

#Co_((0.0207*mol)/(0.0207*mol))O_((0.0309*mol)/(0.0207*mol))=CoO_(1.49)#....

+

But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....

+

#Co_2O_3#.....

+

This is not a good question inasmuch as #Co_2O_3# is poorly characterized and possibly unknown. There are #CoO#, and #Co_3O_4#, a mixed cobalt oxide of #CoO#, and #Co_2O_3#....The person who set the question was not an inorganic chemist.

+
+
+
+
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+ +
+
+
+
+
+ +
+
+ +
+ + Sep 18, 2017 + +
+
+
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+
+
+
+

#Co_2O_3#

+
+
+
+

Explanation:

+
+

Assuming complete reaction of both, convert the masses into moles and normalize. I’ll use a singlet oxygen because we don’t know the ratio yet, even though the actual gas would be diatomic.

+

#(1.216g)/(58.9(g/""mol"") Co) = 0.0206# mole #Co# ; #(0.495g)/(16(g/""mol"" O)) = 0.031# mole O

+

#Co_0.0206O_0.031# ; or #""Co""/O = 0.0206/0.031 = 2/3#
+#Co_2O_3#

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" What is the empirical formula of a oxide of cobalt that contains a #1.216*g# mass of metal, and a #0.495*g# mass of oxygen? nan +424 a8369561-6ddd-11ea-a763-ccda262736ce https://socratic.org/questions/54334fc2581e2a091ddff85f 71.66% start physical_unit 25 28 percent_composition none qc_end physical_unit 0 4 9 10 mass qc_end physical_unit 18 18 16 17 mass qc_end end "[{""type"":""physical unit"",""value"":""Percent composition [OF] aluminum in the sample""}]" "[{""type"":""physical unit"",""value"":""71.66%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] A sample of aluminum oxide [=] \\pu{6.67 g}""},{""type"":""physical unit"",""value"":""Mass [OF] Al [=] \\pu{4.78 g}""}]" "

A sample of aluminum oxide has a mass of #""6.67 g""#. It was found to contain #""4.78 g Al""#. What is the percent composition of aluminum in the sample?

" nan 71.66% "
+

Explanation:

+
+

To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.

+

#(4.78color(red)cancel(color(black)(""g"")))/(6.67color(red)cancel(color(black)(""g"")))xx100 = 71.7%""#

+

The sample of aluminum oxide contains 71.7% aluminum.

+
+
" "
+
+
+

The sample of aluminum oxide contains 71.7% aluminum.

+
+
+
+

Explanation:

+
+

To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.

+

#(4.78color(red)cancel(color(black)(""g"")))/(6.67color(red)cancel(color(black)(""g"")))xx100 = 71.7%""#

+

The sample of aluminum oxide contains 71.7% aluminum.

+
+
+
" "
+

A sample of aluminum oxide has a mass of #""6.67 g""#. It was found to contain #""4.78 g Al""#. What is the percent composition of aluminum in the sample?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Oct 7, 2014 + +
+
+
+
+
+
+
+

The sample of aluminum oxide contains 71.7% aluminum.

+
+
+
+

Explanation:

+
+

To calculate the mass percent composition of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100.

+

#(4.78color(red)cancel(color(black)(""g"")))/(6.67color(red)cancel(color(black)(""g"")))xx100 = 71.7%""#

+

The sample of aluminum oxide contains 71.7% aluminum.

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+
+
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" "A sample of aluminum oxide has a mass of #""6.67 g""#. It was found to contain #""4.78 g Al""#. What is the percent composition of aluminum in the sample?" nan +425 a8369562-6ddd-11ea-b9ec-ccda262736ce https://socratic.org/questions/given-the-value-of-ksp-2-9x10-12-what-would-be-the-solubility-of-mg-oh-2-in-a-0- 2.44 x 10^(-11) M start physical_unit 14 14 solubility mol/l qc_end physical_unit 19 20 17 18 molarity qc_end physical_unit 14 14 5 7 equilibrium_constant_k qc_end end "[{""type"":""physical unit"",""value"":""Solubility [OF] Mg(OH)2 [IN] M""}]" "[{""type"":""physical unit"",""value"":""2.44 x 10^(-11) M""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH solution [=] \\pu{0.345 M}""},{""type"":""physical unit"",""value"":""Ksp [OF] Mg(OH)2 [=] \\pu{2.9 x 10^(-12)}""}]" "

Given the value of #K_(sp) = 2.9 xx 10^(-12)#, what would be the solubility of #""Mg""(""OH"")_2# in a 0.345 M #""NaOH""# solution?

" "
+
+

+

Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution?

+

+
+
" 2.44 x 10^(-11) M "
+

Explanation:

+
+

Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide, #""Mg""(""OH"")_2#, in a solution that contains the soluble sodium hydroxide, #""NaOH""#.

+

Sodium hydroxide dissociates in a #1:1# mole ratio to form hydroxide anions

+
+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#

+
+

Set up your ICE table

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0.345)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaacolor(black)(0.345+color(red)(2)s))#

+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

In your case, this will be equal to

+
+

#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#

+
+

Rearrange to get

+
+

#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#

+
+

This cubic equation will produce one real solution

+
+

#s = 2.44 * 10^(-11)#

+
+

This represents the molar solubility of magnesium hydroxide in a solution that contains #""0.345 M""# hydroxide anions. You can thus say that you have

+
+

#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs.

+

SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.

+

In this case, you would have

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(2)s))#

+

This time, the solubility product constant would be

+
+

#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#

+
+

The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.

+
+
" "
+
+
+

#2.44 * 10^(-11)""M""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide, #""Mg""(""OH"")_2#, in a solution that contains the soluble sodium hydroxide, #""NaOH""#.

+

Sodium hydroxide dissociates in a #1:1# mole ratio to form hydroxide anions

+
+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#

+
+

Set up your ICE table

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0.345)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaacolor(black)(0.345+color(red)(2)s))#

+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

In your case, this will be equal to

+
+

#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#

+
+

Rearrange to get

+
+

#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#

+
+

This cubic equation will produce one real solution

+
+

#s = 2.44 * 10^(-11)#

+
+

This represents the molar solubility of magnesium hydroxide in a solution that contains #""0.345 M""# hydroxide anions. You can thus say that you have

+
+

#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs.

+

SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.

+

In this case, you would have

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(2)s))#

+

This time, the solubility product constant would be

+
+

#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#

+
+

The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.

+
+
+
" "
+

Given the value of #K_(sp) = 2.9 xx 10^(-12)#, what would be the solubility of #""Mg""(""OH"")_2# in a 0.345 M #""NaOH""# solution?

+
+
+

+

Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution?

+

+
+
+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Solubility Equilbria + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 3, 2016 + +
+
+
+
+
+
+
+

#2.44 * 10^(-11)""M""#

+
+
+
+

Explanation:

+
+

Your strategy here will be to use an ICE table to find the solubility of magnesium hydroxide, #""Mg""(""OH"")_2#, in a solution that contains the soluble sodium hydroxide, #""NaOH""#.

+

Sodium hydroxide dissociates in a #1:1# mole ratio to form hydroxide anions

+
+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+
+

Since every mole of sodium hydroxide produces one mole of hydroxide anions, your solution will contain

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.345 M""#

+
+

Set up your ICE table

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0.345)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaacolor(black)(0.345+color(red)(2)s))#

+

By definition, the solubility product constant, #K_(sp)#, will be equal to

+
+

#K_(sp) = [""Mg""^(2+)] * [""OH""^(-)]^color(red)(2)#

+
+

In your case, this will be equal to

+
+

#2.9 * 10^(-12) = s * (0.345 + color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = s * (0.119025 + 1.38s + 4s^2)#

+
+

Rearrange to get

+
+

#4s^3 + 1.38s^2 + 0.119025s - 2.9 * 10^(-12) = 0#

+
+

This cubic equation will produce one real solution

+
+

#s = 2.44 * 10^(-11)#

+
+

This represents the molar solubility of magnesium hydroxide in a solution that contains #""0.345 M""# hydroxide anions. You can thus say that you have

+
+

#""molar solubility""_(""in 0.345M OH""^(-)) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.44 * 10^(-11)""M"")color(white)(a/a)|)))#

+
+

I'll leave the answer rounded to three sig figs.

+

SIDE NOTE You should be able to predict that the solubility of magnesium hydroxide in this solution is significantly lower than the solubility of the salt in pure water.

+

In this case, you would have

+
+

#"" "" ""Mg""(""OH"")_ (color(red)(2)(s)) "" ""rightleftharpoons "" "" ""Mg""_ ((aq))^(2+) "" ""+"" "" color(red)(2)""OH""_ ((aq))^(-)#

+
+

#color(purple)(""I"")color(white)(aaaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)#
+#color(purple)(""C"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)((+s))aaaaaaaacolor(black)((+color(red)(2)s))#
+#color(purple)(""E"")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(color(red)(2)s))#

+

This time, the solubility product constant would be

+
+

#K_(sp) = s * (color(red)(2)s)^color(red)(2)#

+

#2.9 * 10^(-12) = 4s^3 implies s = root(3)( (2.9 * 10^(-12))/4) = 9.0 * 10^(-5)#

+
+

The solubility of the salt decreases in a solution that contains hydroxide anions because of the common-ion effect.

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+
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" "Given the value of #K_(sp) = 2.9 xx 10^(-12)#, what would be the solubility of #""Mg""(""OH"")_2# in a 0.345 M #""NaOH""# solution?" " + + +Given the value of Ksp 2.9x10^-12, what would be the solubility of Mg(OH)2 in a 0.345 M NaOH solution? + + +" +426 a8369563-6ddd-11ea-8114-ccda262736ce https://socratic.org/questions/what-is-the-formula-for-the-compound-made-from-beryllium-and-and-bromine BeBr2 start chemical_formula qc_end substance 9 9 qc_end substance 11 11 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] default""}]" "[{""type"":""chemical equation"",""value"":""BeBr2""}]" "[{""type"":""substance name"",""value"":""Beryllium""},{""type"":""substance name"",""value"":""Bromine""}]" "

What is the formula for the compound made from beryllium and and bromine?

" nan BeBr2 "
+

Explanation:

+
+

#Be(s) + Br_2(l) rarr BeBr_2(s)#

+

Is the synthesis a redox reaction? Why or why not?

+
+
" "
+
+
+

#""Beryllium bromide""#, #BeBr_2#.

+
+
+
+

Explanation:

+
+

#Be(s) + Br_2(l) rarr BeBr_2(s)#

+

Is the synthesis a redox reaction? Why or why not?

+
+
+
" "
+

What is the formula for the compound made from beryllium and and bromine?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Naming Ionic Compounds + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 27, 2017 + +
+
+
+
+
+
+
+

#""Beryllium bromide""#, #BeBr_2#.

+
+
+
+

Explanation:

+
+

#Be(s) + Br_2(l) rarr BeBr_2(s)#

+

Is the synthesis a redox reaction? Why or why not?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
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+ + Creative Commons License + +
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" What is the formula for the compound made from beryllium and and bromine? nan +427 a836bc6e-6ddd-11ea-8778-ccda262736ce https://socratic.org/questions/how-many-moles-of-zn-will-be-needed-to-completely-react-with-0-4-moles-of-hcl-in 0.20 moles start physical_unit 4 4 mole mol qc_end chemical_equation 19 25 qc_end c_other OTHER qc_end physical_unit 15 15 12 13 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] Zn [IN] moles""}]" "[{""type"":""physical unit"",""value"":""0.20 moles""}]" "[{""type"":""chemical equation"",""value"":""Zn + HCl -> ZnCl2 + H2""},{""type"":""other"",""value"":""Completely react.""},{""type"":""physical unit"",""value"":""Mole [OF] HCl [=] \\pu{0.4 moles}""}]" "

How many moles of #Zn# will be needed to completely react with 0.4 moles of #HCl# in the reaction #Zn + HCl -> ZnCl_2 + H_2#?

" nan 0.20 moles "
+

Explanation:

+
+

The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.

+

You started with #0.4# #mol# of #HCl(aq)#; I could have started with 4 atoms, or 40 atoms, or 40 moles. By the stoichiometry of the reaction, each zinc particle requires 2 hydrochloric acid particles (i.e. each equivalent of zinc metal requires 2 hydrochloric acid particles). Thus #0.2# mol of zinc metal were required.

+

Capisce?

+
+
" "
+
+
+

#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#.

+

#0.2# moles of zinc metal are required for reaction with 0.4 moles of hydrochloric acid.

+
+
+
+

Explanation:

+
+

The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.

+

You started with #0.4# #mol# of #HCl(aq)#; I could have started with 4 atoms, or 40 atoms, or 40 moles. By the stoichiometry of the reaction, each zinc particle requires 2 hydrochloric acid particles (i.e. each equivalent of zinc metal requires 2 hydrochloric acid particles). Thus #0.2# mol of zinc metal were required.

+

Capisce?

+
+
+
" "
+

How many moles of #Zn# will be needed to completely react with 0.4 moles of #HCl# in the reaction #Zn + HCl -> ZnCl_2 + H_2#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 23, 2016 + +
+
+
+
+
+
+
+

#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#.

+

#0.2# moles of zinc metal are required for reaction with 0.4 moles of hydrochloric acid.

+
+
+
+

Explanation:

+
+

The equation above is stoichiometrically balanced. What does this mean? It means that for every reactant particle, there is a corresponding product particle. Stoichiometry is also practised in commerce: for every debit item (each charge made to an account), there must be a corresponding credit item (a deposit to made to another account). This principle of ""rob Peter, and pay Paul"" is fundamental to stoichiometry.

+

You started with #0.4# #mol# of #HCl(aq)#; I could have started with 4 atoms, or 40 atoms, or 40 moles. By the stoichiometry of the reaction, each zinc particle requires 2 hydrochloric acid particles (i.e. each equivalent of zinc metal requires 2 hydrochloric acid particles). Thus #0.2# mol of zinc metal were required.

+

Capisce?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 13651 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" How many moles of #Zn# will be needed to completely react with 0.4 moles of #HCl# in the reaction #Zn + HCl -> ZnCl_2 + H_2#? nan +428 a836bc6f-6ddd-11ea-973a-ccda262736ce https://socratic.org/questions/581affb97c01497d8c2e81f0 1.17 × 10^(-2) moles start physical_unit 4 5 mole mol qc_end physical_unit 14 14 10 11 mass qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] copper ion [IN] moles""}]" "[{""type"":""physical unit"",""value"":""1.17 × 10^(-2) moles""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] CuCl2.2H2O [=] \\pu{2.00 g}""}]" "

How many moles of copper ion are present in a #2.00*g# mass of #CuCl_2*2H_2O#?

" nan 1.17 × 10^(-2) moles "
+

Explanation:

+
+

The give quotient is #""mass""/""molar mass""#, and the answer represents the number of moles of #CuCl_2*2H_2O#, of aquated copper chloride.

+
+
" "
+
+
+

#(2.00*g)/(170.48* g*mol^-1)# #=# #??mol#

+
+
+
+

Explanation:

+
+

The give quotient is #""mass""/""molar mass""#, and the answer represents the number of moles of #CuCl_2*2H_2O#, of aquated copper chloride.

+
+
+
" "
+

How many moles of copper ion are present in a #2.00*g# mass of #CuCl_2*2H_2O#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
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+
+ +
+
+ +
+ + Nov 3, 2016 + +
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+

#(2.00*g)/(170.48* g*mol^-1)# #=# #??mol#

+
+
+
+

Explanation:

+
+

The give quotient is #""mass""/""molar mass""#, and the answer represents the number of moles of #CuCl_2*2H_2O#, of aquated copper chloride.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 2657 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many moles of copper ion are present in a #2.00*g# mass of #CuCl_2*2H_2O#? nan +429 a836bc70-6ddd-11ea-a843-ccda262736ce https://socratic.org/questions/what-is-the-oxidation-number-of-carbon-in-na2c2o4 +3 start physical_unit 6 8 oxidation_number none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] carbon in Na2C2O4""}]" "[{""type"":""physical unit"",""value"":""+3""}]" "[{""type"":""chemical equation"",""value"":""Na2C2O4""}]" "

What is the oxidation number of carbon in Na2C2O4?

" nan +3 "
+

Explanation:

+
+

You need to do a little algebra on this one.

+

You have the substance

+

#Na_2C_2O_4# = zero oxidation state

+

Converting this into numbers based on their known oxidation states, you will have:

+

Na = 2 (+1) = +2
+C = 2 (x) = 2x, where x is the unknown
+O = 4 (-2) = -8

+

Hence,

+

(+2) + 2x + (-8) = 0

+

2x + (-6) = 0

+

2x = +6

+

x = +3

+

Therefore, the oxidation state of one Carbon atom in the substance #Na_2C_2O_4# is +3.

+
+
" "
+
+
+

#C^""+3""#

+
+
+
+

Explanation:

+
+

You need to do a little algebra on this one.

+

You have the substance

+

#Na_2C_2O_4# = zero oxidation state

+

Converting this into numbers based on their known oxidation states, you will have:

+

Na = 2 (+1) = +2
+C = 2 (x) = 2x, where x is the unknown
+O = 4 (-2) = -8

+

Hence,

+

(+2) + 2x + (-8) = 0

+

2x + (-6) = 0

+

2x = +6

+

x = +3

+

Therefore, the oxidation state of one Carbon atom in the substance #Na_2C_2O_4# is +3.

+
+
+
" "
+

What is the oxidation number of carbon in Na2C2O4?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+2 Answers +
+
+
+
+
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+ +
+
+ +
+ + Oct 25, 2015 + +
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+

#C^""+3""#

+
+
+
+

Explanation:

+
+

You need to do a little algebra on this one.

+

You have the substance

+

#Na_2C_2O_4# = zero oxidation state

+

Converting this into numbers based on their known oxidation states, you will have:

+

Na = 2 (+1) = +2
+C = 2 (x) = 2x, where x is the unknown
+O = 4 (-2) = -8

+

Hence,

+

(+2) + 2x + (-8) = 0

+

2x + (-6) = 0

+

2x = +6

+

x = +3

+

Therefore, the oxidation state of one Carbon atom in the substance #Na_2C_2O_4# is +3.

+
+
+
+
+
+ +
+
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+ + +
+
+ +
+ + Oct 25, 2015 + +
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+

The oxidation number of each carbon atom in #""Na""_2""C""_2""O""_4""# is +3.

+
+
+
+

Explanation:

+
+

#""Na""_2""C""_2""O""_4""# is the compound sodium oxalate.

+

The sum of the oxidation numbers of the elements in a compound is zero.

+

The oxidation number for sodium is pretty much always #+1#.

+
+

Since there are two sodium atoms, the total oxidation number for sodium is #+2#.

+
+

The oxidation number for oxygen is #-2#, except in peroxides. Oxalate is not a peroxide, so the oxidation number here is still #-2#.

+
+

Since there are four oxygen atoms, the total oxidation number for the oxygen atoms is #-8#.

+

The sum of the oxidation numbers for sodium and oxygen is #+2 - 8 = -6#. Therefore, the total oxidation number for carbon must be #+6# in order for the sum of the oxidation numbers to equal zero.

+

Divide #+6# by two to get the oxidation number of each carbon atom, which is #bb(+3)#.

+
+

Oxidation numbers for all elements in the compound sodium oxalate:

+
+

#stackrel(2 xx +1)(""Na""_2)"" ""stackrel(2 xx +3)(""C""_2)"" ""stackrel(4 xx -2)(""O""_4)#

+

#2xx(+1)+2 xx (+3)+4xx(-2) = 0#

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+ +
+
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+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 26468 views + around the world +
+
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+ + Creative Commons License + +
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+
+
" What is the oxidation number of carbon in Na2C2O4? nan +430 a836bc71-6ddd-11ea-8975-ccda262736ce https://socratic.org/questions/if-the-concentration-of-a-fecl-3-solution-is-2-5-m-what-is-the-concentration-of- 7.50 M start physical_unit 16 17 concentration mol/l qc_end physical_unit 5 6 8 9 concentration qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] chloride ions [IN] M""}]" "[{""type"":""physical unit"",""value"":""7.50 M""}]" "[{""type"":""physical unit"",""value"":""Concentration [OF] FeCl3 solution [=] \\pu{2.5 M}""}]" "

If the concentration of a #FeCl_3# solution is 2.5 M, what is the concentration of just chloride ions?

" nan 7.50 M "
+

Explanation:

+
+

The idea here is that iron(III) chloride, #""FeCl""_3#, is soluble water, which implies that it dissociates completely to form iron(III) cations, #""Fe""^(3+)#, and chloride anions, #""Cl""^(-)#.

+

Now, notice that one formula unit of iron(III) chloride contains

+
+
    +
  • one iron(III) cation, #1 xx ""Fe""^(3+)#
    • +
  • three chloride anions, #3 xx ""Cl""^(-)#
    • +
+
+

This means that when one mole of iron(III) chloride dissolves in water, you get

+
+
    +
  • one mole of iron(III) cations
    • +
  • three moles of chloride anions
    • +
+
+

Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.

+

In your case, you can say that the concentration of chloride anions will be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#

+
+
+
" "
+
+
+

#""7.5 M""#

+
+
+
+

Explanation:

+
+

The idea here is that iron(III) chloride, #""FeCl""_3#, is soluble water, which implies that it dissociates completely to form iron(III) cations, #""Fe""^(3+)#, and chloride anions, #""Cl""^(-)#.

+

Now, notice that one formula unit of iron(III) chloride contains

+
+
    +
  • one iron(III) cation, #1 xx ""Fe""^(3+)#
    • +
  • three chloride anions, #3 xx ""Cl""^(-)#
    • +
+
+

This means that when one mole of iron(III) chloride dissolves in water, you get

+
+
    +
  • one mole of iron(III) cations
    • +
  • three moles of chloride anions
    • +
+
+

Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.

+

In your case, you can say that the concentration of chloride anions will be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#

+
+
+
+
" "
+

If the concentration of a #FeCl_3# solution is 2.5 M, what is the concentration of just chloride ions?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Measuring Concentration + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Oct 28, 2016 + +
+
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#""7.5 M""#

+
+
+
+

Explanation:

+
+

The idea here is that iron(III) chloride, #""FeCl""_3#, is soluble water, which implies that it dissociates completely to form iron(III) cations, #""Fe""^(3+)#, and chloride anions, #""Cl""^(-)#.

+

Now, notice that one formula unit of iron(III) chloride contains

+
+
    +
  • one iron(III) cation, #1 xx ""Fe""^(3+)#
    • +
  • three chloride anions, #3 xx ""Cl""^(-)#
    • +
+
+

This means that when one mole of iron(III) chloride dissolves in water, you get

+
+
    +
  • one mole of iron(III) cations
    • +
  • three moles of chloride anions
    • +
+
+

Therefore, you can say that the concentration of chloride anions in an aqueous solution of iron(III) chloride will be three times higher than the concentration of the salt itself.

+

In your case, you can say that the concentration of chloride anions will be

+
+

#color(green)(bar(ul(|color(white)(a/a)color(black)([""Cl""^(-)] = 3 xx ""2.5 M"" = ""7.5 M"")color(white)(a/a)|)))#

+
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 11423 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" If the concentration of a #FeCl_3# solution is 2.5 M, what is the concentration of just chloride ions? nan +431 a836bc72-6ddd-11ea-bc44-ccda262736ce https://socratic.org/questions/what-is-the-molar-mass-of-nacl-table-salt 58.44 g/mol start physical_unit 6 6 molar_mass g/mol qc_end chemical_equation 6 6 qc_end end "[{""type"":""physical unit"",""value"":""Molar mass [OF] NaCl [IN] g/mol""}]" "[{""type"":""physical unit"",""value"":""58.44 g/mol""}]" "[{""type"":""chemical equation"",""value"":""NaCl""}]" "

What is the molar mass of NaCl (table salt)?

" nan 58.44 g/mol "
+

Explanation:

+
+

Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)# that has the formula unit #""NaCl""#.

+

The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.

+

You know that sodium chloride's formula unit contains

+
+
    +
  • one sodium atom
    • +
  • one chlorine atom
    • +
+
+

This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.

+
+

#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#

+
+

A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are

+
    +
  • #""Na "" -> "" 22.989770 g/mol""#
    • +
  • #""Cl "" -> "" 35.453 g/mol""#
    • +
+

Therefore, the molar mass of sodium chloride will be

+
+

#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#

+
+

In stoichiometric calculations, this value is often used as #""58.44 g/mol""#.

+
+
" "
+
+
+

#""58.44277 g/mol""#

+
+
+
+

Explanation:

+
+

Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)# that has the formula unit #""NaCl""#.

+

The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.

+

You know that sodium chloride's formula unit contains

+
+
    +
  • one sodium atom
    • +
  • one chlorine atom
    • +
+
+

This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.

+
+

#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#

+
+

A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are

+
    +
  • #""Na "" -> "" 22.989770 g/mol""#
    • +
  • #""Cl "" -> "" 35.453 g/mol""#
    • +
+

Therefore, the molar mass of sodium chloride will be

+
+

#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#

+
+

In stoichiometric calculations, this value is often used as #""58.44 g/mol""#.

+
+
+
" "
+

What is the molar mass of NaCl (table salt)?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Oct 29, 2015 + +
+
+
+
+
+
+
+

#""58.44277 g/mol""#

+
+
+
+

Explanation:

+
+

Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations, #""Na""^(+)#, and chloride anions, #""Cl""^(-)# that has the formula unit #""NaCl""#.

+

The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is.

+

You know that sodium chloride's formula unit contains

+
+
    +
  • one sodium atom
    • +
  • one chlorine atom
    • +
+
+

This means that sodium chloride's molar mass will be the sum of the molar masses of those two atoms.

+
+

#M_""M NaCl"" = M_"" MNaCl"" + M_""M Cl""#

+
+

A quick look in the periodic table and you'll see that the molar masses of sodium and chlorine, respectively, are

+
    +
  • #""Na "" -> "" 22.989770 g/mol""#
    • +
  • #""Cl "" -> "" 35.453 g/mol""#
    • +
+

Therefore, the molar mass of sodium chloride will be

+
+

#M_""M"" = ""22.989770 g/mol"" + ""35.453 g/mol"" = color(green)(""58.44277 g/mol"")#

+
+

In stoichiometric calculations, this value is often used as #""58.44 g/mol""#.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Oct 29, 2015 + +
+
+
+
+
+
+
+

The molar mass of sodium chloride is 58.44 g/mol.

+
+
+
+

Explanation:

+
+

#""NaCl""#

+

The molar mass of sodium is 22.990 g/mol. The molar mass of chlorine is 35.45 g/mol.

+

To calculate the molar mass of NaCl, multiply the subscript of each element times its molar mass, then add them together. If there is no subscript, it is understood to be #1#.

+

#(1xx22.990""g/mol"")+(1xx35.45""g/mol"")=""58.44 g/mol""#

+
+
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+
+
+ +
+
+
+
+
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+
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+ + +
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+
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" What is the molar mass of NaCl (table salt)? nan +432 a836bc73-6ddd-11ea-916f-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-0-001-m-naoh-aq-at-298-k 11.00 start physical_unit 7 7 ph none qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 7 7 9 10 temperature qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] NaOH(aq)""}]" "[{""type"":""physical unit"",""value"":""11.00""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] NaOH(aq) [=] \\pu{0.001 M}""},{""type"":""physical unit"",""value"":""Temperature [OF] NaOH(aq) [=] \\pu{298 K}""}]" "

What is the pH of 0.001 M NaOH(aq) at 298 K?

" nan 11.00 "
+

Explanation:

+
+

The problem provides you with the temperature of the solution because at #25^@""C""#, which is about #""298 K""#, an aqueous solution has

+
+

#color(blue)(ul(color(black)(""pH + pOH = 14"")))#

+
+

This means that you can express the #""pH""# of the solution in terms of its #""pOH""#, which, as you know, is defined as

+
+

#""pOH"" = - log([""OH""^(-)])#

+
+

You can thus say that the #""pH""# of the solution is equal to

+
+

#""pH"" = 14 - ""pOH""#

+

#""pH"" = 14 - [ - log([""OH""^(-)])]#

+

#""pH"" = 14 + log([""OH""^(-)])#

+
+

So instead of calculating the #""pH""# of the solution by using the concentration of hydronium cations, #""H""_3""O""^(+)#

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

you can do so indirectly by using the concentration of hydroxide anions.

+

Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.

+

Both ions are produced in #1:1# mole ratio with the strong base, so you can say that

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#

+
+

Now that you know the concentration of hydroxide anions in this solution, you can say that its #""pH""# is equal to

+
+

#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#

+
+

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.

+
+
" "
+
+
+

#""pH"" = 11.0#

+
+
+
+

Explanation:

+
+

The problem provides you with the temperature of the solution because at #25^@""C""#, which is about #""298 K""#, an aqueous solution has

+
+

#color(blue)(ul(color(black)(""pH + pOH = 14"")))#

+
+

This means that you can express the #""pH""# of the solution in terms of its #""pOH""#, which, as you know, is defined as

+
+

#""pOH"" = - log([""OH""^(-)])#

+
+

You can thus say that the #""pH""# of the solution is equal to

+
+

#""pH"" = 14 - ""pOH""#

+

#""pH"" = 14 - [ - log([""OH""^(-)])]#

+

#""pH"" = 14 + log([""OH""^(-)])#

+
+

So instead of calculating the #""pH""# of the solution by using the concentration of hydronium cations, #""H""_3""O""^(+)#

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

you can do so indirectly by using the concentration of hydroxide anions.

+

Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.

+

Both ions are produced in #1:1# mole ratio with the strong base, so you can say that

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#

+
+

Now that you know the concentration of hydroxide anions in this solution, you can say that its #""pH""# is equal to

+
+

#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#

+
+

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.

+
+
+
" "
+

What is the pH of 0.001 M NaOH(aq) at 298 K?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Feb 2, 2018 + +
+
+
+
+
+
+
+

#""pH"" = 11.0#

+
+
+
+

Explanation:

+
+

The problem provides you with the temperature of the solution because at #25^@""C""#, which is about #""298 K""#, an aqueous solution has

+
+

#color(blue)(ul(color(black)(""pH + pOH = 14"")))#

+
+

This means that you can express the #""pH""# of the solution in terms of its #""pOH""#, which, as you know, is defined as

+
+

#""pOH"" = - log([""OH""^(-)])#

+
+

You can thus say that the #""pH""# of the solution is equal to

+
+

#""pH"" = 14 - ""pOH""#

+

#""pH"" = 14 - [ - log([""OH""^(-)])]#

+

#""pH"" = 14 + log([""OH""^(-)])#

+
+

So instead of calculating the #""pH""# of the solution by using the concentration of hydronium cations, #""H""_3""O""^(+)#

+
+

#""pH"" = - log([""H""_3""O""^(+)])#

+
+

you can do so indirectly by using the concentration of hydroxide anions.

+

Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.

+

Both ions are produced in #1:1# mole ratio with the strong base, so you can say that

+
+

#[""OH""^(-)] = [""NaOH""] = ""0.001 M""#

+
+

Now that you know the concentration of hydroxide anions in this solution, you can say that its #""pH""# is equal to

+
+

#""pH"" = 14 + log(0.001) = color(darkgreen)(ul(color(black)(""11.0"")))#

+
+

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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Impact of this question
+
+ 33148 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" What is the pH of 0.001 M NaOH(aq) at 298 K? nan +433 a836bc74-6ddd-11ea-9f65-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-volume-of-25-0-g-of-carbon-monoxide-at-stp 20.17 L start physical_unit 10 11 volume l qc_end physical_unit 10 11 7 8 mass qc_end c_other STP qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] carbon monoxide [IN] L""}]" "[{""type"":""physical unit"",""value"":""20.17 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] carbon monoxide [=] \\pu{25.0 g}""},{""type"":""other"",""value"":""STP""}]" "

How do you calculate the volume of 25.0 g of carbon monoxide at STP?

" nan 20.17 L "
+

Explanation:

+
+
+

We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.

+
+

Moles of #""CO""#

+

#""Moles of CO"" = 25.0 color(red)(cancel(color(black)(""g CO""))) × (""1 mol CO"")/(28.01 color(red)(cancel(color(black)(""g CO"")))) = ""0.8925 mol CO""#

+
+

Volume at STP

+

The Ideal Gas Law is:

+
+

#color(blue)(|bar(ul(PV = nRT)|)#,

+
+

where

+
    +
  • #P# is the pressure
    • +
  • #V# is the volume
    • +
  • #n# is the number of moles
    • +
  • #R# is the gas constant
    • +
  • #T# is the temperature
    • +
+
+

We can rearrange the Ideal Gas Law to get

+
+
+

#V = (nRT)/P#

+
+
+

STP is 1 bar and 0 °C.

+

#n = ""0.8925 mol""#
+#R = ""0.083 14 bar·L·K""^""-1""""mol""^""-1""#
+#T = ""273.15 K""#
+#P = ""1 bar""#

+
+

#V = (nRT)/P = (0.8925 color(red)(cancel(color(black)(""mol""))) × ""0.083 14"" color(red)(cancel(color(black)(""bar"")))·""L""· color(red)(cancel(color(black)(""K""^""-1""""mol""^""-1""))) × 273.15 color(red)(cancel(color(black)(""K""))))/(1 color(red)(cancel(color(black)(""bar"")))) = ""20.3 L""#

+
+
" "
+
+
+

The volume is 20.3 L.

+
+
+
+

Explanation:

+
+
+

We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.

+
+

Moles of #""CO""#

+

#""Moles of CO"" = 25.0 color(red)(cancel(color(black)(""g CO""))) × (""1 mol CO"")/(28.01 color(red)(cancel(color(black)(""g CO"")))) = ""0.8925 mol CO""#

+
+

Volume at STP

+

The Ideal Gas Law is:

+
+

#color(blue)(|bar(ul(PV = nRT)|)#,

+
+

where

+
    +
  • #P# is the pressure
    • +
  • #V# is the volume
    • +
  • #n# is the number of moles
    • +
  • #R# is the gas constant
    • +
  • #T# is the temperature
    • +
+
+

We can rearrange the Ideal Gas Law to get

+
+
+

#V = (nRT)/P#

+
+
+

STP is 1 bar and 0 °C.

+

#n = ""0.8925 mol""#
+#R = ""0.083 14 bar·L·K""^""-1""""mol""^""-1""#
+#T = ""273.15 K""#
+#P = ""1 bar""#

+
+

#V = (nRT)/P = (0.8925 color(red)(cancel(color(black)(""mol""))) × ""0.083 14"" color(red)(cancel(color(black)(""bar"")))·""L""· color(red)(cancel(color(black)(""K""^""-1""""mol""^""-1""))) × 273.15 color(red)(cancel(color(black)(""K""))))/(1 color(red)(cancel(color(black)(""bar"")))) = ""20.3 L""#

+
+
+
" "
+

How do you calculate the volume of 25.0 g of carbon monoxide at STP?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Molar Volume of a Gas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 16, 2016 + +
+
+
+
+
+
+
+

The volume is 20.3 L.

+
+
+
+

Explanation:

+
+
+

We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.

+
+

Moles of #""CO""#

+

#""Moles of CO"" = 25.0 color(red)(cancel(color(black)(""g CO""))) × (""1 mol CO"")/(28.01 color(red)(cancel(color(black)(""g CO"")))) = ""0.8925 mol CO""#

+
+

Volume at STP

+

The Ideal Gas Law is:

+
+

#color(blue)(|bar(ul(PV = nRT)|)#,

+
+

where

+
    +
  • #P# is the pressure
    • +
  • #V# is the volume
    • +
  • #n# is the number of moles
    • +
  • #R# is the gas constant
    • +
  • #T# is the temperature
    • +
+
+

We can rearrange the Ideal Gas Law to get

+
+
+

#V = (nRT)/P#

+
+
+

STP is 1 bar and 0 °C.

+

#n = ""0.8925 mol""#
+#R = ""0.083 14 bar·L·K""^""-1""""mol""^""-1""#
+#T = ""273.15 K""#
+#P = ""1 bar""#

+
+

#V = (nRT)/P = (0.8925 color(red)(cancel(color(black)(""mol""))) × ""0.083 14"" color(red)(cancel(color(black)(""bar"")))·""L""· color(red)(cancel(color(black)(""K""^""-1""""mol""^""-1""))) × 273.15 color(red)(cancel(color(black)(""K""))))/(1 color(red)(cancel(color(black)(""bar"")))) = ""20.3 L""#

+
+
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+
+ +
+
+
+
+
+
+
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+
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+ + Creative Commons License + +
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+
" How do you calculate the volume of 25.0 g of carbon monoxide at STP? nan +434 a836bc75-6ddd-11ea-a4c3-ccda262736ce https://socratic.org/questions/what-is-the-h-if-the-ph-of-a-solution-is-1-65 0.02 mol/L start physical_unit 3 3 molarity mol/l qc_end physical_unit 8 9 11 11 ph qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] H+ [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.02 mol/L""}]" "[{""type"":""physical unit"",""value"":""pH [OF] a solution [=] \\pu{1.65}""}]" "

What is the H+ if the pH of a solution is 1.65?

" nan 0.02 mol/L "
+

Explanation:

+
+
+

The formula for #""pH""# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#

+
+
+

We can rearrange this equation to get

+
+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#

+
+
+

If #""pH"" = 1.65#,

+

#[""H""^+] = 10^""-1.65""color(white)(l)""mol/L"" = ""0.022 mol/L""#

+
+
" "
+
+
+

#[""H""^+] = ""0.022 mol/L""#

+
+
+
+

Explanation:

+
+
+

The formula for #""pH""# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#

+
+
+

We can rearrange this equation to get

+
+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#

+
+
+

If #""pH"" = 1.65#,

+

#[""H""^+] = 10^""-1.65""color(white)(l)""mol/L"" = ""0.022 mol/L""#

+
+
+
" "
+

What is the H+ if the pH of a solution is 1.65?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Sep 7, 2016 + +
+
+
+
+
+
+
+

#[""H""^+] = ""0.022 mol/L""#

+
+
+
+

Explanation:

+
+
+

The formula for #""pH""# is

+
+
+

#color(blue)(|bar(ul(color(white)(a/a) ""pH"" = -log[""H""^+]color(white)(a/a)|)))"" ""#

+
+
+

We can rearrange this equation to get

+
+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""^+] = 10^""-pH""color(white)(a/a)|)))"" ""#

+
+
+

If #""pH"" = 1.65#,

+

#[""H""^+] = 10^""-1.65""color(white)(l)""mol/L"" = ""0.022 mol/L""#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6477 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the H+ if the pH of a solution is 1.65? nan +435 a836bc76-6ddd-11ea-843d-ccda262736ce https://socratic.org/questions/how-can-i-write-the-formula-for-calcium-nitride Ca3N2 start chemical_formula qc_end substance 7 8 qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] calcium nitride [IN] default""}]" "[{""type"":""chemical equation"",""value"":""Ca3N2""}]" "[{""type"":""substance name"",""value"":""Calcium nitride""}]" "

How can I write the formula for calcium nitride?

" nan Ca3N2 "
+

Explanation:

+
+

The correct answer is #Ca_3N_2#.

+

Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.

+

Ca ( Z= 20) has 20 electrons with following electronic configuration. 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#.

+

It loses two electron in its 4s subshell to achieve stability and forms ion #Ca^(2+)#.

+

#Ca^(2+)# = 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#

+

N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion, #N^(3-)# ion.

+

N ( Z=7) = 1#s^2#2#s^2#2#p^3#

+

#N^-3# = 1#s^2#2#s^2#2#p^6#

+

Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes #Ca^(2+)# ion and nitrogen atom after gaining three electrons becomes #N^(3-)# ion.

+

+

so in all we have three #Ca^(2+)# ions and two nitride #N^(3-)# ions.
+so the formula becomes #Ca_3N_2#.

+
+
" "
+
+
+

#""Ca""""""_3""N""_2#

+
+
+
+

Explanation:

+
+

The correct answer is #Ca_3N_2#.

+

Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.

+

Ca ( Z= 20) has 20 electrons with following electronic configuration. 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#.

+

It loses two electron in its 4s subshell to achieve stability and forms ion #Ca^(2+)#.

+

#Ca^(2+)# = 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#

+

N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion, #N^(3-)# ion.

+

N ( Z=7) = 1#s^2#2#s^2#2#p^3#

+

#N^-3# = 1#s^2#2#s^2#2#p^6#

+

Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes #Ca^(2+)# ion and nitrogen atom after gaining three electrons becomes #N^(3-)# ion.

+

+

so in all we have three #Ca^(2+)# ions and two nitride #N^(3-)# ions.
+so the formula becomes #Ca_3N_2#.

+
+
+
" "
+

How can I write the formula for calcium nitride?

+
+
+ + +Chemistry + + + + + +Ionic Bonds + + + + + +Writing Ionic Formulas + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+ +
+
+
+
+
+

#""Ca""""""_3""N""_2#

+
+
+
+

Explanation:

+
+

The correct answer is #Ca_3N_2#.

+

Let us see how we got the answer; Look at the electronic arrangement of Ca and N atom.

+

Ca ( Z= 20) has 20 electrons with following electronic configuration. 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#.

+

It loses two electron in its 4s subshell to achieve stability and forms ion #Ca^(2+)#.

+

#Ca^(2+)# = 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#

+

N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion, #N^(3-)# ion.

+

N ( Z=7) = 1#s^2#2#s^2#2#p^3#

+

#N^-3# = 1#s^2#2#s^2#2#p^6#

+

Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes #Ca^(2+)# ion and nitrogen atom after gaining three electrons becomes #N^(3-)# ion.

+

+

so in all we have three #Ca^(2+)# ions and two nitride #N^(3-)# ions.
+so the formula becomes #Ca_3N_2#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 31170 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
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+
+
" How can I write the formula for calcium nitride? nan +436 a836bc77-6ddd-11ea-8790-ccda262736ce https://socratic.org/questions/a-solution-contains-0-063-g-of-oxalic-acid-h-2c-2o-4-2h-2o-in-250-ml-what-is-the 2.00 × 10^(-3) mol/L start physical_unit 17 18 molarity mol/l qc_end physical_unit 8 8 3 4 mass qc_end physical_unit 1 1 10 11 volume qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] this solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""2.00 × 10^(-3) mol/L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] H2C2O4.2H2O [=] \\pu{0.063 g}""},{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{250 mL}""}]" "

A solution contains 0.063 g of oxalic acid, #H_2C_2O_4 * 2H_2O#, in 250 mL. What is the molarity of this solution?

" nan 2.00 × 10^(-3) mol/L "
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#

+

#=# #(0.063*g)/(126.07*g*mol^-1)xx1/(0.250*L)# #=# #??*mol*L^-1#

+
+
" "
+
+
+

#""Molarity""# #~=# #2xx10^-3*mol*L^-1#

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#

+

#=# #(0.063*g)/(126.07*g*mol^-1)xx1/(0.250*L)# #=# #??*mol*L^-1#

+
+
+
" "
+

A solution contains 0.063 g of oxalic acid, #H_2C_2O_4 * 2H_2O#, in 250 mL. What is the molarity of this solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 7, 2016 + +
+
+
+
+
+
+
+

#""Molarity""# #~=# #2xx10^-3*mol*L^-1#

+
+
+
+

Explanation:

+
+

#""Molarity""# #=# #""Moles of solute""/""Volume of solution""#

+

#=# #(0.063*g)/(126.07*g*mol^-1)xx1/(0.250*L)# #=# #??*mol*L^-1#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 12597 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A solution contains 0.063 g of oxalic acid, #H_2C_2O_4 * 2H_2O#, in 250 mL. What is the molarity of this solution? nan +437 a836bc78-6ddd-11ea-926c-ccda262736ce https://socratic.org/questions/how-do-i-calculate-the-molarity-of-7-24-10-2-ml-of-solution-containing-22-4-g-of 0.19 mol/L start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 7 10 volume qc_end physical_unit 17 18 14 15 mass qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] solution [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.19 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] solution [=] \\pu{7.24 × 10^2 mL}""},{""type"":""physical unit"",""value"":""Mass [OF] potassium iodide [=] \\pu{22.4 g}""}]" "

How do I calculate the molarity of 7.24 * 10^2 mL of solution containing 22.4 g of potassium iodide?

" nan 0.19 mol/L "
+

Explanation:

+
+

And thus.......

+

#""Molarity""=((22.4*g)/(166.0*g*mol^-1))/(7.24xx10^2*mLxx10^-3*L*mL^-1)#

+

#=0.186*mol*L^-1#

+

So what's unusual about this problem? It expresses volume in units of #mL# NOT #L#. But the prefix #m# stands for #""milli""# i.e. #10^-3#, just as #c# stands for #""centi""# i.e. #10^-2#, and #k# stands for #""kilo""# i.e. #10^3#, and #mu# stands for #""micro""# i.e. #10^-6#. These must be simply be learned, but use them first in practice problems.

+

And so when we are quoted a volume of #7.24xx10^2*mL#, we make the conversion......

+

#7.24xx10^2*mL-=7.24xx10^2*cancel(mL)xx10^-3*L*cancel(mL^-1)=0.724*L#

+

Such operations are known as #""dimensional analysis""#, and we did a little example in the problem; we wanted an answer with units of concentration and we got one. It is certainly not straightforward and intuitive, but it does work (eventually!). At your level you simply have to remember that #1*L=1*dm^3=1000*mL=1000*cm^3#. As you do more problems the conversion becomes a bit more familiar.

+
+
" "
+
+
+

#""Molarity""=""Moles of solute""/""Volume of solution""#.........

+

#=0.186*mol*L^-1#

+
+
+
+

Explanation:

+
+

And thus.......

+

#""Molarity""=((22.4*g)/(166.0*g*mol^-1))/(7.24xx10^2*mLxx10^-3*L*mL^-1)#

+

#=0.186*mol*L^-1#

+

So what's unusual about this problem? It expresses volume in units of #mL# NOT #L#. But the prefix #m# stands for #""milli""# i.e. #10^-3#, just as #c# stands for #""centi""# i.e. #10^-2#, and #k# stands for #""kilo""# i.e. #10^3#, and #mu# stands for #""micro""# i.e. #10^-6#. These must be simply be learned, but use them first in practice problems.

+

And so when we are quoted a volume of #7.24xx10^2*mL#, we make the conversion......

+

#7.24xx10^2*mL-=7.24xx10^2*cancel(mL)xx10^-3*L*cancel(mL^-1)=0.724*L#

+

Such operations are known as #""dimensional analysis""#, and we did a little example in the problem; we wanted an answer with units of concentration and we got one. It is certainly not straightforward and intuitive, but it does work (eventually!). At your level you simply have to remember that #1*L=1*dm^3=1000*mL=1000*cm^3#. As you do more problems the conversion becomes a bit more familiar.

+
+
+
" "
+

How do I calculate the molarity of 7.24 * 10^2 mL of solution containing 22.4 g of potassium iodide?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jun 30, 2017 + +
+
+
+
+
+
+
+

#""Molarity""=""Moles of solute""/""Volume of solution""#.........

+

#=0.186*mol*L^-1#

+
+
+
+

Explanation:

+
+

And thus.......

+

#""Molarity""=((22.4*g)/(166.0*g*mol^-1))/(7.24xx10^2*mLxx10^-3*L*mL^-1)#

+

#=0.186*mol*L^-1#

+

So what's unusual about this problem? It expresses volume in units of #mL# NOT #L#. But the prefix #m# stands for #""milli""# i.e. #10^-3#, just as #c# stands for #""centi""# i.e. #10^-2#, and #k# stands for #""kilo""# i.e. #10^3#, and #mu# stands for #""micro""# i.e. #10^-6#. These must be simply be learned, but use them first in practice problems.

+

And so when we are quoted a volume of #7.24xx10^2*mL#, we make the conversion......

+

#7.24xx10^2*mL-=7.24xx10^2*cancel(mL)xx10^-3*L*cancel(mL^-1)=0.724*L#

+

Such operations are known as #""dimensional analysis""#, and we did a little example in the problem; we wanted an answer with units of concentration and we got one. It is certainly not straightforward and intuitive, but it does work (eventually!). At your level you simply have to remember that #1*L=1*dm^3=1000*mL=1000*cm^3#. As you do more problems the conversion becomes a bit more familiar.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4979 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How do I calculate the molarity of 7.24 * 10^2 mL of solution containing 22.4 g of potassium iodide? nan +438 a836bc79-6ddd-11ea-be15-ccda262736ce https://socratic.org/questions/how-much-kbr-should-be-added-to-1-l-of-0-05-m-agno-3-solution-just-to-start-prec 1.19 x 10^(−9) g start physical_unit 2 2 mass g qc_end physical_unit 12 13 7 8 volume qc_end physical_unit 12 13 10 11 molarity qc_end physical_unit 19 19 24 26 equilibrium_constant_k qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] KBr [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.19 x 10^(−9) g""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] AgNO3 solution [=] \\pu{1 L}""},{""type"":""physical unit"",""value"":""Molarity [OF] AgNO3 solution [=] \\pu{0.05 M}""},{""type"":""physical unit"",""value"":""Ksp [OF] AgBr [=] \\pu{5 x 10^(−13)}""}]" "

How much #KBr# should be added to #1# #L# of #0.05# #M# #AgNO_3# solution just to start precipitation of #AgBr#? #K_(sp)# of #AgBr# = #5# x #10^-13#.

" nan 1.19 x 10^(−9) g "
+

Explanation:

+
+

Concentration of #Ag^+=0.05mol//L#
+(one mole of #Ag^+# per mole of #AgNO_3#)
+So #[Ag^+]=5*10^-2#

+

Now we plug in what we know:
+#(5*10^-2)*[Br^-]=5*10^-13#
+#[Br^-]=(cancel5*10^-13)/(cancel5*10^-2)=1*10^-11mol//L#

+

If you have the #KBr# in a not very diluted form, you may assume that you'll only need a drop or two, and that the #1L# won't change significantly.
+So you'll need #1*10^-11mol# of #KBr# and this can easily be converted to grams (or rather milligrams).

+

Extra :
+One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.

+
+
" "
+
+
+

#K_(sp)# means that #[Ag^+]*[Br^-]=5*10^-13#,
+because the 'sp' suffix stands for 'solution product'.

+
+
+
+

Explanation:

+
+

Concentration of #Ag^+=0.05mol//L#
+(one mole of #Ag^+# per mole of #AgNO_3#)
+So #[Ag^+]=5*10^-2#

+

Now we plug in what we know:
+#(5*10^-2)*[Br^-]=5*10^-13#
+#[Br^-]=(cancel5*10^-13)/(cancel5*10^-2)=1*10^-11mol//L#

+

If you have the #KBr# in a not very diluted form, you may assume that you'll only need a drop or two, and that the #1L# won't change significantly.
+So you'll need #1*10^-11mol# of #KBr# and this can easily be converted to grams (or rather milligrams).

+

Extra :
+One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.

+
+
+
" "
+

How much #KBr# should be added to #1# #L# of #0.05# #M# #AgNO_3# solution just to start precipitation of #AgBr#? #K_(sp)# of #AgBr# = #5# x #10^-13#.

+
+
+ + +Chemistry + + + + + +Chemical Equilibrium + + + + + +Ksp + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 30, 2016 + +
+
+
+
+
+
+
+

#K_(sp)# means that #[Ag^+]*[Br^-]=5*10^-13#,
+because the 'sp' suffix stands for 'solution product'.

+
+
+
+

Explanation:

+
+

Concentration of #Ag^+=0.05mol//L#
+(one mole of #Ag^+# per mole of #AgNO_3#)
+So #[Ag^+]=5*10^-2#

+

Now we plug in what we know:
+#(5*10^-2)*[Br^-]=5*10^-13#
+#[Br^-]=(cancel5*10^-13)/(cancel5*10^-2)=1*10^-11mol//L#

+

If you have the #KBr# in a not very diluted form, you may assume that you'll only need a drop or two, and that the #1L# won't change significantly.
+So you'll need #1*10^-11mol# of #KBr# and this can easily be converted to grams (or rather milligrams).

+

Extra :
+One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 5499 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How much #KBr# should be added to #1# #L# of #0.05# #M# #AgNO_3# solution just to start precipitation of #AgBr#? #K_(sp)# of #AgBr# = #5# x #10^-13#. nan +439 a836e37e-6ddd-11ea-8dfc-ccda262736ce https://socratic.org/questions/a-15-75-g-piece-of-iron-absorbs-1086-75-joules-of-heat-energy-and-its-temperatur 0.46 J/(g * ℃) start physical_unit 5 5 specific_heat_capacity j/(°c_·_g) qc_end physical_unit 5 5 1 2 mass qc_end physical_unit 5 5 7 8 heat_energy qc_end physical_unit 5 5 17 18 temperature qc_end physical_unit 5 5 20 21 temperature qc_end end "[{""type"":""physical unit"",""value"":""Specific heat capacity [OF] iron [IN] J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""0.46 J/(g * ℃)""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] iron [=] \\pu{15.75 g}""},{""type"":""physical unit"",""value"":""Heat energy absorbed [OF] iron [=] \\pu{1086.75 joules}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] iron [=] \\pu{25 °C}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] iron [=] \\pu{175 °C}""}]" "

A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. What is the specific heat capacity of iron?

" nan 0.46 J/(g * ℃) "
+

Explanation:

+
+

Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.

+

A substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.

+
+

#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#

+
+

Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.

+

Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.

+
    +
  • Scenario 1
    • +
+

For instance, let's assume that adding #""1086.75 J""# of heat to a #""15.75-g""# sample of iron will only increase its temperature by #1^@""C""#. In this case, the amount of heat needed per gram of iron will be

+
+

#""1086.75 J""/""15.75 g"" = ""69 J/g""#

+
+

In this scenario, adding #""69 J""# of heat for every gram of iron will increase its temperature by #1^@""C""#.

+
    +
  • Scenario 2
    • +
+

Now let's assume that this much heat would increase the temperature of #""1 g""# of iron by

+
+

#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#

+
+

In this case, the amount of heat needed per degree Celsius will be

+
+

#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#

+
+

In this scenario, you will get a #1^@""C""# increase in temperature by adding #""7.245 J""# for every #""1 g""# of iron.

+
    +
  • In real life
    • +
+

But since you know that adding that much heat will increase the temperature of #""15.75 g""# of iron by #150^@""C""#, you can say that

+
+

#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#

+
+

And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.

+

The equation that you'll be using from now on looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added to the sample
+#m# - the mass of the sample
+#c# - its specific heat
+#DeltaT# - the change in temperature

+
+
" "
+
+
+

#0.46""J""/(""g"" """"^@""C"")#

+
+
+
+

Explanation:

+
+

Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.

+

A substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.

+
+

#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#

+
+

Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.

+

Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.

+
    +
  • Scenario 1
    • +
+

For instance, let's assume that adding #""1086.75 J""# of heat to a #""15.75-g""# sample of iron will only increase its temperature by #1^@""C""#. In this case, the amount of heat needed per gram of iron will be

+
+

#""1086.75 J""/""15.75 g"" = ""69 J/g""#

+
+

In this scenario, adding #""69 J""# of heat for every gram of iron will increase its temperature by #1^@""C""#.

+
    +
  • Scenario 2
    • +
+

Now let's assume that this much heat would increase the temperature of #""1 g""# of iron by

+
+

#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#

+
+

In this case, the amount of heat needed per degree Celsius will be

+
+

#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#

+
+

In this scenario, you will get a #1^@""C""# increase in temperature by adding #""7.245 J""# for every #""1 g""# of iron.

+
    +
  • In real life
    • +
+

But since you know that adding that much heat will increase the temperature of #""15.75 g""# of iron by #150^@""C""#, you can say that

+
+

#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#

+
+

And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.

+

The equation that you'll be using from now on looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added to the sample
+#m# - the mass of the sample
+#c# - its specific heat
+#DeltaT# - the change in temperature

+
+
+
" "
+

A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. What is the specific heat capacity of iron?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 25, 2016 + +
+
+
+
+
+
+
+

#0.46""J""/(""g"" """"^@""C"")#

+
+
+
+

Explanation:

+
+

Your task here is to find the specific heat of iron, so take a second to make sure that you understand what it is you're looking for.

+

A substance's specific heat tells you how much heat is needed in order to increase the temperature of #""1 g""# of that substance by #1^@""C""#.

+

So, in essence, you're looking for amount of heat per unit of mass and per unit of temperature.

+
+

#color(blue)(""specific heat"" = ""heat""/(""unit of mass "" xx "" unit of temperature""))#

+
+

Now, let's assume that you don't know the equation that establishes a relationship between heat added, the mass of the sample, the specific heat of the substance, and the resulting increase in temperature.

+

Here's how you could play around with the information provided by the problem to understand how to find the specific heat of iron.

+
    +
  • Scenario 1
    • +
+

For instance, let's assume that adding #""1086.75 J""# of heat to a #""15.75-g""# sample of iron will only increase its temperature by #1^@""C""#. In this case, the amount of heat needed per gram of iron will be

+
+

#""1086.75 J""/""15.75 g"" = ""69 J/g""#

+
+

In this scenario, adding #""69 J""# of heat for every gram of iron will increase its temperature by #1^@""C""#.

+
    +
  • Scenario 2
    • +
+

Now let's assume that this much heat would increase the temperature of #""1 g""# of iron by

+
+

#DeltaT = 175^@""C"" - 25^@""C"" = 150^@""C""#

+
+

In this case, the amount of heat needed per degree Celsius will be

+
+

#""1086.75 J""/(150^@""C"") = ""7.245 J/""""""^@""C""#

+
+

In this scenario, you will get a #1^@""C""# increase in temperature by adding #""7.245 J""# for every #""1 g""# of iron.

+
    +
  • In real life
    • +
+

But since you know that adding that much heat will increase the temperature of #""15.75 g""# of iron by #150^@""C""#, you can say that

+
+

#""1086.75 J""/(""15.75 g"" * 150^@""C"") = 0.46""J""/(""g"" """"^@""C"")#

+
+

And this is the substance's specific heat. So, to find the specific heat of a substance, you need to divide the amount of heat used to produce that increase in temperature for the given sample.

+

The equation that you'll be using from now on looks like this

+
+

#color(blue)(q = m * c * DeltaT)"" ""#, where

+
+

#q# - the amount of heat added to the sample
+#m# - the mass of the sample
+#c# - its specific heat
+#DeltaT# - the change in temperature

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 35249 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. What is the specific heat capacity of iron? nan +440 a836e37f-6ddd-11ea-8ae4-ccda262736ce https://socratic.org/questions/a-gas-held-in-a-25-3-ml-container-is-heated-from-25-c-to-37-c-what-is-the-final- 25.30 mL start physical_unit 22 23 volume ml qc_end physical_unit 7 7 5 6 volume qc_end physical_unit 22 23 11 12 temperature qc_end physical_unit 22 23 14 15 temperature qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] mL""}]" "[{""type"":""physical unit"",""value"":""25.30 mL""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] container [=] \\pu{25.3 mL}""},{""type"":""physical unit"",""value"":""Temperature1 [OF] the gas [=] \\pu{25 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] the gas [=] \\pu{37 ℃}""}]" "

A gas held in a 25.3 mL container is heated from 25°C to 37°C. What is the final volume of the gas?

" nan 25.30 mL "
+

Explanation:

+
+

The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.

+
+
" "
+
+
+

Your question is not well-proposed. The final volume is #25.3*mL#.

+
+
+
+

Explanation:

+
+

The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.

+
+
+
" "
+

A gas held in a 25.3 mL container is heated from 25°C to 37°C. What is the final volume of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Charles' Law + + +
+
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+1 Answer +
+
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+ + Aug 21, 2016 + +
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+
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Your question is not well-proposed. The final volume is #25.3*mL#.

+
+
+
+

Explanation:

+
+

The volume of the container, we must assume, is static. Of course, the pressure will increase, but this is not the question you asked. The gas could have been enclosed in a piston, expanding against external pressure. Please review your question.

+
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+
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+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
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+
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" A gas held in a 25.3 mL container is heated from 25°C to 37°C. What is the final volume of the gas? nan +441 a836e380-6ddd-11ea-97c4-ccda262736ce https://socratic.org/questions/hcl-is-a-strong-acid-what-is-the-ph-of-200-ml-of-0-002-m-hcl 2.70 start physical_unit 0 0 ph none qc_end c_other OTHER qc_end physical_unit 0 0 10 11 volume qc_end physical_unit 0 0 13 14 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] HCl""}]" "[{""type"":""physical unit"",""value"":""2.70""}]" "[{""type"":""other"",""value"":""HCl is a strong acid.""},{""type"":""physical unit"",""value"":""Volume [OF] HCl [=] \\pu{200 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] HCl [=] \\pu{0.002 M}""}]" "

HCl is a strong acid. What is the pH of 200 mL of 0.002 M #HCl#?

" nan 2.70 "
+

Explanation:

+
+

#""Hydrochloric acid""#, as a strong acid, is stoichiometric in #H_3O^+#. And thus:

+

#pH=-log_10[H_3O^+]=-log_10(0.002)=-(-2.70)=2.70#

+

Note that the #pH# thus depends on concentration solely. And thus what would be the #pH# of a #2*L# volume of #2xx10^-3*mol*L^-1# #""hydrochloric acid""#?

+
+
" "
+
+
+

#pH=-log_10[H_3O^+]#

+
+
+
+

Explanation:

+
+

#""Hydrochloric acid""#, as a strong acid, is stoichiometric in #H_3O^+#. And thus:

+

#pH=-log_10[H_3O^+]=-log_10(0.002)=-(-2.70)=2.70#

+

Note that the #pH# thus depends on concentration solely. And thus what would be the #pH# of a #2*L# volume of #2xx10^-3*mol*L^-1# #""hydrochloric acid""#?

+
+
+
" "
+

HCl is a strong acid. What is the pH of 200 mL of 0.002 M #HCl#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Acids and Bases + + +
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+1 Answer +
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+ + Feb 6, 2017 + +
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#pH=-log_10[H_3O^+]#

+
+
+
+

Explanation:

+
+

#""Hydrochloric acid""#, as a strong acid, is stoichiometric in #H_3O^+#. And thus:

+

#pH=-log_10[H_3O^+]=-log_10(0.002)=-(-2.70)=2.70#

+

Note that the #pH# thus depends on concentration solely. And thus what would be the #pH# of a #2*L# volume of #2xx10^-3*mol*L^-1# #""hydrochloric acid""#?

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Related questions
+ + +
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+
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+
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+ + Creative Commons License + +
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+
+
" HCl is a strong acid. What is the pH of 200 mL of 0.002 M #HCl#? nan +442 a836e381-6ddd-11ea-91ed-ccda262736ce https://socratic.org/questions/what-is-the-ph-of-a-solution-with-h-2-3-times-10-3 2.64 start physical_unit 5 6 ph none qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] a solution""}]" "[{""type"":""physical unit"",""value"":""2.64""}]" "[{""type"":""other"",""value"":""[H^+] = 2.3 × 10^(−3)""}]" "

What is the pH of a solution with [H+] = #2.3 times 10^-3#?

" nan 2.64 "
+

Explanation:

+
+

The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.

+

We could represent this reaction by #(i)#:

+

#2H_2O(l) rightleftharpoonsH_3O^+ +HO^-#

+

OR by #(ii)#:

+

#H_2O(l) rightleftharpoonsH^+ +HO^-#

+

Note that #(i)# and #(ii)# ARE EQUIVALENT REPRESENTATIONS, and it really is a matter of preference which equation you decide to use. As far as anyone knows, the actual acidium ion in solution is
+#H_5O_2^+# or #H_7O_3^+#, i.e. a cluster of 2 or 3 or 4 (or more) water molecules with an EXTRA #H^+# tacked on. We can use #H^+#, #""protium ion""# (as is used here) or #H_3O^+#, #""hydronium ion""# equivalently to represent this species.

+

The equilibrium constant for the reaction, under standard conditions, is..........#K_w=[H_3O^+][""""^(-)OH]=10^-14#.

+

And to make the arithmetic a bit easier we can use the #pH# function, where #pH=-log_10[H_3O^+]#, and #pOH=-log_10[HO^-]#. And thus in aqueous solution under the given standard conditions, #pH+pOH=14#.

+

And since we are directly given #[H^+]# or #[H_3O^+]#.....

+

#pH=-log_10[H_3O^+]=-log_10(2.3xx10^-3)=2.64#

+

How do you think #pH# and #pOH# would evolve under non-standard conditions, i.e. at a temperature of say #373*K#? Would #K_w# decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?

+
+
" "
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pH is defined as 'the logarithm (to base 10) of the [H+] with the sign changed'. #Log_10(2.3xx10^-3)=-2.64#. Change the sign to positive and the pH is 2.64.

+
+
+
" "
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What is the pH of a solution with [H+] = #2.3 times 10^-3#?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +pH calculations + + +
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+2 Answers +
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+ + Jun 25, 2017 + +
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pH is defined as 'the logarithm (to base 10) of the [H+] with the sign changed'. #Log_10(2.3xx10^-3)=-2.64#. Change the sign to positive and the pH is 2.64.

+
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+ +
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+ +
+ + Jun 25, 2017 + +
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+

#pH=-log_10[H_3O^+]=-log_10(2.3xx10^-3)=2.64#

+
+
+
+

Explanation:

+
+

The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.

+

We could represent this reaction by #(i)#:

+

#2H_2O(l) rightleftharpoonsH_3O^+ +HO^-#

+

OR by #(ii)#:

+

#H_2O(l) rightleftharpoonsH^+ +HO^-#

+

Note that #(i)# and #(ii)# ARE EQUIVALENT REPRESENTATIONS, and it really is a matter of preference which equation you decide to use. As far as anyone knows, the actual acidium ion in solution is
+#H_5O_2^+# or #H_7O_3^+#, i.e. a cluster of 2 or 3 or 4 (or more) water molecules with an EXTRA #H^+# tacked on. We can use #H^+#, #""protium ion""# (as is used here) or #H_3O^+#, #""hydronium ion""# equivalently to represent this species.

+

The equilibrium constant for the reaction, under standard conditions, is..........#K_w=[H_3O^+][""""^(-)OH]=10^-14#.

+

And to make the arithmetic a bit easier we can use the #pH# function, where #pH=-log_10[H_3O^+]#, and #pOH=-log_10[HO^-]#. And thus in aqueous solution under the given standard conditions, #pH+pOH=14#.

+

And since we are directly given #[H^+]# or #[H_3O^+]#.....

+

#pH=-log_10[H_3O^+]=-log_10(2.3xx10^-3)=2.64#

+

How do you think #pH# and #pOH# would evolve under non-standard conditions, i.e. at a temperature of say #373*K#? Would #K_w# decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?

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Related questions
+ + +
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+
+
Impact of this question
+
+ 14565 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" What is the pH of a solution with [H+] = #2.3 times 10^-3#? nan +443 a836e382-6ddd-11ea-9054-ccda262736ce https://socratic.org/questions/how-much-will-1-2-10-46-atoms-of-carbon-weigh 1.99 × 10^22 g start physical_unit 8 8 mass g qc_end physical_unit 6 8 3 5 number qc_end end "[{""type"":""physical unit"",""value"":""Weight [OF] carbon [IN] g""}]" "[{""type"":""physical unit"",""value"":""1.99 × 10^22 g""}]" "[{""type"":""physical unit"",""value"":""Number [OF] atoms of carbon [=] \\pu{1.2 × 10^46}""}]" "

How much will #1.2 * 10^46# atoms of carbon weigh?

" nan 1.99 × 10^22 g "
+

Explanation:

+
+

A carbon atom has an average mass of #12.0107 \ ""amu""#. Also, one mole of carbon atoms have a mass of #12.0107 \ ""g""#.

+

Recall that one mole of atoms is #6.02*10^23# atoms.

+

So here, we got:

+

#(1.2*10^46color(red)cancelcolor(black)""atoms"")/(6.02*10^23color(red)cancelcolor(black)""atoms""""/mol"")~~2*10^22 \ ""mol""#

+

And so, the total mass will be

+

#2*10^22color(red)cancelcolor(black)""mol""*(12.0107 \ ""g"")/(color(red)cancelcolor(black)""mol"")=2.4*10^23 \ ""g""#

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+
" "
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+
+

#2.4*10^23 \ ""g""#

+
+
+
+

Explanation:

+
+

A carbon atom has an average mass of #12.0107 \ ""amu""#. Also, one mole of carbon atoms have a mass of #12.0107 \ ""g""#.

+

Recall that one mole of atoms is #6.02*10^23# atoms.

+

So here, we got:

+

#(1.2*10^46color(red)cancelcolor(black)""atoms"")/(6.02*10^23color(red)cancelcolor(black)""atoms""""/mol"")~~2*10^22 \ ""mol""#

+

And so, the total mass will be

+

#2*10^22color(red)cancelcolor(black)""mol""*(12.0107 \ ""g"")/(color(red)cancelcolor(black)""mol"")=2.4*10^23 \ ""g""#

+
+
+
" "
+

How much will #1.2 * 10^46# atoms of carbon weigh?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+1 Answer +
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+ + Mar 30, 2018 + +
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#2.4*10^23 \ ""g""#

+
+
+
+

Explanation:

+
+

A carbon atom has an average mass of #12.0107 \ ""amu""#. Also, one mole of carbon atoms have a mass of #12.0107 \ ""g""#.

+

Recall that one mole of atoms is #6.02*10^23# atoms.

+

So here, we got:

+

#(1.2*10^46color(red)cancelcolor(black)""atoms"")/(6.02*10^23color(red)cancelcolor(black)""atoms""""/mol"")~~2*10^22 \ ""mol""#

+

And so, the total mass will be

+

#2*10^22color(red)cancelcolor(black)""mol""*(12.0107 \ ""g"")/(color(red)cancelcolor(black)""mol"")=2.4*10^23 \ ""g""#

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Related questions
+ + +
+
+
+
Impact of this question
+
+ 2945 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
+
" How much will #1.2 * 10^46# atoms of carbon weigh? nan +444 a836e383-6ddd-11ea-823e-ccda262736ce https://socratic.org/questions/596b793e7c014971b5b218b7 C + O2 -> CO2 start chemical_equation qc_end substance 4 4 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the combustion""}]" "[{""type"":""chemical equation"",""value"":""C + O2 -> CO2""}]" "[{""type"":""substance name"",""value"":""Carbon""}]" "

Represent the combustion of carbon?

" nan C + O2 -> CO2 "
+

Explanation:

+
+

Is this reaction exothermic? Could I use this reaction to drive a locomotive?

+
+
" "
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+
+

#C(s) + O_2(g) rarr CO_2(g)#

+
+
+
+

Explanation:

+
+

Is this reaction exothermic? Could I use this reaction to drive a locomotive?

+
+
+
" "
+

Represent the combustion of carbon?

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+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
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+1 Answer +
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+ + Jul 19, 2017 + +
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#C(s) + O_2(g) rarr CO_2(g)#

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Explanation:

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+

Is this reaction exothermic? Could I use this reaction to drive a locomotive?

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+
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+
Related questions
+ + +
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Impact of this question
+
+ 1175 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" Represent the combustion of carbon? nan +445 a836e384-6ddd-11ea-a60a-ccda262736ce https://socratic.org/questions/how-do-you-balance-ag-2o-ag-o-2 2 Ag2O -> 4 Ag + O2 start chemical_equation qc_end chemical_equation 4 8 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""2 Ag2O -> 4 Ag + O2""}]" "[{""type"":""chemical equation"",""value"":""Ag2O -> Ag + O2""}]" "

How do you balance #Ag_2O -> Ag + O_2#?

" nan 2 Ag2O -> 4 Ag + O2 "
+

Explanation:

+
+

Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.

+

Consider the given equation:

+
+

#Ag_2OrarrAg+O_2#

+
+

Notice how there is only one #O# on the reactants side, but two on the products side. This can be fixed by multiplying the reactant by #2#.

+
+

#color(red)(2)Ag_2OrarrAg+O_2#

+
+

Now that the oxygen atoms are balanced, all that remains is the silver. There are now four #Ag# on the reactant side and only one on the product side, which can be remedied with a coefficient of four on the products side.

+
+

#2Ag_2Orarrcolor(blue)(4)Ag+O_2#

+
+

This is completely balanced.

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+
" "
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#2Ag_2Orarr4Ag+O_2#

+
+
+
+

Explanation:

+
+

Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.

+

Consider the given equation:

+
+

#Ag_2OrarrAg+O_2#

+
+

Notice how there is only one #O# on the reactants side, but two on the products side. This can be fixed by multiplying the reactant by #2#.

+
+

#color(red)(2)Ag_2OrarrAg+O_2#

+
+

Now that the oxygen atoms are balanced, all that remains is the silver. There are now four #Ag# on the reactant side and only one on the product side, which can be remedied with a coefficient of four on the products side.

+
+

#2Ag_2Orarrcolor(blue)(4)Ag+O_2#

+
+

This is completely balanced.

+
+
+
" "
+

How do you balance #Ag_2O -> Ag + O_2#?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
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+1 Answer +
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#2Ag_2Orarr4Ag+O_2#

+
+
+
+

Explanation:

+
+

Balancing a chemical equation consists of changing the coefficient of molecules in order to make the number of atoms equal on both side.

+

Consider the given equation:

+
+

#Ag_2OrarrAg+O_2#

+
+

Notice how there is only one #O# on the reactants side, but two on the products side. This can be fixed by multiplying the reactant by #2#.

+
+

#color(red)(2)Ag_2OrarrAg+O_2#

+
+

Now that the oxygen atoms are balanced, all that remains is the silver. There are now four #Ag# on the reactant side and only one on the product side, which can be remedied with a coefficient of four on the products side.

+
+

#2Ag_2Orarrcolor(blue)(4)Ag+O_2#

+
+

This is completely balanced.

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Related questions
+ + +
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+
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Impact of this question
+
+ 1270 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
" How do you balance #Ag_2O -> Ag + O_2#? nan +446 a836e385-6ddd-11ea-a635-ccda262736ce https://socratic.org/questions/how-do-you-calculate-the-amount-of-heat-needed-to-increase-the-temperature-of-25 43.20 kJ start physical_unit 17 17 heat_energy kj qc_end physical_unit 17 17 19 20 temperature qc_end physical_unit 17 17 22 23 temperature qc_end physical_unit 17 17 26 29 cp qc_end physical_unit 17 17 14 15 mass qc_end end "[{""type"":""physical unit"",""value"":""Amount of heat needed [OF] water [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""43.20 kJ""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] water [=] \\pu{20 ℃}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] water [=] \\pu{56 ℃}""},{""type"":""physical unit"",""value"":""Cg [OF] water [=] \\pu{4.8 J/(℃ * g)}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{250 g}""}]" "

How do you calculate the amount of heat needed to increase the temperature of 250g of water from 20 C to 56 C? #C_g = 4.8 J/(C-g)#

" nan 43.20 kJ "
+

Explanation:

+
+
    +
  1. Given the data, the formula to use is
    +#Q=mCpDeltaT#
    +where:
    +#m=250g#
    +#DeltaT=T_2-T_1=56-20=36^oC#
    +#Cp=(4.8J)/(g*C)#
    2. +
  2. This case, just plug in given data to find the heat needed to raise the temperature of the water as provided in the problem.
    +#Q=(250cancel(g))((4.8J)/cancel((g*C)))(36^cancel(oC))#
    +#Q=43200J#
    +#Q=43.2kJ#
    3. +
+
+
" "
+
+
+

#=43.2kJ#

+
+
+
+

Explanation:

+
+
    +
  1. Given the data, the formula to use is
    +#Q=mCpDeltaT#
    +where:
    +#m=250g#
    +#DeltaT=T_2-T_1=56-20=36^oC#
    +#Cp=(4.8J)/(g*C)#
    2. +
  2. This case, just plug in given data to find the heat needed to raise the temperature of the water as provided in the problem.
    +#Q=(250cancel(g))((4.8J)/cancel((g*C)))(36^cancel(oC))#
    +#Q=43200J#
    +#Q=43.2kJ#
    3. +
+
+
+
" "
+

How do you calculate the amount of heat needed to increase the temperature of 250g of water from 20 C to 56 C? #C_g = 4.8 J/(C-g)#

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Calorimetry + + +
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+1 Answer +
+
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+ +
+ + Dec 26, 2017 + +
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+

#=43.2kJ#

+
+
+
+

Explanation:

+
+
    +
  1. Given the data, the formula to use is
    +#Q=mCpDeltaT#
    +where:
    +#m=250g#
    +#DeltaT=T_2-T_1=56-20=36^oC#
    +#Cp=(4.8J)/(g*C)#
    2. +
  2. This case, just plug in given data to find the heat needed to raise the temperature of the water as provided in the problem.
    +#Q=(250cancel(g))((4.8J)/cancel((g*C)))(36^cancel(oC))#
    +#Q=43200J#
    +#Q=43.2kJ#
    3. +
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+
Related questions
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Impact of this question
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" How do you calculate the amount of heat needed to increase the temperature of 250g of water from 20 C to 56 C? #C_g = 4.8 J/(C-g)# nan +447 a836e386-6ddd-11ea-b04b-ccda262736ce https://socratic.org/questions/if-the-k-a-of-a-monoprotic-weak-acid-is-8-1-times-10-6-what-is-the-ph-of-a-0-33- 2.79 start physical_unit 20 23 ph none qc_end physical_unit 4 7 9 11 ka qc_end physical_unit 20 23 18 19 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] solution of this acid""}]" "[{""type"":""physical unit"",""value"":""2.79""}]" "[{""type"":""physical unit"",""value"":""Ka [OF] a monoprotic weak acid [=] \\pu{8.1 × 10^(−6)}""},{""type"":""physical unit"",""value"":""Molarity [OF] solution of this acid [=] \\pu{0.33 M}""}]" "

If the #K_a# of a monoprotic weak acid is #8.1 times 10^-6#, what is the pH of a 0.33 M solution of this acid?

" nan 2.79 "
+

Explanation:

+
+

#(i)# We first write a stoichiometric equation to inform our reasoning:

+

#HA+H_2OrightleftharpoonsH_3O^+ + A^-#

+

#(ii)# And then we write the equilibrium expression:

+

#([H_3O^+][A^-])/([HA])=K_a=8.1xx10^-6#

+

#(iii)# And then we solve the problem using approximations IF APPROPRIATE.

+

If the amount of dissociation is #x*mol*L^-1#, then our expression becomes:

+

#x^2/(0.33-x)=8.1xx10^-6#

+

If #0.33"">>>""x#, then #0.33-x~=0.33#; and we must justify this approx. later. Solving for #x=sqrt(8.1xx10^-6xx0.33)#, we gets....

+

#x_1=1.63xx10^-3#; a value that is indeed small compared to #0.33#. Just to belabour the point, we can input our first approx. back in the calculation, and get a 2nd approx.......

+

#x_2=1.63xx10^-3#

+

Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in #x#).

+

Given #x=1.63xx10^-3*mol*L^-1=[H_3O^+]#, #pH=-log_10[H_3O^+]# #=# #-log_(10)1.63xx10^-3=2.79#.

+

What is #pOH# of this solution?

+
+
" "
+
+
+

#pH=2.79#

+
+
+
+

Explanation:

+
+

#(i)# We first write a stoichiometric equation to inform our reasoning:

+

#HA+H_2OrightleftharpoonsH_3O^+ + A^-#

+

#(ii)# And then we write the equilibrium expression:

+

#([H_3O^+][A^-])/([HA])=K_a=8.1xx10^-6#

+

#(iii)# And then we solve the problem using approximations IF APPROPRIATE.

+

If the amount of dissociation is #x*mol*L^-1#, then our expression becomes:

+

#x^2/(0.33-x)=8.1xx10^-6#

+

If #0.33"">>>""x#, then #0.33-x~=0.33#; and we must justify this approx. later. Solving for #x=sqrt(8.1xx10^-6xx0.33)#, we gets....

+

#x_1=1.63xx10^-3#; a value that is indeed small compared to #0.33#. Just to belabour the point, we can input our first approx. back in the calculation, and get a 2nd approx.......

+

#x_2=1.63xx10^-3#

+

Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in #x#).

+

Given #x=1.63xx10^-3*mol*L^-1=[H_3O^+]#, #pH=-log_10[H_3O^+]# #=# #-log_(10)1.63xx10^-3=2.79#.

+

What is #pOH# of this solution?

+
+
+
" "
+

If the #K_a# of a monoprotic weak acid is #8.1 times 10^-6#, what is the pH of a 0.33 M solution of this acid?

+
+
+ + +Chemistry + + + + + +Acids and Bases + + + + + +Acids and Bases + + +
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+1 Answer +
+
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+ + May 10, 2017 + +
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#pH=2.79#

+
+
+
+

Explanation:

+
+

#(i)# We first write a stoichiometric equation to inform our reasoning:

+

#HA+H_2OrightleftharpoonsH_3O^+ + A^-#

+

#(ii)# And then we write the equilibrium expression:

+

#([H_3O^+][A^-])/([HA])=K_a=8.1xx10^-6#

+

#(iii)# And then we solve the problem using approximations IF APPROPRIATE.

+

If the amount of dissociation is #x*mol*L^-1#, then our expression becomes:

+

#x^2/(0.33-x)=8.1xx10^-6#

+

If #0.33"">>>""x#, then #0.33-x~=0.33#; and we must justify this approx. later. Solving for #x=sqrt(8.1xx10^-6xx0.33)#, we gets....

+

#x_1=1.63xx10^-3#; a value that is indeed small compared to #0.33#. Just to belabour the point, we can input our first approx. back in the calculation, and get a 2nd approx.......

+

#x_2=1.63xx10^-3#

+

Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in #x#).

+

Given #x=1.63xx10^-3*mol*L^-1=[H_3O^+]#, #pH=-log_10[H_3O^+]# #=# #-log_(10)1.63xx10^-3=2.79#.

+

What is #pOH# of this solution?

+
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+ +
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+
+
+
Related questions
+ + +
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+
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" If the #K_a# of a monoprotic weak acid is #8.1 times 10^-6#, what is the pH of a 0.33 M solution of this acid? nan +448 a8370a8c-6ddd-11ea-b8bc-ccda262736ce https://socratic.org/questions/how-many-grams-of-nitrogen-dioxide-must-react-with-water-to-produce-5-00-10-22-m 11.46 grams start physical_unit 4 5 mass g qc_end physical_unit 15 18 12 14 number qc_end substance 9 9 qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] nitrogen dioxide [IN] grams""}]" "[{""type"":""physical unit"",""value"":""11.46 grams""}]" "[{""type"":""physical unit"",""value"":""Number [OF] molecules of nitrogen monoxide [=] \\pu{5.00 × 10^22}""},{""type"":""substance name"",""value"":""Water""}]" "

How many grams of nitrogen dioxide must react with water to produce #5.00*10^22# molecules of nitrogen monoxide?

" nan 11.46 grams "
+

Explanation:

+
+

Consider the balanced chemical equation, which we need to obtain the molar ratios:

+

#3NO_2+H_2O -> NO + 2HNO_3#

+

I see you would like to know the grams #NO_2# required to react with your amount of atoms #NO#.

+

Let's use the molar ratios to calculate the amount of nitrogen dioxide required.

+

Consider the graphic below:
+

+
+
" "
+
+
+

11.46 grams

+
+
+
+

Explanation:

+
+

Consider the balanced chemical equation, which we need to obtain the molar ratios:

+

#3NO_2+H_2O -> NO + 2HNO_3#

+

I see you would like to know the grams #NO_2# required to react with your amount of atoms #NO#.

+

Let's use the molar ratios to calculate the amount of nitrogen dioxide required.

+

Consider the graphic below:
+

+
+
+
" "
+

How many grams of nitrogen dioxide must react with water to produce #5.00*10^22# molecules of nitrogen monoxide?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
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+
+
+1 Answer +
+
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+ +
+
+ +
+ + Dec 6, 2016 + +
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+
+

11.46 grams

+
+
+
+

Explanation:

+
+

Consider the balanced chemical equation, which we need to obtain the molar ratios:

+

#3NO_2+H_2O -> NO + 2HNO_3#

+

I see you would like to know the grams #NO_2# required to react with your amount of atoms #NO#.

+

Let's use the molar ratios to calculate the amount of nitrogen dioxide required.

+

Consider the graphic below:
+

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+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 13358 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" How many grams of nitrogen dioxide must react with water to produce #5.00*10^22# molecules of nitrogen monoxide? nan +449 a8370a8d-6ddd-11ea-8861-ccda262736ce https://socratic.org/questions/what-is-the-balanced-equation-of-so2-o2-so3 2 SO2 + O2 -> 2 SO3 start chemical_equation qc_end chemical_equation 6 10 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the balanced equation""}]" "[{""type"":""chemical equation"",""value"":""2 SO2 + O2 -> 2 SO3""}]" "[{""type"":""chemical equation"",""value"":""SO2 + O2 -> SO3""}]" "

What is the balanced equation of SO2 + O2 = SO3?

" nan 2 SO2 + O2 -> 2 SO3 "
+

Explanation:

+
+
+

You follow a systematic procedure to balance the equation.

+

Start with the unbalanced equation:

+

#""SO""_2 + ""O""_2 → ""SO""_3#

+

A method that often works is to balance everything other than #""O""# and #""H""# first, then balance #""O""#, and finally balance #""H""#.

+

Another useful procedure is to start with what looks like the most complicated formula.

+
+

The most complicated formula looks like #""SO""_3#. We put a 1 in front of it to remind ourselves that the number is now fixed.

+

#""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""S""#:

+

We have #""1 S""# on the right, so we need #""1 S""# on the left. We put a 1 in front of the #""SO""_2#.

+

#color(blue)(1) ""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""O""#:

+

We have #""3 O""# on the right, so we need #""3 O""# on the left. There are already #""4 O""# atoms on the left. We must put a ½ in front of the #""O""_2#.

+

Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.

+

To get rid of the fractions, you multiply all the coefficients by 2 and get

+

#color(blue)(2) ""SO""_2 + ""O""_2 → color(red)(2) ""SO""_3#

+

Now you can balance the equation by putting a 1 in front of the #""O""_2#.

+

#color(blue)(2) ""SO""_2 + color(orange)(1) ""O""_2 → color(red)(1) ""SO""_3#

+

We should now have a balanced equation.

+

Check to make sure:

+

#bb""Atom""color(white)(m)bb ""Left hand side""color(white)(m) bb""Right hand side""#
+#stackrel(————————————————)(color(white)(ml)""S""color(white)(mmmmmm) 2color(white)(mmmmmmmll) 2)#
+#color(white)(ml)""O""color(white)(mmmmmll) 6color(white)(mmmmmmmll) 6#

+

The balanced equation is

+

#""2SO""_2 + ""O""_2 → ""2SO""_3#

+
+
" "
+
+
+

The balanced equation is #""2SO""_2 + ""O""_2 → ""2SO""_3#.

+
+
+
+

Explanation:

+
+
+

You follow a systematic procedure to balance the equation.

+

Start with the unbalanced equation:

+

#""SO""_2 + ""O""_2 → ""SO""_3#

+

A method that often works is to balance everything other than #""O""# and #""H""# first, then balance #""O""#, and finally balance #""H""#.

+

Another useful procedure is to start with what looks like the most complicated formula.

+
+

The most complicated formula looks like #""SO""_3#. We put a 1 in front of it to remind ourselves that the number is now fixed.

+

#""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""S""#:

+

We have #""1 S""# on the right, so we need #""1 S""# on the left. We put a 1 in front of the #""SO""_2#.

+

#color(blue)(1) ""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""O""#:

+

We have #""3 O""# on the right, so we need #""3 O""# on the left. There are already #""4 O""# atoms on the left. We must put a ½ in front of the #""O""_2#.

+

Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.

+

To get rid of the fractions, you multiply all the coefficients by 2 and get

+

#color(blue)(2) ""SO""_2 + ""O""_2 → color(red)(2) ""SO""_3#

+

Now you can balance the equation by putting a 1 in front of the #""O""_2#.

+

#color(blue)(2) ""SO""_2 + color(orange)(1) ""O""_2 → color(red)(1) ""SO""_3#

+

We should now have a balanced equation.

+

Check to make sure:

+

#bb""Atom""color(white)(m)bb ""Left hand side""color(white)(m) bb""Right hand side""#
+#stackrel(————————————————)(color(white)(ml)""S""color(white)(mmmmmm) 2color(white)(mmmmmmmll) 2)#
+#color(white)(ml)""O""color(white)(mmmmmll) 6color(white)(mmmmmmmll) 6#

+

The balanced equation is

+

#""2SO""_2 + ""O""_2 → ""2SO""_3#

+
+
+
" "
+

What is the balanced equation of SO2 + O2 = SO3?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Balancing Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 26, 2014 + +
+
+
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+

The balanced equation is #""2SO""_2 + ""O""_2 → ""2SO""_3#.

+
+
+
+

Explanation:

+
+
+

You follow a systematic procedure to balance the equation.

+

Start with the unbalanced equation:

+

#""SO""_2 + ""O""_2 → ""SO""_3#

+

A method that often works is to balance everything other than #""O""# and #""H""# first, then balance #""O""#, and finally balance #""H""#.

+

Another useful procedure is to start with what looks like the most complicated formula.

+
+

The most complicated formula looks like #""SO""_3#. We put a 1 in front of it to remind ourselves that the number is now fixed.

+

#""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""S""#:

+

We have #""1 S""# on the right, so we need #""1 S""# on the left. We put a 1 in front of the #""SO""_2#.

+

#color(blue)(1) ""SO""_2 + ""O""_2 → color(red)(1) ""SO""_3#

+

Balance #""O""#:

+

We have #""3 O""# on the right, so we need #""3 O""# on the left. There are already #""4 O""# atoms on the left. We must put a ½ in front of the #""O""_2#.

+

Most instructors don't allow fractions in balanced chemical equations, because you can't have a fraction of an atom or a molecule.

+

To get rid of the fractions, you multiply all the coefficients by 2 and get

+

#color(blue)(2) ""SO""_2 + ""O""_2 → color(red)(2) ""SO""_3#

+

Now you can balance the equation by putting a 1 in front of the #""O""_2#.

+

#color(blue)(2) ""SO""_2 + color(orange)(1) ""O""_2 → color(red)(1) ""SO""_3#

+

We should now have a balanced equation.

+

Check to make sure:

+

#bb""Atom""color(white)(m)bb ""Left hand side""color(white)(m) bb""Right hand side""#
+#stackrel(————————————————)(color(white)(ml)""S""color(white)(mmmmmm) 2color(white)(mmmmmmmll) 2)#
+#color(white)(ml)""O""color(white)(mmmmmll) 6color(white)(mmmmmmmll) 6#

+

The balanced equation is

+

#""2SO""_2 + ""O""_2 → ""2SO""_3#

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" What is the balanced equation of SO2 + O2 = SO3? nan +450 a8370a8e-6ddd-11ea-8ad2-ccda262736ce https://socratic.org/questions/what-is-the-oxidation-number-of-hydrogen-in-h2-1 0 start physical_unit 6 8 oxidation_number none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation number [OF] hydrogen in H2""}]" "[{""type"":""physical unit"",""value"":""0""}]" "[{""type"":""chemical equation"",""value"":""H2""}]" "

What is the oxidation number of hydrogen in H2 ??

" nan 0 "
+

Explanation:

+
+

The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give #2xxdotH# radicals; a neutral charge and hence a zerovalent oxidation state. Capisce?

+

+ +

+
+
" "
+
+
+

We has zerovalent hydrogen, i.e. #H(0)#.

+
+
+
+

Explanation:

+
+

The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give #2xxdotH# radicals; a neutral charge and hence a zerovalent oxidation state. Capisce?

+

+ +

+
+
+
" "
+

What is the oxidation number of hydrogen in H2 ??

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+1 Answer +
+
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+ + +
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+ + Jun 12, 2017 + +
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We has zerovalent hydrogen, i.e. #H(0)#.

+
+
+
+

Explanation:

+
+

The oxidation number is the charge assigned when the bonding electrons are distributed to the most electronegative atom. Clearly, for a homonuclear diatomic molecule, the bound atoms have EQUAL electronegativity, and we conceive that the electrons are shared between the two atoms to give #2xxdotH# radicals; a neutral charge and hence a zerovalent oxidation state. Capisce?

+

+ +

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+ +
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+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 6830 views + around the world +
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
+
" What is the oxidation number of hydrogen in H2 ?? nan +451 a8370a8f-6ddd-11ea-8dba-ccda262736ce https://socratic.org/questions/58e68a587c014939ad5f321f 1.12 atm start physical_unit 5 5 pressure atm qc_end physical_unit 5 5 13 14 pressure qc_end end "[{""type"":""physical unit"",""value"":""Pressure2 [OF] atmospheres [IN] atm""}]" "[{""type"":""physical unit"",""value"":""1.12 atm""}]" "[{""type"":""physical unit"",""value"":""Pressure1 [OF] atmospheres [=] \\pu{854 mmHg}""}]" "

What is the pressure in atmospheres, if the pressure is reported to be #854*mm*Hg#?

" nan 1.12 atm "
+

Explanation:

+
+

One atmosphere of pressure will support of column of mercury that is #760*mm# high. And thus we can use a colume on mercury to measure pressures that ARE LESS than #1*atm#, i.e #<760*mm*Hg#.

+

Given your problem, I suppose we could state that #854*""Torr""-=(854*""mm Hg"")/(760 *""mm Hg""*atm^-1)=1.12*atm#. I really find the use of these units objectionable, because if you get a mercury spill in the lab, you have a major clean-up job, and guess who is going to do it?

+

And thus your question has not been consistently proposed. The pressure is LOWER at #1.12*atm#; i.e. it is expanded not compressed. Would you review this problem?

+
+
" "
+
+
+

I do not know why we seem to be getting a lot of questions that quote pressures of #>760*mm*Hg............#

+
+
+
+

Explanation:

+
+

One atmosphere of pressure will support of column of mercury that is #760*mm# high. And thus we can use a colume on mercury to measure pressures that ARE LESS than #1*atm#, i.e #<760*mm*Hg#.

+

Given your problem, I suppose we could state that #854*""Torr""-=(854*""mm Hg"")/(760 *""mm Hg""*atm^-1)=1.12*atm#. I really find the use of these units objectionable, because if you get a mercury spill in the lab, you have a major clean-up job, and guess who is going to do it?

+

And thus your question has not been consistently proposed. The pressure is LOWER at #1.12*atm#; i.e. it is expanded not compressed. Would you review this problem?

+
+
+
" "
+

What is the pressure in atmospheres, if the pressure is reported to be #854*mm*Hg#?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Apr 6, 2017 + +
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+
+
+
+
+
+

I do not know why we seem to be getting a lot of questions that quote pressures of #>760*mm*Hg............#

+
+
+
+

Explanation:

+
+

One atmosphere of pressure will support of column of mercury that is #760*mm# high. And thus we can use a colume on mercury to measure pressures that ARE LESS than #1*atm#, i.e #<760*mm*Hg#.

+

Given your problem, I suppose we could state that #854*""Torr""-=(854*""mm Hg"")/(760 *""mm Hg""*atm^-1)=1.12*atm#. I really find the use of these units objectionable, because if you get a mercury spill in the lab, you have a major clean-up job, and guess who is going to do it?

+

And thus your question has not been consistently proposed. The pressure is LOWER at #1.12*atm#; i.e. it is expanded not compressed. Would you review this problem?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1705 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the pressure in atmospheres, if the pressure is reported to be #854*mm*Hg#? nan +452 a8370a90-6ddd-11ea-9e9d-ccda262736ce https://socratic.org/questions/what-is-the-voltage-of-a-galvanic-cell-made-with-zinc-and-aluminum 0.90 V start physical_unit 5 7 voltage v qc_end c_other OTHER qc_end substance 10 10 qc_end substance 12 12 qc_end end "[{""type"":""physical unit"",""value"":""Voltage [OF] a galvanic cell [IN] V""}]" "[{""type"":""physical unit"",""value"":""0.90 V""}]" "[{""type"":""other"",""value"":""Standard conditions""},{""type"":""substance name"",""value"":""Zinc""},{""type"":""substance name"",""value"":""Aluminum""}]" "

What is the voltage of a galvanic cell made with zinc and aluminum at standard conditions?

" nan 0.90 V "
+

Explanation:

+
+

The standard reduction potentials for zinc and aluminum are listed below:

+

#Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

#Al^(3+)+3e^(-)->Al(s)"" "" "" "" """"xi^@=-1.66V#

+

In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.

+

Since almuninum has the lowest standard reduction potential #xi^@=-1.66V#, therefore, it should be oxidized as follows:

+

#color(red)(""Oxidation"")#: #Al(s)->Al^(3+)+3e^(-)"" "" "" "" """"-xi^@=1.66V#

+

Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.

+

Since zinc will be in the cathode compartment, it will get reduced as follows:

+

#color(blue)(""Reduction"")#: #Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:

+

#color(red)(""Oxidation"")#: #(Al(s)->Al^(3+)+3e^(-))color(purple)(xx2)"" "" "" "" """"-xi^@=1.66V#
+#color(blue)(""Reduction"")#: #(Zn^(2+)+2e^(-)->Zn(s))color(purple)(xx3)"" "" "" ""xi^@=-0.76V#
+#-------------------#
+#color(green)(""RedOx"")#: #color(purple)(2)Al(s)+color(purple)(3)Zn^(2+)->color(purple)(2)Al^(3+)+color(purple)(3)Zn(s)"" "" ""xi_(cell)^@=0.90V#

+

Therefore, the standard cell potential is : #xi_(cell)^@=0.90V#

+

Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.

+

Here is a video that further explains this topic:
+Electrochemistry | The Standard Reduction Potential.
+ +

+
+
" "
+
+
+

#xi_(cell)^@=0.90V#

+
+
+
+

Explanation:

+
+

The standard reduction potentials for zinc and aluminum are listed below:

+

#Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

#Al^(3+)+3e^(-)->Al(s)"" "" "" "" """"xi^@=-1.66V#

+

In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.

+

Since almuninum has the lowest standard reduction potential #xi^@=-1.66V#, therefore, it should be oxidized as follows:

+

#color(red)(""Oxidation"")#: #Al(s)->Al^(3+)+3e^(-)"" "" "" "" """"-xi^@=1.66V#

+

Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.

+

Since zinc will be in the cathode compartment, it will get reduced as follows:

+

#color(blue)(""Reduction"")#: #Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:

+

#color(red)(""Oxidation"")#: #(Al(s)->Al^(3+)+3e^(-))color(purple)(xx2)"" "" "" "" """"-xi^@=1.66V#
+#color(blue)(""Reduction"")#: #(Zn^(2+)+2e^(-)->Zn(s))color(purple)(xx3)"" "" "" ""xi^@=-0.76V#
+#-------------------#
+#color(green)(""RedOx"")#: #color(purple)(2)Al(s)+color(purple)(3)Zn^(2+)->color(purple)(2)Al^(3+)+color(purple)(3)Zn(s)"" "" ""xi_(cell)^@=0.90V#

+

Therefore, the standard cell potential is : #xi_(cell)^@=0.90V#

+

Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.

+

Here is a video that further explains this topic:
+Electrochemistry | The Standard Reduction Potential.
+ +

+
+
+
" "
+

What is the voltage of a galvanic cell made with zinc and aluminum at standard conditions?

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Calculating Energy in Electrochemical Processes + + +
+
+
+
+
+1 Answer +
+
+
+
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+ +
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+ +
+ + Aug 22, 2016 + +
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+

#xi_(cell)^@=0.90V#

+
+
+
+

Explanation:

+
+

The standard reduction potentials for zinc and aluminum are listed below:

+

#Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

#Al^(3+)+3e^(-)->Al(s)"" "" "" "" """"xi^@=-1.66V#

+

In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.

+

Since almuninum has the lowest standard reduction potential #xi^@=-1.66V#, therefore, it should be oxidized as follows:

+

#color(red)(""Oxidation"")#: #Al(s)->Al^(3+)+3e^(-)"" "" "" "" """"-xi^@=1.66V#

+

Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.

+

Since zinc will be in the cathode compartment, it will get reduced as follows:

+

#color(blue)(""Reduction"")#: #Zn^(2+)+2e^(-)->Zn(s)"" "" "" ""xi^@=-0.76V#

+

The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:

+

#color(red)(""Oxidation"")#: #(Al(s)->Al^(3+)+3e^(-))color(purple)(xx2)"" "" "" "" """"-xi^@=1.66V#
+#color(blue)(""Reduction"")#: #(Zn^(2+)+2e^(-)->Zn(s))color(purple)(xx3)"" "" "" ""xi^@=-0.76V#
+#-------------------#
+#color(green)(""RedOx"")#: #color(purple)(2)Al(s)+color(purple)(3)Zn^(2+)->color(purple)(2)Al^(3+)+color(purple)(3)Zn(s)"" "" ""xi_(cell)^@=0.90V#

+

Therefore, the standard cell potential is : #xi_(cell)^@=0.90V#

+

Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.

+

Here is a video that further explains this topic:
+Electrochemistry | The Standard Reduction Potential.
+ +

+
+
+
+
+
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+
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Related questions
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" What is the voltage of a galvanic cell made with zinc and aluminum at standard conditions? nan +453 a8370a91-6ddd-11ea-9335-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-105-mol-of-copper 6.72 g start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] copper [IN] g""}]" "[{""type"":""physical unit"",""value"":""6.72 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] copper [=] \\pu{0.105 mol}""}]" "

What is the mass of .105 mol of copper?

" nan 6.72 g "
+

Explanation:

+
+
    +
  1. You are given the number of mole which will be your data to find the corresponing amount of copper.
    2. +
  2. Find copper from the periodic table and take the atomic mass
    3. +
  3. By dimensional analysis, the amount of copper can be calculated using 0.105 mole of copper your base data.
    4. +
  4. The result will be in grams; the amount of copper at 0.105 mole Cu.
    5. +
+
+
" "
+
+
+

#6.672# grams of Cu

+
+
+
+

Explanation:

+
+
    +
  1. You are given the number of mole which will be your data to find the corresponing amount of copper.
    2. +
  2. Find copper from the periodic table and take the atomic mass
    3. +
  3. By dimensional analysis, the amount of copper can be calculated using 0.105 mole of copper your base data.
    4. +
  4. The result will be in grams; the amount of copper at 0.105 mole Cu.
    5. +
+
+
+
" "
+

What is the mass of .105 mol of copper?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
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+
+1 Answer +
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+ +
+
+ +
+ + Mar 4, 2016 + +
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+
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+
+

#6.672# grams of Cu

+
+
+
+

Explanation:

+
+
    +
  1. You are given the number of mole which will be your data to find the corresponing amount of copper.
    2. +
  2. Find copper from the periodic table and take the atomic mass
    3. +
  3. By dimensional analysis, the amount of copper can be calculated using 0.105 mole of copper your base data.
    4. +
  4. The result will be in grams; the amount of copper at 0.105 mole Cu.
    5. +
+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1421 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the mass of .105 mol of copper? nan +454 a8370a92-6ddd-11ea-8632-ccda262736ce https://socratic.org/questions/a-sample-of-gas-occupies-21-l-under-a-pressure-of-1-3-atm-what-would-the-resulti 7.00 L start physical_unit 0 3 volume l qc_end physical_unit 0 3 11 12 pressure qc_end physical_unit 0 3 5 6 volume qc_end physical_unit 0 3 25 26 pressure qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] A sample of gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""7.00 L""}]" "[{""type"":""physical unit"",""value"":""Pressure1 [OF] A sample of gas [=] \\pu{1.3 atm}""},{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of gas [=] \\pu{21 L}""},{""type"":""physical unit"",""value"":""Pressure2 [OF] A sample of gas [=] \\pu{3.9 atm}""},{""type"":""other"",""value"":""The temperature did not change.""}]" "

A sample of gas occupies 21 L under a pressure of 1.3 atm. What would the resulting volume be if the pressure were increased to 3.9 atm if the temperature did not change?

" nan 7.00 L "
+

Explanation:

+
+

We apply Boyle's Law

+

#P_1V_1=P_2V_2#

+

The initial volume is #V-1=21L#

+

The initial pressure is #P_1=1.3 atm#

+

The final pressure is #P_2=3.9atm#

+

The final volume is

+

#V_2=P_1/P_2*V_1=1.3/3.9*21#

+

#=7 L#

+
+
" "
+
+
+

The volume is #=7L#

+
+
+
+

Explanation:

+
+

We apply Boyle's Law

+

#P_1V_1=P_2V_2#

+

The initial volume is #V-1=21L#

+

The initial pressure is #P_1=1.3 atm#

+

The final pressure is #P_2=3.9atm#

+

The final volume is

+

#V_2=P_1/P_2*V_1=1.3/3.9*21#

+

#=7 L#

+
+
+
" "
+

A sample of gas occupies 21 L under a pressure of 1.3 atm. What would the resulting volume be if the pressure were increased to 3.9 atm if the temperature did not change?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
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+
+1 Answer +
+
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+
+
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+ +
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+ +
+ + Jul 5, 2017 + +
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+

The volume is #=7L#

+
+
+
+

Explanation:

+
+

We apply Boyle's Law

+

#P_1V_1=P_2V_2#

+

The initial volume is #V-1=21L#

+

The initial pressure is #P_1=1.3 atm#

+

The final pressure is #P_2=3.9atm#

+

The final volume is

+

#V_2=P_1/P_2*V_1=1.3/3.9*21#

+

#=7 L#

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+ +
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+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4635 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A sample of gas occupies 21 L under a pressure of 1.3 atm. What would the resulting volume be if the pressure were increased to 3.9 atm if the temperature did not change? nan +455 a8370a93-6ddd-11ea-acc8-ccda262736ce https://socratic.org/questions/5955d17eb72cff185ffec5aa -232726.95 kJ start physical_unit 21 22 reaction_enthalpy kj qc_end chemical_equation 4 10 qc_end c_other OTHER qc_end physical_unit 2 2 24 27 mass qc_end end "[{""type"":""physical unit"",""value"":""Reaction enthalpy [OF] the formation [IN] kJ""}]" "[{""type"":""physical unit"",""value"":""-232726.95 kJ""}]" "[{""type"":""chemical equation"",""value"":""N2(g) + 3 H2(g) -> 2 NH3(g)""},{""type"":""other"",""value"":""DeltaH^{0}_{rxn} = −92.6 kJ/mol""},{""type"":""physical unit"",""value"":""Mass [OF] ammonia [=] \\pu{8.55 × 10^4 g}""}]" "

Given the ammonia synthesis: +#N_2(g) + 3H_2(g) rarr 2NH_3(g)# #DeltaH_""rxn""^@=-92.6*kJ*mol^-1#, + +what reaction enthalpy is associated with the formation of #8.55xx10^4*g# of ammonia?

" nan -232726.95 kJ "
+

Explanation:

+
+

We got...

+

#N_2(g) + 3H_2(g) rarr 2NH_3(g)+92.6*kJ#

+

Which tells that the formation of #34*g# ammonia releases #2xx92.6*kJ#. Agreed?

+

But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.

+

#""Moles of ammonia""=(8.55xx10^4*g)/(17.01*g*mol^-1)=5026.5*mol#.

+

And thus #DeltaH_""rxn""=(-5026.5*molxx92.6*kJ*mol^-1)/2#

+

#=??*kJ#

+

Why is #DeltaH_""rxn""# negative?

+
+
" "
+
+
+

How? We use the enthalpy as a variable in the reaction..........and get #DeltaH_""rxn""=-232725*kJ........#

+
+
+
+

Explanation:

+
+

We got...

+

#N_2(g) + 3H_2(g) rarr 2NH_3(g)+92.6*kJ#

+

Which tells that the formation of #34*g# ammonia releases #2xx92.6*kJ#. Agreed?

+

But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.

+

#""Moles of ammonia""=(8.55xx10^4*g)/(17.01*g*mol^-1)=5026.5*mol#.

+

And thus #DeltaH_""rxn""=(-5026.5*molxx92.6*kJ*mol^-1)/2#

+

#=??*kJ#

+

Why is #DeltaH_""rxn""# negative?

+
+
+
" "
+

Given the ammonia synthesis: +#N_2(g) + 3H_2(g) rarr 2NH_3(g)# #DeltaH_""rxn""^@=-92.6*kJ*mol^-1#, + +what reaction enthalpy is associated with the formation of #8.55xx10^4*g# of ammonia?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Enthalpy + + +
+
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+1 Answer +
+
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+ +
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+ +
+ + Jun 30, 2017 + +
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How? We use the enthalpy as a variable in the reaction..........and get #DeltaH_""rxn""=-232725*kJ........#

+
+
+
+

Explanation:

+
+

We got...

+

#N_2(g) + 3H_2(g) rarr 2NH_3(g)+92.6*kJ#

+

Which tells that the formation of #34*g# ammonia releases #2xx92.6*kJ#. Agreed?

+

But here we have made much more than 2 mol ammonia, and the enthalpy changes appropriately.

+

#""Moles of ammonia""=(8.55xx10^4*g)/(17.01*g*mol^-1)=5026.5*mol#.

+

And thus #DeltaH_""rxn""=(-5026.5*molxx92.6*kJ*mol^-1)/2#

+

#=??*kJ#

+

Why is #DeltaH_""rxn""# negative?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1391 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
" "Given the ammonia synthesis: +#N_2(g) + 3H_2(g) rarr 2NH_3(g)# #DeltaH_""rxn""^@=-92.6*kJ*mol^-1#, + +what reaction enthalpy is associated with the formation of #8.55xx10^4*g# of ammonia?" nan +456 a8370a94-6ddd-11ea-9a3d-ccda262736ce https://socratic.org/questions/a-solution-of-iron-iii-chloride-is-mixed-with-a-solution-of-sodium-hydroxide-and FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq) start chemical_equation qc_end substance 3 4 qc_end substance 11 12 qc_end substance 17 20 qc_end substance 22 24 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this reaction""}]" "[{""type"":""chemical equation"",""value"":""FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq)""}]" "[{""type"":""substance name"",""value"":""Iron(III) chloride""},{""type"":""substance name"",""value"":""Sodium hydroxide""},{""type"":""substance name"",""value"":""solid iron (III) hydroxide""},{""type"":""substance name"",""value"":""aqueous sodium chloride""}]" "

A solution of iron(III)chloride is mixed with a solution of sodium hydroxide and reacts to yield solid iron (III) hydroxide and aqueous sodium chloride. What is the balanced chemical equation for this reaction?

" nan FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq) "
+

Explanation:

+
+

Start by writing the unbalanced chemical equation that describes this double displacement reaction

+
+

#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#

+
+

In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.

+
+

#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#

+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

You can rewrite the unbalanced chemical equation as

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by #color(blue)(3)#.

+

Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by #color(blue)(3)#, you must do the same for the sodium cations

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

Now balance the sodium cations present on the reactants' side by multiplying them by #color(red)(3)#. Once again, the sodium cations are part of the sodium hydroxide, so make sure to multiply the hydroxide anions as well!

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

The hydroxide anions are now balanced because you have #3# hydroxide anions on the products' side as part of the insoluble iron(III) hydroxide. The same can be said about the iron(III) cations, so put all this together

+
+

#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#

+
+

to get the balanced chemical equation

+
+

#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#

+
+

You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.

+
+

#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#

+
+
+
" "
+
+
+

#""FeCl""_ (3(aq)) + 3""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + 3""NaCl""_ ((aq))#

+
+
+
+

Explanation:

+
+

Start by writing the unbalanced chemical equation that describes this double displacement reaction

+
+

#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#

+
+

In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.

+
+

#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#

+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

You can rewrite the unbalanced chemical equation as

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by #color(blue)(3)#.

+

Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by #color(blue)(3)#, you must do the same for the sodium cations

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

Now balance the sodium cations present on the reactants' side by multiplying them by #color(red)(3)#. Once again, the sodium cations are part of the sodium hydroxide, so make sure to multiply the hydroxide anions as well!

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

The hydroxide anions are now balanced because you have #3# hydroxide anions on the products' side as part of the insoluble iron(III) hydroxide. The same can be said about the iron(III) cations, so put all this together

+
+

#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#

+
+

to get the balanced chemical equation

+
+

#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#

+
+

You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.

+
+

#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#

+
+
+
+
" "
+

A solution of iron(III)chloride is mixed with a solution of sodium hydroxide and reacts to yield solid iron (III) hydroxide and aqueous sodium chloride. What is the balanced chemical equation for this reaction?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Jun 14, 2017 + +
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#""FeCl""_ (3(aq)) + 3""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + 3""NaCl""_ ((aq))#

+
+
+
+

Explanation:

+
+

Start by writing the unbalanced chemical equation that describes this double displacement reaction

+
+

#""FeCl""_ (3(aq)) + ""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + ""NaCl""#

+
+

In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.

+
+

#""FeCl""_ (3(aq)) -> ""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)#

+

#""NaOH""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)#

+

#""NaCl""_ ((aq)) -> ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

You can rewrite the unbalanced chemical equation as

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + ""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)#

+
+

The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by #color(blue)(3)#.

+

Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by #color(blue)(3)#, you must do the same for the sodium cations

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + ""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

Now balance the sodium cations present on the reactants' side by multiplying them by #color(red)(3)#. Once again, the sodium cations are part of the sodium hydroxide, so make sure to multiply the hydroxide anions as well!

+
+

#""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-) + color(red)(3)""Na""_ ((aq))^(+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""Na""_ ((aq))^(+) + color(blue)(3)""Cl""_ ((aq))^(-)#

+
+

The hydroxide anions are now balanced because you have #3# hydroxide anions on the products' side as part of the insoluble iron(III) hydroxide. The same can be said about the iron(III) cations, so put all this together

+
+

#[""Fe""_ ((aq))^(3+) + 3""Cl""_ ((aq))^(-)] + color(red)(3) xx [""Na""_ ((aq))^(+) + ""OH""_ ((aq))^(-)] -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)xx[""Na""_ ((aq))^(+) + ""Cl""_ ((aq))^(-)]#

+
+

to get the balanced chemical equation

+
+

#""FeCl""_ (3(aq)) + color(red)(3)""NaOH""_ ((aq)) -> ""Fe""(""OH"")_ (3(s)) darr + color(blue)(3)""NaCl""_ ((aq))#

+
+

You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.

+
+

#""Fe""_ ((aq))^(3+) + color(red)(3)""OH""_ ((aq))^(-) -> ""Fe""(""OH"")_ (3(s)) darr#

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" A solution of iron(III)chloride is mixed with a solution of sodium hydroxide and reacts to yield solid iron (III) hydroxide and aqueous sodium chloride. What is the balanced chemical equation for this reaction? nan +457 a8370a95-6ddd-11ea-8e9c-ccda262736ce https://socratic.org/questions/how-many-liters-of-a-90-acid-solution-must-be-added-to-6-liters-of-a-15-acid-sol 3.00 liters start physical_unit 6 7 volume l qc_end physical_unit 6 6 5 5 percent qc_end physical_unit 6 6 16 16 percent qc_end physical_unit 6 7 12 13 volume qc_end physical_unit 6 6 22 22 percent qc_end end "[{""type"":""physical unit"",""value"":""Volume1 [OF] acid solution [IN] liters""}]" "[{""type"":""physical unit"",""value"":""3.00 liters""}]" "[{""type"":""physical unit"",""value"":""Percent1 [OF] acid [=] \\pu{90%}""},{""type"":""physical unit"",""value"":""Percent2 [OF] acid [=] \\pu{15%}""},{""type"":""physical unit"",""value"":""Volume2 [OF] acid solution [=] \\pu{6 liters}""},{""type"":""physical unit"",""value"":""Percent3 [OF] acid [=] \\pu{40%}""}]" "

How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

" nan 3.00 liters "
+

Explanation:

+
+
+

For this problem, we can use the relation

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#

+
+
+

Let the #""volume of 90 % acid"" = xcolor(white)(l) ""L""#

+

Then, after mixing, we have #(6 + x"") L of 40 % acid""#.

+

This is made up of #""6 L of 15 % acid""# and #xcolor(white)(l) ""L of 90 % acid""#.

+
+

#""Moles before = moles after""#

+

#c_1V_1 = c_2V_2 + c_3V_3#

+

#40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)(""L""))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)(""L""))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)(""L"")))#

+

#240 + 40x = 90 + 90x#

+

#50x = 150#

+

#x = 150/50 = 3#

+

So, we add #""3 L of 90 % acid""# to #""6 L of 15 % acid""# and get #""9 L of 40 % acid""#.

+
+

Check:

+

#3 × 90 + 6 × 15 = 9 × 40#

+

#270 + 90 = 360#

+

#360 = 360#

+

It checks!

+
+
" "
+
+
+

You must add 3 L of the 90 % acid.

+
+
+
+

Explanation:

+
+
+

For this problem, we can use the relation

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#

+
+
+

Let the #""volume of 90 % acid"" = xcolor(white)(l) ""L""#

+

Then, after mixing, we have #(6 + x"") L of 40 % acid""#.

+

This is made up of #""6 L of 15 % acid""# and #xcolor(white)(l) ""L of 90 % acid""#.

+
+

#""Moles before = moles after""#

+

#c_1V_1 = c_2V_2 + c_3V_3#

+

#40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)(""L""))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)(""L""))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)(""L"")))#

+

#240 + 40x = 90 + 90x#

+

#50x = 150#

+

#x = 150/50 = 3#

+

So, we add #""3 L of 90 % acid""# to #""6 L of 15 % acid""# and get #""9 L of 40 % acid""#.

+
+

Check:

+

#3 × 90 + 6 × 15 = 9 × 40#

+

#270 + 90 = 360#

+

#360 = 360#

+

It checks!

+
+
+
" "
+

How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solution Formation + + +
+
+
+
+
+1 Answer +
+
+
+
+
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+ +
+
+ +
+ + Sep 24, 2016 + +
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+
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+

You must add 3 L of the 90 % acid.

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+
+
+

Explanation:

+
+
+

For this problem, we can use the relation

+
+
+

#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = ""amount of solute"" color(white)(a/a)|)))"" ""#

+
+
+

Let the #""volume of 90 % acid"" = xcolor(white)(l) ""L""#

+

Then, after mixing, we have #(6 + x"") L of 40 % acid""#.

+

This is made up of #""6 L of 15 % acid""# and #xcolor(white)(l) ""L of 90 % acid""#.

+
+

#""Moles before = moles after""#

+

#c_1V_1 = c_2V_2 + c_3V_3#

+

#40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)(""L""))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)(""L""))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)(""L"")))#

+

#240 + 40x = 90 + 90x#

+

#50x = 150#

+

#x = 150/50 = 3#

+

So, we add #""3 L of 90 % acid""# to #""6 L of 15 % acid""# and get #""9 L of 40 % acid""#.

+
+

Check:

+

#3 × 90 + 6 × 15 = 9 × 40#

+

#270 + 90 = 360#

+

#360 = 360#

+

It checks!

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
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+
Impact of this question
+
+ 4162 views + around the world +
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+ + Creative Commons License + +
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+
" How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution? nan +458 a8370a96-6ddd-11ea-aacc-ccda262736ce https://socratic.org/questions/if-the-h-3o-in-a-solution-is-2-10-5-m-what-is-the-oh 5.00 × 10^(−10) M start physical_unit 14 14 molarity mol/l qc_end physical_unit 2 2 7 10 molarity qc_end end "[{""type"":""physical unit"",""value"":""Molarity [OF] OH- [IN] M""}]" "[{""type"":""physical unit"",""value"":""5.00 × 10^(−10) M""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] H3O+ [=] \\pu{2 × 10^(−5) M}""}]" "

If the #[""H""_3""O""^(+)]# in a solution is #2*10^-5# #""M""#, what is the #[""OH""^(-)]#?

" nan 5.00 × 10^(−10) M "
+

Explanation:

+
+

For pure water at room temperature, the concentration of hydronium cations, #""H""_3""O""^(+)#, and the concentration of hydroxide anions, #""OH""^(-)#, have the following relationship

+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#

+
+

This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations, #10^(-7)""M""#, of hydronium and hydroxide ions.

+

+

So, you're dealing with an aqueous solution that has

+
+

#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#

+
+

Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.

+

Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.

+

This of course implies that the concentration of hydroxide anions will be lower than #10^(-7)""M""#, what you get in pure water.

+

More specifically, the concentration of hydroxide anions will be

+
+

#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#

+
+

As practice, you can find the pH of this solution by using

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#""pH"" = - log(2 * 10^(-5)) = 4.7#

+
+
+
" "
+
+
+

#[""OH""^(-)] = 5 * 10^(-10)""M""#

+
+
+
+

Explanation:

+
+

For pure water at room temperature, the concentration of hydronium cations, #""H""_3""O""^(+)#, and the concentration of hydroxide anions, #""OH""^(-)#, have the following relationship

+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#

+
+

This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations, #10^(-7)""M""#, of hydronium and hydroxide ions.

+

+

So, you're dealing with an aqueous solution that has

+
+

#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#

+
+

Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.

+

Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.

+

This of course implies that the concentration of hydroxide anions will be lower than #10^(-7)""M""#, what you get in pure water.

+

More specifically, the concentration of hydroxide anions will be

+
+

#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#

+
+

As practice, you can find the pH of this solution by using

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#""pH"" = - log(2 * 10^(-5)) = 4.7#

+
+
+
+
" "
+

If the #[""H""_3""O""^(+)]# in a solution is #2*10^-5# #""M""#, what is the #[""OH""^(-)]#?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Mar 7, 2016 + +
+
+
+
+
+
+
+

#[""OH""^(-)] = 5 * 10^(-10)""M""#

+
+
+
+

Explanation:

+
+

For pure water at room temperature, the concentration of hydronium cations, #""H""_3""O""^(+)#, and the concentration of hydroxide anions, #""OH""^(-)#, have the following relationship

+
+

#color(blue)(|bar(ul(color(white)(a/a)[""H""_3""O""^(+)] * [""OH""^(-)] = 10^(-14)""M""^2color(white)(a/a)|)))#

+
+

This relationship is based on the self-ionization of water, which at room temperature produces equal concentrations, #10^(-7)""M""#, of hydronium and hydroxide ions.

+

+

So, you're dealing with an aqueous solution that has

+
+

#[""H""_3""O""^(+)] = 2 * 10^(-5)""M""#

+
+

Right from the start, you can tell that this solution is acidic, since the concentration of hydronium ions increased compared with that of pure water.

+

Despite the fact that you have more hydronium ions present, the relationship between the hydronium and hydroxide anions remains valid.

+

This of course implies that the concentration of hydroxide anions will be lower than #10^(-7)""M""#, what you get in pure water.

+

More specifically, the concentration of hydroxide anions will be

+
+

#[""OH""^(-)] = (10^(-14)""M""^color(red)(cancel(color(black)(2))))/(2 * 10^(-5)color(red)(cancel(color(black)(""M"")))) = color(green)(|bar(ul(color(white)(a/a)5 * 10^(-10)""M""color(white)(a/a)|)))#

+
+

As practice, you can find the pH of this solution by using

+
+

#color(blue)(|bar(ul(color(white)(a/a)""pH"" = - log([""H""_3""O""^(+)])color(white)(a/a)|)))#

+
+

In this case, you will have

+
+

#""pH"" = - log(2 * 10^(-5)) = 4.7#

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+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 3525 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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" "If the #[""H""_3""O""^(+)]# in a solution is #2*10^-5# #""M""#, what is the #[""OH""^(-)]#?" nan +459 a83731e6-6ddd-11ea-a752-ccda262736ce https://socratic.org/questions/582db73b7c014975f0f79b25 0.24 mol/L start physical_unit 24 26 concentration mol/l qc_end physical_unit 5 6 1 2 volume qc_end physical_unit 17 18 13 14 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Concentration [OF] the nitric acid [IN] mol/L""}]" "[{""type"":""physical unit"",""value"":""0.24 mol/L""}]" "[{""type"":""physical unit"",""value"":""Volume [OF] nitric acid [=] \\pu{33.25 mL}""},{""type"":""physical unit"",""value"":""Mass [OF] sodium carbonate [=] \\pu{0.425 g}""},{""type"":""other"",""value"":""Nitric acid was required for equivalence with sodium carbonate.""}]" "

A #33.25*mL# volume of nitric acid was required for equivalence with a #0.425*g# mass of sodium carbonate. What is the concentration of the nitric acid?

" nan 0.24 mol/L "
+

Explanation:

+
+

With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.

+

#Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + H_2O(l) + CO_2(g)uarr#

+

And then we work out the number of moles of the reagents.

+

#""Moles of sodium carbonate ""=#

+

#(0.425*g)/(105.99*g*mol^-1)=4.01xx10^-3*mol#

+

Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:

+

#[HNO_3]=(2xx4.01xx10^-3*mol)/(33.25*mL)xx10^3*mL*L^-1~=#

+

#0.25*mol*L^-1#.

+
+
" "
+
+
+

We work out the number of moles of each reagent...............and calculate #[HNO_3]# to be a bit over #0.2*mol*L^-1#.

+
+
+
+

Explanation:

+
+

With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.

+

#Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + H_2O(l) + CO_2(g)uarr#

+

And then we work out the number of moles of the reagents.

+

#""Moles of sodium carbonate ""=#

+

#(0.425*g)/(105.99*g*mol^-1)=4.01xx10^-3*mol#

+

Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:

+

#[HNO_3]=(2xx4.01xx10^-3*mol)/(33.25*mL)xx10^3*mL*L^-1~=#

+

#0.25*mol*L^-1#.

+
+
+
" "
+

A #33.25*mL# volume of nitric acid was required for equivalence with a #0.425*g# mass of sodium carbonate. What is the concentration of the nitric acid?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molarity + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Nov 17, 2016 + +
+
+
+
+
+
+
+

We work out the number of moles of each reagent...............and calculate #[HNO_3]# to be a bit over #0.2*mol*L^-1#.

+
+
+
+

Explanation:

+
+

With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.

+

#Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + H_2O(l) + CO_2(g)uarr#

+

And then we work out the number of moles of the reagents.

+

#""Moles of sodium carbonate ""=#

+

#(0.425*g)/(105.99*g*mol^-1)=4.01xx10^-3*mol#

+

Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:

+

#[HNO_3]=(2xx4.01xx10^-3*mol)/(33.25*mL)xx10^3*mL*L^-1~=#

+

#0.25*mol*L^-1#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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+
Impact of this question
+
+ 1333 views + around the world +
+
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+ You can reuse this answer +
+ + Creative Commons License + +
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+
" A #33.25*mL# volume of nitric acid was required for equivalence with a #0.425*g# mass of sodium carbonate. What is the concentration of the nitric acid? nan +460 a83731e7-6ddd-11ea-ae2d-ccda262736ce https://socratic.org/questions/what-is-the-o-n-of-carbonyl-carbon-in-ethyl-4-methylpentanoate-1 +3 start physical_unit 5 9 o.n none qc_end substance 8 9 qc_end end "[{""type"":""physical unit"",""value"":""O.N [OF] carbonyl carbon in Ethyl 4-methylpentanoate""}]" "[{""type"":""physical unit"",""value"":""+3""}]" "[{""type"":""substance name"",""value"":""Ethyl 4-methylpentanoate""}]" "

What is the O.N of carbonyl carbon in Ethyl 4-methylpentanoate ???

" nan +3 "
+

Explanation:

+
+

As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.

+

The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.

+

The bond-line notation for ethyl-4-methylpentanoate looks like this

+

+

A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..

+

When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.

+

So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.

+

Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.

+

This will leave carbon deprived of two electrons, giving it a +2 oxidation state.

+

The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.

+

This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.

+
+
" "
+
+
+

The oxidation number of the carbonyl carbon is +3.

+
+
+
+

Explanation:

+
+

As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.

+

The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.

+

The bond-line notation for ethyl-4-methylpentanoate looks like this

+

+

A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..

+

When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.

+

So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.

+

Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.

+

This will leave carbon deprived of two electrons, giving it a +2 oxidation state.

+

The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.

+

This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.

+
+
+
" "
+

What is the O.N of carbonyl carbon in Ethyl 4-methylpentanoate ???

+
+
+ + +Chemistry + + + + + +Electrochemistry + + + + + +Oxidation Numbers + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 31, 2015 + +
+
+
+
+
+
+
+

The oxidation number of the carbonyl carbon is +3.

+
+
+
+

Explanation:

+
+

As you know, oxidation numbers are assigned by treating all bonds an atom has as ionic.

+

The idea behind that is that the more electronegative atom will take all the bonding electrons from the less electronegative atom.

+

The bond-line notation for ethyl-4-methylpentanoate looks like this

+

+

A carbonyl group is simply a carbon atom double bonded to an oxygen atom. In the case of ethyl-4-methylpentanoate, the carbon that's part of the carbonyl group is also bonded to another carbon atom and to another oxygen atom, th oxygen from the ether group..

+

When a carbon atom is bonded to another carbon atom, the difference in electronegativity between these two atoms is equal to zero.

+

So the carbonyl carbon will neither take, nor lose electrons from its bond to the other carbon atom. Remember that the ""taking"" or ""losing"" of electrons is hypothetical.

+

Since oxygen is more electronegative than carbon, it will take the electrons carbon is contributing to the bond. In the case of the double bond, oxygen will take all four bonding electrons, i.e. both electrons carbon shares to the bond.

+

This will leave carbon deprived of two electrons, giving it a +2 oxidation state.

+

The ether oxygen that is single bonded to the carbonyl atom will take both bonding electrons, i.e. the electron carbon shares to the bond.

+

This means that the carbonyl carbon has ""lost"" an additional 1 electron, bringing the total number of electrons it ""lost"" to 3. As a result, the carbonyl carbon's oxidation state is +3.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
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Impact of this question
+
+ 2094 views + around the world +
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+ + Creative Commons License + +
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" What is the O.N of carbonyl carbon in Ethyl 4-methylpentanoate ??? nan +461 a83731e8-6ddd-11ea-8644-ccda262736ce https://socratic.org/questions/a-student-mixes-50g-of-sugar-with-300g-of-water-and-5g-of-lemon-what-is-the-mass 355.00 g start physical_unit 22 22 mass g qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 16 16 13 14 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] solution [IN] g""}]" "[{""type"":""physical unit"",""value"":""355.00 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] sugar [=] \\pu{50 g}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{300 g}""},{""type"":""physical unit"",""value"":""Mass [OF] lemon [=] \\pu{5 g}""}]" "

A student mixes 50g of sugar with 300g of water and 5g of lemon. What is the mass of solution?

" nan 355.00 g "
+

Explanation:

+
+

Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.

+

Here #355*g# of non-volatile material were used; and necessarily #355*g# of mass remain.

+

In another experiment a punter used the same quantities of material but used #300*g# carbonated water, i.e. fizzy water. After a few hours, the experimenter noticed a slight DROP in mass. How is this possible?

+
+
" "
+
+
+

Why #355*g#.

+
+
+
+

Explanation:

+
+

Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.

+

Here #355*g# of non-volatile material were used; and necessarily #355*g# of mass remain.

+

In another experiment a punter used the same quantities of material but used #300*g# carbonated water, i.e. fizzy water. After a few hours, the experimenter noticed a slight DROP in mass. How is this possible?

+
+
+
" "
+

A student mixes 50g of sugar with 300g of water and 5g of lemon. What is the mass of solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Solution Formation + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 29, 2016 + +
+
+
+
+
+
+
+

Why #355*g#.

+
+
+
+

Explanation:

+
+

Every chemical reaction conserves the component masses; of course such mass may be lost to the experimenter by evolution of gas or thru handling.

+

Here #355*g# of non-volatile material were used; and necessarily #355*g# of mass remain.

+

In another experiment a punter used the same quantities of material but used #300*g# carbonated water, i.e. fizzy water. After a few hours, the experimenter noticed a slight DROP in mass. How is this possible?

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1852 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A student mixes 50g of sugar with 300g of water and 5g of lemon. What is the mass of solution? nan +462 a83731e9-6ddd-11ea-8b0a-ccda262736ce https://socratic.org/questions/what-is-the-percent-water-in-the-compound-barium-chloride-dihydrate 14.51% start physical_unit 4 7 percent none qc_end substance 8 10 qc_end end "[{""type"":""physical unit"",""value"":""Percent [OF] water in the compound""}]" "[{""type"":""physical unit"",""value"":""14.51%""}]" "[{""type"":""substance name"",""value"":""Barium chloride dihydrate""}]" "

What is the percent water in the compound barium chloride dihydrate?

" nan 14.51% "
+

Explanation:

+
+

As its name suggests, barium chloride dihydrate, #""BaCl""_2 * color(red)(2)""H""_2""O""#, is an ionic compound that contains water molecules in its crystal structure, i.e a hydrate.

+

The water that's part of the solid's crystal structure is called water of crystallization.

+

In the case of barium chloride dihydrate, you get #color(red)(2)# moles of water of crystallization for every #1# mole of hydrate.

+

In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with #1# mole of anhydrous barium chloride, #""BaCl""_2#, for every #1# mole of hydrate.

+

The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.

+

Barium chloride dihydrate has a molar mass of #""244.26 g mol""^(-1)#, which means that one mole of hydrate has a mass of #""244.26 g""#.

+

Since one mole of hydrate contains #color(red)(2)# moles of water, and since water has a molar mass of #""18.015 g mol""^(-1)#, it follows that you get #color(red)(2) xx ""18.015 g""# of water for every #""244.26 g""# of hydrate.

+

This means that the percent composition of water is

+
+

#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#

+
+

This means that you get #""14.751 g""# of water for every #""100 g""# of #""BaCl""_2 * 2""H""_2""O""#.

+

Here is a similar lab with analysis conducted using copper (II) sulfate.

+

+ +

+

Hope this helps!

+
+
" "
+
+
+

#14.751%#

+
+
+
+

Explanation:

+
+

As its name suggests, barium chloride dihydrate, #""BaCl""_2 * color(red)(2)""H""_2""O""#, is an ionic compound that contains water molecules in its crystal structure, i.e a hydrate.

+

The water that's part of the solid's crystal structure is called water of crystallization.

+

In the case of barium chloride dihydrate, you get #color(red)(2)# moles of water of crystallization for every #1# mole of hydrate.

+

In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with #1# mole of anhydrous barium chloride, #""BaCl""_2#, for every #1# mole of hydrate.

+

The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.

+

Barium chloride dihydrate has a molar mass of #""244.26 g mol""^(-1)#, which means that one mole of hydrate has a mass of #""244.26 g""#.

+

Since one mole of hydrate contains #color(red)(2)# moles of water, and since water has a molar mass of #""18.015 g mol""^(-1)#, it follows that you get #color(red)(2) xx ""18.015 g""# of water for every #""244.26 g""# of hydrate.

+

This means that the percent composition of water is

+
+

#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#

+
+

This means that you get #""14.751 g""# of water for every #""100 g""# of #""BaCl""_2 * 2""H""_2""O""#.

+

Here is a similar lab with analysis conducted using copper (II) sulfate.

+

+ +

+

Hope this helps!

+
+
+
" "
+

What is the percent water in the compound barium chloride dihydrate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Mar 11, 2016 + +
+
+
+
+
+
+
+

#14.751%#

+
+
+
+

Explanation:

+
+

As its name suggests, barium chloride dihydrate, #""BaCl""_2 * color(red)(2)""H""_2""O""#, is an ionic compound that contains water molecules in its crystal structure, i.e a hydrate.

+

The water that's part of the solid's crystal structure is called water of crystallization.

+

In the case of barium chloride dihydrate, you get #color(red)(2)# moles of water of crystallization for every #1# mole of hydrate.

+

In other words, if you were to drive the water of crystallization off by heating the hydrate, you would be left with #1# mole of anhydrous barium chloride, #""BaCl""_2#, for every #1# mole of hydrate.

+

The find the hydrate's percent composition of water in barium chloride dihydrate, use the mass of one mole of hydrate and of two moles of water.

+

Barium chloride dihydrate has a molar mass of #""244.26 g mol""^(-1)#, which means that one mole of hydrate has a mass of #""244.26 g""#.

+

Since one mole of hydrate contains #color(red)(2)# moles of water, and since water has a molar mass of #""18.015 g mol""^(-1)#, it follows that you get #color(red)(2) xx ""18.015 g""# of water for every #""244.26 g""# of hydrate.

+

This means that the percent composition of water is

+
+

#(color(red)(2) xx 18.015color(red)(cancel(color(black)(""g""))))/(244.26color(red)(cancel(color(black)(""g"")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)""14.751%""color(white)(a/a)|)))#

+
+

This means that you get #""14.751 g""# of water for every #""100 g""# of #""BaCl""_2 * 2""H""_2""O""#.

+

Here is a similar lab with analysis conducted using copper (II) sulfate.

+

+ +

+

Hope this helps!

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 61993 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the percent water in the compound barium chloride dihydrate? nan +463 a83731ea-6ddd-11ea-b303-ccda262736ce https://socratic.org/questions/a-sample-of-gas-occupies-a-volume-of-70-9-ml-as-it-expands-it-does-118-9-j-of-wo 1.21 L start physical_unit 35 36 volume l qc_end physical_unit 0 3 8 9 volume qc_end physical_unit 35 36 15 16 energy qc_end physical_unit 35 36 27 28 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume2 [OF] the gas [IN] L""}]" "[{""type"":""physical unit"",""value"":""1.21 L""}]" "[{""type"":""physical unit"",""value"":""Volume1 [OF] A sample of gas [=] \\pu{70.9 mL}""},{""type"":""physical unit"",""value"":""Energy [OF] the gas [=] \\pu{118.9 J}""},{""type"":""physical unit"",""value"":""Constant pressure [OF] the gas [=] \\pu{783 torr}""}]" "

A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

" nan 1.21 L "
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
" "
+
+
+

#V_2=1.07L#

+
+
+
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
+
" "
+

A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Gas Laws + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ + +
+
+ +
+ + Oct 1, 2016 + +
+
+
+
+
+
+
+

#V_2=1.07L#

+
+
+
+

Explanation:

+
+

The relationship between work (#w#), pressure (#P#) and volume (#V#) is the following:

+

#w=-PDeltaV#

+

where, #DeltaV=V_2-V_1#

+

since the gas is expanding, then the work is done by the system and it is of a negative value .

+

Note that work, in this case, should be expressed in #L*atm#.

+

#1L*atm=101.3J# therefore,
+#w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm#

+

Since work is done by the system: #w=-1.174L*atm#

+

Pressure should then be expressed in #atm#:

+

#P=783cancel(""torr"")xx(1atm)/(760cancel(""torr""))=1.03atm#

+

Thus, replacing every term in its value in the expression #w=-PDeltaV# we get:

+

#cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV#

+

#=>DeltaV=(1.174)/(1.03)=1.14L#

+

Note that #DeltaV=V_2-0.0709L=1.14L#

+

#=>V_2=1.14L+0.0709L=1.21L#

+

Here is a video that further explains this topic:

+

Thermochemistry | The Nature of Energy.
+ +

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 57662 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
+
+
" A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas? nan +464 a83731eb-6ddd-11ea-8d3a-ccda262736ce https://socratic.org/questions/how-do-you-write-calcium-hydroxlde-phosphoric-acid-yield-calcium-phosphate-water 3 Ca(OH)2 + 2 H2PO4 -> Ca3(PO4)2 + 6 H2O start chemical_equation qc_end substance 4 5 qc_end substance 7 8 qc_end substance 10 11 qc_end substance 13 13 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF]""}]" "[{""type"":""chemical equation"",""value"":""3 Ca(OH)2 + 2 H2PO4 -> Ca3(PO4)2 + 6 H2O""}]" "[{""type"":""substance name"",""value"":""Calcium hydroxlde""},{""type"":""substance name"",""value"":""Phosphoric acid""},{""type"":""substance name"",""value"":""Calcium phosphate""},{""type"":""substance name"",""value"":""Water""}]" "

How do you write ""calcium hydroxlde + phosphoric acid yield calcium phosphate + water""?

" nan 3 Ca(OH)2 + 2 H2PO4 -> Ca3(PO4)2 + 6 H2O "
+

Explanation:

+
+

Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
+#Ca(OH)_2 + H_3PO_4 -> Ca_3(PO_4)_2 + H_2O#

+

but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.

+

Balancing this would result to,

+

#3Ca(OH)_2 + 2H_3PO_4 -> Ca_3(PO_4)_2 + 6H_2O#

+
+
" "
+
+
+

#Ca(OH)_2 + H_3PO_4 -> Ca_3(PO_4)_2 + H_2O# (unbalanced)

+
+
+
+

Explanation:

+
+

Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
+#Ca(OH)_2 + H_3PO_4 -> Ca_3(PO_4)_2 + H_2O#

+

but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.

+

Balancing this would result to,

+

#3Ca(OH)_2 + 2H_3PO_4 -> Ca_3(PO_4)_2 + 6H_2O#

+
+
+
" "
+

How do you write ""calcium hydroxlde + phosphoric acid yield calcium phosphate + water""?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jan 18, 2016 + +
+
+
+
+
+
+
+

#Ca(OH)_2 + H_3PO_4 -> Ca_3(PO_4)_2 + H_2O# (unbalanced)

+
+
+
+

Explanation:

+
+

Writing calcium hydroxide + phosphoric acid yield calcium phosphate + water in chemical equation would be,
+#Ca(OH)_2 + H_3PO_4 -> Ca_3(PO_4)_2 + H_2O#

+

but it still needs to be balanced. To balance this, you just have to ensure that there is an equal amount of elements in both the reactant and product side.

+

Balancing this would result to,

+

#3Ca(OH)_2 + 2H_3PO_4 -> Ca_3(PO_4)_2 + 6H_2O#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 72791 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "How do you write ""calcium hydroxlde + phosphoric acid yield calcium phosphate + water""?" nan +465 a8375394-6ddd-11ea-bf24-ccda262736ce https://socratic.org/questions/how-do-you-write-an-equation-for-the-reaction-between-sodium-chloride-sulfur-dio 4 NaCl(s) + SO2(g) + 2 H2O(l) + 2 O2(g) -> 2 NaSO4(aq) + 4 HCl(g) start chemical_equation qc_end substance 10 11 qc_end substance 12 14 qc_end substance 15 15 qc_end substance 17 17 qc_end substance 20 21 qc_end substance 23 25 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] the reaction""}]" "[{""type"":""chemical equation"",""value"":""4 NaCl(s) + SO2(g) + 2 H2O(l) + 2 O2(g) -> 2 NaSO4(aq) + 4 HCl(g)""}]" "[{""type"":""substance name"",""value"":""Sodium chloride""},{""type"":""substance name"",""value"":""Sulfur dioxide gas""},{""type"":""substance name"",""value"":""Steam""},{""type"":""substance name"",""value"":""Oxygen""},{""type"":""substance name"",""value"":""Sodium sulfate""},{""type"":""substance name"",""value"":""Hydrogen chloride gas""}]" "

How do you write an equation for the reaction between sodium chloride, sulfur dioxide gas, steam, and oxygen to give sodium sulfate and hydrogen chloride gas?

" nan 4 NaCl(s) + SO2(g) + 2 H2O(l) + 2 O2(g) -> 2 NaSO4(aq) + 4 HCl(g) "
+

Explanation:

+
+

When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance #Na_2SO_(4(aq)), NaCl_((s)) and HCl_((g))# first.

+

#2NaCl_((s)) + SO_(2(g)) + H_2O_((l)) + O_(2(g)) -> Na_2SO_(4(aq)) + 2HCl_((g))#

+

We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.

+

Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in #SO_4#, we have to make the number of oxygens on the reactants even and then rebalance. The molecule that's making the number of oxygens odd is water, so we add a 2 to make it even and rebalance everything, getting:

+

#4NaCl_((s)) + SO_(2(g)) + 2H_2O_((l)) + 2O_(2(g)) -> 2Na_2SO_(4(aq)) + 4HCl_((g))#

+
+
" "
+
+
+

#4NaCl_((s)) + SO_(2(g)) + 2H_2O_((l)) + 2O_(2(g)) -> 2Na_2SO_(4(aq)) + 4HCl_((g))#

+
+
+
+

Explanation:

+
+

When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance #Na_2SO_(4(aq)), NaCl_((s)) and HCl_((g))# first.

+

#2NaCl_((s)) + SO_(2(g)) + H_2O_((l)) + O_(2(g)) -> Na_2SO_(4(aq)) + 2HCl_((g))#

+

We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.

+

Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in #SO_4#, we have to make the number of oxygens on the reactants even and then rebalance. The molecule that's making the number of oxygens odd is water, so we add a 2 to make it even and rebalance everything, getting:

+

#4NaCl_((s)) + SO_(2(g)) + 2H_2O_((l)) + 2O_(2(g)) -> 2Na_2SO_(4(aq)) + 4HCl_((g))#

+
+
+
" "
+

How do you write an equation for the reaction between sodium chloride, sulfur dioxide gas, steam, and oxygen to give sodium sulfate and hydrogen chloride gas?

+
+
+ + +Chemistry + + + + + +Chemical Reactions + + + + + +Chemical Reactions and Equations + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Oct 5, 2016 + +
+
+
+
+
+
+
+

#4NaCl_((s)) + SO_(2(g)) + 2H_2O_((l)) + 2O_(2(g)) -> 2Na_2SO_(4(aq)) + 4HCl_((g))#

+
+
+
+

Explanation:

+
+

When you balance, start of with the most complicated molecules, and leave the simpler molecules to last. In this case, we want to balance #Na_2SO_(4(aq)), NaCl_((s)) and HCl_((g))# first.

+

#2NaCl_((s)) + SO_(2(g)) + H_2O_((l)) + O_(2(g)) -> Na_2SO_(4(aq)) + 2HCl_((g))#

+

We need two sodiums on the reactants side, so we place a two in front of sodium chloride. Then we need two chlorines as we just added two to sodium chloride, so we place a two in front of hydrogen chloride.

+

Now count your elements and see what's missing. We have 1 sulfur on both sides, 2 hydrogens, but we have 5 oxygens on the reactants side and only 4 on the products. Since the number of oxygens on the products can only be even, as it is in #SO_4#, we have to make the number of oxygens on the reactants even and then rebalance. The molecule that's making the number of oxygens odd is water, so we add a 2 to make it even and rebalance everything, getting:

+

#4NaCl_((s)) + SO_(2(g)) + 2H_2O_((l)) + 2O_(2(g)) -> 2Na_2SO_(4(aq)) + 4HCl_((g))#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6760 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
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+
" How do you write an equation for the reaction between sodium chloride, sulfur dioxide gas, steam, and oxygen to give sodium sulfate and hydrogen chloride gas? nan +466 a8375d90-6ddd-11ea-8030-ccda262736ce https://socratic.org/questions/56c8776111ef6b297543381c 10.00 start physical_unit 17 18 ph none qc_end physical_unit 6 6 4 5 molarity qc_end physical_unit 10 11 8 9 molarity qc_end physical_unit 28 28 23 24 volume qc_end physical_unit 28 28 8 9 molarity qc_end end "[{""type"":""physical unit"",""value"":""pH [OF] the solution""}]" "[{""type"":""physical unit"",""value"":""10.00""}]" "[{""type"":""physical unit"",""value"":""Molarity [OF] ammonia [=] \\pu{0.55 mol/L}""},{""type"":""physical unit"",""value"":""Molarity [OF] ammonium chloride [=] \\pu{0.10 mol/L}""},{""type"":""physical unit"",""value"":""Volume [OF] NaOH [=] \\pu{1.0 mL}""},{""type"":""physical unit"",""value"":""Molarity [OF] NaOH [=] \\pu{0.10 mol/L}""}]" "

A solution P contains 0.55 mol/L ammonia and 0.10 mol/L ammonium chloride. What is the pH of the solution after the addition of 1.0 mL of 0.10 mol/L NaOH?

" nan 10.00 "
+

Explanation:

+
+

This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.

+

The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia, #""NH""_3#, a weak base, and ammonium chloride, the salt of ammonia's conjugate acid, the ammonium ion, #""NH""_4^(+)#.

+

A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.

+

This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.

+

Correct answer: #""pH"" = 10.0#

+

Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.

+

#color(white)(aaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaa)color(red)(""Incorrect answer below"")color(white)(aaaaaaaaa)/color(white)(aaaaaaaaaaaaaa)#

+

The trick here is that the pH of the solution is the only information you actually need here.

+

Don't worry about what solution #""P""# is said to contain, that is not important as long as the pH of the solution is said to be equal to #10.0#.

+

First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide, #""NaOH""#, solution.

+

Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions, #""OH""^(-)#.

+

So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.

+

Prediction:

+
+

#""pH"" > 10.0#

+
+

Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Therefore, you have

+
+

#""pOH"" = 14 - 10.0 = 4.0#

+
+

This means that the concentration of hydroxide anions is

+
+

#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#

+

#[""OH""^(-)] = 10^(-4)""M""#

+
+

Now, you take a #""1-dm""^3# sample of this solution #""P""#. Use the volume and the molarity of the hydroxide anions to calculate how many moles you have present

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#

+
+

Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution #""P""#

+
+

#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#

+
+

Now, the total number of moles of hydroxide anions will be

+
+

#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#

+

#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#

+
+

The total volume of the resulting solution will be

+
+

#V_""total"" = V_""sample P"" + V_""added""#

+

#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#

+
+

The molarity of the hydroxide anions after you add the sodium hydroxide solution will be

+
+

#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#

+
+

The pOH of the resulting solution will be

+
+

#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#

+
+

The pH of this solution will be

+
+

#""pH"" = 14 - 3.70 = color(green)(10.3)#

+
+

As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.

+
+
" "
+
+
+

See explanation.

+
+
+
+

Explanation:

+
+

This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.

+

The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia, #""NH""_3#, a weak base, and ammonium chloride, the salt of ammonia's conjugate acid, the ammonium ion, #""NH""_4^(+)#.

+

A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.

+

This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.

+

Correct answer: #""pH"" = 10.0#

+

Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.

+

#color(white)(aaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaa)color(red)(""Incorrect answer below"")color(white)(aaaaaaaaa)/color(white)(aaaaaaaaaaaaaa)#

+

The trick here is that the pH of the solution is the only information you actually need here.

+

Don't worry about what solution #""P""# is said to contain, that is not important as long as the pH of the solution is said to be equal to #10.0#.

+

First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide, #""NaOH""#, solution.

+

Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions, #""OH""^(-)#.

+

So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.

+

Prediction:

+
+

#""pH"" > 10.0#

+
+

Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Therefore, you have

+
+

#""pOH"" = 14 - 10.0 = 4.0#

+
+

This means that the concentration of hydroxide anions is

+
+

#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#

+

#[""OH""^(-)] = 10^(-4)""M""#

+
+

Now, you take a #""1-dm""^3# sample of this solution #""P""#. Use the volume and the molarity of the hydroxide anions to calculate how many moles you have present

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#

+
+

Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution #""P""#

+
+

#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#

+
+

Now, the total number of moles of hydroxide anions will be

+
+

#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#

+

#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#

+
+

The total volume of the resulting solution will be

+
+

#V_""total"" = V_""sample P"" + V_""added""#

+

#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#

+
+

The molarity of the hydroxide anions after you add the sodium hydroxide solution will be

+
+

#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#

+
+

The pOH of the resulting solution will be

+
+

#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#

+
+

The pH of this solution will be

+
+

#""pH"" = 14 - 3.70 = color(green)(10.3)#

+
+

As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.

+
+
+
" "
+

A solution P contains 0.55 mol/L ammonia and 0.10 mol/L ammonium chloride. What is the pH of the solution after the addition of 1.0 mL of 0.10 mol/L NaOH?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Buffer Calculations + + +
+
+
+
+
+2 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Feb 20, 2016 + +
+
+
+
+
+
+
+

See explanation.

+
+
+
+

Explanation:

+
+

This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.

+

The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia, #""NH""_3#, a weak base, and ammonium chloride, the salt of ammonia's conjugate acid, the ammonium ion, #""NH""_4^(+)#.

+

A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.

+

This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.

+

Correct answer: #""pH"" = 10.0#

+

Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.

+

#color(white)(aaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaa)color(red)(""Incorrect answer below"")color(white)(aaaaaaaaa)/color(white)(aaaaaaaaaaaaaa)#

+

The trick here is that the pH of the solution is the only information you actually need here.

+

Don't worry about what solution #""P""# is said to contain, that is not important as long as the pH of the solution is said to be equal to #10.0#.

+

First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide, #""NaOH""#, solution.

+

Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions, #""OH""^(-)#.

+

So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.

+

Prediction:

+
+

#""pH"" > 10.0#

+
+

Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide

+
+

#color(blue)(""pH"" + ""pOH"" = 14)#

+
+

Therefore, you have

+
+

#""pOH"" = 14 - 10.0 = 4.0#

+
+

This means that the concentration of hydroxide anions is

+
+

#""pOH""= - log([""OH""^(-)]) implies [""OH""^(-)] = 10^(-""pOH"")#

+

#[""OH""^(-)] = 10^(-4)""M""#

+
+

Now, you take a #""1-dm""^3# sample of this solution #""P""#. Use the volume and the molarity of the hydroxide anions to calculate how many moles you have present

+
+

#color(blue)(c = n/V implies n = c * V)#

+

#n_(OH^(-)) = 10^(-4)""mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""moles OH""^(-)#

+
+

Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution #""P""#

+
+

#n_(OH^(-)""added"") = ""0.1 mol"" color(red)(cancel(color(black)(""dm""^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)(""dm""^3))) = 10^(-4)""mol OH""^(-)#

+
+

Now, the total number of moles of hydroxide anions will be

+
+

#n_(OH^(-)""total"") = n_(OH^(-)) + n_(OH^(-)""added"")#

+

#n_(OH^(-)""total"") = 10^(-4)""moles"" + 10^(-4)""moles"" = 2 * 10^(-4)""moles OH""^(-)#

+
+

The total volume of the resulting solution will be

+
+

#V_""total"" = V_""sample P"" + V_""added""#

+

#V_""total"" = ""1 dm""^3 + 1 * 10^(-3)""dm""^3 = ""1.001 dm""^3#

+
+

The molarity of the hydroxide anions after you add the sodium hydroxide solution will be

+
+

#[""OH""^(-)] = (2 * 10^(-4)""moles"")/""1.001 dm""^3 = 1.998 * 10^(-4)""M""#

+
+

The pOH of the resulting solution will be

+
+

#""pOH"" = -log(1.998 * 10^(-4)) = 3.70#

+
+

The pH of this solution will be

+
+

#""pH"" = 14 - 3.70 = color(green)(10.3)#

+
+

As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.

+
+
+
+
+
+ +
+
+
+
+
+ +
+
+ +
+ + Feb 21, 2016 + +
+
+
+
+
+
+
+

#pH=10#

+

i.e unchanged.

+
+
+
+

Explanation:

+
+

This is a recipe for an alkaline buffer. It resists the addition of small amounts of acid and alkali keeping the pH constant.

+

How does it work?

+

You need a solution of a weak base and its salt. In this case we have ammonia solution and the salt ammonium chloride.

+

#NH_4OH# should properly be described as dilute ammonia solution. It forms an alkaline solution in water:

+

#NH_(3(aq))+H_2O_((l))rightleftharpoonsNH_(4(aq))^++OH_((aq))^-"" ""color(red)((1))#

+

Ammonium chloride is an ionic salt and dissociates completely in water:

+

#NH_4Cl_((s))rarrNH_(4(aq))^++Cl_((aq))^-"" ""color(red)((2))#

+

If a small amount of #H^+# ions is added then they will react immediately react with the #OH^-# ions from #color(red)((1))#. The position of equilibrium will then shift to the right to restore the #OH^-# ions thus restoring the pH value.

+

In this question a small amount of #OH^-# ions is added. From #color(red)((2))# you can see that there is a large reserve of #NH_4^+# ions present. These can combine with the #OH^-# thus driving #color(red)((1))# to the left and restoring the pH value.

+

A calculation can show this:

+

Ammonium ions dissociate:

+

#NH_4^+rightleftharpoonsNH_3+H^+#

+

For which:

+

#K_a=([NH_3][H^+])/([NH_4^+])=5.62xx10^(-10)""mol/l""# at 298K #"" ""color(red)((3))#

+

The number of moles of #OH^-# ions added #=1xx0.1/1000=0.0001#

+

These will combine with the #NH_4^+# ions:

+

#NH_4^++OH^(-)rarrNH_3+H_2O#

+

In 1 litre of P we have 0.1 mol of #NH_4^+# ions. (I have ignored the small amount from the dissociation of #NH_3#).

+

So we are left with #0.1-0.0001=0.0999 # moles of #NH_4^+#

+

By the same reasoning you can see that the number of moles of ammonia must have increased by this amount.

+

This gives #0.55+0.0001=0.5501# moles of #NH_3#.

+

To find the pH of a buffer we rearrange #color(red)((3))rArr#

+

#[H^+]=K_axx([NH_4^+])/([NH_3])#

+

Concentration = moles/volume. We can ignore the total volume of the solution since it is common to both #[NH_4^+]# and #[NH_3]#.

+

#:.[H^+]=K_axx(0.0999)/(0.5501)#

+

#[H^+]=5.62xx10^(-10)xx0.181=1.02xx10^(-10)#

+

#:.pH=-log(1.02xx10^(-10))=10#

+

So the pH does not change.

+

This is for illustration only. As you state, this is a multi - choice question so my take on this is that you are required recognise it is the recipe for an alkali buffer so reason that the pH won't change.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 6545 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A solution P contains 0.55 mol/L ammonia and 0.10 mol/L ammonium chloride. What is the pH of the solution after the addition of 1.0 mL of 0.10 mol/L NaOH? nan +467 a8375d91-6ddd-11ea-856b-ccda262736ce https://socratic.org/questions/juan-is-making-spaghetti-and-he-adds-29-grams-of-nacl-to-250-grams-of-water-what 100.50 ℃ start physical_unit 22 23 boiling_point_temperature °c qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 15 15 12 13 mass qc_end end "[{""type"":""physical unit"",""value"":""Boiling point [OF] this solution [IN] ℃""}]" "[{""type"":""physical unit"",""value"":""100.50 ℃""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] NaCl [=] \\pu{29 grams}""},{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{250 grams}""}]" "

Juan is making spaghetti and he adds 29 grams of #NaCl# to 250 grams of water. What is the boiling point of this solution?

" nan 100.50 ℃ "
+

Explanation:

+
+

The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.

+

Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.

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Approximately 214.32 degrees Fahrenheit

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Explanation:

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The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.

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Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.

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Juan is making spaghetti and he adds 29 grams of #NaCl# to 250 grams of water. What is the boiling point of this solution?

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+ + +Chemistry + + + + + +Solutions + + + + + +Colligative Properties + + +
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+2 Answers +
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+ + Aug 22, 2016 + +
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Approximately 214.32 degrees Fahrenheit

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Explanation:

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The salt (NaCl) content of this water is 11.6%. 20% salty water has 216 degrees Fahrenheit boiling point.

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Pure water has a boiling point of 212 degrees Fahrenheit. When you build an equation Delta t = 4 degrees F when salt has a concentration of 20%, you can calculate the boiling point of Juan's water. The answer is 214.32 degrees Fahrenheit.

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+ + Aug 22, 2016 + +
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100.50 degrees celcius.

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Explanation:

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If you dissolve 29.2 g of sodium chloride in 1 kg water, you create what is known as a ""0.5 molal"" solution (as opposed to 0.5 molar solution, which is 29.2 g of sodium chloride dissolved in 1 litre of solution).

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The boiling point of water is raised by 0.5 degrees celcius for each 0.5 molality of the solution, so instead of boiling at 100 degrees C it would boil at 100.50 degrees C.

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" Juan is making spaghetti and he adds 29 grams of #NaCl# to 250 grams of water. What is the boiling point of this solution? nan +468 a8375d92-6ddd-11ea-b515-ccda262736ce https://socratic.org/questions/what-is-the-oxidation-state-of-n-in-nh3 -3 start physical_unit 6 8 oxidation_state none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Oxidation state [OF] N in NH3""}]" "[{""type"":""physical unit"",""value"":""-3""}]" "[{""type"":""chemical equation"",""value"":""NH3""}]" "

What is the oxidation state of N in NH3? +

" nan -3 "
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Explanation:

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Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the #N-H# bonds I am left with #N^(3-)# and #3 xx H^+#.

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What are the oxidation states of nitrogen in #NO# and #NO_2#?

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We could represent the oxidation of ammonia #N(-III)#, to nitrate, #N(V+)#, as an exercise...

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#NH_3 +3H_2O rarr NO_3^(-)+9H^+ +8e^(-)#

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Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...

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#1/2N_2(g)+3H^+ +3e^(-)rarr NH_3(g)#

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The oxidation state of #N# in the ammonia molecule is #-III#.

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Explanation:

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Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the #N-H# bonds I am left with #N^(3-)# and #3 xx H^+#.

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What are the oxidation states of nitrogen in #NO# and #NO_2#?

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We could represent the oxidation of ammonia #N(-III)#, to nitrate, #N(V+)#, as an exercise...

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#NH_3 +3H_2O rarr NO_3^(-)+9H^+ +8e^(-)#

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Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...

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#1/2N_2(g)+3H^+ +3e^(-)rarr NH_3(g)#

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What is the oxidation state of N in NH3? +

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+2 Answers +
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The oxidation state of #N# in the ammonia molecule is #-III#.

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Explanation:

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Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the #N-H# bonds I am left with #N^(3-)# and #3 xx H^+#.

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What are the oxidation states of nitrogen in #NO# and #NO_2#?

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We could represent the oxidation of ammonia #N(-III)#, to nitrate, #N(V+)#, as an exercise...

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#NH_3 +3H_2O rarr NO_3^(-)+9H^+ +8e^(-)#

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Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...

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#1/2N_2(g)+3H^+ +3e^(-)rarr NH_3(g)#

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The oxidation state of Nitrogen in# NH_3# ammonia is -3

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Explanation:

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Nitrogen has a higher electronegativity than Hydrogen. This means that the electron density of the electrons that form the bond between the Nitrogen and the Hydrogen will be pulled closer to the Nitrogen than the Hydrogen.

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There are three sets of bonds between Nitrogen and Hydrogen in #NH_3# The negative electron density in each of these three bonds will be pulled closer to the Nitrogen resulting in a negative three charge on the Nitrogen.

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" "What is the oxidation state of N in NH3? +" nan +469 a8375d93-6ddd-11ea-b789-ccda262736ce https://socratic.org/questions/if-an-ideal-gas-has-a-pressure-of-4-97-atm-a-temperature-of-481-k-and-has-a-volu 3.74 moles start physical_unit 3 3 mole mol qc_end physical_unit 1 3 8 9 pressure qc_end physical_unit 1 3 13 14 temperature qc_end physical_unit 1 3 20 21 volume qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] gas [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.74 moles""}]" "[{""type"":""physical unit"",""value"":""Pressure [OF] an ideal gas [=] \\pu{4.97 atm}""},{""type"":""physical unit"",""value"":""Temperature [OF] an ideal gas [=] \\pu{481 K}""},{""type"":""physical unit"",""value"":""Volume [OF] an ideal gas [=] \\pu{29.83 L}""}]" "

If an ideal gas has a pressure of 4.97 atm, a temperature of 481 K, and has a volume of 29.83 L, how many moles of gas are in the sample?

" nan 3.74 moles "
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Explanation:

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The Ideal Gas Law explicitly states that:

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#n=(PV)/(RT)# #=# #(4.97*cancel(atm)xx29.83*cancelL)/(0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx481*cancelK)# #=# #??*mol#.

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Note that the only problem in solving this problem is choosing an appropriate gas constant, #R#. For chemists, #R=0.0821*L*atm*K^-1*mol^-1#, is very useful because it uses the units that chemists would typically use, i.e. #""atmospheres, litres, etc.""#. Is the equation correct dimensionally?

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#n~=4*mol#

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Explanation:

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The Ideal Gas Law explicitly states that:

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#n=(PV)/(RT)# #=# #(4.97*cancel(atm)xx29.83*cancelL)/(0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx481*cancelK)# #=# #??*mol#.

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Note that the only problem in solving this problem is choosing an appropriate gas constant, #R#. For chemists, #R=0.0821*L*atm*K^-1*mol^-1#, is very useful because it uses the units that chemists would typically use, i.e. #""atmospheres, litres, etc.""#. Is the equation correct dimensionally?

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If an ideal gas has a pressure of 4.97 atm, a temperature of 481 K, and has a volume of 29.83 L, how many moles of gas are in the sample?

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+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+1 Answer +
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#n~=4*mol#

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Explanation:

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The Ideal Gas Law explicitly states that:

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#n=(PV)/(RT)# #=# #(4.97*cancel(atm)xx29.83*cancelL)/(0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx481*cancelK)# #=# #??*mol#.

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Note that the only problem in solving this problem is choosing an appropriate gas constant, #R#. For chemists, #R=0.0821*L*atm*K^-1*mol^-1#, is very useful because it uses the units that chemists would typically use, i.e. #""atmospheres, litres, etc.""#. Is the equation correct dimensionally?

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Related questions
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Impact of this question
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+ 49019 views + around the world +
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" If an ideal gas has a pressure of 4.97 atm, a temperature of 481 K, and has a volume of 29.83 L, how many moles of gas are in the sample? nan +470 a8375d94-6ddd-11ea-ac6b-ccda262736ce https://socratic.org/questions/a-liter-of-air-containing-1-ar-is-repeatedly-passed-over-hot-cu-and-hot-mg-till- 10.00 mL start physical_unit 6 6 volume ml qc_end c_other OTHER qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 6 6 5 5 percent qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] Ar [IN] mL""}]" "[{""type"":""physical unit"",""value"":""10.00 mL""}]" "[{""type"":""other"",""value"":""Air containing Ar is repeatedly passed over hot Cu and hot Mg till no reduction of volume takes place.""},{""type"":""physical unit"",""value"":""Volume [OF] Air [=] \\pu{a liter}""},{""type"":""physical unit"",""value"":""Percent [OF] Ar [=] \\pu{1%}""}]" "

A liter of air containing 1% Ar is repeatedly passed over hot Cu and hot Mg till no reduction of volume takes place. The final volume of Ar will be?

" nan 10.00 mL "
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Explanation:

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The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.

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Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.

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On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide, #""CuO""#, and nitrogen gas will react with the hot magnesium to produce magnesium nitride, #""Mg""_3""N""_2#.

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So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.

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This means that the sample will only contain argon. Since this gas makes up #1%# of the initial volume, you can say that the final volume of argon will be #""10 mL""#, since

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#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#

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The answer is rounded to one significant figure.

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#""10 mL""#

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Explanation:

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The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.

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Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.

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On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide, #""CuO""#, and nitrogen gas will react with the hot magnesium to produce magnesium nitride, #""Mg""_3""N""_2#.

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So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.

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This means that the sample will only contain argon. Since this gas makes up #1%# of the initial volume, you can say that the final volume of argon will be #""10 mL""#, since

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#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#

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The answer is rounded to one significant figure.

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A liter of air containing 1% Ar is repeatedly passed over hot Cu and hot Mg till no reduction of volume takes place. The final volume of Ar will be?

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+1 Answer +
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#""10 mL""#

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Explanation:

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The trick here is to recognize the fact that argon will not react with the hot copper and the hot magnesium, but the oxygen and the nitrogen gas present in the sample will.

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Basically, the fact that argon is a noble gas tells you that you shouldn't expect to see a reaction take place when the sample of air is passed over the two hot metals.

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On the other hand, oxygen gas will react with the hot copper to form copper(II) oxide, #""CuO""#, and nitrogen gas will react with the hot magnesium to produce magnesium nitride, #""Mg""_3""N""_2#.

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So if you pass the sample over the two hot metals until no reduction in volume takes place, you can assume that all the oxygen gas and all the nitrogen gas have reacted.

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This means that the sample will only contain argon. Since this gas makes up #1%# of the initial volume, you can say that the final volume of argon will be #""10 mL""#, since

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#1 color(red)(cancel(color(black)(""%""))) * overbrace(""1000 mL""/(100color(red)(cancel(color(black)(""%"")))))^(color(blue)(""1 L is 100% of the volume"")) = ""10 mL""#

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The answer is rounded to one significant figure.

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Related questions
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" A liter of air containing 1% Ar is repeatedly passed over hot Cu and hot Mg till no reduction of volume takes place. The final volume of Ar will be? nan +471 a8375d95-6ddd-11ea-8e88-ccda262736ce https://socratic.org/questions/548a7796581e2a1c020adbb0 642.86 g start physical_unit 7 8 mass g qc_end physical_unit 7 7 6 6 percent qc_end physical_unit 7 7 11 12 mass qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] HCl solution [IN] g""}]" "[{""type"":""physical unit"",""value"":""642.86 g""}]" "[{""type"":""physical unit"",""value"":""Percent [OF] HCl [=] \\pu{7%}""},{""type"":""physical unit"",""value"":""Mass [OF] HCl [=] \\pu{45.0 g}""}]" "

What is the mass of a #""7% HCl""# solution in which #""45.0 HCl""# has been dissolved?

" nan 642.86 g "
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Explanation:

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In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.

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#""percent concentration""## = ##""mass of solute""/""mass of solution""# x #""100%""#

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Given:
+mass of solute = #""45.0 g HCl""#
+percent concentration = #""7.00%""#

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Unknown:
+mass of solution

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Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.

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#""mass of solution""## = ##""mass of solute""/""percent concentration""# x #""100""#

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#""mass of solution""## = ##""45.0 g HCl""/(7.00color(red)cancel(color(black)(""%"")))# x #100color(red)cancel(color(black)(""%""))# = #""643 g""# (rounded to three significant figures)

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The mass of the solution is #""643 g""#.

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Explanation:

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In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.

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#""percent concentration""## = ##""mass of solute""/""mass of solution""# x #""100%""#

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Given:
+mass of solute = #""45.0 g HCl""#
+percent concentration = #""7.00%""#

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Unknown:
+mass of solution

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Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.

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#""mass of solution""## = ##""mass of solute""/""percent concentration""# x #""100""#

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#""mass of solution""## = ##""45.0 g HCl""/(7.00color(red)cancel(color(black)(""%"")))# x #100color(red)cancel(color(black)(""%""))# = #""643 g""# (rounded to three significant figures)

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What is the mass of a #""7% HCl""# solution in which #""45.0 HCl""# has been dissolved?

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+ + +Chemistry + + + + + +Solutions + + + + + +Percent Concentration + + +
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+1 Answer +
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The mass of the solution is #""643 g""#.

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Explanation:

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In this case, where the mass of the solute is known, and the mass of the solution is desired, the basic equation for percent concentration is the mass of solute divided by the mass of the solution multiplied times 100%.

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#""percent concentration""## = ##""mass of solute""/""mass of solution""# x #""100%""#

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Given:
+mass of solute = #""45.0 g HCl""#
+percent concentration = #""7.00%""#

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Unknown:
+mass of solution

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Solution: Rearrange the equation for percent concentration to isolate mass of solution. Plug in the known values and solve.

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#""mass of solution""## = ##""mass of solute""/""percent concentration""# x #""100""#

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#""mass of solution""## = ##""45.0 g HCl""/(7.00color(red)cancel(color(black)(""%"")))# x #100color(red)cancel(color(black)(""%""))# = #""643 g""# (rounded to three significant figures)

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Impact of this question
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" "What is the mass of a #""7% HCl""# solution in which #""45.0 HCl""# has been dissolved?" nan +472 a8375d96-6ddd-11ea-8452-ccda262736ce https://socratic.org/questions/how-do-you-write-the-equation-for-this-reaction-nitrogen-gas-and-fluorine-gas-co N2(g) + 3 F2(g) -> 2NF3(g) start chemical_equation qc_end substance 9 10 qc_end substance 12 13 qc_end substance 17 19 qc_end end "[{""type"":""other"",""value"":""Chemical Equation [OF] this reaction""}]" "[{""type"":""chemical equation"",""value"":""N2(g) + 3 F2(g) -> 2NF3(g)""}]" "[{""type"":""substance name"",""value"":""Nitrogen gas""},{""type"":""substance name"",""value"":""fluorine gas""},{""type"":""substance name"",""value"":""nitrogen trifluoride gas""}]" "

How do you write the equation for this reaction: Nitrogen gas and fluorine gas combine to produce nitrogen trifluoride gas?

" nan N2(g) + 3 F2(g) -> 2NF3(g) "
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Explanation:

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#NF_3# is in fact not so nasty .... it is much safer than #NCl_3# or #NI_3#.. Of course the direct synthesis with fluorine gas is the province of specialist inorganic chemists...

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#1/2N_2(g) + 3/2F_2(g) rarr NF_3(g)#

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Explanation:

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#NF_3# is in fact not so nasty .... it is much safer than #NCl_3# or #NI_3#.. Of course the direct synthesis with fluorine gas is the province of specialist inorganic chemists...

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How do you write the equation for this reaction: Nitrogen gas and fluorine gas combine to produce nitrogen trifluoride gas?

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#1/2N_2(g) + 3/2F_2(g) rarr NF_3(g)#

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Explanation:

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#NF_3# is in fact not so nasty .... it is much safer than #NCl_3# or #NI_3#.. Of course the direct synthesis with fluorine gas is the province of specialist inorganic chemists...

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#N_2 +3 F_2 -> 2NF_3#

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Explanation:

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Nitrogen molecule is #N_2#

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Flourine molecule is #F_2#

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Nitrogen triflouride is #NF_3#
+(tri- means 3, and #N# usually has 3 covalent bonds, #F# has 1,
+so 3#F# to 1#N#)

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write unbalanced eqn for reaction

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#N_2 + F_2 -> NF_3# (not balanced)

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now balance the number of atoms on each side:
+at least 2 atoms of nitrogen so at least 2 molecules of #NF_3#
+2 molecules of #NF_3# need 6 #F# atoms = 3 #F_2# molecules.

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#N_2 +3 F_2 -> 2NF_3# (now balanced, 2#N# and 6#F# atoms each side)

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Impact of this question
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+
+
" How do you write the equation for this reaction: Nitrogen gas and fluorine gas combine to produce nitrogen trifluoride gas? nan +473 a8375d97-6ddd-11ea-8d9b-ccda262736ce https://socratic.org/questions/how-many-grams-of-barium-hydroxide-required-to-dissolve-in-156g-of-water-to-make 6.60 grams start physical_unit 4 5 mass g qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 19 19 17 18 molality qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] barium hydroxide [IN] grams""}]" "[{""type"":""physical unit"",""value"":""6.60 grams""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] water [=] \\pu{156 g}""},{""type"":""physical unit"",""value"":""Molality [OF] solution [=] \\pu{0.247 m}""}]" "

How many grams of barium hydroxide required to dissolve in 156g of water to make a 0.247m solution?

" nan 6.60 grams "
+

Explanation:

+
+

To answer this question, we need two things:

+
    +
  • The molar mass of barium hydroxide, 171.34 g/mol
    • +
  • The equation for molality:
    +
    • +
+

We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:

+

1000g = 1kg

+

Divide 156g by 1000 to obtain 0.156 kg of solvent.

+

Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:

+

Moles of solute = Molality x Mass of solvent

+

Therefore, #0.247 m xx 0.156kg# = #0.0385 mol Ba(OH)_2#

+

Since we want the mass of #Ba(OH)_2#, the number of moles of solute must be multiplied by the molar mass of the solute:

+

#0.0385 cancel(mol)xx(171.34g)/(1cancelmol)#

+

g #Ba(OH)_2# = 6.60g

+
+
" "
+
+
+

6.60g of #Ba(OH)_2# is required.

+
+
+
+

Explanation:

+
+

To answer this question, we need two things:

+
    +
  • The molar mass of barium hydroxide, 171.34 g/mol
    • +
  • The equation for molality:
    +
    • +
+

We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:

+

1000g = 1kg

+

Divide 156g by 1000 to obtain 0.156 kg of solvent.

+

Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:

+

Moles of solute = Molality x Mass of solvent

+

Therefore, #0.247 m xx 0.156kg# = #0.0385 mol Ba(OH)_2#

+

Since we want the mass of #Ba(OH)_2#, the number of moles of solute must be multiplied by the molar mass of the solute:

+

#0.0385 cancel(mol)xx(171.34g)/(1cancelmol)#

+

g #Ba(OH)_2# = 6.60g

+
+
+
" "
+

How many grams of barium hydroxide required to dissolve in 156g of water to make a 0.247m solution?

+
+
+ + +Chemistry + + + + + +Solutions + + + + + +Molality + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Jul 2, 2016 + +
+
+
+
+
+
+
+

6.60g of #Ba(OH)_2# is required.

+
+
+
+

Explanation:

+
+

To answer this question, we need two things:

+
    +
  • The molar mass of barium hydroxide, 171.34 g/mol
    • +
  • The equation for molality:
    +
    • +
+

We are given the molality and mass of solvent. The mass of solvent is expressed in terms of grams and we need kilograms so the following relationship must be used:

+

1000g = 1kg

+

Divide 156g by 1000 to obtain 0.156 kg of solvent.

+

Next, we determine the number of moles of solute (barium hydroxide), by rearranging the equation:

+

Moles of solute = Molality x Mass of solvent

+

Therefore, #0.247 m xx 0.156kg# = #0.0385 mol Ba(OH)_2#

+

Since we want the mass of #Ba(OH)_2#, the number of moles of solute must be multiplied by the molar mass of the solute:

+

#0.0385 cancel(mol)xx(171.34g)/(1cancelmol)#

+

g #Ba(OH)_2# = 6.60g

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 8476 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
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+
+
+
" How many grams of barium hydroxide required to dissolve in 156g of water to make a 0.247m solution? nan +474 a8375d98-6ddd-11ea-9039-ccda262736ce https://socratic.org/questions/what-is-the-percent-composition-of-water-in-na-2s-9h-2o 67.49% start physical_unit 6 8 percent_composition none qc_end chemical_equation 8 8 qc_end end "[{""type"":""physical unit"",""value"":""Percent composition [OF] water in Na2S.9H2O""}]" "[{""type"":""physical unit"",""value"":""67.49%""}]" "[{""type"":""chemical equation"",""value"":""Na2S.9H2O""}]" "

What is the percent composition of water in #Na_2S • 9H_2O?#

" nan 67.49% "
+

Explanation:

+
+

#%H_2O=""Mass of water""/""Mass of hydrate""=(9xx18.01*g*mol^-1)/(240.18*g*mol^-1)xx100%=??%#

+
+
" "
+
+
+

#%H_2O# #~=# #70%#

+
+
+
+

Explanation:

+
+

#%H_2O=""Mass of water""/""Mass of hydrate""=(9xx18.01*g*mol^-1)/(240.18*g*mol^-1)xx100%=??%#

+
+
+
" "
+

What is the percent composition of water in #Na_2S • 9H_2O?#

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Percent Composition + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 23, 2017 + +
+
+
+
+
+
+
+

#%H_2O# #~=# #70%#

+
+
+
+

Explanation:

+
+

#%H_2O=""Mass of water""/""Mass of hydrate""=(9xx18.01*g*mol^-1)/(240.18*g*mol^-1)xx100%=??%#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1081 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the percent composition of water in #Na_2S • 9H_2O?# nan +475 a8377ff6-6ddd-11ea-ac7e-ccda262736ce https://socratic.org/questions/how-many-molecules-are-in-3-0-moles-of-carbon-dioxide 1.81 × 10^24 start physical_unit 6 9 number none qc_end physical_unit 8 9 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Number [OF] moles of carbon dioxide""}]" "[{""type"":""physical unit"",""value"":""1.81 × 10^24""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] carbon dioxide [=] \\pu{3.0 moles}""}]" "

How many molecules are in 3.0 moles of carbon dioxide?

" nan 1.81 × 10^24 "
+

Explanation:

+
+

Where #N_A-=6.022xx10^23*mol^-1#..

+

And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.

+
+
" "
+
+
+

There are #3xxN_A# molecules...

+
+
+
+

Explanation:

+
+

Where #N_A-=6.022xx10^23*mol^-1#..

+

And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.

+
+
+
" "
+

How many molecules are in 3.0 moles of carbon dioxide?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Aug 1, 2018 + +
+
+
+
+
+
+
+

There are #3xxN_A# molecules...

+
+
+
+

Explanation:

+
+

Where #N_A-=6.022xx10^23*mol^-1#..

+

And what is the mass of this number of molecules? And how many ATOMS? You should be able to answer pdq.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 4030 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" How many molecules are in 3.0 moles of carbon dioxide? nan +476 a8377ff7-6ddd-11ea-88bb-ccda262736ce https://socratic.org/questions/what-is-the-mass-of-0642-mol-ammonium-phosphate 9.57 g start physical_unit 7 8 mass g qc_end physical_unit 7 8 5 6 mole qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] ammonium phosphate [IN] g""}]" "[{""type"":""physical unit"",""value"":""9.57 g""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] ammonium phosphate [=] \\pu{0.0642 mol}""}]" "

What is the mass of .0642 mol ammonium phosphate?

" nan 9.57 g "
+

Explanation:

+
+
    +
  • The molecular formula of Ammonium phosphate is
    +#(NH_4)_3PO_4#
    • +
  • The molar mass of Ammonium phosphate is
    +#(14+4xx1)xx3+31+4xx16=149g/(mol)#
    • +
+

So the mass of .0642 mol ammonium phosphate is
+=#0.0642cancel(mol)xx149g/cancel(mol)=9.5658g#

+
+
" "
+
+
+

#9.5658g#

+
+
+
+

Explanation:

+
+
    +
  • The molecular formula of Ammonium phosphate is
    +#(NH_4)_3PO_4#
    • +
  • The molar mass of Ammonium phosphate is
    +#(14+4xx1)xx3+31+4xx16=149g/(mol)#
    • +
+

So the mass of .0642 mol ammonium phosphate is
+=#0.0642cancel(mol)xx149g/cancel(mol)=9.5658g#

+
+
+
" "
+

What is the mass of .0642 mol ammonium phosphate?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + May 7, 2016 + +
+
+
+
+
+
+
+

#9.5658g#

+
+
+
+

Explanation:

+
+
    +
  • The molecular formula of Ammonium phosphate is
    +#(NH_4)_3PO_4#
    • +
  • The molar mass of Ammonium phosphate is
    +#(14+4xx1)xx3+31+4xx16=149g/(mol)#
    • +
+

So the mass of .0642 mol ammonium phosphate is
+=#0.0642cancel(mol)xx149g/cancel(mol)=9.5658g#

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 3971 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" What is the mass of .0642 mol ammonium phosphate? nan +477 a8377ff8-6ddd-11ea-8d75-ccda262736ce https://socratic.org/questions/584961037c01494eb51125d9 23.18 g start physical_unit 11 11 mass g qc_end physical_unit 7 7 1 2 mass qc_end c_other OTHER qc_end end "[{""type"":""physical unit"",""value"":""Mass [OF] dioxygen [IN] g""}]" "[{""type"":""physical unit"",""value"":""23.18 g""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N2O [=] \\pu{21.3 g}""},{""type"":""other"",""value"":""N2O, is oxidized by dioxygen gas to NO2(g).""}]" "

A #21.3*g# mass of #""nitrous oxide""#, #N_2O#, is oxidized by dioxygen gas to #NO_2(g)#. How much dioxygen is required?

" nan 23.18 g "
+

Explanation:

+
+

#""Moles of""# #N_2O# #=# #(21.3*g)/(44.02*g*mol^-1)=0.483*mol#

+

And for stoichiometric reaction we need,

+

#0.483*molxx3/2xx32.00*g*mol^-1=??*g# #""dioxygen gas""#.

+
+
" "
+
+
+

#N_2O(g) + 3/2O_2(g)rarr2NO_2(g)#

+
+
+
+

Explanation:

+
+

#""Moles of""# #N_2O# #=# #(21.3*g)/(44.02*g*mol^-1)=0.483*mol#

+

And for stoichiometric reaction we need,

+

#0.483*molxx3/2xx32.00*g*mol^-1=??*g# #""dioxygen gas""#.

+
+
+
" "
+

A #21.3*g# mass of #""nitrous oxide""#, #N_2O#, is oxidized by dioxygen gas to #NO_2(g)#. How much dioxygen is required?

+
+
+ + +Chemistry + + + + + +Stoichiometry + + + + + +Stoichiometry + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 14, 2016 + +
+
+
+
+
+
+
+

#N_2O(g) + 3/2O_2(g)rarr2NO_2(g)#

+
+
+
+

Explanation:

+
+

#""Moles of""# #N_2O# #=# #(21.3*g)/(44.02*g*mol^-1)=0.483*mol#

+

And for stoichiometric reaction we need,

+

#0.483*molxx3/2xx32.00*g*mol^-1=??*g# #""dioxygen gas""#.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1391 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" "A #21.3*g# mass of #""nitrous oxide""#, #N_2O#, is oxidized by dioxygen gas to #NO_2(g)#. How much dioxygen is required?" nan +478 a8377ff9-6ddd-11ea-a169-ccda262736ce https://socratic.org/questions/a-gas-mixture-contains-1-19-g-n-2-and-0-79-g-o-2-in-a-1-60-l-container-at-25-deg 63.70% start physical_unit 6 6 mole_fraction none qc_end physical_unit 6 6 4 5 mass qc_end physical_unit 10 10 8 9 mass qc_end physical_unit 15 15 13 14 volume qc_end physical_unit 15 15 17 19 temperature qc_end end "[{""type"":""physical unit"",""value"":""Mole fraction [OF] N2""}]" "[{""type"":""physical unit"",""value"":""63.70%""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] N2 [=] \\pu{1.19 g}""},{""type"":""physical unit"",""value"":""Mass [OF] O2 [=] \\pu{0.79 g}""},{""type"":""physical unit"",""value"":""Volume [OF] container [=] \\pu{1.60 L}""},{""type"":""physical unit"",""value"":""Temperature [OF] container [=] \\pu{25 deg C}""}]" "

A gas mixture contains 1.19 g #N_2# and 0.79 g #O_2# in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of #N_2#?

" nan 63.70% "
+

Explanation:

+
+

#chi_i=(n_i)/(n_""total"")#.

+

Now #n_""total""=(1.19*g)/(28.01*g*mol^-1)+(0.79*g)/(32.00*g*mol^-1)#

+

#=6.67xx10^-2*mol#

+

And #chi_""dinitrogen""# #=# #((1.19*g)/(28.01*g*mol^-1))/(6.67xx10^-2*mol)=0.64#.

+

Given that it is a binary system, what is #chi_""dioxygen""?#

+

Note that the information given with respect to #""volume""# and #""temperature""# was included to distract you.

+
+
" "
+
+
+

#chi_""dinitrogen""# #=# #0.64#

+
+
+
+

Explanation:

+
+

#chi_i=(n_i)/(n_""total"")#.

+

Now #n_""total""=(1.19*g)/(28.01*g*mol^-1)+(0.79*g)/(32.00*g*mol^-1)#

+

#=6.67xx10^-2*mol#

+

And #chi_""dinitrogen""# #=# #((1.19*g)/(28.01*g*mol^-1))/(6.67xx10^-2*mol)=0.64#.

+

Given that it is a binary system, what is #chi_""dioxygen""?#

+

Note that the information given with respect to #""volume""# and #""temperature""# was included to distract you.

+
+
+
" "
+

A gas mixture contains 1.19 g #N_2# and 0.79 g #O_2# in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of #N_2#?

+
+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +The Mole + + +
+
+
+
+
+1 Answer +
+
+
+
+
+
+ +
+
+ +
+ + Dec 10, 2016 + +
+
+
+
+
+
+
+

#chi_""dinitrogen""# #=# #0.64#

+
+
+
+

Explanation:

+
+

#chi_i=(n_i)/(n_""total"")#.

+

Now #n_""total""=(1.19*g)/(28.01*g*mol^-1)+(0.79*g)/(32.00*g*mol^-1)#

+

#=6.67xx10^-2*mol#

+

And #chi_""dinitrogen""# #=# #((1.19*g)/(28.01*g*mol^-1))/(6.67xx10^-2*mol)=0.64#.

+

Given that it is a binary system, what is #chi_""dioxygen""?#

+

Note that the information given with respect to #""volume""# and #""temperature""# was included to distract you.

+
+
+
+
+
+ +
+
+
+
+
+
+
Related questions
+ + +
+
+
+
Impact of this question
+
+ 1624 views + around the world +
+
+
+ +
+ You can reuse this answer +
+ + Creative Commons License + +
+
+
+
+
+
+
" A gas mixture contains 1.19 g #N_2# and 0.79 g #O_2# in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of #N_2#? nan +479 a8377ffa-6ddd-11ea-9914-ccda262736ce https://socratic.org/questions/what-is-the-volume-of-h-2-generated-by-reacting-4-33g-of-zn-with-excess-h-2so-4- 1.69 L start physical_unit 5 5 volume l qc_end physical_unit 12 12 9 10 mass qc_end c_other OTHER qc_end physical_unit 8 8 17 18 temperature qc_end physical_unit 8 8 20 21 pressure qc_end end "[{""type"":""physical unit"",""value"":""Volume [OF] H2 [IN] L""}]" "[{""type"":""physical unit"",""value"":""1.69 L""}]" "[{""type"":""physical unit"",""value"":""Mass [OF] Zn [=] \\pu{4.33 g}""},{""type"":""other"",""value"":""Excess H2SO4.""},{""type"":""physical unit"",""value"":""Temperature [OF] reacting [=] \\pu{38 ℃}""},{""type"":""physical unit"",""value"":""Pressure [OF] reacting [=] \\pu{760 torr}""}]" "

What is the volume of #H_2# generated by reacting 4.33g of #Zn# with excess #H_2SO_4# at 38°C and 760 torr?

" nan 1.69 L "
+

Explanation:

+
+

Zinc metal will react with dilute sulfuric acid to produce zinc sulfate, #""ZnSO""_4#, and hydrogen gas, #""H""_2#, which will bubble out of solution.

+

The balanced chemical equation for this single replacement reaction looks like this

+
+

#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#

+
+

Notice that you have a #1:1# mole ratio between zinc metal and hydrogen gas. This tells you that the reaction will always produce the same number of moles of hydrogen gas as you have moles of zinc metal that take part in the reaction.

+

In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.

+

Use zinc metal's molar mass to determine how many moles you have in that #""4.33-g""# sample

+
+

#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#

+
+

This means that the reaction produced #0.06623# moles of hydrogen gas.

+

Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin!

+

It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.

+

This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.

+

So, rearrange the ideal gas law equation to solve for #V#, the volume of the gas

+
+

#V = (nRT)/P#

+

#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#

+

#V = ""1.692 L""#

+
+

Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be

+
+

#V = color(green)(""1.7 L"")#

+
+

+ +

+
+
" "
+
+
+

#""1.7 L""#

+
+
+
+

Explanation:

+
+

Zinc metal will react with dilute sulfuric acid to produce zinc sulfate, #""ZnSO""_4#, and hydrogen gas, #""H""_2#, which will bubble out of solution.

+

The balanced chemical equation for this single replacement reaction looks like this

+
+

#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#

+
+

Notice that you have a #1:1# mole ratio between zinc metal and hydrogen gas. This tells you that the reaction will always produce the same number of moles of hydrogen gas as you have moles of zinc metal that take part in the reaction.

+

In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.

+

Use zinc metal's molar mass to determine how many moles you have in that #""4.33-g""# sample

+
+

#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#

+
+

This means that the reaction produced #0.06623# moles of hydrogen gas.

+

Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin!

+

It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.

+

This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.

+

So, rearrange the ideal gas law equation to solve for #V#, the volume of the gas

+
+

#V = (nRT)/P#

+

#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#

+

#V = ""1.692 L""#

+
+

Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be

+
+

#V = color(green)(""1.7 L"")#

+
+

+ +

+
+
+
" "
+

What is the volume of #H_2# generated by reacting 4.33g of #Zn# with excess #H_2SO_4# at 38°C and 760 torr?

+
+
+ + +Chemistry + + + + + +Gases + + + + + +Ideal Gas Law + + +
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+2 Answers +
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+ + Dec 29, 2015 + +
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#""1.7 L""#

+
+
+
+

Explanation:

+
+

Zinc metal will react with dilute sulfuric acid to produce zinc sulfate, #""ZnSO""_4#, and hydrogen gas, #""H""_2#, which will bubble out of solution.

+

The balanced chemical equation for this single replacement reaction looks like this

+
+

#""Zn""_text((s]) + ""H""_2""SO""_text(4(aq]) -> ""ZnSO""_text(4(aq]) + ""H""_text(2(g]) uarr#

+
+

Notice that you have a #1:1# mole ratio between zinc metal and hydrogen gas. This tells you that the reaction will always produce the same number of moles of hydrogen gas as you have moles of zinc metal that take part in the reaction.

+

In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.

+

Use zinc metal's molar mass to determine how many moles you have in that #""4.33-g""# sample

+
+

#4.33 color(red)(cancel(color(black)(""g""))) * ""1 mole Zn""/(65.38color(red)(cancel(color(black)(""g"")))) = ""0.06623 moles Zn""#

+
+

This means that the reaction produced #0.06623# moles of hydrogen gas.

+

Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this

+
+

#color(blue)(PV = nRT)"" ""#, where

+
+

#P# - the pressure of the gas
+#V# - the volume it occupies
+#n# - the number of moles of gas
+#R# - the universal gas constant, usually given as #0.0821(""atm"" * ""L"")/(""mol"" * ""K"")#
+#T# - the temperature of the gas, always expressed in Kelvin!

+

It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.

+

This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.

+

So, rearrange the ideal gas law equation to solve for #V#, the volume of the gas

+
+

#V = (nRT)/P#

+

#V = (0.06623 color(red)(cancel(color(black)(""moles""))) * 0.0821(color(red)(cancel(color(black)(""atm""))) * ""L"")/(color(red)(cancel(color(black)(""moles""))) * color(red)(cancel(color(black)(""K"")))) * (273.15 + 38)color(red)(cancel(color(black)(""K""))))/(760/760color(red)(cancel(color(black)(""atm""))))#

+

#V = ""1.692 L""#

+
+

Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be

+
+

#V = color(green)(""1.7 L"")#

+
+

+ +

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+ + Dec 29, 2015 + +
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+
+

#""V""= 1.690""dm""^3#

+
+
+
+

Explanation:

+
+

State the Equation

+

Zinc's only common oxidation state is #""2+""#. When it reacts with sulfuric acid, the result is the following equation:

+

#""Zn""(s) + ""H""_2""SO""_4(aq) -> ""ZnSO""_4 (aq) + ""H""_2 (g)#

+

Calculating Moles of #""H""_2#

+

First, calculate the moles of #""Zn""# reacting:

+

#""mol"" = ""m""/""M""_r#
+#""mol""(""Zn"") = (4.33""g"")/65.4#
+#""mol""(""Zn"") = 0.06620795107"" moles""#

+

Next, calculate the moles of #""H""_2# produced using the mole ratio:

+

#""mol""(""H""_2) = ""mol""(""Zn"") = 0.06620795107"" moles""#

+

Calculating Volume of #""H""_2#

+

In order to find the volume of space occupied by the calculated number of moles of hydrogen gas at the stated conditions, we must consult the ideal gas equation:

+

#""pV "" = "" nRT""#

+

Where:

+
    +
  • #""p""# is pressure in pascals (#""Pa""#)
    • +
  • #""V""# is volume in cubic metres ( #""m""^3#)
    • +
  • #""n""# is the number of moles
    • +
  • #""R""# is the gas constant = #8.314#
    • +
  • #""T""# is the temperature in Kelvin (#""K""#)
    • +
+

First, convert the given data into workable units:

+
    +
  • #1""torr"" ~~ 133.322""Pa"", :. 760""torr"" = 101325""Pa""#
    • +
  • #1^@""C"" = 274.15""K"", :. 38^@""C"" = 311.15""K""#
    • +
+

Next, rearrange the equation to solve for volume:

+

#""pV ""="" nRT "" => "" V ""="" """"nRT""/""p""#

+

Finally, substitute in your values and solve the equation:

+

#""V ""="" ""(0.06620795107*8.314*311.15)/101325#
+#""V ""="" ""1.69 * 10^(-3)""m""^3"" to 3 significant figures""#

+

Another useful unit that is probably suitable to use in this situation is the cubic decimetre ( #""dm""^3# ):

+

#""V ""="" ""1.69 * 10^(-3)""m""^3 * 1000 = 1.690""dm""^3#

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+ +
+
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+
+
+
+
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+
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" What is the volume of #H_2# generated by reacting 4.33g of #Zn# with excess #H_2SO_4# at 38°C and 760 torr? nan +480 a882d200-6ddd-11ea-9acd-ccda262736ce https://socratic.org/questions/molten-iron-is-extremely-hot-averaging-about-1-500-c-the-specific-heat-of-iron-i-1 8.68 × 10^5 J start physical_unit 0 1 heat_energy kj qc_end physical_unit 0 1 7 8 temperature qc_end physical_unit 0 1 36 37 temperature qc_end physical_unit 0 1 28 29 mass qc_end end "[{""type"":""physical unit"",""value"":""Heat released [OF] molten iron [IN] J""}]" "[{""type"":""physical unit"",""value"":""8.68 × 10^5 J""}]" "[{""type"":""physical unit"",""value"":""Temperature1 [OF] molten iron [=] \\pu{1,500 ℃}""},{""type"":""physical unit"",""value"":""Specific heat [OF] iron [=] \\pu{0.46 J/(g * ℃)}""},{""type"":""physical unit"",""value"":""Temperature2 [OF] molten iron [=] \\pu{25 ℃}""},{""type"":""physical unit"",""value"":""Mass [OF] molten iron [=] \\pu{1 kg}""}]" "

Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)?

" nan 8.68 × 10^5 J "
+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

+
+
" "
+
+
+

900 Kj

+
+
+
+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

+
+
+
" "
+

Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)?

+
+
+ + +Chemistry + + + + + +Thermochemistry + + + + + +Specific Heat + + +
+
+
+
+
+3 Answers +
+
+
+
+
+
+ +
+
+ +
+ + Jun 24, 2017 + +
+
+
+
+
+
+
+

900 Kj

+
+
+
+

Explanation:

+
+

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
+*http://www.onlinemetals.com/meltpt.cfm

+

The total heat transfer would be ,,,

+

#Q_""Total = ""Sigma (Q_(""molten"") + Q_(""freezing"") + Q_(""cooling""))#
+..............................................................................................................................
+=> #Q_""molten""=(mcDeltaT)_""molten"" #
+= #(1000gxx0.18""J/g""^oCxx(1500 - 1204)^oC) =53,280# Joules

+

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
+..............................................................................................................................
+=> #Q_""freezing""=(mDeltaH_f)_""freezing""#
+= #(1000gxx272J/g) = 272,000# Joules

+

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

+

...............................................................................................................................
+=> #Q_""cooling""=(mcDeltaT)_""cooling""to""""25^oC""#
+= #(1000gxx0.46J/g^oCxx(1204 - 25)^oC)# = 542,340 Joules

+

(Specific Heat of Iron (s) = 0.46 #J/g^oC# as given in problem data)
+..............................................................................................................................
+#Q_""Total"" = (53,280 + 272,000 + 542,340)J# = 867,620 Joules
+#~~9 xx10^5 ""Joules""# = #900 Kj#

+
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+ + Jun 24, 2017 + +
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+

I got #""1020 kJ""# were RELEASED into the atmosphere, ignoring phase changes between the #alpha#, #delta#, and #gamma# phases and just looking at the temperature changes.

+

You can get more context here:
+https://en.wikipedia.org/wiki/Iron#Phase_diagram_and_allotropes

+

and you can examine the specific heat capacity variations more closely here:
+http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Mask=2&Type=JANAFS&Plot=on#JANAFS

+

On another note, this #""1020 kJ""# is quite a bit higher than what one would normally expect to get, #""655.5 kJ""#, due to taking into account the huge variation in heat capacity across #1475^@ ""C""#.

+

If you simply assume a #C_P# of #""0.46 J/g""cdot""K""# throughout, you would get #""655.5 kJ""# instead (#656# to three sig figs).

+
+

There is a HUGE assumption here that iron's specific heat capacity doesn't change from #25^@ ""C""# to #1500^@ ""C""#, which is clearly not true. Here is the phase diagram of iron:

+

+

Since all these phases at #""1 bar""# are solids, we are safe in assuming there is no major enthalpy of solid-solid phase transitions to worry about.

+

However, the specific heat capacity #C_P# at constant pressure changes drastically as we transition through #alpha#, #gamma#, and #delta# phases:

+

[

+

The wonky curve is the #alpha# and #delta# phase, and the linear curve is the #gamma# phase. Here's how I would treat this:

+
    +
  • #alpha#-phase, from #""298.15 K""# up to #""700 K""# (#426.85^@ ""C""#), using an average of #C_P ~~ ""29.656 J/mol""cdot""K""# (at #~~ ""500 K""#), or #""0.531 J/g""cdot""K""#.
    • +
  • #alpha#-phase, from #""700 K""# to #""935 K""# (#661.85^@ ""C""#) using an average of #C_P ~~ ""40.149 J/mol""cdot""K""# (at #~~ ""816 K""#), or #""0.719 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""935 K""# to #""1042 K""# (#768.85^@ ""C""#) using an average of #C_P ~~ ""59.442 J/mol""cdot""K""# (at #~~ ""1010 K""#), or #""1.064 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""1042 K""# to #""1100 K""# (#826.85^@ ""C""#) using an average of #C_P ~~ ""65.743 J/mol""cdot""K""# (at #~~ ""1068 K""#), or #""1.177 J/g""cdot""K""#
    • +
  • #alpha#-phase, from #""1100 K""# to #""1183.15 K""# (#910^@ ""C""#, the #alpha->gamma# transition temperature) using an average of #C_P ~~ ""43.029 J/mol""cdot""K""# (at #~~ ""1150 K""#), or #""0.770 J/g""cdot""K""#
    • +
  • #gamma#-phase, from #""1183.15 K""# to #""1667.15 K""# (#1394^@ ""C""#, the #gamma->delta# transition temperature) using an average of #C_P ~~ ""35.856 J/mol""cdot""K""# (at #~~ ""1420 K""#), or #""0.642 J/g""cdot""K""#
    • +
  • #delta#-phase, from #""1667.15 K""# to #""1773.15 K""# (#1500^@ ""C""#!), using an average of #C_P ~~ ""41.764 J/mol""cdot""K""# (at #~~ ""1722 K""#), or #""0.748 J/g""cdot""K""#.
    • +
+

Aren't you glad we aren't doing phase changes? :-)

+

So, we would have the heat of cooling as the negative of the heat of heating:

+
+

#q_""cool"" = -(q_1 + . . . + q_7)#

+

#= -m(C_(P1)DeltaT_(0->1) + . . . + C_(P7)DeltaT_(6->7))#

+
+

I'll leave the units out, but you know that they are #""J/g""cdot""K""# for #C_P# and #""K""# for #T#. The mass is in #""g""#.

+
+

#= -1000 cdot [0.531(700 - 298.15) + 0.719(935 - 700) + 1.064(1042 - 935) + 1.177(1100 - 1042) + 0.770(1183.15 - 1100) + 0.642(1667.15 - 1183.15) + 0.748(1773.15 - 1667.15)]#

+
+

Each phase then approximately contributes:

+
+

#= overbrace(-""628487 J"")^(alpha"" phase"") + overbrace(-""310728 J"")^(gamma"" phase"") + overbrace(-""79288 J"")^(delta"" phase"")#

+

#~~# #-1.020 xx 10^(6)# #""J""#,

+
+

or about #color(blue)(-""1020 kJ"")#, to three sig figs.

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+ + Jun 24, 2017 + +
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Thermal history of cooling cast iron

+
+
+
+

Explanation:

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+

Thermal history of cooling cast iron from #1500^oC# to #25^oC# ...
+

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" Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)? nan +481 a9a24eb0-6ddd-11ea-be90-ccda262736ce https://socratic.org/questions/how-many-moles-of-sodium-hydroxide-are-needed-to-neutralize-1-mole-of-phosphoric 3.00 moles start physical_unit 4 5 mole mol qc_end physical_unit 13 14 10 11 mole qc_end end "[{""type"":""physical unit"",""value"":""Mole [OF] sodium hydroxide [IN] moles""}]" "[{""type"":""physical unit"",""value"":""3.00 moles""}]" "[{""type"":""physical unit"",""value"":""Mole [OF] phosphoric acid [=] \\pu{1 mole}""}]" "

How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid?

" nan 3.00 moles "
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

+
+
" "
+
+
+

You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.

+
+
+
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

+
+
+
" "
+

How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid?

+
+
+ + +Chemistry + + + + + +Reactions in Solution + + + + + +Neutralization + + +
+
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+1 Answer +
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+ + May 19, 2014 + +
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+

You need 3 mol of sodium hydroxide to neutralize 1 mol of phosphoric acid.

+
+
+
+

Explanation:

+
+
+

The balanced equation for the reaction is

+

#""3NaOH"" + ""H""_3""PO""_4 → ""Na""_3""PO""_4 + 3""H""_2""O""#

+

Phosphoric acid #""H""_3""PO""_4# is a triprotic acid. It has three acidic hydrogen atoms.

+

Thus, one molecule of phosphoric acid will need three formula units of sodium hydroxide, and

+

One mole of phosphoric acid will need three moles of sodium hydroxide.

+

This is what the balanced equation tells us:

+

3 mol of #""NaOH""# reacts with 1 mol of #""H""_3""PO""_4#.

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+
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+ + Creative Commons License + +
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" How many moles of sodium hydroxide are needed to neutralize 1 mole of phosphoric acid? nan +482 abecac50-6ddd-11ea-8973-ccda262736ce https://socratic.org/questions/an-unknown-compound-is-analyzed-and-found-to-be-composed-of-14-79-nitrogen-50-68 ZnN2O6 start chemical_formula qc_end end "[{""type"":""other"",""value"":""Chemical Formula [OF] the compound [IN] empirical""}]" "[{""type"":""chemical equation"",""value"":""ZnN2O6""}]" "[{""type"":""physical unit"",""value"":""Percent composition [OF] nitrogen [=] \\pu{14.79%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] oxygen [=] \\pu{50.68%}""},{""type"":""physical unit"",""value"":""Percent composition [OF] zinc [=] \\pu{34.53%}""}]" "

An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

" nan ZnN2O6 "
+

Explanation:

+
+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

+

Your values will look like this :

+
    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

+

#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

+

#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

+

#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

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#1.051 / 0.528 = 1.99#

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#3.168 / 0.528 = 6#

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#0.528 / 0.528 = 1#

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To find the smallest whole number ratio, multiply each value by #2#.

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#1.99 * 2 ~~ 4 #

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#6 * 2 = 12#

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#1 * 2 = 2#

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Your formula is now:

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#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

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#""ZnN""_2""O""_6#

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(Remember that if the subscript of an element is #1#, then no subscript is necessary).

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" "
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+

The empirical formula is #""ZnN""_2""O""_6#.

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Explanation:

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+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

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Your values will look like this :

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    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

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#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

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#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

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#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

+

#1.051 / 0.528 = 1.99#

+

#3.168 / 0.528 = 6#

+

#0.528 / 0.528 = 1#

+

To find the smallest whole number ratio, multiply each value by #2#.

+

#1.99 * 2 ~~ 4 #

+

#6 * 2 = 12#

+

#1 * 2 = 2#

+

Your formula is now:

+

#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

+

#""ZnN""_2""O""_6#

+

(Remember that if the subscript of an element is #1#, then no subscript is necessary).

+
+
+
" "
+

An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

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+
+ + +Chemistry + + + + + +The Mole Concept + + + + + +Empirical and Molecular Formulas + + +
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+2 Answers +
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+ + +
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+ +
+ + Jun 18, 2018 + +
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+

The empirical formula is #""ZnN""_2""O""_6#.

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+
+
+

Explanation:

+
+

To begin, using a #""100-g""# sample as a proportion to #100%#, convert each percentage of elements into grams.

+

Your values will look like this :

+
    +
  • #""14.79 g ""# nitrogen
    • +
  • #""50.68 g ""# oxygen
    • +
  • #""34.53 g ""# zinc
    • +
+

Next, convert each value into moles by using molar mass.

+

(Divide your original value by the molar mass to find moles)

+

#""14.79 g N"" / ""14.07 g.mol"" = ""1.051 moles nitrogen""#

+

#""50.68 g O"" / ""16 g"" = ""3.168 moles oxygen""#

+

#""34.53 g Zn"" / ""65.39 g"" = ""0.528 moles zinc""#

+

Take your smallest value of moles and divide every numerical value by it. In this case, the smallest value is #0.528#. The unit of mol will be canceled out automatically by the division, only concern yourself with the numerical values.

+

#1.051 / 0.528 = 1.99#

+

#3.168 / 0.528 = 6#

+

#0.528 / 0.528 = 1#

+

To find the smallest whole number ratio, multiply each value by #2#.

+

#1.99 * 2 ~~ 4 #

+

#6 * 2 = 12#

+

#1 * 2 = 2#

+

Your formula is now:

+

#""Zn""_2""N""_4""O""_12#

+

Remember, empirical formulas are related by a whole number ratios or the least common multiple of all numbers. #4#, #12#, and #2# all share the factor of #2#. Divide each subscript by #2#.

+

The empirical formula is:

+

#""ZnN""_2""O""_6#

+

(Remember that if the subscript of an element is #1#, then no subscript is necessary).

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+ +
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+ +
+ + Jun 18, 2018 + +
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+

#ZnN_2O_6#

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Explanation:

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As always we ASSUME a #100*g# mass of compound, and we assess the EMPIRICAL formula by dividing the elemental masses thru by the ATOMIC masses of each element...

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#""Moles of zinc""-=(34.53*g)/(65.4*g*mol^-1)=0.528*mol#

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#""Moles of nitrogen""-=(14.79*g)/(14.01*g*mol^-1)=1.056*mol#

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#""Moles of oxygen""-=(50.68*g)/(15.999*g*mol^-1)=3.168*mol#

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And so we divide thru by the SMALLEST molar quantity to get a trial empirical formula of...

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#Zn_((0.528*mol)/(0.528*mol))N_((1.056*mol)/(0.528*mol))O_((3.168*mol)/(0.528*mol))-=ZnN_2O_6#...(and I hope you can see those quotients....because I had to magnify them...)

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And of course this is zinc nitrate...#Zn(NO_3)_2#. Note that normally you would NEVER be quoted percentage oxygen. You would be quoted percentage metal, percentage nitrogen, and then you would be expected to get percentage oxygen by difference. Oxygen is a difficult element to analyze...

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" An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound? nan