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3,000 | a9f3a80b-6ddd-11ea-87ea-ccda262736ce | https://socratic.org/questions/how-do-you-balance-albr-3-k-2so-4-kbr-al-2-so-4-3 | 2 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3 | 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 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3"}] | [{"type":"chemical equation","value":"AlBr3 + K2SO4 -> KBr + Al2(SO4)3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#?</h1> | null | 2 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this equation, all you have to do is make sure that there is an equal number of each element in both the reactant and product side.</p>
<p>Step by step balancing:</p>
<p><mathjax>#AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 2 S - 1 O - 4<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> 6KBr + Al_2(SO_4)_3#</mathjax> (balanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2AlBr_3 + 3K_2SO_4 rarr 6KBr + Al_2(SO_4)_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this equation, all you have to do is make sure that there is an equal number of each element in both the reactant and product side.</p>
<p>Step by step balancing:</p>
<p><mathjax>#AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 2 S - 1 O - 4<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> 6KBr + Al_2(SO_4)_3#</mathjax> (balanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#?</h1>
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<div class="markdown"><p><mathjax>#2AlBr_3 + 3K_2SO_4 rarr 6KBr + Al_2(SO_4)_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this equation, all you have to do is make sure that there is an equal number of each element in both the reactant and product side.</p>
<p>Step by step balancing:</p>
<p><mathjax>#AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 2 S - 1 O - 4<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 1 Br - 3 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> KBr + Al_2(SO_4)_3#</mathjax> (unbalanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 1 K - 1 S - 3 O - 12</p>
<p><mathjax>#2AlBr_3 + 3K_2SO_4 -> 6KBr + Al_2(SO_4)_3#</mathjax> (balanced)</p>
<p>reactants:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12<br/>
products:<br/>
Al - 2 Br - 6 K - 6 S - 3 O - 12</p></div>
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</article> | How do you balance #AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3#? | null |
3,001 | a90bb852-6ddd-11ea-aaac-ccda262736ce | https://socratic.org/questions/if-the-h-ion-concentration-is-0-00075-m-what-is-the-oh-ion-concentration | 1.33 × 10^(-11) M | start physical_unit 11 12 concentration mol/l qc_end physical_unit 2 3 6 7 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] OH- ion [IN] M"}] | [{"type":"physical unit","value":"1.33 × 10^(-11) M"}] | [{"type":"physical unit","value":"Concentration [OF] H+ ion [=] \\pu{0.00075 M}"}] | <h1 class="questionTitle" itemprop="name">If the H ion concentration is 0.00075 M, what is the OH ion concentration?</h1> | null | 1.33 × 10^(-11) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that <mathjax>#[HO^-][H_3O^+]=K_w=10^-14#</mathjax></p>
<p>If we take <mathjax>#log_10#</mathjax> of both sides, then:</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]#</mathjax></p>
<p>But by definition, when I say <mathjax>#log_ab=c#</mathjax>, I ask for the power to which I raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>. So since <mathjax>#log_ab=c#</mathjax>, then <mathjax>#a^c=b#</mathjax>.</p>
<p>And likewise, <mathjax>#log_10[10^-14]=-14#</mathjax>, because clearly the exponent of <mathjax>#10^-14=-14#</mathjax>.</p>
<p>And so <mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]=-14#</mathjax></p>
<p>OR,</p>
<p><mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax>,</p>
<p>i.e. <mathjax>#14=pH+pOH#</mathjax>, because that is how we define <mathjax>#pH#</mathjax> etc., i.e. <mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>We are given that <mathjax>#[H_3O^+]=0.00075*mol*L^-1#</mathjax></p>
<p>And <mathjax>#pH=-log_10(0.00075)=-(-3.13)=3.13#</mathjax></p>
<p>So <mathjax>#pOH=10.88#</mathjax>. </p>
<p>And <mathjax>#[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1#</mathjax>.</p>
<p>We assume an aqueous solution under standard conditions. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that <mathjax>#[HO^-][H_3O^+]=K_w=10^-14#</mathjax></p>
<p>If we take <mathjax>#log_10#</mathjax> of both sides, then:</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]#</mathjax></p>
<p>But by definition, when I say <mathjax>#log_ab=c#</mathjax>, I ask for the power to which I raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>. So since <mathjax>#log_ab=c#</mathjax>, then <mathjax>#a^c=b#</mathjax>.</p>
<p>And likewise, <mathjax>#log_10[10^-14]=-14#</mathjax>, because clearly the exponent of <mathjax>#10^-14=-14#</mathjax>.</p>
<p>And so <mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]=-14#</mathjax></p>
<p>OR,</p>
<p><mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax>,</p>
<p>i.e. <mathjax>#14=pH+pOH#</mathjax>, because that is how we define <mathjax>#pH#</mathjax> etc., i.e. <mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>We are given that <mathjax>#[H_3O^+]=0.00075*mol*L^-1#</mathjax></p>
<p>And <mathjax>#pH=-log_10(0.00075)=-(-3.13)=3.13#</mathjax></p>
<p>So <mathjax>#pOH=10.88#</mathjax>. </p>
<p>And <mathjax>#[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">If the H ion concentration is 0.00075 M, what is the OH ion concentration?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1#</mathjax>.</p>
<p>We assume an aqueous solution under standard conditions. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that <mathjax>#[HO^-][H_3O^+]=K_w=10^-14#</mathjax></p>
<p>If we take <mathjax>#log_10#</mathjax> of both sides, then:</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]#</mathjax></p>
<p>But by definition, when I say <mathjax>#log_ab=c#</mathjax>, I ask for the power to which I raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>. So since <mathjax>#log_ab=c#</mathjax>, then <mathjax>#a^c=b#</mathjax>.</p>
<p>And likewise, <mathjax>#log_10[10^-14]=-14#</mathjax>, because clearly the exponent of <mathjax>#10^-14=-14#</mathjax>.</p>
<p>And so <mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]=-14#</mathjax></p>
<p>OR,</p>
<p><mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax>,</p>
<p>i.e. <mathjax>#14=pH+pOH#</mathjax>, because that is how we define <mathjax>#pH#</mathjax> etc., i.e. <mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>We are given that <mathjax>#[H_3O^+]=0.00075*mol*L^-1#</mathjax></p>
<p>And <mathjax>#pH=-log_10(0.00075)=-(-3.13)=3.13#</mathjax></p>
<p>So <mathjax>#pOH=10.88#</mathjax>. </p>
<p>And <mathjax>#[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1#</mathjax>.</p></div>
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</article> | If the H ion concentration is 0.00075 M, what is the OH ion concentration? | null |
3,002 | ab4cf47b-6ddd-11ea-815b-ccda262736ce | https://socratic.org/questions/in-this-reaction-2al-3-2o2-formed-al2o3-9-gm-of-al-will-react-with-how-many-gm-o | 8 gm | start physical_unit 7 7 mass g qc_end physical_unit 4 4 10 11 mass qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 [IN] gm"}] | [{"type":"physical unit","value":"8 gm"}] | [{"type":"physical unit","value":"Mass [OF] Al [=] \\pu{9 gm}"},{"type":"chemical equation","value":"Al2O3"}] | <h1 class="questionTitle" itemprop="name">In this reaction 2Al+3/2O2 formed Al2O3 ,9 gm of Al will react with how many gm of O2?</h1> | null | 8 gm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction (more visibly) is</p>
<p><mathjax>#2Al"(s)" + 3/2O_2"(g)" rarr Al_2O_3"(s)"#</mathjax></p>
<p>To solve this problem, we'll need to convert from mass to moles the <mathjax>#Al#</mathjax>, then use the stoichiometric relationships (the coefficients) to find the moles of <mathjax>#O_2#</mathjax>, then use oxygen's molar mass to find the mass of <mathjax>#O_2#</mathjax> required:</p>
<p><mathjax>#9cancel("g"Al)((1cancel("mol"Al))/(26.98cancel("g"Al)))((3/2cancel("mol"O_2))/(2cancel("mol"Al)))((32.00"g"O_2)/(1cancel("mol"O_2))) = color(red)(8"g"O_2)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#8"g"O_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction (more visibly) is</p>
<p><mathjax>#2Al"(s)" + 3/2O_2"(g)" rarr Al_2O_3"(s)"#</mathjax></p>
<p>To solve this problem, we'll need to convert from mass to moles the <mathjax>#Al#</mathjax>, then use the stoichiometric relationships (the coefficients) to find the moles of <mathjax>#O_2#</mathjax>, then use oxygen's molar mass to find the mass of <mathjax>#O_2#</mathjax> required:</p>
<p><mathjax>#9cancel("g"Al)((1cancel("mol"Al))/(26.98cancel("g"Al)))((3/2cancel("mol"O_2))/(2cancel("mol"Al)))((32.00"g"O_2)/(1cancel("mol"O_2))) = color(red)(8"g"O_2)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In this reaction 2Al+3/2O2 formed Al2O3 ,9 gm of Al will react with how many gm of O2?</h1>
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Nathan L.
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May 22, 2017
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<div class="markdown"><p><mathjax>#8"g"O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction (more visibly) is</p>
<p><mathjax>#2Al"(s)" + 3/2O_2"(g)" rarr Al_2O_3"(s)"#</mathjax></p>
<p>To solve this problem, we'll need to convert from mass to moles the <mathjax>#Al#</mathjax>, then use the stoichiometric relationships (the coefficients) to find the moles of <mathjax>#O_2#</mathjax>, then use oxygen's molar mass to find the mass of <mathjax>#O_2#</mathjax> required:</p>
<p><mathjax>#9cancel("g"Al)((1cancel("mol"Al))/(26.98cancel("g"Al)))((3/2cancel("mol"O_2))/(2cancel("mol"Al)))((32.00"g"O_2)/(1cancel("mol"O_2))) = color(red)(8"g"O_2)#</mathjax></p></div>
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</article> | In this reaction 2Al+3/2O2 formed Al2O3 ,9 gm of Al will react with how many gm of O2? | null |
3,003 | a905eee2-6ddd-11ea-b249-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-oxygen-in-n-2o-5 | 25.93% | start physical_unit 8 10 percent_composition none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Percentage composition [OF] oxygen in N2O5"}] | [{"type":"physical unit","value":"25.93%"}] | [{"type":"chemical equation","value":"N2O5"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the percentage composition of Oxygen in #N_2O_5#?</h1> | null | 25.93% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first need refrence a periodic table to find the weight of <mathjax>#N#</mathjax> and <mathjax>#O#</mathjax></p>
<p><mathjax>#N#</mathjax> = <mathjax>#14.0g#</mathjax><br/>
<mathjax>#O#</mathjax> = <mathjax>#16.0g#</mathjax></p>
<p>Multiply each element based on how many of each element are in the compound. for <mathjax>#N#</mathjax>, it is 2. For <mathjax>#O#</mathjax>, it is 5.</p>
<p><mathjax>#14.0g#</mathjax> <mathjax>#O#</mathjax> * 2 = <mathjax>#28.0g#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#16.0g#</mathjax> <mathjax>#O#</mathjax> * 5 = <mathjax>#80.0g#</mathjax> <mathjax>#O#</mathjax></p>
<p>add the two weights together to get the total weight of <mathjax>#N_2O_5#</mathjax></p>
<p><mathjax>#28.0g#</mathjax> + <mathjax>#80.0g#</mathjax> = <mathjax>#108g#</mathjax> <mathjax>#N_2O_5#</mathjax></p>
<p>Divide each <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> weight by the total weight of the compound and multiply by 100 and you will have the percentage composition</p>
<p><mathjax>#(28.0gN)/(108g)#</mathjax> * 100 = <mathjax>#25.93%#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#(80.0gO)/(108g)#</mathjax> * 100 = <mathjax>#74.07%#</mathjax> <mathjax>#O#</mathjax> </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#25.93%#</mathjax> <mathjax>#N#</mathjax>, <mathjax>#74.07%#</mathjax><mathjax>#O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first need refrence a periodic table to find the weight of <mathjax>#N#</mathjax> and <mathjax>#O#</mathjax></p>
<p><mathjax>#N#</mathjax> = <mathjax>#14.0g#</mathjax><br/>
<mathjax>#O#</mathjax> = <mathjax>#16.0g#</mathjax></p>
<p>Multiply each element based on how many of each element are in the compound. for <mathjax>#N#</mathjax>, it is 2. For <mathjax>#O#</mathjax>, it is 5.</p>
<p><mathjax>#14.0g#</mathjax> <mathjax>#O#</mathjax> * 2 = <mathjax>#28.0g#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#16.0g#</mathjax> <mathjax>#O#</mathjax> * 5 = <mathjax>#80.0g#</mathjax> <mathjax>#O#</mathjax></p>
<p>add the two weights together to get the total weight of <mathjax>#N_2O_5#</mathjax></p>
<p><mathjax>#28.0g#</mathjax> + <mathjax>#80.0g#</mathjax> = <mathjax>#108g#</mathjax> <mathjax>#N_2O_5#</mathjax></p>
<p>Divide each <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> weight by the total weight of the compound and multiply by 100 and you will have the percentage composition</p>
<p><mathjax>#(28.0gN)/(108g)#</mathjax> * 100 = <mathjax>#25.93%#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#(80.0gO)/(108g)#</mathjax> * 100 = <mathjax>#74.07%#</mathjax> <mathjax>#O#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the percentage composition of Oxygen in #N_2O_5#?</h1>
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<div class="markdown"><p><mathjax>#25.93%#</mathjax> <mathjax>#N#</mathjax>, <mathjax>#74.07%#</mathjax><mathjax>#O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You first need refrence a periodic table to find the weight of <mathjax>#N#</mathjax> and <mathjax>#O#</mathjax></p>
<p><mathjax>#N#</mathjax> = <mathjax>#14.0g#</mathjax><br/>
<mathjax>#O#</mathjax> = <mathjax>#16.0g#</mathjax></p>
<p>Multiply each element based on how many of each element are in the compound. for <mathjax>#N#</mathjax>, it is 2. For <mathjax>#O#</mathjax>, it is 5.</p>
<p><mathjax>#14.0g#</mathjax> <mathjax>#O#</mathjax> * 2 = <mathjax>#28.0g#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#16.0g#</mathjax> <mathjax>#O#</mathjax> * 5 = <mathjax>#80.0g#</mathjax> <mathjax>#O#</mathjax></p>
<p>add the two weights together to get the total weight of <mathjax>#N_2O_5#</mathjax></p>
<p><mathjax>#28.0g#</mathjax> + <mathjax>#80.0g#</mathjax> = <mathjax>#108g#</mathjax> <mathjax>#N_2O_5#</mathjax></p>
<p>Divide each <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> weight by the total weight of the compound and multiply by 100 and you will have the percentage composition</p>
<p><mathjax>#(28.0gN)/(108g)#</mathjax> * 100 = <mathjax>#25.93%#</mathjax> <mathjax>#N#</mathjax></p>
<p><mathjax>#(80.0gO)/(108g)#</mathjax> * 100 = <mathjax>#74.07%#</mathjax> <mathjax>#O#</mathjax> </p></div>
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</article> | How do you calculate the percentage composition of Oxygen in #N_2O_5#? | null |
3,004 | ab6e4686-6ddd-11ea-bcf3-ccda262736ce | https://socratic.org/questions/a-gaseous-hydrocarbon-collected-over-water-at-a-temperature-of-21-c-and-a-barome | 58.87 g/mol | start physical_unit 25 26 molecular_weight g/mol qc_end physical_unit 1 2 10 11 temperature qc_end physical_unit 1 2 17 18 barometric_pressure qc_end physical_unit 1 2 23 24 volume qc_end physical_unit 2 2 31 32 mass qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] the hydrocarbon [IN] g/mol"}] | [{"type":"physical unit","value":"58.87 g/mol"}] | [{"type":"physical unit","value":"Temperature [OF] gaseous hydrocarbon [=] \\pu{21 ℃}"},{"type":"physical unit","value":"Barometric pressure [OF] gaseous hydrocarbon [=] \\pu{753 torr}"},{"type":"physical unit","value":"Volume [OF] gaseous hydrocarbon [=] \\pu{48.1 mL}"},{"type":"physical unit","value":"Weight [OF] hydrocarbon [=] \\pu{0.1133 g}"}] | <h1 class="questionTitle" itemprop="name">A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?</h1> | null | 58.87 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before focusing on anything else, use the vapor pressure of water at <mathjax>#21^@"C"#</mathjax> to determine the actual pressure of the gaseous hydrocarbon.</p>
<p>As you know, gases collected <strong>over water</strong> will also contain water vapor. Simply put, the volume of gas collected over water will contain molecules of hydrocarbon <strong>and</strong> molecules of water. </p>
<p>This means that you can use <strong>Dalton's Law of partial pressures</strong> to determine the exact pressure of the hydrocarbon. At <mathjax>#21.0^@"C"#</mathjax>, water vapor has a pressure of about <mathjax>#"18.59 torr"#</mathjax></p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#P_"mixture" = P_"water" + P_"hydrocarbon"#</mathjax></p>
<p><mathjax>#P_"hydrocarbon" = "753 torr" - "18.59 torr" = "734.41 torr"#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to determine how many <strong>moles</strong> of this gaseous hydrocarbon you have in this sample. </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of mole</em> of gas<br/>
<mathjax>#R#</mathjax> - the universal gas constant, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax>, but make sure that the units you have for pressure, temperature, and volume <strong>match</strong> those used in the expression of the universal gas constant</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (734.41/760color(red)(cancel(color(black)("atm"))) * 48.1 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821( color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 21.0)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#n = "0.0019247 moles"#</mathjax></p>
</blockquote>
<p>So, your hydrocarbon sample has a mass of <mathjax>#"0.1133 g"#</mathjax> and it contains <mathjax>#0.0019247#</mathjax> moles, which means that its <strong>molar mass</strong>, which tells you what the mass of <strong>one mole</strong> of a substance is, will be </p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole"))) * "0.1133 g"/(0.0019247color(red)(cancel(color(black)("moles")))) = "58.87 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#M_M = color(green)("58.9 g/mol")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"58.9 g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before focusing on anything else, use the vapor pressure of water at <mathjax>#21^@"C"#</mathjax> to determine the actual pressure of the gaseous hydrocarbon.</p>
<p>As you know, gases collected <strong>over water</strong> will also contain water vapor. Simply put, the volume of gas collected over water will contain molecules of hydrocarbon <strong>and</strong> molecules of water. </p>
<p>This means that you can use <strong>Dalton's Law of partial pressures</strong> to determine the exact pressure of the hydrocarbon. At <mathjax>#21.0^@"C"#</mathjax>, water vapor has a pressure of about <mathjax>#"18.59 torr"#</mathjax></p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#P_"mixture" = P_"water" + P_"hydrocarbon"#</mathjax></p>
<p><mathjax>#P_"hydrocarbon" = "753 torr" - "18.59 torr" = "734.41 torr"#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to determine how many <strong>moles</strong> of this gaseous hydrocarbon you have in this sample. </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of mole</em> of gas<br/>
<mathjax>#R#</mathjax> - the universal gas constant, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax>, but make sure that the units you have for pressure, temperature, and volume <strong>match</strong> those used in the expression of the universal gas constant</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (734.41/760color(red)(cancel(color(black)("atm"))) * 48.1 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821( color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 21.0)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#n = "0.0019247 moles"#</mathjax></p>
</blockquote>
<p>So, your hydrocarbon sample has a mass of <mathjax>#"0.1133 g"#</mathjax> and it contains <mathjax>#0.0019247#</mathjax> moles, which means that its <strong>molar mass</strong>, which tells you what the mass of <strong>one mole</strong> of a substance is, will be </p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole"))) * "0.1133 g"/(0.0019247color(red)(cancel(color(black)("moles")))) = "58.87 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#M_M = color(green)("58.9 g/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-24T19:48:47" itemprop="dateCreated">
Dec 24, 2015
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<div class="markdown"><p><mathjax>#"58.9 g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before focusing on anything else, use the vapor pressure of water at <mathjax>#21^@"C"#</mathjax> to determine the actual pressure of the gaseous hydrocarbon.</p>
<p>As you know, gases collected <strong>over water</strong> will also contain water vapor. Simply put, the volume of gas collected over water will contain molecules of hydrocarbon <strong>and</strong> molecules of water. </p>
<p>This means that you can use <strong>Dalton's Law of partial pressures</strong> to determine the exact pressure of the hydrocarbon. At <mathjax>#21.0^@"C"#</mathjax>, water vapor has a pressure of about <mathjax>#"18.59 torr"#</mathjax></p>
<p><a href="http://www.endmemo.com/chem/vaporpressurewater.php" rel="nofollow" target="_blank">http://www.endmemo.com/chem/vaporpressurewater.php</a></p>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#P_"mixture" = P_"water" + P_"hydrocarbon"#</mathjax></p>
<p><mathjax>#P_"hydrocarbon" = "753 torr" - "18.59 torr" = "734.41 torr"#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to determine how many <strong>moles</strong> of this gaseous hydrocarbon you have in this sample. </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the <em>number of mole</em> of gas<br/>
<mathjax>#R#</mathjax> - the universal gas constant, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax>, but make sure that the units you have for pressure, temperature, and volume <strong>match</strong> those used in the expression of the universal gas constant</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (734.41/760color(red)(cancel(color(black)("atm"))) * 48.1 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821( color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 21.0)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#n = "0.0019247 moles"#</mathjax></p>
</blockquote>
<p>So, your hydrocarbon sample has a mass of <mathjax>#"0.1133 g"#</mathjax> and it contains <mathjax>#0.0019247#</mathjax> moles, which means that its <strong>molar mass</strong>, which tells you what the mass of <strong>one mole</strong> of a substance is, will be </p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole"))) * "0.1133 g"/(0.0019247color(red)(cancel(color(black)("moles")))) = "58.87 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#M_M = color(green)("58.9 g/mol")#</mathjax></p>
</blockquote></div>
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</article> | A gaseous hydrocarbon collected over water at a temperature of 21° C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. What is the molecular mass of the hydrocarbon? | null |
3,005 | ad18f77b-6ddd-11ea-94fc-ccda262736ce | https://socratic.org/questions/how-many-grams-is-3-30-10-25-molecules-of-i-2 | 13908.79 grams | start physical_unit 9 9 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] I2 [IN] grams"}] | [{"type":"physical unit","value":"13908.79 grams"}] | [{"type":"physical unit","value":"Number [OF] I2 molecules [=] \\pu{3.30 × 10^25}"}] | <h1 class="questionTitle" itemprop="name">How many grams is #3.30 * 10^25# molecules of #I_2#?</h1> | null | 13908.79 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants you to go from <em>molecules</em> of iodine, <mathjax>#"I"_2#</mathjax>, to <em>grams</em> of iodine, which means that you're going to have to go through <strong>moles</strong> of iodine first. </p>
<p>More specifically, your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use <strong>Avogadro's number</strong> to determine how many <strong>moles</strong> of iodine you have in your sample</em></li>
<li><em>use iodine's <strong>molar mass</strong> to determine how many <strong>grams</strong> would contain that many moles</em></li>
</ul>
</blockquote>
<p>So, <strong>Avogadro's number</strong> tells you how many molecules you get <strong>in one mole</strong> of a given substance. In order to have one mole of iodine, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> molecules of iodine. </p>
<p>This means that your sample will be equivalent to </p>
<blockquote>
<p><mathjax>#3.30 * 10^(25)color(red)(cancel(color(black)("molecules I"_2))) * overbrace("1 mole I"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2)))))^(color(brown)("Avogadro's number")) = "54.8 moles I"_2#</mathjax></p>
</blockquote>
<p>Now, molecular iodine has a <strong>molar mass</strong> of <mathjax>#"253.81 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of iodine has mass of <mathjax>#"253.81 g"#</mathjax>. </p>
<p>In your case, the sample you're working with will have a mass of </p>
<blockquote>
<p><mathjax>#54.8 color(red)(cancel(color(black)("moles I"_2))) * overbrace("253.81 g"/(1color(red)(cancel(color(black)("mole I"_2)))))^(color(purple)("molar mass of I"_2)) = "13909 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of iodine, the answer will be</p>
<blockquote>
<p><mathjax>#m_(I_2) = color(green)(|bar(ul(color(white)(a/a)"13,900 g I"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"13,900 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants you to go from <em>molecules</em> of iodine, <mathjax>#"I"_2#</mathjax>, to <em>grams</em> of iodine, which means that you're going to have to go through <strong>moles</strong> of iodine first. </p>
<p>More specifically, your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use <strong>Avogadro's number</strong> to determine how many <strong>moles</strong> of iodine you have in your sample</em></li>
<li><em>use iodine's <strong>molar mass</strong> to determine how many <strong>grams</strong> would contain that many moles</em></li>
</ul>
</blockquote>
<p>So, <strong>Avogadro's number</strong> tells you how many molecules you get <strong>in one mole</strong> of a given substance. In order to have one mole of iodine, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> molecules of iodine. </p>
<p>This means that your sample will be equivalent to </p>
<blockquote>
<p><mathjax>#3.30 * 10^(25)color(red)(cancel(color(black)("molecules I"_2))) * overbrace("1 mole I"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2)))))^(color(brown)("Avogadro's number")) = "54.8 moles I"_2#</mathjax></p>
</blockquote>
<p>Now, molecular iodine has a <strong>molar mass</strong> of <mathjax>#"253.81 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of iodine has mass of <mathjax>#"253.81 g"#</mathjax>. </p>
<p>In your case, the sample you're working with will have a mass of </p>
<blockquote>
<p><mathjax>#54.8 color(red)(cancel(color(black)("moles I"_2))) * overbrace("253.81 g"/(1color(red)(cancel(color(black)("mole I"_2)))))^(color(purple)("molar mass of I"_2)) = "13909 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of iodine, the answer will be</p>
<blockquote>
<p><mathjax>#m_(I_2) = color(green)(|bar(ul(color(white)(a/a)"13,900 g I"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams is #3.30 * 10^25# molecules of #I_2#?</h1>
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<div class="markdown"><p><mathjax>#"13,900 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants you to go from <em>molecules</em> of iodine, <mathjax>#"I"_2#</mathjax>, to <em>grams</em> of iodine, which means that you're going to have to go through <strong>moles</strong> of iodine first. </p>
<p>More specifically, your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use <strong>Avogadro's number</strong> to determine how many <strong>moles</strong> of iodine you have in your sample</em></li>
<li><em>use iodine's <strong>molar mass</strong> to determine how many <strong>grams</strong> would contain that many moles</em></li>
</ul>
</blockquote>
<p>So, <strong>Avogadro's number</strong> tells you how many molecules you get <strong>in one mole</strong> of a given substance. In order to have one mole of iodine, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> molecules of iodine. </p>
<p>This means that your sample will be equivalent to </p>
<blockquote>
<p><mathjax>#3.30 * 10^(25)color(red)(cancel(color(black)("molecules I"_2))) * overbrace("1 mole I"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2)))))^(color(brown)("Avogadro's number")) = "54.8 moles I"_2#</mathjax></p>
</blockquote>
<p>Now, molecular iodine has a <strong>molar mass</strong> of <mathjax>#"253.81 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of iodine has mass of <mathjax>#"253.81 g"#</mathjax>. </p>
<p>In your case, the sample you're working with will have a mass of </p>
<blockquote>
<p><mathjax>#54.8 color(red)(cancel(color(black)("moles I"_2))) * overbrace("253.81 g"/(1color(red)(cancel(color(black)("mole I"_2)))))^(color(purple)("molar mass of I"_2)) = "13909 g"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of iodine, the answer will be</p>
<blockquote>
<p><mathjax>#m_(I_2) = color(green)(|bar(ul(color(white)(a/a)"13,900 g I"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
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</article> | How many grams is #3.30 * 10^25# molecules of #I_2#? | null |
3,006 | ac38f753-6ddd-11ea-8827-ccda262736ce | https://socratic.org/questions/how-many-moles-are-equivalent-to-71-1-grams-of-barium-nitride | 0.16 moles | start physical_unit 9 10 mole mol qc_end physical_unit 9 10 6 7 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] barium nitride [IN] moles"}] | [{"type":"physical unit","value":"0.16 moles"}] | [{"type":"physical unit","value":"Mass [OF] barium nitride [=] \\pu{71.1 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are equivalent to 71.1 grams of barium nitride?</h1> | null | 0.16 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to know the molar mass of barium nitride, <mathjax>#("Ba"_3"N"_2")#</mathjax>, which is <mathjax>#"439.995 g/mol"#</mathjax>. <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/15251597" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/15251597</a></p>
<p>Divide the given mass of barium nitride, by its molar mass. Since molar mass is a fraction, g/mol, I prefer to divide by multiplying by its reciprocal, mol/g.</p>
<p><mathjax>#71.1color(red)cancel(color(black)("g Ba"_3"N"_2))xx(1"mol Ba"_3"N"_2)/(439.995color(red)cancel(color(black)("g Ba"_3"N"_2)))="0.162 mol Ba"_3"N"_2"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"71.1 g Ba"_3"N"_2"#</mathjax> is equivalent to <mathjax>#"0.162 mol Ba"_3"N"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to know the molar mass of barium nitride, <mathjax>#("Ba"_3"N"_2")#</mathjax>, which is <mathjax>#"439.995 g/mol"#</mathjax>. <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/15251597" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/15251597</a></p>
<p>Divide the given mass of barium nitride, by its molar mass. Since molar mass is a fraction, g/mol, I prefer to divide by multiplying by its reciprocal, mol/g.</p>
<p><mathjax>#71.1color(red)cancel(color(black)("g Ba"_3"N"_2))xx(1"mol Ba"_3"N"_2)/(439.995color(red)cancel(color(black)("g Ba"_3"N"_2)))="0.162 mol Ba"_3"N"_2"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"71.1 g Ba"_3"N"_2"#</mathjax> is equivalent to <mathjax>#"0.162 mol Ba"_3"N"_2"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>First you need to know the molar mass of barium nitride, <mathjax>#("Ba"_3"N"_2")#</mathjax>, which is <mathjax>#"439.995 g/mol"#</mathjax>. <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/15251597" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/15251597</a></p>
<p>Divide the given mass of barium nitride, by its molar mass. Since molar mass is a fraction, g/mol, I prefer to divide by multiplying by its reciprocal, mol/g.</p>
<p><mathjax>#71.1color(red)cancel(color(black)("g Ba"_3"N"_2))xx(1"mol Ba"_3"N"_2)/(439.995color(red)cancel(color(black)("g Ba"_3"N"_2)))="0.162 mol Ba"_3"N"_2"#</mathjax></p></div>
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</article> | How many moles are equivalent to 71.1 grams of barium nitride? | null |
3,007 | a8fdc136-6ddd-11ea-a98f-ccda262736ce | https://socratic.org/questions/amixture-was-found-to-contain-1-50-g-nh-4cl-0-80-g-nacl-and-1-20-g-sio-2-what-is | 34.29% | start physical_unit 21 24 percent none qc_end physical_unit 8 8 6 7 mass qc_end physical_unit 11 11 9 10 mass qc_end physical_unit 15 15 13 14 mass qc_end end | [{"type":"physical unit","value":"Prcentage [OF] sand in this mixture"}] | [{"type":"physical unit","value":"34.29%"}] | [{"type":"physical unit","value":"Mass [OF] NH4Cl [=] \\pu{1.50 g}"},{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{0.80 g}"},{"type":"physical unit","value":"Mass [OF] SiO2 [=] \\pu{1.20 g}"}] | <h1 class="questionTitle" itemprop="name">Amixture was found to contain 1.50 g #NH_4Cl#, 0.80 g #NaCl#, and 1.20 g #SiO_2#. What is the percentage of sand in this mixture?</h1> | null | 34.29% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Percentage of sand"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*g)/(1.20*g+1.50*g+0.80*g)xx100%#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#30%?#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Percentage of sand"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of sand"/"Mass of mixture"xx100%#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Percentage of sand"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*g)/(1.20*g+1.50*g+0.80*g)xx100%#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#30%?#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Amixture was found to contain 1.50 g #NH_4Cl#, 0.80 g #NaCl#, and 1.20 g #SiO_2#. What is the percentage of sand in this mixture?</h1>
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<div class="markdown"><p><mathjax>#"Percentage of sand"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of sand"/"Mass of mixture"xx100%#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Percentage of sand"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*g)/(1.20*g+1.50*g+0.80*g)xx100%#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#30%?#</mathjax></p></div>
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</article> | Amixture was found to contain 1.50 g #NH_4Cl#, 0.80 g #NaCl#, and 1.20 g #SiO_2#. What is the percentage of sand in this mixture? | null |
3,008 | ac352200-6ddd-11ea-a638-ccda262736ce | https://socratic.org/questions/how-many-liters-of-water-need-to-be-added-to-22-moles-of-kcl-to-make-a-0-2-m-sol | 110.00 liters | start physical_unit 4 4 volume l qc_end physical_unit 13 13 10 11 mole qc_end physical_unit 13 13 17 18 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] liters"}] | [{"type":"physical unit","value":"110.00 liters"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{22 moles}"},{"type":"physical unit","value":"Molarity [OF] KCl solution [=] \\pu{0.2 M}"}] | <h1 class="questionTitle" itemprop="name">How many liters of water need to be added to 22 moles of #KCl# to make a 0.2 M solution? </h1> | null | 110.00 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of soluion"#</mathjax></p>
<p><mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p>And thus, <br/>
<mathjax>#"Volume of solution"=(22*mol)/(0.2*mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Over <mathjax>#100*L#</mathjax> of water must be added.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of soluion"#</mathjax></p>
<p><mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p>And thus, <br/>
<mathjax>#"Volume of solution"=(22*mol)/(0.2*mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of water need to be added to 22 moles of #KCl# to make a 0.2 M solution? </h1>
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<div class="markdown"><p>Over <mathjax>#100*L#</mathjax> of water must be added.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of soluion"#</mathjax></p>
<p><mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p>And thus, <br/>
<mathjax>#"Volume of solution"=(22*mol)/(0.2*mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p></div>
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</article> | How many liters of water need to be added to 22 moles of #KCl# to make a 0.2 M solution? | null |
3,009 | a8b4e001-6ddd-11ea-8fca-ccda262736ce | https://socratic.org/questions/what-is-the-percent-by-mass-of-nahco-3-in-a-solution-containing-50-g-nahco-3-dis | 6.25% | start physical_unit 7 10 mass_percent none qc_end physical_unit 7 7 12 13 mass qc_end physical_unit 20 20 17 18 volume qc_end end | [{"type":"physical unit","value":"Percent by mass [OF] NaHCO3 in a solution"}] | [{"type":"physical unit","value":"6.25%"}] | [{"type":"physical unit","value":"Mass [OF] NaHCO3 [=] \\pu{50 g}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{750 mL}"}] | <h1 class="questionTitle" itemprop="name"> What is the percent by mass of #NaHCO_3# in a solution containing 50 g #NaHCO_3# dissolved in 750 mL of water?</h1> | null | 6.25% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution is a homogeneous mixture consisting of a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. The solution is typically defined by the amount of solute dissolved into an amount of 'solution'. So calculating the percent component in the solution would be ... </p>
<p>Wt% of component = <mathjax>#(("Weight" "of" "Component")/("Weight""of""Solution"))xx100%#</mathjax></p>
<p>where (Solute Wt + Solvent Wt) = Solution Wt.</p>
<p>For this problem ...</p>
<p><mathjax>#Wt% "NaHCO_3 = ((50 g NaHCO_3)/((50g + 750g)Solution))xx100% = #</mathjax>6.25% by weight</p></div>
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<div class="markdown"><p>6.25%</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution is a homogeneous mixture consisting of a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. The solution is typically defined by the amount of solute dissolved into an amount of 'solution'. So calculating the percent component in the solution would be ... </p>
<p>Wt% of component = <mathjax>#(("Weight" "of" "Component")/("Weight""of""Solution"))xx100%#</mathjax></p>
<p>where (Solute Wt + Solvent Wt) = Solution Wt.</p>
<p>For this problem ...</p>
<p><mathjax>#Wt% "NaHCO_3 = ((50 g NaHCO_3)/((50g + 750g)Solution))xx100% = #</mathjax>6.25% by weight</p></div>
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<h1 class="questionTitle" itemprop="name"> What is the percent by mass of #NaHCO_3# in a solution containing 50 g #NaHCO_3# dissolved in 750 mL of water?</h1>
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<div class="markdown"><p>6.25%</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution is a homogeneous mixture consisting of a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. The solution is typically defined by the amount of solute dissolved into an amount of 'solution'. So calculating the percent component in the solution would be ... </p>
<p>Wt% of component = <mathjax>#(("Weight" "of" "Component")/("Weight""of""Solution"))xx100%#</mathjax></p>
<p>where (Solute Wt + Solvent Wt) = Solution Wt.</p>
<p>For this problem ...</p>
<p><mathjax>#Wt% "NaHCO_3 = ((50 g NaHCO_3)/((50g + 750g)Solution))xx100% = #</mathjax>6.25% by weight</p></div>
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</article> | What is the percent by mass of #NaHCO_3# in a solution containing 50 g #NaHCO_3# dissolved in 750 mL of water? | null |
3,010 | ac8ed598-6ddd-11ea-b404-ccda262736ce | https://socratic.org/questions/5930ccef7c0149397b17814d | 1523.86 bar | start physical_unit 1 1 pressure bar qc_end physical_unit 1 1 3 4 volume qc_end physical_unit 1 1 19 20 volume qc_end physical_unit 1 1 25 26 pressure qc_end end | [{"type":"physical unit","value":"Pressure1 [OF] the gas [IN] bar"}] | [{"type":"physical unit","value":"1523.86 bar"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.5650 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{1.204 L}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{715.1 bar}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies #"0.5650 L"# in a closed container. The initial pressure is #"______"# if the volume increases to #"1.204 L"# at a pressure of #"715.1 bar"#?</h1> | null | 1523.86 bar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <strong>Boyle's gas law</strong> , which states that the volume of a given amount of a gas varies inversely with the pressure, as long as temperature is kept constant. This means that when the volume increases, the pressure decreases, and vice-versa. The equation you will need to answer this question is:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>In your question, the blank is the initial pressure, <mathjax>#P_1#</mathjax>, whis is your unknown. Organize the rest of the data.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="0.5650 L"#</mathjax></p>
<p><mathjax>#P_2="715.1 bar"#</mathjax></p>
<p><mathjax>#V_2="1.204 L"#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation above to isolate <mathjax>#P_1#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#P_1=(P_2V_2)/(V_1)#</mathjax></p>
<p><mathjax>#P_1=(715.1"bar"xx1.204color(red)cancel(color(black)("L")))/(0.5650color(red)cancel(color(black)("L")))="1524 bar"#</mathjax> (rounded to four sig figs)</p></div>
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<div class="answerSummary">
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<div class="markdown"><p>The initial pressure, which goes in the blank, was <mathjax>#"1524 bar"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <strong>Boyle's gas law</strong> , which states that the volume of a given amount of a gas varies inversely with the pressure, as long as temperature is kept constant. This means that when the volume increases, the pressure decreases, and vice-versa. The equation you will need to answer this question is:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>In your question, the blank is the initial pressure, <mathjax>#P_1#</mathjax>, whis is your unknown. Organize the rest of the data.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="0.5650 L"#</mathjax></p>
<p><mathjax>#P_2="715.1 bar"#</mathjax></p>
<p><mathjax>#V_2="1.204 L"#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation above to isolate <mathjax>#P_1#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#P_1=(P_2V_2)/(V_1)#</mathjax></p>
<p><mathjax>#P_1=(715.1"bar"xx1.204color(red)cancel(color(black)("L")))/(0.5650color(red)cancel(color(black)("L")))="1524 bar"#</mathjax> (rounded to four sig figs)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas occupies #"0.5650 L"# in a closed container. The initial pressure is #"______"# if the volume increases to #"1.204 L"# at a pressure of #"715.1 bar"#?</h1>
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<div class="markdown"><p>The initial pressure, which goes in the blank, was <mathjax>#"1524 bar"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <strong>Boyle's gas law</strong> , which states that the volume of a given amount of a gas varies inversely with the pressure, as long as temperature is kept constant. This means that when the volume increases, the pressure decreases, and vice-versa. The equation you will need to answer this question is:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>In your question, the blank is the initial pressure, <mathjax>#P_1#</mathjax>, whis is your unknown. Organize the rest of the data.</p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="0.5650 L"#</mathjax></p>
<p><mathjax>#P_2="715.1 bar"#</mathjax></p>
<p><mathjax>#V_2="1.204 L"#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation above to isolate <mathjax>#P_1#</mathjax>. Insert the known data into the equation and solve.</p>
<p><mathjax>#P_1=(P_2V_2)/(V_1)#</mathjax></p>
<p><mathjax>#P_1=(715.1"bar"xx1.204color(red)cancel(color(black)("L")))/(0.5650color(red)cancel(color(black)("L")))="1524 bar"#</mathjax> (rounded to four sig figs)</p></div>
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</article> | A gas occupies #"0.5650 L"# in a closed container. The initial pressure is #"______"# if the volume increases to #"1.204 L"# at a pressure of #"715.1 bar"#? | null |
3,011 | a93dd1be-6ddd-11ea-be09-ccda262736ce | https://socratic.org/questions/how-many-grams-of-sodium-azide-are-needed-to-provide-sufficient-nitrogen-gas-to- | 127.42 grams | start physical_unit 4 5 mass g qc_end physical_unit 22 22 16 21 size qc_end physical_unit 22 22 27 28 pressure qc_end physical_unit 22 22 30 31 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium azide [IN] grams"}] | [{"type":"physical unit","value":"127.42 grams"}] | [{"type":"physical unit","value":"Size [OF] bag [=] \\pu{50.0 × 50.0 × 25.0 cm}"},{"type":"physical unit","value":"Pressure [OF] bag [=] \\pu{1.15 atm}"},{"type":"physical unit","value":"Temperature [OF] bag [=] \\pu{25.0 ℃}"},{"type":"other","value":"Sufficient nitrogen gas."}] | <h1 class="questionTitle" itemprop="name"> How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.15 atm at 25.0 °C?
</h1> | null | 127.42 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of sodium azide (<mathjax>#"NaN"_3#</mathjax>) needed to produce a certain amount of <mathjax>#"N"_2#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal gas equation</strong> to find the moles of nitrogen gas present:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<ul>
<li>
<p><mathjax>#P = 1.15#</mathjax> <mathjax>#"atm"#</mathjax> (given)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> must be in liters, so we can find the volume first in <mathjax>#"cm"^3#</mathjax> and then convert:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#50.0 xx 50.0 xx 25.0 = 62500#</mathjax> <mathjax>#"cm"^3#</mathjax></p>
<p><mathjax>#62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5#</mathjax> <mathjax>#"L"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C"#</mathjax>, which must be in Kelvin:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
</blockquote>
<p>Plugging in known values, and solving for the number of moles, <mathjax>#n#</mathjax>, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((1.15cancel("atm"))(62.5cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(red)(2.94#</mathjax> <mathjax>#color(red)("mol N"_2#</mathjax></p>
<p>Now, we can use the coefficients of the chemical equation to find the relative number of <strong>moles of sodium azide</strong> that must react:</p>
<p><mathjax>#color(red)(2.94)cancel(color(red)("mol N"_2))((2color(white)(l)"mol NaN"_3)/(3cancel("mol N"_2))) = color(green)(1.96#</mathjax> <mathjax>#color(green)("mol NaN"_3#</mathjax></p>
<p>Finally, we can use the <strong>molar mass of sodium azide</strong> (<mathjax>#65.01#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the mass in grams:</p>
<p><mathjax>#color(green)(1.96)cancel(color(green)("mol NaN"_3))((65.01color(white)(l)"g NaN"_3)/(1cancel("mol NaN"_3))) = color(blue)(ul(127color(white)(l)"g NaN"_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#127#</mathjax> <mathjax>#"g NaN"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of sodium azide (<mathjax>#"NaN"_3#</mathjax>) needed to produce a certain amount of <mathjax>#"N"_2#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal gas equation</strong> to find the moles of nitrogen gas present:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<ul>
<li>
<p><mathjax>#P = 1.15#</mathjax> <mathjax>#"atm"#</mathjax> (given)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> must be in liters, so we can find the volume first in <mathjax>#"cm"^3#</mathjax> and then convert:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#50.0 xx 50.0 xx 25.0 = 62500#</mathjax> <mathjax>#"cm"^3#</mathjax></p>
<p><mathjax>#62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5#</mathjax> <mathjax>#"L"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C"#</mathjax>, which must be in Kelvin:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
</blockquote>
<p>Plugging in known values, and solving for the number of moles, <mathjax>#n#</mathjax>, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((1.15cancel("atm"))(62.5cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(red)(2.94#</mathjax> <mathjax>#color(red)("mol N"_2#</mathjax></p>
<p>Now, we can use the coefficients of the chemical equation to find the relative number of <strong>moles of sodium azide</strong> that must react:</p>
<p><mathjax>#color(red)(2.94)cancel(color(red)("mol N"_2))((2color(white)(l)"mol NaN"_3)/(3cancel("mol N"_2))) = color(green)(1.96#</mathjax> <mathjax>#color(green)("mol NaN"_3#</mathjax></p>
<p>Finally, we can use the <strong>molar mass of sodium azide</strong> (<mathjax>#65.01#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the mass in grams:</p>
<p><mathjax>#color(green)(1.96)cancel(color(green)("mol NaN"_3))((65.01color(white)(l)"g NaN"_3)/(1cancel("mol NaN"_3))) = color(blue)(ul(127color(white)(l)"g NaN"_3#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.15 atm at 25.0 °C?
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Nathan L.
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<span class="dateCreated" datetime="2017-07-29T02:16:33" itemprop="dateCreated">
Jul 29, 2017
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<div class="markdown"><p><mathjax>#127#</mathjax> <mathjax>#"g NaN"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of sodium azide (<mathjax>#"NaN"_3#</mathjax>) needed to produce a certain amount of <mathjax>#"N"_2#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal gas equation</strong> to find the moles of nitrogen gas present:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<ul>
<li>
<p><mathjax>#P = 1.15#</mathjax> <mathjax>#"atm"#</mathjax> (given)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> must be in liters, so we can find the volume first in <mathjax>#"cm"^3#</mathjax> and then convert:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#50.0 xx 50.0 xx 25.0 = 62500#</mathjax> <mathjax>#"cm"^3#</mathjax></p>
<p><mathjax>#62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5#</mathjax> <mathjax>#"L"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C"#</mathjax>, which must be in Kelvin:</p>
</li>
</ul>
<blockquote>
<p><mathjax>#T = 25.0#</mathjax> <mathjax>#""^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
</blockquote>
<p>Plugging in known values, and solving for the number of moles, <mathjax>#n#</mathjax>, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((1.15cancel("atm"))(62.5cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(red)(2.94#</mathjax> <mathjax>#color(red)("mol N"_2#</mathjax></p>
<p>Now, we can use the coefficients of the chemical equation to find the relative number of <strong>moles of sodium azide</strong> that must react:</p>
<p><mathjax>#color(red)(2.94)cancel(color(red)("mol N"_2))((2color(white)(l)"mol NaN"_3)/(3cancel("mol N"_2))) = color(green)(1.96#</mathjax> <mathjax>#color(green)("mol NaN"_3#</mathjax></p>
<p>Finally, we can use the <strong>molar mass of sodium azide</strong> (<mathjax>#65.01#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the mass in grams:</p>
<p><mathjax>#color(green)(1.96)cancel(color(green)("mol NaN"_3))((65.01color(white)(l)"g NaN"_3)/(1cancel("mol NaN"_3))) = color(blue)(ul(127color(white)(l)"g NaN"_3#</mathjax></p></div>
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</article> | How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.15 atm at 25.0 °C?
| null |
3,012 | ac817b94-6ddd-11ea-aee0-ccda262736ce | https://socratic.org/questions/56122aac11ef6b16f74201bb | 71.00% | start physical_unit 13 14 mass_percent none qc_end physical_unit 13 14 9 10 molarity qc_end end | [{"type":"physical unit","value":"Percentage concentration by mass [OF] nitric acid in solution"}] | [{"type":"physical unit","value":"71.00%"}] | [{"type":"physical unit","value":"Molarity [OF] nitric acid solution [=] \\pu{16 mol/L}"},{"type":"physical unit","value":"Density [OF] nitric acid solution [=] \\pu{1.42 g/mL}"}] | <h1 class="questionTitle" itemprop="name">What is the percentage concentration by mass of a #16*mol*L^-1# solution of nitric acid, for which density is #1.42*g*mL^-1#?</h1> | null | 71.00% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A <mathjax>#16#</mathjax> <mathjax>#M#</mathjax> concentration refers to the concentration of the solution. That is the nitric acid solution has a concentration of <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax> of <em>solution</em> . We can work out percentages (<mathjax>#w/w#</mathjax> <mathjax>#xx100%#</mathjax>) provided that we know the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution, which we do because you have kindly included it in your question. So <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of solution has a mass of <mathjax>#1420#</mathjax> <mathjax>#g#</mathjax>. In the solution there are <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#HNO_3#</mathjax>, i.e. <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#63.0#</mathjax> <mathjax>#g#</mathjax> <mathjax>#mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1008.2#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>So now we calculate the quotient, 1008.2 g <mathjax>#xx#</mathjax> 1/(1420 g) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#%#</mathjax>. And this gives us our percentage concentration: (mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/mass of solution) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax></p></div>
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<div class="markdown"><p>Around 70% concentration, mass of acid/mass of solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A <mathjax>#16#</mathjax> <mathjax>#M#</mathjax> concentration refers to the concentration of the solution. That is the nitric acid solution has a concentration of <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax> of <em>solution</em> . We can work out percentages (<mathjax>#w/w#</mathjax> <mathjax>#xx100%#</mathjax>) provided that we know the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution, which we do because you have kindly included it in your question. So <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of solution has a mass of <mathjax>#1420#</mathjax> <mathjax>#g#</mathjax>. In the solution there are <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#HNO_3#</mathjax>, i.e. <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#63.0#</mathjax> <mathjax>#g#</mathjax> <mathjax>#mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1008.2#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>So now we calculate the quotient, 1008.2 g <mathjax>#xx#</mathjax> 1/(1420 g) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#%#</mathjax>. And this gives us our percentage concentration: (mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/mass of solution) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the percentage concentration by mass of a #16*mol*L^-1# solution of nitric acid, for which density is #1.42*g*mL^-1#?</h1>
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anor277
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<div class="markdown"><p>Around 70% concentration, mass of acid/mass of solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>A <mathjax>#16#</mathjax> <mathjax>#M#</mathjax> concentration refers to the concentration of the solution. That is the nitric acid solution has a concentration of <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax> of <em>solution</em> . We can work out percentages (<mathjax>#w/w#</mathjax> <mathjax>#xx100%#</mathjax>) provided that we know the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution, which we do because you have kindly included it in your question. So <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of solution has a mass of <mathjax>#1420#</mathjax> <mathjax>#g#</mathjax>. In the solution there are <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#HNO_3#</mathjax>, i.e. <mathjax>#16#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#63.0#</mathjax> <mathjax>#g#</mathjax> <mathjax>#mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1008.2#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>So now we calculate the quotient, 1008.2 g <mathjax>#xx#</mathjax> 1/(1420 g) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#%#</mathjax>. And this gives us our percentage concentration: (mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/mass of solution) <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax></p></div>
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Stefan V.
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<div class="markdown"><p>Your stock solution is <mathjax>#71%"w/w"#</mathjax> nitric acid. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>What you need to do here is pick a sample of this stock nitric acid solution and use its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> to find out how much nitric acid it would contain. </p>
<p>Then use its <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to determine athe sample's mass.</p>
<p>So, to make calculations easier, pick a <mathjax>#"1.00-L"#</mathjax> sample of the stock solution. Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case nitric cid, divided by liters of solution, you get that</p>
<blockquote>
<p><mathjax>#C = n/V implies n = C * V#</mathjax></p>
<p><mathjax>#n_(HNO_3) = "16 M" * "1.00 L" = "16 moles"#</mathjax></p>
</blockquote>
<p>Use nitric acid's molar mass to help you find how many grams of acid would contain this many moles</p>
<blockquote>
<p><mathjax>#16color(red)(cancel(color(black)("moles HNO"""_3))) * "63.013 g"/(1color(red)(cancel(color(black)("mole HNO"""_3)))) = "1008.2 g HNO"""_3#</mathjax></p>
</blockquote>
<p>What would be the mass of the sample? Use its known density!</p>
<blockquote>
<p><mathjax>#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.42 g"/(1color(red)(cancel(color(black)("mL")))) = "1420 g"#</mathjax></p>
</blockquote>
<p>The solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a>, or <mathjax>#"%w/w"#</mathjax>, is defined as </p>
<blockquote>
<p><mathjax>#"%w/w" = "mass of solute"/"mass of solution" xx 100#</mathjax></p>
</blockquote>
<p>In your case, you would have</p>
<blockquote>
<p><mathjax>#"%w/w" = (1008.2color(red)(cancel(color(black)("g"))))/(1420color(red)(cancel(color(black)("g")))) xx 100 = color(green)(71%)#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>You can redo the calculations using any sample of the stock solution, the result will always come out the same.</em></p></div>
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</article> | What is the percentage concentration by mass of a #16*mol*L^-1# solution of nitric acid, for which density is #1.42*g*mL^-1#? | null |
3,013 | aa5f89e7-6ddd-11ea-94e8-ccda262736ce | https://socratic.org/questions/ammonia-and-oxygen-produce-nitrogen-dioxide-and-water-what-volume-of-nitrogen-di | 40.67 L | start physical_unit 11 13 volume l qc_end physical_unit 0 0 18 19 mass qc_end c_other OTHER qc_end substance 7 7 qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen dioxide gas [IN] L"}] | [{"type":"physical unit","value":"40.67 L"}] | [{"type":"physical unit","value":"Mass [OF] ammonia [=] \\pu{30.5 grams}"},{"type":"other","value":"Excess oxygen."},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if 30.5 grams of ammonia is reacted with excess oxygen?</h1> | null | 40.67 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are four steps involved in this <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem:</p>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p><mathjax>#"4NH"_3 + "7O"_2 → "4NO"_2 + "6H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Convert grams of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NH"_3#</mathjax></strong></p>
<p>The molar mass of <mathjax>#"NH"_3#</mathjax> is 17.03 g/mol.</p>
<p>∴ <mathjax>#"moles of NH"_3 = 30.5 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH_3"_3)))) = "1.791 mol NH"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Convert moles of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NO"_2#</mathjax>**</p>
<p>The molar ratio of <mathjax>#"NO"_2#</mathjax> to <mathjax>#"NH"_3#</mathjax> is <mathjax>#"4 mol NO"_2:"4 mol NH"_3#</mathjax>.</p>
<p>∴ <mathjax>#1.791 color(red)(cancel(color(black)("mol NH"_3))) × ("4 mol NO"_2)/(4 color(red)(cancel(color(black)("mol NH"_3)))) = "1.791 mol NO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume of <mathjax>#"NO"_2#</mathjax>.</strong> </p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) PV = nRTcolor(white)(a/a)|)))#</mathjax>, where</p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#P#</mathjax> = the pressure of the gas,</li>
<li><mathjax>#V#</mathjax> = the volume of the gas,</li>
<li><mathjax>#n#</mathjax> = the number of moles of the gas,</li>
<li><mathjax>#R#</mathjax> = the universal gas constant</li>
<li><mathjax>#T#</mathjax> = the temperature of the gas</li>
</ul>
<p>You don't give the pressure or temperature of the gas, so I will assume STP (100 kPa and 0 °C).</p>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V= (nRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#V = (1.791 color(red)(cancel(color(black)("mol"))) ×8.314 color(red)(cancel(color(black)("kPa")))"·L·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(100 color(red)(cancel(color(black)("kPa")))) = "40.7 L"#</mathjax></p>
<p>The volume of <mathjax>#"NO"_2#</mathjax> is 40.7 L.</p>
<blockquote></blockquote>
<p>Here's a useful video on mass-volume calculations.</p>
<p>
<iframe src="https://www.youtube.com/embed/IvzqdYV_UHY?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>40.7 L of <mathjax>#"NO"_2#</mathjax> will be produced at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are four steps involved in this <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem:</p>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p><mathjax>#"4NH"_3 + "7O"_2 → "4NO"_2 + "6H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Convert grams of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NH"_3#</mathjax></strong></p>
<p>The molar mass of <mathjax>#"NH"_3#</mathjax> is 17.03 g/mol.</p>
<p>∴ <mathjax>#"moles of NH"_3 = 30.5 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH_3"_3)))) = "1.791 mol NH"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Convert moles of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NO"_2#</mathjax>**</p>
<p>The molar ratio of <mathjax>#"NO"_2#</mathjax> to <mathjax>#"NH"_3#</mathjax> is <mathjax>#"4 mol NO"_2:"4 mol NH"_3#</mathjax>.</p>
<p>∴ <mathjax>#1.791 color(red)(cancel(color(black)("mol NH"_3))) × ("4 mol NO"_2)/(4 color(red)(cancel(color(black)("mol NH"_3)))) = "1.791 mol NO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume of <mathjax>#"NO"_2#</mathjax>.</strong> </p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) PV = nRTcolor(white)(a/a)|)))#</mathjax>, where</p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#P#</mathjax> = the pressure of the gas,</li>
<li><mathjax>#V#</mathjax> = the volume of the gas,</li>
<li><mathjax>#n#</mathjax> = the number of moles of the gas,</li>
<li><mathjax>#R#</mathjax> = the universal gas constant</li>
<li><mathjax>#T#</mathjax> = the temperature of the gas</li>
</ul>
<p>You don't give the pressure or temperature of the gas, so I will assume STP (100 kPa and 0 °C).</p>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V= (nRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#V = (1.791 color(red)(cancel(color(black)("mol"))) ×8.314 color(red)(cancel(color(black)("kPa")))"·L·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(100 color(red)(cancel(color(black)("kPa")))) = "40.7 L"#</mathjax></p>
<p>The volume of <mathjax>#"NO"_2#</mathjax> is 40.7 L.</p>
<blockquote></blockquote>
<p>Here's a useful video on mass-volume calculations.</p>
<p>
<iframe src="https://www.youtube.com/embed/IvzqdYV_UHY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if 30.5 grams of ammonia is reacted with excess oxygen?</h1>
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<div class="markdown"><p>40.7 L of <mathjax>#"NO"_2#</mathjax> will be produced at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>There are four steps involved in this <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problem:</p>
<p><strong>Step 1. Write the balanced chemical equation.</strong></p>
<p><mathjax>#"4NH"_3 + "7O"_2 → "4NO"_2 + "6H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Convert grams of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NH"_3#</mathjax></strong></p>
<p>The molar mass of <mathjax>#"NH"_3#</mathjax> is 17.03 g/mol.</p>
<p>∴ <mathjax>#"moles of NH"_3 = 30.5 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH_3"_3)))) = "1.791 mol NH"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Convert moles of <mathjax>#"NH"_3#</mathjax> to moles of <mathjax>#"NO"_2#</mathjax>**</p>
<p>The molar ratio of <mathjax>#"NO"_2#</mathjax> to <mathjax>#"NH"_3#</mathjax> is <mathjax>#"4 mol NO"_2:"4 mol NH"_3#</mathjax>.</p>
<p>∴ <mathjax>#1.791 color(red)(cancel(color(black)("mol NH"_3))) × ("4 mol NO"_2)/(4 color(red)(cancel(color(black)("mol NH"_3)))) = "1.791 mol NO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume of <mathjax>#"NO"_2#</mathjax>.</strong> </p>
<p>The Ideal Gas Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) PV = nRTcolor(white)(a/a)|)))#</mathjax>, where</p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#P#</mathjax> = the pressure of the gas,</li>
<li><mathjax>#V#</mathjax> = the volume of the gas,</li>
<li><mathjax>#n#</mathjax> = the number of moles of the gas,</li>
<li><mathjax>#R#</mathjax> = the universal gas constant</li>
<li><mathjax>#T#</mathjax> = the temperature of the gas</li>
</ul>
<p>You don't give the pressure or temperature of the gas, so I will assume STP (100 kPa and 0 °C).</p>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V= (nRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#V = (1.791 color(red)(cancel(color(black)("mol"))) ×8.314 color(red)(cancel(color(black)("kPa")))"·L·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(100 color(red)(cancel(color(black)("kPa")))) = "40.7 L"#</mathjax></p>
<p>The volume of <mathjax>#"NO"_2#</mathjax> is 40.7 L.</p>
<blockquote></blockquote>
<p>Here's a useful video on mass-volume calculations.</p>
<p>
<iframe src="https://www.youtube.com/embed/IvzqdYV_UHY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if 30.5 grams of ammonia is reacted with excess oxygen? | null |
3,014 | aa1674d4-6ddd-11ea-be85-ccda262736ce | https://socratic.org/questions/564eb07a11ef6b270e0433ee | 1555.89 mmHg | start physical_unit 3 5 pressure mmhg qc_end physical_unit 3 5 10 11 pressure qc_end physical_unit 3 5 1 2 volume qc_end physical_unit 3 5 13 14 temperature qc_end physical_unit 3 5 33 34 temperature qc_end physical_unit 3 5 25 26 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] gas sample [IN] mmHg"}] | [{"type":"physical unit","value":"1555.89 mmHg"}] | [{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{570 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{2270 mL}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] gas sample [=] \\pu{175 ℃}"},{"type":"physical unit","value":"Volume2 [OF] gas sample [=] \\pu{1250 mL}"}] | <h1 class="questionTitle" itemprop="name">A #"2270 mL"# sample of gas has a pressure of #"570. mmHg"# at #"25"^@"C"#.
What is the pressure when the volume is decreased to #"1250 mL"# and the temperature is increased to #"175"^@"C"#? </h1> | null | 1555.89 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#V_1/V_2= (T_1P_2)/(T_2P_1)#</mathjax></p>
<p><mathjax>#T_1 = (25 +273.15) K = 298.15 K; T_2 = (175 +273.15) K = 448.15 K#</mathjax></p>
<p><mathjax>#=> P_2 = (V_1T_2P_1)/(V_2T_1)#</mathjax></p>
<p><mathjax>#=> P_2 = (2270 color(red)(cancel(color(black)("mL")))*448.15 color(red)(cancel(color(black)("K")))×"570. mm Hg")/ (1250color(red)(cancel(color(black)("mL")))*298.15 color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#=> P_2 = "1556 mm Hg"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>1556 mm Hg</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#V_1/V_2= (T_1P_2)/(T_2P_1)#</mathjax></p>
<p><mathjax>#T_1 = (25 +273.15) K = 298.15 K; T_2 = (175 +273.15) K = 448.15 K#</mathjax></p>
<p><mathjax>#=> P_2 = (V_1T_2P_1)/(V_2T_1)#</mathjax></p>
<p><mathjax>#=> P_2 = (2270 color(red)(cancel(color(black)("mL")))*448.15 color(red)(cancel(color(black)("K")))×"570. mm Hg")/ (1250color(red)(cancel(color(black)("mL")))*298.15 color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#=> P_2 = "1556 mm Hg"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A #"2270 mL"# sample of gas has a pressure of #"570. mmHg"# at #"25"^@"C"#.
What is the pressure when the volume is decreased to #"1250 mL"# and the temperature is increased to #"175"^@"C"#? </h1>
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<div class="markdown"><p>1556 mm Hg</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#V_1/V_2= (T_1P_2)/(T_2P_1)#</mathjax></p>
<p><mathjax>#T_1 = (25 +273.15) K = 298.15 K; T_2 = (175 +273.15) K = 448.15 K#</mathjax></p>
<p><mathjax>#=> P_2 = (V_1T_2P_1)/(V_2T_1)#</mathjax></p>
<p><mathjax>#=> P_2 = (2270 color(red)(cancel(color(black)("mL")))*448.15 color(red)(cancel(color(black)("K")))×"570. mm Hg")/ (1250color(red)(cancel(color(black)("mL")))*298.15 color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#=> P_2 = "1556 mm Hg"#</mathjax></p></div>
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<div class="markdown"><p>The final pressure is 1560 mmHg.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>, which relates pressure, volume, and temperature.</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/(T_2)#</mathjax></p>
<p><strong>Note:</strong> Temperature must be converted from degrees Celsius to Kelvins.</p>
<p><strong>Known/Given</strong><br/>
<mathjax>#P_1="570. mmHg"#</mathjax><br/>
<mathjax>#V_1="2270 mL"#</mathjax><br/>
<mathjax>#T_1="25"^"o""C"+273.15="298 K"#</mathjax><br/>
<mathjax>#V_2="1250 mL"#</mathjax><br/>
<mathjax>#T_2="175"^"o""C"+273.15="448 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P_2="???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax> and solve.<br/>
<mathjax>#(P_1V_1)/T_1=(P_2V_2)/(T_2)#</mathjax></p>
<p><mathjax>#P_2=(P_1V_1T_2)/(T_1V_2)#</mathjax></p>
<p><mathjax>#P_2=(570."mmHg"xx2270cancel"mL"xx448cancel"K")/(298cancel"K"xx1250cancel"mL")="1560 mmHg"#</mathjax> (rounded to three significant figures)</p></div>
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</article> | A #"2270 mL"# sample of gas has a pressure of #"570. mmHg"# at #"25"^@"C"#.
What is the pressure when the volume is decreased to #"1250 mL"# and the temperature is increased to #"175"^@"C"#? | null |
3,015 | a9a1b1db-6ddd-11ea-bdbd-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-diclofenac-sodium | C14H10Cl2NNaO2 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] Diclofenac Sodium [IN] default"}] | [{"type":"chemical equation","value":"C14H10Cl2NNaO2"}] | [{"type":"substance name","value":"Diclofenac Sodium"}] | <h1 class="questionTitle" itemprop="name">What is the formula for Diclofenac Sodium?</h1> | null | C14H10Cl2NNaO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Its structural formula is</p>
<p><img alt="www.newdruginfo.com" src="http://www.newdruginfo.com/pharmacopeia/usp28/v28230/uspnf/pub/images/v28230/g-1337.gif"/></p>
<p>Its name is sodium {2-[(2,6-dichlorophenyl)amino]phenyl}acetate.</p>
<p>You often see the name written as {2-[(2,6-dichlorophenyl)amino]phenyl}acetic acid, sodium salt</p>
<p>If you mis-spell the name as {2-[(2,6-<mathjax>#bb"dicl"#</mathjax>orophenyl)amino]<mathjax>#bb"f"#</mathjax>enyl}<mathjax>#bb"ac"#</mathjax>etic acid, <mathjax>#bb"Na"#</mathjax> salt, you can get a glimmer of the trade name.</p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>The molecular formula for Diclofenac Sodium is <mathjax>#"C"_14"H"_10"Cl"_2"NNaO"_2#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Its structural formula is</p>
<p><img alt="www.newdruginfo.com" src="http://www.newdruginfo.com/pharmacopeia/usp28/v28230/uspnf/pub/images/v28230/g-1337.gif"/></p>
<p>Its name is sodium {2-[(2,6-dichlorophenyl)amino]phenyl}acetate.</p>
<p>You often see the name written as {2-[(2,6-dichlorophenyl)amino]phenyl}acetic acid, sodium salt</p>
<p>If you mis-spell the name as {2-[(2,6-<mathjax>#bb"dicl"#</mathjax>orophenyl)amino]<mathjax>#bb"f"#</mathjax>enyl}<mathjax>#bb"ac"#</mathjax>etic acid, <mathjax>#bb"Na"#</mathjax> salt, you can get a glimmer of the trade name.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for Diclofenac Sodium?</h1>
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Ernest Z.
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<div class="markdown"><p>The molecular formula for Diclofenac Sodium is <mathjax>#"C"_14"H"_10"Cl"_2"NNaO"_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Its structural formula is</p>
<p><img alt="www.newdruginfo.com" src="http://www.newdruginfo.com/pharmacopeia/usp28/v28230/uspnf/pub/images/v28230/g-1337.gif"/></p>
<p>Its name is sodium {2-[(2,6-dichlorophenyl)amino]phenyl}acetate.</p>
<p>You often see the name written as {2-[(2,6-dichlorophenyl)amino]phenyl}acetic acid, sodium salt</p>
<p>If you mis-spell the name as {2-[(2,6-<mathjax>#bb"dicl"#</mathjax>orophenyl)amino]<mathjax>#bb"f"#</mathjax>enyl}<mathjax>#bb"ac"#</mathjax>etic acid, <mathjax>#bb"Na"#</mathjax> salt, you can get a glimmer of the trade name.</p></div>
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</article> | What is the formula for Diclofenac Sodium? | null |
3,016 | a9904ef8-6ddd-11ea-9863-ccda262736ce | https://socratic.org/questions/at-a-certain-temperature-0-660-mol-of-so3-is-placed-in-a-4-50-l-container-accord | 0.01 | start physical_unit 17 18 equilibrium_constant_k none qc_end physical_unit 7 7 4 5 mole qc_end physical_unit 14 14 12 13 volume qc_end chemical_equation 19 25 qc_end physical_unit 25 25 28 29 mole qc_end end | [{"type":"physical unit","value":"Kc [OF] the reaction"}] | [{"type":"physical unit","value":"0.01"}] | [{"type":"physical unit","value":"Mole [OF] SO3 [=] \\pu{0.660 mol}"},{"type":"physical unit","value":"Volume [OF] container [=] \\pu{4.50 L}"},{"type":"chemical equation","value":"2 SO3(g) -> 2 SO2(g) + O2(g)"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{0.130 mol}"}] | <h1 class="questionTitle" itemprop="name">At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container, according to the reaction:
2SO3(g) --> 2SO2(g) + O2(g),
at equilibrium, 0.130 mol of O2 is present, how would you calculate Kc?</h1> | null | 0.01 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given equilibrium reaction, the <a href="http://socratic.org/chemistry/chemical-equilibrium/equilibrium-constants">equilibrium constant</a> is calculated by taking the ratio between the equilibrium concentrations of the <em>products</em> and the equilibrium concentrations of the <em>reactants</em>, each raised to the power of their respective <strong>stoichiometric coefficients</strong>. </p>
<p>For your equilibrium reaction </p>
<blockquote>
<p><mathjax>#color(red)(2)"SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_text(2(g]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>the equilibrium constant, <mathjax>#K_c#</mathjax>, will take the form </p>
<blockquote>
<p><mathjax>#K_c = ( ["SO"_2]^color(blue)(2) * ["O"_2])/(["SO"_3]^color(red)(2))#</mathjax></p>
</blockquote>
<p>So, you know that the <strong>starting concentration</strong> of sulfur trioxide is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#["SO"_3]_0 = "0.660 moles"/"4.50 L" = "0.1467 M"#</mathjax></p>
</blockquote>
<p>The <strong>starting concentrations</strong> of the two products will be equal to zero, since initially, the reactant is the only one present in the reaction container. </p>
<p>This means that you can use an <strong>ICE table</strong> to help you determine how the concentrations of the species will change once the reaction starts</p>
<blockquote>
<p><mathjax>#" " color(red)(2)"SO"_text(3(g]) " "rightleftharpoons" " color(blue)(2)"SO"_text(2(g]) " "+" " "O"_text(2(g])#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.1467" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-color(red)(2)x)" " " " " "(+color(blue)(2)x)" " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.1467-color(red)(2)x" " " " " "color(blue)(2)x" " " " " " " " " " "x#</mathjax></p>
<p>Now, you know that the reaction container will contain <mathjax>#0.130#</mathjax> moles of oxygen gas <strong>at equilibrium</strong>. This means that the <strong>equilibrium concentration</strong> of oxygen gas will be equal to </p>
<blockquote>
<p><mathjax>#["O"_2] = "0.130 moles"/"4.50 L" = "0.02889 M"#</mathjax></p>
</blockquote>
<p>Now, take a look at the ICE table. The equilibrium concentration of oxygen gas is said to be equal to <mathjax>#x#</mathjax>. This means that the equilibrium concentrations of the other two chemical species involved in the reaction will be </p>
<blockquote>
<p><mathjax>#["SO"_3] = 0.1467 - 2 * 0.02889 = "0.08892 M"#</mathjax></p>
<p><mathjax>#["SO"_2] = 2 * 0.02889 = "0.05778 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant for this reaction <em>at this certain temperature</em> will be </p>
<blockquote>
<p><mathjax>#K_c = ( 0.05778^2 * 0.02889)/0.08892^2 = color(green)(0.0122)#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_c = 0.0122#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given equilibrium reaction, the <a href="http://socratic.org/chemistry/chemical-equilibrium/equilibrium-constants">equilibrium constant</a> is calculated by taking the ratio between the equilibrium concentrations of the <em>products</em> and the equilibrium concentrations of the <em>reactants</em>, each raised to the power of their respective <strong>stoichiometric coefficients</strong>. </p>
<p>For your equilibrium reaction </p>
<blockquote>
<p><mathjax>#color(red)(2)"SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_text(2(g]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>the equilibrium constant, <mathjax>#K_c#</mathjax>, will take the form </p>
<blockquote>
<p><mathjax>#K_c = ( ["SO"_2]^color(blue)(2) * ["O"_2])/(["SO"_3]^color(red)(2))#</mathjax></p>
</blockquote>
<p>So, you know that the <strong>starting concentration</strong> of sulfur trioxide is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#["SO"_3]_0 = "0.660 moles"/"4.50 L" = "0.1467 M"#</mathjax></p>
</blockquote>
<p>The <strong>starting concentrations</strong> of the two products will be equal to zero, since initially, the reactant is the only one present in the reaction container. </p>
<p>This means that you can use an <strong>ICE table</strong> to help you determine how the concentrations of the species will change once the reaction starts</p>
<blockquote>
<p><mathjax>#" " color(red)(2)"SO"_text(3(g]) " "rightleftharpoons" " color(blue)(2)"SO"_text(2(g]) " "+" " "O"_text(2(g])#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.1467" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-color(red)(2)x)" " " " " "(+color(blue)(2)x)" " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.1467-color(red)(2)x" " " " " "color(blue)(2)x" " " " " " " " " " "x#</mathjax></p>
<p>Now, you know that the reaction container will contain <mathjax>#0.130#</mathjax> moles of oxygen gas <strong>at equilibrium</strong>. This means that the <strong>equilibrium concentration</strong> of oxygen gas will be equal to </p>
<blockquote>
<p><mathjax>#["O"_2] = "0.130 moles"/"4.50 L" = "0.02889 M"#</mathjax></p>
</blockquote>
<p>Now, take a look at the ICE table. The equilibrium concentration of oxygen gas is said to be equal to <mathjax>#x#</mathjax>. This means that the equilibrium concentrations of the other two chemical species involved in the reaction will be </p>
<blockquote>
<p><mathjax>#["SO"_3] = 0.1467 - 2 * 0.02889 = "0.08892 M"#</mathjax></p>
<p><mathjax>#["SO"_2] = 2 * 0.02889 = "0.05778 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant for this reaction <em>at this certain temperature</em> will be </p>
<blockquote>
<p><mathjax>#K_c = ( 0.05778^2 * 0.02889)/0.08892^2 = color(green)(0.0122)#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container, according to the reaction:
2SO3(g) --> 2SO2(g) + O2(g),
at equilibrium, 0.130 mol of O2 is present, how would you calculate Kc?</h1>
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Stefan V.
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Dec 6, 2015
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<div class="markdown"><p><mathjax>#K_c = 0.0122#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given equilibrium reaction, the <a href="http://socratic.org/chemistry/chemical-equilibrium/equilibrium-constants">equilibrium constant</a> is calculated by taking the ratio between the equilibrium concentrations of the <em>products</em> and the equilibrium concentrations of the <em>reactants</em>, each raised to the power of their respective <strong>stoichiometric coefficients</strong>. </p>
<p>For your equilibrium reaction </p>
<blockquote>
<p><mathjax>#color(red)(2)"SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_text(2(g]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>the equilibrium constant, <mathjax>#K_c#</mathjax>, will take the form </p>
<blockquote>
<p><mathjax>#K_c = ( ["SO"_2]^color(blue)(2) * ["O"_2])/(["SO"_3]^color(red)(2))#</mathjax></p>
</blockquote>
<p>So, you know that the <strong>starting concentration</strong> of sulfur trioxide is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#["SO"_3]_0 = "0.660 moles"/"4.50 L" = "0.1467 M"#</mathjax></p>
</blockquote>
<p>The <strong>starting concentrations</strong> of the two products will be equal to zero, since initially, the reactant is the only one present in the reaction container. </p>
<p>This means that you can use an <strong>ICE table</strong> to help you determine how the concentrations of the species will change once the reaction starts</p>
<blockquote>
<p><mathjax>#" " color(red)(2)"SO"_text(3(g]) " "rightleftharpoons" " color(blue)(2)"SO"_text(2(g]) " "+" " "O"_text(2(g])#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.1467" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-color(red)(2)x)" " " " " "(+color(blue)(2)x)" " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.1467-color(red)(2)x" " " " " "color(blue)(2)x" " " " " " " " " " "x#</mathjax></p>
<p>Now, you know that the reaction container will contain <mathjax>#0.130#</mathjax> moles of oxygen gas <strong>at equilibrium</strong>. This means that the <strong>equilibrium concentration</strong> of oxygen gas will be equal to </p>
<blockquote>
<p><mathjax>#["O"_2] = "0.130 moles"/"4.50 L" = "0.02889 M"#</mathjax></p>
</blockquote>
<p>Now, take a look at the ICE table. The equilibrium concentration of oxygen gas is said to be equal to <mathjax>#x#</mathjax>. This means that the equilibrium concentrations of the other two chemical species involved in the reaction will be </p>
<blockquote>
<p><mathjax>#["SO"_3] = 0.1467 - 2 * 0.02889 = "0.08892 M"#</mathjax></p>
<p><mathjax>#["SO"_2] = 2 * 0.02889 = "0.05778 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant for this reaction <em>at this certain temperature</em> will be </p>
<blockquote>
<p><mathjax>#K_c = ( 0.05778^2 * 0.02889)/0.08892^2 = color(green)(0.0122)#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
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</article> | At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container, according to the reaction:
2SO3(g) --> 2SO2(g) + O2(g),
at equilibrium, 0.130 mol of O2 is present, how would you calculate Kc? | null |
3,017 | a912be4c-6ddd-11ea-8432-ccda262736ce | https://socratic.org/questions/58d35836b72cff61dbd6d61f | 2.00 L | start physical_unit 6 6 volume l qc_end physical_unit 6 6 4 5 molarity qc_end physical_unit 16 16 12 13 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] solution [IN] L"}] | [{"type":"physical unit","value":"2.00 L"}] | [{"type":"physical unit","value":"Mole [OF] solute [=] \\pu{3 mol}"},{"type":"physical unit","value":"Molarity [OF] solution [=] \\pu{1.5 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What volume of a #1.5*mol*L^-1# solution is required to deliver a #3.0*mol# quantity of solute?</h1> | null | 2.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the problem dimensionally:</p>
<p>we know that <mathjax>#"Moles of solute"/"Volume of solution"=1.5*mol*L^-1#</mathjax></p>
<p>So <mathjax>#"Moles of solute"-="Volume of solution"xx1.5*mol*L^-1#</mathjax></p>
<p>OR <mathjax>#"Volume of solution"="Moles of solute"/(1.5*mol*L^-1)#</mathjax></p>
<p><mathjax>#=(3.0*cancel(mol))/(1.5*cancel(mol)*L^-1)=2*L#</mathjax></p>
<p>We get an answer in <mathjax>#L#</mathjax> because <mathjax>#1/L^-1=1/(1/L)=L#</mathjax> as required. </p>
<p>In order to use these quotients, you have to remember the basic definition:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>; and thus <mathjax>#"concentration"#</mathjax> is commonly quoted with units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.0*L#</mathjax> of solution are required. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the problem dimensionally:</p>
<p>we know that <mathjax>#"Moles of solute"/"Volume of solution"=1.5*mol*L^-1#</mathjax></p>
<p>So <mathjax>#"Moles of solute"-="Volume of solution"xx1.5*mol*L^-1#</mathjax></p>
<p>OR <mathjax>#"Volume of solution"="Moles of solute"/(1.5*mol*L^-1)#</mathjax></p>
<p><mathjax>#=(3.0*cancel(mol))/(1.5*cancel(mol)*L^-1)=2*L#</mathjax></p>
<p>We get an answer in <mathjax>#L#</mathjax> because <mathjax>#1/L^-1=1/(1/L)=L#</mathjax> as required. </p>
<p>In order to use these quotients, you have to remember the basic definition:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>; and thus <mathjax>#"concentration"#</mathjax> is commonly quoted with units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume of a #1.5*mol*L^-1# solution is required to deliver a #3.0*mol# quantity of solute?</h1>
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anor277
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<div class="markdown"><p><mathjax>#2.0*L#</mathjax> of solution are required. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the problem dimensionally:</p>
<p>we know that <mathjax>#"Moles of solute"/"Volume of solution"=1.5*mol*L^-1#</mathjax></p>
<p>So <mathjax>#"Moles of solute"-="Volume of solution"xx1.5*mol*L^-1#</mathjax></p>
<p>OR <mathjax>#"Volume of solution"="Moles of solute"/(1.5*mol*L^-1)#</mathjax></p>
<p><mathjax>#=(3.0*cancel(mol))/(1.5*cancel(mol)*L^-1)=2*L#</mathjax></p>
<p>We get an answer in <mathjax>#L#</mathjax> because <mathjax>#1/L^-1=1/(1/L)=L#</mathjax> as required. </p>
<p>In order to use these quotients, you have to remember the basic definition:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute (mol)"/"Volume of solution (L)"#</mathjax>; and thus <mathjax>#"concentration"#</mathjax> is commonly quoted with units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
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Simon Moore
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<div class="markdown"><p>2 litres.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To actually work it out, use the simple relationship <mathjax>#M = n/V#</mathjax></p>
<p>Where M is <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (or concentration) in moles per <mathjax>#dm^3#</mathjax> (or moles per litre, which is numerically identical), n is the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, and V is the volume of solution in <mathjax>#dm^3#</mathjax> (or litres, which are numerically identical).</p>
<p>Then simply rearrange for V, so <mathjax>#V = n/M#</mathjax> and plug in the numbers:</p>
<p><mathjax>#V = n/M = 3/1.5 = 2#</mathjax> litres</p></div>
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</article> | What volume of a #1.5*mol*L^-1# solution is required to deliver a #3.0*mol# quantity of solute? | null |
3,018 | ace7a5e8-6ddd-11ea-bd10-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-buffer-that-was-prepared-by-adding-3-96-g-of-sodium-benzoate | 4.64 | start physical_unit 6 6 ph none qc_end physical_unit 17 17 12 13 mass qc_end physical_unit 26 26 22 23 molarity qc_end c_other OTHER qc_end physical_unit 24 25 41 43 ka qc_end end | [{"type":"physical unit","value":"pH [OF] the buffer"}] | [{"type":"physical unit","value":"4.64"}] | [{"type":"physical unit","value":"Mass [OF] NaC7H5O2 [=] \\pu{3.96 g}"},{"type":"physical unit","value":"Volume [OF] HC7H5O2 solution [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Molarity [OF] HC7H5O2 solution [=] \\pu{0.0100 M}"},{"type":"other","value":"Assume that there is no change in volume."},{"type":"physical unit","value":"Ka [OF] benzoic acid [=] \\pu{6.3 × 10^(-5)}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, #NaC_7H_5O_2#, to 1.00 L of 0.0100 M benzoic acid, #HC_7H_5O_2#? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Assume that there is no change in volume. The <mathjax>#K_a#</mathjax> for benzoic acid is <mathjax>#6.3 * 10^-5#</mathjax>.</p></div>
</h2>
</div>
</div> | 4.64 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let's represent the benzoate ion, <mathjax>#"C"_7"H"_5"O"_2^"-"#</mathjax>, as <mathjax>#"A"^"-"#</mathjax>.</p>
<p>Then the equation for the dissociation of benzoic acid becomes</p>
<p><mathjax>#"HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-"#</mathjax>; <mathjax>#K_"a" = 6.3 ×10^"-5"#</mathjax></p>
<p>The typical formula for calculating the <mathjax>#"pH"#</mathjax> of a buffer is the <strong>Henderson-Hasselbalch equation</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul("pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]"))|)#</mathjax></p>
</blockquote>
<p>Our job is to calculate the numbers to insert into the formula.</p>
<blockquote></blockquote>
<p><strong>(a)</strong> <mathjax>#"p"K_"a" = "-"logK_"a" = "-"log(6.3 × 10^"-5") = 4.20#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> <mathjax>#[A^"-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O"_2]#</mathjax></p>
<p><mathjax>#3.96 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2))) × ("1 mol NaC"_7"H"_5"O"_2)/( 144.10 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2)))) = "0.027 48 mol NaC"_7"H"_5"O"_2#</mathjax></p>
<p>∴ <mathjax>#[A^"-"] = "0.027 48 mol"/"1 L" = "0.027 48 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c)</strong> <mathjax>#"[HA] = 0.0100 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we insert these values into the Henderson-Hasselbalch equation.</p>
<p><mathjax>#"pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]") = 4.20 + log(("0.027 48" color(red)(cancel(color(black)("mol/L"))))/ (0.0100 color(red)(cancel(color(black)("mol/L"))))) = 4.20 + log(2.748) = 4.20 + 0.439 = 4.64"#</mathjax></p>
<p>The <mathjax>#"pH"#</mathjax> of the buffer is 4.64.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <mathjax>#"pH"#</mathjax> of the buffer is 4.64.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let's represent the benzoate ion, <mathjax>#"C"_7"H"_5"O"_2^"-"#</mathjax>, as <mathjax>#"A"^"-"#</mathjax>.</p>
<p>Then the equation for the dissociation of benzoic acid becomes</p>
<p><mathjax>#"HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-"#</mathjax>; <mathjax>#K_"a" = 6.3 ×10^"-5"#</mathjax></p>
<p>The typical formula for calculating the <mathjax>#"pH"#</mathjax> of a buffer is the <strong>Henderson-Hasselbalch equation</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul("pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]"))|)#</mathjax></p>
</blockquote>
<p>Our job is to calculate the numbers to insert into the formula.</p>
<blockquote></blockquote>
<p><strong>(a)</strong> <mathjax>#"p"K_"a" = "-"logK_"a" = "-"log(6.3 × 10^"-5") = 4.20#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> <mathjax>#[A^"-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O"_2]#</mathjax></p>
<p><mathjax>#3.96 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2))) × ("1 mol NaC"_7"H"_5"O"_2)/( 144.10 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2)))) = "0.027 48 mol NaC"_7"H"_5"O"_2#</mathjax></p>
<p>∴ <mathjax>#[A^"-"] = "0.027 48 mol"/"1 L" = "0.027 48 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c)</strong> <mathjax>#"[HA] = 0.0100 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we insert these values into the Henderson-Hasselbalch equation.</p>
<p><mathjax>#"pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]") = 4.20 + log(("0.027 48" color(red)(cancel(color(black)("mol/L"))))/ (0.0100 color(red)(cancel(color(black)("mol/L"))))) = 4.20 + log(2.748) = 4.20 + 0.439 = 4.64"#</mathjax></p>
<p>The <mathjax>#"pH"#</mathjax> of the buffer is 4.64.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, #NaC_7H_5O_2#, to 1.00 L of 0.0100 M benzoic acid, #HC_7H_5O_2#? </h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Assume that there is no change in volume. The <mathjax>#K_a#</mathjax> for benzoic acid is <mathjax>#6.3 * 10^-5#</mathjax>.</p></div>
</h2>
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<div class="markdown"><p>The <mathjax>#"pH"#</mathjax> of the buffer is 4.64.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let's represent the benzoate ion, <mathjax>#"C"_7"H"_5"O"_2^"-"#</mathjax>, as <mathjax>#"A"^"-"#</mathjax>.</p>
<p>Then the equation for the dissociation of benzoic acid becomes</p>
<p><mathjax>#"HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-"#</mathjax>; <mathjax>#K_"a" = 6.3 ×10^"-5"#</mathjax></p>
<p>The typical formula for calculating the <mathjax>#"pH"#</mathjax> of a buffer is the <strong>Henderson-Hasselbalch equation</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul("pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]"))|)#</mathjax></p>
</blockquote>
<p>Our job is to calculate the numbers to insert into the formula.</p>
<blockquote></blockquote>
<p><strong>(a)</strong> <mathjax>#"p"K_"a" = "-"logK_"a" = "-"log(6.3 × 10^"-5") = 4.20#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> <mathjax>#[A^"-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O"_2]#</mathjax></p>
<p><mathjax>#3.96 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2))) × ("1 mol NaC"_7"H"_5"O"_2)/( 144.10 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2)))) = "0.027 48 mol NaC"_7"H"_5"O"_2#</mathjax></p>
<p>∴ <mathjax>#[A^"-"] = "0.027 48 mol"/"1 L" = "0.027 48 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c)</strong> <mathjax>#"[HA] = 0.0100 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we insert these values into the Henderson-Hasselbalch equation.</p>
<p><mathjax>#"pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]") = 4.20 + log(("0.027 48" color(red)(cancel(color(black)("mol/L"))))/ (0.0100 color(red)(cancel(color(black)("mol/L"))))) = 4.20 + log(2.748) = 4.20 + 0.439 = 4.64"#</mathjax></p>
<p>The <mathjax>#"pH"#</mathjax> of the buffer is 4.64.</p></div>
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</article> | What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, #NaC_7H_5O_2#, to 1.00 L of 0.0100 M benzoic acid, #HC_7H_5O_2#? |
Assume that there is no change in volume. The #K_a# for benzoic acid is #6.3 * 10^-5#.
|
3,019 | abde373a-6ddd-11ea-b02f-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-form-for-sodium-phosphate | Na3PO4 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] sodium phosphate [IN] default"}] | [{"type":"chemical equation","value":"Na3PO4"}] | [{"type":"substance name","value":"Sodium phosphate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical form for sodium phosphate?</h1> | null | Na3PO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In aqueous solution, most of the time, you deal with <mathjax>#Na_2HPO_4, "i.e. sodium biphosphate"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#Na_3PO_4#</mathjax></p></div>
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<div class="markdown"><p>In aqueous solution, most of the time, you deal with <mathjax>#Na_2HPO_4, "i.e. sodium biphosphate"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical form for sodium phosphate?</h1>
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<div class="markdown"><p><mathjax>#Na_3PO_4#</mathjax></p></div>
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<div class="markdown"><p>In aqueous solution, most of the time, you deal with <mathjax>#Na_2HPO_4, "i.e. sodium biphosphate"#</mathjax>.</p></div>
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</article> | What is the chemical form for sodium phosphate? | null |
3,020 | ac9eda1c-6ddd-11ea-9e19-ccda262736ce | https://socratic.org/questions/how-do-you-show-the-reaction-of-sulfurous-acid-with-lithium-hydroxide-to-form-wa | H2SO3(aq) + 2 LiOH(aq) -> Li2SO3(aq) + 2 H2O(l) | start chemical_equation qc_end substance 7 8 qc_end substance 10 11 qc_end substance 14 14 qc_end substance 16 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"H2SO3(aq) + 2 LiOH(aq) -> Li2SO3(aq) + 2 H2O(l)"}] | [{"type":"substance name","value":"Sulfurous acid"},{"type":"substance name","value":"Lithium hydroxide"},{"type":"substance name","value":"Water"},{"type":"substance name","value":"Lithium sulfite"}] | <h1 class="questionTitle" itemprop="name">How do you show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite?</h1> | null | H2SO3(aq) + 2 LiOH(aq) -> Li2SO3(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is charge balanced? Tick. Is mass balanced? Tick.</p>
<p>In aqueous solution, we know that so-called <mathjax>#"sulfurous acid"#</mathjax> is probably solvated <mathjax>#SO_2(aq)#</mathjax>, i.e. <mathjax>#SO_2*H_2O-=H_2SO_3#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#H_2SO_3(aq) + 2LiOH(aq) rarr Li_2SO_3(aq) + 2H_2O(l)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is charge balanced? Tick. Is mass balanced? Tick.</p>
<p>In aqueous solution, we know that so-called <mathjax>#"sulfurous acid"#</mathjax> is probably solvated <mathjax>#SO_2(aq)#</mathjax>, i.e. <mathjax>#SO_2*H_2O-=H_2SO_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite?</h1>
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<div class="markdown"><p><mathjax>#H_2SO_3(aq) + 2LiOH(aq) rarr Li_2SO_3(aq) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is charge balanced? Tick. Is mass balanced? Tick.</p>
<p>In aqueous solution, we know that so-called <mathjax>#"sulfurous acid"#</mathjax> is probably solvated <mathjax>#SO_2(aq)#</mathjax>, i.e. <mathjax>#SO_2*H_2O-=H_2SO_3#</mathjax>.</p></div>
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</article> | How do you show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite? | null |
3,021 | aaac9130-6ddd-11ea-b089-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-has-a-volume-of-240-l-at-4-00-atm-of-pressure-and-a-temperature- | 240.22 L | start physical_unit 1 3 volume l qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 11 12 pressure qc_end physical_unit 1 3 19 20 temperature qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas sample [IN] L"}] | [{"type":"physical unit","value":"240.22 L"}] | [{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{240 L}"},{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{4.00 atm}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{1091 K}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 240. L at 4.00 atm of pressure and a temperature of 1091K. What is the volume at STP?</h1> | null | 240.22 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall the gas law</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>First we need to find moles of gas and then calculate its volume at STP</p>
<p>So moles of gas or n =</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p>Solve the equation using formula</p>
<p><mathjax>#n = "(4atm * 240L)"/"(0.0821L * 1091K)"#</mathjax></p>
<p><mathjax>#n = 960/89.5711"#</mathjax></p>
<p><mathjax>#n = 10.7177426648mol#</mathjax></p>
<p>The volume of 1mol of any gas = 22.4L</p>
<p>Therefore volume of <mathjax>#10.7177426648"moles"#</mathjax><br/>
<mathjax>#= 10.7177426648"moles" * 22.4134L = 240.219981669#</mathjax></p>
<p>But lets still prove it</p>
<p>Volume of 10.7177426648moles in STP<br/>
Temperature at STP = <mathjax>#0^oC or 273K#</mathjax><br/>
Pressure at <mathjax>#STP = 1atm#</mathjax></p>
<p><mathjax>#V = 10.7177426648 * 0.0821 * "273K"/"1atm"#</mathjax></p>
<p><mathjax>#V = 240.219981669/"1atm"#</mathjax></p>
<p><mathjax>#"Volume = 240.219981669L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Volume = 240.219981669L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall the gas law</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>First we need to find moles of gas and then calculate its volume at STP</p>
<p>So moles of gas or n =</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p>Solve the equation using formula</p>
<p><mathjax>#n = "(4atm * 240L)"/"(0.0821L * 1091K)"#</mathjax></p>
<p><mathjax>#n = 960/89.5711"#</mathjax></p>
<p><mathjax>#n = 10.7177426648mol#</mathjax></p>
<p>The volume of 1mol of any gas = 22.4L</p>
<p>Therefore volume of <mathjax>#10.7177426648"moles"#</mathjax><br/>
<mathjax>#= 10.7177426648"moles" * 22.4134L = 240.219981669#</mathjax></p>
<p>But lets still prove it</p>
<p>Volume of 10.7177426648moles in STP<br/>
Temperature at STP = <mathjax>#0^oC or 273K#</mathjax><br/>
Pressure at <mathjax>#STP = 1atm#</mathjax></p>
<p><mathjax>#V = 10.7177426648 * 0.0821 * "273K"/"1atm"#</mathjax></p>
<p><mathjax>#V = 240.219981669/"1atm"#</mathjax></p>
<p><mathjax>#"Volume = 240.219981669L"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 240. L at 4.00 atm of pressure and a temperature of 1091K. What is the volume at STP?</h1>
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<div class="markdown"><p><mathjax>#"Volume = 240.219981669L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall the gas law</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>First we need to find moles of gas and then calculate its volume at STP</p>
<p>So moles of gas or n =</p>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
<p>Solve the equation using formula</p>
<p><mathjax>#n = "(4atm * 240L)"/"(0.0821L * 1091K)"#</mathjax></p>
<p><mathjax>#n = 960/89.5711"#</mathjax></p>
<p><mathjax>#n = 10.7177426648mol#</mathjax></p>
<p>The volume of 1mol of any gas = 22.4L</p>
<p>Therefore volume of <mathjax>#10.7177426648"moles"#</mathjax><br/>
<mathjax>#= 10.7177426648"moles" * 22.4134L = 240.219981669#</mathjax></p>
<p>But lets still prove it</p>
<p>Volume of 10.7177426648moles in STP<br/>
Temperature at STP = <mathjax>#0^oC or 273K#</mathjax><br/>
Pressure at <mathjax>#STP = 1atm#</mathjax></p>
<p><mathjax>#V = 10.7177426648 * 0.0821 * "273K"/"1atm"#</mathjax></p>
<p><mathjax>#V = 240.219981669/"1atm"#</mathjax></p>
<p><mathjax>#"Volume = 240.219981669L"#</mathjax></p></div>
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</article> | A sample of gas has a volume of 240. L at 4.00 atm of pressure and a temperature of 1091K. What is the volume at STP? | null |
3,022 | aa1fdce8-6ddd-11ea-861f-ccda262736ce | https://socratic.org/questions/a-stock-solution-of-hydrochloric-acid-with-1-25-m-25-ml-was-diluted-to-75-ml-wit | 0.42 M | start physical_unit 24 26 concentration mol/l qc_end physical_unit 2 2 9 10 volume qc_end physical_unit 2 2 14 15 volume qc_end substance 17 17 qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] this diluted acid [IN] M"}] | [{"type":"physical unit","value":"0.42 M"}] | [{"type":"physical unit","value":"Concentration1 [OF] hydrochloric acid stock solution [=] \\pu{1.25 M}"},{"type":"physical unit","value":"Volume1 [OF] hydrochloric acid stock solution [=] \\pu{25 mL}"},{"type":"physical unit","value":"Volume2 [OF] hydrochloric acid stock solution [=] \\pu{75 mL}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">A stock solution of hydrochloric acid with 1.25 M, 25 mL was diluted to 75 mL with water. What was the new concentration of this diluted acid?</h1> | null | 0.42 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the new concentration, we'll use the dilution formula</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax></p>
<p><mathjax>#M_1#</mathjax> = Concentration of stock solution = 1.25 M<br/>
<mathjax>#M_2#</mathjax> = Concentration of diluted solution = ?<br/>
<mathjax>#V_1#</mathjax> = Volume of stock solution = 25 mL<br/>
<mathjax>#V_2#</mathjax> = Volume of diluted solution = 75 mL</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax><br/>
<mathjax>#1.25#</mathjax> x <mathjax>#25#</mathjax> = <mathjax>#M_2#</mathjax> x <mathjax>#75#</mathjax><br/>
<mathjax>#M_2#</mathjax> = <mathjax>#0.42 "M"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>0.42 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the new concentration, we'll use the dilution formula</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax></p>
<p><mathjax>#M_1#</mathjax> = Concentration of stock solution = 1.25 M<br/>
<mathjax>#M_2#</mathjax> = Concentration of diluted solution = ?<br/>
<mathjax>#V_1#</mathjax> = Volume of stock solution = 25 mL<br/>
<mathjax>#V_2#</mathjax> = Volume of diluted solution = 75 mL</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax><br/>
<mathjax>#1.25#</mathjax> x <mathjax>#25#</mathjax> = <mathjax>#M_2#</mathjax> x <mathjax>#75#</mathjax><br/>
<mathjax>#M_2#</mathjax> = <mathjax>#0.42 "M"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A stock solution of hydrochloric acid with 1.25 M, 25 mL was diluted to 75 mL with water. What was the new concentration of this diluted acid?</h1>
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<div class="markdown"><p>0.42 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the new concentration, we'll use the dilution formula</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax></p>
<p><mathjax>#M_1#</mathjax> = Concentration of stock solution = 1.25 M<br/>
<mathjax>#M_2#</mathjax> = Concentration of diluted solution = ?<br/>
<mathjax>#V_1#</mathjax> = Volume of stock solution = 25 mL<br/>
<mathjax>#V_2#</mathjax> = Volume of diluted solution = 75 mL</p>
<p><mathjax>#M_1#</mathjax><mathjax>#V_1#</mathjax> = <mathjax>#M_2#</mathjax><mathjax>#V_2#</mathjax><br/>
<mathjax>#1.25#</mathjax> x <mathjax>#25#</mathjax> = <mathjax>#M_2#</mathjax> x <mathjax>#75#</mathjax><br/>
<mathjax>#M_2#</mathjax> = <mathjax>#0.42 "M"#</mathjax></p></div>
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</article> | A stock solution of hydrochloric acid with 1.25 M, 25 mL was diluted to 75 mL with water. What was the new concentration of this diluted acid? | null |
3,023 | a8eef046-6ddd-11ea-9886-ccda262736ce | https://socratic.org/questions/what-mass-of-co-2-contains-4-8-x-10-22-oxygen-atoms | 1.8 g | start physical_unit 3 3 mass g qc_end physical_unit 8 9 5 7 number qc_end end | [{"type":"physical unit","value":"Mass [OF] CO2 [IN] g"}] | [{"type":"physical unit","value":"1.8 g"}] | [{"type":"physical unit","value":"Number [OF] oxygen atoms [=] \\pu{4.8 × 10^22}"}] | <h1 class="questionTitle" itemprop="name">What mass of #CO_2# contains #4.8 xx 10^22# oxygen atoms?</h1> | null | 1.8 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use the chemical formula of carbon dioxide to find the <strong>number of molecules</strong> of</em> <mathjax>#"CO"_2#</mathjax> <em>that would contain that many atoms of oxygen</em></li>
<li><em>use <strong>Avogadro's constant</strong> to convert the number of molecules to <strong>moles</strong> of carbon dioxide</em></li>
<li><em>use the <strong>molar mass</strong> of carbon dioxide to convert the moles to <strong>grams</strong></em></li>
</ul>
</blockquote>
<p>So, you know that <strong>one molecule</strong> of carbon dioxide contains </p>
<blockquote>
<ul>
<li><em><strong>one atom</strong> of carbon</em>, <mathjax>#1 xx "C"#</mathjax></li>
<li><em><strong>two atoms</strong> of oxygen</em>, <mathjax>#2 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the given number of atoms of oxygen would correspond to </p>
<blockquote>
<p><mathjax>#4.8 * 10^(22) color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 2.4 * 10^(22)"molecules CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, <strong>one mole</strong> of any molecular substance contains exactly <mathjax>#6.022 * 10^(22)#</mathjax> <strong>molecules</strong> of that substance -- this is known as <strong>Avogadro's constant</strong>. </p>
<p>In your case, the sample of carbon dioxide molecules contains</p>
<blockquote>
<p><mathjax>#2.4 * 10^(22) color(red)(cancel(color(black)("molecules CO"_2))) * "1 mole CO"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.03985 moles CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, carbon dioxide has a molar mass of <mathjax>#"44.01 g mol"^(-1)#</mathjax>, which means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#0.03985 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(green)(bar(ul(|color(white)(a/a)color(black)("1.8 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</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 sig figs you have for the number of atoms of oxygen present in the sample. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.8 g CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use the chemical formula of carbon dioxide to find the <strong>number of molecules</strong> of</em> <mathjax>#"CO"_2#</mathjax> <em>that would contain that many atoms of oxygen</em></li>
<li><em>use <strong>Avogadro's constant</strong> to convert the number of molecules to <strong>moles</strong> of carbon dioxide</em></li>
<li><em>use the <strong>molar mass</strong> of carbon dioxide to convert the moles to <strong>grams</strong></em></li>
</ul>
</blockquote>
<p>So, you know that <strong>one molecule</strong> of carbon dioxide contains </p>
<blockquote>
<ul>
<li><em><strong>one atom</strong> of carbon</em>, <mathjax>#1 xx "C"#</mathjax></li>
<li><em><strong>two atoms</strong> of oxygen</em>, <mathjax>#2 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the given number of atoms of oxygen would correspond to </p>
<blockquote>
<p><mathjax>#4.8 * 10^(22) color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 2.4 * 10^(22)"molecules CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, <strong>one mole</strong> of any molecular substance contains exactly <mathjax>#6.022 * 10^(22)#</mathjax> <strong>molecules</strong> of that substance -- this is known as <strong>Avogadro's constant</strong>. </p>
<p>In your case, the sample of carbon dioxide molecules contains</p>
<blockquote>
<p><mathjax>#2.4 * 10^(22) color(red)(cancel(color(black)("molecules CO"_2))) * "1 mole CO"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.03985 moles CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, carbon dioxide has a molar mass of <mathjax>#"44.01 g mol"^(-1)#</mathjax>, which means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#0.03985 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(green)(bar(ul(|color(white)(a/a)color(black)("1.8 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</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 sig figs you have for the number of atoms of oxygen present in the sample. </p></div>
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<h1 class="questionTitle" itemprop="name">What mass of #CO_2# contains #4.8 xx 10^22# oxygen atoms?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-10-25T00:13:06" itemprop="dateCreated">
Oct 25, 2016
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<div class="markdown"><p><mathjax>#"1.8 g CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to </p>
<blockquote>
<ul>
<li><em>use the chemical formula of carbon dioxide to find the <strong>number of molecules</strong> of</em> <mathjax>#"CO"_2#</mathjax> <em>that would contain that many atoms of oxygen</em></li>
<li><em>use <strong>Avogadro's constant</strong> to convert the number of molecules to <strong>moles</strong> of carbon dioxide</em></li>
<li><em>use the <strong>molar mass</strong> of carbon dioxide to convert the moles to <strong>grams</strong></em></li>
</ul>
</blockquote>
<p>So, you know that <strong>one molecule</strong> of carbon dioxide contains </p>
<blockquote>
<ul>
<li><em><strong>one atom</strong> of carbon</em>, <mathjax>#1 xx "C"#</mathjax></li>
<li><em><strong>two atoms</strong> of oxygen</em>, <mathjax>#2 xx "O"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the given number of atoms of oxygen would correspond to </p>
<blockquote>
<p><mathjax>#4.8 * 10^(22) color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = 2.4 * 10^(22)"molecules CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, <strong>one mole</strong> of any molecular substance contains exactly <mathjax>#6.022 * 10^(22)#</mathjax> <strong>molecules</strong> of that substance -- this is known as <strong>Avogadro's constant</strong>. </p>
<p>In your case, the sample of carbon dioxide molecules contains</p>
<blockquote>
<p><mathjax>#2.4 * 10^(22) color(red)(cancel(color(black)("molecules CO"_2))) * "1 mole CO"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.03985 moles CO"_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, carbon dioxide has a molar mass of <mathjax>#"44.01 g mol"^(-1)#</mathjax>, which means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#0.03985 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(green)(bar(ul(|color(white)(a/a)color(black)("1.8 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</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 sig figs you have for the number of atoms of oxygen present in the sample. </p></div>
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</article> | What mass of #CO_2# contains #4.8 xx 10^22# oxygen atoms? | null |
3,024 | aabdfbdc-6ddd-11ea-bb93-ccda262736ce | https://socratic.org/questions/how-do-you-balance-hbr-o-2-br-2-h-2o | 4 HBr + O2 -> 2 Br2 + 2 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":"4 HBr + O2 -> 2 Br2 + 2 H2O"}] | [{"type":"chemical equation","value":"HBr + O2 -> Br2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #HBr + O_2 -> Br_2 + H_2O#?</h1> | null | 4 HBr + O2 -> 2 Br2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #HBr + O_2 -> Br_2 + H_2O#?</h1>
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Trevor Ryan.
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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A08
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<div class="markdown"><p><mathjax>#4"HBr"+"O"_2→2"Br"_2+2"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This equation is one of those where you need to balance O at the very beginning. </p>
<p>Otherwise as a thumb-rule one leaves H and O to taken up at the last.</p>
<p>Observe that in the unbalanced equation, on the reactants side there are 2 atoms of oxygen, whereas there is single oxygen atom on the products side.</p>
<p>Therefore to balance O on both sides, we need to put 2 in front of water molecule. This fixes 4 hydrogen atoms on the right side of the equation.</p>
<p>To balance 4H atoms in the products, we need 4 molecules HBr on the reactants side.</p>
<p>Now we are left with balancing of bromine atoms. Clearly to balance 4 Br atoms on the reactants side, we need to put 2 in front of <mathjax>#" Br"_2 ,#</mathjax> to balance the complete equation.</p></div>
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</article> | How do you balance #HBr + O_2 -> Br_2 + H_2O#? | null |
3,025 | abfb50a4-6ddd-11ea-9345-ccda262736ce | https://socratic.org/questions/what-is-the-volume-in-liters-occupied-by-0-723-mol-of-a-gas-at-32-0-c-if-the-pre | 16.2 liters | start physical_unit 12 12 volume l qc_end physical_unit 12 12 8 9 mole qc_end physical_unit 12 12 14 15 temperature qc_end physical_unit 12 12 20 21 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"16.2 liters"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{0.723 mol}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{32.0 ℃}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{852 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg?</h1> | null | 16.2 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for the volume we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> equation </p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=#</mathjax> Pressure in <mathjax>#atm#</mathjax><br/>
<mathjax>#V=#</mathjax> Volume in <mathjax>#L#</mathjax><br/>
<mathjax>#n=#</mathjax> moles<br/>
<mathjax>#R=#</mathjax> the Gas Law Constant <mathjax>#0.0821 (atm*L)/(mol*K)#</mathjax><br/>
<mathjax>#T=#</mathjax> Temperature in <mathjax>#K#</mathjax></p>
<p>When using the Ideal Gas Law Constant we must convert the Pressure to <mathjax>#atm#</mathjax><br/>
<mathjax>#P=852mmHg*(1atm)/(760mmHg)= 1.12 atm#</mathjax></p>
<p><mathjax>#V=?#</mathjax></p>
<p><mathjax>#n=0.723mol#</mathjax></p>
<p><mathjax>#R=0.0821 (atm*L)/(mol*K)#</mathjax> </p>
<p>When using the Ideal Gas Law Constant we must convert the Temperature to <mathjax>#K#</mathjax><br/>
<mathjax>#T=32.0^oC+273 = 305K#</mathjax></p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=((0.723cancel(mol))(0.0821 (cancel(atm)*L)/(cancel(mol)*cancel(K)))(305cancelK))/(1.12cancel(atm))#</mathjax></p>
<p><mathjax>#V = 16.2L#</mathjax></p>
<p>
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<div class="markdown"><p>the volume will be <mathjax>#V = 16.2L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for the volume we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> equation </p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=#</mathjax> Pressure in <mathjax>#atm#</mathjax><br/>
<mathjax>#V=#</mathjax> Volume in <mathjax>#L#</mathjax><br/>
<mathjax>#n=#</mathjax> moles<br/>
<mathjax>#R=#</mathjax> the Gas Law Constant <mathjax>#0.0821 (atm*L)/(mol*K)#</mathjax><br/>
<mathjax>#T=#</mathjax> Temperature in <mathjax>#K#</mathjax></p>
<p>When using the Ideal Gas Law Constant we must convert the Pressure to <mathjax>#atm#</mathjax><br/>
<mathjax>#P=852mmHg*(1atm)/(760mmHg)= 1.12 atm#</mathjax></p>
<p><mathjax>#V=?#</mathjax></p>
<p><mathjax>#n=0.723mol#</mathjax></p>
<p><mathjax>#R=0.0821 (atm*L)/(mol*K)#</mathjax> </p>
<p>When using the Ideal Gas Law Constant we must convert the Temperature to <mathjax>#K#</mathjax><br/>
<mathjax>#T=32.0^oC+273 = 305K#</mathjax></p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=((0.723cancel(mol))(0.0821 (cancel(atm)*L)/(cancel(mol)*cancel(K)))(305cancelK))/(1.12cancel(atm))#</mathjax></p>
<p><mathjax>#V = 16.2L#</mathjax></p>
<p>
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<h1 class="questionTitle" itemprop="name">What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg?</h1>
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<div class="markdown"><p>the volume will be <mathjax>#V = 16.2L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve for the volume we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> equation </p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=#</mathjax> Pressure in <mathjax>#atm#</mathjax><br/>
<mathjax>#V=#</mathjax> Volume in <mathjax>#L#</mathjax><br/>
<mathjax>#n=#</mathjax> moles<br/>
<mathjax>#R=#</mathjax> the Gas Law Constant <mathjax>#0.0821 (atm*L)/(mol*K)#</mathjax><br/>
<mathjax>#T=#</mathjax> Temperature in <mathjax>#K#</mathjax></p>
<p>When using the Ideal Gas Law Constant we must convert the Pressure to <mathjax>#atm#</mathjax><br/>
<mathjax>#P=852mmHg*(1atm)/(760mmHg)= 1.12 atm#</mathjax></p>
<p><mathjax>#V=?#</mathjax></p>
<p><mathjax>#n=0.723mol#</mathjax></p>
<p><mathjax>#R=0.0821 (atm*L)/(mol*K)#</mathjax> </p>
<p>When using the Ideal Gas Law Constant we must convert the Temperature to <mathjax>#K#</mathjax><br/>
<mathjax>#T=32.0^oC+273 = 305K#</mathjax></p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=((0.723cancel(mol))(0.0821 (cancel(atm)*L)/(cancel(mol)*cancel(K)))(305cancelK))/(1.12cancel(atm))#</mathjax></p>
<p><mathjax>#V = 16.2L#</mathjax></p>
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</article> | What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg? | null |
3,026 | a84cbbd8-6ddd-11ea-9069-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-mass-of-a-compound-if-560ml-have-a-mass-of-1-10g-at-stp | 44.59 g/mol | start physical_unit 7 7 molecular_weight g/mol qc_end physical_unit 7 7 9 10 volume qc_end physical_unit 7 7 15 16 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] a compound [IN] g/mol"}] | [{"type":"physical unit","value":"44.59 g/mol"}] | [{"type":"physical unit","value":"Volume [OF] a compound [=] \\pu{560 mL}"},{"type":"physical unit","value":"Mass [OF] a compound [=] \\pu{1.10 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the molecular mass of a compound if 560mL have a mass of 1.10g at STP?
</h1> | null | 44.59 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note: This is going to be a long answer.</p>
<p><strong>First determine the number of moles.</strong></p>
<p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p><strong>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>.</strong></p>
<p><mathjax>#PV=nRT#</mathjax>, <br/>
where <mathjax>#P#</mathjax> is pressure in kiloPascals <mathjax>#("kPa")#</mathjax>, <mathjax>#V#</mathjax> is volume in liters <mathjax>#("L")#</mathjax>, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins <mathjax>#("K")#</mathjax>.</p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><mathjax>#V=560cancel"mL"xx(1"L")/(1000cancel"mL")="0.56 L"#</mathjax></p>
<p><mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax> </p>
<p><strong>Unknown</strong></p>
<p>moles, n</p>
<p><strong>Equation</strong></p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)=#</mathjax></p>
<p><mathjax>#n=((100cancel"kPa")(0.56cancel"L"))/((8.3144598cancel "L" cancel"kPa"cancel" K"^(-1) "mol"^(-1)")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.024658 mol"#</mathjax></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax> (keeping a couple of guard digits)</p>
<p><strong>Determine the molecular (molar) mass.</strong></p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax></p>
<p><mathjax>#"m"="mass"="1.10 g"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>Molar mass: <mathjax>#"MM"#</mathjax></p>
<p><strong>Equation</strong></p>
<p><mathjax>#n=("m")/("MM")#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#"MM"#</mathjax> and solve.</p>
<p><mathjax>#"MM"xx("m")/(n)#</mathjax></p>
<p><mathjax>#"MM"xx(1.10"g")/(0.024658 "mol")=#</mathjax></p>
<p><mathjax>#"MM"="45 g/mol"#</mathjax> (rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to 0.56 L)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molar mass (molecular mass in grams) is <mathjax>#"45 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note: This is going to be a long answer.</p>
<p><strong>First determine the number of moles.</strong></p>
<p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p><strong>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>.</strong></p>
<p><mathjax>#PV=nRT#</mathjax>, <br/>
where <mathjax>#P#</mathjax> is pressure in kiloPascals <mathjax>#("kPa")#</mathjax>, <mathjax>#V#</mathjax> is volume in liters <mathjax>#("L")#</mathjax>, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins <mathjax>#("K")#</mathjax>.</p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><mathjax>#V=560cancel"mL"xx(1"L")/(1000cancel"mL")="0.56 L"#</mathjax></p>
<p><mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax> </p>
<p><strong>Unknown</strong></p>
<p>moles, n</p>
<p><strong>Equation</strong></p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)=#</mathjax></p>
<p><mathjax>#n=((100cancel"kPa")(0.56cancel"L"))/((8.3144598cancel "L" cancel"kPa"cancel" K"^(-1) "mol"^(-1)")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.024658 mol"#</mathjax></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax> (keeping a couple of guard digits)</p>
<p><strong>Determine the molecular (molar) mass.</strong></p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax></p>
<p><mathjax>#"m"="mass"="1.10 g"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>Molar mass: <mathjax>#"MM"#</mathjax></p>
<p><strong>Equation</strong></p>
<p><mathjax>#n=("m")/("MM")#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#"MM"#</mathjax> and solve.</p>
<p><mathjax>#"MM"xx("m")/(n)#</mathjax></p>
<p><mathjax>#"MM"xx(1.10"g")/(0.024658 "mol")=#</mathjax></p>
<p><mathjax>#"MM"="45 g/mol"#</mathjax> (rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to 0.56 L)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molecular mass of a compound if 560mL have a mass of 1.10g at STP?
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<div class="markdown"><p>The molar mass (molecular mass in grams) is <mathjax>#"45 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note: This is going to be a long answer.</p>
<p><strong>First determine the number of moles.</strong></p>
<p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p><strong>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>.</strong></p>
<p><mathjax>#PV=nRT#</mathjax>, <br/>
where <mathjax>#P#</mathjax> is pressure in kiloPascals <mathjax>#("kPa")#</mathjax>, <mathjax>#V#</mathjax> is volume in liters <mathjax>#("L")#</mathjax>, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins <mathjax>#("K")#</mathjax>.</p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><mathjax>#V=560cancel"mL"xx(1"L")/(1000cancel"mL")="0.56 L"#</mathjax></p>
<p><mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax> </p>
<p><strong>Unknown</strong></p>
<p>moles, n</p>
<p><strong>Equation</strong></p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)=#</mathjax></p>
<p><mathjax>#n=((100cancel"kPa")(0.56cancel"L"))/((8.3144598cancel "L" cancel"kPa"cancel" K"^(-1) "mol"^(-1)")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.024658 mol"#</mathjax></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax> (keeping a couple of guard digits)</p>
<p><strong>Determine the molecular (molar) mass.</strong></p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#n"=0.024658 mol"#</mathjax></p>
<p><mathjax>#"m"="mass"="1.10 g"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p>Molar mass: <mathjax>#"MM"#</mathjax></p>
<p><strong>Equation</strong></p>
<p><mathjax>#n=("m")/("MM")#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#"MM"#</mathjax> and solve.</p>
<p><mathjax>#"MM"xx("m")/(n)#</mathjax></p>
<p><mathjax>#"MM"xx(1.10"g")/(0.024658 "mol")=#</mathjax></p>
<p><mathjax>#"MM"="45 g/mol"#</mathjax> (rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to 0.56 L)</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-29T02:20:44" itemprop="dateCreated">
Oct 29, 2015
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<div class="markdown"><p><mathjax>#"45 g/mol"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the fact that at <strong>STP conditions</strong>, <em>one mole</em> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> - this is known as the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. </p>
<p>This means that if you know the volume of the gas, you can determine how many moles the sample contains by using the known molar volume at STP</p>
<blockquote>
<p><mathjax>#560color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(1000color(red)(cancel(color(black)("mL")))) * "1 mole"/(22.7color(red)(cancel(color(black)("L")))) = "0.02467 moles"#</mathjax></p>
</blockquote>
<p>The molar mass of the compound, which tells you what the exact mass of <em>one mole</em> of the gas is, can be determined by using the mass of the sample. </p>
<blockquote>
<p><mathjax>#M_"M" = m/n#</mathjax></p>
<p><mathjax>#M_"M" = "1.10 g"/"0.02467 moles" = "44.59 g/mol"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the nswer will be </p>
<blockquote>
<p><mathjax>#M_"M" = color(green)("45 g/mol")#</mathjax></p>
</blockquote></div>
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</article> | What is the molecular mass of a compound if 560mL have a mass of 1.10g at STP?
| null |
3,027 | ab6abb9d-6ddd-11ea-a228-ccda262736ce | https://socratic.org/questions/the-molecular-formula-for-phtalic-acid-is-is-c-8h-6o-4-what-is-the-empirical-for | C4H3O2 | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] phtalic acid [IN] empirical"}] | [{"type":"chemical equation","value":"C4H3O2"}] | [{"type":"other","value":"The molecular formula for phtalic acid is C8H6O4."}] | <h1 class="questionTitle" itemprop="name">The molecular formula for phtalic acid is is #C_8H_6O_4#. What is the empirical formula?</h1> | null | C4H3O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula is the lowest ratio of kinds of atoms achievable while keeping them all integers. 8, 6, and 4 all have a common factor of two so we can divide this out, but there are no other common factors so we can't reduce it further.</p>
<p>If the compound was being studied experimentally the ratio of different kinds of atoms (so the empirical formula) is usually one of the easiest things to determine, and the molecular formula could be found later by a technique like mass spectrometry to scale up the empirical formula. </p></div>
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<div class="markdown"><p><mathjax>#C_4H_3O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The empirical formula is the lowest ratio of kinds of atoms achievable while keeping them all integers. 8, 6, and 4 all have a common factor of two so we can divide this out, but there are no other common factors so we can't reduce it further.</p>
<p>If the compound was being studied experimentally the ratio of different kinds of atoms (so the empirical formula) is usually one of the easiest things to determine, and the molecular formula could be found later by a technique like mass spectrometry to scale up the empirical formula. </p></div>
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<h1 class="questionTitle" itemprop="name">The molecular formula for phtalic acid is is #C_8H_6O_4#. What is the empirical formula?</h1>
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Rose
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<div class="markdown"><p><mathjax>#C_4H_3O_2#</mathjax></p></div>
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<div class="markdown"><p>The empirical formula is the lowest ratio of kinds of atoms achievable while keeping them all integers. 8, 6, and 4 all have a common factor of two so we can divide this out, but there are no other common factors so we can't reduce it further.</p>
<p>If the compound was being studied experimentally the ratio of different kinds of atoms (so the empirical formula) is usually one of the easiest things to determine, and the molecular formula could be found later by a technique like mass spectrometry to scale up the empirical formula. </p></div>
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</article> | The molecular formula for phtalic acid is is #C_8H_6O_4#. What is the empirical formula? | null |
3,028 | ab4c848d-6ddd-11ea-8367-ccda262736ce | https://socratic.org/questions/how-many-grams-of-mg-oh-2-will-be-needed-to-neutralize-25-ml-of-stomach-acid-if- | 0.07 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 14 10 11 volume qc_end physical_unit 21 21 19 20 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] Mg(OH)2 [IN] grams"}] | [{"type":"physical unit","value":"0.07 grams"}] | [{"type":"physical unit","value":"Volume [OF] stomach acid [=] \\pu{25 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.10 M}"},{"type":"other","value":"Stomach acid is HCl."}] | <h1 class="questionTitle" itemprop="name">How many grams of #Mg(OH)_2# will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M #HCl#? </h1> | null | 0.07 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to keep in mind that you need <mathjax>#2#</mathjax> <strong>moles</strong> of hydrochloric acid in order to neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium hydroxide.</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (2(s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>You know that the hydrochloric acid solution contains <mathjax>#0.10#</mathjax> <strong>moles</strong> of hydrochloric acid <strong>for every</strong> <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, so you can say that your sample contains</p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("mL"))) * "0.10 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.0025 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> will require</p>
<blockquote>
<p><mathjax>#0.0025 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Mg"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.00125 moles Mg"("OH")_2#</mathjax></p>
</blockquote>
<p>Finally, to convert this to <em>grams</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#0.00125 color(red)(cancel(color(black)("moles Mg"("OH")_2))) * "58.32 g"/(color(red)(cancel(color(black)("moles Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.073 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>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.073 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to keep in mind that you need <mathjax>#2#</mathjax> <strong>moles</strong> of hydrochloric acid in order to neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium hydroxide.</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (2(s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>You know that the hydrochloric acid solution contains <mathjax>#0.10#</mathjax> <strong>moles</strong> of hydrochloric acid <strong>for every</strong> <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, so you can say that your sample contains</p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("mL"))) * "0.10 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.0025 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> will require</p>
<blockquote>
<p><mathjax>#0.0025 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Mg"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.00125 moles Mg"("OH")_2#</mathjax></p>
</blockquote>
<p>Finally, to convert this to <em>grams</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#0.00125 color(red)(cancel(color(black)("moles Mg"("OH")_2))) * "58.32 g"/(color(red)(cancel(color(black)("moles Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.073 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>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of #Mg(OH)_2# will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M #HCl#? </h1>
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Stefan V.
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Jun 27, 2017
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<div class="markdown"><p><mathjax>#"0.073 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to keep in mind that you need <mathjax>#2#</mathjax> <strong>moles</strong> of hydrochloric acid in order to neutralize <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium hydroxide.</p>
<blockquote>
<p><mathjax>#"Mg"("OH")_ (2(s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>You know that the hydrochloric acid solution contains <mathjax>#0.10#</mathjax> <strong>moles</strong> of hydrochloric acid <strong>for every</strong> <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, so you can say that your sample contains</p>
<blockquote>
<p><mathjax>#25 color(red)(cancel(color(black)("mL"))) * "0.10 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.0025 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> will require</p>
<blockquote>
<p><mathjax>#0.0025 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Mg"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.00125 moles Mg"("OH")_2#</mathjax></p>
</blockquote>
<p>Finally, to convert this to <em>grams</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#0.00125 color(red)(cancel(color(black)("moles Mg"("OH")_2))) * "58.32 g"/(color(red)(cancel(color(black)("moles Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.073 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>.</p></div>
</div>
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</article> | How many grams of #Mg(OH)_2# will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M #HCl#? | null |
3,029 | aad7cd3a-6ddd-11ea-bf68-ccda262736ce | https://socratic.org/questions/593ce214b72cff475e625101 | 16 M | start physical_unit 5 7 molarity mol/l qc_end physical_unit 5 7 12 13 density qc_end end | [{"type":"physical unit","value":"Molarity [OF] concentrated nitric acid [IN] M"}] | [{"type":"physical unit","value":"16 M"}] | [{"type":"physical unit","value":"m/m [OF] concentrated nitric acid in solution [=] \\pu{70%}"},{"type":"physical unit","value":"ρ [OF] concentrated nitric acid [=] \\pu{1.41 g/mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of concentrated nitric acid that has #rho=1.41*g*mL^-1#, and is #70%# concentrated, #"m/m"#?</h1> | null | 16 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want the expression <mathjax>#"Moles"/"Volume of solution"#</mathjax>, and if we take a <mathjax>#1*mL#</mathjax> volume of nitric acid, with a mass of <mathjax>#1.41*g#</mathjax>, then......</p>
<p><mathjax>#"Concentration"="Moles"/"Volume"#</mathjax> And we use the percentage and <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to find the actual mass of nitric acid <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in (for instance) a <mathjax>#1*mL#</mathjax> volume. </p>
<p><mathjax>#=((70.0%xx1.41*g)/(63.01*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=??*mol*L^-1#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Concentration"~=16*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want the expression <mathjax>#"Moles"/"Volume of solution"#</mathjax>, and if we take a <mathjax>#1*mL#</mathjax> volume of nitric acid, with a mass of <mathjax>#1.41*g#</mathjax>, then......</p>
<p><mathjax>#"Concentration"="Moles"/"Volume"#</mathjax> And we use the percentage and <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to find the actual mass of nitric acid <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in (for instance) a <mathjax>#1*mL#</mathjax> volume. </p>
<p><mathjax>#=((70.0%xx1.41*g)/(63.01*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=??*mol*L^-1#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of concentrated nitric acid that has #rho=1.41*g*mL^-1#, and is #70%# concentrated, #"m/m"#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Concentration"~=16*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want the expression <mathjax>#"Moles"/"Volume of solution"#</mathjax>, and if we take a <mathjax>#1*mL#</mathjax> volume of nitric acid, with a mass of <mathjax>#1.41*g#</mathjax>, then......</p>
<p><mathjax>#"Concentration"="Moles"/"Volume"#</mathjax> And we use the percentage and <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to find the actual mass of nitric acid <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in (for instance) a <mathjax>#1*mL#</mathjax> volume. </p>
<p><mathjax>#=((70.0%xx1.41*g)/(63.01*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=??*mol*L^-1#</mathjax>.</p></div>
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</article> | What is the molarity of concentrated nitric acid that has #rho=1.41*g*mL^-1#, and is #70%# concentrated, #"m/m"#? | null |
3,030 | aa90d840-6ddd-11ea-9966-ccda262736ce | https://socratic.org/questions/a-reaction-is-endothermic-with-h-100-kj-mol-if-the-activation-enthalpy-of-the-fo | 40 kJ/mol | start physical_unit 26 28 activation_barrier kj/mol qc_end physical_unit 1 1 7 8 deltah qc_end physical_unit 15 16 18 19 activation_barrier qc_end end | [{"type":"physical unit","value":"Activation enthalpy [OF] the reverse reaction [IN] kJ/mol"}] | [{"type":"physical unit","value":"40 kJ/mol"}] | [{"type":"physical unit","value":"DeltaH [OF] the reaction [=] \\pu{100 kJ/mol}"},{"type":"physical unit","value":"Activation enthalpy [OF] forward reaction [=] \\pu{140 kJ/mol}"}] | <h1 class="questionTitle" itemprop="name">A reaction is endothermic with H=100 kJ/mol. If the activation enthalpy of the forward reaction is 140 kJ/mol, what is the activation enthalpy of the reverse reaction?
I know the answer is 40 kJ/mol but I’m not quite sure how to get there, thanks!</h1> | null | 40 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Here's a schematic progress of reaction diagram for an endothermic reaction.</p>
<p><img alt="ProgReaction" src="https://useruploads.socratic.org/nGYXchKRziivunMxC61x_ProgReaction.jpg"/> <br/>
(Adapted from BBC)</p>
<p>We see that <mathjax>#Δ_text(‡)H("fwd")#</mathjax> (from reactants to transition state) is <mathjax>#"140 kJ·mol"^"-1"#</mathjax>.</p>
<p>We also see that <mathjax>#Δ_text(rxn)H#</mathjax> (from reactants to products is <mathjax>#"100 kJ·mol"^"-1"#</mathjax>.</p>
<p>Thus, <mathjax>#Δ_text(‡)H("rev")#</mathjax> (products to transition state) must be</p>
<p><mathjax>#Δ_text(‡)H("rev") = "140 kJ·mol"^"-1" - "100 kJ·mol"^"-1" = "40 kJ·mol"^"-1"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#E_text(a) = "40 kJ·mol"^"-1"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Here's a schematic progress of reaction diagram for an endothermic reaction.</p>
<p><img alt="ProgReaction" src="https://useruploads.socratic.org/nGYXchKRziivunMxC61x_ProgReaction.jpg"/> <br/>
(Adapted from BBC)</p>
<p>We see that <mathjax>#Δ_text(‡)H("fwd")#</mathjax> (from reactants to transition state) is <mathjax>#"140 kJ·mol"^"-1"#</mathjax>.</p>
<p>We also see that <mathjax>#Δ_text(rxn)H#</mathjax> (from reactants to products is <mathjax>#"100 kJ·mol"^"-1"#</mathjax>.</p>
<p>Thus, <mathjax>#Δ_text(‡)H("rev")#</mathjax> (products to transition state) must be</p>
<p><mathjax>#Δ_text(‡)H("rev") = "140 kJ·mol"^"-1" - "100 kJ·mol"^"-1" = "40 kJ·mol"^"-1"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A reaction is endothermic with H=100 kJ/mol. If the activation enthalpy of the forward reaction is 140 kJ/mol, what is the activation enthalpy of the reverse reaction?
I know the answer is 40 kJ/mol but I’m not quite sure how to get there, thanks!</h1>
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<div class="markdown"><p><mathjax>#E_text(a) = "40 kJ·mol"^"-1"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Here's a schematic progress of reaction diagram for an endothermic reaction.</p>
<p><img alt="ProgReaction" src="https://useruploads.socratic.org/nGYXchKRziivunMxC61x_ProgReaction.jpg"/> <br/>
(Adapted from BBC)</p>
<p>We see that <mathjax>#Δ_text(‡)H("fwd")#</mathjax> (from reactants to transition state) is <mathjax>#"140 kJ·mol"^"-1"#</mathjax>.</p>
<p>We also see that <mathjax>#Δ_text(rxn)H#</mathjax> (from reactants to products is <mathjax>#"100 kJ·mol"^"-1"#</mathjax>.</p>
<p>Thus, <mathjax>#Δ_text(‡)H("rev")#</mathjax> (products to transition state) must be</p>
<p><mathjax>#Δ_text(‡)H("rev") = "140 kJ·mol"^"-1" - "100 kJ·mol"^"-1" = "40 kJ·mol"^"-1"#</mathjax></p></div>
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</article> | A reaction is endothermic with H=100 kJ/mol. If the activation enthalpy of the forward reaction is 140 kJ/mol, what is the activation enthalpy of the reverse reaction?
I know the answer is 40 kJ/mol but I’m not quite sure how to get there, thanks! | null |
3,031 | ab01a1ee-6ddd-11ea-a085-ccda262736ce | https://socratic.org/questions/what-is-the-mass-o-5-60-mol-naoh | 224.00 grams | start physical_unit 7 7 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] NaOH [IN] grams"}] | [{"type":"physical unit","value":"224.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{5.60 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 5.60 mol #NaOH#?</h1> | null | 224.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> composing the compound;<br/>
<mathjax>#Na#</mathjax>=?<br/>
<mathjax>#O#</mathjax>=?<br/>
<mathjax>#H#</mathjax>=?</li>
<li>Once obtained, multiply to the number of atoms of each of the elements to get the formula mass of <mathjax>#NaOH#</mathjax>;</li>
<li>Apply the conversion method to find the mass of <mathjax>#NaOH#</mathjax> at <mathjax>#5.6#</mathjax> mol <mathjax>#NaOH#</mathjax>;<br/>
<mathjax>#5.6#</mathjax> <mathjax>#cancel(mol NaOH) #</mathjax>x<mathjax># (40 grams NaOH)/ (1 cancel(mol NaOH)#</mathjax>;</li>
<li>The answer is <mathjax>#224#</mathjax> grams <mathjax>#NaOH#</mathjax> per calculation,</li>
</ol></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#224#</mathjax> grams</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> composing the compound;<br/>
<mathjax>#Na#</mathjax>=?<br/>
<mathjax>#O#</mathjax>=?<br/>
<mathjax>#H#</mathjax>=?</li>
<li>Once obtained, multiply to the number of atoms of each of the elements to get the formula mass of <mathjax>#NaOH#</mathjax>;</li>
<li>Apply the conversion method to find the mass of <mathjax>#NaOH#</mathjax> at <mathjax>#5.6#</mathjax> mol <mathjax>#NaOH#</mathjax>;<br/>
<mathjax>#5.6#</mathjax> <mathjax>#cancel(mol NaOH) #</mathjax>x<mathjax># (40 grams NaOH)/ (1 cancel(mol NaOH)#</mathjax>;</li>
<li>The answer is <mathjax>#224#</mathjax> grams <mathjax>#NaOH#</mathjax> per calculation,</li>
</ol></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 5.60 mol #NaOH#?</h1>
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<div class="markdown"><p><mathjax>#224#</mathjax> grams</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Take the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> composing the compound;<br/>
<mathjax>#Na#</mathjax>=?<br/>
<mathjax>#O#</mathjax>=?<br/>
<mathjax>#H#</mathjax>=?</li>
<li>Once obtained, multiply to the number of atoms of each of the elements to get the formula mass of <mathjax>#NaOH#</mathjax>;</li>
<li>Apply the conversion method to find the mass of <mathjax>#NaOH#</mathjax> at <mathjax>#5.6#</mathjax> mol <mathjax>#NaOH#</mathjax>;<br/>
<mathjax>#5.6#</mathjax> <mathjax>#cancel(mol NaOH) #</mathjax>x<mathjax># (40 grams NaOH)/ (1 cancel(mol NaOH)#</mathjax>;</li>
<li>The answer is <mathjax>#224#</mathjax> grams <mathjax>#NaOH#</mathjax> per calculation,</li>
</ol></div>
</div>
</div>
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</article> | What is the mass of 5.60 mol #NaOH#? | null |
3,032 | aceb0eae-6ddd-11ea-bf66-ccda262736ce | https://socratic.org/questions/560ff95f581e2a5c3899aa3d | Fe^3+(aq) + 3 OH-(aq) -> Fe(OH)3(s) v | start chemical_equation qc_end chemical_equation 10 10 qc_end chemical_equation 12 13 qc_end substance 15 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"Fe^3+(aq) + 3 OH-(aq) -> Fe(OH)3(s) v"}] | [{"type":"chemical equation","value":"Fe(OH)3"},{"type":"chemical equation","value":"Ferric nitrate"},{"type":"substance name","value":"Sodium hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for the formation of #Fe(OH)_3# from #"ferric nitrate"# and #"sodium hydroxide"#?</h1> | null | Fe^3+(aq) + 3 OH-(aq) -> Fe(OH)3(s) v | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> that results in the formation of an <em>insoluble solid</em> that <strong>precipitates</strong> out of solution. </p>
<p>More specifically, a solution of iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, will react with a solution of sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, to form <em>iron(III) hydroxide</em>, <mathjax>#"Fe"("OH")_3#</mathjax>, which precipitates out of solution. </p>
<p>So, the <em>molecular equation</em> that describes this reaction is </p>
<blockquote>
<p><mathjax>#"Fe"("NO"_3)_text(3(aq]) + 3"NaOH"_text((aq]) -> "Fe"("OH")_text(3(s]) darr + 3"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <em>complete ionic equation</em> will include the solid and all the ions that are present in solution</p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"NO"_text(3(aq])^(-) + 3"Na"_text((aq])^(+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr + 3"Na"_text((aq])^(+) + 3"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice tha the <em>insoluble solid</em> is not represented as ions! </p>
<p>The <em>net ionic equation</em> will only show the ions that <strong>actually take part in the reaction</strong>. </p>
<p>The reaction between the two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> will result in the formation of the solid, which means that the ions that are <strong>not</strong> a part of the solid will not take part in the reaction. </p>
<p>More specifically, they will act as <strong>spectator ions</strong>. You can tell which ions are spectator ions by their presence on both sides of the equation. </p>
<p>So, the net ionic equation is </p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr#</mathjax></p>
</blockquote>
<p>Here's how this reaction would look like</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000kqCf46dDTeE" src="https://useruploads.socratic.org/3DCUZX6ETwStALMCjGZU_Fphoto-54480206F-2RM.jpg"/> </p>
<p>Here the iron(III) nitrate solution is added to the colorless sodium hydroxide solution</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> that results in the formation of an <em>insoluble solid</em> that <strong>precipitates</strong> out of solution. </p>
<p>More specifically, a solution of iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, will react with a solution of sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, to form <em>iron(III) hydroxide</em>, <mathjax>#"Fe"("OH")_3#</mathjax>, which precipitates out of solution. </p>
<p>So, the <em>molecular equation</em> that describes this reaction is </p>
<blockquote>
<p><mathjax>#"Fe"("NO"_3)_text(3(aq]) + 3"NaOH"_text((aq]) -> "Fe"("OH")_text(3(s]) darr + 3"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <em>complete ionic equation</em> will include the solid and all the ions that are present in solution</p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"NO"_text(3(aq])^(-) + 3"Na"_text((aq])^(+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr + 3"Na"_text((aq])^(+) + 3"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice tha the <em>insoluble solid</em> is not represented as ions! </p>
<p>The <em>net ionic equation</em> will only show the ions that <strong>actually take part in the reaction</strong>. </p>
<p>The reaction between the two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> will result in the formation of the solid, which means that the ions that are <strong>not</strong> a part of the solid will not take part in the reaction. </p>
<p>More specifically, they will act as <strong>spectator ions</strong>. You can tell which ions are spectator ions by their presence on both sides of the equation. </p>
<p>So, the net ionic equation is </p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr#</mathjax></p>
</blockquote>
<p>Here's how this reaction would look like</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000kqCf46dDTeE" src="https://useruploads.socratic.org/3DCUZX6ETwStALMCjGZU_Fphoto-54480206F-2RM.jpg"/> </p>
<p>Here the iron(III) nitrate solution is added to the colorless sodium hydroxide solution</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the net ionic equation for the formation of #Fe(OH)_3# from #"ferric nitrate"# and #"sodium hydroxide"#?</h1>
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Oct 3, 2015
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<div class="markdown"><p><mathjax>#"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a> that results in the formation of an <em>insoluble solid</em> that <strong>precipitates</strong> out of solution. </p>
<p>More specifically, a solution of iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, will react with a solution of sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, to form <em>iron(III) hydroxide</em>, <mathjax>#"Fe"("OH")_3#</mathjax>, which precipitates out of solution. </p>
<p>So, the <em>molecular equation</em> that describes this reaction is </p>
<blockquote>
<p><mathjax>#"Fe"("NO"_3)_text(3(aq]) + 3"NaOH"_text((aq]) -> "Fe"("OH")_text(3(s]) darr + 3"NaNO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The <em>complete ionic equation</em> will include the solid and all the ions that are present in solution</p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"NO"_text(3(aq])^(-) + 3"Na"_text((aq])^(+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr + 3"Na"_text((aq])^(+) + 3"NO"_text(3(aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice tha the <em>insoluble solid</em> is not represented as ions! </p>
<p>The <em>net ionic equation</em> will only show the ions that <strong>actually take part in the reaction</strong>. </p>
<p>The reaction between the two <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> will result in the formation of the solid, which means that the ions that are <strong>not</strong> a part of the solid will not take part in the reaction. </p>
<p>More specifically, they will act as <strong>spectator ions</strong>. You can tell which ions are spectator ions by their presence on both sides of the equation. </p>
<p>So, the net ionic equation is </p>
<blockquote>
<p><mathjax>#"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr#</mathjax></p>
</blockquote>
<p>Here's how this reaction would look like</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000kqCf46dDTeE" src="https://useruploads.socratic.org/3DCUZX6ETwStALMCjGZU_Fphoto-54480206F-2RM.jpg"/> </p>
<p>Here the iron(III) nitrate solution is added to the colorless sodium hydroxide solution</p></div>
</div>
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</article> | What is the net ionic equation for the formation of #Fe(OH)_3# from #"ferric nitrate"# and #"sodium hydroxide"#? | null |
3,033 | ace88aee-6ddd-11ea-adc2-ccda262736ce | https://socratic.org/questions/when-ethane-c-2h-6-reacts-with-chlorine-gas-the-main-product-is-c-2h-5cl-but-sma | 88.8% | start physical_unit 11 11 percent_yield none qc_end physical_unit 1 1 30 31 mass qc_end physical_unit 5 6 35 36 mass qc_end physical_unit 11 11 41 42 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Percent yield [OF] C2H5Cl"}] | [{"type":"physical unit","value":"88.8%"}] | [{"type":"physical unit","value":"Mass [OF] ethane [=] \\pu{125 g}"},{"type":"physical unit","value":"Mass [OF] chlorine gas [=] \\pu{255 g}"},{"type":"physical unit","value":"Mass [OF] C2H5Cl [=] \\pu{206 g}"},{"type":"other","value":"When ethane, C2H6, reacts with chlorine gas the main product is C2H5Cl but small amounts of C2H4Cl2 are produced."}] | <h1 class="questionTitle" itemprop="name">When ethane, #C_2H_6#, reacts with chlorine gas the main product is #C_2H_5Cl# but small amounts of #C_2H_4Cl_2# are produced. What is the percent yield of #C_2H_5Cl# if the reaction of 125g of ethane with 255g of chlorine gas produced 206g of #C2H_5Cl#? </h1> | null | 88.8% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to focus solely on the reaction that produces <em>chloroethane</em>, <mathjax>#"C"_2"H"_5"Cl"#</mathjax>, and ignore the one that produces <em>dichloroethane</em>, <mathjax>#"C"_2"H"_4"Cl"#</mathjax>, the side product of the reaction.</p>
<p>Start by writing the balanced chemical equation for the <em>chlorination of ethane</em></p>
<blockquote>
<p><mathjax>#"C"_2"H"_text(6(g]) + "Cl"_text(2(g]) -> "C"_2"H"_5"Cl"_text((g]) + "HCl"_text((g])#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong> that exists between ethane and chlorine gas to determine whether or not you're dealing with a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>So, you're mixing <mathjax>#"125 g"#</mathjax> of ethane with <mathjax>#"255 g"#</mathjax> of chlorine gas. Convert the masses of the two reactants to <em>moles</em> by using their respective <strong>molar masses</strong></p>
<blockquote>
<p><mathjax>#125 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "4.157 moles C"_2"H"_6#</mathjax></p>
<p><mathjax>#255color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "3.596 moles Cl"_2#</mathjax></p>
</blockquote>
<p>The reaction will <strong>always</strong> consume <em>equal numbers of moles</em> of each reactant, which means that chlorine gas will act as a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> here. </p>
<p>More specifically, out of the <mathjax>#4.157#</mathjax> moles of ethane, only <mathjax>#3.596#</mathjax> will actually take part in the reaction. The rest will be <em>in excess</em>. </p>
<p>Now, you also have a <mathjax>#1:1#</mathjax> mole ratio between the reactants and chloroethane. This means that the reaction will produce <mathjax>#3.596#</mathjax> moles of chloroethane, since that's how many moles of each reactant take part in the reaction. </p>
<p>Use chloroethane's molar mass to figure out how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#3.596 color(red)(cancel(color(black)("moles C"_2"H"_5"Cl"))) * "64.51 g"/(1color(red)(cancel(color(black)("mole C"_2"H"_5"Cl")))) = "231.98 g"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>In <strong>theory</strong>, the reaction should produce <mathjax>#"231.98 g"#</mathjax> of chloroethane. However, you know that the reaction only produced <mathjax>#"206 g"#</mathjax> of chloroethane. </p>
<p>The <strong>difference</strong> between the <strong>theoretical yield</strong> of chloroethane, which is what you get when <mathjax>#100%#</mathjax> of the reactants are converted to chloroethane and hydrogen chloride, and the <strong>actual yield</strong>, which is what you get when some <em>dichloroethane</em> is produced, will represent the <strong><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of chloroethane. </p>
<blockquote>
<p><mathjax>#color(blue)("% yield" = "what you actually get"/"what you should get" xx 100)#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"% yield" = (206 color(red)(cancel(color(black)("g"))))/(231.98color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.8 %")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"% yield" = 88.8%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to focus solely on the reaction that produces <em>chloroethane</em>, <mathjax>#"C"_2"H"_5"Cl"#</mathjax>, and ignore the one that produces <em>dichloroethane</em>, <mathjax>#"C"_2"H"_4"Cl"#</mathjax>, the side product of the reaction.</p>
<p>Start by writing the balanced chemical equation for the <em>chlorination of ethane</em></p>
<blockquote>
<p><mathjax>#"C"_2"H"_text(6(g]) + "Cl"_text(2(g]) -> "C"_2"H"_5"Cl"_text((g]) + "HCl"_text((g])#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong> that exists between ethane and chlorine gas to determine whether or not you're dealing with a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>So, you're mixing <mathjax>#"125 g"#</mathjax> of ethane with <mathjax>#"255 g"#</mathjax> of chlorine gas. Convert the masses of the two reactants to <em>moles</em> by using their respective <strong>molar masses</strong></p>
<blockquote>
<p><mathjax>#125 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "4.157 moles C"_2"H"_6#</mathjax></p>
<p><mathjax>#255color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "3.596 moles Cl"_2#</mathjax></p>
</blockquote>
<p>The reaction will <strong>always</strong> consume <em>equal numbers of moles</em> of each reactant, which means that chlorine gas will act as a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> here. </p>
<p>More specifically, out of the <mathjax>#4.157#</mathjax> moles of ethane, only <mathjax>#3.596#</mathjax> will actually take part in the reaction. The rest will be <em>in excess</em>. </p>
<p>Now, you also have a <mathjax>#1:1#</mathjax> mole ratio between the reactants and chloroethane. This means that the reaction will produce <mathjax>#3.596#</mathjax> moles of chloroethane, since that's how many moles of each reactant take part in the reaction. </p>
<p>Use chloroethane's molar mass to figure out how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#3.596 color(red)(cancel(color(black)("moles C"_2"H"_5"Cl"))) * "64.51 g"/(1color(red)(cancel(color(black)("mole C"_2"H"_5"Cl")))) = "231.98 g"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>In <strong>theory</strong>, the reaction should produce <mathjax>#"231.98 g"#</mathjax> of chloroethane. However, you know that the reaction only produced <mathjax>#"206 g"#</mathjax> of chloroethane. </p>
<p>The <strong>difference</strong> between the <strong>theoretical yield</strong> of chloroethane, which is what you get when <mathjax>#100%#</mathjax> of the reactants are converted to chloroethane and hydrogen chloride, and the <strong>actual yield</strong>, which is what you get when some <em>dichloroethane</em> is produced, will represent the <strong><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of chloroethane. </p>
<blockquote>
<p><mathjax>#color(blue)("% yield" = "what you actually get"/"what you should get" xx 100)#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"% yield" = (206 color(red)(cancel(color(black)("g"))))/(231.98color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.8 %")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When ethane, #C_2H_6#, reacts with chlorine gas the main product is #C_2H_5Cl# but small amounts of #C_2H_4Cl_2# are produced. What is the percent yield of #C_2H_5Cl# if the reaction of 125g of ethane with 255g of chlorine gas produced 206g of #C2H_5Cl#? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"% yield" = 88.8%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to focus solely on the reaction that produces <em>chloroethane</em>, <mathjax>#"C"_2"H"_5"Cl"#</mathjax>, and ignore the one that produces <em>dichloroethane</em>, <mathjax>#"C"_2"H"_4"Cl"#</mathjax>, the side product of the reaction.</p>
<p>Start by writing the balanced chemical equation for the <em>chlorination of ethane</em></p>
<blockquote>
<p><mathjax>#"C"_2"H"_text(6(g]) + "Cl"_text(2(g]) -> "C"_2"H"_5"Cl"_text((g]) + "HCl"_text((g])#</mathjax></p>
</blockquote>
<p>Your next step will be to use the <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong> that exists between ethane and chlorine gas to determine whether or not you're dealing with a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>So, you're mixing <mathjax>#"125 g"#</mathjax> of ethane with <mathjax>#"255 g"#</mathjax> of chlorine gas. Convert the masses of the two reactants to <em>moles</em> by using their respective <strong>molar masses</strong></p>
<blockquote>
<p><mathjax>#125 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "4.157 moles C"_2"H"_6#</mathjax></p>
<p><mathjax>#255color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.91color(red)(cancel(color(black)("g")))) = "3.596 moles Cl"_2#</mathjax></p>
</blockquote>
<p>The reaction will <strong>always</strong> consume <em>equal numbers of moles</em> of each reactant, which means that chlorine gas will act as a <strong><a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> here. </p>
<p>More specifically, out of the <mathjax>#4.157#</mathjax> moles of ethane, only <mathjax>#3.596#</mathjax> will actually take part in the reaction. The rest will be <em>in excess</em>. </p>
<p>Now, you also have a <mathjax>#1:1#</mathjax> mole ratio between the reactants and chloroethane. This means that the reaction will produce <mathjax>#3.596#</mathjax> moles of chloroethane, since that's how many moles of each reactant take part in the reaction. </p>
<p>Use chloroethane's molar mass to figure out how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#3.596 color(red)(cancel(color(black)("moles C"_2"H"_5"Cl"))) * "64.51 g"/(1color(red)(cancel(color(black)("mole C"_2"H"_5"Cl")))) = "231.98 g"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>In <strong>theory</strong>, the reaction should produce <mathjax>#"231.98 g"#</mathjax> of chloroethane. However, you know that the reaction only produced <mathjax>#"206 g"#</mathjax> of chloroethane. </p>
<p>The <strong>difference</strong> between the <strong>theoretical yield</strong> of chloroethane, which is what you get when <mathjax>#100%#</mathjax> of the reactants are converted to chloroethane and hydrogen chloride, and the <strong>actual yield</strong>, which is what you get when some <em>dichloroethane</em> is produced, will represent the <strong><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of chloroethane. </p>
<blockquote>
<p><mathjax>#color(blue)("% yield" = "what you actually get"/"what you should get" xx 100)#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"% yield" = (206 color(red)(cancel(color(black)("g"))))/(231.98color(red)(cancel(color(black)("g")))) xx 100 = color(green)("88.8 %")#</mathjax></p>
</blockquote></div>
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</article> | When ethane, #C_2H_6#, reacts with chlorine gas the main product is #C_2H_5Cl# but small amounts of #C_2H_4Cl_2# are produced. What is the percent yield of #C_2H_5Cl# if the reaction of 125g of ethane with 255g of chlorine gas produced 206g of #C2H_5Cl#? | null |
3,034 | aa0e0f66-6ddd-11ea-93b3-ccda262736ce | https://socratic.org/questions/if-200-ml-of-a-2-50-m-naoh-solution-is-diluted-to-500-ml-what-is-the-new-concent | 1.00 M | start physical_unit 7 8 concentration mol/l qc_end physical_unit 7 8 5 6 concentration qc_end physical_unit 7 8 1 2 volume qc_end physical_unit 7 8 12 13 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] NaOH solution [IN] M"}] | [{"type":"physical unit","value":"1.00 M"}] | [{"type":"physical unit","value":"Concentration1 [OF] NaOH solution [=] \\pu{2.50 M}"},{"type":"physical unit","value":"Volume1 [OF] NaOH solution [=] \\pu{200 mL}"},{"type":"physical unit","value":"Volume2 [OF] NaOH solution [=] \\pu{500 mL}"}] | <h1 class="questionTitle" itemprop="name">If 200 mL of a 2.50 M #NaOH# solution is diluted to 500 mL, what is the new concentration of NaOH?</h1> | null | 1.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We use this relationship to get the initial number of moles, by taking the product,<mathjax>#"concentration"xx"volume"#</mathjax>, and then dividing this molar quantity by the new volume, which we get by simply adding the 2 given volumes. </p>
<p>And thus, <mathjax>#(200xx10^-3cancelLxx2.50*mol*L^-1)/(500xx10^-3cancelL)#</mathjax></p>
<p>And we get an answer with units <mathjax>#mol*L^-1#</mathjax> as we require. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The new concentration is <mathjax>#1.00*mol*L^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We use this relationship to get the initial number of moles, by taking the product,<mathjax>#"concentration"xx"volume"#</mathjax>, and then dividing this molar quantity by the new volume, which we get by simply adding the 2 given volumes. </p>
<p>And thus, <mathjax>#(200xx10^-3cancelLxx2.50*mol*L^-1)/(500xx10^-3cancelL)#</mathjax></p>
<p>And we get an answer with units <mathjax>#mol*L^-1#</mathjax> as we require. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If 200 mL of a 2.50 M #NaOH# solution is diluted to 500 mL, what is the new concentration of NaOH?</h1>
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<div class="markdown"><p>The new concentration is <mathjax>#1.00*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We use this relationship to get the initial number of moles, by taking the product,<mathjax>#"concentration"xx"volume"#</mathjax>, and then dividing this molar quantity by the new volume, which we get by simply adding the 2 given volumes. </p>
<p>And thus, <mathjax>#(200xx10^-3cancelLxx2.50*mol*L^-1)/(500xx10^-3cancelL)#</mathjax></p>
<p>And we get an answer with units <mathjax>#mol*L^-1#</mathjax> as we require. </p></div>
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</article> | If 200 mL of a 2.50 M #NaOH# solution is diluted to 500 mL, what is the new concentration of NaOH? | null |
3,035 | a8c1b0d8-6ddd-11ea-b45a-ccda262736ce | https://socratic.org/questions/in-the-reaction-mg-s-2hcl-aq-h-2-g-mgcl-2-aq-how-many-grams-of-hydrogen-gas-will | 0.76 grams | start physical_unit 15 16 mass g qc_end chemical_equation 3 10 qc_end physical_unit 27 27 21 22 volume qc_end physical_unit 27 27 25 26 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen gas [IN] grams"}] | [{"type":"physical unit","value":"0.76 grams"}] | [{"type":"chemical equation","value":"Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq)"},{"type":"physical unit","value":"Volume [OF] HCI [=] \\pu{125.0 milliliters}"},{"type":"physical unit","value":"Molarity [OF] HCI [=] \\pu{6.0 M}"},{"type":"other","value":"Excess of Mg."}] | <h1 class="questionTitle" itemprop="name">In the reaction# Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (aq)#, how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg? </h1> | null | 0.76 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>I am assuming that <mathjax>#6.0 M HCl#</mathjax> is a typo, and that it should be <mathjax>#6.0 molL^-1 HCl#</mathjax>, since that makes sense in the equation.</em></p>
<p>First we have to find the <strong>amount</strong> of <mathjax>#HCl#</mathjax> in the solution. We use the formula <mathjax>#n = cV#</mathjax> where <mathjax>#n#</mathjax> is the amount of substance in moles, <mathjax>#c#</mathjax> is the concentration of the solution in moles per litre, and <mathjax>#V#</mathjax> is the volume of the substance in litres.</p>
<p><mathjax>#n(HCl) = 6.0molL^-1 xx 0.125L = 0.75mol#</mathjax></p>
<p>Then we find out how many moles of hydrogen gas (<mathjax>#H_2#</mathjax>) are produced. In the formula we see <mathjax>#2HCl#</mathjax>, and <mathjax>#H_2#</mathjax>. This means there is 1 mole of <mathjax>#H_2#</mathjax> for every 2 moles of <mathjax>#HCl#</mathjax> so to find the amount of <mathjax>#H_2#</mathjax> we use:</p>
<p><mathjax>#n(H_2) = 1/2 xx 0.75mol = 0.375mol#</mathjax></p>
<p>Now we find the molar mass of the <mathjax>#H_2#</mathjax> molecules, by adding together the atomic weights of the constituent molecules. In this case: <mathjax>#1.008 + 1.008 = 2.016#</mathjax>. Then we use the formula <mathjax>#m = nM#</mathjax> where <mathjax>#m#</mathjax> is the mass of the substance in grams, and <mathjax>#M#</mathjax> is the molar mass of the substance in grams per mole.</p>
<p><mathjax>#m(H_2) = 0.375mol xx 2.016gmol^-1 = 0.756g#</mathjax></p></div>
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<div class="markdown"><p><mathjax># 0.756g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>I am assuming that <mathjax>#6.0 M HCl#</mathjax> is a typo, and that it should be <mathjax>#6.0 molL^-1 HCl#</mathjax>, since that makes sense in the equation.</em></p>
<p>First we have to find the <strong>amount</strong> of <mathjax>#HCl#</mathjax> in the solution. We use the formula <mathjax>#n = cV#</mathjax> where <mathjax>#n#</mathjax> is the amount of substance in moles, <mathjax>#c#</mathjax> is the concentration of the solution in moles per litre, and <mathjax>#V#</mathjax> is the volume of the substance in litres.</p>
<p><mathjax>#n(HCl) = 6.0molL^-1 xx 0.125L = 0.75mol#</mathjax></p>
<p>Then we find out how many moles of hydrogen gas (<mathjax>#H_2#</mathjax>) are produced. In the formula we see <mathjax>#2HCl#</mathjax>, and <mathjax>#H_2#</mathjax>. This means there is 1 mole of <mathjax>#H_2#</mathjax> for every 2 moles of <mathjax>#HCl#</mathjax> so to find the amount of <mathjax>#H_2#</mathjax> we use:</p>
<p><mathjax>#n(H_2) = 1/2 xx 0.75mol = 0.375mol#</mathjax></p>
<p>Now we find the molar mass of the <mathjax>#H_2#</mathjax> molecules, by adding together the atomic weights of the constituent molecules. In this case: <mathjax>#1.008 + 1.008 = 2.016#</mathjax>. Then we use the formula <mathjax>#m = nM#</mathjax> where <mathjax>#m#</mathjax> is the mass of the substance in grams, and <mathjax>#M#</mathjax> is the molar mass of the substance in grams per mole.</p>
<p><mathjax>#m(H_2) = 0.375mol xx 2.016gmol^-1 = 0.756g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In the reaction# Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (aq)#, how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg? </h1>
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<div class="markdown"><p><mathjax># 0.756g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>I am assuming that <mathjax>#6.0 M HCl#</mathjax> is a typo, and that it should be <mathjax>#6.0 molL^-1 HCl#</mathjax>, since that makes sense in the equation.</em></p>
<p>First we have to find the <strong>amount</strong> of <mathjax>#HCl#</mathjax> in the solution. We use the formula <mathjax>#n = cV#</mathjax> where <mathjax>#n#</mathjax> is the amount of substance in moles, <mathjax>#c#</mathjax> is the concentration of the solution in moles per litre, and <mathjax>#V#</mathjax> is the volume of the substance in litres.</p>
<p><mathjax>#n(HCl) = 6.0molL^-1 xx 0.125L = 0.75mol#</mathjax></p>
<p>Then we find out how many moles of hydrogen gas (<mathjax>#H_2#</mathjax>) are produced. In the formula we see <mathjax>#2HCl#</mathjax>, and <mathjax>#H_2#</mathjax>. This means there is 1 mole of <mathjax>#H_2#</mathjax> for every 2 moles of <mathjax>#HCl#</mathjax> so to find the amount of <mathjax>#H_2#</mathjax> we use:</p>
<p><mathjax>#n(H_2) = 1/2 xx 0.75mol = 0.375mol#</mathjax></p>
<p>Now we find the molar mass of the <mathjax>#H_2#</mathjax> molecules, by adding together the atomic weights of the constituent molecules. In this case: <mathjax>#1.008 + 1.008 = 2.016#</mathjax>. Then we use the formula <mathjax>#m = nM#</mathjax> where <mathjax>#m#</mathjax> is the mass of the substance in grams, and <mathjax>#M#</mathjax> is the molar mass of the substance in grams per mole.</p>
<p><mathjax>#m(H_2) = 0.375mol xx 2.016gmol^-1 = 0.756g#</mathjax></p></div>
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</article> | In the reaction# Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (aq)#, how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg? | null |
3,036 | ab44e165-6ddd-11ea-94ce-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-that-has-a-hydronium-ion-concentration-of-1-00-10-1 | 10.00 | start physical_unit 6 6 ph none qc_end physical_unit 10 11 14 17 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"10.00"}] | [{"type":"physical unit","value":"Concentration [OF] hydronium ion [=] \\pu{1.00 × 10(-10) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution that has a hydronium ion concentration of #1.00*10^-10# M?</h1> | null | 10.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of an aqueous solution can be found simply by taking the negative logarithm of your hydronium ion concentration, or stated mathematically,</p>
<p>pH = <mathjax>#-log[H_3O^+]#</mathjax></p>
<p>So, if we plug in your numbers we'll get:</p>
<p>pH = <mathjax>#-log[1.00*10^-10]#</mathjax></p>
<p>Throw that bad boy in your calculator and you'll find your pH. Or I can just tell you. (It's 10.)</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Take the negative log of your <mathjax>#H_3O^+#</mathjax> <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of an aqueous solution can be found simply by taking the negative logarithm of your hydronium ion concentration, or stated mathematically,</p>
<p>pH = <mathjax>#-log[H_3O^+]#</mathjax></p>
<p>So, if we plug in your numbers we'll get:</p>
<p>pH = <mathjax>#-log[1.00*10^-10]#</mathjax></p>
<p>Throw that bad boy in your calculator and you'll find your pH. Or I can just tell you. (It's 10.)</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a solution that has a hydronium ion concentration of #1.00*10^-10# M?</h1>
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Thomas U.
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<div class="markdown"><p>Take the negative log of your <mathjax>#H_3O^+#</mathjax> <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of an aqueous solution can be found simply by taking the negative logarithm of your hydronium ion concentration, or stated mathematically,</p>
<p>pH = <mathjax>#-log[H_3O^+]#</mathjax></p>
<p>So, if we plug in your numbers we'll get:</p>
<p>pH = <mathjax>#-log[1.00*10^-10]#</mathjax></p>
<p>Throw that bad boy in your calculator and you'll find your pH. Or I can just tell you. (It's 10.)</p></div>
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</article> | What is the pH of a solution that has a hydronium ion concentration of #1.00*10^-10# M? | null |
3,037 | ac189eec-6ddd-11ea-8ad2-ccda262736ce | https://socratic.org/questions/how-many-grams-of-aluminum-are-required-to-produce-8-70-moles-of-aluminum-chlori | 234.90 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 13 9 10 mole qc_end chemical_equation 17 24 qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum [IN] grams"}] | [{"type":"physical unit","value":"234.90 grams"}] | [{"type":"physical unit","value":"Mole [OF] aluminum chloride [=] \\pu{8.70 moles}"},{"type":"chemical equation","value":"2 Al + 3 Cl2 -> 2 AlCl3"}] | <h1 class="questionTitle" itemprop="name">How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#?</h1> | null | 234.90 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which the chemicals combine, and in this case illustrates that <mathjax>#2 mol Al#</mathjax> produces <mathjax>#2molAlCl_3#</mathjax>, hence these 2 chemicals are in a <mathjax>#1:1#</mathjax> ratio.</p>
<p>Thus, to produce <mathjax>#8.70mol#</mathjax> aluminium chloride, it will require <mathjax>#8.70mol#</mathjax> aluminium.</p>
<p>But this quantity of <mathjax>#Al#</mathjax> has a mass in grams of </p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=8.70molxx27g//mol#</mathjax></p>
<p><mathjax>#=234.9g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#234.9g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which the chemicals combine, and in this case illustrates that <mathjax>#2 mol Al#</mathjax> produces <mathjax>#2molAlCl_3#</mathjax>, hence these 2 chemicals are in a <mathjax>#1:1#</mathjax> ratio.</p>
<p>Thus, to produce <mathjax>#8.70mol#</mathjax> aluminium chloride, it will require <mathjax>#8.70mol#</mathjax> aluminium.</p>
<p>But this quantity of <mathjax>#Al#</mathjax> has a mass in grams of </p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=8.70molxx27g//mol#</mathjax></p>
<p><mathjax>#=234.9g#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#?</h1>
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Trevor Ryan.
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<span class="dateCreated" datetime="2016-03-29T20:45:09" itemprop="dateCreated">
Mar 29, 2016
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<div class="markdown"><p><mathjax>#234.9g#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which the chemicals combine, and in this case illustrates that <mathjax>#2 mol Al#</mathjax> produces <mathjax>#2molAlCl_3#</mathjax>, hence these 2 chemicals are in a <mathjax>#1:1#</mathjax> ratio.</p>
<p>Thus, to produce <mathjax>#8.70mol#</mathjax> aluminium chloride, it will require <mathjax>#8.70mol#</mathjax> aluminium.</p>
<p>But this quantity of <mathjax>#Al#</mathjax> has a mass in grams of </p>
<p><mathjax>#m=nxxM_r#</mathjax></p>
<p><mathjax>#=8.70molxx27g//mol#</mathjax></p>
<p><mathjax>#=234.9g#</mathjax></p></div>
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</article> | How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#? | null |
3,038 | a8dd1126-6ddd-11ea-b0c5-ccda262736ce | https://socratic.org/questions/a-36-07-g-sample-of-a-substance-is-initially-at-27-8-c-after-absorbing-2639-j-of | 0.9 J/(g * ℃) | start physical_unit 21 22 specific_heat j/(°c_·_g) qc_end physical_unit 3 6 1 2 mass qc_end physical_unit 3 6 10 11 temperature qc_end physical_unit 21 22 14 15 heat_energy qc_end physical_unit 21 22 24 25 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] the substance [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.9 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] a substance sample [=] \\pu{36.07 g}"},{"type":"physical unit","value":"Temperature1 [OF] a substance sample [=] \\pu{27.8 ℃}"},{"type":"physical unit","value":"Absorbed heat [OF] the substance [=] \\pu{2639 J}"},{"type":"physical unit","value":"Temperature2 [OF] the substance [=] \\pu{109.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 36.07 g sample of a substance is initially at 27.8°C. After absorbing 2639 J of heat, the temperature of the substance is 109.0°C. What is the specific heat of the substance?</h1> | null | 0.9 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the relationship between heat <mathjax>#Q#</mathjax>, mass <mathjax>#m#</mathjax>, <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#c#</mathjax> and change in temperature <mathjax>#DeltaT#</mathjax> as:<br/>
<mathjax>#Q=mcDeltaT#</mathjax><br/>
so:<br/>
<mathjax>#2639=36.07*c*(109.0-27.8)#</mathjax><br/>
rearranging:<br/>
<mathjax>#c=0.9J/(g^@C)#</mathjax></p>
<p>Aluminum perhaps?</p>
<p><img alt="www.eeweb.com" src="https://useruploads.socratic.org/HZk7BYS3TUGWzdraVQbs_Heat-and-Temperature-Change-Specific-Heat-Capacity-4-1336760973_347_251_75.jpg"/></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>I found: <mathjax>#c=0.9J/(g^@C)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the relationship between heat <mathjax>#Q#</mathjax>, mass <mathjax>#m#</mathjax>, <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#c#</mathjax> and change in temperature <mathjax>#DeltaT#</mathjax> as:<br/>
<mathjax>#Q=mcDeltaT#</mathjax><br/>
so:<br/>
<mathjax>#2639=36.07*c*(109.0-27.8)#</mathjax><br/>
rearranging:<br/>
<mathjax>#c=0.9J/(g^@C)#</mathjax></p>
<p>Aluminum perhaps?</p>
<p><img alt="www.eeweb.com" src="https://useruploads.socratic.org/HZk7BYS3TUGWzdraVQbs_Heat-and-Temperature-Change-Specific-Heat-Capacity-4-1336760973_347_251_75.jpg"/></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 36.07 g sample of a substance is initially at 27.8°C. After absorbing 2639 J of heat, the temperature of the substance is 109.0°C. What is the specific heat of the substance?</h1>
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Gió
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<span class="dateCreated" datetime="2016-07-25T21:30:40" itemprop="dateCreated">
Jul 25, 2016
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<div class="markdown"><p>I found: <mathjax>#c=0.9J/(g^@C)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I would use the relationship between heat <mathjax>#Q#</mathjax>, mass <mathjax>#m#</mathjax>, <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#c#</mathjax> and change in temperature <mathjax>#DeltaT#</mathjax> as:<br/>
<mathjax>#Q=mcDeltaT#</mathjax><br/>
so:<br/>
<mathjax>#2639=36.07*c*(109.0-27.8)#</mathjax><br/>
rearranging:<br/>
<mathjax>#c=0.9J/(g^@C)#</mathjax></p>
<p>Aluminum perhaps?</p>
<p><img alt="www.eeweb.com" src="https://useruploads.socratic.org/HZk7BYS3TUGWzdraVQbs_Heat-and-Temperature-Change-Specific-Heat-Capacity-4-1336760973_347_251_75.jpg"/></p></div>
</div>
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</article> | A 36.07 g sample of a substance is initially at 27.8°C. After absorbing 2639 J of heat, the temperature of the substance is 109.0°C. What is the specific heat of the substance? | null |
3,039 | aba205a6-6ddd-11ea-811f-ccda262736ce | https://socratic.org/questions/593c420db72cff4750f36852 | 3.00 L | start physical_unit 5 5 volume l qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 10 11 pressure qc_end physical_unit 5 5 21 22 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"3.00 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{5 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{124 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{206 kPa}"}] | <h1 class="questionTitle" itemprop="name">A #5*L# volume of gas exerts a pressure of #124*kPa#. How should volume evolve if pressure is increased to #206*kPa#?</h1> | null | 3.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity gas, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#P=k/V#</mathjax>, where <mathjax>#k#</mathjax> is some (unspecified!) constant.</p>
<p>And if we solve for <mathjax>#k#</mathjax> under different conditions of pressure and volume (with temperature CONSTANT!), then.......</p>
<p><mathjax>#V_2=(P_1xxV_1)/(P_2)=(5*Lxx124*kPa)/(206*kPa)=3.00*L#</mathjax></p>
<p>Is this answer consistent with the parameters of the problem? Should volume DECREASE with INCREASED pressure?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2#</mathjax>....so <mathjax>#V_2=3.0*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity gas, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#P=k/V#</mathjax>, where <mathjax>#k#</mathjax> is some (unspecified!) constant.</p>
<p>And if we solve for <mathjax>#k#</mathjax> under different conditions of pressure and volume (with temperature CONSTANT!), then.......</p>
<p><mathjax>#V_2=(P_1xxV_1)/(P_2)=(5*Lxx124*kPa)/(206*kPa)=3.00*L#</mathjax></p>
<p>Is this answer consistent with the parameters of the problem? Should volume DECREASE with INCREASED pressure?</p></div>
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<h1 class="questionTitle" itemprop="name">A #5*L# volume of gas exerts a pressure of #124*kPa#. How should volume evolve if pressure is increased to #206*kPa#?</h1>
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<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2#</mathjax>....so <mathjax>#V_2=3.0*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity gas, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#P=k/V#</mathjax>, where <mathjax>#k#</mathjax> is some (unspecified!) constant.</p>
<p>And if we solve for <mathjax>#k#</mathjax> under different conditions of pressure and volume (with temperature CONSTANT!), then.......</p>
<p><mathjax>#V_2=(P_1xxV_1)/(P_2)=(5*Lxx124*kPa)/(206*kPa)=3.00*L#</mathjax></p>
<p>Is this answer consistent with the parameters of the problem? Should volume DECREASE with INCREASED pressure?</p></div>
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</article> | A #5*L# volume of gas exerts a pressure of #124*kPa#. How should volume evolve if pressure is increased to #206*kPa#? | null |
3,040 | ab028cd5-6ddd-11ea-8b58-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-kmno-4 | +7 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] Mn"}] | [{"type":"physical unit","value":"+7"}] | [{"type":"chemical equation","value":"KMnO4"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of #"Mn"# in #"KMnO"_4#? </h1> | null | +7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We assign the oxidation number per a set of rules.</p>
<p>For this question, the important rules are:</p>
<ol>
<li>
<p>The oxidation number of <mathjax>#"O"#</mathjax> in <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is usually -2, but it is -1 in peroxides.</p>
</li>
<li>
<p>The oxidation number of a Group 1 element in a compound is +1.</p>
</li>
<li>
<p>The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of all of the atoms in a neutral compound is 0.</p>
</li>
</ol>
<blockquote></blockquote>
<p><strong>Rule 1</strong> states that the oxidation number of <mathjax>#"O"#</mathjax> is -2.</p>
<p>We write the oxidation number of the element above its symbol and the total for 3 <mathjax>#"O"#</mathjax> atoms below the symbol.</p>
<p>This gives <mathjax>#"KMn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmll)stackrelcolor(blue)("-8"color(white)(mm))#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 2</strong> states that the oxidation number of K is +1.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmll)stackrelcolor(blue)("+1")color(white)(mmmmm)stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 3</strong> states the numbers along the bottom must add up to zero.</p>
<p>The number under <mathjax>#"Mn"#</mathjax> must be +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<p>There is only one <mathjax>#"Mn"#</mathjax> atom, so its oxidation number is +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p></div>
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<div class="markdown"><p>The oxidation number of <mathjax>#"Mn"#</mathjax> is +7.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We assign the oxidation number per a set of rules.</p>
<p>For this question, the important rules are:</p>
<ol>
<li>
<p>The oxidation number of <mathjax>#"O"#</mathjax> in <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is usually -2, but it is -1 in peroxides.</p>
</li>
<li>
<p>The oxidation number of a Group 1 element in a compound is +1.</p>
</li>
<li>
<p>The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of all of the atoms in a neutral compound is 0.</p>
</li>
</ol>
<blockquote></blockquote>
<p><strong>Rule 1</strong> states that the oxidation number of <mathjax>#"O"#</mathjax> is -2.</p>
<p>We write the oxidation number of the element above its symbol and the total for 3 <mathjax>#"O"#</mathjax> atoms below the symbol.</p>
<p>This gives <mathjax>#"KMn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmll)stackrelcolor(blue)("-8"color(white)(mm))#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 2</strong> states that the oxidation number of K is +1.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmll)stackrelcolor(blue)("+1")color(white)(mmmmm)stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 3</strong> states the numbers along the bottom must add up to zero.</p>
<p>The number under <mathjax>#"Mn"#</mathjax> must be +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<p>There is only one <mathjax>#"Mn"#</mathjax> atom, so its oxidation number is +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of #"Mn"# in #"KMnO"_4#? </h1>
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<div class="markdown"><p>The oxidation number of <mathjax>#"Mn"#</mathjax> is +7.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We assign the oxidation number per a set of rules.</p>
<p>For this question, the important rules are:</p>
<ol>
<li>
<p>The oxidation number of <mathjax>#"O"#</mathjax> in <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is usually -2, but it is -1 in peroxides.</p>
</li>
<li>
<p>The oxidation number of a Group 1 element in a compound is +1.</p>
</li>
<li>
<p>The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of all of the atoms in a neutral compound is 0.</p>
</li>
</ol>
<blockquote></blockquote>
<p><strong>Rule 1</strong> states that the oxidation number of <mathjax>#"O"#</mathjax> is -2.</p>
<p>We write the oxidation number of the element above its symbol and the total for 3 <mathjax>#"O"#</mathjax> atoms below the symbol.</p>
<p>This gives <mathjax>#"KMn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmll)stackrelcolor(blue)("-8"color(white)(mm))#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 2</strong> states that the oxidation number of K is +1.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmll)stackrelcolor(blue)("+1")color(white)(mmmmm)stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<blockquote></blockquote>
<p><strong>Rule 3</strong> states the numbers along the bottom must add up to zero.</p>
<p>The number under <mathjax>#"Mn"#</mathjax> must be +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p>
<p>There is only one <mathjax>#"Mn"#</mathjax> atom, so its oxidation number is +7.</p>
<p>This gives <mathjax>#stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O")_4#</mathjax><br/>
<mathjax>#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#</mathjax></p></div>
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<span class="dateCreated" datetime="2017-01-19T18:35:13" itemprop="dateCreated">
Jan 19, 2017
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<div class="markdown"><p>The metal has a <mathjax>#+VII#</mathjax> oxidation state in <mathjax>#MnO_4^-#</mathjax>, <mathjax>#"permanganate ion"#</mathjax>.</p></div>
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<div>
<div class="markdown"><p>See <a href="https://socratic.org/questions/how-do-you-write-oxidation-reduction-half-reactions#278689">this link</a> for more of the same. Remember that oxidation number is a factitious entity, however, it does help us to balance <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>. </p>
<p><a href="https://socratic.org/questions/how-do-you-find-the-oxidation-number-of-the-central-atom">And also here.</a> </p></div>
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</article> | What is the oxidation number of #"Mn"# in #"KMnO"_4#? | null |
3,041 | ac2b2028-6ddd-11ea-b59a-ccda262736ce | https://socratic.org/questions/what-ch-3coona-concentration-is-required-to-prepare-a-buffer-solution-with-a-ph- | 0.34 M | start physical_unit 1 1 concentration mol/l qc_end physical_unit 8 9 14 14 ph qc_end physical_unit 8 9 18 19 [ch3cooh] qc_end physical_unit 16 16 24 26 ka qc_end end | [{"type":"physical unit","value":"Concentration [OF] CH3COO- [IN] M"}] | [{"type":"physical unit","value":"0.34 M"}] | [{"type":"physical unit","value":"pH [OF] buffer solution [=] \\pu{5.10}"},{"type":"physical unit","value":"[CH3COOH] [OF] buffer solution [=] \\pu{0.150 M}"},{"type":"physical unit","value":"Ka [OF] CH3COOH [=] \\pu{1.8 × 10^(-5)}"}] | <h1 class="questionTitle" itemprop="name">What #"CH"_3"COO"^(-)# concentration is required to prepare a buffer solution with a pH of 5.10 if #["CH"_3"COOH"]# is 0.150 M? #K_a# of #"CH"_3"COOH"# is #1.8 xx 10^-5#?</h1> | null | 0.34 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward application of the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak acid</strong> and its conjugate base looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log((["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here you have </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)("p"K_a = - log(K_a))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>where <mathjax>#K_a#</mathjax> - the <em>acid dissociation constant</em> for the weak acid</p>
<p>Before using the H - H equation, calculate the <mathjax>#pK_a#</mathjax> and compare it with the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer. This will help you determine how much conjugate base you need <strong>relative</strong> to the amount of weak acid present in solution. </p>
<p>You will have </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(1.8 * 10^(-5)) = 4.74#</mathjax></p>
</blockquote>
<p>The pH of the buffer is set at <mathjax>#5.10#</mathjax>. Since you have </p>
<blockquote>
<p><mathjax>#"pH " > " p"K_a#</mathjax></p>
</blockquote>
<p>you know for a fact that the buffer must contain <strong>more conjugate base</strong>, which in this case is the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, usually delivered to the solution by <em>sodium acetate</em>, <mathjax>#"CH"_3"COONa"#</mathjax>, than <strong>weak acid</strong>, which in this case is acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>. </p>
<p>Your goal now is to use the H - H equation to find the concentration of the acetate anion. You will have</p>
<blockquote>
<p><mathjax>#5.10 = 4.74 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.36#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.36#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 2.291#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of the acetate anions must be </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx ["CH"_3"COOH"]#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx "0.150 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.344 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>As predicted, the buffer contains more conjugate base than weak acid. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.344 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward application of the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak acid</strong> and its conjugate base looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log((["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here you have </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)("p"K_a = - log(K_a))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>where <mathjax>#K_a#</mathjax> - the <em>acid dissociation constant</em> for the weak acid</p>
<p>Before using the H - H equation, calculate the <mathjax>#pK_a#</mathjax> and compare it with the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer. This will help you determine how much conjugate base you need <strong>relative</strong> to the amount of weak acid present in solution. </p>
<p>You will have </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(1.8 * 10^(-5)) = 4.74#</mathjax></p>
</blockquote>
<p>The pH of the buffer is set at <mathjax>#5.10#</mathjax>. Since you have </p>
<blockquote>
<p><mathjax>#"pH " > " p"K_a#</mathjax></p>
</blockquote>
<p>you know for a fact that the buffer must contain <strong>more conjugate base</strong>, which in this case is the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, usually delivered to the solution by <em>sodium acetate</em>, <mathjax>#"CH"_3"COONa"#</mathjax>, than <strong>weak acid</strong>, which in this case is acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>. </p>
<p>Your goal now is to use the H - H equation to find the concentration of the acetate anion. You will have</p>
<blockquote>
<p><mathjax>#5.10 = 4.74 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.36#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.36#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 2.291#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of the acetate anions must be </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx ["CH"_3"COOH"]#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx "0.150 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.344 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>As predicted, the buffer contains more conjugate base than weak acid. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What #"CH"_3"COO"^(-)# concentration is required to prepare a buffer solution with a pH of 5.10 if #["CH"_3"COOH"]# is 0.150 M? #K_a# of #"CH"_3"COOH"# is #1.8 xx 10^-5#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-21T23:09:30" itemprop="dateCreated">
Jun 21, 2016
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<div class="markdown"><p><mathjax>#"0.344 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward application of the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak acid</strong> and its conjugate base looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = "p"K_a + log((["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here you have </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)("p"K_a = - log(K_a))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>where <mathjax>#K_a#</mathjax> - the <em>acid dissociation constant</em> for the weak acid</p>
<p>Before using the H - H equation, calculate the <mathjax>#pK_a#</mathjax> and compare it with the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer. This will help you determine how much conjugate base you need <strong>relative</strong> to the amount of weak acid present in solution. </p>
<p>You will have </p>
<blockquote>
<p><mathjax>#"p"K_a = - log(1.8 * 10^(-5)) = 4.74#</mathjax></p>
</blockquote>
<p>The pH of the buffer is set at <mathjax>#5.10#</mathjax>. Since you have </p>
<blockquote>
<p><mathjax>#"pH " > " p"K_a#</mathjax></p>
</blockquote>
<p>you know for a fact that the buffer must contain <strong>more conjugate base</strong>, which in this case is the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, usually delivered to the solution by <em>sodium acetate</em>, <mathjax>#"CH"_3"COONa"#</mathjax>, than <strong>weak acid</strong>, which in this case is acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>. </p>
<p>Your goal now is to use the H - H equation to find the concentration of the acetate anion. You will have</p>
<blockquote>
<p><mathjax>#5.10 = 4.74 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#</mathjax></p>
</blockquote>
<p>Rearrange to get</p>
<blockquote>
<p><mathjax>#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.36#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.36#</mathjax></p>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 2.291#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of the acetate anions must be </p>
<blockquote>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx ["CH"_3"COOH"]#</mathjax></p>
<p><mathjax>#["CH"_3"COO"^(-)] = 2.291 xx "0.150 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.344 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>As predicted, the buffer contains more conjugate base than weak acid. </p></div>
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</article> | What #"CH"_3"COO"^(-)# concentration is required to prepare a buffer solution with a pH of 5.10 if #["CH"_3"COOH"]# is 0.150 M? #K_a# of #"CH"_3"COOH"# is #1.8 xx 10^-5#? | null |
3,042 | ac6c499c-6ddd-11ea-92d5-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-composed-of-3-25-hydrogen-h-19-36-ca | CH2O3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O3"}] | [{"type":"physical unit","value":"Percent by mass [OF] H in the compound [=] \\pu{3.25%}"},{"type":"physical unit","value":"Percent by mass [OF] C in the compound [=] \\pu{19.36%}"},{"type":"physical unit","value":"Percent by mass [OF] O in the compound [=] \\pu{77.39%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? </h1> | null | CH2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? </h1>
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anor277
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<span class="dateCreated" datetime="2016-05-07T07:59:02" itemprop="dateCreated">
May 7, 2016
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<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
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<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Assume that there is 100g of the substance, so there would be </p>
<p><mathjax>#3.25g\ H#</mathjax> <mathjax>#19.36g\ C#</mathjax> and <mathjax>#77.39g\ O#</mathjax></p>
<p>Divide each mass by the molar mass of their respective element.</p>
<p><mathjax>#(3.25g)/1 H#</mathjax> <mathjax>#(19.36g)/12 C#</mathjax> <mathjax>#(77.39g) /16 O#</mathjax></p>
<p><mathjax>#=3.25 H#</mathjax> <mathjax>#=1.61 C#</mathjax> <mathjax>#=4.84 O#</mathjax></p>
<p>Divide by the smallest number. In this case, <mathjax>#1.61#</mathjax></p>
<p><mathjax>#=2 H#</mathjax> <mathjax>#=1 C#</mathjax> <mathjax>#=3O#</mathjax></p>
<p>So the empirical formula is <mathjax>#CH_2O_3#</mathjax></p></div>
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</article> | What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? | null |
3,043 | ab81a840-6ddd-11ea-8b0b-ccda262736ce | https://socratic.org/questions/5713343c11ef6b236c1dd4ac | 3.61 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 13 14 8 9 mass qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"3.61 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [=] \\pu{132 g}"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms in a mass of #132*g# with respect to carbon dioxide?</h1> | null | 3.61 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What is <mathjax>#N_A#</mathjax>? It is simply <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>If I have <mathjax>#6.022xx10^23#</mathjax> individual items of stuff, I have a mole of that stuff. <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#CO_2#</mathjax> is equivalent to a mass <mathjax>#44.0*g#</mathjax>; <mathjax>#12.01#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#C#</mathjax>, and <mathjax>#32.0#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#O#</mathjax>.</p></div>
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<div class="markdown"><p>There are <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#CO_2#</mathjax> in such a mass. Thus there are <mathjax>#6xxN_A#</mathjax> individual oxygen atoms. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>What is <mathjax>#N_A#</mathjax>? It is simply <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>If I have <mathjax>#6.022xx10^23#</mathjax> individual items of stuff, I have a mole of that stuff. <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#CO_2#</mathjax> is equivalent to a mass <mathjax>#44.0*g#</mathjax>; <mathjax>#12.01#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#C#</mathjax>, and <mathjax>#32.0#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#O#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many oxygen atoms in a mass of #132*g# with respect to carbon dioxide?</h1>
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<div class="markdown"><p>There are <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#CO_2#</mathjax> in such a mass. Thus there are <mathjax>#6xxN_A#</mathjax> individual oxygen atoms. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>What is <mathjax>#N_A#</mathjax>? It is simply <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>If I have <mathjax>#6.022xx10^23#</mathjax> individual items of stuff, I have a mole of that stuff. <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#CO_2#</mathjax> is equivalent to a mass <mathjax>#44.0*g#</mathjax>; <mathjax>#12.01#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#C#</mathjax>, and <mathjax>#32.0#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#O#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#36.132 times 10^23 #</mathjax> oxygen atoms</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If we work out the relative formula mass of <mathjax>#CO_2#</mathjax></p>
<p>Relative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a>, <mathjax>#A_r#</mathjax> of <mathjax># C= 12#</mathjax><br/>
Relative atomic mass, <mathjax>#A_r#</mathjax> of <mathjax># O=16#</mathjax>, so....</p>
<p><mathjax>#M_r#</mathjax> of <mathjax># CO_2 =44#</mathjax></p>
<p>The number of moles of a substance is given by :</p>
<p><mathjax>#"mass in grams" /(M_r)#</mathjax></p>
<p>So in this case:</p>
<p><mathjax>#132/44=3*"moles"#</mathjax></p>
<p>One mole contains <mathjax>#6.022times 10^23#</mathjax> molecules (Avogadro's number).</p>
<p>So 3 moles contains <mathjax>#18.066 times 10^23#</mathjax> molecules.</p>
<p>Each molecule contains 2 oxygen atoms, so there will be <mathjax>#36.132 times 10^23 #</mathjax>oxygen atoms</p></div>
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</article> | How many oxygen atoms in a mass of #132*g# with respect to carbon dioxide? | null |
3,044 | aade8438-6ddd-11ea-bf9d-ccda262736ce | https://socratic.org/questions/the-deltah-value-for-the-reaction-1-2o-2-g-hg-l-hgo-s-is-90-8-ku-how-many-kj-are | 30.3 kJ | start physical_unit 4 5 heat_energy kj qc_end chemical_equation 6 11 qc_end physical_unit 4 5 13 14 deltah qc_end physical_unit 9 9 21 22 mass qc_end end | [{"type":"physical unit","value":"Released energy [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"30.3 kJ"}] | [{"type":"chemical equation","value":"1/2 O2(g) + Hg(l) -> HgO(s)"},{"type":"physical unit","value":"DeltaH value [OF] the reaction [=] \\pu{-90.8 kJ}"},{"type":"physical unit","value":"Mass [OF] Hg [=] \\pu{66.9 g}"}] | <h1 class="questionTitle" itemprop="name">The #DeltaH# value for the reaction #1/2O_2 (g) + Hg(l) -> HgO (s)# is -90.8 kJ. How many kJ are released when 66.9 g #Hg# is reacted with oxygen?</h1> | null | 30.3 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#1/2"O"_ (2(g)) + "Hg"_ ((l)) -> "HgO"_ ((s))" "DeltaH = -"90.8 kJ"#</mathjax></p>
</blockquote>
<p>This equation tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts with <mathjax>#1/2#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#"90.8 kJ"#</mathjax> of heat are being <strong>given off</strong> by the reaction. </p>
<p>Keep in mind that the <em>minus sign</em> used in the expression of the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction, <mathjax>#DeltaH#</mathjax>, symbolizes <strong>heat given off</strong>. </p>
<p>So, you now want to know how much heat will be released when <mathjax>#"66.9 g"#</mathjax> of mercury take part in the reaction. Since you know how much energy is released when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts, use the element's <strong>molar mass</strong> to convert the mass to <em>moles</em></p>
<blockquote>
<p><mathjax>#66.9 color(red)(cancel(color(black)("g"))) * "1 mole Hg"/(200.6color(red)(cancel(color(black)("g")))) = "0.3335 moles Hg"#</mathjax></p>
</blockquote>
<p>You can now use the known <mathjax>#DeltaH#</mathjax> to determine how much heat is released when <mathjax>#0.3335#</mathjax> <strong>moles</strong> of mercury react</p>
<blockquote>
<p><mathjax>#0.3335 color(red)(cancel(color(black)("moles Hg"))) * "90.8 kJ"/(1color(red)(cancel(color(black)("mole Hg")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("30.3 kJ")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>This is equivalent to saying that when <mathjax>#"66.9 g"#</mathjax> of mercury undergo combustion, the enthalpy change of reaction is </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = -"30.3 kJ"#</mathjax></p>
</blockquote>
<p>Once again, the minus sign symbolizes <strong>heat given off</strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"30.3 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#1/2"O"_ (2(g)) + "Hg"_ ((l)) -> "HgO"_ ((s))" "DeltaH = -"90.8 kJ"#</mathjax></p>
</blockquote>
<p>This equation tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts with <mathjax>#1/2#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#"90.8 kJ"#</mathjax> of heat are being <strong>given off</strong> by the reaction. </p>
<p>Keep in mind that the <em>minus sign</em> used in the expression of the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction, <mathjax>#DeltaH#</mathjax>, symbolizes <strong>heat given off</strong>. </p>
<p>So, you now want to know how much heat will be released when <mathjax>#"66.9 g"#</mathjax> of mercury take part in the reaction. Since you know how much energy is released when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts, use the element's <strong>molar mass</strong> to convert the mass to <em>moles</em></p>
<blockquote>
<p><mathjax>#66.9 color(red)(cancel(color(black)("g"))) * "1 mole Hg"/(200.6color(red)(cancel(color(black)("g")))) = "0.3335 moles Hg"#</mathjax></p>
</blockquote>
<p>You can now use the known <mathjax>#DeltaH#</mathjax> to determine how much heat is released when <mathjax>#0.3335#</mathjax> <strong>moles</strong> of mercury react</p>
<blockquote>
<p><mathjax>#0.3335 color(red)(cancel(color(black)("moles Hg"))) * "90.8 kJ"/(1color(red)(cancel(color(black)("mole Hg")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("30.3 kJ")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>This is equivalent to saying that when <mathjax>#"66.9 g"#</mathjax> of mercury undergo combustion, the enthalpy change of reaction is </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = -"30.3 kJ"#</mathjax></p>
</blockquote>
<p>Once again, the minus sign symbolizes <strong>heat given off</strong>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The #DeltaH# value for the reaction #1/2O_2 (g) + Hg(l) -> HgO (s)# is -90.8 kJ. How many kJ are released when 66.9 g #Hg# is reacted with oxygen?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"30.3 kJ"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the <strong>thermochemical equation</strong> given to you</p>
<blockquote>
<p><mathjax>#1/2"O"_ (2(g)) + "Hg"_ ((l)) -> "HgO"_ ((s))" "DeltaH = -"90.8 kJ"#</mathjax></p>
</blockquote>
<p>This equation tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts with <mathjax>#1/2#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#"90.8 kJ"#</mathjax> of heat are being <strong>given off</strong> by the reaction. </p>
<p>Keep in mind that the <em>minus sign</em> used in the expression of the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction, <mathjax>#DeltaH#</mathjax>, symbolizes <strong>heat given off</strong>. </p>
<p>So, you now want to know how much heat will be released when <mathjax>#"66.9 g"#</mathjax> of mercury take part in the reaction. Since you know how much energy is released when <mathjax>#1#</mathjax> <strong>mole</strong> of mercury reacts, use the element's <strong>molar mass</strong> to convert the mass to <em>moles</em></p>
<blockquote>
<p><mathjax>#66.9 color(red)(cancel(color(black)("g"))) * "1 mole Hg"/(200.6color(red)(cancel(color(black)("g")))) = "0.3335 moles Hg"#</mathjax></p>
</blockquote>
<p>You can now use the known <mathjax>#DeltaH#</mathjax> to determine how much heat is released when <mathjax>#0.3335#</mathjax> <strong>moles</strong> of mercury react</p>
<blockquote>
<p><mathjax>#0.3335 color(red)(cancel(color(black)("moles Hg"))) * "90.8 kJ"/(1color(red)(cancel(color(black)("mole Hg")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("30.3 kJ")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>This is equivalent to saying that when <mathjax>#"66.9 g"#</mathjax> of mercury undergo combustion, the enthalpy change of reaction is </p>
<blockquote>
<p><mathjax>#DeltaH_"rxn" = -"30.3 kJ"#</mathjax></p>
</blockquote>
<p>Once again, the minus sign symbolizes <strong>heat given off</strong>. </p></div>
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</article> | The #DeltaH# value for the reaction #1/2O_2 (g) + Hg(l) -> HgO (s)# is -90.8 kJ. How many kJ are released when 66.9 g #Hg# is reacted with oxygen? | null |
3,045 | a960cfca-6ddd-11ea-9b9b-ccda262736ce | https://socratic.org/questions/what-volume-of-0-250-m-hno3-aq-is-required-to-completely-react-with-10-0-g-of-zi | 1224.00 mL | start physical_unit 5 5 volume ml qc_end chemical_equation 20 27 qc_end physical_unit 15 16 12 13 mass qc_end physical_unit 5 5 3 4 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] HNO3(aq) [IN] mL"}] | [{"type":"physical unit","value":"1224.00 mL"}] | [{"type":"chemical equation","value":"2 HNO3(aq) + Zn(s) -> Zn(NO3)2(aq) + H2(g)"},{"type":"physical unit","value":"Mass [OF] zinc metal [=] \\pu{10.0 g}"},{"type":"physical unit","value":"Molarity [OF] HNO3(aq) [=] \\pu{0.250 M}"}] | <h1 class="questionTitle" itemprop="name">What volume of 0.250 M HNO3(aq) is required to completely react
with 10.0 g of zinc metal according to the equation:
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What volume of 0.250 M HNO3(aq) is required to completely react<br/>
with 10.0 g of zinc metal according to the equation:<br/>
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?</p></div>
</h2>
</div>
</div> | 1224.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation shows a <mathjax>#2:1#</mathjax> equivalence between the acid and zinc respectively.</p>
<p><mathjax>#"Moles of Zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.0*g)/(65.38*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.153*mol#</mathjax>.</p>
<p>Thus we need a <mathjax>#0.306#</mathjax> molar quantity of acid.</p>
<p>We have <mathjax>#0.250*mol*L^-1#</mathjax> nitric acid available, thus,</p>
<p><mathjax>#(0.306*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Zn(s) + 2HNO_3(aq) rarr Zn(NO_3)_2(aq) + H_2(g)uarr#</mathjax></p>
<p>We need over <mathjax>#1*L#</mathjax> of nitric acid. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation shows a <mathjax>#2:1#</mathjax> equivalence between the acid and zinc respectively.</p>
<p><mathjax>#"Moles of Zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.0*g)/(65.38*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.153*mol#</mathjax>.</p>
<p>Thus we need a <mathjax>#0.306#</mathjax> molar quantity of acid.</p>
<p>We have <mathjax>#0.250*mol*L^-1#</mathjax> nitric acid available, thus,</p>
<p><mathjax>#(0.306*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of 0.250 M HNO3(aq) is required to completely react
with 10.0 g of zinc metal according to the equation:
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What volume of 0.250 M HNO3(aq) is required to completely react<br/>
with 10.0 g of zinc metal according to the equation:<br/>
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?</p></div>
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<div class="markdown"><p><mathjax>#Zn(s) + 2HNO_3(aq) rarr Zn(NO_3)_2(aq) + H_2(g)uarr#</mathjax></p>
<p>We need over <mathjax>#1*L#</mathjax> of nitric acid. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation shows a <mathjax>#2:1#</mathjax> equivalence between the acid and zinc respectively.</p>
<p><mathjax>#"Moles of Zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.0*g)/(65.38*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.153*mol#</mathjax>.</p>
<p>Thus we need a <mathjax>#0.306#</mathjax> molar quantity of acid.</p>
<p>We have <mathjax>#0.250*mol*L^-1#</mathjax> nitric acid available, thus,</p>
<p><mathjax>#(0.306*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p></div>
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</article> | What volume of 0.250 M HNO3(aq) is required to completely react
with 10.0 g of zinc metal according to the equation:
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ? |
What volume of 0.250 M HNO3(aq) is required to completely react
with 10.0 g of zinc metal according to the equation:
2 HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g) ?
|
3,046 | a96c719c-6ddd-11ea-951d-ccda262736ce | https://socratic.org/questions/the-pressure-of-a-sample-of-dry-air-is-held-constant-at-2-25-atm-while-the-tempe | 32 L | start physical_unit 29 30 volume l qc_end physical_unit 29 30 32 33 volume qc_end physical_unit 4 7 12 13 pressure qc_end c_other OTHER qc_end physical_unit 4 7 20 21 temperature qc_end physical_unit 4 7 23 24 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the sample [IN] L"}] | [{"type":"physical unit","value":"32 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the sample [=] \\pu{43 L}"},{"type":"physical unit","value":"Pressure1 [OF] dry air sample [=] \\pu{2.25 atm}"},{"type":"other","value":"Constant pressure."},{"type":"physical unit","value":"Temperature1 [OF] dry air sample [=] \\pu{100 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] dry air sample [=] \\pu{7 ℃}"}] | <h1 class="questionTitle" itemprop="name">The pressure of a sample of dry air is held constant at 2.25 atm while the temperature is decreased rom 100 °C to 7 °C. The original volume of the sample is 43 L. What is the final volume of the sample?</h1> | null | 32 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use old <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>.......<mathjax>#VpropT#</mathjax> at constant pressure. </p>
<p><mathjax>#(V_1)/T_1=(V_2)/T_2#</mathjax></p>
<p>And solve for <mathjax>#V_2=(V_1)/(T_1)xxT_2#</mathjax></p>
<p><mathjax>#(43*L)/(373*K)xx280*K=??L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#V_"final"~=32*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use old <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>.......<mathjax>#VpropT#</mathjax> at constant pressure. </p>
<p><mathjax>#(V_1)/T_1=(V_2)/T_2#</mathjax></p>
<p>And solve for <mathjax>#V_2=(V_1)/(T_1)xxT_2#</mathjax></p>
<p><mathjax>#(43*L)/(373*K)xx280*K=??L#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The pressure of a sample of dry air is held constant at 2.25 atm while the temperature is decreased rom 100 °C to 7 °C. The original volume of the sample is 43 L. What is the final volume of the sample?</h1>
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<span class="dateCreated" datetime="2017-09-14T17:45:02" itemprop="dateCreated">
Sep 14, 2017
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<div class="markdown"><p><mathjax>#V_"final"~=32*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We use old <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>.......<mathjax>#VpropT#</mathjax> at constant pressure. </p>
<p><mathjax>#(V_1)/T_1=(V_2)/T_2#</mathjax></p>
<p>And solve for <mathjax>#V_2=(V_1)/(T_1)xxT_2#</mathjax></p>
<p><mathjax>#(43*L)/(373*K)xx280*K=??L#</mathjax></p></div>
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</article> | The pressure of a sample of dry air is held constant at 2.25 atm while the temperature is decreased rom 100 °C to 7 °C. The original volume of the sample is 43 L. What is the final volume of the sample? | null |
3,047 | aa2d6e9e-6ddd-11ea-a428-ccda262736ce | https://socratic.org/questions/at-constant-temperature-the-pressure-on-8-0-l-of-a-gas-increases-from-1-0-atm-to | 2.00 L | start physical_unit 25 26 volume l qc_end c_other constant_temperature qc_end physical_unit 25 26 6 7 volume qc_end physical_unit 25 26 13 14 pressure qc_end physical_unit 25 26 16 17 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"2.00 L"}] | [{"type":"other","value":"ConstantTemperature"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{8.0 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.0 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{4.0 atm}"}] | <h1 class="questionTitle" itemprop="name">At constant temperature, the pressure on 8.0 L of a gas increases from 1.0 atm to 4.0 atm. What will be the new volume of the gas?</h1> | null | 2.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And if <mathjax>#Pprop1/V#</mathjax> (<mathjax>#"temperature"#</mathjax>, and <mathjax>#"moles of gas"#</mathjax> CONSTANT), then <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant.....and so for the same amount of gas UNDER different conditions of <mathjax>#"pressure"#</mathjax> and <mathjax>#"volume"#</mathjax> but constant <mathjax>#"temperature"#</mathjax>, we may write............</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>.........when we solve for <mathjax>#k#</mathjax>.........</p>
<p>And we fill in the blanks, <mathjax>#V_1=8.0*L#</mathjax>, <mathjax>#P_1=1*atm#</mathjax>, and <mathjax>#P_2=4*atm#</mathjax>...........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1*cancel(atm)xx8.0*L)/(4*cancel(atm))=2*L#</mathjax>.</p>
<p>Is it reasonable that volume should be so reduced under higher pressure?</p></div>
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<div class="markdown"><p>Well, according to <a href="https://en.wikipedia.org/wiki/Boyle%27s_law" rel="nofollow">old Boyle's law</a> <mathjax>#Pprop1/V#</mathjax>, and get <mathjax>#V_2=2*L#</mathjax>.........</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And if <mathjax>#Pprop1/V#</mathjax> (<mathjax>#"temperature"#</mathjax>, and <mathjax>#"moles of gas"#</mathjax> CONSTANT), then <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant.....and so for the same amount of gas UNDER different conditions of <mathjax>#"pressure"#</mathjax> and <mathjax>#"volume"#</mathjax> but constant <mathjax>#"temperature"#</mathjax>, we may write............</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>.........when we solve for <mathjax>#k#</mathjax>.........</p>
<p>And we fill in the blanks, <mathjax>#V_1=8.0*L#</mathjax>, <mathjax>#P_1=1*atm#</mathjax>, and <mathjax>#P_2=4*atm#</mathjax>...........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1*cancel(atm)xx8.0*L)/(4*cancel(atm))=2*L#</mathjax>.</p>
<p>Is it reasonable that volume should be so reduced under higher pressure?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At constant temperature, the pressure on 8.0 L of a gas increases from 1.0 atm to 4.0 atm. What will be the new volume of the gas?</h1>
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<div class="markdown"><p>Well, according to <a href="https://en.wikipedia.org/wiki/Boyle%27s_law" rel="nofollow">old Boyle's law</a> <mathjax>#Pprop1/V#</mathjax>, and get <mathjax>#V_2=2*L#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And if <mathjax>#Pprop1/V#</mathjax> (<mathjax>#"temperature"#</mathjax>, and <mathjax>#"moles of gas"#</mathjax> CONSTANT), then <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant.....and so for the same amount of gas UNDER different conditions of <mathjax>#"pressure"#</mathjax> and <mathjax>#"volume"#</mathjax> but constant <mathjax>#"temperature"#</mathjax>, we may write............</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>.........when we solve for <mathjax>#k#</mathjax>.........</p>
<p>And we fill in the blanks, <mathjax>#V_1=8.0*L#</mathjax>, <mathjax>#P_1=1*atm#</mathjax>, and <mathjax>#P_2=4*atm#</mathjax>...........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1*cancel(atm)xx8.0*L)/(4*cancel(atm))=2*L#</mathjax>.</p>
<p>Is it reasonable that volume should be so reduced under higher pressure?</p></div>
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</article> | At constant temperature, the pressure on 8.0 L of a gas increases from 1.0 atm to 4.0 atm. What will be the new volume of the gas? | null |
3,048 | a8f641cc-6ddd-11ea-9204-ccda262736ce | https://socratic.org/questions/what-is-the-mass-in-grams-of-000221-mol-of-carbon | 2.65 × 10^(-3) grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon [IN] grams"}] | [{"type":"physical unit","value":"2.65 × 10^(-3) grams"}] | [{"type":"physical unit","value":"Mole [OF] carbon [=] \\pu{0.000221 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass in grams of .000221 mol of carbon?</h1> | null | 2.65 × 10^(-3) grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall;</p>
<p><mathjax>#"No of moles" = "Mass"/"Molar Mass"#</mathjax></p>
<p>Let;</p>
<p><mathjax>#n = "No of moles"#</mathjax></p>
<p><mathjax>#m = "Mass"#</mathjax></p>
<p><mathjax>#Mm = "Molar Mass"#</mathjax></p>
<p>Given;</p>
<p><mathjax>#n = 0.000221mols#</mathjax></p>
<p><mathjax>#m = ?g#</mathjax></p>
<p><mathjax>#Mm = 12gmol^-1#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p>Making <mathjax>#m#</mathjax> the subject formula;</p>
<p><mathjax>#m = n xx Mm#</mathjax></p>
<p>Plugging in the values;</p>
<p><mathjax>#m = 0.000221mols xx 12gmol^-1#</mathjax></p>
<p><mathjax>#m = 0.000221cancel(mols) xx 12gcancel(mol^-1)#</mathjax></p>
<p>#m = 0.000221 xx 12g</p>
<p><mathjax>#m = 0.002652g#</mathjax></p>
<p><mathjax>#m = 2.652 xx 10^-3g#</mathjax></p>
<p><mathjax>#:. "Mass" = 2.652 xx 10^-3g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Mass" = 2.652 xx 10^-3g#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall;</p>
<p><mathjax>#"No of moles" = "Mass"/"Molar Mass"#</mathjax></p>
<p>Let;</p>
<p><mathjax>#n = "No of moles"#</mathjax></p>
<p><mathjax>#m = "Mass"#</mathjax></p>
<p><mathjax>#Mm = "Molar Mass"#</mathjax></p>
<p>Given;</p>
<p><mathjax>#n = 0.000221mols#</mathjax></p>
<p><mathjax>#m = ?g#</mathjax></p>
<p><mathjax>#Mm = 12gmol^-1#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p>Making <mathjax>#m#</mathjax> the subject formula;</p>
<p><mathjax>#m = n xx Mm#</mathjax></p>
<p>Plugging in the values;</p>
<p><mathjax>#m = 0.000221mols xx 12gmol^-1#</mathjax></p>
<p><mathjax>#m = 0.000221cancel(mols) xx 12gcancel(mol^-1)#</mathjax></p>
<p>#m = 0.000221 xx 12g</p>
<p><mathjax>#m = 0.002652g#</mathjax></p>
<p><mathjax>#m = 2.652 xx 10^-3g#</mathjax></p>
<p><mathjax>#:. "Mass" = 2.652 xx 10^-3g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass in grams of .000221 mol of carbon?</h1>
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<div class="markdown"><p><mathjax>#"Mass" = 2.652 xx 10^-3g#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall;</p>
<p><mathjax>#"No of moles" = "Mass"/"Molar Mass"#</mathjax></p>
<p>Let;</p>
<p><mathjax>#n = "No of moles"#</mathjax></p>
<p><mathjax>#m = "Mass"#</mathjax></p>
<p><mathjax>#Mm = "Molar Mass"#</mathjax></p>
<p>Given;</p>
<p><mathjax>#n = 0.000221mols#</mathjax></p>
<p><mathjax>#m = ?g#</mathjax></p>
<p><mathjax>#Mm = 12gmol^-1#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p>Making <mathjax>#m#</mathjax> the subject formula;</p>
<p><mathjax>#m = n xx Mm#</mathjax></p>
<p>Plugging in the values;</p>
<p><mathjax>#m = 0.000221mols xx 12gmol^-1#</mathjax></p>
<p><mathjax>#m = 0.000221cancel(mols) xx 12gcancel(mol^-1)#</mathjax></p>
<p>#m = 0.000221 xx 12g</p>
<p><mathjax>#m = 0.002652g#</mathjax></p>
<p><mathjax>#m = 2.652 xx 10^-3g#</mathjax></p>
<p><mathjax>#:. "Mass" = 2.652 xx 10^-3g#</mathjax></p></div>
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</article> | What is the mass in grams of .000221 mol of carbon? | null |
3,049 | ad1a1aba-6ddd-11ea-bd6b-ccda262736ce | https://socratic.org/questions/how-many-nitrogen-atoms-are-in-one-molecule-of-the-compound-al-nh-3-2 | 2 | start physical_unit 2 3 number none qc_end end | [{"type":"physical unit","value":"Number [OF] nitrogen atoms"}] | [{"type":"physical unit","value":"2"}] | [{"type":"physical unit","value":"Number [OF] Al(NH3)2 molecule [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How many nitrogen atoms are in one molecule of the compound #Al(NH_3)_2#?</h1> | null | 2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We see that there is one single nitrogen atom in the brackets, but the subscript of 2 affects those in the brackets, so <mathjax>#1xx2=2#</mathjax>.</p>
<p>2 atoms of Nitrogen. </p>
<p>Hope this helps :)</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>2 atoms of Nitrogen. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We see that there is one single nitrogen atom in the brackets, but the subscript of 2 affects those in the brackets, so <mathjax>#1xx2=2#</mathjax>.</p>
<p>2 atoms of Nitrogen. </p>
<p>Hope this helps :)</p></div>
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<div class="markdown"><p>2 atoms of Nitrogen. </p></div>
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<div class="markdown"><p>We see that there is one single nitrogen atom in the brackets, but the subscript of 2 affects those in the brackets, so <mathjax>#1xx2=2#</mathjax>.</p>
<p>2 atoms of Nitrogen. </p>
<p>Hope this helps :)</p></div>
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</article> | How many nitrogen atoms are in one molecule of the compound #Al(NH_3)_2#? | null |
3,050 | aaa2601f-6ddd-11ea-8764-ccda262736ce | https://socratic.org/questions/according-to-the-balanced-equation-below-how-many-moles-of-clo-2-aq-are-needed-t | 3.00 × 10^(-3) moles | start physical_unit 10 10 mole mol qc_end chemical_equation 24 40 qc_end c_other OTHER qc_end physical_unit 22 23 17 18 volume qc_end physical_unit 22 23 20 21 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] ClO2-(aq) [IN] moles"}] | [{"type":"physical unit","value":"3.00 × 10^(-3) moles"}] | [{"type":"chemical equation","value":"2 H2O(l) + 4 MnO4(aq) + 3 ClO2-(aq) -> 4 MnO2(s) + 3 ClO4-(aq) + 4 OH-(aq)"},{"type":"other","value":"React completely."},{"type":"physical unit","value":"Volume [OF] KMnO4 solution [=] \\pu{20 mL}"},{"type":"physical unit","value":"Molarity [OF] KMnO4 solution [=] \\pu{0.20 M}"}] | <h1 class="questionTitle" itemprop="name">According to the balanced equation below, how many moles of #ClO^-2 (aq)# are needed to react completely with 20. mL of 0.20 M #KMnO_4# solution?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#2H_2O(l) + 4MnO_4(aq) + 3ClO_2^(-)(aq) -> 4MnO_2(s) + 3ClO_4^(-)(aq) + 4OH^(-)(aq)#</mathjax></p></div>
</h2>
</div>
</div> | 3.00 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Reduction"#</mathjax><br/>
<mathjax>#MnO_4^(-) +4H^(+) + 3e^(-) rarr MnO_2(s) + 2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Oxidation"#</mathjax><br/>
<mathjax>#ClO_2^(-) +2H_2O rarr ClO_4^(-) + 4H^+ +4e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p><mathjax>#"Overall"#</mathjax> <mathjax>#4xx(i) +3xx(ii)#</mathjax><br/>
<mathjax>#4MnO_4^(-) +3ClO_2^(-) +4H^+rarr 4MnO_2(s) +3ClO_4^(-)+ 2H_2O(l)#</mathjax> </p>
<p><mathjax>#"Moles of "KMnO_4=20xx10^-3Lxx0.20*mol*L^-1=4.0xx10^-3*mol#</mathjax></p>
<p>Given the equation, we need 3/4 equiv of chlorite anion,</p>
<p><mathjax>#3/4xx4.0xx10^-3*mol#</mathjax></p>
<p>What would you observe in this reaction? What was the macroscopic colour change?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#3*mmol#</mathjax> <mathjax>#"chlorite anion"#</mathjax> are required. </p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Reduction"#</mathjax><br/>
<mathjax>#MnO_4^(-) +4H^(+) + 3e^(-) rarr MnO_2(s) + 2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Oxidation"#</mathjax><br/>
<mathjax>#ClO_2^(-) +2H_2O rarr ClO_4^(-) + 4H^+ +4e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p><mathjax>#"Overall"#</mathjax> <mathjax>#4xx(i) +3xx(ii)#</mathjax><br/>
<mathjax>#4MnO_4^(-) +3ClO_2^(-) +4H^+rarr 4MnO_2(s) +3ClO_4^(-)+ 2H_2O(l)#</mathjax> </p>
<p><mathjax>#"Moles of "KMnO_4=20xx10^-3Lxx0.20*mol*L^-1=4.0xx10^-3*mol#</mathjax></p>
<p>Given the equation, we need 3/4 equiv of chlorite anion,</p>
<p><mathjax>#3/4xx4.0xx10^-3*mol#</mathjax></p>
<p>What would you observe in this reaction? What was the macroscopic colour change?</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">According to the balanced equation below, how many moles of #ClO^-2 (aq)# are needed to react completely with 20. mL of 0.20 M #KMnO_4# solution?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#2H_2O(l) + 4MnO_4(aq) + 3ClO_2^(-)(aq) -> 4MnO_2(s) + 3ClO_4^(-)(aq) + 4OH^(-)(aq)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3*mmol#</mathjax> <mathjax>#"chlorite anion"#</mathjax> are required. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Reduction"#</mathjax><br/>
<mathjax>#MnO_4^(-) +4H^(+) + 3e^(-) rarr MnO_2(s) + 2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Oxidation"#</mathjax><br/>
<mathjax>#ClO_2^(-) +2H_2O rarr ClO_4^(-) + 4H^+ +4e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p><mathjax>#"Overall"#</mathjax> <mathjax>#4xx(i) +3xx(ii)#</mathjax><br/>
<mathjax>#4MnO_4^(-) +3ClO_2^(-) +4H^+rarr 4MnO_2(s) +3ClO_4^(-)+ 2H_2O(l)#</mathjax> </p>
<p><mathjax>#"Moles of "KMnO_4=20xx10^-3Lxx0.20*mol*L^-1=4.0xx10^-3*mol#</mathjax></p>
<p>Given the equation, we need 3/4 equiv of chlorite anion,</p>
<p><mathjax>#3/4xx4.0xx10^-3*mol#</mathjax></p>
<p>What would you observe in this reaction? What was the macroscopic colour change?</p></div>
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</article> | According to the balanced equation below, how many moles of #ClO^-2 (aq)# are needed to react completely with 20. mL of 0.20 M #KMnO_4# solution? |
#2H_2O(l) + 4MnO_4(aq) + 3ClO_2^(-)(aq) -> 4MnO_2(s) + 3ClO_4^(-)(aq) + 4OH^(-)(aq)#
|
3,051 | a83de96a-6ddd-11ea-b645-ccda262736ce | https://socratic.org/questions/aluminum-sulfate-al-2-so-4-3-decomposes-to-form-aluminum-oxide-al-2o-3-and-sulfu | Al2(SO4)3 -> Al2O3 + 3 SO3 | start chemical_equation qc_end chemical_equation 2 2 qc_end chemical_equation 8 8 qc_end chemical_equation 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"Al2(SO4)3 -> Al2O3 + 3 SO3"}] | [{"type":"chemical equation","value":"Al2(SO4)3"},{"type":"chemical equation","value":"Al2O3"},{"type":"chemical equation","value":"SO3"}] | <h1 class="questionTitle" itemprop="name">Aluminum sulfate, #Al_2(SO_4)_3#, decomposes to form aluminum oxide, #Al_2O_3#, and sulfur trioxide, #SO_3#. How do you write the balanced equation for this reaction?</h1> | null | Al2(SO4)3 -> Al2O3 + 3 SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory.</p>
<p>Reactant Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=3xx1xx1=3#</mathjax><br/>
<mathjax>#O=3xx4xx1=12#</mathjax></p>
<p>Product Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=1xx3=3#</mathjax><br/>
<mathjax>#O=3xx1 + 3xx3 = 12#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Al_2(SO_4)_3#</mathjax> <mathjax>#->#</mathjax> <mathjax>#Al_2O_3#</mathjax> + <mathjax>#3SO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory.</p>
<p>Reactant Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=3xx1xx1=3#</mathjax><br/>
<mathjax>#O=3xx4xx1=12#</mathjax></p>
<p>Product Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=1xx3=3#</mathjax><br/>
<mathjax>#O=3xx1 + 3xx3 = 12#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Aluminum sulfate, #Al_2(SO_4)_3#, decomposes to form aluminum oxide, #Al_2O_3#, and sulfur trioxide, #SO_3#. How do you write the balanced equation for this reaction?</h1>
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<a class="topContributorPic" href="/users/jarni-renz"><img alt="" class="" src="https://profilepictures.socratic.org/i6OiZQmVRHWljDVlwm8v_WIN_20160301_180231.JPG" title=""/></a>
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Jarni Renz
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<span class="dateCreated" datetime="2016-01-08T14:11:23" itemprop="dateCreated">
Jan 8, 2016
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<div class="markdown"><p><mathjax>#Al_2(SO_4)_3#</mathjax> <mathjax>#->#</mathjax> <mathjax>#Al_2O_3#</mathjax> + <mathjax>#3SO_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory.</p>
<p>Reactant Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=3xx1xx1=3#</mathjax><br/>
<mathjax>#O=3xx4xx1=12#</mathjax></p>
<p>Product Side:<br/>
<mathjax>#Al=2xx1=2#</mathjax><br/>
<mathjax>#S=1xx3=3#</mathjax><br/>
<mathjax>#O=3xx1 + 3xx3 = 12#</mathjax></p></div>
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</article> | Aluminum sulfate, #Al_2(SO_4)_3#, decomposes to form aluminum oxide, #Al_2O_3#, and sulfur trioxide, #SO_3#. How do you write the balanced equation for this reaction? | null |
3,052 | ac2b6e2e-6ddd-11ea-9aaf-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-c-6h-12o-6 | CH2O | start chemical_formula qc_end chemical_equation 8 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O"}] | [{"type":"chemical equation","value":"C6H12O6"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound #C_6H_12O_6#?</h1> | null | CH2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula is a chemical formula written in the simplest form by reducing the subscripts to the smallest possible whole integer. In essence, we just have to find a the smallest common term of each subscript and simplify:</p>
<p>If we look at <mathjax>#C_6H_12O_6#</mathjax> we can see that all subscripts are divisible by <mathjax>#6#</mathjax> so we divide each subscript by <mathjax>#6#</mathjax>. The end result is the empirical formula:</p>
<p><mathjax>#CH_2O#</mathjax> - Empirical formula</p>
<p><mathjax>#C_6H_12O_6#</mathjax> - Molecular formula</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_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula is a chemical formula written in the simplest form by reducing the subscripts to the smallest possible whole integer. In essence, we just have to find a the smallest common term of each subscript and simplify:</p>
<p>If we look at <mathjax>#C_6H_12O_6#</mathjax> we can see that all subscripts are divisible by <mathjax>#6#</mathjax> so we divide each subscript by <mathjax>#6#</mathjax>. The end result is the empirical formula:</p>
<p><mathjax>#CH_2O#</mathjax> - Empirical formula</p>
<p><mathjax>#C_6H_12O_6#</mathjax> - Molecular formula</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound #C_6H_12O_6#?</h1>
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Anthony R.
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May 19, 2017
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<div class="markdown"><p><mathjax>#CH_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula is a chemical formula written in the simplest form by reducing the subscripts to the smallest possible whole integer. In essence, we just have to find a the smallest common term of each subscript and simplify:</p>
<p>If we look at <mathjax>#C_6H_12O_6#</mathjax> we can see that all subscripts are divisible by <mathjax>#6#</mathjax> so we divide each subscript by <mathjax>#6#</mathjax>. The end result is the empirical formula:</p>
<p><mathjax>#CH_2O#</mathjax> - Empirical formula</p>
<p><mathjax>#C_6H_12O_6#</mathjax> - Molecular formula</p></div>
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</article> | What is the empirical formula for the compound #C_6H_12O_6#? | null |
3,053 | a8cde589-6ddd-11ea-85e3-ccda262736ce | https://socratic.org/questions/if-11-grams-of-carbon-dioxide-is-produced-how-much-methane-ch4-must-have-reacted | 0.25 moles | start physical_unit 11 11 mole mol qc_end physical_unit 4 5 1 2 mass qc_end chemical_equation 18 26 qc_end end | [{"type":"physical unit","value":"Mole [OF] CH4 [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [=] \\pu{11 grams}"},{"type":"chemical equation","value":"CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)"}] | <h1 class="questionTitle" itemprop="name">If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#</mathjax></p></div>
</h2>
</div>
</div> | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>From the balanced equation determine the ratio of moles of carbon dioxide produced to methane reacted, then from the molar mass of carbon dioxide get the moles of carbon dioxide and multiply by the methane to <mathjax>#"CO"_2#</mathjax> ratio.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#</mathjax></p></div>
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<div class="markdown"><p>From the balanced equation determine the ratio of moles of carbon dioxide produced to methane reacted, then from the molar mass of carbon dioxide get the moles of carbon dioxide and multiply by the methane to <mathjax>#"CO"_2#</mathjax> ratio.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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Moeletsi M.
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Stefan V.
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<span class="dateCreated" datetime="2018-04-07T15:16:47" itemprop="dateCreated">
Apr 7, 2018
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<div class="markdown"><p>The answer has to be <mathjax>#0.25#</mathjax> mol.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Methane undergoes a combustion reaction as follow:</p>
<p><mathjax>#"CH"_4 +2"O"_2 -> "CO"_2 + 2"H"_2"O"#</mathjax></p>
<p><mathjax>#n_("CO"_2)=m/M_r = "11 g"/((12.01+(2xx16)) \ "g/mol") = "0.25 mol"#</mathjax></p>
<p>According to the balanced equation, <mathjax>#1#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with <mathjax>#2#</mathjax> moles of <mathjax>#"O"_2#</mathjax> to produce <mathjax>#1#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. That means</p>
<p><mathjax>#"1 mol" ("CO"_2) = "1 mol" ("CH"_4)#</mathjax></p>
<p>Therefore </p>
<p><mathjax>#n_("CH"_4) = "0.25 mol"#</mathjax></p></div>
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</article> | If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles? |
#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#
|
3,054 | aa38f908-6ddd-11ea-876d-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-0-202-mol-kcl-in-7-98-l-solutio | 0.03 M | start physical_unit 11 11 molarity mol/l qc_end physical_unit 11 11 9 10 mole qc_end physical_unit 6 6 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] KCl solution [IN] M"}] | [{"type":"physical unit","value":"0.03 M"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{0.202 mol}"},{"type":"physical unit","value":"Volume [OF] KCl solution [=] \\pu{7.98 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 0.202 mol KCl in 7.98 L solution? </h1> | null | 0.03 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is a measure of a solution's <em>concentration</em> in terms of how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> it contains <strong>per liter</strong> of solution. </p>
<p><img alt="http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/cr/m3/s5/" src="https://useruploads.socratic.org/QW9WDL3bTXO0ADTT70qc_crm3s5_1.jpg"/> </p>
<p>So, in essence, if you have the number of moles of solute, like you do here, and the <em>total volume</em> of the solution <strong>expressed in liters</strong>, you can find how many moles of solute you get <em>per liter</em>, i.e. the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution. </p>
<p>Keeping in mind the fact that the moles of solute are <em>distributed evenly</em> in any given volume of the solution, you can say that because <mathjax>#"7.98 L"#</mathjax> of solution contain <mathjax>#0.202#</mathjax> moles of solute, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L"))) * "0.202 moles KCl"/(7.98color(red)(cancel(color(black)("L")))) = "0.0253 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#0.0253#</mathjax> moles of potassium chloride <strong>per liter</strong>, it follows that its molarity is equal to </p>
<blockquote>
<p><mathjax>#c = color(green)("0.0253 M")#</mathjax></p>
</blockquote>
<p>Here <mathjax>#"M"#</mathjax> is equivalent to <mathjax>#"mol/L"#</mathjax>, or <em>molar</em>. </p>
<p>
<iframe src="https://www.youtube.com/embed/SXf9rDnVFao?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.0253 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is a measure of a solution's <em>concentration</em> in terms of how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> it contains <strong>per liter</strong> of solution. </p>
<p><img alt="http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/cr/m3/s5/" src="https://useruploads.socratic.org/QW9WDL3bTXO0ADTT70qc_crm3s5_1.jpg"/> </p>
<p>So, in essence, if you have the number of moles of solute, like you do here, and the <em>total volume</em> of the solution <strong>expressed in liters</strong>, you can find how many moles of solute you get <em>per liter</em>, i.e. the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution. </p>
<p>Keeping in mind the fact that the moles of solute are <em>distributed evenly</em> in any given volume of the solution, you can say that because <mathjax>#"7.98 L"#</mathjax> of solution contain <mathjax>#0.202#</mathjax> moles of solute, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L"))) * "0.202 moles KCl"/(7.98color(red)(cancel(color(black)("L")))) = "0.0253 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#0.0253#</mathjax> moles of potassium chloride <strong>per liter</strong>, it follows that its molarity is equal to </p>
<blockquote>
<p><mathjax>#c = color(green)("0.0253 M")#</mathjax></p>
</blockquote>
<p>Here <mathjax>#"M"#</mathjax> is equivalent to <mathjax>#"mol/L"#</mathjax>, or <em>molar</em>. </p>
<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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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 0.202 mol KCl in 7.98 L solution? </h1>
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<div class="markdown"><p><mathjax>#"0.0253 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is a measure of a solution's <em>concentration</em> in terms of how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> it contains <strong>per liter</strong> of solution. </p>
<p><img alt="http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/cr/m3/s5/" src="https://useruploads.socratic.org/QW9WDL3bTXO0ADTT70qc_crm3s5_1.jpg"/> </p>
<p>So, in essence, if you have the number of moles of solute, like you do here, and the <em>total volume</em> of the solution <strong>expressed in liters</strong>, you can find how many moles of solute you get <em>per liter</em>, i.e. the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution. </p>
<p>Keeping in mind the fact that the moles of solute are <em>distributed evenly</em> in any given volume of the solution, you can say that because <mathjax>#"7.98 L"#</mathjax> of solution contain <mathjax>#0.202#</mathjax> moles of solute, which in your case is potassium chloride, <mathjax>#"KCl"#</mathjax>, <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L"))) * "0.202 moles KCl"/(7.98color(red)(cancel(color(black)("L")))) = "0.0253 moles KCl"#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#0.0253#</mathjax> moles of potassium chloride <strong>per liter</strong>, it follows that its molarity is equal to </p>
<blockquote>
<p><mathjax>#c = color(green)("0.0253 M")#</mathjax></p>
</blockquote>
<p>Here <mathjax>#"M"#</mathjax> is equivalent to <mathjax>#"mol/L"#</mathjax>, or <em>molar</em>. </p>
<p>
<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 that contains 0.202 mol KCl in 7.98 L solution? | null |
3,055 | aadef952-6ddd-11ea-aa48-ccda262736ce | https://socratic.org/questions/when-an-airbag-is-inflated-the-nitrogen-gas-has-a-pressure-of-1-30-atmospheres-a | 47.80 L | start physical_unit 6 7 volume l qc_end physical_unit 6 7 12 13 pressure qc_end physical_unit 6 7 17 18 temperature qc_end physical_unit 6 7 23 24 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen gas [IN] L"}] | [{"type":"physical unit","value":"47.80 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] nitrogen gas [=] \\pu{1.30 atmospheres}"},{"type":"physical unit","value":"Temperature1 [OF] nitrogen gas [=] \\pu{301 K}"},{"type":"physical unit","value":"Volume1 [OF] nitrogen gas [=] \\pu{40.0 liters}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">When an airbag is inflated the nitrogen gas has a pressure of 1.30 atmospheres, a temperature of 301 K, and a volume of 40.0 liters. What is the volume of the nitrogen at STP?</h1> | null | 47.80 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use two approaches to solve this problem. </p>
<p>You can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to find the <em>number of moles</em> of gas present in that sample, then use the known <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> at STP conditions to find the new volume. </p>
<p>As a way to double-check the result you get by using this approach, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> equation to find the new volume of the gas <em>without</em> finding out how many moles it contains. </p>
<p>So, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.30 color(red)(cancel(color(black)("atm"))) * 40.0 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 301color(red)(cancel(color(black)("K")))) = "2.104 moles N"_2#</mathjax></p>
</blockquote>
<p>Now, <strong>STP</strong> conditions are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L" ->#</mathjax> this is the aforementioned <strong>molar volume of a gas</strong> at STP.</p>
<p>In your case, <mathjax>#2.104#</mathjax> <strong>moles</strong> of nitrogen gas will occupy </p>
<blockquote>
<p><mathjax>#2.104 color(red)(cancel(color(black)("moles N"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Now, you can double-check the result by suing the <strong>combined gas law</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>At mentioned before, <strong>STP</strong> conditions are defined as </p>
<blockquote>
<p><mathjax>#P_2 = "100 kPa" = 101/101.325color(white)(a)"atm"#</mathjax></p>
<p><mathjax>#T_2 = 0^@"C" + 273.15 = "273.15 K"#</mathjax></p>
</blockquote>
<p>Rearrange the combined gas law equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (1.30 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/(301color(red)(cancel(color(black)("K")))) * "40.0 L"#</mathjax></p>
<p><mathjax>#V_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>SIDE NOTE</strong> <em>STP conditions are often in accordance to their <strong>old definition</strong> of a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If this is the STP definition given to you, simply redo the calculations using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"47.8 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use two approaches to solve this problem. </p>
<p>You can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to find the <em>number of moles</em> of gas present in that sample, then use the known <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> at STP conditions to find the new volume. </p>
<p>As a way to double-check the result you get by using this approach, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> equation to find the new volume of the gas <em>without</em> finding out how many moles it contains. </p>
<p>So, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.30 color(red)(cancel(color(black)("atm"))) * 40.0 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 301color(red)(cancel(color(black)("K")))) = "2.104 moles N"_2#</mathjax></p>
</blockquote>
<p>Now, <strong>STP</strong> conditions are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L" ->#</mathjax> this is the aforementioned <strong>molar volume of a gas</strong> at STP.</p>
<p>In your case, <mathjax>#2.104#</mathjax> <strong>moles</strong> of nitrogen gas will occupy </p>
<blockquote>
<p><mathjax>#2.104 color(red)(cancel(color(black)("moles N"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Now, you can double-check the result by suing the <strong>combined gas law</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>At mentioned before, <strong>STP</strong> conditions are defined as </p>
<blockquote>
<p><mathjax>#P_2 = "100 kPa" = 101/101.325color(white)(a)"atm"#</mathjax></p>
<p><mathjax>#T_2 = 0^@"C" + 273.15 = "273.15 K"#</mathjax></p>
</blockquote>
<p>Rearrange the combined gas law equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (1.30 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/(301color(red)(cancel(color(black)("K")))) * "40.0 L"#</mathjax></p>
<p><mathjax>#V_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>SIDE NOTE</strong> <em>STP conditions are often in accordance to their <strong>old definition</strong> of a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If this is the STP definition given to you, simply redo the calculations using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When an airbag is inflated the nitrogen gas has a pressure of 1.30 atmospheres, a temperature of 301 K, and a volume of 40.0 liters. What is the volume of the nitrogen at STP?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-27T02:08:49" itemprop="dateCreated">
Jun 27, 2016
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<div class="markdown"><p><mathjax>#"47.8 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use two approaches to solve this problem. </p>
<p>You can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to find the <em>number of moles</em> of gas present in that sample, then use the known <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> at STP conditions to find the new volume. </p>
<p>As a way to double-check the result you get by using this approach, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong> equation to find the new volume of the gas <em>without</em> finding out how many moles it contains. </p>
<p>So, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#n#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.30 color(red)(cancel(color(black)("atm"))) * 40.0 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 301color(red)(cancel(color(black)("K")))) = "2.104 moles N"_2#</mathjax></p>
</blockquote>
<p>Now, <strong>STP</strong> conditions are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these conditions for pressure and temperature, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L" ->#</mathjax> this is the aforementioned <strong>molar volume of a gas</strong> at STP.</p>
<p>In your case, <mathjax>#2.104#</mathjax> <strong>moles</strong> of nitrogen gas will occupy </p>
<blockquote>
<p><mathjax>#2.104 color(red)(cancel(color(black)("moles N"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Now, you can double-check the result by suing the <strong>combined gas law</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)(P_1V_1)/T_1 = (P_2V_2)/T_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>At mentioned before, <strong>STP</strong> conditions are defined as </p>
<blockquote>
<p><mathjax>#P_2 = "100 kPa" = 101/101.325color(white)(a)"atm"#</mathjax></p>
<p><mathjax>#T_2 = 0^@"C" + 273.15 = "273.15 K"#</mathjax></p>
</blockquote>
<p>Rearrange the combined gas law equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2 implies V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (1.30 color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/(301color(red)(cancel(color(black)("K")))) * "40.0 L"#</mathjax></p>
<p><mathjax>#V_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("47.8 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>SIDE NOTE</strong> <em>STP conditions are often in accordance to their <strong>old definition</strong> of a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, <strong>one mole</strong> of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If this is the STP definition given to you, simply redo the calculations using</em> <mathjax>#"22.4 L"#</mathjax> <em>instead of</em> <mathjax>#"22.7 L"#</mathjax>. </p></div>
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</article> | When an airbag is inflated the nitrogen gas has a pressure of 1.30 atmospheres, a temperature of 301 K, and a volume of 40.0 liters. What is the volume of the nitrogen at STP? | null |
3,056 | a9d6509d-6ddd-11ea-b398-ccda262736ce | https://socratic.org/questions/what-mass-in-g-does-4-35-moles-of-ti-titanium-have | 47.87 g | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] Ti [IN] g"}] | [{"type":"physical unit","value":"47.87 g"}] | [{"type":"physical unit","value":"Mole [OF] Ti [=] \\pu{4.35 moles}"}] | <h1 class="questionTitle" itemprop="name">What mass (in g) does 4.35 moles of Ti (titanium) have?</h1> | null | 47.87 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#47.87*g*cancel(mol^-1)xx4.35*cancel(mol)~=210*g#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, the one mole of titanium metal has a mass of <mathjax>#47.87*g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#47.87*g*cancel(mol^-1)xx4.35*cancel(mol)~=210*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What mass (in g) does 4.35 moles of Ti (titanium) have?</h1>
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<div class="markdown"><p>Well, the one mole of titanium metal has a mass of <mathjax>#47.87*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"Mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#47.87*g*cancel(mol^-1)xx4.35*cancel(mol)~=210*g#</mathjax>.</p></div>
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</article> | What mass (in g) does 4.35 moles of Ti (titanium) have? | null |
3,057 | ab1f3e58-6ddd-11ea-b7e5-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-5-3-times-10-23-atoms-of-gold | 0.88 moles | start physical_unit 11 11 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] gold [IN] moles"}] | [{"type":"physical unit","value":"0.88 moles"}] | [{"type":"physical unit","value":"Number [OF] gold atoms [=] \\pu{5.3 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in #5.3 times 10^23# atoms of gold? </h1> | null | 0.88 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>a mol is a quantity of substance that contains an Avogadro's number of molecules (<mathjax>#6.022 xx 10^23#</mathjax>)<br/>
But Gold as the others metals has no a definite molecule. Some authors consider than metals molecule monatomic so you would have<br/>
<mathjax>#n=(5.3 xx 10^23)/(6.022 xx 10^23= 0,88)#</mathjax><br/>
in this case you have the moles of atoms of Gold</p></div>
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<div class="markdown"><p>n= 0,88 </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>a mol is a quantity of substance that contains an Avogadro's number of molecules (<mathjax>#6.022 xx 10^23#</mathjax>)<br/>
But Gold as the others metals has no a definite molecule. Some authors consider than metals molecule monatomic so you would have<br/>
<mathjax>#n=(5.3 xx 10^23)/(6.022 xx 10^23= 0,88)#</mathjax><br/>
in this case you have the moles of atoms of Gold</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are there in #5.3 times 10^23# atoms of gold? </h1>
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<div class="markdown"><p>n= 0,88 </p></div>
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<div class="markdown"><p>a mol is a quantity of substance that contains an Avogadro's number of molecules (<mathjax>#6.022 xx 10^23#</mathjax>)<br/>
But Gold as the others metals has no a definite molecule. Some authors consider than metals molecule monatomic so you would have<br/>
<mathjax>#n=(5.3 xx 10^23)/(6.022 xx 10^23= 0,88)#</mathjax><br/>
in this case you have the moles of atoms of Gold</p></div>
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</article> | How many moles are there in #5.3 times 10^23# atoms of gold? | null |
3,058 | a8b88ad2-6ddd-11ea-bff5-ccda262736ce | https://socratic.org/questions/a-compound-containing-only-carbon-and-hydrogen-is-analyzed-and-is-found-to-conta | CH3 | start chemical_formula qc_end physical_unit 6 6 14 14 mass_percent qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH3"}] | [{"type":"physical unit","value":"Percent by mass [OF] hydrogen [=] \\pu{20.11%}"},{"type":"other","value":"A compound contains only carbon and hydrogen."}] | <h1 class="questionTitle" itemprop="name">A compound containing only carbon and hydrogen is analyzed, and is found to contain 20.11% hydrogen on a mass basis. How can the empirical formula of the compound be calculated? </h1> | null | CH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g hydrocarbon, there are <mathjax>#(20.11*g)/(1.00794*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax>, and <mathjax>#(79.89*g)/(12.011*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#C=6.65*mol#</mathjax> <mathjax>#C#</mathjax>.</p>
<p>We then divide thru by the lowest ratio to give the empirical formula, <mathjax>#CH_3#</mathjax>. </p>
<p>How did I know that <mathjax>#%C#</mathjax> was <mathjax>#79.89#</mathjax>?</p></div>
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<div class="markdown"><p>We assume 100 g of hydrocarbon. The empirical formula is the simplest whole number ratio that defines constituent atoms in a species. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g hydrocarbon, there are <mathjax>#(20.11*g)/(1.00794*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax>, and <mathjax>#(79.89*g)/(12.011*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#C=6.65*mol#</mathjax> <mathjax>#C#</mathjax>.</p>
<p>We then divide thru by the lowest ratio to give the empirical formula, <mathjax>#CH_3#</mathjax>. </p>
<p>How did I know that <mathjax>#%C#</mathjax> was <mathjax>#79.89#</mathjax>?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A compound containing only carbon and hydrogen is analyzed, and is found to contain 20.11% hydrogen on a mass basis. How can the empirical formula of the compound be calculated? </h1>
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<div class="markdown"><p>We assume 100 g of hydrocarbon. The empirical formula is the simplest whole number ratio that defines constituent atoms in a species. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g hydrocarbon, there are <mathjax>#(20.11*g)/(1.00794*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax>, and <mathjax>#(79.89*g)/(12.011*g*mol^(-1)) = 19.95#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#C=6.65*mol#</mathjax> <mathjax>#C#</mathjax>.</p>
<p>We then divide thru by the lowest ratio to give the empirical formula, <mathjax>#CH_3#</mathjax>. </p>
<p>How did I know that <mathjax>#%C#</mathjax> was <mathjax>#79.89#</mathjax>?</p></div>
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</article> | A compound containing only carbon and hydrogen is analyzed, and is found to contain 20.11% hydrogen on a mass basis. How can the empirical formula of the compound be calculated? | null |
3,059 | ab444840-6ddd-11ea-b8ac-ccda262736ce | https://socratic.org/questions/how-many-ml-of-carbon-tetrachloride-are-produced-when-8-0-l-of-chlorine-are-allo | 3.19 mL | start physical_unit 4 5 volume ml qc_end physical_unit 12 12 9 10 volume qc_end physical_unit 21 21 18 19 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon tetrachloride [IN] mL"}] | [{"type":"physical unit","value":"3.19 mL"}] | [{"type":"physical unit","value":"Volume [OF] chlorine [=] \\pu{8.0 L}"},{"type":"physical unit","value":"Volume [OF] methane [=] \\pu{0.75 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many mL of carbon tetrachloride are produced when 8.0 L of chlorine are allowed to react with 0.75 L of methane at STP? </h1> | null | 3.19 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>WARNING! Long answer!</p>
<p>This is a <strong>limiting reactant</strong> problem.</p>
<p>One way to solve it is to figure out how much product you could get from each reactant separately.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation.</strong></p>
<p>The balanced chemical equation is</p>
<p><mathjax>#"CH"_4 + "4Cl"_2 → "CCl"_4 + "4HCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Identify the limiting reactant</strong></p>
<p>We can use <strong>Gay-Lussac's Law of Combining Volumes</strong> to identify the limiting reactant.</p>
<blockquote></blockquote>
<p>From <mathjax>#"CH"_4#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("1 L CH"_4)#</mathjax></p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 0.75 color(red)(cancel(color(black)("L CH"_4))) × ("1 L CCl"4)/(1 color(red)(cancel(color(black)("L CH"_4)))) = "0.75 L CCl"_4#</mathjax></p>
<blockquote></blockquote>
<p>From <mathjax>#"Cl"_2#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("4 L Cl"_2)#</mathjax>.</p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 8.0 color(red)(cancel(color(black)("L Cl"_2))) × ("1 mol CCl"_4)/(4 color(red)(cancel(color(black)("L Cl"_2)))) = "2.0 L CCl"_4#</mathjax></p>
<p><mathjax>#"CH"_4#</mathjax> gives the smaller volume of gaseous <mathjax>#"CCl"_4#</mathjax>, so <mathjax>#"CH"_4#</mathjax> is the limiting reactant.</p>
<p>The theoretical yield of gaseous <mathjax>#"CCl"_4#</mathjax> at STP is 0.75 L.</p>
<blockquote></blockquote>
<p><strong>3. Calculate the mass of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p>To solve this problem, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#m = (PVM)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>STP is defined at 1 bar and 0 °C.</p>
<p><mathjax>#P = "1 bar"#</mathjax><br/>
<mathjax>#V = "0.75 L"#</mathjax><br/>
<mathjax>#M = "153.82 g/mol"#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax></p>
<p>If <mathjax>#"CCl"_4#</mathjax> were an ideal gas at STP, we would calculate that</p>
<p>∴ <mathjax>#m = (1 color(red)(cancel(color(black)("bar"))) × 0.75 color(red)(cancel(color(black)("L"))) × "153.82 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "5.08 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the volume of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p><mathjax>#"CCl"_4#</mathjax> is a <strong>liquid</strong> at STP. Its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> is about 1.59 g/mL.</p>
<p>∴ The volume is</p>
<p><mathjax>#V = 5.08 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.59 color(red)(cancel(color(black)("g")))) = "3.2 mL"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You will get 3.2 mL of <mathjax>#"CCl"_4#</mathjax> at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>WARNING! Long answer!</p>
<p>This is a <strong>limiting reactant</strong> problem.</p>
<p>One way to solve it is to figure out how much product you could get from each reactant separately.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation.</strong></p>
<p>The balanced chemical equation is</p>
<p><mathjax>#"CH"_4 + "4Cl"_2 → "CCl"_4 + "4HCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Identify the limiting reactant</strong></p>
<p>We can use <strong>Gay-Lussac's Law of Combining Volumes</strong> to identify the limiting reactant.</p>
<blockquote></blockquote>
<p>From <mathjax>#"CH"_4#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("1 L CH"_4)#</mathjax></p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 0.75 color(red)(cancel(color(black)("L CH"_4))) × ("1 L CCl"4)/(1 color(red)(cancel(color(black)("L CH"_4)))) = "0.75 L CCl"_4#</mathjax></p>
<blockquote></blockquote>
<p>From <mathjax>#"Cl"_2#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("4 L Cl"_2)#</mathjax>.</p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 8.0 color(red)(cancel(color(black)("L Cl"_2))) × ("1 mol CCl"_4)/(4 color(red)(cancel(color(black)("L Cl"_2)))) = "2.0 L CCl"_4#</mathjax></p>
<p><mathjax>#"CH"_4#</mathjax> gives the smaller volume of gaseous <mathjax>#"CCl"_4#</mathjax>, so <mathjax>#"CH"_4#</mathjax> is the limiting reactant.</p>
<p>The theoretical yield of gaseous <mathjax>#"CCl"_4#</mathjax> at STP is 0.75 L.</p>
<blockquote></blockquote>
<p><strong>3. Calculate the mass of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p>To solve this problem, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#m = (PVM)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>STP is defined at 1 bar and 0 °C.</p>
<p><mathjax>#P = "1 bar"#</mathjax><br/>
<mathjax>#V = "0.75 L"#</mathjax><br/>
<mathjax>#M = "153.82 g/mol"#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax></p>
<p>If <mathjax>#"CCl"_4#</mathjax> were an ideal gas at STP, we would calculate that</p>
<p>∴ <mathjax>#m = (1 color(red)(cancel(color(black)("bar"))) × 0.75 color(red)(cancel(color(black)("L"))) × "153.82 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "5.08 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the volume of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p><mathjax>#"CCl"_4#</mathjax> is a <strong>liquid</strong> at STP. Its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> is about 1.59 g/mL.</p>
<p>∴ The volume is</p>
<p><mathjax>#V = 5.08 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.59 color(red)(cancel(color(black)("g")))) = "3.2 mL"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many mL of carbon tetrachloride are produced when 8.0 L of chlorine are allowed to react with 0.75 L of methane at STP? </h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-07-02T21:15:29" itemprop="dateCreated">
Jul 2, 2016
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<div class="markdown"><p>You will get 3.2 mL of <mathjax>#"CCl"_4#</mathjax> at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>WARNING! Long answer!</p>
<p>This is a <strong>limiting reactant</strong> problem.</p>
<p>One way to solve it is to figure out how much product you could get from each reactant separately.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation.</strong></p>
<p>The balanced chemical equation is</p>
<p><mathjax>#"CH"_4 + "4Cl"_2 → "CCl"_4 + "4HCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Identify the limiting reactant</strong></p>
<p>We can use <strong>Gay-Lussac's Law of Combining Volumes</strong> to identify the limiting reactant.</p>
<blockquote></blockquote>
<p>From <mathjax>#"CH"_4#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("1 L CH"_4)#</mathjax></p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 0.75 color(red)(cancel(color(black)("L CH"_4))) × ("1 L CCl"4)/(1 color(red)(cancel(color(black)("L CH"_4)))) = "0.75 L CCl"_4#</mathjax></p>
<blockquote></blockquote>
<p>From <mathjax>#"Cl"_2#</mathjax>:</p>
<p>The volume ratio is <mathjax>#("1 L CCl"_4)/("4 L Cl"_2)#</mathjax>.</p>
<p>∴ <mathjax>#"Volume of CCl"_4 = 8.0 color(red)(cancel(color(black)("L Cl"_2))) × ("1 mol CCl"_4)/(4 color(red)(cancel(color(black)("L Cl"_2)))) = "2.0 L CCl"_4#</mathjax></p>
<p><mathjax>#"CH"_4#</mathjax> gives the smaller volume of gaseous <mathjax>#"CCl"_4#</mathjax>, so <mathjax>#"CH"_4#</mathjax> is the limiting reactant.</p>
<p>The theoretical yield of gaseous <mathjax>#"CCl"_4#</mathjax> at STP is 0.75 L.</p>
<blockquote></blockquote>
<p><strong>3. Calculate the mass of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p>To solve this problem, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#m = (PVM)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>STP is defined at 1 bar and 0 °C.</p>
<p><mathjax>#P = "1 bar"#</mathjax><br/>
<mathjax>#V = "0.75 L"#</mathjax><br/>
<mathjax>#M = "153.82 g/mol"#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax></p>
<p>If <mathjax>#"CCl"_4#</mathjax> were an ideal gas at STP, we would calculate that</p>
<p>∴ <mathjax>#m = (1 color(red)(cancel(color(black)("bar"))) × 0.75 color(red)(cancel(color(black)("L"))) × "153.82 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "5.08 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the volume of <mathjax>#"CCl"_4#</mathjax></strong></p>
<p><mathjax>#"CCl"_4#</mathjax> is a <strong>liquid</strong> at STP. Its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> is about 1.59 g/mL.</p>
<p>∴ The volume is</p>
<p><mathjax>#V = 5.08 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.59 color(red)(cancel(color(black)("g")))) = "3.2 mL"#</mathjax></p></div>
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</article> | How many mL of carbon tetrachloride are produced when 8.0 L of chlorine are allowed to react with 0.75 L of methane at STP? | null |
3,060 | a9ace051-6ddd-11ea-9b11-ccda262736ce | https://socratic.org/questions/what-is-the-total-number-of-moles-of-nacl-formed-when-2-moles-of-na-2cro-4-react | 4 moles | start physical_unit 8 8 mole mol qc_end c_other OTHER qc_end physical_unit 14 14 11 12 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] NaCl [IN] moles"}] | [{"type":"physical unit","value":"4 moles"}] | [{"type":"other","value":"React completely."},{"type":"physical unit","value":"Mole [OF] Na2CrO4 [=] \\pu{2 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the total number of moles of #NaCl# formed when 2 moles of #Na_2CrO_4# react completely?</h1> | null | 4 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As the formula of <mathjax>#Na_2CrO_4#</mathjax>shows that 1 molecule contains 2 atoms of Na then 2 molecules will contain 4 atoms of Na i.e. 1 mole contains 2 moles Na..Again 1 mole NaCl contains 1 mole Na<br/>
This means 2moles of <mathjax>#Na_2CrO_4#</mathjax> will form 4 moles NaCl</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>It is 4 moles </p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As the formula of <mathjax>#Na_2CrO_4#</mathjax>shows that 1 molecule contains 2 atoms of Na then 2 molecules will contain 4 atoms of Na i.e. 1 mole contains 2 moles Na..Again 1 mole NaCl contains 1 mole Na<br/>
This means 2moles of <mathjax>#Na_2CrO_4#</mathjax> will form 4 moles NaCl</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the total number of moles of #NaCl# formed when 2 moles of #Na_2CrO_4# react completely?</h1>
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<div class="markdown"><p>It is 4 moles </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As the formula of <mathjax>#Na_2CrO_4#</mathjax>shows that 1 molecule contains 2 atoms of Na then 2 molecules will contain 4 atoms of Na i.e. 1 mole contains 2 moles Na..Again 1 mole NaCl contains 1 mole Na<br/>
This means 2moles of <mathjax>#Na_2CrO_4#</mathjax> will form 4 moles NaCl</p></div>
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</article> | What is the total number of moles of #NaCl# formed when 2 moles of #Na_2CrO_4# react completely? | null |
3,061 | a9177931-6ddd-11ea-b6d2-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-salt-solution-made-by-dissolving-240mg-of-nacl-in-4ml- | 1 mol/L | start physical_unit 5 7 molarity mol/l qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 19 19 16 17 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] a salt solution [IN] mol/L"}] | [{"type":"physical unit","value":"1 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{240 mg}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{4 mL}"},{"type":"other","value":"Final volume is the same as the volume of water."}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a salt solution made by dissolving 240mg of NaCl in 4ml of water? Assume final volume is the same as the volume of water.</h1> | null | 1 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you must determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in exactly</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> </p>
</blockquote>
<p>of solution. Notice that you already know the number of milligrams of sodium chloride, the solute, present in <mathjax>#"4 mL"#</mathjax> of solution, so start by calculating the mass of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "240 mg NaCl"/(4color(red)(cancel(color(black)("mL solution")))) = 60 * 10^3#</mathjax> <mathjax>#"mg"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3color(white)(.)"mg"#</mathjax></p>
</blockquote>
<p>you can say that <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution contain</p>
<blockquote>
<p><mathjax>#60 * color(blue)(10^3color(white)(.)"mg") = 60color(white)(.)color(blue)("g")#</mathjax></p>
</blockquote>
<p>of sodium chloride. Now, to convert this to <em>moles</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#60 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.027 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since this represents the number of moles of sodium chloride present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, you can say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1 mol L"^(-1)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for the volume of the solution. </p></div>
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<div class="markdown"><p><mathjax>#"1 mol L"^(-1)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you must determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in exactly</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> </p>
</blockquote>
<p>of solution. Notice that you already know the number of milligrams of sodium chloride, the solute, present in <mathjax>#"4 mL"#</mathjax> of solution, so start by calculating the mass of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "240 mg NaCl"/(4color(red)(cancel(color(black)("mL solution")))) = 60 * 10^3#</mathjax> <mathjax>#"mg"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3color(white)(.)"mg"#</mathjax></p>
</blockquote>
<p>you can say that <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution contain</p>
<blockquote>
<p><mathjax>#60 * color(blue)(10^3color(white)(.)"mg") = 60color(white)(.)color(blue)("g")#</mathjax></p>
</blockquote>
<p>of sodium chloride. Now, to convert this to <em>moles</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#60 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.027 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since this represents the number of moles of sodium chloride present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, you can say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1 mol L"^(-1)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for the volume of the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the molarity of a salt solution made by dissolving 240mg of NaCl in 4ml of water? Assume final volume is the same as the volume of water.</h1>
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<div class="markdown"><p><mathjax>#"1 mol L"^(-1)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you must determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in exactly</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> </p>
</blockquote>
<p>of solution. Notice that you already know the number of milligrams of sodium chloride, the solute, present in <mathjax>#"4 mL"#</mathjax> of solution, so start by calculating the mass of solute present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "240 mg NaCl"/(4color(red)(cancel(color(black)("mL solution")))) = 60 * 10^3#</mathjax> <mathjax>#"mg"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3color(white)(.)"mg"#</mathjax></p>
</blockquote>
<p>you can say that <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution contain</p>
<blockquote>
<p><mathjax>#60 * color(blue)(10^3color(white)(.)"mg") = 60color(white)(.)color(blue)("g")#</mathjax></p>
</blockquote>
<p>of sodium chloride. Now, to convert this to <em>moles</em>, use the <strong>molar mass</strong> of the compound</p>
<blockquote>
<p><mathjax>#60 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.027 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since this represents the number of moles of sodium chloride present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution, you can say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1 mol L"^(-1)))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for the volume of the solution. </p></div>
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<div class="markdown"><p>M=<mathjax>#"1 mol/L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution is the concentration of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per liter of solution, <mathjax>#"mol/L"#</mathjax>. The mass of <mathjax>#"NaCl"#</mathjax> must be converted to grams and then moles. The volume of the solution must be converted to liters.</p>
<p><strong>Convert <mathjax>#"mg NaCl"#</mathjax> to <mathjax>#"g NaCl"#</mathjax>.</strong></p>
<p><mathjax>#"1 g=1000 mg"#</mathjax></p>
<p><mathjax>#240color(red)cancel(color(black)("mg NaCl"))xx(1"g")/(1000color(red)cancel(color(black)("mg")))="0.24 g NaCl"#</mathjax></p>
<p><strong>Determine moles of <mathjax>#"NaCl"#</mathjax>.</strong></p>
<p>Divide the mass of <mathjax>#"NaCl"#</mathjax> by its molar mass by multiplying by its inverse.</p>
<p><strong>Molar mass <mathjax>#"NaCl=58.44 g/mol"#</mathjax></strong></p>
<p><mathjax>#0.24color(red)cancel(color(black)("g NaCl"))xx(1"mol NaCl")/(58.44color(red)cancel(color(black)("g NaCl")))="0.0041 mol NaCl"#</mathjax></p>
<p><strong>Molarity of <mathjax>#"NaCl"#</mathjax>.</strong></p>
<p><strong>Convert <mathjax>#"4 mL"#</mathjax> to liters.</strong></p>
<p><mathjax>#"1 L=1000 mL"#</mathjax></p>
<p><mathjax>#4color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.004 L"#</mathjax></p>
<p><strong>Divide mol NaCl by volume of solution.</strong></p>
<p><mathjax>#"M"=(0.0041"mol NaCl")/(0.004"L solution")="1 mol/L"#</mathjax> rounded to one sig fig due to <mathjax>#4#</mathjax> mL</p></div>
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</article> | What is the molarity of a salt solution made by dissolving 240mg of NaCl in 4ml of water? Assume final volume is the same as the volume of water. | null |
3,062 | a94bfd1a-6ddd-11ea-a5bf-ccda262736ce | https://socratic.org/questions/how-many-moles-of-pcl-5-can-be-produced-from-55-0-g-of-cl-2-and-excess-p-4 | 0.31 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] PCl5 [IN] moles"}] | [{"type":"physical unit","value":"0.31 moles"}] | [{"type":"physical unit","value":"Mass [OF] Cl2 [=] \\pu{55.0 g}"},{"type":"other","value":"Excess P4."}] | <h1 class="questionTitle" itemprop="name">How many moles of #PCl_5# can be produced from 55.0 g of #Cl_2# and excess #P_4#?</h1> | null | 0.31 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the number of <em>moles</em> of <mathjax>#"PCl"_5#</mathjax> formed from excess <mathjax>#"P"_4#</mathjax> and <mathjax>#55.0#</mathjax> <mathjax>#"g CL"_2#</mathjax>.</p>
<p>Let's start by writing the chemical equation for this reaction:</p>
<p><mathjax>#"P"_4 (s) + 10"Cl"_2 (g) rarr 4"PCl"_5(s)#</mathjax></p>
<p>Using the molar mass of <mathjax>#"Cl"_2#</mathjax>, let's convert from grams to moles, and then use the coefficients in the equation to convert to moles of <mathjax>#"PCl"_5#</mathjax>:</p>
<p><mathjax>#55.0#</mathjax> <mathjax>#cancel("g Cl"_2)((1cancel("mol Cl"_2))/(70.90cancel("g Cl"_2)))((4"mol PCl"_5)/(10cancel("mol Cl"_2))) = color(red)(0.310#</mathjax> <mathjax>#color(red)("mol PCl"_5#</mathjax></p>
<p>Thus, with excess phosphorus, <mathjax>#0.310#</mathjax> moles of <mathjax>#"PCl"_5#</mathjax> will form (assuming the reaction goes to completion).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.310#</mathjax> <mathjax>#"mol PCl"_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the number of <em>moles</em> of <mathjax>#"PCl"_5#</mathjax> formed from excess <mathjax>#"P"_4#</mathjax> and <mathjax>#55.0#</mathjax> <mathjax>#"g CL"_2#</mathjax>.</p>
<p>Let's start by writing the chemical equation for this reaction:</p>
<p><mathjax>#"P"_4 (s) + 10"Cl"_2 (g) rarr 4"PCl"_5(s)#</mathjax></p>
<p>Using the molar mass of <mathjax>#"Cl"_2#</mathjax>, let's convert from grams to moles, and then use the coefficients in the equation to convert to moles of <mathjax>#"PCl"_5#</mathjax>:</p>
<p><mathjax>#55.0#</mathjax> <mathjax>#cancel("g Cl"_2)((1cancel("mol Cl"_2))/(70.90cancel("g Cl"_2)))((4"mol PCl"_5)/(10cancel("mol Cl"_2))) = color(red)(0.310#</mathjax> <mathjax>#color(red)("mol PCl"_5#</mathjax></p>
<p>Thus, with excess phosphorus, <mathjax>#0.310#</mathjax> moles of <mathjax>#"PCl"_5#</mathjax> will form (assuming the reaction goes to completion).</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #PCl_5# can be produced from 55.0 g of #Cl_2# and excess #P_4#?</h1>
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Nathan L.
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Jun 12, 2017
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<div class="markdown"><p><mathjax>#0.310#</mathjax> <mathjax>#"mol PCl"_5#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the number of <em>moles</em> of <mathjax>#"PCl"_5#</mathjax> formed from excess <mathjax>#"P"_4#</mathjax> and <mathjax>#55.0#</mathjax> <mathjax>#"g CL"_2#</mathjax>.</p>
<p>Let's start by writing the chemical equation for this reaction:</p>
<p><mathjax>#"P"_4 (s) + 10"Cl"_2 (g) rarr 4"PCl"_5(s)#</mathjax></p>
<p>Using the molar mass of <mathjax>#"Cl"_2#</mathjax>, let's convert from grams to moles, and then use the coefficients in the equation to convert to moles of <mathjax>#"PCl"_5#</mathjax>:</p>
<p><mathjax>#55.0#</mathjax> <mathjax>#cancel("g Cl"_2)((1cancel("mol Cl"_2))/(70.90cancel("g Cl"_2)))((4"mol PCl"_5)/(10cancel("mol Cl"_2))) = color(red)(0.310#</mathjax> <mathjax>#color(red)("mol PCl"_5#</mathjax></p>
<p>Thus, with excess phosphorus, <mathjax>#0.310#</mathjax> moles of <mathjax>#"PCl"_5#</mathjax> will form (assuming the reaction goes to completion).</p></div>
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</article> | How many moles of #PCl_5# can be produced from 55.0 g of #Cl_2# and excess #P_4#? | null |
3,063 | a8d3b1df-6ddd-11ea-8337-ccda262736ce | https://socratic.org/questions/what-is-the-a-balanced-chemical-equation-for-the-combust-of-liquid-methanol-ch-3 | 2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l) | start chemical_equation qc_end chemical_equation 13 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the combust"}] | [{"type":"chemical equation","value":"2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l)"}] | [{"type":"chemical equation","value":"CH3OH"}] | <h1 class="questionTitle" itemprop="name">What is the a balanced chemical equation for the combust of liquid methanol (#CH_3OH#)?</h1> | null | 2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Alcohols combust to give carbon dioxide and water, and the reaction is represented accordingly. </p>
<p>Does garbage out equal garbage in?</p>
<p>How is energy transferred in the reaction?</p>
<p>How can we remove the half-coefficient?</p></div>
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<div class="markdown"><p><mathjax>#H_3COH(l) + 3/2O_2(g) rarrCO_2(g) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Alcohols combust to give carbon dioxide and water, and the reaction is represented accordingly. </p>
<p>Does garbage out equal garbage in?</p>
<p>How is energy transferred in the reaction?</p>
<p>How can we remove the half-coefficient?</p></div>
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anor277
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<div class="markdown"><p><mathjax>#H_3COH(l) + 3/2O_2(g) rarrCO_2(g) + 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Alcohols combust to give carbon dioxide and water, and the reaction is represented accordingly. </p>
<p>Does garbage out equal garbage in?</p>
<p>How is energy transferred in the reaction?</p>
<p>How can we remove the half-coefficient?</p></div>
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</article> | What is the a balanced chemical equation for the combust of liquid methanol (#CH_3OH#)? | null |
3,064 | ab44483f-6ddd-11ea-9315-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-of-li-s-h3po4-aq | 6 Li(s) + H3PO4(aq) -> 2 Li3PO4(aq) + 3 H2(g) ^ | start chemical_equation qc_end chemical_equation 7 7 qc_end chemical_equation 9 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced chemical equation"}] | [{"type":"chemical equation","value":"6 Li(s) + H3PO4(aq) -> 2 Li3PO4(aq) + 3 H2(g) ^"}] | [{"type":"chemical equation","value":"Li(s)"},{"type":"chemical equation","value":"H3PO4(aq)"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation of Li(s) + H3PO4(aq) ?</h1> | null | 6 Li(s) + H3PO4(aq) -> 2 Li3PO4(aq) + 3 H2(g) ^ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Lithium</em> metal will react with phosphoric acid to produce <em>lithium phosphate</em>, <mathjax>#"Li"_3"PO"_4#</mathjax>, and <em>hydrogen gas</em>, <mathjax>#"H"_2#</mathjax>.</p>
<blockquote>
<p><mathjax>#"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>To balance this chemical equation, notice that the only atoms that are not balanced are lithium and hydrogen. </p>
<p>More specifically, you have <strong>one</strong> lithium atom on the reactants' side, and <strong>three</strong> on the products' side. This means that you need to multiply the former by <mathjax>#3#</mathjax></p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>The only atoms that remain unbalanced are the hydrogen atoms. This time you have <strong>three</strong> of these on the reactants's side, and only <strong>two</strong> on the products' side. </p>
<p>This implies that you're going to have to multiply the hydrogen atoms on the reactants' side by <mathjax>#2#</mathjax> and those on the products' side by <mathjax>#3#</mathjax> to get a total of <strong>six</strong> atoms on each side. </p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Now you've unbalanced the number of phosphorus and oxygen atoms. To fix this, you can treat the <em>phosphate anion</em>, <mathjax>#"PO"_4^(3-)#</mathjax>, as <em>a unit</em>.</p>
<p>This means that you will need to multiply the lithium phosphate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Finally, balance the lithium atoms again by multiplying the lithium metal by <mathjax>#6#</mathjax> <strong>instead of</strong> by <mathjax>#3#</mathjax>.</p>
<blockquote>
<p><mathjax>#6"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>And now the chemical equation is balanced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#6"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Lithium</em> metal will react with phosphoric acid to produce <em>lithium phosphate</em>, <mathjax>#"Li"_3"PO"_4#</mathjax>, and <em>hydrogen gas</em>, <mathjax>#"H"_2#</mathjax>.</p>
<blockquote>
<p><mathjax>#"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>To balance this chemical equation, notice that the only atoms that are not balanced are lithium and hydrogen. </p>
<p>More specifically, you have <strong>one</strong> lithium atom on the reactants' side, and <strong>three</strong> on the products' side. This means that you need to multiply the former by <mathjax>#3#</mathjax></p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>The only atoms that remain unbalanced are the hydrogen atoms. This time you have <strong>three</strong> of these on the reactants's side, and only <strong>two</strong> on the products' side. </p>
<p>This implies that you're going to have to multiply the hydrogen atoms on the reactants' side by <mathjax>#2#</mathjax> and those on the products' side by <mathjax>#3#</mathjax> to get a total of <strong>six</strong> atoms on each side. </p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Now you've unbalanced the number of phosphorus and oxygen atoms. To fix this, you can treat the <em>phosphate anion</em>, <mathjax>#"PO"_4^(3-)#</mathjax>, as <em>a unit</em>.</p>
<p>This means that you will need to multiply the lithium phosphate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Finally, balance the lithium atoms again by multiplying the lithium metal by <mathjax>#6#</mathjax> <strong>instead of</strong> by <mathjax>#3#</mathjax>.</p>
<blockquote>
<p><mathjax>#6"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>And now the chemical equation is balanced.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation of Li(s) + H3PO4(aq) ?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-18T11:03:54" itemprop="dateCreated">
Oct 18, 2015
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<div class="markdown"><p><mathjax>#6"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Lithium</em> metal will react with phosphoric acid to produce <em>lithium phosphate</em>, <mathjax>#"Li"_3"PO"_4#</mathjax>, and <em>hydrogen gas</em>, <mathjax>#"H"_2#</mathjax>.</p>
<blockquote>
<p><mathjax>#"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>To balance this chemical equation, notice that the only atoms that are not balanced are lithium and hydrogen. </p>
<p>More specifically, you have <strong>one</strong> lithium atom on the reactants' side, and <strong>three</strong> on the products' side. This means that you need to multiply the former by <mathjax>#3#</mathjax></p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>The only atoms that remain unbalanced are the hydrogen atoms. This time you have <strong>three</strong> of these on the reactants's side, and only <strong>two</strong> on the products' side. </p>
<p>This implies that you're going to have to multiply the hydrogen atoms on the reactants' side by <mathjax>#2#</mathjax> and those on the products' side by <mathjax>#3#</mathjax> to get a total of <strong>six</strong> atoms on each side. </p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Now you've unbalanced the number of phosphorus and oxygen atoms. To fix this, you can treat the <em>phosphate anion</em>, <mathjax>#"PO"_4^(3-)#</mathjax>, as <em>a unit</em>.</p>
<p>This means that you will need to multiply the lithium phosphate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#3"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Finally, balance the lithium atoms again by multiplying the lithium metal by <mathjax>#6#</mathjax> <strong>instead of</strong> by <mathjax>#3#</mathjax>.</p>
<blockquote>
<p><mathjax>#6"Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>And now the chemical equation is balanced.</p></div>
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</article> | What is the balanced chemical equation of Li(s) + H3PO4(aq) ? | null |
3,065 | a8e49e4d-6ddd-11ea-bac8-ccda262736ce | https://socratic.org/questions/how-many-moles-of-mgcl-2-are-there-in-309-g-of-the-compound | 3.25 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 12 8 9 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] MgCl2 [IN] moles"}] | [{"type":"physical unit","value":"3.25 moles"}] | [{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{309 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #MgCl_2# are there in 309 g of the compound?</h1> | null | 3.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 Molecule of MgCl2 = 95.11<br/>
1 Mole in 1 molecule of MgCl2 is <mathjax>#1/95.11#</mathjax> = 0.010514<br/>
Or Total moles in 309 gm of MgCl2 is #0.010514*309 = 3.2488 <br/>
Answer is 3.2488 moles</p></div>
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<div class="markdown"><p>3.2488 moles</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 Molecule of MgCl2 = 95.11<br/>
1 Mole in 1 molecule of MgCl2 is <mathjax>#1/95.11#</mathjax> = 0.010514<br/>
Or Total moles in 309 gm of MgCl2 is #0.010514*309 = 3.2488 <br/>
Answer is 3.2488 moles</p></div>
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<div class="markdown"><p>3.2488 moles</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 Molecule of MgCl2 = 95.11<br/>
1 Mole in 1 molecule of MgCl2 is <mathjax>#1/95.11#</mathjax> = 0.010514<br/>
Or Total moles in 309 gm of MgCl2 is #0.010514*309 = 3.2488 <br/>
Answer is 3.2488 moles</p></div>
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</article> | How many moles of #MgCl_2# are there in 309 g of the compound? | null |
3,066 | a9d594c6-6ddd-11ea-96c2-ccda262736ce | https://socratic.org/questions/how-many-moles-of-oxygen-are-required-to-completely-react-with-16-1-grams-of-mag | 0.33 moles | start physical_unit 4 4 mole mol qc_end chemical_equation 20 26 qc_end physical_unit 14 15 11 12 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"0.33 moles"}] | [{"type":"chemical equation","value":"2 Mg(s) + O2(g) -> 2 MgO(s)"},{"type":"physical unit","value":"Mass [OF] magnesium metal [=] \\pu{16.1 grams}"},{"type":"other","value":"Completely react."}] | <h1 class="questionTitle" itemprop="name"> How many moles of oxygen are required to completely react with 16.1
grams of magnesium metal according to the equation:
2 Mg(s) + O2(g) → 2 MgO(s) ?</h1> | null | 0.33 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Locate <mathjax>#Mg#</mathjax> in <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and take note of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a>;</li>
<li>Find the number of moles of magnesium by dividing the mass as provided in the problem against the atomic mass of magnesium;<br/>
<mathjax>#n Mg=(mass)/(at. mass)#</mathjax><br/>
<mathjax>#n Mg=0.67 mol#</mathjax></li>
<li>Then, compute the number of moles of <mathjax>#O_2#</mathjax> needed to completely react with magnesium through conversion method;<br/>
<mathjax>#0.67 cancel(mol Mg) #</mathjax>x<mathjax># (1 mol O_2)/(2 cancel(mol Mg))#</mathjax></li>
<li>Make sure to cancel out units that are not needed leaving only the desired units;</li>
<li>The final answer is found to be <mathjax>#0.33#</mathjax> moles of <mathjax>#O_2#</mathjax> </li>
</ol></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.33#</mathjax> moles of <mathjax>#O_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Locate <mathjax>#Mg#</mathjax> in <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and take note of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a>;</li>
<li>Find the number of moles of magnesium by dividing the mass as provided in the problem against the atomic mass of magnesium;<br/>
<mathjax>#n Mg=(mass)/(at. mass)#</mathjax><br/>
<mathjax>#n Mg=0.67 mol#</mathjax></li>
<li>Then, compute the number of moles of <mathjax>#O_2#</mathjax> needed to completely react with magnesium through conversion method;<br/>
<mathjax>#0.67 cancel(mol Mg) #</mathjax>x<mathjax># (1 mol O_2)/(2 cancel(mol Mg))#</mathjax></li>
<li>Make sure to cancel out units that are not needed leaving only the desired units;</li>
<li>The final answer is found to be <mathjax>#0.33#</mathjax> moles of <mathjax>#O_2#</mathjax> </li>
</ol></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> How many moles of oxygen are required to completely react with 16.1
grams of magnesium metal according to the equation:
2 Mg(s) + O2(g) → 2 MgO(s) ?</h1>
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<div class="markdown"><p><mathjax>#0.33#</mathjax> moles of <mathjax>#O_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Locate <mathjax>#Mg#</mathjax> in <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and take note of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a>;</li>
<li>Find the number of moles of magnesium by dividing the mass as provided in the problem against the atomic mass of magnesium;<br/>
<mathjax>#n Mg=(mass)/(at. mass)#</mathjax><br/>
<mathjax>#n Mg=0.67 mol#</mathjax></li>
<li>Then, compute the number of moles of <mathjax>#O_2#</mathjax> needed to completely react with magnesium through conversion method;<br/>
<mathjax>#0.67 cancel(mol Mg) #</mathjax>x<mathjax># (1 mol O_2)/(2 cancel(mol Mg))#</mathjax></li>
<li>Make sure to cancel out units that are not needed leaving only the desired units;</li>
<li>The final answer is found to be <mathjax>#0.33#</mathjax> moles of <mathjax>#O_2#</mathjax> </li>
</ol></div>
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</article> | How many moles of oxygen are required to completely react with 16.1
grams of magnesium metal according to the equation:
2 Mg(s) + O2(g) → 2 MgO(s) ? | null |
3,067 | abe233f8-6ddd-11ea-a3ba-ccda262736ce | https://socratic.org/questions/if-24500-is-applied-to-125g-of-water-at-35-c-what-will-the-final-temperature-of- | 81.9 ℃ | start physical_unit 9 9 temperature °c qc_end physical_unit 9 9 1 2 heat_energy qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 9 9 11 12 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"81.9 ℃"}] | [{"type":"physical unit","value":"Applied energy [OF] water [=] \\pu{24500 J}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{125 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{35 ℃}"}] | <h1 class="questionTitle" itemprop="name">If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be?</h1> | null | 81.9 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#delta H = mxxCpxxdeltaT#</mathjax>, where <mathjax>#delta H#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> or change in energy, m is the mass, Cp is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and <mathjax>#delta T#</mathjax> is the change in temperature. </p>
<p>We know <mathjax>#delta H#</mathjax>= 24500J, the specific heat of water is 4.18J/mol*k, and m=125g</p>
<p>So we alter our formula a little bit and substitute in what we know:</p>
<p><mathjax>#deltaT=(deltaH)/(mxxCp)=(24500J)/(125gxx4.18J/(molxxk)) = 46.9k#</mathjax></p>
<p>Celsius and kelvin have the same scale for units, so simply add the change to the original temperature.</p>
<p>We have: Tfinal = 35+46.9=81.9C</p>
<p>Hope that helps! </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>81.9C</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#delta H = mxxCpxxdeltaT#</mathjax>, where <mathjax>#delta H#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> or change in energy, m is the mass, Cp is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and <mathjax>#delta T#</mathjax> is the change in temperature. </p>
<p>We know <mathjax>#delta H#</mathjax>= 24500J, the specific heat of water is 4.18J/mol*k, and m=125g</p>
<p>So we alter our formula a little bit and substitute in what we know:</p>
<p><mathjax>#deltaT=(deltaH)/(mxxCp)=(24500J)/(125gxx4.18J/(molxxk)) = 46.9k#</mathjax></p>
<p>Celsius and kelvin have the same scale for units, so simply add the change to the original temperature.</p>
<p>We have: Tfinal = 35+46.9=81.9C</p>
<p>Hope that helps! </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be?</h1>
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<div class="markdown"><p>81.9C</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#delta H = mxxCpxxdeltaT#</mathjax>, where <mathjax>#delta H#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> or change in energy, m is the mass, Cp is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and <mathjax>#delta T#</mathjax> is the change in temperature. </p>
<p>We know <mathjax>#delta H#</mathjax>= 24500J, the specific heat of water is 4.18J/mol*k, and m=125g</p>
<p>So we alter our formula a little bit and substitute in what we know:</p>
<p><mathjax>#deltaT=(deltaH)/(mxxCp)=(24500J)/(125gxx4.18J/(molxxk)) = 46.9k#</mathjax></p>
<p>Celsius and kelvin have the same scale for units, so simply add the change to the original temperature.</p>
<p>We have: Tfinal = 35+46.9=81.9C</p>
<p>Hope that helps! </p></div>
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</article> | If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be? | null |
3,068 | a8469a4c-6ddd-11ea-865b-ccda262736ce | https://socratic.org/questions/how-many-moles-of-carbon-atoms-are-equivalent-to-38-1-g-carbon | 3.17 moles | start physical_unit 4 5 mole mol qc_end physical_unit 4 4 9 10 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon atoms [IN] moles"}] | [{"type":"physical unit","value":"3.17 moles"}] | [{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{38.1 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of carbon atoms are equivalent to 38.1 g carbon? </h1> | null | 3.17 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>molar mass</strong> of carbon, which you can use as a <em>conversion factor</em> to figure out how many <strong>moles</strong> of carbon would be present in that sample. </p>
<p>You can use the <strong>molar mass</strong> of carbon as a conversion factor because it represents the mass of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("molar mass C " = " mass of 1 mole of C")))#</mathjax></p>
</blockquote>
<p>Now, carbon has a molar mass of <mathjax>#"12.011 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of carbon has a mass of <mathjax>#"12.011 g"#</mathjax>. </p>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#38.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("3.17 moles C")))#</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 mass of the sample. </p></div>
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<div>
<div class="markdown"><p><mathjax>#"3.17 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>molar mass</strong> of carbon, which you can use as a <em>conversion factor</em> to figure out how many <strong>moles</strong> of carbon would be present in that sample. </p>
<p>You can use the <strong>molar mass</strong> of carbon as a conversion factor because it represents the mass of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("molar mass C " = " mass of 1 mole of C")))#</mathjax></p>
</blockquote>
<p>Now, carbon has a molar mass of <mathjax>#"12.011 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of carbon has a mass of <mathjax>#"12.011 g"#</mathjax>. </p>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#38.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("3.17 moles C")))#</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 mass of the sample. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of carbon atoms are equivalent to 38.1 g carbon? </h1>
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Stefan V.
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Dec 27, 2016
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<div class="markdown"><p><mathjax>#"3.17 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>molar mass</strong> of carbon, which you can use as a <em>conversion factor</em> to figure out how many <strong>moles</strong> of carbon would be present in that sample. </p>
<p>You can use the <strong>molar mass</strong> of carbon as a conversion factor because it represents the mass of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("molar mass C " = " mass of 1 mole of C")))#</mathjax></p>
</blockquote>
<p>Now, carbon has a molar mass of <mathjax>#"12.011 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of carbon has a mass of <mathjax>#"12.011 g"#</mathjax>. </p>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#38.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("3.17 moles C")))#</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 mass of the sample. </p></div>
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</article> | How many moles of carbon atoms are equivalent to 38.1 g carbon? | null |
3,069 | acac0de2-6ddd-11ea-a841-ccda262736ce | https://socratic.org/questions/when-glucose-is-consumed-it-reacts-with-oxygen-in-the-body-to-produce-carbon-dio-1 | 65.96 grams | start physical_unit 13 14 mass g qc_end substance 7 7 qc_end substance 15 15 qc_end physical_unit 31 31 28 29 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] grams"}] | [{"type":"physical unit","value":"65.96 grams"}] | [{"type":"substance name","value":"Oxygen"},{"type":"substance name","value":"Water"},{"type":"physical unit","value":"Mass [OF] C6H12O6 [=] \\pu{45 g}"},{"type":"other","value":"Completely react."}] | <h1 class="questionTitle" itemprop="name">When glucose is consumed, it reacts with oxygen in the body to produce carbon dioxide, water, and energy. How many grams of carbon dioxide would be produced if45 g of #C_6H_12O_6# completely reacted with oxygen?</h1> | null | 65.96 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, find the balanced chemical formula:</p>
<p><mathjax>#C_6H_12O_6 + 6O_2 rarr 6CO_2 + 6H_2O#</mathjax></p>
<p>Since glucose forms carbon dioxide in a 1:6 ratio, we know that one mole of glucose yields 6 moles of carbon dioxide. So in order to find the moles of carbon dioxide formed, we need to multiply the moles of glucose by 6.</p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / "molar mass of glucose"#</mathjax></p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / (180.15588 g/(mol))#</mathjax></p>
<p><mathjax>#"moles "C_6H_12O_6 = 0.249783687 mols#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 0.249783687mols * 6#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 1.498702124mols#</mathjax></p>
<p>To find grams of carbon dioxide, just multiply the moles of <mathjax>#CO_2#</mathjax> by <mathjax>#CO_2#</mathjax>'s molar mass.</p>
<p><mathjax>#"grams " CO_2 = 1.498702124mols * 44.0095 g/(mol)#</mathjax></p>
<p><mathjax>#"grams " CO_2 = 65.957131124 grams#</mathjax></p>
<p>Since you only have 2 sig figs, the answer would be 66 grams.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>66 grams <mathjax>#CO_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, find the balanced chemical formula:</p>
<p><mathjax>#C_6H_12O_6 + 6O_2 rarr 6CO_2 + 6H_2O#</mathjax></p>
<p>Since glucose forms carbon dioxide in a 1:6 ratio, we know that one mole of glucose yields 6 moles of carbon dioxide. So in order to find the moles of carbon dioxide formed, we need to multiply the moles of glucose by 6.</p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / "molar mass of glucose"#</mathjax></p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / (180.15588 g/(mol))#</mathjax></p>
<p><mathjax>#"moles "C_6H_12O_6 = 0.249783687 mols#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 0.249783687mols * 6#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 1.498702124mols#</mathjax></p>
<p>To find grams of carbon dioxide, just multiply the moles of <mathjax>#CO_2#</mathjax> by <mathjax>#CO_2#</mathjax>'s molar mass.</p>
<p><mathjax>#"grams " CO_2 = 1.498702124mols * 44.0095 g/(mol)#</mathjax></p>
<p><mathjax>#"grams " CO_2 = 65.957131124 grams#</mathjax></p>
<p>Since you only have 2 sig figs, the answer would be 66 grams.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When glucose is consumed, it reacts with oxygen in the body to produce carbon dioxide, water, and energy. How many grams of carbon dioxide would be produced if45 g of #C_6H_12O_6# completely reacted with oxygen?</h1>
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<div class="markdown"><p>66 grams <mathjax>#CO_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, find the balanced chemical formula:</p>
<p><mathjax>#C_6H_12O_6 + 6O_2 rarr 6CO_2 + 6H_2O#</mathjax></p>
<p>Since glucose forms carbon dioxide in a 1:6 ratio, we know that one mole of glucose yields 6 moles of carbon dioxide. So in order to find the moles of carbon dioxide formed, we need to multiply the moles of glucose by 6.</p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / "molar mass of glucose"#</mathjax></p>
<p><mathjax>#"moles " C_6H_12O_6 = (45g) / (180.15588 g/(mol))#</mathjax></p>
<p><mathjax>#"moles "C_6H_12O_6 = 0.249783687 mols#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 0.249783687mols * 6#</mathjax></p>
<p><mathjax>#"moles "CO_2 = 1.498702124mols#</mathjax></p>
<p>To find grams of carbon dioxide, just multiply the moles of <mathjax>#CO_2#</mathjax> by <mathjax>#CO_2#</mathjax>'s molar mass.</p>
<p><mathjax>#"grams " CO_2 = 1.498702124mols * 44.0095 g/(mol)#</mathjax></p>
<p><mathjax>#"grams " CO_2 = 65.957131124 grams#</mathjax></p>
<p>Since you only have 2 sig figs, the answer would be 66 grams.</p></div>
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</article> | When glucose is consumed, it reacts with oxygen in the body to produce carbon dioxide, water, and energy. How many grams of carbon dioxide would be produced if45 g of #C_6H_12O_6# completely reacted with oxygen? | null |
3,070 | abe20d5f-6ddd-11ea-85c0-ccda262736ce | https://socratic.org/questions/what-mass-of-water-would-release-16700j-of-energy-when-freezing | 50.1 g | start physical_unit 3 3 mass g qc_end physical_unit 3 3 6 7 heat_energy qc_end c_other Freezing qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] g"}] | [{"type":"physical unit","value":"50.1 g"}] | [{"type":"physical unit","value":"Released energy [OF] water [=] \\pu{16700 J}"},{"type":"other","value":"Freezing."}] | <h1 class="questionTitle" itemprop="name">What mass of water would release 16700J of energy when freezing?</h1> | null | 50.1 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</em>, <mathjax>#DeltaH_"fus"#</mathjax>, for water. </p>
<p>For a given substance, the enthalpy of fusion tells you how much heat is either <strong>needed</strong> to melt <mathjax>#"1 g"#</mathjax> of the substance at its melting point or <strong>given off</strong> to freeze <mathjax>#"1 g"#</mathjax> of the substance at its freezing point. </p>
<p>Water has an enthalpy of fusion equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J"#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>This tells you that when <mathjax>#"1 g"#</mathjax> of water goes from <em>liquid</em> at its freezing point to <em>solid</em> at its freezing point, <mathjax>#"333.55 J"#</mathjax> of heat are being <strong>given off</strong>. </p>
<p>In your case, you know that <mathjax>#"16,700 J"#</mathjax> of heat are being <strong>given off</strong> when a mass of water undergoes a liquid <mathjax>#->#</mathjax> solid <em>phase change</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>Use the enthalpy change of fusion as a <strong>conversion factor</strong> to determine how many grams of liquid water would give off this much heat when freezing</p>
<blockquote>
<p><mathjax>#"16,700" color(red)(cancel(color(black)("J"))) * overbrace(("1 g H"_ 2"O")/(333.55 color(red)(cancel(color(black)("J")))))^(color(blue)(= DeltaH_"fus")) = color(green)(|bar(ul(color(white)(a/a)color(black)("50.1 g H"_2"O")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the heat given off. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"50.1 g H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</em>, <mathjax>#DeltaH_"fus"#</mathjax>, for water. </p>
<p>For a given substance, the enthalpy of fusion tells you how much heat is either <strong>needed</strong> to melt <mathjax>#"1 g"#</mathjax> of the substance at its melting point or <strong>given off</strong> to freeze <mathjax>#"1 g"#</mathjax> of the substance at its freezing point. </p>
<p>Water has an enthalpy of fusion equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J"#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>This tells you that when <mathjax>#"1 g"#</mathjax> of water goes from <em>liquid</em> at its freezing point to <em>solid</em> at its freezing point, <mathjax>#"333.55 J"#</mathjax> of heat are being <strong>given off</strong>. </p>
<p>In your case, you know that <mathjax>#"16,700 J"#</mathjax> of heat are being <strong>given off</strong> when a mass of water undergoes a liquid <mathjax>#->#</mathjax> solid <em>phase change</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>Use the enthalpy change of fusion as a <strong>conversion factor</strong> to determine how many grams of liquid water would give off this much heat when freezing</p>
<blockquote>
<p><mathjax>#"16,700" color(red)(cancel(color(black)("J"))) * overbrace(("1 g H"_ 2"O")/(333.55 color(red)(cancel(color(black)("J")))))^(color(blue)(= DeltaH_"fus")) = color(green)(|bar(ul(color(white)(a/a)color(black)("50.1 g H"_2"O")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the heat given off. </p></div>
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<h1 class="questionTitle" itemprop="name">What mass of water would release 16700J of energy when freezing?</h1>
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<div class="markdown"><p><mathjax>#"50.1 g H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <em><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</em>, <mathjax>#DeltaH_"fus"#</mathjax>, for water. </p>
<p>For a given substance, the enthalpy of fusion tells you how much heat is either <strong>needed</strong> to melt <mathjax>#"1 g"#</mathjax> of the substance at its melting point or <strong>given off</strong> to freeze <mathjax>#"1 g"#</mathjax> of the substance at its freezing point. </p>
<p>Water has an enthalpy of fusion equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J"#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>This tells you that when <mathjax>#"1 g"#</mathjax> of water goes from <em>liquid</em> at its freezing point to <em>solid</em> at its freezing point, <mathjax>#"333.55 J"#</mathjax> of heat are being <strong>given off</strong>. </p>
<p>In your case, you know that <mathjax>#"16,700 J"#</mathjax> of heat are being <strong>given off</strong> when a mass of water undergoes a liquid <mathjax>#->#</mathjax> solid <em>phase change</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>Use the enthalpy change of fusion as a <strong>conversion factor</strong> to determine how many grams of liquid water would give off this much heat when freezing</p>
<blockquote>
<p><mathjax>#"16,700" color(red)(cancel(color(black)("J"))) * overbrace(("1 g H"_ 2"O")/(333.55 color(red)(cancel(color(black)("J")))))^(color(blue)(= DeltaH_"fus")) = color(green)(|bar(ul(color(white)(a/a)color(black)("50.1 g H"_2"O")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the heat given off. </p></div>
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</article> | What mass of water would release 16700J of energy when freezing? | null |
3,071 | ac2ac3fe-6ddd-11ea-8d59-ccda262736ce | https://socratic.org/questions/how-many-grams-of-co-2-are-produced-by-the-combustion-of-484g-of-a-mixture-that- | 1406.1 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 11 12 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] CO2 [IN] grams"}] | [{"type":"physical unit","value":"1406.1 grams"}] | [{"type":"physical unit","value":"Mass [OF] the mixture [=] \\pu{484 g}"},{"type":"physical unit","value":"Percent by mass [OF] CH4 in the mixture [=] \\pu{35.1%}"},{"type":"physical unit","value":"Percent by mass [OF] C3H8 in the mixture [=] \\pu{64.9%}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #CO_2# are produced by the combustion of 484g of a mixture that is 35.1% #CH_4# and 64.9% #C_3H_8# by mass?</h1> | null | 1406.1 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that you've got two combustion reactions taking place at the same time. </p>
<p>More specifically, you're going to have to write two balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, one for the combustion of methane and one for the combustion of propane.</p>
<p>The first thing you need to do is figure out exactly how many grams of each hydrocarbon you have. To do that, use the known <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of methane and propane in the mixture.</p>
<p><mathjax>#484cancel("g mixture") * ("35.1 g"CH_4)/(100cancel("g mixture")) = "169.9 g"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>and</p>
<p><mathjax>#484cancel("g mixture") * ("64.9 g" C_3H_8)/(100cancel("g mixture")) = "314.1 g"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>Check to see if the values add up to give <strong>484 g</strong>.</p>
<p><mathjax>#169.9 + 314.1 = "484 g"#</mathjax> <mathjax>#->#</mathjax> so far, so good.</p>
<p>Now focus on the two combustion reactions. Start with the first one</p>
<p><mathjax>#CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between methane and carbon dioxide. This means that, regardless of how many moles of methane eract, the reaction will produce the same exact number of moles of carbon dioxide. </p>
<p>Use methane's molar mass to determine how many moles react</p>
<p><mathjax>#169.9cancel("g") * "1 mole"/(16.04cancel("g")) = "10.59 moles"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>This means that this reaction will produce <strong>10.59 moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p>Now for the second reaction</p>
<p><mathjax>#C_3H_(8(g)) + 5O_(2(g)) -> color(red)(3)CO_(2(g)) + 4H_2O_((l))#</mathjax></p>
<p>This time, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between propane and carbon dioxide is actually equal to <mathjax>#1:color(red)(3)#</mathjax>. So, for every mole of propane that reacts, the reaction produces <strong>3 times more moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p><mathjax>#314.1cancel("g") * "1 mole"/(44.10cancel("g")) = "7.12 moles"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>This means that you have</p>
<p><mathjax>#7.12cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole" C_3H_8)) = "21.36 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>The <em>total number of moles</em> of carbon dioxide will be </p>
<p><mathjax>#n_"total" = 10.59 + 21.36 = "31.95 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Now simply use carbon dioxide's molar mass to determine how many grams would contain this many moles</p>
<p><mathjax>#31.95cancel("moles") * "44.01 g"/(1cancel("mole")) = "1406.1 g"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<p><mathjax>#m_(CO_2) = color(green)("1410 g "CO_2)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This reaction will produce <strong>1410 g</strong> of carbon dioxide.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that you've got two combustion reactions taking place at the same time. </p>
<p>More specifically, you're going to have to write two balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, one for the combustion of methane and one for the combustion of propane.</p>
<p>The first thing you need to do is figure out exactly how many grams of each hydrocarbon you have. To do that, use the known <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of methane and propane in the mixture.</p>
<p><mathjax>#484cancel("g mixture") * ("35.1 g"CH_4)/(100cancel("g mixture")) = "169.9 g"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>and</p>
<p><mathjax>#484cancel("g mixture") * ("64.9 g" C_3H_8)/(100cancel("g mixture")) = "314.1 g"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>Check to see if the values add up to give <strong>484 g</strong>.</p>
<p><mathjax>#169.9 + 314.1 = "484 g"#</mathjax> <mathjax>#->#</mathjax> so far, so good.</p>
<p>Now focus on the two combustion reactions. Start with the first one</p>
<p><mathjax>#CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between methane and carbon dioxide. This means that, regardless of how many moles of methane eract, the reaction will produce the same exact number of moles of carbon dioxide. </p>
<p>Use methane's molar mass to determine how many moles react</p>
<p><mathjax>#169.9cancel("g") * "1 mole"/(16.04cancel("g")) = "10.59 moles"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>This means that this reaction will produce <strong>10.59 moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p>Now for the second reaction</p>
<p><mathjax>#C_3H_(8(g)) + 5O_(2(g)) -> color(red)(3)CO_(2(g)) + 4H_2O_((l))#</mathjax></p>
<p>This time, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between propane and carbon dioxide is actually equal to <mathjax>#1:color(red)(3)#</mathjax>. So, for every mole of propane that reacts, the reaction produces <strong>3 times more moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p><mathjax>#314.1cancel("g") * "1 mole"/(44.10cancel("g")) = "7.12 moles"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>This means that you have</p>
<p><mathjax>#7.12cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole" C_3H_8)) = "21.36 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>The <em>total number of moles</em> of carbon dioxide will be </p>
<p><mathjax>#n_"total" = 10.59 + 21.36 = "31.95 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Now simply use carbon dioxide's molar mass to determine how many grams would contain this many moles</p>
<p><mathjax>#31.95cancel("moles") * "44.01 g"/(1cancel("mole")) = "1406.1 g"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<p><mathjax>#m_(CO_2) = color(green)("1410 g "CO_2)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of #CO_2# are produced by the combustion of 484g of a mixture that is 35.1% #CH_4# and 64.9% #C_3H_8# by mass?</h1>
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<div class="markdown"><p>This reaction will produce <strong>1410 g</strong> of carbon dioxide.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that you've got two combustion reactions taking place at the same time. </p>
<p>More specifically, you're going to have to write two balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a>, one for the combustion of methane and one for the combustion of propane.</p>
<p>The first thing you need to do is figure out exactly how many grams of each hydrocarbon you have. To do that, use the known <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of methane and propane in the mixture.</p>
<p><mathjax>#484cancel("g mixture") * ("35.1 g"CH_4)/(100cancel("g mixture")) = "169.9 g"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>and</p>
<p><mathjax>#484cancel("g mixture") * ("64.9 g" C_3H_8)/(100cancel("g mixture")) = "314.1 g"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>Check to see if the values add up to give <strong>484 g</strong>.</p>
<p><mathjax>#169.9 + 314.1 = "484 g"#</mathjax> <mathjax>#->#</mathjax> so far, so good.</p>
<p>Now focus on the two combustion reactions. Start with the first one</p>
<p><mathjax>#CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))#</mathjax></p>
<p>Notice that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between methane and carbon dioxide. This means that, regardless of how many moles of methane eract, the reaction will produce the same exact number of moles of carbon dioxide. </p>
<p>Use methane's molar mass to determine how many moles react</p>
<p><mathjax>#169.9cancel("g") * "1 mole"/(16.04cancel("g")) = "10.59 moles"#</mathjax> <mathjax>#CH_4#</mathjax></p>
<p>This means that this reaction will produce <strong>10.59 moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p>Now for the second reaction</p>
<p><mathjax>#C_3H_(8(g)) + 5O_(2(g)) -> color(red)(3)CO_(2(g)) + 4H_2O_((l))#</mathjax></p>
<p>This time, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between propane and carbon dioxide is actually equal to <mathjax>#1:color(red)(3)#</mathjax>. So, for every mole of propane that reacts, the reaction produces <strong>3 times more moles</strong> of <mathjax>#CO_2#</mathjax>. </p>
<p><mathjax>#314.1cancel("g") * "1 mole"/(44.10cancel("g")) = "7.12 moles"#</mathjax> <mathjax>#C_3H_8#</mathjax></p>
<p>This means that you have</p>
<p><mathjax>#7.12cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole" C_3H_8)) = "21.36 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>The <em>total number of moles</em> of carbon dioxide will be </p>
<p><mathjax>#n_"total" = 10.59 + 21.36 = "31.95 moles"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Now simply use carbon dioxide's molar mass to determine how many grams would contain this many moles</p>
<p><mathjax>#31.95cancel("moles") * "44.01 g"/(1cancel("mole")) = "1406.1 g"#</mathjax> <mathjax>#CO_2#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<p><mathjax>#m_(CO_2) = color(green)("1410 g "CO_2)#</mathjax></p></div>
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</article> | How many grams of #CO_2# are produced by the combustion of 484g of a mixture that is 35.1% #CH_4# and 64.9% #C_3H_8# by mass? | null |
3,072 | aa07cd3b-6ddd-11ea-8024-ccda262736ce | https://socratic.org/questions/george-is-making-spaghetti-for-dinner-he-places-4-01-kg-of-water-in-a-pan-and-br | 100.25 ℃ | start physical_unit 47 50 temperature °c qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 30 31 27 28 mass qc_end end | [{"type":"physical unit","value":"Temperature [OF] the salty boiling water [IN] ℃"}] | [{"type":"physical unit","value":"100.25 ℃"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{4.01 kg}"},{"type":"physical unit","value":"Mass [OF] table salt [=] \\pu{58 g}"}] | <h1 class="questionTitle" itemprop="name">George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and again brings it to a boil. What is the temperature of the salty boiling water?</h1> | null | 100.25 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that adding salt, <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>, to the pure water will <strong>increase</strong> its boiling point. </p>
<p>In order to determine the boiling point of the solution, you need to use the equation for <strong>boiling-point elevation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = i * K_b * b)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_b#</mathjax> - the boiling-point elevation;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_b#</mathjax> - the ebullioscopic constant 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>A quick search online will reveal that the <em>ebullioscopic constant</em> for water is equal to <mathjax>#0.512^@"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>In order to find the value of <mathjax>#DeltaT_b#</mathjax>, which tells you the boiling point elevation of the solution, you need to know the values of the van't Hoff factor and the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<p>Now, sodium chloride is an <strong>electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations and chloride anions</p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>one mole</strong> of sodium chloride produces <strong>two moles8* of ions in solution, </strong>one mole<strong> of sodium cations and </strong>one mole** of chloride anions. </p>
<p>This means that the van't Hoff factor, which tells you the ratio between the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>added</strong> to the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the number of moles of particles <strong>produced</strong> in solution, will be equal to <mathjax>#2#</mathjax>. </p>
<p>The <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution tells you how many moles of solute you get <strong>per kilogram</strong> of solvent. </p>
<p>Use sodium chloride's molar mass to figure out how many moles are present in that sample </p>
<blockquote>
<p><mathjax>#58 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.9925 moles NaCl"#</mathjax></p>
</blockquote>
<p>This means that the molality of the solution will be </p>
<blockquote>
<p><mathjax>#b = "0.9925 moles"/"4.01 kg" = "0.2475 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in these values into the equation for boiling-point elevation and solve for <mathjax>#DeltaT_b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_b = 2 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2475 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 0.2534^@"C"#</mathjax></p>
</blockquote>
<p>Now, <em>boiling-point elevation</em> is calculated as</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = T_b - T_b^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_b#</mathjax> - the boiling point of the solution<br/>
<mathjax>#T_b^@#</mathjax> - the boiling point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em></p>
<p>Pure water boils at <mathjax>#100^@"C"#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#T_b = 100^@"C" + 0.2534^@"C" = color(green)(100.25^@"C")#</mathjax></p>
</blockquote>
<p>I won't round the answer off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have two sig figs given for the mass of sodium chloride. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#100.25^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that adding salt, <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>, to the pure water will <strong>increase</strong> its boiling point. </p>
<p>In order to determine the boiling point of the solution, you need to use the equation for <strong>boiling-point elevation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = i * K_b * b)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_b#</mathjax> - the boiling-point elevation;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_b#</mathjax> - the ebullioscopic constant 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>A quick search online will reveal that the <em>ebullioscopic constant</em> for water is equal to <mathjax>#0.512^@"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>In order to find the value of <mathjax>#DeltaT_b#</mathjax>, which tells you the boiling point elevation of the solution, you need to know the values of the van't Hoff factor and the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<p>Now, sodium chloride is an <strong>electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations and chloride anions</p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>one mole</strong> of sodium chloride produces <strong>two moles8* of ions in solution, </strong>one mole<strong> of sodium cations and </strong>one mole** of chloride anions. </p>
<p>This means that the van't Hoff factor, which tells you the ratio between the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>added</strong> to the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the number of moles of particles <strong>produced</strong> in solution, will be equal to <mathjax>#2#</mathjax>. </p>
<p>The <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution tells you how many moles of solute you get <strong>per kilogram</strong> of solvent. </p>
<p>Use sodium chloride's molar mass to figure out how many moles are present in that sample </p>
<blockquote>
<p><mathjax>#58 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.9925 moles NaCl"#</mathjax></p>
</blockquote>
<p>This means that the molality of the solution will be </p>
<blockquote>
<p><mathjax>#b = "0.9925 moles"/"4.01 kg" = "0.2475 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in these values into the equation for boiling-point elevation and solve for <mathjax>#DeltaT_b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_b = 2 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2475 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 0.2534^@"C"#</mathjax></p>
</blockquote>
<p>Now, <em>boiling-point elevation</em> is calculated as</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = T_b - T_b^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_b#</mathjax> - the boiling point of the solution<br/>
<mathjax>#T_b^@#</mathjax> - the boiling point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em></p>
<p>Pure water boils at <mathjax>#100^@"C"#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#T_b = 100^@"C" + 0.2534^@"C" = color(green)(100.25^@"C")#</mathjax></p>
</blockquote>
<p>I won't round the answer off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have two sig figs given for the mass of sodium chloride. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and again brings it to a boil. What is the temperature of the salty boiling water?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-02-05T00:02:08" itemprop="dateCreated">
Feb 5, 2016
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<div class="markdown"><p><mathjax>#100.25^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that adding salt, <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>, to the pure water will <strong>increase</strong> its boiling point. </p>
<p>In order to determine the boiling point of the solution, you need to use the equation for <strong>boiling-point elevation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = i * K_b * b)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#DeltaT_b#</mathjax> - the boiling-point elevation;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_b#</mathjax> - the ebullioscopic constant 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>A quick search online will reveal that the <em>ebullioscopic constant</em> for water is equal to <mathjax>#0.512^@"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>In order to find the value of <mathjax>#DeltaT_b#</mathjax>, which tells you the boiling point elevation of the solution, you need to know the values of the van't Hoff factor and the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution.</p>
<p>Now, sodium chloride is an <strong>electrolyte</strong>, which means that it dissociates completely in aqueous solution to form sodium cations and chloride anions</p>
<blockquote>
<p><mathjax>#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>one mole</strong> of sodium chloride produces <strong>two moles8* of ions in solution, </strong>one mole<strong> of sodium cations and </strong>one mole** of chloride anions. </p>
<p>This means that the van't Hoff factor, which tells you the ratio between the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>added</strong> to the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> and the number of moles of particles <strong>produced</strong> in solution, will be equal to <mathjax>#2#</mathjax>. </p>
<p>The <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution tells you how many moles of solute you get <strong>per kilogram</strong> of solvent. </p>
<p>Use sodium chloride's molar mass to figure out how many moles are present in that sample </p>
<blockquote>
<p><mathjax>#58 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.9925 moles NaCl"#</mathjax></p>
</blockquote>
<p>This means that the molality of the solution will be </p>
<blockquote>
<p><mathjax>#b = "0.9925 moles"/"4.01 kg" = "0.2475 mol kg"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in these values into the equation for boiling-point elevation and solve for <mathjax>#DeltaT_b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_b = 2 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2475 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 0.2534^@"C"#</mathjax></p>
</blockquote>
<p>Now, <em>boiling-point elevation</em> is calculated as</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaT_b = T_b - T_b^@)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_b#</mathjax> - the boiling point of the solution<br/>
<mathjax>#T_b^@#</mathjax> - the boiling point of the <em>pure <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em></p>
<p>Pure water boils at <mathjax>#100^@"C"#</mathjax>, which means that you have </p>
<blockquote>
<p><mathjax>#T_b = 100^@"C" + 0.2534^@"C" = color(green)(100.25^@"C")#</mathjax></p>
</blockquote>
<p>I won't round the answer off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have two sig figs given for the mass of sodium chloride. </p></div>
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</article> | George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and again brings it to a boil. What is the temperature of the salty boiling water? | null |
3,073 | ac12f93a-6ddd-11ea-8545-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-chromium-iii-oxide | Cr2O3 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] chromium(III) oxide [IN] default"}] | [{"type":"chemical equation","value":"Cr2O3"}] | [{"type":"substance name","value":"Chromium(III) oxide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for chromium(III) oxide? </h1> | null | Cr2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to notice here is that you're dealing with an <strong>ionic compound</strong> that contains <em>chromium</em>, <mathjax>#"Cr"#</mathjax>, a <strong>transition metal</strong>, and oxygen, <mathjax>#"O"#</mathjax>, a <strong>non-metal</strong>. </p>
<p>The interesting thing about this compound is that it contains a <em>transition metal</em> as its <strong>cation</strong>, i.e. its positively charged ion. As you know, the name of a transition metal cation is written using <strong>Roman numerals</strong>.</p>
<p>This is done to distinguish between the possible oxidation states that transition metals can exhibit. In this case, the name of the compound contains the <strong>(III)</strong> Roman numeral, which implies that the metal has a <mathjax>#+3#</mathjax> oxidation state. </p>
<p>In other words, the cation carries a <mathjax>#color(blue)(3+)#</mathjax> positive charge</p>
<blockquote>
<blockquote>
<p><mathjax>#"Cr"^color(blue)((3+)) ->#</mathjax> <em>the <strong>chromium(III) cation</strong></em></p>
</blockquote>
</blockquote>
<p>The <strong>anion</strong>, i.e. the negatively charged ion, is written using the <strong>-ide</strong> prefix. In this case, you're dealing with the <em>oxide anion</em>, <mathjax>#"O"^color(red)(2-)#</mathjax>.</p>
<p>The anion carries a <mathjax>#color(red)(2-)#</mathjax> charge because oxygen is located in group 16 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and thus requires two more electrons in its outermost shell to complete its <em>octet</em>. </p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^color(red)(2-) ->#</mathjax> <em>the <strong>oxide anion</strong></em></p>
</blockquote>
</blockquote>
<p>Now, an ionic compound must be electrically <strong>neutral</strong>, which implies that the <strong>total positive charge</strong> coming from the cation must be balanced by the <strong>total negative charge</strong> coming from the anion. </p>
<p>In this case, you need <mathjax>#color(red)(2)#</mathjax> chromium(III) cations and <mathjax>#color(blue)(3)#</mathjax> oxide anions in order to have a neutral compound.</p>
<blockquote>
<p><mathjax>#color(red)(2) xx ["Cr"^color(blue)(3+)] + color(blue)(3) xx ["O"^color(red)(2-)] -> "Cr"_ color(red)(2)"O"_ color(blue)(3)#</mathjax></p>
</blockquote>
<p>The chemical formula for <em>chromium(III) oxide</em> will thus be</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Cr"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Cr"_2"O"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to notice here is that you're dealing with an <strong>ionic compound</strong> that contains <em>chromium</em>, <mathjax>#"Cr"#</mathjax>, a <strong>transition metal</strong>, and oxygen, <mathjax>#"O"#</mathjax>, a <strong>non-metal</strong>. </p>
<p>The interesting thing about this compound is that it contains a <em>transition metal</em> as its <strong>cation</strong>, i.e. its positively charged ion. As you know, the name of a transition metal cation is written using <strong>Roman numerals</strong>.</p>
<p>This is done to distinguish between the possible oxidation states that transition metals can exhibit. In this case, the name of the compound contains the <strong>(III)</strong> Roman numeral, which implies that the metal has a <mathjax>#+3#</mathjax> oxidation state. </p>
<p>In other words, the cation carries a <mathjax>#color(blue)(3+)#</mathjax> positive charge</p>
<blockquote>
<blockquote>
<p><mathjax>#"Cr"^color(blue)((3+)) ->#</mathjax> <em>the <strong>chromium(III) cation</strong></em></p>
</blockquote>
</blockquote>
<p>The <strong>anion</strong>, i.e. the negatively charged ion, is written using the <strong>-ide</strong> prefix. In this case, you're dealing with the <em>oxide anion</em>, <mathjax>#"O"^color(red)(2-)#</mathjax>.</p>
<p>The anion carries a <mathjax>#color(red)(2-)#</mathjax> charge because oxygen is located in group 16 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and thus requires two more electrons in its outermost shell to complete its <em>octet</em>. </p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^color(red)(2-) ->#</mathjax> <em>the <strong>oxide anion</strong></em></p>
</blockquote>
</blockquote>
<p>Now, an ionic compound must be electrically <strong>neutral</strong>, which implies that the <strong>total positive charge</strong> coming from the cation must be balanced by the <strong>total negative charge</strong> coming from the anion. </p>
<p>In this case, you need <mathjax>#color(red)(2)#</mathjax> chromium(III) cations and <mathjax>#color(blue)(3)#</mathjax> oxide anions in order to have a neutral compound.</p>
<blockquote>
<p><mathjax>#color(red)(2) xx ["Cr"^color(blue)(3+)] + color(blue)(3) xx ["O"^color(red)(2-)] -> "Cr"_ color(red)(2)"O"_ color(blue)(3)#</mathjax></p>
</blockquote>
<p>The chemical formula for <em>chromium(III) oxide</em> will thus be</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Cr"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for chromium(III) oxide? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-26T01:12:46" itemprop="dateCreated">
Jul 26, 2016
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<div class="markdown"><p><mathjax>#"Cr"_2"O"_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to notice here is that you're dealing with an <strong>ionic compound</strong> that contains <em>chromium</em>, <mathjax>#"Cr"#</mathjax>, a <strong>transition metal</strong>, and oxygen, <mathjax>#"O"#</mathjax>, a <strong>non-metal</strong>. </p>
<p>The interesting thing about this compound is that it contains a <em>transition metal</em> as its <strong>cation</strong>, i.e. its positively charged ion. As you know, the name of a transition metal cation is written using <strong>Roman numerals</strong>.</p>
<p>This is done to distinguish between the possible oxidation states that transition metals can exhibit. In this case, the name of the compound contains the <strong>(III)</strong> Roman numeral, which implies that the metal has a <mathjax>#+3#</mathjax> oxidation state. </p>
<p>In other words, the cation carries a <mathjax>#color(blue)(3+)#</mathjax> positive charge</p>
<blockquote>
<blockquote>
<p><mathjax>#"Cr"^color(blue)((3+)) ->#</mathjax> <em>the <strong>chromium(III) cation</strong></em></p>
</blockquote>
</blockquote>
<p>The <strong>anion</strong>, i.e. the negatively charged ion, is written using the <strong>-ide</strong> prefix. In this case, you're dealing with the <em>oxide anion</em>, <mathjax>#"O"^color(red)(2-)#</mathjax>.</p>
<p>The anion carries a <mathjax>#color(red)(2-)#</mathjax> charge because oxygen is located in group 16 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, and thus requires two more electrons in its outermost shell to complete its <em>octet</em>. </p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^color(red)(2-) ->#</mathjax> <em>the <strong>oxide anion</strong></em></p>
</blockquote>
</blockquote>
<p>Now, an ionic compound must be electrically <strong>neutral</strong>, which implies that the <strong>total positive charge</strong> coming from the cation must be balanced by the <strong>total negative charge</strong> coming from the anion. </p>
<p>In this case, you need <mathjax>#color(red)(2)#</mathjax> chromium(III) cations and <mathjax>#color(blue)(3)#</mathjax> oxide anions in order to have a neutral compound.</p>
<blockquote>
<p><mathjax>#color(red)(2) xx ["Cr"^color(blue)(3+)] + color(blue)(3) xx ["O"^color(red)(2-)] -> "Cr"_ color(red)(2)"O"_ color(blue)(3)#</mathjax></p>
</blockquote>
<p>The chemical formula for <em>chromium(III) oxide</em> will thus be</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Cr"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | What is the chemical formula for chromium(III) oxide? | null |
3,074 | ac6e6892-6ddd-11ea-9e9a-ccda262736ce | https://socratic.org/questions/how-do-you-complete-and-balance-znco-3-s-h-2so-4-aq | ZnCO3(s) + H2SO4(aq) -> ZnSO4(aq) + CO2(g) ^ + H2O(l) | start chemical_equation qc_end chemical_equation 8 8 qc_end chemical_equation 6 6 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"ZnCO3(s) + H2SO4(aq) -> ZnSO4(aq) + CO2(g) ^ + H2O(l)"}] | [{"type":"chemical equation","value":"H2SO4(aq)"},{"type":"chemical equation","value":"ZnCO3(s)"}] | <h1 class="questionTitle" itemprop="name">How do you complete and balance #ZnCO_3(s)+H_2SO_4(aq) ->#?</h1> | null | ZnCO3(s) + H2SO4(aq) -> ZnSO4(aq) + CO2(g) ^ + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Any carbonate will decompose on addition of strong acid to give carbon dioxide, metal salt, and water.</p>
<p>Is the equation above balanced? It MUST be!</p></div>
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<div class="markdown"><p><mathjax>#ZnCO_3(s) + H_2SO_4(aq) rarr ZnSO_4(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Any carbonate will decompose on addition of strong acid to give carbon dioxide, metal salt, and water.</p>
<p>Is the equation above balanced? It MUST be!</p></div>
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<div class="markdown"><p><mathjax>#ZnCO_3(s) + H_2SO_4(aq) rarr ZnSO_4(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Any carbonate will decompose on addition of strong acid to give carbon dioxide, metal salt, and water.</p>
<p>Is the equation above balanced? It MUST be!</p></div>
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</article> | How do you complete and balance #ZnCO_3(s)+H_2SO_4(aq) ->#? | null |
3,075 | ab30db36-6ddd-11ea-92ce-ccda262736ce | https://socratic.org/questions/oxygen-gas-is-at-a-temperature-of-40-c-when-it-occupies-a-volume-of-2-3-liters-t | 884.56 K | start physical_unit 0 1 temperature k qc_end physical_unit 0 1 7 8 temperature qc_end physical_unit 0 1 15 16 volume qc_end physical_unit 0 1 29 30 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] oxygen gas [IN] K"}] | [{"type":"physical unit","value":"884.56 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] oxygen gas [=] \\pu{40 ℃}"},{"type":"physical unit","value":"Volume1 [OF] oxygen gas [=] \\pu{2.3 liters}"},{"type":"physical unit","value":"Volume2 [OF] oxygen gas [=] \\pu{6.5 liters}"}] | <h1 class="questionTitle" itemprop="name">Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters?</h1> | null | 884.56 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1) Convert 40 °C to Kelvin and you get 273 + 40 = 313 K .At this temperature the volume of the gas is 2.3 L.</p>
<p>2) At Temperature <mathjax>#T_2#</mathjax> Kelvin ,this temperature the volume of the gas is 6.5 L.</p>
<p>Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.</p>
<p>Using Charles law equation;</p>
<p><mathjax>#V_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#V_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#V_1#</mathjax> = 2.3 L , <mathjax>#T_1#</mathjax> = 313 K</p>
<p><mathjax>#V_2#</mathjax> = 6.5 L , <mathjax>#T_2#</mathjax> = ?</p>
<p>plug in the values;</p>
<p>2.3 L / 313 K = 6.5 L / <mathjax>#T_2#</mathjax></p>
<p>Cross-multiply and divide:</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 6.5 L x 313 K</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 2034.5 L K</p>
<p><mathjax>#T_2#</mathjax> = 2034.5 LK / 2.3 L</p>
<p><mathjax>#T_2#</mathjax> = 884.5 K</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>884.5 K</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1) Convert 40 °C to Kelvin and you get 273 + 40 = 313 K .At this temperature the volume of the gas is 2.3 L.</p>
<p>2) At Temperature <mathjax>#T_2#</mathjax> Kelvin ,this temperature the volume of the gas is 6.5 L.</p>
<p>Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.</p>
<p>Using Charles law equation;</p>
<p><mathjax>#V_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#V_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#V_1#</mathjax> = 2.3 L , <mathjax>#T_1#</mathjax> = 313 K</p>
<p><mathjax>#V_2#</mathjax> = 6.5 L , <mathjax>#T_2#</mathjax> = ?</p>
<p>plug in the values;</p>
<p>2.3 L / 313 K = 6.5 L / <mathjax>#T_2#</mathjax></p>
<p>Cross-multiply and divide:</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 6.5 L x 313 K</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 2034.5 L K</p>
<p><mathjax>#T_2#</mathjax> = 2034.5 LK / 2.3 L</p>
<p><mathjax>#T_2#</mathjax> = 884.5 K</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters?</h1>
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Jun 27, 2014
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<div class="markdown"><p>884.5 K</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1) Convert 40 °C to Kelvin and you get 273 + 40 = 313 K .At this temperature the volume of the gas is 2.3 L.</p>
<p>2) At Temperature <mathjax>#T_2#</mathjax> Kelvin ,this temperature the volume of the gas is 6.5 L.</p>
<p>Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.</p>
<p>Using Charles law equation;</p>
<p><mathjax>#V_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#V_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#V_1#</mathjax> = 2.3 L , <mathjax>#T_1#</mathjax> = 313 K</p>
<p><mathjax>#V_2#</mathjax> = 6.5 L , <mathjax>#T_2#</mathjax> = ?</p>
<p>plug in the values;</p>
<p>2.3 L / 313 K = 6.5 L / <mathjax>#T_2#</mathjax></p>
<p>Cross-multiply and divide:</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 6.5 L x 313 K</p>
<p>2.3 L x (<mathjax>#T_2#</mathjax>) = 2034.5 L K</p>
<p><mathjax>#T_2#</mathjax> = 2034.5 LK / 2.3 L</p>
<p><mathjax>#T_2#</mathjax> = 884.5 K</p></div>
</div>
</div>
</div>
</div>
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</article> | Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters? | null |
3,076 | a84ac006-6ddd-11ea-b7bf-ccda262736ce | https://socratic.org/questions/57a3637111ef6b5bab57cfcf | Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g) | start chemical_equation qc_end substance 3 4 qc_end substance 8 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)"}] | [{"type":"substance name","value":"Zinc metal"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">What occurs when zinc metal is added to hydrochloric acid?</h1> | null | Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The addition of zinc metal, as a powder, would have produced a vigorous effervescence; heat would also have been evolved. The metal powder would have gone up into solution. We can represent this reaction by the balanced chemical equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>We write the equation in order to establish the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>; the mass equivalence of the reactants and product.</p></div>
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<div class="markdown"><p>I don't know. It has been years since I did that experiment.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The addition of zinc metal, as a powder, would have produced a vigorous effervescence; heat would also have been evolved. The metal powder would have gone up into solution. We can represent this reaction by the balanced chemical equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>We write the equation in order to establish the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>; the mass equivalence of the reactants and product.</p></div>
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<h1 class="questionTitle" itemprop="name">What occurs when zinc metal is added to hydrochloric acid?</h1>
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<div class="markdown"><p>I don't know. It has been years since I did that experiment.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The addition of zinc metal, as a powder, would have produced a vigorous effervescence; heat would also have been evolved. The metal powder would have gone up into solution. We can represent this reaction by the balanced chemical equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>We write the equation in order to establish the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>; the mass equivalence of the reactants and product.</p></div>
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</article> | What occurs when zinc metal is added to hydrochloric acid? | null |
3,077 | a888025a-6ddd-11ea-b306-ccda262736ce | https://socratic.org/questions/how-would-you-complete-and-balance-the-acid-base-neutralizations-reaction-hbr-ba | 2 HBr + Ba(OH)2 -> BaBr2 + 2 H2O | start chemical_equation qc_end chemical_equation 11 15 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the acid base neutralizations reaction"}] | [{"type":"chemical equation","value":"2 HBr + Ba(OH)2 -> BaBr2 + 2 H2O"}] | [{"type":"chemical equation","value":"HBr + Ba(OH)2 -> ?"}] | <h1 class="questionTitle" itemprop="name">How would you complete and balance the acid base neutralizations reaction: #HBr + Ba(OH)_2 ->#?</h1> | null | 2 HBr + Ba(OH)2 -> BaBr2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction, if the base is a hydroxide, then the products formed would be the salt and water.</p>
<blockquote>
<p>acid + base <mathjax>#->#</mathjax> salt + water</p>
</blockquote>
<p>The water (<mathjax>#"H"_2"O"#</mathjax>) is a result of the combination of the <mathjax>#"H"^+#</mathjax> from the acid and the <mathjax>#"OH"^-#</mathjax> from the base.</p>
<p>The salt (<mathjax>#"BaBr"_2#</mathjax>) is simply the remaining anion of the acid (<mathjax>#"Br"^-#</mathjax>) and the cation of the base (<mathjax>#"Ba"^{2+}#</mathjax>). </p>
<p>The <em>unbalanced</em> equation is therefore</p>
<blockquote>
<p><mathjax>#"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>To balance the equation, first notice that the number of Ba on both sides are the same. Ba is balanced.</p>
<p>Next, notice that the LHS is short of 1 x Br. So balance it by adding 1 unit of HBr.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, the RHS is missing 2 x H and 1 x O. Balance that by adding 1 units of <mathjax>#"H"_2"O"#</mathjax>.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2"H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, all the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have equal amounts on both sides of the equation. The chemical equation is now considered balanced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2 "H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction, if the base is a hydroxide, then the products formed would be the salt and water.</p>
<blockquote>
<p>acid + base <mathjax>#->#</mathjax> salt + water</p>
</blockquote>
<p>The water (<mathjax>#"H"_2"O"#</mathjax>) is a result of the combination of the <mathjax>#"H"^+#</mathjax> from the acid and the <mathjax>#"OH"^-#</mathjax> from the base.</p>
<p>The salt (<mathjax>#"BaBr"_2#</mathjax>) is simply the remaining anion of the acid (<mathjax>#"Br"^-#</mathjax>) and the cation of the base (<mathjax>#"Ba"^{2+}#</mathjax>). </p>
<p>The <em>unbalanced</em> equation is therefore</p>
<blockquote>
<p><mathjax>#"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>To balance the equation, first notice that the number of Ba on both sides are the same. Ba is balanced.</p>
<p>Next, notice that the LHS is short of 1 x Br. So balance it by adding 1 unit of HBr.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, the RHS is missing 2 x H and 1 x O. Balance that by adding 1 units of <mathjax>#"H"_2"O"#</mathjax>.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2"H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, all the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have equal amounts on both sides of the equation. The chemical equation is now considered balanced.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you complete and balance the acid base neutralizations reaction: #HBr + Ba(OH)_2 ->#?</h1>
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<div class="markdown"><p><mathjax>#2"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2 "H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction, if the base is a hydroxide, then the products formed would be the salt and water.</p>
<blockquote>
<p>acid + base <mathjax>#->#</mathjax> salt + water</p>
</blockquote>
<p>The water (<mathjax>#"H"_2"O"#</mathjax>) is a result of the combination of the <mathjax>#"H"^+#</mathjax> from the acid and the <mathjax>#"OH"^-#</mathjax> from the base.</p>
<p>The salt (<mathjax>#"BaBr"_2#</mathjax>) is simply the remaining anion of the acid (<mathjax>#"Br"^-#</mathjax>) and the cation of the base (<mathjax>#"Ba"^{2+}#</mathjax>). </p>
<p>The <em>unbalanced</em> equation is therefore</p>
<blockquote>
<p><mathjax>#"HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>To balance the equation, first notice that the number of Ba on both sides are the same. Ba is balanced.</p>
<p>Next, notice that the LHS is short of 1 x Br. So balance it by adding 1 unit of HBr.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + "H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, the RHS is missing 2 x H and 1 x O. Balance that by adding 1 units of <mathjax>#"H"_2"O"#</mathjax>.</p>
<blockquote>
<p><mathjax>#"2HBr" + "Ba(OH)"_2 -> "BaBr"_2 + 2"H"_2"O"#</mathjax></p>
</blockquote>
<p>Now, all the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have equal amounts on both sides of the equation. The chemical equation is now considered balanced.</p></div>
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</article> | How would you complete and balance the acid base neutralizations reaction: #HBr + Ba(OH)_2 ->#? | null |
3,078 | a907ebd2-6ddd-11ea-ba3c-ccda262736ce | https://socratic.org/questions/58d851fc11ef6b3bb4d52051 | 1.24 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 9 10 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] water [IN] moles"}] | [{"type":"physical unit","value":"1.24 moles"}] | [{"type":"physical unit","value":"Mass [OF] B2H6 [=] \\pu{17.2 g}"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion?</h1> | null | 1.24 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p><mathjax>#1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion?</h1>
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<div class="markdown"><p><mathjax>#1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p>The complete combustion of <mathjax>#"17.2 g B"_2"H"_6"#</mathjax> produces <mathjax>#"2.01 mol H"_2"O"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"B"_2"H"_6("g") + "3O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"B"_2"O"_3("s") + "3H"_2"O(g)"#</mathjax></p>
<p>We first need to determine mol <mathjax>#"B"_2"H"_6"#</mathjax> by dividing its given mass by its molar mass <mathjax>#("25.652 g/mol")#</mathjax>. Divide by multiplying by the reciprocal of its molar mass (mol/g).</p>
<p><mathjax>#17.2color(red)cancel(color(black)("g B"_2"H"_6))xx(1"mol B"_2"H"_6)/(25.652color(red)cancel(color(black)("g B"_2"H"_6)))="0.671 mol B"_2"H"_6"#</mathjax></p>
<p>To determine mol <mathjax>#"H"_2"O"#</mathjax> produced, multiply mol <mathjax>#"B"_2"H"_6"#</mathjax> by the mol ratio between <mathjax>#"H"_2"O"#</mathjax> and <mathjax>#"B"_2"H"_6"#</mathjax> from the balanced equation, with <mathjax>#"H"_2"O"#</mathjax> in the numerator.</p>
<p><mathjax>#0.671color(red)cancel(color(black)("mol B"_2"H"_6))xx(3"mol H"_2"O")/(1color(red)cancel(color(black)("mol B"_2"H"_6)))="2.01 mol H"_2"O"#</mathjax></p>
<p>The complete combustion of <mathjax>#"17.2 g B"_2"H"_6"#</mathjax> produces <mathjax>#"2.01 mol H"_2"O"#</mathjax>.</p></div>
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</article> | How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion? | null |
3,079 | abcb10ea-6ddd-11ea-8aca-ccda262736ce | https://socratic.org/questions/578212c87c01497b464b70bc | 12.30 | start physical_unit 9 10 ph none qc_end physical_unit 1 1 3 4 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"12.30"}] | [{"type":"physical unit","value":"Molarity [OF] Ca(OH)2(aq) [=] \\pu{0.01 mol/L}"}] | <h1 class="questionTitle" itemprop="name">If #Ca(OH)_2(aq)=0.01*mol*L^-1#, what is #pH# of the solution?</h1> | null | 12.30 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know (or should know) that in aqueous solution <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus, if we calculate <mathjax>#pOH#</mathjax>, <mathjax>#pH#</mathjax> is directly available.</p>
<p><mathjax>#pOH=-log_10([HO^-])#</mathjax> <mathjax>#=-log_10(0.02)#</mathjax> (why <mathjax>#0.02#</mathjax> here and not <mathjax>#0.01#</mathjax>?) <mathjax>#=#</mathjax> <mathjax>#-(-1.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.70#</mathjax></p>
<p>And from the above, <mathjax>#pH=14-1.70#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH~=12#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know (or should know) that in aqueous solution <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus, if we calculate <mathjax>#pOH#</mathjax>, <mathjax>#pH#</mathjax> is directly available.</p>
<p><mathjax>#pOH=-log_10([HO^-])#</mathjax> <mathjax>#=-log_10(0.02)#</mathjax> (why <mathjax>#0.02#</mathjax> here and not <mathjax>#0.01#</mathjax>?) <mathjax>#=#</mathjax> <mathjax>#-(-1.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.70#</mathjax></p>
<p>And from the above, <mathjax>#pH=14-1.70#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If #Ca(OH)_2(aq)=0.01*mol*L^-1#, what is #pH# of the solution?</h1>
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<div class="markdown"><p><mathjax>#pH~=12#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We know (or should know) that in aqueous solution <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus, if we calculate <mathjax>#pOH#</mathjax>, <mathjax>#pH#</mathjax> is directly available.</p>
<p><mathjax>#pOH=-log_10([HO^-])#</mathjax> <mathjax>#=-log_10(0.02)#</mathjax> (why <mathjax>#0.02#</mathjax> here and not <mathjax>#0.01#</mathjax>?) <mathjax>#=#</mathjax> <mathjax>#-(-1.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.70#</mathjax></p>
<p>And from the above, <mathjax>#pH=14-1.70#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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</article> | If #Ca(OH)_2(aq)=0.01*mol*L^-1#, what is #pH# of the solution? | null |
3,080 | a9eb927a-6ddd-11ea-9c78-ccda262736ce | https://socratic.org/questions/58668e0411ef6b666730610c | 11.10 moles | start physical_unit 2 2 mole mol qc_end physical_unit 12 13 8 9 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] dioxygen [IN] moles"}] | [{"type":"physical unit","value":"11.10 moles"}] | [{"type":"physical unit","value":"Mole [OF] aluminum metal [=] \\pu{4 mol}"}] | <h1 class="questionTitle" itemprop="name">How much dioxygen is required to oxidize a #4*mol# quantity of aluminum metal?</h1> | null | 11.10 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'm going to assume that the reaction you are referring to is</p>
<p><mathjax>#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#</mathjax></p>
<p>In which case, the ratio of <mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4#</mathjax></p>
<p>Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with <mathjax>#x#</mathjax> representing the unknown amount of <mathjax>#O_2#</mathjax>:</p>
<p><mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4 = x/14.8#</mathjax></p>
<p>Solve for <mathjax>#x#</mathjax></p>
<p><mathjax>#x=(3/4) (14.8) = 11.1 " moles of "O_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If the reaction involved is this one</p>
<p><mathjax>#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#</mathjax></p>
<p>the answer is 11.1 moles of <mathjax>#O_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'm going to assume that the reaction you are referring to is</p>
<p><mathjax>#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#</mathjax></p>
<p>In which case, the ratio of <mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4#</mathjax></p>
<p>Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with <mathjax>#x#</mathjax> representing the unknown amount of <mathjax>#O_2#</mathjax>:</p>
<p><mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4 = x/14.8#</mathjax></p>
<p>Solve for <mathjax>#x#</mathjax></p>
<p><mathjax>#x=(3/4) (14.8) = 11.1 " moles of "O_2#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How much dioxygen is required to oxidize a #4*mol# quantity of aluminum metal?</h1>
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Dwight
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<div class="markdown"><p>If the reaction involved is this one</p>
<p><mathjax>#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#</mathjax></p>
<p>the answer is 11.1 moles of <mathjax>#O_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'm going to assume that the reaction you are referring to is</p>
<p><mathjax>#4Al(s) + 3O_2(g) rarr 2 Al_2O_3(s)#</mathjax></p>
<p>In which case, the ratio of <mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4#</mathjax></p>
<p>Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with <mathjax>#x#</mathjax> representing the unknown amount of <mathjax>#O_2#</mathjax>:</p>
<p><mathjax>#("moles of "O_2 " required")/("moles of "Al " used") = 3/4 = x/14.8#</mathjax></p>
<p>Solve for <mathjax>#x#</mathjax></p>
<p><mathjax>#x=(3/4) (14.8) = 11.1 " moles of "O_2#</mathjax></p></div>
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anor277
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Dec 30, 2016
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<div class="markdown"><p>We need <mathjax>#11.1#</mathjax> moles of <mathjax>#O_2(g)#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A stoichiometric equation is required:</p>
<p><mathjax>#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#</mathjax></p>
<p>Given the equation, for each equiv metal <mathjax>#3/4#</mathjax> equiv dioxygen gas are required, i.e. <mathjax>#14.8*molxx3/4=11.1*"mol"#</mathjax> <mathjax>#O_2#</mathjax> gas.</p>
<p>Given standard condtions, how many litres of gas does this represent?</p></div>
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</article> | How much dioxygen is required to oxidize a #4*mol# quantity of aluminum metal? | null |
3,081 | aa7e1293-6ddd-11ea-8e64-ccda262736ce | https://socratic.org/questions/a-21-liter-cylinder-contains-1-5-moles-of-an-ideal-gas-at-311-k-what-is-the-pres | 1.82 atm | start physical_unit 19 20 pressure atm qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 9 10 5 6 mole qc_end physical_unit 9 10 12 13 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"1.82 atm"}] | [{"type":"physical unit","value":"Volume [OF] cylinder [=] \\pu{21 liter}"},{"type":"physical unit","value":"Mole [OF] the ideal gas [=] \\pu{1.5 moles}"},{"type":"physical unit","value":"Temperature [OF] the ideal gas [=] \\pu{311 K}"}] | <h1 class="questionTitle" itemprop="name">A 21-liter cylinder contains 1.5 moles of an ideal gas at 311 K. What is the pressure of the gas?</h1> | null | 1.82 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem.</p>
<p>As you know, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821 ("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, <strong>always</strong> expressed in <em>Kelvin</em></p>
<p>So, the problem provides you with </p>
<blockquote>
<ul>
<li><em>the <strong>volume</strong> of the gas</em>, <mathjax>#V = "21 L"#</mathjax></li>
<li><em>the <strong>number of moles</strong> of gas</em>, <mathjax>#n = 1.5#</mathjax></li>
<li><em>the <strong>absolute temperature</strong> of the gas</em>, <mathjax>#T = "311 K"#</mathjax></li>
</ul>
</blockquote>
<p>Since you know the value of <mathjax>#R#</mathjax>, you can use this information to find the pressure <mathjax>#P#</mathjax>. First, rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax> on one side</p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Next, make sure that the <strong>units</strong> given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<blockquote>
<blockquote>
<p><mathjax># {:(color(red)("Need"), color(white)(aaaaa), color(blue)("Have")), (color(white)(aaaa), color(white)(aaaa), color(white)(aaaa)), (color(white)(aa)"L", color(white)(aaaa), color(white)(aa)"L"color(white)(aaaaaa)color(green)(sqrt())), ("moles", color(white)(aaaa), "moles"color(white)(aaaaa)color(green)(sqrt())), (color(white)(aa)"K", color(white)(aaaa), color(white)(aa)"K"color(white)(aaaaaa)color(green)(sqrt())) :}#</mathjax></p>
</blockquote>
</blockquote>
<p>The units given to you match those used for <mathjax>#R#</mathjax>, so you're good to go. Plug in your values to get</p>
<blockquote>
<p><mathjax>#P = (1.5 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 311color(red)(cancel(color(black)("K"))))/(21color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#P = "1.824 atm"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume and number of moles of gas, the answer will be </p>
<blockquote>
<p><mathjax>#P = color(green)("1.8 atm")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P = "1.8 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem.</p>
<p>As you know, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821 ("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, <strong>always</strong> expressed in <em>Kelvin</em></p>
<p>So, the problem provides you with </p>
<blockquote>
<ul>
<li><em>the <strong>volume</strong> of the gas</em>, <mathjax>#V = "21 L"#</mathjax></li>
<li><em>the <strong>number of moles</strong> of gas</em>, <mathjax>#n = 1.5#</mathjax></li>
<li><em>the <strong>absolute temperature</strong> of the gas</em>, <mathjax>#T = "311 K"#</mathjax></li>
</ul>
</blockquote>
<p>Since you know the value of <mathjax>#R#</mathjax>, you can use this information to find the pressure <mathjax>#P#</mathjax>. First, rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax> on one side</p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Next, make sure that the <strong>units</strong> given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<blockquote>
<blockquote>
<p><mathjax># {:(color(red)("Need"), color(white)(aaaaa), color(blue)("Have")), (color(white)(aaaa), color(white)(aaaa), color(white)(aaaa)), (color(white)(aa)"L", color(white)(aaaa), color(white)(aa)"L"color(white)(aaaaaa)color(green)(sqrt())), ("moles", color(white)(aaaa), "moles"color(white)(aaaaa)color(green)(sqrt())), (color(white)(aa)"K", color(white)(aaaa), color(white)(aa)"K"color(white)(aaaaaa)color(green)(sqrt())) :}#</mathjax></p>
</blockquote>
</blockquote>
<p>The units given to you match those used for <mathjax>#R#</mathjax>, so you're good to go. Plug in your values to get</p>
<blockquote>
<p><mathjax>#P = (1.5 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 311color(red)(cancel(color(black)("K"))))/(21color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#P = "1.824 atm"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume and number of moles of gas, the answer will be </p>
<blockquote>
<p><mathjax>#P = color(green)("1.8 atm")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 21-liter cylinder contains 1.5 moles of an ideal gas at 311 K. What is the pressure of the gas?</h1>
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Jan 23, 2016
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<div class="markdown"><p><mathjax>#P = "1.8 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation practice problem.</p>
<p>As you know, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821 ("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, <strong>always</strong> expressed in <em>Kelvin</em></p>
<p>So, the problem provides you with </p>
<blockquote>
<ul>
<li><em>the <strong>volume</strong> of the gas</em>, <mathjax>#V = "21 L"#</mathjax></li>
<li><em>the <strong>number of moles</strong> of gas</em>, <mathjax>#n = 1.5#</mathjax></li>
<li><em>the <strong>absolute temperature</strong> of the gas</em>, <mathjax>#T = "311 K"#</mathjax></li>
</ul>
</blockquote>
<p>Since you know the value of <mathjax>#R#</mathjax>, you can use this information to find the pressure <mathjax>#P#</mathjax>. First, rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax> on one side</p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Next, make sure that the <strong>units</strong> given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<blockquote>
<blockquote>
<p><mathjax># {:(color(red)("Need"), color(white)(aaaaa), color(blue)("Have")), (color(white)(aaaa), color(white)(aaaa), color(white)(aaaa)), (color(white)(aa)"L", color(white)(aaaa), color(white)(aa)"L"color(white)(aaaaaa)color(green)(sqrt())), ("moles", color(white)(aaaa), "moles"color(white)(aaaaa)color(green)(sqrt())), (color(white)(aa)"K", color(white)(aaaa), color(white)(aa)"K"color(white)(aaaaaa)color(green)(sqrt())) :}#</mathjax></p>
</blockquote>
</blockquote>
<p>The units given to you match those used for <mathjax>#R#</mathjax>, so you're good to go. Plug in your values to get</p>
<blockquote>
<p><mathjax>#P = (1.5 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 311color(red)(cancel(color(black)("K"))))/(21color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#P = "1.824 atm"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume and number of moles of gas, the answer will be </p>
<blockquote>
<p><mathjax>#P = color(green)("1.8 atm")#</mathjax></p>
</blockquote></div>
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</article> | A 21-liter cylinder contains 1.5 moles of an ideal gas at 311 K. What is the pressure of the gas? | null |
3,082 | acbe1ad0-6ddd-11ea-bd40-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-if-you-dissolve-15-g-of-fecl-3-to-make-up-250-ml-solution | 0.37 M | start physical_unit 10 10 molarity mol/l qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 16 16 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] FeCl3 solution [IN] M"}] | [{"type":"physical unit","value":"0.37 M"}] | [{"type":"physical unit","value":"Mass [OF] FeCl3 [=] \\pu{15 g}"},{"type":"physical unit","value":"Volume [OF] FeCl3 solution [=] \\pu{250 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity if you dissolve 15 g of #FeCl_3# to make up 250 mL solution?</h1> | null | 0.37 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the molar mass of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#55.85+3*35.45=162.20 g//mol#</mathjax></p>
<p>Then calculate how many moles are 15 g of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#color(red)(n=m/M#</mathjax></p>
<p><mathjax>#n=15/162.20=0.0925 mol#</mathjax></p>
<p>Then calculate the concentration:</p>
<p><mathjax>#color(red)(c=n/v#</mathjax></p>
<p><mathjax>#n=0.0925mol//0.250L=0.37 mol//L (or 0.37 M)#</mathjax></p>
<p>The concentration is given with two representative digits which are the same ones of 15 grams.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>0.37 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the molar mass of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#55.85+3*35.45=162.20 g//mol#</mathjax></p>
<p>Then calculate how many moles are 15 g of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#color(red)(n=m/M#</mathjax></p>
<p><mathjax>#n=15/162.20=0.0925 mol#</mathjax></p>
<p>Then calculate the concentration:</p>
<p><mathjax>#color(red)(c=n/v#</mathjax></p>
<p><mathjax>#n=0.0925mol//0.250L=0.37 mol//L (or 0.37 M)#</mathjax></p>
<p>The concentration is given with two representative digits which are the same ones of 15 grams.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity if you dissolve 15 g of #FeCl_3# to make up 250 mL solution?</h1>
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José F.
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<div class="markdown"><p>0.37 M</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the molar mass of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#55.85+3*35.45=162.20 g//mol#</mathjax></p>
<p>Then calculate how many moles are 15 g of <mathjax>#FeCl_3#</mathjax>:</p>
<p><mathjax>#color(red)(n=m/M#</mathjax></p>
<p><mathjax>#n=15/162.20=0.0925 mol#</mathjax></p>
<p>Then calculate the concentration:</p>
<p><mathjax>#color(red)(c=n/v#</mathjax></p>
<p><mathjax>#n=0.0925mol//0.250L=0.37 mol//L (or 0.37 M)#</mathjax></p>
<p>The concentration is given with two representative digits which are the same ones of 15 grams.</p></div>
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</article> | What is the molarity if you dissolve 15 g of #FeCl_3# to make up 250 mL solution? | null |
3,083 | ab026562-6ddd-11ea-b6cc-ccda262736ce | https://socratic.org/questions/the-balanced-equation-shows-the-reaction-between-aluminium-hydroxide-and-hydroch | 78.00 g | start physical_unit 7 8 mass g qc_end chemical_equation 18 26 qc_end physical_unit 14 15 27 28 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminium hydroxide [IN] g"}] | [{"type":"physical unit","value":"78.00 g"}] | [{"type":"chemical equation","value":"Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O"},{"type":"physical unit","value":"Mole [OF] aluminium chloride [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">The balanced equation shows the reaction between aluminium hydroxide and hydrochloric acid to form aluminium chloride and water: Al(OH)3 + 3HCl = AlCl3 + 3H2O. 1 mole of aluminium chloride was formed, what mass of aluminium hydroxide was formed?</h1> | null | 78.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'll assume you mean what mass of aluminum hydroxide <strong><em>reacted</em></strong>.</p>
<p>We can use the coefficients of the equation to determine the number of <em>moles</em> of <mathjax>#"Al(OH)"_3#</mathjax> that react, knowing that <mathjax>#1#</mathjax> <mathjax>#"mol AlCl"_3#</mathjax> forms:</p>
<p><mathjax>#1cancel("mol AlCl"_3)((1color(white)(l)"mol Al(OH)"_3)/(1cancel("mol AlCl"_3))) = 1#</mathjax> <mathjax>#"mol Al(OH)"_3#</mathjax></p>
<p>which makes sense given their molar ratio is <mathjax>#1:1#</mathjax>. </p>
<p>Now, we can use the <em>molar mass</em> of aluminum hydroxide (<mathjax>#78.00#</mathjax> <mathjax>#"g/mol"#</mathjax>) to calculate the number of <em>grams</em> that react:</p>
<p><mathjax>#1cancel("mol Al(OH)"_3)((78.00color(white)(l)"g Al(OH)"_3)/(1cancel("mol Al(OH)"_3))) = color(red)(80#</mathjax> <mathjax>#color(red)("g Al(OH)"_3#</mathjax></p>
<p>which I suppose will round to <mathjax>#1#</mathjax> significant figure, the amount given in the problem..</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#80#</mathjax> <mathjax>#"g Al(OH)"_3#</mathjax> (one sig fig)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'll assume you mean what mass of aluminum hydroxide <strong><em>reacted</em></strong>.</p>
<p>We can use the coefficients of the equation to determine the number of <em>moles</em> of <mathjax>#"Al(OH)"_3#</mathjax> that react, knowing that <mathjax>#1#</mathjax> <mathjax>#"mol AlCl"_3#</mathjax> forms:</p>
<p><mathjax>#1cancel("mol AlCl"_3)((1color(white)(l)"mol Al(OH)"_3)/(1cancel("mol AlCl"_3))) = 1#</mathjax> <mathjax>#"mol Al(OH)"_3#</mathjax></p>
<p>which makes sense given their molar ratio is <mathjax>#1:1#</mathjax>. </p>
<p>Now, we can use the <em>molar mass</em> of aluminum hydroxide (<mathjax>#78.00#</mathjax> <mathjax>#"g/mol"#</mathjax>) to calculate the number of <em>grams</em> that react:</p>
<p><mathjax>#1cancel("mol Al(OH)"_3)((78.00color(white)(l)"g Al(OH)"_3)/(1cancel("mol Al(OH)"_3))) = color(red)(80#</mathjax> <mathjax>#color(red)("g Al(OH)"_3#</mathjax></p>
<p>which I suppose will round to <mathjax>#1#</mathjax> significant figure, the amount given in the problem..</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The balanced equation shows the reaction between aluminium hydroxide and hydrochloric acid to form aluminium chloride and water: Al(OH)3 + 3HCl = AlCl3 + 3H2O. 1 mole of aluminium chloride was formed, what mass of aluminium hydroxide was formed?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#80#</mathjax> <mathjax>#"g Al(OH)"_3#</mathjax> (one sig fig)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'll assume you mean what mass of aluminum hydroxide <strong><em>reacted</em></strong>.</p>
<p>We can use the coefficients of the equation to determine the number of <em>moles</em> of <mathjax>#"Al(OH)"_3#</mathjax> that react, knowing that <mathjax>#1#</mathjax> <mathjax>#"mol AlCl"_3#</mathjax> forms:</p>
<p><mathjax>#1cancel("mol AlCl"_3)((1color(white)(l)"mol Al(OH)"_3)/(1cancel("mol AlCl"_3))) = 1#</mathjax> <mathjax>#"mol Al(OH)"_3#</mathjax></p>
<p>which makes sense given their molar ratio is <mathjax>#1:1#</mathjax>. </p>
<p>Now, we can use the <em>molar mass</em> of aluminum hydroxide (<mathjax>#78.00#</mathjax> <mathjax>#"g/mol"#</mathjax>) to calculate the number of <em>grams</em> that react:</p>
<p><mathjax>#1cancel("mol Al(OH)"_3)((78.00color(white)(l)"g Al(OH)"_3)/(1cancel("mol Al(OH)"_3))) = color(red)(80#</mathjax> <mathjax>#color(red)("g Al(OH)"_3#</mathjax></p>
<p>which I suppose will round to <mathjax>#1#</mathjax> significant figure, the amount given in the problem..</p></div>
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</article> | The balanced equation shows the reaction between aluminium hydroxide and hydrochloric acid to form aluminium chloride and water: Al(OH)3 + 3HCl = AlCl3 + 3H2O. 1 mole of aluminium chloride was formed, what mass of aluminium hydroxide was formed? | null |
3,084 | aca836dc-6ddd-11ea-a7c8-ccda262736ce | https://socratic.org/questions/how-would-you-write-the-equation-for-this-reaction-excess-potassium-hydroxide-so | KOU(aq) + KH2PO4(aq) <=> K2HPO4(aq) + H2O(l) | start chemical_equation qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"KOU(aq) + KH2PO4(aq) <=> K2HPO4(aq) + H2O(l)"}] | [{"type":"other","value":"Excess potassium hydroxide solution."},{"type":"substance name","value":"Potassium dihydrogen phosphate solution"}] | <h1 class="questionTitle" itemprop="name">How would you write the equation for this reaction: excess potassium hydroxide solution is added to a solution of potassium dihydrogen phosphate?</h1> | null | KOU(aq) + KH2PO4(aq) <=> K2HPO4(aq) + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid base reaction. It is tempting to speculate that the addition of excess hydroxide would drive the equilibrium to the right, and result in the formation of <mathjax>#K_3PO_4#</mathjax>. Formation of the phosphate trianion is unlikely in aqueous solution. </p>
<p><a href="https://tse4.mm.bing.net/th?id=OIP.M1204f1fcc7b29b190421689ca08f5380o0&pid=15.1&P=0&w=256&h=176" rel="nofollow">See here,</a> </p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/SnQmdGqqRVSl2mRrS0Pu_h3po4_1.jpg"/> </p>
<p>How does the graph illustrate what I have said?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#KOH(aq) + KH_2PO_4(aq) rightleftharpoons K_2HPO_4(aq) + H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid base reaction. It is tempting to speculate that the addition of excess hydroxide would drive the equilibrium to the right, and result in the formation of <mathjax>#K_3PO_4#</mathjax>. Formation of the phosphate trianion is unlikely in aqueous solution. </p>
<p><a href="https://tse4.mm.bing.net/th?id=OIP.M1204f1fcc7b29b190421689ca08f5380o0&pid=15.1&P=0&w=256&h=176" rel="nofollow">See here,</a> </p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/SnQmdGqqRVSl2mRrS0Pu_h3po4_1.jpg"/> </p>
<p>How does the graph illustrate what I have said?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you write the equation for this reaction: excess potassium hydroxide solution is added to a solution of potassium dihydrogen phosphate?</h1>
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<div class="markdown"><p><mathjax>#KOH(aq) + KH_2PO_4(aq) rightleftharpoons K_2HPO_4(aq) + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid base reaction. It is tempting to speculate that the addition of excess hydroxide would drive the equilibrium to the right, and result in the formation of <mathjax>#K_3PO_4#</mathjax>. Formation of the phosphate trianion is unlikely in aqueous solution. </p>
<p><a href="https://tse4.mm.bing.net/th?id=OIP.M1204f1fcc7b29b190421689ca08f5380o0&pid=15.1&P=0&w=256&h=176" rel="nofollow">See here,</a> </p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/SnQmdGqqRVSl2mRrS0Pu_h3po4_1.jpg"/> </p>
<p>How does the graph illustrate what I have said?</p></div>
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<a href="https://socratic.org/answers/346176" itemprop="url">Answer link</a>
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</article> | How would you write the equation for this reaction: excess potassium hydroxide solution is added to a solution of potassium dihydrogen phosphate? | null |
3,085 | a953c47c-6ddd-11ea-baef-ccda262736ce | https://socratic.org/questions/what-is-the-final-concentration-of-a-solution-prepared-by-adding-water-to-50-0-m | 7.5 × 10^(-2) mol/L | start physical_unit 18 18 concentration mol/l qc_end physical_unit 7 7 13 14 volume qc_end physical_unit 18 18 16 17 concentration qc_end physical_unit 7 7 21 22 volume qc_end substance 11 11 qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] NaOH solution [IN] mol/L"}] | [{"type":"physical unit","value":"7.5 × 10^(-2) mol/L"}] | [{"type":"physical unit","value":"Volume1 [OF] NaOH solution [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Concentration1 [OF] NaOH solution [=] \\pu{1.5 M}"},{"type":"physical unit","value":"Volume2 [OF] NaOH solution [=] \\pu{1.00 L}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the final concentration of a solution prepared by adding water to 50.0 mL of 1.5 M NaOH to make 1.00 L of solution?</h1> | null | 7.5 × 10^(-2) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given concentration is dimensionally consistent. The answer has units of <mathjax>#mol*L^-1#</mathjax> as required.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Moles/volume of solution = <mathjax>#(50.0xx10^(-3)cancelLxx1.50*mol*cancel(L^(-1)))/(1.00*L)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given concentration is dimensionally consistent. The answer has units of <mathjax>#mol*L^-1#</mathjax> as required.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the final concentration of a solution prepared by adding water to 50.0 mL of 1.5 M NaOH to make 1.00 L of solution?</h1>
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<div class="markdown"><p>Moles/volume of solution = <mathjax>#(50.0xx10^(-3)cancelLxx1.50*mol*cancel(L^(-1)))/(1.00*L)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given concentration is dimensionally consistent. The answer has units of <mathjax>#mol*L^-1#</mathjax> as required.</p></div>
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</article> | What is the final concentration of a solution prepared by adding water to 50.0 mL of 1.5 M NaOH to make 1.00 L of solution? | null |
3,086 | aa9088ca-6ddd-11ea-844a-ccda262736ce | https://socratic.org/questions/59775fab11ef6b118b24b2eb | 3.22 × 10^22 | start physical_unit 2 3 number none qc_end physical_unit 10 10 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] metal atoms"}] | [{"type":"physical unit","value":"3.22 × 10^22"}] | [{"type":"physical unit","value":"Mass [OF] copper [=] \\pu{3.4 g}"}] | <h1 class="questionTitle" itemprop="name">How many metal atoms in a #3.4*g# mass of copper?</h1> | null | 3.22 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we know this? Because <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> gives us THE MASS of a molar quantity, i.e. <mathjax>#N_A#</mathjax> or <mathjax>#"Avogadro's number"#</mathjax> of ALL the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You got a Periodic Table beside you? You should have one because you are doing your chemistry homework. </p>
<p>And thus to get the number of copper atoms, we just find the molar quantity........</p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xxN_A#</mathjax>, where <mathjax>#N_A="Avocado number"#</mathjax>, </p>
<p><mathjax>#=6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"number of copper atoms....."#</mathjax></p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xx6.022xx10^23*mol^-1-=3.22xx10^22*"copper atoms"#</mathjax>.</p>
<p>Now <mathjax>#"Avogadro's number"#</mathjax> is admittedly a large number. We use it because <mathjax>#N_A#</mathjax> <mathjax>#""^1H#</mathjax> atoms have a mass of <mathjax>#1*g#</mathjax>; alternatively, <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00*g#</mathjax>. It is thus the link between atoms and molecules, what chemists theorize about and of which they conceive, with the macro world of grams, kilograms, and litres, what chemists measure out in the laboratory. </p>
<p>It will be worth your while getting your head around this <mathjax>#"principle of equivalent mass"#</mathjax>, of how a given mass of elements or molecules constitutes a SPECIFIC NUMBER of elements or molecules. So please get it right!</p>
<p>You are on your own with the videos. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We know that a <mathjax>#63.55*g#</mathjax> mass of copper contains <mathjax>#N_A#</mathjax> <mathjax>#"copper atoms......"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we know this? Because <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> gives us THE MASS of a molar quantity, i.e. <mathjax>#N_A#</mathjax> or <mathjax>#"Avogadro's number"#</mathjax> of ALL the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You got a Periodic Table beside you? You should have one because you are doing your chemistry homework. </p>
<p>And thus to get the number of copper atoms, we just find the molar quantity........</p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xxN_A#</mathjax>, where <mathjax>#N_A="Avocado number"#</mathjax>, </p>
<p><mathjax>#=6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"number of copper atoms....."#</mathjax></p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xx6.022xx10^23*mol^-1-=3.22xx10^22*"copper atoms"#</mathjax>.</p>
<p>Now <mathjax>#"Avogadro's number"#</mathjax> is admittedly a large number. We use it because <mathjax>#N_A#</mathjax> <mathjax>#""^1H#</mathjax> atoms have a mass of <mathjax>#1*g#</mathjax>; alternatively, <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00*g#</mathjax>. It is thus the link between atoms and molecules, what chemists theorize about and of which they conceive, with the macro world of grams, kilograms, and litres, what chemists measure out in the laboratory. </p>
<p>It will be worth your while getting your head around this <mathjax>#"principle of equivalent mass"#</mathjax>, of how a given mass of elements or molecules constitutes a SPECIFIC NUMBER of elements or molecules. So please get it right!</p>
<p>You are on your own with the videos. </p></div>
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<h1 class="questionTitle" itemprop="name">How many metal atoms in a #3.4*g# mass of copper?</h1>
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anor277
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<div class="markdown"><p>We know that a <mathjax>#63.55*g#</mathjax> mass of copper contains <mathjax>#N_A#</mathjax> <mathjax>#"copper atoms......"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we know this? Because <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> gives us THE MASS of a molar quantity, i.e. <mathjax>#N_A#</mathjax> or <mathjax>#"Avogadro's number"#</mathjax> of ALL the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You got a Periodic Table beside you? You should have one because you are doing your chemistry homework. </p>
<p>And thus to get the number of copper atoms, we just find the molar quantity........</p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xxN_A#</mathjax>, where <mathjax>#N_A="Avocado number"#</mathjax>, </p>
<p><mathjax>#=6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"number of copper atoms....."#</mathjax></p>
<p><mathjax>#(3.4*g)/(63.55*g*mol^-1)xx6.022xx10^23*mol^-1-=3.22xx10^22*"copper atoms"#</mathjax>.</p>
<p>Now <mathjax>#"Avogadro's number"#</mathjax> is admittedly a large number. We use it because <mathjax>#N_A#</mathjax> <mathjax>#""^1H#</mathjax> atoms have a mass of <mathjax>#1*g#</mathjax>; alternatively, <mathjax>#N_A#</mathjax> <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00*g#</mathjax>. It is thus the link between atoms and molecules, what chemists theorize about and of which they conceive, with the macro world of grams, kilograms, and litres, what chemists measure out in the laboratory. </p>
<p>It will be worth your while getting your head around this <mathjax>#"principle of equivalent mass"#</mathjax>, of how a given mass of elements or molecules constitutes a SPECIFIC NUMBER of elements or molecules. So please get it right!</p>
<p>You are on your own with the videos. </p></div>
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</article> | How many metal atoms in a #3.4*g# mass of copper? | null |
3,087 | aa13409c-6ddd-11ea-a17c-ccda262736ce | https://socratic.org/questions/a-5-0-g-sample-of-a-pure-compound-containing-h-c-and-n-contains-19-g-of-h-and-2- | 1.04 grams | start physical_unit 12 12 mass g qc_end physical_unit 6 7 1 2 mass qc_end physical_unit 9 9 14 15 mass qc_end physical_unit 10 10 19 20 mass qc_end physical_unit 6 7 43 44 mass qc_end end | [{"type":"physical unit","value":"Mass2 [OF] N [IN] grams"}] | [{"type":"physical unit","value":"1.04 grams"}] | [{"type":"physical unit","value":"Mass1 [OF] pure compound sample [=] \\pu{5.0 g}"},{"type":"physical unit","value":"Mass1 [OF] H [=] \\pu{0.19 g}"},{"type":"physical unit","value":"Mass1 [OF] C [=] \\pu{2.22 g}"},{"type":"physical unit","value":"Mass2 [OF] pure compound sample [=] \\pu{2.0 g}"}] | <h1 class="questionTitle" itemprop="name">A 5.0 g sample of a pure compound containing H, C, and N contains .19 g of H and 2.22 g of C. Another smaller sample of this pure compound is analyzed. How many grams of N should be found in the new 2.0 g pure sample?</h1> | null | 1.04 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the amount of hydrogen and the amount of carbon found in the <mathjax>#"5.0-g"#</mathjax> sample of this unknown compound to determine the <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of nitrogen.</p>
<p>Once you know the percent composition of nitrogen, you can determine the amount of nitrogen present in the <mathjax>#"2.0-g"#</mathjax> sample. </p>
<p>So, if your compound contains only hydrogen, <mathjax>#"H"#</mathjax>, carbon, <mathjax>#"C"#</mathjax>, and nitrogen, <mathjax>#"N"#</mathjax>, then you can write the mass of the compound as the <em>sum</em> of the masses of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></p>
<blockquote>
<p><mathjax>#m_"compound" = m_(H) + m_(C) + m_(N)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#"5.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#m_(N) = "5.0 g" - "0.19 g" - "2.22 g" = "2.59 g N"#</mathjax></p>
</blockquote>
<p>If you divide the mass of nitrogen by the <strong>total mass</strong> of the compound and multiply the result by <mathjax>#100#</mathjax>, you will get the compound's <em>percent composition of nitrogen</em></p>
<blockquote>
<p><mathjax>#(2.59 color(red)(cancel(color(black)("g"))))/(5.0color(red)(cancel(color(black)("g")))) xx 100 = "51.8% N"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>For <strong>every</strong> <mathjax>#"100 g"#</mathjax> of this unknown compound, <mathjax>#"51.8 g"#</mathjax> will be nitrogen. </p>
<p>This means that the <mathjax>#"2.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("g compound"))) * overbrace("51.8 g N"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("% composition of N")) = "1.036 g N"#</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>#m_(N) = color(green)("1.0 g")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.0 g N"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the amount of hydrogen and the amount of carbon found in the <mathjax>#"5.0-g"#</mathjax> sample of this unknown compound to determine the <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of nitrogen.</p>
<p>Once you know the percent composition of nitrogen, you can determine the amount of nitrogen present in the <mathjax>#"2.0-g"#</mathjax> sample. </p>
<p>So, if your compound contains only hydrogen, <mathjax>#"H"#</mathjax>, carbon, <mathjax>#"C"#</mathjax>, and nitrogen, <mathjax>#"N"#</mathjax>, then you can write the mass of the compound as the <em>sum</em> of the masses of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></p>
<blockquote>
<p><mathjax>#m_"compound" = m_(H) + m_(C) + m_(N)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#"5.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#m_(N) = "5.0 g" - "0.19 g" - "2.22 g" = "2.59 g N"#</mathjax></p>
</blockquote>
<p>If you divide the mass of nitrogen by the <strong>total mass</strong> of the compound and multiply the result by <mathjax>#100#</mathjax>, you will get the compound's <em>percent composition of nitrogen</em></p>
<blockquote>
<p><mathjax>#(2.59 color(red)(cancel(color(black)("g"))))/(5.0color(red)(cancel(color(black)("g")))) xx 100 = "51.8% N"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>For <strong>every</strong> <mathjax>#"100 g"#</mathjax> of this unknown compound, <mathjax>#"51.8 g"#</mathjax> will be nitrogen. </p>
<p>This means that the <mathjax>#"2.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("g compound"))) * overbrace("51.8 g N"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("% composition of N")) = "1.036 g N"#</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>#m_(N) = color(green)("1.0 g")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 5.0 g sample of a pure compound containing H, C, and N contains .19 g of H and 2.22 g of C. Another smaller sample of this pure compound is analyzed. How many grams of N should be found in the new 2.0 g pure sample?</h1>
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Stefan V.
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Feb 25, 2016
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<div class="markdown"><p><mathjax>#"1.0 g N"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the amount of hydrogen and the amount of carbon found in the <mathjax>#"5.0-g"#</mathjax> sample of this unknown compound to determine the <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of nitrogen.</p>
<p>Once you know the percent composition of nitrogen, you can determine the amount of nitrogen present in the <mathjax>#"2.0-g"#</mathjax> sample. </p>
<p>So, if your compound contains only hydrogen, <mathjax>#"H"#</mathjax>, carbon, <mathjax>#"C"#</mathjax>, and nitrogen, <mathjax>#"N"#</mathjax>, then you can write the mass of the compound as the <em>sum</em> of the masses of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></p>
<blockquote>
<p><mathjax>#m_"compound" = m_(H) + m_(C) + m_(N)#</mathjax></p>
</blockquote>
<p>This means that the <mathjax>#"5.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#m_(N) = "5.0 g" - "0.19 g" - "2.22 g" = "2.59 g N"#</mathjax></p>
</blockquote>
<p>If you divide the mass of nitrogen by the <strong>total mass</strong> of the compound and multiply the result by <mathjax>#100#</mathjax>, you will get the compound's <em>percent composition of nitrogen</em></p>
<blockquote>
<p><mathjax>#(2.59 color(red)(cancel(color(black)("g"))))/(5.0color(red)(cancel(color(black)("g")))) xx 100 = "51.8% N"#</mathjax></p>
</blockquote>
<p><em>So, what does this tell you?</em></p>
<p>For <strong>every</strong> <mathjax>#"100 g"#</mathjax> of this unknown compound, <mathjax>#"51.8 g"#</mathjax> will be nitrogen. </p>
<p>This means that the <mathjax>#"2.0-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("g compound"))) * overbrace("51.8 g N"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("% composition of N")) = "1.036 g N"#</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>#m_(N) = color(green)("1.0 g")#</mathjax></p>
</blockquote></div>
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</article> | A 5.0 g sample of a pure compound containing H, C, and N contains .19 g of H and 2.22 g of C. Another smaller sample of this pure compound is analyzed. How many grams of N should be found in the new 2.0 g pure sample? | null |
3,088 | aae7b180-6ddd-11ea-b3b5-ccda262736ce | https://socratic.org/questions/the-heat-of-vaporization-of-water-is-40-66-kj-mol-how-much-heat-is-absorbed-when | 6.48 kJ | start physical_unit 5 5 heat_energy kj qc_end physical_unit 5 5 7 8 enthalpy_of_vaporization qc_end physical_unit 5 5 15 16 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Absorbed heat [OF] water [IN] kJ"}] | [{"type":"physical unit","value":"6.48 kJ"}] | [{"type":"physical unit","value":"Heat of vaporization [OF] water [=] \\pu{40.66 kJ/mol}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{2.87 g}"},{"type":"other","value":"Water boils at atmospheric pressure."}] | <h1 class="questionTitle" itemprop="name">The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?</h1> | null | 6.48 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>molar heat of vaporization</strong>, <mathjax>#DeltaH_"vap"#</mathjax>, sometimes called the <em>molar <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, tells you how much energy is needed in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of a given substance at its boiling point. </p>
<p>In water's case, a molar heat of vaporization of <mathjax>#"40.66 kJ mol"^(-1)#</mathjax> means that you need to supply <mathjax>#"40.66 kJ"#</mathjax> of heat in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of water at its normal boiling point, i.e. at <mathjax>#100^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1)) #</mathjax></p>
<blockquote>
<blockquote>
<p><em>You need</em> <mathjax>#color(blue)("40.66 kJ")#</mathjax> <em>of heat to boil</em> <mathjax>#color(red)("1 mole")#</mathjax> <em>of water at its normal boiling point</em>.</p>
</blockquote>
</blockquote>
</blockquote>
<p>Now, the first thing to do here is to convert the <em>mass</em> of water to <strong>moles</strong> by using its <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar heat of vaporization as a <strong>conversion factor</strong> to determine how much heat would be needed to boil <mathjax>#0.1593#</mathjax> <strong>moles</strong> of water at its boiling point</p>
<blockquote>
<p><mathjax>#0.1593 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ")))#</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 mass of the sample.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"6.48 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>molar heat of vaporization</strong>, <mathjax>#DeltaH_"vap"#</mathjax>, sometimes called the <em>molar <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, tells you how much energy is needed in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of a given substance at its boiling point. </p>
<p>In water's case, a molar heat of vaporization of <mathjax>#"40.66 kJ mol"^(-1)#</mathjax> means that you need to supply <mathjax>#"40.66 kJ"#</mathjax> of heat in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of water at its normal boiling point, i.e. at <mathjax>#100^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1)) #</mathjax></p>
<blockquote>
<blockquote>
<p><em>You need</em> <mathjax>#color(blue)("40.66 kJ")#</mathjax> <em>of heat to boil</em> <mathjax>#color(red)("1 mole")#</mathjax> <em>of water at its normal boiling point</em>.</p>
</blockquote>
</blockquote>
</blockquote>
<p>Now, the first thing to do here is to convert the <em>mass</em> of water to <strong>moles</strong> by using its <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar heat of vaporization as a <strong>conversion factor</strong> to determine how much heat would be needed to boil <mathjax>#0.1593#</mathjax> <strong>moles</strong> of water at its boiling point</p>
<blockquote>
<p><mathjax>#0.1593 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ")))#</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 mass of the sample.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-11-16T00:03:52" itemprop="dateCreated">
Nov 16, 2016
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<div class="markdown"><p><mathjax>#"6.48 kJ"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>molar heat of vaporization</strong>, <mathjax>#DeltaH_"vap"#</mathjax>, sometimes called the <em>molar <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization</em>, tells you how much energy is needed in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of a given substance at its boiling point. </p>
<p>In water's case, a molar heat of vaporization of <mathjax>#"40.66 kJ mol"^(-1)#</mathjax> means that you need to supply <mathjax>#"40.66 kJ"#</mathjax> of heat in order to boil <mathjax>#1#</mathjax> <strong>mole</strong> of water at its normal boiling point, i.e. at <mathjax>#100^@"C"#</mathjax>. </p>
<blockquote>
<p><mathjax>#DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1)) #</mathjax></p>
<blockquote>
<blockquote>
<p><em>You need</em> <mathjax>#color(blue)("40.66 kJ")#</mathjax> <em>of heat to boil</em> <mathjax>#color(red)("1 mole")#</mathjax> <em>of water at its normal boiling point</em>.</p>
</blockquote>
</blockquote>
</blockquote>
<p>Now, the first thing to do here is to convert the <em>mass</em> of water to <strong>moles</strong> by using its <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar heat of vaporization as a <strong>conversion factor</strong> to determine how much heat would be needed to boil <mathjax>#0.1593#</mathjax> <strong>moles</strong> of water at its boiling point</p>
<blockquote>
<p><mathjax>#0.1593 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ")))#</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 mass of the sample.</p></div>
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</article> | The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure? | null |
3,089 | a8bc5c19-6ddd-11ea-886a-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-aqueous-sodium-chloride-and-aqueous-silver-nit | AgNO3(aq) + NaCl(aq) -> NaNO3(aq) + AgCl(s) v | start chemical_equation qc_end substance 6 8 qc_end substance 10 12 qc_end substance 16 18 qc_end substance 20 22 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"AgNO3(aq) + NaCl(aq) -> NaNO3(aq) + AgCl(s) v"}] | [{"type":"substance name","value":"Aqueous sodium chloride"},{"type":"substance name","value":"Aqueous silver nitrate"},{"type":"substance name","value":"Solid silver chloride"},{"type":"substance name","value":"Aqueous sodium nitrate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for aqueous sodium chloride and aqueous silver nitrate react to form solid silver chloride and aqueous sodium nitrate?</h1> | null | AgNO3(aq) + NaCl(aq) -> NaNO3(aq) + AgCl(s) v | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is commonly used to illustrate basic solubility rules, and solubility equilibria.</p>
<p>All nitrates are soluble, hence silver nitrate is soluble; and all halides are soluble, EXCEPT for <mathjax>#AgX#</mathjax>, <mathjax>#PbX_2#</mathjax>, and <mathjax>#Hg_2X_2#</mathjax>.</p>
<p>Thus silver nitrate is soluble, but silver chloride precipitates from solution as a curdy white solid.</p>
<p>We write the net ionic equation as:</p>
<p><mathjax>#Ag^(+) + X^(-) rarr AgX(s)darr#</mathjax></p>
<p>Silver chloride is a curdy white solid; silver bromide is a creamy yellow; and silver iodide is bright yellow. (On standing the silver halides tend to reduce to silver metal, and the precipitates darken.)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#AgNO_3(aq) + NaCl(aq) rarr NaNO_3(aq) + AgCl(s)darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is commonly used to illustrate basic solubility rules, and solubility equilibria.</p>
<p>All nitrates are soluble, hence silver nitrate is soluble; and all halides are soluble, EXCEPT for <mathjax>#AgX#</mathjax>, <mathjax>#PbX_2#</mathjax>, and <mathjax>#Hg_2X_2#</mathjax>.</p>
<p>Thus silver nitrate is soluble, but silver chloride precipitates from solution as a curdy white solid.</p>
<p>We write the net ionic equation as:</p>
<p><mathjax>#Ag^(+) + X^(-) rarr AgX(s)darr#</mathjax></p>
<p>Silver chloride is a curdy white solid; silver bromide is a creamy yellow; and silver iodide is bright yellow. (On standing the silver halides tend to reduce to silver metal, and the precipitates darken.)</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the chemical equation for aqueous sodium chloride and aqueous silver nitrate react to form solid silver chloride and aqueous sodium nitrate?</h1>
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anor277
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<div class="markdown"><p><mathjax>#AgNO_3(aq) + NaCl(aq) rarr NaNO_3(aq) + AgCl(s)darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This reaction is commonly used to illustrate basic solubility rules, and solubility equilibria.</p>
<p>All nitrates are soluble, hence silver nitrate is soluble; and all halides are soluble, EXCEPT for <mathjax>#AgX#</mathjax>, <mathjax>#PbX_2#</mathjax>, and <mathjax>#Hg_2X_2#</mathjax>.</p>
<p>Thus silver nitrate is soluble, but silver chloride precipitates from solution as a curdy white solid.</p>
<p>We write the net ionic equation as:</p>
<p><mathjax>#Ag^(+) + X^(-) rarr AgX(s)darr#</mathjax></p>
<p>Silver chloride is a curdy white solid; silver bromide is a creamy yellow; and silver iodide is bright yellow. (On standing the silver halides tend to reduce to silver metal, and the precipitates darken.)</p></div>
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</article> | What is the chemical equation for aqueous sodium chloride and aqueous silver nitrate react to form solid silver chloride and aqueous sodium nitrate? | null |
3,090 | ac3bf00c-6ddd-11ea-bbaa-ccda262736ce | https://socratic.org/questions/5995250611ef6b429fd6779c | 10^(-6) M | start physical_unit 5 6 concentration mol/l qc_end physical_unit 9 9 13 13 ph qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydroxyl ions [IN] M"}] | [{"type":"physical unit","value":"10^(-6) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{8}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of hydroxyl ions in a solution of pH #8# ?</h1> | null | 10^(-6) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydrogen ion concentration (actually hydronium <mathjax>#H_3O^+#</mathjax> concentration).</p>
<p><mathjax>#pOH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydroxyl (<mathjax>#OH^-#</mathjax>) ion concentration.</p>
<p>To say <mathjax>#pH + pOH = 14#</mathjax> is to say that the product of the <mathjax>#H_3O^+#</mathjax> and <mathjax>#OH^-#</mathjax> ion concentrations is <mathjax>#10^-14#</mathjax>.</p>
<p>If a solution has <mathjax>#pH = 8#</mathjax>, then its <mathjax>#pOH#</mathjax> is <mathjax>#14-8 = 6#</mathjax> and the corresponding concentration of hydroxyl <mathjax>#OH^-#</mathjax> ions is <mathjax>#10^(-6)#</mathjax> mol per litre.</p></div>
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<div class="markdown"><p><mathjax>#10^(-6)#</mathjax> mol per litre</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydrogen ion concentration (actually hydronium <mathjax>#H_3O^+#</mathjax> concentration).</p>
<p><mathjax>#pOH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydroxyl (<mathjax>#OH^-#</mathjax>) ion concentration.</p>
<p>To say <mathjax>#pH + pOH = 14#</mathjax> is to say that the product of the <mathjax>#H_3O^+#</mathjax> and <mathjax>#OH^-#</mathjax> ion concentrations is <mathjax>#10^-14#</mathjax>.</p>
<p>If a solution has <mathjax>#pH = 8#</mathjax>, then its <mathjax>#pOH#</mathjax> is <mathjax>#14-8 = 6#</mathjax> and the corresponding concentration of hydroxyl <mathjax>#OH^-#</mathjax> ions is <mathjax>#10^(-6)#</mathjax> mol per litre.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of hydroxyl ions in a solution of pH #8# ?</h1>
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<div class="markdown"><p><mathjax>#10^(-6)#</mathjax> mol per litre</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydrogen ion concentration (actually hydronium <mathjax>#H_3O^+#</mathjax> concentration).</p>
<p><mathjax>#pOH#</mathjax> is <mathjax>#-log_10#</mathjax> of the hydroxyl (<mathjax>#OH^-#</mathjax>) ion concentration.</p>
<p>To say <mathjax>#pH + pOH = 14#</mathjax> is to say that the product of the <mathjax>#H_3O^+#</mathjax> and <mathjax>#OH^-#</mathjax> ion concentrations is <mathjax>#10^-14#</mathjax>.</p>
<p>If a solution has <mathjax>#pH = 8#</mathjax>, then its <mathjax>#pOH#</mathjax> is <mathjax>#14-8 = 6#</mathjax> and the corresponding concentration of hydroxyl <mathjax>#OH^-#</mathjax> ions is <mathjax>#10^(-6)#</mathjax> mol per litre.</p></div>
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</article> | What is the concentration of hydroxyl ions in a solution of pH #8# ? | null |
3,091 | ac948d66-6ddd-11ea-8279-ccda262736ce | https://socratic.org/questions/57e02c687c014932132de47b | KClO3 | start chemical_formula qc_end physical_unit 6 7 1 2 mass qc_end c_other OTHER qc_end physical_unit 10 10 15 16 mass qc_end physical_unit 11 11 19 20 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the unknown compound [IN] empirical"}] | [{"type":"chemical equation","value":"KClO3"}] | [{"type":"physical unit","value":"Mass [OF] the unknown compound [=] \\pu{22.35 g}"},{"type":"other","value":"An unknown compound consisting of potassium, chlorine, and oxygen."},{"type":"physical unit","value":"Mass [OF] potassium [=] \\pu{7.13 g}"},{"type":"physical unit","value":"Mass [OF] chlorine [=] \\pu{6.47 g}"}] | <h1 class="questionTitle" itemprop="name">A #22.35*g# mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained #7.13*g# of potassium, #6.47*g# of chlorine. What is its empirical formula?</h1> | null | KClO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always, we divide each constitiuent mass thru by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each constituent atom. </p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.13*g)/(39.10*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.47*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.546*mol#</mathjax>.</p>
<p>And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of <mathjax>#KClO_3#</mathjax>.</p>
<p>How did I know that that there were <mathjax>#(22.35-7.13-6.47)*g#</mathjax> of <mathjax>#O#</mathjax>? The mass of oxygen was not given in the problem.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#KClO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always, we divide each constitiuent mass thru by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each constituent atom. </p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.13*g)/(39.10*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.47*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.546*mol#</mathjax>.</p>
<p>And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of <mathjax>#KClO_3#</mathjax>.</p>
<p>How did I know that that there were <mathjax>#(22.35-7.13-6.47)*g#</mathjax> of <mathjax>#O#</mathjax>? The mass of oxygen was not given in the problem.</p></div>
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<h1 class="questionTitle" itemprop="name">A #22.35*g# mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained #7.13*g# of potassium, #6.47*g# of chlorine. What is its empirical formula?</h1>
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anor277
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Sep 19, 2016
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<div class="markdown"><p><mathjax>#KClO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As always, we divide each constitiuent mass thru by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each constituent atom. </p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.13*g)/(39.10*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.47*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.182*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.546*mol#</mathjax>.</p>
<p>And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of <mathjax>#KClO_3#</mathjax>.</p>
<p>How did I know that that there were <mathjax>#(22.35-7.13-6.47)*g#</mathjax> of <mathjax>#O#</mathjax>? The mass of oxygen was not given in the problem.</p></div>
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</article> | A #22.35*g# mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained #7.13*g# of potassium, #6.47*g# of chlorine. What is its empirical formula? | null |
3,092 | aae6ee07-6ddd-11ea-9754-ccda262736ce | https://socratic.org/questions/a-sample-of-a-gas-has-a-volume-of-2-0-liters-at-a-pressure-of-1-0-atmosphere-wha | 0.50 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 4 4 9 10 volume qc_end physical_unit 4 4 15 16 pressure qc_end physical_unit 4 4 27 28 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas sample [IN] atm"}] | [{"type":"physical unit","value":"0.50 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{2.0 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{1.0 atmosphere}"},{"type":"physical unit","value":"Volume2 [OF] the gas sample [=] \\pu{4.0 liters}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A sample of a gas has a volume of 2.0 liters at a pressure of 1.0 atmosphere. What will the pressure be when the volume increases to 4.0 liters at constant temperature?</h1> | null | 0.50 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the pressure of a gas and the volume it occupies have an <strong>inverse relationship</strong> when the temperature and the number of moles of gas, i.e. the amount of gas present in the sample, remain <strong>constant</strong> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>. </p>
<p><img alt="https://www.pinterest.com/pin/53480314303853411/" src="https://useruploads.socratic.org/1nNzdXHAScun2p8mCDfu_2054720ee0709ddc0f1c43e18fcf9974.jpg"/> </p>
<p>Simply put, when the temperature and the number of moles of gas are constant, a <em>decrease</em> in the volume of the gas will cause an <strong>increase</strong> in its pressure and an <em>increase</em> in the volume of the gas will cause a <strong>decrease</strong> in its pressure. </p>
<p>In your case, you know that the volume is increasing</p>
<blockquote>
<p><mathjax>#"2.0 L " -> " 4.0 L"#</mathjax></p>
</blockquote>
<p>so you should expect the volume to <strong>decrease</strong></p>
<blockquote>
<p><mathjax>#"1.0 atm " > " P"_2#</mathjax></p>
</blockquote>
<p>Your tool of choice here will be this equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (2.0 color(red)(cancel(color(black)("L"))))/(4.0color(red)(cancel(color(black)("L")))) * "1.0 atm" = color(darkgreen)(ul(color(black)("0.50 atm")))#</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>#"0.50 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the pressure of a gas and the volume it occupies have an <strong>inverse relationship</strong> when the temperature and the number of moles of gas, i.e. the amount of gas present in the sample, remain <strong>constant</strong> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>. </p>
<p><img alt="https://www.pinterest.com/pin/53480314303853411/" src="https://useruploads.socratic.org/1nNzdXHAScun2p8mCDfu_2054720ee0709ddc0f1c43e18fcf9974.jpg"/> </p>
<p>Simply put, when the temperature and the number of moles of gas are constant, a <em>decrease</em> in the volume of the gas will cause an <strong>increase</strong> in its pressure and an <em>increase</em> in the volume of the gas will cause a <strong>decrease</strong> in its pressure. </p>
<p>In your case, you know that the volume is increasing</p>
<blockquote>
<p><mathjax>#"2.0 L " -> " 4.0 L"#</mathjax></p>
</blockquote>
<p>so you should expect the volume to <strong>decrease</strong></p>
<blockquote>
<p><mathjax>#"1.0 atm " > " P"_2#</mathjax></p>
</blockquote>
<p>Your tool of choice here will be this equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (2.0 color(red)(cancel(color(black)("L"))))/(4.0color(red)(cancel(color(black)("L")))) * "1.0 atm" = color(darkgreen)(ul(color(black)("0.50 atm")))#</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> | <article>
<h1 class="questionTitle" itemprop="name">A sample of a gas has a volume of 2.0 liters at a pressure of 1.0 atmosphere. What will the pressure be when the volume increases to 4.0 liters at constant temperature?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-06-12T02:07:38" itemprop="dateCreated">
Jun 12, 2017
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<div>
<div class="markdown"><p><mathjax>#"0.50 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the pressure of a gas and the volume it occupies have an <strong>inverse relationship</strong> when the temperature and the number of moles of gas, i.e. the amount of gas present in the sample, remain <strong>constant</strong> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>. </p>
<p><img alt="https://www.pinterest.com/pin/53480314303853411/" src="https://useruploads.socratic.org/1nNzdXHAScun2p8mCDfu_2054720ee0709ddc0f1c43e18fcf9974.jpg"/> </p>
<p>Simply put, when the temperature and the number of moles of gas are constant, a <em>decrease</em> in the volume of the gas will cause an <strong>increase</strong> in its pressure and an <em>increase</em> in the volume of the gas will cause a <strong>decrease</strong> in its pressure. </p>
<p>In your case, you know that the volume is increasing</p>
<blockquote>
<p><mathjax>#"2.0 L " -> " 4.0 L"#</mathjax></p>
</blockquote>
<p>so you should expect the volume to <strong>decrease</strong></p>
<blockquote>
<p><mathjax>#"1.0 atm " > " P"_2#</mathjax></p>
</blockquote>
<p>Your tool of choice here will be this equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (2.0 color(red)(cancel(color(black)("L"))))/(4.0color(red)(cancel(color(black)("L")))) * "1.0 atm" = color(darkgreen)(ul(color(black)("0.50 atm")))#</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> | A sample of a gas has a volume of 2.0 liters at a pressure of 1.0 atmosphere. What will the pressure be when the volume increases to 4.0 liters at constant temperature? | null |
3,093 | aad476e6-6ddd-11ea-8260-ccda262736ce | https://socratic.org/questions/58505bb3b72cff163fd9a3e9 | SO2 + 2 HNO2 -> H2SO4 + 2 NO | start chemical_equation qc_end chemical_equation 7 7 qc_end substance 9 10 qc_end substance 13 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"SO2 + 2 HNO2 -> H2SO4 + 2 NO"}] | [{"type":"chemical equation","value":"SO2(g)"},{"type":"substance name","value":"Nitrous acid"},{"type":"substance name","value":"Sulfuric acid"}] | <h1 class="questionTitle" itemprop="name">How do you represent the oxidation of #SO_2(g)# by #"nitrous acid"# to give #"sulfuric acid"#?</h1> | null | SO2 + 2 HNO2 -> H2SO4 + 2 NO | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_2 +2H_2O rarr H_2SO_4+2H^(+) +2e^-#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#HNO_2 +H^(+) + e^(-) rarr NO+H_2O#</mathjax></p>
<p><mathjax>#"Overall:"#</mathjax></p>
<p><mathjax>#SO_2 +2 HNO_2 rarr H_2SO_4 +2NO#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#S(+IV)rarrS(+VI)#</mathjax>; <mathjax>#N(+III)rarrN(+II)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_2 +2H_2O rarr H_2SO_4+2H^(+) +2e^-#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#HNO_2 +H^(+) + e^(-) rarr NO+H_2O#</mathjax></p>
<p><mathjax>#"Overall:"#</mathjax></p>
<p><mathjax>#SO_2 +2 HNO_2 rarr H_2SO_4 +2NO#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you represent the oxidation of #SO_2(g)# by #"nitrous acid"# to give #"sulfuric acid"#?</h1>
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anor277
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Dec 13, 2016
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<div class="markdown"><p><mathjax>#S(+IV)rarrS(+VI)#</mathjax>; <mathjax>#N(+III)rarrN(+II)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_2 +2H_2O rarr H_2SO_4+2H^(+) +2e^-#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#HNO_2 +H^(+) + e^(-) rarr NO+H_2O#</mathjax></p>
<p><mathjax>#"Overall:"#</mathjax></p>
<p><mathjax>#SO_2 +2 HNO_2 rarr H_2SO_4 +2NO#</mathjax></p></div>
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</article> | How do you represent the oxidation of #SO_2(g)# by #"nitrous acid"# to give #"sulfuric acid"#? | null |
3,094 | a99f19f5-6ddd-11ea-8ebc-ccda262736ce | https://socratic.org/questions/what-is-the-standard-enthalpy-of-formation-for-ethanol-c-2h-5oh | -277.7 kJ/mol | start physical_unit 9 9 standard_enthalpy_formation kj/mol qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Standard enthalpy of formation [OF] C2H5OH [IN] kJ/mol"}] | [{"type":"physical unit","value":"-277.7 kJ/mol"}] | [{"type":"chemical equation","value":"C2H5OH"}] | <h1 class="questionTitle" itemprop="name">What is the standard enthalpy of formation for ethanol #C_2H_5OH#?</h1> | null | -277.7 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this specifies the reaction....</p>
<p><mathjax>#2C(s) + 3H_2(g) + 1/2O_2(g) rarr C_2H_5OH(l)+DeltaH_f^@#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>Well, <a href="http://wps.prenhall.com/wps/media/objects/3311/3390593/blb0507.html" rel="nofollow">this site quotes</a> <mathjax>#DeltaH_f^@("ethanol")=-277.7*kJ*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this specifies the reaction....</p>
<p><mathjax>#2C(s) + 3H_2(g) + 1/2O_2(g) rarr C_2H_5OH(l)+DeltaH_f^@#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the standard enthalpy of formation for ethanol #C_2H_5OH#?</h1>
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anor277
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Jan 15, 2018
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<div class="markdown"><p>Well, <a href="http://wps.prenhall.com/wps/media/objects/3311/3390593/blb0507.html" rel="nofollow">this site quotes</a> <mathjax>#DeltaH_f^@("ethanol")=-277.7*kJ*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And this specifies the reaction....</p>
<p><mathjax>#2C(s) + 3H_2(g) + 1/2O_2(g) rarr C_2H_5OH(l)+DeltaH_f^@#</mathjax></p></div>
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</article> | What is the standard enthalpy of formation for ethanol #C_2H_5OH#? | null |
3,095 | ad05f32f-6ddd-11ea-98e7-ccda262736ce | https://socratic.org/questions/a-solution-contains-0-3-liters-of-sulfuric-acid-if-this-represents-12-of-the-tot | 2.50 liters | start physical_unit 1 1 mass l qc_end physical_unit 6 7 3 4 volume qc_end end | [{"type":"physical unit","value":"Total amount [OF] the solution [IN] liters"}] | [{"type":"physical unit","value":"2.50 liters"}] | [{"type":"physical unit","value":"Percent [OF] sulfuric acid in the solution [=] \\pu{12%}"},{"type":"physical unit","value":"Volume [OF] sulfuric acid [=] \\pu{0.3 liters}"}] | <h1 class="questionTitle" itemprop="name">A solution contains 0.3 liters of sulfuric acid. If this represents 12% of the total amount, how do you find the total amount?</h1> | null | 2.50 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is more of an algebra problem than a chemistry problem.</p>
<p>0.3 liters = 12%</p>
<p>total volume = 100 % </p>
<p>Set up a ratio proportion. </p>
<p><mathjax># 0.3/x = 12/100#</mathjax> Solve for x </p>
<p>First multiply both sides by 100 x. This gives </p>
<p><mathjax># 100 xx x xx( 0.3/x) = 100 xx x xx ( 12/100) #</mathjax> This gives</p>
<p><mathjax>#100 xx 0.3 = x xx 12 #</mathjax> This gives </p>
<p>30 = 12 x divide both sides by 12 </p>
<p><mathjax># 30/12 = 12 x/12#</mathjax> This gives </p>
<p>2.5 = x </p>
<p>x = the total volume of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>.</p></div>
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<div class="markdown"><p>The total amount of solution is 2.5 liters. Use a ratio proportion method. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is more of an algebra problem than a chemistry problem.</p>
<p>0.3 liters = 12%</p>
<p>total volume = 100 % </p>
<p>Set up a ratio proportion. </p>
<p><mathjax># 0.3/x = 12/100#</mathjax> Solve for x </p>
<p>First multiply both sides by 100 x. This gives </p>
<p><mathjax># 100 xx x xx( 0.3/x) = 100 xx x xx ( 12/100) #</mathjax> This gives</p>
<p><mathjax>#100 xx 0.3 = x xx 12 #</mathjax> This gives </p>
<p>30 = 12 x divide both sides by 12 </p>
<p><mathjax># 30/12 = 12 x/12#</mathjax> This gives </p>
<p>2.5 = x </p>
<p>x = the total volume of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>.</p></div>
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<h1 class="questionTitle" itemprop="name">A solution contains 0.3 liters of sulfuric acid. If this represents 12% of the total amount, how do you find the total amount?</h1>
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<div class="markdown"><p>The total amount of solution is 2.5 liters. Use a ratio proportion method. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is more of an algebra problem than a chemistry problem.</p>
<p>0.3 liters = 12%</p>
<p>total volume = 100 % </p>
<p>Set up a ratio proportion. </p>
<p><mathjax># 0.3/x = 12/100#</mathjax> Solve for x </p>
<p>First multiply both sides by 100 x. This gives </p>
<p><mathjax># 100 xx x xx( 0.3/x) = 100 xx x xx ( 12/100) #</mathjax> This gives</p>
<p><mathjax>#100 xx 0.3 = x xx 12 #</mathjax> This gives </p>
<p>30 = 12 x divide both sides by 12 </p>
<p><mathjax># 30/12 = 12 x/12#</mathjax> This gives </p>
<p>2.5 = x </p>
<p>x = the total volume of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>.</p></div>
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</article> | A solution contains 0.3 liters of sulfuric acid. If this represents 12% of the total amount, how do you find the total amount? | null |
3,096 | a92cd87e-6ddd-11ea-ba6a-ccda262736ce | https://socratic.org/questions/what-is-the-steric-number-of-nh-3 | 4 | start physical_unit 6 6 steric_number none qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Steric number [OF] NH3"}] | [{"type":"physical unit","value":"4"}] | [{"type":"chemical equation","value":"NH3"}] | <h1 class="questionTitle" itemprop="name">What is the steric number of #NH_3#?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given molecule, the <strong>steric number</strong> is calculated by counting how many <strong>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> surround the central atom. </p>
<p>A <em>region</em> of electron density can be</p>
<blockquote>
<ul>
<li><em>a single, double, <strong>or</strong> triple bond -- all three count as <strong>one region</strong> of electron density</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>In order to calculate the steric number of nitrogen, which is the central atom in ammonia, <mathjax>#"NH"_3#</mathjax>, you must first draw the molecule's <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong>. </p>
<p>The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong>, <mathjax>#5#</mathjax> coming from nitrogen and <mathjax>#1#</mathjax> coming from each of the three hydrogen atoms. </p>
<p>The nitrogen atom will form <strong>single bonds</strong> with the three hydrogen atoms, with each of these bonds accounting for <mathjax>#2#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>. The remaining <mathjax>#2#</mathjax> valence electrons will be placed on the nitrogen atom as a <strong>lone pair of electrons</strong>. </p>
<p>The Lewis structure for ammonia will thus looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps" src="https://useruploads.socratic.org/SxEs655MQuOiht5iAY8b_%3D768e2d063688200c5482e54c7905ab81.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Notice that the central atom is surrounded by a total of <mathjax>#4#</mathjax> <strong>regions of electron density</strong></p>
<blockquote>
<ul>
<li><em><strong>three single bonds</strong></em></li>
<li><em><strong>one lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>You can thus say that the nitrogen atom has a <strong>steric number</strong> equal to <mathjax>#4#</mathjax>. The steric number tells you the <strong>hybridization</strong> of the central atom.</p>
<p>In this case, a steric number equal to <mathjax>#4#</mathjax> implies that the nitrogen atom is <mathjax>#"sp"^3#</mathjax> hybridized. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given molecule, the <strong>steric number</strong> is calculated by counting how many <strong>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> surround the central atom. </p>
<p>A <em>region</em> of electron density can be</p>
<blockquote>
<ul>
<li><em>a single, double, <strong>or</strong> triple bond -- all three count as <strong>one region</strong> of electron density</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>In order to calculate the steric number of nitrogen, which is the central atom in ammonia, <mathjax>#"NH"_3#</mathjax>, you must first draw the molecule's <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong>. </p>
<p>The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong>, <mathjax>#5#</mathjax> coming from nitrogen and <mathjax>#1#</mathjax> coming from each of the three hydrogen atoms. </p>
<p>The nitrogen atom will form <strong>single bonds</strong> with the three hydrogen atoms, with each of these bonds accounting for <mathjax>#2#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>. The remaining <mathjax>#2#</mathjax> valence electrons will be placed on the nitrogen atom as a <strong>lone pair of electrons</strong>. </p>
<p>The Lewis structure for ammonia will thus looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps" src="https://useruploads.socratic.org/SxEs655MQuOiht5iAY8b_%3D768e2d063688200c5482e54c7905ab81.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Notice that the central atom is surrounded by a total of <mathjax>#4#</mathjax> <strong>regions of electron density</strong></p>
<blockquote>
<ul>
<li><em><strong>three single bonds</strong></em></li>
<li><em><strong>one lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>You can thus say that the nitrogen atom has a <strong>steric number</strong> equal to <mathjax>#4#</mathjax>. The steric number tells you the <strong>hybridization</strong> of the central atom.</p>
<p>In this case, a steric number equal to <mathjax>#4#</mathjax> implies that the nitrogen atom is <mathjax>#"sp"^3#</mathjax> hybridized. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the steric number of #NH_3#?</h1>
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<div>
<div class="markdown"><p><mathjax>#4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a given molecule, the <strong>steric number</strong> is calculated by counting how many <strong>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> surround the central atom. </p>
<p>A <em>region</em> of electron density can be</p>
<blockquote>
<ul>
<li><em>a single, double, <strong>or</strong> triple bond -- all three count as <strong>one region</strong> of electron density</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>In order to calculate the steric number of nitrogen, which is the central atom in ammonia, <mathjax>#"NH"_3#</mathjax>, you must first draw the molecule's <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong>. </p>
<p>The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong>, <mathjax>#5#</mathjax> coming from nitrogen and <mathjax>#1#</mathjax> coming from each of the three hydrogen atoms. </p>
<p>The nitrogen atom will form <strong>single bonds</strong> with the three hydrogen atoms, with each of these bonds accounting for <mathjax>#2#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>. The remaining <mathjax>#2#</mathjax> valence electrons will be placed on the nitrogen atom as a <strong>lone pair of electrons</strong>. </p>
<p>The Lewis structure for ammonia will thus looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps" src="https://useruploads.socratic.org/SxEs655MQuOiht5iAY8b_%3D768e2d063688200c5482e54c7905ab81.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Notice that the central atom is surrounded by a total of <mathjax>#4#</mathjax> <strong>regions of electron density</strong></p>
<blockquote>
<ul>
<li><em><strong>three single bonds</strong></em></li>
<li><em><strong>one lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>You can thus say that the nitrogen atom has a <strong>steric number</strong> equal to <mathjax>#4#</mathjax>. The steric number tells you the <strong>hybridization</strong> of the central atom.</p>
<p>In this case, a steric number equal to <mathjax>#4#</mathjax> implies that the nitrogen atom is <mathjax>#"sp"^3#</mathjax> hybridized. </p></div>
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</article> | What is the steric number of #NH_3#? | null |
3,097 | ace60dd0-6ddd-11ea-9bbc-ccda262736ce | https://socratic.org/questions/how-many-moles-of-nitrogen-dioxide-are-produced-from-4-754-mol-no-given-the-equa | 4.75 moles | start physical_unit 4 5 mole mol qc_end physical_unit 11 11 9 10 mole qc_end chemical_equation 15 19 qc_end end | [{"type":"physical unit","value":"Mole [OF] nitrogen dioxide [IN] moles"}] | [{"type":"physical unit","value":"4.75 moles"}] | [{"type":"physical unit","value":"Mole [OF] NO [=] \\pu{4.754 mol}"},{"type":"chemical equation","value":"NO(g) + O2(g) -> NO2(g)"}] | <h1 class="questionTitle" itemprop="name">How many moles of nitrogen dioxide are produced from 4.754 mol NO? Given the equation: NO(g) + O2(g) --> NO2(g).</h1> | null | 4.75 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above is stoichiometric. What does this mean? It means that for every product particle, there is a corresponding reactant particle; indeed there must be because masses are conserved in every chemical reaction.</p>
<p>Now of course I cannot have 1/2 an oxygen molecule, but I can certainly have 16 g of oxygen, which represents 1/2 a mole (i.e. Avogadro's number) of oxygen, <mathjax>#O_2#</mathjax>, molecules.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, 1 mole of nitrous oxide, <mathjax>#NO#</mathjax>, combines with stoichiometic oxygen, to give stoichiometric nitric oxide, <mathjax>#NO_2#</mathjax>. So there is a 1:1 equivalence between nitrous oxide and its oxidation product.</p>
<p>You have started with 4.754 moles of nitrous oxide. Given the 1:1 stoichiometry, with how many moles of nitric oxide will you finish given sufficient oxygen gas? I think the answer is pretty clear. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#NO(g) + 1/2O_2(g) rarr NO_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above is stoichiometric. What does this mean? It means that for every product particle, there is a corresponding reactant particle; indeed there must be because masses are conserved in every chemical reaction.</p>
<p>Now of course I cannot have 1/2 an oxygen molecule, but I can certainly have 16 g of oxygen, which represents 1/2 a mole (i.e. Avogadro's number) of oxygen, <mathjax>#O_2#</mathjax>, molecules.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, 1 mole of nitrous oxide, <mathjax>#NO#</mathjax>, combines with stoichiometic oxygen, to give stoichiometric nitric oxide, <mathjax>#NO_2#</mathjax>. So there is a 1:1 equivalence between nitrous oxide and its oxidation product.</p>
<p>You have started with 4.754 moles of nitrous oxide. Given the 1:1 stoichiometry, with how many moles of nitric oxide will you finish given sufficient oxygen gas? I think the answer is pretty clear. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of nitrogen dioxide are produced from 4.754 mol NO? Given the equation: NO(g) + O2(g) --> NO2(g).</h1>
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anor277
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<div class="markdown"><p><mathjax>#NO(g) + 1/2O_2(g) rarr NO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above is stoichiometric. What does this mean? It means that for every product particle, there is a corresponding reactant particle; indeed there must be because masses are conserved in every chemical reaction.</p>
<p>Now of course I cannot have 1/2 an oxygen molecule, but I can certainly have 16 g of oxygen, which represents 1/2 a mole (i.e. Avogadro's number) of oxygen, <mathjax>#O_2#</mathjax>, molecules.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, 1 mole of nitrous oxide, <mathjax>#NO#</mathjax>, combines with stoichiometic oxygen, to give stoichiometric nitric oxide, <mathjax>#NO_2#</mathjax>. So there is a 1:1 equivalence between nitrous oxide and its oxidation product.</p>
<p>You have started with 4.754 moles of nitrous oxide. Given the 1:1 stoichiometry, with how many moles of nitric oxide will you finish given sufficient oxygen gas? I think the answer is pretty clear. </p></div>
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</article> | How many moles of nitrogen dioxide are produced from 4.754 mol NO? Given the equation: NO(g) + O2(g) --> NO2(g). | null |
3,098 | a8fa60c8-6ddd-11ea-a0a3-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-the-gas-after-the-explosion-in-the-following-problem | 200 m^3 | start physical_unit 5 6 volume m^3 qc_end physical_unit 5 6 21 22 volume qc_end physical_unit 5 6 30 33 pressure qc_end physical_unit 5 6 56 57 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] m^3"}] | [{"type":"physical unit","value":"200 m^3"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.050 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{4.0 × 10^6 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{1.00 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of the gas after the explosion in the following problem?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>In a thermonuclear device the pressure of 0.050 liters of gas within the bomb casing reaches <mathjax>#4.0 x 10^6#</mathjax> atm. When the bomb casing is destroyed by the explosion. the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. </p></div>
</h2>
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</div> | 200 m^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = 4.0 × 10^6color(white)(l) "atm"; V_1 = "0.050 L"#</mathjax><br/>
<mathjax>#P_2 ="1.00 atm"; color(white)(mmll) V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "0.050 L" × (4.00 ×10^6 color(red)(cancel(color(black)("atm"))))/(1.00 color(red)(cancel(color(black)("atm")))) = "200 000 L" = "200 m"^3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume of the gas is <mathjax>#"200 m"^3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = 4.0 × 10^6color(white)(l) "atm"; V_1 = "0.050 L"#</mathjax><br/>
<mathjax>#P_2 ="1.00 atm"; color(white)(mmll) V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "0.050 L" × (4.00 ×10^6 color(red)(cancel(color(black)("atm"))))/(1.00 color(red)(cancel(color(black)("atm")))) = "200 000 L" = "200 m"^3#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the volume of the gas after the explosion in the following problem?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>In a thermonuclear device the pressure of 0.050 liters of gas within the bomb casing reaches <mathjax>#4.0 x 10^6#</mathjax> atm. When the bomb casing is destroyed by the explosion. the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. </p></div>
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<div class="markdown"><p>The volume of the gas is <mathjax>#"200 m"^3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<p>This appears to be a <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
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<p>We can rearrange Boyle's Law to get</p>
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<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
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</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = 4.0 × 10^6color(white)(l) "atm"; V_1 = "0.050 L"#</mathjax><br/>
<mathjax>#P_2 ="1.00 atm"; color(white)(mmll) V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "0.050 L" × (4.00 ×10^6 color(red)(cancel(color(black)("atm"))))/(1.00 color(red)(cancel(color(black)("atm")))) = "200 000 L" = "200 m"^3#</mathjax></p></div>
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</article> | What is the volume of the gas after the explosion in the following problem? |
In a thermonuclear device the pressure of 0.050 liters of gas within the bomb casing reaches #4.0 x 10^6# atm. When the bomb casing is destroyed by the explosion. the gas is released into the atmosphere where it reaches a pressure of 1.00 atm.
|
3,099 | aaaf461c-6ddd-11ea-b47a-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-equation-for-the-combustion-of-octane-c8h18-l-o2-g-co2 | 2 C8H18(l) + O2(g) -> 12 CO2(g) + 14 H2O(l) | start chemical_equation qc_end chemical_equation 11 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the combustion of octane"}] | [{"type":"chemical equation","value":"2 C8H18(l) + O2(g) -> 12 CO2(g) + 14 H2O(l)"}] | [{"type":"chemical equation","value":"C8H18(l) + O2(g) -> CO2(g) + H2O(l)"}] | <h1 class="questionTitle" itemprop="name">How would you balance the equation for the combustion of octane: C8H18(l)+O2(g)---->CO2(g)+H2O(l)?</h1> | null | 2 C8H18(l) + O2(g) -> 12 CO2(g) + 14 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_6H_14(g) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(g)#</mathjax></p>
<p>Is this equation balanced? How do you know? How is the complete combustion of octane, <mathjax>#C_8H_18#</mathjax>, to be represented? I balanced the carbons, and then the hydrogens, and then the oxygens. The order I used is unimportant, it is important that I balance the equation.</p></div>
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<div class="markdown"><p>The complete combustion of any hydrocarbon gives carbon dioxide and water. I will represent the combustion of hexane.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_6H_14(g) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(g)#</mathjax></p>
<p>Is this equation balanced? How do you know? How is the complete combustion of octane, <mathjax>#C_8H_18#</mathjax>, to be represented? I balanced the carbons, and then the hydrogens, and then the oxygens. The order I used is unimportant, it is important that I balance the equation.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance the equation for the combustion of octane: C8H18(l)+O2(g)---->CO2(g)+H2O(l)?</h1>
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anor277
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<div class="markdown"><p>The complete combustion of any hydrocarbon gives carbon dioxide and water. I will represent the combustion of hexane.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_6H_14(g) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(g)#</mathjax></p>
<p>Is this equation balanced? How do you know? How is the complete combustion of octane, <mathjax>#C_8H_18#</mathjax>, to be represented? I balanced the carbons, and then the hydrogens, and then the oxygens. The order I used is unimportant, it is important that I balance the equation.</p></div>
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Nikka C.
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Nov 11, 2015
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<div class="markdown"><p><mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#25/2O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8CO_2 (g)#</mathjax> + <mathjax>#9H_2O (l)#</mathjax></p>
<p>or</p>
<p><mathjax>#2C_8H_18 (l)#</mathjax> + <mathjax>#25 O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#16CO_2 (g)#</mathjax> + <mathjax>#18H_2O (l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><strong>First</strong>, you need to tally all the atoms.</p>
<p><mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2 (g)#</mathjax> + <mathjax>#H_2O (l)#</mathjax> (unbalanced)</p>
<p>Based on the subscripts, you have</p>
<p><em>left side:</em><br/>
C = 8<br/>
H = 18<br/>
O = 2</p>
<p><em>right side:</em><br/>
C = 1<br/>
H = 2<br/>
O = 2 + 1 (do not add this up yet)</p>
<p><strong>Second</strong>, find the easiest atom to balance. In this case, the <mathjax>#C#</mathjax> atom. Always remember that in balancing, you are NOT SUPPOSED TO CHANGE THE SUBSCRIPTS, only put coefficients before the substance (as changing the subscripts means that you are changing the molecular structure instead).</p>
<p><mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (red) 8CO_2 (g)#</mathjax> + <mathjax>#H_2O (l)#</mathjax> </p>
<p><em>left side:</em><br/>
C = <strong>8</strong><br/>
H = 18<br/>
O = 2</p>
<p><em>right side:</em><br/>
C = (1 x <mathjax>#color (red) 8#</mathjax>) = <strong>8</strong><br/>
H = 2<br/>
O = (2 x <mathjax>#color (red) 8#</mathjax>) + 1 </p>
<p>Since <mathjax>#CO_2#</mathjax> is a substance, you have to apply the coefficient to both <mathjax>#C#</mathjax> and two <mathjax>#O#</mathjax> atoms as they are all bonded to each other.</p>
<p><strong>Third</strong>, balance the next easiest atom.</p>
<p><mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8CO_2 (g)#</mathjax> + <mathjax>#color (blue) 9H_2O (l)#</mathjax> </p>
<p><em>left side:</em><br/>
C = <strong>8</strong><br/>
H = <strong>18</strong><br/>
O = 2</p>
<p><em>right side:</em><br/>
C = (1 x 8) = <strong>8</strong><br/>
H = (2 x <mathjax>#color (blue) 9#</mathjax>) = <strong>18</strong><br/>
O = (2 x 8) + (1 x <mathjax>#color (blue) 9#</mathjax>) = 25</p>
<p>Now all that is left is to balance are the <mathjax>#O#</mathjax> atoms. Since the sum of <mathjax>#O#</mathjax> atoms on the right side is an odd number, I can use my knowledge in fractions to balance the left side of the equation. </p>
<p>Thus,</p>
<p><mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#color (green) (25/2)O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8CO_2 (g)#</mathjax> + <mathjax>#9H_2O (l)#</mathjax> (balance)</p>
<p><em>left side:</em><br/>
C = <strong>8</strong><br/>
H = <strong>18</strong><br/>
O = (2 x <mathjax>#color (green) (25/2)#</mathjax>) = <strong>25</strong></p>
<p><em>right side:</em><br/>
C = (1 x 8) = <strong>8</strong><br/>
H = (2 x 9) = <strong>18</strong><br/>
O = (2 x 8) + (1 x 9) = <strong>25</strong></p>
<p>But if you don't want fractions as coefficients, you can always multiply the WHOLE equation by 2.</p>
<p><mathjax>#cancel 2#</mathjax> [<mathjax>#C_8H_18 (l)#</mathjax> + <mathjax>#25/ cancel 2O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8CO_2 (g)#</mathjax> + <mathjax>#9H_2O (l)#</mathjax>]</p>
<p>=</p>
<p><mathjax>#2C_8H_18 (l)#</mathjax> + <mathjax>#25 O_2 (g)#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#16CO_2 (g)#</mathjax> + <mathjax>#18H_2O (l)#</mathjax> (balance)</p>
<p>Both answers are considered correct.</p></div>
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</article> | How would you balance the equation for the combustion of octane: C8H18(l)+O2(g)---->CO2(g)+H2O(l)? | null |
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