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2,300 | aa87d6ca-6ddd-11ea-86d7-ccda262736ce | https://socratic.org/questions/nitrogen-dioxide-is-a-covalent-compound-how-many-oxygen-atoms-are-in-a-nitrogen- | 2 | start physical_unit 8 9 number none qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"2"}] | [{"type":"other","value":"Nitrogen dioxide is a covalent compound."},{"type":"physical unit","value":"Number [OF] nitrogen dioxide molecule [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">Nitrogen dioxide is a covalent compound. How many oxygen atoms are in a nitrogen dioxide molecule?</h1> | null | 2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NO_2#</mathjax> is an interesting molecule; because it is a radical species with a formal, lone electron on the central nitrogen:</p>
<p>i.e. <mathjax>#O=N^(+)-O^(-)#</mathjax></p>
<p>On the central nitrogen, there are FOUR <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, i.e. there is a nitrogen-centred single electron. That there are 4 <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a> rather than 5 justifies our distribution of a formal positive charge to this site. Because of this single electron, we can rationalize the formation of <mathjax>#N_2O_4#</mathjax>, where the single electrons couple to form a <mathjax>#N-N#</mathjax> bond:</p>
<p><mathjax>#2NO_2 rarr N_2O_4#</mathjax></p>
<p>A Lewis structure of <mathjax>#""^(-)O(O=)N^(+)-""^(+)N(=O)O^-#</mathjax> is usually invoked. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>In <mathjax>#NO_2#</mathjax>? Well I think you can count them yourself. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NO_2#</mathjax> is an interesting molecule; because it is a radical species with a formal, lone electron on the central nitrogen:</p>
<p>i.e. <mathjax>#O=N^(+)-O^(-)#</mathjax></p>
<p>On the central nitrogen, there are FOUR <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, i.e. there is a nitrogen-centred single electron. That there are 4 <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a> rather than 5 justifies our distribution of a formal positive charge to this site. Because of this single electron, we can rationalize the formation of <mathjax>#N_2O_4#</mathjax>, where the single electrons couple to form a <mathjax>#N-N#</mathjax> bond:</p>
<p><mathjax>#2NO_2 rarr N_2O_4#</mathjax></p>
<p>A Lewis structure of <mathjax>#""^(-)O(O=)N^(+)-""^(+)N(=O)O^-#</mathjax> is usually invoked. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Nitrogen dioxide is a covalent compound. How many oxygen atoms are in a nitrogen dioxide molecule?</h1>
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anor277
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Nov 28, 2016
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<div class="markdown"><p>In <mathjax>#NO_2#</mathjax>? Well I think you can count them yourself. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NO_2#</mathjax> is an interesting molecule; because it is a radical species with a formal, lone electron on the central nitrogen:</p>
<p>i.e. <mathjax>#O=N^(+)-O^(-)#</mathjax></p>
<p>On the central nitrogen, there are FOUR <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>, i.e. there is a nitrogen-centred single electron. That there are 4 <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a> rather than 5 justifies our distribution of a formal positive charge to this site. Because of this single electron, we can rationalize the formation of <mathjax>#N_2O_4#</mathjax>, where the single electrons couple to form a <mathjax>#N-N#</mathjax> bond:</p>
<p><mathjax>#2NO_2 rarr N_2O_4#</mathjax></p>
<p>A Lewis structure of <mathjax>#""^(-)O(O=)N^(+)-""^(+)N(=O)O^-#</mathjax> is usually invoked. </p></div>
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</article> | Nitrogen dioxide is a covalent compound. How many oxygen atoms are in a nitrogen dioxide molecule? | null |
2,301 | a94c4bdd-6ddd-11ea-a606-ccda262736ce | https://socratic.org/questions/the-reaction-2-mg-s-o2-g-2-mgo-s-has-a-free-energy-grxn-597-kj-mol-under-standar | -298.5 kJ/mol | start physical_unit 8 8 standard_free_energy kj/mol qc_end chemical_equation 2 8 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Free energy of formation [OF] MgO [IN] kJ/mol"}] | [{"type":"physical unit","value":"-298.5 kJ/mol"}] | [{"type":"physical unit","value":"DeltaG [OF] rxn [=] \\pu{-597 kJ/mol}"},{"type":"chemical equation","value":"2 Mg(s) + O2(g) -> 2 MgO(s)"},{"type":"other","value":"Standard conditions."}] | <h1 class="questionTitle" itemprop="name">The reaction #2 "Mg"(s) + "O"_2(g) -> 2 "MgO"(s)# has a free energy
#DeltaG_"rxn" = -"597 kJ/mol"# under standard conditions. What is the free energy of formation of #"MgO"#?</h1> | null | -298.5 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>-298.5 kJ/mol<br/>
The given reaction provides the <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs free energy</a> of formation for the formation of two moles of MgO. Therefore, the <mathjax>#G_f#</mathjax> for one mole is half that amount.</p></div>
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<div class="markdown"><p>-298.5 kJ/mol</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>-298.5 kJ/mol<br/>
The given reaction provides the <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs free energy</a> of formation for the formation of two moles of MgO. Therefore, the <mathjax>#G_f#</mathjax> for one mole is half that amount.</p></div>
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<h1 class="questionTitle" itemprop="name">The reaction #2 "Mg"(s) + "O"_2(g) -> 2 "MgO"(s)# has a free energy
#DeltaG_"rxn" = -"597 kJ/mol"# under standard conditions. What is the free energy of formation of #"MgO"#?</h1>
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<div class="markdown"><p>-298.5 kJ/mol</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>-298.5 kJ/mol<br/>
The given reaction provides the <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs free energy</a> of formation for the formation of two moles of MgO. Therefore, the <mathjax>#G_f#</mathjax> for one mole is half that amount.</p></div>
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</article> | The reaction #2 "Mg"(s) + "O"_2(g) -> 2 "MgO"(s)# has a free energy
#DeltaG_"rxn" = -"597 kJ/mol"# under standard conditions. What is the free energy of formation of #"MgO"#? | null |
2,302 | aa7effac-6ddd-11ea-b4bd-ccda262736ce | https://socratic.org/questions/5921e6497c0149508b46647c | 1043.74 grams | start physical_unit 5 5 mass g qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] NaCl [IN] grams"}] | [{"type":"physical unit","value":"1043.74 grams"}] | [{"type":"physical unit","value":"Volume [OF] Cl2 gas [=] \\pu{200 L}"},{"type":"other","value":"Excess sodium metal."}] | <h1 class="questionTitle" itemprop="name">What mass of sodium chloride, #NaCl#, is made when #200# #L# of chlorine gas, #Cl_2#, reacts with excess sodium metal?</h1> | null | 1043.74 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of an ideal gas has a volume of <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure. </p>
<p>The number of moles of gas, then, is given by <mathjax>#n=V/22.4 = 200/22.4 = 8.93#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Looking at the balanced equation, each <mathjax>#1#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2#</mathjax> mol of <mathjax>#NaCl#</mathjax>, so <mathjax>#8.93#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2xx8.93 =17.86#</mathjax> mol of <mathjax>#NaCl#</mathjax>.</p>
<p>The molar mass of <mathjax>#NaCl#</mathjax> is <mathjax>#58.44#</mathjax> <mathjax>#g#</mathjax>. To find the mass of the product, we take <mathjax>#n=m/M#</mathjax> and rearrange to make <mathjax>#m#</mathjax> the product. <mathjax>#m=nM = 17.86xx58.44 = 1044#</mathjax> <mathjax>#g#</mathjax> (with some rounding)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1044#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#NaCl#</mathjax> are produced, since there are <mathjax>#8.93#</mathjax> mol of <mathjax>#Cl_2#</mathjax> gas and each mole of <mathjax>#Cl_2#</mathjax> produces <mathjax>#2#</mathjax> mol of <mathjax>#NaCl#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of an ideal gas has a volume of <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure. </p>
<p>The number of moles of gas, then, is given by <mathjax>#n=V/22.4 = 200/22.4 = 8.93#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Looking at the balanced equation, each <mathjax>#1#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2#</mathjax> mol of <mathjax>#NaCl#</mathjax>, so <mathjax>#8.93#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2xx8.93 =17.86#</mathjax> mol of <mathjax>#NaCl#</mathjax>.</p>
<p>The molar mass of <mathjax>#NaCl#</mathjax> is <mathjax>#58.44#</mathjax> <mathjax>#g#</mathjax>. To find the mass of the product, we take <mathjax>#n=m/M#</mathjax> and rearrange to make <mathjax>#m#</mathjax> the product. <mathjax>#m=nM = 17.86xx58.44 = 1044#</mathjax> <mathjax>#g#</mathjax> (with some rounding)</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass of sodium chloride, #NaCl#, is made when #200# #L# of chlorine gas, #Cl_2#, reacts with excess sodium metal?</h1>
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David G.
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<span class="dateCreated" datetime="2017-05-21T22:52:44" itemprop="dateCreated">
May 21, 2017
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<div class="markdown"><p><mathjax>#1044#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#NaCl#</mathjax> are produced, since there are <mathjax>#8.93#</mathjax> mol of <mathjax>#Cl_2#</mathjax> gas and each mole of <mathjax>#Cl_2#</mathjax> produces <mathjax>#2#</mathjax> mol of <mathjax>#NaCl#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of an ideal gas has a volume of <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure. </p>
<p>The number of moles of gas, then, is given by <mathjax>#n=V/22.4 = 200/22.4 = 8.93#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Looking at the balanced equation, each <mathjax>#1#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2#</mathjax> mol of <mathjax>#NaCl#</mathjax>, so <mathjax>#8.93#</mathjax> mol of <mathjax>#Cl_2#</mathjax> yields <mathjax>#2xx8.93 =17.86#</mathjax> mol of <mathjax>#NaCl#</mathjax>.</p>
<p>The molar mass of <mathjax>#NaCl#</mathjax> is <mathjax>#58.44#</mathjax> <mathjax>#g#</mathjax>. To find the mass of the product, we take <mathjax>#n=m/M#</mathjax> and rearrange to make <mathjax>#m#</mathjax> the product. <mathjax>#m=nM = 17.86xx58.44 = 1044#</mathjax> <mathjax>#g#</mathjax> (with some rounding)</p></div>
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</article> | What mass of sodium chloride, #NaCl#, is made when #200# #L# of chlorine gas, #Cl_2#, reacts with excess sodium metal? | null |
2,303 | ac27e7a4-6ddd-11ea-a23c-ccda262736ce | https://socratic.org/questions/what-is-the-final-concentration-of-50-0-ml-of-a-2-00m-solution-diluted-to-500-0- | 0.20 M | start physical_unit 12 12 concentration mol/l qc_end physical_unit 12 12 10 11 concentration qc_end physical_unit 12 12 6 7 volume qc_end physical_unit 12 12 15 16 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] solution [IN] M"}] | [{"type":"physical unit","value":"0.20 M"}] | [{"type":"physical unit","value":"Concentration1 [OF] solution [=] \\pu{2.00 M}"},{"type":"physical unit","value":"Volume1 [OF] solution [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Volume2 [OF] solution [=] \\pu{500.0 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the final concentration of 50.0 mL of a 2.00M solution, diluted to 500.0 mL?</h1> | null | 0.20 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(50.0xx10^(-3)cancel(L)xx2.00*mol*cancel(L^-1))/(500.0xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that the answer is reasonable. If you read the question you have diluted the starting solution from <mathjax>#50.0*mL#</mathjax> to <mathjax>#500.0*mL#</mathjax>, a tenfold dilution, so the final solution shoould be <mathjax>#10xx#</mathjax> as dilute. </p></div>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of stuff"/"Volume of solution"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#(50.0xx10^(-3)cancel(L)xx2.00*mol*cancel(L^-1))/(500.0xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that the answer is reasonable. If you read the question you have diluted the starting solution from <mathjax>#50.0*mL#</mathjax> to <mathjax>#500.0*mL#</mathjax>, a tenfold dilution, so the final solution shoould be <mathjax>#10xx#</mathjax> as dilute. </p></div>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of stuff"/"Volume of solution"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(50.0xx10^(-3)cancel(L)xx2.00*mol*cancel(L^-1))/(500.0xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that the answer is reasonable. If you read the question you have diluted the starting solution from <mathjax>#50.0*mL#</mathjax> to <mathjax>#500.0*mL#</mathjax>, a tenfold dilution, so the final solution shoould be <mathjax>#10xx#</mathjax> as dilute. </p></div>
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</article> | What is the final concentration of 50.0 mL of a 2.00M solution, diluted to 500.0 mL? | null |
2,304 | ab81cfb1-6ddd-11ea-815a-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-volume-of-3-03-g-of-h2-g-at-a-pressure-of-560-torr-a | 47 L | start physical_unit 10 10 volume l qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 15 16 pressure qc_end physical_unit 10 10 20 21 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] H2(g) [IN] L"}] | [{"type":"physical unit","value":"47 L"}] | [{"type":"physical unit","value":"Mass [OF] H2(g) [=] \\pu{3.03 g}"},{"type":"physical unit","value":"Pressure [OF] H2(g) [=] \\pu{560 torr}"},{"type":"physical unit","value":"Temperature [OF] H2(g) [=] \\pu{139 K}"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the volume of 3.03 g of H2(g) at a pressure of 560 torr and temp of 139 k?</h1> | null | 47 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>and rearrange to solve for the volume of the gas, <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#V = (nRT)/P" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax> - the number of moles of the gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given in as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin<br/>
<mathjax>#P#</mathjax> - the pressure of the gas</p>
<p>Now, you have everything you need to calculate the volume of the gas <em>except</em> the number of moles. </p>
<p>To find how many moles of gas you have in that sample, use hydrogen's molar mass</p>
<blockquote>
<p><mathjax>#3.03color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(1.00794color(red)(cancel(color(black)("g")))) = "3.006 moles H"_2#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V#</mathjax> - do not forget to convert the pressure from <em>torr</em> to <em>atm</em></p>
<blockquote>
<p><mathjax>#V = (3.006color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 139color(red)(cancel(color(black)("K"))))/(560/760color(red)(cancel(color(black)("atm")))) = color(green)("47 L")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the pressure of the gas. </p>
<p>
<iframe src="https://www.youtube.com/embed/TqLlfHBFY08?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p><mathjax>#"47 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>and rearrange to solve for the volume of the gas, <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#V = (nRT)/P" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax> - the number of moles of the gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given in as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin<br/>
<mathjax>#P#</mathjax> - the pressure of the gas</p>
<p>Now, you have everything you need to calculate the volume of the gas <em>except</em> the number of moles. </p>
<p>To find how many moles of gas you have in that sample, use hydrogen's molar mass</p>
<blockquote>
<p><mathjax>#3.03color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(1.00794color(red)(cancel(color(black)("g")))) = "3.006 moles H"_2#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V#</mathjax> - do not forget to convert the pressure from <em>torr</em> to <em>atm</em></p>
<blockquote>
<p><mathjax>#V = (3.006color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 139color(red)(cancel(color(black)("K"))))/(560/760color(red)(cancel(color(black)("atm")))) = color(green)("47 L")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the pressure of the gas. </p>
<p>
<iframe src="https://www.youtube.com/embed/TqLlfHBFY08?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">How would you calculate the volume of 3.03 g of H2(g) at a pressure of 560 torr and temp of 139 k?</h1>
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Nov 1, 2015
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<div class="markdown"><p><mathjax>#"47 L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>and rearrange to solve for the volume of the gas, <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#V = (nRT)/P" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#n#</mathjax> - the number of moles of the gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given in as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin<br/>
<mathjax>#P#</mathjax> - the pressure of the gas</p>
<p>Now, you have everything you need to calculate the volume of the gas <em>except</em> the number of moles. </p>
<p>To find how many moles of gas you have in that sample, use hydrogen's molar mass</p>
<blockquote>
<p><mathjax>#3.03color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(1.00794color(red)(cancel(color(black)("g")))) = "3.006 moles H"_2#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V#</mathjax> - do not forget to convert the pressure from <em>torr</em> to <em>atm</em></p>
<blockquote>
<p><mathjax>#V = (3.006color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 139color(red)(cancel(color(black)("K"))))/(560/760color(red)(cancel(color(black)("atm")))) = color(green)("47 L")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the pressure of the gas. </p>
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</article> | How would you calculate the volume of 3.03 g of H2(g) at a pressure of 560 torr and temp of 139 k? | null |
2,305 | ab422414-6ddd-11ea-9657-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-for-the-synthesis-of-sodium-bromide-from- | 2 Na(s) + Br2(l) -> 2 NaBr(s) | 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 equation"}] | [{"type":"chemical equation","value":"2 Na(s) + Br2(l) -> 2 NaBr(s)"}] | [{"type":"substance name","value":"Sodium bromide"},{"type":"substance name","value":"Sodium"},{"type":"substance name","value":"Bromine"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for the synthesis of sodium bromide from sodium and bromine?</h1> | null | 2 Na(s) + Br2(l) -> 2 NaBr(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Alternatively.......</p>
<p><mathjax>#2Na(s) + Br_2(l) rarr 2NaBr#</mathjax></p>
<p>Is this a redox equation? Why?</p></div>
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<div class="markdown"><p><mathjax>#Na(s) + 1/2Br_2(l) rarr NaBr(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Alternatively.......</p>
<p><mathjax>#2Na(s) + Br_2(l) rarr 2NaBr#</mathjax></p>
<p>Is this a redox equation? Why?</p></div>
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<div class="markdown"><p><mathjax>#Na(s) + 1/2Br_2(l) rarr NaBr(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Alternatively.......</p>
<p><mathjax>#2Na(s) + Br_2(l) rarr 2NaBr#</mathjax></p>
<p>Is this a redox equation? Why?</p></div>
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</article> | What is the balanced chemical equation for the synthesis of sodium bromide from sodium and bromine? | null |
2,306 | aae6ee08-6ddd-11ea-826d-ccda262736ce | https://socratic.org/questions/how-many-liters-of-oxygen-are-required-to-react-completely-with-2-4-liters-of-hy-1 | 1.20 liters | start physical_unit 4 4 volume l qc_end physical_unit 14 14 11 12 volume qc_end chemical_equation 18 24 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] liters"}] | [{"type":"physical unit","value":"1.20 liters"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{2.4 liters}"},{"type":"chemical equation","value":"2 H2(g) + O2(g) -> 2 H2O(g)"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#?</h1> | null | 1.20 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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<div class="markdown"><p>Why <mathjax>#1.2*L#</mathjax> precisely..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#?</h1>
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<div class="markdown"><p>Why <mathjax>#1.2*L#</mathjax> precisely..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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<div class="markdown"><p><mathjax>#1.2 L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Know that one mole of any gas at STP takes up 22.4 liters. So 2.4 L of hydrogen is <mathjax>#0.107142857 mols H_2#</mathjax>.</p>
<p>Since <mathjax>#H_2#</mathjax> reacts with <mathjax>#O_2#</mathjax> in a 2:1 ratio, to completely react the two you will need half as much oxygen as hydrogen.</p>
<p><mathjax>#"mols " O_2= (0.107142857 mols H_2)/2#</mathjax></p>
<p><mathjax>#"mols " O_2 = 0.053571429 mols#</mathjax></p>
<p>To find liters of oxygen, just multiply the moles above by 22.4 liters.</p>
<p><mathjax>#"liters "O_2 = 0.053571429 mols * 22.4 L/(mols)#</mathjax></p>
<p><mathjax>#"liters " O_2 = 1.2L#</mathjax></p></div>
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</article> | How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#? | null |
2,307 | a8e16e94-6ddd-11ea-a4ff-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-4-65-times-10-24-molecules-of-no-2 | 7.72 moles | start physical_unit 10 10 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] NO2 [IN] moles"}] | [{"type":"physical unit","value":"7.72 moles"}] | [{"type":"physical unit","value":"Number [OF] NO2 molecules [=] \\pu{4.65 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in #4.65 times 10^24# molecules of #NO_2#?</h1> | null | 7.72 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we take the quotient.....</p>
<p><mathjax>#(4.65xx10^24*NO_2* "molecules")/(6.022xx10^23*NO_2* "molecules"*mol^-1)=7.72*mol#</mathjax>...</p>
<p>What is the mass of molar quantity?</p></div>
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<div class="markdown"><p>Well, one mole contains <mathjax>#6.022xx10^23#</mathjax> <mathjax>#NO_2#</mathjax> molecules....</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we take the quotient.....</p>
<p><mathjax>#(4.65xx10^24*NO_2* "molecules")/(6.022xx10^23*NO_2* "molecules"*mol^-1)=7.72*mol#</mathjax>...</p>
<p>What is the mass of molar quantity?</p></div>
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anor277
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<div class="markdown"><p>Well, one mole contains <mathjax>#6.022xx10^23#</mathjax> <mathjax>#NO_2#</mathjax> molecules....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we take the quotient.....</p>
<p><mathjax>#(4.65xx10^24*NO_2* "molecules")/(6.022xx10^23*NO_2* "molecules"*mol^-1)=7.72*mol#</mathjax>...</p>
<p>What is the mass of molar quantity?</p></div>
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</article> | How many moles are in #4.65 times 10^24# molecules of #NO_2#? | null |
2,308 | ad27b9f5-6ddd-11ea-90ec-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-0605-m-solution-of-sodium-hydroxide-naoh | 12.78 | start physical_unit 8 8 ph none qc_end physical_unit 12 12 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NaOH solution"}] | [{"type":"physical unit","value":"12.78"}] | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.0605 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.0605 M solution of sodium hydroxide, #NaOH#? </h1> | null | 12.78 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a solution that contains a <strong>strong base</strong>, so right from the start you should know that sodium hydroxide <strong>dissociates completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html" src="https://useruploads.socratic.org/unebI1DJQmSgtVxkuScI_FG14_13.JPG"/> </p>
<p>This means that <strong>for every mole</strong> of sodium hydroxide that you're dissolving in water, you will get <strong>one mole</strong> of hydroxide anions in aqueous solution. </p>
<p>Therefore, for a given volume of your solution, the <em>concentration</em> of the hydroxide anions will be <strong>equal</strong> to that of the strong base</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = ["NaOH"] = "0.0605 M"#</mathjax></p>
</blockquote>
<p>Now that you know the concentration of hydroxide anions, you can calculate the pOH of the solution by using </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pOH" = - log(0.0605) = 1.22#</mathjax></p>
</blockquote>
<p>You know that for aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature you have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH " + " pH" = 14color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the pOH of the solution to get</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 1.22 = color(green)(|bar(ul(color(white)(a/a)color(black)(12.78)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#"pH" = 12.78#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a solution that contains a <strong>strong base</strong>, so right from the start you should know that sodium hydroxide <strong>dissociates completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html" src="https://useruploads.socratic.org/unebI1DJQmSgtVxkuScI_FG14_13.JPG"/> </p>
<p>This means that <strong>for every mole</strong> of sodium hydroxide that you're dissolving in water, you will get <strong>one mole</strong> of hydroxide anions in aqueous solution. </p>
<p>Therefore, for a given volume of your solution, the <em>concentration</em> of the hydroxide anions will be <strong>equal</strong> to that of the strong base</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = ["NaOH"] = "0.0605 M"#</mathjax></p>
</blockquote>
<p>Now that you know the concentration of hydroxide anions, you can calculate the pOH of the solution by using </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pOH" = - log(0.0605) = 1.22#</mathjax></p>
</blockquote>
<p>You know that for aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature you have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH " + " pH" = 14color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the pOH of the solution to get</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 1.22 = color(green)(|bar(ul(color(white)(a/a)color(black)(12.78)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.0605 M solution of sodium hydroxide, #NaOH#? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 12.78#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a solution that contains a <strong>strong base</strong>, so right from the start you should know that sodium hydroxide <strong>dissociates completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html" src="https://useruploads.socratic.org/unebI1DJQmSgtVxkuScI_FG14_13.JPG"/> </p>
<p>This means that <strong>for every mole</strong> of sodium hydroxide that you're dissolving in water, you will get <strong>one mole</strong> of hydroxide anions in aqueous solution. </p>
<p>Therefore, for a given volume of your solution, the <em>concentration</em> of the hydroxide anions will be <strong>equal</strong> to that of the strong base</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = ["NaOH"] = "0.0605 M"#</mathjax></p>
</blockquote>
<p>Now that you know the concentration of hydroxide anions, you can calculate the pOH of the solution by using </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pOH" = - log(0.0605) = 1.22#</mathjax></p>
</blockquote>
<p>You know that for aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature you have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH " + " pH" = 14color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the pOH of the solution to get</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 1.22 = color(green)(|bar(ul(color(white)(a/a)color(black)(12.78)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | What is the pH of a 0.0605 M solution of sodium hydroxide, #NaOH#? | null |
2,309 | a901d014-6ddd-11ea-ba73-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-of-a-sample-of-ch-4-gas-6-022-g-in-a-30-0-l-vessel-at-402-k | 106.27 atm | start physical_unit 6 9 pressure atm qc_end physical_unit 6 9 10 11 mass qc_end physical_unit 16 16 14 15 volume qc_end physical_unit 6 9 18 19 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] CH4 gas sample [IN] atm"}] | [{"type":"physical unit","value":"106.27 atm"}] | [{"type":"physical unit","value":"Mass [OF] CH4 gas sample [=] \\pu{6.022 g}"},{"type":"physical unit","value":"Volume [OF] vessel [=] \\pu{30.0 L}"},{"type":"physical unit","value":"Temperature [OF] CH4 gas sample [=] \\pu{402 K}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure of a sample of #CH_4# gas (6.022 g) in a 30.0 L vessel at 402 K?</h1> | null | 106.27 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#" Lets recall the gas law"#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#"so "P = "(nRT)"/V"#</mathjax></p>
<p><mathjax>#"Where P = pressure in atm"#</mathjax><br/>
<mathjax>#"R = the gas constant 0.0821L"#</mathjax><br/>
<mathjax>#"T = temperature in kelvin"#</mathjax><br/>
<mathjax>#"n = moles of substance "#</mathjax><br/>
<mathjax>#"V = volume in litres"#</mathjax></p>
<p>Lets now plug in the variables <br/>
R = 0.0821L<br/>
T = 402K<br/>
now for <mathjax>#n#</mathjax> we have to calculate</p>
<p><mathjax>#"n" = "amount of substance in grams"/"molar mass"#</mathjax></p>
<p><mathjax>#"n" = "6.022g"/ "16.04 g/mol" = "96.6moles"#</mathjax></p>
<p>V = 30.0L</p>
<p><mathjax>#"(96.6g * 0.0821 * 402)" / "30.0L" = Pressure#</mathjax></p>
<p><mathjax># 3188.20572 / "30.0L" = "106.273524 atm"#</mathjax></p></div>
</div>
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<div>
<div class="markdown"><p><mathjax>#"106.273524 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#" Lets recall the gas law"#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#"so "P = "(nRT)"/V"#</mathjax></p>
<p><mathjax>#"Where P = pressure in atm"#</mathjax><br/>
<mathjax>#"R = the gas constant 0.0821L"#</mathjax><br/>
<mathjax>#"T = temperature in kelvin"#</mathjax><br/>
<mathjax>#"n = moles of substance "#</mathjax><br/>
<mathjax>#"V = volume in litres"#</mathjax></p>
<p>Lets now plug in the variables <br/>
R = 0.0821L<br/>
T = 402K<br/>
now for <mathjax>#n#</mathjax> we have to calculate</p>
<p><mathjax>#"n" = "amount of substance in grams"/"molar mass"#</mathjax></p>
<p><mathjax>#"n" = "6.022g"/ "16.04 g/mol" = "96.6moles"#</mathjax></p>
<p>V = 30.0L</p>
<p><mathjax>#"(96.6g * 0.0821 * 402)" / "30.0L" = Pressure#</mathjax></p>
<p><mathjax># 3188.20572 / "30.0L" = "106.273524 atm"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pressure of a sample of #CH_4# gas (6.022 g) in a 30.0 L vessel at 402 K?</h1>
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<div class="markdown"><p><mathjax>#"106.273524 atm"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#" Lets recall the gas law"#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#"so "P = "(nRT)"/V"#</mathjax></p>
<p><mathjax>#"Where P = pressure in atm"#</mathjax><br/>
<mathjax>#"R = the gas constant 0.0821L"#</mathjax><br/>
<mathjax>#"T = temperature in kelvin"#</mathjax><br/>
<mathjax>#"n = moles of substance "#</mathjax><br/>
<mathjax>#"V = volume in litres"#</mathjax></p>
<p>Lets now plug in the variables <br/>
R = 0.0821L<br/>
T = 402K<br/>
now for <mathjax>#n#</mathjax> we have to calculate</p>
<p><mathjax>#"n" = "amount of substance in grams"/"molar mass"#</mathjax></p>
<p><mathjax>#"n" = "6.022g"/ "16.04 g/mol" = "96.6moles"#</mathjax></p>
<p>V = 30.0L</p>
<p><mathjax>#"(96.6g * 0.0821 * 402)" / "30.0L" = Pressure#</mathjax></p>
<p><mathjax># 3188.20572 / "30.0L" = "106.273524 atm"#</mathjax></p></div>
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</article> | What is the pressure of a sample of #CH_4# gas (6.022 g) in a 30.0 L vessel at 402 K? | null |
2,310 | a98d6918-6ddd-11ea-810b-ccda262736ce | https://socratic.org/questions/how-many-joules-does-it-take-to-melt-35-g-of-ice-at-0-c | 1.17 × 10^4 joules | start physical_unit 11 11 energy j qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 11 11 13 14 temperature qc_end end | [{"type":"physical unit","value":"Needed energy [OF] ice [IN] joules"}] | [{"type":"physical unit","value":"1.17 × 10^4 joules"}] | [{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{35 g}"},{"type":"physical unit","value":"Temperature [OF] ice [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">How many joules does it take to melt 35 g of ice at 0° C? </h1> | null | 1.17 × 10^4 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:</p>
<p><mathjax>#q=nxxDeltaH#</mathjax></p>
<p>where:<br/>
<mathjax>#q#</mathjax> = heat<br/>
<mathjax>#n#</mathjax> = moles <br/>
<mathjax>#DeltaH#</mathjax> = enthalpy</p>
<p>In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:<br/>
1. Calculate moles of water<br/>
2. multiply by the enthalpy of fusion<br/>
3. Convert kJ to J</p>
<p><strong>Step 1: Calculating moles of water</strong></p>
<p><mathjax>#35 g xx ((1 mol)/(18.02g)) = 1.94 mols#</mathjax></p>
<p><strong>Step 2: Multiply by enthalpy of fusion</strong></p>
<p><mathjax>#q=nxxDeltaH= 1.94xx6.02= 11.678 kJ#</mathjax></p>
<p><strong>Step 3: Convert kJ to J</strong></p>
<p><mathjax>#11.678 kJxx((1000 J)/(1 kJ)) = 11,678 J#</mathjax></p>
<p>Finally rounding to 2 sig figs (since 34°C has two sig figs) we get</p>
<p><mathjax>#q=12,000J#</mathjax></p>
<p>For more examples on <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase changes</a> and enthalpy, see the video below: </p>
<p>
<iframe src="https://www.youtube.com/embed/tEms2L6eKJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>One last note, if the temperature were not 0 °C then the ice would have to be heated in addition to melted. This would be a phase change problem combined with a heat capacity problem.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>It takes 12,000 Joules of energy to melt 35 grams of ice at 0 °C</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:</p>
<p><mathjax>#q=nxxDeltaH#</mathjax></p>
<p>where:<br/>
<mathjax>#q#</mathjax> = heat<br/>
<mathjax>#n#</mathjax> = moles <br/>
<mathjax>#DeltaH#</mathjax> = enthalpy</p>
<p>In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:<br/>
1. Calculate moles of water<br/>
2. multiply by the enthalpy of fusion<br/>
3. Convert kJ to J</p>
<p><strong>Step 1: Calculating moles of water</strong></p>
<p><mathjax>#35 g xx ((1 mol)/(18.02g)) = 1.94 mols#</mathjax></p>
<p><strong>Step 2: Multiply by enthalpy of fusion</strong></p>
<p><mathjax>#q=nxxDeltaH= 1.94xx6.02= 11.678 kJ#</mathjax></p>
<p><strong>Step 3: Convert kJ to J</strong></p>
<p><mathjax>#11.678 kJxx((1000 J)/(1 kJ)) = 11,678 J#</mathjax></p>
<p>Finally rounding to 2 sig figs (since 34°C has two sig figs) we get</p>
<p><mathjax>#q=12,000J#</mathjax></p>
<p>For more examples on <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase changes</a> and enthalpy, see the video below: </p>
<p>
<iframe src="https://www.youtube.com/embed/tEms2L6eKJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>One last note, if the temperature were not 0 °C then the ice would have to be heated in addition to melted. This would be a phase change problem combined with a heat capacity problem.</p></div>
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<h1 class="questionTitle" itemprop="name">How many joules does it take to melt 35 g of ice at 0° C? </h1>
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<div class="markdown"><p>It takes 12,000 Joules of energy to melt 35 grams of ice at 0 °C</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:</p>
<p><mathjax>#q=nxxDeltaH#</mathjax></p>
<p>where:<br/>
<mathjax>#q#</mathjax> = heat<br/>
<mathjax>#n#</mathjax> = moles <br/>
<mathjax>#DeltaH#</mathjax> = enthalpy</p>
<p>In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:<br/>
1. Calculate moles of water<br/>
2. multiply by the enthalpy of fusion<br/>
3. Convert kJ to J</p>
<p><strong>Step 1: Calculating moles of water</strong></p>
<p><mathjax>#35 g xx ((1 mol)/(18.02g)) = 1.94 mols#</mathjax></p>
<p><strong>Step 2: Multiply by enthalpy of fusion</strong></p>
<p><mathjax>#q=nxxDeltaH= 1.94xx6.02= 11.678 kJ#</mathjax></p>
<p><strong>Step 3: Convert kJ to J</strong></p>
<p><mathjax>#11.678 kJxx((1000 J)/(1 kJ)) = 11,678 J#</mathjax></p>
<p>Finally rounding to 2 sig figs (since 34°C has two sig figs) we get</p>
<p><mathjax>#q=12,000J#</mathjax></p>
<p>For more examples on <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase changes</a> and enthalpy, see the video below: </p>
<p>
<iframe src="https://www.youtube.com/embed/tEms2L6eKJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>One last note, if the temperature were not 0 °C then the ice would have to be heated in addition to melted. This would be a phase change problem combined with a heat capacity problem.</p></div>
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</article> | How many joules does it take to melt 35 g of ice at 0° C? | null |
2,311 | aa1515d8-6ddd-11ea-9100-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-2-0-moles-of-an-ideal-gas-at-290-k-and-2-8-atm | 1.70 × 10^(-2) L | start physical_unit 9 10 volume l qc_end physical_unit 9 10 5 6 mole qc_end physical_unit 9 10 12 13 temperature qc_end physical_unit 9 10 15 16 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] the ideal gas [IN] L"}] | [{"type":"physical unit","value":"1.70 × 10^(-2) L"}] | [{"type":"physical unit","value":"Mole [OF] the ideal gas [=] \\pu{2.0 moles}"},{"type":"physical unit","value":"Temperature [OF] the ideal gas [=] \\pu{290 K}"},{"type":"physical unit","value":"Pressure [OF] the ideal gas [=] \\pu{2.8 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm? </h1> | null | 1.70 × 10^(-2) L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here,</p>
<p><mathjax>#temperature,T=color(red)(290K)#</mathjax></p>
<p><mathjax>#pressure=2.8atm=2.8xx101325pa=color(red)(283710pa)#</mathjax> </p>
<p><mathjax>#n=color(red)(2.0mol es)#</mathjax> </p>
<p>for ideal gas, we know,</p>
<p><mathjax>#PV=nRT#</mathjax> </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(2.0mol exx8.314Nmmol e^(-1)K^(-1)xx290K)/(283710Nm^(-2)#</mathjax> </p>
<p><mathjax>#=0.016996651m^3#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.016996651m^3#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here,</p>
<p><mathjax>#temperature,T=color(red)(290K)#</mathjax></p>
<p><mathjax>#pressure=2.8atm=2.8xx101325pa=color(red)(283710pa)#</mathjax> </p>
<p><mathjax>#n=color(red)(2.0mol es)#</mathjax> </p>
<p>for ideal gas, we know,</p>
<p><mathjax>#PV=nRT#</mathjax> </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(2.0mol exx8.314Nmmol e^(-1)K^(-1)xx290K)/(283710Nm^(-2)#</mathjax> </p>
<p><mathjax>#=0.016996651m^3#</mathjax> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm? </h1>
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<div class="markdown"><p><mathjax>#0.016996651m^3#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>here,</p>
<p><mathjax>#temperature,T=color(red)(290K)#</mathjax></p>
<p><mathjax>#pressure=2.8atm=2.8xx101325pa=color(red)(283710pa)#</mathjax> </p>
<p><mathjax>#n=color(red)(2.0mol es)#</mathjax> </p>
<p>for ideal gas, we know,</p>
<p><mathjax>#PV=nRT#</mathjax> </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(2.0mol exx8.314Nmmol e^(-1)K^(-1)xx290K)/(283710Nm^(-2)#</mathjax> </p>
<p><mathjax>#=0.016996651m^3#</mathjax> </p></div>
</div>
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</article> | What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm? | null |
2,312 | a9ca0180-6ddd-11ea-b46c-ccda262736ce | https://socratic.org/questions/how-much-volume-would-2-00-moles-of-gas-take-up-under-standard-temperature-and-p | 44.8 dm^3 | start physical_unit 7 7 volume dm^3 qc_end physical_unit 7 7 4 5 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] gas [IN] dm^3"}] | [{"type":"physical unit","value":"44.8 dm^3"}] | [{"type":"physical unit","value":"Mole [OF] gas [=] \\pu{2.00 moles}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How much volume would 2.00 moles of gas take up, under standard temperature and pressure conditions? </h1> | null | 44.8 dm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of gas would take up 22.4 liters of volume, which is equivalent to 22.4<mathjax>#dm^3#</mathjax> of volume. </p>
<p>Therefore, multiply answer by 2 (since there are 2 moles), and hence you would get an answer of <strong><mathjax># 44.8dm^3#</mathjax></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>44.8 <mathjax>#dm^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of gas would take up 22.4 liters of volume, which is equivalent to 22.4<mathjax>#dm^3#</mathjax> of volume. </p>
<p>Therefore, multiply answer by 2 (since there are 2 moles), and hence you would get an answer of <strong><mathjax># 44.8dm^3#</mathjax></strong>. </p></div>
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<h1 class="questionTitle" itemprop="name">How much volume would 2.00 moles of gas take up, under standard temperature and pressure conditions? </h1>
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<div class="markdown"><p>44.8 <mathjax>#dm^3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of gas would take up 22.4 liters of volume, which is equivalent to 22.4<mathjax>#dm^3#</mathjax> of volume. </p>
<p>Therefore, multiply answer by 2 (since there are 2 moles), and hence you would get an answer of <strong><mathjax># 44.8dm^3#</mathjax></strong>. </p></div>
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</article> | How much volume would 2.00 moles of gas take up, under standard temperature and pressure conditions? | null |
2,313 | a8b6c28a-6ddd-11ea-9d27-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-for-this-reaction-gaseous-chlorine-reacts | Cl2(g) + 2 KBr(aq) -> Br2(l) + 2 KCl(aq) | start chemical_equation qc_end substance 9 10 qc_end substance 21 22 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"Cl2(g) + 2 KBr(aq) -> Br2(l) + 2 KCl(aq)"}] | [{"type":"substance name","value":"Gaseous chlorine"},{"type":"substance name","value":"Potassium bromide aqueous solution"},{"type":"substance name","value":"Liquid bromine"},{"type":"substance name","value":"Potassium chloride aqueous solution"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for this reaction: Gaseous chlorine reacts with an aqueous solution of potassium bromide to form liquid bromine and an aqueous solution of potassium chloride?</h1> | null | Cl2(g) + 2 KBr(aq) -> Br2(l) + 2 KCl(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement reaction in which a more reactive halogen (group 17/VIIA) replaces a less reactive halogen in a compound.</p>
<p>The reactivity of the halogens relative to one another decreases down the group. Since chlorine is above bromine, it is more reactive and will replace the bromine in the potassium bromide compound.</p>
<p>However, the reverse reaction would not occur, because bromine is less reactive than chlorine, so it would not be able to replace the chlorine in the potassium chloride compound.</p>
<p><img alt="http://mizbanan.com/most-and-least-reactive-elements-on-the-periodic-table/" src="https://useruploads.socratic.org/3Y18anuRQBCEQTmJGsCb_Organization%2Bof%2Bthe%2BPeriodic%2BTable%2BChemical%2BFamily%2B%25E2%2580%2593%2BHalogens.jpg"/> </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Cl"_2("g") + "2KBr(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Br"_2(l) + "2KCl(aq)"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement reaction in which a more reactive halogen (group 17/VIIA) replaces a less reactive halogen in a compound.</p>
<p>The reactivity of the halogens relative to one another decreases down the group. Since chlorine is above bromine, it is more reactive and will replace the bromine in the potassium bromide compound.</p>
<p>However, the reverse reaction would not occur, because bromine is less reactive than chlorine, so it would not be able to replace the chlorine in the potassium chloride compound.</p>
<p><img alt="http://mizbanan.com/most-and-least-reactive-elements-on-the-periodic-table/" src="https://useruploads.socratic.org/3Y18anuRQBCEQTmJGsCb_Organization%2Bof%2Bthe%2BPeriodic%2BTable%2BChemical%2BFamily%2B%25E2%2580%2593%2BHalogens.jpg"/> </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for this reaction: Gaseous chlorine reacts with an aqueous solution of potassium bromide to form liquid bromine and an aqueous solution of potassium chloride?</h1>
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<div class="markdown"><p><mathjax>#"Cl"_2("g") + "2KBr(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Br"_2(l) + "2KCl(aq)"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement reaction in which a more reactive halogen (group 17/VIIA) replaces a less reactive halogen in a compound.</p>
<p>The reactivity of the halogens relative to one another decreases down the group. Since chlorine is above bromine, it is more reactive and will replace the bromine in the potassium bromide compound.</p>
<p>However, the reverse reaction would not occur, because bromine is less reactive than chlorine, so it would not be able to replace the chlorine in the potassium chloride compound.</p>
<p><img alt="http://mizbanan.com/most-and-least-reactive-elements-on-the-periodic-table/" src="https://useruploads.socratic.org/3Y18anuRQBCEQTmJGsCb_Organization%2Bof%2Bthe%2BPeriodic%2BTable%2BChemical%2BFamily%2B%25E2%2580%2593%2BHalogens.jpg"/> </p></div>
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</article> | What is the balanced chemical equation for this reaction: Gaseous chlorine reacts with an aqueous solution of potassium bromide to form liquid bromine and an aqueous solution of potassium chloride? | null |
2,314 | ac206e78-6ddd-11ea-8c3d-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-containing-13-5-g-ca-10-8-g-o-and | Ca(OH)2 | 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] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Ca(OH)2"}] | [{"type":"physical unit","value":"Mass [OF] Ca [=] \\pu{13.5 g}"},{"type":"physical unit","value":"Mass [OF] O [=] \\pu{10.8 g}"},{"type":"physical unit","value":"Mass [OF] H [=] \\pu{0.675 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound containing 13.5 g Ca, 10.8 g O, and 0.675 g H? </h1> | null | Ca(OH)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide each given mass by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each individual element:</p>
<p><mathjax>#"Calcium:"#</mathjax> <mathjax>#(13.5*g)/(40.1*g*mol^-1)=0.337*mol.#</mathjax></p>
<p><mathjax>#"Oxygen:"#</mathjax> <mathjax>#(10.8*g)/(15.999*g*mol^-1)=0.675*mol.#</mathjax></p>
<p><mathjax>#"Hydrogen:"#</mathjax> <mathjax>#(0.675*g)/(1.008*g*mol^-1)=0.675*mol.#</mathjax></p>
<p>If we divide thru by the smallest atomic quantity, we get an empirical formula of, <mathjax>#CaH_2O_2#</mathjax>, i.e. <mathjax>#Ca(OH)_2#</mathjax>. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Ca(OH)_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide each given mass by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each individual element:</p>
<p><mathjax>#"Calcium:"#</mathjax> <mathjax>#(13.5*g)/(40.1*g*mol^-1)=0.337*mol.#</mathjax></p>
<p><mathjax>#"Oxygen:"#</mathjax> <mathjax>#(10.8*g)/(15.999*g*mol^-1)=0.675*mol.#</mathjax></p>
<p><mathjax>#"Hydrogen:"#</mathjax> <mathjax>#(0.675*g)/(1.008*g*mol^-1)=0.675*mol.#</mathjax></p>
<p>If we divide thru by the smallest atomic quantity, we get an empirical formula of, <mathjax>#CaH_2O_2#</mathjax>, i.e. <mathjax>#Ca(OH)_2#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound containing 13.5 g Ca, 10.8 g O, and 0.675 g H? </h1>
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<div class="markdown"><p><mathjax>#Ca(OH)_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide each given mass by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each individual element:</p>
<p><mathjax>#"Calcium:"#</mathjax> <mathjax>#(13.5*g)/(40.1*g*mol^-1)=0.337*mol.#</mathjax></p>
<p><mathjax>#"Oxygen:"#</mathjax> <mathjax>#(10.8*g)/(15.999*g*mol^-1)=0.675*mol.#</mathjax></p>
<p><mathjax>#"Hydrogen:"#</mathjax> <mathjax>#(0.675*g)/(1.008*g*mol^-1)=0.675*mol.#</mathjax></p>
<p>If we divide thru by the smallest atomic quantity, we get an empirical formula of, <mathjax>#CaH_2O_2#</mathjax>, i.e. <mathjax>#Ca(OH)_2#</mathjax>. </p></div>
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</article> | What is the empirical formula for the compound containing 13.5 g Ca, 10.8 g O, and 0.675 g H? | null |
2,315 | aa31090c-6ddd-11ea-b449-ccda262736ce | https://socratic.org/questions/58f51e85b72cff7edc888fad | 1.80 L | start physical_unit 3 5 volume l qc_end physical_unit 3 5 7 8 concentration qc_end physical_unit 3 4 15 16 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] lithium sulfite solution [IN] L"}] | [{"type":"physical unit","value":"1.80 L"}] | [{"type":"physical unit","value":"Concentration [OF] lithium sulfite solution [=] \\pu{2.50 mol/L}"},{"type":"physical unit","value":"Mass [OF] lithium sulfite [=] \\pu{422.55 g}"}] | <h1 class="questionTitle" itemprop="name">What volume of #"lithium sulfite solution"# of #2.50*mol*L^-1# concentration would contain a mass of #422.55*g#?</h1> | null | 1.80 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#=(422.55*g)/(93.94*g*mol^-1)=4.50*mol#</mathjax>.</p>
<p>And since <mathjax>#"concentration"="moles"/"volume"#</mathjax>, we need the quotient..........</p>
<p><mathjax>#"volume"="moles"/"concentration"=(4.50*mol)/(2.50*mol*L^-1)=1.80*1/L^-1#</mathjax></p>
<p><mathjax>#1.80*1/(1/L)=1.80*L#</mathjax></p>
<p>In practice, water is quite an involatile material. You would not evaporate this quantity of water. </p></div>
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<div class="markdown"><p><mathjax>#"Moles of lithium sulfite req'd..........."=(422.55*g)/(93.94*g*mol^-1)=#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#=(422.55*g)/(93.94*g*mol^-1)=4.50*mol#</mathjax>.</p>
<p>And since <mathjax>#"concentration"="moles"/"volume"#</mathjax>, we need the quotient..........</p>
<p><mathjax>#"volume"="moles"/"concentration"=(4.50*mol)/(2.50*mol*L^-1)=1.80*1/L^-1#</mathjax></p>
<p><mathjax>#1.80*1/(1/L)=1.80*L#</mathjax></p>
<p>In practice, water is quite an involatile material. You would not evaporate this quantity of water. </p></div>
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<h1 class="questionTitle" itemprop="name">What volume of #"lithium sulfite solution"# of #2.50*mol*L^-1# concentration would contain a mass of #422.55*g#?</h1>
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<div class="markdown"><p><mathjax>#"Moles of lithium sulfite req'd..........."=(422.55*g)/(93.94*g*mol^-1)=#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#=(422.55*g)/(93.94*g*mol^-1)=4.50*mol#</mathjax>.</p>
<p>And since <mathjax>#"concentration"="moles"/"volume"#</mathjax>, we need the quotient..........</p>
<p><mathjax>#"volume"="moles"/"concentration"=(4.50*mol)/(2.50*mol*L^-1)=1.80*1/L^-1#</mathjax></p>
<p><mathjax>#1.80*1/(1/L)=1.80*L#</mathjax></p>
<p>In practice, water is quite an involatile material. You would not evaporate this quantity of water. </p></div>
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</article> | What volume of #"lithium sulfite solution"# of #2.50*mol*L^-1# concentration would contain a mass of #422.55*g#? | null |
2,316 | ac26fc5e-6ddd-11ea-b711-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-has-a-volume-of-25-l-at-a-pressure-of-200-kpa-and-a-temperature- | 18.32 L | start physical_unit 1 3 volume l qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 14 15 pressure qc_end physical_unit 1 3 20 21 temperature qc_end physical_unit 1 3 33 34 pressure qc_end physical_unit 1 3 41 42 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas sample [IN] L"}] | [{"type":"physical unit","value":"18.32 L"}] | [{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{25 L}"},{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{200 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] gas sample [=] \\pu{250 kPa}"},{"type":"physical unit","value":"Temperature2 [OF] gas sample [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 25 L at a pressure of 200 kPa and a temperature of 25°C. What would be the volume if the pressure were increased to 250 kPa and the temperature were decreased to 0°C?</h1> | null | 18.32 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>You need to convert degC to Kelvin by adding 273:</p>
<p><mathjax>#:.(200xx25)/298=(250xxV_2)/273#</mathjax></p>
<p><mathjax>#:.V_2=(200xx25xx273)/(298xx250)#</mathjax></p>
<p><mathjax>#:.V_2=18.32"L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#18.32"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>You need to convert degC to Kelvin by adding 273:</p>
<p><mathjax>#:.(200xx25)/298=(250xxV_2)/273#</mathjax></p>
<p><mathjax>#:.V_2=(200xx25xx273)/(298xx250)#</mathjax></p>
<p><mathjax>#:.V_2=18.32"L"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 25 L at a pressure of 200 kPa and a temperature of 25°C. What would be the volume if the pressure were increased to 250 kPa and the temperature were decreased to 0°C?</h1>
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<a class="topContributorPic" href="/users/michael-2"><img alt="" class="" src="https://lh3.googleusercontent.com/-gCv6FQRhls0/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7oOJS05Ylqn3KuDSW0LfnbOk7FezQ/mo/photo.jpg?sz=50" title=""/></a>
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Michael
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<span class="dateCreated" datetime="2015-11-22T17:49:08" itemprop="dateCreated">
Nov 22, 2015
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<div class="markdown"><p><mathjax>#18.32"L"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>You need to convert degC to Kelvin by adding 273:</p>
<p><mathjax>#:.(200xx25)/298=(250xxV_2)/273#</mathjax></p>
<p><mathjax>#:.V_2=(200xx25xx273)/(298xx250)#</mathjax></p>
<p><mathjax>#:.V_2=18.32"L"#</mathjax></p></div>
</div>
</div>
</div>
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<a href="https://socratic.org/answers/191938" itemprop="url">Answer link</a>
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</article> | A sample of gas has a volume of 25 L at a pressure of 200 kPa and a temperature of 25°C. What would be the volume if the pressure were increased to 250 kPa and the temperature were decreased to 0°C? | null |
2,317 | a9a0ee9c-6ddd-11ea-9915-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-in-mass-m-v-of-a-solution-if-you-dissolve-1-25-g-nacl- | 5% | start physical_unit 9 10 density none qc_end physical_unit 16 16 14 15 mass qc_end physical_unit 20 21 18 19 volume qc_end end | [{"type":"physical unit","value":"concentration by % (m/v) [OF] a solution"}] | [{"type":"physical unit","value":"5%"}] | [{"type":"physical unit","value":"mass [OF] NaCl [=] \\pu{1.25 g}"},{"type":"physical unit","value":"Volume [OF] DI H2O [=] \\pu{25.00 mL}"}] | <h1 class="questionTitle" itemprop="name"> What is the concentration in % mass (m/v) of a solution if you dissolve 1.25 g NaCl in 25.00 mL DI H2O?
</h1> | null | 5% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<blockquote>
<p><mathjax>#%("m/v") = "Mass of solute (grams)"/"Volume of solution (mL)" × 100#</mathjax></p>
</blockquote>
</blockquote>
<p>For given solution,</p>
<ul>
<li>Mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = 1.25 g</li>
<li>Volume of solution = 25.00 mL</li>
</ul>
<p><mathjax>#%("m/v") = "1.25 g"/"25.00 mL" × 100 = color(blue)"5% (m/v)"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5% ("m/v")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<blockquote>
<p><mathjax>#%("m/v") = "Mass of solute (grams)"/"Volume of solution (mL)" × 100#</mathjax></p>
</blockquote>
</blockquote>
<p>For given solution,</p>
<ul>
<li>Mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = 1.25 g</li>
<li>Volume of solution = 25.00 mL</li>
</ul>
<p><mathjax>#%("m/v") = "1.25 g"/"25.00 mL" × 100 = color(blue)"5% (m/v)"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> What is the concentration in % mass (m/v) of a solution if you dissolve 1.25 g NaCl in 25.00 mL DI H2O?
</h1>
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<a class="topContributorPic" href="/users/junaid-mirza"><img alt="" class="" src="https://profilepictures.socratic.org/BBeNx7NTQ7WmDQKll8wN_BDC9EDF4-5DCB-440D-8ADC-ADACE2966216.jpeg" title=""/></a>
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Junaid Mirza
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Mar 28, 2018
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<div class="markdown"><p><mathjax>#5% ("m/v")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<blockquote>
<p><mathjax>#%("m/v") = "Mass of solute (grams)"/"Volume of solution (mL)" × 100#</mathjax></p>
</blockquote>
</blockquote>
<p>For given solution,</p>
<ul>
<li>Mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = 1.25 g</li>
<li>Volume of solution = 25.00 mL</li>
</ul>
<p><mathjax>#%("m/v") = "1.25 g"/"25.00 mL" × 100 = color(blue)"5% (m/v)"#</mathjax></p></div>
</div>
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</article> | What is the concentration in % mass (m/v) of a solution if you dissolve 1.25 g NaCl in 25.00 mL DI H2O?
| null |
2,318 | a926676e-6ddd-11ea-baa5-ccda262736ce | https://socratic.org/questions/how-many-grams-of-potassium-chloride-are-in-45-0-g-of-a-5-00-by-mass-solution | 2.25 grams | start physical_unit 4 5 mass g qc_end physical_unit 15 15 8 9 mass qc_end physical_unit 4 5 12 12 mass_percent qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium chloride [IN] grams"}] | [{"type":"physical unit","value":"2.25 grams"}] | [{"type":"physical unit","value":"Mass [OF] solution [=] \\pu{45.0 g}"},{"type":"physical unit","value":"Percent by mass [OF] potassium chloride in solution [=] \\pu{5.00%}"}] | <h1 class="questionTitle" itemprop="name">How many grams of potassium chloride are in 45.0 g of a 5.00% (by mass) solution? </h1> | null | 2.25 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, essentially tells you how many <em>grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You're dealing with a <mathjax>#"5.00% m/m"#</mathjax> solution of potassium chloride, so right from the start you should be able to look at this solution and say that it contains <mathjax>#"5.00 g"#</mathjax> of solute <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>So, if <mathjax>#"100 g"#</mathjax> of solution will contain <mathjax>#"5.00 g"#</mathjax> of solute, it follows that <mathjax>#"45.0 g"#</mathjax> of solution will contain</p>
<blockquote>
<p><mathjax>#45.0 color(red)(cancel(color(black)("g solution"))) * overbrace("5.00 g KCl"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 5.00% m/m")) = color(green)(|bar(ul(color(white)(a/a)"2.25 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.25 g KCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, essentially tells you how many <em>grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You're dealing with a <mathjax>#"5.00% m/m"#</mathjax> solution of potassium chloride, so right from the start you should be able to look at this solution and say that it contains <mathjax>#"5.00 g"#</mathjax> of solute <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>So, if <mathjax>#"100 g"#</mathjax> of solution will contain <mathjax>#"5.00 g"#</mathjax> of solute, it follows that <mathjax>#"45.0 g"#</mathjax> of solution will contain</p>
<blockquote>
<p><mathjax>#45.0 color(red)(cancel(color(black)("g solution"))) * overbrace("5.00 g KCl"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 5.00% m/m")) = color(green)(|bar(ul(color(white)(a/a)"2.25 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of potassium chloride are in 45.0 g of a 5.00% (by mass) solution? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"2.25 g KCl"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, essentially tells you how many <em>grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You're dealing with a <mathjax>#"5.00% m/m"#</mathjax> solution of potassium chloride, so right from the start you should be able to look at this solution and say that it contains <mathjax>#"5.00 g"#</mathjax> of solute <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>So, if <mathjax>#"100 g"#</mathjax> of solution will contain <mathjax>#"5.00 g"#</mathjax> of solute, it follows that <mathjax>#"45.0 g"#</mathjax> of solution will contain</p>
<blockquote>
<p><mathjax>#45.0 color(red)(cancel(color(black)("g solution"))) * overbrace("5.00 g KCl"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 5.00% m/m")) = color(green)(|bar(ul(color(white)(a/a)"2.25 g"color(white)(a/a)|)))#</mathjax></p>
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<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | How many grams of potassium chloride are in 45.0 g of a 5.00% (by mass) solution? | null |
2,319 | acf50e0d-6ddd-11ea-b8bd-ccda262736ce | https://socratic.org/questions/589d1c9611ef6b2c1aee4bd2 | 39.93 g/mol | start physical_unit 19 20 molecular_weight g/mol qc_end physical_unit 19 20 9 10 density qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] this gas [IN] g/mol"}] | [{"type":"physical unit","value":"39.93 g/mol"}] | [{"type":"physical unit","value":"Density [OF] this gas [=] \\pu{1.7824 g/L}"},{"type":"other","value":"STP"},{"type":"other","value":"A molar volume of gas."}] | <h1 class="questionTitle" itemprop="name">A molar volume of gas has a density of #1.7824*g*L^-1# at #"STP"#. What is the molecular mass of this gas? </h1> | null | 39.93 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assume Ideal Gas behaviour..........</p>
<p>And thus, at <mathjax>#STP#</mathjax>, we have gots a molar mass of........</p>
<p><mathjax>#22.414*L*mol^-1xx1.7824*g*L^-1=39.95*g*mol^-1#</mathjax>......</p>
<p>You should be able to identify this gas <mathjax>#"pdq"#</mathjax>.</p></div>
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<div class="markdown"><p>Well the molar volume of an Ideal Gas at <mathjax>#STP#</mathjax> is <mathjax>#22.4*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assume Ideal Gas behaviour..........</p>
<p>And thus, at <mathjax>#STP#</mathjax>, we have gots a molar mass of........</p>
<p><mathjax>#22.414*L*mol^-1xx1.7824*g*L^-1=39.95*g*mol^-1#</mathjax>......</p>
<p>You should be able to identify this gas <mathjax>#"pdq"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A molar volume of gas has a density of #1.7824*g*L^-1# at #"STP"#. What is the molecular mass of this gas? </h1>
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<div class="markdown"><p>Well the molar volume of an Ideal Gas at <mathjax>#STP#</mathjax> is <mathjax>#22.4*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assume Ideal Gas behaviour..........</p>
<p>And thus, at <mathjax>#STP#</mathjax>, we have gots a molar mass of........</p>
<p><mathjax>#22.414*L*mol^-1xx1.7824*g*L^-1=39.95*g*mol^-1#</mathjax>......</p>
<p>You should be able to identify this gas <mathjax>#"pdq"#</mathjax>.</p></div>
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<div class="markdown"><p>we know , 22.4 L of any gas gas at STP is 1 mol of the gas.</p>
<p>We have been given that the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of an elemental gas is 1.7824g/L at STP.</p>
<p>Hence mass of 22.4 L of the elemental gas gas at STP is its molar mass.<br/>
Hence molar mass <mathjax>#=22.4L/"mol"xx1.7824g/L~~39.9258g"/"mol#</mathjax></p></div>
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</article> | A molar volume of gas has a density of #1.7824*g*L^-1# at #"STP"#. What is the molecular mass of this gas? | null |
2,320 | a963e16c-6ddd-11ea-9083-ccda262736ce | https://socratic.org/questions/592f6199b72cff557473ef74 | 0.26 M | start physical_unit 21 23 molarity mol/l qc_end physical_unit 21 23 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity2 [OF] the diluted solution [IN] M"}] | [{"type":"physical unit","value":"0.26 M"}] | [{"type":"physical unit","value":"Volume1 [OF] sulfuric acid stock soluition [=] \\pu{60.0 mL}"},{"type":"physical unit","value":"Molarity1 [OF] sulfuric acid stock solution [=] \\pu{6.07 M}"},{"type":"physical unit","value":"Volume2 [OF] the diluted solution [=] \\pu{1.42 L}"}] | <h1 class="questionTitle" itemprop="name">#"60.0 mL"# of a #"6.07 M"# stock solution of sulfuric acid is diluted to #"1.42 L"#. What is the molarity of the diluted solution?</h1> | null | 0.26 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When diluting a solution, you can determine the molarity of the diluted solution using the formula:</p>
<p><mathjax>#M_1V_1=M_2V_2#</mathjax>,</p>
<p>where <mathjax>#M#</mathjax> is molarity in <mathjax>#"mol/L"#</mathjax> and <mathjax>#V#</mathjax> is the volume of the solution in liters.</p>
<p>The initial volume is given in <mathjax>#"mL"#</mathjax>, but it needs to be in liters. Convert <mathjax>#"60.0 mL"#</mathjax> into liters. <mathjax>#"1 L = 1000 mL"#</mathjax></p>
<p><mathjax>#60.0color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.0600 L"#</mathjax></p>
<p><strong>Organize your data.</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#M_1="6.07 M"="6.07 mol/L"#</mathjax></p>
<p><mathjax>#V_1="0.0600 L"#</mathjax></p>
<p><mathjax>#V_2="1.42 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#M_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the formula above to isolate <mathjax>#M_2#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=((6.07"mol")/("1L")xx0.0600color(red)cancel(color(black)("L")))/(1.42color(red)cancel(color(black)("L")))="0.256 mol/L"#</mathjax> or <mathjax>#"0.256 M"#</mathjax> (rounded to three sig figs)</p></div>
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<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the diluted sulfuric acid solution will be <mathjax>#"0.256 mol/L"#</mathjax>, or <mathjax>#"0.256 M"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When diluting a solution, you can determine the molarity of the diluted solution using the formula:</p>
<p><mathjax>#M_1V_1=M_2V_2#</mathjax>,</p>
<p>where <mathjax>#M#</mathjax> is molarity in <mathjax>#"mol/L"#</mathjax> and <mathjax>#V#</mathjax> is the volume of the solution in liters.</p>
<p>The initial volume is given in <mathjax>#"mL"#</mathjax>, but it needs to be in liters. Convert <mathjax>#"60.0 mL"#</mathjax> into liters. <mathjax>#"1 L = 1000 mL"#</mathjax></p>
<p><mathjax>#60.0color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.0600 L"#</mathjax></p>
<p><strong>Organize your data.</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#M_1="6.07 M"="6.07 mol/L"#</mathjax></p>
<p><mathjax>#V_1="0.0600 L"#</mathjax></p>
<p><mathjax>#V_2="1.42 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#M_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the formula above to isolate <mathjax>#M_2#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=((6.07"mol")/("1L")xx0.0600color(red)cancel(color(black)("L")))/(1.42color(red)cancel(color(black)("L")))="0.256 mol/L"#</mathjax> or <mathjax>#"0.256 M"#</mathjax> (rounded to three sig figs)</p></div>
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<h1 class="questionTitle" itemprop="name">#"60.0 mL"# of a #"6.07 M"# stock solution of sulfuric acid is diluted to #"1.42 L"#. What is the molarity of the diluted solution?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the diluted sulfuric acid solution will be <mathjax>#"0.256 mol/L"#</mathjax>, or <mathjax>#"0.256 M"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When diluting a solution, you can determine the molarity of the diluted solution using the formula:</p>
<p><mathjax>#M_1V_1=M_2V_2#</mathjax>,</p>
<p>where <mathjax>#M#</mathjax> is molarity in <mathjax>#"mol/L"#</mathjax> and <mathjax>#V#</mathjax> is the volume of the solution in liters.</p>
<p>The initial volume is given in <mathjax>#"mL"#</mathjax>, but it needs to be in liters. Convert <mathjax>#"60.0 mL"#</mathjax> into liters. <mathjax>#"1 L = 1000 mL"#</mathjax></p>
<p><mathjax>#60.0color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.0600 L"#</mathjax></p>
<p><strong>Organize your data.</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#M_1="6.07 M"="6.07 mol/L"#</mathjax></p>
<p><mathjax>#V_1="0.0600 L"#</mathjax></p>
<p><mathjax>#V_2="1.42 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#M_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the formula above to isolate <mathjax>#M_2#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=((6.07"mol")/("1L")xx0.0600color(red)cancel(color(black)("L")))/(1.42color(red)cancel(color(black)("L")))="0.256 mol/L"#</mathjax> or <mathjax>#"0.256 M"#</mathjax> (rounded to three sig figs)</p></div>
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</article> | #"60.0 mL"# of a #"6.07 M"# stock solution of sulfuric acid is diluted to #"1.42 L"#. What is the molarity of the diluted solution? | null |
2,321 | ac7b1259-6ddd-11ea-bf40-ccda262736ce | https://socratic.org/questions/5647948311ef6b5c1ccd6e0d | 23.82 atm | start physical_unit 9 9 pressure atm qc_end physical_unit 9 9 6 7 pressure qc_end physical_unit 9 9 11 12 temperature qc_end physical_unit 9 9 18 19 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"23.82 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{50 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{300 ℃}"},{"type":"other","value":"Constant volume."}] | <h1 class="questionTitle" itemprop="name">Given constant volume, what pressure would #50*atm# of gas at #0# #""^@C#, develop at a temperature of #300# #""^@C#?</h1> | null | 23.82 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, <mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50.0*atmxx273K)/(573K)#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#?? atmospheres#</mathjax></p>
<p>This is reasonable on the basis that we would expect a given quantity of gas to exert less pressure at lower temperature. Note that I converted the temperature to the Kelvin scale in order to avoid multiplying by <mathjax>#0#</mathjax>.</p></div>
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<div class="markdown"><p>Gay Lusaac's law holds that at constant volume, <mathjax>#((P_1)/T_1 = (P_2)/T_2)#</mathjax> (i.e. pressure is proportional to temperature at constant volume).</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, <mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50.0*atmxx273K)/(573K)#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#?? atmospheres#</mathjax></p>
<p>This is reasonable on the basis that we would expect a given quantity of gas to exert less pressure at lower temperature. Note that I converted the temperature to the Kelvin scale in order to avoid multiplying by <mathjax>#0#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Given constant volume, what pressure would #50*atm# of gas at #0# #""^@C#, develop at a temperature of #300# #""^@C#?</h1>
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<div class="markdown"><p>Gay Lusaac's law holds that at constant volume, <mathjax>#((P_1)/T_1 = (P_2)/T_2)#</mathjax> (i.e. pressure is proportional to temperature at constant volume).</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, <mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(50.0*atmxx273K)/(573K)#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#?? atmospheres#</mathjax></p>
<p>This is reasonable on the basis that we would expect a given quantity of gas to exert less pressure at lower temperature. Note that I converted the temperature to the Kelvin scale in order to avoid multiplying by <mathjax>#0#</mathjax>.</p></div>
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</article> | Given constant volume, what pressure would #50*atm# of gas at #0# #""^@C#, develop at a temperature of #300# #""^@C#? | null |
2,322 | ab263393-6ddd-11ea-b3f3-ccda262736ce | https://socratic.org/questions/what-is-the-final-volume-when-2-50-ml-of-a-11-0-m-hcl-solution-is-diluted-to-0-1 | 275.00 mL | start physical_unit 12 13 volume ml qc_end physical_unit 12 13 6 7 volume qc_end physical_unit 12 13 10 11 molarity qc_end physical_unit 12 13 17 18 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] HCl solution [IN] mL"}] | [{"type":"physical unit","value":"275.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] HCl solution [=] \\pu{2.50 mL}"},{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{11.0 M}"},{"type":"physical unit","value":"Molarity2 [OF] HCl solution [=] \\pu{0.100 M}"}] | <h1 class="questionTitle" itemprop="name">What is the final volume when 2.50 mL of a 11.0 M #HCl# solution is diluted to 0.100 M #HCl# solution? </h1> | null | 275.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember when diluting a solution is that you can determine the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> by using the <em>concentrations <strong>or</strong> the volumes</em> of the stock and target <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p>
<p>The underlying principle of a <strong>dilution</strong> is that the concentration of the solution is <strong>decreased</strong> by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/PhMo7ioS8RX3MJGvksJw_6243113_orig.jpg"/> </p>
<p><mathjax>#color(white)(a)#</mathjax><br/>
You can express this by using molarities and volumes </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the diluted solution </p>
<p>Now, you can find a solution's <em>dilution factor</em> be rearranging the above equation to get </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This will be the solution's dilution factor </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In essence, the dilution factor tells you how much higher the concentration of the stock solution was compared with that of the <em>diluted solution</em>. </p>
<p>In your case, you know that the stock solution had a concentration of <mathjax>#"11.0 M"#</mathjax>, and that the diluted solution has a concentration of <mathjax>#"0.100 M"#</mathjax>. </p>
<p>This means that your solution was diluted by a factor of </p>
<blockquote>
<p><mathjax>#"D.F." = (11.0 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = 110#</mathjax></p>
</blockquote>
<p>If the concentration of the stock solution was <mathjax>#110#</mathjax> <strong>times higher</strong> than that of the diluted solution, it follows that the <em>volume</em> of the diluted solution must be <mathjax>#110#</mathjax> <strong>times bigger</strong> than that of the stock solution</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("D.F." = V_2/V_1 implies V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#V_2 = 110 * "2.50 mL" = color(green)(|bar(ul(color(white)(a/a)"275 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you'd start with <mathjax>#"2.50 mL"#</mathjax> of <mathjax>#"11.0 M"#</mathjax> hydrochloric acid solution and add enough water to get the total volume to <mathjax>#"275 mL"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"275 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember when diluting a solution is that you can determine the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> by using the <em>concentrations <strong>or</strong> the volumes</em> of the stock and target <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p>
<p>The underlying principle of a <strong>dilution</strong> is that the concentration of the solution is <strong>decreased</strong> by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/PhMo7ioS8RX3MJGvksJw_6243113_orig.jpg"/> </p>
<p><mathjax>#color(white)(a)#</mathjax><br/>
You can express this by using molarities and volumes </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the diluted solution </p>
<p>Now, you can find a solution's <em>dilution factor</em> be rearranging the above equation to get </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This will be the solution's dilution factor </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In essence, the dilution factor tells you how much higher the concentration of the stock solution was compared with that of the <em>diluted solution</em>. </p>
<p>In your case, you know that the stock solution had a concentration of <mathjax>#"11.0 M"#</mathjax>, and that the diluted solution has a concentration of <mathjax>#"0.100 M"#</mathjax>. </p>
<p>This means that your solution was diluted by a factor of </p>
<blockquote>
<p><mathjax>#"D.F." = (11.0 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = 110#</mathjax></p>
</blockquote>
<p>If the concentration of the stock solution was <mathjax>#110#</mathjax> <strong>times higher</strong> than that of the diluted solution, it follows that the <em>volume</em> of the diluted solution must be <mathjax>#110#</mathjax> <strong>times bigger</strong> than that of the stock solution</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("D.F." = V_2/V_1 implies V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#V_2 = 110 * "2.50 mL" = color(green)(|bar(ul(color(white)(a/a)"275 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you'd start with <mathjax>#"2.50 mL"#</mathjax> of <mathjax>#"11.0 M"#</mathjax> hydrochloric acid solution and add enough water to get the total volume to <mathjax>#"275 mL"#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the final volume when 2.50 mL of a 11.0 M #HCl# solution is diluted to 0.100 M #HCl# solution? </h1>
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<div class="markdown"><p><mathjax>#"275 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember when diluting a solution is that you can determine the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> by using the <em>concentrations <strong>or</strong> the volumes</em> of the stock and target <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p>
<p>The underlying principle of a <strong>dilution</strong> is that the concentration of the solution is <strong>decreased</strong> by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/PhMo7ioS8RX3MJGvksJw_6243113_orig.jpg"/> </p>
<p><mathjax>#color(white)(a)#</mathjax><br/>
You can express this by using molarities and volumes </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the diluted solution </p>
<p>Now, you can find a solution's <em>dilution factor</em> be rearranging the above equation to get </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This will be the solution's dilution factor </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In essence, the dilution factor tells you how much higher the concentration of the stock solution was compared with that of the <em>diluted solution</em>. </p>
<p>In your case, you know that the stock solution had a concentration of <mathjax>#"11.0 M"#</mathjax>, and that the diluted solution has a concentration of <mathjax>#"0.100 M"#</mathjax>. </p>
<p>This means that your solution was diluted by a factor of </p>
<blockquote>
<p><mathjax>#"D.F." = (11.0 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = 110#</mathjax></p>
</blockquote>
<p>If the concentration of the stock solution was <mathjax>#110#</mathjax> <strong>times higher</strong> than that of the diluted solution, it follows that the <em>volume</em> of the diluted solution must be <mathjax>#110#</mathjax> <strong>times bigger</strong> than that of the stock solution</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("D.F." = V_2/V_1 implies V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you'll have </p>
<blockquote>
<p><mathjax>#V_2 = 110 * "2.50 mL" = color(green)(|bar(ul(color(white)(a/a)"275 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you'd start with <mathjax>#"2.50 mL"#</mathjax> of <mathjax>#"11.0 M"#</mathjax> hydrochloric acid solution and add enough water to get the total volume to <mathjax>#"275 mL"#</mathjax>. </p></div>
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</article> | What is the final volume when 2.50 mL of a 11.0 M #HCl# solution is diluted to 0.100 M #HCl# solution? | null |
2,323 | aa94879c-6ddd-11ea-a56c-ccda262736ce | https://socratic.org/questions/a-sample-of-chlorine-gas-is-confined-in-a-3-93-l-container-at-407-torr-and-21-0- | 0.09 moles | start physical_unit 1 4 mole mol qc_end physical_unit 1 4 9 10 volume qc_end physical_unit 1 4 13 14 pressure qc_end physical_unit 1 4 16 18 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] chlorine gas sample [IN] moles"}] | [{"type":"physical unit","value":"0.09 moles"}] | [{"type":"physical unit","value":"Volume [OF] chlorine gas sample [=] \\pu{3.93 L}"},{"type":"physical unit","value":"Pressure [OF] chlorine gas sample [=] \\pu{407 torr }"},{"type":"physical unit","value":"Temperature [OF] chlorine gas sample [=] \\pu{21.0 degrees Celsius}"}] | <h1 class="questionTitle" itemprop="name">A sample of chlorine gas is confined in a 3.93-L container at 407 torr and 21.0 degrees Celsius. How many moles of gas are in the sample?</h1> | null | 0.09 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old Ideal Gas equation.....</p>
<p><mathjax>#PV=nRT#</mathjax>, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, <mathjax>#"force per unit area"#</mathjax> is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that <mathjax>#"1 atmosphere"#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high (and <mathjax>#"1 Torr"-="1 mm Hg"#</mathjax>). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?</p>
<p>So here <mathjax>#P=(407*mm*Hg)/(760*mm*Hg*atm^-1)=0.536*atm#</mathjax>.</p>
<p>The problem is almost done now.........</p>
<p><mathjax>#PV=nRT#</mathjax>; and so <mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(0.536*atmxx3.93*L)/(294.15*Kxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p><mathjax>#=??*atm#</mathjax></p>
<p>Note the dimensional consistency of the answer.....</p>
<p><mathjax>#n=(0.536*cancel(atm)xx3.93*cancelL)/(294.15*cancelKxx0.0821*cancelL*cancel(atm*K^-1)*mol^-1)#</mathjax></p>
<p><mathjax>#=1/(mol^-1)=1/(1/(mol))=mol#</mathjax> as required...............this is what we want, and the consistent unit is an internal check on our calculation......</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, here we measure pressure in <mathjax>#"Torr"#</mathjax>, and of course there is assumed knowledge........I make <mathjax>#P=0.09*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old Ideal Gas equation.....</p>
<p><mathjax>#PV=nRT#</mathjax>, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, <mathjax>#"force per unit area"#</mathjax> is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that <mathjax>#"1 atmosphere"#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high (and <mathjax>#"1 Torr"-="1 mm Hg"#</mathjax>). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?</p>
<p>So here <mathjax>#P=(407*mm*Hg)/(760*mm*Hg*atm^-1)=0.536*atm#</mathjax>.</p>
<p>The problem is almost done now.........</p>
<p><mathjax>#PV=nRT#</mathjax>; and so <mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(0.536*atmxx3.93*L)/(294.15*Kxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p><mathjax>#=??*atm#</mathjax></p>
<p>Note the dimensional consistency of the answer.....</p>
<p><mathjax>#n=(0.536*cancel(atm)xx3.93*cancelL)/(294.15*cancelKxx0.0821*cancelL*cancel(atm*K^-1)*mol^-1)#</mathjax></p>
<p><mathjax>#=1/(mol^-1)=1/(1/(mol))=mol#</mathjax> as required...............this is what we want, and the consistent unit is an internal check on our calculation......</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of chlorine gas is confined in a 3.93-L container at 407 torr and 21.0 degrees Celsius. How many moles of gas are in the sample?</h1>
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<div class="markdown"><p>Well, here we measure pressure in <mathjax>#"Torr"#</mathjax>, and of course there is assumed knowledge........I make <mathjax>#P=0.09*atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old Ideal Gas equation.....</p>
<p><mathjax>#PV=nRT#</mathjax>, and now all the given measurements are kosher except for the PRESSURE measurement. Pressure, <mathjax>#"force per unit area"#</mathjax> is something that is hard and non-intuitive to measure. Physical scientists, however, have long known that <mathjax>#"1 atmosphere"#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high (and <mathjax>#"1 Torr"-="1 mm Hg"#</mathjax>). And thus we can use a measurement of length to give the pressure in atmospheres. Are you with me.....?</p>
<p>So here <mathjax>#P=(407*mm*Hg)/(760*mm*Hg*atm^-1)=0.536*atm#</mathjax>.</p>
<p>The problem is almost done now.........</p>
<p><mathjax>#PV=nRT#</mathjax>; and so <mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(0.536*atmxx3.93*L)/(294.15*Kxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p><mathjax>#=??*atm#</mathjax></p>
<p>Note the dimensional consistency of the answer.....</p>
<p><mathjax>#n=(0.536*cancel(atm)xx3.93*cancelL)/(294.15*cancelKxx0.0821*cancelL*cancel(atm*K^-1)*mol^-1)#</mathjax></p>
<p><mathjax>#=1/(mol^-1)=1/(1/(mol))=mol#</mathjax> as required...............this is what we want, and the consistent unit is an internal check on our calculation......</p></div>
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</article> | A sample of chlorine gas is confined in a 3.93-L container at 407 torr and 21.0 degrees Celsius. How many moles of gas are in the sample? | null |
2,324 | a9245e39-6ddd-11ea-bb36-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-4-10-2-m-hcl-solution | 1.40 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] HCl solution"}] | [{"type":"physical unit","value":"1.40"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{4 × 10^(-2) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #4*10^-2# #M# #HCl# solution? </h1> | null | 1.40 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)(4.0xx10^-2)#</mathjax></p>
<p>We assume (reasonably) that hydrochloric acid will undergo complete dissociation in water to give stoichiometric <mathjax>#H_3O^+#</mathjax> ion, according to the given reaction:</p>
<p><mathjax>#HCl(aq) + H_2O(l) rarr H_3O^(+) + Cl^-#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)(4.0xx10^-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.40#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)(4.0xx10^-2)#</mathjax></p>
<p>We assume (reasonably) that hydrochloric acid will undergo complete dissociation in water to give stoichiometric <mathjax>#H_3O^+#</mathjax> ion, according to the given reaction:</p>
<p><mathjax>#HCl(aq) + H_2O(l) rarr H_3O^(+) + Cl^-#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #4*10^-2# #M# #HCl# solution? </h1>
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anor277
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<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)(4.0xx10^-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.40#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)(4.0xx10^-2)#</mathjax></p>
<p>We assume (reasonably) that hydrochloric acid will undergo complete dissociation in water to give stoichiometric <mathjax>#H_3O^+#</mathjax> ion, according to the given reaction:</p>
<p><mathjax>#HCl(aq) + H_2O(l) rarr H_3O^(+) + Cl^-#</mathjax></p></div>
</div>
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</article> | What is the pH of a #4*10^-2# #M# #HCl# solution? | null |
2,325 | a9b44076-6ddd-11ea-91c4-ccda262736ce | https://socratic.org/questions/56070a93581e2a2b0ec57420 | 2 | start physical_unit 2 4 number none qc_end end | [{"type":"physical unit","value":"Number [OF] double bond equivalents"}] | [{"type":"physical unit","value":"2"}] | [{"type":"substance name","value":" 2,2,3,3-tetramethyl-4-octyne"}] | <h1 class="questionTitle" itemprop="name">How many double bond equivalents (degrees of unsaturation) are in 2,2,3,3-tetramethyl-4-octyne?</h1> | null | 2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of 2,2,3,3-tetramethyl-4-octyne is <mathjax>#C_12H_22#</mathjax></p>
<p><img alt="" src="https://useruploads.socratic.org/iLpclAY3RZ6p34EYNKLu_2%2C2%2C3%2C3-tetramethyl-4-octyne.PNG"/> </p>
<p>The general formula for calculating the double bond equivalent (DBE) is:</p>
<p><mathjax>#DBE=C-H/2+N/2+1#</mathjax></p>
<p>where, <mathjax>#C#</mathjax> is the number of carbon atoms, <mathjax>#H#</mathjax> is the number of hydrogen and halogen atoms, and <mathjax>#N#</mathjax> is the number of nitrogen atoms if there is any in the molecule. </p>
<p>Therefore, in this case:</p>
<p><mathjax>#DBE=12-(22)/2+(0)/2+1=2#</mathjax></p>
<p>It is logic to find <mathjax>#DBE=2#</mathjax> since we have two <mathjax>#pi#</mathjax> bonds in a triple bond.</p></div>
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<div class="markdown"><p><mathjax>#DBE=2#</mathjax></p>
<p>In other words, the molecule has 2 degrees of unsaturation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of 2,2,3,3-tetramethyl-4-octyne is <mathjax>#C_12H_22#</mathjax></p>
<p><img alt="" src="https://useruploads.socratic.org/iLpclAY3RZ6p34EYNKLu_2%2C2%2C3%2C3-tetramethyl-4-octyne.PNG"/> </p>
<p>The general formula for calculating the double bond equivalent (DBE) is:</p>
<p><mathjax>#DBE=C-H/2+N/2+1#</mathjax></p>
<p>where, <mathjax>#C#</mathjax> is the number of carbon atoms, <mathjax>#H#</mathjax> is the number of hydrogen and halogen atoms, and <mathjax>#N#</mathjax> is the number of nitrogen atoms if there is any in the molecule. </p>
<p>Therefore, in this case:</p>
<p><mathjax>#DBE=12-(22)/2+(0)/2+1=2#</mathjax></p>
<p>It is logic to find <mathjax>#DBE=2#</mathjax> since we have two <mathjax>#pi#</mathjax> bonds in a triple bond.</p></div>
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<h1 class="questionTitle" itemprop="name">How many double bond equivalents (degrees of unsaturation) are in 2,2,3,3-tetramethyl-4-octyne?</h1>
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<div class="markdown"><p><mathjax>#DBE=2#</mathjax></p>
<p>In other words, the molecule has 2 degrees of unsaturation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of 2,2,3,3-tetramethyl-4-octyne is <mathjax>#C_12H_22#</mathjax></p>
<p><img alt="" src="https://useruploads.socratic.org/iLpclAY3RZ6p34EYNKLu_2%2C2%2C3%2C3-tetramethyl-4-octyne.PNG"/> </p>
<p>The general formula for calculating the double bond equivalent (DBE) is:</p>
<p><mathjax>#DBE=C-H/2+N/2+1#</mathjax></p>
<p>where, <mathjax>#C#</mathjax> is the number of carbon atoms, <mathjax>#H#</mathjax> is the number of hydrogen and halogen atoms, and <mathjax>#N#</mathjax> is the number of nitrogen atoms if there is any in the molecule. </p>
<p>Therefore, in this case:</p>
<p><mathjax>#DBE=12-(22)/2+(0)/2+1=2#</mathjax></p>
<p>It is logic to find <mathjax>#DBE=2#</mathjax> since we have two <mathjax>#pi#</mathjax> bonds in a triple bond.</p></div>
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</article> | How many double bond equivalents (degrees of unsaturation) are in 2,2,3,3-tetramethyl-4-octyne? | null |
2,326 | aa7e1292-6ddd-11ea-ba02-ccda262736ce | https://socratic.org/questions/800-ml-of-solution-has-a-concentration-of-3-m-how-much-water-needs-to-be-added-t | 464 mL | start physical_unit 12 12 volume ml qc_end physical_unit 19 20 8 9 concentration qc_end physical_unit 19 20 22 23 concentration qc_end physical_unit 19 20 0 1 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] mL"}] | [{"type":"physical unit","value":"464 mL"}] | [{"type":"physical unit","value":"Concentration1 [OF] the solution [=] \\pu{3.0 M}"},{"type":"physical unit","value":"Concentration2 [OF] the solution [=] \\pu{1.9 M}"},{"type":"physical unit","value":"Volume1 [OF] the solution [=] \\pu{800 mL}"}] | <h1 class="questionTitle" itemprop="name">800. mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the solution to 1.9 M?</h1> | null | 464 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <strong>dilution</strong> is used to <strong>decrease</strong> the concentration of a given solution. </p>
<p>In order to <em>dilute</em> a given solution, you must make sure that you keep the <em>number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> present <strong>constant</strong> and that you <strong>increase</strong> the total volume of the solution, usually by adding more <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/4PxpNIFcQNSU3n0ocsAo_6243113_orig.jpg"/></p>
<p>Since <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is calculated by taking the <em>number of moles of solute</em> present <strong>per liter</strong> of solution, increasing the volume while keeping the number of moles of solute constant will result in a <strong>decrease</strong> in concentration. </p>
<p>Mathematically, <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> can be performed using the following equation</p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Now, you can rearrange this equation to find </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This gives you the solution's <strong>dilution factor</strong>, <mathjax>#"D.F."#</mathjax>, which tells you how concentrated the stock solution was compared with the <em>diluted solution</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution decreases from <mathjax>#"3.0 M"#</mathjax> to <mathjax>#'1.9 M"#</mathjax>, which means that the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.0 color(red)(cancel(color(black)("M"))))/(1.9color(red)(cancel(color(black)("M")))) = 1.58#</mathjax></p>
</blockquote>
<p>So, if the stock solution was <mathjax>#1.58#</mathjax> <strong>times more concentrated</strong> than the diluted solution, it follows that the <em>volume</em> of the diluted solution <strong>must have increased</strong> by a factor of <mathjax>#1.58#</mathjax></p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Therefore, the total volume of the diluted solution must be equal to </p>
<blockquote>
<p><mathjax>#V_2 = 1.58 * "800. mL" = "1264 mL"#</mathjax></p>
</blockquote>
<p>The volume of water needed will thus be </p>
<blockquote>
<p><mathjax>#V_2 = V_1 + V_"water"#</mathjax></p>
<p><mathjax>#V_"water" = "1264 mL" - "800. mL" = "464 mL"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, despite the fact that you only have two sig figs for the concentration of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a></p>
<blockquote>
<p><mathjax>#V_"water" = color(green)(|bar(ul(color(white)(a/a)"464 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"464 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <strong>dilution</strong> is used to <strong>decrease</strong> the concentration of a given solution. </p>
<p>In order to <em>dilute</em> a given solution, you must make sure that you keep the <em>number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> present <strong>constant</strong> and that you <strong>increase</strong> the total volume of the solution, usually by adding more <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/4PxpNIFcQNSU3n0ocsAo_6243113_orig.jpg"/></p>
<p>Since <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is calculated by taking the <em>number of moles of solute</em> present <strong>per liter</strong> of solution, increasing the volume while keeping the number of moles of solute constant will result in a <strong>decrease</strong> in concentration. </p>
<p>Mathematically, <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> can be performed using the following equation</p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Now, you can rearrange this equation to find </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This gives you the solution's <strong>dilution factor</strong>, <mathjax>#"D.F."#</mathjax>, which tells you how concentrated the stock solution was compared with the <em>diluted solution</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution decreases from <mathjax>#"3.0 M"#</mathjax> to <mathjax>#'1.9 M"#</mathjax>, which means that the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.0 color(red)(cancel(color(black)("M"))))/(1.9color(red)(cancel(color(black)("M")))) = 1.58#</mathjax></p>
</blockquote>
<p>So, if the stock solution was <mathjax>#1.58#</mathjax> <strong>times more concentrated</strong> than the diluted solution, it follows that the <em>volume</em> of the diluted solution <strong>must have increased</strong> by a factor of <mathjax>#1.58#</mathjax></p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Therefore, the total volume of the diluted solution must be equal to </p>
<blockquote>
<p><mathjax>#V_2 = 1.58 * "800. mL" = "1264 mL"#</mathjax></p>
</blockquote>
<p>The volume of water needed will thus be </p>
<blockquote>
<p><mathjax>#V_2 = V_1 + V_"water"#</mathjax></p>
<p><mathjax>#V_"water" = "1264 mL" - "800. mL" = "464 mL"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, despite the fact that you only have two sig figs for the concentration of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a></p>
<blockquote>
<p><mathjax>#V_"water" = color(green)(|bar(ul(color(white)(a/a)"464 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">800. mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the solution to 1.9 M?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-04-17T12:23:01" itemprop="dateCreated">
Apr 17, 2016
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<div class="markdown"><p><mathjax>#"464 mL"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <strong>dilution</strong> is used to <strong>decrease</strong> the concentration of a given solution. </p>
<p>In order to <em>dilute</em> a given solution, you must make sure that you keep the <em>number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> present <strong>constant</strong> and that you <strong>increase</strong> the total volume of the solution, usually by adding more <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://acidsandbasesfordummieschem.weebly.com/molarity.html" src="https://useruploads.socratic.org/4PxpNIFcQNSU3n0ocsAo_6243113_orig.jpg"/></p>
<p>Since <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is calculated by taking the <em>number of moles of solute</em> present <strong>per liter</strong> of solution, increasing the volume while keeping the number of moles of solute constant will result in a <strong>decrease</strong> in concentration. </p>
<p>Mathematically, <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> can be performed using the following equation</p>
<p>The equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))#</mathjax></p>
</blockquote>
<p>Here </p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Now, you can rearrange this equation to find </p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#</mathjax></p>
</blockquote>
<p>This gives you the solution's <strong>dilution factor</strong>, <mathjax>#"D.F."#</mathjax>, which tells you how concentrated the stock solution was compared with the <em>diluted solution</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution decreases from <mathjax>#"3.0 M"#</mathjax> to <mathjax>#'1.9 M"#</mathjax>, which means that the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.0 color(red)(cancel(color(black)("M"))))/(1.9color(red)(cancel(color(black)("M")))) = 1.58#</mathjax></p>
</blockquote>
<p>So, if the stock solution was <mathjax>#1.58#</mathjax> <strong>times more concentrated</strong> than the diluted solution, it follows that the <em>volume</em> of the diluted solution <strong>must have increased</strong> by a factor of <mathjax>#1.58#</mathjax></p>
<blockquote>
<p><mathjax>#"D.F." = V_2/V_1 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(V_2 = "D.F." xx V_1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Therefore, the total volume of the diluted solution must be equal to </p>
<blockquote>
<p><mathjax>#V_2 = 1.58 * "800. mL" = "1264 mL"#</mathjax></p>
</blockquote>
<p>The volume of water needed will thus be </p>
<blockquote>
<p><mathjax>#V_2 = V_1 + V_"water"#</mathjax></p>
<p><mathjax>#V_"water" = "1264 mL" - "800. mL" = "464 mL"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, despite the fact that you only have two sig figs for the concentration of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a></p>
<blockquote>
<p><mathjax>#V_"water" = color(green)(|bar(ul(color(white)(a/a)"464 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | 800. mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the solution to 1.9 M? | null |
2,327 | a90db558-6ddd-11ea-89ed-ccda262736ce | https://socratic.org/questions/a-breathing-mixture-used-by-deep-sea-divers-contains-helium-oxygen-and-carbon-di | 18.5 kPa | start physical_unit 9 9 partial_pressure kpa qc_end physical_unit 1 2 21 22 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] oxygen [IN] kPa"}] | [{"type":"physical unit","value":"18.5 kPa"}] | [{"type":"physical unit","value":"Total pressure [OF] breathing mixture [=] \\pu{101.4 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] He [=] \\pu{82.5 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] CO2 [=] \\pu{0.4 kPa}"}] | <h1 class="questionTitle" itemprop="name">A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide What is the partial pressure of oxygen at #101.4# #kPa# if #P_(He)# = #82.5# #kPa# and #P_(CO2)# = #.4# #kPa#?</h1> | null | 18.5 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure is the sum of the partial pressures of the given gases.</p>
<p><mathjax>#P_"Total"=P_"He"+P_"CO2"+P_"O2"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#P_"O2"#</mathjax>, substitute the given values into the equation and solve.</p>
<p><mathjax>#P_"O2"=P_"Total"-(P_"He"+P_"CO2")#</mathjax></p>
<p><mathjax>#P_"O2"=101.4"kPa"+(82.5"kPa"+0.4"kPa")#</mathjax></p>
<p><mathjax>#P_"O2"=18.5"kPa"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen in the mixture is 18.5 kPa.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure is the sum of the partial pressures of the given gases.</p>
<p><mathjax>#P_"Total"=P_"He"+P_"CO2"+P_"O2"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#P_"O2"#</mathjax>, substitute the given values into the equation and solve.</p>
<p><mathjax>#P_"O2"=P_"Total"-(P_"He"+P_"CO2")#</mathjax></p>
<p><mathjax>#P_"O2"=101.4"kPa"+(82.5"kPa"+0.4"kPa")#</mathjax></p>
<p><mathjax>#P_"O2"=18.5"kPa"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide What is the partial pressure of oxygen at #101.4# #kPa# if #P_(He)# = #82.5# #kPa# and #P_(CO2)# = #.4# #kPa#?</h1>
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Meave60
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen in the mixture is 18.5 kPa.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure is the sum of the partial pressures of the given gases.</p>
<p><mathjax>#P_"Total"=P_"He"+P_"CO2"+P_"O2"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#P_"O2"#</mathjax>, substitute the given values into the equation and solve.</p>
<p><mathjax>#P_"O2"=P_"Total"-(P_"He"+P_"CO2")#</mathjax></p>
<p><mathjax>#P_"O2"=101.4"kPa"+(82.5"kPa"+0.4"kPa")#</mathjax></p>
<p><mathjax>#P_"O2"=18.5"kPa"#</mathjax></p></div>
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David G.
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<span class="dateCreated" datetime="2016-02-22T01:45:29" itemprop="dateCreated">
Feb 22, 2016
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<div class="markdown"><p>The total pressure of <mathjax>#101.4#</mathjax> <mathjax>#kPa#</mathjax> is made up of the sum of the partial pressures of the three gases, so <mathjax>#P_(O2)=101.4-82.5-0.4=18.5#</mathjax> <mathjax>#kPa#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a little more detail, the total pressure is the sum of the partial pressures of the gases present:</p>
<p><mathjax>#P_("tot")=P_(O2)+P_(He)+P_(CO2)#</mathjax></p>
<p>We want to know the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen, <mathjax>#P_(O2)#</mathjax>, so we rearrange:</p>
<p><mathjax>#P_(O2)=P_("tot")-P_(He)-P_(CO2)#</mathjax></p>
<p>Substituting in the known values gives the answer shown above.</p></div>
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</article> | A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide What is the partial pressure of oxygen at #101.4# #kPa# if #P_(He)# = #82.5# #kPa# and #P_(CO2)# = #.4# #kPa#? | null |
2,328 | a893247a-6ddd-11ea-9cbe-ccda262736ce | https://socratic.org/questions/using-the-equation-n-2-3h-2-2nh-3-how-many-grams-of-hydrogen-must-react-if-the-r | 22.58 grams | start physical_unit 14 14 mass g qc_end chemical_equation 3 9 qc_end physical_unit 9 9 21 22 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen [IN] grams"}] | [{"type":"physical unit","value":"22.58 grams"}] | [{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"},{"type":"physical unit","value":"Mass [OF] NH3 [=] \\pu{127 grams}"}] | <h1 class="questionTitle" itemprop="name">Using the equation #N_2 + 3H_2 -> 2NH_3#, how many grams of hydrogen must react if the reaction needs 127 grams of #NH_3#?</h1> | null | 22.58 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"3H"_2::"2NH"_3#</mathjax></p>
<p><mathjax>#3/2"H"_2::"1NH"_3#</mathjax></p>
<p><mathjax>#"Moles of NH"_3=127/17.0#</mathjax></p>
<p>So <mathjax>#"moles of H"_2=3/2*127/17.0 = 11.2#</mathjax></p>
<p>Mass of <mathjax>#"H"_2 = "11.2 mol" × "2.016 g/mol" = "22.6 g"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>22.6 g <mathjax>#"H"_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"3H"_2::"2NH"_3#</mathjax></p>
<p><mathjax>#3/2"H"_2::"1NH"_3#</mathjax></p>
<p><mathjax>#"Moles of NH"_3=127/17.0#</mathjax></p>
<p>So <mathjax>#"moles of H"_2=3/2*127/17.0 = 11.2#</mathjax></p>
<p>Mass of <mathjax>#"H"_2 = "11.2 mol" × "2.016 g/mol" = "22.6 g"#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Using the equation #N_2 + 3H_2 -> 2NH_3#, how many grams of hydrogen must react if the reaction needs 127 grams of #NH_3#?</h1>
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Ernest Z.
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<div class="markdown"><p>22.6 g <mathjax>#"H"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"3H"_2::"2NH"_3#</mathjax></p>
<p><mathjax>#3/2"H"_2::"1NH"_3#</mathjax></p>
<p><mathjax>#"Moles of NH"_3=127/17.0#</mathjax></p>
<p>So <mathjax>#"moles of H"_2=3/2*127/17.0 = 11.2#</mathjax></p>
<p>Mass of <mathjax>#"H"_2 = "11.2 mol" × "2.016 g/mol" = "22.6 g"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"22.6 g of H"_2#</mathjax> must react.</p>
<blockquote></blockquote>
<p><strong>Given:</strong> Mass of <mathjax>#"NH"_3#</mathjax> and the chemical equation.</p>
<blockquote></blockquote>
<p><strong>Find:</strong> Mass of <mathjax>#"H"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Strategy:</strong></p>
<p>The central part of any <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem is to <strong>convert moles of something to moles of something else</strong>.</p>
<p><strong>(a)</strong> We start with the balanced chemical equation for the reaction.</p>
<p><strong>(b)</strong> We can use the molar mass of <mathjax>#"NH"_3#</mathjax> to find the moles of <mathjax>#"NH"_3#</mathjax>.</p>
<p><strong>(c)</strong> We can use the <strong>molar ratio</strong> from the equation to convert moles of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#"moles of NH"_3 stackrelcolor(blue)("molar ratio"color(white)(Xl))( →) "moles of H"_2#</mathjax></p>
<p><strong>(d)</strong> Then we can use the molar mass to convert the moles of <mathjax>#"H"_2#</mathjax> to mass of <mathjax>#"H"_2#</mathjax>.</p>
<blockquote></blockquote>
<p>Our complete strategy is:</p>
<p><mathjax>#"Mass of NH"_3stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of NH"_3stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of H"_2 stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "mass of H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Solution</strong></p>
<p><strong>(a)</strong> The balanced equation is</p>
<p><mathjax>#"N"_2 + "3H"_2 → "2NH"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> Calculate moles of <mathjax>#"NH"_3#</mathjax></p>
<p><mathjax>#127 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "7.457 mol NH"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c)</strong> Calculate moles of <mathjax>#"H"_2#</mathjax></p>
<p>The molar ratio of <mathjax>#"H"_2#</mathjax> to <mathjax>#"NH"_3#</mathjax> is <mathjax>#("3 mol H"_2)/("2 mol NH"_3)"#</mathjax>.</p>
<p><mathjax>#"Moles of H"_2 = 7.457 color(red)(cancel(color(black)("mol NH"_3))) × ("3 mol H"_2)/(2 color(red)(cancel(color(black)("mol NH"_3)))) = "11.19 mol H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(d)</strong> Calculate the mass of <mathjax>#"H"_2#</mathjax></p>
<p><mathjax>#11.19 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "22.6 g H"_2 #</mathjax></p>
<p><strong>Answer: </strong> <mathjax>#"22.6 g of H"_2 #</mathjax> must react.</p></div>
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<div class="markdown"><p><mathjax>#22.58 g H_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#N_2 + 3H_2 -> 2NH_3#</mathjax></p>
<p>If <mathjax>#127#</mathjax> grams of necessary we can use <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> conversions to find the mass of Hydrogen required.</p>
<p><mathjax>#127g NH_3 * (1 mol NH_3)/(17.04 g NH_3)*(3 mol H_2)/(2 mol NH_3)*(2.02 g H_2)/(1 mol H_2) = #</mathjax> </p>
<p><mathjax>#N = 1 * 14.01 g = 14.01#</mathjax><br/>
<mathjax>#H= 3 * 1.01 g = 3.03#</mathjax></p>
<p><mathjax>#14.01+3.03=17.04g#</mathjax></p>
<p><mathjax>#127cancel(g NH_3) * (1 cancel(mol NH_3))/(17.04 cancel(g NH_3))*(3 cancel(mol H_2))/(2 cancel(mol NH_3))*(2.02 g H_2)/(1 cancel(mol H_2)) = #</mathjax> </p>
<p><mathjax>#=22.58 g H_2#</mathjax></p>
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</article> | Using the equation #N_2 + 3H_2 -> 2NH_3#, how many grams of hydrogen must react if the reaction needs 127 grams of #NH_3#? | null |
2,329 | a9c22066-6ddd-11ea-98d1-ccda262736ce | https://socratic.org/questions/how-many-moles-of-silver-nitrate-do-2-888-x-10-15-formula-units-equal | 4.80 x 10^(-10) moles | start physical_unit 4 5 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] silver nitrate [IN] moles"}] | [{"type":"physical unit","value":"4.80 x 10^(-10) moles"}] | [{"type":"physical unit","value":"Number [OF] silver nitrate formula units [=] \\pu{2.888 x 10^15}"}] | <h1 class="questionTitle" itemprop="name">How many moles of silver nitrate do 2.888 x #10^15# formula units equal? </h1> | null | 4.80 x 10^(-10) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"No. of moles"="Number of molecules"/"Avogadro's Number"#</mathjax></p>
<p><mathjax>#=(2.888xx10^15)/(6.022xx10^23*mol^-1)=??*mol#</mathjax></p></div>
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<div class="markdown"><p>If I have 4 eggs, then how many dozen eggs does I have; what about 3 eggses? Anyway you have a very small molar quantity here. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"No. of moles"="Number of molecules"/"Avogadro's Number"#</mathjax></p>
<p><mathjax>#=(2.888xx10^15)/(6.022xx10^23*mol^-1)=??*mol#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of silver nitrate do 2.888 x #10^15# formula units equal? </h1>
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<div class="markdown"><p>If I have 4 eggs, then how many dozen eggs does I have; what about 3 eggses? Anyway you have a very small molar quantity here. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"No. of moles"="Number of molecules"/"Avogadro's Number"#</mathjax></p>
<p><mathjax>#=(2.888xx10^15)/(6.022xx10^23*mol^-1)=??*mol#</mathjax></p></div>
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</article> | How many moles of silver nitrate do 2.888 x #10^15# formula units equal? | null |
2,330 | a8b4e000-6ddd-11ea-a4dd-ccda262736ce | https://socratic.org/questions/59f0bf3f11ef6b6a90f9e202 | 3.01 × 10^22 | start physical_unit 2 3 number none qc_end physical_unit 2 2 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] water molecules"}] | [{"type":"physical unit","value":"3.01 × 10^22"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{0.90 g}"}] | <h1 class="questionTitle" itemprop="name">How many water molecules in a #0.90*g# mass of water?</h1> | null | 3.01 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The mass of one mole of water is <mathjax>#"18 g"#</mathjax>. Therefore, <mathjax>#"0.9 g"#</mathjax> of water is </p>
<p><mathjax>#"0.9 g"/"18 g/mol" = "0.05 moles"#</mathjax> </p>
<p>of water. One mole of water contains Avagadro's constant, ie, <mathjax>#6.022 xx 10^23#</mathjax> molecules of water, therefore, <mathjax>#0.05#</mathjax> moles of water contain</p>
<p><mathjax>#"0.05 mol" xx (6.022^23"molec.")/("1 mol") = 3.01xx10^22#</mathjax> <mathjax>#"molecules"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3.01xx10^22#</mathjax> molecules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The mass of one mole of water is <mathjax>#"18 g"#</mathjax>. Therefore, <mathjax>#"0.9 g"#</mathjax> of water is </p>
<p><mathjax>#"0.9 g"/"18 g/mol" = "0.05 moles"#</mathjax> </p>
<p>of water. One mole of water contains Avagadro's constant, ie, <mathjax>#6.022 xx 10^23#</mathjax> molecules of water, therefore, <mathjax>#0.05#</mathjax> moles of water contain</p>
<p><mathjax>#"0.05 mol" xx (6.022^23"molec.")/("1 mol") = 3.01xx10^22#</mathjax> <mathjax>#"molecules"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many water molecules in a #0.90*g# mass of water?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#3.01xx10^22#</mathjax> molecules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The mass of one mole of water is <mathjax>#"18 g"#</mathjax>. Therefore, <mathjax>#"0.9 g"#</mathjax> of water is </p>
<p><mathjax>#"0.9 g"/"18 g/mol" = "0.05 moles"#</mathjax> </p>
<p>of water. One mole of water contains Avagadro's constant, ie, <mathjax>#6.022 xx 10^23#</mathjax> molecules of water, therefore, <mathjax>#0.05#</mathjax> moles of water contain</p>
<p><mathjax>#"0.05 mol" xx (6.022^23"molec.")/("1 mol") = 3.01xx10^22#</mathjax> <mathjax>#"molecules"#</mathjax></p></div>
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<div class="markdown"><p>We find the molar quantity.....and get <mathjax>#"number of water molecules"#</mathjax> <mathjax>#=3.01xx10^22#</mathjax>...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of water"=(0.9*g)/(18.01*g*mol^-1)=0.050*mol#</mathjax></p>
<p>And we know that there are <mathjax>#6.022xx10^23#</mathjax> molecules in one mole of any substance.....</p>
<p>And so we take the product......</p>
<p><mathjax>#0.050*molxx6.022xx10^23*mol^-1=3.01xx10^22#</mathjax> <mathjax>#"water molecules"#</mathjax>. (And thus the product gives a dimensionless number as required....)</p>
<p>(i) how many oxygen atoms in this molar quantity; and (ii) how many hydrogen atoms......?</p></div>
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<div class="markdown"><p><mathjax>#30.1#</mathjax> x <mathjax>#10^21 mlcs#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#Mass of water(H_2O)#</mathjax> <mathjax>#= 0.9grams#</mathjax></p>
<p><mathjax>#Molar mass of (H_2O)#</mathjax> <mathjax>#= 1.008#</mathjax> x <mathjax># 2#</mathjax> + 1<mathjax>#6#</mathjax> <br/>
<mathjax># = 18.016gmol^-#</mathjax></p>
<p><strong> </strong>Number of molecules<strong> </strong> = Mass in grams<mathjax>#/#</mathjax>Molar mass x <mathjax>#N_A#</mathjax></p>
<p><mathjax>#= 0.9/18.016#</mathjax> x <mathjax>#6.02#</mathjax> x <mathjax>#10^23#</mathjax></p>
<p><mathjax># = 30.1#</mathjax> x <mathjax>#10^21 #</mathjax> <mathjax>#mlcs #</mathjax></p></div>
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</article> | How many water molecules in a #0.90*g# mass of water? | null |
2,331 | aa5fd5f8-6ddd-11ea-8e3b-ccda262736ce | https://socratic.org/questions/how-much-energy-is-needed-to-raise-the-temperature-of-2-0-g-of-water-5-00-c | 41.84 J | start physical_unit 13 13 energy j qc_end physical_unit 13 13 14 15 temperature qc_end physical_unit 13 13 10 11 mass qc_end end | [{"type":"physical unit","value":"Needed energy [OF] water [IN] J"}] | [{"type":"physical unit","value":"41.84 J"}] | [{"type":"physical unit","value":"Raised temperature [OF] water [=] \\pu{5.00 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{2.0 g}"}] | <h1 class="questionTitle" itemprop="name">How much energy is needed to raise the temperature of 2.0 g of water 5.00°C?</h1> | null | 41.84 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree Celsius (°C).<br/>
<a href="https://water.usgs.gov/edu/heat-capacity.html" rel="nofollow" target="_blank">https://water.usgs.gov/edu/heat-capacity.html</a></p>
<p><mathjax>#"Energy" = 2.0g xx 5.00^oC xx 4.184 J/(g**^oC) = 41.84J#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#41.84J#</mathjax></p></div>
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<div>
<div class="markdown"><p>Water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree Celsius (°C).<br/>
<a href="https://water.usgs.gov/edu/heat-capacity.html" rel="nofollow" target="_blank">https://water.usgs.gov/edu/heat-capacity.html</a></p>
<p><mathjax>#"Energy" = 2.0g xx 5.00^oC xx 4.184 J/(g**^oC) = 41.84J#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#41.84J#</mathjax></p></div>
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<div class="markdown"><p>Water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree Celsius (°C).<br/>
<a href="https://water.usgs.gov/edu/heat-capacity.html" rel="nofollow" target="_blank">https://water.usgs.gov/edu/heat-capacity.html</a></p>
<p><mathjax>#"Energy" = 2.0g xx 5.00^oC xx 4.184 J/(g**^oC) = 41.84J#</mathjax></p></div>
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2,332 | ac7f57a8-6ddd-11ea-a1fd-ccda262736ce | https://socratic.org/questions/what-is-the-solubility-of-mns-in-grams-per-liter-in-a-buffer-solution-that-is-0- | 1.81 grams per liter | start physical_unit 5 5 solubility g/l qc_end physical_unit 18 18 16 17 molarity qc_end physical_unit 22 22 20 21 molarity qc_end end | [{"type":"physical unit","value":"Solubility [OF] MnS [IN] grams per liter"}] | [{"type":"physical unit","value":"1.81 grams per liter"}] | [{"type":"physical unit","value":"Molarity [OF] CH3COOH solution [=] \\pu{0.105 M}"},{"type":"physical unit","value":"Molarity [OF] NaCH3COO solution [=] \\pu{0.490 M}"}] | <h1 class="questionTitle" itemprop="name">What is the solubility of #MnS#, in grams per liter, in a buffer solution that is 0.105 M #CH_3COOH# - 0.490 M #NaCH_3COO#? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What is the solubility of <mathjax>#MnS#</mathjax>, in grams per liter, in a buffer solution that is 0.105 M <mathjax>#CH_3COOH#</mathjax> - 0.490 M <mathjax>#NaCH_3COO#</mathjax>?</p>
<hr/>
<p><mathjax>#MnS#</mathjax> has a <mathjax>#K_(\color(red)(spa))#</mathjax> of <mathjax>#3\cdot10^7#</mathjax></p></div>
</h2>
</div>
</div> | 1.81 grams per liter | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! EXTREMELY LONG ANSWER !!</strong></p>
<p>The idea here is that the solubility of manganese(II) sulfide will <strong>increase</strong> in an <em>acidic solution</em> because the sulfide anions will act <strong>as a base</strong> and react with the hydronium cations to form hydrogen sulfide, a <em>weak acid</em>.</p>
<p>In pure water, manganese(II) sulfide will only partially dissociate to produce manganese(II) cations and sulfide anions</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) rightleftharpoons "Mn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for this equilibrium reaction is</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that the sulfide anion <strong>cannot</strong> exist in water. Once it dissociates from the solid, it will react with water to form <em>bisulfide anions</em>, <mathjax>#"HS"^(-)#</mathjax>. </p>
<p>In other words, the sulfide anions act as a <strong>strong base</strong> in aqueous solution. </p>
<blockquote>
<p><mathjax>#"S"_ ((aq))^(2-) + "H"_ 2"O"_ ((l)) -> "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that a more <strong>accurate depiction</strong> of what happens when manganese(II) sulfide partially dissociates in water would be </p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + "H"_ 2"O"_ ((l)) rightleftharpoons "Mn"_ ((aq))^(2+) + "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The <mathjax>#K_(sp)#</mathjax> would thus take the form</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["HS"^(-)] * ["OH"^(-)]#</mathjax></p>
</blockquote>
<p>Here comes the cool part. If you have an <strong>acidic solution</strong>, the excess hydronium cations will <strong>neutralize</strong> the hydroxide anions produced when the solid partially dissociates. </p>
<p>Moreover, the excess hydronium cations will <strong>protonate</strong> the bisulfide anions to hydrogen sulfide, <mathjax>#"H"_ 2"S"#</mathjax>.</p>
<p>As a result, the equilibrium will <strong>shift to the right</strong> and <strong>more solid will dissolve</strong> <mathjax>#->#</mathjax> as always, <strong><a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a></strong> governs equilibrium reactions. </p>
<p>This means that in <strong>acidic solution</strong>, you will have</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + 2"H"_ 3"O"_ ((aq))^(+) rightleftharpoons "Mn"_ ((aq))^(2+) + "H"_ 2"S"_ ((aq)) + 2"H"_ 2"O"_ ((l))" "color(purple)("(!)")#</mathjax></p>
</blockquote>
<p>This time, the solubility product constant is called the <strong>acid solubility product constant</strong>, <mathjax>#K_(spa)#</mathjax>, and it takes the form -- remember, pure liquids or solids are <strong>not</strong> included in the expression of the equilibrium constant!</p>
<blockquote>
<p><mathjax>#K_(spa) = (["H"_2"S"] * ["Mn"^(2+)])/(["H"_3"O"^(+)]^2)" "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>In your case, you know that the <mathjax>#K_(spa)#</mathjax> for manganese(II) sulfide is </p>
<blockquote>
<p><mathjax>#K_(spa) = 3 * 10^7#</mathjax></p>
</blockquote>
<p>So, know that we've figured out how the balanced chemical equation should look like, focus on determining the concentration of hydronium cations in the buffer.</p>
<p>The <mathjax>#"p"K_a#</mathjax> of acetic acid is </p>
<blockquote>
<p><mathjax>#"p"K_a = 4.75#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>Use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 4.75 + log( (0.490 color(red)(cancel(color(black)("M"))))/(0.105color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 5.42#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"pH" = - log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>which implies</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>The concentration of hydronium cations in this solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-5.42) = 3.80 * 10^(-6)#</mathjax></p>
</blockquote>
<p>Now, take a look at the balanced chemical equation <mathjax>#color(purple)("(!)")#</mathjax>. Notice that <strong>every mole</strong> of manganese(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of manganese(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen sulfide. </p>
<p>This means that the two chemical species will have <strong>equal concentrations</strong> at equilibrium. </p>
<p>The amount of manganese(II) sulfide that dissociates produces manganese(II) cations in a <mathjax>#1:1#</mathjax> mole ratio, which means that if you take <mathjax>#s#</mathjax> to be this equilibrium concentration, you can say that this value represents the <strong>molar solubility</strong> of the salt and that equation <mathjax>#color(darkorange)("(*)")#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#K_(spa) = (s * s)/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#3 * 10^7 = s^2/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#s#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#s^2 = 3 * 10^7 * (3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#s = sqrt(3 * 10^7 * (3.80 * 10^(-6))^2) = 0.0208#</mathjax></p>
</blockquote>
<p>You can thus say that in this buffer, the molar solubility of the salt is equal to </p>
<blockquote>
<p><mathjax>#s = "0.0208 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams per liter</em>, use the <strong>molar mass</strong> of manganese(II) sulfide</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("solubility in g/L"))) = 0.0208 color(white)(.)color(red)(cancel(color(black)("mol")))/"L" * "87.003 g"/(1color(red)(cancel(color(black)("mol MnS")))) = color(darkgreen)(ul(color(black)("1.81 g L"^(-1)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the concentrations of acetic acid and acetate anions. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.81 g L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! EXTREMELY LONG ANSWER !!</strong></p>
<p>The idea here is that the solubility of manganese(II) sulfide will <strong>increase</strong> in an <em>acidic solution</em> because the sulfide anions will act <strong>as a base</strong> and react with the hydronium cations to form hydrogen sulfide, a <em>weak acid</em>.</p>
<p>In pure water, manganese(II) sulfide will only partially dissociate to produce manganese(II) cations and sulfide anions</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) rightleftharpoons "Mn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for this equilibrium reaction is</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that the sulfide anion <strong>cannot</strong> exist in water. Once it dissociates from the solid, it will react with water to form <em>bisulfide anions</em>, <mathjax>#"HS"^(-)#</mathjax>. </p>
<p>In other words, the sulfide anions act as a <strong>strong base</strong> in aqueous solution. </p>
<blockquote>
<p><mathjax>#"S"_ ((aq))^(2-) + "H"_ 2"O"_ ((l)) -> "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that a more <strong>accurate depiction</strong> of what happens when manganese(II) sulfide partially dissociates in water would be </p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + "H"_ 2"O"_ ((l)) rightleftharpoons "Mn"_ ((aq))^(2+) + "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The <mathjax>#K_(sp)#</mathjax> would thus take the form</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["HS"^(-)] * ["OH"^(-)]#</mathjax></p>
</blockquote>
<p>Here comes the cool part. If you have an <strong>acidic solution</strong>, the excess hydronium cations will <strong>neutralize</strong> the hydroxide anions produced when the solid partially dissociates. </p>
<p>Moreover, the excess hydronium cations will <strong>protonate</strong> the bisulfide anions to hydrogen sulfide, <mathjax>#"H"_ 2"S"#</mathjax>.</p>
<p>As a result, the equilibrium will <strong>shift to the right</strong> and <strong>more solid will dissolve</strong> <mathjax>#->#</mathjax> as always, <strong><a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a></strong> governs equilibrium reactions. </p>
<p>This means that in <strong>acidic solution</strong>, you will have</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + 2"H"_ 3"O"_ ((aq))^(+) rightleftharpoons "Mn"_ ((aq))^(2+) + "H"_ 2"S"_ ((aq)) + 2"H"_ 2"O"_ ((l))" "color(purple)("(!)")#</mathjax></p>
</blockquote>
<p>This time, the solubility product constant is called the <strong>acid solubility product constant</strong>, <mathjax>#K_(spa)#</mathjax>, and it takes the form -- remember, pure liquids or solids are <strong>not</strong> included in the expression of the equilibrium constant!</p>
<blockquote>
<p><mathjax>#K_(spa) = (["H"_2"S"] * ["Mn"^(2+)])/(["H"_3"O"^(+)]^2)" "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>In your case, you know that the <mathjax>#K_(spa)#</mathjax> for manganese(II) sulfide is </p>
<blockquote>
<p><mathjax>#K_(spa) = 3 * 10^7#</mathjax></p>
</blockquote>
<p>So, know that we've figured out how the balanced chemical equation should look like, focus on determining the concentration of hydronium cations in the buffer.</p>
<p>The <mathjax>#"p"K_a#</mathjax> of acetic acid is </p>
<blockquote>
<p><mathjax>#"p"K_a = 4.75#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>Use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 4.75 + log( (0.490 color(red)(cancel(color(black)("M"))))/(0.105color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 5.42#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"pH" = - log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>which implies</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>The concentration of hydronium cations in this solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-5.42) = 3.80 * 10^(-6)#</mathjax></p>
</blockquote>
<p>Now, take a look at the balanced chemical equation <mathjax>#color(purple)("(!)")#</mathjax>. Notice that <strong>every mole</strong> of manganese(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of manganese(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen sulfide. </p>
<p>This means that the two chemical species will have <strong>equal concentrations</strong> at equilibrium. </p>
<p>The amount of manganese(II) sulfide that dissociates produces manganese(II) cations in a <mathjax>#1:1#</mathjax> mole ratio, which means that if you take <mathjax>#s#</mathjax> to be this equilibrium concentration, you can say that this value represents the <strong>molar solubility</strong> of the salt and that equation <mathjax>#color(darkorange)("(*)")#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#K_(spa) = (s * s)/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#3 * 10^7 = s^2/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#s#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#s^2 = 3 * 10^7 * (3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#s = sqrt(3 * 10^7 * (3.80 * 10^(-6))^2) = 0.0208#</mathjax></p>
</blockquote>
<p>You can thus say that in this buffer, the molar solubility of the salt is equal to </p>
<blockquote>
<p><mathjax>#s = "0.0208 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams per liter</em>, use the <strong>molar mass</strong> of manganese(II) sulfide</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("solubility in g/L"))) = 0.0208 color(white)(.)color(red)(cancel(color(black)("mol")))/"L" * "87.003 g"/(1color(red)(cancel(color(black)("mol MnS")))) = color(darkgreen)(ul(color(black)("1.81 g L"^(-1)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the concentrations of acetic acid and acetate anions. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the solubility of #MnS#, in grams per liter, in a buffer solution that is 0.105 M #CH_3COOH# - 0.490 M #NaCH_3COO#? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What is the solubility of <mathjax>#MnS#</mathjax>, in grams per liter, in a buffer solution that is 0.105 M <mathjax>#CH_3COOH#</mathjax> - 0.490 M <mathjax>#NaCH_3COO#</mathjax>?</p>
<hr/>
<p><mathjax>#MnS#</mathjax> has a <mathjax>#K_(\color(red)(spa))#</mathjax> of <mathjax>#3\cdot10^7#</mathjax></p></div>
</h2>
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Stefan V.
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<span class="dateCreated" datetime="2017-04-03T20:05:06" itemprop="dateCreated">
Apr 3, 2017
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<div class="markdown"><p><mathjax>#"1.81 g L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! EXTREMELY LONG ANSWER !!</strong></p>
<p>The idea here is that the solubility of manganese(II) sulfide will <strong>increase</strong> in an <em>acidic solution</em> because the sulfide anions will act <strong>as a base</strong> and react with the hydronium cations to form hydrogen sulfide, a <em>weak acid</em>.</p>
<p>In pure water, manganese(II) sulfide will only partially dissociate to produce manganese(II) cations and sulfide anions</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) rightleftharpoons "Mn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for this equilibrium reaction is</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that the sulfide anion <strong>cannot</strong> exist in water. Once it dissociates from the solid, it will react with water to form <em>bisulfide anions</em>, <mathjax>#"HS"^(-)#</mathjax>. </p>
<p>In other words, the sulfide anions act as a <strong>strong base</strong> in aqueous solution. </p>
<blockquote>
<p><mathjax>#"S"_ ((aq))^(2-) + "H"_ 2"O"_ ((l)) -> "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that a more <strong>accurate depiction</strong> of what happens when manganese(II) sulfide partially dissociates in water would be </p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + "H"_ 2"O"_ ((l)) rightleftharpoons "Mn"_ ((aq))^(2+) + "HS"_ ((aq))^(-) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The <mathjax>#K_(sp)#</mathjax> would thus take the form</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mn"^(2+)] * ["HS"^(-)] * ["OH"^(-)]#</mathjax></p>
</blockquote>
<p>Here comes the cool part. If you have an <strong>acidic solution</strong>, the excess hydronium cations will <strong>neutralize</strong> the hydroxide anions produced when the solid partially dissociates. </p>
<p>Moreover, the excess hydronium cations will <strong>protonate</strong> the bisulfide anions to hydrogen sulfide, <mathjax>#"H"_ 2"S"#</mathjax>.</p>
<p>As a result, the equilibrium will <strong>shift to the right</strong> and <strong>more solid will dissolve</strong> <mathjax>#->#</mathjax> as always, <strong><a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a></strong> governs equilibrium reactions. </p>
<p>This means that in <strong>acidic solution</strong>, you will have</p>
<blockquote>
<p><mathjax>#"MnS"_ ((s)) + 2"H"_ 3"O"_ ((aq))^(+) rightleftharpoons "Mn"_ ((aq))^(2+) + "H"_ 2"S"_ ((aq)) + 2"H"_ 2"O"_ ((l))" "color(purple)("(!)")#</mathjax></p>
</blockquote>
<p>This time, the solubility product constant is called the <strong>acid solubility product constant</strong>, <mathjax>#K_(spa)#</mathjax>, and it takes the form -- remember, pure liquids or solids are <strong>not</strong> included in the expression of the equilibrium constant!</p>
<blockquote>
<p><mathjax>#K_(spa) = (["H"_2"S"] * ["Mn"^(2+)])/(["H"_3"O"^(+)]^2)" "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>In your case, you know that the <mathjax>#K_(spa)#</mathjax> for manganese(II) sulfide is </p>
<blockquote>
<p><mathjax>#K_(spa) = 3 * 10^7#</mathjax></p>
</blockquote>
<p>So, know that we've figured out how the balanced chemical equation should look like, focus on determining the concentration of hydronium cations in the buffer.</p>
<p>The <mathjax>#"p"K_a#</mathjax> of acetic acid is </p>
<blockquote>
<p><mathjax>#"p"K_a = 4.75#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>Use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 4.75 + log( (0.490 color(red)(cancel(color(black)("M"))))/(0.105color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = 5.42#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"pH" = - log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>which implies</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
</blockquote>
<p>The concentration of hydronium cations in this solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-5.42) = 3.80 * 10^(-6)#</mathjax></p>
</blockquote>
<p>Now, take a look at the balanced chemical equation <mathjax>#color(purple)("(!)")#</mathjax>. Notice that <strong>every mole</strong> of manganese(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of manganese(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen sulfide. </p>
<p>This means that the two chemical species will have <strong>equal concentrations</strong> at equilibrium. </p>
<p>The amount of manganese(II) sulfide that dissociates produces manganese(II) cations in a <mathjax>#1:1#</mathjax> mole ratio, which means that if you take <mathjax>#s#</mathjax> to be this equilibrium concentration, you can say that this value represents the <strong>molar solubility</strong> of the salt and that equation <mathjax>#color(darkorange)("(*)")#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#K_(spa) = (s * s)/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which is </p>
<blockquote>
<p><mathjax>#3 * 10^7 = s^2/(3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#s#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#s^2 = 3 * 10^7 * (3.80 * 10^(-6))^2#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#s = sqrt(3 * 10^7 * (3.80 * 10^(-6))^2) = 0.0208#</mathjax></p>
</blockquote>
<p>You can thus say that in this buffer, the molar solubility of the salt is equal to </p>
<blockquote>
<p><mathjax>#s = "0.0208 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams per liter</em>, use the <strong>molar mass</strong> of manganese(II) sulfide</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("solubility in g/L"))) = 0.0208 color(white)(.)color(red)(cancel(color(black)("mol")))/"L" * "87.003 g"/(1color(red)(cancel(color(black)("mol MnS")))) = color(darkgreen)(ul(color(black)("1.81 g L"^(-1)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the concentrations of acetic acid and acetate anions. </p></div>
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</article> | What is the solubility of #MnS#, in grams per liter, in a buffer solution that is 0.105 M #CH_3COOH# - 0.490 M #NaCH_3COO#? |
What is the solubility of #MnS#, in grams per liter, in a buffer solution that is 0.105 M #CH_3COOH# - 0.490 M #NaCH_3COO#?
#MnS# has a #K_(\color(red)(spa))# of #3\cdot10^7#
|
2,333 | ab6627b6-6ddd-11ea-a9d2-ccda262736ce | https://socratic.org/questions/5925f54711ef6b72af01fa47 | 16.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 13 8 9 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] hydrogen [IN] moles"}] | [{"type":"physical unit","value":"16.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] ammonium hydroxide [=] \\pu{3.2 mol}"}] | <h1 class="questionTitle" itemprop="name">How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#?</h1> | null | 16.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
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<div>
<div class="markdown"><p><mathjax>#3.2xxN_Axx5=??#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#3.2xxN_Axx5=??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
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Nathan L.
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<div class="markdown"><p><mathjax>#9.6xx10^24#</mathjax> <mathjax>#"atoms H"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We're asked to calculate the number of atoms of <mathjax>#"H"#</mathjax> in <mathjax>#3.2#</mathjax> <mathjax>#"mol NH"_4"OH"#</mathjax>.</p>
<p>We can do this two different ways. The first is to calculate the number of molecules of <mathjax>#"NH"_4"OH"#</mathjax>, and then atoms of <mathjax>#"H"#</mathjax>, and the second method is to first calculate the number of moles of <mathjax>#"H"#</mathjax>, and then atoms of <mathjax>#"H"#</mathjax>.</p>
<p>Method 1:</p>
<p>Using <em>Avogadro's number</em> (<mathjax>#6.022xx10^23"particles"/"mol"#</mathjax>), let's calculate the number of <em>molecules</em> of <mathjax>#"NH"_4"OH"#</mathjax>:</p>
<p><mathjax>#3.2cancel("mol NH"_4"OH")((6.022xx10^23"molecules NH"_4"OH")/(1 cancel("mol NH"_4"OH")))#</mathjax></p>
<p><mathjax>#= 1.927xx10^24#</mathjax> <mathjax>#"molecules NH"_4"OH"#</mathjax></p>
<p>Now, using the fact that there are <mathjax>#5#</mathjax> atoms of <mathjax>#"H"#</mathjax> per molecule of <mathjax>#"NH"_4"OH"#</mathjax>, we have</p>
<p><mathjax>#"atoms H" =#</mathjax><br/>
<mathjax>#1.927xx10^24cancel("molecules NH"_4"OH")((5"atoms H")/(1 cancel("molecule NH"_4"OH")))#</mathjax></p>
<p><mathjax>#= color(red)(9.6xx10^24#</mathjax> <mathjax>#color(red)("atoms H"#</mathjax></p>
<p>rounded to <mathjax>#2#</mathjax> <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, the amount given in the problem.</p>
<blockquote></blockquote>
<p>Method 2:</p>
<p>For the second method, we can calculate the number of moles of <mathjax>#"H"#</mathjax> per mole of <mathjax>#"NH"_4"OH"#</mathjax>, and then atoms <mathjax>#"H"#</mathjax> from moles:</p>
<p><mathjax>#3.2#</mathjax> <mathjax>#cancel("mol NH"_4"OH")((5cancel("mol H"))/(1cancel("mol NH"_4"OH")))((6.022xx10^23"atoms H")/(1cancel("mol H")))#</mathjax></p>
<p><mathjax>#= color(red)(9.6xx10^24#</mathjax> <mathjax>#color(red)("atoms H"#</mathjax></p>
<p>rounded to <mathjax>#2#</mathjax> significant figures, the amount given in the problem.</p>
<p>As you can see, both methods yield the same result, so either one can be used (I would probably prefer the second method; I find it to be a little more exact).</p></div>
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</article> | How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#? | null |
2,334 | a90b6cc0-6ddd-11ea-b2ba-ccda262736ce | https://socratic.org/questions/if-29-4-grams-of-hydrogen-and-3701-3-grams-of-iodine-combine-to-form-hydrogen-io | 3730.33 grams | start physical_unit 13 14 mass g qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 9 10 6 7 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen iodide [IN] grams"}] | [{"type":"physical unit","value":"3730.33 grams"}] | [{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{29.4 grams}"},{"type":"physical unit","value":"Mass [OF] iodine combine [=] \\pu{3701.3 grams}"}] | <h1 class="questionTitle" itemprop="name">If 29.4 grams of hydrogen and 3701.3 grams of iodine combine to form hydrogen iodide, how many grams of hydrogen iodide must form?</h1> | null | 3730.33 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write down a balanced chemical equation for the reaction :</p>
<p><mathjax>#H_2+I_2->2HI#</mathjax></p>
<p>This represents <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Now calculate the number of moles of each reactant present :</p>
<p><mathjax>#n_(H_2)=m/M_r=29.4/2=14.7mol#</mathjax></p>
<p><mathjax>#n_(I_2)=m/M_r=3701.3/253.8=14.583mol#</mathjax></p>
<p>Since the balanced chemical equation illustrates that the reactants combine in a <mathjax>#1:1#</mathjax> ratio, it implies that iodine is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> and decides how much product is formed. Hydrogen is in excess.</p>
<p>So hence, since <mathjax>#1mol I_2#</mathjax> produces <mathjax>#2molHI#</mathjax>, it follows that <mathjax>#14.583molI_2#</mathjax> will produce <mathjax>#29.166molHI#</mathjax></p>
<p>This corresponds to a mass of product of</p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=29.166xx127.9#</mathjax></p>
<p><mathjax>#=3730.33g#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#3.730kg#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write down a balanced chemical equation for the reaction :</p>
<p><mathjax>#H_2+I_2->2HI#</mathjax></p>
<p>This represents <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Now calculate the number of moles of each reactant present :</p>
<p><mathjax>#n_(H_2)=m/M_r=29.4/2=14.7mol#</mathjax></p>
<p><mathjax>#n_(I_2)=m/M_r=3701.3/253.8=14.583mol#</mathjax></p>
<p>Since the balanced chemical equation illustrates that the reactants combine in a <mathjax>#1:1#</mathjax> ratio, it implies that iodine is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> and decides how much product is formed. Hydrogen is in excess.</p>
<p>So hence, since <mathjax>#1mol I_2#</mathjax> produces <mathjax>#2molHI#</mathjax>, it follows that <mathjax>#14.583molI_2#</mathjax> will produce <mathjax>#29.166molHI#</mathjax></p>
<p>This corresponds to a mass of product of</p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=29.166xx127.9#</mathjax></p>
<p><mathjax>#=3730.33g#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If 29.4 grams of hydrogen and 3701.3 grams of iodine combine to form hydrogen iodide, how many grams of hydrogen iodide must form?</h1>
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Trevor Ryan.
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Apr 5, 2016
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<div class="markdown"><p><mathjax>#3.730kg#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write down a balanced chemical equation for the reaction :</p>
<p><mathjax>#H_2+I_2->2HI#</mathjax></p>
<p>This represents <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Now calculate the number of moles of each reactant present :</p>
<p><mathjax>#n_(H_2)=m/M_r=29.4/2=14.7mol#</mathjax></p>
<p><mathjax>#n_(I_2)=m/M_r=3701.3/253.8=14.583mol#</mathjax></p>
<p>Since the balanced chemical equation illustrates that the reactants combine in a <mathjax>#1:1#</mathjax> ratio, it implies that iodine is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> and decides how much product is formed. Hydrogen is in excess.</p>
<p>So hence, since <mathjax>#1mol I_2#</mathjax> produces <mathjax>#2molHI#</mathjax>, it follows that <mathjax>#14.583molI_2#</mathjax> will produce <mathjax>#29.166molHI#</mathjax></p>
<p>This corresponds to a mass of product of</p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=29.166xx127.9#</mathjax></p>
<p><mathjax>#=3730.33g#</mathjax></p></div>
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</article> | If 29.4 grams of hydrogen and 3701.3 grams of iodine combine to form hydrogen iodide, how many grams of hydrogen iodide must form? | null |
2,335 | a9ada4fe-6ddd-11ea-bec9-ccda262736ce | https://socratic.org/questions/55d9b7a911ef6b17f380ca20 | 136.65 L | start physical_unit 22 23 volume l qc_end physical_unit 3 4 0 1 mass qc_end physical_unit 8 9 6 7 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] the piston [IN] L"}] | [{"type":"physical unit","value":"136.65 L"}] | [{"type":"physical unit","value":"Mass [OF] ammonia gas [=] \\pu{60 g}"},{"type":"physical unit","value":"Mole [OF] dihydrogen gas [=] \\pu{2.5 mol}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">#60*g# of ammonia gas, and #2.5*mol# dihydrogen gas, are confined in a piston at #"STP"#. What is the volume of the piston?</h1> | null | 136.65 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V = (nRT)/P#</mathjax>. Use an appropriate gas constant, <mathjax>#R#</mathjax>, i.e. <mathjax>#8.314#</mathjax> <mathjax>#J#</mathjax> <mathjax># K^-1#</mathjax> <mathjax>#mol^-1#</mathjax>. All you need to do is calculate the moles of each gas, dihydrogen, and of ammonia, and plug the numbers in. Remember that you have to use absolute temperature, that is Celsius temperature + 273.15.</p></div>
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<div class="markdown"><p>Treat both gases as ideal and use the ideal gas equation. Note that idealization of gases is not always justified, but here, with moderate pressures (1 atm) and moderate temperatures, it is entirely reasonable.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V = (nRT)/P#</mathjax>. Use an appropriate gas constant, <mathjax>#R#</mathjax>, i.e. <mathjax>#8.314#</mathjax> <mathjax>#J#</mathjax> <mathjax># K^-1#</mathjax> <mathjax>#mol^-1#</mathjax>. All you need to do is calculate the moles of each gas, dihydrogen, and of ammonia, and plug the numbers in. Remember that you have to use absolute temperature, that is Celsius temperature + 273.15.</p></div>
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<h1 class="questionTitle" itemprop="name">#60*g# of ammonia gas, and #2.5*mol# dihydrogen gas, are confined in a piston at #"STP"#. What is the volume of the piston?</h1>
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anor277
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Aug 23, 2015
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<div class="markdown"><p>Treat both gases as ideal and use the ideal gas equation. Note that idealization of gases is not always justified, but here, with moderate pressures (1 atm) and moderate temperatures, it is entirely reasonable.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V = (nRT)/P#</mathjax>. Use an appropriate gas constant, <mathjax>#R#</mathjax>, i.e. <mathjax>#8.314#</mathjax> <mathjax>#J#</mathjax> <mathjax># K^-1#</mathjax> <mathjax>#mol^-1#</mathjax>. All you need to do is calculate the moles of each gas, dihydrogen, and of ammonia, and plug the numbers in. Remember that you have to use absolute temperature, that is Celsius temperature + 273.15.</p></div>
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<div class="markdown"><p>Volume of hydrogen: <strong>57 L</strong><br/>
Volume of ammonia: <strong>80 L</strong></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that your gases are at <strong>STP</strong> conditions, which means that you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to make your life easier. </p>
<p><strong>STP</strong> conditions imply a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these condtions for pressure and temperature, <strong>one mole</strong> of any idea lgas aoccupies exactly <strong>22.7 L</strong>. </p>
<p>So basically all you need to do is figure out how many moles of each gas you have. You already know how many moles of hydrogen gas you have, so you can calculate its volume to be </p>
<blockquote>
<p><mathjax>#2.5color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "56.75 L"#</mathjax></p>
</blockquote>
<p>To get the number of moles of ammonia, use its molar mass </p>
<blockquote>
<p><mathjax>#60color(red)(cancel(color(black)("g"))) * ("1 mole NH"""_3)/(17.03color(red)(cancel(color(black)("g")))) = "3.52 moles NH"""_3#</mathjax></p>
</blockquote>
<p>This means that the volume of the ammonia sample will be </p>
<blockquote>
<p><mathjax>#3.52color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "79.9 L"#</mathjax></p>
</blockquote>
<p>You need to round the answers to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a> for hydrogen and to one sig fig for ammonia</p>
<blockquote>
<p><mathjax>#V_(H_2) = color(green)("57 L")" "#</mathjax> and <mathjax>#" "V_(NH_3) = color(green)("80 L")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP as being equal to 22.4 L.</em></p>
<p><em>That value isbased on the old definition of STP, which implied a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>If this was the value you were supposed to use, simply redo the calculations using 22.4 instead of 22.7 L.</em></p></div>
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</article> | #60*g# of ammonia gas, and #2.5*mol# dihydrogen gas, are confined in a piston at #"STP"#. What is the volume of the piston? | null |
2,336 | acf55fc9-6ddd-11ea-99bf-ccda262736ce | https://socratic.org/questions/how-many-moles-of-sodium-hydroxide-are-in-500-ml-of-a-2-0-m-naoh-solution | 1.00 moles | start physical_unit 4 5 mole mol qc_end physical_unit 14 15 8 9 volume qc_end physical_unit 14 15 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium hydroxide [IN] moles"}] | [{"type":"physical unit","value":"1.00 moles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{500 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{2.0 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of sodium hydroxide are in 500 mL of a 2.0 M #NaOH# solution?</h1> | null | 1.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We calculate the mass of <mathjax>#"NaOH"#</mathjax>, using the molarity formula.</p>
<blockquote>
<p><mathjax>#["NaOH"] = "mol solute"/"L solution"#</mathjax></p>
<p><mathjax>#"2 M" = (x " mol NaOH")/"0.500 L"#</mathjax> </p>
<p><mathjax>#2 xx "0.500 mol" = x " mol NaOH"#</mathjax></p>
<p><mathjax>#x = "1 mol NaOH"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1 mol"#</mathjax> of <mathjax>#"NaOH"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We calculate the mass of <mathjax>#"NaOH"#</mathjax>, using the molarity formula.</p>
<blockquote>
<p><mathjax>#["NaOH"] = "mol solute"/"L solution"#</mathjax></p>
<p><mathjax>#"2 M" = (x " mol NaOH")/"0.500 L"#</mathjax> </p>
<p><mathjax>#2 xx "0.500 mol" = x " mol NaOH"#</mathjax></p>
<p><mathjax>#x = "1 mol NaOH"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of sodium hydroxide are in 500 mL of a 2.0 M #NaOH# solution?</h1>
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<span class="dateCreated" datetime="2016-04-04T12:03:04" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"1 mol"#</mathjax> of <mathjax>#"NaOH"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We calculate the mass of <mathjax>#"NaOH"#</mathjax>, using the molarity formula.</p>
<blockquote>
<p><mathjax>#["NaOH"] = "mol solute"/"L solution"#</mathjax></p>
<p><mathjax>#"2 M" = (x " mol NaOH")/"0.500 L"#</mathjax> </p>
<p><mathjax>#2 xx "0.500 mol" = x " mol NaOH"#</mathjax></p>
<p><mathjax>#x = "1 mol NaOH"#</mathjax></p>
</blockquote></div>
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</article> | How many moles of sodium hydroxide are in 500 mL of a 2.0 M #NaOH# solution? | null |
2,337 | ab457a5d-6ddd-11ea-8ca2-ccda262736ce | https://socratic.org/questions/5912429f7c01497b7a3986c6 | 1.60 | start physical_unit 5 5 poh none qc_end physical_unit 14 15 10 11 mass qc_end physical_unit 5 5 17 18 volume qc_end end | [{"type":"physical unit","value":"pOH [OF] the solution"}] | [{"type":"physical unit","value":"1.60"}] | [{"type":"physical unit","value":"Mass [OF] potassium hydroxide [=] \\pu{15.8 mg}"},{"type":"physical unit","value":"Volume [OF] the solution [=] \\pu{11 mL}"}] | <h1 class="questionTitle" itemprop="name">What is #pOH# of a solution prepared by adding a #15.8*mg# mass of potassium hydroxide to #11*mL# of solution?</h1> | null | 1.60 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now..........<mathjax>#pOH=-log_10[HO^-]#</mathjax>...........</p>
<p>And we have <mathjax>#15.8*mg#</mathjax> <mathjax>#KOH#</mathjax> dissolved in <mathjax>#11*mL#</mathjax> of solution.</p>
<p><mathjax>#[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)#</mathjax></p>
<p><mathjax>#=2.56xx10^-2*mol*L^-1#</mathjax>.</p>
<p>But by definition, <mathjax>#pOH=-log_10(2.56xx10^-2)=1.60#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution? You should be able to answer immediately. Why? Because in aqueous solution, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Also see this <a href="https://socratic.org/questions/ph-value-definition">old problem here for more treatment.</a> </p></div>
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<div class="markdown"><p><mathjax>#pOH=1.60#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now..........<mathjax>#pOH=-log_10[HO^-]#</mathjax>...........</p>
<p>And we have <mathjax>#15.8*mg#</mathjax> <mathjax>#KOH#</mathjax> dissolved in <mathjax>#11*mL#</mathjax> of solution.</p>
<p><mathjax>#[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)#</mathjax></p>
<p><mathjax>#=2.56xx10^-2*mol*L^-1#</mathjax>.</p>
<p>But by definition, <mathjax>#pOH=-log_10(2.56xx10^-2)=1.60#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution? You should be able to answer immediately. Why? Because in aqueous solution, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Also see this <a href="https://socratic.org/questions/ph-value-definition">old problem here for more treatment.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">What is #pOH# of a solution prepared by adding a #15.8*mg# mass of potassium hydroxide to #11*mL# of solution?</h1>
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anor277
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<div class="markdown"><p><mathjax>#pOH=1.60#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now..........<mathjax>#pOH=-log_10[HO^-]#</mathjax>...........</p>
<p>And we have <mathjax>#15.8*mg#</mathjax> <mathjax>#KOH#</mathjax> dissolved in <mathjax>#11*mL#</mathjax> of solution.</p>
<p><mathjax>#[HO^-]=((15.8xx10^-3*g)/(56.11*g*mol^-1))/(11*mLxx10^-3*mL*L^-1)#</mathjax></p>
<p><mathjax>#=2.56xx10^-2*mol*L^-1#</mathjax>.</p>
<p>But by definition, <mathjax>#pOH=-log_10(2.56xx10^-2)=1.60#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution? You should be able to answer immediately. Why? Because in aqueous solution, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Also see this <a href="https://socratic.org/questions/ph-value-definition">old problem here for more treatment.</a> </p></div>
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</article> | What is #pOH# of a solution prepared by adding a #15.8*mg# mass of potassium hydroxide to #11*mL# of solution? | null |
2,338 | a90a5b52-6ddd-11ea-b854-ccda262736ce | https://socratic.org/questions/using-the-equation-n-2-3h-2-2nh-3-if-28g-n-2-react-how-many-grams-of-nh-3-will-b | 34 grams | start physical_unit 9 9 mass g qc_end physical_unit 3 3 11 12 mass qc_end chemical_equation 3 9 qc_end end | [{"type":"physical unit","value":"Mass [OF] NH3 [IN] grams"}] | [{"type":"physical unit","value":"34 grams"}] | [{"type":"physical unit","value":"Mass [OF] N2 [=] \\pu{28 g}"},{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"}] | <h1 class="questionTitle" itemprop="name">Using the equation #"N"_2 + 3"H"_2 -> 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?</h1> | null | 34 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to any <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem is the <strong>mole ratio</strong> that exists in the <em>balanced</em> chemical equation between the two chemical species that are of interest.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>The balanced chemical equation tells you that the reaction consumes <mathjax>#3#</mathjax> <strong>moles</strong> of hydrogen gas, <mathjax>#"H"_2#</mathjax>, and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia, <mathjax>#"NH"_3#</mathjax>, <strong>for every mole</strong> of nitrogen gas, <mathjax>#"N"_2#</mathjax>, that takes part in the reaction. </p>
<p>Since the problem doesn't mention the amount of hydrogen gas available for the reaction, you can assume that it's <em>in excess</em>, which means that the reaction will consume <strong>all the grams</strong> of nitrogen gas present. </p>
<p>Now, you can convert <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of the two species</p>
<blockquote>
<p><mathjax>#M_("M N"_2) ~~ "28 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M NH"_3) ~~ "17 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#1:color(red0(2)#</mathjax> mole ratio that exists between nitrogen gas and ammonia can be written as </p>
<blockquote>
<p><mathjax>#(1 color(gray)(cancel(color(black)("mole"))) * "28 g" color(gray)(cancel(color(black)("mol"^(-1)))))/(color(red)(2) color(gray)(cancel(color(black)("moles"))) * "17 g" color(gray)(cancel(color(black)("mol"^(-1))))) = (28 color(gray)(cancel(color(black)("g"))))/(34color(gray)(cancel(color(black)("g")))) = 14/17 ->#</mathjax> <em><strong>gram ratio</strong></em></p>
</blockquote>
<p>So, if the reaction consumes <mathjax>#"28 g"#</mathjax> of nitrogen gas, it follows that it will produce </p>
<blockquote>
<p><mathjax>#28 color(red)(cancel(color(black)("g N"_2))) * "17 g NH"_3/(14color(red)(cancel(color(black)("g N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("34 g NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In other words, <mathjax>#1#</mathjax> <strong>mole</strong> of nitrogen gas will produce <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"34 g NH"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to any <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem is the <strong>mole ratio</strong> that exists in the <em>balanced</em> chemical equation between the two chemical species that are of interest.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>The balanced chemical equation tells you that the reaction consumes <mathjax>#3#</mathjax> <strong>moles</strong> of hydrogen gas, <mathjax>#"H"_2#</mathjax>, and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia, <mathjax>#"NH"_3#</mathjax>, <strong>for every mole</strong> of nitrogen gas, <mathjax>#"N"_2#</mathjax>, that takes part in the reaction. </p>
<p>Since the problem doesn't mention the amount of hydrogen gas available for the reaction, you can assume that it's <em>in excess</em>, which means that the reaction will consume <strong>all the grams</strong> of nitrogen gas present. </p>
<p>Now, you can convert <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of the two species</p>
<blockquote>
<p><mathjax>#M_("M N"_2) ~~ "28 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M NH"_3) ~~ "17 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#1:color(red0(2)#</mathjax> mole ratio that exists between nitrogen gas and ammonia can be written as </p>
<blockquote>
<p><mathjax>#(1 color(gray)(cancel(color(black)("mole"))) * "28 g" color(gray)(cancel(color(black)("mol"^(-1)))))/(color(red)(2) color(gray)(cancel(color(black)("moles"))) * "17 g" color(gray)(cancel(color(black)("mol"^(-1))))) = (28 color(gray)(cancel(color(black)("g"))))/(34color(gray)(cancel(color(black)("g")))) = 14/17 ->#</mathjax> <em><strong>gram ratio</strong></em></p>
</blockquote>
<p>So, if the reaction consumes <mathjax>#"28 g"#</mathjax> of nitrogen gas, it follows that it will produce </p>
<blockquote>
<p><mathjax>#28 color(red)(cancel(color(black)("g N"_2))) * "17 g NH"_3/(14color(red)(cancel(color(black)("g N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("34 g NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In other words, <mathjax>#1#</mathjax> <strong>mole</strong> of nitrogen gas will produce <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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<h1 class="questionTitle" itemprop="name">Using the equation #"N"_2 + 3"H"_2 -> 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"34 g NH"_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to any <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem is the <strong>mole ratio</strong> that exists in the <em>balanced</em> chemical equation between the two chemical species that are of interest.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>The balanced chemical equation tells you that the reaction consumes <mathjax>#3#</mathjax> <strong>moles</strong> of hydrogen gas, <mathjax>#"H"_2#</mathjax>, and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia, <mathjax>#"NH"_3#</mathjax>, <strong>for every mole</strong> of nitrogen gas, <mathjax>#"N"_2#</mathjax>, that takes part in the reaction. </p>
<p>Since the problem doesn't mention the amount of hydrogen gas available for the reaction, you can assume that it's <em>in excess</em>, which means that the reaction will consume <strong>all the grams</strong> of nitrogen gas present. </p>
<p>Now, you can convert <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of the two species</p>
<blockquote>
<p><mathjax>#M_("M N"_2) ~~ "28 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M NH"_3) ~~ "17 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#1:color(red0(2)#</mathjax> mole ratio that exists between nitrogen gas and ammonia can be written as </p>
<blockquote>
<p><mathjax>#(1 color(gray)(cancel(color(black)("mole"))) * "28 g" color(gray)(cancel(color(black)("mol"^(-1)))))/(color(red)(2) color(gray)(cancel(color(black)("moles"))) * "17 g" color(gray)(cancel(color(black)("mol"^(-1))))) = (28 color(gray)(cancel(color(black)("g"))))/(34color(gray)(cancel(color(black)("g")))) = 14/17 ->#</mathjax> <em><strong>gram ratio</strong></em></p>
</blockquote>
<p>So, if the reaction consumes <mathjax>#"28 g"#</mathjax> of nitrogen gas, it follows that it will produce </p>
<blockquote>
<p><mathjax>#28 color(red)(cancel(color(black)("g N"_2))) * "17 g NH"_3/(14color(red)(cancel(color(black)("g N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("34 g NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In other words, <mathjax>#1#</mathjax> <strong>mole</strong> of nitrogen gas will produce <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | Using the equation #"N"_2 + 3"H"_2 -> 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced? | null |
2,339 | a85876c0-6ddd-11ea-872d-ccda262736ce | https://socratic.org/questions/50-ml-of-0-10-m-naoh-completely-neutralizes-30-ml-of-hydrochloric-acid-how-would | 0.17 M | start physical_unit 20 20 concentration mol/l qc_end c_other OTHER qc_end physical_unit 5 5 3 4 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] HCl solution [IN] M"}] | [{"type":"physical unit","value":"0.17 M"}] | [{"type":"other","value":"Completely neutralize."},{"type":"physical unit","value":"Concentration [OF] NaOH solution [=] \\pu{0.10 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{50.0 ml}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{30.0 ml}"}] | <h1 class="questionTitle" itemprop="name">50.0 ml of 0.10 M NaOH completely neutralizes 30.0 ml of hydrochloric acid. How would you calculate the concentration of HCl?</h1> | null | 0.17 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by making sure that you have a clear understanding of what a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> actually means. </p>
<p>In this case, you're dealing with sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <em>strong base</em>, and hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a strong acid. Both these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> will completely dissociate in aqueous solution to form </p>
<blockquote>
<p><mathjax>#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydroxide ions, <mathjax>#"OH"^(-)#</mathjax>, that can be found in the sodium hydroxide solution will react with the hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>, to form water. </p>
<p>This is what a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> is all about. </p>
<p>The balanced chemical equation for the reaction looks like this - I'll show you the <em>net ionic equation</em></p>
<blockquote>
<p><mathjax>#"OH"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>This tells you that in order for the neutalization to be complete, you need <strong>equal numbers of moles</strong> of hydroxide and hydronium ions. </p>
<p>Use the volume and <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the sodium hydroxide solution to find how many moles of hydroxide you have</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.10 M" * 50.0 * 10^(-3)"L" = "0.0050 moles OH"^(-)#</mathjax></p>
</blockquote>
<p>This means that the hydrochloic acid solution you used must provide <mathjax>#0.0050#</mathjax> moles of hydrochloric acid to the mix. This means that its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> will be</p>
<blockquote>
<p><mathjax>#c = "0.0050 moles"/(30.0 * 10^(-3)"L") = "0.1667 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#c_"HCl" = color(green)("0.17 M")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.17 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by making sure that you have a clear understanding of what a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> actually means. </p>
<p>In this case, you're dealing with sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <em>strong base</em>, and hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a strong acid. Both these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> will completely dissociate in aqueous solution to form </p>
<blockquote>
<p><mathjax>#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydroxide ions, <mathjax>#"OH"^(-)#</mathjax>, that can be found in the sodium hydroxide solution will react with the hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>, to form water. </p>
<p>This is what a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> is all about. </p>
<p>The balanced chemical equation for the reaction looks like this - I'll show you the <em>net ionic equation</em></p>
<blockquote>
<p><mathjax>#"OH"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>This tells you that in order for the neutalization to be complete, you need <strong>equal numbers of moles</strong> of hydroxide and hydronium ions. </p>
<p>Use the volume and <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the sodium hydroxide solution to find how many moles of hydroxide you have</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.10 M" * 50.0 * 10^(-3)"L" = "0.0050 moles OH"^(-)#</mathjax></p>
</blockquote>
<p>This means that the hydrochloic acid solution you used must provide <mathjax>#0.0050#</mathjax> moles of hydrochloric acid to the mix. This means that its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> will be</p>
<blockquote>
<p><mathjax>#c = "0.0050 moles"/(30.0 * 10^(-3)"L") = "0.1667 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#c_"HCl" = color(green)("0.17 M")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">50.0 ml of 0.10 M NaOH completely neutralizes 30.0 ml of hydrochloric acid. How would you calculate the concentration of HCl?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-31T01:24:58" itemprop="dateCreated">
Oct 31, 2015
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<div class="markdown"><p><mathjax>#"0.17 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by making sure that you have a clear understanding of what a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> actually means. </p>
<p>In this case, you're dealing with sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <em>strong base</em>, and hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a strong acid. Both these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> will completely dissociate in aqueous solution to form </p>
<blockquote>
<p><mathjax>#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydroxide ions, <mathjax>#"OH"^(-)#</mathjax>, that can be found in the sodium hydroxide solution will react with the hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>, to form water. </p>
<p>This is what a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> is all about. </p>
<p>The balanced chemical equation for the reaction looks like this - I'll show you the <em>net ionic equation</em></p>
<blockquote>
<p><mathjax>#"OH"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>This tells you that in order for the neutalization to be complete, you need <strong>equal numbers of moles</strong> of hydroxide and hydronium ions. </p>
<p>Use the volume and <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the sodium hydroxide solution to find how many moles of hydroxide you have</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.10 M" * 50.0 * 10^(-3)"L" = "0.0050 moles OH"^(-)#</mathjax></p>
</blockquote>
<p>This means that the hydrochloic acid solution you used must provide <mathjax>#0.0050#</mathjax> moles of hydrochloric acid to the mix. This means that its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> will be</p>
<blockquote>
<p><mathjax>#c = "0.0050 moles"/(30.0 * 10^(-3)"L") = "0.1667 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
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<p><mathjax>#c_"HCl" = color(green)("0.17 M")#</mathjax></p>
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</article> | 50.0 ml of 0.10 M NaOH completely neutralizes 30.0 ml of hydrochloric acid. How would you calculate the concentration of HCl? | null |
2,340 | a8d2797f-6ddd-11ea-b5fc-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-ph-of-acid-rain | 4 | start physical_unit 7 8 ph none qc_end substance 7 8 qc_end end | [{"type":"physical unit","value":"pH [OF] acid rain"}] | [{"type":"physical unit","value":"4"}] | [{"type":"substance name","value":"Acid rain"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the pH of acid rain?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>pH is an expression of the hydrogen ion concentration within a solution.</p>
<p>Mathematically, the pH will equal <mathjax>#-log_10[H^+]#</mathjax>, or the negative logarithm of the molar concentration of hydrogen ion about the solution.</p>
<p>We would need to make this calculation for any relevant solution in question to determine its pH value. </p>
<p>Rainwater itself is known to be slightly acidic, and its pH will vary about the sampled solution. A normal pH value for rain water would be somewhere around <mathjax>#5-5.5#</mathjax>.</p>
<p>For acid rain, we can expect the pH value to be lower, around <mathjax>#4#</mathjax>.</p></div>
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<div class="markdown"><p>Acid rain will have a <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> <mathjax>#≈4#</mathjax>, see below</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>pH is an expression of the hydrogen ion concentration within a solution.</p>
<p>Mathematically, the pH will equal <mathjax>#-log_10[H^+]#</mathjax>, or the negative logarithm of the molar concentration of hydrogen ion about the solution.</p>
<p>We would need to make this calculation for any relevant solution in question to determine its pH value. </p>
<p>Rainwater itself is known to be slightly acidic, and its pH will vary about the sampled solution. A normal pH value for rain water would be somewhere around <mathjax>#5-5.5#</mathjax>.</p>
<p>For acid rain, we can expect the pH value to be lower, around <mathjax>#4#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the pH of acid rain?</h1>
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<div class="markdown"><p>Acid rain will have a <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> <mathjax>#≈4#</mathjax>, see below</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>pH is an expression of the hydrogen ion concentration within a solution.</p>
<p>Mathematically, the pH will equal <mathjax>#-log_10[H^+]#</mathjax>, or the negative logarithm of the molar concentration of hydrogen ion about the solution.</p>
<p>We would need to make this calculation for any relevant solution in question to determine its pH value. </p>
<p>Rainwater itself is known to be slightly acidic, and its pH will vary about the sampled solution. A normal pH value for rain water would be somewhere around <mathjax>#5-5.5#</mathjax>.</p>
<p>For acid rain, we can expect the pH value to be lower, around <mathjax>#4#</mathjax>.</p></div>
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</article> | How do you calculate the pH of acid rain? | null |
2,341 | acafb992-6ddd-11ea-92f0-ccda262736ce | https://socratic.org/questions/when-7-25-g-of-al-reacts-with-excess-h-2so-4-in-the-reaction-2al-s-3-h-2so-4-aq- | 5.95 liters | start physical_unit 21 21 volume l qc_end physical_unit 4 4 1 2 mass qc_end c_other OTHER qc_end chemical_equation 12 21 qc_end physical_unit 10 11 30 31 temperature qc_end physical_unit 10 11 36 37 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] H2 [IN] liters"}] | [{"type":"physical unit","value":"5.95 liters"}] | [{"type":"physical unit","value":"Mass [OF] Al [=] \\pu{7.25 g}"},{"type":"other","value":"Excess H2SO4."},{"type":"chemical equation","value":"2 Al(s) + 3 H2SO4(aq) -> Al2(SO4)3(aq) + 3 H2(g)"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure [OF] the reaction [=] \\pu{1.650 atm}"}] | <h1 class="questionTitle" itemprop="name">When 7.25 g of #Al# reacts with excess #H_2SO_4# in the reaction #2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g)# how many liters of #H_2# is formed at 25° C and a pressure of 1.650 atm? </h1> | null | 5.95 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2#</mathjax> <mathjax>#Al(s)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2SO_4(aq) to #</mathjax> <mathjax>#Al_2(SO_4)_3(aq)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2(g)#</mathjax></p>
<p>First calculate the number of moles of <mathjax>#Al#</mathjax> in <mathjax>#7.25#</mathjax> <mathjax>#g#</mathjax>, knowing the molar mass of <mathjax>#Al#</mathjax> is <mathjax>#27#</mathjax> <mathjax>#gmol^-1#</mathjax>:</p>
<p><mathjax>#n=m/M=7.25/27=0.27#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the equation we can see that each <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#Al#</mathjax> yields <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#H_2#</mathjax>, so multiply this by <mathjax>#3/2#</mathjax> to find the number of moles of <mathjax>#H_2#</mathjax> released.</p>
<p><mathjax>#n=0.40#</mathjax> <mathjax>#mol#</mathjax></p>
<p>We know that a mole of an ideal gas occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at standard temperature and pressure (STP), which is <mathjax>#0^o#</mathjax> <mathjax>#C#</mathjax> or <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so this number of moles will produce <mathjax>#0.4xx22.4=9.0#</mathjax> <mathjax>#L#</mathjax> of gas at STP.</p>
<p>We are not asked about STP, though, but <mathjax>#25^o#</mathjax> <mathjax>#C#</mathjax> (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax>) and <mathjax>#1.65#</mathjax> <mathjax>#atm#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>Rearranging to make <mathjax>#V_2#</mathjax> the subject:</p>
<p><mathjax>#V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95#</mathjax> <mathjax>#L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The first step is to calculate the number of moles of <mathjax>#Al#</mathjax> present, <mathjax>#n=0.27#</mathjax> <mathjax>#mol#</mathjax>, then use <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in the equation to determine that this releases <mathjax>#0.40#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#H_2#</mathjax>, with a volume of <mathjax>#9.0#</mathjax> <mathjax>#L#</mathjax> at STP. Next convert to the given temperature and pressure, and <mathjax>#V=5.95#</mathjax> <mathjax>#L#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2#</mathjax> <mathjax>#Al(s)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2SO_4(aq) to #</mathjax> <mathjax>#Al_2(SO_4)_3(aq)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2(g)#</mathjax></p>
<p>First calculate the number of moles of <mathjax>#Al#</mathjax> in <mathjax>#7.25#</mathjax> <mathjax>#g#</mathjax>, knowing the molar mass of <mathjax>#Al#</mathjax> is <mathjax>#27#</mathjax> <mathjax>#gmol^-1#</mathjax>:</p>
<p><mathjax>#n=m/M=7.25/27=0.27#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the equation we can see that each <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#Al#</mathjax> yields <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#H_2#</mathjax>, so multiply this by <mathjax>#3/2#</mathjax> to find the number of moles of <mathjax>#H_2#</mathjax> released.</p>
<p><mathjax>#n=0.40#</mathjax> <mathjax>#mol#</mathjax></p>
<p>We know that a mole of an ideal gas occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at standard temperature and pressure (STP), which is <mathjax>#0^o#</mathjax> <mathjax>#C#</mathjax> or <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so this number of moles will produce <mathjax>#0.4xx22.4=9.0#</mathjax> <mathjax>#L#</mathjax> of gas at STP.</p>
<p>We are not asked about STP, though, but <mathjax>#25^o#</mathjax> <mathjax>#C#</mathjax> (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax>) and <mathjax>#1.65#</mathjax> <mathjax>#atm#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>Rearranging to make <mathjax>#V_2#</mathjax> the subject:</p>
<p><mathjax>#V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95#</mathjax> <mathjax>#L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When 7.25 g of #Al# reacts with excess #H_2SO_4# in the reaction #2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g)# how many liters of #H_2# is formed at 25° C and a pressure of 1.650 atm? </h1>
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<div class="markdown"><p>The first step is to calculate the number of moles of <mathjax>#Al#</mathjax> present, <mathjax>#n=0.27#</mathjax> <mathjax>#mol#</mathjax>, then use <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in the equation to determine that this releases <mathjax>#0.40#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#H_2#</mathjax>, with a volume of <mathjax>#9.0#</mathjax> <mathjax>#L#</mathjax> at STP. Next convert to the given temperature and pressure, and <mathjax>#V=5.95#</mathjax> <mathjax>#L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2#</mathjax> <mathjax>#Al(s)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2SO_4(aq) to #</mathjax> <mathjax>#Al_2(SO_4)_3(aq)#</mathjax> + <mathjax>#3#</mathjax> <mathjax>#H_2(g)#</mathjax></p>
<p>First calculate the number of moles of <mathjax>#Al#</mathjax> in <mathjax>#7.25#</mathjax> <mathjax>#g#</mathjax>, knowing the molar mass of <mathjax>#Al#</mathjax> is <mathjax>#27#</mathjax> <mathjax>#gmol^-1#</mathjax>:</p>
<p><mathjax>#n=m/M=7.25/27=0.27#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the equation we can see that each <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#Al#</mathjax> yields <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#H_2#</mathjax>, so multiply this by <mathjax>#3/2#</mathjax> to find the number of moles of <mathjax>#H_2#</mathjax> released.</p>
<p><mathjax>#n=0.40#</mathjax> <mathjax>#mol#</mathjax></p>
<p>We know that a mole of an ideal gas occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> at standard temperature and pressure (STP), which is <mathjax>#0^o#</mathjax> <mathjax>#C#</mathjax> or <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so this number of moles will produce <mathjax>#0.4xx22.4=9.0#</mathjax> <mathjax>#L#</mathjax> of gas at STP.</p>
<p>We are not asked about STP, though, but <mathjax>#25^o#</mathjax> <mathjax>#C#</mathjax> (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax>) and <mathjax>#1.65#</mathjax> <mathjax>#atm#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>Rearranging to make <mathjax>#V_2#</mathjax> the subject:</p>
<p><mathjax>#V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95#</mathjax> <mathjax>#L#</mathjax></p></div>
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</article> | When 7.25 g of #Al# reacts with excess #H_2SO_4# in the reaction #2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g)# how many liters of #H_2# is formed at 25° C and a pressure of 1.650 atm? | null |
2,342 | a99bc09c-6ddd-11ea-ae53-ccda262736ce | https://socratic.org/questions/oxalic-acid-is-26-68-carbon-2-24-hydrogen-and-the-remainder-oxygen-what-is-its-e | CHO2 | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] oxalic acid [IN] empirical"}] | [{"type":"chemical equation","value":"CHO2"}] | [{"type":"physical unit","value":"Percentage [OF] carbon in oxalic acid [=] \\pu{26.68%}"},{"type":"physical unit","value":"Percentage [OF] hydrogen in oxalic acid [=] \\pu{2.24%}"},{"type":"other","value":"Oxalic acid is carbon, hydrogen, and the remainder oxygen."}] | <h1 class="questionTitle" itemprop="name">Oxalic acid is 26.68% carbon, 2.24% hydrogen, and the remainder oxygen, what is its empirical formula?</h1> | null | CHO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume 100 g of acid. Breaking it down into its components, there are <mathjax>#(26.68*g)/(12.01 *g*mol^-1*C); (2.24*g)/(1.00794 *g *mol^-1*H); (71.08*g)/(16.00*g*mol^(-1)*O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.22:2.22:4.44#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1:1:2#</mathjax>.</p>
<p>Therefore, empirical formula of oxalic acid is <mathjax>#CHO_2#</mathjax>. If its molecular mass is <mathjax>#90.03*g*mol^-1#</mathjax>, what is its molecular formula?</p></div>
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<div class="markdown"><p><mathjax>#CHO_2#</mathjax> is the empirical formula for oxalic acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Assume 100 g of acid. Breaking it down into its components, there are <mathjax>#(26.68*g)/(12.01 *g*mol^-1*C); (2.24*g)/(1.00794 *g *mol^-1*H); (71.08*g)/(16.00*g*mol^(-1)*O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.22:2.22:4.44#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1:1:2#</mathjax>.</p>
<p>Therefore, empirical formula of oxalic acid is <mathjax>#CHO_2#</mathjax>. If its molecular mass is <mathjax>#90.03*g*mol^-1#</mathjax>, what is its molecular formula?</p></div>
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<div class="markdown"><p><mathjax>#CHO_2#</mathjax> is the empirical formula for oxalic acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Assume 100 g of acid. Breaking it down into its components, there are <mathjax>#(26.68*g)/(12.01 *g*mol^-1*C); (2.24*g)/(1.00794 *g *mol^-1*H); (71.08*g)/(16.00*g*mol^(-1)*O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.22:2.22:4.44#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1:1:2#</mathjax>.</p>
<p>Therefore, empirical formula of oxalic acid is <mathjax>#CHO_2#</mathjax>. If its molecular mass is <mathjax>#90.03*g*mol^-1#</mathjax>, what is its molecular formula?</p></div>
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</article> | Oxalic acid is 26.68% carbon, 2.24% hydrogen, and the remainder oxygen, what is its empirical formula? | null |
2,343 | a9db0bba-6ddd-11ea-bab5-ccda262736ce | https://socratic.org/questions/a-50-m-solution-of-sodium-thiosulfate-na-2s-2o-3-is-used-to-create-a-more-dilute | 0.05 M | start physical_unit 7 7 concentration mol/l qc_end physical_unit 7 7 1 2 concentration qc_end physical_unit 3 3 17 18 volume qc_end physical_unit 3 3 29 30 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] Na2S2O3 solution [IN] M"}] | [{"type":"physical unit","value":"0.05 M"}] | [{"type":"physical unit","value":"Concentration1 [OF] Na2S2O3 solution [=] \\pu{0.50 M}"},{"type":"physical unit","value":"Volume1 [OF] Na2S2O3 solution [=] \\pu{250 mL}"},{"type":"physical unit","value":"Volume2 [OF] Na2S2O3 solution [=] \\pu{2.5 L}"}] | <h1 class="questionTitle" itemprop="name">A .50 M solution of sodium thiosulfate, #Na_2S_2O_3#, is used to create a more dilute solution. If 250 mL of the concentrated solution is diluted to a volume of 2.5 L, determine the concentration of the new solution?</h1> | null | 0.05 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.50*mol*L^-1xx0.250*L)/(2.50*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.050*mol*L^-1#</mathjax> with respect to sodium thiosulate. What is the concentration with respect to sodium ion?</p></div>
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<div class="markdown"><p>Th initial solution is diluted 10 fold, thus it is <mathjax>#0.050*mol*L^-1#</mathjax> after dilution. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.50*mol*L^-1xx0.250*L)/(2.50*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.050*mol*L^-1#</mathjax> with respect to sodium thiosulate. What is the concentration with respect to sodium ion?</p></div>
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<h1 class="questionTitle" itemprop="name">A .50 M solution of sodium thiosulfate, #Na_2S_2O_3#, is used to create a more dilute solution. If 250 mL of the concentrated solution is diluted to a volume of 2.5 L, determine the concentration of the new solution?</h1>
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<div class="markdown"><p>Th initial solution is diluted 10 fold, thus it is <mathjax>#0.050*mol*L^-1#</mathjax> after dilution. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.50*mol*L^-1xx0.250*L)/(2.50*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.050*mol*L^-1#</mathjax> with respect to sodium thiosulate. What is the concentration with respect to sodium ion?</p></div>
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</article> | A .50 M solution of sodium thiosulfate, #Na_2S_2O_3#, is used to create a more dilute solution. If 250 mL of the concentrated solution is diluted to a volume of 2.5 L, determine the concentration of the new solution? | null |
2,344 | aad203ee-6ddd-11ea-bf97-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-capacity-of-a-metal-if-it-requires-178-1-j-to-change-t-1 | 1.70 J/(g * ℃） | start physical_unit 22 23 specific_heat_capacity j/(g_·_°c） qc_end physical_unit 22 23 12 13 energy qc_end physical_unit 22 23 25 26 temperature qc_end physical_unit 22 23 28 29 temperature qc_end physical_unit 22 23 19 20 mass qc_end end | [{"type":"physical unit","value":"Specific heat capacity [OF] the metal [IN] J/(g * ℃）"}] | [{"type":"physical unit","value":"1.70 J/(g * ℃）"}] | [{"type":"physical unit","value":"Required energy [OF] the metal [=] \\pu{178.1 J}"},{"type":"physical unit","value":"Temperature1 [OF] the metal [=] \\pu{25.00 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the metal [=] \\pu{32.00 ℃}"},{"type":"physical unit","value":"Mass [OF] the metal [=] \\pu{15.0 g}"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat capacity of a metal if it requires 178.1 J to change the temperature to 15.0 g of the metal from 25.00 C to 32.00 C?</h1> | null | 1.70 J/(g * ℃） | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When a problem asks you to find a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong>, <mathjax>#c#</mathjax>, it's essentially telling you to find how much heat is needed in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, you know that you need <mathjax>#"178.1 J"#</mathjax> in order to increase the temperature of <mathjax>#"15.0 g"#</mathjax> of your unknown metal by <mathjax>#7^@"C"#</mathjax>, since</p>
<blockquote>
<p><mathjax>#DeltaT = 32.00^@"C" - 25.00^@"C" = 7.00^@"C"#</mathjax></p>
</blockquote>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#7.00^@"C ?"#</mathjax></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * "178.1 J"/(15.0color(red)(cancel(color(black)("g")))) = "11.873 J"#</mathjax></p>
</blockquote>
<p>Since this much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> by <mathjax>#7.00^@"C"#</mathjax>, it follows that you can increase temperature by <mathjax>#1^@"C"#</mathjax> by adding</p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)(""^@"C"))) * "11.873 J"/(7.00color(red)(cancel(color(black)(""^@"C")))) = "1.70 J"#</mathjax></p>
</blockquote>
<p>Since you need <mathjax>#"1.70 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#1^@"C"#</mathjax>, you can say that the metal's <strong>specific heat</strong> will be </p>
<blockquote>
<p><mathjax>#c = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|))) -> #</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p><strong>ALTERNATIVELY</strong></p>
<p>You can also use the following equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat released<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Rearrange to solve for <mathjax>#c#</mathjax></p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#c = "178.1 J"/("15.0 g" * 7.00^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.70 J g"^(-1)""^@"C"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When a problem asks you to find a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong>, <mathjax>#c#</mathjax>, it's essentially telling you to find how much heat is needed in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, you know that you need <mathjax>#"178.1 J"#</mathjax> in order to increase the temperature of <mathjax>#"15.0 g"#</mathjax> of your unknown metal by <mathjax>#7^@"C"#</mathjax>, since</p>
<blockquote>
<p><mathjax>#DeltaT = 32.00^@"C" - 25.00^@"C" = 7.00^@"C"#</mathjax></p>
</blockquote>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#7.00^@"C ?"#</mathjax></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * "178.1 J"/(15.0color(red)(cancel(color(black)("g")))) = "11.873 J"#</mathjax></p>
</blockquote>
<p>Since this much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> by <mathjax>#7.00^@"C"#</mathjax>, it follows that you can increase temperature by <mathjax>#1^@"C"#</mathjax> by adding</p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)(""^@"C"))) * "11.873 J"/(7.00color(red)(cancel(color(black)(""^@"C")))) = "1.70 J"#</mathjax></p>
</blockquote>
<p>Since you need <mathjax>#"1.70 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#1^@"C"#</mathjax>, you can say that the metal's <strong>specific heat</strong> will be </p>
<blockquote>
<p><mathjax>#c = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|))) -> #</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p><strong>ALTERNATIVELY</strong></p>
<p>You can also use the following equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat released<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Rearrange to solve for <mathjax>#c#</mathjax></p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#c = "178.1 J"/("15.0 g" * 7.00^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the specific heat capacity of a metal if it requires 178.1 J to change the temperature to 15.0 g of the metal from 25.00 C to 32.00 C?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-01T10:29:36" itemprop="dateCreated">
Jul 1, 2016
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<div class="markdown"><p><mathjax>#"1.70 J g"^(-1)""^@"C"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When a problem asks you to find a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong>, <mathjax>#c#</mathjax>, it's essentially telling you to find how much heat is needed in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, you know that you need <mathjax>#"178.1 J"#</mathjax> in order to increase the temperature of <mathjax>#"15.0 g"#</mathjax> of your unknown metal by <mathjax>#7^@"C"#</mathjax>, since</p>
<blockquote>
<p><mathjax>#DeltaT = 32.00^@"C" - 25.00^@"C" = 7.00^@"C"#</mathjax></p>
</blockquote>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#7.00^@"C ?"#</mathjax></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * "178.1 J"/(15.0color(red)(cancel(color(black)("g")))) = "11.873 J"#</mathjax></p>
</blockquote>
<p>Since this much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> by <mathjax>#7.00^@"C"#</mathjax>, it follows that you can increase temperature by <mathjax>#1^@"C"#</mathjax> by adding</p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)(""^@"C"))) * "11.873 J"/(7.00color(red)(cancel(color(black)(""^@"C")))) = "1.70 J"#</mathjax></p>
</blockquote>
<p>Since you need <mathjax>#"1.70 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of this metal by <mathjax>#1^@"C"#</mathjax>, you can say that the metal's <strong>specific heat</strong> will be </p>
<blockquote>
<p><mathjax>#c = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|))) -> #</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p><strong>ALTERNATIVELY</strong></p>
<p>You can also use the following equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat released<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Rearrange to solve for <mathjax>#c#</mathjax></p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#c = "178.1 J"/("15.0 g" * 7.00^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | What is the specific heat capacity of a metal if it requires 178.1 J to change the temperature to 15.0 g of the metal from 25.00 C to 32.00 C? | null |
2,345 | aca9bd75-6ddd-11ea-9f33-ccda262736ce | https://socratic.org/questions/595e5f9f7c014916daab07c5 | Al2S3 | start chemical_formula qc_end physical_unit 7 8 14 15 mass qc_end physical_unit 7 7 18 19 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] aluminum sulfide [IN] empirical"}] | [{"type":"chemical equation","value":"Al2S3"}] | [{"type":"physical unit","value":"Mass [OF] aluminum sulfide [=] \\pu{2.03 g}"},{"type":"physical unit","value":"Mass [OF] aluminum [=] \\pu{0.73 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of an #"aluminum sulfide"#, for which a mass of #2.03*g# contains a #0.73*g# mass of #"aluminum"#?</h1> | null | Al2S3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we work out moles of aluminum and moles of sulfur.</p>
<p><mathjax>#"Moles of Al"=(0.73*g)/(26.98*g*mol^-1)=0.0271*mol#</mathjax></p>
<p><mathjax>#"Moles of S"=(2.03*g-0.73*g)/(32.06*g*mol^-1)=0.0406*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity...........</p>
<p>And get <mathjax>#AlS_(1.498)#</mathjax>. But the <mathjax>#"empirical formula"#</mathjax> is specified to be the <mathjax>#"simplest WHOLE number ratio"#</mathjax>.........and thus we get an empirical formula of <mathjax>#Al_2S_3#</mathjax>.........</p>
<p>How did I know that there were <mathjax>#1.30*g#</mathjax> of sulfur? </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We interrogate the empirical formula, and get <mathjax>#Al_2S_3#</mathjax>....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we work out moles of aluminum and moles of sulfur.</p>
<p><mathjax>#"Moles of Al"=(0.73*g)/(26.98*g*mol^-1)=0.0271*mol#</mathjax></p>
<p><mathjax>#"Moles of S"=(2.03*g-0.73*g)/(32.06*g*mol^-1)=0.0406*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity...........</p>
<p>And get <mathjax>#AlS_(1.498)#</mathjax>. But the <mathjax>#"empirical formula"#</mathjax> is specified to be the <mathjax>#"simplest WHOLE number ratio"#</mathjax>.........and thus we get an empirical formula of <mathjax>#Al_2S_3#</mathjax>.........</p>
<p>How did I know that there were <mathjax>#1.30*g#</mathjax> of sulfur? </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of an #"aluminum sulfide"#, for which a mass of #2.03*g# contains a #0.73*g# mass of #"aluminum"#?</h1>
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<span class="dateCreated" datetime="2017-07-06T16:51:50" itemprop="dateCreated">
Jul 6, 2017
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<div class="markdown"><p>We interrogate the empirical formula, and get <mathjax>#Al_2S_3#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so we work out moles of aluminum and moles of sulfur.</p>
<p><mathjax>#"Moles of Al"=(0.73*g)/(26.98*g*mol^-1)=0.0271*mol#</mathjax></p>
<p><mathjax>#"Moles of S"=(2.03*g-0.73*g)/(32.06*g*mol^-1)=0.0406*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity...........</p>
<p>And get <mathjax>#AlS_(1.498)#</mathjax>. But the <mathjax>#"empirical formula"#</mathjax> is specified to be the <mathjax>#"simplest WHOLE number ratio"#</mathjax>.........and thus we get an empirical formula of <mathjax>#Al_2S_3#</mathjax>.........</p>
<p>How did I know that there were <mathjax>#1.30*g#</mathjax> of sulfur? </p></div>
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</article> | What is the empirical formula of an #"aluminum sulfide"#, for which a mass of #2.03*g# contains a #0.73*g# mass of #"aluminum"#? | null |
2,346 | a95bed9a-6ddd-11ea-b76b-ccda262736ce | https://socratic.org/questions/what-are-the-grams-of-sodium-chloride-produced-when-10-0-g-of-sodium-react-with- | 16.48 grams | start physical_unit 5 6 mass g qc_end physical_unit 5 5 9 10 mass qc_end physical_unit 18 19 9 10 mass qc_end chemical_equation 23 29 qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium chloride [IN] grams"}] | [{"type":"physical unit","value":"16.48 grams"}] | [{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{10.0 g}"},{"type":"physical unit","value":"Mass [OF] chlorine gas [=] \\pu{10.0 g}"},{"type":"chemical equation","value":"2 Na + Cl2 -> 2 NaCl"}] | <h1 class="questionTitle" itemprop="name">What are the grams of sodium chloride produced when 10.0 g of sodium react with 10.0 g of chlorine gas in the equation #2Na+Cl_2 -> 2NaCl#?</h1> | null | 16.48 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1) We need to know which of reagents is excess,<br/>
so we calculate it by reagents' mass.<br/>
<mathjax>#M (Na)=23#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Na)=n×M=2×23=46#</mathjax> g</p>
<p><mathjax>#M (Cl_2)=35.5×2=71#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Cl_2)=n×M=1×71=71#</mathjax> g</p>
<p><mathjax>#\frac{10}{46}:\frac{x}{71}#</mathjax></p>
<p><mathjax>#x=\frac {10×71}{46}=15.43#</mathjax></p>
<p>This means the sodium is excess.<br/>
So, we're continuing solving the problem by chlorine. </p>
<p>2) <mathjax>#M (NaCl)=23+35.5=58.5#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (NaCl)=n×M=2×58.5=117#</mathjax>g</p>
<p>3) <mathjax>#10g (Cl_2) - xg (NaCl)#</mathjax><br/>
<mathjax>#71g (Cl_2) - 117g (NaCl)#</mathjax> </p>
<p><mathjax>#x=\frac {10×117}{71}=16.479#</mathjax> g</p></div>
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<div class="markdown"><p><mathjax>#m (NaCl)=16.479#</mathjax>g</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1) We need to know which of reagents is excess,<br/>
so we calculate it by reagents' mass.<br/>
<mathjax>#M (Na)=23#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Na)=n×M=2×23=46#</mathjax> g</p>
<p><mathjax>#M (Cl_2)=35.5×2=71#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Cl_2)=n×M=1×71=71#</mathjax> g</p>
<p><mathjax>#\frac{10}{46}:\frac{x}{71}#</mathjax></p>
<p><mathjax>#x=\frac {10×71}{46}=15.43#</mathjax></p>
<p>This means the sodium is excess.<br/>
So, we're continuing solving the problem by chlorine. </p>
<p>2) <mathjax>#M (NaCl)=23+35.5=58.5#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (NaCl)=n×M=2×58.5=117#</mathjax>g</p>
<p>3) <mathjax>#10g (Cl_2) - xg (NaCl)#</mathjax><br/>
<mathjax>#71g (Cl_2) - 117g (NaCl)#</mathjax> </p>
<p><mathjax>#x=\frac {10×117}{71}=16.479#</mathjax> g</p></div>
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<h1 class="questionTitle" itemprop="name">What are the grams of sodium chloride produced when 10.0 g of sodium react with 10.0 g of chlorine gas in the equation #2Na+Cl_2 -> 2NaCl#?</h1>
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<div class="markdown"><p><mathjax>#m (NaCl)=16.479#</mathjax>g</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>1) We need to know which of reagents is excess,<br/>
so we calculate it by reagents' mass.<br/>
<mathjax>#M (Na)=23#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Na)=n×M=2×23=46#</mathjax> g</p>
<p><mathjax>#M (Cl_2)=35.5×2=71#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (Cl_2)=n×M=1×71=71#</mathjax> g</p>
<p><mathjax>#\frac{10}{46}:\frac{x}{71}#</mathjax></p>
<p><mathjax>#x=\frac {10×71}{46}=15.43#</mathjax></p>
<p>This means the sodium is excess.<br/>
So, we're continuing solving the problem by chlorine. </p>
<p>2) <mathjax>#M (NaCl)=23+35.5=58.5#</mathjax> <mathjax>#frac\{g}{mol}#</mathjax><br/>
<mathjax>#m (NaCl)=n×M=2×58.5=117#</mathjax>g</p>
<p>3) <mathjax>#10g (Cl_2) - xg (NaCl)#</mathjax><br/>
<mathjax>#71g (Cl_2) - 117g (NaCl)#</mathjax> </p>
<p><mathjax>#x=\frac {10×117}{71}=16.479#</mathjax> g</p></div>
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</article> | What are the grams of sodium chloride produced when 10.0 g of sodium react with 10.0 g of chlorine gas in the equation #2Na+Cl_2 -> 2NaCl#? | null |
2,347 | aa8454ae-6ddd-11ea-acd7-ccda262736ce | https://socratic.org/questions/59809aef11ef6b5b1036c3b8 | +3 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 12 12 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] carbon"}] | [{"type":"physical unit","value":"+3"}] | [{"type":"chemical equation","value":"Na2+C2O4^2-"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#?</h1> | null | +3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3#</mathjax>. See below.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#?</h1>
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Room 204
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<div class="markdown"><p><mathjax>#3#</mathjax>. See below.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
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<div class="markdown"><p>We have <mathjax>#C(+III)#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, the oxidation state is the charge left on the central atom, when all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> atoms are removed with the charge (the electrons) devolved to the most electronegative atom.....</p>
<p>We gots <mathjax>#""^(-)O(O=)C-C(=O)O^(-)#</mathjax>. The oxidation number of oxygen is usually <mathjax>#-II#</mathjax> and it is here. The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> equals the charge on the ion.....here <mathjax>#-2#</mathjax>.</p>
<p>And so <mathjax>#2xxC_"oxidation number"+4xx(-2)=-2#</mathjax></p>
<p>And <mathjax>#2xxC_"oxidation number"=+6#</mathjax></p>
<p>And <mathjax>#C_"oxidation number"=+III#</mathjax> if you will forgive me mixing Roman and Arabic numerals.........of course you will. </p>
<p>Another way we could look at this is to split the <mathjax>#C-C#</mathjax> bond in oxalate ion to give <mathjax>#2xx""^(-)O(O=)C*#</mathjax>, <mathjax>#C(+III)#</mathjax> as required. And thus the carbon in oxalic acid is almost fully oxidized. </p></div>
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<div class="markdown"><p>answer = 3 [oxidation state]</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Oxygen is one of the most electronegative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> after flourine so it shows 2 as oxidation state for most <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> [exception - hydrogen peroxide <mathjax>#H2O2#</mathjax>]<br/>
SO IN <mathjax>#C2O4^(2-)#</mathjax> Oxidation state of oxygen must be -2 . <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> on compound is (-2) . <br/>
oxidation state of carbon will be positive .<br/>
.°. 2(oxidation state of carbon) + 4 ( oxidation state of oxygen) = -2<br/>
2 (C) + 4 (-2) = -2<br/>
2(C) = 6<br/>
OXIDATION STATE OF CARBON IS <mathjax>#6/2#</mathjax> = 3.</p></div>
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</article> | What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#? | null |
2,348 | aaf0b23b-6ddd-11ea-9274-ccda262736ce | https://socratic.org/questions/how-many-moles-of-nh-3-can-be-produced-from-21-0-mol-of-h-2-and-excess-n-2 | 14 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] NH3 [IN] moles"}] | [{"type":"physical unit","value":"14 moles"}] | [{"type":"physical unit","value":"Mole [OF] H2 [=] \\pu{21.0 mol}"},{"type":"other","value":"Excess N2."}] | <h1 class="questionTitle" itemprop="name">How many moles of #NH_3# can be produced from 21.0 mol of #H_2# and excess #N_2#? </h1> | null | 14 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichimetric equation:</p>
<p><mathjax>#N_2(g) + 3H_2(g)rarr2NH_3(g)#</mathjax></p>
<p>The stoichiometric equation explicitly states that each mol, each equiv, dinitrogen, requires 3 mol of dihydrogen.</p>
<p>Given that we have <mathjax>#21.0*mol" dihydrogen"#</mathjax>, <mathjax>#14*mol#</mathjax> of ammonia could be produced given stoichiometric reaction. Of course, in this reaction, stoichiometric yields are unthinkable, but industrial procedures can recycle the unreacted dinitrogen and dihydrogen and achieve acceptable turnovers. That ammonia is condensable, whereas the reactant gases are much less so, allows recycling. </p></div>
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<div class="markdown"><p>Clearly, <mathjax>#14*mol#</mathjax> of ammonia could be produced. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichimetric equation:</p>
<p><mathjax>#N_2(g) + 3H_2(g)rarr2NH_3(g)#</mathjax></p>
<p>The stoichiometric equation explicitly states that each mol, each equiv, dinitrogen, requires 3 mol of dihydrogen.</p>
<p>Given that we have <mathjax>#21.0*mol" dihydrogen"#</mathjax>, <mathjax>#14*mol#</mathjax> of ammonia could be produced given stoichiometric reaction. Of course, in this reaction, stoichiometric yields are unthinkable, but industrial procedures can recycle the unreacted dinitrogen and dihydrogen and achieve acceptable turnovers. That ammonia is condensable, whereas the reactant gases are much less so, allows recycling. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #NH_3# can be produced from 21.0 mol of #H_2# and excess #N_2#? </h1>
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<div class="markdown"><p>Clearly, <mathjax>#14*mol#</mathjax> of ammonia could be produced. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The stoichimetric equation:</p>
<p><mathjax>#N_2(g) + 3H_2(g)rarr2NH_3(g)#</mathjax></p>
<p>The stoichiometric equation explicitly states that each mol, each equiv, dinitrogen, requires 3 mol of dihydrogen.</p>
<p>Given that we have <mathjax>#21.0*mol" dihydrogen"#</mathjax>, <mathjax>#14*mol#</mathjax> of ammonia could be produced given stoichiometric reaction. Of course, in this reaction, stoichiometric yields are unthinkable, but industrial procedures can recycle the unreacted dinitrogen and dihydrogen and achieve acceptable turnovers. That ammonia is condensable, whereas the reactant gases are much less so, allows recycling. </p></div>
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</article> | How many moles of #NH_3# can be produced from 21.0 mol of #H_2# and excess #N_2#? | null |
2,349 | a9611dd0-6ddd-11ea-830a-ccda262736ce | https://socratic.org/questions/a-gas-at-pressure-of-5-0-bar-from-0-to-546-c-and-simultaneously-compressed-to-on | 45 bar | start physical_unit 1 1 pressure bar qc_end physical_unit 1 1 5 6 pressure qc_end physical_unit 1 1 8 9 temperature qc_end physical_unit 1 1 11 12 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] bar"}] | [{"type":"physical unit","value":"45 bar"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{5.0 bar}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{546 ℃}"},{"type":"other","value":"A gas simultaneously compressed to one third of its original volume."}] | <h1 class="questionTitle" itemprop="name">A gas at pressure of 5.0 bar from 0 to 546°c and simultaneously compressed to one third of its original volume.What will be the final pressure in bar?</h1> | null | 45 bar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Initial Pressure of the gas"=P_1=5" bar"#</mathjax></p>
<p><mathjax>#"Initial Volume of the gas"=V_1 L#</mathjax></p>
<p><mathjax>#"Initial Temperature of the gas"=T_1=(0+273)K=273K#</mathjax></p>
<p><mathjax>#"Final pressure of the gas"=P_2=?#</mathjax></p>
<p><mathjax>#"Final Volume of the gas"=V_2 L#</mathjax></p>
<p><mathjax>#"Final Temperature of the gas"=T_2=(546+273)K=819K#</mathjax></p>
<p>Given condition is the is compressed to one third of its original volume. </p>
<p>So <mathjax>#V_1=3V_2#</mathjax></p>
<p>We know from gas law </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#(P_2V_2)/T_2=(P_1V_1)/T_1#</mathjax></p>
<p><mathjax>#(P_2V_2)/819=(5xx3V_2)/273#</mathjax></p>
<p><mathjax>#P_2=(5xx3xxcancel819^3)/cancel273" bar"=45" bar"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote></div>
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<div class="markdown"><p>Final pressure =5 bar</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Initial Pressure of the gas"=P_1=5" bar"#</mathjax></p>
<p><mathjax>#"Initial Volume of the gas"=V_1 L#</mathjax></p>
<p><mathjax>#"Initial Temperature of the gas"=T_1=(0+273)K=273K#</mathjax></p>
<p><mathjax>#"Final pressure of the gas"=P_2=?#</mathjax></p>
<p><mathjax>#"Final Volume of the gas"=V_2 L#</mathjax></p>
<p><mathjax>#"Final Temperature of the gas"=T_2=(546+273)K=819K#</mathjax></p>
<p>Given condition is the is compressed to one third of its original volume. </p>
<p>So <mathjax>#V_1=3V_2#</mathjax></p>
<p>We know from gas law </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#(P_2V_2)/T_2=(P_1V_1)/T_1#</mathjax></p>
<p><mathjax>#(P_2V_2)/819=(5xx3V_2)/273#</mathjax></p>
<p><mathjax>#P_2=(5xx3xxcancel819^3)/cancel273" bar"=45" bar"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A gas at pressure of 5.0 bar from 0 to 546°c and simultaneously compressed to one third of its original volume.What will be the final pressure in bar?</h1>
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<span class="dateCreated" datetime="2016-10-01T08:29:23" itemprop="dateCreated">
Oct 1, 2016
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<div class="markdown"><p>Final pressure =5 bar</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Initial Pressure of the gas"=P_1=5" bar"#</mathjax></p>
<p><mathjax>#"Initial Volume of the gas"=V_1 L#</mathjax></p>
<p><mathjax>#"Initial Temperature of the gas"=T_1=(0+273)K=273K#</mathjax></p>
<p><mathjax>#"Final pressure of the gas"=P_2=?#</mathjax></p>
<p><mathjax>#"Final Volume of the gas"=V_2 L#</mathjax></p>
<p><mathjax>#"Final Temperature of the gas"=T_2=(546+273)K=819K#</mathjax></p>
<p>Given condition is the is compressed to one third of its original volume. </p>
<p>So <mathjax>#V_1=3V_2#</mathjax></p>
<p>We know from gas law </p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#(P_2V_2)/T_2=(P_1V_1)/T_1#</mathjax></p>
<p><mathjax>#(P_2V_2)/819=(5xx3V_2)/273#</mathjax></p>
<p><mathjax>#P_2=(5xx3xxcancel819^3)/cancel273" bar"=45" bar"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote></div>
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</article> | A gas at pressure of 5.0 bar from 0 to 546°c and simultaneously compressed to one third of its original volume.What will be the final pressure in bar? | null |
2,350 | ab88386c-6ddd-11ea-86b3-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-is-made-by-diluting-50-00-ml-of-a-4-74-m | 0.95 M | start physical_unit 20 20 molarity mol/l qc_end physical_unit 20 20 16 17 molarity qc_end physical_unit 6 6 12 13 volume qc_end physical_unit 6 6 22 23 volume qc_end end | [{"type":"physical unit","value":"Molarity2 [OF] HCl solution [IN] M"}] | [{"type":"physical unit","value":"0.95 M"}] | [{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{4.74 M}"},{"type":"physical unit","value":"Volume1 [OF] HCl solution [=] \\pu{50.00 mL}"},{"type":"physical unit","value":"Volume2 [OF] HCl solution [=] \\pu{250.00 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution that is made by diluting
50.00 mL of a 4.74 M solution of HCl to 250.00 mL?</h1> | null | 0.95 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use this equation</p>
<blockquote>
<p><mathjax>#"M"_1"V"_1 = "M"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"M"_2 = ("M"_1"V"_1)/("V"_2) = ("4.74 M"× 50.00 cancel"mL")/(250.00 cancel"mL") = "0.948 M"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"0.948 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use this equation</p>
<blockquote>
<p><mathjax>#"M"_1"V"_1 = "M"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"M"_2 = ("M"_1"V"_1)/("V"_2) = ("4.74 M"× 50.00 cancel"mL")/(250.00 cancel"mL") = "0.948 M"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution that is made by diluting
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<div class="markdown"><p><mathjax>#"0.948 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use this equation</p>
<blockquote>
<p><mathjax>#"M"_1"V"_1 = "M"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"M"_2 = ("M"_1"V"_1)/("V"_2) = ("4.74 M"× 50.00 cancel"mL")/(250.00 cancel"mL") = "0.948 M"#</mathjax></p></div>
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</article> | What is the molarity of a solution that is made by diluting
50.00 mL of a 4.74 M solution of HCl to 250.00 mL? | null |
2,351 | a9866468-6ddd-11ea-8123-ccda262736ce | https://socratic.org/questions/the-reaction-between-cerium-iv-ions-and-nitride-ions-can-be-expressed-in-the-for | 2Ce^4+ + NO2- + H2O -> NO3- + 2 H+ 2 Ce^3+ | start chemical_equation qc_end chemical_equation 25 29 qc_end chemical_equation 31 40 qc_end end | [{"type":"other","value":"Chemical Equation [OF] complete stoichiometric equation"}] | [{"type":"chemical equation","value":"2Ce^4+ + NO2- + H2O -> NO3- + 2 H+ 2 Ce^3+"}] | [{"type":"chemical equation","value":"Ce^4+ + e- -> Ce^3+"},{"type":"chemical equation","value":"NO2- + H2O -> NO3- + 2 H+ 2 e-"}] | <h1 class="questionTitle" itemprop="name">Write the complete stoichiometric equation for the reaction?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction between cerium(IV) ions and nitride ions can be expressed in the form of two half-equations:<br/>
<mathjax>#"Ce"^(4+) + e^- to "Ce"^(3+)#</mathjax><br/>
<mathjax>#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#</mathjax></p></div>
</h2>
</div>
</div> | 2Ce^4+ + NO2- + H2O -> NO3- + 2 H+ 2 Ce^3+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Half-reactions:</p>
<p>Reduction: <mathjax>#"Ce"^(4+) + e^- to "Ce"^(3+)#</mathjax><br/>
Oxidation: <mathjax>#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#</mathjax></p>
<p>For each mole of the half-reaction:<br/>
the reduction half <em>consumes</em> one mole of electrons, whereas<br/>
the oxidation half <em>produces</em> two moles of electrons.</p>
<p>Adding both halves of the reaction shall give the complete equation for the redox reaction; however, since electrons are neither eliminated nor generated in any chemical process, a balanced equation shall contain no electrons. </p>
<p>Multiplying the first half-reaction with a coefficient of two shall eliminate <mathjax>#e^-#</mathjax> from the net equation, giving </p>
<p><mathjax>#color(blue)(2)"Ce"^(4+) + color(green)(cancel[color(blue)(2)color(black)(e^(-))])+"NO"_2^(-) + "H"_2"O" to color(blue)(2)"Ce"^(3+)+"NO"_3^(-) + 2"H"^(+) + color(green)(cancel[color(black)(2)color(black)(e^(-))])#</mathjax><br/>
<mathjax>#2"Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H"^(+)#</mathjax></p>
<p>Double-check the equation:<br/>
Make sure that on the two sides of the equation</p>
<ul>
<li>electrons cancel out;</li>
<li>charges balance;</li>
<li>number of atoms balance.</li>
</ul></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H"^(+)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Half-reactions:</p>
<p>Reduction: <mathjax>#"Ce"^(4+) + e^- to "Ce"^(3+)#</mathjax><br/>
Oxidation: <mathjax>#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#</mathjax></p>
<p>For each mole of the half-reaction:<br/>
the reduction half <em>consumes</em> one mole of electrons, whereas<br/>
the oxidation half <em>produces</em> two moles of electrons.</p>
<p>Adding both halves of the reaction shall give the complete equation for the redox reaction; however, since electrons are neither eliminated nor generated in any chemical process, a balanced equation shall contain no electrons. </p>
<p>Multiplying the first half-reaction with a coefficient of two shall eliminate <mathjax>#e^-#</mathjax> from the net equation, giving </p>
<p><mathjax>#color(blue)(2)"Ce"^(4+) + color(green)(cancel[color(blue)(2)color(black)(e^(-))])+"NO"_2^(-) + "H"_2"O" to color(blue)(2)"Ce"^(3+)+"NO"_3^(-) + 2"H"^(+) + color(green)(cancel[color(black)(2)color(black)(e^(-))])#</mathjax><br/>
<mathjax>#2"Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H"^(+)#</mathjax></p>
<p>Double-check the equation:<br/>
Make sure that on the two sides of the equation</p>
<ul>
<li>electrons cancel out;</li>
<li>charges balance;</li>
<li>number of atoms balance.</li>
</ul></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Write the complete stoichiometric equation for the reaction?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction between cerium(IV) ions and nitride ions can be expressed in the form of two half-equations:<br/>
<mathjax>#"Ce"^(4+) + e^- to "Ce"^(3+)#</mathjax><br/>
<mathjax>#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#</mathjax></p></div>
</h2>
</div>
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<div class="markdown"><p><mathjax>#2"Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H"^(+)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Half-reactions:</p>
<p>Reduction: <mathjax>#"Ce"^(4+) + e^- to "Ce"^(3+)#</mathjax><br/>
Oxidation: <mathjax>#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#</mathjax></p>
<p>For each mole of the half-reaction:<br/>
the reduction half <em>consumes</em> one mole of electrons, whereas<br/>
the oxidation half <em>produces</em> two moles of electrons.</p>
<p>Adding both halves of the reaction shall give the complete equation for the redox reaction; however, since electrons are neither eliminated nor generated in any chemical process, a balanced equation shall contain no electrons. </p>
<p>Multiplying the first half-reaction with a coefficient of two shall eliminate <mathjax>#e^-#</mathjax> from the net equation, giving </p>
<p><mathjax>#color(blue)(2)"Ce"^(4+) + color(green)(cancel[color(blue)(2)color(black)(e^(-))])+"NO"_2^(-) + "H"_2"O" to color(blue)(2)"Ce"^(3+)+"NO"_3^(-) + 2"H"^(+) + color(green)(cancel[color(black)(2)color(black)(e^(-))])#</mathjax><br/>
<mathjax>#2"Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H"^(+)#</mathjax></p>
<p>Double-check the equation:<br/>
Make sure that on the two sides of the equation</p>
<ul>
<li>electrons cancel out;</li>
<li>charges balance;</li>
<li>number of atoms balance.</li>
</ul></div>
</div>
</div>
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</article> | Write the complete stoichiometric equation for the reaction? |
The reaction between cerium(IV) ions and nitride ions can be expressed in the form of two half-equations:
#"Ce"^(4+) + e^- to "Ce"^(3+)#
#"NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H"^(+) + 2e^-#
|
2,352 | ac40f06c-6ddd-11ea-a02e-ccda262736ce | https://socratic.org/questions/how-many-moles-of-h-2so-4-are-present-in-1-63-liters-of-a-954-m-solution | 1.56 moles | start physical_unit 4 4 mole mol qc_end physical_unit 14 14 8 9 volume qc_end physical_unit 4 4 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] H2SO4 [IN] moles"}] | [{"type":"physical unit","value":"1.56 moles"}] | [{"type":"physical unit","value":"Volume [OF] H2SO4 solution [=] \\pu{1.63 liters}"},{"type":"physical unit","value":"Molarity [OF] H2SO4 solution [=] \\pu{0.954 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #H_2SO_4# are present in 1.63 liters of a .954 M solution?</h1> | null | 1.56 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#"Concentration"="Moles"/"Volume"#</mathjax>, and thus by manipulating this expression, we get, <mathjax>#"Volume"="Moles"/"Concentration"#</mathjax>, or, as we require here..............</p>
<p><mathjax>#"Moles"="Volume"xx"Concentration"#</mathjax>; and so we calculate the product.........</p>
<p><mathjax>#1.63*Lxx0.954*mol*L^-1=1.63*cancelLxx0.954*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=1.56*mol#</mathjax> of <mathjax>#H_2SO_4#</mathjax>. </p>
<p>Such a product is dimensionally consistent; we wanted an answer in moles and the product gave us one, which persuades us that we have done the right thing arithmetically. </p>
<p>What is the concentration with respect to <mathjax>#H^+#</mathjax> or <mathjax>#H_3O^+#</mathjax>? </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Here, there are <mathjax>#1.56*mol#</mathjax> with respect to <mathjax>#H_2SO_4#</mathjax>.........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#"Concentration"="Moles"/"Volume"#</mathjax>, and thus by manipulating this expression, we get, <mathjax>#"Volume"="Moles"/"Concentration"#</mathjax>, or, as we require here..............</p>
<p><mathjax>#"Moles"="Volume"xx"Concentration"#</mathjax>; and so we calculate the product.........</p>
<p><mathjax>#1.63*Lxx0.954*mol*L^-1=1.63*cancelLxx0.954*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=1.56*mol#</mathjax> of <mathjax>#H_2SO_4#</mathjax>. </p>
<p>Such a product is dimensionally consistent; we wanted an answer in moles and the product gave us one, which persuades us that we have done the right thing arithmetically. </p>
<p>What is the concentration with respect to <mathjax>#H^+#</mathjax> or <mathjax>#H_3O^+#</mathjax>? </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #H_2SO_4# are present in 1.63 liters of a .954 M solution?</h1>
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<div class="markdown"><p>Here, there are <mathjax>#1.56*mol#</mathjax> with respect to <mathjax>#H_2SO_4#</mathjax>.........</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#"Concentration"="Moles"/"Volume"#</mathjax>, and thus by manipulating this expression, we get, <mathjax>#"Volume"="Moles"/"Concentration"#</mathjax>, or, as we require here..............</p>
<p><mathjax>#"Moles"="Volume"xx"Concentration"#</mathjax>; and so we calculate the product.........</p>
<p><mathjax>#1.63*Lxx0.954*mol*L^-1=1.63*cancelLxx0.954*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=1.56*mol#</mathjax> of <mathjax>#H_2SO_4#</mathjax>. </p>
<p>Such a product is dimensionally consistent; we wanted an answer in moles and the product gave us one, which persuades us that we have done the right thing arithmetically. </p>
<p>What is the concentration with respect to <mathjax>#H^+#</mathjax> or <mathjax>#H_3O^+#</mathjax>? </p></div>
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</article> | How many moles of #H_2SO_4# are present in 1.63 liters of a .954 M solution? | null |
2,353 | a9e504da-6ddd-11ea-b682-ccda262736ce | https://socratic.org/questions/how-many-moles-of-copper-are-equivalent-to-3-44-10-23-atoms-of-copper | 0.57 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] copper [IN] moles"}] | [{"type":"physical unit","value":"0.57 moles"}] | [{"type":"physical unit","value":"Number [OF] copper atoms [=] \\pu{3.44 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles of copper are equivalent to #3.44*10^23# atoms of copper?</h1> | null | 0.57 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to find the conversion factor. We know that 1 mol. contains Avogadro's number of atoms and therefore our conversion factor will be: </p>
<p><mathjax># \ 1 " mol." = 6.02xx10^23 " atoms"#</mathjax></p>
<p><mathjax>#3.44xx10^23 \ "atoms" xx (1 \ "mol.")/(6.02xx10^23 \ "atoms")#</mathjax></p>
<p><mathjax>#3.44xx10^23\ cancel("atoms") xx (1 \ "mol.")/(6.02xx10^23 \ cancel("atoms"))#</mathjax></p>
<p><mathjax>#0.571 " mol."#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#0.571 " mol."#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to find the conversion factor. We know that 1 mol. contains Avogadro's number of atoms and therefore our conversion factor will be: </p>
<p><mathjax># \ 1 " mol." = 6.02xx10^23 " atoms"#</mathjax></p>
<p><mathjax>#3.44xx10^23 \ "atoms" xx (1 \ "mol.")/(6.02xx10^23 \ "atoms")#</mathjax></p>
<p><mathjax>#3.44xx10^23\ cancel("atoms") xx (1 \ "mol.")/(6.02xx10^23 \ cancel("atoms"))#</mathjax></p>
<p><mathjax>#0.571 " mol."#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of copper are equivalent to #3.44*10^23# atoms of copper?</h1>
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<div class="markdown"><p><mathjax>#0.571 " mol."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to find the conversion factor. We know that 1 mol. contains Avogadro's number of atoms and therefore our conversion factor will be: </p>
<p><mathjax># \ 1 " mol." = 6.02xx10^23 " atoms"#</mathjax></p>
<p><mathjax>#3.44xx10^23 \ "atoms" xx (1 \ "mol.")/(6.02xx10^23 \ "atoms")#</mathjax></p>
<p><mathjax>#3.44xx10^23\ cancel("atoms") xx (1 \ "mol.")/(6.02xx10^23 \ cancel("atoms"))#</mathjax></p>
<p><mathjax>#0.571 " mol."#</mathjax></p></div>
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</article> | How many moles of copper are equivalent to #3.44*10^23# atoms of copper? | null |
2,354 | acf1649c-6ddd-11ea-b41e-ccda262736ce | https://socratic.org/questions/a-substance-contains-36-1-percent-calcium-and-63-9-percent-chlorine-by-weight-wh | CaCl2 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the given compound [IN] empirical"}] | [{"type":"chemical equation","value":"CaCl2"}] | [{"type":"physical unit","value":"Percent by weight [OF] calcium in the given compound [=] \\pu{36.1 percent}"},{"type":"physical unit","value":"Percent by weight [OF] chlorine in the given compound [=] \\pu{63.9 percent}"}] | <h1 class="questionTitle" itemprop="name">A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compound?</h1> | null | CaCl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! SHORT ANSWER !!</strong></p>
<p>Grab a periodic table and look at the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You have</p>
<blockquote>
<p><mathjax>#"For Ca: " "40.078 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Cl: " "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Notice that the two molar masses are relatively close to each other. Now, the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound tells you that you get about <strong>one part</strong> calcium <strong>per two parts</strong> chlorine <em>by mass</em>. </p>
<p>You can thus say that for <strong>every mole</strong> of this compound, you will get <strong>one mole</strong> of calcium and <strong>two moles</strong> of chlorine. Therefore, the empirical formula for this compound will be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)()#</mathjax><br/>
<strong>!! LONG ANSWER !!</strong></p>
<p>Here's how you can double-check your result. Let's assume that you have a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li>
<p><mathjax>#"36.1 g"#</mathjax> <em>of calcium</em>, <mathjax>#"Ca"#</mathjax></p>
</li>
<li>
<p><mathjax>#"63.9 g"#</mathjax> <em>of chlorine</em>, <mathjax>#"Cl"#</mathjax></p>
</li>
</ul>
</blockquote>
<p>Use the molar masses of the two elements to figure out how many <strong>moles</strong> of each you have in this sample </p>
<blockquote>
<p><mathjax>#"For Ca: " 36.1 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.901 moles Ca"#</mathjax></p>
<p><mathjax>#"For Cl: " 63.9 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "1.802 moles Cl"#</mathjax></p>
</blockquote>
<p>To get <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between the two elements in the compound, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For Ca: " (0.901 color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (1.802color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 2#</mathjax></p>
</blockquote>
<p>Once again, the empirical formula of the compound, which tells you the <strong>smallest whole number ratio</strong> that exists between the elements that make up the compound, comes out to be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"CaCl"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! SHORT ANSWER !!</strong></p>
<p>Grab a periodic table and look at the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You have</p>
<blockquote>
<p><mathjax>#"For Ca: " "40.078 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Cl: " "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Notice that the two molar masses are relatively close to each other. Now, the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound tells you that you get about <strong>one part</strong> calcium <strong>per two parts</strong> chlorine <em>by mass</em>. </p>
<p>You can thus say that for <strong>every mole</strong> of this compound, you will get <strong>one mole</strong> of calcium and <strong>two moles</strong> of chlorine. Therefore, the empirical formula for this compound will be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)()#</mathjax><br/>
<strong>!! LONG ANSWER !!</strong></p>
<p>Here's how you can double-check your result. Let's assume that you have a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li>
<p><mathjax>#"36.1 g"#</mathjax> <em>of calcium</em>, <mathjax>#"Ca"#</mathjax></p>
</li>
<li>
<p><mathjax>#"63.9 g"#</mathjax> <em>of chlorine</em>, <mathjax>#"Cl"#</mathjax></p>
</li>
</ul>
</blockquote>
<p>Use the molar masses of the two elements to figure out how many <strong>moles</strong> of each you have in this sample </p>
<blockquote>
<p><mathjax>#"For Ca: " 36.1 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.901 moles Ca"#</mathjax></p>
<p><mathjax>#"For Cl: " 63.9 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "1.802 moles Cl"#</mathjax></p>
</blockquote>
<p>To get <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between the two elements in the compound, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For Ca: " (0.901 color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (1.802color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 2#</mathjax></p>
</blockquote>
<p>Once again, the empirical formula of the compound, which tells you the <strong>smallest whole number ratio</strong> that exists between the elements that make up the compound, comes out to be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compound?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-05-21T16:11:26" itemprop="dateCreated">
May 21, 2016
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<div class="markdown"><p><mathjax>#"CaCl"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! SHORT ANSWER !!</strong></p>
<p>Grab a periodic table and look at the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You have</p>
<blockquote>
<p><mathjax>#"For Ca: " "40.078 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Cl: " "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Notice that the two molar masses are relatively close to each other. Now, the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of the compound tells you that you get about <strong>one part</strong> calcium <strong>per two parts</strong> chlorine <em>by mass</em>. </p>
<p>You can thus say that for <strong>every mole</strong> of this compound, you will get <strong>one mole</strong> of calcium and <strong>two moles</strong> of chlorine. Therefore, the empirical formula for this compound will be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)()#</mathjax><br/>
<strong>!! LONG ANSWER !!</strong></p>
<p>Here's how you can double-check your result. Let's assume that you have a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li>
<p><mathjax>#"36.1 g"#</mathjax> <em>of calcium</em>, <mathjax>#"Ca"#</mathjax></p>
</li>
<li>
<p><mathjax>#"63.9 g"#</mathjax> <em>of chlorine</em>, <mathjax>#"Cl"#</mathjax></p>
</li>
</ul>
</blockquote>
<p>Use the molar masses of the two elements to figure out how many <strong>moles</strong> of each you have in this sample </p>
<blockquote>
<p><mathjax>#"For Ca: " 36.1 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.901 moles Ca"#</mathjax></p>
<p><mathjax>#"For Cl: " 63.9 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "1.802 moles Cl"#</mathjax></p>
</blockquote>
<p>To get <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between the two elements in the compound, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For Ca: " (0.901 color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (1.802color(red)(cancel(color(black)("moles"))))/(0.901color(red)(cancel(color(black)("moles")))) = 2#</mathjax></p>
</blockquote>
<p>Once again, the empirical formula of the compound, which tells you the <strong>smallest whole number ratio</strong> that exists between the elements that make up the compound, comes out to be </p>
<blockquote>
<p><mathjax>#"Ca"_1"Cl"_2 implies color(green)(|bar(ul(color(white)(a/a)"CaCl"_2color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compound? | null |
2,355 | a9e089dc-6ddd-11ea-8e81-ccda262736ce | https://socratic.org/questions/you-know-that-a-gas-in-a-sealed-container-has-a-pressure-of-111-kpa-at-231-c-wha | 164.74 kPa | start physical_unit 4 4 pressure kpa qc_end physical_unit 4 4 13 14 pressure qc_end physical_unit 4 4 16 17 temperature qc_end physical_unit 4 4 28 29 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] kPa"}] | [{"type":"physical unit","value":"164.74 kPa"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{111 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{231 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{475 ℃}"}] | <h1 class="questionTitle" itemprop="name">You know that a gas in a sealed container has a pressure of 111 kPa at 231°C. What will the pressure be if the temperature rises to 475°C?</h1> | null | 164.74 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, this is a sealed container which means that the volume is constant. Since the only change happening is the raise of temperature, therefore, the number of mole is also constant.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can say that at <mathjax>#V=cte#</mathjax> and <mathjax>#n=cte#</mathjax> the pressure <mathjax>#P#</mathjax> is directly proportional to the temperature <mathjax>#T#</mathjax>.</p>
<p>Therefore, <mathjax>#P=kT or P/T=k#</mathjax>, where <mathjax>#k#</mathjax> is a constant <mathjax>#k=(nR)/V#</mathjax>.</p>
<p>Thus, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>P_2=(P_1xxT_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(111kPaxx748cancel(K))/(504cancel(K))=165kPa#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_2=165kPa#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, this is a sealed container which means that the volume is constant. Since the only change happening is the raise of temperature, therefore, the number of mole is also constant.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can say that at <mathjax>#V=cte#</mathjax> and <mathjax>#n=cte#</mathjax> the pressure <mathjax>#P#</mathjax> is directly proportional to the temperature <mathjax>#T#</mathjax>.</p>
<p>Therefore, <mathjax>#P=kT or P/T=k#</mathjax>, where <mathjax>#k#</mathjax> is a constant <mathjax>#k=(nR)/V#</mathjax>.</p>
<p>Thus, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>P_2=(P_1xxT_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(111kPaxx748cancel(K))/(504cancel(K))=165kPa#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">You know that a gas in a sealed container has a pressure of 111 kPa at 231°C. What will the pressure be if the temperature rises to 475°C?</h1>
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Dr. Hayek
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<div class="markdown"><p><mathjax>#P_2=165kPa#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, this is a sealed container which means that the volume is constant. Since the only change happening is the raise of temperature, therefore, the number of mole is also constant.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can say that at <mathjax>#V=cte#</mathjax> and <mathjax>#n=cte#</mathjax> the pressure <mathjax>#P#</mathjax> is directly proportional to the temperature <mathjax>#T#</mathjax>.</p>
<p>Therefore, <mathjax>#P=kT or P/T=k#</mathjax>, where <mathjax>#k#</mathjax> is a constant <mathjax>#k=(nR)/V#</mathjax>.</p>
<p>Thus, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>P_2=(P_1xxT_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(111kPaxx748cancel(K))/(504cancel(K))=165kPa#</mathjax></p></div>
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</article> | You know that a gas in a sealed container has a pressure of 111 kPa at 231°C. What will the pressure be if the temperature rises to 475°C? | null |
2,356 | a8b3cee8-6ddd-11ea-b843-ccda262736ce | https://socratic.org/questions/in-the-reaction-h-2s-2o-2-h-2so-4-the-law-of-definite-proportions-predicts-that- | 2.00 moles | start physical_unit 6 6 mole mol qc_end chemical_equation 3 8 qc_end end | [{"type":"physical unit","value":"Mole [OF] O2 [IN] moles"}] | [{"type":"physical unit","value":"2.00 moles"}] | [{"type":"chemical equation","value":"H2S + 2 O2 -> H2SO4"},{"type":"physical unit","value":"Mole [OF] H2S [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">In the reaction #H_2S + 2O_2 -> H_2SO_4#, the law of definite proportions predicts that for every mole of #H_2S#, you will need how many moles of #O_2#?</h1> | null | 2.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would look at your equation again. You have even balanced and formatted it properly. What does it say? It says 1 molecule of hydrogen sulfide reacts with 2 molecules of oxygen to make 1 molecule of sulfuric acid. </p>
<p>What is the relationship between molecules and moles? A mole is simply Avogadro's number of molecules, a very large number, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>. In the reaction you have given above, I could CORRECTLY say that 1 molecule of hydrogen sulfide reacts, or 1 MOLE of hydrogen sulfide reacts. </p>
<p>Please note that sulfuric acid is not produced in this way (from so-called conc. sour gas which sends the environmentalists screaming), and I doubt hydrogen sulfide can be oxidized up to to sulfuric acid easily. </p></div>
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<div class="markdown"><p>Clearly it is 2 moles. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would look at your equation again. You have even balanced and formatted it properly. What does it say? It says 1 molecule of hydrogen sulfide reacts with 2 molecules of oxygen to make 1 molecule of sulfuric acid. </p>
<p>What is the relationship between molecules and moles? A mole is simply Avogadro's number of molecules, a very large number, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>. In the reaction you have given above, I could CORRECTLY say that 1 molecule of hydrogen sulfide reacts, or 1 MOLE of hydrogen sulfide reacts. </p>
<p>Please note that sulfuric acid is not produced in this way (from so-called conc. sour gas which sends the environmentalists screaming), and I doubt hydrogen sulfide can be oxidized up to to sulfuric acid easily. </p></div>
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<h1 class="questionTitle" itemprop="name">In the reaction #H_2S + 2O_2 -> H_2SO_4#, the law of definite proportions predicts that for every mole of #H_2S#, you will need how many moles of #O_2#?</h1>
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<div class="markdown"><p>Clearly it is 2 moles. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I would look at your equation again. You have even balanced and formatted it properly. What does it say? It says 1 molecule of hydrogen sulfide reacts with 2 molecules of oxygen to make 1 molecule of sulfuric acid. </p>
<p>What is the relationship between molecules and moles? A mole is simply Avogadro's number of molecules, a very large number, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>. In the reaction you have given above, I could CORRECTLY say that 1 molecule of hydrogen sulfide reacts, or 1 MOLE of hydrogen sulfide reacts. </p>
<p>Please note that sulfuric acid is not produced in this way (from so-called conc. sour gas which sends the environmentalists screaming), and I doubt hydrogen sulfide can be oxidized up to to sulfuric acid easily. </p></div>
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</article> | In the reaction #H_2S + 2O_2 -> H_2SO_4#, the law of definite proportions predicts that for every mole of #H_2S#, you will need how many moles of #O_2#? | null |
2,357 | a860d112-6ddd-11ea-acc9-ccda262736ce | https://socratic.org/questions/58a24ddf11ef6b74754025e6 | 2 kPa | start physical_unit 44 45 pressure kpa qc_end physical_unit 44 45 14 15 pressure qc_end physical_unit 44 45 1 2 volume qc_end physical_unit 44 45 20 21 temperature qc_end physical_unit 44 45 28 29 temperature qc_end physical_unit 44 45 36 37 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] kPa"}] | [{"type":"physical unit","value":"2 kPa"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{100 kPa}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{73 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{80 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{4530 mL}"}] | <h1 class="questionTitle" itemprop="name">A #73*mL# volume of gas enclosed in a piston at a pressure of #100*kPa#, and a temperature of #0# #""^@C# was heated to a temperature of #80# #""^@C# and the new volume occupied was #4530*mL#. What is the new pressure that the gas exerts?</h1> | null | 2 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature must be quoted on the <mathjax>#"Kelvin scale"#</mathjax>, and so...</p>
<p><mathjax>#P_2=(P_1V_1)/T_1xxT_2/V_2#</mathjax>, which will certainly give us an answer with the required units of pressure. Why so?</p>
<p>And thus...........</p>
<p><mathjax>#P_2=(100*kPaxx73.0*mL)/(273.15*K)xx(353.15*K)/(4530*mL)#</mathjax></p>
<p><mathjax>#~=2*kPa#</mathjax></p>
<p>The pressure change is marked because we increase the volume GREATLY........i.e. more than 50 times the original volume, whereas the temperature is increased only marginally. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well for a given molar quantity of gas, <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature must be quoted on the <mathjax>#"Kelvin scale"#</mathjax>, and so...</p>
<p><mathjax>#P_2=(P_1V_1)/T_1xxT_2/V_2#</mathjax>, which will certainly give us an answer with the required units of pressure. Why so?</p>
<p>And thus...........</p>
<p><mathjax>#P_2=(100*kPaxx73.0*mL)/(273.15*K)xx(353.15*K)/(4530*mL)#</mathjax></p>
<p><mathjax>#~=2*kPa#</mathjax></p>
<p>The pressure change is marked because we increase the volume GREATLY........i.e. more than 50 times the original volume, whereas the temperature is increased only marginally. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A #73*mL# volume of gas enclosed in a piston at a pressure of #100*kPa#, and a temperature of #0# #""^@C# was heated to a temperature of #80# #""^@C# and the new volume occupied was #4530*mL#. What is the new pressure that the gas exerts?</h1>
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<div class="markdown"><p>Well for a given molar quantity of gas, <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature must be quoted on the <mathjax>#"Kelvin scale"#</mathjax>, and so...</p>
<p><mathjax>#P_2=(P_1V_1)/T_1xxT_2/V_2#</mathjax>, which will certainly give us an answer with the required units of pressure. Why so?</p>
<p>And thus...........</p>
<p><mathjax>#P_2=(100*kPaxx73.0*mL)/(273.15*K)xx(353.15*K)/(4530*mL)#</mathjax></p>
<p><mathjax>#~=2*kPa#</mathjax></p>
<p>The pressure change is marked because we increase the volume GREATLY........i.e. more than 50 times the original volume, whereas the temperature is increased only marginally. </p></div>
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</article> | A #73*mL# volume of gas enclosed in a piston at a pressure of #100*kPa#, and a temperature of #0# #""^@C# was heated to a temperature of #80# #""^@C# and the new volume occupied was #4530*mL#. What is the new pressure that the gas exerts? | null |
2,358 | a8abe60a-6ddd-11ea-9e3c-ccda262736ce | https://socratic.org/questions/59488c8b11ef6b2bcad62087 | 1.71 g | start physical_unit 3 4 mass g qc_end physical_unit 13 14 9 10 volume qc_end physical_unit 13 14 16 17 concentration qc_end physical_unit 27 28 23 24 volume qc_end physical_unit 27 28 30 31 concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] lead chloride [IN] g"}] | [{"type":"physical unit","value":"1.71 g"}] | [{"type":"physical unit","value":"Volume [OF] lead nitrate [=] \\pu{21.5 mL}"},{"type":"physical unit","value":"Concentration [OF] lead nitrate [=] \\pu{0.0806 mol/L}"},{"type":"physical unit","value":"Volume [OF] potassium chloride [=] \\pu{18.00 mL}"},{"type":"physical unit","value":"Concentration [OF] potassium chloride [=] \\pu{0.683 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What mass of #"lead chloride"# would result if a #21.5*mL# volume of #"lead nitrate"# at #0.0806*mol*L^-1# concentration were mixed with a #18.00*mL# volume of #"potassium chloride"# at #0.683*mol*L^-1# concentration?</h1> | null | 1.71 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)#</mathjax></p>
<p>Lead chloride is one of the few water-insoluble halides. It crashes out of water solution with alacrity.</p>
<p>And (ii) we need equivalent quantities of each reagent......</p>
<p><mathjax>#"Moles of lead nitrate"=21.5xx10^-3*Lxx0.806*mol*L^-1=0.0173*mol#</mathjax></p>
<p><mathjax>#"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*mol#</mathjax>.</p>
<p>And thus (clearly) chloride anion is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> predicts that half an equiv of plumbous chloride will precipitate.</p>
<p><mathjax>#"Mass of"#</mathjax> <mathjax>#PbCl_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.0123*mol)/2xx278.10*g*mol^-1#</mathjax></p>
<p><mathjax>#=1.71*g#</mathjax> </p>
<p>It is unusual that the chloride ion is in stoichiometric deficiency in that you would think we want to salt out the lead content......</p></div>
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<div class="markdown"><p>Under a <mathjax>#2*g#</mathjax> mass................</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)#</mathjax></p>
<p>Lead chloride is one of the few water-insoluble halides. It crashes out of water solution with alacrity.</p>
<p>And (ii) we need equivalent quantities of each reagent......</p>
<p><mathjax>#"Moles of lead nitrate"=21.5xx10^-3*Lxx0.806*mol*L^-1=0.0173*mol#</mathjax></p>
<p><mathjax>#"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*mol#</mathjax>.</p>
<p>And thus (clearly) chloride anion is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> predicts that half an equiv of plumbous chloride will precipitate.</p>
<p><mathjax>#"Mass of"#</mathjax> <mathjax>#PbCl_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.0123*mol)/2xx278.10*g*mol^-1#</mathjax></p>
<p><mathjax>#=1.71*g#</mathjax> </p>
<p>It is unusual that the chloride ion is in stoichiometric deficiency in that you would think we want to salt out the lead content......</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass of #"lead chloride"# would result if a #21.5*mL# volume of #"lead nitrate"# at #0.0806*mol*L^-1# concentration were mixed with a #18.00*mL# volume of #"potassium chloride"# at #0.683*mol*L^-1# concentration?</h1>
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<div class="markdown"><p>Under a <mathjax>#2*g#</mathjax> mass................</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation.....</p>
<p><mathjax>#Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)#</mathjax></p>
<p>Lead chloride is one of the few water-insoluble halides. It crashes out of water solution with alacrity.</p>
<p>And (ii) we need equivalent quantities of each reagent......</p>
<p><mathjax>#"Moles of lead nitrate"=21.5xx10^-3*Lxx0.806*mol*L^-1=0.0173*mol#</mathjax></p>
<p><mathjax>#"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*mol#</mathjax>.</p>
<p>And thus (clearly) chloride anion is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> predicts that half an equiv of plumbous chloride will precipitate.</p>
<p><mathjax>#"Mass of"#</mathjax> <mathjax>#PbCl_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.0123*mol)/2xx278.10*g*mol^-1#</mathjax></p>
<p><mathjax>#=1.71*g#</mathjax> </p>
<p>It is unusual that the chloride ion is in stoichiometric deficiency in that you would think we want to salt out the lead content......</p></div>
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</article> | What mass of #"lead chloride"# would result if a #21.5*mL# volume of #"lead nitrate"# at #0.0806*mol*L^-1# concentration were mixed with a #18.00*mL# volume of #"potassium chloride"# at #0.683*mol*L^-1# concentration? | null |
2,359 | a8c35e8b-6ddd-11ea-9651-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-potassium-phosphide | K3P | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] potassium phosphide [IN] default"}] | [{"type":"chemical equation","value":"K3P"}] | [{"type":"substance name","value":"Potassium phosphide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for potassium phosphide?</h1> | null | K3P | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium almost always has a charge of + 1<br/>
Phosphorus has charges of + 3, -3, + 5. <br/>
In this case Phosphorus will act as an Oxidizing agent with Potassium. Meaning that it will take electrons from Potassium <br/>
resulting in a negative three charge. </p>
<p><mathjax># 3 xx + 1 = 1 xx -3#</mathjax> so it will take three potassium ions to balance one Phosphorus ion. So the formula is </p>
<p><mathjax># K_3P#</mathjax></p></div>
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<div class="markdown"><p><mathjax># K_3P#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium almost always has a charge of + 1<br/>
Phosphorus has charges of + 3, -3, + 5. <br/>
In this case Phosphorus will act as an Oxidizing agent with Potassium. Meaning that it will take electrons from Potassium <br/>
resulting in a negative three charge. </p>
<p><mathjax># 3 xx + 1 = 1 xx -3#</mathjax> so it will take three potassium ions to balance one Phosphorus ion. So the formula is </p>
<p><mathjax># K_3P#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for potassium phosphide?</h1>
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<div class="markdown"><p><mathjax># K_3P#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium almost always has a charge of + 1<br/>
Phosphorus has charges of + 3, -3, + 5. <br/>
In this case Phosphorus will act as an Oxidizing agent with Potassium. Meaning that it will take electrons from Potassium <br/>
resulting in a negative three charge. </p>
<p><mathjax># 3 xx + 1 = 1 xx -3#</mathjax> so it will take three potassium ions to balance one Phosphorus ion. So the formula is </p>
<p><mathjax># K_3P#</mathjax></p></div>
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</article> | What is the formula for potassium phosphide? | null |
2,360 | ab32414c-6ddd-11ea-9fab-ccda262736ce | https://socratic.org/questions/if-i-start-with-5-grams-of-c-3h-8-what-is-the-theoretical-yield-of-water-in-the- | 8.18 g | start physical_unit 14 14 theoretical_yield g qc_end chemical_equation 18 27 qc_end physical_unit 7 7 4 5 mass qc_end end | [{"type":"physical unit","value":"Theoretical yield [OF] water [IN] g"}] | [{"type":"physical unit","value":"8.18 g"}] | [{"type":"chemical equation","value":"C3H8 + 5 O2 -> 3 CO2 + 4 H2O"},{"type":"physical unit","value":"Mass [OF] C3H8 [=] \\pu{5 grams}"}] | <h1 class="questionTitle" itemprop="name">If I start with 5 grams of #C_3H_8#, what is the theoretical yield of water in the equation #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#?</h1> | null | 8.18 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>the molar mass of<mathjax>#C_3H_8=3xx12+8xx1=44gmmol^=1#</mathjax><br/>
molar mass of water=<mathjax>#18gmmol^-1#</mathjax></p>
<p>As per the equation 44 gm<mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72gm#</mathjax> water<br/>
1gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44gm#</mathjax> water<br/>
5gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44xx5~~8.18gm#</mathjax> water</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#~~8.18gm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>the molar mass of<mathjax>#C_3H_8=3xx12+8xx1=44gmmol^=1#</mathjax><br/>
molar mass of water=<mathjax>#18gmmol^-1#</mathjax></p>
<p>As per the equation 44 gm<mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72gm#</mathjax> water<br/>
1gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44gm#</mathjax> water<br/>
5gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44xx5~~8.18gm#</mathjax> water</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If I start with 5 grams of #C_3H_8#, what is the theoretical yield of water in the equation #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#?</h1>
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Mar 11, 2016
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<div class="markdown"><p><mathjax>#~~8.18gm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>the molar mass of<mathjax>#C_3H_8=3xx12+8xx1=44gmmol^=1#</mathjax><br/>
molar mass of water=<mathjax>#18gmmol^-1#</mathjax></p>
<p>As per the equation 44 gm<mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72gm#</mathjax> water<br/>
1gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44gm#</mathjax> water<br/>
5gm <mathjax>#C_3H_8#</mathjax> gives <mathjax>#4xx18=72/44xx5~~8.18gm#</mathjax> water</p></div>
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</article> | If I start with 5 grams of #C_3H_8#, what is the theoretical yield of water in the equation #C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O#? | null |
2,361 | ac097f71-6ddd-11ea-bc96-ccda262736ce | https://socratic.org/questions/to-what-volume-should-you-dilute-20-ml-of-a-12-m-h-2so-4-solution-to-obtain-a-0- | 1500 mL | start physical_unit 12 13 volume ml qc_end physical_unit 12 13 6 7 volume qc_end physical_unit 12 13 10 11 molarity qc_end physical_unit 12 13 17 18 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] H2SO4 solution [IN] mL"}] | [{"type":"physical unit","value":"1500 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] H2SO4 solution [=] \\pu{20 mL}"},{"type":"physical unit","value":"Molarity1 [OF] H2SO4 solution [=] \\pu{12 M}"},{"type":"physical unit","value":"Molarity2 [OF] H2SO4 solution [=] \\pu{0.16 M}"}] | <h1 class="questionTitle" itemprop="name"> To what volume should you dilute 20 mL of a 12 M #H_2SO_4# solution to obtain a 0.16 M #H_2SO_4# solution?</h1> | null | 1500 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> is by definition the number of moles of stuff, which is fixed according to the volume and the concentration of your starting material.</p>
<p>We started with <mathjax>#20*mL#</mathjax> of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#"sulfuric acid"#</mathjax>.</p>
<p>And thus <mathjax>#(20*mLxx12*cancel(mol*L^-1))/(0.16*cancel(mol*L^-1))=V_2=1500*mL#</mathjax></p>
<p>PS When you do the dilution, do you add the acid to the water? Why?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#C_1V_1=C_2V_2#</mathjax>.</p>
<p>A final volume of <mathjax>#1.500*L#</mathjax> is required. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> is by definition the number of moles of stuff, which is fixed according to the volume and the concentration of your starting material.</p>
<p>We started with <mathjax>#20*mL#</mathjax> of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#"sulfuric acid"#</mathjax>.</p>
<p>And thus <mathjax>#(20*mLxx12*cancel(mol*L^-1))/(0.16*cancel(mol*L^-1))=V_2=1500*mL#</mathjax></p>
<p>PS When you do the dilution, do you add the acid to the water? Why?</p></div>
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<h1 class="questionTitle" itemprop="name"> To what volume should you dilute 20 mL of a 12 M #H_2SO_4# solution to obtain a 0.16 M #H_2SO_4# solution?</h1>
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<div class="markdown"><p><mathjax>#C_1V_1=C_2V_2#</mathjax>.</p>
<p>A final volume of <mathjax>#1.500*L#</mathjax> is required. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> is by definition the number of moles of stuff, which is fixed according to the volume and the concentration of your starting material.</p>
<p>We started with <mathjax>#20*mL#</mathjax> of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#"sulfuric acid"#</mathjax>.</p>
<p>And thus <mathjax>#(20*mLxx12*cancel(mol*L^-1))/(0.16*cancel(mol*L^-1))=V_2=1500*mL#</mathjax></p>
<p>PS When you do the dilution, do you add the acid to the water? Why?</p></div>
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</article> | To what volume should you dilute 20 mL of a 12 M #H_2SO_4# solution to obtain a 0.16 M #H_2SO_4# solution? | null |
2,362 | a8e111cc-6ddd-11ea-a91e-ccda262736ce | https://socratic.org/questions/how-do-you-balance-ircl-3-ca-oh-2-ir-2o-3-hcl-cacl-2 | 2 IrCl3 + 3 Ca(OH)2 -> Ir2O3 + 3 H2O + 3 CaCl2 | start chemical_equation qc_end chemical_equation 4 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 IrCl3 + 3 Ca(OH)2 -> Ir2O3 + 3 H2O + 3 CaCl2"}] | [{"type":"chemical equation","value":"IrCl3 + Ca(OH)2 -> Ir2O3 + H2O + CaCl2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #IrCl_3 + Ca(OH)_2 -> Ir_2O_3 + HCl + CaCl_2#?</h1> | null | 2 IrCl3 + 3 Ca(OH)2 -> Ir2O3 + 3 H2O + 3 CaCl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As Ernest wrote, <mathjax>#HCl#</mathjax> will not result from the reaction. While the reaction product would be exceedingly difficult to characterize, an <mathjax>#Ir(OH)_3#</mathjax> species would probably precipitate to give a hydrous oxide of <mathjax>#Ir(III)#</mathjax>:</p>
<p><mathjax>#Ir^(3+) + 3HO^(-) rarr 2Ir(OH)_3 rarr Ir_2O_3 + 3H_2O#</mathjax></p>
<p>Of course if you did this reaction you would have to collect all of the solid. Not only is iridium a heavy metal, it is also a precious metal, that is worth a lot of money per gram. </p></div>
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<div class="markdown"><p><mathjax>#2IrCl_3 + 3Ca(OH)_2 rarr Ir_2O_3 + 3CaCl_2 + 3H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As Ernest wrote, <mathjax>#HCl#</mathjax> will not result from the reaction. While the reaction product would be exceedingly difficult to characterize, an <mathjax>#Ir(OH)_3#</mathjax> species would probably precipitate to give a hydrous oxide of <mathjax>#Ir(III)#</mathjax>:</p>
<p><mathjax>#Ir^(3+) + 3HO^(-) rarr 2Ir(OH)_3 rarr Ir_2O_3 + 3H_2O#</mathjax></p>
<p>Of course if you did this reaction you would have to collect all of the solid. Not only is iridium a heavy metal, it is also a precious metal, that is worth a lot of money per gram. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #IrCl_3 + Ca(OH)_2 -> Ir_2O_3 + HCl + CaCl_2#?</h1>
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<div class="markdown"><p><mathjax>#2IrCl_3 + 3Ca(OH)_2 rarr Ir_2O_3 + 3CaCl_2 + 3H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As Ernest wrote, <mathjax>#HCl#</mathjax> will not result from the reaction. While the reaction product would be exceedingly difficult to characterize, an <mathjax>#Ir(OH)_3#</mathjax> species would probably precipitate to give a hydrous oxide of <mathjax>#Ir(III)#</mathjax>:</p>
<p><mathjax>#Ir^(3+) + 3HO^(-) rarr 2Ir(OH)_3 rarr Ir_2O_3 + 3H_2O#</mathjax></p>
<p>Of course if you did this reaction you would have to collect all of the solid. Not only is iridium a heavy metal, it is also a precious metal, that is worth a lot of money per gram. </p></div>
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</article> | How do you balance #IrCl_3 + Ca(OH)_2 -> Ir_2O_3 + HCl + CaCl_2#? | null |
2,363 | a9177930-6ddd-11ea-bed6-ccda262736ce | https://socratic.org/questions/aballoon-that-had-a-volume-of-3-50-l-at-25-0-c-is-placed-in-a-hot-roomat-40-0-c- | 3.68 L | start physical_unit 35 36 volume l qc_end physical_unit 35 36 7 8 volume qc_end physical_unit 35 36 10 11 temperature qc_end physical_unit 35 36 27 28 pressure qc_end physical_unit 35 36 19 20 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] L"}] | [{"type":"physical unit","value":"3.68 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{3.50 L}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{25.0 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the balloon [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the balloon [=] \\pu{40.0 ℃}"},{"type":"other","value":"constant pressure"}] | <h1 class="questionTitle" itemprop="name">A balloon that had a volume of 3.50 L at 25.0 °C is placed in a hot room at 40.0 °C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?</h1> | null | 3.68 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a Charles Law problem </p>
<blockquote>
<p><mathjax># V_1/T_1 = V_2/T_2#</mathjax> </p>
<p><mathjax># V_1 = "3.5 L"#</mathjax>,</p>
<p><mathjax># T_1 = 25 + 273 = "298 K"#</mathjax> ,</p>
<p><mathjax>#V_2 = "unknown L"#</mathjax>,</p>
<p><mathjax># T_2 = 40 + 273 = "313 K"#</mathjax>,</p>
</blockquote>
<p>so </p>
<blockquote>
<p><mathjax># ("3.5 L")/("298 K") = V_2/("313 K")#</mathjax> .</p>
</blockquote>
<p>This gives </p>
<blockquote>
<p><mathjax># ("3.5 L") xx ("313 K")/("298 K") = V_2#</mathjax>.</p>
<p><mathjax>#=> "3.68 L" = V_2#</mathjax> </p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>3.68 liters to three significant numbers. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a Charles Law problem </p>
<blockquote>
<p><mathjax># V_1/T_1 = V_2/T_2#</mathjax> </p>
<p><mathjax># V_1 = "3.5 L"#</mathjax>,</p>
<p><mathjax># T_1 = 25 + 273 = "298 K"#</mathjax> ,</p>
<p><mathjax>#V_2 = "unknown L"#</mathjax>,</p>
<p><mathjax># T_2 = 40 + 273 = "313 K"#</mathjax>,</p>
</blockquote>
<p>so </p>
<blockquote>
<p><mathjax># ("3.5 L")/("298 K") = V_2/("313 K")#</mathjax> .</p>
</blockquote>
<p>This gives </p>
<blockquote>
<p><mathjax># ("3.5 L") xx ("313 K")/("298 K") = V_2#</mathjax>.</p>
<p><mathjax>#=> "3.68 L" = V_2#</mathjax> </p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A balloon that had a volume of 3.50 L at 25.0 °C is placed in a hot room at 40.0 °C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?</h1>
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<span class="dateCreated" datetime="2017-06-08T20:14:24" itemprop="dateCreated">
Jun 8, 2017
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<div class="markdown"><p>3.68 liters to three significant numbers. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a Charles Law problem </p>
<blockquote>
<p><mathjax># V_1/T_1 = V_2/T_2#</mathjax> </p>
<p><mathjax># V_1 = "3.5 L"#</mathjax>,</p>
<p><mathjax># T_1 = 25 + 273 = "298 K"#</mathjax> ,</p>
<p><mathjax>#V_2 = "unknown L"#</mathjax>,</p>
<p><mathjax># T_2 = 40 + 273 = "313 K"#</mathjax>,</p>
</blockquote>
<p>so </p>
<blockquote>
<p><mathjax># ("3.5 L")/("298 K") = V_2/("313 K")#</mathjax> .</p>
</blockquote>
<p>This gives </p>
<blockquote>
<p><mathjax># ("3.5 L") xx ("313 K")/("298 K") = V_2#</mathjax>.</p>
<p><mathjax>#=> "3.68 L" = V_2#</mathjax> </p>
</blockquote></div>
</div>
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<a href="https://socratic.org/answers/436740" itemprop="url">Answer link</a>
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</article> | A balloon that had a volume of 3.50 L at 25.0 °C is placed in a hot room at 40.0 °C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room? | null |
2,364 | aa843b40-6ddd-11ea-a039-ccda262736ce | https://socratic.org/questions/how-many-grams-of-20-zinc-oxide-ointment-would-contain-10-g-of-zinc-oxide | 50.00 grams | start physical_unit 5 7 mass g qc_end physical_unit 5 6 10 11 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] zinc oxide ointment [IN] grams"}] | [{"type":"physical unit","value":"50.00 grams"}] | [{"type":"physical unit","value":"Mass [OF] zinc oxide [=] \\pu{10 g}"},{"type":"physical unit","value":"Percent [OF] zinc oxide in ointment [=] \\pu{20%}"}] | <h1 class="questionTitle" itemprop="name"> How many grams of 20% zinc oxide ointment would contain 10 g of zinc oxide?</h1> | null | 50.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's check, <mathjax>#50*gxx20%(m_"ZnO"/m_"cream")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10*g#</mathjax> <mathjax>#ZnO#</mathjax>.</p>
<p>Do you know what the major use of this cream is?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#10*g#</mathjax> of zinc oxide, are contained in <mathjax>#50*g#</mathjax> of <mathjax>#20%#</mathjax> ointment. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Let's check, <mathjax>#50*gxx20%(m_"ZnO"/m_"cream")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10*g#</mathjax> <mathjax>#ZnO#</mathjax>.</p>
<p>Do you know what the major use of this cream is?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name"> How many grams of 20% zinc oxide ointment would contain 10 g of zinc oxide?</h1>
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anor277
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<div class="markdown"><p><mathjax>#10*g#</mathjax> of zinc oxide, are contained in <mathjax>#50*g#</mathjax> of <mathjax>#20%#</mathjax> ointment. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Let's check, <mathjax>#50*gxx20%(m_"ZnO"/m_"cream")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10*g#</mathjax> <mathjax>#ZnO#</mathjax>.</p>
<p>Do you know what the major use of this cream is?</p></div>
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</article> | How many grams of 20% zinc oxide ointment would contain 10 g of zinc oxide? | null |
2,365 | a99d6c1e-6ddd-11ea-9468-ccda262736ce | https://socratic.org/questions/salt-nacl-has-a-solubility-of-35-7g-per-100g-of-water-at-0-degrees-c-what-is-the | -22.7 ℃ | start physical_unit 25 25 melting_point_temperature °c qc_end physical_unit 1 1 6 7 solubility qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 14 16 temperature qc_end physical_unit 40 40 42 43 kf qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Lowest possible melting point [OF] ice [IN] ℃"}] | [{"type":"physical unit","value":"-22.7 ℃"}] | [{"type":"physical unit","value":"Solubility [OF] NaCl [=] \\pu{35.7 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 g}"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{0 degrees C}"},{"type":"physical unit","value":"Kf [OF] H2O [=] \\pu{1.86 C/m}"},{"type":"other","value":"Considering the solubility of NaCl in water."}] | <h1 class="questionTitle" itemprop="name">Salt, NaCl, has a solubility of 35.7g per 100g of water at 0 degrees C. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCl in water? The Kf for H2O is 1.86 C/m</h1> | null | -22.7 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the equation that expresses the <strong>freezing-point depression</strong> of a solution, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<p>In your case, the cryoscopic constant of water is said to be </p>
<blockquote>
<p><mathjax>#K_f = 1.86^@"C kg mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, the idea here is that you can only dissolve <mathjax>#"35.7 g"#</mathjax> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, in water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The thing to remember about sodium chloride is that the compound is a <strong>strong electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"^(-)#</mathjax></p>
</blockquote>
<p>This means that the <em>van't Hoff factor</em>, which tells you the ratio that exists between the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that are <em>dissolved</em> and the number of moles of <strong>solute particles</strong> produced in solution, will be equal to </p>
<blockquote>
<p><mathjax>#i = 2#</mathjax></p>
</blockquote>
<p>That is the case because <strong>one mole</strong> of sodium chloride will dissociate completely to form <strong>one mole</strong> of sodium cations and <strong>one mole</strong> of chloride anions, hence <strong>two moles</strong> of particles. </p>
<p>Now, let's assume that you're working with <mathjax>#"100 g"#</mathjax> of water at <mathjax>#0^@"C"#</mathjax>. Use sodium chloride's <strong>molar mass</strong> to determine how many <em>moles</em> of sodium chloride you'd get in that <mathjax>#"35.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#35.7 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443 color(red)(cancel(color(black)("g")))) = "0.61085 moles NaCl"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of a solution tells you how many moles of solute you get <strong>per kilogram of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have </p>
<blockquote>
<p><mathjax>#b = "0.61085 moles"/(100 * 10^(-3)"kg") = "6.1085 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>The freezing-point depression of this solution would be</p>
<blockquote>
<p><mathjax>#DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 6.1085 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_f = 22.72^@"C"#</mathjax><br/>
The freezing-point depression is defined as </p>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"f sol"#</mathjax> - the freezing point of the solution</p>
<p>The freezing point of the solution will thus be </p>
<blockquote>
<p><mathjax>#T_"f sol" = T_f^@ - DeltaT_f#</mathjax></p>
<p><mathjax>#T_"f sol" = 0^@"C" - 22.72^@"C" = color(green)(|bar(ul(color(white)(a/a)-22.7^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#-22.7^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the equation that expresses the <strong>freezing-point depression</strong> of a solution, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<p>In your case, the cryoscopic constant of water is said to be </p>
<blockquote>
<p><mathjax>#K_f = 1.86^@"C kg mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, the idea here is that you can only dissolve <mathjax>#"35.7 g"#</mathjax> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, in water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The thing to remember about sodium chloride is that the compound is a <strong>strong electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"^(-)#</mathjax></p>
</blockquote>
<p>This means that the <em>van't Hoff factor</em>, which tells you the ratio that exists between the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that are <em>dissolved</em> and the number of moles of <strong>solute particles</strong> produced in solution, will be equal to </p>
<blockquote>
<p><mathjax>#i = 2#</mathjax></p>
</blockquote>
<p>That is the case because <strong>one mole</strong> of sodium chloride will dissociate completely to form <strong>one mole</strong> of sodium cations and <strong>one mole</strong> of chloride anions, hence <strong>two moles</strong> of particles. </p>
<p>Now, let's assume that you're working with <mathjax>#"100 g"#</mathjax> of water at <mathjax>#0^@"C"#</mathjax>. Use sodium chloride's <strong>molar mass</strong> to determine how many <em>moles</em> of sodium chloride you'd get in that <mathjax>#"35.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#35.7 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443 color(red)(cancel(color(black)("g")))) = "0.61085 moles NaCl"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of a solution tells you how many moles of solute you get <strong>per kilogram of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have </p>
<blockquote>
<p><mathjax>#b = "0.61085 moles"/(100 * 10^(-3)"kg") = "6.1085 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>The freezing-point depression of this solution would be</p>
<blockquote>
<p><mathjax>#DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 6.1085 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_f = 22.72^@"C"#</mathjax><br/>
The freezing-point depression is defined as </p>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"f sol"#</mathjax> - the freezing point of the solution</p>
<p>The freezing point of the solution will thus be </p>
<blockquote>
<p><mathjax>#T_"f sol" = T_f^@ - DeltaT_f#</mathjax></p>
<p><mathjax>#T_"f sol" = 0^@"C" - 22.72^@"C" = color(green)(|bar(ul(color(white)(a/a)-22.7^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Salt, NaCl, has a solubility of 35.7g per 100g of water at 0 degrees C. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCl in water? The Kf for H2O is 1.86 C/m</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-05-07T20:11:07" itemprop="dateCreated">
May 7, 2016
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<div class="markdown"><p><mathjax>#-22.7^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the equation that expresses the <strong>freezing-point depression</strong> of a solution, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<p>In your case, the cryoscopic constant of water is said to be </p>
<blockquote>
<p><mathjax>#K_f = 1.86^@"C kg mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, the idea here is that you can only dissolve <mathjax>#"35.7 g"#</mathjax> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, in water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The thing to remember about sodium chloride is that the compound is a <strong>strong electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"^(-)#</mathjax></p>
</blockquote>
<p>This means that the <em>van't Hoff factor</em>, which tells you the ratio that exists between the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that are <em>dissolved</em> and the number of moles of <strong>solute particles</strong> produced in solution, will be equal to </p>
<blockquote>
<p><mathjax>#i = 2#</mathjax></p>
</blockquote>
<p>That is the case because <strong>one mole</strong> of sodium chloride will dissociate completely to form <strong>one mole</strong> of sodium cations and <strong>one mole</strong> of chloride anions, hence <strong>two moles</strong> of particles. </p>
<p>Now, let's assume that you're working with <mathjax>#"100 g"#</mathjax> of water at <mathjax>#0^@"C"#</mathjax>. Use sodium chloride's <strong>molar mass</strong> to determine how many <em>moles</em> of sodium chloride you'd get in that <mathjax>#"35.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#35.7 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443 color(red)(cancel(color(black)("g")))) = "0.61085 moles NaCl"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of a solution tells you how many moles of solute you get <strong>per kilogram of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have </p>
<blockquote>
<p><mathjax>#b = "0.61085 moles"/(100 * 10^(-3)"kg") = "6.1085 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>The freezing-point depression of this solution would be</p>
<blockquote>
<p><mathjax>#DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 6.1085 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_f = 22.72^@"C"#</mathjax><br/>
The freezing-point depression is defined as </p>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"f sol"#</mathjax> - the freezing point of the solution</p>
<p>The freezing point of the solution will thus be </p>
<blockquote>
<p><mathjax>#T_"f sol" = T_f^@ - DeltaT_f#</mathjax></p>
<p><mathjax>#T_"f sol" = 0^@"C" - 22.72^@"C" = color(green)(|bar(ul(color(white)(a/a)-22.7^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | Salt, NaCl, has a solubility of 35.7g per 100g of water at 0 degrees C. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCl in water? The Kf for H2O is 1.86 C/m | null |
2,366 | aa58470c-6ddd-11ea-97d7-ccda262736ce | https://socratic.org/questions/how-to-determine-the-molecular-formula-of-this-organic-compound | C4H8O2 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 30 31 27 28 mass qc_end physical_unit 36 37 33 34 mass qc_end physical_unit 42 42 39 40 mass qc_end physical_unit 54 54 52 53 volume qc_end physical_unit 54 54 56 57 temperature qc_end physical_unit 54 54 62 63 pressure qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] this organic compound [IN] molecular"}] | [{"type":"chemical equation","value":"C4H8O2"}] | [{"type":"other","value":"Completely oxidized in air."},{"type":"physical unit","value":"Mass1 [OF] the compound [=] \\pu{0.900 g}"},{"type":"physical unit","value":"Mass [OF] carbon dioxide [=] \\pu{1.8 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{0.736 g}"},{"type":"physical unit","value":"Mass2 [OF] the separate sample [=] \\pu{2.279 g}"},{"type":"physical unit","value":"Volume [OF] vessel [=] \\pu{1.00 dm^3}"},{"type":"physical unit","value":"Temperature [OF] vessel [=] \\pu{100 ℃}"},{"type":"physical unit","value":"Pressure [OF] vessel [=] \\pu{84 kPa}"},{"type":"other","value":"An organic compound containing only carbon, hydrogen and oxygen was analysed gravimetrically."}] | <h1 class="questionTitle" itemprop="name">How to determine the molecular formula of this organic compound?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>An organic compound containing only carbon, hydrogen and oxygen was analysed gravimetrically. When completely oxidized in air, <mathjax>#0.900 g#</mathjax> of the compound produced <mathjax>#1.80 g#</mathjax> of carbon dioxide and <mathjax>#0.736 g#</mathjax> of water. A separate <mathjax>#2.279 g#</mathjax> sample, when vaporized in a <mathjax>#1.00 dm3#</mathjax> vessel at <mathjax>#100°C#</mathjax>, had a pressure of <mathjax>#84 kPa#</mathjax>. Determine the molecular formula of the compound.</p></div>
</h2>
</div>
</div> | C4H8O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Calculate the empirical formula.</strong></p>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 1.80 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.4912 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.736 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0823 g H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "0.900 g - 0.4912 g - 0.0823 g" = "0.3265 g"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.4912 color(white)(mll)"0.040 90"
color(white)(Xll)2.004color(white)(mm)2)#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXl)0.0823 color(white)(mll)"0.0816" color(white)(mml)4.00 color(white)(Xmll)4#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)0.3265color(white)(mll)"0.020 41"color(white)(mll)1color(white)(mmmll)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_2"H"_4"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the molar mass</strong></p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (mRT)/(PV)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#m = "2.279 g"#</mathjax><br/>
<mathjax>#R = "8.314 kPa·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(100 + 273.15) K" = "373.15 K"#</mathjax><br/>
<mathjax>#P = "84 kPa"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("2.279 g" × 8.314 color(red)(cancel(color(black)("kPa·dm"^3"·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/(84 color(red)(cancel(color(black)("kPa")))) = "84.2 g/mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the molecular formula</strong></p>
<p>The empirical formula mass of <mathjax>#"C"_2"H"_4"O"#</mathjax> is 44.05 u.</p>
<p>The molecular mass is 84.2 u.</p>
<p>The molecular mass must be an integral multiple of the empirical formula mass.</p>
<p><mathjax>#"MM"/"EFM" = (84.2 color(red)(cancel(color(black)("u"))))/(44.05 color(red)(cancel(color(black)("u")))) = 1.91 ≈ 2#</mathjax></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p><mathjax>#"Molecular formula" = ("C"_2"H"_4"O")_2 = "C"_4"H"_8"O"_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>WARNING! Long answer!</strong> The molecular formula is <mathjax>#"C"_4"H"_8"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Calculate the empirical formula.</strong></p>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 1.80 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.4912 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.736 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0823 g H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "0.900 g - 0.4912 g - 0.0823 g" = "0.3265 g"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.4912 color(white)(mll)"0.040 90"
color(white)(Xll)2.004color(white)(mm)2)#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXl)0.0823 color(white)(mll)"0.0816" color(white)(mml)4.00 color(white)(Xmll)4#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)0.3265color(white)(mll)"0.020 41"color(white)(mll)1color(white)(mmmll)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_2"H"_4"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the molar mass</strong></p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (mRT)/(PV)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#m = "2.279 g"#</mathjax><br/>
<mathjax>#R = "8.314 kPa·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(100 + 273.15) K" = "373.15 K"#</mathjax><br/>
<mathjax>#P = "84 kPa"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("2.279 g" × 8.314 color(red)(cancel(color(black)("kPa·dm"^3"·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/(84 color(red)(cancel(color(black)("kPa")))) = "84.2 g/mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the molecular formula</strong></p>
<p>The empirical formula mass of <mathjax>#"C"_2"H"_4"O"#</mathjax> is 44.05 u.</p>
<p>The molecular mass is 84.2 u.</p>
<p>The molecular mass must be an integral multiple of the empirical formula mass.</p>
<p><mathjax>#"MM"/"EFM" = (84.2 color(red)(cancel(color(black)("u"))))/(44.05 color(red)(cancel(color(black)("u")))) = 1.91 ≈ 2#</mathjax></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p><mathjax>#"Molecular formula" = ("C"_2"H"_4"O")_2 = "C"_4"H"_8"O"_2#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How to determine the molecular formula of this organic compound?</h1>
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<div class="markdown"><p>An organic compound containing only carbon, hydrogen and oxygen was analysed gravimetrically. When completely oxidized in air, <mathjax>#0.900 g#</mathjax> of the compound produced <mathjax>#1.80 g#</mathjax> of carbon dioxide and <mathjax>#0.736 g#</mathjax> of water. A separate <mathjax>#2.279 g#</mathjax> sample, when vaporized in a <mathjax>#1.00 dm3#</mathjax> vessel at <mathjax>#100°C#</mathjax>, had a pressure of <mathjax>#84 kPa#</mathjax>. Determine the molecular formula of the compound.</p></div>
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Ernest Z.
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<div class="markdown"><p><strong>WARNING! Long answer!</strong> The molecular formula is <mathjax>#"C"_4"H"_8"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Calculate the empirical formula.</strong></p>
<p>We can calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 1.80 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.4912 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 0.736 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0823 g H"#</mathjax></p>
<p><mathjax>#"Mass of O" = "Mass of compound - mass of C - mass of O" = "0.900 g - 0.4912 g - 0.0823 g" = "0.3265 g"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.4912 color(white)(mll)"0.040 90"
color(white)(Xll)2.004color(white)(mm)2)#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXl)0.0823 color(white)(mll)"0.0816" color(white)(mml)4.00 color(white)(Xmll)4#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(mmmml)0.3265color(white)(mll)"0.020 41"color(white)(mll)1color(white)(mmmll)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_2"H"_4"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the molar mass</strong></p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (mRT)/(PV)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#m = "2.279 g"#</mathjax><br/>
<mathjax>#R = "8.314 kPa·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(100 + 273.15) K" = "373.15 K"#</mathjax><br/>
<mathjax>#P = "84 kPa"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("2.279 g" × 8.314 color(red)(cancel(color(black)("kPa·dm"^3"·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/(84 color(red)(cancel(color(black)("kPa")))) = "84.2 g/mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculate the molecular formula</strong></p>
<p>The empirical formula mass of <mathjax>#"C"_2"H"_4"O"#</mathjax> is 44.05 u.</p>
<p>The molecular mass is 84.2 u.</p>
<p>The molecular mass must be an integral multiple of the empirical formula mass.</p>
<p><mathjax>#"MM"/"EFM" = (84.2 color(red)(cancel(color(black)("u"))))/(44.05 color(red)(cancel(color(black)("u")))) = 1.91 ≈ 2#</mathjax></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p><mathjax>#"Molecular formula" = ("C"_2"H"_4"O")_2 = "C"_4"H"_8"O"_2#</mathjax></p></div>
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</article> | How to determine the molecular formula of this organic compound? |
An organic compound containing only carbon, hydrogen and oxygen was analysed gravimetrically. When completely oxidized in air, #0.900 g# of the compound produced #1.80 g# of carbon dioxide and #0.736 g# of water. A separate #2.279 g# sample, when vaporized in a #1.00 dm3# vessel at #100°C#, had a pressure of #84 kPa#. Determine the molecular formula of the compound.
|
2,367 | a8b11024-6ddd-11ea-8fc2-ccda262736ce | https://socratic.org/questions/how-many-moles-of-n-are-in-0-193-g-of-n-2o | 8.78 × 10^(−3) moles | start physical_unit 4 4 mole mol qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] N [IN] moles"}] | [{"type":"physical unit","value":"8.78 × 10^(−3) moles"}] | [{"type":"physical unit","value":"Mass [OF] N2O [=] \\pu{0.193 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #N# are in 0.193 g of #N_2O#? </h1> | null | 8.78 × 10^(−3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.193*g)/(44.01*g*mol^-1)#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*mol.#</mathjax></p>
<p>And thus <mathjax>#"moles of nitrogen atoms"#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*molxx2=??#</mathjax></p>
<p>How can I convert this molar quantity into a number of nitrogen atoms?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>There are <mathjax>#=8.77xx10^-3*mol" of nitrogen atoms.........."#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.193*g)/(44.01*g*mol^-1)#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*mol.#</mathjax></p>
<p>And thus <mathjax>#"moles of nitrogen atoms"#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*molxx2=??#</mathjax></p>
<p>How can I convert this molar quantity into a number of nitrogen atoms?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #N# are in 0.193 g of #N_2O#? </h1>
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anor277
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<div class="markdown"><p>There are <mathjax>#=8.77xx10^-3*mol" of nitrogen atoms.........."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.193*g)/(44.01*g*mol^-1)#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*mol.#</mathjax></p>
<p>And thus <mathjax>#"moles of nitrogen atoms"#</mathjax></p>
<p><mathjax>#=4.39xx10^-3*molxx2=??#</mathjax></p>
<p>How can I convert this molar quantity into a number of nitrogen atoms?</p></div>
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</article> | How many moles of #N# are in 0.193 g of #N_2O#? | null |
2,368 | a8d4ea98-6ddd-11ea-bd57-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-5-7-mol-potassium-permanganate | 900.79 grams | 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] potassium permanganate [IN] grams"}] | [{"type":"physical unit","value":"900.79 grams"}] | [{"type":"physical unit","value":"Mole [OF] potassium permanganate [=] \\pu{5.7 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in 5.7 mol potassium permanganate?</h1> | null | 900.79 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The required mass is simply the product, <mathjax>#"moles "xx" molar mass"#</mathjax>, i.e. <mathjax>#5.7*cancel(mol)xx158.034*g*cancel(mol^-1)=??*g#</mathjax>.</p></div>
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<div class="markdown"><p>There are about <mathjax>#900*g#</mathjax> in a <mathjax>#5.7#</mathjax> molar quantity of <mathjax>#KMnO_4#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The required mass is simply the product, <mathjax>#"moles "xx" molar mass"#</mathjax>, i.e. <mathjax>#5.7*cancel(mol)xx158.034*g*cancel(mol^-1)=??*g#</mathjax>.</p></div>
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<div class="markdown"><p>There are about <mathjax>#900*g#</mathjax> in a <mathjax>#5.7#</mathjax> molar quantity of <mathjax>#KMnO_4#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The required mass is simply the product, <mathjax>#"moles "xx" molar mass"#</mathjax>, i.e. <mathjax>#5.7*cancel(mol)xx158.034*g*cancel(mol^-1)=??*g#</mathjax>.</p></div>
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</article> | How many grams are in 5.7 mol potassium permanganate? | null |
2,369 | ab9f6d6e-6ddd-11ea-8134-ccda262736ce | https://socratic.org/questions/a-mixture-of-helium-and-neon-gases-is-collected-over-water-at-28-0-c-and-745-mmh | 359.00 mmHg | start physical_unit 5 5 partial_pressure mmhg qc_end physical_unit 10 10 12 13 temperature qc_end physical_unit 10 10 15 16 pressure qc_end physical_unit 3 3 24 25 partial_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] neon [IN] mmHg"}] | [{"type":"physical unit","value":"359.00 mmHg"}] | [{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{28.0 ℃}"},{"type":"physical unit","value":"Pressure [OF] water [=] \\pu{745 mmHg}"},{"type":"physical unit","value":"Partial pressure [OF] helium [=] \\pu{368 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A mixture of helium and neon gases is collected over water at 28.0°C and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon?</h1> | null | 359.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The partial pressures are additive; but to solve the problem we need 1 more parameter, <mathjax>#P_"SVP"#</mathjax>, the <mathjax>#"saturated vapour pressure"#</mathjax>, which is the <mathjax>#"vapour pressure"#</mathjax> of water at the given temperature. <mathjax>#P_"SVP"=18*mm*Hg#</mathjax> at <mathjax>#301*K#</mathjax>.</p>
<p>From above, <mathjax>#P_"Ne"=P_"Total"-P_"He"-P_"SVP"#</mathjax></p>
<p><mathjax>#=(745-368-18)*mm*Hg#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mm*Hg#</mathjax>.</p></div>
</div>
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<div class="markdown"><p><mathjax>#P_"Total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#P_"He"+P_"Ne"+P_"SVP"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The partial pressures are additive; but to solve the problem we need 1 more parameter, <mathjax>#P_"SVP"#</mathjax>, the <mathjax>#"saturated vapour pressure"#</mathjax>, which is the <mathjax>#"vapour pressure"#</mathjax> of water at the given temperature. <mathjax>#P_"SVP"=18*mm*Hg#</mathjax> at <mathjax>#301*K#</mathjax>.</p>
<p>From above, <mathjax>#P_"Ne"=P_"Total"-P_"He"-P_"SVP"#</mathjax></p>
<p><mathjax>#=(745-368-18)*mm*Hg#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mm*Hg#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A mixture of helium and neon gases is collected over water at 28.0°C and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon?</h1>
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<div class="markdown"><p><mathjax>#P_"Total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#P_"He"+P_"Ne"+P_"SVP"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The partial pressures are additive; but to solve the problem we need 1 more parameter, <mathjax>#P_"SVP"#</mathjax>, the <mathjax>#"saturated vapour pressure"#</mathjax>, which is the <mathjax>#"vapour pressure"#</mathjax> of water at the given temperature. <mathjax>#P_"SVP"=18*mm*Hg#</mathjax> at <mathjax>#301*K#</mathjax>.</p>
<p>From above, <mathjax>#P_"Ne"=P_"Total"-P_"He"-P_"SVP"#</mathjax></p>
<p><mathjax>#=(745-368-18)*mm*Hg#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mm*Hg#</mathjax>.</p></div>
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</article> | A mixture of helium and neon gases is collected over water at 28.0°C and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon? | null |
2,370 | a8d2047a-6ddd-11ea-9dfe-ccda262736ce | https://socratic.org/questions/lead-ii-nitrate-reacts-with-potassium-iodide-to-form-lead-ii-iodide-and-potassiu | Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3 | start chemical_equation qc_end substance 0 2 qc_end substance 5 6 qc_end substance 9 11 qc_end substance 13 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3"}] | [{"type":"substance name","value":"Lead (II) nitrate"},{"type":"substance name","value":"Potassium iodide"},{"type":"substance name","value":"Lead (II) iodide"},{"type":"substance name","value":"Potassium nitrate"}] | <h1 class="questionTitle" itemprop="name">Lead (II) nitrate reacts with potassium iodide to form lead (II) iodide and potassium nitrate. How do you write the balanced equation for this?</h1> | null | Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We must first convert from a word equation to a symbol equation:</p>
<p><mathjax>#"Lead (II) Nitrate + Potassium Iodide "->" Lead (II) Iodide + Potassium Nitrate"#</mathjax></p>
<p>The lead (II) ion is represented as <mathjax>#"Pb"^(2+)#</mathjax>, whilst the nitrate ion is <mathjax>#"NO"_3^-#</mathjax>. To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is <mathjax>#"Pb(NO"_3")"_2#</mathjax> .</p>
<p>The potassium ion is <mathjax>#"K"^+#</mathjax> and the iodide ion is <mathjax>#"I"^-#</mathjax>. The two charges balance in a <mathjax>#1 : 1#</mathjax> ratio, so potassium iodide is simply <mathjax>#"KI"#</mathjax>.</p>
<p>In lead (II) iodide, the charges balance in a <mathjax>#1 : 2#</mathjax> ratio, so the formula is <mathjax>#"PbI"_2#</mathjax>.</p>
<p>Finally, in potassium nitrate, the charges balance in another <mathjax>#1 : 1#</mathjax> ratio, giving a formula of <mathjax>#"KNO"_3#</mathjax> .</p>
<p>The symbol equation is as follows:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + "KNO"_3#</mathjax></p>
<p>The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can do this by placing a coefficient of <mathjax>#2#</mathjax> before the potassium nitrate:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of <mathjax>#2#</mathjax>, this time before the potassium iodide:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing our potassium ions. Now, our equation is balanced.</p>
<p>And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would have arrived at the same answer simply by working backwards.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We must first convert from a word equation to a symbol equation:</p>
<p><mathjax>#"Lead (II) Nitrate + Potassium Iodide "->" Lead (II) Iodide + Potassium Nitrate"#</mathjax></p>
<p>The lead (II) ion is represented as <mathjax>#"Pb"^(2+)#</mathjax>, whilst the nitrate ion is <mathjax>#"NO"_3^-#</mathjax>. To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is <mathjax>#"Pb(NO"_3")"_2#</mathjax> .</p>
<p>The potassium ion is <mathjax>#"K"^+#</mathjax> and the iodide ion is <mathjax>#"I"^-#</mathjax>. The two charges balance in a <mathjax>#1 : 1#</mathjax> ratio, so potassium iodide is simply <mathjax>#"KI"#</mathjax>.</p>
<p>In lead (II) iodide, the charges balance in a <mathjax>#1 : 2#</mathjax> ratio, so the formula is <mathjax>#"PbI"_2#</mathjax>.</p>
<p>Finally, in potassium nitrate, the charges balance in another <mathjax>#1 : 1#</mathjax> ratio, giving a formula of <mathjax>#"KNO"_3#</mathjax> .</p>
<p>The symbol equation is as follows:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + "KNO"_3#</mathjax></p>
<p>The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can do this by placing a coefficient of <mathjax>#2#</mathjax> before the potassium nitrate:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of <mathjax>#2#</mathjax>, this time before the potassium iodide:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing our potassium ions. Now, our equation is balanced.</p>
<p>And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would have arrived at the same answer simply by working backwards.</p></div>
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<h1 class="questionTitle" itemprop="name">Lead (II) nitrate reacts with potassium iodide to form lead (II) iodide and potassium nitrate. How do you write the balanced equation for this?</h1>
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Owen Bell
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<span class="dateCreated" datetime="2015-12-28T19:23:07" itemprop="dateCreated">
Dec 28, 2015
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<div class="markdown"><p><mathjax>#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We must first convert from a word equation to a symbol equation:</p>
<p><mathjax>#"Lead (II) Nitrate + Potassium Iodide "->" Lead (II) Iodide + Potassium Nitrate"#</mathjax></p>
<p>The lead (II) ion is represented as <mathjax>#"Pb"^(2+)#</mathjax>, whilst the nitrate ion is <mathjax>#"NO"_3^-#</mathjax>. To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is <mathjax>#"Pb(NO"_3")"_2#</mathjax> .</p>
<p>The potassium ion is <mathjax>#"K"^+#</mathjax> and the iodide ion is <mathjax>#"I"^-#</mathjax>. The two charges balance in a <mathjax>#1 : 1#</mathjax> ratio, so potassium iodide is simply <mathjax>#"KI"#</mathjax>.</p>
<p>In lead (II) iodide, the charges balance in a <mathjax>#1 : 2#</mathjax> ratio, so the formula is <mathjax>#"PbI"_2#</mathjax>.</p>
<p>Finally, in potassium nitrate, the charges balance in another <mathjax>#1 : 1#</mathjax> ratio, giving a formula of <mathjax>#"KNO"_3#</mathjax> .</p>
<p>The symbol equation is as follows:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + "KNO"_3#</mathjax></p>
<p>The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can do this by placing a coefficient of <mathjax>#2#</mathjax> before the potassium nitrate:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + "KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of <mathjax>#2#</mathjax>, this time before the potassium iodide:</p>
<p><mathjax>#"Pb(NO"_3")"_2 + 2"KI" -> "PbI"_2 + 2"KNO"_3#</mathjax></p>
<p>Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing our potassium ions. Now, our equation is balanced.</p>
<p>And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would have arrived at the same answer simply by working backwards.</p></div>
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</article> | Lead (II) nitrate reacts with potassium iodide to form lead (II) iodide and potassium nitrate. How do you write the balanced equation for this? | null |
2,371 | ac9615e4-6ddd-11ea-be16-ccda262736ce | https://socratic.org/questions/a-student-dissolves-106-g-of-na-2co-3-in-enough-water-to-make-a-6-00-l-solution- | 0.17 M | start physical_unit 21 22 molarity mol/l qc_end physical_unit 6 6 3 4 mass qc_end c_other OTHER qc_end physical_unit 15 15 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"0.17 M"}] | [{"type":"physical unit","value":"Mass [OF] Na2CO3 [=] \\pu{106 g}"},{"type":"other","value":"Enough water."},{"type":"physical unit","value":"Volume [OF] Na2CO3 solution [=] \\pu{6.00 L}"}] | <h1 class="questionTitle" itemprop="name">A student dissolves 106 g of #Na_2CO_3# in enough water to make a 6.00 L solution. What is the molarity of the solution?</h1> | null | 0.17 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the formula mass of <mathjax>#Na_2CO_3#</mathjax></li>
<li>Find the number of moles by dividing the amount of <mathjax>#Na_2CO_3#</mathjax> as provided in the problem by its formulas mass;</li>
<li>Compute the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> (M) using the formula; <br/>
M=number of mol/L solution</li>
<li>Make sure to cancel out units leaving only the desired unit</li>
<li>Per calculation, M=0.1667 mol/L</li>
</ol></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#0.1667M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the formula mass of <mathjax>#Na_2CO_3#</mathjax></li>
<li>Find the number of moles by dividing the amount of <mathjax>#Na_2CO_3#</mathjax> as provided in the problem by its formulas mass;</li>
<li>Compute the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> (M) using the formula; <br/>
M=number of mol/L solution</li>
<li>Make sure to cancel out units leaving only the desired unit</li>
<li>Per calculation, M=0.1667 mol/L</li>
</ol></div>
</div>
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<h1 class="questionTitle" itemprop="name">A student dissolves 106 g of #Na_2CO_3# in enough water to make a 6.00 L solution. What is the molarity of the solution?</h1>
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Jarni Renz
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<div class="markdown"><p><mathjax>#0.1667M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the formula mass of <mathjax>#Na_2CO_3#</mathjax></li>
<li>Find the number of moles by dividing the amount of <mathjax>#Na_2CO_3#</mathjax> as provided in the problem by its formulas mass;</li>
<li>Compute the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> (M) using the formula; <br/>
M=number of mol/L solution</li>
<li>Make sure to cancel out units leaving only the desired unit</li>
<li>Per calculation, M=0.1667 mol/L</li>
</ol></div>
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</article> | A student dissolves 106 g of #Na_2CO_3# in enough water to make a 6.00 L solution. What is the molarity of the solution? | null |
2,372 | a9ca2830-6ddd-11ea-8b09-ccda262736ce | https://socratic.org/questions/for-the-reaction-h2-i2-2hi-kc-is-49-calculate-the-concentration-of-hi-at-equilib | 0.78 M | start physical_unit 8 8 concentration mol/l qc_end chemical_equation 3 8 qc_end physical_unit 1 2 11 11 equilibrium_constant_k qc_end physical_unit 3 3 21 22 mole qc_end physical_unit 5 5 21 22 mole qc_end physical_unit 35 35 33 34 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] HI [IN] M"}] | [{"type":"physical unit","value":"0.78 M"}] | [{"type":"chemical equation","value":"H2 + I2 <=> 2 HI"},{"type":"physical unit","value":"Kc [OF] the reaction [=] \\pu{49}"},{"type":"physical unit","value":"Mole [OF] H2 [=] \\pu{1 mole}"},{"type":"physical unit","value":"Mole [OF] I2 [=] \\pu{1 mole}"},{"type":"physical unit","value":"Volume [OF] flask [=] \\pu{2 litre}"}] | <h1 class="questionTitle" itemprop="name">For the reaction H2+I2----2HI
Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?</h1> | null | 0.78 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the equilibrium concentration of <mathjax>#"HI"#</mathjax> with an initial concentration of <mathjax>#1#</mathjax> <mathjax>#"mol H"_2#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#"mol I"_2#</mathjax>.</p>
<p>The equilibrium constant expression for this reaction is</p>
<p><mathjax>#K_c = (["HI"]^2)/(["H"_2]["I"_2]) = 49#</mathjax></p>
<p>The initial concentrations of both <mathjax>#"H"_2#</mathjax> and <mathjax>#"I"_2#</mathjax> are</p>
<p><mathjax>#(1color(white)(l)"mol")/(2color(white)(l)"L") = 0.5M#</mathjax></p>
<p>So our initial concentrations for each species are</p>
<p><strong>Initial</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#0#</mathjax></p>
</li>
</ul>
<p>From the coefficients of the chemical equation, we can predict the <em>changes</em> in concentration with the quantity <mathjax>#x#</mathjax>:</p>
<p><strong>Change</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#+2x#</mathjax></p>
</li>
</ul>
<p>and the <em>final</em> concentrations are the sum of the initial and change:</p>
<p><strong>Final</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#2x#</mathjax></p>
</li>
</ul>
<p>It's algebra time! Let's plug these into our equilibrium constant expression to start solving for <mathjax>#x#</mathjax>:</p>
<p><mathjax>#K_c = ((2x)^2)/((0.50-x)(0.50-x)) = 49#</mathjax></p>
<p><mathjax>#4x^2 = 49(0.25-x+x^2)#</mathjax></p>
<p><mathjax>#49x^2 - 49x + 12.25 = 4x^2#</mathjax></p>
<p><mathjax>#45x^2 - 49x + 12.25 = 0#</mathjax></p>
<p><mathjax>#x = (49 +-sqrt((-49)^2-4(45)(12.25)))/(2(45)) = 0.7#</mathjax> or <mathjax>#0.389#</mathjax></p>
<p>We can neglect the solution that is greater than <mathjax>#0.5#</mathjax>, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is <mathjax>#color(blue)(0.389#</mathjax>.</p>
<p>Therefore, the <strong>final</strong> equilibrium concentration of <mathjax>#"HI"#</mathjax> is</p>
<p><mathjax>#["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#["HI"] = 0.778M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the equilibrium concentration of <mathjax>#"HI"#</mathjax> with an initial concentration of <mathjax>#1#</mathjax> <mathjax>#"mol H"_2#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#"mol I"_2#</mathjax>.</p>
<p>The equilibrium constant expression for this reaction is</p>
<p><mathjax>#K_c = (["HI"]^2)/(["H"_2]["I"_2]) = 49#</mathjax></p>
<p>The initial concentrations of both <mathjax>#"H"_2#</mathjax> and <mathjax>#"I"_2#</mathjax> are</p>
<p><mathjax>#(1color(white)(l)"mol")/(2color(white)(l)"L") = 0.5M#</mathjax></p>
<p>So our initial concentrations for each species are</p>
<p><strong>Initial</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#0#</mathjax></p>
</li>
</ul>
<p>From the coefficients of the chemical equation, we can predict the <em>changes</em> in concentration with the quantity <mathjax>#x#</mathjax>:</p>
<p><strong>Change</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#+2x#</mathjax></p>
</li>
</ul>
<p>and the <em>final</em> concentrations are the sum of the initial and change:</p>
<p><strong>Final</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#2x#</mathjax></p>
</li>
</ul>
<p>It's algebra time! Let's plug these into our equilibrium constant expression to start solving for <mathjax>#x#</mathjax>:</p>
<p><mathjax>#K_c = ((2x)^2)/((0.50-x)(0.50-x)) = 49#</mathjax></p>
<p><mathjax>#4x^2 = 49(0.25-x+x^2)#</mathjax></p>
<p><mathjax>#49x^2 - 49x + 12.25 = 4x^2#</mathjax></p>
<p><mathjax>#45x^2 - 49x + 12.25 = 0#</mathjax></p>
<p><mathjax>#x = (49 +-sqrt((-49)^2-4(45)(12.25)))/(2(45)) = 0.7#</mathjax> or <mathjax>#0.389#</mathjax></p>
<p>We can neglect the solution that is greater than <mathjax>#0.5#</mathjax>, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is <mathjax>#color(blue)(0.389#</mathjax>.</p>
<p>Therefore, the <strong>final</strong> equilibrium concentration of <mathjax>#"HI"#</mathjax> is</p>
<p><mathjax>#["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">For the reaction H2+I2----2HI
Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?</h1>
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Nathan L.
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Jun 25, 2017
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<div class="markdown"><p><mathjax>#["HI"] = 0.778M#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the equilibrium concentration of <mathjax>#"HI"#</mathjax> with an initial concentration of <mathjax>#1#</mathjax> <mathjax>#"mol H"_2#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#"mol I"_2#</mathjax>.</p>
<p>The equilibrium constant expression for this reaction is</p>
<p><mathjax>#K_c = (["HI"]^2)/(["H"_2]["I"_2]) = 49#</mathjax></p>
<p>The initial concentrations of both <mathjax>#"H"_2#</mathjax> and <mathjax>#"I"_2#</mathjax> are</p>
<p><mathjax>#(1color(white)(l)"mol")/(2color(white)(l)"L") = 0.5M#</mathjax></p>
<p>So our initial concentrations for each species are</p>
<p><strong>Initial</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#0#</mathjax></p>
</li>
</ul>
<p>From the coefficients of the chemical equation, we can predict the <em>changes</em> in concentration with the quantity <mathjax>#x#</mathjax>:</p>
<p><strong>Change</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#+2x#</mathjax></p>
</li>
</ul>
<p>and the <em>final</em> concentrations are the sum of the initial and change:</p>
<p><strong>Final</strong>:</p>
<ul>
<li>
<p><mathjax>#"H"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"I"_2#</mathjax>: <mathjax>#0.5M-x#</mathjax></p>
</li>
<li>
<p><mathjax>#"HI"#</mathjax>: <mathjax>#2x#</mathjax></p>
</li>
</ul>
<p>It's algebra time! Let's plug these into our equilibrium constant expression to start solving for <mathjax>#x#</mathjax>:</p>
<p><mathjax>#K_c = ((2x)^2)/((0.50-x)(0.50-x)) = 49#</mathjax></p>
<p><mathjax>#4x^2 = 49(0.25-x+x^2)#</mathjax></p>
<p><mathjax>#49x^2 - 49x + 12.25 = 4x^2#</mathjax></p>
<p><mathjax>#45x^2 - 49x + 12.25 = 0#</mathjax></p>
<p><mathjax>#x = (49 +-sqrt((-49)^2-4(45)(12.25)))/(2(45)) = 0.7#</mathjax> or <mathjax>#0.389#</mathjax></p>
<p>We can neglect the solution that is greater than <mathjax>#0.5#</mathjax>, because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is <mathjax>#color(blue)(0.389#</mathjax>.</p>
<p>Therefore, the <strong>final</strong> equilibrium concentration of <mathjax>#"HI"#</mathjax> is</p>
<p><mathjax>#["HI"] = 2x = 2(color(blue)(0.389)) = color(red)(0.778M#</mathjax></p></div>
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</article> | For the reaction H2+I2----2HI
Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.? | null |
2,373 | a8c02a1c-6ddd-11ea-bbe9-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-one-mole-of-any-gas-at-stp | 22.4 L | start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] any gas [IN] L"}] | [{"type":"physical unit","value":"22.4 L"}] | [{"type":"physical unit","value":"Mole [OF] any gas [=] \\pu{1 mole}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of one mole of any gas at STP?</h1> | null | 22.4 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The volume of one mole of an ideal gas at STP can be found as follows:</p>
<p>First, What is STP?</p>
<p>STP is the standard conditions of Temperature and Pressure, where, <mathjax>#T=273K#</mathjax> and <mathjax>#P=1atm#</mathjax>.</p>
<p>Since we are looking for the volume of one mole, then <mathjax>#n=1mol#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can find the volume by:</p>
<p><mathjax>#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#22.4L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The volume of one mole of an ideal gas at STP can be found as follows:</p>
<p>First, What is STP?</p>
<p>STP is the standard conditions of Temperature and Pressure, where, <mathjax>#T=273K#</mathjax> and <mathjax>#P=1atm#</mathjax>.</p>
<p>Since we are looking for the volume of one mole, then <mathjax>#n=1mol#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can find the volume by:</p>
<p><mathjax>#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of one mole of any gas at STP?</h1>
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Dr. Hayek
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Nov 19, 2015
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<div class="markdown"><p><mathjax>#22.4L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The volume of one mole of an ideal gas at STP can be found as follows:</p>
<p>First, What is STP?</p>
<p>STP is the standard conditions of Temperature and Pressure, where, <mathjax>#T=273K#</mathjax> and <mathjax>#P=1atm#</mathjax>.</p>
<p>Since we are looking for the volume of one mole, then <mathjax>#n=1mol#</mathjax>.</p>
<p>Using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> we can find the volume by:</p>
<p><mathjax>#V=(nRT)/P =(1cancel(mol)xx0.08206(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))/(1cancel(atm))=22.4L#</mathjax></p></div>
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Meave60
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Nov 19, 2015
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<div class="markdown"><p>The volume of one mole of a gas is either <mathjax>#"22.414 L"#</mathjax>, or <mathjax>#"22.711 L"#</mathjax>, depending on the pressure used for <mathjax>#"STP"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It depends on what values you use for <mathjax>#"STP"#</mathjax>. If you use the values of <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the volume of one mole of a gas is <mathjax>#"22.414 L"#</mathjax>.</p>
<p>However, the IUPAC and NIST have updated the values for <mathjax>#"STP"#</mathjax> to <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"10"^5 "Pa"#</mathjax> or <mathjax>#"100 kPa"#</mathjax>. With these values, the volume of one mole of a gas is <mathjax>#"22.711 L"#</mathjax>.</p>
<p><a href="http://en.citizendium.org/wiki/Reference_conditions_of_gas_temperature_and_pressure#Molar_volume_of_a_gas" rel="nofollow" target="_blank">http://en.citizendium.org/wiki/Reference_conditions_of_gas_temperature_and_pressure#Molar_volume_of_a_gas</a></p></div>
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</article> | What is the volume of one mole of any gas at STP? | null |
2,374 | aa0b03e7-6ddd-11ea-bd3d-ccda262736ce | https://socratic.org/questions/what-volume-of-nitrogen-gas-can-be-produced-by-the-decomposition-of-50-0-g-of-am | 7.88 L | start physical_unit 3 4 volume l qc_end physical_unit 15 16 12 13 mass qc_end physical_unit 21 22 24 24 percent_yield qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [IN] L"}] | [{"type":"physical unit","value":"7.88 L"}] | [{"type":"physical unit","value":"Mass [OF] ammonium nitrite [=] \\pu{50.0 g}"},{"type":"physical unit","value":"Yield [OF] this reaction [=] \\pu{45.0%}"},{"type":"other","value":"Water is the other product."}] | <h1 class="questionTitle" itemprop="name">What volume of nitrogen gas can be produced by the decomposition of 50.0 g of ammonium nitrite if the yield for this reaction is 45.0%? Water is the other product.</h1> | null | 7.88 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"NH"_4"NO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"N"_2" + 2H"_2"O"#</mathjax></p>
<p><strong>Determine moles <mathjax>#"NH"_4"NO"_2"#</mathjax>. The molar mass of ammonium nitrite is <mathjax>#"64.044 g/mol"#</mathjax>.</strong></p>
<p>We will need the molar volume of a gas for these calculations. Molar volume at the most commonly used values for STP, <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> (for gases), and <mathjax>#"1 atm"#</mathjax>. The molar volume of a gas under these conditions is <mathjax>#"22.414 L/mol"#</mathjax>.</p>
<p><mathjax>#50.0color(red)cancel(color(black)("g NH"_4"NO"_2))xx(1"mol NH"_4"NO"_2)/(64.044color(red)cancel(color(black)("g NH"_4"NO"_2)))="0.7807 mol NH"_4"NO"_2#</mathjax></p>
<p><strong>I am keeping an extra digit to reduce rounding errors. I will round to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</strong></p>
<p>Determine moles <mathjax>#"N"_2"#</mathjax> produced by multiplying the moles <mathjax>#"NH"_4"NO"_2#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"N"_2#</mathjax> and <mathjax>#"NH"_4"NO"_2"#</mathjax> from the balanced equation. <mathjax>#"1 mol N"_2":#</mathjax> <mathjax>#"1 mol NH"_4"NO"_2"#</mathjax></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol NH"_4"NO"_2))xx(1"mol N"_2)/(1color(red)cancel(color(black)("mol NH"_4"NO"_2)))="0.7807 mol N"_2"#</mathjax></p>
<p><strong>Determine the theoretical volume of <mathjax>#"N"_2"#</mathjax> produced by multiplying by the mole <mathjax>#"N"_2"#</mathjax> by the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> <mathjax>#("22.414 mol/L")#</mathjax>.</strong></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol N"_2))xx(22.414"L N"_2)/(1color(red)cancel(color(black)("mol N"_2)))="17.5 L N"_2"#</mathjax></p>
<p><strong>Convert <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> to decimal form.</strong></p>
<p><mathjax>#"Percent yield"=45.0%=45.0/100=0.450#</mathjax></p>
<p><strong>Determine actual yield.</strong></p>
<p><strong>Multiply the theoretical volume of <mathjax>#"N"_2#</mathjax> by the percent yield.</strong></p>
<p><mathjax>#17.5" L N"_2xx0.450="7.88 L N"_2"#</mathjax></p>
<p>The actual yield is <mathjax>#"7.88 L N"_2"#</mathjax>.</p>
<p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</p>
<p>The most current values for STP, recommended by the International Pure and Applied Chemistry (IUPAC) in 1982, are <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#10^5" Pa"#</mathjax>. Under these conditions, The molar volume of a gas is <mathjax>#"22.71 L/mol"#</mathjax>.</p>
<p><strong>Theoretical yield</strong></p>
<p><mathjax>#0.7807"mol N"_2xx(22.71"L N"_2)/(1"mol N"_2)="17.7 L N"_2"#</mathjax></p>
<p><strong>Actual yield</strong></p>
<p><mathjax>#17.7"L N"_2xx0.450="7.97 L N"_2"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The actual yield using the molar volume <mathjax>#"22.414 mol/L"#</mathjax> is <mathjax>#"7.88 L N"_2"#</mathjax>.</p>
<p>The actual yield using the molar volume <mathjax>#"22.71 mol/L"#</mathjax> is <mathjax>#"7.97 L N"_2"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"NH"_4"NO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"N"_2" + 2H"_2"O"#</mathjax></p>
<p><strong>Determine moles <mathjax>#"NH"_4"NO"_2"#</mathjax>. The molar mass of ammonium nitrite is <mathjax>#"64.044 g/mol"#</mathjax>.</strong></p>
<p>We will need the molar volume of a gas for these calculations. Molar volume at the most commonly used values for STP, <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> (for gases), and <mathjax>#"1 atm"#</mathjax>. The molar volume of a gas under these conditions is <mathjax>#"22.414 L/mol"#</mathjax>.</p>
<p><mathjax>#50.0color(red)cancel(color(black)("g NH"_4"NO"_2))xx(1"mol NH"_4"NO"_2)/(64.044color(red)cancel(color(black)("g NH"_4"NO"_2)))="0.7807 mol NH"_4"NO"_2#</mathjax></p>
<p><strong>I am keeping an extra digit to reduce rounding errors. I will round to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</strong></p>
<p>Determine moles <mathjax>#"N"_2"#</mathjax> produced by multiplying the moles <mathjax>#"NH"_4"NO"_2#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"N"_2#</mathjax> and <mathjax>#"NH"_4"NO"_2"#</mathjax> from the balanced equation. <mathjax>#"1 mol N"_2":#</mathjax> <mathjax>#"1 mol NH"_4"NO"_2"#</mathjax></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol NH"_4"NO"_2))xx(1"mol N"_2)/(1color(red)cancel(color(black)("mol NH"_4"NO"_2)))="0.7807 mol N"_2"#</mathjax></p>
<p><strong>Determine the theoretical volume of <mathjax>#"N"_2"#</mathjax> produced by multiplying by the mole <mathjax>#"N"_2"#</mathjax> by the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> <mathjax>#("22.414 mol/L")#</mathjax>.</strong></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol N"_2))xx(22.414"L N"_2)/(1color(red)cancel(color(black)("mol N"_2)))="17.5 L N"_2"#</mathjax></p>
<p><strong>Convert <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> to decimal form.</strong></p>
<p><mathjax>#"Percent yield"=45.0%=45.0/100=0.450#</mathjax></p>
<p><strong>Determine actual yield.</strong></p>
<p><strong>Multiply the theoretical volume of <mathjax>#"N"_2#</mathjax> by the percent yield.</strong></p>
<p><mathjax>#17.5" L N"_2xx0.450="7.88 L N"_2"#</mathjax></p>
<p>The actual yield is <mathjax>#"7.88 L N"_2"#</mathjax>.</p>
<p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</p>
<p>The most current values for STP, recommended by the International Pure and Applied Chemistry (IUPAC) in 1982, are <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#10^5" Pa"#</mathjax>. Under these conditions, The molar volume of a gas is <mathjax>#"22.71 L/mol"#</mathjax>.</p>
<p><strong>Theoretical yield</strong></p>
<p><mathjax>#0.7807"mol N"_2xx(22.71"L N"_2)/(1"mol N"_2)="17.7 L N"_2"#</mathjax></p>
<p><strong>Actual yield</strong></p>
<p><mathjax>#17.7"L N"_2xx0.450="7.97 L N"_2"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of nitrogen gas can be produced by the decomposition of 50.0 g of ammonium nitrite if the yield for this reaction is 45.0%? Water is the other product.</h1>
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<div class="markdown"><p>The actual yield using the molar volume <mathjax>#"22.414 mol/L"#</mathjax> is <mathjax>#"7.88 L N"_2"#</mathjax>.</p>
<p>The actual yield using the molar volume <mathjax>#"22.71 mol/L"#</mathjax> is <mathjax>#"7.97 L N"_2"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"NH"_4"NO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"N"_2" + 2H"_2"O"#</mathjax></p>
<p><strong>Determine moles <mathjax>#"NH"_4"NO"_2"#</mathjax>. The molar mass of ammonium nitrite is <mathjax>#"64.044 g/mol"#</mathjax>.</strong></p>
<p>We will need the molar volume of a gas for these calculations. Molar volume at the most commonly used values for STP, <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> (for gases), and <mathjax>#"1 atm"#</mathjax>. The molar volume of a gas under these conditions is <mathjax>#"22.414 L/mol"#</mathjax>.</p>
<p><mathjax>#50.0color(red)cancel(color(black)("g NH"_4"NO"_2))xx(1"mol NH"_4"NO"_2)/(64.044color(red)cancel(color(black)("g NH"_4"NO"_2)))="0.7807 mol NH"_4"NO"_2#</mathjax></p>
<p><strong>I am keeping an extra digit to reduce rounding errors. I will round to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</strong></p>
<p>Determine moles <mathjax>#"N"_2"#</mathjax> produced by multiplying the moles <mathjax>#"NH"_4"NO"_2#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"N"_2#</mathjax> and <mathjax>#"NH"_4"NO"_2"#</mathjax> from the balanced equation. <mathjax>#"1 mol N"_2":#</mathjax> <mathjax>#"1 mol NH"_4"NO"_2"#</mathjax></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol NH"_4"NO"_2))xx(1"mol N"_2)/(1color(red)cancel(color(black)("mol NH"_4"NO"_2)))="0.7807 mol N"_2"#</mathjax></p>
<p><strong>Determine the theoretical volume of <mathjax>#"N"_2"#</mathjax> produced by multiplying by the mole <mathjax>#"N"_2"#</mathjax> by the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> <mathjax>#("22.414 mol/L")#</mathjax>.</strong></p>
<p><mathjax>#0.7807color(red)cancel(color(black)("mol N"_2))xx(22.414"L N"_2)/(1color(red)cancel(color(black)("mol N"_2)))="17.5 L N"_2"#</mathjax></p>
<p><strong>Convert <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> to decimal form.</strong></p>
<p><mathjax>#"Percent yield"=45.0%=45.0/100=0.450#</mathjax></p>
<p><strong>Determine actual yield.</strong></p>
<p><strong>Multiply the theoretical volume of <mathjax>#"N"_2#</mathjax> by the percent yield.</strong></p>
<p><mathjax>#17.5" L N"_2xx0.450="7.88 L N"_2"#</mathjax></p>
<p>The actual yield is <mathjax>#"7.88 L N"_2"#</mathjax>.</p>
<p>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</p>
<p>The most current values for STP, recommended by the International Pure and Applied Chemistry (IUPAC) in 1982, are <mathjax>#0^@"C"#</mathjax>, or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#10^5" Pa"#</mathjax>. Under these conditions, The molar volume of a gas is <mathjax>#"22.71 L/mol"#</mathjax>.</p>
<p><strong>Theoretical yield</strong></p>
<p><mathjax>#0.7807"mol N"_2xx(22.71"L N"_2)/(1"mol N"_2)="17.7 L N"_2"#</mathjax></p>
<p><strong>Actual yield</strong></p>
<p><mathjax>#17.7"L N"_2xx0.450="7.97 L N"_2"#</mathjax></p></div>
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</article> | What volume of nitrogen gas can be produced by the decomposition of 50.0 g of ammonium nitrite if the yield for this reaction is 45.0%? Water is the other product. | null |
2,375 | a8ecd070-6ddd-11ea-8abc-ccda262736ce | https://socratic.org/questions/a-balloon-has-a-volume-of-253-2-ml-at-356-k-the-volume-of-the-balloon-is-decreas | 233 K | start physical_unit 14 15 temperature k qc_end physical_unit 14 15 6 7 volume qc_end physical_unit 14 15 19 20 volume qc_end physical_unit 14 15 9 10 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the balloon [IN] K"}] | [{"type":"physical unit","value":"233 K"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{253.2 mL}"},{"type":"physical unit","value":"Volume2 [OF] the balloon [=] \\pu{165.4 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{356 K}"}] | <h1 class="questionTitle" itemprop="name">A balloon has a volume of 253.2 mL at 356 K. The volume of the balloon is decreased to 165.4 mL. What is the new temperature?</h1> | null | 233 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> states that when pressure is held constant, temperature and volume are directly proportional. The equation to use is <mathjax>#V_1/T_1V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1=252.3cancel"mL"xx(1"L")/(1000cancel"mL")="0.2523 L"#</mathjax><br/>
<mathjax>#T_1="356 K"#</mathjax><br/>
<mathjax>#V_2=165.4cancel"mL"xx(1"L")/(1000cancel"mL")="0.1654 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#V_1/T_1V_2/T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.<br/>
<mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((356"K")xx(0.1654cancel"L"))/(0.2523cancel"L")="233 K"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new temperature is <mathjax>#"233 K"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> states that when pressure is held constant, temperature and volume are directly proportional. The equation to use is <mathjax>#V_1/T_1V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1=252.3cancel"mL"xx(1"L")/(1000cancel"mL")="0.2523 L"#</mathjax><br/>
<mathjax>#T_1="356 K"#</mathjax><br/>
<mathjax>#V_2=165.4cancel"mL"xx(1"L")/(1000cancel"mL")="0.1654 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#V_1/T_1V_2/T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.<br/>
<mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((356"K")xx(0.1654cancel"L"))/(0.2523cancel"L")="233 K"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A balloon has a volume of 253.2 mL at 356 K. The volume of the balloon is decreased to 165.4 mL. What is the new temperature?</h1>
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<span class="dateCreated" datetime="2015-11-19T00:54:25" itemprop="dateCreated">
Nov 19, 2015
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<div class="markdown"><p>The new temperature is <mathjax>#"233 K"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> states that when pressure is held constant, temperature and volume are directly proportional. The equation to use is <mathjax>#V_1/T_1V_2/T_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#V_1=252.3cancel"mL"xx(1"L")/(1000cancel"mL")="0.2523 L"#</mathjax><br/>
<mathjax>#T_1="356 K"#</mathjax><br/>
<mathjax>#V_2=165.4cancel"mL"xx(1"L")/(1000cancel"mL")="0.1654 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#V_1/T_1V_2/T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>.<br/>
<mathjax>#T_2=(T_1V_2)/V_1#</mathjax></p>
<p><mathjax>#T_2=((356"K")xx(0.1654cancel"L"))/(0.2523cancel"L")="233 K"#</mathjax></p></div>
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</article> | A balloon has a volume of 253.2 mL at 356 K. The volume of the balloon is decreased to 165.4 mL. What is the new temperature? | null |
2,376 | a9572036-6ddd-11ea-a7d1-ccda262736ce | https://socratic.org/questions/you-want-to-produce-579-ml-of-a-0-440-m-solution-of-nacl-you-have-a-1-40-m-solut | 181.97 mL | start physical_unit 10 10 volume ml qc_end physical_unit 10 10 4 5 volume qc_end physical_unit 12 12 8 9 molarity qc_end physical_unit 12 12 16 17 molarity qc_end substance 30 30 qc_end end | [{"type":"physical unit","value":"Volume1 [OF] NaCl solution [IN] mL"}] | [{"type":"physical unit","value":"181.97 mL"}] | [{"type":"physical unit","value":"Volume2 [OF] NaCl solution [=] \\pu{579 mL}"},{"type":"physical unit","value":"Molarity2 [OF] NaCl solution [=] \\pu{0.440 M}"},{"type":"physical unit","value":"Molarity1 [OF] NaCl solution [=] \\pu{1.40 M}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">You want to produce 579 mL of a 0.440 M solution of NaCl. You have a 1.40 M solution. How many mL of it should you use and dilute with water?</h1> | null | 181.97 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by <strong>assuming</strong> that you're not familiar with the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, so that you can't just plug in your values and do a quick calculation. </p>
<p>So, the idea with <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> is that the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong> when diluting a given sample. </p>
<p>In essence, to dilute a solution you need to <strong>decrease</strong> the concentration of the solute by keeping the number of moles of solute <em>constant</em> and <strong>Increasing</strong> the volume. </p>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute, which in your case will be sodium chloride, <mathjax>#"NaCl"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>Use the molarity and volume of the target solution to determine how many moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.440 M" * 579 * 10^(-3)"L" = "0.25476 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute must be present in the <strong>sample of stock solution</strong> used to prepare the target solution. </p>
<p>This means that you can use the definition of molarity to determine what <em>volume</em> of the stock solution would contain this many moles of sodium chloride</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V = (0.25476 color(red)(cancel(color(black)("moles"))))/(1.40 color(red)(cancel(color(black)("moles")))/"L") = "0.182 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("182 mL")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can use the formula for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which expresses the exact same concept </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(color(purple)(stackrel("moles of solute")("in stock solution"))) = overbrace(c_2V_2)^(color(red)(stackrel("moles of solute")("in target solution")))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the stock solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the target solution</p>
<p>In this case, you would get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.440 color(red)(cancel(color(black)("M"))))/(1.40color(red)(cancel(color(black)("M")))) * "579 mL" = color(green)("182 mL")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"182 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by <strong>assuming</strong> that you're not familiar with the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, so that you can't just plug in your values and do a quick calculation. </p>
<p>So, the idea with <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> is that the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong> when diluting a given sample. </p>
<p>In essence, to dilute a solution you need to <strong>decrease</strong> the concentration of the solute by keeping the number of moles of solute <em>constant</em> and <strong>Increasing</strong> the volume. </p>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute, which in your case will be sodium chloride, <mathjax>#"NaCl"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>Use the molarity and volume of the target solution to determine how many moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.440 M" * 579 * 10^(-3)"L" = "0.25476 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute must be present in the <strong>sample of stock solution</strong> used to prepare the target solution. </p>
<p>This means that you can use the definition of molarity to determine what <em>volume</em> of the stock solution would contain this many moles of sodium chloride</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V = (0.25476 color(red)(cancel(color(black)("moles"))))/(1.40 color(red)(cancel(color(black)("moles")))/"L") = "0.182 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("182 mL")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can use the formula for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which expresses the exact same concept </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(color(purple)(stackrel("moles of solute")("in stock solution"))) = overbrace(c_2V_2)^(color(red)(stackrel("moles of solute")("in target solution")))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the stock solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the target solution</p>
<p>In this case, you would get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.440 color(red)(cancel(color(black)("M"))))/(1.40color(red)(cancel(color(black)("M")))) * "579 mL" = color(green)("182 mL")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">You want to produce 579 mL of a 0.440 M solution of NaCl. You have a 1.40 M solution. How many mL of it should you use and dilute with water?</h1>
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<div class="answer" id="198024" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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Stefan V.
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<span class="dateCreated" datetime="2015-12-09T11:29:24" itemprop="dateCreated">
Dec 9, 2015
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<div class="markdown"><p><mathjax>#"182 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by <strong>assuming</strong> that you're not familiar with the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, so that you can't just plug in your values and do a quick calculation. </p>
<p>So, the idea with <a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a> is that the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong> when diluting a given sample. </p>
<p>In essence, to dilute a solution you need to <strong>decrease</strong> the concentration of the solute by keeping the number of moles of solute <em>constant</em> and <strong>Increasing</strong> the volume. </p>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute, which in your case will be sodium chloride, <mathjax>#"NaCl"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>Use the molarity and volume of the target solution to determine how many moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.440 M" * 579 * 10^(-3)"L" = "0.25476 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute must be present in the <strong>sample of stock solution</strong> used to prepare the target solution. </p>
<p>This means that you can use the definition of molarity to determine what <em>volume</em> of the stock solution would contain this many moles of sodium chloride</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V = (0.25476 color(red)(cancel(color(black)("moles"))))/(1.40 color(red)(cancel(color(black)("moles")))/"L") = "0.182 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("182 mL")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can use the formula for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which expresses the exact same concept </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(color(purple)(stackrel("moles of solute")("in stock solution"))) = overbrace(c_2V_2)^(color(red)(stackrel("moles of solute")("in target solution")))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the stock solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the target solution</p>
<p>In this case, you would get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.440 color(red)(cancel(color(black)("M"))))/(1.40color(red)(cancel(color(black)("M")))) * "579 mL" = color(green)("182 mL")#</mathjax></p>
</blockquote></div>
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<a href="https://socratic.org/answers/198024" itemprop="url">Answer link</a>
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</article> | You want to produce 579 mL of a 0.440 M solution of NaCl. You have a 1.40 M solution. How many mL of it should you use and dilute with water? | null |
2,377 | a9a60646-6ddd-11ea-b85b-ccda262736ce | https://socratic.org/questions/how-many-grams-of-aluminum-are-required-to-react-with-35-ml-of-2-0-m-hydrochlori | 0.63 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 16 10 11 volume qc_end physical_unit 15 16 13 14 molarity qc_end chemical_equation 18 24 qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum [IN] grams"}] | [{"type":"physical unit","value":"0.63 grams"}] | [{"type":"physical unit","value":"Volume [OF] hydrochloric acid [=] \\pu{35 mL}"},{"type":"physical unit","value":"Molarity [OF] hydrochloric acid [=] \\pu{2.0 M}"},{"type":"chemical equation","value":"HCl + Al -> AlCl3 + H2"}] | <h1 class="questionTitle" itemprop="name">How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl? __ HCl + __ Al __ AlCl3 + __ H2 </h1> | null | 0.63 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thanks for your <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> question.</p>
<p>In looking at the balanced equation, we see:</p>
<p><mathjax>#6 HCl + 2 Al -> 2 AlCl_3 +3H_2#</mathjax></p>
<p>The first thing we need to do is to determine the number of moles of HCl that are reacting.</p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles/liters of solution</p>
<p>2.0 M = X/.035 L X = .070 moles of HCl</p>
<p>Now, we can determine the grams of Al that will be needed to react with .070 moles of HCl by using the balanced equation.</p>
<p><mathjax>#".070 moles HCl" * ( "2 mol Al"/"6 mol HCl")("26.98g Al"/"1 mol Al")#</mathjax></p>
<p>= .63 g Al</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>.63 g Al is required to react with 35 mL of 2.0 M hydrochloric acid.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thanks for your <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> question.</p>
<p>In looking at the balanced equation, we see:</p>
<p><mathjax>#6 HCl + 2 Al -> 2 AlCl_3 +3H_2#</mathjax></p>
<p>The first thing we need to do is to determine the number of moles of HCl that are reacting.</p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles/liters of solution</p>
<p>2.0 M = X/.035 L X = .070 moles of HCl</p>
<p>Now, we can determine the grams of Al that will be needed to react with .070 moles of HCl by using the balanced equation.</p>
<p><mathjax>#".070 moles HCl" * ( "2 mol Al"/"6 mol HCl")("26.98g Al"/"1 mol Al")#</mathjax></p>
<p>= .63 g Al</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl? __ HCl + __ Al __ AlCl3 + __ H2 </h1>
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Christopher P.
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Jul 18, 2014
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<div class="markdown"><p>.63 g Al is required to react with 35 mL of 2.0 M hydrochloric acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thanks for your <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> question.</p>
<p>In looking at the balanced equation, we see:</p>
<p><mathjax>#6 HCl + 2 Al -> 2 AlCl_3 +3H_2#</mathjax></p>
<p>The first thing we need to do is to determine the number of moles of HCl that are reacting.</p>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles/liters of solution</p>
<p>2.0 M = X/.035 L X = .070 moles of HCl</p>
<p>Now, we can determine the grams of Al that will be needed to react with .070 moles of HCl by using the balanced equation.</p>
<p><mathjax>#".070 moles HCl" * ( "2 mol Al"/"6 mol HCl")("26.98g Al"/"1 mol Al")#</mathjax></p>
<p>= .63 g Al</p></div>
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</article> | How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl? __ HCl + __ Al __ AlCl3 + __ H2 | null |
2,378 | a86f1671-6ddd-11ea-ad84-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-of-an-unknown-substance-if-2000-j-of-energy-are-requir | 0.11 J/(g * ℃) | start physical_unit 6 8 specific_heat j/(°c_·_g) qc_end physical_unit 24 25 10 11 energy qc_end physical_unit 24 25 26 28 temperature qc_end physical_unit 24 25 21 22 mass qc_end end | [{"type":"physical unit","value":"Specific heat [OF] an unknown substance [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.11 J/(g * ℃)"}] | [{"type":"physical unit","value":"required energy [OF] the substance [=] \\pu{2000 J}"},{"type":"physical unit","value":"raised temperature [OF] the substance [=] \\pu{45 degrees Celsius}"},{"type":"physical unit","value":"mass [OF] the substance [=] \\pu{400 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat of an unknown substance if 2000 J of energy are required to raise the temperature of 400 grams of the substance 45 degrees Celsius? </h1> | null | 0.11 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> tells you how much heat is needed in order to raise the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Your tool of choice here will be this equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>In your case, the sample is said to have a mass of <mathjax>#"400 g"#</mathjax>. The change in temperature is said to be equal to <mathjax>#45^@"C"#</mathjax>. </p>
<p>You know that in order to increase the temperature of this sample by <mathjax>#45^@"C"#</mathjax>, you need to provide it with <mathjax>#"2000 J"#</mathjax> of energy, which means that its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> will be</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = "2000 J"/("400 g" * 45^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.1 J g"^(-1)""^@"C"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> tells you how much heat is needed in order to raise the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Your tool of choice here will be this equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>In your case, the sample is said to have a mass of <mathjax>#"400 g"#</mathjax>. The change in temperature is said to be equal to <mathjax>#45^@"C"#</mathjax>. </p>
<p>You know that in order to increase the temperature of this sample by <mathjax>#45^@"C"#</mathjax>, you need to provide it with <mathjax>#"2000 J"#</mathjax> of energy, which means that its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> will be</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = "2000 J"/("400 g" * 45^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the specific heat of an unknown substance if 2000 J of energy are required to raise the temperature of 400 grams of the substance 45 degrees Celsius? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-05-28T23:50:18" itemprop="dateCreated">
May 28, 2016
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<div class="markdown"><p><mathjax>#"0.1 J g"^(-1)""^@"C"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> tells you how much heat is needed in order to raise the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Your tool of choice here will be this equation </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>In your case, the sample is said to have a mass of <mathjax>#"400 g"#</mathjax>. The change in temperature is said to be equal to <mathjax>#45^@"C"#</mathjax>. </p>
<p>You know that in order to increase the temperature of this sample by <mathjax>#45^@"C"#</mathjax>, you need to provide it with <mathjax>#"2000 J"#</mathjax> of energy, which means that its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> will be</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = "2000 J"/("400 g" * 45^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>.</p></div>
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</article> | What is the specific heat of an unknown substance if 2000 J of energy are required to raise the temperature of 400 grams of the substance 45 degrees Celsius? | null |
2,379 | a91f2dec-6ddd-11ea-a0bc-ccda262736ce | https://socratic.org/questions/150ml-of-water-at-25oc-is-mixed-325ml-of-80oc-water-in-a-sealed-vacuum-insulated | 64.72 ℃ | start physical_unit 3 3 temperature °c qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 3 3 5 6 temperature qc_end physical_unit 3 3 12 13 temperature qc_end physical_unit 3 3 9 10 volume qc_end physical_unit 3 3 28 29 volume qc_end physical_unit 3 3 37 38 density qc_end end | [{"type":"physical unit","value":"Temperature3 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"64.72 ℃"}] | [{"type":"physical unit","value":"Volume1 [OF] water [=] \\pu{150 mL}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{80 ℃}"},{"type":"physical unit","value":"Volume2 [OF] water [=] \\pu{325 mL}"},{"type":"physical unit","value":"Volume3 [OF] water [=] \\pu{475 mL}"},{"type":"physical unit","value":"Density [OF] water [=] \\pu{1 g/mL}"}] | <h1 class="questionTitle" itemprop="name">150mL of water at 25oC is mixed 325mL of 80oC water in a sealed vacuum insulated vessel. Determine the final temp of the combined 475mL of water. The density of water is 1g/mL. What's the final temperature??</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I got 127.23 degrees celsius. Am I correct?</p></div>
</h2>
</div>
</div> | 64.72 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Method of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">Mixtures</a> <mathjax>#=> T_(f"i"nal) = 64.7^oC#</mathjax> see written explanation below.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/H2SZAeAKQ5uweoKOQErS_Method%20of%20Mixtures%20Calculation%20.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final mix temp is <mathjax>#64.7^oC#</mathjax>. Note: Generally, for mixing two quantities of water at different temperatures, the final temperature will be between the initial temp and the final temp. Such is a good way to determine if your calculation is in the proper range. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Method of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">Mixtures</a> <mathjax>#=> T_(f"i"nal) = 64.7^oC#</mathjax> see written explanation below.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/H2SZAeAKQ5uweoKOQErS_Method%20of%20Mixtures%20Calculation%20.jpg"/> </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">150mL of water at 25oC is mixed 325mL of 80oC water in a sealed vacuum insulated vessel. Determine the final temp of the combined 475mL of water. The density of water is 1g/mL. What's the final temperature??</h1>
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<div class="markdown"><p>The final mix temp is <mathjax>#64.7^oC#</mathjax>. Note: Generally, for mixing two quantities of water at different temperatures, the final temperature will be between the initial temp and the final temp. Such is a good way to determine if your calculation is in the proper range. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Method of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">Mixtures</a> <mathjax>#=> T_(f"i"nal) = 64.7^oC#</mathjax> see written explanation below.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/H2SZAeAKQ5uweoKOQErS_Method%20of%20Mixtures%20Calculation%20.jpg"/> </p></div>
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</article> | 150mL of water at 25oC is mixed 325mL of 80oC water in a sealed vacuum insulated vessel. Determine the final temp of the combined 475mL of water. The density of water is 1g/mL. What's the final temperature?? |
I got 127.23 degrees celsius. Am I correct?
|
2,380 | ac0fec0d-6ddd-11ea-81f7-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-0-7891-mol-of-ferric-oxide-fe-2o-3 | 128 g | start physical_unit 10 10 mass g qc_end physical_unit 10 10 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] Fe2O3 [IN] g"}] | [{"type":"physical unit","value":"128 g"}] | [{"type":"physical unit","value":"Mole [OF] Fe2O3 [=] \\pu{0.7891 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 0.7891 mol of ferric oxide (#Fe_2O_3#)?</h1> | null | 128 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the mass of a specific molar quantity, mulitply the number of moles by the molar mass to give an answer in grams:</p>
<p><mathjax>#159.69*g*mol^-1xx0.7891*mol#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#128*g#</mathjax>.</p>
<p>What is the oxidation state of iron in ferric oxide?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Fe_2O_3#</mathjax> has a formula mass of <mathjax>#159.69*g*mol^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the mass of a specific molar quantity, mulitply the number of moles by the molar mass to give an answer in grams:</p>
<p><mathjax>#159.69*g*mol^-1xx0.7891*mol#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#128*g#</mathjax>.</p>
<p>What is the oxidation state of iron in ferric oxide?</p></div>
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May 15, 2016
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<div class="markdown"><p><mathjax>#Fe_2O_3#</mathjax> has a formula mass of <mathjax>#159.69*g*mol^-1#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the mass of a specific molar quantity, mulitply the number of moles by the molar mass to give an answer in grams:</p>
<p><mathjax>#159.69*g*mol^-1xx0.7891*mol#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#128*g#</mathjax>.</p>
<p>What is the oxidation state of iron in ferric oxide?</p></div>
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</article> | What is the mass of 0.7891 mol of ferric oxide (#Fe_2O_3#)? | null |
2,381 | a9243c6a-6ddd-11ea-a7ae-ccda262736ce | https://socratic.org/questions/how-many-moles-is-in-8-g-ca | 0.02 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] Ca [IN] moles"}] | [{"type":"physical unit","value":"0.02 moles"}] | [{"type":"physical unit","value":"Mass [OF] Ca [=] \\pu{0.8 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in .8 g #Ca#?</h1> | null | 0.02 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>no. of moles <mathjax>#=(no of gram)/(molar mass)#</mathjax> =<mathjax>#0.8/40#</mathjax>= 0.02 </p></div>
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<div class="markdown"><p>0.02 moles</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>no. of moles <mathjax>#=(no of gram)/(molar mass)#</mathjax> =<mathjax>#0.8/40#</mathjax>= 0.02 </p></div>
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<div class="markdown"><p>0.02 moles</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>no. of moles <mathjax>#=(no of gram)/(molar mass)#</mathjax> =<mathjax>#0.8/40#</mathjax>= 0.02 </p></div>
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</article> | How many moles are in .8 g #Ca#? | null |
2,382 | ac2e4428-6ddd-11ea-9449-ccda262736ce | https://socratic.org/questions/calculate-ph-of-a-solution-that-resulted-when-40-ml-of-0-1-m-ammonia-nh3-liquid- | 11.0 | start physical_unit 5 5 ph none qc_end physical_unit 15 16 9 10 volume qc_end physical_unit 15 16 13 14 molarity qc_end physical_unit 15 16 20 21 volume qc_end physical_unit 26 27 29 31 ka qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"11.0"}] | [{"type":"physical unit","value":"Volume1 [OF] ammonia solution [=] \\pu{40 mL}"},{"type":"physical unit","value":"Molarity1 [OF] ammonia solution [=] \\pu{0.1 M}"},{"type":"physical unit","value":"Volume2 [OF] ammonia solution [=] \\pu{60 mL}"},{"type":"physical unit","value":"Ka [OF] ammonium cation [=] \\pu{5.7 × 10^(-10)}"}] | <h1 class="questionTitle" itemprop="name">Calculate the #"pH"# of a solution that resulted when #"40 mL"# of a #"0.1-M"# ammonia solution was diluted to #"60 mL"# ?
</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The <mathjax>#K_a#</mathjax> of the ammonium cation is <mathjax>#5.7 * 10^-10#</mathjax></p></div>
</h2>
</div>
</div> | 11.0 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the diluted ammonia solution.</p>
<p>You know that your stock solution contains</p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3#</mathjax></p>
</blockquote>
<p>After you dilute this solution by adding enough water to increase its volume from <mathjax>#"40 mL"#</mathjax> to <mathjax>#"60 mL"#</mathjax>, its molarity will be--remember to use the volume of the solution <strong>in liters</strong>!</p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"#</mathjax></p>
</blockquote>
<p>Now, ammonia will act as a weak base in aqueous solution.</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the expression of the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is given by</p>
<blockquote>
<p><mathjax>#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3^(+)])#</mathjax></p>
</blockquote>
<p>As you know, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))#</mathjax></p>
</blockquote>
<p>Notice that the problem gives you the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, of the ammonium cation, ammonia's conjugate acid. This means that the base dissociation constant for ammonia is equal to</p>
<blockquote>
<p><mathjax>#K_b= (1 * 10^(-14))/(5.7 * 10^(-10)) = 1.75 * 10^(-5)#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the <strong>equilibrium concentration</strong> of the ammonium cations and of the hydroxide anions, you can say that, at equilibrium, the solution will also contain </p>
<blockquote>
<p><mathjax>#["NH"_3] = (0.0667 - x) quad "M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to produce <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of ammonium cations and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydroxide anions, the concentration of ammonia must <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>. </p>
</blockquote>
</blockquote>
<p>Plug this back into the expression of the base dissociation constant to find</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = (x * x)/(0.0667 - x)#</mathjax></p>
<p><mathjax>#1.75 * 10^(-5) = x^2/(0.0667 - x)#</mathjax></p>
</blockquote>
<p>The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation</p>
<blockquote>
<p><mathjax>#0.0667 - x ~~ 0.0667#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = x^2/0.0667#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.0667 * 1.75 * 10^(-5)) = 1.08 * 10^(-3)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of the hydroxide anions, you can say that your solution contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 1.08 * 10^(-3) quad "M"#</mathjax></p>
</blockquote>
<p>Consequently, the <mathjax>#"pH"#</mathjax> of the solution, which can be found using the fact that at <mathjax>#25^@"C"#</mathjax>, an aqueous solution has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#</mathjax></p>
</blockquote>
<p>you can say that your solution has</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [- log(1.08 * 10^(-3))] = color(darkgreen)(ul(color(black)(11.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 11.0#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the diluted ammonia solution.</p>
<p>You know that your stock solution contains</p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3#</mathjax></p>
</blockquote>
<p>After you dilute this solution by adding enough water to increase its volume from <mathjax>#"40 mL"#</mathjax> to <mathjax>#"60 mL"#</mathjax>, its molarity will be--remember to use the volume of the solution <strong>in liters</strong>!</p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"#</mathjax></p>
</blockquote>
<p>Now, ammonia will act as a weak base in aqueous solution.</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the expression of the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is given by</p>
<blockquote>
<p><mathjax>#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3^(+)])#</mathjax></p>
</blockquote>
<p>As you know, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))#</mathjax></p>
</blockquote>
<p>Notice that the problem gives you the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, of the ammonium cation, ammonia's conjugate acid. This means that the base dissociation constant for ammonia is equal to</p>
<blockquote>
<p><mathjax>#K_b= (1 * 10^(-14))/(5.7 * 10^(-10)) = 1.75 * 10^(-5)#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the <strong>equilibrium concentration</strong> of the ammonium cations and of the hydroxide anions, you can say that, at equilibrium, the solution will also contain </p>
<blockquote>
<p><mathjax>#["NH"_3] = (0.0667 - x) quad "M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to produce <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of ammonium cations and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydroxide anions, the concentration of ammonia must <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>. </p>
</blockquote>
</blockquote>
<p>Plug this back into the expression of the base dissociation constant to find</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = (x * x)/(0.0667 - x)#</mathjax></p>
<p><mathjax>#1.75 * 10^(-5) = x^2/(0.0667 - x)#</mathjax></p>
</blockquote>
<p>The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation</p>
<blockquote>
<p><mathjax>#0.0667 - x ~~ 0.0667#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = x^2/0.0667#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.0667 * 1.75 * 10^(-5)) = 1.08 * 10^(-3)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of the hydroxide anions, you can say that your solution contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 1.08 * 10^(-3) quad "M"#</mathjax></p>
</blockquote>
<p>Consequently, the <mathjax>#"pH"#</mathjax> of the solution, which can be found using the fact that at <mathjax>#25^@"C"#</mathjax>, an aqueous solution has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#</mathjax></p>
</blockquote>
<p>you can say that your solution has</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [- log(1.08 * 10^(-3))] = color(darkgreen)(ul(color(black)(11.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Calculate the #"pH"# of a solution that resulted when #"40 mL"# of a #"0.1-M"# ammonia solution was diluted to #"60 mL"# ?
</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The <mathjax>#K_a#</mathjax> of the ammonium cation is <mathjax>#5.7 * 10^-10#</mathjax></p></div>
</h2>
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Stefan V.
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Feb 19, 2018
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<div class="markdown"><p><mathjax>#"pH" = 11.0#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the diluted ammonia solution.</p>
<p>You know that your stock solution contains</p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3#</mathjax></p>
</blockquote>
<p>After you dilute this solution by adding enough water to increase its volume from <mathjax>#"40 mL"#</mathjax> to <mathjax>#"60 mL"#</mathjax>, its molarity will be--remember to use the volume of the solution <strong>in liters</strong>!</p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"#</mathjax></p>
</blockquote>
<p>Now, ammonia will act as a weak base in aqueous solution.</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the expression of the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is given by</p>
<blockquote>
<p><mathjax>#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3^(+)])#</mathjax></p>
</blockquote>
<p>As you know, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))#</mathjax></p>
</blockquote>
<p>Notice that the problem gives you the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, of the ammonium cation, ammonia's conjugate acid. This means that the base dissociation constant for ammonia is equal to</p>
<blockquote>
<p><mathjax>#K_b= (1 * 10^(-14))/(5.7 * 10^(-10)) = 1.75 * 10^(-5)#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the <strong>equilibrium concentration</strong> of the ammonium cations and of the hydroxide anions, you can say that, at equilibrium, the solution will also contain </p>
<blockquote>
<p><mathjax>#["NH"_3] = (0.0667 - x) quad "M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to produce <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of ammonium cations and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydroxide anions, the concentration of ammonia must <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>. </p>
</blockquote>
</blockquote>
<p>Plug this back into the expression of the base dissociation constant to find</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = (x * x)/(0.0667 - x)#</mathjax></p>
<p><mathjax>#1.75 * 10^(-5) = x^2/(0.0667 - x)#</mathjax></p>
</blockquote>
<p>The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation</p>
<blockquote>
<p><mathjax>#0.0667 - x ~~ 0.0667#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-5) = x^2/0.0667#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.0667 * 1.75 * 10^(-5)) = 1.08 * 10^(-3)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of the hydroxide anions, you can say that your solution contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 1.08 * 10^(-3) quad "M"#</mathjax></p>
</blockquote>
<p>Consequently, the <mathjax>#"pH"#</mathjax> of the solution, which can be found using the fact that at <mathjax>#25^@"C"#</mathjax>, an aqueous solution has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#</mathjax></p>
</blockquote>
<p>you can say that your solution has</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [- log(1.08 * 10^(-3))] = color(darkgreen)(ul(color(black)(11.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
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</article> | Calculate the #"pH"# of a solution that resulted when #"40 mL"# of a #"0.1-M"# ammonia solution was diluted to #"60 mL"# ?
|
The #K_a# of the ammonium cation is #5.7 * 10^-10#
|
2,383 | abff6d40-6ddd-11ea-98c3-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-the-solution-32-x-10-2-moles-of-ch-3cooh-in-20-0-liters- | 1.60 M | start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 7 10 mole qc_end physical_unit 17 17 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] CH3COOH solution [IN] M"}] | [{"type":"physical unit","value":"1.60 M"}] | [{"type":"physical unit","value":"Mole [OF] CH3COOH [=] \\pu{0.32 × 10^2 moles}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{20.0 liters}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of the solution #.32 x 10^2 # moles of #CH_3COOH# in 20.0 liters of water?</h1> | null | 1.60 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>0.32 x <mathjax>#10^2#</mathjax> is 32 moles.</p>
<p>You have 32 moles in 20 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles / litre</p>
<p>Molarity = 32 / 20 = 1.6 M</p>
<p>If you want more help with moles and molarity then have a look at this blog post: <a href="http://maths4biosciences.com/blog/concentrations/molarity-formula/" rel="nofollow">Moles and Molarity</a> </p></div>
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<div class="markdown"><p>1.6 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>0.32 x <mathjax>#10^2#</mathjax> is 32 moles.</p>
<p>You have 32 moles in 20 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles / litre</p>
<p>Molarity = 32 / 20 = 1.6 M</p>
<p>If you want more help with moles and molarity then have a look at this blog post: <a href="http://maths4biosciences.com/blog/concentrations/molarity-formula/" rel="nofollow">Moles and Molarity</a> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of the solution #.32 x 10^2 # moles of #CH_3COOH# in 20.0 liters of water?</h1>
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Nick
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<div class="markdown"><p>1.6 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>0.32 x <mathjax>#10^2#</mathjax> is 32 moles.</p>
<p>You have 32 moles in 20 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = moles / litre</p>
<p>Molarity = 32 / 20 = 1.6 M</p>
<p>If you want more help with moles and molarity then have a look at this blog post: <a href="http://maths4biosciences.com/blog/concentrations/molarity-formula/" rel="nofollow">Moles and Molarity</a> </p></div>
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</article> | What is the molarity of the solution #.32 x 10^2 # moles of #CH_3COOH# in 20.0 liters of water? | null |
2,384 | abe679aa-6ddd-11ea-bf6b-ccda262736ce | https://socratic.org/questions/a-sample-of-hydrogen-has-a-volume-of-1107-ml-when-the-temperature-is-101-9-degc--1 | 708.11 mL | start physical_unit 29 30 volume ml qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 14 16 temperature qc_end physical_unit 1 3 21 22 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"708.11 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] hydrogen sample [=] \\pu{1107 mL}"},{"type":"physical unit","value":"Temperature1 [OF] hydrogen sample [=] \\pu{101.9 deg C}"},{"type":"physical unit","value":"Pressure1 [OF] hydrogen sample [=] \\pu{0.867 atm}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of hydrogen has a volume of 1107 mL when the temperature is 101.9 degC and the pressure is 0.867 atm. What will be the volume of the gas at STP?</h1> | null | 708.11 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So what is <mathjax>#"STP"#</mathjax>? Standards differ across curricula. I use the <a href="http://chemistry.about.com/od/gas2/f/What-Is-Stp-In-Chemistry.htm" rel="nofollow">IUPAC</a> standard of <mathjax>#273.15*K#</mathjax>, and <mathjax>#0.98692*atm#</mathjax>. (The pressure is this absurd value because this represents <mathjax>#100*kPa#</mathjax>. Chemists usually prefer to work in <mathjax>#"atmospheres"#</mathjax>, as this may be directly measured in a laboratory with a mercury column.)</p>
<p>For the calculation, <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1*V_1*T_2)/(T_1*P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.867*atmxx1107*mLxx273.15*K)/(375.1*Kxx0.987*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p>
<p>Clearly, the expression gives an answer in <mathjax>#mL#</mathjax>, as is required for a volume. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So what is <mathjax>#"STP"#</mathjax>? Standards differ across curricula. I use the <a href="http://chemistry.about.com/od/gas2/f/What-Is-Stp-In-Chemistry.htm" rel="nofollow">IUPAC</a> standard of <mathjax>#273.15*K#</mathjax>, and <mathjax>#0.98692*atm#</mathjax>. (The pressure is this absurd value because this represents <mathjax>#100*kPa#</mathjax>. Chemists usually prefer to work in <mathjax>#"atmospheres"#</mathjax>, as this may be directly measured in a laboratory with a mercury column.)</p>
<p>For the calculation, <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1*V_1*T_2)/(T_1*P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.867*atmxx1107*mLxx273.15*K)/(375.1*Kxx0.987*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p>
<p>Clearly, the expression gives an answer in <mathjax>#mL#</mathjax>, as is required for a volume. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of hydrogen has a volume of 1107 mL when the temperature is 101.9 degC and the pressure is 0.867 atm. What will be the volume of the gas at STP?</h1>
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anor277
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<span class="dateCreated" datetime="2016-06-20T16:55:31" itemprop="dateCreated">
Jun 20, 2016
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So what is <mathjax>#"STP"#</mathjax>? Standards differ across curricula. I use the <a href="http://chemistry.about.com/od/gas2/f/What-Is-Stp-In-Chemistry.htm" rel="nofollow">IUPAC</a> standard of <mathjax>#273.15*K#</mathjax>, and <mathjax>#0.98692*atm#</mathjax>. (The pressure is this absurd value because this represents <mathjax>#100*kPa#</mathjax>. Chemists usually prefer to work in <mathjax>#"atmospheres"#</mathjax>, as this may be directly measured in a laboratory with a mercury column.)</p>
<p>For the calculation, <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1*V_1*T_2)/(T_1*P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.867*atmxx1107*mLxx273.15*K)/(375.1*Kxx0.987*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p>
<p>Clearly, the expression gives an answer in <mathjax>#mL#</mathjax>, as is required for a volume. </p></div>
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</article> | A sample of hydrogen has a volume of 1107 mL when the temperature is 101.9 degC and the pressure is 0.867 atm. What will be the volume of the gas at STP? | null |
2,385 | a8e138d2-6ddd-11ea-97b7-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-solubility-moles-l-of-mg-oh-2-in-water-the-ksp-5-0-10-1 | 2.32 × 10^(-4) moles/L | start physical_unit 8 8 solubility mol/l qc_end physical_unit 8 8 13 15 equilibrium_constant_k qc_end substance 10 10 qc_end end | [{"type":"physical unit","value":"Solubility [OF] Mg(OH)2 [IN] moles/L"}] | [{"type":"physical unit","value":"2.32 × 10^(-4) moles/L"}] | [{"type":"physical unit","value":"Ksp [OF] Mg(OH)2 [=] \\pu{5.0 × 10^(-11)}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the solubility (moles/L) of Mg(OH)2 in water.The Ksp=#5.0*10^-11#?</h1> | null | 2.32 × 10^(-4) moles/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium..........</p>
<p><mathjax>#Mg(OH)_2(s) rightleftharpoons Mg^(2+) + 2HO^(-)#</mathjax></p>
<p>And <mathjax>#K_"sp"=[Mg^(2+)][""^(-)OH]^2#</mathjax></p>
<p>And now if we call the solubility of <mathjax>#Mg(OH)_2#</mathjax> <mathjax>#S#</mathjax>, then by definition, <mathjax>#S=[Mg^(2+)]#</mathjax>, and <mathjax>#2S=""^(-)OH#</mathjax>.</p>
<p>And thus <mathjax>#K_"sp"=Sxx(2S)^2=4S^3#</mathjax></p>
<p>And so (finally!), <mathjax>#S=""^(3)sqrt(K_"sp"/4)#</mathjax></p>
<p><mathjax>#S=""^(3)sqrt((5.0xx10^-11)/4)=2.32xx10^-4*mol*L^-1#</mathjax></p>
<p>And of course we can convert this to a <mathjax>#"mass solubility"#</mathjax>, by multiplying <mathjax>#S#</mathjax> by the molar mass..........i.e. </p>
<p><mathjax>#58.32*g*mol^-1xx2.32xx10^-4*mol*L^-1=1.35xx10^-2*g*L^-1.#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of a saturated solution? This is a common follow up question.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We interrogate the equilibrium..........and get,</p>
<p><mathjax>#S=2.32xx10^-4*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium..........</p>
<p><mathjax>#Mg(OH)_2(s) rightleftharpoons Mg^(2+) + 2HO^(-)#</mathjax></p>
<p>And <mathjax>#K_"sp"=[Mg^(2+)][""^(-)OH]^2#</mathjax></p>
<p>And now if we call the solubility of <mathjax>#Mg(OH)_2#</mathjax> <mathjax>#S#</mathjax>, then by definition, <mathjax>#S=[Mg^(2+)]#</mathjax>, and <mathjax>#2S=""^(-)OH#</mathjax>.</p>
<p>And thus <mathjax>#K_"sp"=Sxx(2S)^2=4S^3#</mathjax></p>
<p>And so (finally!), <mathjax>#S=""^(3)sqrt(K_"sp"/4)#</mathjax></p>
<p><mathjax>#S=""^(3)sqrt((5.0xx10^-11)/4)=2.32xx10^-4*mol*L^-1#</mathjax></p>
<p>And of course we can convert this to a <mathjax>#"mass solubility"#</mathjax>, by multiplying <mathjax>#S#</mathjax> by the molar mass..........i.e. </p>
<p><mathjax>#58.32*g*mol^-1xx2.32xx10^-4*mol*L^-1=1.35xx10^-2*g*L^-1.#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of a saturated solution? This is a common follow up question.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the solubility (moles/L) of Mg(OH)2 in water.The Ksp=#5.0*10^-11#?</h1>
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<div class="markdown"><p>We interrogate the equilibrium..........and get,</p>
<p><mathjax>#S=2.32xx10^-4*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We interrogate the equilibrium..........</p>
<p><mathjax>#Mg(OH)_2(s) rightleftharpoons Mg^(2+) + 2HO^(-)#</mathjax></p>
<p>And <mathjax>#K_"sp"=[Mg^(2+)][""^(-)OH]^2#</mathjax></p>
<p>And now if we call the solubility of <mathjax>#Mg(OH)_2#</mathjax> <mathjax>#S#</mathjax>, then by definition, <mathjax>#S=[Mg^(2+)]#</mathjax>, and <mathjax>#2S=""^(-)OH#</mathjax>.</p>
<p>And thus <mathjax>#K_"sp"=Sxx(2S)^2=4S^3#</mathjax></p>
<p>And so (finally!), <mathjax>#S=""^(3)sqrt(K_"sp"/4)#</mathjax></p>
<p><mathjax>#S=""^(3)sqrt((5.0xx10^-11)/4)=2.32xx10^-4*mol*L^-1#</mathjax></p>
<p>And of course we can convert this to a <mathjax>#"mass solubility"#</mathjax>, by multiplying <mathjax>#S#</mathjax> by the molar mass..........i.e. </p>
<p><mathjax>#58.32*g*mol^-1xx2.32xx10^-4*mol*L^-1=1.35xx10^-2*g*L^-1.#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of a saturated solution? This is a common follow up question.</p></div>
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</article> | How do you calculate the solubility (moles/L) of Mg(OH)2 in water.The Ksp=#5.0*10^-11#? | null |
2,386 | abe456c0-6ddd-11ea-a637-ccda262736ce | https://socratic.org/questions/56fafbc211ef6b535d702dbb | 3.88 moles | start physical_unit 3 3 mole mol qc_end physical_unit 3 3 8 9 density qc_end physical_unit 3 3 5 6 temperature qc_end physical_unit 3 3 16 17 volume qc_end physical_unit 3 3 24 25 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] CO2 [IN] moles"}] | [{"type":"physical unit","value":"3.88 moles"}] | [{"type":"physical unit","value":"Density [OF] CO2 [=] \\pu{1.98 g/L}"},{"type":"physical unit","value":"Temperature [OF] CO2 [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Volume [OF] CO2 [=] \\pu{88.0 L}"},{"type":"physical unit","value":"Pressure [OF] CO2 [=] \\pu{1 atm}"}] | <h1 class="questionTitle" itemprop="name">The density of #"CO"_2# at #0^@ "C"# is #"1.98 g/L"#. How many mols are found in #"88.0 L"# of #"CO"_2# at this temperature and #"1 atm"#?</h1> | null | 3.88 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <strong>Standard Temperature and Pressure</strong>, <strong>STP</strong>, is that when an ideal gas is being kept under these conditions for <em>pressure</em> and <em>temperature</em>, <strong>one mole</strong> of this gas occupies <mathjax>#"22.7 L"#</mathjax>. </p>
<p>The volume occupied by one mole of an ideal gas at <em>any</em> conditions for pressure and temperature is called the <strong>molar volume</strong>. </p>
<p>Under STP conditions, which are <strong>currently</strong> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, you will be dealing with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>, which is equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>So, you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP as a <em>conversion factor</em> to figure out how many moles of carbon dioxide, <mathjax>#"CO"_2#</mathjax>< would occupy <mathjax>#"88.0 L"#</mathjax></p>
<blockquote>
<p><mathjax>#88.0 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas aat STP")) = "3.87665 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"volume of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"3.88 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>More often than not, you'll see STP conditions being defined as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>.</p>
<p><em>Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. </p>
<p><em>If this is the value given to you for the molar volume of a gas at STP, simply redo the calculation using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
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<div>
<div class="markdown"><p><mathjax>#"3.88 moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <strong>Standard Temperature and Pressure</strong>, <strong>STP</strong>, is that when an ideal gas is being kept under these conditions for <em>pressure</em> and <em>temperature</em>, <strong>one mole</strong> of this gas occupies <mathjax>#"22.7 L"#</mathjax>. </p>
<p>The volume occupied by one mole of an ideal gas at <em>any</em> conditions for pressure and temperature is called the <strong>molar volume</strong>. </p>
<p>Under STP conditions, which are <strong>currently</strong> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, you will be dealing with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>, which is equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>So, you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP as a <em>conversion factor</em> to figure out how many moles of carbon dioxide, <mathjax>#"CO"_2#</mathjax>< would occupy <mathjax>#"88.0 L"#</mathjax></p>
<blockquote>
<p><mathjax>#88.0 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas aat STP")) = "3.87665 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"volume of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"3.88 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>More often than not, you'll see STP conditions being defined as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>.</p>
<p><em>Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. </p>
<p><em>If this is the value given to you for the molar volume of a gas at STP, simply redo the calculation using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The density of #"CO"_2# at #0^@ "C"# is #"1.98 g/L"#. How many mols are found in #"88.0 L"# of #"CO"_2# at this temperature and #"1 atm"#?</h1>
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<div class="markdown"><p><mathjax>#"3.88 moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <strong>Standard Temperature and Pressure</strong>, <strong>STP</strong>, is that when an ideal gas is being kept under these conditions for <em>pressure</em> and <em>temperature</em>, <strong>one mole</strong> of this gas occupies <mathjax>#"22.7 L"#</mathjax>. </p>
<p>The volume occupied by one mole of an ideal gas at <em>any</em> conditions for pressure and temperature is called the <strong>molar volume</strong>. </p>
<p>Under STP conditions, which are <strong>currently</strong> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>, you will be dealing with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>, which is equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>So, you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP as a <em>conversion factor</em> to figure out how many moles of carbon dioxide, <mathjax>#"CO"_2#</mathjax>< would occupy <mathjax>#"88.0 L"#</mathjax></p>
<blockquote>
<p><mathjax>#88.0 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas aat STP")) = "3.87665 moles"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"volume of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"3.88 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>More often than not, you'll see STP conditions being defined as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>.</p>
<p><em>Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. </p>
<p><em>If this is the value given to you for the molar volume of a gas at STP, simply redo the calculation using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
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Truong-Son N.
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<span class="dateCreated" datetime="2016-04-08T01:10:34" itemprop="dateCreated">
Apr 8, 2016
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<div class="markdown"><p>Here's an alternative way to do this, without requiring the assumption that <mathjax>#"CO"_2#</mathjax> is ideal (though it is quite close).</p>
<p>If you use the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"CO"_2#</mathjax> (which is given as <mathjax>#"1.98 g/L"#</mathjax> at <mathjax>#0^@ "C"#</mathjax>), it is possible to solve for the exact number of <mathjax>#"mol"#</mathjax>s. </p>
<p>The main catch with using the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> is that it assumes all gases have the <em>same</em> <strong>molar volume</strong> at STP, and of course, the "more realistic" gases don't.</p>
<p>When you use the experimental density and <em>convert</em> that to the <strong>molar density</strong>, it circumvents that assumption because you'd be using all experimental data. Here's how you do it:</p>
<blockquote>
<p><mathjax>#1/(barV) = barrho = n/V = m/(VM_r) = \mathbf((rho)/M_r)#</mathjax></p>
<p>where:</p>
</blockquote>
<ul>
<li><mathjax>#V#</mathjax> is volume in <mathjax>#"L"#</mathjax></li>
<li><mathjax>#n#</mathjax> is <mathjax>#"mol"#</mathjax>s</li>
<li><mathjax>#M_r#</mathjax> is relative molar mass in <mathjax>#"g/mol"#</mathjax></li>
<li><mathjax>#m#</mathjax> is mass in <mathjax>#"g"#</mathjax></li>
<li><mathjax>#rho#</mathjax> is the mass density in <mathjax>#"g/mol"#</mathjax></li>
</ul>
<p>The molar density has units of <mathjax>#"mol/L"#</mathjax>. Thus, simply multiply by the volume to get:</p>
<blockquote>
<p><mathjax>#"mol CO"_2 = ("1.98" cancel"g")/(cancel"L")xx("mol")/(44.009 cancel"g") xx 88.0 cancel"L"#</mathjax></p>
<p><mathjax># = color(blue)("3.92 mol CO"_2)#</mathjax></p>
</blockquote>
<p>And this is larger but somewhat close to Stefan's answer of <mathjax>#"3.88 mol"#</mathjax> by about <mathjax>#1.2%#</mathjax>. </p>
<p><mathjax>#Z = (PV)/(nRT) = 0.99435 ~~ 1#</mathjax> at <mathjax>#15^@ "C"#</mathjax> suggests that <mathjax>#"CO"_2#</mathjax> is actually easier to compress than an ideal gas. </p>
<p>This answer agrees with that statement because there are a greater number of <mathjax>#"mol"#</mathjax>s of <mathjax>#"CO"_2#</mathjax> than expected for an ideal gas for the same volume, meaning that it is more compressed and thus more dense than an ideal gas.</p></div>
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</article> | The density of #"CO"_2# at #0^@ "C"# is #"1.98 g/L"#. How many mols are found in #"88.0 L"# of #"CO"_2# at this temperature and #"1 atm"#? | null |
2,387 | a8c077f1-6ddd-11ea-9b87-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-458-grams-of-na-2so-4 | 3.22 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] Na2SO4 [IN] moles"}] | [{"type":"physical unit","value":"3.22 moles"}] | [{"type":"physical unit","value":"Mass [OF] Na2SO4 [=] \\pu{458 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in 458 grams of #Na_2SO_4#?</h1> | null | 3.22 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of <mathjax>#"Na"_2"SO"_4 = "142.04 g/mol"#</mathjax></p>
<p>Number of moles in <mathjax>#"458 g"#</mathjax> of it is<mathjax>#= "458 g"/"142.04 g/mol" = "3.22 mol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"3.22 mol"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of <mathjax>#"Na"_2"SO"_4 = "142.04 g/mol"#</mathjax></p>
<p>Number of moles in <mathjax>#"458 g"#</mathjax> of it is<mathjax>#= "458 g"/"142.04 g/mol" = "3.22 mol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"3.22 mol"#</mathjax></p></div>
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<div class="markdown"><p>Molar mass of <mathjax>#"Na"_2"SO"_4 = "142.04 g/mol"#</mathjax></p>
<p>Number of moles in <mathjax>#"458 g"#</mathjax> of it is<mathjax>#= "458 g"/"142.04 g/mol" = "3.22 mol"#</mathjax></p></div>
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</article> | How many moles are there in 458 grams of #Na_2SO_4#? | null |
2,388 | acbe4870-6ddd-11ea-94ef-ccda262736ce | https://socratic.org/questions/the-formula-for-the-oxalate-ion-is-c-2o-4-2-what-would-you-predict-the-formula-f | C2H2O4 | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] oxalic acid [IN] default"}] | [{"type":"chemical equation","value":"C2H2O4"}] | [{"type":"other","value":"The formula for the oxalate ion is C2O4^2-."}] | <h1 class="questionTitle" itemprop="name">The formula for the oxalate ion is #C_2O_4^(2-)#. What would you predict the formula for oxalic acid to be?</h1> | null | C2H2O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2O_4^(2-)#</mathjax> is the dianion; <mathjax>#HO(O=)C-C(=O)H#</mathjax> is the parent diacid. </p></div>
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<div class="markdown"><p>Add two protons.............?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2O_4^(2-)#</mathjax> is the dianion; <mathjax>#HO(O=)C-C(=O)H#</mathjax> is the parent diacid. </p></div>
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<h1 class="questionTitle" itemprop="name">The formula for the oxalate ion is #C_2O_4^(2-)#. What would you predict the formula for oxalic acid to be?</h1>
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anor277
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<div class="markdown"><p>Add two protons.............?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#C_2O_4^(2-)#</mathjax> is the dianion; <mathjax>#HO(O=)C-C(=O)H#</mathjax> is the parent diacid. </p></div>
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<div class="markdown"><p><mathjax>#C_(2)H_(2)O_(4)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An acid consists of positively-charged hydrogen ion(s) and an anion. This can be symbolized in a chemical equation: <mathjax>#nH^(+)+A^(-n)->H_(n)A#</mathjax></p>
<p>Also, we can use the <a href="https://en.wikipedia.org/wiki/Chemical_formula#Hill_system" rel="nofollow"><em>Hill System</em></a> to determine the order to put <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the chemical formula.</p>
<p>"The Hill system (or Hill notation) is a system of writing empirical chemical formulas, molecular chemical formulas and components of a condensed formula such that the number of <strong><em>carbon atoms</em></strong> in a molecule is indicated first, the number of <strong><em>hydrogen atoms</em></strong> next, and then the number of all other chemical elements subsequently, in alphabetical order of the chemical symbols. When the formula contains no carbon, all the elements, including hydrogen, are listed alphabetically."</p>
<p>Therefore the formula should be <mathjax>#C_(2)H_(2)O_(4)#</mathjax>.</p>
<p>Reference: </p>
<ul>
<li><a href="http://www.ck12.org/chemistry/Naming-Acids/lesson/Naming-Acids-CHEM/" rel="nofollow" target="_blank">http://www.ck12.org/chemistry/Naming-Acids/lesson/Naming-Acids-CHEM/</a></li>
<li><a href="https://chemistry.stackexchange.com/questions/1239/order-of-elements-in-a-formula" rel="nofollow"></a><a href="https://chemistry.stackexchange.com/questions/1239/order-of-elements-in-a-formula" rel="nofollow" target="_blank">https://chemistry.stackexchange.com/questions/1239/order-of-elements-in-a-formula</a> </li>
<li><a href="https://en.wikipedia.org/wiki/Chemical_formula#Hill_system" rel="nofollow"></a><a href="https://en.wikipedia.org/wiki/Chemical_formula" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Chemical_formula</a> </li>
</ul></div>
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</article> | The formula for the oxalate ion is #C_2O_4^(2-)#. What would you predict the formula for oxalic acid to be? | null |
2,389 | a8a7e3a9-6ddd-11ea-afe2-ccda262736ce | https://socratic.org/questions/how-many-moles-of-co-2-are-formed-from-5-mole-of-c-8h-18-with-ample-o-2 | 4.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 8 9 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] CO2 [IN] moles"}] | [{"type":"physical unit","value":"4.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] C8H18 [=] \\pu{0.5 mole}"},{"type":"other","value":"Ample O2."}] | <h1 class="questionTitle" itemprop="name">How many moles of #CO_2# are formed from #.5# mole of #C_8H_18# with ample #O_2#?</h1> | null | 4.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So how many moles of product if only <mathjax>#1/2#</mathjax> a mole of octane were used? </p></div>
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<div class="markdown"><p>Complete combustion of octane:<br/>
<mathjax>#C_8H_18(g) + 17/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So how many moles of product if only <mathjax>#1/2#</mathjax> a mole of octane were used? </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #CO_2# are formed from #.5# mole of #C_8H_18# with ample #O_2#?</h1>
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<div class="markdown"><p>Complete combustion of octane:<br/>
<mathjax>#C_8H_18(g) + 17/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p></div>
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<div class="markdown"><p>So how many moles of product if only <mathjax>#1/2#</mathjax> a mole of octane were used? </p></div>
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</article> | How many moles of #CO_2# are formed from #.5# mole of #C_8H_18# with ample #O_2#? | null |
2,390 | ac6a21f4-6ddd-11ea-ab11-ccda262736ce | https://socratic.org/questions/if-25-21-ml-of-naoh-solution-is-required-to-react-completely-with-550-g-khp-what | 0.11 M | start physical_unit 4 5 molarity mol/l qc_end physical_unit 4 5 1 2 volume qc_end physical_unit 14 14 12 13 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [IN] M"}] | [{"type":"physical unit","value":"0.11 M"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{25.21 mL}"},{"type":"physical unit","value":"Mass [OF] KHP [=] \\pu{550 g}"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">If 25.21 mL of #NaOH# solution is required to react completely with .550 g #KHP#, what is the molarity of the #NaOH# solution?</h1> | null | 0.11 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide. </p>
<p><mathjax>#"Moles of KHP"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.550*g)/(204.22*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.69xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of NaOH(aq) solution:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.69xx10^-3*mol)/(25.21xx10^-3*L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.100*mol*L^-1#</mathjax>, but do the calculation for an exact value. </p>
<p><mathjax>#"KHP"#</mathjax> is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution. </p></div>
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<div class="markdown"><p>Rxn: <mathjax>#C_6H_4(CO_2H)(CO_2^(-)K^(+)) + NaOH(aq) rarr C_6H_4(CO_2^(-)Na^+)(CO_2^(-)K^(+)) + H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide. </p>
<p><mathjax>#"Moles of KHP"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.550*g)/(204.22*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.69xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of NaOH(aq) solution:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.69xx10^-3*mol)/(25.21xx10^-3*L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.100*mol*L^-1#</mathjax>, but do the calculation for an exact value. </p>
<p><mathjax>#"KHP"#</mathjax> is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution. </p></div>
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<h1 class="questionTitle" itemprop="name">If 25.21 mL of #NaOH# solution is required to react completely with .550 g #KHP#, what is the molarity of the #NaOH# solution?</h1>
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<div class="markdown"><p>Rxn: <mathjax>#C_6H_4(CO_2H)(CO_2^(-)K^(+)) + NaOH(aq) rarr C_6H_4(CO_2^(-)Na^+)(CO_2^(-)K^(+)) + H_2O#</mathjax></p></div>
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<div class="markdown"><p>Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide. </p>
<p><mathjax>#"Moles of KHP"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.550*g)/(204.22*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.69xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of NaOH(aq) solution:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.69xx10^-3*mol)/(25.21xx10^-3*L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.100*mol*L^-1#</mathjax>, but do the calculation for an exact value. </p>
<p><mathjax>#"KHP"#</mathjax> is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution. </p></div>
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</article> | If 25.21 mL of #NaOH# solution is required to react completely with .550 g #KHP#, what is the molarity of the #NaOH# solution? | null |
2,391 | ad21c698-6ddd-11ea-878e-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-prepared-by-dissolving-141-6-g-of-citric-acid | 0.21 M | start physical_unit 15 15 molarity mol/l qc_end physical_unit 15 15 10 11 mass qc_end physical_unit 22 23 25 26 volume qc_end substance 17 17 qc_end end | [{"type":"physical unit","value":"Molarity [OF] C3H5O(COOH)3 solution [IN] M"}] | [{"type":"physical unit","value":"0.21 M"}] | [{"type":"physical unit","value":"Mass [OF] C3H5O(COOH)3 [=] \\pu{141.6 g}"},{"type":"physical unit","value":"Volume [OF] resulting solution [=] \\pu{3500.0 mL}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by dissolving 141.6 g of citric acid, #C_3H_5O(COOH)_3#, in water and then diluting the resulting solution to 3500.0 mL?</h1> | null | 0.21 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the mols of a compound/element over 1 liter of solution</p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p>you first need to convert 141.6g of citric acid into mols </p>
<p><mathjax>#C#</mathjax> = 12.01g * 6 = 72.06g<br/>
<mathjax>#H#</mathjax> = 1.0079g * 8 = 8.0632g<br/>
<mathjax>#O#</mathjax> = 16.00g * 7 = 112.0g<br/>
<mathjax>#C_3H_5O(COOH)_3#</mathjax>= 192.12g</p>
<p>141.6g <mathjax>#C_3H_5O(COOH)_3#</mathjax> <mathjax>#(1mol C_3H_5O(COOH)_3)/(192.12g)#</mathjax>=.7370 mol <mathjax>#C_3H_5O(COOH)_3#</mathjax></p>
<p>We not have the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax>, and we are given it is diluted to 3500.0 mL</p>
<p>3500.0 mL <mathjax>#(1Liter)/(1000mL)#</mathjax>=3.5000<mathjax>#L#</mathjax></p>
<p>plug both the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax> and <mathjax>#L#</mathjax> of <mathjax>#C_3H_5O(COOH)_3#</mathjax> into </p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(.7370 mol)/(3.5000L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>= .2106</p>
<p>.2106<mathjax>#M#</mathjax> <mathjax>#C_3H_5O(COOH)_3#</mathjax></p></div>
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<div class="markdown"><p>.2106<mathjax>#M#</mathjax> <mathjax>#C_3H_5O(COOH)_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the mols of a compound/element over 1 liter of solution</p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p>you first need to convert 141.6g of citric acid into mols </p>
<p><mathjax>#C#</mathjax> = 12.01g * 6 = 72.06g<br/>
<mathjax>#H#</mathjax> = 1.0079g * 8 = 8.0632g<br/>
<mathjax>#O#</mathjax> = 16.00g * 7 = 112.0g<br/>
<mathjax>#C_3H_5O(COOH)_3#</mathjax>= 192.12g</p>
<p>141.6g <mathjax>#C_3H_5O(COOH)_3#</mathjax> <mathjax>#(1mol C_3H_5O(COOH)_3)/(192.12g)#</mathjax>=.7370 mol <mathjax>#C_3H_5O(COOH)_3#</mathjax></p>
<p>We not have the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax>, and we are given it is diluted to 3500.0 mL</p>
<p>3500.0 mL <mathjax>#(1Liter)/(1000mL)#</mathjax>=3.5000<mathjax>#L#</mathjax></p>
<p>plug both the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax> and <mathjax>#L#</mathjax> of <mathjax>#C_3H_5O(COOH)_3#</mathjax> into </p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(.7370 mol)/(3.5000L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>= .2106</p>
<p>.2106<mathjax>#M#</mathjax> <mathjax>#C_3H_5O(COOH)_3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by dissolving 141.6 g of citric acid, #C_3H_5O(COOH)_3#, in water and then diluting the resulting solution to 3500.0 mL?</h1>
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<div class="markdown"><p>.2106<mathjax>#M#</mathjax> <mathjax>#C_3H_5O(COOH)_3#</mathjax></p></div>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the mols of a compound/element over 1 liter of solution</p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p>you first need to convert 141.6g of citric acid into mols </p>
<p><mathjax>#C#</mathjax> = 12.01g * 6 = 72.06g<br/>
<mathjax>#H#</mathjax> = 1.0079g * 8 = 8.0632g<br/>
<mathjax>#O#</mathjax> = 16.00g * 7 = 112.0g<br/>
<mathjax>#C_3H_5O(COOH)_3#</mathjax>= 192.12g</p>
<p>141.6g <mathjax>#C_3H_5O(COOH)_3#</mathjax> <mathjax>#(1mol C_3H_5O(COOH)_3)/(192.12g)#</mathjax>=.7370 mol <mathjax>#C_3H_5O(COOH)_3#</mathjax></p>
<p>We not have the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax>, and we are given it is diluted to 3500.0 mL</p>
<p>3500.0 mL <mathjax>#(1Liter)/(1000mL)#</mathjax>=3.5000<mathjax>#L#</mathjax></p>
<p>plug both the mols of <mathjax>#C_3H_5O(COOH)_3#</mathjax> and <mathjax>#L#</mathjax> of <mathjax>#C_3H_5O(COOH)_3#</mathjax> into </p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(Mols)/(1L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>(Molarity)= <mathjax>#(.7370 mol)/(3.5000L solution)#</mathjax></p>
<p><mathjax>#M#</mathjax>= .2106</p>
<p>.2106<mathjax>#M#</mathjax> <mathjax>#C_3H_5O(COOH)_3#</mathjax></p></div>
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</article> | What is the molarity of a solution prepared by dissolving 141.6 g of citric acid, #C_3H_5O(COOH)_3#, in water and then diluting the resulting solution to 3500.0 mL? | null |
2,392 | ab2e6a66-6ddd-11ea-b02d-ccda262736ce | https://socratic.org/questions/what-is-the-poh-of-a-4-8x10-10-m-h-solution | 4.68 | start physical_unit 11 11 poh none qc_end end | [{"type":"physical unit","value":"pOH [OF] the solution"}] | [{"type":"physical unit","value":"4.68"}] | [{"type":"physical unit","value":"Molarity [OF] H+ solutiom [=] \\pu{4.8 × 10^(-10) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pOH of a #4.8xx10^(-10) M# #H^+# solution?</h1> | null | 4.68 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>For aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> we know that <mathjax>#pH+pOH=14#</mathjax>...</p>
<p>And so <mathjax>#pOH=14-pH#</mathjax>...</p>
<p>And since ........... </p>
<p><mathjax>#pH=-log_10[H^+]=-log_10(4.8xx10^-10)=-(-9.32)=+9.32....#</mathjax></p>
<p><mathjax>#pOH=14-9.32=4.68#</mathjax>...</p></div>
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<div class="markdown"><p><mathjax>#pOH-=4.68#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> we know that <mathjax>#pH+pOH=14#</mathjax>...</p>
<p>And so <mathjax>#pOH=14-pH#</mathjax>...</p>
<p>And since ........... </p>
<p><mathjax>#pH=-log_10[H^+]=-log_10(4.8xx10^-10)=-(-9.32)=+9.32....#</mathjax></p>
<p><mathjax>#pOH=14-9.32=4.68#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pOH of a #4.8xx10^(-10) M# #H^+# solution?</h1>
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<div class="markdown"><p><mathjax>#pOH-=4.68#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>For aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> we know that <mathjax>#pH+pOH=14#</mathjax>...</p>
<p>And so <mathjax>#pOH=14-pH#</mathjax>...</p>
<p>And since ........... </p>
<p><mathjax>#pH=-log_10[H^+]=-log_10(4.8xx10^-10)=-(-9.32)=+9.32....#</mathjax></p>
<p><mathjax>#pOH=14-9.32=4.68#</mathjax>...</p></div>
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</article> | What is the pOH of a #4.8xx10^(-10) M# #H^+# solution? | null |
2,393 | aab67264-6ddd-11ea-8540-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-3-3-l-at-25-c-and-1-3-atm-at-what-temperature-would-the-ga | 174 K | start physical_unit 18 19 temperature k qc_end physical_unit 18 19 9 10 temperature qc_end physical_unit 18 19 12 13 pressure qc_end physical_unit 18 19 6 7 volume qc_end physical_unit 18 19 21 22 volume qc_end physical_unit 18 19 24 25 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"174 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.3 atm}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{3.3 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{2.5 L}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{1 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 3.3 L at 25°C and 1.3 atm; at what temperature would the gas occupy 2.5 L at 1 atm? </h1> | null | 174 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1 / T_1 = P_2V_2 / T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (P_2V_2 T_1)//(P_1 V_1#</mathjax>)</p>
<p><mathjax>#T_2#</mathjax> =(1.0 atm ) (2.5 L) <mathjax>#298 K#</mathjax> // 1.3 atm x 3.3 L</p>
<p><mathjax>#T_2#</mathjax> = 745 atm L K // 4.29 L atm</p>
<p><mathjax>#T_2#</mathjax> = 174 K</p></div>
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<div class="markdown"><p>174 K</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1 / T_1 = P_2V_2 / T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (P_2V_2 T_1)//(P_1 V_1#</mathjax>)</p>
<p><mathjax>#T_2#</mathjax> =(1.0 atm ) (2.5 L) <mathjax>#298 K#</mathjax> // 1.3 atm x 3.3 L</p>
<p><mathjax>#T_2#</mathjax> = 745 atm L K // 4.29 L atm</p>
<p><mathjax>#T_2#</mathjax> = 174 K</p></div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 3.3 L at 25°C and 1.3 atm; at what temperature would the gas occupy 2.5 L at 1 atm? </h1>
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<div class="markdown"><p>174 K</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1 / T_1 = P_2V_2 / T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (P_2V_2 T_1)//(P_1 V_1#</mathjax>)</p>
<p><mathjax>#T_2#</mathjax> =(1.0 atm ) (2.5 L) <mathjax>#298 K#</mathjax> // 1.3 atm x 3.3 L</p>
<p><mathjax>#T_2#</mathjax> = 745 atm L K // 4.29 L atm</p>
<p><mathjax>#T_2#</mathjax> = 174 K</p></div>
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</article> | A gas has a volume of 3.3 L at 25°C and 1.3 atm; at what temperature would the gas occupy 2.5 L at 1 atm? | null |
2,394 | abf3afd3-6ddd-11ea-9cb6-ccda262736ce | https://socratic.org/questions/at-stp-what-is-the-volume-of-7-08-mol-of-nitrogen-gas | 158.59 L | start physical_unit 10 11 volume l qc_end c_other STP qc_end physical_unit 10 11 7 8 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [IN] L"}] | [{"type":"physical unit","value":"158.59 L"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Mole [OF] nitrogen gas [=] \\pu{7.08 mol}"}] | <h1 class="questionTitle" itemprop="name">At STP, what is the volume of 7.08 mol of nitrogen gas?</h1> | null | 158.59 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP the molar volume of all gases is <mathjax>#22.4dm^3//mol.#</mathjax></p>
<p>So hence <mathjax>#7.08mol#</mathjax> of any gas at STP would occupy a volume of <mathjax>#7.08xx22.4=158.592dm^3#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#158.592dm^3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP the molar volume of all gases is <mathjax>#22.4dm^3//mol.#</mathjax></p>
<p>So hence <mathjax>#7.08mol#</mathjax> of any gas at STP would occupy a volume of <mathjax>#7.08xx22.4=158.592dm^3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">At STP, what is the volume of 7.08 mol of nitrogen gas?</h1>
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<div class="markdown"><p><mathjax>#158.592dm^3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At STP the molar volume of all gases is <mathjax>#22.4dm^3//mol.#</mathjax></p>
<p>So hence <mathjax>#7.08mol#</mathjax> of any gas at STP would occupy a volume of <mathjax>#7.08xx22.4=158.592dm^3#</mathjax>.</p></div>
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</article> | At STP, what is the volume of 7.08 mol of nitrogen gas? | null |
2,395 | ac2734dc-6ddd-11ea-a036-ccda262736ce | https://socratic.org/questions/how-many-silver-atoms-are-there-in-2-00-moles-of-silver | 1.20 × 10^24 | 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] silver atoms"}] | [{"type":"physical unit","value":"1.20 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] silver [=] \\pu{2.00 moles}"}] | <h1 class="questionTitle" itemprop="name">How many silver atoms are there in 2.00 moles of silver? </h1> | null | 1.20 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*mol#</mathjax> simply specifies a number, i.e. <mathjax>#N_A=6.0222xx10^23*mol^-1#</mathjax>.</p>
<p>And, thus in <mathjax>#2.00*mol#</mathjax> of silver, there are <mathjax>#2*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*cancel(mol^-1)xx2*cancel(mol)=1.20xx10^24#</mathjax> <mathjax>#"individual silver atoms."#</mathjax></p>
<p>What is the mass of this number, this quantity of silver atoms?</p></div>
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<div class="markdown"><p>How many eggses in 2 dozen eggs?</p></div>
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<div class="markdown"><p><mathjax>#1*mol#</mathjax> simply specifies a number, i.e. <mathjax>#N_A=6.0222xx10^23*mol^-1#</mathjax>.</p>
<p>And, thus in <mathjax>#2.00*mol#</mathjax> of silver, there are <mathjax>#2*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*cancel(mol^-1)xx2*cancel(mol)=1.20xx10^24#</mathjax> <mathjax>#"individual silver atoms."#</mathjax></p>
<p>What is the mass of this number, this quantity of silver atoms?</p></div>
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<h1 class="questionTitle" itemprop="name">How many silver atoms are there in 2.00 moles of silver? </h1>
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<div class="markdown"><p><mathjax>#1*mol#</mathjax> simply specifies a number, i.e. <mathjax>#N_A=6.0222xx10^23*mol^-1#</mathjax>.</p>
<p>And, thus in <mathjax>#2.00*mol#</mathjax> of silver, there are <mathjax>#2*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*cancel(mol^-1)xx2*cancel(mol)=1.20xx10^24#</mathjax> <mathjax>#"individual silver atoms."#</mathjax></p>
<p>What is the mass of this number, this quantity of silver atoms?</p></div>
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</article> | How many silver atoms are there in 2.00 moles of silver? | null |
2,396 | acde06ec-6ddd-11ea-8935-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-methane-given-methane-can-be-decomposed-into-4- | CH4 | start chemical_formula qc_end physical_unit 16 16 13 14 mass qc_end physical_unit 21 21 18 19 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] methane [IN] empirical"}] | [{"type":"chemical equation","value":"CH4"}] | [{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{4.5 g}"},{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{1.5 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of methane given methane can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen?</h1> | null | CH4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal when trying to figure out a compound's <strong>empirical formula</strong> is to find the <strong>smallest whole number ratio</strong> that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>The problem tells you that a sample of methane underwent combustion and produced</p>
<blockquote>
<ul>
<li><mathjax>#"4.5 g carbon, C"#</mathjax></li>
<li><mathjax>#"1.5 g hydrogen, H"#</mathjax></li>
</ul>
</blockquote>
<p>The first thing to do here is to convert these masses to <em>moles</em> by using the <strong>molar masses</strong> of the two elements. You will have</p>
<blockquote>
<p><mathjax>#"For C: " 4.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.3747 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 1.5 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "1.4881 moles H"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong>mole ratio</strong> that existed between these two elements in the sample that underwent combustion, you must divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " (0.3747 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.4881 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 3.97 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist between these two elements, you can say that the empirical formula for methane is </p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("C"_1"H"_4 implies "CH"_4)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"CH"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal when trying to figure out a compound's <strong>empirical formula</strong> is to find the <strong>smallest whole number ratio</strong> that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>The problem tells you that a sample of methane underwent combustion and produced</p>
<blockquote>
<ul>
<li><mathjax>#"4.5 g carbon, C"#</mathjax></li>
<li><mathjax>#"1.5 g hydrogen, H"#</mathjax></li>
</ul>
</blockquote>
<p>The first thing to do here is to convert these masses to <em>moles</em> by using the <strong>molar masses</strong> of the two elements. You will have</p>
<blockquote>
<p><mathjax>#"For C: " 4.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.3747 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 1.5 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "1.4881 moles H"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong>mole ratio</strong> that existed between these two elements in the sample that underwent combustion, you must divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " (0.3747 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.4881 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 3.97 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist between these two elements, you can say that the empirical formula for methane is </p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("C"_1"H"_4 implies "CH"_4)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of methane given methane can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen?</h1>
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<div class="markdown"><p><mathjax>#"CH"_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal when trying to figure out a compound's <strong>empirical formula</strong> is to find the <strong>smallest whole number ratio</strong> that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>The problem tells you that a sample of methane underwent combustion and produced</p>
<blockquote>
<ul>
<li><mathjax>#"4.5 g carbon, C"#</mathjax></li>
<li><mathjax>#"1.5 g hydrogen, H"#</mathjax></li>
</ul>
</blockquote>
<p>The first thing to do here is to convert these masses to <em>moles</em> by using the <strong>molar masses</strong> of the two elements. You will have</p>
<blockquote>
<p><mathjax>#"For C: " 4.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.3747 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 1.5 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "1.4881 moles H"#</mathjax></p>
</blockquote>
<p>Now, in order to find the <strong>mole ratio</strong> that existed between these two elements in the sample that underwent combustion, you must divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " (0.3747 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.4881 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 3.97 ~~ 4#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1:4#</mathjax> is already the <strong>smallest whole number ratio</strong> that can exist between these two elements, you can say that the empirical formula for methane is </p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("C"_1"H"_4 implies "CH"_4)color(white)(a/a)|)))#</mathjax></p>
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</article> | What is the empirical formula of methane given methane can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen? | null |
2,397 | abfc1400-6ddd-11ea-a1d5-ccda262736ce | https://socratic.org/questions/5960e5277c014916cb7082b7 | 4.00 moles | start physical_unit 4 5 mole mol qc_end physical_unit 11 11 7 8 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] hydrogen atoms [IN] moles"}] | [{"type":"physical unit","value":"4.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] CH4 [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How many moles of hydrogen atoms in one mole of methane, #CH_4#?</h1> | null | 4.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of methane specifies <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's number"#</mathjax> of methane molecules, i.e. <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#CH_4#</mathjax> molecules. And thus it contains <mathjax>#4xx6.022xx10^23#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>. </p>
<p>I would have used precisely the same procedure if I were asked the number of <mathjax>#"hydrogen atoms"#</mathjax> in <mathjax>#"4 dozen methane molecules"#</mathjax>; i.e. there are <mathjax>#4xx12=48#</mathjax>, <mathjax>#"four dozen"#</mathjax>, <mathjax>#"hydrogen atoms....."#</mathjax> <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a collective number like a <mathjax>#"dozen"#</mathjax> or a <mathjax>#"gross"#</mathjax>.....admittedly, it is a much larger quantity.</p>
<p>The <mathjax>#"mole"#</mathjax>, however, has a special property: <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.0*g#</mathjax> precisely, and thus we use it as a means to equate the number of atoms, which we can estimate but not individually count, with a given mass of stuff. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well <mathjax>#4*"moles"#</mathjax> of hydrogen atoms.......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of methane specifies <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's number"#</mathjax> of methane molecules, i.e. <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#CH_4#</mathjax> molecules. And thus it contains <mathjax>#4xx6.022xx10^23#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>. </p>
<p>I would have used precisely the same procedure if I were asked the number of <mathjax>#"hydrogen atoms"#</mathjax> in <mathjax>#"4 dozen methane molecules"#</mathjax>; i.e. there are <mathjax>#4xx12=48#</mathjax>, <mathjax>#"four dozen"#</mathjax>, <mathjax>#"hydrogen atoms....."#</mathjax> <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a collective number like a <mathjax>#"dozen"#</mathjax> or a <mathjax>#"gross"#</mathjax>.....admittedly, it is a much larger quantity.</p>
<p>The <mathjax>#"mole"#</mathjax>, however, has a special property: <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.0*g#</mathjax> precisely, and thus we use it as a means to equate the number of atoms, which we can estimate but not individually count, with a given mass of stuff. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of hydrogen atoms in one mole of methane, #CH_4#?</h1>
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anor277
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<div class="markdown"><p>Well <mathjax>#4*"moles"#</mathjax> of hydrogen atoms.......</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of methane specifies <mathjax>#N_A#</mathjax>, <mathjax>#"Avogadro's number"#</mathjax> of methane molecules, i.e. <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#CH_4#</mathjax> molecules. And thus it contains <mathjax>#4xx6.022xx10^23#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>. </p>
<p>I would have used precisely the same procedure if I were asked the number of <mathjax>#"hydrogen atoms"#</mathjax> in <mathjax>#"4 dozen methane molecules"#</mathjax>; i.e. there are <mathjax>#4xx12=48#</mathjax>, <mathjax>#"four dozen"#</mathjax>, <mathjax>#"hydrogen atoms....."#</mathjax> <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a collective number like a <mathjax>#"dozen"#</mathjax> or a <mathjax>#"gross"#</mathjax>.....admittedly, it is a much larger quantity.</p>
<p>The <mathjax>#"mole"#</mathjax>, however, has a special property: <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.0*g#</mathjax> precisely, and thus we use it as a means to equate the number of atoms, which we can estimate but not individually count, with a given mass of stuff. </p></div>
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</article> | How many moles of hydrogen atoms in one mole of methane, #CH_4#? | null |
2,398 | a9fdb58a-6ddd-11ea-bc24-ccda262736ce | https://socratic.org/questions/write-the-balanced-equation-of-the-complete-combustion-of-butane-c4h10 | 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O | start chemical_equation qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the complete combustion"}] | [{"type":"chemical equation","value":"2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O"}] | [{"type":"chemical equation","value":"C4H10"}] | <h1 class="questionTitle" itemprop="name">Write the balanced equation of the complete combustion of butane C4H10?</h1> | null | 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_4"H"_10 + "O"_2 -> "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Whenever it is said 'complete combustion', what must be remembered is just to write <mathjax>#"CO"_2#</mathjax> instead of <mathjax>#"CO"#</mathjax> (this is for 'incomplete combustion') </p>
<p>Now write coefficients:</p>
<p><mathjax>#"C"_4"H"_10+ 13/2"O"_2 -> 4"CO"_2 + 5"H"_2"O"#</mathjax></p>
<p>or we can multiply each side by <mathjax>#2#</mathjax>:</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 -> 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#2"C"_4"H"_10 + 13"O"_2 -> 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_4"H"_10 + "O"_2 -> "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Whenever it is said 'complete combustion', what must be remembered is just to write <mathjax>#"CO"_2#</mathjax> instead of <mathjax>#"CO"#</mathjax> (this is for 'incomplete combustion') </p>
<p>Now write coefficients:</p>
<p><mathjax>#"C"_4"H"_10+ 13/2"O"_2 -> 4"CO"_2 + 5"H"_2"O"#</mathjax></p>
<p>or we can multiply each side by <mathjax>#2#</mathjax>:</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 -> 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Write the balanced equation of the complete combustion of butane C4H10?</h1>
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<div class="markdown"><p><mathjax>#2"C"_4"H"_10 + 13"O"_2 -> 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_4"H"_10 + "O"_2 -> "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Whenever it is said 'complete combustion', what must be remembered is just to write <mathjax>#"CO"_2#</mathjax> instead of <mathjax>#"CO"#</mathjax> (this is for 'incomplete combustion') </p>
<p>Now write coefficients:</p>
<p><mathjax>#"C"_4"H"_10+ 13/2"O"_2 -> 4"CO"_2 + 5"H"_2"O"#</mathjax></p>
<p>or we can multiply each side by <mathjax>#2#</mathjax>:</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 -> 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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</article> | Write the balanced equation of the complete combustion of butane C4H10? | null |
2,399 | a840a7da-6ddd-11ea-a434-ccda262736ce | https://socratic.org/questions/how-many-grams-of-ki-are-required-to-make-550-0-ml-of-a-1-50-m-ki-solution | 137 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 16 9 10 volume qc_end physical_unit 15 16 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] KI [IN] grams"}] | [{"type":"physical unit","value":"137 grams"}] | [{"type":"physical unit","value":"Volume [OF] KI solution [=] \\pu{550.0 mL}"},{"type":"physical unit","value":"Molarity [OF] KI solution [=] \\pu{1.50 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of KI are required to make 550.0 mL of a 1.50 M KI solution ? </h1> | null | 137 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>More specifically, you know that in order for the solution to have a molarity of <mathjax>#"1.50 M"#</mathjax>, it must contain <mathjax>#1.50#</mathjax> <strong>moles</strong> of potassium iodide, the solute, for every <mathjax>#"1.00 L"#</mathjax> of the solution.</p>
<p>Since</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3 quad "mL"#</mathjax></p>
</blockquote>
<p>you can say that your sample must contain </p>
<blockquote>
<p><mathjax>#550.0 color(red)(cancel(color(black)("mL solution"))) * "1.50 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.825 moles KI"#</mathjax></p>
</blockquote>
<p>To convert the number of moles of potassium iodide to <em>grams</em>, use the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.825 color(red)(cancel(color(black)("moles KI"))) * "166.003 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("137 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"137 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>More specifically, you know that in order for the solution to have a molarity of <mathjax>#"1.50 M"#</mathjax>, it must contain <mathjax>#1.50#</mathjax> <strong>moles</strong> of potassium iodide, the solute, for every <mathjax>#"1.00 L"#</mathjax> of the solution.</p>
<p>Since</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3 quad "mL"#</mathjax></p>
</blockquote>
<p>you can say that your sample must contain </p>
<blockquote>
<p><mathjax>#550.0 color(red)(cancel(color(black)("mL solution"))) * "1.50 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.825 moles KI"#</mathjax></p>
</blockquote>
<p>To convert the number of moles of potassium iodide to <em>grams</em>, use the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.825 color(red)(cancel(color(black)("moles KI"))) * "166.003 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("137 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of KI are required to make 550.0 mL of a 1.50 M KI solution ? </h1>
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<a class="topContributorPic" href="/users/stefan-zdre"><img alt="" class="" src="https://profilepictures.socratic.org/LrguokJzR9yQlbiWbCvr_proba_1.jpg" title=""/></a>
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Stefan V.
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<span class="dateCreated" datetime="2018-04-11T23:50:22" itemprop="dateCreated">
Apr 11, 2018
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<div class="markdown"><p><mathjax>#"137 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>More specifically, you know that in order for the solution to have a molarity of <mathjax>#"1.50 M"#</mathjax>, it must contain <mathjax>#1.50#</mathjax> <strong>moles</strong> of potassium iodide, the solute, for every <mathjax>#"1.00 L"#</mathjax> of the solution.</p>
<p>Since</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3 quad "mL"#</mathjax></p>
</blockquote>
<p>you can say that your sample must contain </p>
<blockquote>
<p><mathjax>#550.0 color(red)(cancel(color(black)("mL solution"))) * "1.50 moles KI"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.825 moles KI"#</mathjax></p>
</blockquote>
<p>To convert the number of moles of potassium iodide to <em>grams</em>, use the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.825 color(red)(cancel(color(black)("moles KI"))) * "166.003 g"/(1color(red)(cancel(color(black)("mole KI")))) = color(darkgreen)(ul(color(black)("137 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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</article> | How many grams of KI are required to make 550.0 mL of a 1.50 M KI solution ? | null |
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