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3,100 | ac06ed0d-6ddd-11ea-b879-ccda262736ce | https://socratic.org/questions/how-many-moles-are-present-in-25cm3-of-0-2m-sodium-hydroxide-solution | 5.00 × 10^(−3) moles | start physical_unit 11 12 mole mol qc_end physical_unit 11 13 6 7 volume qc_end physical_unit 11 13 9 10 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium hydroxide [IN] moles"}] | [{"type":"physical unit","value":"5.00 × 10^(−3) moles"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide solution [=] \\pu{25 cm^3}"},{"type":"physical unit","value":"Molarity [OF] sodium hydroxide solution [=] \\pu{0.2 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution?</h1> | null | 5.00 × 10^(−3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5.0xx10^-3*mol#</mathjax> with respect to <mathjax>#NaOH#</mathjax>....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution?</h1>
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Apr 23, 2018
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<div class="markdown"><p><mathjax>#5.0xx10^-3*mol#</mathjax> with respect to <mathjax>#NaOH#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
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Dominic B.
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Stefan V.
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Apr 23, 2018
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<div class="markdown"><p><mathjax>#"0.005 moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Base equation </p>
<p><mathjax>#"concentration" = "moles"/("volume" ("in dm"^3)#</mathjax></p>
<p>Here</p>
<ul>
<li>concentration <mathjax>#= \ "0.2 M"#</mathjax></li>
<li>volume <mathjax>#= "25 cm"^3 = (25/1000) = "0.025 dm"^3#</mathjax></li>
<li>moles <mathjax>#=x#</mathjax></li>
</ul>
<p>Therefore </p>
<p><mathjax>#0.2= x/0.025#</mathjax></p>
<p>Multiply by <mathjax>#0.025#</mathjax> on both sides to leave <mathjax>#x#</mathjax> by itself</p>
<p><mathjax>#- 0.2xx0.025=x#</mathjax></p>
<p><mathjax>#x = "0.005 moles" = 5xx 10^(-3) \ "moles"#</mathjax></p></div>
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</article> | How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution? | null |
3,101 | a865a0e4-6ddd-11ea-9fe5-ccda262736ce | https://socratic.org/questions/58228f9511ef6b527f8c2aba | HI(aq) + KOH(aq) -> KI(aq) + H2O(l) | start chemical_equation qc_end substance 14 15 qc_end substance 17 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"HI(aq) + KOH(aq) -> KI(aq) + H2O(l)"}] | [{"type":"substance name","value":"Hydrogen iodide"},{"type":"substance name","value":"Potassium hydroxide"}] | <h1 class="questionTitle" itemprop="name">How do we represent the complete, and net ionic equations for the reaction between hydrogen iodide, and potassium hydroxide?</h1> | null | HI(aq) + KOH(aq) -> KI(aq) + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the net ionic equation is rather simpler:</p>
<p><mathjax>#H_3O^+ + HO^(-) rarr 2H_2O(l)#</mathjax></p>
<p>But what do we mean by <mathjax>#HI(aq)#</mathjax>? The <mathjax>#(aq)#</mathjax> designation means the socalled <mathjax>#"aquated ion"#</mathjax>. In water, hydrogen iodide speciates to <mathjax>#H_3O^+#</mathjax> and <mathjax>#I^-#</mathjax>. So-called hydronium ion, <mathjax>#H_3O^+#</mathjax>, as far as anyone knows, is a cluster of water molecules, with an EXTRA proton, <mathjax>#H_9O_4^+#</mathjax> or <mathjax>#H_11O_5^+#</mathjax> or something similar.</p>
<p>This water cluster not only holds for the hydronium ion, but potassium ion, <mathjax>#K^(+)(aq)#</mathjax>, is probably <mathjax>#[K(OH_2)_6]^+#</mathjax>, or <mathjax>#Cl(aq)^-#</mathjax> <mathjax>#[Cl(H_2O)_(4-6)]^-#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#HI(aq) + KOH(aq) rarr KI(aq) + H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the net ionic equation is rather simpler:</p>
<p><mathjax>#H_3O^+ + HO^(-) rarr 2H_2O(l)#</mathjax></p>
<p>But what do we mean by <mathjax>#HI(aq)#</mathjax>? The <mathjax>#(aq)#</mathjax> designation means the socalled <mathjax>#"aquated ion"#</mathjax>. In water, hydrogen iodide speciates to <mathjax>#H_3O^+#</mathjax> and <mathjax>#I^-#</mathjax>. So-called hydronium ion, <mathjax>#H_3O^+#</mathjax>, as far as anyone knows, is a cluster of water molecules, with an EXTRA proton, <mathjax>#H_9O_4^+#</mathjax> or <mathjax>#H_11O_5^+#</mathjax> or something similar.</p>
<p>This water cluster not only holds for the hydronium ion, but potassium ion, <mathjax>#K^(+)(aq)#</mathjax>, is probably <mathjax>#[K(OH_2)_6]^+#</mathjax>, or <mathjax>#Cl(aq)^-#</mathjax> <mathjax>#[Cl(H_2O)_(4-6)]^-#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do we represent the complete, and net ionic equations for the reaction between hydrogen iodide, and potassium hydroxide?</h1>
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<div class="markdown"><p><mathjax>#HI(aq) + KOH(aq) rarr KI(aq) + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the net ionic equation is rather simpler:</p>
<p><mathjax>#H_3O^+ + HO^(-) rarr 2H_2O(l)#</mathjax></p>
<p>But what do we mean by <mathjax>#HI(aq)#</mathjax>? The <mathjax>#(aq)#</mathjax> designation means the socalled <mathjax>#"aquated ion"#</mathjax>. In water, hydrogen iodide speciates to <mathjax>#H_3O^+#</mathjax> and <mathjax>#I^-#</mathjax>. So-called hydronium ion, <mathjax>#H_3O^+#</mathjax>, as far as anyone knows, is a cluster of water molecules, with an EXTRA proton, <mathjax>#H_9O_4^+#</mathjax> or <mathjax>#H_11O_5^+#</mathjax> or something similar.</p>
<p>This water cluster not only holds for the hydronium ion, but potassium ion, <mathjax>#K^(+)(aq)#</mathjax>, is probably <mathjax>#[K(OH_2)_6]^+#</mathjax>, or <mathjax>#Cl(aq)^-#</mathjax> <mathjax>#[Cl(H_2O)_(4-6)]^-#</mathjax>.</p></div>
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</article> | How do we represent the complete, and net ionic equations for the reaction between hydrogen iodide, and potassium hydroxide? | null |
3,102 | a8c30fe4-6ddd-11ea-89e2-ccda262736ce | https://socratic.org/questions/how-many-liters-of-a-500m-h-2so-4-must-react-with-excess-cu-to-produce-15-0-g-cu | 0.376 liters | start physical_unit 6 6 volume l qc_end physical_unit 6 6 4 5 molarity qc_end c_other OTHER qc_end physical_unit 16 16 14 15 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] H2SO4 [IN] liters"}] | [{"type":"physical unit","value":"0.376 liters"}] | [{"type":"physical unit","value":"Molarity [OF] H2SO4 [=] \\pu{0.500 M}"},{"type":"other","value":"Excess Cu."},{"type":"physical unit","value":"Mass [OF] CuSO4 [=] \\pu{15.0 g}"}] | <h1 class="questionTitle" itemprop="name">How many liters of a .500M #H_2SO_4# must react with excess #Cu# to produce 15.0 g #CuSO_4#?</h1> | null | 0.376 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper is too stable to react with dilute aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> of sulfuric acid (such as this one of concentration <mathjax>#0.500 color(white)(l) "mol" * "dm"^(-3)#</mathjax>) under standard conditions.</p>
<p>It takes concentrated sulfuric acid (which is indeed oxidizing but only at high concentrations) to directly react with copper and produce <mathjax>#"CuSO"_4#</mathjax>. Some orange-colored <mathjax>#"SO"_2(g)#</mathjax> gas evolves in this process. The reaction takes place as follows:</p>
<p><mathjax>#stackrel(color(navy)(0))("Cu")(s) + 2 color(white)(l) "H"_2 stackrel(color(purple)(+6))("S") "O"_4("conc." ,aq ) #</mathjax><br/>
<mathjax>#to stackrel(color(navy)(+2))("Cu") stackrel(color(grey)(+6))("S") "O"_4 (aq) + stackrel(color(purple)(+4))("S")"O"_2(g) + "H"_2"O"(l)#</mathjax></p>
<p>This equation can be balanced by tracking changes in the oxidation state of <mathjax>#"Cu"#</mathjax> and <mathjax>#"S"#</mathjax>. From the reaction equation:</p>
<p><mathjax>#(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = color(purple)(2)/color(blue)(1)#</mathjax></p>
<p><mathjax>#n(color(blue)("CuSO"_4)) = (m(color(blue)("CuSO"_4)))/(M(color(blue)("CuSO"_4))) = (15.0 color(white)(l) "g")/ (159.6 color(white)(l) "g" * "mol"^(-1)) = 0.0940 color(white)(l) "mol"#</mathjax></p>
<p><mathjax>#n(color(purple)("H"_2"SO"_4)) = n(color(blue)("CuSO"_4)) *(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = 0.0940 color(white)(l) "mol" * 2 = 0.188 color(white)(l) "mol"#</mathjax></p>
<p>Meaning that it takes <mathjax>#0.188 color(white)(l) "mol"#</mathjax> of (concentrated) sulfuric acid to produce <mathjax>#15.0 color(white)(l) "g"#</mathjax> of <mathjax>#"CuSO"_4#</mathjax> through direct oxidation of copper metal <mathjax>#"Cu"#</mathjax>.</p></div>
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<div class="markdown"><p>As @ernest-z and @anor277 have pointed out, a dilute solution of <mathjax>#"H"_2 "SO"_4#</mathjax> is unable to oxidize <mathjax>#"Cu"#</mathjax> under standard conditions. </p>
<p>This answer demonstrates the calculation of the number of moles of <mathjax>#ul("concentrated")#</mathjax> <mathjax>#"H"_2 "SO"_4 #</mathjax> necessary to produce the same amount of <mathjax>#"Cu" "SO"_4#</mathjax> from excess <mathjax>#"Cu"#</mathjax> metal.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper is too stable to react with dilute aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> of sulfuric acid (such as this one of concentration <mathjax>#0.500 color(white)(l) "mol" * "dm"^(-3)#</mathjax>) under standard conditions.</p>
<p>It takes concentrated sulfuric acid (which is indeed oxidizing but only at high concentrations) to directly react with copper and produce <mathjax>#"CuSO"_4#</mathjax>. Some orange-colored <mathjax>#"SO"_2(g)#</mathjax> gas evolves in this process. The reaction takes place as follows:</p>
<p><mathjax>#stackrel(color(navy)(0))("Cu")(s) + 2 color(white)(l) "H"_2 stackrel(color(purple)(+6))("S") "O"_4("conc." ,aq ) #</mathjax><br/>
<mathjax>#to stackrel(color(navy)(+2))("Cu") stackrel(color(grey)(+6))("S") "O"_4 (aq) + stackrel(color(purple)(+4))("S")"O"_2(g) + "H"_2"O"(l)#</mathjax></p>
<p>This equation can be balanced by tracking changes in the oxidation state of <mathjax>#"Cu"#</mathjax> and <mathjax>#"S"#</mathjax>. From the reaction equation:</p>
<p><mathjax>#(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = color(purple)(2)/color(blue)(1)#</mathjax></p>
<p><mathjax>#n(color(blue)("CuSO"_4)) = (m(color(blue)("CuSO"_4)))/(M(color(blue)("CuSO"_4))) = (15.0 color(white)(l) "g")/ (159.6 color(white)(l) "g" * "mol"^(-1)) = 0.0940 color(white)(l) "mol"#</mathjax></p>
<p><mathjax>#n(color(purple)("H"_2"SO"_4)) = n(color(blue)("CuSO"_4)) *(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = 0.0940 color(white)(l) "mol" * 2 = 0.188 color(white)(l) "mol"#</mathjax></p>
<p>Meaning that it takes <mathjax>#0.188 color(white)(l) "mol"#</mathjax> of (concentrated) sulfuric acid to produce <mathjax>#15.0 color(white)(l) "g"#</mathjax> of <mathjax>#"CuSO"_4#</mathjax> through direct oxidation of copper metal <mathjax>#"Cu"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of a .500M #H_2SO_4# must react with excess #Cu# to produce 15.0 g #CuSO_4#?</h1>
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<div class="markdown"><p>As @ernest-z and @anor277 have pointed out, a dilute solution of <mathjax>#"H"_2 "SO"_4#</mathjax> is unable to oxidize <mathjax>#"Cu"#</mathjax> under standard conditions. </p>
<p>This answer demonstrates the calculation of the number of moles of <mathjax>#ul("concentrated")#</mathjax> <mathjax>#"H"_2 "SO"_4 #</mathjax> necessary to produce the same amount of <mathjax>#"Cu" "SO"_4#</mathjax> from excess <mathjax>#"Cu"#</mathjax> metal.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper is too stable to react with dilute aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> of sulfuric acid (such as this one of concentration <mathjax>#0.500 color(white)(l) "mol" * "dm"^(-3)#</mathjax>) under standard conditions.</p>
<p>It takes concentrated sulfuric acid (which is indeed oxidizing but only at high concentrations) to directly react with copper and produce <mathjax>#"CuSO"_4#</mathjax>. Some orange-colored <mathjax>#"SO"_2(g)#</mathjax> gas evolves in this process. The reaction takes place as follows:</p>
<p><mathjax>#stackrel(color(navy)(0))("Cu")(s) + 2 color(white)(l) "H"_2 stackrel(color(purple)(+6))("S") "O"_4("conc." ,aq ) #</mathjax><br/>
<mathjax>#to stackrel(color(navy)(+2))("Cu") stackrel(color(grey)(+6))("S") "O"_4 (aq) + stackrel(color(purple)(+4))("S")"O"_2(g) + "H"_2"O"(l)#</mathjax></p>
<p>This equation can be balanced by tracking changes in the oxidation state of <mathjax>#"Cu"#</mathjax> and <mathjax>#"S"#</mathjax>. From the reaction equation:</p>
<p><mathjax>#(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = color(purple)(2)/color(blue)(1)#</mathjax></p>
<p><mathjax>#n(color(blue)("CuSO"_4)) = (m(color(blue)("CuSO"_4)))/(M(color(blue)("CuSO"_4))) = (15.0 color(white)(l) "g")/ (159.6 color(white)(l) "g" * "mol"^(-1)) = 0.0940 color(white)(l) "mol"#</mathjax></p>
<p><mathjax>#n(color(purple)("H"_2"SO"_4)) = n(color(blue)("CuSO"_4)) *(n(color(purple)("H"_2"SO"_4))) /(n(color(blue)("CuSO"_4))) = 0.0940 color(white)(l) "mol" * 2 = 0.188 color(white)(l) "mol"#</mathjax></p>
<p>Meaning that it takes <mathjax>#0.188 color(white)(l) "mol"#</mathjax> of (concentrated) sulfuric acid to produce <mathjax>#15.0 color(white)(l) "g"#</mathjax> of <mathjax>#"CuSO"_4#</mathjax> through direct oxidation of copper metal <mathjax>#"Cu"#</mathjax>.</p></div>
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</article> | How many liters of a .500M #H_2SO_4# must react with excess #Cu# to produce 15.0 g #CuSO_4#? | null |
3,103 | a8e50838-6ddd-11ea-9d3a-ccda262736ce | https://socratic.org/questions/how-many-neutrons-are-found-in-one-atom-of-36cl | 392.60 kJ | start physical_unit 14 14 heat_energy kj qc_end physical_unit 14 14 8 11 mass qc_end physical_unit 14 14 26 27 enthalpy_of_sublimation qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] CO2 [IN] kJ"}] | [{"type":"physical unit","value":"392.60 kJ"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{2.0 × 10^3 g }"},{"type":"physical unit","value":"Enthalpy of sublimation [OF] CO2 [=] \\pu{196.3 J/g}"},{"type":"other","value":"Ice (CO2) sublimate at the normal dry sublimation point."}] | <h1 class="questionTitle" itemprop="name">How many neutrons are found in one atom of #""^36Cl#?</h1> | null | 392.60 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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<div class="markdown"><p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> is <mathjax>#19#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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<div class="markdown"><p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> is <mathjax>#19#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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</article> | How many neutrons are found in one atom of #""^36Cl#? | null |
3,104 | ad18ce50-6ddd-11ea-9aed-ccda262736ce | https://socratic.org/questions/5874980f11ef6b50c729fa48 | 1.67 moles | start physical_unit 11 12 mole mol qc_end physical_unit 11 12 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] acetic acid [IN] moles"}] | [{"type":"physical unit","value":"1.67 moles"}] | [{"type":"physical unit","value":"Mass [OF] acetic acid [=] \\pu{100 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are present in a #100*g# mass of acetic acid?</h1> | null | 1.67 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of acetic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#(100*cancelg)/(60.05*cancelg*mol^-1)=??mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Number of moles of acetic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5/3*mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of acetic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#(100*cancelg)/(60.05*cancelg*mol^-1)=??mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Number of moles of acetic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5/3*mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of acetic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#(100*cancelg)/(60.05*cancelg*mol^-1)=??mol#</mathjax></p></div>
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</article> | How many moles are present in a #100*g# mass of acetic acid? | null |
3,105 | a98f1998-6ddd-11ea-a61a-ccda262736ce | https://socratic.org/questions/the-total-pressure-is-600-mmhg-after-0-50-moles-of-nitrogen-gas-n-2-and-1-0-mole | 400 mmHg | start physical_unit 17 18 partial_pressure mmhg qc_end physical_unit 25 26 4 5 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] nitrogen dioxide [IN] mmHg"}] | [{"type":"physical unit","value":"400 mmHg"}] | [{"type":"physical unit","value":"Total pressure [OF] the closed container [=] \\pu{600 mmHg}"},{"type":"physical unit","value":"Mole [OF] N2 gas [=] \\pu{0.50 moles}"},{"type":"physical unit","value":"Mole [OF] NO2 gas [=] \\pu{1.0 mole}"}] | <h1 class="questionTitle" itemprop="name">The total pressure is 600 mmHg after 0.50 moles of nitrogen gas, #N_2#, and 1.0 mole of nitrogen dioxide, #NO_2#, gas are mixed in a closed container. There is no reaction between the gases. What is the partial pressure of the nitrogen dioxide? </h1> | null | 400 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Dalton's Law of partial pressures"#</mathjax> tells us that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.</p>
<p>The total pressure is the sum of the individual partial pressures. </p>
<p>Thus <mathjax>#P_"Total"=P_(NO_2)+P_(N_2)=600*mm*Hg#</mathjax></p>
<p>But <mathjax>#P_(NO_2)=(1.0*molxxRT)/V#</mathjax></p>
<p>And <mathjax>#P_(N_2)=(0.50*molxxRT)/V=1/2xxP_(NO_2)#</mathjax></p>
<p>And so <mathjax>#P_"Total"=600*mm*Hg=3/2xxP_(NO_2)#</mathjax></p>
<p>And finally, <mathjax>#P_(NO_2)=400*mm*Hg#</mathjax>, and <mathjax>#P_(N_2)=200*mm*Hg#</mathjax>; the which satisfies Dalton's Law....</p></div>
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<div class="markdown"><p><mathjax>#P_(NO_2)=400*mm*Hg#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Dalton's Law of partial pressures"#</mathjax> tells us that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.</p>
<p>The total pressure is the sum of the individual partial pressures. </p>
<p>Thus <mathjax>#P_"Total"=P_(NO_2)+P_(N_2)=600*mm*Hg#</mathjax></p>
<p>But <mathjax>#P_(NO_2)=(1.0*molxxRT)/V#</mathjax></p>
<p>And <mathjax>#P_(N_2)=(0.50*molxxRT)/V=1/2xxP_(NO_2)#</mathjax></p>
<p>And so <mathjax>#P_"Total"=600*mm*Hg=3/2xxP_(NO_2)#</mathjax></p>
<p>And finally, <mathjax>#P_(NO_2)=400*mm*Hg#</mathjax>, and <mathjax>#P_(N_2)=200*mm*Hg#</mathjax>; the which satisfies Dalton's Law....</p></div>
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<h1 class="questionTitle" itemprop="name">The total pressure is 600 mmHg after 0.50 moles of nitrogen gas, #N_2#, and 1.0 mole of nitrogen dioxide, #NO_2#, gas are mixed in a closed container. There is no reaction between the gases. What is the partial pressure of the nitrogen dioxide? </h1>
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<div class="markdown"><p><mathjax>#P_(NO_2)=400*mm*Hg#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Dalton's Law of partial pressures"#</mathjax> tells us that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.</p>
<p>The total pressure is the sum of the individual partial pressures. </p>
<p>Thus <mathjax>#P_"Total"=P_(NO_2)+P_(N_2)=600*mm*Hg#</mathjax></p>
<p>But <mathjax>#P_(NO_2)=(1.0*molxxRT)/V#</mathjax></p>
<p>And <mathjax>#P_(N_2)=(0.50*molxxRT)/V=1/2xxP_(NO_2)#</mathjax></p>
<p>And so <mathjax>#P_"Total"=600*mm*Hg=3/2xxP_(NO_2)#</mathjax></p>
<p>And finally, <mathjax>#P_(NO_2)=400*mm*Hg#</mathjax>, and <mathjax>#P_(N_2)=200*mm*Hg#</mathjax>; the which satisfies Dalton's Law....</p></div>
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</article> | The total pressure is 600 mmHg after 0.50 moles of nitrogen gas, #N_2#, and 1.0 mole of nitrogen dioxide, #NO_2#, gas are mixed in a closed container. There is no reaction between the gases. What is the partial pressure of the nitrogen dioxide? | null |
3,106 | ac0587ee-6ddd-11ea-8ac4-ccda262736ce | https://socratic.org/questions/a-0-25-g-sample-of-a-pretzel-is-burned-the-heat-it-gives-off-is-used-to-heat-50- | 4.8 kcal/g | start physical_unit 34 35 energy kcal/g qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 21 21 18 19 mass qc_end physical_unit 21 21 23 24 temperature qc_end physical_unit 21 21 26 27 temperature qc_end end | [{"type":"physical unit","value":"Energy value [OF] the pretzel [IN] kcal/g"}] | [{"type":"physical unit","value":"4.8 kcal/g"}] | [{"type":"physical unit","value":"Mass [OF] pretzel sample [=] \\pu{0.25 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{50 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{18 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{42 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 0.25 g sample of a pretzel is burned. The heat it gives off is used to heat 50 g of water from 18°C to 42°C. What is the energy value of the pretzel, in kcal/g?</h1> | null | 4.8 kcal/g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The calculation assumes that all of the heat generated went into heating up the water; in reality that will never happen, although in a very well made calorimeter any losses will be reduced by efficient thermal lagging.</p>
<p>The energy used, E is equal to m.c.<mathjax>#theta#</mathjax> where m is the mass of water, c is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of water, and <mathjax>#theta#</mathjax> is the temperature change.</p>
<p>Ths specific heat capacity of water under the conditions described is 1 kcal/K.kg , where K is the temperature in Kelvins, and kg is the mass.</p>
<p>So now we can work out the figures.</p>
<p>E = 0.05 x 1 x 24 = 1.2 kcal.</p>
<p>So we know that 0.25 g of pretzel gives off (assuming no losses) 1.2 kcal of heat energy, so per gram the energy would be (1/0.25) x 1.2 = 4.8 kcal.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>4.8 kcal per g of pretzel (assuming no heat losses during combustion)</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The calculation assumes that all of the heat generated went into heating up the water; in reality that will never happen, although in a very well made calorimeter any losses will be reduced by efficient thermal lagging.</p>
<p>The energy used, E is equal to m.c.<mathjax>#theta#</mathjax> where m is the mass of water, c is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of water, and <mathjax>#theta#</mathjax> is the temperature change.</p>
<p>Ths specific heat capacity of water under the conditions described is 1 kcal/K.kg , where K is the temperature in Kelvins, and kg is the mass.</p>
<p>So now we can work out the figures.</p>
<p>E = 0.05 x 1 x 24 = 1.2 kcal.</p>
<p>So we know that 0.25 g of pretzel gives off (assuming no losses) 1.2 kcal of heat energy, so per gram the energy would be (1/0.25) x 1.2 = 4.8 kcal.</p></div>
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<h1 class="questionTitle" itemprop="name">A 0.25 g sample of a pretzel is burned. The heat it gives off is used to heat 50 g of water from 18°C to 42°C. What is the energy value of the pretzel, in kcal/g?</h1>
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<div class="markdown"><p>4.8 kcal per g of pretzel (assuming no heat losses during combustion)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The calculation assumes that all of the heat generated went into heating up the water; in reality that will never happen, although in a very well made calorimeter any losses will be reduced by efficient thermal lagging.</p>
<p>The energy used, E is equal to m.c.<mathjax>#theta#</mathjax> where m is the mass of water, c is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of water, and <mathjax>#theta#</mathjax> is the temperature change.</p>
<p>Ths specific heat capacity of water under the conditions described is 1 kcal/K.kg , where K is the temperature in Kelvins, and kg is the mass.</p>
<p>So now we can work out the figures.</p>
<p>E = 0.05 x 1 x 24 = 1.2 kcal.</p>
<p>So we know that 0.25 g of pretzel gives off (assuming no losses) 1.2 kcal of heat energy, so per gram the energy would be (1/0.25) x 1.2 = 4.8 kcal.</p></div>
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</article> | A 0.25 g sample of a pretzel is burned. The heat it gives off is used to heat 50 g of water from 18°C to 42°C. What is the energy value of the pretzel, in kcal/g? | null |
3,107 | a8b72b06-6ddd-11ea-b0a0-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-nitroglycerin | C3H5N3O9 | start chemical_formula qc_end substance 6 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] nitroglycerin [IN] empirical"}] | [{"type":"chemical equation","value":"C3H5N3O9 "}] | [{"type":"substance name","value":"Nitroglycerin"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for nitroglycerin?</h1> | null | C3H5N3O9 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound. In the case of nitroglycerin, its formula <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax> cannot be reduced further and keep the ratios whole numbers. Therefore, its molecular formula is the same as its empirical formula.</p></div>
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<div class="markdown"><p>The molecular formula for nitroglycerin is <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax>. <br/>
<a href="https://en.wikipedia.org/wiki/Nitroglycerin" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Nitroglycerin</a><br/>
This is also the empirical formula.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound. In the case of nitroglycerin, its formula <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax> cannot be reduced further and keep the ratios whole numbers. Therefore, its molecular formula is the same as its empirical formula.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for nitroglycerin?</h1>
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<div class="markdown"><p>The molecular formula for nitroglycerin is <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax>. <br/>
<a href="https://en.wikipedia.org/wiki/Nitroglycerin" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Nitroglycerin</a><br/>
This is also the empirical formula.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound. In the case of nitroglycerin, its formula <mathjax>#"C"_3"H"_5"N"_3"O"_9#</mathjax> cannot be reduced further and keep the ratios whole numbers. Therefore, its molecular formula is the same as its empirical formula.</p></div>
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</article> | What is the empirical formula for nitroglycerin? | null |
3,108 | ab3e9300-6ddd-11ea-b4d6-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-9-461-10-24-formula-units-of-manganese-iii-dichromate | 15.71 moles | start physical_unit 11 12 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] manganese(III) dichromate [IN] moles"}] | [{"type":"physical unit","value":"15.71 moles"}] | [{"type":"physical unit","value":"Number [OF] manganese(III) dichromate formula units [=] \\pu{9.461 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in #9.461 * 10^24# formula units of manganese(III) dichromate?</h1> | null | 15.71 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> form giant lattice structures in which the positively charge ions, or cations, and the negatively charged ions, or anions, are held together by <em>electrostatic attraction</em>. </p>
<p>This of course implies that you cannot speak of <strong>molecules</strong> of an ionic compound. Instead, the <strong>formula unit</strong> is used to express the smallest ratio that exists between the cations and anions in the ionic compound. </p>
<p>Simply put, the formula unit is the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of an ionic compound. </p>
<p>This means that <strong>one mole</strong> of <em>manganese(II) dichromate</em>, <mathjax>#"MnCr"_2"O"_7#</mathjax>, will contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of manganese(II) dichromate - this is known as <strong>Avogadro's number</strong>.</p>
<p>So, you know that you're dealing with <mathjax>#9.461 * 10^(24)#</mathjax> formula units of manganese(II) dichromate. Since you have more than <mathjax>#6.022 * 10^(23)#</mathjax> formula units, you can predict that you will have <em>more than one mole</em> of the compound. </p>
<p>In fact, looking at the exponents, you can say that you'll have <em>more than ten moles</em> of the compound. </p>
<p>Confirm this by doing</p>
<blockquote>
<p><mathjax>#9.461 * 10^(24)color(red)(cancel(color(black)("form. units"))) * ("1 mole MnCr"_2"O"_7)/(6.022 * 10^(23)color(red)(cancel(color(black)("form. units")))) = color(green)("15.71 moles MnCr"_2"O"_7)#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the number of formula units. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"15.71 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> form giant lattice structures in which the positively charge ions, or cations, and the negatively charged ions, or anions, are held together by <em>electrostatic attraction</em>. </p>
<p>This of course implies that you cannot speak of <strong>molecules</strong> of an ionic compound. Instead, the <strong>formula unit</strong> is used to express the smallest ratio that exists between the cations and anions in the ionic compound. </p>
<p>Simply put, the formula unit is the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of an ionic compound. </p>
<p>This means that <strong>one mole</strong> of <em>manganese(II) dichromate</em>, <mathjax>#"MnCr"_2"O"_7#</mathjax>, will contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of manganese(II) dichromate - this is known as <strong>Avogadro's number</strong>.</p>
<p>So, you know that you're dealing with <mathjax>#9.461 * 10^(24)#</mathjax> formula units of manganese(II) dichromate. Since you have more than <mathjax>#6.022 * 10^(23)#</mathjax> formula units, you can predict that you will have <em>more than one mole</em> of the compound. </p>
<p>In fact, looking at the exponents, you can say that you'll have <em>more than ten moles</em> of the compound. </p>
<p>Confirm this by doing</p>
<blockquote>
<p><mathjax>#9.461 * 10^(24)color(red)(cancel(color(black)("form. units"))) * ("1 mole MnCr"_2"O"_7)/(6.022 * 10^(23)color(red)(cancel(color(black)("form. units")))) = color(green)("15.71 moles MnCr"_2"O"_7)#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the number of formula units. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles are in #9.461 * 10^24# formula units of manganese(III) dichromate?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"15.71 moles"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> form giant lattice structures in which the positively charge ions, or cations, and the negatively charged ions, or anions, are held together by <em>electrostatic attraction</em>. </p>
<p>This of course implies that you cannot speak of <strong>molecules</strong> of an ionic compound. Instead, the <strong>formula unit</strong> is used to express the smallest ratio that exists between the cations and anions in the ionic compound. </p>
<p>Simply put, the formula unit is the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of an ionic compound. </p>
<p>This means that <strong>one mole</strong> of <em>manganese(II) dichromate</em>, <mathjax>#"MnCr"_2"O"_7#</mathjax>, will contain <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of manganese(II) dichromate - this is known as <strong>Avogadro's number</strong>.</p>
<p>So, you know that you're dealing with <mathjax>#9.461 * 10^(24)#</mathjax> formula units of manganese(II) dichromate. Since you have more than <mathjax>#6.022 * 10^(23)#</mathjax> formula units, you can predict that you will have <em>more than one mole</em> of the compound. </p>
<p>In fact, looking at the exponents, you can say that you'll have <em>more than ten moles</em> of the compound. </p>
<p>Confirm this by doing</p>
<blockquote>
<p><mathjax>#9.461 * 10^(24)color(red)(cancel(color(black)("form. units"))) * ("1 mole MnCr"_2"O"_7)/(6.022 * 10^(23)color(red)(cancel(color(black)("form. units")))) = color(green)("15.71 moles MnCr"_2"O"_7)#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the number of formula units. </p></div>
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</article> | How many moles are in #9.461 * 10^24# formula units of manganese(III) dichromate? | null |
3,109 | acbe68d0-6ddd-11ea-b434-ccda262736ce | https://socratic.org/questions/what-is-the-net-ionic-equation-for-the-reaction-na-aq-c-2h-3o-2-aq-h-aq-cl-aq-na | C2H3O2-(aq) + H+(aq) -> HC2H3O2(aq) | start chemical_equation qc_end chemical_equation 9 21 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"C2H3O2-(aq) + H+(aq) -> HC2H3O2(aq)"}] | [{"type":"chemical equation","value":"Na+(aq) + C2H3O2-(aq) + H+(aq) + Cl-(aq) -> Na+(aq) + Cl-(aq) + HC2H3O2(aq)"}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction #Na^+(aq) + C_2H_3O_2^-(aq) + H^+ (aq) + Cl^- (aq) -> Na^+(aq) + Cl^- (aq) + HC_2H_3O_2(aq)#?</h1> | null | C2H3O2-(aq) + H+(aq) -> HC2H3O2(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For determining net-ionic reaction from an ionic reaction you must cut out the ions which are same. One of the product is (gaseous), (liquid) but it is not (aq) <br/>
For instance <mathjax>#HC_2H_3O_2#</mathjax> is the (l) here </p>
<p><mathjax>#Na^+(aq) + C_2H_3O_2^-)(aq) + H^+(aq) + Cl^-) = Na^+(aq) + Cl^-)(aq) + HC_2H_3O_2(l)#</mathjax></p>
<p>Cut out the same ions </p>
<p><mathjax>#cancel(Na^+(aq)) + C_2H_3O_2^-)(aq) + H^+(aq) +cancel( Cl^-) = cancel(Na^+(aq)) + cancel(Cl^-)(aq) = HC_2H_3O_2(l)#</mathjax></p>
<p><mathjax>#C_2H_3O_2^-)(aq) + H^+(aq) = HC_2H_3O_2(l)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#C_2H_3O_2^-)(aq) + H^+(aq) = HC_2H_3O_2(l)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For determining net-ionic reaction from an ionic reaction you must cut out the ions which are same. One of the product is (gaseous), (liquid) but it is not (aq) <br/>
For instance <mathjax>#HC_2H_3O_2#</mathjax> is the (l) here </p>
<p><mathjax>#Na^+(aq) + C_2H_3O_2^-)(aq) + H^+(aq) + Cl^-) = Na^+(aq) + Cl^-)(aq) + HC_2H_3O_2(l)#</mathjax></p>
<p>Cut out the same ions </p>
<p><mathjax>#cancel(Na^+(aq)) + C_2H_3O_2^-)(aq) + H^+(aq) +cancel( Cl^-) = cancel(Na^+(aq)) + cancel(Cl^-)(aq) = HC_2H_3O_2(l)#</mathjax></p>
<p><mathjax>#C_2H_3O_2^-)(aq) + H^+(aq) = HC_2H_3O_2(l)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction #Na^+(aq) + C_2H_3O_2^-(aq) + H^+ (aq) + Cl^- (aq) -> Na^+(aq) + Cl^- (aq) + HC_2H_3O_2(aq)#?</h1>
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<div class="markdown"><p><mathjax>#C_2H_3O_2^-)(aq) + H^+(aq) = HC_2H_3O_2(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For determining net-ionic reaction from an ionic reaction you must cut out the ions which are same. One of the product is (gaseous), (liquid) but it is not (aq) <br/>
For instance <mathjax>#HC_2H_3O_2#</mathjax> is the (l) here </p>
<p><mathjax>#Na^+(aq) + C_2H_3O_2^-)(aq) + H^+(aq) + Cl^-) = Na^+(aq) + Cl^-)(aq) + HC_2H_3O_2(l)#</mathjax></p>
<p>Cut out the same ions </p>
<p><mathjax>#cancel(Na^+(aq)) + C_2H_3O_2^-)(aq) + H^+(aq) +cancel( Cl^-) = cancel(Na^+(aq)) + cancel(Cl^-)(aq) = HC_2H_3O_2(l)#</mathjax></p>
<p><mathjax>#C_2H_3O_2^-)(aq) + H^+(aq) = HC_2H_3O_2(l)#</mathjax></p></div>
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</article> | What is the net ionic equation for the reaction #Na^+(aq) + C_2H_3O_2^-(aq) + H^+ (aq) + Cl^- (aq) -> Na^+(aq) + Cl^- (aq) + HC_2H_3O_2(aq)#? | null |
3,110 | aaf0364c-6ddd-11ea-94e0-ccda262736ce | https://socratic.org/questions/what-is-the-molar-volume-of-3-00-moles-of-hydrogen-gas | 68.10 L | start physical_unit 9 10 molar_volume l qc_end physical_unit 9 10 6 7 mole qc_end end | [{"type":"physical unit","value":"Molar volume [OF] hydrogen gas [IN] L"}] | [{"type":"physical unit","value":"68.10 L"}] | [{"type":"physical unit","value":"Mole [OF] hydrogen gas [=] \\pu{3.00 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the molar volume of 3.00 moles of hydrogen gas?</h1> | null | 68.10 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, assuming that the hydrogen gas is contained at <mathjax>#"STP"#</mathjax>, then the molar volume would be <mathjax>#22.7*3=68.1 \ "L"#</mathjax>.</p>
<p>Due to the <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure" rel="nofollow">new conditions</a> of <mathjax>#"STP"#</mathjax>, the new molar volume using the ideal gas equation <mathjax>#(PV=nRT)#</mathjax> comes out as <mathjax>#22.7 \ "L"#</mathjax>.</p>
<p>So, <mathjax>#3#</mathjax> moles of hydrogen gas at <mathjax>#"STP"#</mathjax> will have a molar volume of</p>
<p><mathjax>#3color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=68.1 \ "L"#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume is <mathjax>#68.1 \ "L"#</mathjax> at <mathjax>#"STP"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, assuming that the hydrogen gas is contained at <mathjax>#"STP"#</mathjax>, then the molar volume would be <mathjax>#22.7*3=68.1 \ "L"#</mathjax>.</p>
<p>Due to the <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure" rel="nofollow">new conditions</a> of <mathjax>#"STP"#</mathjax>, the new molar volume using the ideal gas equation <mathjax>#(PV=nRT)#</mathjax> comes out as <mathjax>#22.7 \ "L"#</mathjax>.</p>
<p>So, <mathjax>#3#</mathjax> moles of hydrogen gas at <mathjax>#"STP"#</mathjax> will have a molar volume of</p>
<p><mathjax>#3color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=68.1 \ "L"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar volume of 3.00 moles of hydrogen gas?</h1>
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<div class="markdown"><p>The volume is <mathjax>#68.1 \ "L"#</mathjax> at <mathjax>#"STP"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, assuming that the hydrogen gas is contained at <mathjax>#"STP"#</mathjax>, then the molar volume would be <mathjax>#22.7*3=68.1 \ "L"#</mathjax>.</p>
<p>Due to the <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure" rel="nofollow">new conditions</a> of <mathjax>#"STP"#</mathjax>, the new molar volume using the ideal gas equation <mathjax>#(PV=nRT)#</mathjax> comes out as <mathjax>#22.7 \ "L"#</mathjax>.</p>
<p>So, <mathjax>#3#</mathjax> moles of hydrogen gas at <mathjax>#"STP"#</mathjax> will have a molar volume of</p>
<p><mathjax>#3color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=68.1 \ "L"#</mathjax>.</p></div>
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</article> | What is the molar volume of 3.00 moles of hydrogen gas? | null |
3,111 | a986b7a2-6ddd-11ea-a7ef-ccda262736ce | https://socratic.org/questions/a-sample-of-h-2o-with-a-mass-of-46-0-grams-has-a-temperature-of-58-0-c-how-many--1 | 5496.1 joules | start physical_unit 25 26 heat_energy j qc_end physical_unit 1 3 8 9 mass qc_end physical_unit 1 3 14 15 temperature qc_end physical_unit 25 26 14 15 temperature qc_end physical_unit 26 26 31 34 specific_heat qc_end end | [{"type":"physical unit","value":"Necessary energy [OF] the ice [IN] joules"}] | [{"type":"physical unit","value":"5496.1 joules"}] | [{"type":"physical unit","value":"Mass [OF] H2O sample [=] \\pu{46.0 grams}"},{"type":"physical unit","value":"Temperature1 [OF] H2O sample [=] \\pu{-58.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the ice [=] \\pu{0 ℃}"},{"type":"physical unit","value":"specific heat [OF] ice [=] \\pu{2.1 J/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">A sample of #H_2O# with a mass of 46.0 grams has a temperature of -58.0°C. How many joules of energy are necessary to heat the ice to 0°C? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Use 2.1 J/g°C for the specific heat of ice.</p></div>
</h2>
</div>
</div> | 5496.1 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:</p>
<p><mathjax>#"m = 46 g"#</mathjax></p>
<p><mathjax>#"T = -58"^@"C"#</mathjax></p>
<p><mathjax>#c = "2.06 J/g"cdot""^@"C"#</mathjax> (You have to look this up)</p>
<p>I recommend drawing the diagram of ice melting.</p>
<blockquote>
<p><mathjax>#q = (46.0cancel("g")) xx [0 - (-58)cancel(""^@"C")] xx (2.06 J/(cancel("g") cdot cancel(""^@"C"))) = "5496.1 J"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>5496.1 J</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:</p>
<p><mathjax>#"m = 46 g"#</mathjax></p>
<p><mathjax>#"T = -58"^@"C"#</mathjax></p>
<p><mathjax>#c = "2.06 J/g"cdot""^@"C"#</mathjax> (You have to look this up)</p>
<p>I recommend drawing the diagram of ice melting.</p>
<blockquote>
<p><mathjax>#q = (46.0cancel("g")) xx [0 - (-58)cancel(""^@"C")] xx (2.06 J/(cancel("g") cdot cancel(""^@"C"))) = "5496.1 J"#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A sample of #H_2O# with a mass of 46.0 grams has a temperature of -58.0°C. How many joules of energy are necessary to heat the ice to 0°C? </h1>
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<div class="markdown"><p><strong>5496.1 J</strong></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:</p>
<p><mathjax>#"m = 46 g"#</mathjax></p>
<p><mathjax>#"T = -58"^@"C"#</mathjax></p>
<p><mathjax>#c = "2.06 J/g"cdot""^@"C"#</mathjax> (You have to look this up)</p>
<p>I recommend drawing the diagram of ice melting.</p>
<blockquote>
<p><mathjax>#q = (46.0cancel("g")) xx [0 - (-58)cancel(""^@"C")] xx (2.06 J/(cancel("g") cdot cancel(""^@"C"))) = "5496.1 J"#</mathjax></p>
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</article> | A sample of #H_2O# with a mass of 46.0 grams has a temperature of -58.0°C. How many joules of energy are necessary to heat the ice to 0°C? |
Use 2.1 J/g°C for the specific heat of ice.
|
3,112 | aadeab3f-6ddd-11ea-857b-ccda262736ce | https://socratic.org/questions/5631d10b11ef6b4609a78ee2 | 29.53 grams | start physical_unit 17 18 mass g qc_end physical_unit 3 4 0 1 mass qc_end physical_unit 10 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] silver chloride [IN] grams"}] | [{"type":"physical unit","value":"29.53 grams"}] | [{"type":"physical unit","value":"Mass [OF] sodium chloride [=] \\pu{25 g}"},{"type":"physical unit","value":"Mass [OF] silver nitrate [=] \\pu{35.0 g}"}] | <h1 class="questionTitle" itemprop="name">#25*g# of sodium chloride are mixed with #35.0*g# silver nitrate in aqueous solution. How much silver chloride can be prepared?</h1> | null | 29.53 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#Ag^+#</mathjax>, <mathjax>#(35.0*g)/(169.9*g*mol^(-1)) = ??#</mathjax></p>
<p>Moles of <mathjax>#Cl^-#</mathjax>, <mathjax>#(25.0*g)/(58.4*g*mol^(-1)) = ??#</mathjax></p>
<p>The <mathjax>#Na^+(aq)#</mathjax> and <mathjax>#NO_3^(-)(aq)#</mathjax> ions are spectators and do not participate in the reaction. Chloride anion is clearly in excess; precipitation <mathjax>#AgCl#</mathjax> is (almost) quantitative. You are free to do the math.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Reaction: <mathjax>#Ag^+(aq) + Cl^(-)(aq) rarr AgCl(s)darr#</mathjax></p>
<p>Note that I assume you are using silver nitrate, <mathjax>#AgNO_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#Ag^+#</mathjax>, <mathjax>#(35.0*g)/(169.9*g*mol^(-1)) = ??#</mathjax></p>
<p>Moles of <mathjax>#Cl^-#</mathjax>, <mathjax>#(25.0*g)/(58.4*g*mol^(-1)) = ??#</mathjax></p>
<p>The <mathjax>#Na^+(aq)#</mathjax> and <mathjax>#NO_3^(-)(aq)#</mathjax> ions are spectators and do not participate in the reaction. Chloride anion is clearly in excess; precipitation <mathjax>#AgCl#</mathjax> is (almost) quantitative. You are free to do the math.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">#25*g# of sodium chloride are mixed with #35.0*g# silver nitrate in aqueous solution. How much silver chloride can be prepared?</h1>
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<div class="markdown"><p>Reaction: <mathjax>#Ag^+(aq) + Cl^(-)(aq) rarr AgCl(s)darr#</mathjax></p>
<p>Note that I assume you are using silver nitrate, <mathjax>#AgNO_3#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#Ag^+#</mathjax>, <mathjax>#(35.0*g)/(169.9*g*mol^(-1)) = ??#</mathjax></p>
<p>Moles of <mathjax>#Cl^-#</mathjax>, <mathjax>#(25.0*g)/(58.4*g*mol^(-1)) = ??#</mathjax></p>
<p>The <mathjax>#Na^+(aq)#</mathjax> and <mathjax>#NO_3^(-)(aq)#</mathjax> ions are spectators and do not participate in the reaction. Chloride anion is clearly in excess; precipitation <mathjax>#AgCl#</mathjax> is (almost) quantitative. You are free to do the math.</p></div>
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<div class="markdown"><p>The maximum yield of <mathjax>#"AgCl"#</mathjax> is <mathjax>#"29.5 g"#</mathjax>. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"NaCl(aq)"+"AgNO"_3("aq")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaNO"_3("aq")"+AgCl(s)"#</mathjax></p>
<p>This is a limiting reagent question. The reactant that produces the least amount of silver chloride is the limiting reagent, and it determines the amount of silver chloride that can be produced.</p>
<p>For each reactant, determine how much silver chloride can be produced.</p>
<ol>
<li>
<p>Determine moles of reactant by dividing given mass of the reactant by its molar mass.</p>
</li>
<li>
<p>Determine moles silver chloride by multiplying moles of the reactant by the mole ratio between the reactant and silver chloride in the balanced equation, with silver chloride in the numerator.</p>
</li>
<li>
<p>Determine mass silver chloride by multiplying moles of silver chloride by its molar mass.</p>
</li>
</ol>
<p><strong>Sodium chloride</strong></p>
<p><mathjax>#(25.0color(red)cancel(color(black)("g NaCl")))/(58.44color(red)cancel(color(black)("g NaCl"))/color(red)cancel(color(black)("mol NaCl")))xx(1color(red)cancel(color(black)("mol AgCl")))/(1color(red)cancel(color(black)("mol NaCl")))xx(143.32"g AgCl")/(1color(red)cancel(color(black)("mol AgCl")))="61.3 g AgCl"#</mathjax></p>
<p><strong>Silver nitrate</strong></p>
<p><mathjax>#(35.0color(red)cancel(color(black)("g AgNO"_3)))/(169.87color(red)cancel(color(black)("g AgNO"_3))/color(red)cancel(color(black)("mol AgNO"_3)))xx(1color(red)cancel(color(black)("mol AgCl")))/(1color(red)cancel(color(black)("mol AgNO3")))xx(143.32"g AgCl")/(1color(red)cancel(color(black)("mol AgCl")))="29.5 g AgCl"#</mathjax></p>
<p><mathjax>#"AgNO"_3"#</mathjax> is the limiting reagent. The maximum yield of <mathjax>#"AgCl"#</mathjax> is <mathjax>#"29.5 g"#</mathjax>. </p>
<p>Refer to the following website for more information.<br/>
<a href="https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm" rel="nofollow" target="_blank">https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm</a></p></div>
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</article> | #25*g# of sodium chloride are mixed with #35.0*g# silver nitrate in aqueous solution. How much silver chloride can be prepared? | null |
3,113 | a9b2494d-6ddd-11ea-ae3f-ccda262736ce | https://socratic.org/questions/how-do-you-balance-pbso-4-pbso-3-o-2 | 2 PbSO4 -> 2 PbSO3 + O2 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 PbSO4 -> 2 PbSO3 + O2"}] | [{"type":"chemical equation","value":"PbSO4 -> PbSO3 + O2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #PbSO_4 → PbSO_3 + O_2#?</h1> | null | 2 PbSO4 -> 2 PbSO3 + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With problems like this, I would do an atom inventory to see how many atoms of each element are present on both sides of the reaction arrow. </p>
<p>Initially, you have 1 atom of Pb on both the reactants and products side, followed by 1 atom of sulfur on both sides, and 4 atoms of oxygen on the reactants side and 5 atoms of oxygen on the products side. </p>
<p>Place a coefficient of two in front of the PbSO4 to obtain 2 atoms of Pb and S, and 8 atoms of O.</p>
<p>I started with 2 because I based it off the number of oxygen atoms on both sides of the reaction arrow. I knew that if there's a 2 in front of PbSO₄ , there will be 8 oxygen atoms on the left hand side. And if you place 2 on the right hand side in front of <mathjax>#PbSO_3#</mathjax>, you'll end up with 6 oxygen atoms, plus the two that you already have from <br/>
<mathjax>#O_2#</mathjax>, which happens to balance all the atoms.</p>
<p>I basically focused on the oxygen atoms since lead and sulfur has a coefficient of 1. Next to balance the Pb and S on the products side, place a coefficient of two in front of <mathjax>#PbSO_3#</mathjax> to obtain 2 atoms of Pb and S, and 6 atoms of O. The last two oxygen atoms come from the <mathjax>#O_2#</mathjax>, which gives you a total of 8 oxygen atoms.</p>
<p>Now it's balanced! Hope I helped.</p>
<p><a href="https://www.youtube.com/watch?v=yA3TZJ2em6g" rel="nofollow">Balancing Chemical Equations Help</a> </p></div>
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<div class="markdown"><p>2 PbSO4→2 PbSO3 + O2</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With problems like this, I would do an atom inventory to see how many atoms of each element are present on both sides of the reaction arrow. </p>
<p>Initially, you have 1 atom of Pb on both the reactants and products side, followed by 1 atom of sulfur on both sides, and 4 atoms of oxygen on the reactants side and 5 atoms of oxygen on the products side. </p>
<p>Place a coefficient of two in front of the PbSO4 to obtain 2 atoms of Pb and S, and 8 atoms of O.</p>
<p>I started with 2 because I based it off the number of oxygen atoms on both sides of the reaction arrow. I knew that if there's a 2 in front of PbSO₄ , there will be 8 oxygen atoms on the left hand side. And if you place 2 on the right hand side in front of <mathjax>#PbSO_3#</mathjax>, you'll end up with 6 oxygen atoms, plus the two that you already have from <br/>
<mathjax>#O_2#</mathjax>, which happens to balance all the atoms.</p>
<p>I basically focused on the oxygen atoms since lead and sulfur has a coefficient of 1. Next to balance the Pb and S on the products side, place a coefficient of two in front of <mathjax>#PbSO_3#</mathjax> to obtain 2 atoms of Pb and S, and 6 atoms of O. The last two oxygen atoms come from the <mathjax>#O_2#</mathjax>, which gives you a total of 8 oxygen atoms.</p>
<p>Now it's balanced! Hope I helped.</p>
<p><a href="https://www.youtube.com/watch?v=yA3TZJ2em6g" rel="nofollow">Balancing Chemical Equations Help</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #PbSO_4 → PbSO_3 + O_2#?</h1>
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<div class="markdown"><p>2 PbSO4→2 PbSO3 + O2</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With problems like this, I would do an atom inventory to see how many atoms of each element are present on both sides of the reaction arrow. </p>
<p>Initially, you have 1 atom of Pb on both the reactants and products side, followed by 1 atom of sulfur on both sides, and 4 atoms of oxygen on the reactants side and 5 atoms of oxygen on the products side. </p>
<p>Place a coefficient of two in front of the PbSO4 to obtain 2 atoms of Pb and S, and 8 atoms of O.</p>
<p>I started with 2 because I based it off the number of oxygen atoms on both sides of the reaction arrow. I knew that if there's a 2 in front of PbSO₄ , there will be 8 oxygen atoms on the left hand side. And if you place 2 on the right hand side in front of <mathjax>#PbSO_3#</mathjax>, you'll end up with 6 oxygen atoms, plus the two that you already have from <br/>
<mathjax>#O_2#</mathjax>, which happens to balance all the atoms.</p>
<p>I basically focused on the oxygen atoms since lead and sulfur has a coefficient of 1. Next to balance the Pb and S on the products side, place a coefficient of two in front of <mathjax>#PbSO_3#</mathjax> to obtain 2 atoms of Pb and S, and 6 atoms of O. The last two oxygen atoms come from the <mathjax>#O_2#</mathjax>, which gives you a total of 8 oxygen atoms.</p>
<p>Now it's balanced! Hope I helped.</p>
<p><a href="https://www.youtube.com/watch?v=yA3TZJ2em6g" rel="nofollow">Balancing Chemical Equations Help</a> </p></div>
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</article> | How do you balance #PbSO_4 → PbSO_3 + O_2#? | null |
3,114 | a8b1f876-6ddd-11ea-942d-ccda262736ce | https://socratic.org/questions/how-much-heat-is-added-to-10-0-g-of-ice-at-200-c-to-convert-it-to-steam-at-120-0 | 34561.50 J | start physical_unit 9 9 heat_energy j qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 9 9 11 11 temperature qc_end end | [{"type":"physical unit","value":"Added heat [OF] ice [IN] J"}] | [{"type":"physical unit","value":"34561.50 J"}] | [{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{10.0 g}"},{"type":"physical unit","value":"Temperature [OF] ice [=] \\pu{-200 ℃}"},{"type":"physical unit","value":"Temperature [OF] steam [=] \\pu{120 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much heat is added to 10.0 g of ice at -200°C to convert it to steam at 120.00°C?</h1> | null | 34561.50 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The standard heating/cooling curve goes through four phases. All temperature changes in a single phase (solid, liquid or gas) depend on the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> (oC/g) of the compound in that phase, and the inter-phase transition energies. </p>
<p>So, in this case we have to heat solid water from -200oC to 0oC, then change phase from solid to liquid at 0oC, followed by heating liquid water from 0oC to 100oC. Then the water must change phase again from liquid to gas, and then finally heat the gas phase from 100oC to 120oC.</p>
<p>Solid Specific Heat: 2.05 J/g-oK<br/>
Liquid Specific Heat: 4.178 J/g-oK<br/>
Vapor Specific Heat: 1.89 J/g-oK<br/>
Heat of Fusion: 333.55 J/g<br/>
Heat of Vaporization: 2257 J/g</p>
<p>10.0g * (200oC) * 2.05 J/g-oC + 10.0g * 333.55 J/g + 10.0g * (100oC) * 4.178 J/g-oC + 10.0g * 2257 J/g + 10.0g * (20oC) * 1.89 J/g-oC</p>
<p>10.0j * ((200oC * 2.05 J/g-oC) + 333.55 J/g + (100oC * 4.178 J/g-oC) + 2257 J/g + (20oC * 1.89 J/g-oC))</p>
<p>10.0g * (410J/g + 333.55 J/g + 417.8 J/g + 2257 J/g + 37.8 J/g)</p>
<p><strong>= 34561.5 Joules</strong></p>
<p>A good example of this type of problem is explained at the Khan Academy here:<br/>
<a href="https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization" rel="nofollow" target="_blank">https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization</a></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><strong>34561.5 Joules</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The standard heating/cooling curve goes through four phases. All temperature changes in a single phase (solid, liquid or gas) depend on the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> (oC/g) of the compound in that phase, and the inter-phase transition energies. </p>
<p>So, in this case we have to heat solid water from -200oC to 0oC, then change phase from solid to liquid at 0oC, followed by heating liquid water from 0oC to 100oC. Then the water must change phase again from liquid to gas, and then finally heat the gas phase from 100oC to 120oC.</p>
<p>Solid Specific Heat: 2.05 J/g-oK<br/>
Liquid Specific Heat: 4.178 J/g-oK<br/>
Vapor Specific Heat: 1.89 J/g-oK<br/>
Heat of Fusion: 333.55 J/g<br/>
Heat of Vaporization: 2257 J/g</p>
<p>10.0g * (200oC) * 2.05 J/g-oC + 10.0g * 333.55 J/g + 10.0g * (100oC) * 4.178 J/g-oC + 10.0g * 2257 J/g + 10.0g * (20oC) * 1.89 J/g-oC</p>
<p>10.0j * ((200oC * 2.05 J/g-oC) + 333.55 J/g + (100oC * 4.178 J/g-oC) + 2257 J/g + (20oC * 1.89 J/g-oC))</p>
<p>10.0g * (410J/g + 333.55 J/g + 417.8 J/g + 2257 J/g + 37.8 J/g)</p>
<p><strong>= 34561.5 Joules</strong></p>
<p>A good example of this type of problem is explained at the Khan Academy here:<br/>
<a href="https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization" rel="nofollow" target="_blank">https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization</a></p></div>
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<h1 class="questionTitle" itemprop="name">How much heat is added to 10.0 g of ice at -200°C to convert it to steam at 120.00°C?</h1>
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<div class="markdown"><p><strong>34561.5 Joules</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The standard heating/cooling curve goes through four phases. All temperature changes in a single phase (solid, liquid or gas) depend on the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> (oC/g) of the compound in that phase, and the inter-phase transition energies. </p>
<p>So, in this case we have to heat solid water from -200oC to 0oC, then change phase from solid to liquid at 0oC, followed by heating liquid water from 0oC to 100oC. Then the water must change phase again from liquid to gas, and then finally heat the gas phase from 100oC to 120oC.</p>
<p>Solid Specific Heat: 2.05 J/g-oK<br/>
Liquid Specific Heat: 4.178 J/g-oK<br/>
Vapor Specific Heat: 1.89 J/g-oK<br/>
Heat of Fusion: 333.55 J/g<br/>
Heat of Vaporization: 2257 J/g</p>
<p>10.0g * (200oC) * 2.05 J/g-oC + 10.0g * 333.55 J/g + 10.0g * (100oC) * 4.178 J/g-oC + 10.0g * 2257 J/g + 10.0g * (20oC) * 1.89 J/g-oC</p>
<p>10.0j * ((200oC * 2.05 J/g-oC) + 333.55 J/g + (100oC * 4.178 J/g-oC) + 2257 J/g + (20oC * 1.89 J/g-oC))</p>
<p>10.0g * (410J/g + 333.55 J/g + 417.8 J/g + 2257 J/g + 37.8 J/g)</p>
<p><strong>= 34561.5 Joules</strong></p>
<p>A good example of this type of problem is explained at the Khan Academy here:<br/>
<a href="https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization" rel="nofollow" target="_blank">https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization</a></p></div>
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</article> | How much heat is added to 10.0 g of ice at -200°C to convert it to steam at 120.00°C? | null |
3,115 | ac5f0d6f-6ddd-11ea-a408-ccda262736ce | https://socratic.org/questions/a-sample-of-oxygen-gas-occupies-a-volume-of-250-ml-at-740-mmhg-what-volume-will- | 231.25 mL | start physical_unit 1 4 volume ml qc_end physical_unit 1 4 9 10 volume qc_end physical_unit 1 4 12 13 pressure qc_end physical_unit 1 4 20 21 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] oxygen gas sample [IN] mL"}] | [{"type":"physical unit","value":"231.25 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] oxygen gas sample [=] \\pu{250 mL}"},{"type":"physical unit","value":"Pressure1 [OF] oxygen gas sample [=] \\pu{740 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] oxygen gas sample [=] \\pu{800 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A sample of oxygen gas occupies a volume of 250 mL at 740 mmHg. What volume will it occupy at 800 mmHg? </h1> | null | 231.25 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I don't like this question very much, because it suggests that you can get a column of mercury that is <mathjax>#800*mm#</mathjax> high. Try this, and you'll get mercury all over your laboratory, where it will inhabit every crack, and every cranny. Cleaners also will not touch your lab until you clean it up.</p>
<p>One atmosphere will support a column of mercury <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> high, and this is as high as you go. Would you care to use new units of pressure? Your teacher should know this. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I don't like this question very much, because it suggests that you can get a column of mercury that is <mathjax>#800*mm#</mathjax> high. Try this, and you'll get mercury all over your laboratory, where it will inhabit every crack, and every cranny. Cleaners also will not touch your lab until you clean it up.</p>
<p>One atmosphere will support a column of mercury <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> high, and this is as high as you go. Would you care to use new units of pressure? Your teacher should know this. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of oxygen gas occupies a volume of 250 mL at 740 mmHg. What volume will it occupy at 800 mmHg? </h1>
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anor277
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Dec 14, 2016
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I don't like this question very much, because it suggests that you can get a column of mercury that is <mathjax>#800*mm#</mathjax> high. Try this, and you'll get mercury all over your laboratory, where it will inhabit every crack, and every cranny. Cleaners also will not touch your lab until you clean it up.</p>
<p>One atmosphere will support a column of mercury <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> high, and this is as high as you go. Would you care to use new units of pressure? Your teacher should know this. </p></div>
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</article> | A sample of oxygen gas occupies a volume of 250 mL at 740 mmHg. What volume will it occupy at 800 mmHg? | null |
3,116 | abcc16b6-6ddd-11ea-8c61-ccda262736ce | https://socratic.org/questions/an-ideal-gas-in-a-sealed-container-has-an-initial-volume-of-2-70-l-at-constant-p | 450.51 K | start physical_unit 1 2 temperature k qc_end physical_unit 1 2 12 13 volume qc_end physical_unit 1 2 21 22 temperature qc_end physical_unit 1 2 28 29 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature1 [OF] the ideal gas [IN] K"}] | [{"type":"physical unit","value":"450.51 K"}] | [{"type":"physical unit","value":"Volume1 [OF] the ideal gas [=] \\pu{2.70 L}"},{"type":"physical unit","value":"Temperature2 [OF] the ideal gas [=] \\pu{19 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the ideal gas [=] \\pu{1.75 L}"},{"type":"other","value":"Constant pressure."}] | <h1 class="questionTitle" itemprop="name">An ideal gas in a sealed container has an initial volume of 2.70 L. At constant pressure, it is cooled to 19° C where its final volume is 1.75 L. What was the initial temperature? </h1> | null | 450.51 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/V_2xxT_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.70*L)/(1.75*L)xx292*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p>
<p>The question SHOULD have proposed a piston rather than a sealed container, whose volume should reasonably remain constant. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This question is ill-proposed, but <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> applies, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, temperatures are absolute values. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/V_2xxT_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.70*L)/(1.75*L)xx292*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p>
<p>The question SHOULD have proposed a piston rather than a sealed container, whose volume should reasonably remain constant. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An ideal gas in a sealed container has an initial volume of 2.70 L. At constant pressure, it is cooled to 19° C where its final volume is 1.75 L. What was the initial temperature? </h1>
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anor277
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<div class="markdown"><p>This question is ill-proposed, but <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> applies, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, temperatures are absolute values. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/V_2xxT_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.70*L)/(1.75*L)xx292*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax></p>
<p>The question SHOULD have proposed a piston rather than a sealed container, whose volume should reasonably remain constant. </p></div>
</div>
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</article> | An ideal gas in a sealed container has an initial volume of 2.70 L. At constant pressure, it is cooled to 19° C where its final volume is 1.75 L. What was the initial temperature? | null |
3,117 | ad0eab9d-6ddd-11ea-8a29-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-01-m-solution-of-the-strong-acid-hno-3-in-water | 2.00 | start physical_unit 8 8 ph none qc_end physical_unit 13 13 6 7 molarity qc_end substance 15 15 qc_end end | [{"type":"physical unit","value":"pH [OF] HNO3 solution"}] | [{"type":"physical unit","value":"2.00"}] | [{"type":"physical unit","value":"Molarity [OF] HNO3 solution [=] \\pu{0.01 M}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.01 M solution of the strong acid #HNO_3# in water?</h1> | null | 2.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind when dealing with <strong>strong acids</strong> is that they dissociate completely in aqueous solution to form <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and their <a href="https://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>. </p>
<p><em>Complete dissociation</em> means that <strong>every molecule</strong> of acid will ionize, i.e. donate its acidic proton to a water molecule. </p>
<p>That tells you that a strong <strong>monoprotic acid</strong> will produce hydronium ions in a <mathjax>#1:1#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>, that is, every mole of acid will produce one mole of hydronium ions. </p>
<p><img alt="http://cikguwong.blogspot.ro/2011/08/chemistry-form-4-chapter-7-strength-of.html" src="https://useruploads.socratic.org/s4eLRTKURXOlIQmTTdSW_16-22.jpg"/> </p>
<p>In this case, nitric acid will dissociate to form </p>
<blockquote>
<p><mathjax>#"HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that the concentration of hydronium ions will be <strong>equal</strong> to that of the nitric acid </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HNO"_3] = "0.01 M"#</mathjax></p>
</blockquote>
<p>As you know, a solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is simply a measure of its concentration of hydronium ions </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log( ["H"_3"O"^(+)]))#</mathjax></p>
</blockquote>
<p>In this case, the pH of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.01) = color(green)(2)#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"pH" = 2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind when dealing with <strong>strong acids</strong> is that they dissociate completely in aqueous solution to form <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and their <a href="https://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>. </p>
<p><em>Complete dissociation</em> means that <strong>every molecule</strong> of acid will ionize, i.e. donate its acidic proton to a water molecule. </p>
<p>That tells you that a strong <strong>monoprotic acid</strong> will produce hydronium ions in a <mathjax>#1:1#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>, that is, every mole of acid will produce one mole of hydronium ions. </p>
<p><img alt="http://cikguwong.blogspot.ro/2011/08/chemistry-form-4-chapter-7-strength-of.html" src="https://useruploads.socratic.org/s4eLRTKURXOlIQmTTdSW_16-22.jpg"/> </p>
<p>In this case, nitric acid will dissociate to form </p>
<blockquote>
<p><mathjax>#"HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that the concentration of hydronium ions will be <strong>equal</strong> to that of the nitric acid </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HNO"_3] = "0.01 M"#</mathjax></p>
</blockquote>
<p>As you know, a solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is simply a measure of its concentration of hydronium ions </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log( ["H"_3"O"^(+)]))#</mathjax></p>
</blockquote>
<p>In this case, the pH of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.01) = color(green)(2)#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.01 M solution of the strong acid #HNO_3# in water?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind when dealing with <strong>strong acids</strong> is that they dissociate completely in aqueous solution to form <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and their <a href="https://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>. </p>
<p><em>Complete dissociation</em> means that <strong>every molecule</strong> of acid will ionize, i.e. donate its acidic proton to a water molecule. </p>
<p>That tells you that a strong <strong>monoprotic acid</strong> will produce hydronium ions in a <mathjax>#1:1#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a>, that is, every mole of acid will produce one mole of hydronium ions. </p>
<p><img alt="http://cikguwong.blogspot.ro/2011/08/chemistry-form-4-chapter-7-strength-of.html" src="https://useruploads.socratic.org/s4eLRTKURXOlIQmTTdSW_16-22.jpg"/> </p>
<p>In this case, nitric acid will dissociate to form </p>
<blockquote>
<p><mathjax>#"HNO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>This means that the concentration of hydronium ions will be <strong>equal</strong> to that of the nitric acid </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HNO"_3] = "0.01 M"#</mathjax></p>
</blockquote>
<p>As you know, a solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is simply a measure of its concentration of hydronium ions </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log( ["H"_3"O"^(+)]))#</mathjax></p>
</blockquote>
<p>In this case, the pH of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.01) = color(green)(2)#</mathjax></p>
</blockquote></div>
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</article> | What is the pH of a 0.01 M solution of the strong acid #HNO_3# in water? | null |
3,118 | ac151c4a-6ddd-11ea-88d7-ccda262736ce | https://socratic.org/questions/how-many-moles-of-sodium-hydroxide-contained-in-300-ml-of-a-100m-naoh-solution | 3.00 × 10^(-2) 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":"3.00 × 10^(-2) moles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{300 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.100 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of sodium hydroxide contained in 300 mL of a .100M NaOH solution?</h1> | null | 3.00 × 10^(-2) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#=300*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)xx0.100*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=3.00xx10^-2*mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Number of moles of sodium hydroxide"#</mathjax></p>
<p><mathjax>#=300*mLxx10^-3L*mL^-1xx0.100*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#=300*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)xx0.100*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=3.00xx10^-2*mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Number of moles of sodium hydroxide"#</mathjax></p>
<p><mathjax>#=300*mLxx10^-3L*mL^-1xx0.100*mol*L^-1#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#=300*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)xx0.100*mol*cancel(L^-1)#</mathjax></p>
<p><mathjax>#=3.00xx10^-2*mol#</mathjax></p></div>
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</article> | How many moles of sodium hydroxide contained in 300 mL of a .100M NaOH solution? | null |
3,119 | aae2a2ee-6ddd-11ea-9881-ccda262736ce | https://socratic.org/questions/ais-found-to-be-12-7-carbon-3-2-hydrogen-and-84-1-bromine-what-is-its-compound-e | CH3Br | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound A [IN] empirical"}] | [{"type":"chemical equation","value":"CH3Br"}] | [{"type":"physical unit","value":"Percentage [OF] carbon in the compound A [=] \\pu{12.7%}"},{"type":"physical unit","value":"Percentage [OF] hydrogen in the compound A [=] \\pu{3.2%}"},{"type":"physical unit","value":"Percentage [OF] bromine in the compound A [=] \\pu{84.1%}"}] | <h1 class="questionTitle" itemprop="name">A is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine What is its compound empirical formula?</h1> | null | CH3Br | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we assumed the percentage to be in grams (like mass)<br/>
then we take the percentage and divide by the molar masses of each element to get the no. of mol of each element<br/>
Afterwards, divide all of the results by the smallest mol value in the table, which gives us the simplest molar ratio<img alt="my own workings" src="https://useruploads.socratic.org/Fyv4PLh2QeKFRYaKvTPe_Photo%20on%201-5-17%20at%203.28+PM.jpg"/> </p></div>
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<div class="markdown"><p>This may be simplified into a table to help answer this question</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we assumed the percentage to be in grams (like mass)<br/>
then we take the percentage and divide by the molar masses of each element to get the no. of mol of each element<br/>
Afterwards, divide all of the results by the smallest mol value in the table, which gives us the simplest molar ratio<img alt="my own workings" src="https://useruploads.socratic.org/Fyv4PLh2QeKFRYaKvTPe_Photo%20on%201-5-17%20at%203.28+PM.jpg"/> </p></div>
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<h1 class="questionTitle" itemprop="name">A is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine What is its compound empirical formula?</h1>
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Constance
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<div class="markdown"><p>This may be simplified into a table to help answer this question</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we assumed the percentage to be in grams (like mass)<br/>
then we take the percentage and divide by the molar masses of each element to get the no. of mol of each element<br/>
Afterwards, divide all of the results by the smallest mol value in the table, which gives us the simplest molar ratio<img alt="my own workings" src="https://useruploads.socratic.org/Fyv4PLh2QeKFRYaKvTPe_Photo%20on%201-5-17%20at%203.28+PM.jpg"/> </p></div>
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<div class="markdown"><p><mathjax>#"Empirical formula"=CH_3Br#</mathjax>.</p></div>
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<div class="markdown"><p>As is usual in these problems, we ASSUME a mass of <mathjax>#100*g#</mathjax>, and then work out the atomic composition:</p>
<p><mathjax>#"Moles of carbon"=(12.7*g)/(12.01*g*mol^-1)=1.05*mol#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"=(3.2*g)/(1.00794*g*mol^-1)=3.175*mol#</mathjax>.</p>
<p><mathjax>#"Moles of bromine"=(84.1*g)/(79.9*g*mol^-1)=1.05*mol#</mathjax>.</p>
<p>And then we divide thru by the smallest molar quantity, to get an empirical formula, the simplest whole number ratio defining constitutent atoms in a species as, <mathjax>#CH_3Br#</mathjax>.</p>
<p>Just to add, that we calculated the empirical formula. The molecule COULD be methyl bromide, in which case the question is highly unreasonable. You do not perform microanalysis, i.e. <mathjax>#%C,%H,%N#</mathjax> composition on LIQUIDS; you can do so on solids. </p></div>
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</article> | A is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine What is its compound empirical formula? | null |
3,120 | aabe6ab4-6ddd-11ea-a16a-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>
<div>
<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>
<div>
<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 |
3,121 | abcfdbd2-6ddd-11ea-ba91-ccda262736ce | https://socratic.org/questions/how-would-you-make-250-ml-of-0-150-m-sodium-thiosulfate-from-solid-na-2-s-2-o-3- | 9.31 g | start physical_unit 13 13 mass g qc_end physical_unit 9 10 7 8 molarity qc_end physical_unit 13 13 17 18 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Na2S2O3.5H2O [IN] g"}] | [{"type":"physical unit","value":"9.31 g"}] | [{"type":"physical unit","value":"Volume [OF] sodium thiosulfate solution [=] \\pu{250 ml}"},{"type":"physical unit","value":"Molarity [OF] sodium thiosulfate solution [=] \\pu{0.150 M}"},{"type":"physical unit","value":"Molar mass [OF] Na2S2O3.5H2O [=] \\pu{248.2 g/mole}"}] | <h1 class="questionTitle" itemprop="name">How would you make 250 ml of 0.150 M sodium thiosulfate from solid Na2S2O3 *
5H2O?
(molar mass = 248.2 g/mole)</h1> | null | 9.31 g | <div class="answerDescription">
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<div class="markdown"><p>Strength of the sodium thiosulphate solution to be produced is <mathjax>#0.150M=0.150"mol/"Lxx248.2g"/"mol=37.23g"/"L#</mathjax></p>
<p>So 1L or <mathjax>#1000mL#</mathjax> solution contains <mathjax>#37.23g#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>Hence 250mL wYill contain <mathjax>#37.23/1000*250 =9.3075g#</mathjax> solute</p>
<p>This mass of solute is to be taken in a 250mL volumetric flask by weighing in an electronic balance and then it is disolved in water of to make a solution of volume less than 250mL and finally required volume of water is added to it to make the volume 250mL.</p></div>
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<div class="markdown"><p>Strength of the sodium thiosulphate solution to be produced is <mathjax>#0.150M=0.150"mol/"Lxx248.2g"/"mol=37.23g"/"L#</mathjax></p>
<p>So 1L or <mathjax>#1000mL#</mathjax> solution contains <mathjax>#37.23g#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>Hence 250mL wYill contain <mathjax>#37.23/1000*250 =9.3075g#</mathjax> solute</p>
<p>This mass of solute is to be taken in a 250mL volumetric flask by weighing in an electronic balance and then it is disolved in water of to make a solution of volume less than 250mL and finally required volume of water is added to it to make the volume 250mL.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you make 250 ml of 0.150 M sodium thiosulfate from solid Na2S2O3 *
5H2O?
(molar mass = 248.2 g/mole)</h1>
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<div class="markdown"><p>Strength of the sodium thiosulphate solution to be produced is <mathjax>#0.150M=0.150"mol/"Lxx248.2g"/"mol=37.23g"/"L#</mathjax></p>
<p>So 1L or <mathjax>#1000mL#</mathjax> solution contains <mathjax>#37.23g#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>Hence 250mL wYill contain <mathjax>#37.23/1000*250 =9.3075g#</mathjax> solute</p>
<p>This mass of solute is to be taken in a 250mL volumetric flask by weighing in an electronic balance and then it is disolved in water of to make a solution of volume less than 250mL and finally required volume of water is added to it to make the volume 250mL.</p></div>
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<div class="markdown"><p>Dissolve 9.31 g of <mathjax>#Na_2S_2O_3 * 5H_2O#</mathjax> into 250mL water.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dissolve 9.31 g of <mathjax>#Na_2S_2O_3 * 5H_2O#</mathjax> into 250mL water.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is moles/Liter. 0.150 M in 250mL is ¼ of 0.150 moles/L, or 0.0375 moles/0.25L. </p>
<p><mathjax>#Na_2S_2O_3 * 5H_2O#</mathjax> just has the extra mass (moles) of water to consider. That is irrelevant to this problem because we are already given the molar mass for the compound. It doesn’t matter that there are a few additional water molecules hanging on, the moles of sodium thiosuflate in the compound are the same.</p>
<p>0.0375 moles * 248.2 g/mol = 9.3075g <mathjax>#Na_2S_2O_3 * 5H_2O#</mathjax> </p></div>
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</article> | How would you make 250 ml of 0.150 M sodium thiosulfate from solid Na2S2O3 *
5H2O?
(molar mass = 248.2 g/mole) | null |
3,122 | aacbff1a-6ddd-11ea-975b-ccda262736ce | https://socratic.org/questions/582dfa7511ef6b1ec925a3c6 | 288.32 g | start physical_unit 3 3 mass g qc_end physical_unit 13 14 9 10 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] g"}] | [{"type":"physical unit","value":"288.32 g"}] | [{"type":"physical unit","value":"Mole [OF] methane gas [=] \\pu{8.0 mol}"}] | <h1 class="questionTitle" itemprop="name">What mass of water results from combustion of an #8.0*mol# quantity of #"methane gas"#?</h1> | null | 288.32 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need a stoichiometrically balanced equation to represent the combustion:</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> dictates that for each mole of methane, 2 moles of water are evolved. </p>
<p>So if <mathjax>#8.0*mol#</mathjax> methane gas were combusted, <mathjax>#16*mol#</mathjax> of water were evolved: <mathjax>#16*molxx18.02*g*mol^-1=???g#</mathjax></p></div>
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<div class="markdown"><p>Under <mathjax>#300*g#</mathjax> water. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need a stoichiometrically balanced equation to represent the combustion:</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> dictates that for each mole of methane, 2 moles of water are evolved. </p>
<p>So if <mathjax>#8.0*mol#</mathjax> methane gas were combusted, <mathjax>#16*mol#</mathjax> of water were evolved: <mathjax>#16*molxx18.02*g*mol^-1=???g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of water results from combustion of an #8.0*mol# quantity of #"methane gas"#?</h1>
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<div class="markdown"><p>Under <mathjax>#300*g#</mathjax> water. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need a stoichiometrically balanced equation to represent the combustion:</p>
<p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> dictates that for each mole of methane, 2 moles of water are evolved. </p>
<p>So if <mathjax>#8.0*mol#</mathjax> methane gas were combusted, <mathjax>#16*mol#</mathjax> of water were evolved: <mathjax>#16*molxx18.02*g*mol^-1=???g#</mathjax></p></div>
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</article> | What mass of water results from combustion of an #8.0*mol# quantity of #"methane gas"#? | null |
3,123 | abe456bf-6ddd-11ea-9d60-ccda262736ce | https://socratic.org/questions/a-solution-is-a-mixture-of-0-05m-nacl-and-0-05m-nai-the-concentration-of-iodide- | 2 × 10^(-7) M | start physical_unit 15 16 concentration mol/l qc_end physical_unit 8 8 6 7 molarity qc_end physical_unit 12 12 6 7 molarity qc_end c_other OTHER qc_end physical_unit 21 21 31 34 equilibrium_constant_k qc_end physical_unit 36 36 38 41 equilibrium_constant_k qc_end end | [{"type":"physical unit","value":"Concentration [OF] iodide ion [IN] M"}] | [{"type":"physical unit","value":"2 × 10^(-7) M"}] | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [=] \\pu{0.05 M}"},{"type":"physical unit","value":"Molarity [OF] NaI solution [=] \\pu{0.05 M}"},{"type":"other","value":"When AgCl just starts precipitating."},{"type":"physical unit","value":"Ksp [OF] AgCl [=] \\pu{1 × 10^(-10) M^2}"},{"type":"physical unit","value":"Ksp [OF] AgI [=] \\pu{4 × 10^(-16) M^2}"}] | <h1 class="questionTitle" itemprop="name">A solution is a mixture of #"0.05M NaCl"# and #"0.05M NaI"#.The concentration of iodide ion in the solution when #"AgCl"# just starts precipitating is equal to?
</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#K_("sp AgCl") = 1 * 10^(-10)#</mathjax> <mathjax>#"M"^2#</mathjax></p>
<p><mathjax>#K_("sp AgI") = 4* 10^(-16)#</mathjax> <mathjax>#"M"^2#</mathjax> </p></div>
</h2>
</div>
</div> | 2 × 10^(-7) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>After you mix the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, the resulting solution will contain two anions that can combine with the silver(I) cations to form an insoluble precipitate, the chloride anions and the iodide anions, respectively. </p>
<p>For silver chloride, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>solubility product constant</strong> for this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgCl") = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you know that the solution contains</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 1" = K_("sp AgCl")/(["Cl"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_ "min 1" = (1 * 10^(-10) "M"^color(red)(cancel(color(black)(2))))/(0.05color(red)(cancel(color(black)("M")))) = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(a)#</mathjax><br/>
Now, for silver iodide, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This time, the <strong>solubility product constant</strong> is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgI") = ["Ag"^(+)] * ["I"^(-)]#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver iodide to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" = K_("sp AgI")/(["I"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_"min 2" = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(0.05 color(red)(cancel(color(black)("M")))) = 8 * 10^(-15) quad "M"#</mathjax></p>
</blockquote>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" " << " ["Ag"^(+)]_"min 1"#</mathjax></p>
</blockquote>
<p>This tells you that as you start to add silver(I) cations to the solution, the silver iodide will precipitate <strong>first</strong>, consuming some of the iodide ions present in the solution in the process. </p>
<p>This tells you that when silver chloride precipitate, the solution will <strong>not</strong> contain</p>
<blockquote>
<p><mathjax>#["I"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>because some of the iodide ions already precipitated as silver iodide. In order for the silver(I) cations to precipitate the silver chloride when </p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>you need a concentration of silver(I) cations equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min1" = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p>present in the solution. This means that at that point, the corresponding concentration of iodide anions must be at a <em>maximum</em>--which corresponds to the solubility equilibrium condition--value of</p>
<blockquote>
<p><mathjax>#["I"^(-)] = K_("sp AgI")/(["Ag"^(+)]_ "min 1")#</mathjax></p>
<p><mathjax>#["I"^(-)] = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(2 * 10^(-9) color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(2 * 10^(-7) quad "M")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#["I"^(-)] = 2 * 10^(-7)#</mathjax> <mathjax>#"M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>After you mix the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, the resulting solution will contain two anions that can combine with the silver(I) cations to form an insoluble precipitate, the chloride anions and the iodide anions, respectively. </p>
<p>For silver chloride, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>solubility product constant</strong> for this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgCl") = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you know that the solution contains</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 1" = K_("sp AgCl")/(["Cl"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_ "min 1" = (1 * 10^(-10) "M"^color(red)(cancel(color(black)(2))))/(0.05color(red)(cancel(color(black)("M")))) = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(a)#</mathjax><br/>
Now, for silver iodide, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This time, the <strong>solubility product constant</strong> is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgI") = ["Ag"^(+)] * ["I"^(-)]#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver iodide to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" = K_("sp AgI")/(["I"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_"min 2" = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(0.05 color(red)(cancel(color(black)("M")))) = 8 * 10^(-15) quad "M"#</mathjax></p>
</blockquote>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" " << " ["Ag"^(+)]_"min 1"#</mathjax></p>
</blockquote>
<p>This tells you that as you start to add silver(I) cations to the solution, the silver iodide will precipitate <strong>first</strong>, consuming some of the iodide ions present in the solution in the process. </p>
<p>This tells you that when silver chloride precipitate, the solution will <strong>not</strong> contain</p>
<blockquote>
<p><mathjax>#["I"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>because some of the iodide ions already precipitated as silver iodide. In order for the silver(I) cations to precipitate the silver chloride when </p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>you need a concentration of silver(I) cations equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min1" = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p>present in the solution. This means that at that point, the corresponding concentration of iodide anions must be at a <em>maximum</em>--which corresponds to the solubility equilibrium condition--value of</p>
<blockquote>
<p><mathjax>#["I"^(-)] = K_("sp AgI")/(["Ag"^(+)]_ "min 1")#</mathjax></p>
<p><mathjax>#["I"^(-)] = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(2 * 10^(-9) color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(2 * 10^(-7) quad "M")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">A solution is a mixture of #"0.05M NaCl"# and #"0.05M NaI"#.The concentration of iodide ion in the solution when #"AgCl"# just starts precipitating is equal to?
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<div class="markdown"><p><mathjax>#K_("sp AgCl") = 1 * 10^(-10)#</mathjax> <mathjax>#"M"^2#</mathjax></p>
<p><mathjax>#K_("sp AgI") = 4* 10^(-16)#</mathjax> <mathjax>#"M"^2#</mathjax> </p></div>
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Stefan V.
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Feb 10, 2018
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<div class="markdown"><p><mathjax>#["I"^(-)] = 2 * 10^(-7)#</mathjax> <mathjax>#"M"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>After you mix the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, the resulting solution will contain two anions that can combine with the silver(I) cations to form an insoluble precipitate, the chloride anions and the iodide anions, respectively. </p>
<p>For silver chloride, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>solubility product constant</strong> for this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgCl") = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you know that the solution contains</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver chloride to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 1" = K_("sp AgCl")/(["Cl"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_ "min 1" = (1 * 10^(-10) "M"^color(red)(cancel(color(black)(2))))/(0.05color(red)(cancel(color(black)("M")))) = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(a)#</mathjax><br/>
Now, for silver iodide, the solubility equilibrium looks like this</p>
<blockquote>
<p><mathjax>#"AgI"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "I"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This time, the <strong>solubility product constant</strong> is equal to</p>
<blockquote>
<p><mathjax>#K_("sp AgI") = ["Ag"^(+)] * ["I"^(-)]#</mathjax></p>
</blockquote>
<p>This means that the <em>minimum</em> concentration of silver(I) cations that will cause the silver iodide to precipitate is equal to</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" = K_("sp AgI")/(["I"^(-)])#</mathjax></p>
<p><mathjax>#["Ag"^(+)]_"min 2" = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(0.05 color(red)(cancel(color(black)("M")))) = 8 * 10^(-15) quad "M"#</mathjax></p>
</blockquote>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min 2" " << " ["Ag"^(+)]_"min 1"#</mathjax></p>
</blockquote>
<p>This tells you that as you start to add silver(I) cations to the solution, the silver iodide will precipitate <strong>first</strong>, consuming some of the iodide ions present in the solution in the process. </p>
<p>This tells you that when silver chloride precipitate, the solution will <strong>not</strong> contain</p>
<blockquote>
<p><mathjax>#["I"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>because some of the iodide ions already precipitated as silver iodide. In order for the silver(I) cations to precipitate the silver chloride when </p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = "0.05 M"#</mathjax></p>
</blockquote>
<p>you need a concentration of silver(I) cations equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)]_ "min1" = 2 * 10^(-9) quad "M"#</mathjax></p>
</blockquote>
<p>present in the solution. This means that at that point, the corresponding concentration of iodide anions must be at a <em>maximum</em>--which corresponds to the solubility equilibrium condition--value of</p>
<blockquote>
<p><mathjax>#["I"^(-)] = K_("sp AgI")/(["Ag"^(+)]_ "min 1")#</mathjax></p>
<p><mathjax>#["I"^(-)] = (4 * 10^(-16) "M"^color(red)(cancel(color(black)(2))))/(2 * 10^(-9) color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(2 * 10^(-7) quad "M")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p></div>
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</article> | A solution is a mixture of #"0.05M NaCl"# and #"0.05M NaI"#.The concentration of iodide ion in the solution when #"AgCl"# just starts precipitating is equal to?
|
#K_("sp AgCl") = 1 * 10^(-10)# #"M"^2#
#K_("sp AgI") = 4* 10^(-16)# #"M"^2#
|
3,124 | a94daa12-6ddd-11ea-a093-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-3ch4-g-o2-co2-h2o-g-energy | CH4(g) + 2O2 -> CO2 + 2H2O(g) | start chemical_equation qc_end chemical_equation 7 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"CH4(g) + 2O2 -> CO2 + 2H2O(g) "}] | [{"type":"chemical equation","value":"3 CH4(g) + O2 -> CO2 + H2O(g) + energy"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation:
3CH4(g) + O2 --> CO2 + H2O(g) + energy?</h1> | null | CH4(g) + 2O2 -> CO2 + 2H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the equation balanced? How do you know. Where would you place the energy term?</p>
<p>How would I balance the following word equation?</p>
<p>Ethane + oxygen <mathjax>#rarr#</mathjax> carbon dioxide + water.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2 rarr CO_2(g) + 2H_2O(g)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the equation balanced? How do you know. Where would you place the energy term?</p>
<p>How would I balance the following word equation?</p>
<p>Ethane + oxygen <mathjax>#rarr#</mathjax> carbon dioxide + water.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance the following equation:
3CH4(g) + O2 --> CO2 + H2O(g) + energy?</h1>
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<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2 rarr CO_2(g) + 2H_2O(g)#</mathjax></p></div>
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<div class="markdown"><p>Is the equation balanced? How do you know. Where would you place the energy term?</p>
<p>How would I balance the following word equation?</p>
<p>Ethane + oxygen <mathjax>#rarr#</mathjax> carbon dioxide + water.</p></div>
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</article> | How would you balance the following equation:
3CH4(g) + O2 --> CO2 + H2O(g) + energy? | null |
3,125 | a8c077ee-6ddd-11ea-a4cc-ccda262736ce | https://socratic.org/questions/the-freezing-point-of-an-aqueous-solution-is-2-79-c-what-is-the-boiling-point-of | 100.77 ℃ | start physical_unit 16 17 boiling_point_temperature °c qc_end physical_unit 4 6 8 9 freezing_point_temperature qc_end end | [{"type":"physical unit","value":"Boiling point [OF] this solution [IN] ℃"}] | [{"type":"physical unit","value":"100.77 ℃"}] | [{"type":"physical unit","value":"Freezing point [OF] an aqueous solution [=] \\pu{-2.79 ℃}"}] | <h1 class="questionTitle" itemprop="name">The freezing point of an aqueous solution is -2.79°C. What is the boiling point of this solution?</h1> | null | 100.77 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you have to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of particles in the solution using the following formula:</p>
<p><mathjax>#Delta T= K_fxxmxx i#</mathjax> </p>
<p>where:</p>
<p><mathjax>#m -> " is the molality of the solute expressed in " (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T ->" is the difference between the freezing point of the "#</mathjax><br/>
<mathjax>#"solution"\ (T_f)" and that of the pure solvent "(T_i) . #</mathjax><br/>
<mathjax># \Delta T = T_f- T_i #</mathjax></p>
<p><mathjax># i -> " is the Van'to Hoff factor. It is related to the number of"#</mathjax><br/>
<mathjax>#"particles ( or ions) of the solute dissolved in the solution."#</mathjax></p>
<p><mathjax># K_f->"is the freezing point depression constant of the solvent."#</mathjax><br/>
<mathjax>#"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_("molality of particles")#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_(m')#</mathjax></p>
<p><mathjax>#Delta T= - K_fxx\ m'#</mathjax></p>
<p><mathjax>#m'=- (Delta T)/ K_f#</mathjax></p>
<p><mathjax>#m'=- ((T_f-T_i))/ K_f#</mathjax></p>
<p><mathjax>#m'= -((-2.79 °C-0°C))/(1.86°C.Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= (2.79 cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= 1.5 \ (mol)/(Kg)#</mathjax></p>
<p>Once the molality of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is determined, the boiling point of the solution could be determined.</p>
<p>The change in the boiling point is determined using the following formula: </p>
<p><mathjax>#Delta T= K_b\ xx mxx i#</mathjax></p>
<p>Where <mathjax>#K_b#</mathjax> is the boiling point elevation constant for water.</p>
<p><mathjax>#K_b = 0.510 \°C* Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxunderbrace (mxx i)_(m')#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxm'#</mathjax></p>
<p><mathjax>#Delta T= 0.510 \°C* Kg*mol^-1 xx1.5 \ (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T= 0.765 °C#</mathjax></p>
<p><mathjax>#Delta T = T_f- T_i #</mathjax></p>
<p>Note that <mathjax>#T_i#</mathjax> is the normal boiling point of water and <mathjax>#T_f#</mathjax> is the temperature at which the solution boils.</p>
<p><mathjax>#0.765 °C= T_f- 100.000 °C #</mathjax></p>
<p><mathjax># T_f=100.765 °C #</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax># 100.765 °C #</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you have to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of particles in the solution using the following formula:</p>
<p><mathjax>#Delta T= K_fxxmxx i#</mathjax> </p>
<p>where:</p>
<p><mathjax>#m -> " is the molality of the solute expressed in " (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T ->" is the difference between the freezing point of the "#</mathjax><br/>
<mathjax>#"solution"\ (T_f)" and that of the pure solvent "(T_i) . #</mathjax><br/>
<mathjax># \Delta T = T_f- T_i #</mathjax></p>
<p><mathjax># i -> " is the Van'to Hoff factor. It is related to the number of"#</mathjax><br/>
<mathjax>#"particles ( or ions) of the solute dissolved in the solution."#</mathjax></p>
<p><mathjax># K_f->"is the freezing point depression constant of the solvent."#</mathjax><br/>
<mathjax>#"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_("molality of particles")#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_(m')#</mathjax></p>
<p><mathjax>#Delta T= - K_fxx\ m'#</mathjax></p>
<p><mathjax>#m'=- (Delta T)/ K_f#</mathjax></p>
<p><mathjax>#m'=- ((T_f-T_i))/ K_f#</mathjax></p>
<p><mathjax>#m'= -((-2.79 °C-0°C))/(1.86°C.Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= (2.79 cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= 1.5 \ (mol)/(Kg)#</mathjax></p>
<p>Once the molality of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is determined, the boiling point of the solution could be determined.</p>
<p>The change in the boiling point is determined using the following formula: </p>
<p><mathjax>#Delta T= K_b\ xx mxx i#</mathjax></p>
<p>Where <mathjax>#K_b#</mathjax> is the boiling point elevation constant for water.</p>
<p><mathjax>#K_b = 0.510 \°C* Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxunderbrace (mxx i)_(m')#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxm'#</mathjax></p>
<p><mathjax>#Delta T= 0.510 \°C* Kg*mol^-1 xx1.5 \ (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T= 0.765 °C#</mathjax></p>
<p><mathjax>#Delta T = T_f- T_i #</mathjax></p>
<p>Note that <mathjax>#T_i#</mathjax> is the normal boiling point of water and <mathjax>#T_f#</mathjax> is the temperature at which the solution boils.</p>
<p><mathjax>#0.765 °C= T_f- 100.000 °C #</mathjax></p>
<p><mathjax># T_f=100.765 °C #</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The freezing point of an aqueous solution is -2.79°C. What is the boiling point of this solution?</h1>
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Jun 16, 2016
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<div class="markdown"><p><mathjax># 100.765 °C #</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you have to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of particles in the solution using the following formula:</p>
<p><mathjax>#Delta T= K_fxxmxx i#</mathjax> </p>
<p>where:</p>
<p><mathjax>#m -> " is the molality of the solute expressed in " (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T ->" is the difference between the freezing point of the "#</mathjax><br/>
<mathjax>#"solution"\ (T_f)" and that of the pure solvent "(T_i) . #</mathjax><br/>
<mathjax># \Delta T = T_f- T_i #</mathjax></p>
<p><mathjax># i -> " is the Van'to Hoff factor. It is related to the number of"#</mathjax><br/>
<mathjax>#"particles ( or ions) of the solute dissolved in the solution."#</mathjax></p>
<p><mathjax># K_f->"is the freezing point depression constant of the solvent."#</mathjax><br/>
<mathjax>#"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_("molality of particles")#</mathjax></p>
<p><mathjax>#Delta T= - K_f\ xxunderbrace (mxxi)_(m')#</mathjax></p>
<p><mathjax>#Delta T= - K_fxx\ m'#</mathjax></p>
<p><mathjax>#m'=- (Delta T)/ K_f#</mathjax></p>
<p><mathjax>#m'=- ((T_f-T_i))/ K_f#</mathjax></p>
<p><mathjax>#m'= -((-2.79 °C-0°C))/(1.86°C.Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= (2.79 cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)#</mathjax></p>
<p><mathjax>#m'= 1.5 \ (mol)/(Kg)#</mathjax></p>
<p>Once the molality of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is determined, the boiling point of the solution could be determined.</p>
<p>The change in the boiling point is determined using the following formula: </p>
<p><mathjax>#Delta T= K_b\ xx mxx i#</mathjax></p>
<p>Where <mathjax>#K_b#</mathjax> is the boiling point elevation constant for water.</p>
<p><mathjax>#K_b = 0.510 \°C* Kg*mol^-1#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxunderbrace (mxx i)_(m')#</mathjax></p>
<p><mathjax>#Delta T= K_b\ xxm'#</mathjax></p>
<p><mathjax>#Delta T= 0.510 \°C* Kg*mol^-1 xx1.5 \ (mol)/(Kg)#</mathjax></p>
<p><mathjax>#Delta T= 0.765 °C#</mathjax></p>
<p><mathjax>#Delta T = T_f- T_i #</mathjax></p>
<p>Note that <mathjax>#T_i#</mathjax> is the normal boiling point of water and <mathjax>#T_f#</mathjax> is the temperature at which the solution boils.</p>
<p><mathjax>#0.765 °C= T_f- 100.000 °C #</mathjax></p>
<p><mathjax># T_f=100.765 °C #</mathjax></p></div>
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</article> | The freezing point of an aqueous solution is -2.79°C. What is the boiling point of this solution? | null |
3,126 | abefe55e-6ddd-11ea-a124-ccda262736ce | https://socratic.org/questions/how-do-you-balance-hcl-l-zn-s-zncl-h-2-g-110kj | 2 HCl(l) + Zn(s) -> ZnCl + H2(g) + 110 kJ | start chemical_equation qc_end chemical_equation 4 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 HCl(l) + Zn(s) -> ZnCl + H2(g) + 110 kJ"}] | [{"type":"chemical equation","value":"HCl(l) + Zn(s) -> ZnCl + H2(g) + 110 kJ"}] | <h1 class="questionTitle" itemprop="name">How do you balance #HCl(l) + Zn(s) -> ZnCl + H_2(g) + 110kJ#?</h1> | null | 2 HCl(l) + Zn(s) -> ZnCl + H2(g) + 110 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ul>
<li>
<p>When creating a chemical equation, the energy produced from the shouldn't be placed with the rest of the equation. However the energy produced shouldn't be forgotten and should be placed bellow.</p>
</li>
<li>
<p>Make sure that the number of atoms for each element on the left side of the equation adds up to the the number of atoms for each element on the right side.</p>
</li>
</ul></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#</mathjax></p>
<p>Energy produced from exothermic reaction: 110kJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ul>
<li>
<p>When creating a chemical equation, the energy produced from the shouldn't be placed with the rest of the equation. However the energy produced shouldn't be forgotten and should be placed bellow.</p>
</li>
<li>
<p>Make sure that the number of atoms for each element on the left side of the equation adds up to the the number of atoms for each element on the right side.</p>
</li>
</ul></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance #HCl(l) + Zn(s) -> ZnCl + H_2(g) + 110kJ#?</h1>
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<div class="markdown"><p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#</mathjax></p>
<p>Energy produced from exothermic reaction: 110kJ</p></div>
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<li>
<p>When creating a chemical equation, the energy produced from the shouldn't be placed with the rest of the equation. However the energy produced shouldn't be forgotten and should be placed bellow.</p>
</li>
<li>
<p>Make sure that the number of atoms for each element on the left side of the equation adds up to the the number of atoms for each element on the right side.</p>
</li>
</ul></div>
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</article> | How do you balance #HCl(l) + Zn(s) -> ZnCl + H_2(g) + 110kJ#? | null |
3,127 | acab45ba-6ddd-11ea-a441-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-sulfur-in-magnesium-sulfate | +6 | start physical_unit 6 6 oxidation_number none qc_end substance 8 9 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] sulfur"}] | [{"type":"physical unit","value":"+6"}] | [{"type":"substance name","value":"Magnesium sulfate"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of sulfur in magnesium sulfate?</h1> | null | +6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur in the sulfate ion achieves the group oxidation number: i.e.<mathjax>#S_(ON)=(-2-4xx-2)=-2-(-8)=+6#</mathjax>.</p>
<p>Can you tell me the oxidation number of the sulfurs in thiosulfate, <mathjax>#S_2O_3^(2-)?#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>In sulfate? <mathjax>#S(VI+)#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur in the sulfate ion achieves the group oxidation number: i.e.<mathjax>#S_(ON)=(-2-4xx-2)=-2-(-8)=+6#</mathjax>.</p>
<p>Can you tell me the oxidation number of the sulfurs in thiosulfate, <mathjax>#S_2O_3^(2-)?#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of sulfur in magnesium sulfate?</h1>
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<div class="markdown"><p>In sulfate? <mathjax>#S(VI+)#</mathjax>.</p></div>
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<div class="markdown"><p>Sulfur in the sulfate ion achieves the group oxidation number: i.e.<mathjax>#S_(ON)=(-2-4xx-2)=-2-(-8)=+6#</mathjax>.</p>
<p>Can you tell me the oxidation number of the sulfurs in thiosulfate, <mathjax>#S_2O_3^(2-)?#</mathjax></p></div>
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</article> | What is the oxidation number of sulfur in magnesium sulfate? | null |
3,128 | aa909936-6ddd-11ea-8c99-ccda262736ce | https://socratic.org/questions/if-2-0-moles-of-a-substance-are-dissolved-in-2-0-liters-of-solution-what-is-the- | 1.00 M | start physical_unit 18 19 molarity mol/l qc_end physical_unit 5 5 1 2 mole qc_end physical_unit 18 19 9 10 volume qc_end end | [{"type":"physical unit","value":"molarity [OF] this solution [IN] M"}] | [{"type":"physical unit","value":"1.00 M"}] | [{"type":"physical unit","value":"Mole [OF] the substance [=] \\pu{2.0 moles}"},{"type":"physical unit","value":"Volume [OF] this solution [=] \\pu{2.0 liters}"}] | <h1 class="questionTitle" itemprop="name">If 2.0 moles of a substance are dissolved in 2.0 liters of solution, what is the molarity of this solution?</h1> | null | 1.00 M | <div class="answerDescription">
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<div class="markdown"><p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) equals moles of substance over liters of solution, that would be <mathjax>#(2 mols)/(2 L)#</mathjax>, which makes the molarity 1 M.</p></div>
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<div class="markdown"><p>1 M.`</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) equals moles of substance over liters of solution, that would be <mathjax>#(2 mols)/(2 L)#</mathjax>, which makes the molarity 1 M.</p></div>
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<h1 class="questionTitle" itemprop="name">If 2.0 moles of a substance are dissolved in 2.0 liters of solution, what is the molarity of this solution?</h1>
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<div class="markdown"><p>1 M.`</p></div>
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<div class="markdown"><p>Since <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) equals moles of substance over liters of solution, that would be <mathjax>#(2 mols)/(2 L)#</mathjax>, which makes the molarity 1 M.</p></div>
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</article> | If 2.0 moles of a substance are dissolved in 2.0 liters of solution, what is the molarity of this solution? | null |
3,129 | aa8b0a5c-6ddd-11ea-9a4c-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-manganese-iv-bicarbonate | Mn(HCO3)4 | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] manganese iV bicarbonate [IN] default"}] | [{"type":"chemical equation","value":"Mn(HCO3)4"}] | [{"type":"substance name","value":"Manganese iV bicarbonate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for manganese iV bicarbonate?</h1> | null | Mn(HCO3)4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Manganese (IV) bicarbonate is an ionic compound, composed of two ions, Manganese (IV), and bicarbonate, also called hydrogen carbonate.</p>
<p><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">Ionic compounds</a> are neutral, meaning that the total positive charge and total negative charge are equal.</p>
<p>The name manganese (IV) tells us that the manganese ion has a charge of <mathjax>#4+#</mathjax>. The formula for manganese (IV) is <mathjax>#"Mn"^(4+)"#</mathjax>, and bicarbonate has the formula <mathjax>#"HCO"_3"^(-)#</mathjax> Since the bicarbonate ion has a <mathjax>#1-#</mathjax> charge, and the manganese ion has a <mathjax>#4+#</mathjax> charge, four bicarbonate ions will be needed to form a neutral compound.</p>
<p><mathjax>#"Mn"^(4+)#</mathjax><mathjax>#+#</mathjax><mathjax>#"4HCO"_3"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mn(HCO"_3)_4#</mathjax></p>
<p>Another way to determine the formula of an ionic compound is by crisscrossing. The size of the charge on the positive ion becomes the subscript on the negative ion, and the size of the charge on the negative ion becomes the subscript of the positive ion.</p>
<p>The examples below show how the process works.</p>
<p><img alt="http://www.ck12.org/section/Ionic-Compounds/" src="https://d2jmvrsizmvf4x.cloudfront.net/7s33hN7sQZiSfKs4Tdo9_images"/> </p>
<p><img alt="http://faculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch100_OL/aMy_FileLec/04OL_LecNotes_Ch100/05_CompoundBonding/501_IonicCompounds/501_IonicCompounds.htm" src="https://useruploads.socratic.org/1MZkF4ySEC6W2np2k6q0_AlScrisscross.gif"/> </p></div>
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<div class="markdown"><p><mathjax>#"Mn(HCO"_3)_4#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Manganese (IV) bicarbonate is an ionic compound, composed of two ions, Manganese (IV), and bicarbonate, also called hydrogen carbonate.</p>
<p><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">Ionic compounds</a> are neutral, meaning that the total positive charge and total negative charge are equal.</p>
<p>The name manganese (IV) tells us that the manganese ion has a charge of <mathjax>#4+#</mathjax>. The formula for manganese (IV) is <mathjax>#"Mn"^(4+)"#</mathjax>, and bicarbonate has the formula <mathjax>#"HCO"_3"^(-)#</mathjax> Since the bicarbonate ion has a <mathjax>#1-#</mathjax> charge, and the manganese ion has a <mathjax>#4+#</mathjax> charge, four bicarbonate ions will be needed to form a neutral compound.</p>
<p><mathjax>#"Mn"^(4+)#</mathjax><mathjax>#+#</mathjax><mathjax>#"4HCO"_3"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mn(HCO"_3)_4#</mathjax></p>
<p>Another way to determine the formula of an ionic compound is by crisscrossing. The size of the charge on the positive ion becomes the subscript on the negative ion, and the size of the charge on the negative ion becomes the subscript of the positive ion.</p>
<p>The examples below show how the process works.</p>
<p><img alt="http://www.ck12.org/section/Ionic-Compounds/" src="https://d2jmvrsizmvf4x.cloudfront.net/7s33hN7sQZiSfKs4Tdo9_images"/> </p>
<p><img alt="http://faculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch100_OL/aMy_FileLec/04OL_LecNotes_Ch100/05_CompoundBonding/501_IonicCompounds/501_IonicCompounds.htm" src="https://useruploads.socratic.org/1MZkF4ySEC6W2np2k6q0_AlScrisscross.gif"/> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for manganese iV bicarbonate?</h1>
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<div class="markdown"><p><mathjax>#"Mn(HCO"_3)_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Manganese (IV) bicarbonate is an ionic compound, composed of two ions, Manganese (IV), and bicarbonate, also called hydrogen carbonate.</p>
<p><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">Ionic compounds</a> are neutral, meaning that the total positive charge and total negative charge are equal.</p>
<p>The name manganese (IV) tells us that the manganese ion has a charge of <mathjax>#4+#</mathjax>. The formula for manganese (IV) is <mathjax>#"Mn"^(4+)"#</mathjax>, and bicarbonate has the formula <mathjax>#"HCO"_3"^(-)#</mathjax> Since the bicarbonate ion has a <mathjax>#1-#</mathjax> charge, and the manganese ion has a <mathjax>#4+#</mathjax> charge, four bicarbonate ions will be needed to form a neutral compound.</p>
<p><mathjax>#"Mn"^(4+)#</mathjax><mathjax>#+#</mathjax><mathjax>#"4HCO"_3"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mn(HCO"_3)_4#</mathjax></p>
<p>Another way to determine the formula of an ionic compound is by crisscrossing. The size of the charge on the positive ion becomes the subscript on the negative ion, and the size of the charge on the negative ion becomes the subscript of the positive ion.</p>
<p>The examples below show how the process works.</p>
<p><img alt="http://www.ck12.org/section/Ionic-Compounds/" src="https://d2jmvrsizmvf4x.cloudfront.net/7s33hN7sQZiSfKs4Tdo9_images"/> </p>
<p><img alt="http://faculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch100_OL/aMy_FileLec/04OL_LecNotes_Ch100/05_CompoundBonding/501_IonicCompounds/501_IonicCompounds.htm" src="https://useruploads.socratic.org/1MZkF4ySEC6W2np2k6q0_AlScrisscross.gif"/> </p></div>
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</article> | What is the formula for manganese iV bicarbonate? | null |
3,130 | abf314d1-6ddd-11ea-b05b-ccda262736ce | https://socratic.org/questions/5906898c7c014934fbabb98a | 7.98 grams | start physical_unit 10 10 mass g qc_end physical_unit 4 4 1 2 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Cl2 [IN] grams"}] | [{"type":"physical unit","value":"7.98 grams"}] | [{"type":"physical unit","value":"Mass [OF] AlCl3 [=] \\pu{10 g}"}] | <h1 class="questionTitle" itemprop="name">If #"10 g"# of #"AlCl"_3"# decomposes, how many grams of #"Cl"_2"# will be produced?</h1> | null | 7.98 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you need to find the number of moles of aluminum chloride. </p>
<p>moles = mass/rmm</p>
<p><mathjax>#(10g)/ (26.98+3(35.453))#</mathjax> = 0.075 moles</p>
<p>Since the equation states that for every 3 molecules of aluminum chloride, you produce 2 molecules of chlorine, by cross-multiplying the amount of moles, you obtain 0.1125 moles of chlorine produced. </p>
<p>Now, just convert moles to grams by rearranging the formula earlier, and you should obtain 7.98 grams (rounded to 2 decimal places)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you need to find the number of moles of aluminum chloride. </p>
<p>moles = mass/rmm</p>
<p><mathjax>#(10g)/ (26.98+3(35.453))#</mathjax> = 0.075 moles</p>
<p>Since the equation states that for every 3 molecules of aluminum chloride, you produce 2 molecules of chlorine, by cross-multiplying the amount of moles, you obtain 0.1125 moles of chlorine produced. </p>
<p>Now, just convert moles to grams by rearranging the formula earlier, and you should obtain 7.98 grams (rounded to 2 decimal places)</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If #"10 g"# of #"AlCl"_3"# decomposes, how many grams of #"Cl"_2"# will be produced?</h1>
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<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you need to find the number of moles of aluminum chloride. </p>
<p>moles = mass/rmm</p>
<p><mathjax>#(10g)/ (26.98+3(35.453))#</mathjax> = 0.075 moles</p>
<p>Since the equation states that for every 3 molecules of aluminum chloride, you produce 2 molecules of chlorine, by cross-multiplying the amount of moles, you obtain 0.1125 moles of chlorine produced. </p>
<p>Now, just convert moles to grams by rearranging the formula earlier, and you should obtain 7.98 grams (rounded to 2 decimal places)</p></div>
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Meave60
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<div class="markdown"><p><mathjax>#~~"8 g Cl"_2"#</mathjax> will be produced</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"2AlCl"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2Al + 3Cl"_2#</mathjax></p>
<p>The process for solving this problem is:</p>
<p><mathjax>#color(red)("given mass AlCl"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("mol AlCl"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)("mol Cl"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)
("mass Cl"_2"#</mathjax></p>
<p>The molar masses of aluminum chloride and chlorine gas must be determined. Molar mass is determined by multiplying the subscript of each element by its atomic weight (relative atomic mass) from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole, or g/mol.</p>
<p><strong>Molar Masses</strong><br/>
<mathjax>#"AlCl"_3:#</mathjax><mathjax>#(1xx26.982"g/mol Al")+(3xx35.45"g/mol Cl")="133.332 g/mol AlCl"_3"#</mathjax><br/>
(I will round to the correct number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</p>
<p><mathjax>#"Cl"_2:#</mathjax><mathjax>#(2xx35.45"g/mol Cl")="70.9 g/mol Cl"_2"#</mathjax></p>
<p><strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">Mole Ratios</a></strong><br/>
Write <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratios between <mathjax>#"AlCl"_3"#</mathjax> and <mathjax>#"Cl"_2"#</mathjax> from the balanced equation. The mole ratio allows us to convert between moles and mass and vice-versa.</p>
<p><mathjax>#(2"mol AlCl"_3)/(3"mol Cl"_2)#</mathjax> <mathjax>#color(white)(....)"and"#</mathjax> <mathjax>#color(white)(....)(3"mol Cl"_2)/(2"mol AlCl"_3)#</mathjax></p>
<p><mathjax>#color(red)("Given Mass"#</mathjax> <mathjax>#color(red)("AlCl"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("Mol AlCl"_3"#</mathjax></p>
<p>Multiply the given mass of <mathjax>#"AlCl"_3"#</mathjax> by the reciprocal of the molar mass of <mathjax>#"Cl"_2"#</mathjax>.</p>
<p><mathjax>#10color(red)cancel(color(black)("g AlCl"_3))xx(1"mol AlCl"_3)/(133.332color(red)cancel(color(black)("g AlCl"_3)))="0.0750008 mol AlCl"_3"#</mathjax></p>
<p><mathjax>#color(blue)("Mol AlCl"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)("Mol Cl"_2"#</mathjax></p>
<p>Multiply mol <mathjax>#"AlCl"_3"#</mathjax> by the mole ratio with <mathjax>#"Cl"_2"#</mathjax> in the numerator.</p>
<p><mathjax>#0.0750008color(red)cancel(color(black)("mol AlCl"_3))xx(3"mol Cl"_2)/(2color(red)cancel(color(black)("mol AlCl"_3)))="0.11250 mol Cl"_2"#</mathjax></p>
<p><mathjax>#color(green)("Mole Cl"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)"Mass Cl"_2"#</mathjax></p>
<p>Multiply mol <mathjax>#"Cl"_2"#</mathjax> by its molar mass.</p>
<p><mathjax>#0.11250color(red)cancel(color(black)("mol Cl"_2))xx(70.9"g Cl"_2)/(1color(red)cancel(color(black)("mol Cl"_2)))="7.976 g Cl"_2~~"8 g Cl"_2"#</mathjax> (rounded to one significant figure due to 10 g <mathjax>#"AlCl"_3"#</mathjax></p></div>
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</article> | If #"10 g"# of #"AlCl"_3"# decomposes, how many grams of #"Cl"_2"# will be produced? | null |
3,131 | aa3b19fb-6ddd-11ea-b88a-ccda262736ce | https://socratic.org/questions/what-would-be-the-formula-between-iron-iii-and-sulfur | Fe2S3 | start chemical_formula qc_end substance 6 7 qc_end substance 9 9 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"Fe2S3"}] | [{"type":"substance name","value":"Iron III"},{"type":"substance name","value":"Sulfur"}] | <h1 class="questionTitle" itemprop="name">What would be the formula between Iron III and Sulfur?</h1> | null | Fe2S3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Iron III I has a valency of <mathjax>#3+#</mathjax> and sulphur, being in group 6< has a valency of <mathjax>#2-#</mathjax>.</p>
<p>A molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero and so we would need 3 sulphur atoms and 2 iron atoms in order to form a neutral molecule of Iron(III)sulphide. (<mathjax>#6+ and 6-#</mathjax>).</p>
<p>Hence the formula is <mathjax>#Fe_2S_3#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Fe_2S_3#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Iron III I has a valency of <mathjax>#3+#</mathjax> and sulphur, being in group 6< has a valency of <mathjax>#2-#</mathjax>.</p>
<p>A molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero and so we would need 3 sulphur atoms and 2 iron atoms in order to form a neutral molecule of Iron(III)sulphide. (<mathjax>#6+ and 6-#</mathjax>).</p>
<p>Hence the formula is <mathjax>#Fe_2S_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What would be the formula between Iron III and Sulfur?</h1>
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<div class="markdown"><p><mathjax>#Fe_2S_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Iron III I has a valency of <mathjax>#3+#</mathjax> and sulphur, being in group 6< has a valency of <mathjax>#2-#</mathjax>.</p>
<p>A molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero and so we would need 3 sulphur atoms and 2 iron atoms in order to form a neutral molecule of Iron(III)sulphide. (<mathjax>#6+ and 6-#</mathjax>).</p>
<p>Hence the formula is <mathjax>#Fe_2S_3#</mathjax>.</p></div>
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</article> | What would be the formula between Iron III and Sulfur? | null |
3,132 | ab028cd4-6ddd-11ea-9f7f-ccda262736ce | https://socratic.org/questions/the-ph-of-a-solution-is-10-what-is-its-h-concentration | 1.00 × 10^(-10) M | start physical_unit 10 10 concentration mol/l qc_end physical_unit 4 4 6 6 ph qc_end end | [{"type":"physical unit","value":"Concentration [OF] [H+] [IN] M"}] | [{"type":"physical unit","value":"1.00 × 10^(-10) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{10}"}] | <h1 class="questionTitle" itemprop="name">The pH of a solution is 10. What is its #[H^+]# concentration?</h1> | null | 1.00 × 10^(-10) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H^+]#</mathjax>.</p>
<p>In water, under standard conditions, <mathjax>#pH#</mathjax> characterizes the autoprotolysis reaction.......</p>
<p><mathjax>#2H_2OrightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p><mathjax>#K_w=[HO^-][H_3O^+]=10^-14#</mathjax>, we take <mathjax>#log_10#</mathjax> of BOTH SIDES...</p>
<p><mathjax>#log_10K_w=log_10[HO^-]+log_10[H_3O^+]#</mathjax></p>
<p>And on rearrangement, </p>
<p><mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[HO^-])_(pOH)underbrace(-log_10[H_3O^+])_(pH)#</mathjax></p>
<p>And thus <mathjax>#14=pOH+pH#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#[H^+]"/"[H_3O^+]=10^(-10)*mol*L^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H^+]#</mathjax>.</p>
<p>In water, under standard conditions, <mathjax>#pH#</mathjax> characterizes the autoprotolysis reaction.......</p>
<p><mathjax>#2H_2OrightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p><mathjax>#K_w=[HO^-][H_3O^+]=10^-14#</mathjax>, we take <mathjax>#log_10#</mathjax> of BOTH SIDES...</p>
<p><mathjax>#log_10K_w=log_10[HO^-]+log_10[H_3O^+]#</mathjax></p>
<p>And on rearrangement, </p>
<p><mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[HO^-])_(pOH)underbrace(-log_10[H_3O^+])_(pH)#</mathjax></p>
<p>And thus <mathjax>#14=pOH+pH#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">The pH of a solution is 10. What is its #[H^+]# concentration?</h1>
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<div class="markdown"><p><mathjax>#[H^+]"/"[H_3O^+]=10^(-10)*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H^+]#</mathjax>.</p>
<p>In water, under standard conditions, <mathjax>#pH#</mathjax> characterizes the autoprotolysis reaction.......</p>
<p><mathjax>#2H_2OrightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p><mathjax>#K_w=[HO^-][H_3O^+]=10^-14#</mathjax>, we take <mathjax>#log_10#</mathjax> of BOTH SIDES...</p>
<p><mathjax>#log_10K_w=log_10[HO^-]+log_10[H_3O^+]#</mathjax></p>
<p>And on rearrangement, </p>
<p><mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[HO^-])_(pOH)underbrace(-log_10[H_3O^+])_(pH)#</mathjax></p>
<p>And thus <mathjax>#14=pOH+pH#</mathjax>.</p></div>
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</article> | The pH of a solution is 10. What is its #[H^+]# concentration? | null |
3,133 | ab693586-6ddd-11ea-bf5a-ccda262736ce | https://socratic.org/questions/5938df87b72cff2d818df41c | 1 mg | start physical_unit 6 7 mass mg qc_end physical_unit 6 7 3 5 number qc_end end | [{"type":"physical unit","value":"Mass [OF] lead atoms [IN] mg"}] | [{"type":"physical unit","value":"1 mg"}] | [{"type":"physical unit","value":"Number [OF] lead atoms [=] \\pu{2.50 × 10^18}"}] | <h1 class="questionTitle" itemprop="name">If there are #2.50xx10^18# lead atoms, what is the mass of this quantity?</h1> | null | 1 mg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I don't know the mass of a ream of lead, and I have never heard of lead being measured with this quantity.</p>
<p>But the number of atoms of any elemental quantity is <mathjax>#"Number of moles "xx" Molar mass"#</mathjax>, where a <mathjax>#"mole"#</mathjax> specifies <mathjax>#"Avogadro's Number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus if there were <mathjax>#2.50xx10^18#</mathjax> lead atoms........we have a mass of..........</p>
<p><mathjax>#(2.50xx10^18*cancel"lead atoms")/(6.022xx10^23*cancel"lead atoms"*cancel(mol^-1))xx207.2*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#~=1*mg#</mathjax></p></div>
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<div class="markdown"><p>Isn't a <mathjax>#"ream"#</mathjax> a quantity of paper, i.e. 480 or 500 individual sheets?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I don't know the mass of a ream of lead, and I have never heard of lead being measured with this quantity.</p>
<p>But the number of atoms of any elemental quantity is <mathjax>#"Number of moles "xx" Molar mass"#</mathjax>, where a <mathjax>#"mole"#</mathjax> specifies <mathjax>#"Avogadro's Number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus if there were <mathjax>#2.50xx10^18#</mathjax> lead atoms........we have a mass of..........</p>
<p><mathjax>#(2.50xx10^18*cancel"lead atoms")/(6.022xx10^23*cancel"lead atoms"*cancel(mol^-1))xx207.2*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#~=1*mg#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If there are #2.50xx10^18# lead atoms, what is the mass of this quantity?</h1>
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<div class="markdown"><p>Isn't a <mathjax>#"ream"#</mathjax> a quantity of paper, i.e. 480 or 500 individual sheets?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I don't know the mass of a ream of lead, and I have never heard of lead being measured with this quantity.</p>
<p>But the number of atoms of any elemental quantity is <mathjax>#"Number of moles "xx" Molar mass"#</mathjax>, where a <mathjax>#"mole"#</mathjax> specifies <mathjax>#"Avogadro's Number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus if there were <mathjax>#2.50xx10^18#</mathjax> lead atoms........we have a mass of..........</p>
<p><mathjax>#(2.50xx10^18*cancel"lead atoms")/(6.022xx10^23*cancel"lead atoms"*cancel(mol^-1))xx207.2*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#~=1*mg#</mathjax></p></div>
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<div class="markdown"><p>I'm not quite sure how much lead you have but this is how to calculate the number of atoms:</p>
<p><mathjax>#N=n*N_A#</mathjax></p>
<p>Where:</p>
<p><mathjax>#N="number of atoms"#</mathjax></p>
<p><mathjax>#N_A="Avogadro constant"=6.02*10^23" mol"^-1#</mathjax></p>
<p><mathjax>#n="mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"molar mass of lead"=207.2" g/mol"#</mathjax></p>
<p><mathjax>#N_A#</mathjax> is a constant and is the number of atoms in one mole of any substance. So if you have <mathjax>#n=2" mol"#</mathjax>, you have <mathjax>#2*N_A#</mathjax> atoms.</p>
<p>As an example, if you have <mathjax>#2.50*10^18#</mathjax> grams of lead, you have this many atoms:</p>
<p><mathjax>#n=(2.50*10^18)/207.2=1.21*10^16" mol"#</mathjax></p>
<p><mathjax>#N=n*N_A=1.21*10^16*6.02*10^23=7.28*10^39" atoms"#</mathjax></p>
<p>This is a lot of lead!</p></div>
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</article> | If there are #2.50xx10^18# lead atoms, what is the mass of this quantity? | null |
3,134 | a9658c18-6ddd-11ea-acac-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-aluminum-replacing-hydrogen-when-the-metal-is- | 2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq) | start chemical_equation qc_end substance 6 6 qc_end substance 8 8 qc_end substance 11 11 qc_end substance 15 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the replacing"}] | [{"type":"chemical equation","value":"2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq)"}] | [{"type":"substance name","value":"aluminum"},{"type":"substance name","value":"hydrogen"},{"type":"substance name","value":"metal"},{"type":"substance name","value":"hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for aluminum replacing hydrogen when the metal is placed in hydrochloric acid?</h1> | null | 2 Al(s) + 6 HCl(aq) -> 3 H2(g) + 2 AlCl3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement (single displacement) reaction in which the aluminum replaces the hydrogen in hydrochloric acid, which produces aluminum chloride and hydrogen gas.</p>
<p>The general form for a single replacement reaction is:</p>
<p><mathjax>#color(red)("A") + color(blue)("B")color(teal)("C")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("B") + color(red)("A")color(teal)("C")#</mathjax></p>
<p>Not all possible <a href="https://socratic.org/chemistry/chemical-reactions/single-replacement-reactions">single replacement reactions</a> will occur. The element that replaces the element in the compound must be more reactive than the one it is replacing. We use an activity series to determine whether a single replacement reaction will occur.</p>
<p>You can see in the activity series below, aluminum is above hydrogen, so it is more reactive and will replace it in the reaction.</p>
<p><img alt="http://yeahchemistry.com/downloads/activity-series" src="https://useruploads.socratic.org/FcDMDdlRKaHpcJ7AtfJG_activityseries_1.png"/> </p></div>
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<div class="markdown"><p><mathjax>#"2Al(s)" + "6HCl(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"3H"_2("g")"+2AlCl"_3("aq")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement (single displacement) reaction in which the aluminum replaces the hydrogen in hydrochloric acid, which produces aluminum chloride and hydrogen gas.</p>
<p>The general form for a single replacement reaction is:</p>
<p><mathjax>#color(red)("A") + color(blue)("B")color(teal)("C")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("B") + color(red)("A")color(teal)("C")#</mathjax></p>
<p>Not all possible <a href="https://socratic.org/chemistry/chemical-reactions/single-replacement-reactions">single replacement reactions</a> will occur. The element that replaces the element in the compound must be more reactive than the one it is replacing. We use an activity series to determine whether a single replacement reaction will occur.</p>
<p>You can see in the activity series below, aluminum is above hydrogen, so it is more reactive and will replace it in the reaction.</p>
<p><img alt="http://yeahchemistry.com/downloads/activity-series" src="https://useruploads.socratic.org/FcDMDdlRKaHpcJ7AtfJG_activityseries_1.png"/> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for aluminum replacing hydrogen when the metal is placed in hydrochloric acid?</h1>
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<div class="markdown"><p><mathjax>#"2Al(s)" + "6HCl(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"3H"_2("g")"+2AlCl"_3("aq")#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a single replacement (single displacement) reaction in which the aluminum replaces the hydrogen in hydrochloric acid, which produces aluminum chloride and hydrogen gas.</p>
<p>The general form for a single replacement reaction is:</p>
<p><mathjax>#color(red)("A") + color(blue)("B")color(teal)("C")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("B") + color(red)("A")color(teal)("C")#</mathjax></p>
<p>Not all possible <a href="https://socratic.org/chemistry/chemical-reactions/single-replacement-reactions">single replacement reactions</a> will occur. The element that replaces the element in the compound must be more reactive than the one it is replacing. We use an activity series to determine whether a single replacement reaction will occur.</p>
<p>You can see in the activity series below, aluminum is above hydrogen, so it is more reactive and will replace it in the reaction.</p>
<p><img alt="http://yeahchemistry.com/downloads/activity-series" src="https://useruploads.socratic.org/FcDMDdlRKaHpcJ7AtfJG_activityseries_1.png"/> </p></div>
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</article> | What is the chemical equation for aluminum replacing hydrogen when the metal is placed in hydrochloric acid? | null |
3,135 | aae73c40-6ddd-11ea-a7dd-ccda262736ce | https://socratic.org/questions/what-volume-in-ml-of-200-m-hydrochloric-acid-is-needed-to-react-with-1-62-grams- | 123.84 mL | start physical_unit 7 8 volume ml qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 17 17 14 15 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrochloric acid [IN] mL"}] | [{"type":"physical unit","value":"123.84 mL"}] | [{"type":"physical unit","value":"Molarity [OF] hydrochloric acid [=] \\pu{0.200 M}"},{"type":"physical unit","value":"Mass [OF] zinc [=] \\pu{1.62 grams}"}] | <h1 class="questionTitle" itemprop="name">What volume in mL of .200 M hydrochloric acid is needed to react with 1.62 grams of zinc?</h1> | null | 123.84 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Zinc will react with 2 equiv hydrochloric acid, according to the given stoichiometric equation. How many moles of metal do you have? Can you convert twice that molar quantity to a given volume of hydrochloric acid?</p></div>
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<div class="markdown"><p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + 2H_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Zinc will react with 2 equiv hydrochloric acid, according to the given stoichiometric equation. How many moles of metal do you have? Can you convert twice that molar quantity to a given volume of hydrochloric acid?</p></div>
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<h1 class="questionTitle" itemprop="name">What volume in mL of .200 M hydrochloric acid is needed to react with 1.62 grams of zinc?</h1>
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<div class="markdown"><p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + 2H_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Zinc will react with 2 equiv hydrochloric acid, according to the given stoichiometric equation. How many moles of metal do you have? Can you convert twice that molar quantity to a given volume of hydrochloric acid?</p></div>
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</article> | What volume in mL of .200 M hydrochloric acid is needed to react with 1.62 grams of zinc? | null |
3,136 | a8e3f69a-6ddd-11ea-a3b9-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-manganese-ill-fluoride | MnF3 | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] manganese (Ill) fluoride [IN] default"}] | [{"type":"chemical equation","value":"MnF3"}] | [{"type":"substance name","value":"Manganese (Ill) fluoride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for manganese (Ill) fluoride?</h1> | null | MnF3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Manganese%28III%29_fluoride" rel="nofollow">site.</a></p></div>
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<div class="markdown"><p><mathjax>#MnF_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Manganese%28III%29_fluoride" rel="nofollow">site.</a></p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for manganese (Ill) fluoride?</h1>
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<div class="markdown"><p><mathjax>#MnF_3#</mathjax></p></div>
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<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Manganese%28III%29_fluoride" rel="nofollow">site.</a></p></div>
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</article> | What is the formula for manganese (Ill) fluoride? | null |
3,137 | ac2f2e8a-6ddd-11ea-97b5-ccda262736ce | https://socratic.org/questions/what-is-the-total-number-of-grams-of-nai-needed-to-make-1-0-liter-of-a-0-010-m-s | 1.50 grams | start physical_unit 8 8 mass g qc_end physical_unit 18 18 12 13 volume qc_end physical_unit 8 8 16 17 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] NaI [IN] grams"}] | [{"type":"physical unit","value":"1.50 grams"}] | [{"type":"physical unit","value":"Volume [OF] NaI solution [=] \\pu{1.0 liter}"},{"type":"physical unit","value":"Molarity [OF] NaI solution [=] \\pu{0.010 M}"}] | <h1 class="questionTitle" itemprop="name">What is the total number of grams of #NaI# needed to make 1.0 liter of a 0.010 M solution?</h1> | null | 1.50 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:<br/>
<img alt="www.chemistrytutorials.org" src="https://useruploads.socratic.org/J53EfLgVTdGDrxn6hzDX_molarity.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution, both of which have the appropriate units.</p>
<p>Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = <mathjax>#liters of solution xxMolarity#</mathjax></p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (1.0 L)<mathjax>#xx#</mathjax>(0.010 M) = 0.010 moles of NaI</p>
<p>Now we have to convert the 0.010 moles of NaI into grams of NaI. This can be done by multiplying 0.010 moles by the molecular weight of NaI, which is <mathjax>#149.89 g/(mol)#</mathjax>. </p>
<p><mathjax>#0.010 #</mathjax>moles<mathjax># #</mathjax><mathjax>#xx#</mathjax><mathjax>#149.89 g/(mol)#</mathjax> = 1.50g NaI</p></div>
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<div class="markdown"><p>1.50g of NaI</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:<br/>
<img alt="www.chemistrytutorials.org" src="https://useruploads.socratic.org/J53EfLgVTdGDrxn6hzDX_molarity.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution, both of which have the appropriate units.</p>
<p>Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = <mathjax>#liters of solution xxMolarity#</mathjax></p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (1.0 L)<mathjax>#xx#</mathjax>(0.010 M) = 0.010 moles of NaI</p>
<p>Now we have to convert the 0.010 moles of NaI into grams of NaI. This can be done by multiplying 0.010 moles by the molecular weight of NaI, which is <mathjax>#149.89 g/(mol)#</mathjax>. </p>
<p><mathjax>#0.010 #</mathjax>moles<mathjax># #</mathjax><mathjax>#xx#</mathjax><mathjax>#149.89 g/(mol)#</mathjax> = 1.50g NaI</p></div>
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<h1 class="questionTitle" itemprop="name">What is the total number of grams of #NaI# needed to make 1.0 liter of a 0.010 M solution?</h1>
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<div class="markdown"><p>1.50g of NaI</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:<br/>
<img alt="www.chemistrytutorials.org" src="https://useruploads.socratic.org/J53EfLgVTdGDrxn6hzDX_molarity.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution, both of which have the appropriate units.</p>
<p>Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = <mathjax>#liters of solution xxMolarity#</mathjax></p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (1.0 L)<mathjax>#xx#</mathjax>(0.010 M) = 0.010 moles of NaI</p>
<p>Now we have to convert the 0.010 moles of NaI into grams of NaI. This can be done by multiplying 0.010 moles by the molecular weight of NaI, which is <mathjax>#149.89 g/(mol)#</mathjax>. </p>
<p><mathjax>#0.010 #</mathjax>moles<mathjax># #</mathjax><mathjax>#xx#</mathjax><mathjax>#149.89 g/(mol)#</mathjax> = 1.50g NaI</p></div>
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</article> | What is the total number of grams of #NaI# needed to make 1.0 liter of a 0.010 M solution? | null |
3,138 | a861ba68-6ddd-11ea-8788-ccda262736ce | https://socratic.org/questions/a-solution-of-naoh-has-a-concentration-of-5-51-10-3-m-what-is-the-ph-of-this-sol | 11.74 | start physical_unit 17 18 ph none qc_end physical_unit 3 3 8 11 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] this solution"}] | [{"type":"physical unit","value":"11.74"}] | [{"type":"physical unit","value":"Concentration [OF] NaOH solution [=] \\pu{5.51 × 10^(−3) M}"}] | <h1 class="questionTitle" itemprop="name">A solution of #NaOH# has a concentration of #5.51*10^-3# #M#. What is the pH of this solution?</h1> | null | 11.74 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that <mathjax>#pH+pOH=14#</mathjax> for water at <mathjax>#298K#</mathjax>.</p>
<p><mathjax>#pOH=-log_(10)5.51xx10^(-3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.26#</mathjax></p>
<p><mathjax>#pH=14-2.26=??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=11.74#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that <mathjax>#pH+pOH=14#</mathjax> for water at <mathjax>#298K#</mathjax>.</p>
<p><mathjax>#pOH=-log_(10)5.51xx10^(-3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.26#</mathjax></p>
<p><mathjax>#pH=14-2.26=??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A solution of #NaOH# has a concentration of #5.51*10^-3# #M#. What is the pH of this solution?</h1>
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<div class="markdown"><p><mathjax>#pH=11.74#</mathjax></p></div>
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<div class="markdown"><p>We know that <mathjax>#pH+pOH=14#</mathjax> for water at <mathjax>#298K#</mathjax>.</p>
<p><mathjax>#pOH=-log_(10)5.51xx10^(-3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.26#</mathjax></p>
<p><mathjax>#pH=14-2.26=??#</mathjax></p></div>
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</article> | A solution of #NaOH# has a concentration of #5.51*10^-3# #M#. What is the pH of this solution? | null |
3,139 | aacfbe48-6ddd-11ea-ba92-ccda262736ce | https://socratic.org/questions/the-reaction-of-methane-and-water-is-one-way-to-prepare-hydrogen-ch-4-g-h-2o-g-c | 376 g | start physical_unit 19 19 maximum_possible_yield g qc_end chemical_equation 12 19 qc_end physical_unit 12 12 30 31 mass qc_end physical_unit 5 5 35 36 mass qc_end end | [{"type":"physical unit","value":"maximum possible yield [OF] H2 [IN] g"}] | [{"type":"physical unit","value":"376 g"}] | [{"type":"chemical equation","value":"CH4(g) + H2O(g) -> CO(g) + 3 H2(g)"},{"type":"physical unit","value":"Molar mass [OF] CH4 [=] \\pu{16.04 g/mol}"},{"type":"physical unit","value":"Molar mass [OF] H2O [=] \\pu{18.02 g/mol}"},{"type":"physical unit","value":"Molar mass [OF] CO [=] \\pu{28.01 g/mol}"},{"type":"physical unit","value":"Molar mass [OF] H2 [=] \\pu{2.02 g/mol}"},{"type":"physical unit","value":"Mass [OF] CH4 [=] \\pu{995 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{2510 g}"}] | <h1 class="questionTitle" itemprop="name">The reaction of methane and water is one way to prepare hydrogen: #CH_4(g) + H_2O (g) -> CO (g) + 3H_2(g)# [Molar masses 16.04 18.02 28.01 2.02]. If you begin with 995 g of #CH_4# and 2510 g of water, what is the maximum possible yield of #H_2#?</h1> | null | 376 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"CH"_4("g") + "H"_2"O(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO(g)" + "3H"_2("g")"#</mathjax></p>
<p>This is a limiting reactant <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem. We will determine whether methane or water is the limiting reactant. The limiting reactant will give us the maximum possible yield of hydrogen gas.</p>
<p><mathjax>#color(blue)("Maximum Yield of")#</mathjax> <mathjax>#color(blue)("H"_2"#</mathjax> <mathjax>#color(blue)("Produced by"#</mathjax> <mathjax>#color(blue)("995 g CH"_4#</mathjax></p>
<p>Determine the moles <mathjax>#"CH"_4#</mathjax> in <mathjax>#"995 g CH"_4#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#995color(red)cancel(color(black)("g CH"_4))xx(1"mol CH"_4)/(16.04color(red)cancel(color(black)("g CH"_4)))="62.0 mol CH"_4#</mathjax></p>
<p>Multiply the moles <mathjax>#"CH"_4#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"CH"_4#</mathjax>, <mathjax>#"3 mol H"_2:#</mathjax><mathjax>#"1 mol CH"_4#</mathjax>.</p>
<p><mathjax>#62.0color(red)cancel(color(black)("mol CH"_4))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol CH"_4)))="186 mol H"_2"#</mathjax></p>
<p><mathjax>#color(teal)"Maximum Yield of"#</mathjax> <mathjax>#color(teal)("H"_2)#</mathjax> <mathjax>#color(teal)("produced by"#</mathjax> <mathjax>#color(teal)(2510"g H"_2"O"#</mathjax></p>
<p>Determine the moles <mathjax>#"H"_2"O"#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#2510color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.02color(red)cancel(color(black)("g H"_2"O")))="139 mol H"_2"O"#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_2"O"#</mathjax> by the mole ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>, <mathjax>#"3 mole H"_2:#</mathjax><mathjax>#1"mol H"_2"O"#</mathjax>.</p>
<p><mathjax>#139color(red)cancel(color(black)("mol H"_2"O"))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol H"_2"O")))="417 mol H"_2"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"CH"_4#</mathjax> since it produced the least number of moles of <mathjax>#"H"_2#</mathjax>. In order to determine the mass <mathjax>#"H"_2"#</mathjax> that is possible, multiply the moles <mathjax>#"H"_2#</mathjax> by the molar mass of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(magenta)("Maximum Possible Yield of H"_2"#</mathjax></p>
<p><mathjax>#186color(red)cancel(color(black)("mol H"_2))xx(2.02"g H"_2)/(1color(red)cancel(color(black)("mol H"_2)))="376 g H"_2"#</mathjax> (rounded to three sig figs)</p></div>
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<div class="markdown"><p>Methane is the limiting reactant, therefore the maximum possible yield of hydrogen gas is 376 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"CH"_4("g") + "H"_2"O(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO(g)" + "3H"_2("g")"#</mathjax></p>
<p>This is a limiting reactant <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem. We will determine whether methane or water is the limiting reactant. The limiting reactant will give us the maximum possible yield of hydrogen gas.</p>
<p><mathjax>#color(blue)("Maximum Yield of")#</mathjax> <mathjax>#color(blue)("H"_2"#</mathjax> <mathjax>#color(blue)("Produced by"#</mathjax> <mathjax>#color(blue)("995 g CH"_4#</mathjax></p>
<p>Determine the moles <mathjax>#"CH"_4#</mathjax> in <mathjax>#"995 g CH"_4#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#995color(red)cancel(color(black)("g CH"_4))xx(1"mol CH"_4)/(16.04color(red)cancel(color(black)("g CH"_4)))="62.0 mol CH"_4#</mathjax></p>
<p>Multiply the moles <mathjax>#"CH"_4#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"CH"_4#</mathjax>, <mathjax>#"3 mol H"_2:#</mathjax><mathjax>#"1 mol CH"_4#</mathjax>.</p>
<p><mathjax>#62.0color(red)cancel(color(black)("mol CH"_4))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol CH"_4)))="186 mol H"_2"#</mathjax></p>
<p><mathjax>#color(teal)"Maximum Yield of"#</mathjax> <mathjax>#color(teal)("H"_2)#</mathjax> <mathjax>#color(teal)("produced by"#</mathjax> <mathjax>#color(teal)(2510"g H"_2"O"#</mathjax></p>
<p>Determine the moles <mathjax>#"H"_2"O"#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#2510color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.02color(red)cancel(color(black)("g H"_2"O")))="139 mol H"_2"O"#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_2"O"#</mathjax> by the mole ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>, <mathjax>#"3 mole H"_2:#</mathjax><mathjax>#1"mol H"_2"O"#</mathjax>.</p>
<p><mathjax>#139color(red)cancel(color(black)("mol H"_2"O"))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol H"_2"O")))="417 mol H"_2"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"CH"_4#</mathjax> since it produced the least number of moles of <mathjax>#"H"_2#</mathjax>. In order to determine the mass <mathjax>#"H"_2"#</mathjax> that is possible, multiply the moles <mathjax>#"H"_2#</mathjax> by the molar mass of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(magenta)("Maximum Possible Yield of H"_2"#</mathjax></p>
<p><mathjax>#186color(red)cancel(color(black)("mol H"_2))xx(2.02"g H"_2)/(1color(red)cancel(color(black)("mol H"_2)))="376 g H"_2"#</mathjax> (rounded to three sig figs)</p></div>
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<h1 class="questionTitle" itemprop="name">The reaction of methane and water is one way to prepare hydrogen: #CH_4(g) + H_2O (g) -> CO (g) + 3H_2(g)# [Molar masses 16.04 18.02 28.01 2.02]. If you begin with 995 g of #CH_4# and 2510 g of water, what is the maximum possible yield of #H_2#?</h1>
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<div class="markdown"><p>Methane is the limiting reactant, therefore the maximum possible yield of hydrogen gas is 376 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"CH"_4("g") + "H"_2"O(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO(g)" + "3H"_2("g")"#</mathjax></p>
<p>This is a limiting reactant <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem. We will determine whether methane or water is the limiting reactant. The limiting reactant will give us the maximum possible yield of hydrogen gas.</p>
<p><mathjax>#color(blue)("Maximum Yield of")#</mathjax> <mathjax>#color(blue)("H"_2"#</mathjax> <mathjax>#color(blue)("Produced by"#</mathjax> <mathjax>#color(blue)("995 g CH"_4#</mathjax></p>
<p>Determine the moles <mathjax>#"CH"_4#</mathjax> in <mathjax>#"995 g CH"_4#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#995color(red)cancel(color(black)("g CH"_4))xx(1"mol CH"_4)/(16.04color(red)cancel(color(black)("g CH"_4)))="62.0 mol CH"_4#</mathjax></p>
<p>Multiply the moles <mathjax>#"CH"_4#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"CH"_4#</mathjax>, <mathjax>#"3 mol H"_2:#</mathjax><mathjax>#"1 mol CH"_4#</mathjax>.</p>
<p><mathjax>#62.0color(red)cancel(color(black)("mol CH"_4))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol CH"_4)))="186 mol H"_2"#</mathjax></p>
<p><mathjax>#color(teal)"Maximum Yield of"#</mathjax> <mathjax>#color(teal)("H"_2)#</mathjax> <mathjax>#color(teal)("produced by"#</mathjax> <mathjax>#color(teal)(2510"g H"_2"O"#</mathjax></p>
<p>Determine the moles <mathjax>#"H"_2"O"#</mathjax> by dividing the given mass by its molar mass by multiplying by the inverse of the molar mass.</p>
<p><mathjax>#2510color(red)cancel(color(black)("g H"_2"O"))xx(1"mol H"_2"O")/(18.02color(red)cancel(color(black)("g H"_2"O")))="139 mol H"_2"O"#</mathjax></p>
<p>Multiply the moles <mathjax>#"H"_2"O"#</mathjax> by the mole ratio between <mathjax>#"H"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>, <mathjax>#"3 mole H"_2:#</mathjax><mathjax>#1"mol H"_2"O"#</mathjax>.</p>
<p><mathjax>#139color(red)cancel(color(black)("mol H"_2"O"))xx(3"mol H"_2)/(1color(red)cancel(color(black)("mol H"_2"O")))="417 mol H"_2"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"CH"_4#</mathjax> since it produced the least number of moles of <mathjax>#"H"_2#</mathjax>. In order to determine the mass <mathjax>#"H"_2"#</mathjax> that is possible, multiply the moles <mathjax>#"H"_2#</mathjax> by the molar mass of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(magenta)("Maximum Possible Yield of H"_2"#</mathjax></p>
<p><mathjax>#186color(red)cancel(color(black)("mol H"_2))xx(2.02"g H"_2)/(1color(red)cancel(color(black)("mol H"_2)))="376 g H"_2"#</mathjax> (rounded to three sig figs)</p></div>
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</article> | The reaction of methane and water is one way to prepare hydrogen: #CH_4(g) + H_2O (g) -> CO (g) + 3H_2(g)# [Molar masses 16.04 18.02 28.01 2.02]. If you begin with 995 g of #CH_4# and 2510 g of water, what is the maximum possible yield of #H_2#? | null |
3,140 | ac7bfcc2-6ddd-11ea-94ce-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-in-which-58-g-of-nacl-are-dissolved-in-1-0-l- | 0.99 M | start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 6 6 16 17 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [IN] M"}] | [{"type":"physical unit","value":"0.99 M"}] | [{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{58 g}"},{"type":"physical unit","value":"Volume [OF] NaCl solution [=] \\pu{1.0 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution in which 58 g of #NaCl# are dissolved in 1.0 L of solution?</h1> | null | 0.99 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium chloride has an equivalent mass of <mathjax>#58.44*g*mol^-1#</mathjax>.</p>
<p>Since <mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>, the concentration of sodium chloride <mathjax>#=#</mathjax> <mathjax>#(58*g)/(58.44*g*mol^-1)xx1/(1.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p>A tad under <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium chloride has an equivalent mass of <mathjax>#58.44*g*mol^-1#</mathjax>.</p>
<p>Since <mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>, the concentration of sodium chloride <mathjax>#=#</mathjax> <mathjax>#(58*g)/(58.44*g*mol^-1)xx1/(1.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution in which 58 g of #NaCl# are dissolved in 1.0 L of solution?</h1>
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<div class="markdown"><p>A tad under <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Sodium chloride has an equivalent mass of <mathjax>#58.44*g*mol^-1#</mathjax>.</p>
<p>Since <mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>, the concentration of sodium chloride <mathjax>#=#</mathjax> <mathjax>#(58*g)/(58.44*g*mol^-1)xx1/(1.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1#</mathjax></p></div>
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</article> | What is the molarity of a solution in which 58 g of #NaCl# are dissolved in 1.0 L of solution? | null |
3,141 | ace5db5c-6ddd-11ea-9617-ccda262736ce | https://socratic.org/questions/how-many-moles-of-argon-are-in-2-2x1020-molecules-of-argon | 3.65 × 10^(-4) moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] argon [IN] moles"}] | [{"type":"physical unit","value":"3.65 × 10^(-4) moles"}] | [{"type":"physical unit","value":"Number [OF] argon molecules [=] \\pu{2.2 × 10^20}"}] | <h1 class="questionTitle" itemprop="name">How many moles of argon are in #2.2xx10^20# molecules of argon?</h1> | null | 3.65 × 10^(-4) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The relationship here is very simple to understand, I hope this helps! </p>
<p>A very useful constant known as Avogadro's number was created to demonstrate the relationship between molecules and moles. </p>
<p>It was found that for every 1 mole of a substance, <mathjax>#6.02*10^23#</mathjax> molecules were in that substance. In other words, There is <mathjax>#6.02*10^23#</mathjax> molecules/mole. </p>
<p>Therefore, </p>
<p>Looking at our amount of Argon, we see that we have <mathjax>#2.2*10^20#</mathjax> molecules. We can determine our amount of moles by dividing by Avogadro's number: </p>
<p><mathjax>#(2.2*10^20 "molecules") /(6.02*10^23 "molecules/mole")#</mathjax> </p>
<p>which cancels out molecules and leaves:</p>
<p><mathjax>#"0.00037 moles"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>0.00037</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The relationship here is very simple to understand, I hope this helps! </p>
<p>A very useful constant known as Avogadro's number was created to demonstrate the relationship between molecules and moles. </p>
<p>It was found that for every 1 mole of a substance, <mathjax>#6.02*10^23#</mathjax> molecules were in that substance. In other words, There is <mathjax>#6.02*10^23#</mathjax> molecules/mole. </p>
<p>Therefore, </p>
<p>Looking at our amount of Argon, we see that we have <mathjax>#2.2*10^20#</mathjax> molecules. We can determine our amount of moles by dividing by Avogadro's number: </p>
<p><mathjax>#(2.2*10^20 "molecules") /(6.02*10^23 "molecules/mole")#</mathjax> </p>
<p>which cancels out molecules and leaves:</p>
<p><mathjax>#"0.00037 moles"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of argon are in #2.2xx10^20# molecules of argon?</h1>
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Stefan V.
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<div class="markdown"><p>0.00037</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The relationship here is very simple to understand, I hope this helps! </p>
<p>A very useful constant known as Avogadro's number was created to demonstrate the relationship between molecules and moles. </p>
<p>It was found that for every 1 mole of a substance, <mathjax>#6.02*10^23#</mathjax> molecules were in that substance. In other words, There is <mathjax>#6.02*10^23#</mathjax> molecules/mole. </p>
<p>Therefore, </p>
<p>Looking at our amount of Argon, we see that we have <mathjax>#2.2*10^20#</mathjax> molecules. We can determine our amount of moles by dividing by Avogadro's number: </p>
<p><mathjax>#(2.2*10^20 "molecules") /(6.02*10^23 "molecules/mole")#</mathjax> </p>
<p>which cancels out molecules and leaves:</p>
<p><mathjax>#"0.00037 moles"#</mathjax></p></div>
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</article> | How many moles of argon are in #2.2xx10^20# molecules of argon? | null |
3,142 | ac038c14-6ddd-11ea-953c-ccda262736ce | https://socratic.org/questions/if-500-000g-of-limestone-is-heated-up-what-is-the-mass-of-cao-caco3-cao-co2 | 280.1 kg | start physical_unit 13 13 mass kg qc_end physical_unit 4 4 1 2 mass qc_end chemical_equation 14 18 qc_end end | [{"type":"physical unit","value":"Mass [OF] CaO [IN] kg"}] | [{"type":"physical unit","value":"280.1 kg"}] | [{"type":"physical unit","value":"Mass [OF] limestone [=] \\pu{500,000 g}"},{"type":"chemical equation","value":"CaCO3 -> CaO + CO2"}] | <h1 class="questionTitle" itemprop="name">If 500,000g of limestone is heated up, what is the mass of CaO?
CaCO3 = CaO + CO2</h1> | null | 280.1 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)#</mathjax></p>
<p>And, clearly, there is 1:1 molar relationship between the limestone and calcium oxide product........</p>
<p><mathjax>#"Moles of calcium carbonate"=(500xx10^3*g)/(100.09*g*mol^-1)=5.00xx10^2*mol#</mathjax></p>
<p><mathjax>#"Mass of calcium oxide"=5.00xx10^3*molxx56.08*g*mol^-1#</mathjax></p>
<p><mathjax>#=280.1xx10^3*g-=280#</mathjax> <mathjax>#"kg"#</mathjax></p>
<p>For what would we typically use this quantity of <mathjax>#"quicklime"#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Well we have the stoichiometric equation........and we get over <mathjax>#280#</mathjax> <mathjax>#"kg"#</mathjax> of calcium oxide..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)#</mathjax></p>
<p>And, clearly, there is 1:1 molar relationship between the limestone and calcium oxide product........</p>
<p><mathjax>#"Moles of calcium carbonate"=(500xx10^3*g)/(100.09*g*mol^-1)=5.00xx10^2*mol#</mathjax></p>
<p><mathjax>#"Mass of calcium oxide"=5.00xx10^3*molxx56.08*g*mol^-1#</mathjax></p>
<p><mathjax>#=280.1xx10^3*g-=280#</mathjax> <mathjax>#"kg"#</mathjax></p>
<p>For what would we typically use this quantity of <mathjax>#"quicklime"#</mathjax>?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If 500,000g of limestone is heated up, what is the mass of CaO?
CaCO3 = CaO + CO2</h1>
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<div class="markdown"><p>Well we have the stoichiometric equation........and we get over <mathjax>#280#</mathjax> <mathjax>#"kg"#</mathjax> of calcium oxide..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)#</mathjax></p>
<p>And, clearly, there is 1:1 molar relationship between the limestone and calcium oxide product........</p>
<p><mathjax>#"Moles of calcium carbonate"=(500xx10^3*g)/(100.09*g*mol^-1)=5.00xx10^2*mol#</mathjax></p>
<p><mathjax>#"Mass of calcium oxide"=5.00xx10^3*molxx56.08*g*mol^-1#</mathjax></p>
<p><mathjax>#=280.1xx10^3*g-=280#</mathjax> <mathjax>#"kg"#</mathjax></p>
<p>For what would we typically use this quantity of <mathjax>#"quicklime"#</mathjax>?</p></div>
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</article> | If 500,000g of limestone is heated up, what is the mass of CaO?
CaCO3 = CaO + CO2 | null |
3,143 | ac703edb-6ddd-11ea-b6e3-ccda262736ce | https://socratic.org/questions/saccharomyces-cerevisiae-produces-ethanol-by-fermentation-the-theoretical-yield- | 1.28 g | start physical_unit 3 3 theoretical_yield g qc_end physical_unit 15 15 12 13 mass qc_end end | [{"type":"physical unit","value":"Theoretical yield [OF] ethanol [IN] g"}] | [{"type":"physical unit","value":"1.28 g"}] | [{"type":"physical unit","value":"Mass [OF] glucose [=] \\pu{2.5 g}"}] | <h1 class="questionTitle" itemprop="name">Saccharomyces cerevisiae produces ethanol by fermentation. The theoretical yield of ethanol from 2.5 g of glucose is ______________ g ?</h1> | null | 1.28 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by finding the balanced chemical reaction:</p>
<p><mathjax>#C_6H_12O_6 → 2 C_2H_5OH + 2 CO_2#</mathjax></p>
<p>Remember that the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> is the actual yield divided by the theoretical yield multiplied by 100.</p>
<p>See if you can follow the graphic below, with this information you have provided, this is all I can do:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/ulvEDUc0TVKh54P2Hg1l_Percent%20Yield%20Glucose.png"/> </p></div>
</div>
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<div class="markdown"><p>50.3% yield</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by finding the balanced chemical reaction:</p>
<p><mathjax>#C_6H_12O_6 → 2 C_2H_5OH + 2 CO_2#</mathjax></p>
<p>Remember that the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> is the actual yield divided by the theoretical yield multiplied by 100.</p>
<p>See if you can follow the graphic below, with this information you have provided, this is all I can do:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/ulvEDUc0TVKh54P2Hg1l_Percent%20Yield%20Glucose.png"/> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Saccharomyces cerevisiae produces ethanol by fermentation. The theoretical yield of ethanol from 2.5 g of glucose is ______________ g ?</h1>
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Dec 5, 2016
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<div class="markdown"><p>50.3% yield</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by finding the balanced chemical reaction:</p>
<p><mathjax>#C_6H_12O_6 → 2 C_2H_5OH + 2 CO_2#</mathjax></p>
<p>Remember that the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> is the actual yield divided by the theoretical yield multiplied by 100.</p>
<p>See if you can follow the graphic below, with this information you have provided, this is all I can do:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/ulvEDUc0TVKh54P2Hg1l_Percent%20Yield%20Glucose.png"/> </p></div>
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</article> | Saccharomyces cerevisiae produces ethanol by fermentation. The theoretical yield of ethanol from 2.5 g of glucose is ______________ g ? | null |
3,144 | a94f579c-6ddd-11ea-89f4-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-in-which-6-9mol-of-potassium-chloride-is-diss | 3.83 mol/L | start physical_unit 12 13 molarity mol/l qc_end physical_unit 12 13 9 10 mole qc_end physical_unit 6 6 23 24 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] potassium chloride solution [IN] mol/L"}] | [{"type":"physical unit","value":"3.83 mol/L"}] | [{"type":"physical unit","value":"Mole [OF] potassium chloride [=] \\pu{6.9 mol}"},{"type":"physical unit","value":"Volume [OF] potassium chloride solution [=] \\pu{1.8 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution in which 6.9mol of potassium chloride is dissolved in water to a final volume of 1.8L? </h1> | null | 3.83 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> is a measure of a solution's concentration in terms of how many <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter of solution</strong>. </p>
<p>The unit of volume selected for molarity is <em>one liter</em>, <mathjax>#"1 L"#</mathjax>. This means that your ultimate goal when calculating a solution's molarity is to figure out how many moles of solute are present in <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>In your case, the solution is said to contain <mathjax>#6.9#</mathjax> <strong>moles</strong> of potassium chloride, <mathjax>#"KCl"#</mathjax>, your solute, in <mathjax>#"1.8 L"#</mathjax> of solution. </p>
<p>Now, a solution is characterized by the fact that the solute and the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, are <strong>evenly mixed</strong>. This implies that <strong>any sample</strong> of this solution will have <strong>the same concentration</strong>, since scaling down the volume will also scale down the number of moles of solute it contains. </p>
<p>You can thus use the volume of the solution to figure out how many moles of solute you'd get in <mathjax>#"1 L"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#1.0 color(red)(cancel(color(black)("L solution"))) * "6.9 moles KCl"/(1.8 color(red)(cancel(color(black)("L solution")))) = "3.833 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 L"#</mathjax> of solution contains <mathjax>#3.833#</mathjax> <strong>moles</strong> of solute, it follows that the <em>molarity</em> of the solution will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("3.8 mol L"^(-1))color(white)(a/a)|))) ->#</mathjax> <em>rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/SXf9rDnVFao?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.8 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> is a measure of a solution's concentration in terms of how many <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter of solution</strong>. </p>
<p>The unit of volume selected for molarity is <em>one liter</em>, <mathjax>#"1 L"#</mathjax>. This means that your ultimate goal when calculating a solution's molarity is to figure out how many moles of solute are present in <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>In your case, the solution is said to contain <mathjax>#6.9#</mathjax> <strong>moles</strong> of potassium chloride, <mathjax>#"KCl"#</mathjax>, your solute, in <mathjax>#"1.8 L"#</mathjax> of solution. </p>
<p>Now, a solution is characterized by the fact that the solute and the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, are <strong>evenly mixed</strong>. This implies that <strong>any sample</strong> of this solution will have <strong>the same concentration</strong>, since scaling down the volume will also scale down the number of moles of solute it contains. </p>
<p>You can thus use the volume of the solution to figure out how many moles of solute you'd get in <mathjax>#"1 L"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#1.0 color(red)(cancel(color(black)("L solution"))) * "6.9 moles KCl"/(1.8 color(red)(cancel(color(black)("L solution")))) = "3.833 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 L"#</mathjax> of solution contains <mathjax>#3.833#</mathjax> <strong>moles</strong> of solute, it follows that the <em>molarity</em> of the solution will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("3.8 mol L"^(-1))color(white)(a/a)|))) ->#</mathjax> <em>rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/SXf9rDnVFao?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution in which 6.9mol of potassium chloride is dissolved in water to a final volume of 1.8L? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-01T11:59:30" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"3.8 mol L"^(-1)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> is a measure of a solution's concentration in terms of how many <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter of solution</strong>. </p>
<p>The unit of volume selected for molarity is <em>one liter</em>, <mathjax>#"1 L"#</mathjax>. This means that your ultimate goal when calculating a solution's molarity is to figure out how many moles of solute are present in <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>In your case, the solution is said to contain <mathjax>#6.9#</mathjax> <strong>moles</strong> of potassium chloride, <mathjax>#"KCl"#</mathjax>, your solute, in <mathjax>#"1.8 L"#</mathjax> of solution. </p>
<p>Now, a solution is characterized by the fact that the solute and the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, are <strong>evenly mixed</strong>. This implies that <strong>any sample</strong> of this solution will have <strong>the same concentration</strong>, since scaling down the volume will also scale down the number of moles of solute it contains. </p>
<p>You can thus use the volume of the solution to figure out how many moles of solute you'd get in <mathjax>#"1 L"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#1.0 color(red)(cancel(color(black)("L solution"))) * "6.9 moles KCl"/(1.8 color(red)(cancel(color(black)("L solution")))) = "3.833 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 L"#</mathjax> of solution contains <mathjax>#3.833#</mathjax> <strong>moles</strong> of solute, it follows that the <em>molarity</em> of the solution will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("3.8 mol L"^(-1))color(white)(a/a)|))) ->#</mathjax> <em>rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
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<iframe src="https://www.youtube.com/embed/SXf9rDnVFao?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | What is the molarity of a solution in which 6.9mol of potassium chloride is dissolved in water to a final volume of 1.8L? | null |
3,145 | a9d8a7da-6ddd-11ea-9a69-ccda262736ce | https://socratic.org/questions/57b2171b7c0149321faa50ca | 17.3 grams | start physical_unit 3 3 mass g qc_end physical_unit 3 3 8 9 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mass [OF] ammonia [IN] grams"}] | [{"type":"physical unit","value":"17.3 grams"}] | [{"type":"physical unit","value":"Volume [OF] ammonia [=] \\pu{22.4 dm^3}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What mass of ammonia constitutes a volume of #22.4*dm^3# at #"STP"#? </h1> | null | 17.3 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So if we treat ammonia as an ideal gas, there are <mathjax>#(14.01+3xx1.00794)*g#</mathjax> of <mathjax>#NH_3(g)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#22.4*dm^3#</mathjax> of ammonia at <mathjax>#"STP"#</mathjax> specifies a mole of ideal gas............</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So if we treat ammonia as an ideal gas, there are <mathjax>#(14.01+3xx1.00794)*g#</mathjax> of <mathjax>#NH_3(g)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of ammonia constitutes a volume of #22.4*dm^3# at #"STP"#? </h1>
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anor277
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<div class="markdown"><p><mathjax>#22.4*dm^3#</mathjax> of ammonia at <mathjax>#"STP"#</mathjax> specifies a mole of ideal gas............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So if we treat ammonia as an ideal gas, there are <mathjax>#(14.01+3xx1.00794)*g#</mathjax> of <mathjax>#NH_3(g)#</mathjax></p></div>
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</article> | What mass of ammonia constitutes a volume of #22.4*dm^3# at #"STP"#? | null |
3,146 | ab7cd5b6-6ddd-11ea-b34c-ccda262736ce | https://socratic.org/questions/58b8d414b72cff4c1b489d15 | 0.04 grams | start physical_unit 3 4 mass g qc_end physical_unit 12 13 10 11 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"0.04 grams"}] | [{"type":"physical unit","value":"Volume [OF] chloric acid [=] \\pu{60.00 mL}"},{"type":"physical unit","value":"Molarity [OF] chloric acid [=] \\pu{0.015 mol/L }"}] | <h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid?</h1> | null | 0.04 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.036 g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid?</h1>
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Stefan V.
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Mar 4, 2017
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<div class="markdown"><p><mathjax>#"0.036 g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
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Ernest Z.
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<span class="dateCreated" datetime="2017-03-04T00:58:22" itemprop="dateCreated">
Mar 4, 2017
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<div class="markdown"><p>The mass of sodium hydroxide is 36 mg.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equation for the reaction is</p>
<p><mathjax>#"HClO"_3 + "NaOH" → "NaClO"_3 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>1.</strong> Calculate the moles of <mathjax>#"HClO"_3#</mathjax>.</p>
<p><mathjax>#"Moles of HClO"_3 = "0.060 00" color(red)(cancel(color(black)("L NaOH"))) × "0.015 mol HClO"_3/(1 color(red)(cancel(color(black)("L HClO"_3)))) = "0.000 900 mol HClO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"NaOH"#</mathjax>.</p>
<p><mathjax>#"Moles of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol HClO"_3))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol HClO"_3)))) = "0.000 900 mol NaOH"#</mathjax> </p>
<blockquote></blockquote>
<p><strong>3.</strong> Calculate the mass of the <mathjax>#"NaOH"#</mathjax>.</p>
<p><mathjax>#"Mass of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.036 g NaOH" = "36 mg NaOH"#</mathjax></p></div>
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</article> | What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid? | null |
3,147 | a8d4791b-6ddd-11ea-bae0-ccda262736ce | https://socratic.org/questions/58f99abcb72cff3516912215 | 3 | start physical_unit 6 7 ph none qc_end physical_unit 7 7 20 21 volume qc_end physical_unit 24 24 26 27 concentration qc_end physical_unit 7 7 34 35 volume qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"3"}] | [{"type":"physical unit","value":"Volume1 [OF] HCl(aq) solution [=] \\pu{10 mL}"},{"type":"physical unit","value":"Concentration [OF] HCl(aq) solution [=] \\pu{0.10 mol/L}"},{"type":"physical unit","value":"Volume2 [OF] HCl(aq) solution [=] \\pu{1 L}"}] | <h1 class="questionTitle" itemprop="name">What is the resultant #pH# of the solution.....?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>What is the resultant <mathjax>#pH#</mathjax> of the solution prepared by taking a <mathjax>#10*mL#</mathjax> volume of <mathjax>#HCl(aq)#</mathjax> of <mathjax>#0.10*mol*L^-1#</mathjax> concentration, and diluting this to a <mathjax>#1*L#</mathjax> volume?</p></div>
</h2>
</div>
</div> | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)#</mathjax></p>
<p><mathjax>#10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#-log_(10)(10^-3)=3...........#</mathjax></p>
<p>And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#pH=3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)#</mathjax></p>
<p><mathjax>#10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#-log_(10)(10^-3)=3...........#</mathjax></p>
<p>And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the resultant #pH# of the solution.....?</h1>
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<div class="markdown"><p>What is the resultant <mathjax>#pH#</mathjax> of the solution prepared by taking a <mathjax>#10*mL#</mathjax> volume of <mathjax>#HCl(aq)#</mathjax> of <mathjax>#0.10*mol*L^-1#</mathjax> concentration, and diluting this to a <mathjax>#1*L#</mathjax> volume?</p></div>
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anor277
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<div class="markdown"><p><mathjax>#pH=3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)#</mathjax></p>
<p><mathjax>#10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p><mathjax>#-log_(10)(10^-3)=3...........#</mathjax></p>
<p>And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition. </p></div>
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</article> | What is the resultant #pH# of the solution.....? |
What is the resultant #pH# of the solution prepared by taking a #10*mL# volume of #HCl(aq)# of #0.10*mol*L^-1# concentration, and diluting this to a #1*L# volume?
|
3,148 | ab818150-6ddd-11ea-9d54-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-of-zn-pb-no3-2-pb-zn-no3-2 | Zn + Pb(NO3)2 -> Pb + Zn(NO3)2 | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced equation"}] | [{"type":"chemical equation","value":"Zn + Pb(NO3)2 -> Pb + Zn(NO3)2"}] | [{"type":"chemical equation","value":"Zn + Pb(NO3)2 -> Pb + Zn(NO3)2"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation of Zn+Pb(NO3)2---Pb+Zn(NO3)2?</h1> | null | Zn + Pb(NO3)2 -> Pb + Zn(NO3)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the given equation:</p>
<p><mathjax>#"Zn" + "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<p>A method that often works is to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax> first, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"Pb(NO"_3)_2#</mathjax> or <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's choose <mathjax>#"Pb(NO"_3)_2#</mathjax>. We put a 1 in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Pb"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Pb"#</mathjax> on the left, so we need 1 <mathjax>#"Pb"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Pb"#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"N"#</mathjax>:</strong></p>
<p>We have fixed 2 <mathjax>#"N"#</mathjax> on the left, so we need 2 <mathjax>#"N"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Zn"#</mathjax> atom on the right so we need 1 <mathjax>#"Zn"#</mathjax> atom on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(ml)"Zn"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"Pb"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"N"color(white)(mmmmmml)2color(white)(mmmmmmmll)2#</mathjax><br/>
<mathjax>#color(white)(ml)"O"color(white)(mmmmmml)6color(white)(mmmmmmmll)6#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#color(red)("Zn" + "Pb"("NO"_3)_2 → "Pb" + "Zn"("NO"_3)_2#</mathjax></p>
<p>Surprise! The original equation was already balanced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The balanced equation is <mathjax>#"Zn" + "Pb"("NO"_3)_2 → "Pb" + "Zn"("NO"_3)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the given equation:</p>
<p><mathjax>#"Zn" + "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<p>A method that often works is to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax> first, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"Pb(NO"_3)_2#</mathjax> or <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's choose <mathjax>#"Pb(NO"_3)_2#</mathjax>. We put a 1 in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Pb"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Pb"#</mathjax> on the left, so we need 1 <mathjax>#"Pb"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Pb"#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"N"#</mathjax>:</strong></p>
<p>We have fixed 2 <mathjax>#"N"#</mathjax> on the left, so we need 2 <mathjax>#"N"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Zn"#</mathjax> atom on the right so we need 1 <mathjax>#"Zn"#</mathjax> atom on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(ml)"Zn"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"Pb"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"N"color(white)(mmmmmml)2color(white)(mmmmmmmll)2#</mathjax><br/>
<mathjax>#color(white)(ml)"O"color(white)(mmmmmml)6color(white)(mmmmmmmll)6#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#color(red)("Zn" + "Pb"("NO"_3)_2 → "Pb" + "Zn"("NO"_3)_2#</mathjax></p>
<p>Surprise! The original equation was already balanced.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation of Zn+Pb(NO3)2---Pb+Zn(NO3)2?</h1>
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Ernest Z.
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May 26, 2014
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<div class="markdown"><p>The balanced equation is <mathjax>#"Zn" + "Pb"("NO"_3)_2 → "Pb" + "Zn"("NO"_3)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the given equation:</p>
<p><mathjax>#"Zn" + "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<p>A method that often works is to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax> first, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"Pb(NO"_3)_2#</mathjax> or <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's choose <mathjax>#"Pb(NO"_3)_2#</mathjax>. We put a 1 in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Pb"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Pb"#</mathjax> on the left, so we need 1 <mathjax>#"Pb"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Pb"#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"N"#</mathjax>:</strong></p>
<p>We have fixed 2 <mathjax>#"N"#</mathjax> on the left, so we need 2 <mathjax>#"N"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"Zn(NO"_3)_2#</mathjax>.</p>
<p><mathjax>#"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have fixed 1 <mathjax>#"Zn"#</mathjax> atom on the right so we need 1 <mathjax>#"Zn"#</mathjax> atom on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(red)(1) "Pb(NO"_3)_2 → color(green)(1) "Pb" + color(orange)(1) "Zn(NO"_3)_2#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(ml)"Zn"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"Pb"color(white)(mmmmmll)1color(white)(mmmmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(ml)"N"color(white)(mmmmmml)2color(white)(mmmmmmmll)2#</mathjax><br/>
<mathjax>#color(white)(ml)"O"color(white)(mmmmmml)6color(white)(mmmmmmmll)6#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#color(red)("Zn" + "Pb"("NO"_3)_2 → "Pb" + "Zn"("NO"_3)_2#</mathjax></p>
<p>Surprise! The original equation was already balanced.</p></div>
</div>
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</article> | What is the balanced equation of Zn+Pb(NO3)2---Pb+Zn(NO3)2? | null |
3,149 | ac974ef0-6ddd-11ea-b304-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-937-g-ca-c-2h-3o-2-2 | 5.92 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(C2H3O2)2 [IN] moles"}] | [{"type":"physical unit","value":"5.92 moles"}] | [{"type":"physical unit","value":"Mass [OF] Ca(C2H3O2)2 [=] \\pu{937 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 937 g #Ca(C_2H_3O_2)_2#?</h1> | null | 5.92 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Calcium acetate has a formula weight of <mathjax>#158.17#</mathjax> <mathjax>#g*mol^-1#</mathjax>. To get the the number of moles, in fact the NUMBER of calcium acetate formula units, we divide the given mass by the molar mass:</p>
<p><mathjax>#(937*cancelg)/(158.17*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that this answer is dimensionally consistent. We seek an answer in <mathjax>#mol#</mathjax>, and we get an answer in <mathjax>#1/(mol^-1)="moles"#</mathjax> as required. Capisce?</p>
<p>What number of calcium acetate formula units does this molar quantity represent?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>About <mathjax>#6#</mathjax> <mathjax>#mol#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Calcium acetate has a formula weight of <mathjax>#158.17#</mathjax> <mathjax>#g*mol^-1#</mathjax>. To get the the number of moles, in fact the NUMBER of calcium acetate formula units, we divide the given mass by the molar mass:</p>
<p><mathjax>#(937*cancelg)/(158.17*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that this answer is dimensionally consistent. We seek an answer in <mathjax>#mol#</mathjax>, and we get an answer in <mathjax>#1/(mol^-1)="moles"#</mathjax> as required. Capisce?</p>
<p>What number of calcium acetate formula units does this molar quantity represent?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 937 g #Ca(C_2H_3O_2)_2#?</h1>
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<div class="markdown"><p>About <mathjax>#6#</mathjax> <mathjax>#mol#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Calcium acetate has a formula weight of <mathjax>#158.17#</mathjax> <mathjax>#g*mol^-1#</mathjax>. To get the the number of moles, in fact the NUMBER of calcium acetate formula units, we divide the given mass by the molar mass:</p>
<p><mathjax>#(937*cancelg)/(158.17*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that this answer is dimensionally consistent. We seek an answer in <mathjax>#mol#</mathjax>, and we get an answer in <mathjax>#1/(mol^-1)="moles"#</mathjax> as required. Capisce?</p>
<p>What number of calcium acetate formula units does this molar quantity represent?</p></div>
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</article> | How many moles are in 937 g #Ca(C_2H_3O_2)_2#? | null |
3,150 | a93b3014-6ddd-11ea-b5fd-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-has-a-volume-of-425-ml-at-25-c-and-760-torr-what-volume-would-th | 389 mL | start physical_unit 19 21 volume ml qc_end c_other STP qc_end physical_unit 20 21 8 9 volume qc_end physical_unit 20 21 11 12 temperature qc_end physical_unit 20 21 14 15 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] this gas sample [IN] mL"}] | [{"type":"physical unit","value":"389 mL"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{425 mL}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{760 torr}"}] | <h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 425 mL at 25°C and 760 torr. What volume would this gas sample have at STP?</h1> | null | 389 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP: P=760 torr and T=273K.</p>
<p>To solve this question, we could use <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> since both the number of mole and pressure are constant:</p>
<p><mathjax>#(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=T_2xx(V_1)/(T_1)#</mathjax></p>
<p><mathjax>#=>V_2=273 cancel(K)xx(425mL)/(298cancel(K))=389mL#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_2=389mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP: P=760 torr and T=273K.</p>
<p>To solve this question, we could use <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> since both the number of mole and pressure are constant:</p>
<p><mathjax>#(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=T_2xx(V_1)/(T_1)#</mathjax></p>
<p><mathjax>#=>V_2=273 cancel(K)xx(425mL)/(298cancel(K))=389mL#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 425 mL at 25°C and 760 torr. What volume would this gas sample have at STP?</h1>
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<div class="markdown"><p><mathjax>#V_2=389mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At STP: P=760 torr and T=273K.</p>
<p>To solve this question, we could use <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> since both the number of mole and pressure are constant:</p>
<p><mathjax>#(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=T_2xx(V_1)/(T_1)#</mathjax></p>
<p><mathjax>#=>V_2=273 cancel(K)xx(425mL)/(298cancel(K))=389mL#</mathjax></p></div>
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</article> | A sample of gas has a volume of 425 mL at 25°C and 760 torr. What volume would this gas sample have at STP? | null |
3,151 | aadf249e-6ddd-11ea-bdac-ccda262736ce | https://socratic.org/questions/suppose-you-have-125-0-ml-of-0-400-m-nacl-solution-this-is-too-salty-so-you-dilu | 0.29 molarity | start physical_unit 8 9 concentration mol/l qc_end physical_unit 8 9 3 4 volume qc_end physical_unit 8 9 6 7 concentration qc_end physical_unit 24 24 21 22 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] NaCl solution [IN] molarity"}] | [{"type":"physical unit","value":"0.29 molarity"}] | [{"type":"physical unit","value":"Volume1 [OF] NaCl solution [=] \\pu{125.0 mL}"},{"type":"physical unit","value":"Concentration1 [OF] NaCl solution [=] \\pu{0.400 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{50.0 mL}"}] | <h1 class="questionTitle" itemprop="name">Suppose you have 125.0 mL of 0.400 M #NaCl# solution. This is too salty, so you dilute it by adding another 50.0 mL of water. What is the new concentration of #NaCl# in units of molarity? </h1> | null | 0.29 molarity | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a great example of a practice problem that requires you to use the formula for <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong>. </p>
<p>To make it more interesting, let's assume that you're not familiar with that problem.</p>
<p>The most important thing to remember about a <strong>dilution</strong> is the the <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>must remain constant</strong>.</p>
<p>That is, the number of moles of solute present in the initial solution <strong>must be equal</strong> to the number of moles of solute present in the diluted solution. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liter of solution, keeping the number of moles of solute constant allows you to <strong>decrease</strong> concentration by <strong>increasing</strong> the total volume of the solution. </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://pharmrx.yolasite.com/concentration-and-dilutions.php" src="https://useruploads.socratic.org/mYO9qa8FTkCHOARoBWtg_solution.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>So, use the molarity and volume of the initial solution to determine how many moles of sodium chloride it contains</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(c = n/V))| implies n = c * V#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.400 M" * 125.0 * 10^(-3)"L" = "0.0500 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute you must have in the diluted solution. </p>
<p>The new volume of the solution will be </p>
<blockquote>
<p><mathjax>#V_"new" = "125.0 mL" + "50.0 mL" = "175.0 mL"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the diluted solution will be</p>
<blockquote>
<p><mathjax>#c = "0.0500 moles"/(175.0 * 10^(-3)"L") = color(green)(|bar(ul("0.286 M")|)#</mathjax></p>
</blockquote>
<p>Now, this is exactly what the formula for dilutions calculations allows you to do</p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(purple)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>Rearrange to solve for <mathjax>#c_2#</mathjax></p>
<blockquote>
<p><mathjax>#c_2 = V_1/V_2 xx c_1#</mathjax></p>
<p><mathjax>#c_2 = (125.0 color(red)(cancel(color(black)("mL"))))/(175.0color(red)(cancel(color(black)("mL")))) * "0.400 M" = color(green)(|bar(ul("0.286 M")|)#</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>#"0.286 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a great example of a practice problem that requires you to use the formula for <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong>. </p>
<p>To make it more interesting, let's assume that you're not familiar with that problem.</p>
<p>The most important thing to remember about a <strong>dilution</strong> is the the <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>must remain constant</strong>.</p>
<p>That is, the number of moles of solute present in the initial solution <strong>must be equal</strong> to the number of moles of solute present in the diluted solution. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liter of solution, keeping the number of moles of solute constant allows you to <strong>decrease</strong> concentration by <strong>increasing</strong> the total volume of the solution. </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://pharmrx.yolasite.com/concentration-and-dilutions.php" src="https://useruploads.socratic.org/mYO9qa8FTkCHOARoBWtg_solution.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>So, use the molarity and volume of the initial solution to determine how many moles of sodium chloride it contains</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(c = n/V))| implies n = c * V#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.400 M" * 125.0 * 10^(-3)"L" = "0.0500 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute you must have in the diluted solution. </p>
<p>The new volume of the solution will be </p>
<blockquote>
<p><mathjax>#V_"new" = "125.0 mL" + "50.0 mL" = "175.0 mL"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the diluted solution will be</p>
<blockquote>
<p><mathjax>#c = "0.0500 moles"/(175.0 * 10^(-3)"L") = color(green)(|bar(ul("0.286 M")|)#</mathjax></p>
</blockquote>
<p>Now, this is exactly what the formula for dilutions calculations allows you to do</p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(purple)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>Rearrange to solve for <mathjax>#c_2#</mathjax></p>
<blockquote>
<p><mathjax>#c_2 = V_1/V_2 xx c_1#</mathjax></p>
<p><mathjax>#c_2 = (125.0 color(red)(cancel(color(black)("mL"))))/(175.0color(red)(cancel(color(black)("mL")))) * "0.400 M" = color(green)(|bar(ul("0.286 M")|)#</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">Suppose you have 125.0 mL of 0.400 M #NaCl# solution. This is too salty, so you dilute it by adding another 50.0 mL of water. What is the new concentration of #NaCl# in units of molarity? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-04T22:28:15" itemprop="dateCreated">
Mar 4, 2016
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<div class="markdown"><p><mathjax>#"0.286 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a great example of a practice problem that requires you to use the formula for <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong>. </p>
<p>To make it more interesting, let's assume that you're not familiar with that problem.</p>
<p>The most important thing to remember about a <strong>dilution</strong> is the the <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>must remain constant</strong>.</p>
<p>That is, the number of moles of solute present in the initial solution <strong>must be equal</strong> to the number of moles of solute present in the diluted solution. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liter of solution, keeping the number of moles of solute constant allows you to <strong>decrease</strong> concentration by <strong>increasing</strong> the total volume of the solution. </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://pharmrx.yolasite.com/concentration-and-dilutions.php" src="https://useruploads.socratic.org/mYO9qa8FTkCHOARoBWtg_solution.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>So, use the molarity and volume of the initial solution to determine how many moles of sodium chloride it contains</p>
<blockquote>
<p><mathjax>#color(blue)(| bar(ul(c = n/V))| implies n = c * V#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.400 M" * 125.0 * 10^(-3)"L" = "0.0500 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of solute you must have in the diluted solution. </p>
<p>The new volume of the solution will be </p>
<blockquote>
<p><mathjax>#V_"new" = "125.0 mL" + "50.0 mL" = "175.0 mL"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the diluted solution will be</p>
<blockquote>
<p><mathjax>#c = "0.0500 moles"/(175.0 * 10^(-3)"L") = color(green)(|bar(ul("0.286 M")|)#</mathjax></p>
</blockquote>
<p>Now, this is exactly what the formula for dilutions calculations allows you to do</p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(purple)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution</p>
<p>Rearrange to solve for <mathjax>#c_2#</mathjax></p>
<blockquote>
<p><mathjax>#c_2 = V_1/V_2 xx c_1#</mathjax></p>
<p><mathjax>#c_2 = (125.0 color(red)(cancel(color(black)("mL"))))/(175.0color(red)(cancel(color(black)("mL")))) * "0.400 M" = color(green)(|bar(ul("0.286 M")|)#</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>
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</article> | Suppose you have 125.0 mL of 0.400 M #NaCl# solution. This is too salty, so you dilute it by adding another 50.0 mL of water. What is the new concentration of #NaCl# in units of molarity? | null |
3,152 | a9a1d95b-6ddd-11ea-9624-ccda262736ce | https://socratic.org/questions/what-is-the-maximum-number-of-moles-of-mg-oh-2-that-can-remain-dissolved-in-a-5- | 0.04 moles | start physical_unit 8 8 mole mol qc_end physical_unit 17 17 15 16 volume qc_end physical_unit 17 17 21 21 ph qc_end end | [{"type":"physical unit","value":"Mole [OF] Mg(OH)2 [IN] moles"}] | [{"type":"physical unit","value":"0.04 moles"}] | [{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{5.0 L}"},{"type":"physical unit","value":"pH [OF] solution [=] \\pu{9.70}"}] | <h1 class="questionTitle" itemprop="name">What is the maximum number of moles of #"Mg"("OH")_2#, that can remain dissolved in a 5.0 L solution whose pH is 9.70? </h1> | null | 0.04 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! CORRECT SOLUTION !!</strong></p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax><br/>
<strong>SIDE NOTE</strong> <em>My first attempt at solving this problem was incorrect, to put it mildly. I will add the correct answer here, and leave the incorrect one below to serve as a <strong>'How not to solve this problem'</strong> example.</em><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax></p>
<p>The idea here is that magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution. This means that when you dissolve this salt in water, the following dissociation equilibrium is established</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this equilibrium looks like this </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>The value listed here is </p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-11)#</mathjax></p>
</blockquote>
<p><a href="http://www.wiredchemist.com/chemistry/data/solubility-product-constants" rel="nofollow" target="_blank">http://www.wiredchemist.com/chemistry/data/solubility-product-constants</a></p>
<p>Assuming that the solution is <strong>buffered</strong> at a pH of <mathjax>#9.70#</mathjax>, i.e. the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>, is <strong>constant</strong>, your first goal here will be figure out the concentration of <strong>magnesium cations</strong>, <mathjax>#"Mg"^(2+)#</mathjax>. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, according to the expression you have for the solution's solubility product constant, the equilibrium concentration of magnesium cations will be equal to </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = K_(sp)/(["OH"^(-)])^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = (1.8 * 10^(-11))/(5.01 * 10^(-5))^color(red)(2) = 7.17 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that when you place some solid magnesium hydroxide in a buffered solution that has a pH equal to <mathjax>#9.70#</mathjax>, the salt will dissolve until the equilibrium concentration of magnesium cations is equal to <mathjax>#7.17 * 10^(-3)"M"#</mathjax>. </p>
<p>Use the <strong>volume</strong> of the solution to find the <em>number of moles</em> of magnesium cations </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 7.17 * 10^(-3)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = "0.03585 moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Since magnesium hydroxide dissociates in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce magnesium cations, you can say that the number of moles of magnesium hydroxide that dissolve in this solution will be </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.036 moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><mathjax>#color(red)(color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)())#</mathjax></p>
<p><strong>!! INCORRECT SOLUTION !!</strong></p>
<p>The problem wants you to use the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to find the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Since magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution, its dissociation equilibrium will be affected by the pH of the solution. More specifically, a <strong>higher</strong> concentration of hydroxide anions will <strong>reduce</strong> the solubility of magnesium hydroxide. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Magnesium hydroxide's dissociation equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of magnesium hydroxide <strong>that dissociates</strong> in aqueous solution produces </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of magnesium cations</em>, <mathjax>#1 xx "Mg"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of hydroxide anions</em>, <mathjax>#color(red)(2) xx "OH"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that the equilibrium concentration of hydroxide anions will be <strong>twice as high</strong> as the equilibrium concentration of magnesium cations. </p>
<p>You can thus say that this solution will contain </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx ["OH"^(-)]#</mathjax></p>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx 5.01 * 10^(-5)"M" = 2.5 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Use the <strong>volume</strong> of the solution to calculate how many <em>moles</em> of magnesium cations would be present </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 2.5 * 10^(-5)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = 1.25 * 10^(-4)"moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Now, you already know that <strong>for every mole</strong> of magnesium hydroxide that dissociates you get <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium cations in solution. </p>
<p>This means that in a solution that has a pH equal to <mathjax>#9.70#</mathjax> you can only hope to dissolve </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.3 * 10^(-4)"moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<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>#"0.036 moles Mg"("OH")_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! CORRECT SOLUTION !!</strong></p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax><br/>
<strong>SIDE NOTE</strong> <em>My first attempt at solving this problem was incorrect, to put it mildly. I will add the correct answer here, and leave the incorrect one below to serve as a <strong>'How not to solve this problem'</strong> example.</em><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax></p>
<p>The idea here is that magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution. This means that when you dissolve this salt in water, the following dissociation equilibrium is established</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this equilibrium looks like this </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>The value listed here is </p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-11)#</mathjax></p>
</blockquote>
<p><a href="http://www.wiredchemist.com/chemistry/data/solubility-product-constants" rel="nofollow" target="_blank">http://www.wiredchemist.com/chemistry/data/solubility-product-constants</a></p>
<p>Assuming that the solution is <strong>buffered</strong> at a pH of <mathjax>#9.70#</mathjax>, i.e. the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>, is <strong>constant</strong>, your first goal here will be figure out the concentration of <strong>magnesium cations</strong>, <mathjax>#"Mg"^(2+)#</mathjax>. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, according to the expression you have for the solution's solubility product constant, the equilibrium concentration of magnesium cations will be equal to </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = K_(sp)/(["OH"^(-)])^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = (1.8 * 10^(-11))/(5.01 * 10^(-5))^color(red)(2) = 7.17 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that when you place some solid magnesium hydroxide in a buffered solution that has a pH equal to <mathjax>#9.70#</mathjax>, the salt will dissolve until the equilibrium concentration of magnesium cations is equal to <mathjax>#7.17 * 10^(-3)"M"#</mathjax>. </p>
<p>Use the <strong>volume</strong> of the solution to find the <em>number of moles</em> of magnesium cations </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 7.17 * 10^(-3)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = "0.03585 moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Since magnesium hydroxide dissociates in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce magnesium cations, you can say that the number of moles of magnesium hydroxide that dissolve in this solution will be </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.036 moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><mathjax>#color(red)(color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)())#</mathjax></p>
<p><strong>!! INCORRECT SOLUTION !!</strong></p>
<p>The problem wants you to use the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to find the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Since magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution, its dissociation equilibrium will be affected by the pH of the solution. More specifically, a <strong>higher</strong> concentration of hydroxide anions will <strong>reduce</strong> the solubility of magnesium hydroxide. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Magnesium hydroxide's dissociation equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of magnesium hydroxide <strong>that dissociates</strong> in aqueous solution produces </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of magnesium cations</em>, <mathjax>#1 xx "Mg"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of hydroxide anions</em>, <mathjax>#color(red)(2) xx "OH"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that the equilibrium concentration of hydroxide anions will be <strong>twice as high</strong> as the equilibrium concentration of magnesium cations. </p>
<p>You can thus say that this solution will contain </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx ["OH"^(-)]#</mathjax></p>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx 5.01 * 10^(-5)"M" = 2.5 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Use the <strong>volume</strong> of the solution to calculate how many <em>moles</em> of magnesium cations would be present </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 2.5 * 10^(-5)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = 1.25 * 10^(-4)"moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Now, you already know that <strong>for every mole</strong> of magnesium hydroxide that dissociates you get <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium cations in solution. </p>
<p>This means that in a solution that has a pH equal to <mathjax>#9.70#</mathjax> you can only hope to dissolve </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.3 * 10^(-4)"moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<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>
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<h1 class="questionTitle" itemprop="name">What is the maximum number of moles of #"Mg"("OH")_2#, that can remain dissolved in a 5.0 L solution whose pH is 9.70? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.036 moles Mg"("OH")_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! CORRECT SOLUTION !!</strong></p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax><br/>
<strong>SIDE NOTE</strong> <em>My first attempt at solving this problem was incorrect, to put it mildly. I will add the correct answer here, and leave the incorrect one below to serve as a <strong>'How not to solve this problem'</strong> example.</em><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#</mathjax></p>
<p>The idea here is that magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution. This means that when you dissolve this salt in water, the following dissociation equilibrium is established</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this equilibrium looks like this </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>The value listed here is </p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-11)#</mathjax></p>
</blockquote>
<p><a href="http://www.wiredchemist.com/chemistry/data/solubility-product-constants" rel="nofollow" target="_blank">http://www.wiredchemist.com/chemistry/data/solubility-product-constants</a></p>
<p>Assuming that the solution is <strong>buffered</strong> at a pH of <mathjax>#9.70#</mathjax>, i.e. the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>, is <strong>constant</strong>, your first goal here will be figure out the concentration of <strong>magnesium cations</strong>, <mathjax>#"Mg"^(2+)#</mathjax>. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, according to the expression you have for the solution's solubility product constant, the equilibrium concentration of magnesium cations will be equal to </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = K_(sp)/(["OH"^(-)])^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = (1.8 * 10^(-11))/(5.01 * 10^(-5))^color(red)(2) = 7.17 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that when you place some solid magnesium hydroxide in a buffered solution that has a pH equal to <mathjax>#9.70#</mathjax>, the salt will dissolve until the equilibrium concentration of magnesium cations is equal to <mathjax>#7.17 * 10^(-3)"M"#</mathjax>. </p>
<p>Use the <strong>volume</strong> of the solution to find the <em>number of moles</em> of magnesium cations </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 7.17 * 10^(-3)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = "0.03585 moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Since magnesium hydroxide dissociates in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce magnesium cations, you can say that the number of moles of magnesium hydroxide that dissolve in this solution will be </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.036 moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><mathjax>#color(red)(color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)())#</mathjax></p>
<p><strong>!! INCORRECT SOLUTION !!</strong></p>
<p>The problem wants you to use the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to find the concentration of hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Since magnesium hydroxide, <mathjax>#"Mg"("OH")_2#</mathjax>, is <strong>insoluble</strong> in aqueous solution, its dissociation equilibrium will be affected by the pH of the solution. More specifically, a <strong>higher</strong> concentration of hydroxide anions will <strong>reduce</strong> the solubility of magnesium hydroxide. </p>
<p>Start by calculating the concentration of hydroxide anions. 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(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))" "#</mathjax> and <mathjax>#" "color(purple)(|bar(ul(color(white)(a/a)color(black)("poH" = - log(["OH"^(-)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pOH" = 14 - 9.70 = 4.30#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 10^(-4.30) = 5.01 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Magnesium hydroxide's dissociation equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (color(red)(2)(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of magnesium hydroxide <strong>that dissociates</strong> in aqueous solution produces </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of magnesium cations</em>, <mathjax>#1 xx "Mg"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of hydroxide anions</em>, <mathjax>#color(red)(2) xx "OH"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>This tells you that the equilibrium concentration of hydroxide anions will be <strong>twice as high</strong> as the equilibrium concentration of magnesium cations. </p>
<p>You can thus say that this solution will contain </p>
<blockquote>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx ["OH"^(-)]#</mathjax></p>
<p><mathjax>#["Mg"^(2+)] = 1/color(red)(2) xx 5.01 * 10^(-5)"M" = 2.5 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Use the <strong>volume</strong> of the solution to calculate how many <em>moles</em> of magnesium cations would be present </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#n_("Mg"^(2+)) = 2.5 * 10^(-5)"mol" color(red)(cancel(color(black)("L"^(-1)))) * 5.0 color(red)(cancel(color(black)("L"))#</mathjax></p>
<p><mathjax>#n_("Mg"^(2+)) = 1.25 * 10^(-4)"moles Mg"^(2+)#</mathjax></p>
</blockquote>
<p>Now, you already know that <strong>for every mole</strong> of magnesium hydroxide that dissociates you get <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium cations in solution. </p>
<p>This means that in a solution that has a pH equal to <mathjax>#9.70#</mathjax> you can only hope to dissolve </p>
<blockquote>
<p><mathjax>#n_("Mg"("OH")_2) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.3 * 10^(-4)"moles Mg"("OH")_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<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> | What is the maximum number of moles of #"Mg"("OH")_2#, that can remain dissolved in a 5.0 L solution whose pH is 9.70? | null |
3,153 | ab87c328-6ddd-11ea-a714-ccda262736ce | https://socratic.org/questions/563c165c11ef6b3f75083e0c | K2MnO4 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"K2MnO4"}] | [{"type":"physical unit","value":"Percent [OF] K in the compound [=] \\pu{39%}"},{"type":"physical unit","value":"Percent [OF] Mn in the compound [=] \\pu{27.9%}"},{"type":"physical unit","value":"Percent [OF] O in the compound [=] \\pu{32.5%}"}] | <h1 class="questionTitle" itemprop="name">A compound was found to contain #"39% K"#, #"27.9% Mn"#, and #"32.5% O"#. What is the empirical formula for this compound?</h1> | null | K2MnO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the percentages add up to 100, we can assume that we have 100.0 g of the compound, which will convert the percentages to grams.</p>
<p>Determine the moles of each element by dividing their masses in the compound by their molar masses (atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole, or g/mol). I will be saving some guard digits to reduce rounding errors. I will round to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</p>
<p><mathjax>#39.6cancel"g K"xx(1"mol K")/(39.0983cancel"g K")="1.0128 mol K"#</mathjax></p>
<p><mathjax>#27.9cancel"g Mn"xx(1"mol Mn")/(54.938049cancel"g Mn")="0.50784 mol Mn"#</mathjax></p>
<p><mathjax>#32.5cancel"g O"xx(1"mol O")/(15.999cancel"g O")="2.0314 mol O"#</mathjax></p>
<p>Next divide the moles of each element by the least number of moles. This will give you the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> for the formula.</p>
<p><mathjax>#"K":#</mathjax><mathjax>#(1.0128)/(0.50784)#</mathjax><mathjax>#=#</mathjax><mathjax>#"1.99#</mathjax><mathjax>#~~#</mathjax><mathjax>#2#</mathjax></p>
<p><mathjax>#"Mn":#</mathjax><mathjax>#(0.50784)/(0.50784)="1.00"#</mathjax></p>
<p><mathjax>#"O":#</mathjax><mathjax>#(2.0314)/(0.50784)="4.00"#</mathjax></p>
<p>The empirical formula of this compound is <mathjax>#"K"_2"MnO"_4"#</mathjax>, which is potassium manganate.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula of this compound is <mathjax>#"K"_2"MnO"_4"#</mathjax>.<br/>
This compound is potassium manganate.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the percentages add up to 100, we can assume that we have 100.0 g of the compound, which will convert the percentages to grams.</p>
<p>Determine the moles of each element by dividing their masses in the compound by their molar masses (atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole, or g/mol). I will be saving some guard digits to reduce rounding errors. I will round to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</p>
<p><mathjax>#39.6cancel"g K"xx(1"mol K")/(39.0983cancel"g K")="1.0128 mol K"#</mathjax></p>
<p><mathjax>#27.9cancel"g Mn"xx(1"mol Mn")/(54.938049cancel"g Mn")="0.50784 mol Mn"#</mathjax></p>
<p><mathjax>#32.5cancel"g O"xx(1"mol O")/(15.999cancel"g O")="2.0314 mol O"#</mathjax></p>
<p>Next divide the moles of each element by the least number of moles. This will give you the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> for the formula.</p>
<p><mathjax>#"K":#</mathjax><mathjax>#(1.0128)/(0.50784)#</mathjax><mathjax>#=#</mathjax><mathjax>#"1.99#</mathjax><mathjax>#~~#</mathjax><mathjax>#2#</mathjax></p>
<p><mathjax>#"Mn":#</mathjax><mathjax>#(0.50784)/(0.50784)="1.00"#</mathjax></p>
<p><mathjax>#"O":#</mathjax><mathjax>#(2.0314)/(0.50784)="4.00"#</mathjax></p>
<p>The empirical formula of this compound is <mathjax>#"K"_2"MnO"_4"#</mathjax>, which is potassium manganate.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound was found to contain #"39% K"#, #"27.9% Mn"#, and #"32.5% O"#. What is the empirical formula for this compound?</h1>
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<div class="markdown"><p>The empirical formula of this compound is <mathjax>#"K"_2"MnO"_4"#</mathjax>.<br/>
This compound is potassium manganate.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the percentages add up to 100, we can assume that we have 100.0 g of the compound, which will convert the percentages to grams.</p>
<p>Determine the moles of each element by dividing their masses in the compound by their molar masses (atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole, or g/mol). I will be saving some guard digits to reduce rounding errors. I will round to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> at the end.</p>
<p><mathjax>#39.6cancel"g K"xx(1"mol K")/(39.0983cancel"g K")="1.0128 mol K"#</mathjax></p>
<p><mathjax>#27.9cancel"g Mn"xx(1"mol Mn")/(54.938049cancel"g Mn")="0.50784 mol Mn"#</mathjax></p>
<p><mathjax>#32.5cancel"g O"xx(1"mol O")/(15.999cancel"g O")="2.0314 mol O"#</mathjax></p>
<p>Next divide the moles of each element by the least number of moles. This will give you the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> for the formula.</p>
<p><mathjax>#"K":#</mathjax><mathjax>#(1.0128)/(0.50784)#</mathjax><mathjax>#=#</mathjax><mathjax>#"1.99#</mathjax><mathjax>#~~#</mathjax><mathjax>#2#</mathjax></p>
<p><mathjax>#"Mn":#</mathjax><mathjax>#(0.50784)/(0.50784)="1.00"#</mathjax></p>
<p><mathjax>#"O":#</mathjax><mathjax>#(2.0314)/(0.50784)="4.00"#</mathjax></p>
<p>The empirical formula of this compound is <mathjax>#"K"_2"MnO"_4"#</mathjax>, which is potassium manganate.</p></div>
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</article> | A compound was found to contain #"39% K"#, #"27.9% Mn"#, and #"32.5% O"#. What is the empirical formula for this compound? | null |
3,154 | a9a22bd4-6ddd-11ea-a49c-ccda262736ce | https://socratic.org/questions/the-ph-of-a-5-95-times-10-1-m-solution-of-a-weak-base-is-11-54-what-is-the-pk-b- | 4.69 | start physical_unit 20 21 pkb none qc_end physical_unit 10 12 4 7 molarity qc_end physical_unit 8 8 14 14 ph qc_end end | [{"type":"physical unit","value":"pKb [OF] this base"}] | [{"type":"physical unit","value":"4.69"}] | [{"type":"physical unit","value":"Molarity [OF] a weak base solution [=] \\pu{5.95 × 10^(−1) M}"},{"type":"physical unit","value":"pH [OF] a weak base solution [=] \\pu{11.54}"}] | <h1 class="questionTitle" itemprop="name">The pH of a #5.95\times 10^-1# M solution of a weak base is 11.54. What is the #pK_b# for this base to two decimal places?</h1> | null | 4.69 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From an ICE table you can use this expression which applies to a weak base:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>b is the concentration of the base.</p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pOH=14-11.54=2.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(2.46=1/2[pK_b-log(0.595)])#</mathjax></p>
<p><mathjax>#sf(2.46=1/2pK_b+0.1127)#</mathjax></p>
<p><mathjax>#sf(1/2pK_b=2.3473)#</mathjax></p>
<p><mathjax>#sf(pK_b=4.69)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#sf(pK_b=4.69)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From an ICE table you can use this expression which applies to a weak base:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>b is the concentration of the base.</p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pOH=14-11.54=2.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(2.46=1/2[pK_b-log(0.595)])#</mathjax></p>
<p><mathjax>#sf(2.46=1/2pK_b+0.1127)#</mathjax></p>
<p><mathjax>#sf(1/2pK_b=2.3473)#</mathjax></p>
<p><mathjax>#sf(pK_b=4.69)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The pH of a #5.95\times 10^-1# M solution of a weak base is 11.54. What is the #pK_b# for this base to two decimal places?</h1>
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<div class="markdown"><p><mathjax>#sf(pK_b=4.69)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>From an ICE table you can use this expression which applies to a weak base:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>b is the concentration of the base.</p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pOH=14-11.54=2.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(2.46=1/2[pK_b-log(0.595)])#</mathjax></p>
<p><mathjax>#sf(2.46=1/2pK_b+0.1127)#</mathjax></p>
<p><mathjax>#sf(1/2pK_b=2.3473)#</mathjax></p>
<p><mathjax>#sf(pK_b=4.69)#</mathjax></p></div>
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</article> | The pH of a #5.95\times 10^-1# M solution of a weak base is 11.54. What is the #pK_b# for this base to two decimal places? | null |
3,155 | a8ebbed3-6ddd-11ea-b196-ccda262736ce | https://socratic.org/questions/how-to-convert-3-0-moles-of-zncl-2-into-grams | 408.95 grams | start physical_unit 6 6 mass g qc_end physical_unit 6 6 3 4 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] ZnCl2 [IN] grams"}] | [{"type":"physical unit","value":"408.95 grams"}] | [{"type":"physical unit","value":"Mole [OF] ZnCl2 [=] \\pu{3.0 moles}"}] | <h1 class="questionTitle" itemprop="name">How to convert 3.0 moles of #"ZnCl"_2 #into grams?
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<div class="markdown"><p>I answered 405 g but apparently I was wrong. </p></div>
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</div> | 408.95 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>136.315 g/mol = ZnCl2</p>
<p><mathjax>#136.315*3=408.945#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>408.945g or 410g based off your <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>136.315 g/mol = ZnCl2</p>
<p><mathjax>#136.315*3=408.945#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How to convert 3.0 moles of #"ZnCl"_2 #into grams?
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<div class="markdown"><p>408.945g or 410g based off your <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>136.315 g/mol = ZnCl2</p>
<p><mathjax>#136.315*3=408.945#</mathjax></p></div>
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</article> | How to convert 3.0 moles of #"ZnCl"_2 #into grams?
|
I answered 405 g but apparently I was wrong.
|
3,156 | ab30664e-6ddd-11ea-a259-ccda262736ce | https://socratic.org/questions/how-would-you-complete-and-balance-the-acid-base-neutralizations-reaction-hno-3- | 3 HNO3 + Al(OH3) -> Al(NO)3 + 3 H2O | start chemical_equation qc_end chemical_equation 11 11 qc_end chemical_equation 13 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the acid base neutralizations reaction"}] | [{"type":"chemical equation","value":"3 HNO3 + Al(OH3) -> Al(NO)3 + 3 H2O"}] | [{"type":"chemical equation","value":"HNO3"},{"type":"chemical equation","value":"Al(OH3)"}] | <h1 class="questionTitle" itemprop="name">How would you complete and balance the acid base neutralizations reaction: #HNO_3 + Al(OH_3) ->#?</h1> | null | 3 HNO3 + Al(OH3) -> Al(NO)3 + 3 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The 3 hydroxide ions from the aluminum hydroxide need 3 hydrogen ions to form 3 water molecules. Since the nitric acid is the only reactant that will give off hydrogen ions, there must be 3 of them. This also checks out when noting that the nitrate ion has a charge of -1, so 3 of them must join with the aluminum ion (charge of +3) when forming the product, which must be aluminum nitrate.</p></div>
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<div class="markdown"><p><mathjax>#3"HNO"_3+"Al"("OH")_3rarr"Al"("NO")_3+3"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The 3 hydroxide ions from the aluminum hydroxide need 3 hydrogen ions to form 3 water molecules. Since the nitric acid is the only reactant that will give off hydrogen ions, there must be 3 of them. This also checks out when noting that the nitrate ion has a charge of -1, so 3 of them must join with the aluminum ion (charge of +3) when forming the product, which must be aluminum nitrate.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you complete and balance the acid base neutralizations reaction: #HNO_3 + Al(OH_3) ->#?</h1>
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<div class="markdown"><p><mathjax>#3"HNO"_3+"Al"("OH")_3rarr"Al"("NO")_3+3"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The 3 hydroxide ions from the aluminum hydroxide need 3 hydrogen ions to form 3 water molecules. Since the nitric acid is the only reactant that will give off hydrogen ions, there must be 3 of them. This also checks out when noting that the nitrate ion has a charge of -1, so 3 of them must join with the aluminum ion (charge of +3) when forming the product, which must be aluminum nitrate.</p></div>
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</article> | How would you complete and balance the acid base neutralizations reaction: #HNO_3 + Al(OH_3) ->#? | null |
3,157 | a837f719-6ddd-11ea-b435-ccda262736ce | https://socratic.org/questions/in-the-reaction-fecl-2-2naoh-fe-oh-2-s-2nacl-if-6-moles-of-fecl-2-are-added-to-6 | 3 moles | start physical_unit 3 3 mole mol qc_end chemical_equation 3 11 qc_end physical_unit 3 3 13 14 mole qc_end physical_unit 6 6 13 14 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] FeCl2 [IN] moles"}] | [{"type":"physical unit","value":"3 moles"}] | [{"type":"chemical equation","value":" FeCl2 + 2 NaOH -> Fe(OH)2(s) + 2 NaCl"},{"type":"physical unit","value":"Mole [OF] FeCl2 [=] \\pu{6 moles}"},{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{6 moles}"}] | <h1 class="questionTitle" itemprop="name">In the reaction #"FeC"l_2 + 2"NaOH" -> "Fe(OH)"_2"(s)" + 2"NaC"l#, if 6 moles of #"FeC"l_2# are added to 6 moles of #"NaOH"#, how many moles of #"FeC"l_2# would be used up in the reaction?</h1> | null | 3 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For any chemical reaction, the <strong>balanced chemical equation</strong> tells you the ratio that must <strong>always</strong> exist between the reactants. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"FeCl"_ (2(aq)) + color(blue)(2)"NaOH"_ ((aq)) -> "Fe"("OH")_ (2(s)) darr + 2"NaCl"_ ((aq))#</mathjax></p>
</blockquote>
<p>The <em>stoichiometric coefficients</em> that belong to iron(II) chloride and to sodium hydroxide, respectively, tell you the <strong>mole ratio</strong> that must exist between the two reactants when this reaction takes place.</p>
<p>Notice that you have a <mathjax>#1:color(blue)(2)#</mathjax> <strong>mole ratio</strong> between the two reactants, so you can say that the reaction will <strong>always</strong> consume <em>twice as many</em> moles of sodium hydroxide than moles of iron(II) chloride.</p>
<p>Now, you know that <mathjax>#6#</mathjax> <strong>moles</strong> of iron(II) chloride are added to <mathjax>#6#</mathjax> <strong>moles</strong> of sodium hydroxide. </p>
<p>Use the aforementioned mole ratio to determine how many moles of iron(II) chloride will <strong>react</strong> with the moles of sodium hydroxide</p>
<blockquote>
<p><mathjax>#6 color(red)(cancel(color(black)("moles NaOH"))) * "1 mole FeCl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles NaOH")))) = "3 moles FeCl"_2#</mathjax></p>
</blockquote>
<p>This tells you that in order for <strong>all the moles</strong> of sodium hydroxide to react, you need <mathjax>#3#</mathjax> <strong>moles</strong> of iron(II) chloride. The other <mathjax>#3#</mathjax> <strong>moles</strong> will not take part in the reaction, i.e. they are <em>in excess</em>.</p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#"3 moles of FeCl"_2 -> "will react"#</mathjax></p>
<p><mathjax>#"3 moles of FeCl"_2 -> "will not react"#</mathjax></p>
</blockquote>
<p>Notice that sodium hydroxide <strong>is completely consumed</strong> before <em>all the moles</em> of iron(II) chloride get the chance to take part in the reaction. </p>
<p>This tells you that sodium hydroxide acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from <mathjax>#6#</mathjax> <strong>moles</strong> to <mathjax>#3#</mathjax> <strong>moles</strong>. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"3 moles FeCl"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For any chemical reaction, the <strong>balanced chemical equation</strong> tells you the ratio that must <strong>always</strong> exist between the reactants. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"FeCl"_ (2(aq)) + color(blue)(2)"NaOH"_ ((aq)) -> "Fe"("OH")_ (2(s)) darr + 2"NaCl"_ ((aq))#</mathjax></p>
</blockquote>
<p>The <em>stoichiometric coefficients</em> that belong to iron(II) chloride and to sodium hydroxide, respectively, tell you the <strong>mole ratio</strong> that must exist between the two reactants when this reaction takes place.</p>
<p>Notice that you have a <mathjax>#1:color(blue)(2)#</mathjax> <strong>mole ratio</strong> between the two reactants, so you can say that the reaction will <strong>always</strong> consume <em>twice as many</em> moles of sodium hydroxide than moles of iron(II) chloride.</p>
<p>Now, you know that <mathjax>#6#</mathjax> <strong>moles</strong> of iron(II) chloride are added to <mathjax>#6#</mathjax> <strong>moles</strong> of sodium hydroxide. </p>
<p>Use the aforementioned mole ratio to determine how many moles of iron(II) chloride will <strong>react</strong> with the moles of sodium hydroxide</p>
<blockquote>
<p><mathjax>#6 color(red)(cancel(color(black)("moles NaOH"))) * "1 mole FeCl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles NaOH")))) = "3 moles FeCl"_2#</mathjax></p>
</blockquote>
<p>This tells you that in order for <strong>all the moles</strong> of sodium hydroxide to react, you need <mathjax>#3#</mathjax> <strong>moles</strong> of iron(II) chloride. The other <mathjax>#3#</mathjax> <strong>moles</strong> will not take part in the reaction, i.e. they are <em>in excess</em>.</p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#"3 moles of FeCl"_2 -> "will react"#</mathjax></p>
<p><mathjax>#"3 moles of FeCl"_2 -> "will not react"#</mathjax></p>
</blockquote>
<p>Notice that sodium hydroxide <strong>is completely consumed</strong> before <em>all the moles</em> of iron(II) chloride get the chance to take part in the reaction. </p>
<p>This tells you that sodium hydroxide acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from <mathjax>#6#</mathjax> <strong>moles</strong> to <mathjax>#3#</mathjax> <strong>moles</strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">In the reaction #"FeC"l_2 + 2"NaOH" -> "Fe(OH)"_2"(s)" + 2"NaC"l#, if 6 moles of #"FeC"l_2# are added to 6 moles of #"NaOH"#, how many moles of #"FeC"l_2# would be used up in the reaction?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"3 moles FeCl"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For any chemical reaction, the <strong>balanced chemical equation</strong> tells you the ratio that must <strong>always</strong> exist between the reactants. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"FeCl"_ (2(aq)) + color(blue)(2)"NaOH"_ ((aq)) -> "Fe"("OH")_ (2(s)) darr + 2"NaCl"_ ((aq))#</mathjax></p>
</blockquote>
<p>The <em>stoichiometric coefficients</em> that belong to iron(II) chloride and to sodium hydroxide, respectively, tell you the <strong>mole ratio</strong> that must exist between the two reactants when this reaction takes place.</p>
<p>Notice that you have a <mathjax>#1:color(blue)(2)#</mathjax> <strong>mole ratio</strong> between the two reactants, so you can say that the reaction will <strong>always</strong> consume <em>twice as many</em> moles of sodium hydroxide than moles of iron(II) chloride.</p>
<p>Now, you know that <mathjax>#6#</mathjax> <strong>moles</strong> of iron(II) chloride are added to <mathjax>#6#</mathjax> <strong>moles</strong> of sodium hydroxide. </p>
<p>Use the aforementioned mole ratio to determine how many moles of iron(II) chloride will <strong>react</strong> with the moles of sodium hydroxide</p>
<blockquote>
<p><mathjax>#6 color(red)(cancel(color(black)("moles NaOH"))) * "1 mole FeCl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles NaOH")))) = "3 moles FeCl"_2#</mathjax></p>
</blockquote>
<p>This tells you that in order for <strong>all the moles</strong> of sodium hydroxide to react, you need <mathjax>#3#</mathjax> <strong>moles</strong> of iron(II) chloride. The other <mathjax>#3#</mathjax> <strong>moles</strong> will not take part in the reaction, i.e. they are <em>in excess</em>.</p>
<p>So, you can say that</p>
<blockquote>
<p><mathjax>#"3 moles of FeCl"_2 -> "will react"#</mathjax></p>
<p><mathjax>#"3 moles of FeCl"_2 -> "will not react"#</mathjax></p>
</blockquote>
<p>Notice that sodium hydroxide <strong>is completely consumed</strong> before <em>all the moles</em> of iron(II) chloride get the chance to take part in the reaction. </p>
<p>This tells you that sodium hydroxide acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from <mathjax>#6#</mathjax> <strong>moles</strong> to <mathjax>#3#</mathjax> <strong>moles</strong>. </p></div>
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</article> | In the reaction #"FeC"l_2 + 2"NaOH" -> "Fe(OH)"_2"(s)" + 2"NaC"l#, if 6 moles of #"FeC"l_2# are added to 6 moles of #"NaOH"#, how many moles of #"FeC"l_2# would be used up in the reaction? | null |
3,158 | aab38bba-6ddd-11ea-8cbe-ccda262736ce | https://socratic.org/questions/glycerin-c-3h-8o-3-is-a-nonvolatile-liquid-what-is-the-vapor-pressure-of-a-solut | 49.98 mmHg | start physical_unit 13 13 vapor_pressure mmhg qc_end physical_unit 23 23 21 22 volume qc_end physical_unit 13 13 25 26 temperature qc_end end | [{"type":"physical unit","value":"Vapor pressure [OF] the solution [IN] mmHg"}] | [{"type":"physical unit","value":"49.98 mmHg"}] | [{"type":"physical unit","value":"Mass [OF] C3H8O3 [=] \\pu{164 g}"},{"type":"physical unit","value":"Volume [OF] H2O [=] \\pu{338 mL}"},{"type":"physical unit","value":"Temperature [OF] the solution [=] \\pu{39.8 ℃}"}] | <h1 class="questionTitle" itemprop="name">Glycerin, #C_3H_8O_3#, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL #H_2O# at 39.8°C?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The vapor pressure of pure water at 39.8°C is 54.74 torr and its density is .992 <mathjax>#g#</mathjax><mathjax>#/#</mathjax><mathjax>#cm^3#</mathjax>.</p></div>
</h2>
</div>
</div> | 49.98 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so the vapour pressure will be somewhat reduced with respect to PURE water...we calculate the mole fractions of each component...</p>
<p><mathjax>#chi_"water"="moles of water"/"moles of water + moles of glycerol"#</mathjax></p>
<p><mathjax>#chi_"water"=((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(18.62*mol)/(18.62*mol+1.781*mol)=0.913#</mathjax></p>
<p><mathjax>#chi_"glycerol"=((164*g)/(92.09*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(1.781*mol)/(18.62*mol+1.781*mol)=0.0873#</mathjax></p>
<p>I did this the long way, but of course we know that the SUM of the mole fractions in a BINARY solution is UNITY...i.e. <mathjax>#chi_"water"+chi_"glycerol"-=1#</mathjax></p>
<p>And so <mathjax>#"vapour pressure"=chi_"water"xx54.74*mm*Hg#</mathjax></p>
<p><mathjax>#=underbrace0.913_(chi_"water")xx54.74*mm*Hg=??*mm*Hg#</mathjax></p>
<p>As is typical for aqueous solutions...<mathjax>#chi_"water"#</mathjax> DOMINATES....the molar mass of water is so low...</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The vapour pressure of the solution is proportional to <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of the volatile component...</p>
<p>Here...<mathjax>#P_"solution"=50*mm*Hg#</mathjax>...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so the vapour pressure will be somewhat reduced with respect to PURE water...we calculate the mole fractions of each component...</p>
<p><mathjax>#chi_"water"="moles of water"/"moles of water + moles of glycerol"#</mathjax></p>
<p><mathjax>#chi_"water"=((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(18.62*mol)/(18.62*mol+1.781*mol)=0.913#</mathjax></p>
<p><mathjax>#chi_"glycerol"=((164*g)/(92.09*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(1.781*mol)/(18.62*mol+1.781*mol)=0.0873#</mathjax></p>
<p>I did this the long way, but of course we know that the SUM of the mole fractions in a BINARY solution is UNITY...i.e. <mathjax>#chi_"water"+chi_"glycerol"-=1#</mathjax></p>
<p>And so <mathjax>#"vapour pressure"=chi_"water"xx54.74*mm*Hg#</mathjax></p>
<p><mathjax>#=underbrace0.913_(chi_"water")xx54.74*mm*Hg=??*mm*Hg#</mathjax></p>
<p>As is typical for aqueous solutions...<mathjax>#chi_"water"#</mathjax> DOMINATES....the molar mass of water is so low...</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Glycerin, #C_3H_8O_3#, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL #H_2O# at 39.8°C?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The vapor pressure of pure water at 39.8°C is 54.74 torr and its density is .992 <mathjax>#g#</mathjax><mathjax>#/#</mathjax><mathjax>#cm^3#</mathjax>.</p></div>
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<div class="markdown"><p>The vapour pressure of the solution is proportional to <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of the volatile component...</p>
<p>Here...<mathjax>#P_"solution"=50*mm*Hg#</mathjax>...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so the vapour pressure will be somewhat reduced with respect to PURE water...we calculate the mole fractions of each component...</p>
<p><mathjax>#chi_"water"="moles of water"/"moles of water + moles of glycerol"#</mathjax></p>
<p><mathjax>#chi_"water"=((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(18.62*mol)/(18.62*mol+1.781*mol)=0.913#</mathjax></p>
<p><mathjax>#chi_"glycerol"=((164*g)/(92.09*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(1.781*mol)/(18.62*mol+1.781*mol)=0.0873#</mathjax></p>
<p>I did this the long way, but of course we know that the SUM of the mole fractions in a BINARY solution is UNITY...i.e. <mathjax>#chi_"water"+chi_"glycerol"-=1#</mathjax></p>
<p>And so <mathjax>#"vapour pressure"=chi_"water"xx54.74*mm*Hg#</mathjax></p>
<p><mathjax>#=underbrace0.913_(chi_"water")xx54.74*mm*Hg=??*mm*Hg#</mathjax></p>
<p>As is typical for aqueous solutions...<mathjax>#chi_"water"#</mathjax> DOMINATES....the molar mass of water is so low...</p></div>
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</article> | Glycerin, #C_3H_8O_3#, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL #H_2O# at 39.8°C? |
The vapor pressure of pure water at 39.8°C is 54.74 torr and its density is .992 #g##/##cm^3#.
|
3,159 | abfc13ff-6ddd-11ea-93ed-ccda262736ce | https://socratic.org/questions/if-the-temperature-of-34-4-g-of-ethanol-increases-from-25-c-to-78-8-c-how-much-h | 4500 J | start physical_unit 7 7 heat_energy j qc_end physical_unit 7 7 4 5 mass qc_end physical_unit 7 7 10 11 temperature qc_end physical_unit 7 7 13 14 temperature qc_end physical_unit 7 7 30 33 specific_heat qc_end end | [{"type":"physical unit","value":"Absorbed heat [OF] ethanol [IN] J"}] | [{"type":"physical unit","value":"4500 J"}] | [{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{34.4 g}"},{"type":"physical unit","value":"Temperature1 [OF] ethanol [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] ethanol [=] \\pu{78.8 ℃}"},{"type":"physical unit","value":"Specific heat [OF] ethanol [=] \\pu{2.44 J/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">If the temperature of 34.4 g of ethanol increases from 25 °C to 78.8 °C, how much heat has been absorbed by the ethanol? The specific heat of ethanol is 2.44 J/(gC)?</h1> | null | 4500 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the following formula:</p>
<p><mathjax>#q=mcDeltat#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is energy in Joules, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Deltat#</mathjax> is change in temperature: <mathjax>#Deltat=(T_"final"-T_"initial")#</mathjax></p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#m="34.4 g"#</mathjax></p>
<p><mathjax>#c=("2.44 J")/("g"*"C"^@")#</mathjax></p>
<p><mathjax>#T_"initial"="25"^@"C"#</mathjax></p>
<p><mathjax>#T_"final"="78.8"^@"C"#</mathjax></p>
<p><mathjax>#Deltat=78.8^@"C"-25^@"C"="54"^@"C"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#q#</mathjax></p>
<p>Insert you data into the formula and solve.</p>
<p><mathjax>#q=(34.4color(red)cancel(color(black)("g")))xx((2.44"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))xx(54^@color(red)cancel(color(black)("C")))="4500 J"#</mathjax> (rounded to two significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The ethanol has absorbed <mathjax>#"4500 J"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the following formula:</p>
<p><mathjax>#q=mcDeltat#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is energy in Joules, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Deltat#</mathjax> is change in temperature: <mathjax>#Deltat=(T_"final"-T_"initial")#</mathjax></p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#m="34.4 g"#</mathjax></p>
<p><mathjax>#c=("2.44 J")/("g"*"C"^@")#</mathjax></p>
<p><mathjax>#T_"initial"="25"^@"C"#</mathjax></p>
<p><mathjax>#T_"final"="78.8"^@"C"#</mathjax></p>
<p><mathjax>#Deltat=78.8^@"C"-25^@"C"="54"^@"C"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#q#</mathjax></p>
<p>Insert you data into the formula and solve.</p>
<p><mathjax>#q=(34.4color(red)cancel(color(black)("g")))xx((2.44"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))xx(54^@color(red)cancel(color(black)("C")))="4500 J"#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">If the temperature of 34.4 g of ethanol increases from 25 °C to 78.8 °C, how much heat has been absorbed by the ethanol? The specific heat of ethanol is 2.44 J/(gC)?</h1>
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<div class="markdown"><p>The ethanol has absorbed <mathjax>#"4500 J"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the following formula:</p>
<p><mathjax>#q=mcDeltat#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is energy in Joules, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#Deltat#</mathjax> is change in temperature: <mathjax>#Deltat=(T_"final"-T_"initial")#</mathjax></p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#m="34.4 g"#</mathjax></p>
<p><mathjax>#c=("2.44 J")/("g"*"C"^@")#</mathjax></p>
<p><mathjax>#T_"initial"="25"^@"C"#</mathjax></p>
<p><mathjax>#T_"final"="78.8"^@"C"#</mathjax></p>
<p><mathjax>#Deltat=78.8^@"C"-25^@"C"="54"^@"C"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#q#</mathjax></p>
<p>Insert you data into the formula and solve.</p>
<p><mathjax>#q=(34.4color(red)cancel(color(black)("g")))xx((2.44"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))xx(54^@color(red)cancel(color(black)("C")))="4500 J"#</mathjax> (rounded to two significant figures)</p></div>
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</article> | If the temperature of 34.4 g of ethanol increases from 25 °C to 78.8 °C, how much heat has been absorbed by the ethanol? The specific heat of ethanol is 2.44 J/(gC)? | null |
3,160 | acb26ffa-6ddd-11ea-8213-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-chemical-equation-n-2-h-2-nh-3 | N2 + 3 H2 -> 2 NH3 | start chemical_equation qc_end chemical_equation 7 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this chemical equation"}] | [{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"}] | [{"type":"chemical equation","value":"N2 + H2 -> NH3"}] | <h1 class="questionTitle" itemprop="name">How do you balance this chemical equation: #N_2 + H_2 -> NH_3#?</h1> | null | N2 + 3 H2 -> 2 NH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since there are <mathjax>#3#</mathjax> hydrogen atoms in <mathjax>#NH_3#</mathjax> and <mathjax>#2#</mathjax> in <mathjax>#H_2#</mathjax>, we will need a multiple of <mathjax>#3H_2#</mathjax> on the left to balance the right hand side. Then notice that <mathjax>#N_2+3H_2#</mathjax> does balance <mathjax>#2NH_3#</mathjax>, so we're done.</p></div>
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<div class="markdown"><p><mathjax>#N_2+3H_2 -> 2NH_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since there are <mathjax>#3#</mathjax> hydrogen atoms in <mathjax>#NH_3#</mathjax> and <mathjax>#2#</mathjax> in <mathjax>#H_2#</mathjax>, we will need a multiple of <mathjax>#3H_2#</mathjax> on the left to balance the right hand side. Then notice that <mathjax>#N_2+3H_2#</mathjax> does balance <mathjax>#2NH_3#</mathjax>, so we're done.</p></div>
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<div class="markdown"><p><mathjax>#N_2+3H_2 -> 2NH_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since there are <mathjax>#3#</mathjax> hydrogen atoms in <mathjax>#NH_3#</mathjax> and <mathjax>#2#</mathjax> in <mathjax>#H_2#</mathjax>, we will need a multiple of <mathjax>#3H_2#</mathjax> on the left to balance the right hand side. Then notice that <mathjax>#N_2+3H_2#</mathjax> does balance <mathjax>#2NH_3#</mathjax>, so we're done.</p></div>
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</article> | How do you balance this chemical equation: #N_2 + H_2 -> NH_3#? | null |
3,161 | abcbd41e-6ddd-11ea-8dcc-ccda262736ce | https://socratic.org/questions/58e5332e7c014964302c2864 | 18.00 molar | start physical_unit 9 9 mole mol/l qc_end physical_unit 9 10 6 8 number qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] molar"}] | [{"type":"physical unit","value":"18.00 molar"}] | [{"type":"physical unit","value":"Number [OF] oxygen molecules [=] \\pu{1.084 × 10^25}"}] | <h1 class="questionTitle" itemprop="name">What molar quantity is represented by #1.084xx10^25# oxygen molecules?</h1> | null | 18.00 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I could ask the same question question using <mathjax>#"dozens"#</mathjax> in place of <mathjax>#"moles"#</mathjax>. I have <mathjax>#36#</mathjax> <mathjax>#C_6H_12O_6#</mathjax> molecules or <mathjax>#"3 dozen"#</mathjax> sugar molecules.....clearly I have</p>
<p>....<mathjax>#(6xx36*"oxygen atoms")/(12*"oxygen atoms"*"dozen"^-1)#</mathjax> </p>
<p><mathjax>#=18*"dozen"*"oxygen atoms"#</mathjax></p>
<p>And in the question, we replace <mathjax>#"dozen"#</mathjax> with <mathjax>#"mole"#</mathjax>, where <mathjax>#1*mol-=6.022xx10^23#</mathjax> individual items.....</p>
<p>What is the mass of this quantity of oxygen atoms?</p></div>
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<div class="markdown"><p>Well, 18 moles......with respect to dioxygen...</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I could ask the same question question using <mathjax>#"dozens"#</mathjax> in place of <mathjax>#"moles"#</mathjax>. I have <mathjax>#36#</mathjax> <mathjax>#C_6H_12O_6#</mathjax> molecules or <mathjax>#"3 dozen"#</mathjax> sugar molecules.....clearly I have</p>
<p>....<mathjax>#(6xx36*"oxygen atoms")/(12*"oxygen atoms"*"dozen"^-1)#</mathjax> </p>
<p><mathjax>#=18*"dozen"*"oxygen atoms"#</mathjax></p>
<p>And in the question, we replace <mathjax>#"dozen"#</mathjax> with <mathjax>#"mole"#</mathjax>, where <mathjax>#1*mol-=6.022xx10^23#</mathjax> individual items.....</p>
<p>What is the mass of this quantity of oxygen atoms?</p></div>
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<h1 class="questionTitle" itemprop="name">What molar quantity is represented by #1.084xx10^25# oxygen molecules?</h1>
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<div class="markdown"><p>Well, 18 moles......with respect to dioxygen...</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I could ask the same question question using <mathjax>#"dozens"#</mathjax> in place of <mathjax>#"moles"#</mathjax>. I have <mathjax>#36#</mathjax> <mathjax>#C_6H_12O_6#</mathjax> molecules or <mathjax>#"3 dozen"#</mathjax> sugar molecules.....clearly I have</p>
<p>....<mathjax>#(6xx36*"oxygen atoms")/(12*"oxygen atoms"*"dozen"^-1)#</mathjax> </p>
<p><mathjax>#=18*"dozen"*"oxygen atoms"#</mathjax></p>
<p>And in the question, we replace <mathjax>#"dozen"#</mathjax> with <mathjax>#"mole"#</mathjax>, where <mathjax>#1*mol-=6.022xx10^23#</mathjax> individual items.....</p>
<p>What is the mass of this quantity of oxygen atoms?</p></div>
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</article> | What molar quantity is represented by #1.084xx10^25# oxygen molecules? | null |
3,162 | aab0f3af-6ddd-11ea-847e-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-a-zinc-atom-with-an-electron-con-guration-of-ar- | +2 | start physical_unit 7 8 oxidation_number none qc_end chemical_equation 14 14 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] zinc atom"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"[Ar]3d10"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of a zinc atom with an electron configuration of #[Ar]3d^10#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The electron configuration for a neutral zinc atom is <mathjax>#"[Ar]3d"^10"4s"^2"#</mathjax>.</p>
<p>A zinc 2+ ion has the electron configuration of <mathjax>#"[Ar]3d"^10"#</mathjax>, as it has lost its two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> <mathjax>#"4s"^2"#</mathjax>. The oxidation number of an ion is the same as its charge.</p></div>
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<div class="answerSummary">
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<div class="markdown"><p>A zinc ion with the <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> <mathjax>#"[Ar]3d"^10"#</mathjax> has an oxidation number of +2.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The electron configuration for a neutral zinc atom is <mathjax>#"[Ar]3d"^10"4s"^2"#</mathjax>.</p>
<p>A zinc 2+ ion has the electron configuration of <mathjax>#"[Ar]3d"^10"#</mathjax>, as it has lost its two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> <mathjax>#"4s"^2"#</mathjax>. The oxidation number of an ion is the same as its charge.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of a zinc atom with an electron configuration of #[Ar]3d^10#?</h1>
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<div class="markdown"><p>A zinc ion with the <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> <mathjax>#"[Ar]3d"^10"#</mathjax> has an oxidation number of +2.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The electron configuration for a neutral zinc atom is <mathjax>#"[Ar]3d"^10"4s"^2"#</mathjax>.</p>
<p>A zinc 2+ ion has the electron configuration of <mathjax>#"[Ar]3d"^10"#</mathjax>, as it has lost its two <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> <mathjax>#"4s"^2"#</mathjax>. The oxidation number of an ion is the same as its charge.</p></div>
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<div class="markdown"><p>The 'oxidation number' is just the difference (+ or -) in the number of electrons in a zinc atom with its normal complement of electrons to that of the defined change.</p></div>
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<div class="markdown"><p>Oxidation states are the changes from a 'neutral' atom containing balanced protons and electrons to those states in which it either loses electrons or gains them. So, you must first write out the standard, or stable state <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> of zinc, then compare it to [Ar]3d^10. Fewer electrons will give it a positive oxidation number, and more electrons will result in a negative oxidation number.</p></div>
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</article> | What is the oxidation number of a zinc atom with an electron configuration of #[Ar]3d^10#? | null |
3,163 | ac63e23b-6ddd-11ea-a1fb-ccda262736ce | https://socratic.org/questions/58e5d403b72cff40094bd9c8 | 15.00 moles | start physical_unit 11 11 mole mol qc_end physical_unit 8 8 4 5 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] dihydrogen [IN] moles"}] | [{"type":"physical unit","value":"15.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] N2 [=] \\pu{5 mol}"}] | <h1 class="questionTitle" itemprop="name">If there is a #5*mol# quantity of #N_2#, how much dihydrogen is required to make ammonia?</h1> | null | 15.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the stoichiometric reaction:</p>
<p><mathjax>#N_2(g) + 3H_2(g) rarr 2NH_3(g)#</mathjax></p>
<p>And thus, if we have a <mathjax>#5#</mathjax> <mathjax>#mol#</mathjax> quantity of dinitrogen, the reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> INSISTS that we must have a <mathjax>#15#</mathjax> <mathjax>#mol#</mathjax> quantity of dihydrogen gas. Are you satisfied?</p></div>
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<div class="markdown"><p>To make ammonia, <mathjax>#NH_3#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assess the stoichiometric reaction:</p>
<p><mathjax>#N_2(g) + 3H_2(g) rarr 2NH_3(g)#</mathjax></p>
<p>And thus, if we have a <mathjax>#5#</mathjax> <mathjax>#mol#</mathjax> quantity of dinitrogen, the reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> INSISTS that we must have a <mathjax>#15#</mathjax> <mathjax>#mol#</mathjax> quantity of dihydrogen gas. Are you satisfied?</p></div>
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<h1 class="questionTitle" itemprop="name">If there is a #5*mol# quantity of #N_2#, how much dihydrogen is required to make ammonia?</h1>
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<div class="markdown"><p>To make ammonia, <mathjax>#NH_3#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assess the stoichiometric reaction:</p>
<p><mathjax>#N_2(g) + 3H_2(g) rarr 2NH_3(g)#</mathjax></p>
<p>And thus, if we have a <mathjax>#5#</mathjax> <mathjax>#mol#</mathjax> quantity of dinitrogen, the reaction <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> INSISTS that we must have a <mathjax>#15#</mathjax> <mathjax>#mol#</mathjax> quantity of dihydrogen gas. Are you satisfied?</p></div>
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</article> | If there is a #5*mol# quantity of #N_2#, how much dihydrogen is required to make ammonia? | null |
3,164 | ac28aa0b-6ddd-11ea-91c2-ccda262736ce | https://socratic.org/questions/exactly-5-00-l-of-air-at-50-0-c-is-warmed-to-100-0-c-what-is-the-new-volume-if-t | 8 L | start physical_unit 4 4 volume l qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 4 4 6 7 temperature qc_end physical_unit 4 4 11 12 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] air [IN] L"}] | [{"type":"physical unit","value":"8 L"}] | [{"type":"physical unit","value":"Volume1 [OF] air [=] \\pu{5.00 L}"},{"type":"physical unit","value":"Temperature1 [OF] air [=] \\pu{-50.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] air [=] \\pu{100.0 ℃}"},{"type":"other","value":"The pressure remains constant."}] | <h1 class="questionTitle" itemprop="name">Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?</h1> | null | 8 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you have asked this question before; it is still suspect because we must assume that the gas is enclosed in a container whose volume can change, i.e. a piston.</p>
<p><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that <mathjax>#"Volume "prop" Temperature"#</mathjax>, i.e. <mathjax>#VpropT#</mathjax>, or <mathjax>#V=kT#</mathjax>.</p>
<p>Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and here we must use <mathjax>#" Absolute Temperature"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#""^@C + 273.15*K#</mathjax>.</p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.00*Lxx373.15*K)/(223.15*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p>Given your reaction conditions, your volume of gas is in a piston. The gas should expand to a <mathjax>#~=8*L#</mathjax> volume.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you have asked this question before; it is still suspect because we must assume that the gas is enclosed in a container whose volume can change, i.e. a piston.</p>
<p><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that <mathjax>#"Volume "prop" Temperature"#</mathjax>, i.e. <mathjax>#VpropT#</mathjax>, or <mathjax>#V=kT#</mathjax>.</p>
<p>Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and here we must use <mathjax>#" Absolute Temperature"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#""^@C + 273.15*K#</mathjax>.</p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.00*Lxx373.15*K)/(223.15*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?</h1>
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<div class="markdown"><p>Given your reaction conditions, your volume of gas is in a piston. The gas should expand to a <mathjax>#~=8*L#</mathjax> volume.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you have asked this question before; it is still suspect because we must assume that the gas is enclosed in a container whose volume can change, i.e. a piston.</p>
<p><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that <mathjax>#"Volume "prop" Temperature"#</mathjax>, i.e. <mathjax>#VpropT#</mathjax>, or <mathjax>#V=kT#</mathjax>.</p>
<p>Solving for <mathjax>#k#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, and here we must use <mathjax>#" Absolute Temperature"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#""^@C + 273.15*K#</mathjax>.</p>
<p><mathjax>#V_2=(V_1T_2)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.00*Lxx373.15*K)/(223.15*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</article> | Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant? | null |
3,165 | ab021736-6ddd-11ea-b982-ccda262736ce | https://socratic.org/questions/57a099ad11ef6b5baa1dd7f7 | BaCl2(aq) + H2SO4(aq) -> BaSO4(s) + 2 HCl(aq) | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"BaCl2(aq) + H2SO4(aq) -> BaSO4(s) + 2 HCl(aq)"}] | [{"type":"substance name","value":"Barium chloride"},{"type":"substance name","value":"Sulfuric acid"}] | <h1 class="questionTitle" itemprop="name">How does barium chloride react with sulfuric acid?</h1> | null | BaCl2(aq) + H2SO4(aq) -> BaSO4(s) + 2 HCl(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is the equation stoichiometrically balanced? Note that barium sulfate is quite insoluble and will precipitate from aqueous solution. </p></div>
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<div class="markdown"><p><mathjax>#BaCl_2(aq) + H_2SO_4(aq) rarr BaSO_4(s)darr + 2HCl(aq)#</mathjax></p></div>
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<div class="markdown"><p>Is the equation stoichiometrically balanced? Note that barium sulfate is quite insoluble and will precipitate from aqueous solution. </p></div>
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<div class="markdown"><p><mathjax>#BaCl_2(aq) + H_2SO_4(aq) rarr BaSO_4(s)darr + 2HCl(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is the equation stoichiometrically balanced? Note that barium sulfate is quite insoluble and will precipitate from aqueous solution. </p></div>
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</article> | How does barium chloride react with sulfuric acid? | null |
3,166 | aaec4530-6ddd-11ea-b2cb-ccda262736ce | https://socratic.org/questions/595fedd511ef6b614dd0404d | 20.00 g | start physical_unit 10 11 mass g qc_end physical_unit 10 11 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium hydroxide [IN] g"}] | [{"type":"physical unit","value":"20.00 g"}] | [{"type":"physical unit","value":"Mole [OF] sodium hydroxide [=] \\pu{1/2 mol}"}] | <h1 class="questionTitle" itemprop="name">What mass is associated with a #1/2*mol# quantity of sodium hydroxide?</h1> | null | 20.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus if we have a <mathjax>#1/2*mol#</mathjax> of <mathjax>#"caustic soda"#</mathjax>, we have a mass of..........</p>
<p><mathjax>#1/2*cancel(mol)xx40*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Note the DIMENSIONAL consistency of the answer. We want an answer with units of mass, and the product gives us an answer with units of mass. This is a check on my calculations, and built in detection of possible error on my part in framing the problem......</p></div>
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<div class="markdown"><p>Well........<mathjax>#"sodium hydroxide"#</mathjax> has a molar mass of <mathjax>#40.0*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus if we have a <mathjax>#1/2*mol#</mathjax> of <mathjax>#"caustic soda"#</mathjax>, we have a mass of..........</p>
<p><mathjax>#1/2*cancel(mol)xx40*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Note the DIMENSIONAL consistency of the answer. We want an answer with units of mass, and the product gives us an answer with units of mass. This is a check on my calculations, and built in detection of possible error on my part in framing the problem......</p></div>
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<div class="markdown"><p>Well........<mathjax>#"sodium hydroxide"#</mathjax> has a molar mass of <mathjax>#40.0*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus if we have a <mathjax>#1/2*mol#</mathjax> of <mathjax>#"caustic soda"#</mathjax>, we have a mass of..........</p>
<p><mathjax>#1/2*cancel(mol)xx40*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Note the DIMENSIONAL consistency of the answer. We want an answer with units of mass, and the product gives us an answer with units of mass. This is a check on my calculations, and built in detection of possible error on my part in framing the problem......</p></div>
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</article> | What mass is associated with a #1/2*mol# quantity of sodium hydroxide? | null |
3,167 | a8ebbed4-6ddd-11ea-a2c7-ccda262736ce | https://socratic.org/questions/magnesium-chloride-has-an-atomic-mass-of-94-amu-what-is-the-mass-of-one-mole-of- | 94 g | start physical_unit 0 1 mass g qc_end physical_unit 0 1 7 8 atomic_mass qc_end physical_unit 0 1 14 15 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] magnesium chloride [IN] g"}] | [{"type":"physical unit","value":"94 g"}] | [{"type":"physical unit","value":"Atomic mass [OF] magnesium chloride [=] \\pu{94 amu}"},{"type":"physical unit","value":"Mole [OF] magnesium chloride [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">Magnesium chloride has an atomic mass of 94 amu. What is the mass of one mole of magnesium chloride? </h1> | null | 94 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#1#</mathjax> particle of <mathjax>#MgCl_2#</mathjax> has a mass of <mathjax>#"94 amu"#</mathjax>, then <mathjax>#1#</mathjax> mole of <mathjax>#MgCl_2#</mathjax> will have a mass of <mathjax>#"94 g"#</mathjax>. </p>
<p>It might seem weirdly convenient that these are the same value at first, but <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> was actually defined for this exact purpose!—to easily relate things from the macroscopic and atomic worlds. </p>
<p>One mole is defined as the number of atoms in <mathjax>#"12 g"#</mathjax> of <mathjax>#"carbon-12"#</mathjax>, one atom of which has a mass of <mathjax>#"12 amu"#</mathjax>. </p>
<p><mathjax>#"1 mole of atoms" = "12 g"/"12 amu"#</mathjax></p>
<p>This definition effectively establishes this ratio:</p>
<p><mathjax>#"amu"/"atom" = "grams"/"mole"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"94 g"#</mathjax>!</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#1#</mathjax> particle of <mathjax>#MgCl_2#</mathjax> has a mass of <mathjax>#"94 amu"#</mathjax>, then <mathjax>#1#</mathjax> mole of <mathjax>#MgCl_2#</mathjax> will have a mass of <mathjax>#"94 g"#</mathjax>. </p>
<p>It might seem weirdly convenient that these are the same value at first, but <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> was actually defined for this exact purpose!—to easily relate things from the macroscopic and atomic worlds. </p>
<p>One mole is defined as the number of atoms in <mathjax>#"12 g"#</mathjax> of <mathjax>#"carbon-12"#</mathjax>, one atom of which has a mass of <mathjax>#"12 amu"#</mathjax>. </p>
<p><mathjax>#"1 mole of atoms" = "12 g"/"12 amu"#</mathjax></p>
<p>This definition effectively establishes this ratio:</p>
<p><mathjax>#"amu"/"atom" = "grams"/"mole"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Magnesium chloride has an atomic mass of 94 amu. What is the mass of one mole of magnesium chloride? </h1>
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<div class="markdown"><p><mathjax>#"94 g"#</mathjax>!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#1#</mathjax> particle of <mathjax>#MgCl_2#</mathjax> has a mass of <mathjax>#"94 amu"#</mathjax>, then <mathjax>#1#</mathjax> mole of <mathjax>#MgCl_2#</mathjax> will have a mass of <mathjax>#"94 g"#</mathjax>. </p>
<p>It might seem weirdly convenient that these are the same value at first, but <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> was actually defined for this exact purpose!—to easily relate things from the macroscopic and atomic worlds. </p>
<p>One mole is defined as the number of atoms in <mathjax>#"12 g"#</mathjax> of <mathjax>#"carbon-12"#</mathjax>, one atom of which has a mass of <mathjax>#"12 amu"#</mathjax>. </p>
<p><mathjax>#"1 mole of atoms" = "12 g"/"12 amu"#</mathjax></p>
<p>This definition effectively establishes this ratio:</p>
<p><mathjax>#"amu"/"atom" = "grams"/"mole"#</mathjax></p></div>
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</article> | Magnesium chloride has an atomic mass of 94 amu. What is the mass of one mole of magnesium chloride? | null |
3,168 | ab5710ee-6ddd-11ea-b5a5-ccda262736ce | https://socratic.org/questions/a-pharmacist-wishes-to-mix-a-solution-that-is-4-minoxidil-she-has-on-hand-60-ml- | 30.00 mL | start physical_unit 6 6 volume ml qc_end physical_unit 10 10 19 19 percent qc_end physical_unit 10 10 26 26 percent qc_end physical_unit 10 10 9 9 percent qc_end physical_unit 6 6 15 16 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] solution [IN] mL"}] | [{"type":"physical unit","value":"30.00 mL"}] | [{"type":"physical unit","value":"Percent1 [OF] minoxidil in solution [=] \\pu{3%}"},{"type":"physical unit","value":"Percent2 [OF] minoxidil in solution [=] \\pu{6%}"},{"type":"physical unit","value":"Percent3 [OF] minoxidil in solution [=] \\pu{4%}"},{"type":"physical unit","value":"Volume1 [OF] solution [=] \\pu{60 ml}"}] | <h1 class="questionTitle" itemprop="name">A pharmacist wishes to mix a solution that is 4% Minoxidil. She has on hand 60 ml of a 3% solution and wishes to add some 6% solution to obtain the desired 4% solution. How much 6% solution should she add?</h1> | null | 30.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can express concentration as amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p>So the amount of substance <mathjax>#n#</mathjax> is given by:</p>
<p><mathjax>#n=cxxv#</mathjax></p>
<p>In this problem the total amount of <mathjax>#"Minoxidil"^("tm")#</mathjax> is the same before and after mixing, so we can set up this equation where <mathjax>#V#</mathjax> is the volume of the 6% solution:</p>
<p><mathjax>#(60+V)4=(60xx3)+6V#</mathjax></p>
<p><mathjax>#:.240+4V=180+6V#</mathjax></p>
<p><mathjax>#:.6V-4V=240-180#</mathjax></p>
<p><mathjax>#:.2V=60#</mathjax></p>
<p><mathjax>#V=60/2=30"ml"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>She should add 30ml.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can express concentration as amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p>So the amount of substance <mathjax>#n#</mathjax> is given by:</p>
<p><mathjax>#n=cxxv#</mathjax></p>
<p>In this problem the total amount of <mathjax>#"Minoxidil"^("tm")#</mathjax> is the same before and after mixing, so we can set up this equation where <mathjax>#V#</mathjax> is the volume of the 6% solution:</p>
<p><mathjax>#(60+V)4=(60xx3)+6V#</mathjax></p>
<p><mathjax>#:.240+4V=180+6V#</mathjax></p>
<p><mathjax>#:.6V-4V=240-180#</mathjax></p>
<p><mathjax>#:.2V=60#</mathjax></p>
<p><mathjax>#V=60/2=30"ml"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A pharmacist wishes to mix a solution that is 4% Minoxidil. She has on hand 60 ml of a 3% solution and wishes to add some 6% solution to obtain the desired 4% solution. How much 6% solution should she add?</h1>
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<div class="markdown"><p>She should add 30ml.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can express concentration as amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p>So the amount of substance <mathjax>#n#</mathjax> is given by:</p>
<p><mathjax>#n=cxxv#</mathjax></p>
<p>In this problem the total amount of <mathjax>#"Minoxidil"^("tm")#</mathjax> is the same before and after mixing, so we can set up this equation where <mathjax>#V#</mathjax> is the volume of the 6% solution:</p>
<p><mathjax>#(60+V)4=(60xx3)+6V#</mathjax></p>
<p><mathjax>#:.240+4V=180+6V#</mathjax></p>
<p><mathjax>#:.6V-4V=240-180#</mathjax></p>
<p><mathjax>#:.2V=60#</mathjax></p>
<p><mathjax>#V=60/2=30"ml"#</mathjax></p></div>
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</article> | A pharmacist wishes to mix a solution that is 4% Minoxidil. She has on hand 60 ml of a 3% solution and wishes to add some 6% solution to obtain the desired 4% solution. How much 6% solution should she add? | null |
3,169 | aa15b1a7-6ddd-11ea-b9c1-ccda262736ce | https://socratic.org/questions/at-7-0-c-the-volume-of-a-gas-is-49-ml-at-the-same-pressure-its-volume-is-74-ml-a | 422.86 K | start physical_unit 7 7 temperature k qc_end physical_unit 7 7 1 2 temperature qc_end physical_unit 7 7 9 10 volume qc_end physical_unit 7 7 18 19 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"422.86 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{7.0 ℃}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{49 mL}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{74 mL}"},{"type":"other","value":"At the same pressure."}] | <h1 class="questionTitle" itemprop="name">At 7.0°C, the volume of a gas is 49 mL. At the same pressure, its volume is 74 mL at what temperature?</h1> | null | 422.86 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chalres' law states that at constant pressure, <mathjax>#VpropT#</mathjax>, i.e. <mathjax>#V_1/T_1=k#</mathjax> or <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, units of absolute temperature are (reasonably) used, and we could use any units of volume (<mathjax>#"pints, bushels, pecks?"#</mathjax>), because the volumes cancel out.</p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(280*Kxx74*mL)/(49*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax> <mathjax>#=??""^@C#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use <a href="https://en.wikipedia.org/wiki/Charles%27s_law" rel="nofollow">Charles' Law</a> to give <mathjax>#T_2~=423*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chalres' law states that at constant pressure, <mathjax>#VpropT#</mathjax>, i.e. <mathjax>#V_1/T_1=k#</mathjax> or <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, units of absolute temperature are (reasonably) used, and we could use any units of volume (<mathjax>#"pints, bushels, pecks?"#</mathjax>), because the volumes cancel out.</p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(280*Kxx74*mL)/(49*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax> <mathjax>#=??""^@C#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">At 7.0°C, the volume of a gas is 49 mL. At the same pressure, its volume is 74 mL at what temperature?</h1>
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<span class="dateCreated" datetime="2016-06-07T04:33:42" itemprop="dateCreated">
Jun 7, 2016
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<div class="markdown"><p>We use <a href="https://en.wikipedia.org/wiki/Charles%27s_law" rel="nofollow">Charles' Law</a> to give <mathjax>#T_2~=423*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chalres' law states that at constant pressure, <mathjax>#VpropT#</mathjax>, i.e. <mathjax>#V_1/T_1=k#</mathjax> or <mathjax>#V_1/T_1=V_2/T_2#</mathjax>, units of absolute temperature are (reasonably) used, and we could use any units of volume (<mathjax>#"pints, bushels, pecks?"#</mathjax>), because the volumes cancel out.</p>
<p><mathjax>#T_2=(T_1V_2)/V_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(280*Kxx74*mL)/(49*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??K#</mathjax> <mathjax>#=??""^@C#</mathjax></p></div>
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</article> | At 7.0°C, the volume of a gas is 49 mL. At the same pressure, its volume is 74 mL at what temperature? | null |
3,170 | aa951d1c-6ddd-11ea-b229-ccda262736ce | https://socratic.org/questions/you-have-1-25-l-of-oxygen-gas-at-stp-what-is-its-mass | 1.78 g | start physical_unit 5 6 mass g qc_end physical_unit 5 6 2 3 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen gas [IN] g"}] | [{"type":"physical unit","value":"1.78 g"}] | [{"type":"physical unit","value":"Volume [OF] oxygen gas [=] \\pu{1.25 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">You have 1.25 L of oxygen gas at STP. What is its mass? </h1> | null | 1.78 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP and that means we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>!<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx1.25cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 0.05577 mol#</mathjax></p>
<p>To get the mass of <mathjax>#O_2#</mathjax>, we just have to use the molar mass of oxygen as a conversion factor:</p>
<p><mathjax>#0.0577cancel"mol" O_2 xx (32.00g)/(1cancel "mol")#</mathjax> = 1.78g <mathjax>#O_2#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#O_2#</mathjax> has a mass of 1.78g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP and that means we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>!<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx1.25cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 0.05577 mol#</mathjax></p>
<p>To get the mass of <mathjax>#O_2#</mathjax>, we just have to use the molar mass of oxygen as a conversion factor:</p>
<p><mathjax>#0.0577cancel"mol" O_2 xx (32.00g)/(1cancel "mol")#</mathjax> = 1.78g <mathjax>#O_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">You have 1.25 L of oxygen gas at STP. What is its mass? </h1>
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Jun 19, 2016
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<div class="markdown"><p><mathjax>#O_2#</mathjax> has a mass of 1.78g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP and that means we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>!<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx1.25cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 0.05577 mol#</mathjax></p>
<p>To get the mass of <mathjax>#O_2#</mathjax>, we just have to use the molar mass of oxygen as a conversion factor:</p>
<p><mathjax>#0.0577cancel"mol" O_2 xx (32.00g)/(1cancel "mol")#</mathjax> = 1.78g <mathjax>#O_2#</mathjax></p></div>
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</article> | You have 1.25 L of oxygen gas at STP. What is its mass? | null |
3,171 | acf55fca-6ddd-11ea-96b3-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-to-show-the-combustion-reaction-of-ethanol | C2H6O(l) + 3 O2(g) -> 2 CO2(g) + 3 H2O(l) | start chemical_equation qc_end substance 11 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the combustion reaction"}] | [{"type":"chemical equation","value":"C2H6O(l) + 3 O2(g) -> 2 CO2(g) + 3 H2O(l)"}] | [{"type":"substance name","value":"Ethanol"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation to show the combustion reaction of ethanol?</h1> | null | C2H6O(l) + 3 O2(g) -> 2 CO2(g) + 3 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2("g") + "H"_2"O(l")#</mathjax></p>
<p><strong>Balance the Equation:</strong></p>
<p>Balance the <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> first. There are <mathjax>#2#</mathjax> carbon atoms on the left side, and <mathjax>#1#</mathjax> carbon atom on the right side. There are <mathjax>#6#</mathjax> hydrogen atoms on the left side and <mathjax>#2#</mathjax> hydrogen atoms on the right side.</p>
<p>Add a coefficient of <mathjax>#color(red)2#</mathjax> in front of the carbon dioxide, and a coefficient of <mathjax>#color(purple)3#</mathjax> in front of the water.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")"#</mathjax></p>
<p>We now have equal numbers of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> atoms on both sides. However, the <mathjax>#"O"#</mathjax> is not balanced.</p>
<p>We have <mathjax>#7#</mathjax> <mathjax>#"O"#</mathjax> atoms on the right side, and <mathjax>#3#</mathjax> on the left side. Place a coefficient of <mathjax>#color(orange)3#</mathjax> in front of the <mathjax>#"O"_2#</mathjax> on the left side.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")#</mathjax></p>
<p>We now have a balanced equation, with <mathjax>#2#</mathjax> carbon atoms on both sides, <mathjax>#6#</mathjax> hydrogen atoms on both sides, and <mathjax>#7#</mathjax> oxygen atoms on both sides.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2("g") + "H"_2"O(l")#</mathjax></p>
<p><strong>Balance the Equation:</strong></p>
<p>Balance the <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> first. There are <mathjax>#2#</mathjax> carbon atoms on the left side, and <mathjax>#1#</mathjax> carbon atom on the right side. There are <mathjax>#6#</mathjax> hydrogen atoms on the left side and <mathjax>#2#</mathjax> hydrogen atoms on the right side.</p>
<p>Add a coefficient of <mathjax>#color(red)2#</mathjax> in front of the carbon dioxide, and a coefficient of <mathjax>#color(purple)3#</mathjax> in front of the water.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")"#</mathjax></p>
<p>We now have equal numbers of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> atoms on both sides. However, the <mathjax>#"O"#</mathjax> is not balanced.</p>
<p>We have <mathjax>#7#</mathjax> <mathjax>#"O"#</mathjax> atoms on the right side, and <mathjax>#3#</mathjax> on the left side. Place a coefficient of <mathjax>#color(orange)3#</mathjax> in front of the <mathjax>#"O"_2#</mathjax> on the left side.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")#</mathjax></p>
<p>We now have a balanced equation, with <mathjax>#2#</mathjax> carbon atoms on both sides, <mathjax>#6#</mathjax> hydrogen atoms on both sides, and <mathjax>#7#</mathjax> oxygen atoms on both sides.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation to show the combustion reaction of ethanol?</h1>
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Nov 2, 2015
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<div class="markdown"><p><mathjax>#"C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2("g") + "H"_2"O(l")#</mathjax></p>
<p><strong>Balance the Equation:</strong></p>
<p>Balance the <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> first. There are <mathjax>#2#</mathjax> carbon atoms on the left side, and <mathjax>#1#</mathjax> carbon atom on the right side. There are <mathjax>#6#</mathjax> hydrogen atoms on the left side and <mathjax>#2#</mathjax> hydrogen atoms on the right side.</p>
<p>Add a coefficient of <mathjax>#color(red)2#</mathjax> in front of the carbon dioxide, and a coefficient of <mathjax>#color(purple)3#</mathjax> in front of the water.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + "O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")"#</mathjax></p>
<p>We now have equal numbers of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> atoms on both sides. However, the <mathjax>#"O"#</mathjax> is not balanced.</p>
<p>We have <mathjax>#7#</mathjax> <mathjax>#"O"#</mathjax> atoms on the right side, and <mathjax>#3#</mathjax> on the left side. Place a coefficient of <mathjax>#color(orange)3#</mathjax> in front of the <mathjax>#"O"_2#</mathjax> on the left side.</p>
<p><mathjax>#"C"_2"H"_6"O(l)" + color(orange)3"O"_2"(g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)2"CO"_2("g") + color(purple)3"H"_2"O(l")#</mathjax></p>
<p>We now have a balanced equation, with <mathjax>#2#</mathjax> carbon atoms on both sides, <mathjax>#6#</mathjax> hydrogen atoms on both sides, and <mathjax>#7#</mathjax> oxygen atoms on both sides.</p></div>
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</article> | What is the chemical equation to show the combustion reaction of ethanol? | null |
3,172 | aa15b1a8-6ddd-11ea-b714-ccda262736ce | https://socratic.org/questions/how-many-liters-of-4-00-m-solution-can-be-made-using-exactly-100-0-grams-of-lith-1 | 0.29 liters | start physical_unit 6 6 volume l qc_end physical_unit 15 16 4 5 molarity qc_end physical_unit 15 16 12 13 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] lithum bromide solution [IN] liters"}] | [{"type":"physical unit","value":"0.29 liters"}] | [{"type":"physical unit","value":"Molarity [OF] lithum bromide solution [=] \\pu{4.00 M}"},{"type":"physical unit","value":"Mass [OF] lithum bromide [=] \\pu{100.0 grams}"}] | <h1 class="questionTitle" itemprop="name">How many liters of 4.00 M solution can be made using exactly 100.0 grams of lithum bromide?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>the answer is given i just don'y know how to do the work to get the answer?</p></div>
</h2>
</div>
</div> | 0.29 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to convert the mass of lithium bromide to <em>moles</em> by using the compound's <strong>molar mass</strong>. </p>
<blockquote>
<p><mathjax>#100.0 color(red)(cancel(color(black)("g"))) * "1 mole LiBr"/(86.845color(red)(cancel(color(black)("g")))) = "1.1515 moles LiBr"#</mathjax></p>
</blockquote>
<p>Now, the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution is simply a measure of the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is lithium bromide, present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In order to have a <mathjax>#"4.00-M"#</mathjax> solution of lithium bromide, you need to have <mathjax>#4.00#</mathjax> <strong>moles</strong> of lithium bromide <strong>for every</strong> <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>You know that your sample contains <mathjax>#1.1515#</mathjax> <strong>moles</strong> of lithium bromide, so you can use the molarity of the solution as a <em>conversion factor</em> to determine how many liters of this solution can be made.</p>
<blockquote>
<p><mathjax>#1.1515 color(red)(cancel(color(black)("moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution")))#</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>#"0.288 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to convert the mass of lithium bromide to <em>moles</em> by using the compound's <strong>molar mass</strong>. </p>
<blockquote>
<p><mathjax>#100.0 color(red)(cancel(color(black)("g"))) * "1 mole LiBr"/(86.845color(red)(cancel(color(black)("g")))) = "1.1515 moles LiBr"#</mathjax></p>
</blockquote>
<p>Now, the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution is simply a measure of the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is lithium bromide, present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In order to have a <mathjax>#"4.00-M"#</mathjax> solution of lithium bromide, you need to have <mathjax>#4.00#</mathjax> <strong>moles</strong> of lithium bromide <strong>for every</strong> <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>You know that your sample contains <mathjax>#1.1515#</mathjax> <strong>moles</strong> of lithium bromide, so you can use the molarity of the solution as a <em>conversion factor</em> to determine how many liters of this solution can be made.</p>
<blockquote>
<p><mathjax>#1.1515 color(red)(cancel(color(black)("moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution")))#</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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<h1 class="questionTitle" itemprop="name">How many liters of 4.00 M solution can be made using exactly 100.0 grams of lithum bromide?</h1>
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<div class="markdown"><p>the answer is given i just don'y know how to do the work to get the answer?</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-04-17T23:49:32" itemprop="dateCreated">
Apr 17, 2018
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<div class="markdown"><p><mathjax>#"0.288 L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to convert the mass of lithium bromide to <em>moles</em> by using the compound's <strong>molar mass</strong>. </p>
<blockquote>
<p><mathjax>#100.0 color(red)(cancel(color(black)("g"))) * "1 mole LiBr"/(86.845color(red)(cancel(color(black)("g")))) = "1.1515 moles LiBr"#</mathjax></p>
</blockquote>
<p>Now, the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution is simply a measure of the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is lithium bromide, present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In order to have a <mathjax>#"4.00-M"#</mathjax> solution of lithium bromide, you need to have <mathjax>#4.00#</mathjax> <strong>moles</strong> of lithium bromide <strong>for every</strong> <mathjax>#"1.00 L"#</mathjax> of this solution. </p>
<p>You know that your sample contains <mathjax>#1.1515#</mathjax> <strong>moles</strong> of lithium bromide, so you can use the molarity of the solution as a <em>conversion factor</em> to determine how many liters of this solution can be made.</p>
<blockquote>
<p><mathjax>#1.1515 color(red)(cancel(color(black)("moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution")))#</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 liters of 4.00 M solution can be made using exactly 100.0 grams of lithum bromide? |
the answer is given i just don'y know how to do the work to get the answer?
|
3,173 | ad2792a9-6ddd-11ea-9dfb-ccda262736ce | https://socratic.org/questions/a-vessel-contains-a-mixture-of-nitrogen-and-oxygen-if-the-partial-pressure-of-ni | 62.80 atm | start physical_unit 33 34 total_pressure atm qc_end physical_unit 6 6 16 17 partial_pressure qc_end physical_unit 8 8 25 26 partial_pressure qc_end end | [{"type":"physical unit","value":"Total pressure [OF] the vessel [IN] atm"}] | [{"type":"physical unit","value":"62.80 atm"}] | [{"type":"physical unit","value":"Partial pressure [OF] nitrogen [=] \\pu{20 atmospheres}"},{"type":"physical unit","value":"Partial pressure [OF] oxygen [=] \\pu{42.8 atmospheres}"}] | <h1 class="questionTitle" itemprop="name">A vessel contains a mixture of nitrogen and oxygen. If the partial pressure of nitrogen is 20 atmospheres and the partial pressure of oxygen is 42.8 atmospheres, what is the total pressure inside the vessel?</h1> | null | 62.80 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Datlon's law of Partial Pressures states that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the SUM of the partial pressures.</p>
<p>For your problem <mathjax>#P_"Total"=P_"dinitrogen"+P_"dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(20+42.8)*atm#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_"Total"=(20+42.8)*atm=62.8*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Datlon's law of Partial Pressures states that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the SUM of the partial pressures.</p>
<p>For your problem <mathjax>#P_"Total"=P_"dinitrogen"+P_"dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(20+42.8)*atm#</mathjax>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A vessel contains a mixture of nitrogen and oxygen. If the partial pressure of nitrogen is 20 atmospheres and the partial pressure of oxygen is 42.8 atmospheres, what is the total pressure inside the vessel?</h1>
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anor277
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<div class="markdown"><p><mathjax>#P_"Total"=(20+42.8)*atm=62.8*atm#</mathjax></p></div>
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<div class="markdown"><p>Datlon's law of Partial Pressures states that in a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the SUM of the partial pressures.</p>
<p>For your problem <mathjax>#P_"Total"=P_"dinitrogen"+P_"dioxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(20+42.8)*atm#</mathjax>.</p></div>
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</article> | A vessel contains a mixture of nitrogen and oxygen. If the partial pressure of nitrogen is 20 atmospheres and the partial pressure of oxygen is 42.8 atmospheres, what is the total pressure inside the vessel? | null |
3,174 | ab04da9a-6ddd-11ea-9a05-ccda262736ce | https://socratic.org/questions/how-many-moles-of-gas-occupy-98-l-at-a-pressure-of-2-8-atmospheres-and-a-tempera | 11.45 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 6 7 volume qc_end physical_unit 4 4 12 13 pressure qc_end physical_unit 4 4 18 19 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] moles"}] | [{"type":"physical unit","value":"11.45 moles"}] | [{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{98 L}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{2.8 atmospheres }"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{292 K}"}] | <h1 class="questionTitle" itemprop="name">How many moles of gas occupy 98 L at a pressure of 2.8 atmospheres and a temperature of 292°K?</h1> | null | 11.45 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve for <mathjax>#n#</mathjax> in the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(2.8*atmxx98*L)/(0.0821*L*atm*K^-1*mol^-1xx292*K)#</mathjax></p>
<p><mathjax>#~=11*mol#</mathjax>...</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#n_"gas"~=11*mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve for <mathjax>#n#</mathjax> in the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(2.8*atmxx98*L)/(0.0821*L*atm*K^-1*mol^-1xx292*K)#</mathjax></p>
<p><mathjax>#~=11*mol#</mathjax>...</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of gas occupy 98 L at a pressure of 2.8 atmospheres and a temperature of 292°K?</h1>
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<div class="markdown"><p><mathjax>#n_"gas"~=11*mol#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve for <mathjax>#n#</mathjax> in the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(2.8*atmxx98*L)/(0.0821*L*atm*K^-1*mol^-1xx292*K)#</mathjax></p>
<p><mathjax>#~=11*mol#</mathjax>...</p></div>
</div>
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</article> | How many moles of gas occupy 98 L at a pressure of 2.8 atmospheres and a temperature of 292°K? | null |
3,175 | ac81c96c-6ddd-11ea-8b8a-ccda262736ce | https://socratic.org/questions/how-liters-of-1-5-m-hcl-solution-would-react-completely-with-25-moles-ca-oh-2 | 33.33 L | start physical_unit 5 6 volume l qc_end physical_unit 5 6 3 4 molarity qc_end c_other OTHER qc_end physical_unit 13 13 11 12 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] HCl solution [IN] L"}] | [{"type":"physical unit","value":"33.33 L"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{1.5 M}"},{"type":"other","value":"React completely."},{"type":"physical unit","value":"Mole [OF] Ca(OH)2 [=] \\pu{25 moles}"}] | <h1 class="questionTitle" itemprop="name">How liters of 1.5 M #HCl# solution would react completely with 25 moles #Ca(OH)_2#?</h1> | null | 33.33 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write the balanced equation:</p>
<blockquote>
<p><mathjax>#Ca(OH)_2 (s) + 2HCl (aq) -> CaCl_2 (aq) + 2H_2 O (l)#</mathjax></p>
</blockquote>
<p>From this you can see that each mole of calcium hydroxide reacts with 2 moles of hydrochloric acid. </p>
<p>You have 25 moles of calcium hydroxide, so it means that 50 moles of hydrochloric acid is required.</p>
<p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid is 1.5 M, and </p>
<blockquote>
<p><mathjax>#"number of litres" = "number of moles"/"molarity"#</mathjax></p>
</blockquote>
<p>This is </p>
<blockquote>
<p><mathjax>#"50 moles"/"1.5 mol/L" = "33.3 litres"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Just over 33 litres</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write the balanced equation:</p>
<blockquote>
<p><mathjax>#Ca(OH)_2 (s) + 2HCl (aq) -> CaCl_2 (aq) + 2H_2 O (l)#</mathjax></p>
</blockquote>
<p>From this you can see that each mole of calcium hydroxide reacts with 2 moles of hydrochloric acid. </p>
<p>You have 25 moles of calcium hydroxide, so it means that 50 moles of hydrochloric acid is required.</p>
<p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid is 1.5 M, and </p>
<blockquote>
<p><mathjax>#"number of litres" = "number of moles"/"molarity"#</mathjax></p>
</blockquote>
<p>This is </p>
<blockquote>
<p><mathjax>#"50 moles"/"1.5 mol/L" = "33.3 litres"#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How liters of 1.5 M #HCl# solution would react completely with 25 moles #Ca(OH)_2#?</h1>
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<div class="markdown"><p>Just over 33 litres</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First write the balanced equation:</p>
<blockquote>
<p><mathjax>#Ca(OH)_2 (s) + 2HCl (aq) -> CaCl_2 (aq) + 2H_2 O (l)#</mathjax></p>
</blockquote>
<p>From this you can see that each mole of calcium hydroxide reacts with 2 moles of hydrochloric acid. </p>
<p>You have 25 moles of calcium hydroxide, so it means that 50 moles of hydrochloric acid is required.</p>
<p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid is 1.5 M, and </p>
<blockquote>
<p><mathjax>#"number of litres" = "number of moles"/"molarity"#</mathjax></p>
</blockquote>
<p>This is </p>
<blockquote>
<p><mathjax>#"50 moles"/"1.5 mol/L" = "33.3 litres"#</mathjax></p>
</blockquote></div>
</div>
</div>
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</article> | How liters of 1.5 M #HCl# solution would react completely with 25 moles #Ca(OH)_2#? | null |
3,176 | ac8b8e7a-6ddd-11ea-945d-ccda262736ce | https://socratic.org/questions/5947a0d8b72cff162ba58670 | 6.02 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 10 11 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"6.02 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] sulfuric acid [=] \\pu{10 mol}"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms in a #10*mol# quantity of sulfuric acid?</h1> | null | 6.02 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of sulfuric acid there are <mathjax>#4xxN_A#</mathjax>, <mathjax>#4xx6.022xx10^23#</mathjax> individual oxygen atoms.........</p>
<p>And thus in <mathjax>#10*mol#</mathjax> of same stuff, there are <mathjax>#4xx10xxN_A~=#</mathjax></p>
<p><mathjax>#24xx10^24#</mathjax> individual oxygen atoms........ </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, I count <mathjax>#40xxN_A#</mathjax> individual oxygen atoms.......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of sulfuric acid there are <mathjax>#4xxN_A#</mathjax>, <mathjax>#4xx6.022xx10^23#</mathjax> individual oxygen atoms.........</p>
<p>And thus in <mathjax>#10*mol#</mathjax> of same stuff, there are <mathjax>#4xx10xxN_A~=#</mathjax></p>
<p><mathjax>#24xx10^24#</mathjax> individual oxygen atoms........ </p></div>
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<h1 class="questionTitle" itemprop="name">How many oxygen atoms in a #10*mol# quantity of sulfuric acid?</h1>
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<div class="markdown"><p>Well, I count <mathjax>#40xxN_A#</mathjax> individual oxygen atoms.......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In one mole of sulfuric acid there are <mathjax>#4xxN_A#</mathjax>, <mathjax>#4xx6.022xx10^23#</mathjax> individual oxygen atoms.........</p>
<p>And thus in <mathjax>#10*mol#</mathjax> of same stuff, there are <mathjax>#4xx10xxN_A~=#</mathjax></p>
<p><mathjax>#24xx10^24#</mathjax> individual oxygen atoms........ </p></div>
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</article> | How many oxygen atoms in a #10*mol# quantity of sulfuric acid? | null |
3,177 | a8c6b980-6ddd-11ea-91b4-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-291-moles-of-arsenic | 21801.72 grams | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] arsenic [IN] grams"}] | [{"type":"physical unit","value":"21801.72 grams"}] | [{"type":"physical unit","value":"Mole [OF] arsenic [=] \\pu{291 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in 291 moles of arsenic? </h1> | null | 21801.72 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The molar mass of arsenic is <mathjax>#74.92*g*mol^-1#</mathjax>.</p>
<p>The mass of 291 moles is: <mathjax>#74.92*g*mol^-1xx291*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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<div class="markdown"><p>About <mathjax>#22xx10^3*g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of arsenic is <mathjax>#74.92*g*mol^-1#</mathjax>.</p>
<p>The mass of 291 moles is: <mathjax>#74.92*g*mol^-1xx291*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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<div class="markdown"><p>About <mathjax>#22xx10^3*g#</mathjax></p></div>
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<div class="markdown"><p>The molar mass of arsenic is <mathjax>#74.92*g*mol^-1#</mathjax>.</p>
<p>The mass of 291 moles is: <mathjax>#74.92*g*mol^-1xx291*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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</article> | How many grams are in 291 moles of arsenic? | null |
3,178 | a8a80ab8-6ddd-11ea-ba4c-ccda262736ce | https://socratic.org/questions/how-many-molecules-are-in-10-0-g-of-o-2-gas-at-stp | 1.51 × 10^23 | start physical_unit 2 2 number none qc_end physical_unit 8 9 5 6 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Number [OF] molecules"}] | [{"type":"physical unit","value":"1.51 × 10^23"}] | [{"type":"physical unit","value":"Mass [OF] O2 gas [=] \\pu{10.0 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many molecules are in 10.0 g of #O_2# gas at STP? </h1> | null | 1.51 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specified a mass of 10.0 g. Clearly, this is a finite number of dioxygen molecules. This same finite number (which is <mathjax>#N_Axx(10.0*g)/(32.0*g*mol^-1)#</mathjax>, <mathjax>#N_A = "Avogadro's number"#</mathjax>) is the same mass whatever the state, solid, liquid, or gas. </p>
<p>Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#xx#</mathjax> its molar quantity, approx. <mathjax>#8*L#</mathjax>. </p></div>
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<div class="markdown"><p>How many molecules at STP? The same number of molecules in 10.0 g of LIQUID OXYGEN at LOW TEMPERATURES.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specified a mass of 10.0 g. Clearly, this is a finite number of dioxygen molecules. This same finite number (which is <mathjax>#N_Axx(10.0*g)/(32.0*g*mol^-1)#</mathjax>, <mathjax>#N_A = "Avogadro's number"#</mathjax>) is the same mass whatever the state, solid, liquid, or gas. </p>
<p>Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#xx#</mathjax> its molar quantity, approx. <mathjax>#8*L#</mathjax>. </p></div>
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<div class="markdown"><p>How many molecules at STP? The same number of molecules in 10.0 g of LIQUID OXYGEN at LOW TEMPERATURES.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You specified a mass of 10.0 g. Clearly, this is a finite number of dioxygen molecules. This same finite number (which is <mathjax>#N_Axx(10.0*g)/(32.0*g*mol^-1)#</mathjax>, <mathjax>#N_A = "Avogadro's number"#</mathjax>) is the same mass whatever the state, solid, liquid, or gas. </p>
<p>Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#xx#</mathjax> its molar quantity, approx. <mathjax>#8*L#</mathjax>. </p></div>
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<div class="markdown"><p>There are <mathjax>#1.51xx10^23"molecules O"_2"#</mathjax> in <mathjax>#"10.0 g O"_2"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>STP is irrelevant - it's a detractor. Mass is not dependent on temperature or pressure. The number of molecules of <mathjax>#"O"_2"#</mathjax> are dependent only on the moles of <mathjax>#"O"_2"#</mathjax> in <mathjax>#"10.0 g O"_2"#</mathjax>.</p>
<p>Determine the molar mass of <mathjax>#"O"_2"#</mathjax>, which is <mathjax>#(2xx15.998"g/mol")="39.998 g/mol"#</mathjax>.</p>
<p><mathjax>#1"mol of molecules"=6.022xx10^23"molecules"#</mathjax></p>
<p>You will make the following conversions:</p>
<p><mathjax>#color(red)("mass O"_2)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)("mol O"_2)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)("molecules O"_2)#</mathjax>** <br/>
by 1) dividing the given mass <mathjax>#"O"_2"#</mathjax> by its molar mass, and 2) multiplying mol <mathjax>#"O"_2"#</mathjax> by <mathjax>#6.022xx10^23"molecules".#</mathjax></p>
<p><mathjax>#color(red)10.0cancel(color(red)("g O"_2))xx(color(green)(1)cancelcolor(green)("mol O"_2))/(color(green)(39.998)cancelcolor(green)("g O"_2))xx(color(blue)(6.022xx10^23"molecules O"_2))/(cancelcolor(blue)("mol O"_2))=color(blue)(1.51xx10^23"molecules O"_2")#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | How many molecules are in 10.0 g of #O_2# gas at STP? | null |
3,179 | aa405da8-6ddd-11ea-8651-ccda262736ce | https://socratic.org/questions/how-do-you-determine-the-maximum-number-of-moles-of-product-that-can-be-produced | 5.33 moles | start physical_unit 10 10 mole mol qc_end physical_unit 18 18 16 17 mole qc_end physical_unit 22 22 20 21 mole qc_end chemical_equation 27 34 qc_end end | [{"type":"physical unit","value":"Mole [OF] the product [IN] moles"}] | [{"type":"physical unit","value":"5.33 moles"}] | [{"type":"physical unit","value":"Mole [OF] Al [=] \\pu{7.0 mol}"},{"type":"physical unit","value":"Mole [OF] Cl2 [=] \\pu{8.0 mol}"},{"type":"chemical equation","value":"2 Al + 3 Cl2 -> 2 AlCl3"}] | <h1 class="questionTitle" itemprop="name">How do you determine the maximum number of moles of product that can be produced from 7.0 mol #Al# and 8.0 mol #Cl_2# according to the equation #2Al + 3Cl_2 -> 2AlCl_3#? </h1> | null | 5.33 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of a limiting reactant problem.</p>
<p>The first step is to write down the balanced chemical equation. That is,</p>
<p><mathjax>#2Al(s)+3Cl_2(g)->2AlCl_3(s)#</mathjax></p>
<p>We got seven moles of aluminium and eight moles of chlorine. So, we find the limiting reactant.</p>
<p><mathjax>#7color(red)cancelcolor(black)("mol of" \ Al)*(2 \ "mol of" \ AlCl_3)/(2color(red)cancelcolor(black)("mol of" \ Al))=7 \ "mol of" \ AlCl_3#</mathjax></p>
<p><mathjax>#8color(red)cancelcolor(black)("mol of" \ Cl_2)*(2 \ "mol of" \ AlCl_3)/(3color(red)cancelcolor(black)("mol of" \ Cl_2))=5 1/3 \ "mol of" \ AlCl_3#</mathjax></p>
<p>So, this shows that chlorine gas is the <strong>limiting reactant</strong> here, and only the maximum amount of aluminium chloride will form is dependent on the amount of chlorine gas.</p>
<p>Since there are eight moles of chlorine gas, then <mathjax>#5 1/3 \ "mol"#</mathjax> of aluminium chloride is formed.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5 1/3 \ "mol of" \ AlCl_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of a limiting reactant problem.</p>
<p>The first step is to write down the balanced chemical equation. That is,</p>
<p><mathjax>#2Al(s)+3Cl_2(g)->2AlCl_3(s)#</mathjax></p>
<p>We got seven moles of aluminium and eight moles of chlorine. So, we find the limiting reactant.</p>
<p><mathjax>#7color(red)cancelcolor(black)("mol of" \ Al)*(2 \ "mol of" \ AlCl_3)/(2color(red)cancelcolor(black)("mol of" \ Al))=7 \ "mol of" \ AlCl_3#</mathjax></p>
<p><mathjax>#8color(red)cancelcolor(black)("mol of" \ Cl_2)*(2 \ "mol of" \ AlCl_3)/(3color(red)cancelcolor(black)("mol of" \ Cl_2))=5 1/3 \ "mol of" \ AlCl_3#</mathjax></p>
<p>So, this shows that chlorine gas is the <strong>limiting reactant</strong> here, and only the maximum amount of aluminium chloride will form is dependent on the amount of chlorine gas.</p>
<p>Since there are eight moles of chlorine gas, then <mathjax>#5 1/3 \ "mol"#</mathjax> of aluminium chloride is formed.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you determine the maximum number of moles of product that can be produced from 7.0 mol #Al# and 8.0 mol #Cl_2# according to the equation #2Al + 3Cl_2 -> 2AlCl_3#? </h1>
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<div class="markdown"><p><mathjax>#5 1/3 \ "mol of" \ AlCl_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of a limiting reactant problem.</p>
<p>The first step is to write down the balanced chemical equation. That is,</p>
<p><mathjax>#2Al(s)+3Cl_2(g)->2AlCl_3(s)#</mathjax></p>
<p>We got seven moles of aluminium and eight moles of chlorine. So, we find the limiting reactant.</p>
<p><mathjax>#7color(red)cancelcolor(black)("mol of" \ Al)*(2 \ "mol of" \ AlCl_3)/(2color(red)cancelcolor(black)("mol of" \ Al))=7 \ "mol of" \ AlCl_3#</mathjax></p>
<p><mathjax>#8color(red)cancelcolor(black)("mol of" \ Cl_2)*(2 \ "mol of" \ AlCl_3)/(3color(red)cancelcolor(black)("mol of" \ Cl_2))=5 1/3 \ "mol of" \ AlCl_3#</mathjax></p>
<p>So, this shows that chlorine gas is the <strong>limiting reactant</strong> here, and only the maximum amount of aluminium chloride will form is dependent on the amount of chlorine gas.</p>
<p>Since there are eight moles of chlorine gas, then <mathjax>#5 1/3 \ "mol"#</mathjax> of aluminium chloride is formed.</p></div>
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</article> | How do you determine the maximum number of moles of product that can be produced from 7.0 mol #Al# and 8.0 mol #Cl_2# according to the equation #2Al + 3Cl_2 -> 2AlCl_3#? | null |
3,180 | a97cb7fb-6ddd-11ea-8b59-ccda262736ce | https://socratic.org/questions/what-is-the-change-in-the-freezing-point-of-water-when-35-0-g-of-sucrose-is-diss | -0.63 ℃ | start physical_unit 9 9 freezing_point_temperature °c qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 9 9 18 19 mass qc_end physical_unit 9 9 26 27 kf qc_end physical_unit 14 14 32 33 molar_mass qc_end physical_unit 37 37 26 26 van_t_hoff_factor qc_end end | [{"type":"physical unit","value":"Change in the freezing point [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"-0.63 ℃"}] | [{"type":"physical unit","value":"Mass [OF] sucrose [=] \\pu{35.0 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{300.0 g}"},{"type":"physical unit","value":"Kf [OF] water [=] \\pu{-1.86 degC/mol}"},{"type":"physical unit","value":"Molar mass [OF] sucrose [=] \\pu{342.30 g/mol}"},{"type":"physical unit","value":"i value [OF] sugar [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">What is the change in the freezing point of water when 35.0 g of sucrose is dissolved in 300.0 g of water?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Kf of water = -1.86 degC/mol<br/>
molar mass sucrose = 342.30 g/mol<br/>
i value of sugar = 1</p></div>
</h2>
</div>
</div> | -0.63 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>freezing point depression</strong> of a solution.</p>
<p>To do this, we use the equation</p>
<p><mathjax>#DeltaT_f = i·m·K_f#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_f#</mathjax> is the <strong>change in freezing point temperature</strong> (what we're trying to find)</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong>, which is given as <mathjax>#1#</mathjax> (and usually is <mathjax>#1#</mathjax> in the case of nonelectrolytes)</p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#"molality" = "mol solute"/"kg solvent"#</mathjax></p>
<p>Convert the given mass of sucrose to moles using its <em>molar mass</em>:</p>
<p><mathjax>#35.0cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.102#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>The molality is thus</p>
<p><mathjax>#"molality" = color(red)(0.120color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg water") = color(green)(0.341m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#K_f#</mathjax> is the <strong>molal freezing point constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (water), which is given as <mathjax>#-1.86#</mathjax> <mathjax>#""^"o""C/"m#</mathjax><blockquote></blockquote>
</li>
</ul>
<p>Plugging in known values, we have</p>
<p><mathjax>#DeltaT_f = (1)(color(green)(0.341)cancel(color(green)(m)))(-1.86color(white)(l)""^"o""C/"cancel(m))#</mathjax></p>
<p><mathjax>#= color(blue)(ul(-0.634color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
<p>This represents by how much the freezing point <strong>decreases</strong>. The new freezing <em>point</em> of the solution is found by adding this value from the normal freezing point of the solvent (<mathjax>#0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> for water):</p>
<p><mathjax>#"new f.p." = 0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- color(blue)(0.634#</mathjax> <mathjax>#color(blue)(""^"o""C"#</mathjax> <mathjax>#= ul(-0.634color(white)(l)""^"o""C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaT_f = -0.634#</mathjax> <mathjax>#""^"o""C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>freezing point depression</strong> of a solution.</p>
<p>To do this, we use the equation</p>
<p><mathjax>#DeltaT_f = i·m·K_f#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_f#</mathjax> is the <strong>change in freezing point temperature</strong> (what we're trying to find)</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong>, which is given as <mathjax>#1#</mathjax> (and usually is <mathjax>#1#</mathjax> in the case of nonelectrolytes)</p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#"molality" = "mol solute"/"kg solvent"#</mathjax></p>
<p>Convert the given mass of sucrose to moles using its <em>molar mass</em>:</p>
<p><mathjax>#35.0cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.102#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>The molality is thus</p>
<p><mathjax>#"molality" = color(red)(0.120color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg water") = color(green)(0.341m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#K_f#</mathjax> is the <strong>molal freezing point constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (water), which is given as <mathjax>#-1.86#</mathjax> <mathjax>#""^"o""C/"m#</mathjax><blockquote></blockquote>
</li>
</ul>
<p>Plugging in known values, we have</p>
<p><mathjax>#DeltaT_f = (1)(color(green)(0.341)cancel(color(green)(m)))(-1.86color(white)(l)""^"o""C/"cancel(m))#</mathjax></p>
<p><mathjax>#= color(blue)(ul(-0.634color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
<p>This represents by how much the freezing point <strong>decreases</strong>. The new freezing <em>point</em> of the solution is found by adding this value from the normal freezing point of the solvent (<mathjax>#0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> for water):</p>
<p><mathjax>#"new f.p." = 0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- color(blue)(0.634#</mathjax> <mathjax>#color(blue)(""^"o""C"#</mathjax> <mathjax>#= ul(-0.634color(white)(l)""^"o""C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the change in the freezing point of water when 35.0 g of sucrose is dissolved in 300.0 g of water?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Kf of water = -1.86 degC/mol<br/>
molar mass sucrose = 342.30 g/mol<br/>
i value of sugar = 1</p></div>
</h2>
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Nathan L.
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<div class="markdown"><p><mathjax>#DeltaT_f = -0.634#</mathjax> <mathjax>#""^"o""C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>freezing point depression</strong> of a solution.</p>
<p>To do this, we use the equation</p>
<p><mathjax>#DeltaT_f = i·m·K_f#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_f#</mathjax> is the <strong>change in freezing point temperature</strong> (what we're trying to find)</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong>, which is given as <mathjax>#1#</mathjax> (and usually is <mathjax>#1#</mathjax> in the case of nonelectrolytes)</p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#"molality" = "mol solute"/"kg solvent"#</mathjax></p>
<p>Convert the given mass of sucrose to moles using its <em>molar mass</em>:</p>
<p><mathjax>#35.0cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.102#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>The molality is thus</p>
<p><mathjax>#"molality" = color(red)(0.120color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg water") = color(green)(0.341m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#K_f#</mathjax> is the <strong>molal freezing point constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (water), which is given as <mathjax>#-1.86#</mathjax> <mathjax>#""^"o""C/"m#</mathjax><blockquote></blockquote>
</li>
</ul>
<p>Plugging in known values, we have</p>
<p><mathjax>#DeltaT_f = (1)(color(green)(0.341)cancel(color(green)(m)))(-1.86color(white)(l)""^"o""C/"cancel(m))#</mathjax></p>
<p><mathjax>#= color(blue)(ul(-0.634color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
<p>This represents by how much the freezing point <strong>decreases</strong>. The new freezing <em>point</em> of the solution is found by adding this value from the normal freezing point of the solvent (<mathjax>#0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> for water):</p>
<p><mathjax>#"new f.p." = 0.0#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- color(blue)(0.634#</mathjax> <mathjax>#color(blue)(""^"o""C"#</mathjax> <mathjax>#= ul(-0.634color(white)(l)""^"o""C"#</mathjax></p></div>
</div>
</div>
</div>
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</article> | What is the change in the freezing point of water when 35.0 g of sucrose is dissolved in 300.0 g of water? |
Kf of water = -1.86 degC/mol
molar mass sucrose = 342.30 g/mol
i value of sugar = 1
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3,181 | a91815c6-6ddd-11ea-a16a-ccda262736ce | https://socratic.org/questions/how-many-joules-of-heat-energy-are-required-to-change-64-grams-of-methane-from-a | 36708 joules | start physical_unit 13 13 heat_energy j qc_end physical_unit 13 13 10 11 mass qc_end c_other ConstantTemperaturePressue qc_end c_other OTHER qc_end physical_unit 13 13 32 33 enthalpy_of_vaporization qc_end end | [{"type":"physical unit","value":"Required heat energy [OF] methane [IN] joules"}] | [{"type":"physical unit","value":"36708 joules"}] | [{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{64 grams}"},{"type":"other","value":"ConstantTemperaturePressue"},{"type":"other","value":"Change methane from a liquid to a gas."},{"type":"physical unit","value":"Heat of vaporization [OF] methane [=] \\pu{9.2 kJ/mol}"}] | <h1 class="questionTitle" itemprop="name">How many joules of heat energy are required to change 64 grams of methane from a liquid to a gas at constant temperature and pressure? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The heat of vaporization of methane is 9.2 kJ/mol?</p></div>
</h2>
</div>
</div> | 36708 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem is the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, <mathjax>#DeltaH_"vap"#</mathjax>, for methane. </p>
<p>In essence, the <em>enthalpy change of vaporization</em> tells you how much heat is <strong>needed</strong> in order to convert <strong>one mole</strong> of a substance from <em>liquid</em> at its boiling point to <em>vapor</em> at its boiling point. </p>
<p>The enthalpy change of vaporization expresses the amount of heat needed for <strong>one mole</strong> of a substance to undergo a <em>liquid</em> <mathjax>#->#</mathjax> <em>vapor</em> <strong><a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a></strong>. </p>
<p>In this case, you know that</p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = "9.2 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>one mole</strong> of liquid methane at its boiling point must absorb <mathjax>#"9.2 kJ"#</mathjax> of heat to go to vapor at the <em>same temperature</em>. </p>
<p>Use methane's <strong>molar mass</strong> to determine how many <em>moles</em> of methane you have in that <mathjax>#"64-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "3.99 moles CH"_4#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> requires <mathjax>#"9.2 kJ"#</mathjax> of heat, it follows that this many moles would require</p>
<blockquote>
<p><mathjax>#3.99color(red)(cancel(color(black)("moles CH"_4))) * overbrace("9.2 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))))^(color(purple)(DeltaH_text(vap)color(white)(a) "for methane")) = "36.708 kJ"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"heat required" = color(green)(|bar(ul(color(white)(a/a)"37 kJ"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"37 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem is the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, <mathjax>#DeltaH_"vap"#</mathjax>, for methane. </p>
<p>In essence, the <em>enthalpy change of vaporization</em> tells you how much heat is <strong>needed</strong> in order to convert <strong>one mole</strong> of a substance from <em>liquid</em> at its boiling point to <em>vapor</em> at its boiling point. </p>
<p>The enthalpy change of vaporization expresses the amount of heat needed for <strong>one mole</strong> of a substance to undergo a <em>liquid</em> <mathjax>#->#</mathjax> <em>vapor</em> <strong><a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a></strong>. </p>
<p>In this case, you know that</p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = "9.2 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>one mole</strong> of liquid methane at its boiling point must absorb <mathjax>#"9.2 kJ"#</mathjax> of heat to go to vapor at the <em>same temperature</em>. </p>
<p>Use methane's <strong>molar mass</strong> to determine how many <em>moles</em> of methane you have in that <mathjax>#"64-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "3.99 moles CH"_4#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> requires <mathjax>#"9.2 kJ"#</mathjax> of heat, it follows that this many moles would require</p>
<blockquote>
<p><mathjax>#3.99color(red)(cancel(color(black)("moles CH"_4))) * overbrace("9.2 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))))^(color(purple)(DeltaH_text(vap)color(white)(a) "for methane")) = "36.708 kJ"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"heat required" = color(green)(|bar(ul(color(white)(a/a)"37 kJ"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many joules of heat energy are required to change 64 grams of methane from a liquid to a gas at constant temperature and pressure? </h1>
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Stefan V.
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Mar 11, 2016
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<div class="markdown"><p><mathjax>#"37 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to this problem is the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, <mathjax>#DeltaH_"vap"#</mathjax>, for methane. </p>
<p>In essence, the <em>enthalpy change of vaporization</em> tells you how much heat is <strong>needed</strong> in order to convert <strong>one mole</strong> of a substance from <em>liquid</em> at its boiling point to <em>vapor</em> at its boiling point. </p>
<p>The enthalpy change of vaporization expresses the amount of heat needed for <strong>one mole</strong> of a substance to undergo a <em>liquid</em> <mathjax>#->#</mathjax> <em>vapor</em> <strong><a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a></strong>. </p>
<p>In this case, you know that</p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = "9.2 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>one mole</strong> of liquid methane at its boiling point must absorb <mathjax>#"9.2 kJ"#</mathjax> of heat to go to vapor at the <em>same temperature</em>. </p>
<p>Use methane's <strong>molar mass</strong> to determine how many <em>moles</em> of methane you have in that <mathjax>#"64-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "3.99 moles CH"_4#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> requires <mathjax>#"9.2 kJ"#</mathjax> of heat, it follows that this many moles would require</p>
<blockquote>
<p><mathjax>#3.99color(red)(cancel(color(black)("moles CH"_4))) * overbrace("9.2 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))))^(color(purple)(DeltaH_text(vap)color(white)(a) "for methane")) = "36.708 kJ"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"heat required" = color(green)(|bar(ul(color(white)(a/a)"37 kJ"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | How many joules of heat energy are required to change 64 grams of methane from a liquid to a gas at constant temperature and pressure? |
The heat of vaporization of methane is 9.2 kJ/mol?
|
3,182 | ab0153ee-6ddd-11ea-9cf7-ccda262736ce | https://socratic.org/questions/a-calorimeter-contains-15-grams-of-water-the-water-s-temperature-increases-by-10 | 150.00 cal | start physical_unit 6 6 heat_energy cal qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 6 6 12 13 temperature qc_end physical_unit 6 6 28 31 specific_heat qc_end end | [{"type":"physical unit","value":"Added heat energy [OF] water [IN] cal"}] | [{"type":"physical unit","value":"150.00 cal"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{15 grams}"},{"type":"physical unit","value":"Increased temperature [OF] water [=] \\pu{10 ℃}"},{"type":"physical unit","value":"Specific heat [OF] water [=] \\pu{1 cal/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">A calorimeter contains 15 grams of water. The water's temperature increases by 10 C°. How much heat energy was added to the water? (Hint: water's specific heat is 1 cal/g°C)?</h1> | null | 150.00 cal | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Unsurprisingly, the key to this problem is the hint. </p>
<p>The thing to remember about a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> is that it tells you the amount of energy needed in order increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, water is said to have a specific heat of <mathjax>#"1 cal g"^(-1)""^@"C"^(-1)#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you must provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>The thing to remember here is that you need <mathjax>#"1 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of water</strong> and <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>in temperature</strong>. </p>
<p>So, you have <mathjax>#"15 g"#</mathjax> of water to work with. You can use the specific heat of water to figure out how much heat is needed to increase the temperature of this sample by <mathjax>#1^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#15 color(red)(cancel(color(black)("g"))) * "1 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "15 cal" ""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>increase</strong> in the temperature of the sample, <mathjax>#"15 cal"#</mathjax> of heat are needed. Since you want the temperature to increase by <mathjax>#10^@"C"#</mathjax>, you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * overbrace("15 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 15 g of water")) = color(darkgreen)(ul(color(black)("150 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig ifgs</a></strong>, but keep in mind that you only have one significant figure for the increase in temperature. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"150 cal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Unsurprisingly, the key to this problem is the hint. </p>
<p>The thing to remember about a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> is that it tells you the amount of energy needed in order increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, water is said to have a specific heat of <mathjax>#"1 cal g"^(-1)""^@"C"^(-1)#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you must provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>The thing to remember here is that you need <mathjax>#"1 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of water</strong> and <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>in temperature</strong>. </p>
<p>So, you have <mathjax>#"15 g"#</mathjax> of water to work with. You can use the specific heat of water to figure out how much heat is needed to increase the temperature of this sample by <mathjax>#1^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#15 color(red)(cancel(color(black)("g"))) * "1 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "15 cal" ""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>increase</strong> in the temperature of the sample, <mathjax>#"15 cal"#</mathjax> of heat are needed. Since you want the temperature to increase by <mathjax>#10^@"C"#</mathjax>, you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * overbrace("15 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 15 g of water")) = color(darkgreen)(ul(color(black)("150 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig ifgs</a></strong>, but keep in mind that you only have one significant figure for the increase in temperature. </p></div>
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<h1 class="questionTitle" itemprop="name">A calorimeter contains 15 grams of water. The water's temperature increases by 10 C°. How much heat energy was added to the water? (Hint: water's specific heat is 1 cal/g°C)?</h1>
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<div class="markdown"><p><mathjax>#"150 cal"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Unsurprisingly, the key to this problem is the hint. </p>
<p>The thing to remember about a substance's <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> is that it tells you the amount of energy needed in order increase the temperature of <mathjax>#"1 g"#</mathjax> of said substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, water is said to have a specific heat of <mathjax>#"1 cal g"^(-1)""^@"C"^(-1)#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you must provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>The thing to remember here is that you need <mathjax>#"1 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of water</strong> and <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>in temperature</strong>. </p>
<p>So, you have <mathjax>#"15 g"#</mathjax> of water to work with. You can use the specific heat of water to figure out how much heat is needed to increase the temperature of this sample by <mathjax>#1^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#15 color(red)(cancel(color(black)("g"))) * "1 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "15 cal" ""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that <strong>for every</strong> <mathjax>#1^@"C"#</mathjax> <strong>increase</strong> in the temperature of the sample, <mathjax>#"15 cal"#</mathjax> of heat are needed. Since you want the temperature to increase by <mathjax>#10^@"C"#</mathjax>, you will need</p>
<blockquote>
<p><mathjax>#10color(red)(cancel(color(black)(""^@"C"))) * overbrace("15 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 15 g of water")) = color(darkgreen)(ul(color(black)("150 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig ifgs</a></strong>, but keep in mind that you only have one significant figure for the increase in temperature. </p></div>
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</article> | A calorimeter contains 15 grams of water. The water's temperature increases by 10 C°. How much heat energy was added to the water? (Hint: water's specific heat is 1 cal/g°C)? | null |
3,183 | a9b29d0c-6ddd-11ea-87b9-ccda262736ce | https://socratic.org/questions/what-is-the-van-t-hoff-factor-of-na-3po-4-in-a-0-40-m-solution-whose-freezing-po | 3.5 | start physical_unit 7 7 van_t_hoff_factor none qc_end physical_unit 7 7 10 11 molarity qc_end end | [{"type":"physical unit","value":"Van't Hoff factor [OF] Na3PO4"}] | [{"type":"physical unit","value":"3.5"}] | [{"type":"physical unit","value":"Molarity [OF] Na3PO4 solution [=] \\pu{0.40 M}"},{"type":"physical unit","value":"Freezing point [OF] Na3PO4 solution [=] \\pu{-2.6 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the van't Hoff factor of #Na_3PO_4# in a 0.40 m solution whose freezing point is -2.6°C ?</h1> | null | 3.5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's start by figuring out what you would <em>expect</em> the van't Hoff factor, <mathjax>#i#</mathjax>, to be for <em>sodium phosphate</em>, <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>As you know, the <strong>van't Hoff factor</strong> tells you what the ratio between the number of particles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the number of particles <em>produced in solution</em>* after dissolving the solute. </p>
<p>For <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, this comes down to how many <strong>ions</strong> will be produced <em>per formula unit</em> of solute. </p>
<p>Sodium phosphate will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>So, if <em>every</em> formula unit of sodium phosphate should produce <mathjax>#4#</mathjax> ions in solution, you should expect the van't Hoff factor to be equal to <mathjax>#4#</mathjax>. </p>
<p>Now, the equation that describes <em>freezing-point depression</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = i * K_f * b)" "#</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>The cryoscopic constant of water is equal to <mathjax>#1.86 ""^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>The freezing-point depression is defined as </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = T_f^@ - T_f)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em><br/>
<mathjax>#T_f#</mathjax> - the freezing point of the solution</p>
<p>In your case, you would have </p>
<blockquote>
<p><mathjax>#DeltaT_f = 0^@"C" - (-2.6^@"C") = 2.6^@"C"#</mathjax></p>
</blockquote>
<p>Plug in your values into the equation for freezing-point depression and solve for <mathjax>#i#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies i = (DeltaT_f)/(K_f * b)#</mathjax></p>
<p><mathjax>#i = (2.6 color(red)(cancel(color(black)(""^@"C"))))/(1.86 color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.40 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))) = color(green)(3.5)#</mathjax></p>
</blockquote>
<p>Notice that the <em>observed</em> van't Hoff factor is <strong>smaller</strong> than the <em>expected</em> van't Hoff factor. </p>
<p>This happens because some of the ions will actually bind to form <em>solvation cells</em>, which is why the number of ions produced per formula unit is smaller than the expected number. </p>
<p>In other words, some of the sodium cations will bind to the phosphate anions and exist either as <mathjax>#"Na"^(+)"PO"_4^(3-)#</mathjax>, or as <mathjax>#"Na"_2^(+)"PO"_4^(3-)#</mathjax> solvation cells. </p>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "Na"^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> "Na"_text((aq])^(+) + "Na"_2^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#i = 3.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's start by figuring out what you would <em>expect</em> the van't Hoff factor, <mathjax>#i#</mathjax>, to be for <em>sodium phosphate</em>, <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>As you know, the <strong>van't Hoff factor</strong> tells you what the ratio between the number of particles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the number of particles <em>produced in solution</em>* after dissolving the solute. </p>
<p>For <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, this comes down to how many <strong>ions</strong> will be produced <em>per formula unit</em> of solute. </p>
<p>Sodium phosphate will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>So, if <em>every</em> formula unit of sodium phosphate should produce <mathjax>#4#</mathjax> ions in solution, you should expect the van't Hoff factor to be equal to <mathjax>#4#</mathjax>. </p>
<p>Now, the equation that describes <em>freezing-point depression</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = i * K_f * b)" "#</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>The cryoscopic constant of water is equal to <mathjax>#1.86 ""^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>The freezing-point depression is defined as </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = T_f^@ - T_f)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em><br/>
<mathjax>#T_f#</mathjax> - the freezing point of the solution</p>
<p>In your case, you would have </p>
<blockquote>
<p><mathjax>#DeltaT_f = 0^@"C" - (-2.6^@"C") = 2.6^@"C"#</mathjax></p>
</blockquote>
<p>Plug in your values into the equation for freezing-point depression and solve for <mathjax>#i#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies i = (DeltaT_f)/(K_f * b)#</mathjax></p>
<p><mathjax>#i = (2.6 color(red)(cancel(color(black)(""^@"C"))))/(1.86 color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.40 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))) = color(green)(3.5)#</mathjax></p>
</blockquote>
<p>Notice that the <em>observed</em> van't Hoff factor is <strong>smaller</strong> than the <em>expected</em> van't Hoff factor. </p>
<p>This happens because some of the ions will actually bind to form <em>solvation cells</em>, which is why the number of ions produced per formula unit is smaller than the expected number. </p>
<p>In other words, some of the sodium cations will bind to the phosphate anions and exist either as <mathjax>#"Na"^(+)"PO"_4^(3-)#</mathjax>, or as <mathjax>#"Na"_2^(+)"PO"_4^(3-)#</mathjax> solvation cells. </p>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "Na"^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> "Na"_text((aq])^(+) + "Na"_2^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the van't Hoff factor of #Na_3PO_4# in a 0.40 m solution whose freezing point is -2.6°C ?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#i = 3.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's start by figuring out what you would <em>expect</em> the van't Hoff factor, <mathjax>#i#</mathjax>, to be for <em>sodium phosphate</em>, <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>As you know, the <strong>van't Hoff factor</strong> tells you what the ratio between the number of particles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the number of particles <em>produced in solution</em>* after dissolving the solute. </p>
<p>For <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, this comes down to how many <strong>ions</strong> will be produced <em>per formula unit</em> of solute. </p>
<p>Sodium phosphate will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>So, if <em>every</em> formula unit of sodium phosphate should produce <mathjax>#4#</mathjax> ions in solution, you should expect the van't Hoff factor to be equal to <mathjax>#4#</mathjax>. </p>
<p>Now, the equation that describes <em>freezing-point depression</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = i * K_f * b)" "#</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>The cryoscopic constant of water is equal to <mathjax>#1.86 ""^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>The freezing-point depression is defined as </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_f = T_f^@ - T_f)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_f^@#</mathjax> - the freezing point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em><br/>
<mathjax>#T_f#</mathjax> - the freezing point of the solution</p>
<p>In your case, you would have </p>
<blockquote>
<p><mathjax>#DeltaT_f = 0^@"C" - (-2.6^@"C") = 2.6^@"C"#</mathjax></p>
</blockquote>
<p>Plug in your values into the equation for freezing-point depression and solve for <mathjax>#i#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies i = (DeltaT_f)/(K_f * b)#</mathjax></p>
<p><mathjax>#i = (2.6 color(red)(cancel(color(black)(""^@"C"))))/(1.86 color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.40 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))) = color(green)(3.5)#</mathjax></p>
</blockquote>
<p>Notice that the <em>observed</em> van't Hoff factor is <strong>smaller</strong> than the <em>expected</em> van't Hoff factor. </p>
<p>This happens because some of the ions will actually bind to form <em>solvation cells</em>, which is why the number of ions produced per formula unit is smaller than the expected number. </p>
<p>In other words, some of the sodium cations will bind to the phosphate anions and exist either as <mathjax>#"Na"^(+)"PO"_4^(3-)#</mathjax>, or as <mathjax>#"Na"_2^(+)"PO"_4^(3-)#</mathjax> solvation cells. </p>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "Na"^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
<p><mathjax>#"Na"_3"PO"_text(4(aq]) -> "Na"_text((aq])^(+) + "Na"_2^(+) "PO"_text(4(aq])^(3-)#</mathjax></p>
</blockquote></div>
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</article> | What is the van't Hoff factor of #Na_3PO_4# in a 0.40 m solution whose freezing point is -2.6°C ? | null |
3,184 | aaded79c-6ddd-11ea-b680-ccda262736ce | https://socratic.org/questions/575773f37c0149375b2ae30a | 2.76 L | start physical_unit 11 11 volume l qc_end physical_unit 11 11 7 8 mole qc_end physical_unit 11 11 16 17 temperature qc_end physical_unit 11 11 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"2.76 L"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{0.1 mol}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{310 K}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{700 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What volume would be occupied by a #0.1*mol# quantity of gas at a temperature of #310*K#, and a pressure of #700*mm*Hg#?</h1> | null | 2.76 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You must simply KNOW that <mathjax>#1*atm#</mathjax> pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high; and thus <mathjax>#"Torr"-=1*mm*Hg#</mathjax> stands in for a pressure measurement. You must also know the relationship between <mathjax>#"degrees Kelvin"#</mathjax>, and <mathjax>#"degrees centigrade"#</mathjax>: <mathjax>#0#</mathjax> <mathjax>#""^@C-=+273.15*K#</mathjax></p>
<p><mathjax>#V=(nRT)/P=(0.1*molxx0.0821*(L*atm)/(K*mol)xx310*K)/((700*"Torr")/(760*"Torr"*atm^-1))#</mathjax></p>
<p><mathjax>#=??*L#</mathjax>. The question would have been better proposed had it been specified that <mathjax>#0.1*mol#</mathjax> quantity of gas had been enclosed in a piston at the given temperature and pressure. </p></div>
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<div class="markdown"><p>Well, <mathjax>#PV=nRT#</mathjax>.........and so.......<mathjax>#V~=3*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You must simply KNOW that <mathjax>#1*atm#</mathjax> pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high; and thus <mathjax>#"Torr"-=1*mm*Hg#</mathjax> stands in for a pressure measurement. You must also know the relationship between <mathjax>#"degrees Kelvin"#</mathjax>, and <mathjax>#"degrees centigrade"#</mathjax>: <mathjax>#0#</mathjax> <mathjax>#""^@C-=+273.15*K#</mathjax></p>
<p><mathjax>#V=(nRT)/P=(0.1*molxx0.0821*(L*atm)/(K*mol)xx310*K)/((700*"Torr")/(760*"Torr"*atm^-1))#</mathjax></p>
<p><mathjax>#=??*L#</mathjax>. The question would have been better proposed had it been specified that <mathjax>#0.1*mol#</mathjax> quantity of gas had been enclosed in a piston at the given temperature and pressure. </p></div>
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<h1 class="questionTitle" itemprop="name">What volume would be occupied by a #0.1*mol# quantity of gas at a temperature of #310*K#, and a pressure of #700*mm*Hg#?</h1>
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<div class="markdown"><p>Well, <mathjax>#PV=nRT#</mathjax>.........and so.......<mathjax>#V~=3*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You must simply KNOW that <mathjax>#1*atm#</mathjax> pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high; and thus <mathjax>#"Torr"-=1*mm*Hg#</mathjax> stands in for a pressure measurement. You must also know the relationship between <mathjax>#"degrees Kelvin"#</mathjax>, and <mathjax>#"degrees centigrade"#</mathjax>: <mathjax>#0#</mathjax> <mathjax>#""^@C-=+273.15*K#</mathjax></p>
<p><mathjax>#V=(nRT)/P=(0.1*molxx0.0821*(L*atm)/(K*mol)xx310*K)/((700*"Torr")/(760*"Torr"*atm^-1))#</mathjax></p>
<p><mathjax>#=??*L#</mathjax>. The question would have been better proposed had it been specified that <mathjax>#0.1*mol#</mathjax> quantity of gas had been enclosed in a piston at the given temperature and pressure. </p></div>
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</article> | What volume would be occupied by a #0.1*mol# quantity of gas at a temperature of #310*K#, and a pressure of #700*mm*Hg#? | null |
3,185 | ace68fd8-6ddd-11ea-a76a-ccda262736ce | https://socratic.org/questions/what-is-the-total-number-of-molecules-represented-by-the-equation-2c-4h-10-g-13o | 33 | start physical_unit 6 6 number none qc_end chemical_equation 11 21 qc_end end | [{"type":"physical unit","value":"Total number [OF] molecules"}] | [{"type":"physical unit","value":"33"}] | [{"type":"chemical equation","value":"2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)"}] | <h1 class="questionTitle" itemprop="name">What is the total number of molecules represented by the equation #2C_4H_10(g) + 13O_2(g) -> 8CO_2(g) + 10H_2O(g)#?</h1> | null | 33 | <div class="answerDescription">
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<div>
<div class="markdown"><p>Adding up the coefficients of all molecular species mentioned in the balanced equation, we can get total number of molecules in the reaction represented by the equation and this becomes <mathjax>#2+13+8+10=33#</mathjax></p></div>
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<div class="markdown"><p>33</p></div>
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<div class="markdown"><p>Adding up the coefficients of all molecular species mentioned in the balanced equation, we can get total number of molecules in the reaction represented by the equation and this becomes <mathjax>#2+13+8+10=33#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the total number of molecules represented by the equation #2C_4H_10(g) + 13O_2(g) -> 8CO_2(g) + 10H_2O(g)#?</h1>
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<div class="markdown"><p>33</p></div>
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<div class="markdown"><p>Adding up the coefficients of all molecular species mentioned in the balanced equation, we can get total number of molecules in the reaction represented by the equation and this becomes <mathjax>#2+13+8+10=33#</mathjax></p></div>
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</article> | What is the total number of molecules represented by the equation #2C_4H_10(g) + 13O_2(g) -> 8CO_2(g) + 10H_2O(g)#? | null |
3,186 | a9fc07ca-6ddd-11ea-bec6-ccda262736ce | https://socratic.org/questions/phosphoric-acid-is-neutralized-by-a-solution-of-sodium-hydroxide-what-is-the-nam | Na3PO4 | start chemical_formula qc_end substance 0 1 qc_end substance 8 9 qc_end c_other Neutralization qc_end end | [{"type":"other","value":"Chemical Formula [OF] the salt [IN] default"}] | [{"type":"chemical equation","value":"Na3PO4"}] | [{"type":"substance name","value":"Phosphoric acid"},{"type":"substance name","value":"Sodium hydroxide"},{"type":"other","value":"Neutralization."}] | <h1 class="questionTitle" itemprop="name">Phosphoric acid is neutralized by a solution of sodium hydroxide. What is the name of the salt formed from the neutralization?</h1> | null | Na3PO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_3PO_4 + 3NaOH -> Na_3PO_4 + 3H_2O#</mathjax></p></div>
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<div class="markdown"><p>Trisodium Phosphate (<mathjax>#Na_3PO_4#</mathjax>)</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_3PO_4 + 3NaOH -> Na_3PO_4 + 3H_2O#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Phosphoric acid is neutralized by a solution of sodium hydroxide. What is the name of the salt formed from the neutralization?</h1>
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Neh Pandya
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Grace
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Stefan V.
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<div class="markdown"><p>Trisodium Phosphate (<mathjax>#Na_3PO_4#</mathjax>)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_3PO_4 + 3NaOH -> Na_3PO_4 + 3H_2O#</mathjax></p></div>
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</article> | Phosphoric acid is neutralized by a solution of sodium hydroxide. What is the name of the salt formed from the neutralization? | null |
3,187 | ac689e9a-6ddd-11ea-983f-ccda262736ce | https://socratic.org/questions/56f3438b11ef6b1ea0655f82 | CS2 | start chemical_formula qc_end physical_unit 1 1 6 6 electron_configuration qc_end physical_unit 1 1 20 21 atomic_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"CS2"}] | [{"type":"physical unit","value":"Electronic configuration [OF] the element [=] \\pu{1s^22s^22p^6}"},{"type":"physical unit","value":"Atomic mass [OF] the element [=] \\pu{32.06 g/mol}"},{"type":"physical unit","value":"Number [OF] element atoms [=] \\pu{2}"}] | <h1 class="questionTitle" itemprop="name">An element whose electronic configuration is #1s^(2)2s^(2)2p^6# is chemically bound to TWO atoms of an element whose atomic mass is #32.06*g*mol^-1#. What is the compound described?</h1> | null | CS2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All the bumf about electronic configuration was there to confuse and distract you. You have a Group 14 element with 6 electrons.....This means that you got 6 protons. And if there are 6 protons, 6 positively charged nuclear particles, then <mathjax>#Z=6#</mathjax>. And if <mathjax>#Z=6#</mathjax> the merest glance at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> will tell you that you gots the element carbon.</p>
<p>But we are not finished there. We were further told that this carbon element was bound to a Group 16 element, whose <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> was <mathjax>#32.06*"amu"#</mathjax>. This CLEARLY specified the element sulfur. How do I know? Because I have a Periodic Table in front of me. Is there one in front of you? If you are doing your chemistry homework, and there is not one accessible you are wasting your time.</p>
<p>And the standard Lewis structure for carbon disulfide is <mathjax>#:stackrel(ddot)S=C=stackrel(ddot)S:#</mathjax>; this is a volatile, foul-smelling liquid, that poses a significant fire hazard in that it is dead easy to ignite. </p></div>
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<div class="markdown"><p>You have specified the <mathjax>#CS_2#</mathjax> molecule........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All the bumf about electronic configuration was there to confuse and distract you. You have a Group 14 element with 6 electrons.....This means that you got 6 protons. And if there are 6 protons, 6 positively charged nuclear particles, then <mathjax>#Z=6#</mathjax>. And if <mathjax>#Z=6#</mathjax> the merest glance at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> will tell you that you gots the element carbon.</p>
<p>But we are not finished there. We were further told that this carbon element was bound to a Group 16 element, whose <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> was <mathjax>#32.06*"amu"#</mathjax>. This CLEARLY specified the element sulfur. How do I know? Because I have a Periodic Table in front of me. Is there one in front of you? If you are doing your chemistry homework, and there is not one accessible you are wasting your time.</p>
<p>And the standard Lewis structure for carbon disulfide is <mathjax>#:stackrel(ddot)S=C=stackrel(ddot)S:#</mathjax>; this is a volatile, foul-smelling liquid, that poses a significant fire hazard in that it is dead easy to ignite. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">An element whose electronic configuration is #1s^(2)2s^(2)2p^6# is chemically bound to TWO atoms of an element whose atomic mass is #32.06*g*mol^-1#. What is the compound described?</h1>
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<div class="markdown"><p>You have specified the <mathjax>#CS_2#</mathjax> molecule........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All the bumf about electronic configuration was there to confuse and distract you. You have a Group 14 element with 6 electrons.....This means that you got 6 protons. And if there are 6 protons, 6 positively charged nuclear particles, then <mathjax>#Z=6#</mathjax>. And if <mathjax>#Z=6#</mathjax> the merest glance at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> will tell you that you gots the element carbon.</p>
<p>But we are not finished there. We were further told that this carbon element was bound to a Group 16 element, whose <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> was <mathjax>#32.06*"amu"#</mathjax>. This CLEARLY specified the element sulfur. How do I know? Because I have a Periodic Table in front of me. Is there one in front of you? If you are doing your chemistry homework, and there is not one accessible you are wasting your time.</p>
<p>And the standard Lewis structure for carbon disulfide is <mathjax>#:stackrel(ddot)S=C=stackrel(ddot)S:#</mathjax>; this is a volatile, foul-smelling liquid, that poses a significant fire hazard in that it is dead easy to ignite. </p></div>
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</article> | An element whose electronic configuration is #1s^(2)2s^(2)2p^6# is chemically bound to TWO atoms of an element whose atomic mass is #32.06*g*mol^-1#. What is the compound described? | null |
3,188 | a9acba48-6ddd-11ea-b8b8-ccda262736ce | https://socratic.org/questions/58b8d414b72cff4c1b489d15 | 0.04 grams | start physical_unit 3 4 mass g qc_end physical_unit 12 13 7 8 volume qc_end physical_unit 12 13 10 11 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"0.04 grams"}] | [{"type":"physical unit","value":"Volume [OF] chloric acid [=] \\pu{60.00 mL}"},{"type":"physical unit","value":"Molarity [OF] chloric acid [=] \\pu{0.015 mol/L }"}] | <h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid?</h1> | null | 0.04 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.036 g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
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<h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.036 g NaOH"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium hydroxide and chloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce water and aqueous sodium chlorate, as shown by the chemical equation</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>This tells you that the reaction will consume <strong>equal numbers of moles</strong> of sodium hydroxide and of chloric acid. </p>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the chloric acid solution to calculate how many <em>moles</em> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> it contained</p>
<blockquote>
<p><mathjax>#60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 9.00 * 10^(-4)color(white)(.)"moles HClO"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>So, if the reaction consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of chloric acid, then it must have also consumed <mathjax>#9.00 * 10^(-4)#</mathjax> <strong>moles</strong> of sodium hydroxide, since the two reactants react in a <mathjax>#1:1#</mathjax> mole ratio. </p>
<p>To convert the number of moles of sodium hydroxide to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the molarity of the chloric acid solution. </p></div>
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Ernest Z.
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<div class="markdown"><p>The mass of sodium hydroxide is 36 mg.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equation for the reaction is</p>
<p><mathjax>#"HClO"_3 + "NaOH" → "NaClO"_3 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>1.</strong> Calculate the moles of <mathjax>#"HClO"_3#</mathjax>.</p>
<p><mathjax>#"Moles of HClO"_3 = "0.060 00" color(red)(cancel(color(black)("L NaOH"))) × "0.015 mol HClO"_3/(1 color(red)(cancel(color(black)("L HClO"_3)))) = "0.000 900 mol HClO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"NaOH"#</mathjax>.</p>
<p><mathjax>#"Moles of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol HClO"_3))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol HClO"_3)))) = "0.000 900 mol NaOH"#</mathjax> </p>
<blockquote></blockquote>
<p><strong>3.</strong> Calculate the mass of the <mathjax>#"NaOH"#</mathjax>.</p>
<p><mathjax>#"Mass of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.036 g NaOH" = "36 mg NaOH"#</mathjax></p></div>
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</article> | What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid? | null |
3,189 | aa87fef8-6ddd-11ea-b301-ccda262736ce | https://socratic.org/questions/mg-s-1-2-o2-g-mgo-s-what-is-the-h-f-of-mgo | -601.6 kJ/mol | start physical_unit 10 10 standard_enthalpy_change kj/mol qc_end chemical_equation 14 19 qc_end end | [{"type":"physical unit","value":"Standard change in enthalpy of formation [OF] MgO(s) [IN] kJ/mol"}] | [{"type":"physical unit","value":"-601.6 kJ/mol"}] | [{"type":"chemical equation","value":"Mg(s) + 1/2 O2(g) -> MgO(s)"}] | <h1 class="questionTitle" itemprop="name">What is the standard change in enthalpy of formation of #"MgO"(s)#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is:</p>
<blockquote>
<p><mathjax>#"Mg"(s) + 1/2 "O"_2(g) -> "MgO"(s)#</mathjax></p>
</blockquote></div>
</h2>
</div>
</div> | -601.6 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction you're using represents the formation of magnesium oxide from its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable forms at standard state, <mathjax>#25^@"C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>. </p>
<p>In the case of formation reactions, the standard enthalpy change of formation for a compound is measured, not calculated; usually, you'll find this value listed either in books or online.</p>
<p><a href="http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>So, in your case, the standard enthalpy change of formation for magnesium oxide was measured to be </p>
<p><mathjax>#DeltaH_f^@ = "-601.6 kJ/mol"#</mathjax>.</p>
<p>In every day language, the large negative value indicates that this process releases quite a bit of energy as can be observed in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <strong>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of formation</strong>, or <mathjax>#DeltaH_f^@#</mathjax>, of magnesium oxide will be <strong>-601.6 kJ/mol</strong>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction you're using represents the formation of magnesium oxide from its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable forms at standard state, <mathjax>#25^@"C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>. </p>
<p>In the case of formation reactions, the standard enthalpy change of formation for a compound is measured, not calculated; usually, you'll find this value listed either in books or online.</p>
<p><a href="http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>So, in your case, the standard enthalpy change of formation for magnesium oxide was measured to be </p>
<p><mathjax>#DeltaH_f^@ = "-601.6 kJ/mol"#</mathjax>.</p>
<p>In every day language, the large negative value indicates that this process releases quite a bit of energy as can be observed in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the standard change in enthalpy of formation of #"MgO"(s)#?</h1>
<div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is:</p>
<blockquote>
<p><mathjax>#"Mg"(s) + 1/2 "O"_2(g) -> "MgO"(s)#</mathjax></p>
</blockquote></div>
</h2>
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<div class="markdown"><p>The <strong>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of formation</strong>, or <mathjax>#DeltaH_f^@#</mathjax>, of magnesium oxide will be <strong>-601.6 kJ/mol</strong>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction you're using represents the formation of magnesium oxide from its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable forms at standard state, <mathjax>#25^@"C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>. </p>
<p>In the case of formation reactions, the standard enthalpy change of formation for a compound is measured, not calculated; usually, you'll find this value listed either in books or online.</p>
<p><a href="http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>So, in your case, the standard enthalpy change of formation for magnesium oxide was measured to be </p>
<p><mathjax>#DeltaH_f^@ = "-601.6 kJ/mol"#</mathjax>.</p>
<p>In every day language, the large negative value indicates that this process releases quite a bit of energy as can be observed in the video below.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | What is the standard change in enthalpy of formation of #"MgO"(s)#? |
The reaction is:
#"Mg"(s) + 1/2 "O"_2(g) -> "MgO"(s)#
|
3,190 | a8c0ed08-6ddd-11ea-ad61-ccda262736ce | https://socratic.org/questions/5a8095feb72cff7df9896019 | 4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s) | start chemical_equation qc_end chemical_equation 6 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)"}] | [{"type":"chemical equation","value":"4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)"}] | <h1 class="questionTitle" itemprop="name">How do you balance the equation 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ?</h1> | null | 4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>write the equation of the reaction and balnce it <br/>
4 <mathjax>#Fe#</mathjax>(s) + 3 <mathjax>#O_2#</mathjax>(g) <mathjax>#->#</mathjax> 2 <mathjax>#Fe_2O_3#</mathjax>(s)<br/>
If you have problems with balancing this chemical equation, here is the whole process explained: <br/>
<iframe src="https://www.youtube.com/embed/Jm1I9wJISrQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</li>
<li>put iron and oxygen [the reactants] in ratio to extract the number of moles for oxygen <br/>
On the picture below you can see how to work it out.</li>
</ol>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/el3ZAKoiRgGUDzWqLfyZ_New%20Doc%202018-02-11%20%281%29_1%20%281%29.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>I got 0,666 mol</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>write the equation of the reaction and balnce it <br/>
4 <mathjax>#Fe#</mathjax>(s) + 3 <mathjax>#O_2#</mathjax>(g) <mathjax>#->#</mathjax> 2 <mathjax>#Fe_2O_3#</mathjax>(s)<br/>
If you have problems with balancing this chemical equation, here is the whole process explained: <br/>
<iframe src="https://www.youtube.com/embed/Jm1I9wJISrQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</li>
<li>put iron and oxygen [the reactants] in ratio to extract the number of moles for oxygen <br/>
On the picture below you can see how to work it out.</li>
</ol>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/el3ZAKoiRgGUDzWqLfyZ_New%20Doc%202018-02-11%20%281%29_1%20%281%29.jpg"/> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance the equation 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ?</h1>
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Marta O.
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Feb 11, 2018
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<div class="markdown"><p>I got 0,666 mol</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>write the equation of the reaction and balnce it <br/>
4 <mathjax>#Fe#</mathjax>(s) + 3 <mathjax>#O_2#</mathjax>(g) <mathjax>#->#</mathjax> 2 <mathjax>#Fe_2O_3#</mathjax>(s)<br/>
If you have problems with balancing this chemical equation, here is the whole process explained: <br/>
<iframe src="https://www.youtube.com/embed/Jm1I9wJISrQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</li>
<li>put iron and oxygen [the reactants] in ratio to extract the number of moles for oxygen <br/>
On the picture below you can see how to work it out.</li>
</ol>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/el3ZAKoiRgGUDzWqLfyZ_New%20Doc%202018-02-11%20%281%29_1%20%281%29.jpg"/> </p></div>
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</article> | How do you balance the equation 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ? | null |
3,191 | ac201f14-6ddd-11ea-823a-ccda262736ce | https://socratic.org/questions/how-many-moles-of-a-pure-gas-are-in-a-30-liter-scuba-canister-if-the-temperature | 0.49 moles | start physical_unit 5 6 mole mol qc_end physical_unit 5 6 10 11 volume qc_end physical_unit 5 6 21 22 temperature qc_end physical_unit 5 6 27 28 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] pure gas [IN] moles"}] | [{"type":"physical unit","value":"0.49 moles"}] | [{"type":"physical unit","value":"Volume [OF] pure gas [=] \\pu{30 liters}"},{"type":"physical unit","value":"Temperature [OF] pure gas [=] \\pu{27 ℃}"},{"type":"physical unit","value":"Pressure [OF] pure gas [=] \\pu{0.4 atmospheres}"}] | <h1 class="questionTitle" itemprop="name">How many moles of a pure gas are in a 30 liter scuba canister if the temperature of the canister is 27°C and the pressure is 0.4 atmospheres? </h1> | null | 0.49 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n=(PV)/(RT)=(0.4*cancel(atm)xx30*cancelL)/(0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)#</mathjax></p>
<p><mathjax>#=0.487*mol#</mathjax></p>
<p>This is not a very realistic problem given that few scuba tanks would have LESS than <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>. Scuba tanks are also filled with air, and NOT a pure gas. For the purposes of this problem, we can still find the molar quantity. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#n=0.487*mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n=(PV)/(RT)=(0.4*cancel(atm)xx30*cancelL)/(0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)#</mathjax></p>
<p><mathjax>#=0.487*mol#</mathjax></p>
<p>This is not a very realistic problem given that few scuba tanks would have LESS than <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>. Scuba tanks are also filled with air, and NOT a pure gas. For the purposes of this problem, we can still find the molar quantity. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of a pure gas are in a 30 liter scuba canister if the temperature of the canister is 27°C and the pressure is 0.4 atmospheres? </h1>
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anor277
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Feb 24, 2017
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<div class="markdown"><p><mathjax>#n=0.487*mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#n=(PV)/(RT)=(0.4*cancel(atm)xx30*cancelL)/(0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)#</mathjax></p>
<p><mathjax>#=0.487*mol#</mathjax></p>
<p>This is not a very realistic problem given that few scuba tanks would have LESS than <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>. Scuba tanks are also filled with air, and NOT a pure gas. For the purposes of this problem, we can still find the molar quantity. </p></div>
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</article> | How many moles of a pure gas are in a 30 liter scuba canister if the temperature of the canister is 27°C and the pressure is 0.4 atmospheres? | null |
3,192 | a87a8e26-6ddd-11ea-bbd1-ccda262736ce | https://socratic.org/questions/the-composition-of-dry-air-is-approximately-78-nitrogen-21-oxygen-and-1-other-ga-1 | 79.03 kPa | start physical_unit 8 8 partial_pressure kpa qc_end physical_unit 3 4 26 27 standard_atmospheric_pressure qc_end end | [{"type":"physical unit","value":"partial pressure [OF] nitrogen [IN] kPa"}] | [{"type":"physical unit","value":"79.03 kPa"}] | [{"type":"physical unit","value":"percent composition [OF] nitrogen in dry air [=] \\pu{78%}"},{"type":"physical unit","value":"percent composition [OF] oxygen in dry air [=] \\pu{21%}"},{"type":"physical unit","value":"percent composition [OF] other gases in dry air [=] \\pu{1%}"},{"type":"physical unit","value":"standard atmospheric pressure [OF] dry air [=] \\pu{101.3 kPa}"}] | <h1 class="questionTitle" itemprop="name">The composition of dry air is approximately 78% nitrogen, 21% oxygen, and 1% other gases. What is the partial pressure of nitrogen at standard atmospheric pressure (101.3 kPa)? </h1> | null | 79.03 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Dalton's law of partial pressures, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas is equal to <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of the gas multiplied by the total pressure. Thus, the partial pressure of <mathjax>#N_2#</mathjax> (in <mathjax>#kPa#</mathjax>)is</p>
<p><mathjax>#P_(N_2) = (0.78)(101.325 kPa) = 79.0335 kPa#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#79.0335 kPa#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to Dalton's law of partial pressures, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas is equal to <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of the gas multiplied by the total pressure. Thus, the partial pressure of <mathjax>#N_2#</mathjax> (in <mathjax>#kPa#</mathjax>)is</p>
<p><mathjax>#P_(N_2) = (0.78)(101.325 kPa) = 79.0335 kPa#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The composition of dry air is approximately 78% nitrogen, 21% oxygen, and 1% other gases. What is the partial pressure of nitrogen at standard atmospheric pressure (101.3 kPa)? </h1>
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Nathan L.
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May 19, 2017
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<div class="markdown"><p><mathjax>#79.0335 kPa#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>According to Dalton's law of partial pressures, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas is equal to <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of the gas multiplied by the total pressure. Thus, the partial pressure of <mathjax>#N_2#</mathjax> (in <mathjax>#kPa#</mathjax>)is</p>
<p><mathjax>#P_(N_2) = (0.78)(101.325 kPa) = 79.0335 kPa#</mathjax></p></div>
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</article> | The composition of dry air is approximately 78% nitrogen, 21% oxygen, and 1% other gases. What is the partial pressure of nitrogen at standard atmospheric pressure (101.3 kPa)? | null |
3,193 | a977d61c-6ddd-11ea-be73-ccda262736ce | https://socratic.org/questions/59a04b217c01495b712b0c8d | 4.82 × 10^24 | start physical_unit 2 3 number none qc_end end | [{"type":"physical unit","value":"Number [OF] valence electrons"}] | [{"type":"physical unit","value":"4.82 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] chloride anions [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How many valence electrons in a mole of chloride anions?</h1> | null | 4.82 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus in a mole of chloride anions, there are <mathjax>#N_Axx8*"electrons"#</mathjax>, i.e. approx. <mathjax>#48xx10^23*"valence electrons."#</mathjax></p></div>
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<div class="markdown"><p>Per isolated chloride ion, there are 8 <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus in a mole of chloride anions, there are <mathjax>#N_Axx8*"electrons"#</mathjax>, i.e. approx. <mathjax>#48xx10^23*"valence electrons."#</mathjax></p></div>
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<div class="markdown"><p>Per isolated chloride ion, there are 8 <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>.........</p></div>
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<div class="markdown"><p>And thus in a mole of chloride anions, there are <mathjax>#N_Axx8*"electrons"#</mathjax>, i.e. approx. <mathjax>#48xx10^23*"valence electrons."#</mathjax></p></div>
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</article> | How many valence electrons in a mole of chloride anions? | null |
3,194 | ab03555e-6ddd-11ea-8edf-ccda262736ce | https://socratic.org/questions/how-do-you-find-the-volume-of-concentrated-18m-sulfuric-acid-to-prepare-250ml-of | 83.33 mL | start physical_unit 20 20 volume ml qc_end physical_unit 10 11 18 19 molarity qc_end physical_unit 20 20 14 15 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] concentrated sulfuric acid solution [IN] mL"}] | [{"type":"physical unit","value":"83.33 mL"}] | [{"type":"physical unit","value":"Molarity2 [OF] concentrated sulfuric acid solution [=] \\pu{18 M}"},{"type":"physical unit","value":"Molarity1 [OF] sulfuric acid solution [=] \\pu{6.0 M}"},{"type":"physical unit","value":"Volume1 [OF] sulfuric acid solution [=] \\pu{250 mL}"}] | <h1 class="questionTitle" itemprop="name">How do you find the volume of concentrated 18M sulfuric acid to prepare 250mL of a 6.0M solution?</h1> | null | 83.33 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We require <mathjax>#250*mL#</mathjax> of a <mathjax>#6.0*mol*L^-1#</mathjax> solution (with respect to <mathjax>#H_2SO_4)#</mathjax>.</p>
<p>With respect to <mathjax>#H_2SO_4#</mathjax>, this represents a molar quantity of <mathjax>#0.250*Lxx6.0*mol*L^-1=1.50*mol#</mathjax>.</p>
<p>If the mother solution has a <mathjax>#18*mol*L^-1#</mathjax> concentration, then we solve the quotient, <mathjax>#(1.50*cancel(mol))/(18*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)=??*mL#</mathjax>.</p>
<p>Alternatively we could use the following formula for the molar equivalence:</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>.............</p>
<p><mathjax>#"Important, so get it right................."#</mathjax></p>
<p><mathjax>#"WHEN YOU DO THESE DILUTIONS YOU ALWAYS ADD"#</mathjax> <br/>
<mathjax>#"ACID TO WATER!!!!"#</mathjax></p>
<p><mathjax>#"Why?"#</mathjax> BECAUSE IF YOU SPIT IN ACID IT SPITS BACK AT YOU. I am not joking. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>We use the relationship: <mathjax>#"concentration"="Moles of solute"/"Volume of solution"#</mathjax>, and get under <mathjax>#100*mL#</mathjax> of conc. acid. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We require <mathjax>#250*mL#</mathjax> of a <mathjax>#6.0*mol*L^-1#</mathjax> solution (with respect to <mathjax>#H_2SO_4)#</mathjax>.</p>
<p>With respect to <mathjax>#H_2SO_4#</mathjax>, this represents a molar quantity of <mathjax>#0.250*Lxx6.0*mol*L^-1=1.50*mol#</mathjax>.</p>
<p>If the mother solution has a <mathjax>#18*mol*L^-1#</mathjax> concentration, then we solve the quotient, <mathjax>#(1.50*cancel(mol))/(18*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)=??*mL#</mathjax>.</p>
<p>Alternatively we could use the following formula for the molar equivalence:</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>.............</p>
<p><mathjax>#"Important, so get it right................."#</mathjax></p>
<p><mathjax>#"WHEN YOU DO THESE DILUTIONS YOU ALWAYS ADD"#</mathjax> <br/>
<mathjax>#"ACID TO WATER!!!!"#</mathjax></p>
<p><mathjax>#"Why?"#</mathjax> BECAUSE IF YOU SPIT IN ACID IT SPITS BACK AT YOU. I am not joking. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you find the volume of concentrated 18M sulfuric acid to prepare 250mL of a 6.0M solution?</h1>
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<div class="markdown"><p>We use the relationship: <mathjax>#"concentration"="Moles of solute"/"Volume of solution"#</mathjax>, and get under <mathjax>#100*mL#</mathjax> of conc. acid. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We require <mathjax>#250*mL#</mathjax> of a <mathjax>#6.0*mol*L^-1#</mathjax> solution (with respect to <mathjax>#H_2SO_4)#</mathjax>.</p>
<p>With respect to <mathjax>#H_2SO_4#</mathjax>, this represents a molar quantity of <mathjax>#0.250*Lxx6.0*mol*L^-1=1.50*mol#</mathjax>.</p>
<p>If the mother solution has a <mathjax>#18*mol*L^-1#</mathjax> concentration, then we solve the quotient, <mathjax>#(1.50*cancel(mol))/(18*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)=??*mL#</mathjax>.</p>
<p>Alternatively we could use the following formula for the molar equivalence:</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>.............</p>
<p><mathjax>#"Important, so get it right................."#</mathjax></p>
<p><mathjax>#"WHEN YOU DO THESE DILUTIONS YOU ALWAYS ADD"#</mathjax> <br/>
<mathjax>#"ACID TO WATER!!!!"#</mathjax></p>
<p><mathjax>#"Why?"#</mathjax> BECAUSE IF YOU SPIT IN ACID IT SPITS BACK AT YOU. I am not joking. </p></div>
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</article> | How do you find the volume of concentrated 18M sulfuric acid to prepare 250mL of a 6.0M solution? | null |
3,195 | a913d09c-6ddd-11ea-98e5-ccda262736ce | https://socratic.org/questions/a-solution-is-prepared-by-dissolving-40-0-g-of-sucrose-in-250-g-of-water-at-25-c | 23.55 mmHg | start physical_unit 24 25 vapor_pressure mmhg qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 14 14 36 37 vapor_pressure qc_end end | [{"type":"physical unit","value":"Vapor pressure [OF] the solution [IN] mmHg"}] | [{"type":"physical unit","value":"23.55 mmHg"}] | [{"type":"physical unit","value":"Mass [OF] sucrose [=] \\pu{40.0 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{250 g}"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Vapor pressure [OF] water [=] \\pu{23.76 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A solution is prepared by dissolving 40.0 g of sucrose, in 250 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?</h1> | null | 23.55 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Raoults Law</p>
<p><mathjax># p_{i}=p_{i}^{\star }x_{i},#</mathjax></p>
<p><mathjax>#" Calculate number of moles in each of the substance"#</mathjax></p>
<p><mathjax>#"Mol of H20 = 250 g H2O * 1 mol/18.02 g H2O = 13.9 mol H2O"#</mathjax></p>
<p><mathjax>#"Mol C12H22O11 = 40.0 g C12H22O11 * 1 mol/342.296 g = 0.12 mol sucrose"#</mathjax></p>
<p>Now calculate vapour pressure</p>
<p><mathjax>#"Mole fraction H2O = 13.9 mol H_2O/(0.12 mol+ 13.9 mol) = 0.991"#</mathjax> </p>
<p><mathjax>#"Partial pressure of soln = 0.991 * 23.76 = 23.546 mm Hg"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#23.6mmHg#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Raoults Law</p>
<p><mathjax># p_{i}=p_{i}^{\star }x_{i},#</mathjax></p>
<p><mathjax>#" Calculate number of moles in each of the substance"#</mathjax></p>
<p><mathjax>#"Mol of H20 = 250 g H2O * 1 mol/18.02 g H2O = 13.9 mol H2O"#</mathjax></p>
<p><mathjax>#"Mol C12H22O11 = 40.0 g C12H22O11 * 1 mol/342.296 g = 0.12 mol sucrose"#</mathjax></p>
<p>Now calculate vapour pressure</p>
<p><mathjax>#"Mole fraction H2O = 13.9 mol H_2O/(0.12 mol+ 13.9 mol) = 0.991"#</mathjax> </p>
<p><mathjax>#"Partial pressure of soln = 0.991 * 23.76 = 23.546 mm Hg"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A solution is prepared by dissolving 40.0 g of sucrose, in 250 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?</h1>
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<div class="markdown"><p><mathjax>#23.6mmHg#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Raoults Law</p>
<p><mathjax># p_{i}=p_{i}^{\star }x_{i},#</mathjax></p>
<p><mathjax>#" Calculate number of moles in each of the substance"#</mathjax></p>
<p><mathjax>#"Mol of H20 = 250 g H2O * 1 mol/18.02 g H2O = 13.9 mol H2O"#</mathjax></p>
<p><mathjax>#"Mol C12H22O11 = 40.0 g C12H22O11 * 1 mol/342.296 g = 0.12 mol sucrose"#</mathjax></p>
<p>Now calculate vapour pressure</p>
<p><mathjax>#"Mole fraction H2O = 13.9 mol H_2O/(0.12 mol+ 13.9 mol) = 0.991"#</mathjax> </p>
<p><mathjax>#"Partial pressure of soln = 0.991 * 23.76 = 23.546 mm Hg"#</mathjax></p></div>
</div>
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</article> | A solution is prepared by dissolving 40.0 g of sucrose, in 250 g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg? | null |
3,196 | ad1bcd80-6ddd-11ea-996e-ccda262736ce | https://socratic.org/questions/how-would-you-balance-c8-h18-o2-co2-h20 | 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O"}] | [{"type":"chemical equation","value":"C8H18 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How would you balance
C8 H18 + O2 = CO2 +H20?</h1> | null | 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I were to combust heptane, I would formulate the combustion rxn this way:</p>
<p><mathjax>#C_7H_16 + 11O_2 rarr 7CO_2 + 8H_2O#</mathjax></p>
<p>You have octane. How are you going to modify the equation?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>You have octane. Under complete combustion there MUST be 8 equivalents carbon dioxide as a consequence. Water is the other product; how many equiv water result? And how many equiv oxygen gas are used?</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I were to combust heptane, I would formulate the combustion rxn this way:</p>
<p><mathjax>#C_7H_16 + 11O_2 rarr 7CO_2 + 8H_2O#</mathjax></p>
<p>You have octane. How are you going to modify the equation?</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance
C8 H18 + O2 = CO2 +H20?</h1>
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<div class="markdown"><p>You have octane. Under complete combustion there MUST be 8 equivalents carbon dioxide as a consequence. Water is the other product; how many equiv water result? And how many equiv oxygen gas are used?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I were to combust heptane, I would formulate the combustion rxn this way:</p>
<p><mathjax>#C_7H_16 + 11O_2 rarr 7CO_2 + 8H_2O#</mathjax></p>
<p>You have octane. How are you going to modify the equation?</p></div>
</div>
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</article> | How would you balance
C8 H18 + O2 = CO2 +H20? | null |
3,197 | ac6dcb91-6ddd-11ea-b0b8-ccda262736ce | https://socratic.org/questions/when-435-j-of-heat-is-added-to-3-40-g-of-olive-oil-at-21-0-c-the-temperature-inc | 1.51 J/(g * ℃) | start physical_unit 28 30 specific_heat j/(°c_·_g) qc_end physical_unit 28 30 1 2 heat_energy qc_end physical_unit 28 30 8 9 mass qc_end physical_unit 28 30 14 15 temperature qc_end physical_unit 28 30 20 21 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] the olive oil [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"1.51 J/(g * ℃)"}] | [{"type":"physical unit","value":"Added heat [OF] the olive oil [=] \\pu{435 J}"},{"type":"physical unit","value":"Mass [OF] the olive oil [=] \\pu{3.40 g}"},{"type":"physical unit","value":"Temperature [OF] the olive oil [=] \\pu{21.0 ℃}"},{"type":"physical unit","value":"Increased temperature [OF] the olive oil [=] \\pu{85 ℃}"}] | <h1 class="questionTitle" itemprop="name">When 435 J of heat is added to 3.40 g of olive oil at 21.0°C, the temperature increases by 85°C. What is the specific heat of the olive oil?</h1> | null | 1.51 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> (<mathjax>#"S"#</mathjax>) is given by the formula</p>
<blockquote>
<p><mathjax>#"S" = "Q"/"mΔT"#</mathjax></p>
</blockquote>
<p>Where</p>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat added/released </li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in Temperature</li>
</ul>
<p><mathjax>#"S" = "435 J"/"3.40 g × 85°C" = "1.51 J/g°C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.51 J/g°C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> (<mathjax>#"S"#</mathjax>) is given by the formula</p>
<blockquote>
<p><mathjax>#"S" = "Q"/"mΔT"#</mathjax></p>
</blockquote>
<p>Where</p>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat added/released </li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in Temperature</li>
</ul>
<p><mathjax>#"S" = "435 J"/"3.40 g × 85°C" = "1.51 J/g°C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When 435 J of heat is added to 3.40 g of olive oil at 21.0°C, the temperature increases by 85°C. What is the specific heat of the olive oil?</h1>
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<div class="markdown"><p><mathjax>#"1.51 J/g°C"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> (<mathjax>#"S"#</mathjax>) is given by the formula</p>
<blockquote>
<p><mathjax>#"S" = "Q"/"mΔT"#</mathjax></p>
</blockquote>
<p>Where</p>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat added/released </li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in Temperature</li>
</ul>
<p><mathjax>#"S" = "435 J"/"3.40 g × 85°C" = "1.51 J/g°C"#</mathjax></p></div>
</div>
</div>
</div>
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</article> | When 435 J of heat is added to 3.40 g of olive oil at 21.0°C, the temperature increases by 85°C. What is the specific heat of the olive oil? | null |
3,198 | aa1738e2-6ddd-11ea-9d13-ccda262736ce | https://socratic.org/questions/a-gas-at-155-kpa-and-25-c-has-an-initial-volume-of-1-00-l-the-pressure-of-the-ga-1 | 0.34 L | start physical_unit 18 19 volume l qc_end physical_unit 18 19 3 4 pressure qc_end physical_unit 18 19 6 7 temperature qc_end physical_unit 18 19 13 14 volume qc_end physical_unit 18 19 22 23 pressure qc_end physical_unit 18 19 30 31 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.34 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{155 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 °C}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{605 kPa}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{125 °C}"}] | <h1 class="questionTitle" itemprop="name">A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?</h1> | null | 0.34 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V_2=0.342*L#</mathjax>...the gas is compressed....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?</h1>
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<div class="markdown"><p><mathjax>#V_2=0.342*L#</mathjax>...the gas is compressed....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
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</article> | A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? | null |
3,199 | ac4704cb-6ddd-11ea-8ca2-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-lithium-sulfide | Li2S | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] lithium sulfide [IN] default"}] | [{"type":"chemical equation","value":"Li2S"}] | [{"type":"substance name","value":"Lithium sulfide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for lithium sulfide?</h1> | null | Li2S | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium is a Group 1 metal .... it commonly forms a <mathjax>#Li^+#</mathjax> ion... Sulfur is a Group 16 non-metal...upon reduction it commonly forms a <mathjax>#S^(2-)#</mathjax> ion...</p>
<p>And so when we make the NEUTRAL SALT we gets...</p>
<p><mathjax>#S^(2-)+2xxLi^+ -=Li_2S#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Li_2S#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium is a Group 1 metal .... it commonly forms a <mathjax>#Li^+#</mathjax> ion... Sulfur is a Group 16 non-metal...upon reduction it commonly forms a <mathjax>#S^(2-)#</mathjax> ion...</p>
<p>And so when we make the NEUTRAL SALT we gets...</p>
<p><mathjax>#S^(2-)+2xxLi^+ -=Li_2S#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for lithium sulfide?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Li_2S#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium is a Group 1 metal .... it commonly forms a <mathjax>#Li^+#</mathjax> ion... Sulfur is a Group 16 non-metal...upon reduction it commonly forms a <mathjax>#S^(2-)#</mathjax> ion...</p>
<p>And so when we make the NEUTRAL SALT we gets...</p>
<p><mathjax>#S^(2-)+2xxLi^+ -=Li_2S#</mathjax></p></div>
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</article> | What is the formula for lithium sulfide? | null |
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