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2,500 | ac1f0d66-6ddd-11ea-b19c-ccda262736ce | https://socratic.org/questions/how-much-heat-will-be-released-when-12-0-g-of-h-2-reacts-with-76-0-g-of-o-2-acco | 1.3 × 10^3 kJ | start physical_unit 19 21 heat_energy kj qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 16 16 13 14 mass qc_end chemical_equation 22 28 qc_end physical_unit 19 21 31 32 deltah qc_end end | [{"type":"physical unit","value":"Released heat [OF] the following equation [IN] kJ"}] | [{"type":"physical unit","value":"1.3 × 10^3 kJ"}] | [{"type":"physical unit","value":"Mass [OF] H2 [=] \\pu{12.0 g}"},{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{76.0 g}"},{"type":"chemical equation","value":"2 H2 + O2 -> 2 H2O"},{"type":"physical unit","value":"DeltaH [OF] the following equation [=] \\pu{-571.6 kJ}"}] | <h1 class="questionTitle" itemprop="name">How much heat will be released when 12.0 g of #H_2# reacts with 76.0 g of #O_2# according to the following equation? #2H_2 + O_2 -> 2H_2O# #DeltaH#= -571.6 kJ?</h1> | null | 1.3 × 10^3 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax> <mathjax>#DeltaH=-576.6*kJ*mol^-1.#</mathjax></p>
<p><mathjax>#"Moles of dihydrogen"=(12.0*g)/(2.016*g*mol^-1)=5.95*mol#</mathjax>.</p>
<p><mathjax>#"Moles of dioxygen"=(76.0*g)/(32.0*g*mol^-1)=2.38*mol#</mathjax>.</p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, clearly, there is a insufficient molar quantity of dioxygen for complete combustion. At most <mathjax>#4.75*mol#</mathjax> dihydrogen can react (i.e. <mathjax>#2xx2.38*mol)#</mathjax> according to the given equation.</p>
<p>And thus energy released can based on the molar quantity of <mathjax>#"dioxygen gas"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.38*molxx-576.6*kJ*mol^-1=??kJ#</mathjax> </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Dioxygen gas"#</mathjax> is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. Over <mathjax>#1.3xx10^3*kJ#</mathjax> are evolved.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax> <mathjax>#DeltaH=-576.6*kJ*mol^-1.#</mathjax></p>
<p><mathjax>#"Moles of dihydrogen"=(12.0*g)/(2.016*g*mol^-1)=5.95*mol#</mathjax>.</p>
<p><mathjax>#"Moles of dioxygen"=(76.0*g)/(32.0*g*mol^-1)=2.38*mol#</mathjax>.</p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, clearly, there is a insufficient molar quantity of dioxygen for complete combustion. At most <mathjax>#4.75*mol#</mathjax> dihydrogen can react (i.e. <mathjax>#2xx2.38*mol)#</mathjax> according to the given equation.</p>
<p>And thus energy released can based on the molar quantity of <mathjax>#"dioxygen gas"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.38*molxx-576.6*kJ*mol^-1=??kJ#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">How much heat will be released when 12.0 g of #H_2# reacts with 76.0 g of #O_2# according to the following equation? #2H_2 + O_2 -> 2H_2O# #DeltaH#= -571.6 kJ?</h1>
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anor277
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Nov 19, 2016
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<div class="markdown"><p><mathjax>#"Dioxygen gas"#</mathjax> is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. Over <mathjax>#1.3xx10^3*kJ#</mathjax> are evolved.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax> <mathjax>#DeltaH=-576.6*kJ*mol^-1.#</mathjax></p>
<p><mathjax>#"Moles of dihydrogen"=(12.0*g)/(2.016*g*mol^-1)=5.95*mol#</mathjax>.</p>
<p><mathjax>#"Moles of dioxygen"=(76.0*g)/(32.0*g*mol^-1)=2.38*mol#</mathjax>.</p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, clearly, there is a insufficient molar quantity of dioxygen for complete combustion. At most <mathjax>#4.75*mol#</mathjax> dihydrogen can react (i.e. <mathjax>#2xx2.38*mol)#</mathjax> according to the given equation.</p>
<p>And thus energy released can based on the molar quantity of <mathjax>#"dioxygen gas"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.38*molxx-576.6*kJ*mol^-1=??kJ#</mathjax> </p></div>
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</article> | How much heat will be released when 12.0 g of #H_2# reacts with 76.0 g of #O_2# according to the following equation? #2H_2 + O_2 -> 2H_2O# #DeltaH#= -571.6 kJ? | null |
2,501 | a99c5d3a-6ddd-11ea-b98f-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-19-87-mol-of-ammonium-chloride-nh-4cl-at-stp | 445.35 L | start physical_unit 10 10 volume l qc_end physical_unit 10 10 5 6 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] NH4Cl [IN] L"}] | [{"type":"physical unit","value":"445.35 L"}] | [{"type":"physical unit","value":"Mole [OF] NH4Cl [=] \\pu{19.87 mol}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 19.87 mol of ammonium chloride (#NH_4Cl#) at STP?</h1> | null | 445.35 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP, we would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a><br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (must have units of atm)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the proportionality constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the volume of <mathjax>#NH_4Cl#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given moles and the proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax><br/>
Now all we have to do is rearrange the equation and solve for V like so: <br/>
<mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (19.87molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#</mathjax><br/>
<mathjax>#V = 445.3 L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>445.3 L of <mathjax>#NH_4Cl#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP, we would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a><br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (must have units of atm)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the proportionality constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the volume of <mathjax>#NH_4Cl#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given moles and the proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax><br/>
Now all we have to do is rearrange the equation and solve for V like so: <br/>
<mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (19.87molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#</mathjax><br/>
<mathjax>#V = 445.3 L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 19.87 mol of ammonium chloride (#NH_4Cl#) at STP?</h1>
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<div class="markdown"><p>445.3 L of <mathjax>#NH_4Cl#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP, we would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a><br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (must have units of atm)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the proportionality constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the volume of <mathjax>#NH_4Cl#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given moles and the proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax><br/>
Now all we have to do is rearrange the equation and solve for V like so: <br/>
<mathjax>#V = (nxxRxxT)/P#</mathjax><br/>
<mathjax>#V = (19.87molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#</mathjax><br/>
<mathjax>#V = 445.3 L#</mathjax></p></div>
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</article> | What is the volume of 19.87 mol of ammonium chloride (#NH_4Cl#) at STP? | null |
2,502 | a8e3cf7b-6ddd-11ea-9716-ccda262736ce | https://socratic.org/questions/a-chemist-reacted-0-05-moles-of-solid-sodium-with-water-to-form-sodium-hydroxide | 1.15 g | start physical_unit 7 7 mass g qc_end chemical_equation 21 27 qc_end physical_unit 6 7 3 4 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium [IN] g"}] | [{"type":"physical unit","value":"1.15 g"}] | [{"type":"chemical equation","value":"Na + H2O -> NaOH + H2"},{"type":"physical unit","value":"Mole [OF] solid sodium [=] \\pu{0.05 moles}"}] | <h1 class="questionTitle" itemprop="name">A chemist reacted 0.05 moles of solid sodium with water to form sodium hydroxide solution. The chemical reaction for this is: Na+H2O = NaOH +H2
What mass of sodium was reacted?</h1> | null | 1.15 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium has an <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of <mathjax>#22.99*g*mol^-1#</mathjax>. How do I know this? Well, for a start it is printed on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>, a copy of which will be made available in every examination of chemistry and physics you sit........</p>
<p>And so we take the (dimensional) product..........</p>
<p><mathjax>#22.99*g*cancel(mol^-1)xx0.05*cancel(mol)=??*g#</mathjax>, a bit over one gram...........</p>
<p>Typically, questions like these would also ask you to calculate the volume of dihydrogen gas evolved ......... Can you make this assessment?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Got a Periodic Table........?</p>
<p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium has an <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of <mathjax>#22.99*g*mol^-1#</mathjax>. How do I know this? Well, for a start it is printed on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>, a copy of which will be made available in every examination of chemistry and physics you sit........</p>
<p>And so we take the (dimensional) product..........</p>
<p><mathjax>#22.99*g*cancel(mol^-1)xx0.05*cancel(mol)=??*g#</mathjax>, a bit over one gram...........</p>
<p>Typically, questions like these would also ask you to calculate the volume of dihydrogen gas evolved ......... Can you make this assessment?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A chemist reacted 0.05 moles of solid sodium with water to form sodium hydroxide solution. The chemical reaction for this is: Na+H2O = NaOH +H2
What mass of sodium was reacted?</h1>
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anor277
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<span class="dateCreated" datetime="2017-07-08T21:40:40" itemprop="dateCreated">
Jul 8, 2017
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<div class="markdown"><p>Got a Periodic Table........?</p>
<p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium has an <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of <mathjax>#22.99*g*mol^-1#</mathjax>. How do I know this? Well, for a start it is printed on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>, a copy of which will be made available in every examination of chemistry and physics you sit........</p>
<p>And so we take the (dimensional) product..........</p>
<p><mathjax>#22.99*g*cancel(mol^-1)xx0.05*cancel(mol)=??*g#</mathjax>, a bit over one gram...........</p>
<p>Typically, questions like these would also ask you to calculate the volume of dihydrogen gas evolved ......... Can you make this assessment?</p></div>
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<a href="https://socratic.org/answers/449424" itemprop="url">Answer link</a>
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</article> | A chemist reacted 0.05 moles of solid sodium with water to form sodium hydroxide solution. The chemical reaction for this is: Na+H2O = NaOH +H2
What mass of sodium was reacted? | null |
2,503 | acad7109-6ddd-11ea-b464-ccda262736ce | https://socratic.org/questions/if-22-5-l-of-nitrogen-at-748-mm-hg-are-compressed-to-725-mm-hg-at-constant-tempe | 23.2 L | start physical_unit 4 4 volume l qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 11 12 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen [IN] L"}] | [{"type":"physical unit","value":"23.2 L"}] | [{"type":"physical unit","value":"Volume1 [OF] nitrogen [=] \\pu{22.5 L}"},{"type":"physical unit","value":"Pressure1 [OF] nitrogen [=] \\pu{748 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] nitrogen [=] \\pu{725 mmHg}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature, what is the new volume?</h1> | null | 23.2 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=748 mm Hg#</mathjax></p>
<p><mathjax>#V_1=22.5 L#</mathjax></p>
<p><mathjax>#P_2=725 mm Hg#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(748*22.5)/725=23.2 L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The new volume <mathjax>#=23.2 L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=748 mm Hg#</mathjax></p>
<p><mathjax>#V_1=22.5 L#</mathjax></p>
<p><mathjax>#P_2=725 mm Hg#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(748*22.5)/725=23.2 L#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature, what is the new volume?</h1>
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Narad T.
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<div class="markdown"><p>The new volume <mathjax>#=23.2 L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1=748 mm Hg#</mathjax></p>
<p><mathjax>#V_1=22.5 L#</mathjax></p>
<p><mathjax>#P_2=725 mm Hg#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(748*22.5)/725=23.2 L#</mathjax></p></div>
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</article> | If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature, what is the new volume? | null |
2,504 | ad06239d-6ddd-11ea-94e5-ccda262736ce | https://socratic.org/questions/59d4e5deb72cff5f7b40ffe0 | 32 | start physical_unit 2 2 number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"32"}] | [{"type":"chemical equation","value":"CO3^2-"}] | <h1 class="questionTitle" itemprop="name">How many electrons are there in carbonate ion, #CO_3^(2-)#?</h1> | null | 32 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now, clearly, the ion must contain 2 more electrons, than it does nucular charges.....in one formula unit of <mathjax>#CO_3^(2-)#</mathjax> there are <mathjax>#6_C+3xx8_O=30*"positively charged nuclear particles"#</mathjax>. And thus there MUST be 32 electrons conceived to whizz about the nuclei. </p>
<p>And their presence is clear from the Lewis structure of carbonate anion.....there are 8 inner core electrons, i.e. formally the <mathjax>#1s^2#</mathjax> electrons, and for <mathjax>#O=C(-O^(-))_2#</mathjax> there 24 electrons (i.e. 12 electron pairs) involved in <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> or present as lone pairs. Thirty-two electrons as required. </p>
<p>And all I have done is considered the atomic numbers of each element....</p></div>
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<div class="markdown"><p>The carbonate dianion is <mathjax>#CO_3^(2-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now, clearly, the ion must contain 2 more electrons, than it does nucular charges.....in one formula unit of <mathjax>#CO_3^(2-)#</mathjax> there are <mathjax>#6_C+3xx8_O=30*"positively charged nuclear particles"#</mathjax>. And thus there MUST be 32 electrons conceived to whizz about the nuclei. </p>
<p>And their presence is clear from the Lewis structure of carbonate anion.....there are 8 inner core electrons, i.e. formally the <mathjax>#1s^2#</mathjax> electrons, and for <mathjax>#O=C(-O^(-))_2#</mathjax> there 24 electrons (i.e. 12 electron pairs) involved in <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> or present as lone pairs. Thirty-two electrons as required. </p>
<p>And all I have done is considered the atomic numbers of each element....</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many electrons are there in carbonate ion, #CO_3^(2-)#?</h1>
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<div class="markdown"><p>The carbonate dianion is <mathjax>#CO_3^(2-)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now, clearly, the ion must contain 2 more electrons, than it does nucular charges.....in one formula unit of <mathjax>#CO_3^(2-)#</mathjax> there are <mathjax>#6_C+3xx8_O=30*"positively charged nuclear particles"#</mathjax>. And thus there MUST be 32 electrons conceived to whizz about the nuclei. </p>
<p>And their presence is clear from the Lewis structure of carbonate anion.....there are 8 inner core electrons, i.e. formally the <mathjax>#1s^2#</mathjax> electrons, and for <mathjax>#O=C(-O^(-))_2#</mathjax> there 24 electrons (i.e. 12 electron pairs) involved in <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> or present as lone pairs. Thirty-two electrons as required. </p>
<p>And all I have done is considered the atomic numbers of each element....</p></div>
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</article> | How many electrons are there in carbonate ion, #CO_3^(2-)#? | null |
2,505 | ac1fe12e-6ddd-11ea-8b6a-ccda262736ce | https://socratic.org/questions/in-the-reaction-nh-4no-2-s-n-2-g-2h-2o-g-how-many-liters-of-nitrogen-gas-is-prod | 25.00 liters | start physical_unit 13 14 volume l qc_end chemical_equation 3 8 qc_end physical_unit 21 21 18 19 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [IN] liters"}] | [{"type":"physical unit","value":"25.00 liters"}] | [{"type":"chemical equation","value":"NH4NO2(s) -> N2(g) + 2 H2O(g)"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{50.0 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">In the reaction #NH_4NO_2(s) -> N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP?</h1> | null | 25.00 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I follow the reaction as written:</p>
<p><mathjax>#NH_4NO_2(s) rarr N_2(g) + 2H_2O(g)#</mathjax>.</p>
<p>Both products are gaseous. Because <mathjax>#Vpropn#</mathjax> at constant temperature and pressure, the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> is easy to assess. Of course if <mathjax>#50*L#</mathjax> of <mathjax>#H_2O(l)#</mathjax> were evolved, it would be a different story. </p></div>
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<div class="answerSummary">
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<div class="markdown"><p>If <mathjax>#50*L#</mathjax> of water gas were evolved, then we can immediately see that a <mathjax>#25*L#</mathjax> volume of dinitrogen was likewise evolved. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I follow the reaction as written:</p>
<p><mathjax>#NH_4NO_2(s) rarr N_2(g) + 2H_2O(g)#</mathjax>.</p>
<p>Both products are gaseous. Because <mathjax>#Vpropn#</mathjax> at constant temperature and pressure, the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> is easy to assess. Of course if <mathjax>#50*L#</mathjax> of <mathjax>#H_2O(l)#</mathjax> were evolved, it would be a different story. </p></div>
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<h1 class="questionTitle" itemprop="name">In the reaction #NH_4NO_2(s) -> N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP?</h1>
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<div class="markdown"><p>If <mathjax>#50*L#</mathjax> of water gas were evolved, then we can immediately see that a <mathjax>#25*L#</mathjax> volume of dinitrogen was likewise evolved. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I follow the reaction as written:</p>
<p><mathjax>#NH_4NO_2(s) rarr N_2(g) + 2H_2O(g)#</mathjax>.</p>
<p>Both products are gaseous. Because <mathjax>#Vpropn#</mathjax> at constant temperature and pressure, the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> is easy to assess. Of course if <mathjax>#50*L#</mathjax> of <mathjax>#H_2O(l)#</mathjax> were evolved, it would be a different story. </p></div>
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</article> | In the reaction #NH_4NO_2(s) -> N_2(g) + 2H_2O(g)#, how many liters of nitrogen gas is produced if 50.0 L of water is produced at STP? | null |
2,506 | ab1b974c-6ddd-11ea-a392-ccda262736ce | https://socratic.org/questions/how-do-you-balance-no-2-h-2o-hno-3-no | 3 NO2 + H2O -> 2 HNO3 + NO | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 NO2 + H2O -> 2 HNO3 + NO"}] | [{"type":"chemical equation","value":"NO2 + H2O -> HNO3 + NO"}] | <h1 class="questionTitle" itemprop="name">How do you balance #NO_2 + H_2O → HNO_3 + NO#? </h1> | null | 3 NO2 + H2O -> 2 HNO3 + NO | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#NO_2 + H_2O -> HNO_3 + NO#</mathjax></p>
</blockquote>
<p>We are looking for something like:</p>
<blockquote>
<p><mathjax>#aNO_2 + bH_2O -> cHNO_3 + d NO#</mathjax></p>
</blockquote>
<p>for some integers <mathjax>#a, b, c, d#</mathjax>.</p>
<p>Equating the amounts of <mathjax>#N#</mathjax>, <mathjax>#O#</mathjax> and <mathjax>#H#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#{ (a = c + d), (2a+b = 3c+d), (2b = c) :}#</mathjax></p>
</blockquote>
<p>Subtracting the first equation from the second to eliminate <mathjax>#d#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#a+b = 2c#</mathjax></p>
</blockquote>
<p>Then subtracting twice the third equation, we find:</p>
<blockquote>
<p><mathjax>#a-3b = 0#</mathjax></p>
</blockquote>
<p>So let's put <mathjax>#b=1#</mathjax>, <mathjax>#a=3#</mathjax>, <mathjax>#c=2#</mathjax> and <mathjax>#d=1#</mathjax> to find:</p>
<blockquote>
<p><mathjax>#3NO_2+H_2O -> 2HNO_3+NO#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3NO_2+H_2O -> 2HNO_3+NO#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#NO_2 + H_2O -> HNO_3 + NO#</mathjax></p>
</blockquote>
<p>We are looking for something like:</p>
<blockquote>
<p><mathjax>#aNO_2 + bH_2O -> cHNO_3 + d NO#</mathjax></p>
</blockquote>
<p>for some integers <mathjax>#a, b, c, d#</mathjax>.</p>
<p>Equating the amounts of <mathjax>#N#</mathjax>, <mathjax>#O#</mathjax> and <mathjax>#H#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#{ (a = c + d), (2a+b = 3c+d), (2b = c) :}#</mathjax></p>
</blockquote>
<p>Subtracting the first equation from the second to eliminate <mathjax>#d#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#a+b = 2c#</mathjax></p>
</blockquote>
<p>Then subtracting twice the third equation, we find:</p>
<blockquote>
<p><mathjax>#a-3b = 0#</mathjax></p>
</blockquote>
<p>So let's put <mathjax>#b=1#</mathjax>, <mathjax>#a=3#</mathjax>, <mathjax>#c=2#</mathjax> and <mathjax>#d=1#</mathjax> to find:</p>
<blockquote>
<p><mathjax>#3NO_2+H_2O -> 2HNO_3+NO#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance #NO_2 + H_2O → HNO_3 + NO#? </h1>
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George C.
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Nov 25, 2017
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<div class="markdown"><p><mathjax>#3NO_2+H_2O -> 2HNO_3+NO#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#NO_2 + H_2O -> HNO_3 + NO#</mathjax></p>
</blockquote>
<p>We are looking for something like:</p>
<blockquote>
<p><mathjax>#aNO_2 + bH_2O -> cHNO_3 + d NO#</mathjax></p>
</blockquote>
<p>for some integers <mathjax>#a, b, c, d#</mathjax>.</p>
<p>Equating the amounts of <mathjax>#N#</mathjax>, <mathjax>#O#</mathjax> and <mathjax>#H#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#{ (a = c + d), (2a+b = 3c+d), (2b = c) :}#</mathjax></p>
</blockquote>
<p>Subtracting the first equation from the second to eliminate <mathjax>#d#</mathjax>, we find:</p>
<blockquote>
<p><mathjax>#a+b = 2c#</mathjax></p>
</blockquote>
<p>Then subtracting twice the third equation, we find:</p>
<blockquote>
<p><mathjax>#a-3b = 0#</mathjax></p>
</blockquote>
<p>So let's put <mathjax>#b=1#</mathjax>, <mathjax>#a=3#</mathjax>, <mathjax>#c=2#</mathjax> and <mathjax>#d=1#</mathjax> to find:</p>
<blockquote>
<p><mathjax>#3NO_2+H_2O -> 2HNO_3+NO#</mathjax></p>
</blockquote></div>
</div>
</div>
</div>
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</article> | How do you balance #NO_2 + H_2O → HNO_3 + NO#? | null |
2,507 | a8ef1600-6ddd-11ea-9d2b-ccda262736ce | https://socratic.org/questions/how-to-calculate-the-ph-of-0-180-g-of-potassium-biphthalate-pka-5-4-in-50-0-ml-o | 3.58 | start physical_unit 9 10 ph none qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 9 10 6 7 mass qc_end physical_unit 9 10 13 13 pka qc_end end | [{"type":"physical unit","value":"pH [OF] potassium biphthalate"}] | [{"type":"physical unit","value":"3.58"}] | [{"type":"physical unit","value":"Volume [OF] water [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Mass [OF] potassium biphthalate [=] \\pu{0.180 g}"},{"type":"physical unit","value":"pKa [OF] potassium biphthalate [=] \\pu{5.4}"}] | <h1 class="questionTitle" itemprop="name">How to Calculate the pH of 0.180 g of potassium biphthalate (pKa = 5.4) in 50.0 mL of water?
</h1> | null | 3.58 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydrogen phthalate, or simply <mathjax>#"KHP"#</mathjax>, is actually an acidic salt because its anion, the hydrogen phthalate anion, <mathjax>#"HP"^(-)#</mathjax>, acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"KHP"_ ((aq)) -> "K"_ ((aq))^(+) + "HP"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the hydrogen phthalate will partially ionize to produce phthalate anions and hydronium cations.</p>
<blockquote>
<p><mathjax>#"HP"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "P"_ ((aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong> for the hydrogen phthalate anion is given by </p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of potassium hydrogen phthalate to calculate the number of moles present in the sample</p>
<blockquote>
<p><mathjax>#0.180 color(red)(cancel(color(black)("g"))) * "1 mole KHP"/(204.22 color(red)(cancel(color(black)("g")))) = "0.0008814 moles KHP"#</mathjax></p>
</blockquote>
<p>This means that the solution will contain <mathjax>#0.0008814#</mathjax> <strong>moles</strong> of hydrogen phthalate anions, which would have an initial concentration of--you can <em>assume</em> that the volume of the solution will be <strong>equal</strong> to the volume of water.</p>
<blockquote>
<p><mathjax>#["HP"^(-)] = "0.0008814 moles"/(50.0 * 10^(-3) quad "L") = "0.01763 M"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydrogen phthalate anion ionize, you will get <mathjax>#x#</mathjax> <mathjax>#"M#</mathjax>" of phthalate anions and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations.</p>
<p>Moreover, the concentration of the hydrogen phthalate anions will <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>, so you can say that, at equilibrium, the solution will contain </p>
<blockquote>
<p><mathjax>#["P"^(2-)] = ["H"_3"O"^(+)] = x quad "M"#</mathjax></p>
<p><mathjax>#["HP"^(-)] = (0.01763 - x) quad "M"#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the acid dissociation constant</p>
<blockquote>
<p><mathjax>#K_a = (["P"^(2-)] * ["H"_3"O"^(+)])/(["HP"^(-)])#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = (x * x)/(0.01763 - x)#</mathjax></p>
<p><mathjax>#10^(-"p"K_a) = x^2/(0.01763 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the acid dissociation constant is significantly smaller than the initial concentration of the hydrogen phthalate anions, you can use the approximation</p>
<blockquote>
<p><mathjax>#0.01763 - x ~~ 0.01763#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = x^2/0.01763#</mathjax></p>
</blockquote>
<p>which will get you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.01763 * 10^(-5.4)) = 0.00026493#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of hydronium cations, you can say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.00026493 M"#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>In your case, the <mathjax>#"pH"#</mathjax> of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.00026493) = color(darkgreen)(ul(color(black)(3.577)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the mass of potassium hydrogen phthalate and the volume of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 3.577#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydrogen phthalate, or simply <mathjax>#"KHP"#</mathjax>, is actually an acidic salt because its anion, the hydrogen phthalate anion, <mathjax>#"HP"^(-)#</mathjax>, acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"KHP"_ ((aq)) -> "K"_ ((aq))^(+) + "HP"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the hydrogen phthalate will partially ionize to produce phthalate anions and hydronium cations.</p>
<blockquote>
<p><mathjax>#"HP"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "P"_ ((aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong> for the hydrogen phthalate anion is given by </p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of potassium hydrogen phthalate to calculate the number of moles present in the sample</p>
<blockquote>
<p><mathjax>#0.180 color(red)(cancel(color(black)("g"))) * "1 mole KHP"/(204.22 color(red)(cancel(color(black)("g")))) = "0.0008814 moles KHP"#</mathjax></p>
</blockquote>
<p>This means that the solution will contain <mathjax>#0.0008814#</mathjax> <strong>moles</strong> of hydrogen phthalate anions, which would have an initial concentration of--you can <em>assume</em> that the volume of the solution will be <strong>equal</strong> to the volume of water.</p>
<blockquote>
<p><mathjax>#["HP"^(-)] = "0.0008814 moles"/(50.0 * 10^(-3) quad "L") = "0.01763 M"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydrogen phthalate anion ionize, you will get <mathjax>#x#</mathjax> <mathjax>#"M#</mathjax>" of phthalate anions and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations.</p>
<p>Moreover, the concentration of the hydrogen phthalate anions will <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>, so you can say that, at equilibrium, the solution will contain </p>
<blockquote>
<p><mathjax>#["P"^(2-)] = ["H"_3"O"^(+)] = x quad "M"#</mathjax></p>
<p><mathjax>#["HP"^(-)] = (0.01763 - x) quad "M"#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the acid dissociation constant</p>
<blockquote>
<p><mathjax>#K_a = (["P"^(2-)] * ["H"_3"O"^(+)])/(["HP"^(-)])#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = (x * x)/(0.01763 - x)#</mathjax></p>
<p><mathjax>#10^(-"p"K_a) = x^2/(0.01763 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the acid dissociation constant is significantly smaller than the initial concentration of the hydrogen phthalate anions, you can use the approximation</p>
<blockquote>
<p><mathjax>#0.01763 - x ~~ 0.01763#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = x^2/0.01763#</mathjax></p>
</blockquote>
<p>which will get you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.01763 * 10^(-5.4)) = 0.00026493#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of hydronium cations, you can say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.00026493 M"#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>In your case, the <mathjax>#"pH"#</mathjax> of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.00026493) = color(darkgreen)(ul(color(black)(3.577)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the mass of potassium hydrogen phthalate and the volume of the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How to Calculate the pH of 0.180 g of potassium biphthalate (pKa = 5.4) in 50.0 mL of water?
</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-16T00:37:32" itemprop="dateCreated">
Feb 16, 2018
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<div class="markdown"><p><mathjax>#"pH" = 3.577#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydrogen phthalate, or simply <mathjax>#"KHP"#</mathjax>, is actually an acidic salt because its anion, the hydrogen phthalate anion, <mathjax>#"HP"^(-)#</mathjax>, acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>So when you dissolve the salt in water, you get potassium cations and hydrogen phthalate anions in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"KHP"_ ((aq)) -> "K"_ ((aq))^(+) + "HP"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, the hydrogen phthalate will partially ionize to produce phthalate anions and hydronium cations.</p>
<blockquote>
<p><mathjax>#"HP"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "P"_ ((aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong> for the hydrogen phthalate anion is given by </p>
<blockquote>
<p><mathjax>#K_a = 10^(-"p"K_a)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of potassium hydrogen phthalate to calculate the number of moles present in the sample</p>
<blockquote>
<p><mathjax>#0.180 color(red)(cancel(color(black)("g"))) * "1 mole KHP"/(204.22 color(red)(cancel(color(black)("g")))) = "0.0008814 moles KHP"#</mathjax></p>
</blockquote>
<p>This means that the solution will contain <mathjax>#0.0008814#</mathjax> <strong>moles</strong> of hydrogen phthalate anions, which would have an initial concentration of--you can <em>assume</em> that the volume of the solution will be <strong>equal</strong> to the volume of water.</p>
<blockquote>
<p><mathjax>#["HP"^(-)] = "0.0008814 moles"/(50.0 * 10^(-3) quad "L") = "0.01763 M"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydrogen phthalate anion ionize, you will get <mathjax>#x#</mathjax> <mathjax>#"M#</mathjax>" of phthalate anions and <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of hydronium cations.</p>
<p>Moreover, the concentration of the hydrogen phthalate anions will <strong>decrease</strong> by <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax>, so you can say that, at equilibrium, the solution will contain </p>
<blockquote>
<p><mathjax>#["P"^(2-)] = ["H"_3"O"^(+)] = x quad "M"#</mathjax></p>
<p><mathjax>#["HP"^(-)] = (0.01763 - x) quad "M"#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the acid dissociation constant</p>
<blockquote>
<p><mathjax>#K_a = (["P"^(2-)] * ["H"_3"O"^(+)])/(["HP"^(-)])#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = (x * x)/(0.01763 - x)#</mathjax></p>
<p><mathjax>#10^(-"p"K_a) = x^2/(0.01763 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the acid dissociation constant is significantly smaller than the initial concentration of the hydrogen phthalate anions, you can use the approximation</p>
<blockquote>
<p><mathjax>#0.01763 - x ~~ 0.01763#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#10^(-"p"K_a) = x^2/0.01763#</mathjax></p>
</blockquote>
<p>which will get you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.01763 * 10^(-5.4)) = 0.00026493#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> represents the equilibrium concentration of hydronium cations, you can say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.00026493 M"#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>In your case, the <mathjax>#"pH"#</mathjax> of the solution will be </p>
<blockquote>
<p><mathjax>#"pH" = - log(0.00026493) = color(darkgreen)(ul(color(black)(3.577)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the mass of potassium hydrogen phthalate and the volume of the solution. </p></div>
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</article> | How to Calculate the pH of 0.180 g of potassium biphthalate (pKa = 5.4) in 50.0 mL of water?
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2,508 | ab5d0be4-6ddd-11ea-a783-ccda262736ce | https://socratic.org/questions/how-many-grams-of-k-2co-3-are-needed-in-order-to-produce-2-0-mol-h-2o | 276.42 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 11 12 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] K2CO3 [IN] grams"}] | [{"type":"physical unit","value":"276.42 grams"}] | [{"type":"physical unit","value":"Mole [OF] K2O [=] \\pu{2.0 mol}"}] | <h1 class="questionTitle" itemprop="name"> How many grams of #K_2CO_3# are needed in order to produce 2.0 mol #K_2O#?</h1> | null | 276.42 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><h2></h2>
<p>When heated, <mathjax>#K_2CO_3#</mathjax> decomposes according to the following equation:</p>
<p><mathjax>#K_2CO_3 -> K_2O + CO_2#</mathjax></p>
<p>The molar mass of <mathjax># K_2CO_3 = 138.21 \ g. mol^-1#</mathjax></p>
<p><mathjax>#2.0 \ mol. K_2O xx (1 \ mol. K_2CO_3)/(1 \ mol. K_2O) xx(138.21 \ g \ K_2CO_3) / (1 \ mol. K_2CO_3) #</mathjax></p>
<p><mathjax>#2.0 \cancel( mol. K_2O) xx ( 1 \color(red) cancel( mol. K_2CO_3))/(1 \ cancel(mol. K_2O)) xx(138.21 \ g \ K_2CO_3) / (1 color(red) cancel(\ mol. K_2CO_3) ) #</mathjax></p>
<p><mathjax>#= 2.8xx10^2 g#</mathjax></p></div>
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<div class="markdown"><p><mathjax># Mass_(\ K_2CO_3)= 2.8xx10^2 g#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><h2></h2>
<p>When heated, <mathjax>#K_2CO_3#</mathjax> decomposes according to the following equation:</p>
<p><mathjax>#K_2CO_3 -> K_2O + CO_2#</mathjax></p>
<p>The molar mass of <mathjax># K_2CO_3 = 138.21 \ g. mol^-1#</mathjax></p>
<p><mathjax>#2.0 \ mol. K_2O xx (1 \ mol. K_2CO_3)/(1 \ mol. K_2O) xx(138.21 \ g \ K_2CO_3) / (1 \ mol. K_2CO_3) #</mathjax></p>
<p><mathjax>#2.0 \cancel( mol. K_2O) xx ( 1 \color(red) cancel( mol. K_2CO_3))/(1 \ cancel(mol. K_2O)) xx(138.21 \ g \ K_2CO_3) / (1 color(red) cancel(\ mol. K_2CO_3) ) #</mathjax></p>
<p><mathjax>#= 2.8xx10^2 g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name"> How many grams of #K_2CO_3# are needed in order to produce 2.0 mol #K_2O#?</h1>
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<div class="markdown"><p><mathjax># Mass_(\ K_2CO_3)= 2.8xx10^2 g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><h2></h2>
<p>When heated, <mathjax>#K_2CO_3#</mathjax> decomposes according to the following equation:</p>
<p><mathjax>#K_2CO_3 -> K_2O + CO_2#</mathjax></p>
<p>The molar mass of <mathjax># K_2CO_3 = 138.21 \ g. mol^-1#</mathjax></p>
<p><mathjax>#2.0 \ mol. K_2O xx (1 \ mol. K_2CO_3)/(1 \ mol. K_2O) xx(138.21 \ g \ K_2CO_3) / (1 \ mol. K_2CO_3) #</mathjax></p>
<p><mathjax>#2.0 \cancel( mol. K_2O) xx ( 1 \color(red) cancel( mol. K_2CO_3))/(1 \ cancel(mol. K_2O)) xx(138.21 \ g \ K_2CO_3) / (1 color(red) cancel(\ mol. K_2CO_3) ) #</mathjax></p>
<p><mathjax>#= 2.8xx10^2 g#</mathjax></p></div>
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</article> | How many grams of #K_2CO_3# are needed in order to produce 2.0 mol #K_2O#? | null |
2,509 | ab213a3a-6ddd-11ea-b263-ccda262736ce | https://socratic.org/questions/57270ea211ef6b018904dd2e | 5.01 L | start physical_unit 15 16 volume l qc_end physical_unit 8 8 4 5 mass qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] chlorine gas [IN] L"}] | [{"type":"physical unit","value":"5.01 L"}] | [{"type":"physical unit","value":"Mass [OF] MnO2 [=] \\pu{14.75 g}"},{"type":"other","value":"Excess HCl(aq)."},{"type":"other","value":"Under standard conditions."}] | <h1 class="questionTitle" itemprop="name">Upon treatment of a #14.75*g# mass of #MnO_2# with excess #HCl(aq)#, what VOLUME of chlorine gas is generated under standard conditions...?</h1> | null | 5.01 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MnO_2(s) + 4HCl(aq) rarr MnCl_2(aq) + Cl_2(g)uarr + 2H_2O(l)#</mathjax></p>
<p><mathjax>#"Moles of "MnO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.75*g)/(86.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.170*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction, one equiv of chlorine gas is evolved per equiv of manganese oxide consumed. Thus <mathjax>#0.170*mol#</mathjax> of <mathjax>#Cl_2(g)#</mathjax>, whose behaviour we can assume to be ideal, expresses the following volume under the given conditions of temperature and pressure. </p>
<p><mathjax>#V=(nRT)/P=#</mathjax> </p>
<p><mathjax>#((0.170*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/(646/760*atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? L#</mathjax></p>
<p>Here, I know for a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> pressure will support a column of mercury <mathjax>#760*mm#</mathjax> high. I converted the given mercury column to atmospheres in the denominator. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Very close to <mathjax>#5*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MnO_2(s) + 4HCl(aq) rarr MnCl_2(aq) + Cl_2(g)uarr + 2H_2O(l)#</mathjax></p>
<p><mathjax>#"Moles of "MnO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.75*g)/(86.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.170*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction, one equiv of chlorine gas is evolved per equiv of manganese oxide consumed. Thus <mathjax>#0.170*mol#</mathjax> of <mathjax>#Cl_2(g)#</mathjax>, whose behaviour we can assume to be ideal, expresses the following volume under the given conditions of temperature and pressure. </p>
<p><mathjax>#V=(nRT)/P=#</mathjax> </p>
<p><mathjax>#((0.170*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/(646/760*atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? L#</mathjax></p>
<p>Here, I know for a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> pressure will support a column of mercury <mathjax>#760*mm#</mathjax> high. I converted the given mercury column to atmospheres in the denominator. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Upon treatment of a #14.75*g# mass of #MnO_2# with excess #HCl(aq)#, what VOLUME of chlorine gas is generated under standard conditions...?</h1>
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anor277
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<div class="markdown"><p>Very close to <mathjax>#5*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#MnO_2(s) + 4HCl(aq) rarr MnCl_2(aq) + Cl_2(g)uarr + 2H_2O(l)#</mathjax></p>
<p><mathjax>#"Moles of "MnO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.75*g)/(86.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.170*mol#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction, one equiv of chlorine gas is evolved per equiv of manganese oxide consumed. Thus <mathjax>#0.170*mol#</mathjax> of <mathjax>#Cl_2(g)#</mathjax>, whose behaviour we can assume to be ideal, expresses the following volume under the given conditions of temperature and pressure. </p>
<p><mathjax>#V=(nRT)/P=#</mathjax> </p>
<p><mathjax>#((0.170*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/(646/760*atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? L#</mathjax></p>
<p>Here, I know for a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> pressure will support a column of mercury <mathjax>#760*mm#</mathjax> high. I converted the given mercury column to atmospheres in the denominator. </p></div>
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</article> | Upon treatment of a #14.75*g# mass of #MnO_2# with excess #HCl(aq)#, what VOLUME of chlorine gas is generated under standard conditions...? | null |
2,510 | ad0cfa68-6ddd-11ea-8734-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-2-14-10-10-molecules-of-water | 6.39 × 10^(-13) grams | start physical_unit 10 10 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] grams"}] | [{"type":"physical unit","value":"6.39 × 10^(-13) grams"}] | [{"type":"physical unit","value":"Number [OF] water molecules [=] \\pu{2.14 × 10^10}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #2.14 * 10^10# molecules of water?</h1> | null | 6.39 × 10^(-13) grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>molar mass of water = <mathjax>#18gmol^-1#</mathjax><br/>
So <mathjax>#6.022xx10^23#</mathjax>molecules have mass 18g<br/>
mass of <mathjax>#2.14xx10^10#</mathjax> molecules will have mass<br/>
<mathjax>#(2.14xx10^10)xx(18gcancel(mol^-1))/(6.022xx10^23cancel(mol^-1))~~6.39xx10^-13g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#~~6.39xx10^-13g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>molar mass of water = <mathjax>#18gmol^-1#</mathjax><br/>
So <mathjax>#6.022xx10^23#</mathjax>molecules have mass 18g<br/>
mass of <mathjax>#2.14xx10^10#</mathjax> molecules will have mass<br/>
<mathjax>#(2.14xx10^10)xx(18gcancel(mol^-1))/(6.022xx10^23cancel(mol^-1))~~6.39xx10^-13g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#~~6.39xx10^-13g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>molar mass of water = <mathjax>#18gmol^-1#</mathjax><br/>
So <mathjax>#6.022xx10^23#</mathjax>molecules have mass 18g<br/>
mass of <mathjax>#2.14xx10^10#</mathjax> molecules will have mass<br/>
<mathjax>#(2.14xx10^10)xx(18gcancel(mol^-1))/(6.022xx10^23cancel(mol^-1))~~6.39xx10^-13g#</mathjax></p></div>
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</article> | What is the mass of #2.14 * 10^10# molecules of water? | null |
2,511 | a89f7fe4-6ddd-11ea-bd48-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-molecule-containing-24-carbon-6-hydrogen-and- | CH3Cl | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] a molecule [IN] empirical"}] | [{"type":"chemical equation","value":"CH3Cl"}] | [{"type":"physical unit","value":"percent [OF] carbon in molecule [=] \\pu{24%}"},{"type":"physical unit","value":"percent [OF] hydrogen in molecule [=] \\pu{6%}"},{"type":"physical unit","value":"percent [OF] chlorine in molecule [=] \\pu{70%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a molecule containing 24% carbon, 6% hydrogen, and 70% chlorine?</h1> | null | CH3Cl | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step we need to do is to pick our base, i.e.: a random value of our choosing that will let us change this from percentages to actual mass. For convenience's sake we can choose something like <mathjax>#100 g#</mathjax> or <mathjax>#100 kg#</mathjax> or something of the like, using <mathjax>#100 g#</mathjax>:</p>
<p>In <mathjax>#100 g#</mathjax>, since any percentage of <mathjax>#100#</mathjax> is that number, we have</p>
<p><mathjax>#70% 100g = m_Cl = 70g#</mathjax><br/>
<mathjax>#24%100g = m_C = 24g#</mathjax><br/>
<mathjax>#6%100g = m_H = 6g#</mathjax></p>
<p>Now we need to divide these values by their molar masses, because then we'll know how much atoms we can find in a molecule. For practical purposes like this we say <mathjax>#MM_Cl = 35.5 g*mol^-1#</mathjax>, <mathjax>#MM_C = 12 g* mol^-1#</mathjax>, <mathjax>#MM_H = 1g*mol^-1#</mathjax></p>
<p>So we have</p>
<p><mathjax>#n_Cl = m_Cl/(MM_Cl) = 70/35.5 = 1.97 ~= 2#</mathjax><br/>
<mathjax>#n_C = m_C/(MM_C) = 24/12 = 2#</mathjax><br/>
<mathjax>#n_H = m_H/(MM_H) = 6/1 = 6#</mathjax></p>
<p>The empirical formula wants us to have all the values as indivisible integers, and since all of these values can divide by 2, we can reduce them from <mathjax>#C_2H_6Cl#</mathjax> to <mathjax>#CH_3Cl#</mathjax>, which conveniently enough gave us a molecule that could exist from one that couldn't. (Why the first molecule can't exist and the last can is a matter for another time though, and if you have some grounding in organical chemistry, should be easy enough to see why, if not and you're curious just send me a message and I'll happy to explain why.)</p></div>
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<div class="markdown"><p><mathjax>#CH_3Cl#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step we need to do is to pick our base, i.e.: a random value of our choosing that will let us change this from percentages to actual mass. For convenience's sake we can choose something like <mathjax>#100 g#</mathjax> or <mathjax>#100 kg#</mathjax> or something of the like, using <mathjax>#100 g#</mathjax>:</p>
<p>In <mathjax>#100 g#</mathjax>, since any percentage of <mathjax>#100#</mathjax> is that number, we have</p>
<p><mathjax>#70% 100g = m_Cl = 70g#</mathjax><br/>
<mathjax>#24%100g = m_C = 24g#</mathjax><br/>
<mathjax>#6%100g = m_H = 6g#</mathjax></p>
<p>Now we need to divide these values by their molar masses, because then we'll know how much atoms we can find in a molecule. For practical purposes like this we say <mathjax>#MM_Cl = 35.5 g*mol^-1#</mathjax>, <mathjax>#MM_C = 12 g* mol^-1#</mathjax>, <mathjax>#MM_H = 1g*mol^-1#</mathjax></p>
<p>So we have</p>
<p><mathjax>#n_Cl = m_Cl/(MM_Cl) = 70/35.5 = 1.97 ~= 2#</mathjax><br/>
<mathjax>#n_C = m_C/(MM_C) = 24/12 = 2#</mathjax><br/>
<mathjax>#n_H = m_H/(MM_H) = 6/1 = 6#</mathjax></p>
<p>The empirical formula wants us to have all the values as indivisible integers, and since all of these values can divide by 2, we can reduce them from <mathjax>#C_2H_6Cl#</mathjax> to <mathjax>#CH_3Cl#</mathjax>, which conveniently enough gave us a molecule that could exist from one that couldn't. (Why the first molecule can't exist and the last can is a matter for another time though, and if you have some grounding in organical chemistry, should be easy enough to see why, if not and you're curious just send me a message and I'll happy to explain why.)</p></div>
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<div class="markdown"><p><mathjax>#CH_3Cl#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step we need to do is to pick our base, i.e.: a random value of our choosing that will let us change this from percentages to actual mass. For convenience's sake we can choose something like <mathjax>#100 g#</mathjax> or <mathjax>#100 kg#</mathjax> or something of the like, using <mathjax>#100 g#</mathjax>:</p>
<p>In <mathjax>#100 g#</mathjax>, since any percentage of <mathjax>#100#</mathjax> is that number, we have</p>
<p><mathjax>#70% 100g = m_Cl = 70g#</mathjax><br/>
<mathjax>#24%100g = m_C = 24g#</mathjax><br/>
<mathjax>#6%100g = m_H = 6g#</mathjax></p>
<p>Now we need to divide these values by their molar masses, because then we'll know how much atoms we can find in a molecule. For practical purposes like this we say <mathjax>#MM_Cl = 35.5 g*mol^-1#</mathjax>, <mathjax>#MM_C = 12 g* mol^-1#</mathjax>, <mathjax>#MM_H = 1g*mol^-1#</mathjax></p>
<p>So we have</p>
<p><mathjax>#n_Cl = m_Cl/(MM_Cl) = 70/35.5 = 1.97 ~= 2#</mathjax><br/>
<mathjax>#n_C = m_C/(MM_C) = 24/12 = 2#</mathjax><br/>
<mathjax>#n_H = m_H/(MM_H) = 6/1 = 6#</mathjax></p>
<p>The empirical formula wants us to have all the values as indivisible integers, and since all of these values can divide by 2, we can reduce them from <mathjax>#C_2H_6Cl#</mathjax> to <mathjax>#CH_3Cl#</mathjax>, which conveniently enough gave us a molecule that could exist from one that couldn't. (Why the first molecule can't exist and the last can is a matter for another time though, and if you have some grounding in organical chemistry, should be easy enough to see why, if not and you're curious just send me a message and I'll happy to explain why.)</p></div>
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</article> | What is the empirical formula of a molecule containing 24% carbon, 6% hydrogen, and 70% chlorine? | null |
2,512 | a8e9ef6e-6ddd-11ea-bce7-ccda262736ce | https://socratic.org/questions/what-volume-of-18-0-m-h-2so-4-is-needed-to-contain-2-45-g-h-2so-4 | 1.5 mL | start physical_unit 5 5 volume ml qc_end physical_unit 5 5 3 4 molarity qc_end physical_unit 5 5 10 11 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] H2SO4 [IN] mL"}] | [{"type":"physical unit","value":"1.5 mL"}] | [{"type":"physical unit","value":"Molarity [OF] H2SO4 [=] \\pu{18.0 M}"},{"type":"physical unit","value":"Mass [OF] H2SO4 [=] \\pu{2.45 g}"}] | <h1 class="questionTitle" itemprop="name">What volume of 18.0 M #H_2SO_4# is needed to contain 2.45 g #H_2SO_4#?</h1> | null | 1.5 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(2.45*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0250*mol" sulfuric acid"#</mathjax>.</p>
<p>We have <mathjax>#18.0*mol*L^-1#</mathjax> <mathjax>#H_2SO_4#</mathjax> available. </p>
<p>Volume required <mathjax>#=#</mathjax> <mathjax>#(0.0250*mol)/(18.0*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p>
<p>PS Have you ever lifted a <mathjax>#2.5*L#</mathjax> bottle of conc. sulfuric? Why is it so heavy?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Under <mathjax>#1.5*mL#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(2.45*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0250*mol" sulfuric acid"#</mathjax>.</p>
<p>We have <mathjax>#18.0*mol*L^-1#</mathjax> <mathjax>#H_2SO_4#</mathjax> available. </p>
<p>Volume required <mathjax>#=#</mathjax> <mathjax>#(0.0250*mol)/(18.0*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p>
<p>PS Have you ever lifted a <mathjax>#2.5*L#</mathjax> bottle of conc. sulfuric? Why is it so heavy?</p></div>
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<h1 class="questionTitle" itemprop="name">What volume of 18.0 M #H_2SO_4# is needed to contain 2.45 g #H_2SO_4#?</h1>
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<span class="dateCreated" datetime="2016-05-26T06:03:57" itemprop="dateCreated">
May 26, 2016
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<div class="markdown"><p>Under <mathjax>#1.5*mL#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(2.45*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0250*mol" sulfuric acid"#</mathjax>.</p>
<p>We have <mathjax>#18.0*mol*L^-1#</mathjax> <mathjax>#H_2SO_4#</mathjax> available. </p>
<p>Volume required <mathjax>#=#</mathjax> <mathjax>#(0.0250*mol)/(18.0*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mL#</mathjax></p>
<p>PS Have you ever lifted a <mathjax>#2.5*L#</mathjax> bottle of conc. sulfuric? Why is it so heavy?</p></div>
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</article> | What volume of 18.0 M #H_2SO_4# is needed to contain 2.45 g #H_2SO_4#? | null |
2,513 | a865792c-6ddd-11ea-a7b9-ccda262736ce | https://socratic.org/questions/58e1c0337c014928b75a8e8f | 75.5 g/mol | start physical_unit 1 1 molecular_weight g/mol qc_end physical_unit 1 1 6 7 density qc_end physical_unit 1 1 12 13 temperature qc_end physical_unit 1 1 18 19 pressure qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"75.5 g/mol"}] | [{"type":"physical unit","value":"Density [OF] the gas [=] \\pu{3.18 g/L}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{347 K}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{1.2 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas has a density of #3.18*g*L^-1# at a temperature of #347*K#, and a pressure of #1.2*atm#. What is its molecular mass?</h1> | null | 75.5 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume ideality, and so <mathjax>#PV=nRT#</mathjax>, and thus,</p>
<p><mathjax>#PV=("mass"/"molar mass")RT#</mathjax> because <mathjax>#n="mass"/"molar mass"#</mathjax></p>
<p>On rearrangement, <mathjax>#"molar mass"="mass"/Vxx(RT)/P#</mathjax></p>
<p>But <mathjax>#"mass"/V=rho, "density"#</mathjax>, and thus...........</p>
<p><mathjax>#"molar mass"=(rhoRT)/P=#</mathjax></p>
<p><mathjax>#(3.18*g*L^-1xx0.0821*(L*atm)/(K*mol)xx347*K)/(1.2*atm)#</mathjax></p>
<p>Let's just cancel out the units to see if we have got this right. I am not immune to mistakes.............</p>
<p><mathjax>#"Molar mass"=(3.18*g*cancel(L^-1)xx0.0821*cancel(L*atm)/(cancelK*mol)xx347*cancelK)/(1.2*cancel"atm")#</mathjax></p>
<p>And this gives, I think, an answer of........</p>
<p><mathjax>#75.5*g*mol^-1#</mathjax>. Because we have got consistent units, I think our order of operations is correct. </p></div>
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<div class="markdown"><p>We use the Ideal Gas equation to get <mathjax>#"molecular mass"=75.5*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume ideality, and so <mathjax>#PV=nRT#</mathjax>, and thus,</p>
<p><mathjax>#PV=("mass"/"molar mass")RT#</mathjax> because <mathjax>#n="mass"/"molar mass"#</mathjax></p>
<p>On rearrangement, <mathjax>#"molar mass"="mass"/Vxx(RT)/P#</mathjax></p>
<p>But <mathjax>#"mass"/V=rho, "density"#</mathjax>, and thus...........</p>
<p><mathjax>#"molar mass"=(rhoRT)/P=#</mathjax></p>
<p><mathjax>#(3.18*g*L^-1xx0.0821*(L*atm)/(K*mol)xx347*K)/(1.2*atm)#</mathjax></p>
<p>Let's just cancel out the units to see if we have got this right. I am not immune to mistakes.............</p>
<p><mathjax>#"Molar mass"=(3.18*g*cancel(L^-1)xx0.0821*cancel(L*atm)/(cancelK*mol)xx347*cancelK)/(1.2*cancel"atm")#</mathjax></p>
<p>And this gives, I think, an answer of........</p>
<p><mathjax>#75.5*g*mol^-1#</mathjax>. Because we have got consistent units, I think our order of operations is correct. </p></div>
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<h1 class="questionTitle" itemprop="name">A gas has a density of #3.18*g*L^-1# at a temperature of #347*K#, and a pressure of #1.2*atm#. What is its molecular mass?</h1>
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<div class="markdown"><p>We use the Ideal Gas equation to get <mathjax>#"molecular mass"=75.5*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume ideality, and so <mathjax>#PV=nRT#</mathjax>, and thus,</p>
<p><mathjax>#PV=("mass"/"molar mass")RT#</mathjax> because <mathjax>#n="mass"/"molar mass"#</mathjax></p>
<p>On rearrangement, <mathjax>#"molar mass"="mass"/Vxx(RT)/P#</mathjax></p>
<p>But <mathjax>#"mass"/V=rho, "density"#</mathjax>, and thus...........</p>
<p><mathjax>#"molar mass"=(rhoRT)/P=#</mathjax></p>
<p><mathjax>#(3.18*g*L^-1xx0.0821*(L*atm)/(K*mol)xx347*K)/(1.2*atm)#</mathjax></p>
<p>Let's just cancel out the units to see if we have got this right. I am not immune to mistakes.............</p>
<p><mathjax>#"Molar mass"=(3.18*g*cancel(L^-1)xx0.0821*cancel(L*atm)/(cancelK*mol)xx347*cancelK)/(1.2*cancel"atm")#</mathjax></p>
<p>And this gives, I think, an answer of........</p>
<p><mathjax>#75.5*g*mol^-1#</mathjax>. Because we have got consistent units, I think our order of operations is correct. </p></div>
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</article> | A gas has a density of #3.18*g*L^-1# at a temperature of #347*K#, and a pressure of #1.2*atm#. What is its molecular mass? | null |
2,514 | aacc4890-6ddd-11ea-b8d2-ccda262736ce | https://socratic.org/questions/58d7f5a4b72cff483e8d02d7 | 24.6 g/mol | start physical_unit 23 24 molar_mass g/mol qc_end physical_unit 23 24 1 2 mass qc_end physical_unit 23 24 8 9 volume qc_end physical_unit 23 24 12 13 pressure qc_end physical_unit 23 24 15 16 temperature qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"24.6 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{0.20 g}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{0.200 L}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{1 atmosphere}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{300 K}"}] | <h1 class="questionTitle" itemprop="name">A #0.20*g# mass of gas occupies a #0.200*L# volume at one atmosphere, and #300*K#. What is the molar mass of the gas?</h1> | null | 24.6 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation.......</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>And thus <mathjax>#n="Mass"/"Molar Mass"=(PV)/(RT)#</mathjax></p>
<p>And so, <mathjax>#"Molar Mass"=("Mass"xxRT)/(PV)#</mathjax></p>
<p><mathjax>#=(0.20*gxx0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)/(1*cancel(atm)xx0.200*cancelL)#</mathjax></p>
<p><mathjax>#=24.6*g*mol^-1#</mathjax>. I think this is dimensionally correct, however, it does not correspond to a gas I can think of right now.........</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#24.6*g*mol^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation.......</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>And thus <mathjax>#n="Mass"/"Molar Mass"=(PV)/(RT)#</mathjax></p>
<p>And so, <mathjax>#"Molar Mass"=("Mass"xxRT)/(PV)#</mathjax></p>
<p><mathjax>#=(0.20*gxx0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)/(1*cancel(atm)xx0.200*cancelL)#</mathjax></p>
<p><mathjax>#=24.6*g*mol^-1#</mathjax>. I think this is dimensionally correct, however, it does not correspond to a gas I can think of right now.........</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A #0.20*g# mass of gas occupies a #0.200*L# volume at one atmosphere, and #300*K#. What is the molar mass of the gas?</h1>
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<div class="markdown"><p><mathjax>#"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#24.6*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation.......</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>And thus <mathjax>#n="Mass"/"Molar Mass"=(PV)/(RT)#</mathjax></p>
<p>And so, <mathjax>#"Molar Mass"=("Mass"xxRT)/(PV)#</mathjax></p>
<p><mathjax>#=(0.20*gxx0.0821*cancelL*cancel(atm)*cancel(K^-1)*mol^-1xx300*cancelK)/(1*cancel(atm)xx0.200*cancelL)#</mathjax></p>
<p><mathjax>#=24.6*g*mol^-1#</mathjax>. I think this is dimensionally correct, however, it does not correspond to a gas I can think of right now.........</p></div>
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</article> | A #0.20*g# mass of gas occupies a #0.200*L# volume at one atmosphere, and #300*K#. What is the molar mass of the gas? | null |
2,515 | aad1dcf6-6ddd-11ea-842d-ccda262736ce | https://socratic.org/questions/a-mass-of-134-g-of-manganese-dibromide-is-dissolved-in-225-g-of-water-what-is-th | 2.77 mol/kg | start physical_unit 20 21 molality mol/kg qc_end physical_unit 6 7 3 4 mass qc_end physical_unit 14 14 11 12 mass qc_end end | [{"type":"physical unit","value":"molality [OF] the solution [IN] mol/kg"}] | [{"type":"physical unit","value":"2.77 mol/kg"}] | [{"type":"physical unit","value":"Mass [OF] manganese dibromide [=] \\pu{134 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{225 g}"}] | <h1 class="questionTitle" itemprop="name">A mass of 134 g of manganese dibromide is dissolved in 225 g of water. What is the molality of the solution?</h1> | null | 2.77 mol/kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#"molality"="moles of solute"/"kilograms of solvent"#</mathjax>, and thus for this scenario, we work out the quotient,</p>
<p><mathjax>#"Molality"=((134*g)/(214.75*g*mol^-1))/(225*gxx10^-3*kg*g^-1)=??*mol*kg^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Molality"~=2.8*mol*kg^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#"molality"="moles of solute"/"kilograms of solvent"#</mathjax>, and thus for this scenario, we work out the quotient,</p>
<p><mathjax>#"Molality"=((134*g)/(214.75*g*mol^-1))/(225*gxx10^-3*kg*g^-1)=??*mol*kg^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A mass of 134 g of manganese dibromide is dissolved in 225 g of water. What is the molality of the solution?</h1>
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<div class="markdown"><p><mathjax>#"Molality"~=2.8*mol*kg^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, <mathjax>#"molality"="moles of solute"/"kilograms of solvent"#</mathjax>, and thus for this scenario, we work out the quotient,</p>
<p><mathjax>#"Molality"=((134*g)/(214.75*g*mol^-1))/(225*gxx10^-3*kg*g^-1)=??*mol*kg^-1#</mathjax></p></div>
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</article> | A mass of 134 g of manganese dibromide is dissolved in 225 g of water. What is the molality of the solution? | null |
2,516 | aa3b414d-6ddd-11ea-a049-ccda262736ce | https://socratic.org/questions/calculate-the-ph-of-a-solution-when-1-g-of-acetic-acid-and-4-g-of-sodium-hydroxi | 12.6 | start physical_unit 5 5 ph none qc_end physical_unit 10 11 7 8 mass qc_end physical_unit 16 17 13 14 mass qc_end physical_unit 24 24 21 22 volume qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"12.6"}] | [{"type":"physical unit","value":"Mass [OF] acetic acid [=] \\pu{1 g}"},{"type":"physical unit","value":"Mass [OF] sodium hydroxide [=] \\pu{4 g}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{2 L}"}] | <h1 class="questionTitle" itemprop="name"> Calculate the #"pH"# of a solution when #"1 g"# of acetic acid and #"4 g"# of sodium hydroxide are dissolved in #"2 L"# of water?</h1> | null | 12.6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine how many <em>moles</em> of each you are adding.</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05 color(red)(cancel(color(black)("g")))) = "0.0167 moles CH"_3"COOH"#</mathjax></p>
<p><mathjax>#4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.10 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, acetic acid and sodium hydroxide react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetate and water.</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ( (aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, you have more moles of sodium hydroxide than of acetic acid, so right from the start, you can say that sodium hydroxide is <em>in excess</em>, which implies that acetic acid is the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>This means that acetic acid will be <strong>completely consumed</strong> by the reaction. After the reaction is complete, the solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of acetic acid and </p>
<blockquote>
<p><mathjax>#"0.10 moles " - " 0.0167 moles" = "0.0833 moles NaOH"#</mathjax></p>
</blockquote>
<p>Sodium hydroxide is a <strong>strong base</strong>, so it dissociates completely in aqueous solution to produce hydroxide anions in a <mathjax>#1:1#</mathjax> mole ratio. This means that the resulting solution will contain <mathjax>#0.0833#</mathjax> <strong>moles</strong> of hydroxide anions.</p>
<p>You can thus say that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydroxide anions in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.0833 moles"/"2 L" = "0.04165 M"#</mathjax></p>
</blockquote>
<p>Finally, you know that an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#"pH + pOH = 14"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>you can say that the <mathjax>#"pH"#</mathjax> of the solution is given by the equation</p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(0.04165) = color(darkgreen)(ul(color(black)(12.6)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 12.6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine how many <em>moles</em> of each you are adding.</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05 color(red)(cancel(color(black)("g")))) = "0.0167 moles CH"_3"COOH"#</mathjax></p>
<p><mathjax>#4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.10 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, acetic acid and sodium hydroxide react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetate and water.</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ( (aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, you have more moles of sodium hydroxide than of acetic acid, so right from the start, you can say that sodium hydroxide is <em>in excess</em>, which implies that acetic acid is the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>This means that acetic acid will be <strong>completely consumed</strong> by the reaction. After the reaction is complete, the solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of acetic acid and </p>
<blockquote>
<p><mathjax>#"0.10 moles " - " 0.0167 moles" = "0.0833 moles NaOH"#</mathjax></p>
</blockquote>
<p>Sodium hydroxide is a <strong>strong base</strong>, so it dissociates completely in aqueous solution to produce hydroxide anions in a <mathjax>#1:1#</mathjax> mole ratio. This means that the resulting solution will contain <mathjax>#0.0833#</mathjax> <strong>moles</strong> of hydroxide anions.</p>
<p>You can thus say that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydroxide anions in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.0833 moles"/"2 L" = "0.04165 M"#</mathjax></p>
</blockquote>
<p>Finally, you know that an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#"pH + pOH = 14"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>you can say that the <mathjax>#"pH"#</mathjax> of the solution is given by the equation</p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(0.04165) = color(darkgreen)(ul(color(black)(12.6)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> Calculate the #"pH"# of a solution when #"1 g"# of acetic acid and #"4 g"# of sodium hydroxide are dissolved in #"2 L"# of water?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 12.6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by using the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine how many <em>moles</em> of each you are adding.</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05 color(red)(cancel(color(black)("g")))) = "0.0167 moles CH"_3"COOH"#</mathjax></p>
<p><mathjax>#4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "0.10 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, acetic acid and sodium hydroxide react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous sodium acetate and water.</p>
<blockquote>
<p><mathjax>#"CH"_ 3"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "CH"_ 3"COONa"_ ( (aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, you have more moles of sodium hydroxide than of acetic acid, so right from the start, you can say that sodium hydroxide is <em>in excess</em>, which implies that acetic acid is the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>. </p>
<p>This means that acetic acid will be <strong>completely consumed</strong> by the reaction. After the reaction is complete, the solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of acetic acid and </p>
<blockquote>
<p><mathjax>#"0.10 moles " - " 0.0167 moles" = "0.0833 moles NaOH"#</mathjax></p>
</blockquote>
<p>Sodium hydroxide is a <strong>strong base</strong>, so it dissociates completely in aqueous solution to produce hydroxide anions in a <mathjax>#1:1#</mathjax> mole ratio. This means that the resulting solution will contain <mathjax>#0.0833#</mathjax> <strong>moles</strong> of hydroxide anions.</p>
<p>You can thus say that the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydroxide anions in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.0833 moles"/"2 L" = "0.04165 M"#</mathjax></p>
</blockquote>
<p>Finally, you know that an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#"pH + pOH = 14"#</mathjax></p>
</blockquote>
<p>Since</p>
<blockquote>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>you can say that the <mathjax>#"pH"#</mathjax> of the solution is given by the equation</p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = 14 + log(0.04165) = color(darkgreen)(ul(color(black)(12.6)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for your values. </p></div>
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</article> | Calculate the #"pH"# of a solution when #"1 g"# of acetic acid and #"4 g"# of sodium hydroxide are dissolved in #"2 L"# of water? | null |
2,517 | aa3b013a-6ddd-11ea-8790-ccda262736ce | https://socratic.org/questions/562fd27311ef6b390f1bf22c | 0.35 g | start physical_unit 7 8 mass g qc_end physical_unit 7 8 4 5 volume qc_end c_other STP qc_end physical_unit 7 8 15 16 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] helium gas [IN] g"}] | [{"type":"physical unit","value":"0.35 g"}] | [{"type":"physical unit","value":"Volume [OF] helium gas [=] \\pu{2.0 L}"},{"type":"other","value":"STP"},{"type":"physical unit","value":"Molar mass [OF] helium gas [=] \\pu{4.002602 g/mol}"}] | <h1 class="questionTitle" itemprop="name">Find the mass of #"2.0 L"# of helium gas at STP? The molar mass is #"4.002602 g/mol"#. </h1> | null | 0.35 g | <div class="answerDescription">
<div>
<div class="markdown"><p>To do this problem, we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, knowing that we are at STP:</p>
<blockquote>
<p><mathjax>#PbarV = RT#</mathjax></p>
</blockquote>
<p>where <mathjax>#barV = V/n#</mathjax>, the molar volume.</p>
<p>For an ideal gas at STP:</p>
<ul>
<li><mathjax>#P = "1 bar"#</mathjax> (standard pressure)</li>
<li><mathjax>#barV_"ideal" = ?#</mathjax></li>
<li><mathjax>#R = "0.083145 L" cdot "bar/mol" cdot "K"#</mathjax></li>
<li><mathjax>#T = "273.15 K"#</mathjax> (standard temperature)</li>
</ul>
<p>We can solve for the ideal molar volume <mathjax>#barV_"ideal"#</mathjax>, and then from that determine the <mathjax>#"mol"#</mathjax>s of helium and thus the mass.</p>
<blockquote>
<p><mathjax>#barV_"ideal" = (RT)/P#</mathjax></p>
<p><mathjax>#= (0.083145*273.15)/(1) = color(green)"22.711 L/mol"#</mathjax></p>
</blockquote>
<p>Then, we know that when assuming helium is <em>ideal</em>:</p>
<blockquote>
<p><mathjax>#barV_"ideal" = barV_"He"#</mathjax></p>
<p><mathjax>#"22.711 L"/"1 mol ideal gas" = "2.0 L"/(n_"He")#</mathjax></p>
<p><mathjax>#n_"He" = "0.0881 mol"#</mathjax></p>
</blockquote>
<p>Lastly, converting to grams:</p>
<blockquote>
<p><mathjax>#0.0881 cancel"mol He" xx ("4.002602 g")/(cancel"mol He")#</mathjax></p>
<p><mathjax>#color(blue)("= 0.3525 g He")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerDescription">
<div>
<div class="markdown"><p>To do this problem, we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, knowing that we are at STP:</p>
<blockquote>
<p><mathjax>#PbarV = RT#</mathjax></p>
</blockquote>
<p>where <mathjax>#barV = V/n#</mathjax>, the molar volume.</p>
<p>For an ideal gas at STP:</p>
<ul>
<li><mathjax>#P = "1 bar"#</mathjax> (standard pressure)</li>
<li><mathjax>#barV_"ideal" = ?#</mathjax></li>
<li><mathjax>#R = "0.083145 L" cdot "bar/mol" cdot "K"#</mathjax></li>
<li><mathjax>#T = "273.15 K"#</mathjax> (standard temperature)</li>
</ul>
<p>We can solve for the ideal molar volume <mathjax>#barV_"ideal"#</mathjax>, and then from that determine the <mathjax>#"mol"#</mathjax>s of helium and thus the mass.</p>
<blockquote>
<p><mathjax>#barV_"ideal" = (RT)/P#</mathjax></p>
<p><mathjax>#= (0.083145*273.15)/(1) = color(green)"22.711 L/mol"#</mathjax></p>
</blockquote>
<p>Then, we know that when assuming helium is <em>ideal</em>:</p>
<blockquote>
<p><mathjax>#barV_"ideal" = barV_"He"#</mathjax></p>
<p><mathjax>#"22.711 L"/"1 mol ideal gas" = "2.0 L"/(n_"He")#</mathjax></p>
<p><mathjax>#n_"He" = "0.0881 mol"#</mathjax></p>
</blockquote>
<p>Lastly, converting to grams:</p>
<blockquote>
<p><mathjax>#0.0881 cancel"mol He" xx ("4.002602 g")/(cancel"mol He")#</mathjax></p>
<p><mathjax>#color(blue)("= 0.3525 g He")#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Find the mass of #"2.0 L"# of helium gas at STP? The molar mass is #"4.002602 g/mol"#. </h1>
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Oct 27, 2015
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<div class="markdown"><p>To do this problem, we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, knowing that we are at STP:</p>
<blockquote>
<p><mathjax>#PbarV = RT#</mathjax></p>
</blockquote>
<p>where <mathjax>#barV = V/n#</mathjax>, the molar volume.</p>
<p>For an ideal gas at STP:</p>
<ul>
<li><mathjax>#P = "1 bar"#</mathjax> (standard pressure)</li>
<li><mathjax>#barV_"ideal" = ?#</mathjax></li>
<li><mathjax>#R = "0.083145 L" cdot "bar/mol" cdot "K"#</mathjax></li>
<li><mathjax>#T = "273.15 K"#</mathjax> (standard temperature)</li>
</ul>
<p>We can solve for the ideal molar volume <mathjax>#barV_"ideal"#</mathjax>, and then from that determine the <mathjax>#"mol"#</mathjax>s of helium and thus the mass.</p>
<blockquote>
<p><mathjax>#barV_"ideal" = (RT)/P#</mathjax></p>
<p><mathjax>#= (0.083145*273.15)/(1) = color(green)"22.711 L/mol"#</mathjax></p>
</blockquote>
<p>Then, we know that when assuming helium is <em>ideal</em>:</p>
<blockquote>
<p><mathjax>#barV_"ideal" = barV_"He"#</mathjax></p>
<p><mathjax>#"22.711 L"/"1 mol ideal gas" = "2.0 L"/(n_"He")#</mathjax></p>
<p><mathjax>#n_"He" = "0.0881 mol"#</mathjax></p>
</blockquote>
<p>Lastly, converting to grams:</p>
<blockquote>
<p><mathjax>#0.0881 cancel"mol He" xx ("4.002602 g")/(cancel"mol He")#</mathjax></p>
<p><mathjax>#color(blue)("= 0.3525 g He")#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p>The mass of helium present is <mathjax>#"0.36 g"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"STP"#</mathjax> is <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>.</p>
<p>You will use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> in order to determine moles of helium. Then you will multiply the moles of helium times its molar mass to determine the mass of helium in grams.</p>
<p><strong>Ideal gas law</strong></p>
<p><mathjax>#PV=nRT#</mathjax>, where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, a <mathjax>#T#</mathjax> is the temperature.</p>
<p><strong>Known/Given</strong><br/>
<mathjax>#P="1 atm"#</mathjax><br/>
<mathjax>#V="2.0 L"#</mathjax><br/>
<mathjax>#R="0.08206 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="273.15 K"#</mathjax><br/>
Molar mass of helium: <mathjax>#"4.002602 g/mol"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Moles of helium:<mathjax>#n_"He"#</mathjax><br/>
Mass of helium in grams</p>
<p><strong>Equations</strong></p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n_"He"xx(4.002602"g He")/(1"mol He")#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the ideal gas law equation to isolate <mathjax>#n#</mathjax>, then solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(1cancel"atm"*2.0cancel"L")/(0.08206cancel"L atm K"^(-1)"mol"^(-1)*273.15cancel"K")="0.089 mol He"#</mathjax></p>
<p><mathjax>#0.089cancel"mol He"xx(4.002602"g He")/(1cancel"mol He")="0.36 g He"#</mathjax></p></div>
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</article> | Find the mass of #"2.0 L"# of helium gas at STP? The molar mass is #"4.002602 g/mol"#. | null |
2,518 | ab19e729-6ddd-11ea-97cb-ccda262736ce | https://socratic.org/questions/how-would-you-balance-n2o5-no2-o2 | 2 N2O5 -> 4 NO2 + O2 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 N2O5 -> 4 NO2 + O2"}] | [{"type":"chemical equation","value":"N2O5 -> NO2 + O2"}] | <h1 class="questionTitle" itemprop="name">How would you balance
_N2O5 -->_NO2+_O2?</h1> | null | 2 N2O5 -> 4 NO2 + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Can I have 1/2 a molecule of <mathjax>#O_2(g)#</mathjax>? How do I get rid of the <mathjax>#1/2#</mathjax> coefficient? Is this a redox reaction?</p></div>
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<div class="markdown"><p><mathjax>#N_2O_5(g) rarr 2NO_2(g) + 1/2O_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Can I have 1/2 a molecule of <mathjax>#O_2(g)#</mathjax>? How do I get rid of the <mathjax>#1/2#</mathjax> coefficient? Is this a redox reaction?</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance
_N2O5 -->_NO2+_O2?</h1>
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<div class="markdown"><p><mathjax>#N_2O_5(g) rarr 2NO_2(g) + 1/2O_2(g)#</mathjax></p></div>
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<div class="markdown"><p>Can I have 1/2 a molecule of <mathjax>#O_2(g)#</mathjax>? How do I get rid of the <mathjax>#1/2#</mathjax> coefficient? Is this a redox reaction?</p></div>
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<span class="dateCreated" datetime="2015-10-21T11:33:24" itemprop="dateCreated">
Oct 21, 2015
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<div class="markdown"><p><mathjax>#"2N"_2"O"_5#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NO"_2+"O"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"N"_2"O"_5#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NO"_2+"O"_2#</mathjax></p>
<p>Add coefficients in front of the formulas in order to balance the equation.</p>
<p><mathjax>#"2N"_2"O"_5#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NO"_2+"O"_2#</mathjax></p>
<p>There are now 4 atoms of N and 10 atoms of oxygen on both sides, and the equation is balanced.</p></div>
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</article> | How would you balance
_N2O5 -->_NO2+_O2? | null |
2,519 | aa90fe43-6ddd-11ea-9b78-ccda262736ce | https://socratic.org/questions/5926653d7c0149508b4fe969 | 2.30 grams | start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium metal [IN] grams"}] | [{"type":"physical unit","value":"2.30 grams"}] | [{"type":"physical unit","value":"Mole [OF] sodium metal [=] \\pu{0.1 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #0.1*mol# of sodium metal?</h1> | null | 2.30 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do I know this? Well, I have got a Periodic Table beside me, and you should have one beside you too.</p>
<p>And so we take the product............</p>
<p><mathjax>#0.1*cancel(mol)xx22.99*g*cancel(mol^-1)=2.30*g..........#</mathjax></p>
<p>This is an example or what is called dimensional analysis. We wanted an answer in <mathjax>#"grams"#</mathjax>, and the product gave us an answer with dimensions of <mathjax>#"grams"#</mathjax>. This persuades us that we got the order of operations right. It is all too easy to divide when you should multiply, or vice versa. </p></div>
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<div class="markdown"><p>Well, the mass of one mole of sodium is <mathjax>#22.99*g#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>How do I know this? Well, I have got a Periodic Table beside me, and you should have one beside you too.</p>
<p>And so we take the product............</p>
<p><mathjax>#0.1*cancel(mol)xx22.99*g*cancel(mol^-1)=2.30*g..........#</mathjax></p>
<p>This is an example or what is called dimensional analysis. We wanted an answer in <mathjax>#"grams"#</mathjax>, and the product gave us an answer with dimensions of <mathjax>#"grams"#</mathjax>. This persuades us that we got the order of operations right. It is all too easy to divide when you should multiply, or vice versa. </p></div>
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<div class="markdown"><p>Well, the mass of one mole of sodium is <mathjax>#22.99*g#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>How do I know this? Well, I have got a Periodic Table beside me, and you should have one beside you too.</p>
<p>And so we take the product............</p>
<p><mathjax>#0.1*cancel(mol)xx22.99*g*cancel(mol^-1)=2.30*g..........#</mathjax></p>
<p>This is an example or what is called dimensional analysis. We wanted an answer in <mathjax>#"grams"#</mathjax>, and the product gave us an answer with dimensions of <mathjax>#"grams"#</mathjax>. This persuades us that we got the order of operations right. It is all too easy to divide when you should multiply, or vice versa. </p></div>
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</article> | What is the mass of #0.1*mol# of sodium metal? | null |
2,520 | a8a9b862-6ddd-11ea-aaf2-ccda262736ce | https://socratic.org/questions/what-volume-in-ml-s-of-a-6-m-stock-solution-of-hcl-should-be-used-to-prepare-6-0 | 250.00 mL | start physical_unit 9 9 volume ml qc_end physical_unit 23 24 21 22 molarity qc_end physical_unit 23 24 17 18 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] HCl stock solution [IN] mL"}] | [{"type":"physical unit","value":"250.00 mL"}] | [{"type":"physical unit","value":"Molarity [OF] HCl stock solution [=] \\pu{6.0 M}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.250 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{6.00 L}"}] | <h1 class="questionTitle" itemprop="name">What volume (in mL's) of a 6.0 M stock solution of HCl should be used to prepare 6.00 L of a 0.250 M HCl solution?</h1> | null | 250.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume for a second that you're not familiar with the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>.</p>
<p>What would your approach be here? </p>
<p>As you know, <em>diluting</em> a solution means keeping the number of moles of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>constant</strong> while <strong>increasing</strong> the total volume of the solution by adding more <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://slideplayer.com/slide/2557276/" src="https://useruploads.socratic.org/ADMgZ3mSQmrDL0e8987w_slide_29.jpg"/> </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liters of solution, diluting a solution will result in a <strong>decrease</strong> in its concentration. </p>
<p>So, if the number of moles of solute <strong>must be constant</strong> in order for a dilution to be performed, you can use the molarity and volume of the <em>target solution</em> to figure out how many moles of solute <strong>must be present</strong> in the sample of stock solution you're diluting. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.250 M" * "6.00 L" = "1.5 moles HCl"#</mathjax></p>
</blockquote>
<p>Now all you have to do is figure out what volume of <mathjax>#"6.0 M"#</mathjax> stock solution will contain <mathjax>#1.5#</mathjax> moles of hydrochloric acid </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V_"stock" = (1.5 color(red)(cancel(color(black)("moles"))))/(6.0 color(red)(cancel(color(black)("moles")))/"L") = "0.25 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("250 mL") ->#</mathjax> <em>rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>This is the principle behind the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(c_1V_1 = c_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the dilute solution</p>
<p>Notice that you actually have </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(stackrel( color(brown)("moles of solute"))(color(brown)("in stock solution"))) = overbrace(c_2V_2)^(stackrel(color(purple)("moles of solute"))(color(purple)("in target solution")))#</mathjax></p>
</blockquote>
<p>In your case, you'll once again get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.250 color(red)(cancel(color(black)("M"))))/(6.0color(red)(cancel(color(black)("M")))) * "6.00 L" = "0.25 L" = color(green)("250 L")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"250 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume for a second that you're not familiar with the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>.</p>
<p>What would your approach be here? </p>
<p>As you know, <em>diluting</em> a solution means keeping the number of moles of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>constant</strong> while <strong>increasing</strong> the total volume of the solution by adding more <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://slideplayer.com/slide/2557276/" src="https://useruploads.socratic.org/ADMgZ3mSQmrDL0e8987w_slide_29.jpg"/> </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liters of solution, diluting a solution will result in a <strong>decrease</strong> in its concentration. </p>
<p>So, if the number of moles of solute <strong>must be constant</strong> in order for a dilution to be performed, you can use the molarity and volume of the <em>target solution</em> to figure out how many moles of solute <strong>must be present</strong> in the sample of stock solution you're diluting. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.250 M" * "6.00 L" = "1.5 moles HCl"#</mathjax></p>
</blockquote>
<p>Now all you have to do is figure out what volume of <mathjax>#"6.0 M"#</mathjax> stock solution will contain <mathjax>#1.5#</mathjax> moles of hydrochloric acid </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V_"stock" = (1.5 color(red)(cancel(color(black)("moles"))))/(6.0 color(red)(cancel(color(black)("moles")))/"L") = "0.25 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("250 mL") ->#</mathjax> <em>rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>This is the principle behind the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(c_1V_1 = c_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the dilute solution</p>
<p>Notice that you actually have </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(stackrel( color(brown)("moles of solute"))(color(brown)("in stock solution"))) = overbrace(c_2V_2)^(stackrel(color(purple)("moles of solute"))(color(purple)("in target solution")))#</mathjax></p>
</blockquote>
<p>In your case, you'll once again get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.250 color(red)(cancel(color(black)("M"))))/(6.0color(red)(cancel(color(black)("M")))) * "6.00 L" = "0.25 L" = color(green)("250 L")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume (in mL's) of a 6.0 M stock solution of HCl should be used to prepare 6.00 L of a 0.250 M HCl solution?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-11T00:17:28" itemprop="dateCreated">
Jan 11, 2016
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<div class="markdown"><p><mathjax>#"250 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume for a second that you're not familiar with the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>.</p>
<p>What would your approach be here? </p>
<p>As you know, <em>diluting</em> a solution means keeping the number of moles of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>constant</strong> while <strong>increasing</strong> the total volume of the solution by adding more <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p><img alt="http://slideplayer.com/slide/2557276/" src="https://useruploads.socratic.org/ADMgZ3mSQmrDL0e8987w_slide_29.jpg"/> </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of solute per liters of solution, diluting a solution will result in a <strong>decrease</strong> in its concentration. </p>
<p>So, if the number of moles of solute <strong>must be constant</strong> in order for a dilution to be performed, you can use the molarity and volume of the <em>target solution</em> to figure out how many moles of solute <strong>must be present</strong> in the sample of stock solution you're diluting. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HCl) = "0.250 M" * "6.00 L" = "1.5 moles HCl"#</mathjax></p>
</blockquote>
<p>Now all you have to do is figure out what volume of <mathjax>#"6.0 M"#</mathjax> stock solution will contain <mathjax>#1.5#</mathjax> moles of hydrochloric acid </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies V = n/c)#</mathjax></p>
<p><mathjax>#V_"stock" = (1.5 color(red)(cancel(color(black)("moles"))))/(6.0 color(red)(cancel(color(black)("moles")))/"L") = "0.25 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)("250 mL") ->#</mathjax> <em>rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>This is the principle behind the equation for <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(c_1V_1 = c_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the concentrated solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the dilute solution</p>
<p>Notice that you actually have </p>
<blockquote>
<p><mathjax>#overbrace(c_1V_1)^(stackrel( color(brown)("moles of solute"))(color(brown)("in stock solution"))) = overbrace(c_2V_2)^(stackrel(color(purple)("moles of solute"))(color(purple)("in target solution")))#</mathjax></p>
</blockquote>
<p>In your case, you'll once again get </p>
<blockquote>
<p><mathjax>#V_1 = c_2/c_1 * V_2#</mathjax></p>
<p><mathjax>#V_1 = (0.250 color(red)(cancel(color(black)("M"))))/(6.0color(red)(cancel(color(black)("M")))) * "6.00 L" = "0.25 L" = color(green)("250 L")#</mathjax></p>
</blockquote></div>
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</article> | What volume (in mL's) of a 6.0 M stock solution of HCl should be used to prepare 6.00 L of a 0.250 M HCl solution? | null |
2,521 | a9da2152-6ddd-11ea-94ef-ccda262736ce | https://socratic.org/questions/what-volume-will-3-4-g-of-co-2-occupy-at-stp | 1.76 L | start physical_unit 6 6 volume l qc_end physical_unit 6 6 3 4 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 [IN] L"}] | [{"type":"physical unit","value":"1.76 L"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{3.4 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What volume will 3.4 g of #CO_2# occupy at STP?</h1> | null | 1.76 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At <mathjax>#"STP"#</mathjax>, a gas will contain approximately a volume of <mathjax>#22.7 \ "L"#</mathjax>. Click <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure#Molar_volume_of_a_gas" rel="nofollow">here</a> for the explanation.</p>
<p>So here, we need to convert <mathjax>#3.4 \ "g"#</mathjax> of carbon dioxide into moles.</p>
<p>Carbon dioxide has a molar mass of <mathjax>#44 \ "g/mol"#</mathjax>. So here, there exist,</p>
<p><mathjax>#(3.4color(red)cancelcolor(black)"g")/(44color(red)cancelcolor(black)"g""/mol")=0.0772727273 \ "mol"#</mathjax></p>
<p>And so, the molar volume is:</p>
<p><mathjax>#0.0772727273color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=1.75409090971 \ "L"#</mathjax></p>
<p><mathjax>#~~1.755 \ "L"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#~~1.755 \ "L"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At <mathjax>#"STP"#</mathjax>, a gas will contain approximately a volume of <mathjax>#22.7 \ "L"#</mathjax>. Click <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure#Molar_volume_of_a_gas" rel="nofollow">here</a> for the explanation.</p>
<p>So here, we need to convert <mathjax>#3.4 \ "g"#</mathjax> of carbon dioxide into moles.</p>
<p>Carbon dioxide has a molar mass of <mathjax>#44 \ "g/mol"#</mathjax>. So here, there exist,</p>
<p><mathjax>#(3.4color(red)cancelcolor(black)"g")/(44color(red)cancelcolor(black)"g""/mol")=0.0772727273 \ "mol"#</mathjax></p>
<p>And so, the molar volume is:</p>
<p><mathjax>#0.0772727273color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=1.75409090971 \ "L"#</mathjax></p>
<p><mathjax>#~~1.755 \ "L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume will 3.4 g of #CO_2# occupy at STP?</h1>
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<div class="markdown"><p><mathjax>#~~1.755 \ "L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At <mathjax>#"STP"#</mathjax>, a gas will contain approximately a volume of <mathjax>#22.7 \ "L"#</mathjax>. Click <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure#Molar_volume_of_a_gas" rel="nofollow">here</a> for the explanation.</p>
<p>So here, we need to convert <mathjax>#3.4 \ "g"#</mathjax> of carbon dioxide into moles.</p>
<p>Carbon dioxide has a molar mass of <mathjax>#44 \ "g/mol"#</mathjax>. So here, there exist,</p>
<p><mathjax>#(3.4color(red)cancelcolor(black)"g")/(44color(red)cancelcolor(black)"g""/mol")=0.0772727273 \ "mol"#</mathjax></p>
<p>And so, the molar volume is:</p>
<p><mathjax>#0.0772727273color(red)cancelcolor(black)"mol"*(22.7 \ "L")/(color(red)cancelcolor(black)"mol")=1.75409090971 \ "L"#</mathjax></p>
<p><mathjax>#~~1.755 \ "L"#</mathjax></p></div>
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</article> | What volume will 3.4 g of #CO_2# occupy at STP? | null |
2,522 | a99c5d3b-6ddd-11ea-9dde-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-equation-nahco-3-naoh-co-2 | NaHCO3 -> NaOH + CO2 | start chemical_equation qc_end chemical_equation 6 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"NaHCO3 -> NaOH + CO2"}] | [{"type":"chemical equation","value":"? NaHCO3 -> ? NaOH + ? CO2"}] | <h1 class="questionTitle" itemprop="name">How do you balance this equation: #?NaHCO_3 → ?NaOH + ?CO_2#?</h1> | null | NaHCO3 -> NaOH + CO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Na is sodium,<br/>
H is hydrogen,<br/>
C is carbon and <br/>
O is oxygen.</p>
<p>Sodium is balanced, there's 1 on each side.<br/>
Oxygen is also balanced, there's 3 on each side. (For the RHS, add up the O in <mathjax>#NaOH#</mathjax> and <mathjax>#CO_2#</mathjax>. <br/>
Hydrogen is balanced, there's 1 on each side.<br/>
Carbon is balanced, there's 1 on each side. </p></div>
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<div class="markdown"><p>It is already balanced.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Na is sodium,<br/>
H is hydrogen,<br/>
C is carbon and <br/>
O is oxygen.</p>
<p>Sodium is balanced, there's 1 on each side.<br/>
Oxygen is also balanced, there's 3 on each side. (For the RHS, add up the O in <mathjax>#NaOH#</mathjax> and <mathjax>#CO_2#</mathjax>. <br/>
Hydrogen is balanced, there's 1 on each side.<br/>
Carbon is balanced, there's 1 on each side. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance this equation: #?NaHCO_3 → ?NaOH + ?CO_2#?</h1>
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<div class="markdown"><p>It is already balanced.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Na is sodium,<br/>
H is hydrogen,<br/>
C is carbon and <br/>
O is oxygen.</p>
<p>Sodium is balanced, there's 1 on each side.<br/>
Oxygen is also balanced, there's 3 on each side. (For the RHS, add up the O in <mathjax>#NaOH#</mathjax> and <mathjax>#CO_2#</mathjax>. <br/>
Hydrogen is balanced, there's 1 on each side.<br/>
Carbon is balanced, there's 1 on each side. </p></div>
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<div class="markdown"><p>All question marks get 1.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NaHCO_3 -> NaOH + CO_2#</mathjax></p>
<p>If you compute reactants and products, check it:</p>
<p>Reactants: <mathjax>#1 Na, 1 H, 1 C, and 3 O#</mathjax> <br/>
Products: <mathjax>#1 Na, 1 H, C, and (1+2) O#</mathjax></p>
<p>This reaction is a balanced reaction:</p>
<p><mathjax>#NaHCO_3 -> NaOH + CO_2#</mathjax></p></div>
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</article> | How do you balance this equation: #?NaHCO_3 → ?NaOH + ?CO_2#? | null |
2,523 | a940397a-6ddd-11ea-a39f-ccda262736ce | https://socratic.org/questions/how-many-millliters-of-275-m-h-2so-4-aq-are-needed-to-react-completely-with-58-5 | 1254.55 millliters | start physical_unit 6 6 volume ml qc_end physical_unit 6 6 4 5 molarity qc_end physical_unit 16 16 13 14 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] H2SO4(aq) [IN] millliters"}] | [{"type":"physical unit","value":"1254.55 millliters"}] | [{"type":"physical unit","value":"Molarity [OF] H2SO4(aq) [=] \\pu{275 M}"},{"type":"physical unit","value":"Mass [OF] BaO2(s) [=] \\pu{58.5 g}"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">How many millliters of 275 M #H_2SO_4(aq)# are needed to react completely with 58.5 g of #BaO_2(s)#?</h1> | null | 1254.55 millliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#BaO_2(s) + H_2SO_4(aq) rarr BaSO_4darr + H_2O_2(aq)#</mathjax></p>
<p>This used to be used as an industrial method to make hydrogen peroxide. </p>
<p>So <mathjax>#"moles of barium peroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(58.5*g)/(169.33*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.345*mol#</mathjax>.</p>
<p>We need <mathjax>#1#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid.</p>
<p>If <mathjax>#0.275*mol*L^-1#</mathjax> sulfuric acid is available, we need <mathjax>#(0.345*mol)/(0.275*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1255*mL#</mathjax>.</p></div>
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<div class="markdown"><p>Are you sure mean barium peroxide, <mathjax>#BaO_2#</mathjax>, and not barium oxide, <mathjax>#BaO#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#BaO_2(s) + H_2SO_4(aq) rarr BaSO_4darr + H_2O_2(aq)#</mathjax></p>
<p>This used to be used as an industrial method to make hydrogen peroxide. </p>
<p>So <mathjax>#"moles of barium peroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(58.5*g)/(169.33*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.345*mol#</mathjax>.</p>
<p>We need <mathjax>#1#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid.</p>
<p>If <mathjax>#0.275*mol*L^-1#</mathjax> sulfuric acid is available, we need <mathjax>#(0.345*mol)/(0.275*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1255*mL#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many millliters of 275 M #H_2SO_4(aq)# are needed to react completely with 58.5 g of #BaO_2(s)#?</h1>
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<div class="markdown"><p>Are you sure mean barium peroxide, <mathjax>#BaO_2#</mathjax>, and not barium oxide, <mathjax>#BaO#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#BaO_2(s) + H_2SO_4(aq) rarr BaSO_4darr + H_2O_2(aq)#</mathjax></p>
<p>This used to be used as an industrial method to make hydrogen peroxide. </p>
<p>So <mathjax>#"moles of barium peroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(58.5*g)/(169.33*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.345*mol#</mathjax>.</p>
<p>We need <mathjax>#1#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid.</p>
<p>If <mathjax>#0.275*mol*L^-1#</mathjax> sulfuric acid is available, we need <mathjax>#(0.345*mol)/(0.275*mol*L^-1)xx1000*mL*L^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1255*mL#</mathjax>.</p></div>
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</article> | How many millliters of 275 M #H_2SO_4(aq)# are needed to react completely with 58.5 g of #BaO_2(s)#? | null |
2,524 | a9ca0181-6ddd-11ea-89de-ccda262736ce | https://socratic.org/questions/5886a6b8b72cff3c781c03eb | 2.71 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 2 2 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] water molecules"}] | [{"type":"physical unit","value":"2.71 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] water [=] \\pu{4.5 mol}"}] | <h1 class="questionTitle" itemprop="name">How many water molecules in a #4.5*mol# quantity of water?</h1> | null | 2.71 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#4.5xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number"=6.022xx10^23*mol^-1#</mathjax>.</p>
<p><mathjax>#"The mole"#</mathjax> is simply a number, like a dozen, or a score, or a gross. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> has a special property. <mathjax>#1*mol#</mathjax> of <mathjax>#""^1H#</mathjax> atoms has a mass of <mathjax>#1*g#</mathjax>.</p>
<p>What is the mass of this quantity of water molecules?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#4.5xxN_A#</mathjax>, where............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#4.5xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number"=6.022xx10^23*mol^-1#</mathjax>.</p>
<p><mathjax>#"The mole"#</mathjax> is simply a number, like a dozen, or a score, or a gross. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> has a special property. <mathjax>#1*mol#</mathjax> of <mathjax>#""^1H#</mathjax> atoms has a mass of <mathjax>#1*g#</mathjax>.</p>
<p>What is the mass of this quantity of water molecules?</p></div>
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<h1 class="questionTitle" itemprop="name">How many water molecules in a #4.5*mol# quantity of water?</h1>
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<div class="markdown"><p><mathjax>#4.5xxN_A#</mathjax>, where............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#4.5xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number"=6.022xx10^23*mol^-1#</mathjax>.</p>
<p><mathjax>#"The mole"#</mathjax> is simply a number, like a dozen, or a score, or a gross. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> has a special property. <mathjax>#1*mol#</mathjax> of <mathjax>#""^1H#</mathjax> atoms has a mass of <mathjax>#1*g#</mathjax>.</p>
<p>What is the mass of this quantity of water molecules?</p></div>
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</article> | How many water molecules in a #4.5*mol# quantity of water? | null |
2,525 | aa5b69ba-6ddd-11ea-b29c-ccda262736ce | https://socratic.org/questions/the-concentration-of-h-ions-in-a-solution-is-1-0-x-10-12-m-what-is-the-ph-of-the | 12.00 | start physical_unit 18 19 ph none qc_end physical_unit 3 4 9 12 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"12.00"}] | [{"type":"physical unit","value":"Concentration [OF] H+ ions [=] \\pu{1.0 × 10^(-12) M}"}] | <h1 class="questionTitle" itemprop="name">The concentration of H+ ions in a solution is #1.0 xx 10^-12# M. What is the pH of the solution? </h1> | null | 12.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the <mathjax>#"pH"#</mathjax> of a solution with a known hydrogen (hydronium) ion concentration (<mathjax>#["H"^+]#</mathjax> or <mathjax>#["H"_3"O"^+]#</mathjax>; they both represent the same thing in this case).</p>
<p>To calculate the <mathjax>#"pH"#</mathjax> of a solution from a known <mathjax>#["H"^+]#</mathjax>, we can use the formula </p>
<p><mathjax>#"pH" = -log["H"^+]#</mathjax></p>
<p>That is, the <mathjax>#"pH"#</mathjax> of a solution is the negative of the logarithm of the hydronium/hydrogen ion concentration.</p>
<p>Using this formula, we have</p>
<p><mathjax>#"pH" = -log(1.0 xx 10^-12M) = color(red)(12.00#</mathjax></p>
<p>The number of <em><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em> in the <mathjax>#["H"^+]#</mathjax> is equal to the number of <em>decimal places</em> of the <mathjax>#"pH"#</mathjax>, so the <mathjax>#"pH"#</mathjax> is <mathjax>#12.00#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#12.00#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the <mathjax>#"pH"#</mathjax> of a solution with a known hydrogen (hydronium) ion concentration (<mathjax>#["H"^+]#</mathjax> or <mathjax>#["H"_3"O"^+]#</mathjax>; they both represent the same thing in this case).</p>
<p>To calculate the <mathjax>#"pH"#</mathjax> of a solution from a known <mathjax>#["H"^+]#</mathjax>, we can use the formula </p>
<p><mathjax>#"pH" = -log["H"^+]#</mathjax></p>
<p>That is, the <mathjax>#"pH"#</mathjax> of a solution is the negative of the logarithm of the hydronium/hydrogen ion concentration.</p>
<p>Using this formula, we have</p>
<p><mathjax>#"pH" = -log(1.0 xx 10^-12M) = color(red)(12.00#</mathjax></p>
<p>The number of <em><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em> in the <mathjax>#["H"^+]#</mathjax> is equal to the number of <em>decimal places</em> of the <mathjax>#"pH"#</mathjax>, so the <mathjax>#"pH"#</mathjax> is <mathjax>#12.00#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The concentration of H+ ions in a solution is #1.0 xx 10^-12# M. What is the pH of the solution? </h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-05-30T21:20:42" itemprop="dateCreated">
May 30, 2017
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<div class="markdown"><p><mathjax>#12.00#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the <mathjax>#"pH"#</mathjax> of a solution with a known hydrogen (hydronium) ion concentration (<mathjax>#["H"^+]#</mathjax> or <mathjax>#["H"_3"O"^+]#</mathjax>; they both represent the same thing in this case).</p>
<p>To calculate the <mathjax>#"pH"#</mathjax> of a solution from a known <mathjax>#["H"^+]#</mathjax>, we can use the formula </p>
<p><mathjax>#"pH" = -log["H"^+]#</mathjax></p>
<p>That is, the <mathjax>#"pH"#</mathjax> of a solution is the negative of the logarithm of the hydronium/hydrogen ion concentration.</p>
<p>Using this formula, we have</p>
<p><mathjax>#"pH" = -log(1.0 xx 10^-12M) = color(red)(12.00#</mathjax></p>
<p>The number of <em><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em> in the <mathjax>#["H"^+]#</mathjax> is equal to the number of <em>decimal places</em> of the <mathjax>#"pH"#</mathjax>, so the <mathjax>#"pH"#</mathjax> is <mathjax>#12.00#</mathjax>.</p></div>
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</article> | The concentration of H+ ions in a solution is #1.0 xx 10^-12# M. What is the pH of the solution? | null |
2,526 | a967b6a4-6ddd-11ea-8726-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-hydroxide-ions-in-a-solution-that-has-a-ph-of-6-0 | 10^(-8) M | start physical_unit 5 6 concentration mol/l qc_end physical_unit 9 9 15 15 ph qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydroxide ions [IN] M"}] | [{"type":"physical unit","value":"10^(-8) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{6.0}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of hydroxide ions in a solution that has a pH of 6.0?</h1> | null | 10^(-8) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that water undergoes self-ionization, which we may represent by the following reaction:</p>
<p><mathjax>#H_2O rightleftharpoons H^+ + HO^-#</mathjax></p>
<p>This equilibrium has been carefully measured, and at <mathjax>#298*K#</mathjax>, we may write,</p>
<p><mathjax>#[H^+][HO^-]=10^-14#</mathjax>, and taking <mathjax>#log_10#</mathjax> of both sides,</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p><mathjax>#-log_10[H_3O^+]-log_10[HO^-]=+14#</mathjax></p>
<p>Given the definition of <mathjax>#pH=-log_10[H^+]#</mathjax>, then...........</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>So if <mathjax>#pH=6#</mathjax>, then <mathjax>#pOH=8#</mathjax> and <mathjax>#[HO^-]=10^-8*mol*L^-1#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#[HO^-]=10^-8*mol*L^-1#</mathjax>..........</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that water undergoes self-ionization, which we may represent by the following reaction:</p>
<p><mathjax>#H_2O rightleftharpoons H^+ + HO^-#</mathjax></p>
<p>This equilibrium has been carefully measured, and at <mathjax>#298*K#</mathjax>, we may write,</p>
<p><mathjax>#[H^+][HO^-]=10^-14#</mathjax>, and taking <mathjax>#log_10#</mathjax> of both sides,</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p><mathjax>#-log_10[H_3O^+]-log_10[HO^-]=+14#</mathjax></p>
<p>Given the definition of <mathjax>#pH=-log_10[H^+]#</mathjax>, then...........</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>So if <mathjax>#pH=6#</mathjax>, then <mathjax>#pOH=8#</mathjax> and <mathjax>#[HO^-]=10^-8*mol*L^-1#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of hydroxide ions in a solution that has a pH of 6.0?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[HO^-]=10^-8*mol*L^-1#</mathjax>..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that water undergoes self-ionization, which we may represent by the following reaction:</p>
<p><mathjax>#H_2O rightleftharpoons H^+ + HO^-#</mathjax></p>
<p>This equilibrium has been carefully measured, and at <mathjax>#298*K#</mathjax>, we may write,</p>
<p><mathjax>#[H^+][HO^-]=10^-14#</mathjax>, and taking <mathjax>#log_10#</mathjax> of both sides,</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=-14#</mathjax></p>
<p><mathjax>#-log_10[H_3O^+]-log_10[HO^-]=+14#</mathjax></p>
<p>Given the definition of <mathjax>#pH=-log_10[H^+]#</mathjax>, then...........</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>So if <mathjax>#pH=6#</mathjax>, then <mathjax>#pOH=8#</mathjax> and <mathjax>#[HO^-]=10^-8*mol*L^-1#</mathjax>.</p></div>
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</article> | What is the concentration of hydroxide ions in a solution that has a pH of 6.0? | null |
2,527 | a8cab111-6ddd-11ea-bf3f-ccda262736ce | https://socratic.org/questions/what-is-the-percent-by-mass-of-water-in-cuso-4-5h-2o | 36% | start physical_unit 7 9 mass_percent none qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Percent by mass [OF] water in CuSO4.5H2O"}] | [{"type":"physical unit","value":"36%"}] | [{"type":"chemical equation","value":"CuSO4.5H2O"}] | <h1 class="questionTitle" itemprop="name">What is the percent by mass of water in #CuSO_4*5H_2O#?</h1> | null | 36% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%"water"=(5xx18.01*g*mol^-1)/(249.68*g*mol^-1)xx100%~=36%#</mathjax></p>
<p>This video provides a detailed explanation of how to calculate this solution.</p>
<p>
<iframe src="https://www.youtube.com/embed/RPWkndwt2FA?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#%"water"="Mass of water"/"Mass of copper sulfate pentahydrate"~=36%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%"water"=(5xx18.01*g*mol^-1)/(249.68*g*mol^-1)xx100%~=36%#</mathjax></p>
<p>This video provides a detailed explanation of how to calculate this solution.</p>
<p>
<iframe src="https://www.youtube.com/embed/RPWkndwt2FA?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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<div class="markdown"><p><mathjax>#%"water"="Mass of water"/"Mass of copper sulfate pentahydrate"~=36%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%"water"=(5xx18.01*g*mol^-1)/(249.68*g*mol^-1)xx100%~=36%#</mathjax></p>
<p>This video provides a detailed explanation of how to calculate this solution.</p>
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<iframe src="https://www.youtube.com/embed/RPWkndwt2FA?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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</article> | What is the percent by mass of water in #CuSO_4*5H_2O#? | null |
2,528 | ac0f01c8-6ddd-11ea-abbb-ccda262736ce | https://socratic.org/questions/597cae97b72cff300c577dac | 8.44 | start physical_unit 2 2 ph none qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 12 15 9 10 molarity qc_end physical_unit 6 7 23 23 pkb qc_end end | [{"type":"physical unit","value":"pH [OF] the buffer"}] | [{"type":"physical unit","value":"8.44"}] | [{"type":"physical unit","value":"Molarity [OF] weak base [=] \\pu{0.110 M}"},{"type":"physical unit","value":"Molarity [OF] the weak conjugate acid [=] \\pu{0.440 M}"},{"type":"physical unit","value":"pKb [OF] weak base [=] \\pu{4.96}"}] | <h1 class="questionTitle" itemprop="name">If a buffer contains #"0.110 M"# weak base and #"0.440 M"# of the weak conjugate acid, what is the #"pH"#? The #"pK"_b# is #4.96#.</h1> | null | 8.44 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong>Henderson - Hasselbalch equation</strong>, which for a weak base/conjugate acid buffer looks like this</p>
<blockquote>
<p><mathjax>#"pH" = 14 - overbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])^(color(blue)("the pOH of the buffer solution"))#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"p"K_b = - log(K_b)#</mathjax></p>
</blockquote>
<p>In your case, you know that the buffer contains <mathjax>#"0.110 M"#</mathjax> of the weak base and <mathjax>#"0.440 M"#</mathjax> of its conjugate acid, so even without doing any calculations, you should be able to say that the <mathjax>#"pOH"#</mathjax> of the buffer will be <strong>higher</strong> than the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>In other words, the <mathjax>#"pH"#</mathjax> of the buffer will be <strong>lower</strong> than <mathjax>#14 - "p"K_b#</mathjax>, what you would get for a <mathjax>#"pOH"#</mathjax> equal to the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>Plug in your values into the Henderson - Hasselbalch equation to find</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(1.1 * 10^(-5)) + log((0.440 color(red)(cancel(color(black)("M"))))/(0.110color(red)(cancel(color(black)("M")))))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.44)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the base dissociation constant.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 8.44#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong>Henderson - Hasselbalch equation</strong>, which for a weak base/conjugate acid buffer looks like this</p>
<blockquote>
<p><mathjax>#"pH" = 14 - overbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])^(color(blue)("the pOH of the buffer solution"))#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"p"K_b = - log(K_b)#</mathjax></p>
</blockquote>
<p>In your case, you know that the buffer contains <mathjax>#"0.110 M"#</mathjax> of the weak base and <mathjax>#"0.440 M"#</mathjax> of its conjugate acid, so even without doing any calculations, you should be able to say that the <mathjax>#"pOH"#</mathjax> of the buffer will be <strong>higher</strong> than the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>In other words, the <mathjax>#"pH"#</mathjax> of the buffer will be <strong>lower</strong> than <mathjax>#14 - "p"K_b#</mathjax>, what you would get for a <mathjax>#"pOH"#</mathjax> equal to the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>Plug in your values into the Henderson - Hasselbalch equation to find</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(1.1 * 10^(-5)) + log((0.440 color(red)(cancel(color(black)("M"))))/(0.110color(red)(cancel(color(black)("M")))))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.44)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the base dissociation constant.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If a buffer contains #"0.110 M"# weak base and #"0.440 M"# of the weak conjugate acid, what is the #"pH"#? The #"pK"_b# is #4.96#.</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-07-30T00:02:30" itemprop="dateCreated">
Jul 30, 2017
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<div class="markdown"><p><mathjax>#"pH" = 8.44#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong>Henderson - Hasselbalch equation</strong>, which for a weak base/conjugate acid buffer looks like this</p>
<blockquote>
<p><mathjax>#"pH" = 14 - overbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])^(color(blue)("the pOH of the buffer solution"))#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#"p"K_b = - log(K_b)#</mathjax></p>
</blockquote>
<p>In your case, you know that the buffer contains <mathjax>#"0.110 M"#</mathjax> of the weak base and <mathjax>#"0.440 M"#</mathjax> of its conjugate acid, so even without doing any calculations, you should be able to say that the <mathjax>#"pOH"#</mathjax> of the buffer will be <strong>higher</strong> than the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>In other words, the <mathjax>#"pH"#</mathjax> of the buffer will be <strong>lower</strong> than <mathjax>#14 - "p"K_b#</mathjax>, what you would get for a <mathjax>#"pOH"#</mathjax> equal to the <mathjax>#"p"K_b#</mathjax> of the weak base. </p>
<p>Plug in your values into the Henderson - Hasselbalch equation to find</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(1.1 * 10^(-5)) + log((0.440 color(red)(cancel(color(black)("M"))))/(0.110color(red)(cancel(color(black)("M")))))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.44)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the base dissociation constant.</p></div>
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Truong-Son N.
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<div class="markdown"><p>Stefan has a good answer, but I thought I'd give another approach to this. For buffers (i.e. weak acid + conjugate base, weak base + conjugate acid), the <strong>Henderson-Hasselbalch equation</strong> applies.</p>
<p>To make it so I only have to know one Henderson-Hasselbalch equation, I use the <mathjax>#"pK"_a#</mathjax> one and recall the relationships to interconvert between <mathjax>#"pK"_a#</mathjax>, <mathjax>#"pK"_b#</mathjax>, <mathjax>#"pH"#</mathjax>, and <mathjax>#"pOH"#</mathjax>.</p>
<blockquote>
<p><mathjax>#"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])#</mathjax></p>
</blockquote>
<p>Using the idea that <mathjax>#"pK"_a + "pK"_b = 14#</mathjax>, we get:</p>
<blockquote>
<p><mathjax>#-log(K_b) = "pK"_b = -log(1.1 xx 10^(-5)) = 4.96#</mathjax></p>
<p><mathjax>#=> "pK"_a = 14 - 4.96 = 9.04#</mathjax></p>
</blockquote>
<p>And thus, noting the difference in notation (treating the base as <mathjax>#"A"^(-)#</mathjax> or <mathjax>#"B"#</mathjax> and the conjugate acid as <mathjax>#"HA"#</mathjax> or <mathjax>#"BH"^(+)#</mathjax>), the <mathjax>#"pH"#</mathjax> is alternatively found as:</p>
<blockquote>
<p><mathjax>#color(blue)("pH") = 9.04 + log (("0.110 M")/("0.440 M"))#</mathjax></p>
<p><mathjax>#= 9.04 - log 4 = color(blue)(8.44)#</mathjax></p>
</blockquote>
<p>And this makes physical sense, as we started with a weak base, whose conjugate acid dissociates less (<mathjax>#K_a < K_b#</mathjax> if <mathjax>#K_b > 10^(-7)#</mathjax> at <mathjax>#25^@ "C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>). </p>
<p>Furthermore, there is a higher concentration of conjugate acid than the weak base. So, we should expect the <mathjax>#"pH"#</mathjax> to be basic, but also more acidic than the <mathjax>#"pK"_a#</mathjax> of the conjugate acid. </p>
<hr/>
<p><strong>APPENDIX</strong></p>
<p>And just so you see, this gives the same equation Stefan has. Recall that:</p>
<ul>
<li>At <mathjax>#25^@ "C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, <mathjax>#"pK"_a + "pK"_b = 14 = "pK"_w#</mathjax></li>
<li><mathjax>#log(a/b) = -log(b/a)#</mathjax></li>
</ul>
<p>Therefore:</p>
<blockquote>
<p><mathjax>#"pH" = (14 - "pK"_b) + log\frac(["A"^(-)])(["HA"])#</mathjax></p>
<p><mathjax>#= (14 - "pK"_b) - log\frac(["HA"])(["A"^(-)])#</mathjax></p>
<p><mathjax>#= 14 - ["pK"_b + log\frac(["HA"])(["A"^(-)])]#</mathjax></p>
<p>which is what Stefan used. </p>
</blockquote>
<p>With slightly changed notation, and knowing that <mathjax>#"pH" + "pOH" = 14#</mathjax> at <mathjax>#25^@ "C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, we can get the other form of the Henderson-Hasselbalch equation:</p>
<blockquote>
<p><mathjax>#"pH" = 14 - overbrace(["pK"_b + log\frac(["BH"^(+)])(["B"])])^("pOH")#</mathjax></p>
<p><mathjax>#= 14 - "pOH"#</mathjax></p>
</blockquote>
<p>Thus, </p>
<blockquote>
<p><mathjax>#barul(|stackrel(" ")(" " "pOH" = "pK"_b + log\frac(["BH"^(+)])(["B"])" ")|)#</mathjax></p>
</blockquote></div>
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</article> | If a buffer contains #"0.110 M"# weak base and #"0.440 M"# of the weak conjugate acid, what is the #"pH"#? The #"pK"_b# is #4.96#. | null |
2,529 | a91df6e2-6ddd-11ea-a36c-ccda262736ce | https://socratic.org/questions/56c386f37c01491c30f00e61 | 2.15 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 9 10 mole qc_end chemical_equation 15 24 qc_end end | [{"type":"physical unit","value":"Mole [OF] NiCl2 [IN] moles"}] | [{"type":"physical unit","value":"2.15 moles"}] | [{"type":"physical unit","value":"Mole [OF] Ni3(PO4)2 [=] \\pu{0.715 moles}"},{"type":"chemical equation","value":"3 NiCl2 + 2 Na3PO4 -> Ni3(PO4)2 + 6 NaCl"}] | <h1 class="questionTitle" itemprop="name">How many moles of #"NiCl"_2# are required to produce #"0.715 moles Ni"_3"(PO"_4)_2"#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The equation is:</p>
<p><mathjax>#"3NiCl"_2 + "2Na"_3"PO"_4"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Ni"_3"(PO"_4)_2 + "6NaCl"#</mathjax></p></div>
</h2>
</div>
</div> | 2.15 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NiCl"_2+"2Na"_3"PO"_4#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Ni"_3("PO"_4")"_2+"6NaCl"#</mathjax></p>
<p>Since we are starting with moles of <mathjax>#"Ni"_3("PO"_4")#</mathjax> and ending with moles of <mathjax>#"NiCl"_2"#</mathjax>, we need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> from the balanced equation.</p>
<p><strong>Mole Ratio</strong></p>
<p><mathjax>#"3 mol NiCl"_2":#</mathjax><mathjax>#"1 mol Ni"_3("PO"_4")"_2"#</mathjax></p>
<p>Multiply the given moles of <mathjax>#"Ni"_3("PO"_4")"_2"#</mathjax> times the mole ratio with <mathjax>#"NiCl"_2"#</mathjax> in the numerator.</p>
<p><mathjax>#0.715cancel("mol Ni"_3("PO"_4")"_2")#</mathjax><mathjax>#xx#</mathjax><mathjax>#(3"mol NiCl"_2)/((1cancel("mol Ni"_3("PO"_4")"_2"))#</mathjax><mathjax>#=#</mathjax><mathjax>#"2.15 mol NiCl"_2"#</mathjax></p>
<p>You will need <mathjax>#"2.15 mol NiCl"_2"#</mathjax> to produce <mathjax>#"0.715 mol Ni"_3("PO"_4")"_2"#</mathjax>.</p>
<p>Since it's easier to work with mass, you can convert moles <mathjax>#"NiCl"_2"#</mathjax> to mass in grams by multiplying the calculated moles <mathjax>#"NiCl"_2"#</mathjax> times its molar mass, <mathjax>#"129.5994 g/mol"#</mathjax>. <a href="https://pubchem.ncbi.nlm.nih.gov/compound/24385" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/24385</a></p>
<p><mathjax>#2.15cancel"mol NiCl"_2xx(129.5994"g NiCl"_2)/(1cancel"mol NiCl"_2)="279 g NiCl"_2"#</mathjax> rounded to three significant figures. </p>
<p>So in order to obtain <mathjax>#"2.15 mol NiCl"_2"#</mathjax>, you would use <mathjax>#"279 g NiCl"_2"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You will need <mathjax>#"2.15 mol NiCl"_2"#</mathjax> to produce <mathjax>#"0.715 mol Ni"_3("PO"_4")"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NiCl"_2+"2Na"_3"PO"_4#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Ni"_3("PO"_4")"_2+"6NaCl"#</mathjax></p>
<p>Since we are starting with moles of <mathjax>#"Ni"_3("PO"_4")#</mathjax> and ending with moles of <mathjax>#"NiCl"_2"#</mathjax>, we need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> from the balanced equation.</p>
<p><strong>Mole Ratio</strong></p>
<p><mathjax>#"3 mol NiCl"_2":#</mathjax><mathjax>#"1 mol Ni"_3("PO"_4")"_2"#</mathjax></p>
<p>Multiply the given moles of <mathjax>#"Ni"_3("PO"_4")"_2"#</mathjax> times the mole ratio with <mathjax>#"NiCl"_2"#</mathjax> in the numerator.</p>
<p><mathjax>#0.715cancel("mol Ni"_3("PO"_4")"_2")#</mathjax><mathjax>#xx#</mathjax><mathjax>#(3"mol NiCl"_2)/((1cancel("mol Ni"_3("PO"_4")"_2"))#</mathjax><mathjax>#=#</mathjax><mathjax>#"2.15 mol NiCl"_2"#</mathjax></p>
<p>You will need <mathjax>#"2.15 mol NiCl"_2"#</mathjax> to produce <mathjax>#"0.715 mol Ni"_3("PO"_4")"_2"#</mathjax>.</p>
<p>Since it's easier to work with mass, you can convert moles <mathjax>#"NiCl"_2"#</mathjax> to mass in grams by multiplying the calculated moles <mathjax>#"NiCl"_2"#</mathjax> times its molar mass, <mathjax>#"129.5994 g/mol"#</mathjax>. <a href="https://pubchem.ncbi.nlm.nih.gov/compound/24385" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/24385</a></p>
<p><mathjax>#2.15cancel"mol NiCl"_2xx(129.5994"g NiCl"_2)/(1cancel"mol NiCl"_2)="279 g NiCl"_2"#</mathjax> rounded to three significant figures. </p>
<p>So in order to obtain <mathjax>#"2.15 mol NiCl"_2"#</mathjax>, you would use <mathjax>#"279 g NiCl"_2"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of #"NiCl"_2# are required to produce #"0.715 moles Ni"_3"(PO"_4)_2"#?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The equation is:</p>
<p><mathjax>#"3NiCl"_2 + "2Na"_3"PO"_4"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Ni"_3"(PO"_4)_2 + "6NaCl"#</mathjax></p></div>
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<div class="markdown"><p>You will need <mathjax>#"2.15 mol NiCl"_2"#</mathjax> to produce <mathjax>#"0.715 mol Ni"_3("PO"_4")"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NiCl"_2+"2Na"_3"PO"_4#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Ni"_3("PO"_4")"_2+"6NaCl"#</mathjax></p>
<p>Since we are starting with moles of <mathjax>#"Ni"_3("PO"_4")#</mathjax> and ending with moles of <mathjax>#"NiCl"_2"#</mathjax>, we need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> from the balanced equation.</p>
<p><strong>Mole Ratio</strong></p>
<p><mathjax>#"3 mol NiCl"_2":#</mathjax><mathjax>#"1 mol Ni"_3("PO"_4")"_2"#</mathjax></p>
<p>Multiply the given moles of <mathjax>#"Ni"_3("PO"_4")"_2"#</mathjax> times the mole ratio with <mathjax>#"NiCl"_2"#</mathjax> in the numerator.</p>
<p><mathjax>#0.715cancel("mol Ni"_3("PO"_4")"_2")#</mathjax><mathjax>#xx#</mathjax><mathjax>#(3"mol NiCl"_2)/((1cancel("mol Ni"_3("PO"_4")"_2"))#</mathjax><mathjax>#=#</mathjax><mathjax>#"2.15 mol NiCl"_2"#</mathjax></p>
<p>You will need <mathjax>#"2.15 mol NiCl"_2"#</mathjax> to produce <mathjax>#"0.715 mol Ni"_3("PO"_4")"_2"#</mathjax>.</p>
<p>Since it's easier to work with mass, you can convert moles <mathjax>#"NiCl"_2"#</mathjax> to mass in grams by multiplying the calculated moles <mathjax>#"NiCl"_2"#</mathjax> times its molar mass, <mathjax>#"129.5994 g/mol"#</mathjax>. <a href="https://pubchem.ncbi.nlm.nih.gov/compound/24385" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/24385</a></p>
<p><mathjax>#2.15cancel"mol NiCl"_2xx(129.5994"g NiCl"_2)/(1cancel"mol NiCl"_2)="279 g NiCl"_2"#</mathjax> rounded to three significant figures. </p>
<p>So in order to obtain <mathjax>#"2.15 mol NiCl"_2"#</mathjax>, you would use <mathjax>#"279 g NiCl"_2"#</mathjax>.</p></div>
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</article> | How many moles of #"NiCl"_2# are required to produce #"0.715 moles Ni"_3"(PO"_4)_2"#? |
The equation is:
#"3NiCl"_2 + "2Na"_3"PO"_4"##rarr##"Ni"_3"(PO"_4)_2 + "6NaCl"#
|
2,530 | aa16b2c6-6ddd-11ea-bd19-ccda262736ce | https://socratic.org/questions/in-a-mixture-of-and-nitrogen-gas-80-0-percent-of-the-total-gas-pressure-is-exert | 0.4 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 2 2 24 25 total_pressure qc_end end | [{"type":"physical unit","value":"Pressure [OF] oxygen [IN] atm"}] | [{"type":"physical unit","value":"0.4 atm"}] | [{"type":"physical unit","value":"Pressure percent [OF] nitrogen in total gas [=] \\pu{80.0%}"},{"type":"physical unit","value":"Total pressure [OF] the mixture [=] \\pu{2.0 atm}"}] | <h1 class="questionTitle" itemprop="name">In a mixture of oxygen and nitrogen gas, #80.0%# of the total gas pressure is exerted by the nitrogen. If the total pressure is #"2.0 atm"#, what pressure does the oxygen exert? </h1> | null | 0.4 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#80%#</mathjax>of pressure is exerted by nitrogen gas. So, <mathjax>#20%#</mathjax>of pressure will be exerted by oxygen gas.</p>
<p><mathjax>#"p"_("O"_2) = 20% "of 2 atm" = 20/100 × "2 atm" = "0.4 atm"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"0.4 atm"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#80%#</mathjax>of pressure is exerted by nitrogen gas. So, <mathjax>#20%#</mathjax>of pressure will be exerted by oxygen gas.</p>
<p><mathjax>#"p"_("O"_2) = 20% "of 2 atm" = 20/100 × "2 atm" = "0.4 atm"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In a mixture of oxygen and nitrogen gas, #80.0%# of the total gas pressure is exerted by the nitrogen. If the total pressure is #"2.0 atm"#, what pressure does the oxygen exert? </h1>
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<div class="markdown"><p><mathjax>#"0.4 atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#80%#</mathjax>of pressure is exerted by nitrogen gas. So, <mathjax>#20%#</mathjax>of pressure will be exerted by oxygen gas.</p>
<p><mathjax>#"p"_("O"_2) = 20% "of 2 atm" = 20/100 × "2 atm" = "0.4 atm"#</mathjax></p></div>
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</article> | In a mixture of oxygen and nitrogen gas, #80.0%# of the total gas pressure is exerted by the nitrogen. If the total pressure is #"2.0 atm"#, what pressure does the oxygen exert? | null |
2,531 | a8d00836-6ddd-11ea-8291-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-3-9-10-8-m-oh-solution | 7.08 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NaOH solution"}] | [{"type":"physical unit","value":"7.08"}] | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{3.9 × 10^(−8) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #3.9*10^-8# #M# #NaOH# solution? </h1> | null | 7.08 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is 7.08.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #3.9*10^-8# #M# #NaOH# solution? </h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is 7.08.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#sf(pH=7.08)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We would predict that as this is such a low concentration the <mathjax>#sf(pH)#</mathjax> will be close to 7.</p>
<p>This is an incredibly low concentration, so we must take into account the <mathjax>#sf(OH^-)#</mathjax> ions which arise from the auto - ionisation of water:</p>
<p><mathjax>#sf(H_2O_((l))rightleftharpoonsH_((aq))^(+)+OH_((aq))^-)#</mathjax></p>
<p>For which:</p>
<p><mathjax>#sf(K_w=1xx10^(-14)" ""mol"^(2)."l"^(-2) #</mathjax> at <mathjax>#sf(25^@"C"#</mathjax>.</p>
<p>This means that the concentration of <mathjax>#sf(OH^-#</mathjax> ions is <mathjax>#sf(10^(-7)"M")#</mathjax> arising from the water.</p>
<p>A tiny amount of <mathjax>#sf(OH^-)#</mathjax> ions is added such that the concentration is <mathjax>#sf(3.9xx10^(-8)"M"#</mathjax>.</p>
<p>It might be reasoned, therefore, that the total <mathjax>#sf(OH^-)#</mathjax> ion concentration will be <mathjax>#sf((3.9xx10^(-8))+10^(-7)"M"#</mathjax>.</p>
<p>However this will not be the <mathjax>#sf(OH^-)#</mathjax> ion concentration <strong>at equilibrium.</strong></p>
<p>By adding those ions we have disturbed a system which is at equilibrium.</p>
<p>Le Chatelier's Principle tells us that the system will respond by opposing that change, thus restoring the equilibrium which represents the lowest energy state.</p>
<p>The reaction quotient <mathjax>#sf(Q)#</mathjax> is given by:</p>
<p><mathjax>#sf(Q=[H_((aq))^(+)][OH_((aq))^-]=10^(-7)xx[(3.9xx10^(-8)+10^(-7))])#</mathjax></p>
<p><mathjax>#=sf(1.39xx10^(-14)" ""mol"^2."l"^(-2))#</mathjax></p>
<p>This shows that <mathjax>#sf(Q>K)#</mathjax> which confirms that the equilibrium will shift to the left. More <mathjax>#sf(H^+)#</mathjax> will be consumed by the extra <mathjax>#sf(OH^-)#</mathjax> ions to produce more water.</p>
<p>To calculate the <mathjax>#sf(pH)#</mathjax> we need to get the concentration of <mathjax>#sf(H^+)#</mathjax> ions at equilibrium which we can do by setting up an <strong>ICE</strong> table:</p>
<p><mathjax>#" "sf(H_2O" "rightleftharpoons" "H^(+)" "+" "OH^(-))#</mathjax></p>
<p><mathjax>#sf(color(red)(I)" "--" "10^(-7)" "1.39xx10^(-7))#</mathjax></p>
<p><mathjax>#sf(color(red)(C)" "--" " -x" " -x)#</mathjax></p>
<p><mathjax>#sf(color(red)(E)" "--" "(10^(-7)-x)" "(1.39xx10^(-7)-x))#</mathjax></p>
<p>So:</p>
<p><mathjax>#sf(K_(w)=(10^(-7)-x)(1.39xx10^(-7))=10^(-14))#</mathjax></p>
<p>This simplifies to:</p>
<p><mathjax>#sf(x^2-(2.39xx10^(-7))x+0.39xx10^(-14)=0)#</mathjax></p>
<p>This is a quadratic equation which can be solved using the quadratic formula.</p>
<p>Discarding the absurd root we get:</p>
<p><mathjax>#sf(x=0.1765xx10^(-7)"mol/l")#</mathjax></p>
<p>So:</p>
<p><mathjax>#sf([H_(eqm)^(+)]=(1xx10^(-7)-0.1765xx10^(-7))=0.824xx10^(-7)"mol/l")#</mathjax></p>
<p><mathjax>#sf(pH=-log[H_(eqm)^+]=-log[0.824xx10^(-7)]#</mathjax></p>
<p><mathjax>#sf(color(red)(pH=7.08))#</mathjax></p>
<p>You may see this more generally described as "The Common Ion Effect" . <mathjax>#sf(OH^-)#</mathjax>being the common ion in question. </p></div>
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</article> | What is the pH of a #3.9*10^-8# #M# #NaOH# solution? | null |
2,532 | abc2d438-6ddd-11ea-8b40-ccda262736ce | https://socratic.org/questions/how-many-moles-of-na-ions-are-in-150-ml-of-a-0-200-m-na-3po-4-solution | 0.09 moles | start physical_unit 4 5 mole mol qc_end physical_unit 14 15 8 9 volume qc_end physical_unit 14 15 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] Na+ ions [IN] moles"}] | [{"type":"physical unit","value":"0.09 moles"}] | [{"type":"physical unit","value":"Volume [OF] Na3PO4 solution [=] \\pu{150 mL}"},{"type":"physical unit","value":"Molarity [OF] Na3PO4 solution [=] \\pu{0.200 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #Na^+# ions are in 150 mL of a 0.200 M #Na_3PO_4# solution?</h1> | null | 0.09 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#"Na"_3"PO"_4#</mathjax> in the solution is</p>
<p><mathjax>#"n" =\ "Molarity × Volume"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.200 M × 0.150 L"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.03 mol"#</mathjax></p>
<p>There are <mathjax>#3#</mathjax> atoms in each molecule of <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>So, Moles of <mathjax>#"Na"^+#</mathjax> ions <mathjax>#"= 3 × 0.03 mol = 0.09 mol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"0.09 mol"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#"Na"_3"PO"_4#</mathjax> in the solution is</p>
<p><mathjax>#"n" =\ "Molarity × Volume"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.200 M × 0.150 L"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.03 mol"#</mathjax></p>
<p>There are <mathjax>#3#</mathjax> atoms in each molecule of <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>So, Moles of <mathjax>#"Na"^+#</mathjax> ions <mathjax>#"= 3 × 0.03 mol = 0.09 mol"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #Na^+# ions are in 150 mL of a 0.200 M #Na_3PO_4# solution?</h1>
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<div class="markdown"><p><mathjax>#"0.09 mol"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#"Na"_3"PO"_4#</mathjax> in the solution is</p>
<p><mathjax>#"n" =\ "Molarity × Volume"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.200 M × 0.150 L"#</mathjax></p>
<p><mathjax>#color(white)("n") =\ "0.03 mol"#</mathjax></p>
<p>There are <mathjax>#3#</mathjax> atoms in each molecule of <mathjax>#"Na"_3"PO"_4#</mathjax>. </p>
<p>So, Moles of <mathjax>#"Na"^+#</mathjax> ions <mathjax>#"= 3 × 0.03 mol = 0.09 mol"#</mathjax></p></div>
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</article> | How many moles of #Na^+# ions are in 150 mL of a 0.200 M #Na_3PO_4# solution? | null |
2,533 | aa7b2be3-6ddd-11ea-8eaa-ccda262736ce | https://socratic.org/questions/5950d977b72cff192abedb85 | 17.5 g/mol | start physical_unit 22 23 molar_mass g/mol qc_end physical_unit 22 23 1 2 mass qc_end physical_unit 22 23 11 12 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"17.5 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{3.49 g}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{4.48 dm^3}"},{"type":"other","value":"Standard conditions."}] | <h1 class="questionTitle" itemprop="name">A #3.49*g# mass of a gas occupies a volume of #4.48*dm^3# under standard conditions, what is the molar mass of the gas?</h1> | null | 17.5 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Depending on your syllabus, at <mathjax>#"STP"#</mathjax> <mathjax>#1*mol#</mathjax> of Ideal Gas occupies <mathjax>#22.4*dm^3#</mathjax>, i.e. <mathjax>#22.4*dm^3*mol^-1#</mathjax>.......and note that <mathjax>#1*dm^3=1xx(10^-1*m)^3=1/1000*m^3=1*L#</mathjax>, cos' there are <mathjax>#1000*L#</mathjax> in a <mathjax>#"cubic metre"#</mathjax>.</p>
<p>And thus the <mathjax>#"molar mass"="mass"/"molar quantity"#</mathjax></p>
<p><mathjax>#=(3.49*g)/((4.48*dm^3)/(22.4*dm^3*mol^-1))=17.5*g*mol^-1#</mathjax>.......</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Molar mass"=17.5*g*mol^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Depending on your syllabus, at <mathjax>#"STP"#</mathjax> <mathjax>#1*mol#</mathjax> of Ideal Gas occupies <mathjax>#22.4*dm^3#</mathjax>, i.e. <mathjax>#22.4*dm^3*mol^-1#</mathjax>.......and note that <mathjax>#1*dm^3=1xx(10^-1*m)^3=1/1000*m^3=1*L#</mathjax>, cos' there are <mathjax>#1000*L#</mathjax> in a <mathjax>#"cubic metre"#</mathjax>.</p>
<p>And thus the <mathjax>#"molar mass"="mass"/"molar quantity"#</mathjax></p>
<p><mathjax>#=(3.49*g)/((4.48*dm^3)/(22.4*dm^3*mol^-1))=17.5*g*mol^-1#</mathjax>.......</p></div>
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<h1 class="questionTitle" itemprop="name">A #3.49*g# mass of a gas occupies a volume of #4.48*dm^3# under standard conditions, what is the molar mass of the gas?</h1>
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<div class="markdown"><p><mathjax>#"Molar mass"=17.5*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Depending on your syllabus, at <mathjax>#"STP"#</mathjax> <mathjax>#1*mol#</mathjax> of Ideal Gas occupies <mathjax>#22.4*dm^3#</mathjax>, i.e. <mathjax>#22.4*dm^3*mol^-1#</mathjax>.......and note that <mathjax>#1*dm^3=1xx(10^-1*m)^3=1/1000*m^3=1*L#</mathjax>, cos' there are <mathjax>#1000*L#</mathjax> in a <mathjax>#"cubic metre"#</mathjax>.</p>
<p>And thus the <mathjax>#"molar mass"="mass"/"molar quantity"#</mathjax></p>
<p><mathjax>#=(3.49*g)/((4.48*dm^3)/(22.4*dm^3*mol^-1))=17.5*g*mol^-1#</mathjax>.......</p></div>
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</article> | A #3.49*g# mass of a gas occupies a volume of #4.48*dm^3# under standard conditions, what is the molar mass of the gas? | null |
2,534 | ab0a6914-6ddd-11ea-bcb1-ccda262736ce | https://socratic.org/questions/color-red-sf-last-question-for-today-find-k-c-for-the-reaction-below-see-details | 3 × 10^(-4) | start physical_unit 3 4 equilibrium_constant_k none qc_end chemical_equation 8 13 qc_end physical_unit 17 19 15 16 mass qc_end physical_unit 26 27 24 25 volume qc_end physical_unit 17 19 29 30 temperature qc_end physical_unit 11 11 43 44 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Kc [OF] the reaction"}] | [{"type":"physical unit","value":"3 × 10^(-4)"}] | [{"type":"chemical equation","value":"2 ICl(g) <=> I2(g) + Cl2(g)"},{"type":"physical unit","value":"Mass [OF] ICl(g) sample [=] \\pu{0.0682 grams}"},{"type":"physical unit","value":"Volume [OF] reaction vessel [=] \\pu{625 mL}"},{"type":"physical unit","value":"Temperature [OF] ICl(g) sample [=] \\pu{628 K}"},{"type":"physical unit","value":"Mass [OF] I2 [=] \\pu{0.0383 grams}"},{"type":"other","value":"Equilibrium is reached between the ICl(g) and I2(g) formed by dissociation."}] | <h1 class="questionTitle" itemprop="name"> Find #K_c# for the reaction below (see details)?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#\sf{2ICl(g)\harrI_2(g)+Cl_2(g)#</mathjax></p>
<p>A 0.0682-gram sample of <mathjax>#"ICl (g)"#</mathjax> is placed in a 625-mL reaction vessel at 628 K. When equilibrium is reached between the <mathjax>#"ICl (g)"#</mathjax> and <mathjax>#I_2" (g)"#</mathjax> formed by dissociation, 0.0383-grams of <mathjax>#I_2#</mathjax> are present.<br/>
What is <mathjax>#K_c#</mathjax> for this reaction?</p>
<hr/>
<p>Sorry if this has been asked before!</p></div>
</h2>
</div>
</div> | 3 × 10^(-4) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation of the given reversible gaseous reaction</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#\sf{2ICl(g)rightleftharpoonsI_2(g)+Cl_2(g)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Molar masses of reactant and products</p>
<p><mathjax>#ICl->162.5" "g*mol^-1#</mathjax></p>
<p><mathjax>#I_2->254" "g*mol^-1#</mathjax></p>
<p><mathjax>#Cl_2->71" "g*mol^-1#</mathjax></p>
<p>Volume of reaction vessel <mathjax>#V=625mL=0.625L#</mathjax></p>
<p><strong>ICE Table</strong></p>
<p><mathjax>#" "" "\sf{2ICl(g)" "" "rightleftharpoons" "I_2(g)" "+" "Cl_2(g)#</mathjax></p>
<p><mathjax>#I" " " "\alpha" "mol" "" "" "" "0" "mol" "" "0" "mol#</mathjax></p>
<p><mathjax>#C" "-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p><mathjax>#E" "\alpha-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p>By the problem initial amount of <mathjax>#ICl(g)#</mathjax></p>
<p><mathjax>#\alpha=(0.0682" g "ICl)/(162.5" g/mol "ICl)~~4.2xx10^(-4)mol#</mathjax></p>
<p>Amount of <mathjax>#I_2(g)#</mathjax> as well as <mathjax>#Cl_2(g)#</mathjax> in equilibrium mixture </p>
<p><mathjax>#x=(0.0383" g")/(254" g/mol")~~1.5xx10^-4mol#</mathjax></p>
<p>The equilibrium constant <mathjax>#K_c#</mathjax> of the reaction</p>
<p><mathjax>#K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]#</mathjax></p>
<blockquote>
<p><mathjax>#=(x/V*x/V)/((\alpha-2x)/V)#</mathjax></p>
<p><mathjax>#=x^2/(V(\alpha-2x))#</mathjax></p>
<p><mathjax>#=(1.5xx10^-4)^2/(0.625(4.2xx10^(-4)-2*1.5xx10^-4))#</mathjax></p>
<p><mathjax>#=(2.25xx10^-8)/(0.625xx1.2xx10^-4)=3xx10^-4#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#k=3xx10^-4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation of the given reversible gaseous reaction</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#\sf{2ICl(g)rightleftharpoonsI_2(g)+Cl_2(g)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Molar masses of reactant and products</p>
<p><mathjax>#ICl->162.5" "g*mol^-1#</mathjax></p>
<p><mathjax>#I_2->254" "g*mol^-1#</mathjax></p>
<p><mathjax>#Cl_2->71" "g*mol^-1#</mathjax></p>
<p>Volume of reaction vessel <mathjax>#V=625mL=0.625L#</mathjax></p>
<p><strong>ICE Table</strong></p>
<p><mathjax>#" "" "\sf{2ICl(g)" "" "rightleftharpoons" "I_2(g)" "+" "Cl_2(g)#</mathjax></p>
<p><mathjax>#I" " " "\alpha" "mol" "" "" "" "0" "mol" "" "0" "mol#</mathjax></p>
<p><mathjax>#C" "-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p><mathjax>#E" "\alpha-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p>By the problem initial amount of <mathjax>#ICl(g)#</mathjax></p>
<p><mathjax>#\alpha=(0.0682" g "ICl)/(162.5" g/mol "ICl)~~4.2xx10^(-4)mol#</mathjax></p>
<p>Amount of <mathjax>#I_2(g)#</mathjax> as well as <mathjax>#Cl_2(g)#</mathjax> in equilibrium mixture </p>
<p><mathjax>#x=(0.0383" g")/(254" g/mol")~~1.5xx10^-4mol#</mathjax></p>
<p>The equilibrium constant <mathjax>#K_c#</mathjax> of the reaction</p>
<p><mathjax>#K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]#</mathjax></p>
<blockquote>
<p><mathjax>#=(x/V*x/V)/((\alpha-2x)/V)#</mathjax></p>
<p><mathjax>#=x^2/(V(\alpha-2x))#</mathjax></p>
<p><mathjax>#=(1.5xx10^-4)^2/(0.625(4.2xx10^(-4)-2*1.5xx10^-4))#</mathjax></p>
<p><mathjax>#=(2.25xx10^-8)/(0.625xx1.2xx10^-4)=3xx10^-4#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> Find #K_c# for the reaction below (see details)?</h1>
<div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#\sf{2ICl(g)\harrI_2(g)+Cl_2(g)#</mathjax></p>
<p>A 0.0682-gram sample of <mathjax>#"ICl (g)"#</mathjax> is placed in a 625-mL reaction vessel at 628 K. When equilibrium is reached between the <mathjax>#"ICl (g)"#</mathjax> and <mathjax>#I_2" (g)"#</mathjax> formed by dissociation, 0.0383-grams of <mathjax>#I_2#</mathjax> are present.<br/>
What is <mathjax>#K_c#</mathjax> for this reaction?</p>
<hr/>
<p>Sorry if this has been asked before!</p></div>
</h2>
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<span class="dateCreated" datetime="2018-08-05T15:24:46" itemprop="dateCreated">
Aug 5, 2018
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#k=3xx10^-4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation of the given reversible gaseous reaction</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#\sf{2ICl(g)rightleftharpoonsI_2(g)+Cl_2(g)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Molar masses of reactant and products</p>
<p><mathjax>#ICl->162.5" "g*mol^-1#</mathjax></p>
<p><mathjax>#I_2->254" "g*mol^-1#</mathjax></p>
<p><mathjax>#Cl_2->71" "g*mol^-1#</mathjax></p>
<p>Volume of reaction vessel <mathjax>#V=625mL=0.625L#</mathjax></p>
<p><strong>ICE Table</strong></p>
<p><mathjax>#" "" "\sf{2ICl(g)" "" "rightleftharpoons" "I_2(g)" "+" "Cl_2(g)#</mathjax></p>
<p><mathjax>#I" " " "\alpha" "mol" "" "" "" "0" "mol" "" "0" "mol#</mathjax></p>
<p><mathjax>#C" "-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p><mathjax>#E" "\alpha-2x" "mol" "" "" "x" "mol" "" "x" "mol#</mathjax></p>
<p>By the problem initial amount of <mathjax>#ICl(g)#</mathjax></p>
<p><mathjax>#\alpha=(0.0682" g "ICl)/(162.5" g/mol "ICl)~~4.2xx10^(-4)mol#</mathjax></p>
<p>Amount of <mathjax>#I_2(g)#</mathjax> as well as <mathjax>#Cl_2(g)#</mathjax> in equilibrium mixture </p>
<p><mathjax>#x=(0.0383" g")/(254" g/mol")~~1.5xx10^-4mol#</mathjax></p>
<p>The equilibrium constant <mathjax>#K_c#</mathjax> of the reaction</p>
<p><mathjax>#K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]#</mathjax></p>
<blockquote>
<p><mathjax>#=(x/V*x/V)/((\alpha-2x)/V)#</mathjax></p>
<p><mathjax>#=x^2/(V(\alpha-2x))#</mathjax></p>
<p><mathjax>#=(1.5xx10^-4)^2/(0.625(4.2xx10^(-4)-2*1.5xx10^-4))#</mathjax></p>
<p><mathjax>#=(2.25xx10^-8)/(0.625xx1.2xx10^-4)=3xx10^-4#</mathjax></p>
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</article> | Find #K_c# for the reaction below (see details)? |
#\sf{2ICl(g)\harrI_2(g)+Cl_2(g)#
A 0.0682-gram sample of #"ICl (g)"# is placed in a 625-mL reaction vessel at 628 K. When equilibrium is reached between the #"ICl (g)"# and #I_2" (g)"# formed by dissociation, 0.0383-grams of #I_2# are present.
What is #K_c# for this reaction?
Sorry if this has been asked before!
|
2,535 | aa476252-6ddd-11ea-9456-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-of-ice-in-joules | 2.03 J/(g * ℃) | start physical_unit 6 6 specific_heat j/(°c_·_g) qc_end substance 6 6 qc_end end | [{"type":"physical unit","value":"Specific heat [OF] ice [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"2.03 J/(g * ℃)"}] | [{"type":"substance name","value":"Ice"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat of ice in joules?</h1> | null | 2.03 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Specific heat is not measured in joules, it is measured in <mathjax>#J/(g.^oC)#</mathjax> </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The number is 2.03 but the units of <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> are not joules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Specific heat is not measured in joules, it is measured in <mathjax>#J/(g.^oC)#</mathjax> </p></div>
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<div class="markdown"><p>The number is 2.03 but the units of <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> are not joules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Specific heat is not measured in joules, it is measured in <mathjax>#J/(g.^oC)#</mathjax> </p></div>
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</article> | What is the specific heat of ice in joules? | null |
2,536 | acaec677-6ddd-11ea-b302-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-1-5-moles-of-co-2 | 66.00 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] CO2 [IN] g"}] | [{"type":"physical unit","value":"66.00 g"}] | [{"type":"physical unit","value":"Mole [OF] CO2 [=] \\pu{1.5 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 1.5 moles of #CO_2#?</h1> | null | 66.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let:<br/>
<mathjax>#n#</mathjax> be the number of moles<br/>
<mathjax>#m#</mathjax> be the mass <br/>
<mathjax>#M#</mathjax> be the molecular weight</p>
<p><mathjax>#n = 1.5mol#</mathjax></p>
<p>By definition:</p>
<p><mathjax>#n = m/M#</mathjax></p>
<p><mathjax>#m = n*M#</mathjax></p>
<p><mathjax>#M(C) = 12g//mol#</mathjax><br/>
<mathjax>#M(O_2) = 2*16g//mol= 32g//mol#</mathjax><br/>
<mathjax>#M = 12g//mol + 32g//mol#</mathjax><br/>
<mathjax>#M = 44g//mol#</mathjax></p>
<p><mathjax>#m = 1.5mol*44g//mol#</mathjax><br/>
<mathjax>#m = 66g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass of 1.5 moles of <mathjax>#CO_2#</mathjax> is 66 grams</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let:<br/>
<mathjax>#n#</mathjax> be the number of moles<br/>
<mathjax>#m#</mathjax> be the mass <br/>
<mathjax>#M#</mathjax> be the molecular weight</p>
<p><mathjax>#n = 1.5mol#</mathjax></p>
<p>By definition:</p>
<p><mathjax>#n = m/M#</mathjax></p>
<p><mathjax>#m = n*M#</mathjax></p>
<p><mathjax>#M(C) = 12g//mol#</mathjax><br/>
<mathjax>#M(O_2) = 2*16g//mol= 32g//mol#</mathjax><br/>
<mathjax>#M = 12g//mol + 32g//mol#</mathjax><br/>
<mathjax>#M = 44g//mol#</mathjax></p>
<p><mathjax>#m = 1.5mol*44g//mol#</mathjax><br/>
<mathjax>#m = 66g#</mathjax></p></div>
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<div class="markdown"><p>The mass of 1.5 moles of <mathjax>#CO_2#</mathjax> is 66 grams</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let:<br/>
<mathjax>#n#</mathjax> be the number of moles<br/>
<mathjax>#m#</mathjax> be the mass <br/>
<mathjax>#M#</mathjax> be the molecular weight</p>
<p><mathjax>#n = 1.5mol#</mathjax></p>
<p>By definition:</p>
<p><mathjax>#n = m/M#</mathjax></p>
<p><mathjax>#m = n*M#</mathjax></p>
<p><mathjax>#M(C) = 12g//mol#</mathjax><br/>
<mathjax>#M(O_2) = 2*16g//mol= 32g//mol#</mathjax><br/>
<mathjax>#M = 12g//mol + 32g//mol#</mathjax><br/>
<mathjax>#M = 44g//mol#</mathjax></p>
<p><mathjax>#m = 1.5mol*44g//mol#</mathjax><br/>
<mathjax>#m = 66g#</mathjax></p></div>
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</article> | What is the mass of 1.5 moles of #CO_2#? | null |
2,537 | acaa3966-6ddd-11ea-b645-ccda262736ce | https://socratic.org/questions/how-many-grams-is-25-mol-of-co-2-molecules | 11.00 grams | start physical_unit 7 8 mass g qc_end physical_unit 7 8 4 5 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] CO2 molecules [IN] grams"}] | [{"type":"physical unit","value":"11.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] CO2 molecules [=] \\pu{0.25 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams is .25 mol of #CO_2# molecules?</h1> | null | 11.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specified 0.25 mol.</p>
<p>So <mathjax>#1/4#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#44.01*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> carbon dioxide has a mass of <mathjax>#44.01#</mathjax> <mathjax>#g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specified 0.25 mol.</p>
<p>So <mathjax>#1/4#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#44.01*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams is .25 mol of #CO_2# molecules?</h1>
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anor277
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<div class="markdown"><p>Well <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> carbon dioxide has a mass of <mathjax>#44.01#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You specified 0.25 mol.</p>
<p>So <mathjax>#1/4#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#44.01*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? g#</mathjax></p></div>
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</article> | How many grams is .25 mol of #CO_2# molecules? | null |
2,538 | ad1a2e3a-6ddd-11ea-a026-ccda262736ce | https://socratic.org/questions/in-the-equation-2kclo-3-2kci-3o-2-how-many-moles-of-oxygen-are-produced-when-5-0 | 7.58 moles | start physical_unit 15 15 mole mol qc_end chemical_equation 3 10 qc_end physical_unit 4 4 19 20 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"7.58 moles"}] | [{"type":"chemical equation","value":"2 KClO3 -> 2 KCI + 3 O2"},{"type":"physical unit","value":"Mole [OF] KClO3 [=] \\pu{5.05 moles}"},{"type":"other","value":"Decompose completely."}] | <h1 class="questionTitle" itemprop="name">In the equation #2KClO_3 -> 2KCI + 3O_2#, how many moles of oxygen are produced when 5.05 m of #KClO_3# decompose completely?</h1> | null | 7.58 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are multiple lines of reasoning for solving this. Let me give you the most detailed:</p>
<p>You start with a plan: we will convert moles of <mathjax>#KClO_3#</mathjax> to moles of <mathjax>#O_2#</mathjax> using the balanced chemical equation:</p>
<p>You will get 3 moles of <mathjax>#O_2#</mathjax> for every 2 moles of <mathjax>#KClO_3#</mathjax> you decompose.</p>
<p>Now let's substitute the number of moles of <mathjax>#KClO_3#</mathjax> that you actually started with:</p>
<p><mathjax>#5.05 " moles "KClO_3 xx (3 " moles " O_2)/(2 " moles " KClO_3) = 7.575 " moles " O_2#</mathjax></p>
<p>Finally, since we started with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we have to end up with three significant figures since all we did was multiply and divide, so...</p>
<p>7.58 moles <mathjax>#O_2#</mathjax></p></div>
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<div class="markdown"><p>7.58 moles of <mathjax>#O_2#</mathjax> are produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are multiple lines of reasoning for solving this. Let me give you the most detailed:</p>
<p>You start with a plan: we will convert moles of <mathjax>#KClO_3#</mathjax> to moles of <mathjax>#O_2#</mathjax> using the balanced chemical equation:</p>
<p>You will get 3 moles of <mathjax>#O_2#</mathjax> for every 2 moles of <mathjax>#KClO_3#</mathjax> you decompose.</p>
<p>Now let's substitute the number of moles of <mathjax>#KClO_3#</mathjax> that you actually started with:</p>
<p><mathjax>#5.05 " moles "KClO_3 xx (3 " moles " O_2)/(2 " moles " KClO_3) = 7.575 " moles " O_2#</mathjax></p>
<p>Finally, since we started with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we have to end up with three significant figures since all we did was multiply and divide, so...</p>
<p>7.58 moles <mathjax>#O_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In the equation #2KClO_3 -> 2KCI + 3O_2#, how many moles of oxygen are produced when 5.05 m of #KClO_3# decompose completely?</h1>
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<div class="markdown"><p>7.58 moles of <mathjax>#O_2#</mathjax> are produced.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are multiple lines of reasoning for solving this. Let me give you the most detailed:</p>
<p>You start with a plan: we will convert moles of <mathjax>#KClO_3#</mathjax> to moles of <mathjax>#O_2#</mathjax> using the balanced chemical equation:</p>
<p>You will get 3 moles of <mathjax>#O_2#</mathjax> for every 2 moles of <mathjax>#KClO_3#</mathjax> you decompose.</p>
<p>Now let's substitute the number of moles of <mathjax>#KClO_3#</mathjax> that you actually started with:</p>
<p><mathjax>#5.05 " moles "KClO_3 xx (3 " moles " O_2)/(2 " moles " KClO_3) = 7.575 " moles " O_2#</mathjax></p>
<p>Finally, since we started with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we have to end up with three significant figures since all we did was multiply and divide, so...</p>
<p>7.58 moles <mathjax>#O_2#</mathjax></p></div>
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</article> | In the equation #2KClO_3 -> 2KCI + 3O_2#, how many moles of oxygen are produced when 5.05 m of #KClO_3# decompose completely? | null |
2,539 | aa621fc0-6ddd-11ea-bb0c-ccda262736ce | https://socratic.org/questions/58e3251f7c01491e20de705d | +1.50 V | start physical_unit 10 10 value v qc_end chemical_equation 4 15 qc_end physical_unit 19 19 21 22 value qc_end end | [{"type":"physical unit","value":"Value [OF] E [IN] V"}] | [{"type":"physical unit","value":"+1.50 V"}] | [{"type":"chemical equation","value":"MnO4- + 8 H+ + 5 e- <=> Mn^2+ + 7 H2O"},{"type":"physical unit","value":"Value [OF] E^0 [=] \\pu{+1.51 V}"},{"type":"physical unit","value":"[MnO4-] [OF] the reaction [=] \\pu{0.0001 M}"},{"type":"physical unit","value":"[Mn^2+] [OF] the reaction [=] \\pu{0.0005 M}"}] | <h1 class="questionTitle" itemprop="name">For the 1/2 cell #sf(MnO_4^(-)+8H^(+)+5erightleftharpoonsMn^(2+)+7H_2O)# the value of #sf(E^(@)=+1.51V)#. What is the value for E if #sf([MnO_4^(-)]=0.0001M)# and #sf([Mn^(2+)]=0.0005M)# ?</h1> | null | +1.50 V | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before the calculation it is helpful to make a prediction for the electrode potential using <a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a>:</p>
<p>The 1/2 cell reaction is:</p>
<p><mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#</mathjax></p>
<p><mathjax>#sf(MnO_4^(-)+8H^++5erightleftharpoonsMn^(2+)+4H_2O)#</mathjax></p>
<p><mathjax>#sf(color(red)(0.0001Mcolor(white)(xxxxxxxxx)0.0005M)#</mathjax></p>
<p><mathjax>#sf(E^@=+1.51color(white)(x)V)#</mathjax></p>
<p>You can see that the concentration of <mathjax>#sf(MnO_4^-)#</mathjax> has been <strong>reduced</strong> relative to the concentration of <mathjax>#sf(Mn^(2+))#</mathjax>.</p>
<p>According to Le Chatelier we would predict that the position of equilibrium will shift to the <strong>left</strong> to oppose that change, as shown by the blue arrow.</p>
<p>You can see from the 1/2 equation that this will tend to push out more electrons so we would expect the electrode potential to be <strong>less</strong> positive.</p>
<p>We can calculate this using The Nernst Equation:</p>
<p><mathjax>#sf(E=E^(@)-(RT)/(zF)ln(([red])/(["ox"]))#</mathjax></p>
<p>At 298K this simplifies to:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(z)log((["ox"])/([red])))#</mathjax></p>
<p>Where <strong>z</strong> is the number of moles of electrons transferred.</p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-][H^+]^8)/([Mn^(2+)]))#</mathjax></p>
<p>Since we are at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 0 we can say that <mathjax>#sf([H^+]=1color(white)(x)M)#</mathjax></p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-])/([Mn^(2+)]))#</mathjax></p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(E=+1.51+(0.05916)/(5)log((0.0001)/(0.0005))#</mathjax></p>
<p><mathjax>#sf(E=+1.51-0.00827=+1.50color(white)(x)V)#</mathjax></p>
<p>As you can see, this is in accordance with our prediction. The potential of the electrode has been made slightly less positive.</p>
<p>Put simply, reducing the concentration of Mn(VII) relative to Mn(II) has made the Mn(VII) slightly <strong>less</strong> effective as an oxidising agent.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#sf(E=+1.50color(white)(x)V)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before the calculation it is helpful to make a prediction for the electrode potential using <a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a>:</p>
<p>The 1/2 cell reaction is:</p>
<p><mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#</mathjax></p>
<p><mathjax>#sf(MnO_4^(-)+8H^++5erightleftharpoonsMn^(2+)+4H_2O)#</mathjax></p>
<p><mathjax>#sf(color(red)(0.0001Mcolor(white)(xxxxxxxxx)0.0005M)#</mathjax></p>
<p><mathjax>#sf(E^@=+1.51color(white)(x)V)#</mathjax></p>
<p>You can see that the concentration of <mathjax>#sf(MnO_4^-)#</mathjax> has been <strong>reduced</strong> relative to the concentration of <mathjax>#sf(Mn^(2+))#</mathjax>.</p>
<p>According to Le Chatelier we would predict that the position of equilibrium will shift to the <strong>left</strong> to oppose that change, as shown by the blue arrow.</p>
<p>You can see from the 1/2 equation that this will tend to push out more electrons so we would expect the electrode potential to be <strong>less</strong> positive.</p>
<p>We can calculate this using The Nernst Equation:</p>
<p><mathjax>#sf(E=E^(@)-(RT)/(zF)ln(([red])/(["ox"]))#</mathjax></p>
<p>At 298K this simplifies to:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(z)log((["ox"])/([red])))#</mathjax></p>
<p>Where <strong>z</strong> is the number of moles of electrons transferred.</p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-][H^+]^8)/([Mn^(2+)]))#</mathjax></p>
<p>Since we are at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 0 we can say that <mathjax>#sf([H^+]=1color(white)(x)M)#</mathjax></p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-])/([Mn^(2+)]))#</mathjax></p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(E=+1.51+(0.05916)/(5)log((0.0001)/(0.0005))#</mathjax></p>
<p><mathjax>#sf(E=+1.51-0.00827=+1.50color(white)(x)V)#</mathjax></p>
<p>As you can see, this is in accordance with our prediction. The potential of the electrode has been made slightly less positive.</p>
<p>Put simply, reducing the concentration of Mn(VII) relative to Mn(II) has made the Mn(VII) slightly <strong>less</strong> effective as an oxidising agent.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">For the 1/2 cell #sf(MnO_4^(-)+8H^(+)+5erightleftharpoonsMn^(2+)+7H_2O)# the value of #sf(E^(@)=+1.51V)#. What is the value for E if #sf([MnO_4^(-)]=0.0001M)# and #sf([Mn^(2+)]=0.0005M)# ?</h1>
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Michael
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<span class="dateCreated" datetime="2017-04-07T16:30:47" itemprop="dateCreated">
Apr 7, 2017
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<div class="markdown"><p><mathjax>#sf(E=+1.50color(white)(x)V)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Before the calculation it is helpful to make a prediction for the electrode potential using <a href="https://socratic.org/chemistry/chemical-equilibrium/le-chatelier-s-principle">Le Chatelier's Principle</a>:</p>
<p>The 1/2 cell reaction is:</p>
<p><mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#</mathjax></p>
<p><mathjax>#sf(MnO_4^(-)+8H^++5erightleftharpoonsMn^(2+)+4H_2O)#</mathjax></p>
<p><mathjax>#sf(color(red)(0.0001Mcolor(white)(xxxxxxxxx)0.0005M)#</mathjax></p>
<p><mathjax>#sf(E^@=+1.51color(white)(x)V)#</mathjax></p>
<p>You can see that the concentration of <mathjax>#sf(MnO_4^-)#</mathjax> has been <strong>reduced</strong> relative to the concentration of <mathjax>#sf(Mn^(2+))#</mathjax>.</p>
<p>According to Le Chatelier we would predict that the position of equilibrium will shift to the <strong>left</strong> to oppose that change, as shown by the blue arrow.</p>
<p>You can see from the 1/2 equation that this will tend to push out more electrons so we would expect the electrode potential to be <strong>less</strong> positive.</p>
<p>We can calculate this using The Nernst Equation:</p>
<p><mathjax>#sf(E=E^(@)-(RT)/(zF)ln(([red])/(["ox"]))#</mathjax></p>
<p>At 298K this simplifies to:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(z)log((["ox"])/([red])))#</mathjax></p>
<p>Where <strong>z</strong> is the number of moles of electrons transferred.</p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-][H^+]^8)/([Mn^(2+)]))#</mathjax></p>
<p>Since we are at <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 0 we can say that <mathjax>#sf([H^+]=1color(white)(x)M)#</mathjax></p>
<p>This becomes:</p>
<p><mathjax>#sf(E=E^@+(0.05916)/(5)log(([MnO_4^-])/([Mn^(2+)]))#</mathjax></p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(E=+1.51+(0.05916)/(5)log((0.0001)/(0.0005))#</mathjax></p>
<p><mathjax>#sf(E=+1.51-0.00827=+1.50color(white)(x)V)#</mathjax></p>
<p>As you can see, this is in accordance with our prediction. The potential of the electrode has been made slightly less positive.</p>
<p>Put simply, reducing the concentration of Mn(VII) relative to Mn(II) has made the Mn(VII) slightly <strong>less</strong> effective as an oxidising agent.</p></div>
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</article> | For the 1/2 cell #sf(MnO_4^(-)+8H^(+)+5erightleftharpoonsMn^(2+)+7H_2O)# the value of #sf(E^(@)=+1.51V)#. What is the value for E if #sf([MnO_4^(-)]=0.0001M)# and #sf([Mn^(2+)]=0.0005M)# ? | null |
2,540 | a9ab3a26-6ddd-11ea-bdb2-ccda262736ce | https://socratic.org/questions/the-compression-and-expansion-of-gases-form-the-basis-of-how-air-is-cooled-by-ai | -67.2 ℃ | start physical_unit 22 23 temperature °c qc_end physical_unit 22 23 30 31 temperature qc_end physical_unit 22 23 25 26 pressure qc_end physical_unit 22 23 38 39 pressure qc_end physical_unit 22 23 18 19 volume qc_end physical_unit 22 23 35 36 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the ideal gas [IN] ℃"}] | [{"type":"physical unit","value":"-67.2 ℃"}] | [{"type":"physical unit","value":"Temperature1 [OF] the ideal gas [=] \\pu{20.5 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the ideal gas [=] \\pu{6.38 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the ideal gas [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Volume1 [OF] the ideal gas [=] \\pu{1.55 L}"},{"type":"physical unit","value":"Volume2 [OF] the ideal gas [=] \\pu{6.95 L}"}] | <h1 class="questionTitle" itemprop="name">The compression and expansion of gases form the basis of how air is cooled by air conditioners. Suppose 1.55 L of an ideal gas under 6.38 atm of pressure at 20.5°C is expanded to 6.95 L at 1.00 atm. What is the new temperature?</h1> | null | -67.2 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The pressure, temperature, and volume of the gas will change when going from its initial state to its final state, which tells you that you must use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> to find the new <em>temperature</em> of the gas. </p>
<p>As you know, when the <em>number of moles</em> of gas is <strong>kept constant</strong>, pressure, volume, and temperature have the following relationship</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<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>From this point on, the important thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>, so don't forget to convert it from the given degrees Celsius. </p>
<p>So, rearrange that equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#T_2 = (1.00 color(red)(cancel(color(black)("atm"))))/(6.38color(red)(cancel(color(black)("atm")))) * (6.95 color(red)(cancel(color(black)("L"))))/(1.55color(red)(cancel(color(black)("L")))) * (273.15 + 20.5)"K"#</mathjax></p>
<p><mathjax>#T_2 = "206.38 K"#</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>#T_2 = color(green)("206 K")#</mathjax></p>
</blockquote>
<p>If you want, you can express the answer in degrees Celsius</p>
<blockquote>
<p><mathjax>#T_2[""^@"C"] = 206 - 273.15 = -67.2^@"C"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"206 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The pressure, temperature, and volume of the gas will change when going from its initial state to its final state, which tells you that you must use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> to find the new <em>temperature</em> of the gas. </p>
<p>As you know, when the <em>number of moles</em> of gas is <strong>kept constant</strong>, pressure, volume, and temperature have the following relationship</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<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>From this point on, the important thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>, so don't forget to convert it from the given degrees Celsius. </p>
<p>So, rearrange that equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#T_2 = (1.00 color(red)(cancel(color(black)("atm"))))/(6.38color(red)(cancel(color(black)("atm")))) * (6.95 color(red)(cancel(color(black)("L"))))/(1.55color(red)(cancel(color(black)("L")))) * (273.15 + 20.5)"K"#</mathjax></p>
<p><mathjax>#T_2 = "206.38 K"#</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>#T_2 = color(green)("206 K")#</mathjax></p>
</blockquote>
<p>If you want, you can express the answer in degrees Celsius</p>
<blockquote>
<p><mathjax>#T_2[""^@"C"] = 206 - 273.15 = -67.2^@"C"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The compression and expansion of gases form the basis of how air is cooled by air conditioners. Suppose 1.55 L of an ideal gas under 6.38 atm of pressure at 20.5°C is expanded to 6.95 L at 1.00 atm. What is the new temperature?</h1>
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Stefan V.
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Dec 25, 2015
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<div class="markdown"><p><mathjax>#"206 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The pressure, temperature, and volume of the gas will change when going from its initial state to its final state, which tells you that you must use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> to find the new <em>temperature</em> of the gas. </p>
<p>As you know, when the <em>number of moles</em> of gas is <strong>kept constant</strong>, pressure, volume, and temperature have the following relationship</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<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>From this point on, the important thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>, so don't forget to convert it from the given degrees Celsius. </p>
<p>So, rearrange that equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#T_2 = (1.00 color(red)(cancel(color(black)("atm"))))/(6.38color(red)(cancel(color(black)("atm")))) * (6.95 color(red)(cancel(color(black)("L"))))/(1.55color(red)(cancel(color(black)("L")))) * (273.15 + 20.5)"K"#</mathjax></p>
<p><mathjax>#T_2 = "206.38 K"#</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>#T_2 = color(green)("206 K")#</mathjax></p>
</blockquote>
<p>If you want, you can express the answer in degrees Celsius</p>
<blockquote>
<p><mathjax>#T_2[""^@"C"] = 206 - 273.15 = -67.2^@"C"#</mathjax></p>
</blockquote></div>
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</article> | The compression and expansion of gases form the basis of how air is cooled by air conditioners. Suppose 1.55 L of an ideal gas under 6.38 atm of pressure at 20.5°C is expanded to 6.95 L at 1.00 atm. What is the new temperature? | null |
2,541 | aaac6a26-6ddd-11ea-89cd-ccda262736ce | https://socratic.org/questions/a-gas-has-a-pressure-of-710-kpa-at-227-c-what-will-its-pressure-be-at-27-0-c-if- | 213 kPa | start physical_unit 1 1 pressure kpa qc_end physical_unit 1 1 6 7 pressure qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 17 18 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] kPa"}] | [{"type":"physical unit","value":"213 kPa"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{710 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{227 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{27.0 ℃}"},{"type":"other","value":"Volume doubles."}] | <h1 class="questionTitle" itemprop="name">A gas has a pressure of 710 kPa at 227°C. What will its pressure be at 27.0°C, if the volume doubles?</h1> | null | 213 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the final pressure of the gas, given its change in temperature and the fact that its volume doubled.</p>
<p>We can solve this problem using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>We must convert our given temperatures to Kelvin, which we <em>always</em> must do for gas equations:</p>
<p><mathjax>#T_1 = 227^"o""C" + 273 = 500#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#T_2 = 27.0^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>We're not given any values for the volume, but we know that <mathjax>#V_2#</mathjax> is twice that of <mathjax>#V_1#</mathjax>, let's make things as simple as possible by making <mathjax>#V_1#</mathjax> <mathjax>#1#</mathjax> liter and <mathjax>#V_2#</mathjax> <mathjax>#2#</mathjax> liters.</p>
<p>Plugging in our known values, and rearranging the equation to solve for the final pressure, <mathjax>#P_2#</mathjax>, we have</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2) = ((710"kPa")(1cancel("L"))(300cancel("K")))/((500cancel("K"))(2cancel("L"))) = color(red)(213#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p>
<p>The final pressure is thus <mathjax>#color(red)(213#</mathjax> kilopascals</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Final Pressure" = 213#</mathjax> <mathjax>#"kPa"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the final pressure of the gas, given its change in temperature and the fact that its volume doubled.</p>
<p>We can solve this problem using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>We must convert our given temperatures to Kelvin, which we <em>always</em> must do for gas equations:</p>
<p><mathjax>#T_1 = 227^"o""C" + 273 = 500#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#T_2 = 27.0^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>We're not given any values for the volume, but we know that <mathjax>#V_2#</mathjax> is twice that of <mathjax>#V_1#</mathjax>, let's make things as simple as possible by making <mathjax>#V_1#</mathjax> <mathjax>#1#</mathjax> liter and <mathjax>#V_2#</mathjax> <mathjax>#2#</mathjax> liters.</p>
<p>Plugging in our known values, and rearranging the equation to solve for the final pressure, <mathjax>#P_2#</mathjax>, we have</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2) = ((710"kPa")(1cancel("L"))(300cancel("K")))/((500cancel("K"))(2cancel("L"))) = color(red)(213#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p>
<p>The final pressure is thus <mathjax>#color(red)(213#</mathjax> kilopascals</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas has a pressure of 710 kPa at 227°C. What will its pressure be at 27.0°C, if the volume doubles?</h1>
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Nathan L.
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Jun 15, 2017
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<div class="markdown"><p><mathjax>#"Final Pressure" = 213#</mathjax> <mathjax>#"kPa"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the final pressure of the gas, given its change in temperature and the fact that its volume doubled.</p>
<p>We can solve this problem using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>We must convert our given temperatures to Kelvin, which we <em>always</em> must do for gas equations:</p>
<p><mathjax>#T_1 = 227^"o""C" + 273 = 500#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#T_2 = 27.0^"o""C" + 273 = 300#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>We're not given any values for the volume, but we know that <mathjax>#V_2#</mathjax> is twice that of <mathjax>#V_1#</mathjax>, let's make things as simple as possible by making <mathjax>#V_1#</mathjax> <mathjax>#1#</mathjax> liter and <mathjax>#V_2#</mathjax> <mathjax>#2#</mathjax> liters.</p>
<p>Plugging in our known values, and rearranging the equation to solve for the final pressure, <mathjax>#P_2#</mathjax>, we have</p>
<p><mathjax>#P_2 = (P_1V_1T_2)/(T_1V_2) = ((710"kPa")(1cancel("L"))(300cancel("K")))/((500cancel("K"))(2cancel("L"))) = color(red)(213#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p>
<p>The final pressure is thus <mathjax>#color(red)(213#</mathjax> kilopascals</p></div>
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</article> | A gas has a pressure of 710 kPa at 227°C. What will its pressure be at 27.0°C, if the volume doubles? | null |
2,542 | ac9ca6cc-6ddd-11ea-a6cf-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-chemical-equation-for-the-fermentation-of-sucrose-c- | C12H22O11 + H2O -> 4 C2H5OH + 4 CO2 | start chemical_equation qc_end chemical_equation 13 13 qc_end chemical_equation 29 29 qc_end substance 31 33 qc_end substance 23 23 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the fermentation"}] | [{"type":"chemical equation","value":"C12H22O11 + H2O -> 4 C2H5OH + 4 CO2"}] | [{"type":"chemical equation","value":"C12H22O11"},{"type":"chemical equation","value":"C2H5OH"},{"type":"substance name","value":"Carbon dioxide gas"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced chemical equation for the fermentation of sucrose (#C_12H_22O_11#) by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol (#C_2H_5OH#) and carbon dioxide gas?</h1> | null | C12H22O11 + H2O -> 4 C2H5OH + 4 CO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a <strong>reduction-oxidation reaction</strong>.</p>
<p>You can find the general technique for balancing redox equations <a href="https://socratic.org/questions/how-do-you-balance-redox-equations-in-acidic-solutions">here</a>.</p>
<blockquote></blockquote>
<p>We can use the method of <strong><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to balance this equation.</p>
<p>We start with the unbalanced equation:</p>
<p><mathjax>#"C"_12"H"_22"O"_11 + "H"_2"O" → "C"_2"H"_5"OH" + "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Identify the atoms that change oxidation number</strong></p>
<p>Determine the oxidation numbers of every atom in the equation.</p>
<p><mathjax>#stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+1")("H")_22stackrelcolor(blue)("-2")("O")_11 + color(white)(l)stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O")_2#</mathjax><br/>
<mathjax>#color(white)(stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+22")("H")_22stackrelcolor(blue)("-22")("O")_11 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-4")("C")_2stackrelcolor(blue)("+5")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-4")("O")_2)#</mathjax></p>
<p>We see that the oxidation number of <mathjax>#"C"#</mathjax> in sucrose is reduced to -2 in <mathjax>#"C"_2"H"_5"OH"#</mathjax> and increased to +4 in <mathjax>#"CO"_2#</mathjax>.</p>
<p>This is a <strong>disproportionation</strong> reaction.</p>
<blockquote></blockquote>
<p>The changes in oxidation number are:</p>
<p><mathjax>#"C: 0 → -2";color(white)(l) "Change ="color(white)(m) "-2 (oxidation)"#</mathjax><br/>
<mathjax>#"C: 0 → +4"; "Change ="color(white)(l) "+4 (reduction)"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Equalize the changes in oxidation number</strong></p>
<p>We need 2 atoms of <mathjax>#"C"#</mathjax> that become ethanol for every 1 atom of <mathjax>#"C"#</mathjax> that becomes <mathjax>#"CO"_2#</mathjax>.</p>
<p>We must also have a total of 12 <mathjax>#"C"#</mathjax> atoms from the sucrose.</p>
<p>That means we need 1 molecule of sucrose, 4 of ethanol, and 4 of <mathjax>#"CO"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Insert coefficients to get these numbers</strong></p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + "H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Balance <mathjax>#"O"#</mathjax></strong></p>
<p>We have fixed 12 <mathjax>#"O"#</mathjax> atoms on the right and 11 <mathjax>#"O"#</mathjax> atoms on the right, so we need 1 more <mathjax>#"O"#</mathjax> atoms on the left. Put a 1 before <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + color(blue)(1)"H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<p>Every formula now has a coefficient. The equation should be balanced.</p>
<blockquote></blockquote>
<p><strong>Step 5. Check that all atoms are balanced.</strong></p>
<p><mathjax>#bb("On the left"color(white)(l) "On the right")#</mathjax><br/>
<mathjax>#color(white)(mm)"12 C"color(white)(mmmml) "12 C"#</mathjax><br/>
<mathjax>#color(white)(mm)"24 H"color(white)(mmmml) "24 H"#</mathjax><br/>
<mathjax>#color(white)(mm)"12 O"color(white)(mmmml) "12 O"#</mathjax></p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#color(blue)("C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>WARNING! Long answer!</strong> The balanced equation is </p>
<p><mathjax>#"C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a <strong>reduction-oxidation reaction</strong>.</p>
<p>You can find the general technique for balancing redox equations <a href="https://socratic.org/questions/how-do-you-balance-redox-equations-in-acidic-solutions">here</a>.</p>
<blockquote></blockquote>
<p>We can use the method of <strong><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to balance this equation.</p>
<p>We start with the unbalanced equation:</p>
<p><mathjax>#"C"_12"H"_22"O"_11 + "H"_2"O" → "C"_2"H"_5"OH" + "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Identify the atoms that change oxidation number</strong></p>
<p>Determine the oxidation numbers of every atom in the equation.</p>
<p><mathjax>#stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+1")("H")_22stackrelcolor(blue)("-2")("O")_11 + color(white)(l)stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O")_2#</mathjax><br/>
<mathjax>#color(white)(stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+22")("H")_22stackrelcolor(blue)("-22")("O")_11 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-4")("C")_2stackrelcolor(blue)("+5")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-4")("O")_2)#</mathjax></p>
<p>We see that the oxidation number of <mathjax>#"C"#</mathjax> in sucrose is reduced to -2 in <mathjax>#"C"_2"H"_5"OH"#</mathjax> and increased to +4 in <mathjax>#"CO"_2#</mathjax>.</p>
<p>This is a <strong>disproportionation</strong> reaction.</p>
<blockquote></blockquote>
<p>The changes in oxidation number are:</p>
<p><mathjax>#"C: 0 → -2";color(white)(l) "Change ="color(white)(m) "-2 (oxidation)"#</mathjax><br/>
<mathjax>#"C: 0 → +4"; "Change ="color(white)(l) "+4 (reduction)"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Equalize the changes in oxidation number</strong></p>
<p>We need 2 atoms of <mathjax>#"C"#</mathjax> that become ethanol for every 1 atom of <mathjax>#"C"#</mathjax> that becomes <mathjax>#"CO"_2#</mathjax>.</p>
<p>We must also have a total of 12 <mathjax>#"C"#</mathjax> atoms from the sucrose.</p>
<p>That means we need 1 molecule of sucrose, 4 of ethanol, and 4 of <mathjax>#"CO"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Insert coefficients to get these numbers</strong></p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + "H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Balance <mathjax>#"O"#</mathjax></strong></p>
<p>We have fixed 12 <mathjax>#"O"#</mathjax> atoms on the right and 11 <mathjax>#"O"#</mathjax> atoms on the right, so we need 1 more <mathjax>#"O"#</mathjax> atoms on the left. Put a 1 before <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + color(blue)(1)"H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<p>Every formula now has a coefficient. The equation should be balanced.</p>
<blockquote></blockquote>
<p><strong>Step 5. Check that all atoms are balanced.</strong></p>
<p><mathjax>#bb("On the left"color(white)(l) "On the right")#</mathjax><br/>
<mathjax>#color(white)(mm)"12 C"color(white)(mmmml) "12 C"#</mathjax><br/>
<mathjax>#color(white)(mm)"24 H"color(white)(mmmml) "24 H"#</mathjax><br/>
<mathjax>#color(white)(mm)"12 O"color(white)(mmmml) "12 O"#</mathjax></p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#color(blue)("C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2)#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you write a balanced chemical equation for the fermentation of sucrose (#C_12H_22O_11#) by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol (#C_2H_5OH#) and carbon dioxide gas?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2017-02-19T15:27:39" itemprop="dateCreated">
Feb 19, 2017
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<div class="markdown"><p><strong>WARNING! Long answer!</strong> The balanced equation is </p>
<p><mathjax>#"C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is a <strong>reduction-oxidation reaction</strong>.</p>
<p>You can find the general technique for balancing redox equations <a href="https://socratic.org/questions/how-do-you-balance-redox-equations-in-acidic-solutions">here</a>.</p>
<blockquote></blockquote>
<p>We can use the method of <strong><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to balance this equation.</p>
<p>We start with the unbalanced equation:</p>
<p><mathjax>#"C"_12"H"_22"O"_11 + "H"_2"O" → "C"_2"H"_5"OH" + "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Identify the atoms that change oxidation number</strong></p>
<p>Determine the oxidation numbers of every atom in the equation.</p>
<p><mathjax>#stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+1")("H")_22stackrelcolor(blue)("-2")("O")_11 + color(white)(l)stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O")_2#</mathjax><br/>
<mathjax>#color(white)(stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+22")("H")_22stackrelcolor(blue)("-22")("O")_11 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-4")("C")_2stackrelcolor(blue)("+5")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-4")("O")_2)#</mathjax></p>
<p>We see that the oxidation number of <mathjax>#"C"#</mathjax> in sucrose is reduced to -2 in <mathjax>#"C"_2"H"_5"OH"#</mathjax> and increased to +4 in <mathjax>#"CO"_2#</mathjax>.</p>
<p>This is a <strong>disproportionation</strong> reaction.</p>
<blockquote></blockquote>
<p>The changes in oxidation number are:</p>
<p><mathjax>#"C: 0 → -2";color(white)(l) "Change ="color(white)(m) "-2 (oxidation)"#</mathjax><br/>
<mathjax>#"C: 0 → +4"; "Change ="color(white)(l) "+4 (reduction)"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Equalize the changes in oxidation number</strong></p>
<p>We need 2 atoms of <mathjax>#"C"#</mathjax> that become ethanol for every 1 atom of <mathjax>#"C"#</mathjax> that becomes <mathjax>#"CO"_2#</mathjax>.</p>
<p>We must also have a total of 12 <mathjax>#"C"#</mathjax> atoms from the sucrose.</p>
<p>That means we need 1 molecule of sucrose, 4 of ethanol, and 4 of <mathjax>#"CO"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Insert coefficients to get these numbers</strong></p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + "H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Balance <mathjax>#"O"#</mathjax></strong></p>
<p>We have fixed 12 <mathjax>#"O"#</mathjax> atoms on the right and 11 <mathjax>#"O"#</mathjax> atoms on the right, so we need 1 more <mathjax>#"O"#</mathjax> atoms on the left. Put a 1 before <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_12"H"_22"O"_11 + color(blue)(1)"H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#</mathjax></p>
<p>Every formula now has a coefficient. The equation should be balanced.</p>
<blockquote></blockquote>
<p><strong>Step 5. Check that all atoms are balanced.</strong></p>
<p><mathjax>#bb("On the left"color(white)(l) "On the right")#</mathjax><br/>
<mathjax>#color(white)(mm)"12 C"color(white)(mmmml) "12 C"#</mathjax><br/>
<mathjax>#color(white)(mm)"24 H"color(white)(mmmml) "24 H"#</mathjax><br/>
<mathjax>#color(white)(mm)"12 O"color(white)(mmmml) "12 O"#</mathjax></p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#color(blue)("C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2)#</mathjax></p></div>
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</article> | How do you write a balanced chemical equation for the fermentation of sucrose (#C_12H_22O_11#) by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol (#C_2H_5OH#) and carbon dioxide gas? | null |
2,543 | a973c470-6ddd-11ea-a995-ccda262736ce | https://socratic.org/questions/how-many-moles-of-he-are-in-16-g-of-the-element | 4.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] He [IN] moles"}] | [{"type":"physical unit","value":"4.00 moles"}] | [{"type":"physical unit","value":"Mass [OF] He [=] \\pu{16 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of He are in 16 g of the element?</h1> | null | 4.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and find the molar mass of Helium.</p>
<p><img alt="http://www.keywordsuggests.com/lIkG0YEvhgjVW3F|bqpmjOC|3Q1OTcRgHkZMdVAh0AdlIqP07Ctda35xSJ|*ylhNwi0bnkXaASomScIDUvTrAg/" src="https://useruploads.socratic.org/zS9HFPMDTtac5V3uB0Ks_255431-efb0297173e9a65aeaf68002aa6eed62.jpg"/> </p>
<p>This tells us that <mathjax>#4.00#</mathjax> grams of <mathjax>#"Helium"#</mathjax> equates to <mathjax>#1" mole"#</mathjax>. We write it as follows:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaa)(4.00"g He")/(1" mole")#</mathjax></p>
<p>If we are given <mathjax>#16" grams"#</mathjax> of <mathjax>#"He"#</mathjax>, then we can use what we wrote above to figure out number of <mathjax>#"moles"#</mathjax> of <mathjax>#"He"#</mathjax>.</p>
<p><mathjax>#(16cancel"g He")/(1) * (1" mole")/(4.00cancel"g He") = 4" moles of He"#</mathjax></p>
<p><mathjax>#"Answer": 4" moles of He"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#4" moles of He"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and find the molar mass of Helium.</p>
<p><img alt="http://www.keywordsuggests.com/lIkG0YEvhgjVW3F|bqpmjOC|3Q1OTcRgHkZMdVAh0AdlIqP07Ctda35xSJ|*ylhNwi0bnkXaASomScIDUvTrAg/" src="https://useruploads.socratic.org/zS9HFPMDTtac5V3uB0Ks_255431-efb0297173e9a65aeaf68002aa6eed62.jpg"/> </p>
<p>This tells us that <mathjax>#4.00#</mathjax> grams of <mathjax>#"Helium"#</mathjax> equates to <mathjax>#1" mole"#</mathjax>. We write it as follows:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaa)(4.00"g He")/(1" mole")#</mathjax></p>
<p>If we are given <mathjax>#16" grams"#</mathjax> of <mathjax>#"He"#</mathjax>, then we can use what we wrote above to figure out number of <mathjax>#"moles"#</mathjax> of <mathjax>#"He"#</mathjax>.</p>
<p><mathjax>#(16cancel"g He")/(1) * (1" mole")/(4.00cancel"g He") = 4" moles of He"#</mathjax></p>
<p><mathjax>#"Answer": 4" moles of He"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of He are in 16 g of the element?</h1>
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<div class="markdown"><p><mathjax>#4" moles of He"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> and find the molar mass of Helium.</p>
<p><img alt="http://www.keywordsuggests.com/lIkG0YEvhgjVW3F|bqpmjOC|3Q1OTcRgHkZMdVAh0AdlIqP07Ctda35xSJ|*ylhNwi0bnkXaASomScIDUvTrAg/" src="https://useruploads.socratic.org/zS9HFPMDTtac5V3uB0Ks_255431-efb0297173e9a65aeaf68002aa6eed62.jpg"/> </p>
<p>This tells us that <mathjax>#4.00#</mathjax> grams of <mathjax>#"Helium"#</mathjax> equates to <mathjax>#1" mole"#</mathjax>. We write it as follows:</p>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaa)(4.00"g He")/(1" mole")#</mathjax></p>
<p>If we are given <mathjax>#16" grams"#</mathjax> of <mathjax>#"He"#</mathjax>, then we can use what we wrote above to figure out number of <mathjax>#"moles"#</mathjax> of <mathjax>#"He"#</mathjax>.</p>
<p><mathjax>#(16cancel"g He")/(1) * (1" mole")/(4.00cancel"g He") = 4" moles of He"#</mathjax></p>
<p><mathjax>#"Answer": 4" moles of He"#</mathjax></p></div>
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</article> | How many moles of He are in 16 g of the element? | null |
2,544 | a8e8b4e2-6ddd-11ea-be10-ccda262736ce | https://socratic.org/questions/how-many-grams-of-ki-are-in-25-0-ml-of-a-3-0-m-v-ki-solution | 0.75 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 14 7 8 volume qc_end physical_unit 13 14 11 11 mass_concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] KI [IN] grams"}] | [{"type":"physical unit","value":"0.75 grams"}] | [{"type":"physical unit","value":"Volume [OF] KI solution [=] \\pu{25.0 mL}"},{"type":"physical unit","value":"m/v [OF] KI solution [=] \\pu{3.0% }"}] | <h1 class="questionTitle" itemprop="name">How many grams of KI are in 25.0 mL of a 3.0% (m/v) KI solution? </h1> | null | 0.75 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you were given the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">mass by volume percent concentration</a>, which is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by volume of solution, and multiplied by <mathjax>#100#</mathjax>.</p>
<p>In your case, a <mathjax>#"3.0% w/v"#</mathjax> potassium iodide solution will contain <mathjax>#"3.0 g"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of solution. </p>
<p>You know that the volume of the sample is <em>four times smaller</em> than this value, so you can say right from the start that it will contain <em>four times less</em> potassium iodide. </p>
<blockquote>
<p><mathjax>#25.0color(red)(cancel(color(black)("mL solution"))) * "3.0 g KI"/(100color(red)(cancel(color(black)("mL solution")))) = color(green)("0.75 g KI")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the given concentration.</p>
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<div class="markdown"><p><mathjax>#"0.75 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you were given the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">mass by volume percent concentration</a>, which is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by volume of solution, and multiplied by <mathjax>#100#</mathjax>.</p>
<p>In your case, a <mathjax>#"3.0% w/v"#</mathjax> potassium iodide solution will contain <mathjax>#"3.0 g"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of solution. </p>
<p>You know that the volume of the sample is <em>four times smaller</em> than this value, so you can say right from the start that it will contain <em>four times less</em> potassium iodide. </p>
<blockquote>
<p><mathjax>#25.0color(red)(cancel(color(black)("mL solution"))) * "3.0 g KI"/(100color(red)(cancel(color(black)("mL solution")))) = color(green)("0.75 g KI")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the given concentration.</p>
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<h1 class="questionTitle" itemprop="name">How many grams of KI are in 25.0 mL of a 3.0% (m/v) KI solution? </h1>
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<div class="markdown"><p><mathjax>#"0.75 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you were given the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">mass by volume percent concentration</a>, which is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by volume of solution, and multiplied by <mathjax>#100#</mathjax>.</p>
<p>In your case, a <mathjax>#"3.0% w/v"#</mathjax> potassium iodide solution will contain <mathjax>#"3.0 g"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of solution. </p>
<p>You know that the volume of the sample is <em>four times smaller</em> than this value, so you can say right from the start that it will contain <em>four times less</em> potassium iodide. </p>
<blockquote>
<p><mathjax>#25.0color(red)(cancel(color(black)("mL solution"))) * "3.0 g KI"/(100color(red)(cancel(color(black)("mL solution")))) = color(green)("0.75 g KI")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the given concentration.</p>
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</article> | How many grams of KI are in 25.0 mL of a 3.0% (m/v) KI solution? | null |
2,545 | ad18ad0d-6ddd-11ea-8092-ccda262736ce | https://socratic.org/questions/how-do-you-find-the-ph-of-a-solution-whose-h-3-44-times-10-12 | 11.46 | start physical_unit 8 8 ph none qc_end physical_unit 8 8 12 15 [h+] qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"11.46"}] | [{"type":"physical unit","value":"[H+] [OF] the solution [=] \\pu{3.44 × 10^(-12) M}"}] | <h1 class="questionTitle" itemprop="name">How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#?</h1> | null | 11.46 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<div class="markdown"><p>Well, by definition <mathjax>#pH=-log_10[H_3O^+].............#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#?</h1>
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anor277
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<div class="markdown"><p>Well, by definition <mathjax>#pH=-log_10[H_3O^+].............#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 11.5</p></div>
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<div class="markdown"><p><img alt="My own work (Mert METİN)" src="https://useruploads.socratic.org/82fX22inTL6sAJvvNYmo_qwdsa.png"/> </p></div>
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</article> | How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#? | null |
2,546 | aa74d1c8-6ddd-11ea-9d5f-ccda262736ce | https://socratic.org/questions/what-is-the-mass-percent-of-a-solution-that-contains-45-6-grams-ethanol-c-2h-6o- | 15.25% | start physical_unit 7 7 mass_percent none qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 18 18 15 16 mass qc_end end | [{"type":"physical unit","value":"Mass percent [OF] C2H6O in solution"}] | [{"type":"physical unit","value":"15.25%"}] | [{"type":"physical unit","value":"Mass [OF] C2H6O [=] \\pu{45.6 grams}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{254 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the mass percent of a solution that contains 45.6 grams ethanol (#C_2H_6O#) in 254 grams of water?</h1> | null | 15.25% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>mass percent</em> of a solution is simply the ratio between the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case ethanol, and the mass of the <strong>entire solution</strong>, which contains ethanol and water. </p>
<p>This is what's known as the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a>, or <strong>%m/m</strong>.</p>
<p>So, the mass of the solution will be </p>
<p><mathjax>#m_"solution" = m_"ethanol" + m_"water"#</mathjax></p>
<p><mathjax>#m_"solution" = 45.6 + 254 = "299.6 g"#</mathjax></p>
<p>This means that the mass percent of the solution is </p>
<p><mathjax>#"%m/m" = m_"solute"/m_"solution" * 100#</mathjax></p>
<p><mathjax>#"%m/m" = (45.6cancel("g"))/(299.6cancel("g")) * 100 = color(green)("15.2%")#</mathjax></p></div>
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<div class="markdown"><p>The solution is <strong>15.2% m/m</strong>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>mass percent</em> of a solution is simply the ratio between the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case ethanol, and the mass of the <strong>entire solution</strong>, which contains ethanol and water. </p>
<p>This is what's known as the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a>, or <strong>%m/m</strong>.</p>
<p>So, the mass of the solution will be </p>
<p><mathjax>#m_"solution" = m_"ethanol" + m_"water"#</mathjax></p>
<p><mathjax>#m_"solution" = 45.6 + 254 = "299.6 g"#</mathjax></p>
<p>This means that the mass percent of the solution is </p>
<p><mathjax>#"%m/m" = m_"solute"/m_"solution" * 100#</mathjax></p>
<p><mathjax>#"%m/m" = (45.6cancel("g"))/(299.6cancel("g")) * 100 = color(green)("15.2%")#</mathjax></p></div>
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Jul 7, 2015
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<div class="markdown"><p>The solution is <strong>15.2% m/m</strong>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>mass percent</em> of a solution is simply the ratio between the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case ethanol, and the mass of the <strong>entire solution</strong>, which contains ethanol and water. </p>
<p>This is what's known as the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a>, or <strong>%m/m</strong>.</p>
<p>So, the mass of the solution will be </p>
<p><mathjax>#m_"solution" = m_"ethanol" + m_"water"#</mathjax></p>
<p><mathjax>#m_"solution" = 45.6 + 254 = "299.6 g"#</mathjax></p>
<p>This means that the mass percent of the solution is </p>
<p><mathjax>#"%m/m" = m_"solute"/m_"solution" * 100#</mathjax></p>
<p><mathjax>#"%m/m" = (45.6cancel("g"))/(299.6cancel("g")) * 100 = color(green)("15.2%")#</mathjax></p></div>
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</article> | What is the mass percent of a solution that contains 45.6 grams ethanol (#C_2H_6O#) in 254 grams of water? | null |
2,547 | ac049e0a-6ddd-11ea-b79d-ccda262736ce | https://socratic.org/questions/the-solubility-of-lead-ii-iodate-pb-io-3-2-is-0-76-g-l-at-25-c-how-do-you-calcul | 1.03 × 10^(-8) | start physical_unit 6 6 equilibrium_constant_k none qc_end physical_unit 6 6 8 9 solubility qc_end physical_unit 6 6 11 12 temperature qc_end end | [{"type":"physical unit","value":"Ksp [OF] Pb(IO3)2"}] | [{"type":"physical unit","value":"1.03 × 10^(-8)"}] | [{"type":"physical unit","value":"Solubility [OF] Pb(IO3)2 [=] \\pu{0.76 g/L}"},{"type":"physical unit","value":"Temperature [OF] Pb(IO3)2 [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">The solubility of lead (II) Iodate, #Pb(IO_3)_2#, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?</h1> | null | 1.03 × 10^(-8) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can represent the solubility of <mathjax>#Pb(IO_3)_2#</mathjax> as:</p>
<p><mathjax>#Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-#</mathjax></p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[Pb^(2+)][IO_3^-]^2#</mathjax></p>
<p>And if we let <mathjax>#S="solubility of lead iodate"#</mathjax>, then, </p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>.</p>
<p>So now we work out the solubility of <mathjax>#Pb(IO_3)_2#</mathjax>.</p>
<p>We are given <mathjax>#S_("mass")=0.76*g*L^-1#</mathjax></p>
<p><mathjax>#S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#1.37xx10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_(sp)=4xx(1.37xx10^-3)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Lead iodate is thus quite an insoluble beast. </p>
<p>See <a href="https://socratic.org/questions/what-is-ksp-in-chemistry">here</a> and <a href="https://socratic.org/questions/how-can-i-calculate-solubility-equilibrium">and here</a> for other examples.</p></div>
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<div class="markdown"><p><mathjax>#K_(sp)#</mathjax> <mathjax>#Pb(IO_3)_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can represent the solubility of <mathjax>#Pb(IO_3)_2#</mathjax> as:</p>
<p><mathjax>#Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-#</mathjax></p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[Pb^(2+)][IO_3^-]^2#</mathjax></p>
<p>And if we let <mathjax>#S="solubility of lead iodate"#</mathjax>, then, </p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>.</p>
<p>So now we work out the solubility of <mathjax>#Pb(IO_3)_2#</mathjax>.</p>
<p>We are given <mathjax>#S_("mass")=0.76*g*L^-1#</mathjax></p>
<p><mathjax>#S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#1.37xx10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_(sp)=4xx(1.37xx10^-3)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Lead iodate is thus quite an insoluble beast. </p>
<p>See <a href="https://socratic.org/questions/what-is-ksp-in-chemistry">here</a> and <a href="https://socratic.org/questions/how-can-i-calculate-solubility-equilibrium">and here</a> for other examples.</p></div>
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<h1 class="questionTitle" itemprop="name">The solubility of lead (II) Iodate, #Pb(IO_3)_2#, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature?</h1>
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<div class="markdown"><p><mathjax>#K_(sp)#</mathjax> <mathjax>#Pb(IO_3)_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We can represent the solubility of <mathjax>#Pb(IO_3)_2#</mathjax> as:</p>
<p><mathjax>#Pb(IO_3)_2(s) rightleftharpoons Pb^(2+) + 2IO_3^-#</mathjax></p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[Pb^(2+)][IO_3^-]^2#</mathjax></p>
<p>And if we let <mathjax>#S="solubility of lead iodate"#</mathjax>, then, </p>
<p><mathjax>#K_(sp)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>.</p>
<p>So now we work out the solubility of <mathjax>#Pb(IO_3)_2#</mathjax>.</p>
<p>We are given <mathjax>#S_("mass")=0.76*g*L^-1#</mathjax></p>
<p><mathjax>#S_("molar")=(0.76*g)/(557.00*g*mol^-1)xx1/(1*L)#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#1.37xx10^-3*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_(sp)=4xx(1.37xx10^-3)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Lead iodate is thus quite an insoluble beast. </p>
<p>See <a href="https://socratic.org/questions/what-is-ksp-in-chemistry">here</a> and <a href="https://socratic.org/questions/how-can-i-calculate-solubility-equilibrium">and here</a> for other examples.</p></div>
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</article> | The solubility of lead (II) Iodate, #Pb(IO_3)_2#, is 0.76 g/L at 25*C. How do you calculate the Value of Ksp at this same temperature? | null |
2,548 | ac887f6e-6ddd-11ea-bb28-ccda262736ce | https://socratic.org/questions/how-much-heat-energy-joules-is-required-to-to-raise-the-temperature-of-120-0-g-o | 21501.6 Joules | start physical_unit 16 16 heat_energy j qc_end physical_unit 16 16 13 14 mass qc_end physical_unit 16 16 18 19 temperature qc_end physical_unit 16 16 21 22 temperature qc_end end | [{"type":"physical unit","value":"Required heat energy [OF] water [IN] Joules"}] | [{"type":"physical unit","value":"21501.6 Joules"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{120.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{-90 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{-5 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much heat energy (Joules) is required to to raise the temperature of 120.0 g of water from -90 C to -5 C?</h1> | null | 21501.6 Joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use this equation </p>
<blockquote>
<p><mathjax>#"Q = mC"Δ"T"#</mathjax></p>
</blockquote>
<p>where</p>
<blockquote>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat</li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"C ="#</mathjax> <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> of sample (<mathjax>#"2.108 J/g°C"#</mathjax> for ice)</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in temperature</li>
</ul>
</blockquote>
<p><mathjax>#"Q" = 120 cancel"g" × "2.108 J"/(cancel"g" cancel"°C") × [-5 - (-90)] cancel"°C" = "21501.6 J" ≈ "21.5 kJ"#</mathjax></p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"21.5 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use this equation </p>
<blockquote>
<p><mathjax>#"Q = mC"Δ"T"#</mathjax></p>
</blockquote>
<p>where</p>
<blockquote>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat</li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"C ="#</mathjax> <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> of sample (<mathjax>#"2.108 J/g°C"#</mathjax> for ice)</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in temperature</li>
</ul>
</blockquote>
<p><mathjax>#"Q" = 120 cancel"g" × "2.108 J"/(cancel"g" cancel"°C") × [-5 - (-90)] cancel"°C" = "21501.6 J" ≈ "21.5 kJ"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How much heat energy (Joules) is required to to raise the temperature of 120.0 g of water from -90 C to -5 C?</h1>
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<div class="markdown"><p><mathjax>#"21.5 kJ"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use this equation </p>
<blockquote>
<p><mathjax>#"Q = mC"Δ"T"#</mathjax></p>
</blockquote>
<p>where</p>
<blockquote>
<ul>
<li><mathjax>#"Q ="#</mathjax> Heat</li>
<li><mathjax>#"m ="#</mathjax> Mass of sample</li>
<li><mathjax>#"C ="#</mathjax> <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> of sample (<mathjax>#"2.108 J/g°C"#</mathjax> for ice)</li>
<li><mathjax>#"ΔT ="#</mathjax> Change in temperature</li>
</ul>
</blockquote>
<p><mathjax>#"Q" = 120 cancel"g" × "2.108 J"/(cancel"g" cancel"°C") × [-5 - (-90)] cancel"°C" = "21501.6 J" ≈ "21.5 kJ"#</mathjax></p></div>
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</article> | How much heat energy (Joules) is required to to raise the temperature of 120.0 g of water from -90 C to -5 C? | null |
2,549 | aa9320de-6ddd-11ea-ab1a-ccda262736ce | https://socratic.org/questions/propane-and-oxygen-react-according-to-the-equation-c-3h-8-g-5o-2-g-3co-2-g-4h-2o | 144 grams | start physical_unit 21 21 mass g qc_end chemical_equation 8 16 qc_end physical_unit 8 8 30 31 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] grams"}] | [{"type":"physical unit","value":"144 grams"}] | [{"type":"chemical equation","value":"C3H8(g) + 5O2(g) -> 3 CO2(g) + 4 H2O(g)"},{"type":"physical unit","value":"Mole [OF] C3H8(g) [=] \\pu{2.0 moles }"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">Propane and oxygen react according to the equation #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#. How many grams of water can be produced from the complete combustion of 2.0 moles of #C_3H_8(g)#?</h1> | null | 144 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the molar mass of water <mathjax>#M(H_2O)=2xx1+16=18g//mol#</mathjax></p>
<p>2 mol propane will generate <mathjax>#2xx4xx18=144g#</mathjax> of water</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>From the equation you will see that 1 mol of propane generates 4 mols of water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the molar mass of water <mathjax>#M(H_2O)=2xx1+16=18g//mol#</mathjax></p>
<p>2 mol propane will generate <mathjax>#2xx4xx18=144g#</mathjax> of water</p></div>
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<h1 class="questionTitle" itemprop="name">Propane and oxygen react according to the equation #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#. How many grams of water can be produced from the complete combustion of 2.0 moles of #C_3H_8(g)#?</h1>
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<div class="markdown"><p>From the equation you will see that 1 mol of propane generates 4 mols of water.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the molar mass of water <mathjax>#M(H_2O)=2xx1+16=18g//mol#</mathjax></p>
<p>2 mol propane will generate <mathjax>#2xx4xx18=144g#</mathjax> of water</p></div>
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</article> | Propane and oxygen react according to the equation #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#. How many grams of water can be produced from the complete combustion of 2.0 moles of #C_3H_8(g)#? | null |
2,550 | ab71b22c-6ddd-11ea-b615-ccda262736ce | https://socratic.org/questions/how-many-grams-of-solute-are-present-in-935-ml-of-0-720-m-kbr | 50.81 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 8 9 volume qc_end physical_unit 13 13 11 12 molarity qc_end end | [{"type":"physical unit","value":"mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"50.81 grams"}] | [{"type":"physical unit","value":"Volume [OF] KBr [=] \\pu{935 mL}"},{"type":"physical unit","value":"Molarity [OF] KBr [=] \\pu{0.720 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of solute are present in 935 mL of 0.720 M #KBr#?</h1> | null | 50.81 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>. This gives units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"="volume"xx"concentration".=#</mathjax></p>
<p><mathjax>#935*cancel(mL)xx10^-3cancelL*cancel(mL^-1)xx0.729*cancel(mol)*cancel(L^-1)xx74.55*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Grams of KCl"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#50*g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>. This gives units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"="volume"xx"concentration".=#</mathjax></p>
<p><mathjax>#935*cancel(mL)xx10^-3cancelL*cancel(mL^-1)xx0.729*cancel(mol)*cancel(L^-1)xx74.55*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of solute are present in 935 mL of 0.720 M #KBr#?</h1>
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<div class="markdown"><p><mathjax>#"Grams of KCl"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#50*g#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>. This gives units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"="volume"xx"concentration".=#</mathjax></p>
<p><mathjax>#935*cancel(mL)xx10^-3cancelL*cancel(mL^-1)xx0.729*cancel(mol)*cancel(L^-1)xx74.55*g*cancel(mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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</article> | How many grams of solute are present in 935 mL of 0.720 M #KBr#? | null |
2,551 | aa86cb52-6ddd-11ea-8bd9-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-1-532-moles-of-a-compound-with-a-molar-mass-of-44-g-mol | 67.41 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 5 6 mole qc_end physical_unit 9 9 15 16 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] the compound [IN] grams"}] | [{"type":"physical unit","value":"67.41 grams"}] | [{"type":"physical unit","value":"Mole [OF] the compound [=] \\pu{1.532 moles}"},{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{44 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 1.532 moles of a compound with a molar mass of 44 g/mol?</h1> | null | 67.41 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <strong>molar mass</strong> of a compound can be used as a <em>conversion factor</em> between <strong>grams</strong> and <strong>moles</strong>.</p>
<p>As you know, the molar mass tells you the mass of <strong>one mole</strong> of a given compound. In your case, the compound is said to have a molar mass equal to <mathjax>#"44 g mol"^(-1)#</mathjax>.</p>
<p>This means that if you were to measure out exactly <strong>one mole</strong> of this compound, its mass would be equal to <mathjax>#"44 g"#</mathjax>. </p>
<p>Now, your sample contains <mathjax>#1.532#</mathjax> <strong>moles</strong> of this unknown compound. To use the molar mass a conversion factor, place the unit that you <strong>have</strong> on the bottom of the fraction and the unit that you <strong>need</strong> on the top of the fraction</p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have moles" * "44 grams"/"1 mole" = "get grams"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>or, when you have grams and need moles</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have grams" * "1 mole"/"44 grams" = "get moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#1.532 color(red)(cancel(color(black)("moles"))) * "44 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("67.41 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> because you can say that the molar mass of the compound is a <em>constant</em>, which implies that it has an "infinite" number of sig figs. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"67.41 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <strong>molar mass</strong> of a compound can be used as a <em>conversion factor</em> between <strong>grams</strong> and <strong>moles</strong>.</p>
<p>As you know, the molar mass tells you the mass of <strong>one mole</strong> of a given compound. In your case, the compound is said to have a molar mass equal to <mathjax>#"44 g mol"^(-1)#</mathjax>.</p>
<p>This means that if you were to measure out exactly <strong>one mole</strong> of this compound, its mass would be equal to <mathjax>#"44 g"#</mathjax>. </p>
<p>Now, your sample contains <mathjax>#1.532#</mathjax> <strong>moles</strong> of this unknown compound. To use the molar mass a conversion factor, place the unit that you <strong>have</strong> on the bottom of the fraction and the unit that you <strong>need</strong> on the top of the fraction</p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have moles" * "44 grams"/"1 mole" = "get grams"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>or, when you have grams and need moles</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have grams" * "1 mole"/"44 grams" = "get moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#1.532 color(red)(cancel(color(black)("moles"))) * "44 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("67.41 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> because you can say that the molar mass of the compound is a <em>constant</em>, which implies that it has an "infinite" number of sig figs. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 1.532 moles of a compound with a molar mass of 44 g/mol?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"67.41 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <strong>molar mass</strong> of a compound can be used as a <em>conversion factor</em> between <strong>grams</strong> and <strong>moles</strong>.</p>
<p>As you know, the molar mass tells you the mass of <strong>one mole</strong> of a given compound. In your case, the compound is said to have a molar mass equal to <mathjax>#"44 g mol"^(-1)#</mathjax>.</p>
<p>This means that if you were to measure out exactly <strong>one mole</strong> of this compound, its mass would be equal to <mathjax>#"44 g"#</mathjax>. </p>
<p>Now, your sample contains <mathjax>#1.532#</mathjax> <strong>moles</strong> of this unknown compound. To use the molar mass a conversion factor, place the unit that you <strong>have</strong> on the bottom of the fraction and the unit that you <strong>need</strong> on the top of the fraction</p>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have moles" * "44 grams"/"1 mole" = "get grams"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>or, when you have grams and need moles</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"have grams" * "1 mole"/"44 grams" = "get moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#1.532 color(red)(cancel(color(black)("moles"))) * "44 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("67.41 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> because you can say that the molar mass of the compound is a <em>constant</em>, which implies that it has an "infinite" number of sig figs. </p></div>
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</article> | What is the mass of 1.532 moles of a compound with a molar mass of 44 g/mol? | null |
2,552 | a980619e-6ddd-11ea-9030-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-2000-ml-at-253-c-and-a-pressure-of-101-3-kpa-what-will-be- | 2.55 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 15 16 pressure qc_end physical_unit 1 1 29 30 temperature qc_end physical_unit 1 1 37 38 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"2.55 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2000 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{-253 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{101.3 kPa}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{-233 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{198.5 kPa}"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 2000 mL at -253 °C and a pressure of 101.3 kPa. What will be the new volume when the temperature is changed to -233 °C and the pressure is changed to 198.5 kPa?</h1> | null | 2.55 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <mathjax>#"degrees Kelvin"#</mathjax>, and so <mathjax>#V_2=(P_1*V_1*T_2)/(P_2*T_1)#</mathjax></p>
<p><mathjax>#=(101.3*kPaxx2000*mLxx50*K)/(198.5*kPaxx20*K)#</mathjax></p>
<p><mathjax>#=??mL#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, <mathjax>#V_2
~=2.5*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <mathjax>#"degrees Kelvin"#</mathjax>, and so <mathjax>#V_2=(P_1*V_1*T_2)/(P_2*T_1)#</mathjax></p>
<p><mathjax>#=(101.3*kPaxx2000*mLxx50*K)/(198.5*kPaxx20*K)#</mathjax></p>
<p><mathjax>#=??mL#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas has a volume of 2000 mL at -253 °C and a pressure of 101.3 kPa. What will be the new volume when the temperature is changed to -233 °C and the pressure is changed to 198.5 kPa?</h1>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, <mathjax>#V_2
~=2.5*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <mathjax>#"degrees Kelvin"#</mathjax>, and so <mathjax>#V_2=(P_1*V_1*T_2)/(P_2*T_1)#</mathjax></p>
<p><mathjax>#=(101.3*kPaxx2000*mLxx50*K)/(198.5*kPaxx20*K)#</mathjax></p>
<p><mathjax>#=??mL#</mathjax></p></div>
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</article> | A gas has a volume of 2000 mL at -253 °C and a pressure of 101.3 kPa. What will be the new volume when the temperature is changed to -233 °C and the pressure is changed to 198.5 kPa? | null |
2,553 | ab1f6574-6ddd-11ea-b8db-ccda262736ce | https://socratic.org/questions/a-sample-of-helium-gas-occupies-14-7-l-at-23-c-and-956-atm-what-volume-will-it-o | 12.22 L | start physical_unit 1 4 volume l qc_end physical_unit 1 4 6 7 volume qc_end physical_unit 1 4 9 10 temperature qc_end physical_unit 1 4 12 13 pressure qc_end physical_unit 1 4 20 21 temperature qc_end physical_unit 1 4 23 24 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] helium gas sample [IN] L"}] | [{"type":"physical unit","value":"12.22 L"}] | [{"type":"physical unit","value":"Volume1 [OF] helium gas sample [=] \\pu{14.7 L}"},{"type":"physical unit","value":"Temperature1 [OF] helium gas sample [=] \\pu{23 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] helium gas sample [=] \\pu{0.956 atm}"},{"type":"physical unit","value":"Temperature2 [OF] helium gas sample [=] \\pu{40 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] helium gas sample [=] \\pu{1.20 atm}"}] | <h1 class="questionTitle" itemprop="name">A sample of helium gas occupies 14.7 L at 23°C and .956 atm. What volume will it occupy at 40°C and 1.20 atm?</h1> | null | 12.22 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, which is the <a href="https://en.wikipedia.org/wiki/Combined_gas_law" rel="nofollow">combined gas law?</a> Temperature is specified to be on the <mathjax>#"Absolute scale"#</mathjax>.</p>
<p>So, <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.956*atmxx14.7*Lxx313*K)/(300*Kxx1.20*atm)#</mathjax></p>
<p>I get an answer (clearly, why?) in <mathjax>#"litres"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, which is the <a href="https://en.wikipedia.org/wiki/Combined_gas_law" rel="nofollow">combined gas law?</a> Temperature is specified to be on the <mathjax>#"Absolute scale"#</mathjax>.</p>
<p>So, <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.956*atmxx14.7*Lxx313*K)/(300*Kxx1.20*atm)#</mathjax></p>
<p>I get an answer (clearly, why?) in <mathjax>#"litres"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of helium gas occupies 14.7 L at 23°C and .956 atm. What volume will it occupy at 40°C and 1.20 atm?</h1>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, which is the <a href="https://en.wikipedia.org/wiki/Combined_gas_law" rel="nofollow">combined gas law?</a> Temperature is specified to be on the <mathjax>#"Absolute scale"#</mathjax>.</p>
<p>So, <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.956*atmxx14.7*Lxx313*K)/(300*Kxx1.20*atm)#</mathjax></p>
<p>I get an answer (clearly, why?) in <mathjax>#"litres"#</mathjax>.</p></div>
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</article> | A sample of helium gas occupies 14.7 L at 23°C and .956 atm. What volume will it occupy at 40°C and 1.20 atm? | null |
2,554 | a92b5249-6ddd-11ea-91c5-ccda262736ce | https://socratic.org/questions/59809aef11ef6b5b1036c3b8 | 3 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 12 12 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] carbon"}] | [{"type":"physical unit","value":"3"}] | [{"type":"chemical equation","value":"Na2+C2O4^2−"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#?</h1> | null | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3#</mathjax>. See below.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#?</h1>
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Room 204
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<div class="markdown"><p><mathjax>#3#</mathjax>. See below.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers"> <strong>oxidation states</strong> </a> to atoms in a molecule by basically using one premise: <em>the more electronegative atom gets all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons</em>.</p>
<p>Here we are considering the oxalate ion: <br/>
<img alt="https://upload.wikimedia.org/wikipedia/commons/d/db/Oxalate-ion-2D-skeletal.png" src="https://useruploads.socratic.org/fNjNFETJRVbUM5QL18jU_Oxalate-ion-2D-skeletal.png"/></p>
<p>We know that oxygen is more <strong>electronegative</strong> than carbon and it is typically assigned an oxidation number of <mathjax>#"-2"#</mathjax>.</p>
<p>So<br/>
<mathjax>#stackrel(color(red)(2x))("C")#</mathjax> <mathjax>#stackrel(color(blue)(4(-2)))("O")_2#</mathjax> </p>
<p>As the whole complex ion has a charge of <mathjax>#-2#</mathjax></p>
<p><mathjax>#:.#</mathjax> <mathjax>#color(red)(2x)+color(blue)(4(-2)=color(green)(-2)#</mathjax></p>
<p><mathjax>#color(white)(344)#</mathjax><mathjax>#color(red)(2x)-color(blue)(8)=color(green)(-2)#</mathjax> </p>
<p><mathjax>#=>#</mathjax> <mathjax>#color(red)(2x)=color(green)(6)#</mathjax></p>
<p>i.e. <mathjax>#color(white)(34)#</mathjax> <mathjax>#x=3#</mathjax></p></div>
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<div class="markdown"><p>We have <mathjax>#C(+III)#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, the oxidation state is the charge left on the central atom, when all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> atoms are removed with the charge (the electrons) devolved to the most electronegative atom.....</p>
<p>We gots <mathjax>#""^(-)O(O=)C-C(=O)O^(-)#</mathjax>. The oxidation number of oxygen is usually <mathjax>#-II#</mathjax> and it is here. The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> equals the charge on the ion.....here <mathjax>#-2#</mathjax>.</p>
<p>And so <mathjax>#2xxC_"oxidation number"+4xx(-2)=-2#</mathjax></p>
<p>And <mathjax>#2xxC_"oxidation number"=+6#</mathjax></p>
<p>And <mathjax>#C_"oxidation number"=+III#</mathjax> if you will forgive me mixing Roman and Arabic numerals.........of course you will. </p>
<p>Another way we could look at this is to split the <mathjax>#C-C#</mathjax> bond in oxalate ion to give <mathjax>#2xx""^(-)O(O=)C*#</mathjax>, <mathjax>#C(+III)#</mathjax> as required. And thus the carbon in oxalic acid is almost fully oxidized. </p></div>
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<div class="markdown"><p>answer = 3 [oxidation state]</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Oxygen is one of the most electronegative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> after flourine so it shows 2 as oxidation state for most <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> [exception - hydrogen peroxide <mathjax>#H2O2#</mathjax>]<br/>
SO IN <mathjax>#C2O4^(2-)#</mathjax> Oxidation state of oxygen must be -2 . <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> on compound is (-2) . <br/>
oxidation state of carbon will be positive .<br/>
.°. 2(oxidation state of carbon) + 4 ( oxidation state of oxygen) = -2<br/>
2 (C) + 4 (-2) = -2<br/>
2(C) = 6<br/>
OXIDATION STATE OF CARBON IS <mathjax>#6/2#</mathjax> = 3.</p></div>
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</article> | What is the oxidation number of carbon in the salt sodium oxalate, #Na_2^(+)C_2O_4^(2-)#? | null |
2,555 | a95efaca-6ddd-11ea-93a7-ccda262736ce | https://socratic.org/questions/56a087b37c01494d912f21fc | 70.00 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 5 6 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] NaNO2 [IN] grams"}] | [{"type":"physical unit","value":"70.00 grams"}] | [{"type":"physical unit","value":"Molarity [OF] NaNO2 solution [=] \\pu{1 mol/L}"}] | <h1 class="questionTitle" itemprop="name">In order to make a #1*mol*L^-1# solution of #NaNO_2# in water, how many grams of sodium nitrite are necessary?</h1> | null | 70.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We require, <mathjax>#(5*mmol)/(5*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5xx10^(-3)*mol)/(5xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax></p>
<p>Note that the <mathjax>#m#</mathjax> prefix simply means <mathjax>#xx10^-3#</mathjax>. So <mathjax>#mmol*mL^-1#</mathjax> is precisely <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax>.</p>
<p>So to make this volume we simply take a <mathjax>#5#</mathjax> <mathjax>#mL#</mathjax> volume of <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax> sodium nitrite.</p>
<p>It would be best to make a stock solution. Accurately weigh <mathjax>#70.00#</mathjax> <mathjax>#g#</mathjax> of sodium nitrite, and quantitatively transfer this to a <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> volumetric flask. Fill up to the mark with distilled water, and make sure all the solid is dissolved. </p></div>
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<div class="markdown"><p>Concentration <mathjax>#=#</mathjax> <mathjax>#("moles")/("volume of solution")#</mathjax>. Here you need a <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax> solution. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We require, <mathjax>#(5*mmol)/(5*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5xx10^(-3)*mol)/(5xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax></p>
<p>Note that the <mathjax>#m#</mathjax> prefix simply means <mathjax>#xx10^-3#</mathjax>. So <mathjax>#mmol*mL^-1#</mathjax> is precisely <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax>.</p>
<p>So to make this volume we simply take a <mathjax>#5#</mathjax> <mathjax>#mL#</mathjax> volume of <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax> sodium nitrite.</p>
<p>It would be best to make a stock solution. Accurately weigh <mathjax>#70.00#</mathjax> <mathjax>#g#</mathjax> of sodium nitrite, and quantitatively transfer this to a <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> volumetric flask. Fill up to the mark with distilled water, and make sure all the solid is dissolved. </p></div>
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<h1 class="questionTitle" itemprop="name">In order to make a #1*mol*L^-1# solution of #NaNO_2# in water, how many grams of sodium nitrite are necessary?</h1>
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<div class="markdown"><p>Concentration <mathjax>#=#</mathjax> <mathjax>#("moles")/("volume of solution")#</mathjax>. Here you need a <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax> solution. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We require, <mathjax>#(5*mmol)/(5*mL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5xx10^(-3)*mol)/(5xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#L^-1#</mathjax></p>
<p>Note that the <mathjax>#m#</mathjax> prefix simply means <mathjax>#xx10^-3#</mathjax>. So <mathjax>#mmol*mL^-1#</mathjax> is precisely <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax>.</p>
<p>So to make this volume we simply take a <mathjax>#5#</mathjax> <mathjax>#mL#</mathjax> volume of <mathjax>#1#</mathjax> <mathjax>#mol*L^-1#</mathjax> sodium nitrite.</p>
<p>It would be best to make a stock solution. Accurately weigh <mathjax>#70.00#</mathjax> <mathjax>#g#</mathjax> of sodium nitrite, and quantitatively transfer this to a <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> volumetric flask. Fill up to the mark with distilled water, and make sure all the solid is dissolved. </p></div>
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</article> | In order to make a #1*mol*L^-1# solution of #NaNO_2# in water, how many grams of sodium nitrite are necessary? | null |
2,556 | a853e790-6ddd-11ea-9cc7-ccda262736ce | https://socratic.org/questions/how-many-electrons-does-the-iron-ion-have-when-it-forms-the-ionic-compound-fecl- | 23 | start physical_unit 2 2 number none qc_end substance 5 6 qc_end chemical_equation 14 14 qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"23"}] | [{"type":"substance name","value":"Iron ion"},{"type":"chemical equation","value":"FeCl3"}] | <h1 class="questionTitle" itemprop="name">How many electrons does the iron ion have when it forms the ionic compound #FeCl_3#?</h1> | null | 23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The iron atom has 26 electrons in total; it must have if <mathjax>#Z=26#</mathjax>; why <mathjax>#"must?"#</mathjax> And thus upon oxidation.......</p>
<p><mathjax>#FerarrFe^(3+) + 3e^-#</mathjax></p>
<p>....the resultant ion has 23 electrons........</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Fe^(3+)#</mathjax> has <mathjax>#"5 d electrons"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The iron atom has 26 electrons in total; it must have if <mathjax>#Z=26#</mathjax>; why <mathjax>#"must?"#</mathjax> And thus upon oxidation.......</p>
<p><mathjax>#FerarrFe^(3+) + 3e^-#</mathjax></p>
<p>....the resultant ion has 23 electrons........</p></div>
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<h1 class="questionTitle" itemprop="name">How many electrons does the iron ion have when it forms the ionic compound #FeCl_3#?</h1>
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<div class="markdown"><p><mathjax>#Fe^(3+)#</mathjax> has <mathjax>#"5 d electrons"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The iron atom has 26 electrons in total; it must have if <mathjax>#Z=26#</mathjax>; why <mathjax>#"must?"#</mathjax> And thus upon oxidation.......</p>
<p><mathjax>#FerarrFe^(3+) + 3e^-#</mathjax></p>
<p>....the resultant ion has 23 electrons........</p></div>
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</article> | How many electrons does the iron ion have when it forms the ionic compound #FeCl_3#? | null |
2,557 | a83a660a-6ddd-11ea-80f8-ccda262736ce | https://socratic.org/questions/8-000x10-1-g-of-a-gas-is-collected-in-the-lab-and-found-to-have-a-volume-of-8-90 | 23.4 g/mol | start physical_unit 50 51 molar_mass g/mol qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"23.4 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{8.000 × 10^(-1) g}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{8.9000 × 10^(-1) L}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{9.600 × 10^1 kPa}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{3.00 × 10^2 K}"}] | <h1 class="questionTitle" itemprop="name">(8.000x10^-1) g of a gas is collected in the lab and found to have a volume of (8.9000x10^-1) L when the pressure is (9.600x10^1) kPa and the temperature is (3.00x10^2) K. According to this collected data what is the molar mass of the gas?</h1> | null | 23.4 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#m = 8.000 × 10^"-1"color(white)(l) "g" #</mathjax><br/>
<mathjax>#R = "8.314 kPa·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3.00 × 10^2color(white)(l) "K"#</mathjax><br/>
<mathjax>#P = 9.600 × 10^1 color(white)(l) "kPa "#</mathjax><br/>
<mathjax>#V = 8.9000 × 10^"-1"color(white)(l) "L "#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (8.000 × 10^"-1"color(white)(l)"g" × 8.314 color(red)(cancel(color(black)("kPa·L·K"^"-1")))"mol"^"-1" × 3.00 × 10^2 color(red)(cancel(color(black)("K"))))/(9.600 × 10^1 color(red)(cancel(color(black)("kPa"))) ×8.900 × 10^"-1" color(red)(cancel(color(black)("L")))) = "23.4 g/mol"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>According to these data, the molar mass of the gas is 23.4 g/mol.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#m = 8.000 × 10^"-1"color(white)(l) "g" #</mathjax><br/>
<mathjax>#R = "8.314 kPa·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3.00 × 10^2color(white)(l) "K"#</mathjax><br/>
<mathjax>#P = 9.600 × 10^1 color(white)(l) "kPa "#</mathjax><br/>
<mathjax>#V = 8.9000 × 10^"-1"color(white)(l) "L "#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (8.000 × 10^"-1"color(white)(l)"g" × 8.314 color(red)(cancel(color(black)("kPa·L·K"^"-1")))"mol"^"-1" × 3.00 × 10^2 color(red)(cancel(color(black)("K"))))/(9.600 × 10^1 color(red)(cancel(color(black)("kPa"))) ×8.900 × 10^"-1" color(red)(cancel(color(black)("L")))) = "23.4 g/mol"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">(8.000x10^-1) g of a gas is collected in the lab and found to have a volume of (8.9000x10^-1) L when the pressure is (9.600x10^1) kPa and the temperature is (3.00x10^2) K. According to this collected data what is the molar mass of the gas?</h1>
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<div class="markdown"><p>According to these data, the molar mass of the gas is 23.4 g/mol.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#m = 8.000 × 10^"-1"color(white)(l) "g" #</mathjax><br/>
<mathjax>#R = "8.314 kPa·L·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3.00 × 10^2color(white)(l) "K"#</mathjax><br/>
<mathjax>#P = 9.600 × 10^1 color(white)(l) "kPa "#</mathjax><br/>
<mathjax>#V = 8.9000 × 10^"-1"color(white)(l) "L "#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (8.000 × 10^"-1"color(white)(l)"g" × 8.314 color(red)(cancel(color(black)("kPa·L·K"^"-1")))"mol"^"-1" × 3.00 × 10^2 color(red)(cancel(color(black)("K"))))/(9.600 × 10^1 color(red)(cancel(color(black)("kPa"))) ×8.900 × 10^"-1" color(red)(cancel(color(black)("L")))) = "23.4 g/mol"#</mathjax></p></div>
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</article> | (8.000x10^-1) g of a gas is collected in the lab and found to have a volume of (8.9000x10^-1) L when the pressure is (9.600x10^1) kPa and the temperature is (3.00x10^2) K. According to this collected data what is the molar mass of the gas? | null |
2,558 | ab0069bb-6ddd-11ea-8c5a-ccda262736ce | https://socratic.org/questions/5976514d11ef6b7fde83e6bf | 1.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mole qc_end chemical_equation 15 23 qc_end end | [{"type":"physical unit","value":"Mole [OF] C6H12O6 [IN] moles"}] | [{"type":"physical unit","value":"1.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{6 moles}"},{"type":"chemical equation","value":"C6H12O6 + 6O2 -> 6 CO2 + 6 H2O"}] | <h1 class="questionTitle" itemprop="name">How many moles of #"C"_6"H"_12"O"_6"# will be consumed if 6 moles of #"O"_2"# are consumed?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"C"_6"H"_12"O"_6 + "6O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"6CO"_2 +"6H"_2"O"#</mathjax></p></div>
</h2>
</div>
</div> | 1.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"_6"H"_12"O"_6 + "6O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"6CO"_2 +"6H"_2"O"#</mathjax></p>
<p>The coefficients in a chemical equation indicate the number of moles of each element, molecule, or formula unit. In this equation, all of the substances are molecules. If there is no coefficient, it is understood to be <mathjax>#1#</mathjax>.</p>
<p>The balanced equation shows us that 1 mole <mathjax>#"C"_12"H"_2"O"_6#</mathjax> will be consumed if 6 moles <mathjax>#"O"_2#</mathjax> are consumed. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>One mole of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> will be consumed if <mathjax>#"6 mol O"_2#</mathjax> are consumed.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"_6"H"_12"O"_6 + "6O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"6CO"_2 +"6H"_2"O"#</mathjax></p>
<p>The coefficients in a chemical equation indicate the number of moles of each element, molecule, or formula unit. In this equation, all of the substances are molecules. If there is no coefficient, it is understood to be <mathjax>#1#</mathjax>.</p>
<p>The balanced equation shows us that 1 mole <mathjax>#"C"_12"H"_2"O"_6#</mathjax> will be consumed if 6 moles <mathjax>#"O"_2#</mathjax> are consumed. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #"C"_6"H"_12"O"_6"# will be consumed if 6 moles of #"O"_2"# are consumed?</h1>
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<div class="markdown"><p><mathjax>#"C"_6"H"_12"O"_6 + "6O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"6CO"_2 +"6H"_2"O"#</mathjax></p></div>
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<div class="markdown"><p>One mole of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> will be consumed if <mathjax>#"6 mol O"_2#</mathjax> are consumed.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"_6"H"_12"O"_6 + "6O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"6CO"_2 +"6H"_2"O"#</mathjax></p>
<p>The coefficients in a chemical equation indicate the number of moles of each element, molecule, or formula unit. In this equation, all of the substances are molecules. If there is no coefficient, it is understood to be <mathjax>#1#</mathjax>.</p>
<p>The balanced equation shows us that 1 mole <mathjax>#"C"_12"H"_2"O"_6#</mathjax> will be consumed if 6 moles <mathjax>#"O"_2#</mathjax> are consumed. </p></div>
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</article> | How many moles of #"C"_6"H"_12"O"_6"# will be consumed if 6 moles of #"O"_2"# are consumed? |
#"C"_6"H"_12"O"_6 + "6O"_2"##rarr##"6CO"_2 +"6H"_2"O"#
|
2,559 | aa6f1e0b-6ddd-11ea-bb19-ccda262736ce | https://socratic.org/questions/elemental-sulfur-occurs-as-octatomic-molecules-s-8-what-mass-g-of-fluorine-gas-i | 63.29 g | start physical_unit 11 12 mass g qc_end physical_unit 1 1 19 20 mass qc_end c_other OTHER qc_end substance 25 26 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] fluorine gas [IN] g"}] | [{"type":"physical unit","value":"63.29 g"}] | [{"type":"physical unit","value":"Mass [OF] sulfur [=] \\pu{17.8 grams}"},{"type":"other","value":"Elemental sulfur occurs as octatomic molecules, S8."},{"type":"substance name","value":"Sulfur hexafluoride"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">Elemental sulfur occurs as octatomic molecules, #S_8#. What mass (g) of fluorine gas is needed to react completely with 17.8 grams of sulfur to form sulfur hexafluoride?</h1> | null | 63.29 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Start with a balanced equation.</strong></p>
<p><mathjax>#"S"_8+"24F"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"8SF"_6"#</mathjax></p>
<p>Then determine the molar masses of <mathjax>#"S"_8"#</mathjax> and <mathjax>#"F"_2"#</mathjax> by multiplying the subscript of each element by its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> from <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><mathjax>#"S"_8":#</mathjax> <mathjax>#(8xx32.06 "g/mol")="256.48 g/mol"#</mathjax><br/>
<mathjax>#"F"_2":#</mathjax> <mathjax>#(2xx18.998 "g/mol")="37.996 g/mol"#</mathjax></p>
<p><strong>Determine the mass of <mathjax>#"F"_2"#</mathjax> needed to react completely with <mathjax>#"S"_8"#</mathjax> to form <mathjax>#"SF"_6"#</mathjax>.</strong></p>
<ol>
<li>Divide the given mass <mathjax>#"S"_8"#</mathjax> by its molar mass. </li>
<li>Multiply <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"F"_2"#</mathjax> and <mathjax>#"S"_8"#</mathjax> from the balanced equation with the moles <mathjax>#"F"_2"#</mathjax> in the numerator: <mathjax>#(24"mol F"_2)/(1"mol S"_8)#</mathjax></li>
<li>Multiply moles <mathjax>#"F"_2"#</mathjax> times its molar mass to get grams <mathjax>#"F"_2"#</mathjax>.</li>
</ol>
<p><mathjax>#17.8cancel"g S"_8xx(1cancel"mol S"_8)/(256.48cancel"g S"_8)xx(24cancel"mol F"_2)/(1cancel"mol S"_8)xx(37.996"g F"_2)/(1cancel"mol F"_2)="63.2 g F"_2#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"63.2 g F"_2"#</mathjax> are needed to react completely with <mathjax>#"17.8 g S"_8"#</mathjax> to produce <mathjax>#"SF"_6"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Start with a balanced equation.</strong></p>
<p><mathjax>#"S"_8+"24F"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"8SF"_6"#</mathjax></p>
<p>Then determine the molar masses of <mathjax>#"S"_8"#</mathjax> and <mathjax>#"F"_2"#</mathjax> by multiplying the subscript of each element by its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> from <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><mathjax>#"S"_8":#</mathjax> <mathjax>#(8xx32.06 "g/mol")="256.48 g/mol"#</mathjax><br/>
<mathjax>#"F"_2":#</mathjax> <mathjax>#(2xx18.998 "g/mol")="37.996 g/mol"#</mathjax></p>
<p><strong>Determine the mass of <mathjax>#"F"_2"#</mathjax> needed to react completely with <mathjax>#"S"_8"#</mathjax> to form <mathjax>#"SF"_6"#</mathjax>.</strong></p>
<ol>
<li>Divide the given mass <mathjax>#"S"_8"#</mathjax> by its molar mass. </li>
<li>Multiply <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"F"_2"#</mathjax> and <mathjax>#"S"_8"#</mathjax> from the balanced equation with the moles <mathjax>#"F"_2"#</mathjax> in the numerator: <mathjax>#(24"mol F"_2)/(1"mol S"_8)#</mathjax></li>
<li>Multiply moles <mathjax>#"F"_2"#</mathjax> times its molar mass to get grams <mathjax>#"F"_2"#</mathjax>.</li>
</ol>
<p><mathjax>#17.8cancel"g S"_8xx(1cancel"mol S"_8)/(256.48cancel"g S"_8)xx(24cancel"mol F"_2)/(1cancel"mol S"_8)xx(37.996"g F"_2)/(1cancel"mol F"_2)="63.2 g F"_2#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Elemental sulfur occurs as octatomic molecules, #S_8#. What mass (g) of fluorine gas is needed to react completely with 17.8 grams of sulfur to form sulfur hexafluoride?</h1>
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<div class="markdown"><p><mathjax>#"63.2 g F"_2"#</mathjax> are needed to react completely with <mathjax>#"17.8 g S"_8"#</mathjax> to produce <mathjax>#"SF"_6"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Start with a balanced equation.</strong></p>
<p><mathjax>#"S"_8+"24F"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"8SF"_6"#</mathjax></p>
<p>Then determine the molar masses of <mathjax>#"S"_8"#</mathjax> and <mathjax>#"F"_2"#</mathjax> by multiplying the subscript of each element by its <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> from <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><mathjax>#"S"_8":#</mathjax> <mathjax>#(8xx32.06 "g/mol")="256.48 g/mol"#</mathjax><br/>
<mathjax>#"F"_2":#</mathjax> <mathjax>#(2xx18.998 "g/mol")="37.996 g/mol"#</mathjax></p>
<p><strong>Determine the mass of <mathjax>#"F"_2"#</mathjax> needed to react completely with <mathjax>#"S"_8"#</mathjax> to form <mathjax>#"SF"_6"#</mathjax>.</strong></p>
<ol>
<li>Divide the given mass <mathjax>#"S"_8"#</mathjax> by its molar mass. </li>
<li>Multiply <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"F"_2"#</mathjax> and <mathjax>#"S"_8"#</mathjax> from the balanced equation with the moles <mathjax>#"F"_2"#</mathjax> in the numerator: <mathjax>#(24"mol F"_2)/(1"mol S"_8)#</mathjax></li>
<li>Multiply moles <mathjax>#"F"_2"#</mathjax> times its molar mass to get grams <mathjax>#"F"_2"#</mathjax>.</li>
</ol>
<p><mathjax>#17.8cancel"g S"_8xx(1cancel"mol S"_8)/(256.48cancel"g S"_8)xx(24cancel"mol F"_2)/(1cancel"mol S"_8)xx(37.996"g F"_2)/(1cancel"mol F"_2)="63.2 g F"_2#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | Elemental sulfur occurs as octatomic molecules, #S_8#. What mass (g) of fluorine gas is needed to react completely with 17.8 grams of sulfur to form sulfur hexafluoride? | null |
2,560 | a8d4c7ae-6ddd-11ea-982f-ccda262736ce | https://socratic.org/questions/what-is-the-maximum-number-of-grams-of-nh-4cl-that-will-dissolve-in-200-grams-of | 120 grams | start physical_unit 8 8 mass g qc_end physical_unit 16 16 18 19 temperature qc_end physical_unit 16 16 13 14 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] NH4Cl [IN] grams"}] | [{"type":"physical unit","value":"120 grams"}] | [{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{70 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{200 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the maximum number of grams of #"NH"_4"Cl"# that will dissolve in 200 grams of water at 70°C?</h1> | null | 120 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a></strong> for ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, which looks like this</p>
<p><img alt="http://www.kentchemistry.com/links/Kinetics/SolubilityCurves.htm" src="https://useruploads.socratic.org/dIdVaFuQSxWIjjGwFEgn_Solubi1.gif"/> </p>
<p>The solubility graph provides you with information on the <em>solubility</em> of a given salt at various temperatures. In this case, the graph shows you the solubility of ammonium chloride <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of water</strong> at temperatures ranging from <mathjax>#0^@"C"#</mathjax> to <mathjax>#100^@"C"#</mathjax>. </p>
<p>The actual curve represents the <strong>solubility</strong> of the salt, i.e. the <em>maximum</em> amount that can be dissolved in <mathjax>#"100 g"#</mathjax> of water to form a <strong>saturated solution</strong>. </p>
<p>At <mathjax>#70^@"C"#</mathjax>, ammonium chloride has a solubility of about <mathjax>#"62 g / 100 g H"_2"O"#</mathjax>. This tells you that can only hope to dissolve <mathjax>#"62 g"#</mathjax> of ammonium chloride for every <mathjax>#"100 g"#</mathjax> of water before the solution becomes <strong>saturated</strong>.</p>
<p>After this point, any extra salt added would <em>remain undissolved</em>. </p>
<p>So, you know how much salt can be dissolved in <mathjax>#"100 g"#</mathjax> of water, so use this as a <em>conversion factor</em> to find the solubility of the salt in <mathjax>#"200 g"#</mathjax> of water</p>
<blockquote>
<p><mathjax>#200 color(red)(cancel(color(black)("g H"_2"O"))) * ("62 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g H"_2"O")))) = "124 g NH"_4"Cl"#</mathjax></p>
</blockquote>
<p>I will leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the mass of water</p>
<blockquote>
<p><mathjax>#"solubility at 70"""^@"C in 200 g H"_2"O" = color(green)(|bar(ul(color(white)(a/a)color(black)("120 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#"120 g NH"_4"Cl"#</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><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a></strong> for ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, which looks like this</p>
<p><img alt="http://www.kentchemistry.com/links/Kinetics/SolubilityCurves.htm" src="https://useruploads.socratic.org/dIdVaFuQSxWIjjGwFEgn_Solubi1.gif"/> </p>
<p>The solubility graph provides you with information on the <em>solubility</em> of a given salt at various temperatures. In this case, the graph shows you the solubility of ammonium chloride <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of water</strong> at temperatures ranging from <mathjax>#0^@"C"#</mathjax> to <mathjax>#100^@"C"#</mathjax>. </p>
<p>The actual curve represents the <strong>solubility</strong> of the salt, i.e. the <em>maximum</em> amount that can be dissolved in <mathjax>#"100 g"#</mathjax> of water to form a <strong>saturated solution</strong>. </p>
<p>At <mathjax>#70^@"C"#</mathjax>, ammonium chloride has a solubility of about <mathjax>#"62 g / 100 g H"_2"O"#</mathjax>. This tells you that can only hope to dissolve <mathjax>#"62 g"#</mathjax> of ammonium chloride for every <mathjax>#"100 g"#</mathjax> of water before the solution becomes <strong>saturated</strong>.</p>
<p>After this point, any extra salt added would <em>remain undissolved</em>. </p>
<p>So, you know how much salt can be dissolved in <mathjax>#"100 g"#</mathjax> of water, so use this as a <em>conversion factor</em> to find the solubility of the salt in <mathjax>#"200 g"#</mathjax> of water</p>
<blockquote>
<p><mathjax>#200 color(red)(cancel(color(black)("g H"_2"O"))) * ("62 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g H"_2"O")))) = "124 g NH"_4"Cl"#</mathjax></p>
</blockquote>
<p>I will leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the mass of water</p>
<blockquote>
<p><mathjax>#"solubility at 70"""^@"C in 200 g H"_2"O" = color(green)(|bar(ul(color(white)(a/a)color(black)("120 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the maximum number of grams of #"NH"_4"Cl"# that will dissolve in 200 grams of water at 70°C?</h1>
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Jul 5, 2016
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<div class="markdown"><p><mathjax>#"120 g NH"_4"Cl"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a></strong> for ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, which looks like this</p>
<p><img alt="http://www.kentchemistry.com/links/Kinetics/SolubilityCurves.htm" src="https://useruploads.socratic.org/dIdVaFuQSxWIjjGwFEgn_Solubi1.gif"/> </p>
<p>The solubility graph provides you with information on the <em>solubility</em> of a given salt at various temperatures. In this case, the graph shows you the solubility of ammonium chloride <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of water</strong> at temperatures ranging from <mathjax>#0^@"C"#</mathjax> to <mathjax>#100^@"C"#</mathjax>. </p>
<p>The actual curve represents the <strong>solubility</strong> of the salt, i.e. the <em>maximum</em> amount that can be dissolved in <mathjax>#"100 g"#</mathjax> of water to form a <strong>saturated solution</strong>. </p>
<p>At <mathjax>#70^@"C"#</mathjax>, ammonium chloride has a solubility of about <mathjax>#"62 g / 100 g H"_2"O"#</mathjax>. This tells you that can only hope to dissolve <mathjax>#"62 g"#</mathjax> of ammonium chloride for every <mathjax>#"100 g"#</mathjax> of water before the solution becomes <strong>saturated</strong>.</p>
<p>After this point, any extra salt added would <em>remain undissolved</em>. </p>
<p>So, you know how much salt can be dissolved in <mathjax>#"100 g"#</mathjax> of water, so use this as a <em>conversion factor</em> to find the solubility of the salt in <mathjax>#"200 g"#</mathjax> of water</p>
<blockquote>
<p><mathjax>#200 color(red)(cancel(color(black)("g H"_2"O"))) * ("62 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g H"_2"O")))) = "124 g NH"_4"Cl"#</mathjax></p>
</blockquote>
<p>I will leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the mass of water</p>
<blockquote>
<p><mathjax>#"solubility at 70"""^@"C in 200 g H"_2"O" = color(green)(|bar(ul(color(white)(a/a)color(black)("120 g")color(white)(a/a)|)))#</mathjax></p>
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</article> | What is the maximum number of grams of #"NH"_4"Cl"# that will dissolve in 200 grams of water at 70°C? | null |
2,561 | abafe050-6ddd-11ea-98f5-ccda262736ce | https://socratic.org/questions/a-sample-of-seawater-sodium-chloride-liter-solution-how-many-mg-of-of-per-of-sod | 94.16 mg | start physical_unit 8 9 mass mg qc_end physical_unit 13 13 24 25 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium chloride [IN] mg"}] | [{"type":"physical unit","value":"94.16 mg"}] | [{"type":"physical unit","value":"Density [OF] sodium chloride solution [=] \\pu{6.277 g/L}"},{"type":"physical unit","value":"Volume [OF] sodium chloride solution [=] \\pu{15.0 mL}"}] | <h1 class="questionTitle" itemprop="name">A sample of seawater contains 6.277 g of sodium chloride per liter of solution. How many mg of sodium chloride would be contained in 15.0 mL of this solution?</h1> | null | 94.16 mg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use multiple approaches to solve this problem, but they revolve around a couple of <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a>.</p>
<p>One way to approach this is to use the given concentration to figure out how many <em>milligrams</em> of sodium chloride you get <strong>per milliliter</strong> of solution. </p>
<p>This will allow you to find the number of <em>milligrams</em> of sodium chloride present in <mathjax>#"15.0 mL"#</mathjax> of this solution, provided that you use the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3"mg") " "#</mathjax> and <mathjax>#" " color(blue)("1 L" = 10^3"mL")#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#"6.277 g"#</mathjax> of sodium chloride <strong>per liter</strong>, you can say that it contains </p>
<blockquote>
<p><mathjax>#6.277 "g"/color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3"mL") = 6.277 * 10^(_3)"g/mL"#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#6.277 * color(purple)(cancel(color(black)(10^(-3)))) color(red)(cancel(color(black)("g")))/"mL" * (color(purple)(cancel(color(black)(10^3)))"mg")/(1color(red)(cancel(color(black)("g")))) = "6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>You now know that </p>
<blockquote>
<p><mathjax>#"6.277 g/L " = " 6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 mL"#</mathjax> of this solution contains <mathjax>#"6.277 mg"#</mathjax> of sodium chloride, <mathjax>#"15.0 mL"#</mathjax> will contain </p>
<blockquote>
<p><mathjax>#15.0 color(red)(cancel(color(black)("mL solution"))) * overbrace( ("6.277 mg NaCl")/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the given concentration")) = "94.155 mg NaCl"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the sample, the answer will be </p>
<blockquote>
<p><mathjax>#m_(NaCl) = color(green)("94.2 mg NaCl")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"94.2 mg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use multiple approaches to solve this problem, but they revolve around a couple of <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a>.</p>
<p>One way to approach this is to use the given concentration to figure out how many <em>milligrams</em> of sodium chloride you get <strong>per milliliter</strong> of solution. </p>
<p>This will allow you to find the number of <em>milligrams</em> of sodium chloride present in <mathjax>#"15.0 mL"#</mathjax> of this solution, provided that you use the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3"mg") " "#</mathjax> and <mathjax>#" " color(blue)("1 L" = 10^3"mL")#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#"6.277 g"#</mathjax> of sodium chloride <strong>per liter</strong>, you can say that it contains </p>
<blockquote>
<p><mathjax>#6.277 "g"/color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3"mL") = 6.277 * 10^(_3)"g/mL"#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#6.277 * color(purple)(cancel(color(black)(10^(-3)))) color(red)(cancel(color(black)("g")))/"mL" * (color(purple)(cancel(color(black)(10^3)))"mg")/(1color(red)(cancel(color(black)("g")))) = "6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>You now know that </p>
<blockquote>
<p><mathjax>#"6.277 g/L " = " 6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 mL"#</mathjax> of this solution contains <mathjax>#"6.277 mg"#</mathjax> of sodium chloride, <mathjax>#"15.0 mL"#</mathjax> will contain </p>
<blockquote>
<p><mathjax>#15.0 color(red)(cancel(color(black)("mL solution"))) * overbrace( ("6.277 mg NaCl")/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the given concentration")) = "94.155 mg NaCl"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the sample, the answer will be </p>
<blockquote>
<p><mathjax>#m_(NaCl) = color(green)("94.2 mg NaCl")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of seawater contains 6.277 g of sodium chloride per liter of solution. How many mg of sodium chloride would be contained in 15.0 mL of this solution?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-22T12:51:24" itemprop="dateCreated">
Jan 22, 2016
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<div class="markdown"><p><mathjax>#"94.2 mg"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can actually use multiple approaches to solve this problem, but they revolve around a couple of <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a>.</p>
<p>One way to approach this is to use the given concentration to figure out how many <em>milligrams</em> of sodium chloride you get <strong>per milliliter</strong> of solution. </p>
<p>This will allow you to find the number of <em>milligrams</em> of sodium chloride present in <mathjax>#"15.0 mL"#</mathjax> of this solution, provided that you use the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)("1 g" = 10^3"mg") " "#</mathjax> and <mathjax>#" " color(blue)("1 L" = 10^3"mL")#</mathjax></p>
</blockquote>
<p>So, if your solution contains <mathjax>#"6.277 g"#</mathjax> of sodium chloride <strong>per liter</strong>, you can say that it contains </p>
<blockquote>
<p><mathjax>#6.277 "g"/color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3"mL") = 6.277 * 10^(_3)"g/mL"#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#6.277 * color(purple)(cancel(color(black)(10^(-3)))) color(red)(cancel(color(black)("g")))/"mL" * (color(purple)(cancel(color(black)(10^3)))"mg")/(1color(red)(cancel(color(black)("g")))) = "6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>You now know that </p>
<blockquote>
<p><mathjax>#"6.277 g/L " = " 6.277 mg/mL"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"1 mL"#</mathjax> of this solution contains <mathjax>#"6.277 mg"#</mathjax> of sodium chloride, <mathjax>#"15.0 mL"#</mathjax> will contain </p>
<blockquote>
<p><mathjax>#15.0 color(red)(cancel(color(black)("mL solution"))) * overbrace( ("6.277 mg NaCl")/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the given concentration")) = "94.155 mg NaCl"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the sample, the answer will be </p>
<blockquote>
<p><mathjax>#m_(NaCl) = color(green)("94.2 mg NaCl")#</mathjax></p>
</blockquote></div>
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</article> | A sample of seawater contains 6.277 g of sodium chloride per liter of solution. How many mg of sodium chloride would be contained in 15.0 mL of this solution? | null |
2,562 | aa0498e8-6ddd-11ea-ae55-ccda262736ce | https://socratic.org/questions/59bc0d13b72cff515ff88474 | 1.96 × 10^(-3) g/mL | start physical_unit 10 10 density g/ml qc_end physical_unit 10 10 6 7 mass qc_end physical_unit 10 10 14 15 volume qc_end end | [{"type":"physical unit","value":"Density [OF] the gas [IN] g/mL"}] | [{"type":"physical unit","value":"1.96 × 10^(-3) g/mL"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{0.196 g}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{100 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the density of a #0.196*g# mass of gas, confined to a #100*mL# volume?</h1> | null | 1.96 × 10^(-3) g/mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">Density</a> is mass over volume. (Mass per unit volume).</p>
<p>So...</p>
<p><mathjax>#(0.196g)/(100 ml) or .00196g/(ml)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#.00196g/(ml)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">Density</a> is mass over volume. (Mass per unit volume).</p>
<p>So...</p>
<p><mathjax>#(0.196g)/(100 ml) or .00196g/(ml)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the density of a #0.196*g# mass of gas, confined to a #100*mL# volume?</h1>
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jim p.
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Sep 15, 2017
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<div class="markdown"><p><mathjax>#.00196g/(ml)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">Density</a> is mass over volume. (Mass per unit volume).</p>
<p>So...</p>
<p><mathjax>#(0.196g)/(100 ml) or .00196g/(ml)#</mathjax></p></div>
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anor277
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Sep 15, 2017
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<div class="markdown"><p>Well, <mathjax>#rho="Mass"/"Volume"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we simply work out the quotient.....</p>
<p><mathjax>#rho=(0.196*g)/(100*mL)=1.96xx10^-3*g*mL^-1#</mathjax>.....</p></div>
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</article> | What is the density of a #0.196*g# mass of gas, confined to a #100*mL# volume? | null |
2,563 | ac86abd9-6ddd-11ea-aeff-ccda262736ce | https://socratic.org/questions/a78-0-g-sample-of-an-unknown-compound-contains-12-4-g-of-hydrogen-what-is-the-pe | 15.90% | start physical_unit 20 23 mass_percent none qc_end physical_unit 6 7 1 2 mass qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Percent by mass [OF] hydrogen in the compound"}] | [{"type":"physical unit","value":"15.90%"}] | [{"type":"physical unit","value":"Mass [OF] unknown compound sample [=] \\pu{78.0 g}"},{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{12.4 g}"}] | <h1 class="questionTitle" itemprop="name">A78.0-g sample of an unknown compound contains 12.4 g of hydrogen. What is the percent by mass of hydrogen in the compound?</h1> | null | 15.90% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have done is take the quotient:</p>
<p><mathjax>#"Mass of hydrogen"/"Mass of stuff"xx100%#</mathjax></p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"%hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.4*g)/(78.0*g)xx100%#</mathjax> <mathjax>#~=15%#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have done is take the quotient:</p>
<p><mathjax>#"Mass of hydrogen"/"Mass of stuff"xx100%#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A78.0-g sample of an unknown compound contains 12.4 g of hydrogen. What is the percent by mass of hydrogen in the compound?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"%hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.4*g)/(78.0*g)xx100%#</mathjax> <mathjax>#~=15%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So all we have done is take the quotient:</p>
<p><mathjax>#"Mass of hydrogen"/"Mass of stuff"xx100%#</mathjax></p></div>
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</article> | A78.0-g sample of an unknown compound contains 12.4 g of hydrogen. What is the percent by mass of hydrogen in the compound? | null |
2,564 | aa6bc282-6ddd-11ea-a35a-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-moles-in-500-l-of-he-gas-at-stp | 22.03 moles | start physical_unit 10 11 mole mol qc_end physical_unit 10 11 7 8 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mole [OF] He gas [IN] moles"}] | [{"type":"physical unit","value":"22.03 moles"}] | [{"type":"physical unit","value":"Volume [OF] He gas [=] \\pu{500 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the number of moles in 500 L of #He# gas at STP? </h1> | null | 22.03 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that you're working under <strong>STP conditions</strong>, which implies that you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to find how many <em>moles</em> of helium will occupy that volume. </p>
<p>Now, the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> represents the volume occupied by <strong>one mole</strong> of a gas under some specific conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>Starting from the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>you can say that the molar volume of gas at a pressure <mathjax>#P#</mathjax> and a temperature <mathjax>#T#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (RT)/P#</mathjax></p>
</blockquote>
<p>Now, <strong>Standard Temperature and Pressure</strong> conditions are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these specific conditions, the molar volume of a gas will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (0.0821 * (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V/n = "22.7 L/mol"#</mathjax></p>
</blockquote>
<p>This of course implies that <strong>one mole</strong> of <strong>any</strong> ideal gas will occupy <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In your case, the volume of the gas is said to be equal to <mathjax>#"500 L"#</mathjax>. This means that you will have </p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("L"))) * "1 mole He"/(22.7color(red)(cancel(color(black)("L")))) = "22.026 moles He"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the volume of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#n_(He) = color(green)("20 moles")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions for pressure and temperature, one mole of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If these are the values for STP given to you by your instructor, make sure to 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>#"20 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that you're working under <strong>STP conditions</strong>, which implies that you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to find how many <em>moles</em> of helium will occupy that volume. </p>
<p>Now, the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> represents the volume occupied by <strong>one mole</strong> of a gas under some specific conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>Starting from the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>you can say that the molar volume of gas at a pressure <mathjax>#P#</mathjax> and a temperature <mathjax>#T#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (RT)/P#</mathjax></p>
</blockquote>
<p>Now, <strong>Standard Temperature and Pressure</strong> conditions are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these specific conditions, the molar volume of a gas will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (0.0821 * (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V/n = "22.7 L/mol"#</mathjax></p>
</blockquote>
<p>This of course implies that <strong>one mole</strong> of <strong>any</strong> ideal gas will occupy <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In your case, the volume of the gas is said to be equal to <mathjax>#"500 L"#</mathjax>. This means that you will have </p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("L"))) * "1 mole He"/(22.7color(red)(cancel(color(black)("L")))) = "22.026 moles He"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the volume of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#n_(He) = color(green)("20 moles")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions for pressure and temperature, one mole of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If these are the values for STP given to you by your instructor, make sure to 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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<h1 class="questionTitle" itemprop="name">What is the number of moles in 500 L of #He# gas at STP? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"20 moles"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that you're working under <strong>STP conditions</strong>, which implies that you can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to find how many <em>moles</em> of helium will occupy that volume. </p>
<p>Now, the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong> represents the volume occupied by <strong>one mole</strong> of a gas under some specific conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>Starting from the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>you can say that the molar volume of gas at a pressure <mathjax>#P#</mathjax> and a temperature <mathjax>#T#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (RT)/P#</mathjax></p>
</blockquote>
<p>Now, <strong>Standard Temperature and Pressure</strong> conditions are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Under these specific conditions, the molar volume of a gas will be equal to </p>
<blockquote>
<p><mathjax>#V/n = (0.0821 * (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V/n = "22.7 L/mol"#</mathjax></p>
</blockquote>
<p>This of course implies that <strong>one mole</strong> of <strong>any</strong> ideal gas will occupy <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In your case, the volume of the gas is said to be equal to <mathjax>#"500 L"#</mathjax>. This means that you will have </p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("L"))) * "1 mole He"/(22.7color(red)(cancel(color(black)("L")))) = "22.026 moles He"#</mathjax></p>
</blockquote>
<p>Rounded to one <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the volume of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#n_(He) = color(green)("20 moles")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions for pressure and temperature, one mole of any ideal gas occupies</em> <mathjax>#"22.4 L"#</mathjax>. <em>If these are the values for STP given to you by your instructor, make sure to 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> | What is the number of moles in 500 L of #He# gas at STP? | null |
2,565 | ac2c587e-6ddd-11ea-9433-ccda262736ce | https://socratic.org/questions/a-balloon-contains-2-58-10-24-molecules-of-methane-gas-how-many-moles-of-methane | 4.28 moles | start physical_unit 8 8 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] methane [IN] moles"}] | [{"type":"physical unit","value":"4.28 moles"}] | [{"type":"physical unit","value":"Number [OF] methane molecules [=] \\pu{2.58 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon?</h1> | null | 4.28 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>A farmer collects 150 eggs; how many dozen eggs does he collect? </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon?</h1>
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anor277
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Mar 24, 2016
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<div class="markdown"><p>A farmer collects 150 eggs; how many dozen eggs does he collect? </p></div>
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<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
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<span class="dateCreated" datetime="2016-03-25T00:59:06" itemprop="dateCreated">
Mar 25, 2016
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<div class="markdown"><p><mathjax>#2.58xx10^24"molecules CH"_4"#</mathjax> contains <mathjax>#4.28"mol CH"_4"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#1 "mole CH"_4=6.022xx10^23 "molecules CH"_4"#</mathjax></p>
<p>In order to convert molecules to moles of a substance, divide the number of given molecules by <mathjax>#6.022xx10^23 "molecules/mol"#</mathjax>.</p>
<p><mathjax>#2.58xx10^24cancel"molecules CH"_4xx(1"mol CH"_4)/(6.022xx10^23cancel"molecules CH"_4)="4.28 moles CH"_4"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon? | null |
2,566 | aa49371e-6ddd-11ea-aba7-ccda262736ce | https://socratic.org/questions/a-20-00-ml-sample-of-a-koh-solution-required-31-32-ml-of-0-118-m-hcl-for-neutral | 0.21 g | start physical_unit 6 6 mass g qc_end physical_unit 6 7 1 2 volume qc_end physical_unit 7 7 9 10 volume qc_end physical_unit 14 14 12 13 molarity qc_end physical_unit 6 6 37 38 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] KOH [IN] g"}] | [{"type":"physical unit","value":"0.21 g"}] | [{"type":"physical unit","value":"Volume [OF] KOH solution sample [=] \\pu{20.00 mL}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{31.32 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.118 M}"},{"type":"physical unit","value":"Molar mass [OF] KOH [=] \\pu{56.11 g/mol}"}] | <h1 class="questionTitle" itemprop="name">A 20.00 mL sample of a KOH solution required 31.32 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was present in the 20.00 mL of the sample of KOH - (Molar mass of KOH 56.11 g/mol)?</h1> | null | 0.21 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to </p>
<ul>
<li><em>calculate the <strong>number of moles</strong> of hydrochloric acid</em>, <mathjax>#"HCl#</mathjax>, <em>that were consumed in the reaction</em></li>
<li><em>use the <strong>mole ratio</strong> that exists between the two reactants to calculate the <strong>number of moles</strong> of potassium hydroxide, <mathjax>#"KOH"#</mathjax>, present in the sample</em></li>
<li><em>use the <strong>molar mass</strong> of potassium hydroxide to convert the number of moles to <strong>grams</strong></em></li>
</ul>
<p>So, you know that a <mathjax>#"0.118 M"#</mathjax> solution of hydrochloric acid will contain <mathjax>#0.118#</mathjax> <strong>moles</strong> of acid <strong>per liter of solution</strong>. This means that your </p>
<blockquote>
<p><mathjax>#31.32 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.03132 L"#</mathjax></p>
</blockquote>
<p>sample will contain </p>
<blockquote>
<p><mathjax>#0.03132 color(red)(cancel(color(black)("L solution"))) * "0.118 moles HCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.003696 moles HCl"#</mathjax></p>
</blockquote>
<p>Potassium hydroxide and hydrochloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous potassium chloride and water</p>
<blockquote>
<p><mathjax>#"KOH"_ ((aq)) + "HCl"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> requires <strong>equal numbers of moles</strong> of acid and of base. You can thus say that the sample of potassium hydroxide must have contained <mathjax>#0.003696#</mathjax> <strong>moles</strong> of potassium hydroxide, since that's how many moles of hydrochloric acid were consumed in the reaction. </p>
<p>Finally, potassium hydroxide has a <strong>molar mass</strong> of <mathjax>#"56.11 g mol"^(-1)#</mathjax>, i.e. <strong>one mole</strong> of this compound has a mass of <mathjax>#"56.11 g"#</mathjax>. </p>
<p>This means that the mass of potassium hydroxide dissolved in solution was</p>
<blockquote>
<p><mathjax>#0.003696 color(red)(cancel(color(black)("moles KOH"))) * "56.11 g"/(1color(red)(cancel(color(black)("mole KOH")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.207 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.207 g KOH"#</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>
<ul>
<li><em>calculate the <strong>number of moles</strong> of hydrochloric acid</em>, <mathjax>#"HCl#</mathjax>, <em>that were consumed in the reaction</em></li>
<li><em>use the <strong>mole ratio</strong> that exists between the two reactants to calculate the <strong>number of moles</strong> of potassium hydroxide, <mathjax>#"KOH"#</mathjax>, present in the sample</em></li>
<li><em>use the <strong>molar mass</strong> of potassium hydroxide to convert the number of moles to <strong>grams</strong></em></li>
</ul>
<p>So, you know that a <mathjax>#"0.118 M"#</mathjax> solution of hydrochloric acid will contain <mathjax>#0.118#</mathjax> <strong>moles</strong> of acid <strong>per liter of solution</strong>. This means that your </p>
<blockquote>
<p><mathjax>#31.32 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.03132 L"#</mathjax></p>
</blockquote>
<p>sample will contain </p>
<blockquote>
<p><mathjax>#0.03132 color(red)(cancel(color(black)("L solution"))) * "0.118 moles HCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.003696 moles HCl"#</mathjax></p>
</blockquote>
<p>Potassium hydroxide and hydrochloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous potassium chloride and water</p>
<blockquote>
<p><mathjax>#"KOH"_ ((aq)) + "HCl"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> requires <strong>equal numbers of moles</strong> of acid and of base. You can thus say that the sample of potassium hydroxide must have contained <mathjax>#0.003696#</mathjax> <strong>moles</strong> of potassium hydroxide, since that's how many moles of hydrochloric acid were consumed in the reaction. </p>
<p>Finally, potassium hydroxide has a <strong>molar mass</strong> of <mathjax>#"56.11 g mol"^(-1)#</mathjax>, i.e. <strong>one mole</strong> of this compound has a mass of <mathjax>#"56.11 g"#</mathjax>. </p>
<p>This means that the mass of potassium hydroxide dissolved in solution was</p>
<blockquote>
<p><mathjax>#0.003696 color(red)(cancel(color(black)("moles KOH"))) * "56.11 g"/(1color(red)(cancel(color(black)("mole KOH")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.207 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid solution. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A 20.00 mL sample of a KOH solution required 31.32 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was present in the 20.00 mL of the sample of KOH - (Molar mass of KOH 56.11 g/mol)?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.207 g KOH"#</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>
<ul>
<li><em>calculate the <strong>number of moles</strong> of hydrochloric acid</em>, <mathjax>#"HCl#</mathjax>, <em>that were consumed in the reaction</em></li>
<li><em>use the <strong>mole ratio</strong> that exists between the two reactants to calculate the <strong>number of moles</strong> of potassium hydroxide, <mathjax>#"KOH"#</mathjax>, present in the sample</em></li>
<li><em>use the <strong>molar mass</strong> of potassium hydroxide to convert the number of moles to <strong>grams</strong></em></li>
</ul>
<p>So, you know that a <mathjax>#"0.118 M"#</mathjax> solution of hydrochloric acid will contain <mathjax>#0.118#</mathjax> <strong>moles</strong> of acid <strong>per liter of solution</strong>. This means that your </p>
<blockquote>
<p><mathjax>#31.32 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.03132 L"#</mathjax></p>
</blockquote>
<p>sample will contain </p>
<blockquote>
<p><mathjax>#0.03132 color(red)(cancel(color(black)("L solution"))) * "0.118 moles HCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.003696 moles HCl"#</mathjax></p>
</blockquote>
<p>Potassium hydroxide and hydrochloric acid react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> to produce aqueous potassium chloride and water</p>
<blockquote>
<p><mathjax>#"KOH"_ ((aq)) + "HCl"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This means that a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> requires <strong>equal numbers of moles</strong> of acid and of base. You can thus say that the sample of potassium hydroxide must have contained <mathjax>#0.003696#</mathjax> <strong>moles</strong> of potassium hydroxide, since that's how many moles of hydrochloric acid were consumed in the reaction. </p>
<p>Finally, potassium hydroxide has a <strong>molar mass</strong> of <mathjax>#"56.11 g mol"^(-1)#</mathjax>, i.e. <strong>one mole</strong> of this compound has a mass of <mathjax>#"56.11 g"#</mathjax>. </p>
<p>This means that the mass of potassium hydroxide dissolved in solution was</p>
<blockquote>
<p><mathjax>#0.003696 color(red)(cancel(color(black)("moles KOH"))) * "56.11 g"/(1color(red)(cancel(color(black)("mole KOH")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.207 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid solution. </p></div>
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</article> | A 20.00 mL sample of a KOH solution required 31.32 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was present in the 20.00 mL of the sample of KOH - (Molar mass of KOH 56.11 g/mol)? | null |
2,567 | aacae742-6ddd-11ea-aa63-ccda262736ce | https://socratic.org/questions/548c0cb9581e2a6c68b7cad7 | 525.47 kg | start physical_unit 3 3 mass kg qc_end physical_unit 10 10 13 13 ph qc_end physical_unit 10 10 18 21 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] limestone [IN] kg"}] | [{"type":"physical unit","value":"525.47 kg"}] | [{"type":"physical unit","value":"pH [OF] the lake [=] \\pu{5.6}"},{"type":"physical unit","value":"Volume [OF] the lake [=] \\pu{4.2 × 10^9 L}"}] | <h1 class="questionTitle" itemprop="name">What mass of limestone do I need to neutralize a lake with pH 5.6 and a volume of #4.2 × 10# L?</h1> | null | 525.47 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1.</strong> Calculate the concentration of H⁺</p>
<p><mathjax>#["H"^+] = 10^("-pH")" mol/L" =10^-5.6" mol/L" = 2.5 × 10^-6" mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Calculate the moles of H⁺</p>
<p>Moles = <mathjax>#4.2 ×10^9" L" × (2.5 × 10^-6" mol")/"1 L" = 1.05 × 10^4" mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Calculate the mass of CaCO₃</p>
<p>CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂</p>
<p>Mass of CaCO₃ = <mathjax>#1.05 × 10^4" mol H"^+ × ("1 mol CaCO"_3)/("2 mol H"^+) × ("100.09 g CaCO"_3)/("1 mol CaCO"_3) × ("1 kg CaCO"_3)/("1000 g CaCO"_3) = "530 kg CaCO"_3#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You need 530 kg of limestone to neutralize the lake.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1.</strong> Calculate the concentration of H⁺</p>
<p><mathjax>#["H"^+] = 10^("-pH")" mol/L" =10^-5.6" mol/L" = 2.5 × 10^-6" mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Calculate the moles of H⁺</p>
<p>Moles = <mathjax>#4.2 ×10^9" L" × (2.5 × 10^-6" mol")/"1 L" = 1.05 × 10^4" mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Calculate the mass of CaCO₃</p>
<p>CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂</p>
<p>Mass of CaCO₃ = <mathjax>#1.05 × 10^4" mol H"^+ × ("1 mol CaCO"_3)/("2 mol H"^+) × ("100.09 g CaCO"_3)/("1 mol CaCO"_3) × ("1 kg CaCO"_3)/("1000 g CaCO"_3) = "530 kg CaCO"_3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of limestone do I need to neutralize a lake with pH 5.6 and a volume of #4.2 × 10# L?</h1>
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<div class="markdown"><p>You need 530 kg of limestone to neutralize the lake.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1.</strong> Calculate the concentration of H⁺</p>
<p><mathjax>#["H"^+] = 10^("-pH")" mol/L" =10^-5.6" mol/L" = 2.5 × 10^-6" mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Calculate the moles of H⁺</p>
<p>Moles = <mathjax>#4.2 ×10^9" L" × (2.5 × 10^-6" mol")/"1 L" = 1.05 × 10^4" mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Calculate the mass of CaCO₃</p>
<p>CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂</p>
<p>Mass of CaCO₃ = <mathjax>#1.05 × 10^4" mol H"^+ × ("1 mol CaCO"_3)/("2 mol H"^+) × ("100.09 g CaCO"_3)/("1 mol CaCO"_3) × ("1 kg CaCO"_3)/("1000 g CaCO"_3) = "530 kg CaCO"_3#</mathjax></p></div>
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</article> | What mass of limestone do I need to neutralize a lake with pH 5.6 and a volume of #4.2 × 10# L? | null |
2,568 | a9baee1e-6ddd-11ea-8056-ccda262736ce | https://socratic.org/questions/58d882ec11ef6b7d8f23fa4f | 8.24 L | start physical_unit 3 4 volume l qc_end c_other OTHER qc_end physical_unit 21 22 17 18 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] dihydrogen gas [IN] L"}] | [{"type":"physical unit","value":"8.24 L"}] | [{"type":"other","value":"Excess hydrochloric acid."},{"type":"physical unit","value":"Mass [OF] zinc metal [=] \\pu{21 g}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{13 ℃}"},{"type":"physical unit","value":"Pressure [OF] the reaction [=] \\pu{704 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What volume of dihydrogen gas will be generated by the action of excess hydrochloric acid on a #21*g# mass of zinc metal?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>A temperature of <mathjax>#13#</mathjax> <mathjax>#""^@C#</mathjax>, and a pressure of <mathjax>#704*mm*Hg#</mathjax> are specified...</p></div>
</h2>
</div>
</div> | 8.24 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>That is <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. This is a convenient laboratory measurement that I hope has been shown to you in the lab. The difference between mercury columns can thus be related to differences between absolute pressures. A mercury column is thus useful for pressures BELOW <mathjax>#1*atm#</mathjax>. IT IS NOT USED FOR PRESSURES ABOVE <mathjax>#1* atm#</mathjax>. (Why not? Because you will get mercury all over the lab, and guess who is going to clean it up.)</p>
<p>And of course we need (ii) a stoichiometrically balanced equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>So for each equiv metal, 1 equiv of dihydrogen gas results. </p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#(21*g)/(65.38*g*mol^-1)=0.321*mol#</mathjax>.</p>
<p>And thus, by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#0.321*mol#</mathjax> <mathjax>#H_2(g)#</mathjax> will result. </p>
<p>And so now, this is an Ideal Gas Equation, where we solve for volume:</p>
<p><mathjax>#V=(nRT)/P=(0.321*cancel(mol)xx0.0821*(L*cancel"atm")/(cancel(K^-1*mol^-1))xx286*cancelK)/((704cancel(*mm*Hg))/(760cancel(*mm*Hg*atm^-1))#</mathjax></p>
<p>You can do the math. I get an answer of approx. <mathjax>#8*L#</mathjax> at this pressure. And this is consistent with the known molar volume of an Ideal Gas under standard conditions, i.e. approx. <mathjax>#25*L#</mathjax>.</p>
<p>See <a href="https://socratic.org/questions/how-do-you-convert-atm-to-torr">here</a> for more on the use of mercury to measure moderate pressure. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>To answer this question we need (i) to know that:</p>
<p><mathjax>#"760 mm Hg"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#1*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>That is <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. This is a convenient laboratory measurement that I hope has been shown to you in the lab. The difference between mercury columns can thus be related to differences between absolute pressures. A mercury column is thus useful for pressures BELOW <mathjax>#1*atm#</mathjax>. IT IS NOT USED FOR PRESSURES ABOVE <mathjax>#1* atm#</mathjax>. (Why not? Because you will get mercury all over the lab, and guess who is going to clean it up.)</p>
<p>And of course we need (ii) a stoichiometrically balanced equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>So for each equiv metal, 1 equiv of dihydrogen gas results. </p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#(21*g)/(65.38*g*mol^-1)=0.321*mol#</mathjax>.</p>
<p>And thus, by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#0.321*mol#</mathjax> <mathjax>#H_2(g)#</mathjax> will result. </p>
<p>And so now, this is an Ideal Gas Equation, where we solve for volume:</p>
<p><mathjax>#V=(nRT)/P=(0.321*cancel(mol)xx0.0821*(L*cancel"atm")/(cancel(K^-1*mol^-1))xx286*cancelK)/((704cancel(*mm*Hg))/(760cancel(*mm*Hg*atm^-1))#</mathjax></p>
<p>You can do the math. I get an answer of approx. <mathjax>#8*L#</mathjax> at this pressure. And this is consistent with the known molar volume of an Ideal Gas under standard conditions, i.e. approx. <mathjax>#25*L#</mathjax>.</p>
<p>See <a href="https://socratic.org/questions/how-do-you-convert-atm-to-torr">here</a> for more on the use of mercury to measure moderate pressure. </p></div>
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<h1 class="questionTitle" itemprop="name">What volume of dihydrogen gas will be generated by the action of excess hydrochloric acid on a #21*g# mass of zinc metal?</h1>
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<div class="markdown"><p>A temperature of <mathjax>#13#</mathjax> <mathjax>#""^@C#</mathjax>, and a pressure of <mathjax>#704*mm*Hg#</mathjax> are specified...</p></div>
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<div class="markdown"><p>To answer this question we need (i) to know that:</p>
<p><mathjax>#"760 mm Hg"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#1*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>That is <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. This is a convenient laboratory measurement that I hope has been shown to you in the lab. The difference between mercury columns can thus be related to differences between absolute pressures. A mercury column is thus useful for pressures BELOW <mathjax>#1*atm#</mathjax>. IT IS NOT USED FOR PRESSURES ABOVE <mathjax>#1* atm#</mathjax>. (Why not? Because you will get mercury all over the lab, and guess who is going to clean it up.)</p>
<p>And of course we need (ii) a stoichiometrically balanced equation:</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>So for each equiv metal, 1 equiv of dihydrogen gas results. </p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#(21*g)/(65.38*g*mol^-1)=0.321*mol#</mathjax>.</p>
<p>And thus, by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, <mathjax>#0.321*mol#</mathjax> <mathjax>#H_2(g)#</mathjax> will result. </p>
<p>And so now, this is an Ideal Gas Equation, where we solve for volume:</p>
<p><mathjax>#V=(nRT)/P=(0.321*cancel(mol)xx0.0821*(L*cancel"atm")/(cancel(K^-1*mol^-1))xx286*cancelK)/((704cancel(*mm*Hg))/(760cancel(*mm*Hg*atm^-1))#</mathjax></p>
<p>You can do the math. I get an answer of approx. <mathjax>#8*L#</mathjax> at this pressure. And this is consistent with the known molar volume of an Ideal Gas under standard conditions, i.e. approx. <mathjax>#25*L#</mathjax>.</p>
<p>See <a href="https://socratic.org/questions/how-do-you-convert-atm-to-torr">here</a> for more on the use of mercury to measure moderate pressure. </p></div>
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</article> | What volume of dihydrogen gas will be generated by the action of excess hydrochloric acid on a #21*g# mass of zinc metal? |
A temperature of #13# #""^@C#, and a pressure of #704*mm*Hg# are specified...
|
2,569 | a9f6b100-6ddd-11ea-984a-ccda262736ce | https://socratic.org/questions/what-is-the-charge-on-the-theoretical-element-x-that-forms-a-cation-with-a-oxyge | +1 | start physical_unit 6 8 charge none qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Charge [OF] theoretical element X"}] | [{"type":"physical unit","value":"+1"}] | [{"type":"other","value":"Element X forms a cation with a oxygen and the formula of its oxide is X2O."}] | <h1 class="questionTitle" itemprop="name">What is the charge on the theoretical element #X# that forms a cation with a oxygen and the formula of its oxide is #X_2O#?</h1> | null | +1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to remember about <strong><a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they are <strong>electrically neutral</strong>. </p>
<p>This means that the <em>overall</em> positive charge coming from the cations <strong>must be balanced</strong> by the <em>overall</em> negative charge coming from the anions. </p>
<p>Now, your theoretical element <mathjax>#"X"#</mathjax> forms an ionic compound with the <em>oxide anion</em>, <mathjax>#"O"^(2-)#</mathjax>. This means that you need an <em>overall</em> positive charge of <mathjax>#2+#</mathjax> in order to balance the negative <mathjax>#2-#</mathjax> charge present on the oxide anion. </p>
<p>Notice that the formula of the compound is <mathjax>#"X"_2"O"#</mathjax>. This tells you that one <em>formula unit</em> of this oxide will contain </p>
<blockquote>
<ul>
<li><em><strong>two</strong></em> <mathjax>#"X"#</mathjax> <em>cations</em></li>
<li><em><strong>one</strong></em> <mathjax>#"O"^(2-)#</mathjax> <em>anion</em></li>
</ul>
</blockquote>
<p>Well, if you have <strong>two</strong> <mathjax>#"X"#</mathjax> cations that must have an overall charge of <mathjax>#2+#</mathjax>, it follows that <strong>each</strong> <mathjax>#"X"#</mathjax> cation will have a charge of <mathjax>#1+#</mathjax>. </p>
<p>So, one formula unit for this oxide will contain <strong>two</strong> <mathjax>#"X"^(+)#</mathjax> cations and <strong>one</strong> <mathjax>#"O"^(2-)#</mathjax> anion. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1+#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to remember about <strong><a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they are <strong>electrically neutral</strong>. </p>
<p>This means that the <em>overall</em> positive charge coming from the cations <strong>must be balanced</strong> by the <em>overall</em> negative charge coming from the anions. </p>
<p>Now, your theoretical element <mathjax>#"X"#</mathjax> forms an ionic compound with the <em>oxide anion</em>, <mathjax>#"O"^(2-)#</mathjax>. This means that you need an <em>overall</em> positive charge of <mathjax>#2+#</mathjax> in order to balance the negative <mathjax>#2-#</mathjax> charge present on the oxide anion. </p>
<p>Notice that the formula of the compound is <mathjax>#"X"_2"O"#</mathjax>. This tells you that one <em>formula unit</em> of this oxide will contain </p>
<blockquote>
<ul>
<li><em><strong>two</strong></em> <mathjax>#"X"#</mathjax> <em>cations</em></li>
<li><em><strong>one</strong></em> <mathjax>#"O"^(2-)#</mathjax> <em>anion</em></li>
</ul>
</blockquote>
<p>Well, if you have <strong>two</strong> <mathjax>#"X"#</mathjax> cations that must have an overall charge of <mathjax>#2+#</mathjax>, it follows that <strong>each</strong> <mathjax>#"X"#</mathjax> cation will have a charge of <mathjax>#1+#</mathjax>. </p>
<p>So, one formula unit for this oxide will contain <strong>two</strong> <mathjax>#"X"^(+)#</mathjax> cations and <strong>one</strong> <mathjax>#"O"^(2-)#</mathjax> anion. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the charge on the theoretical element #X# that forms a cation with a oxygen and the formula of its oxide is #X_2O#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-02-09T23:16:19" itemprop="dateCreated">
Feb 9, 2016
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<div class="markdown"><p><mathjax>#1+#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to remember about <strong><a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a></strong> is that they are <strong>electrically neutral</strong>. </p>
<p>This means that the <em>overall</em> positive charge coming from the cations <strong>must be balanced</strong> by the <em>overall</em> negative charge coming from the anions. </p>
<p>Now, your theoretical element <mathjax>#"X"#</mathjax> forms an ionic compound with the <em>oxide anion</em>, <mathjax>#"O"^(2-)#</mathjax>. This means that you need an <em>overall</em> positive charge of <mathjax>#2+#</mathjax> in order to balance the negative <mathjax>#2-#</mathjax> charge present on the oxide anion. </p>
<p>Notice that the formula of the compound is <mathjax>#"X"_2"O"#</mathjax>. This tells you that one <em>formula unit</em> of this oxide will contain </p>
<blockquote>
<ul>
<li><em><strong>two</strong></em> <mathjax>#"X"#</mathjax> <em>cations</em></li>
<li><em><strong>one</strong></em> <mathjax>#"O"^(2-)#</mathjax> <em>anion</em></li>
</ul>
</blockquote>
<p>Well, if you have <strong>two</strong> <mathjax>#"X"#</mathjax> cations that must have an overall charge of <mathjax>#2+#</mathjax>, it follows that <strong>each</strong> <mathjax>#"X"#</mathjax> cation will have a charge of <mathjax>#1+#</mathjax>. </p>
<p>So, one formula unit for this oxide will contain <strong>two</strong> <mathjax>#"X"^(+)#</mathjax> cations and <strong>one</strong> <mathjax>#"O"^(2-)#</mathjax> anion. </p></div>
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</article> | What is the charge on the theoretical element #X# that forms a cation with a oxygen and the formula of its oxide is #X_2O#? | null |
2,570 | a94bae86-6ddd-11ea-a047-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-occupies-17-ml-at-112-c-what-volume-does-the-sample-occupy-at-70-1 | 29 mL | start physical_unit 1 3 volume ml qc_end physical_unit 1 3 5 6 volume qc_end physical_unit 1 3 8 9 temperature qc_end physical_unit 1 3 17 18 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas sample [IN] mL"}] | [{"type":"physical unit","value":"29 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{17 mL}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{-112 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] gas sample [=] \\pu{70 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of gas occupies 17 mL at -112 C. What volume does the sample occupy at 70°C?</h1> | null | 29 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Charle's - GayLussac's law,</p>
<p><mathjax>#V_1/T_1 = V_2/T_2#</mathjax></p>
<p>manipulating this,<br/>
<mathjax>#V_2 = (V_1 T_2) / T_1#</mathjax><br/>
<mathjax>#V_2 = (17 mL * (70^oC + 273.15 K)) / (-112^oC + 273.15 K)#</mathjax><br/>
<mathjax>#V_2 =29 mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>29 mL</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Charle's - GayLussac's law,</p>
<p><mathjax>#V_1/T_1 = V_2/T_2#</mathjax></p>
<p>manipulating this,<br/>
<mathjax>#V_2 = (V_1 T_2) / T_1#</mathjax><br/>
<mathjax>#V_2 = (17 mL * (70^oC + 273.15 K)) / (-112^oC + 273.15 K)#</mathjax><br/>
<mathjax>#V_2 =29 mL#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A sample of gas occupies 17 mL at -112 C. What volume does the sample occupy at 70°C?</h1>
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<div class="markdown"><p>29 mL</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Using Charle's - GayLussac's law,</p>
<p><mathjax>#V_1/T_1 = V_2/T_2#</mathjax></p>
<p>manipulating this,<br/>
<mathjax>#V_2 = (V_1 T_2) / T_1#</mathjax><br/>
<mathjax>#V_2 = (17 mL * (70^oC + 273.15 K)) / (-112^oC + 273.15 K)#</mathjax><br/>
<mathjax>#V_2 =29 mL#</mathjax></p></div>
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</article> | A sample of gas occupies 17 mL at -112 C. What volume does the sample occupy at 70°C? | null |
2,571 | ac9ccf2f-6ddd-11ea-b2c6-ccda262736ce | https://socratic.org/questions/55dae6cc581e2a35831c3721 | 2.22 | start physical_unit 8 8 ph none qc_end physical_unit 10 11 14 14 pka qc_end physical_unit 10 11 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] formic acid solution"}] | [{"type":"physical unit","value":"2.22"}] | [{"type":"physical unit","value":"pKa [OF] formic acid [=] \\pu{3.75}"},{"type":"physical unit","value":"Molarity [OF] formic acid solution [=] \\pu{0.20 M}"}] | <h1 class="questionTitle" itemprop="name">What is the #"pH"# of a #"0.20-M"# solution of formic acid? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#pK_a = 3.75#</mathjax></p></div>
</h2>
</div>
</div> | 2.22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will show you how to solve part <strong>(a)</strong>, so that you can use this example to solve part <strong>(b)</strong> on your own.</p>
<p>So, you're dealing with <em>formic acid</em>, <mathjax>#"HCOOH"#</mathjax>, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid. </p>
<p>You can use an <strong>ICE table</strong> and the initial <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">concentration</a> ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes</p>
<blockquote>
<p><mathjax>#"HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "HCOO"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.20" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "(0.20-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x#</mathjax></p>
<p>You need to use the acid's <mathjax>#pK_a#</mathjax> to determine its <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(K_a = 10^(-pK_a))#</mathjax></p>
</blockquote>
<p>In your case, you have </p>
<blockquote>
<p><mathjax>#K_a = 10^(-3.75) = 1.78 * 10^(-4)#</mathjax></p>
</blockquote>
<p>You know that the equilibrium constant is also equal to </p>
<blockquote>
<p><mathjax>#K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])#</mathjax></p>
<p><mathjax>#K_a = (x * x)/(0.20 - x)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#K_a#</mathjax> is small compared with the initial concentration of the acid, you can approximate <mathjax>#(0.20-x)#</mathjax> with <mathjax>#0.20#</mathjax>. This will give you </p>
<blockquote>
<p><mathjax>#K_a = x^2/0.2 impliesx = sqrt(0.2 * 1.78 * 10^(-4)) = 5.97 * 10^(-3)#</mathjax></p>
</blockquote>
<p>The concentration of the hydronium ions will thus be</p>
<blockquote>
<p><mathjax>#x = ["H"_3"O"^(+)] = 5.97 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that the solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> will be </p>
<blockquote>
<p><mathjax>#pH_"sol" = -log(["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#pH_"sol" = -log(5.97 * 10^(-3)) = color(green)(2.22)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For part <strong>(a)</strong>: <mathjax>#pH_"sol" = 2.22#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will show you how to solve part <strong>(a)</strong>, so that you can use this example to solve part <strong>(b)</strong> on your own.</p>
<p>So, you're dealing with <em>formic acid</em>, <mathjax>#"HCOOH"#</mathjax>, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid. </p>
<p>You can use an <strong>ICE table</strong> and the initial <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">concentration</a> ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes</p>
<blockquote>
<p><mathjax>#"HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "HCOO"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.20" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "(0.20-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x#</mathjax></p>
<p>You need to use the acid's <mathjax>#pK_a#</mathjax> to determine its <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(K_a = 10^(-pK_a))#</mathjax></p>
</blockquote>
<p>In your case, you have </p>
<blockquote>
<p><mathjax>#K_a = 10^(-3.75) = 1.78 * 10^(-4)#</mathjax></p>
</blockquote>
<p>You know that the equilibrium constant is also equal to </p>
<blockquote>
<p><mathjax>#K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])#</mathjax></p>
<p><mathjax>#K_a = (x * x)/(0.20 - x)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#K_a#</mathjax> is small compared with the initial concentration of the acid, you can approximate <mathjax>#(0.20-x)#</mathjax> with <mathjax>#0.20#</mathjax>. This will give you </p>
<blockquote>
<p><mathjax>#K_a = x^2/0.2 impliesx = sqrt(0.2 * 1.78 * 10^(-4)) = 5.97 * 10^(-3)#</mathjax></p>
</blockquote>
<p>The concentration of the hydronium ions will thus be</p>
<blockquote>
<p><mathjax>#x = ["H"_3"O"^(+)] = 5.97 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that the solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> will be </p>
<blockquote>
<p><mathjax>#pH_"sol" = -log(["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#pH_"sol" = -log(5.97 * 10^(-3)) = color(green)(2.22)#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the #"pH"# of a #"0.20-M"# solution of formic acid? </h1>
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<div class="markdown"><p><mathjax>#pK_a = 3.75#</mathjax></p></div>
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Stefan V.
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Aug 30, 2015
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<div class="markdown"><p>For part <strong>(a)</strong>: <mathjax>#pH_"sol" = 2.22#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will show you how to solve part <strong>(a)</strong>, so that you can use this example to solve part <strong>(b)</strong> on your own.</p>
<p>So, you're dealing with <em>formic acid</em>, <mathjax>#"HCOOH"#</mathjax>, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid. </p>
<p>You can use an <strong>ICE table</strong> and the initial <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">concentration</a> ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes</p>
<blockquote>
<p><mathjax>#"HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "HCOO"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.20" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "(0.20-x)" " " " " " " " " " " " " "x" " " " " " " " " " " "x#</mathjax></p>
<p>You need to use the acid's <mathjax>#pK_a#</mathjax> to determine its <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(K_a = 10^(-pK_a))#</mathjax></p>
</blockquote>
<p>In your case, you have </p>
<blockquote>
<p><mathjax>#K_a = 10^(-3.75) = 1.78 * 10^(-4)#</mathjax></p>
</blockquote>
<p>You know that the equilibrium constant is also equal to </p>
<blockquote>
<p><mathjax>#K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])#</mathjax></p>
<p><mathjax>#K_a = (x * x)/(0.20 - x)#</mathjax></p>
</blockquote>
<p>Since <mathjax>#K_a#</mathjax> is small compared with the initial concentration of the acid, you can approximate <mathjax>#(0.20-x)#</mathjax> with <mathjax>#0.20#</mathjax>. This will give you </p>
<blockquote>
<p><mathjax>#K_a = x^2/0.2 impliesx = sqrt(0.2 * 1.78 * 10^(-4)) = 5.97 * 10^(-3)#</mathjax></p>
</blockquote>
<p>The concentration of the hydronium ions will thus be</p>
<blockquote>
<p><mathjax>#x = ["H"_3"O"^(+)] = 5.97 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>This means that the solution's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> will be </p>
<blockquote>
<p><mathjax>#pH_"sol" = -log(["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#pH_"sol" = -log(5.97 * 10^(-3)) = color(green)(2.22)#</mathjax></p>
</blockquote></div>
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</article> | What is the #"pH"# of a #"0.20-M"# solution of formic acid? |
#pK_a = 3.75#
|
2,572 | ac106154-6ddd-11ea-b979-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-prepared-by-dissolving-0-450-of-hno-3-in-300- | 0.02 M | start physical_unit 6 6 molarity mol/l qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 18 18 15 16 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"0.02 M"}] | [{"type":"physical unit","value":"Mass [OF] HNO3 [=] \\pu{0.450 g}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{300 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by dissolving 0.450 of #HNO_3# in 300 mL of water?</h1> | null | 0.02 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#("moles of solute")/("volume of solution")#</mathjax>. </p>
<p><mathjax>#((0.450*g)/(63.01*g*mol^(-1)))/(0.300*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^(-1)#</mathjax></p>
<p>Please do the math, and report the answer in <mathjax>#mol*L^(-1)#</mathjax>.</p></div>
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<div class="markdown"><p>Moles of nitric acid: <mathjax>#(0.450*g)/(63.01*g*mol^(-1))#</mathjax>. This is dissolved in <mathjax>#300#</mathjax> <mathjax>#mL#</mathjax> of water. I have assumed that you mean <mathjax>#0.450* g#</mathjax> of nitric acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#("moles of solute")/("volume of solution")#</mathjax>. </p>
<p><mathjax>#((0.450*g)/(63.01*g*mol^(-1)))/(0.300*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^(-1)#</mathjax></p>
<p>Please do the math, and report the answer in <mathjax>#mol*L^(-1)#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by dissolving 0.450 of #HNO_3# in 300 mL of water?</h1>
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<div class="markdown"><p>Moles of nitric acid: <mathjax>#(0.450*g)/(63.01*g*mol^(-1))#</mathjax>. This is dissolved in <mathjax>#300#</mathjax> <mathjax>#mL#</mathjax> of water. I have assumed that you mean <mathjax>#0.450* g#</mathjax> of nitric acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#("moles of solute")/("volume of solution")#</mathjax>. </p>
<p><mathjax>#((0.450*g)/(63.01*g*mol^(-1)))/(0.300*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^(-1)#</mathjax></p>
<p>Please do the math, and report the answer in <mathjax>#mol*L^(-1)#</mathjax>.</p></div>
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</article> | What is the molarity of a solution prepared by dissolving 0.450 of #HNO_3# in 300 mL of water? | null |
2,573 | abcae99e-6ddd-11ea-9913-ccda262736ce | https://socratic.org/questions/for-a-0-15-m-solution-of-k-2so-4-what-is-the-concentration-of-potassium | 0.30 M | start physical_unit 12 12 concentration mol/l qc_end physical_unit 6 6 2 3 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] potassium [IN] M"}] | [{"type":"physical unit","value":"0.30 M"}] | [{"type":"physical unit","value":"Concentration [OF] K2SO4 solution [=] \\pu{0.15 M}"}] | <h1 class="questionTitle" itemprop="name">For a 0.15 M solution of #K_2SO_4#, what is the concentration of potassium?</h1> | null | 0.30 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For each mol of <mathjax>#K_2SO_4#</mathjax> you have 2 mols of potassium, so the concentration of potassium will be also the double.</p></div>
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<div class="markdown"><p>0.30 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For each mol of <mathjax>#K_2SO_4#</mathjax> you have 2 mols of potassium, so the concentration of potassium will be also the double.</p></div>
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José F.
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<div class="markdown"><p>0.30 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>For each mol of <mathjax>#K_2SO_4#</mathjax> you have 2 mols of potassium, so the concentration of potassium will be also the double.</p></div>
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</article> | For a 0.15 M solution of #K_2SO_4#, what is the concentration of potassium? | null |
2,574 | ab5fa982-6ddd-11ea-acda-ccda262736ce | https://socratic.org/questions/how-many-grams-of-solute-are-present-in-795-ml-of-0-870-m-kbr | 82.3 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"82.3 grams"}] | [{"type":"physical unit","value":"Volume [OF] KBr solution [=] \\pu{795 mL}"},{"type":"physical unit","value":"Molarity [OF] KBr solution [=] \\pu{0.870 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of solute are present in 795 mL of 0.870 M #"KBr"#? </h1> | null | 82.3 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the solution to determine the <strong>number of moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium bromide, <mathjax>#"KBr"#</mathjax>, it contains.</p>
<p>Once you know that, you can use the compound's <strong>molar mass</strong> to convert to <em>grams</em>. </p>
<p>So, you know that a <mathjax>#"0.870 M"#</mathjax> potassium bromide solution will contain <mathjax>#0.870#</mathjax> <strong>moles</strong> of potassium bromide <strong>for every liter</strong> of solution. </p>
<p>In your case, the solution is said to have a volume of</p>
<blockquote>
<p><mathjax>#795 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.795 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.795 color(red)(cancel(color(black)("L solution"))) * "0.870 moles KBr"/(1color(red)(cancel(color(black)("L solution")))) = "0.69165 moles KBr"#</mathjax></p>
</blockquote>
<p>Now, potassium bromide has a molar mass of <mathjax>#"119.002 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of potassium bromide has a mass of <mathjax>#"119.002 g"#</mathjax>. </p>
<p>Your solution will thus contain </p>
<blockquote>
<p><mathjax>#0.69165 color(red)(cancel(color(black)("moles KBr"))) * "119.002 g"/(1color(red)(cancel(color(black)("mole KBr")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("82.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div>
<div class="markdown"><p><mathjax>#"82.3 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the solution to determine the <strong>number of moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium bromide, <mathjax>#"KBr"#</mathjax>, it contains.</p>
<p>Once you know that, you can use the compound's <strong>molar mass</strong> to convert to <em>grams</em>. </p>
<p>So, you know that a <mathjax>#"0.870 M"#</mathjax> potassium bromide solution will contain <mathjax>#0.870#</mathjax> <strong>moles</strong> of potassium bromide <strong>for every liter</strong> of solution. </p>
<p>In your case, the solution is said to have a volume of</p>
<blockquote>
<p><mathjax>#795 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.795 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.795 color(red)(cancel(color(black)("L solution"))) * "0.870 moles KBr"/(1color(red)(cancel(color(black)("L solution")))) = "0.69165 moles KBr"#</mathjax></p>
</blockquote>
<p>Now, potassium bromide has a molar mass of <mathjax>#"119.002 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of potassium bromide has a mass of <mathjax>#"119.002 g"#</mathjax>. </p>
<p>Your solution will thus contain </p>
<blockquote>
<p><mathjax>#0.69165 color(red)(cancel(color(black)("moles KBr"))) * "119.002 g"/(1color(red)(cancel(color(black)("mole KBr")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("82.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of solute are present in 795 mL of 0.870 M #"KBr"#? </h1>
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<div class="markdown"><p><mathjax>#"82.3 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the solution to determine the <strong>number of moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium bromide, <mathjax>#"KBr"#</mathjax>, it contains.</p>
<p>Once you know that, you can use the compound's <strong>molar mass</strong> to convert to <em>grams</em>. </p>
<p>So, you know that a <mathjax>#"0.870 M"#</mathjax> potassium bromide solution will contain <mathjax>#0.870#</mathjax> <strong>moles</strong> of potassium bromide <strong>for every liter</strong> of solution. </p>
<p>In your case, the solution is said to have a volume of</p>
<blockquote>
<p><mathjax>#795 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.795 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.795 color(red)(cancel(color(black)("L solution"))) * "0.870 moles KBr"/(1color(red)(cancel(color(black)("L solution")))) = "0.69165 moles KBr"#</mathjax></p>
</blockquote>
<p>Now, potassium bromide has a molar mass of <mathjax>#"119.002 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of potassium bromide has a mass of <mathjax>#"119.002 g"#</mathjax>. </p>
<p>Your solution will thus contain </p>
<blockquote>
<p><mathjax>#0.69165 color(red)(cancel(color(black)("moles KBr"))) * "119.002 g"/(1color(red)(cancel(color(black)("mole KBr")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("82.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many grams of solute are present in 795 mL of 0.870 M #"KBr"#? | null |
2,575 | a8aa588c-6ddd-11ea-bc88-ccda262736ce | https://socratic.org/questions/in-a-2-5-m-cacl-2-solution-what-is-the-molarity-of-the-calcium-ions | 2.50 M | start physical_unit 11 13 molarity mol/l qc_end physical_unit 4 5 2 3 molarity qc_end end | [{"type":"physical unit","value":"Molarity [OF] the calcium ions [IN] M"}] | [{"type":"physical unit","value":"2.50 M"}] | [{"type":"physical unit","value":"Molarity [OF] CaCl2 solution [=] \\pu{2.5 M}"}] | <h1 class="questionTitle" itemprop="name">In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions?</h1> | null | 2.50 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<div class="markdown"><p>AS the name says on the tin, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of calcium ion is <mathjax>#2.5*mol*L^-1#</mathjax>....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<h1 class="questionTitle" itemprop="name">In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions?</h1>
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<div class="markdown"><p>AS the name says on the tin, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of calcium ion is <mathjax>#2.5*mol*L^-1#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<div class="markdown"><p><mathjax>#2.5M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the number of moles in a liter. If the solution is <mathjax>#2.5M#</mathjax>, the molarity of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> is the same basic amount, modified by the number of atoms of each element in the original compound.</p>
<p>In this case, we have one <mathjax>#Ca#</mathjax> atom and two <mathjax>#Cl#</mathjax> atoms in the compound. So, we will have <mathjax>#2.5M Ca^(+2)#</mathjax> and <mathjax>#2.5 xx 2 = 5M Cl^(-1)#</mathjax> ions in solution.</p></div>
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</article> | In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions? | null |
2,576 | a8605a50-6ddd-11ea-926d-ccda262736ce | https://socratic.org/questions/how-many-grams-of-o-2-are-needed-to-react-with-84-5-g-of-nh-3 | 198 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 10 11 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 [IN] grams"}] | [{"type":"physical unit","value":"198 grams"}] | [{"type":"physical unit","value":"Mass [OF] NH3 [=] \\pu{84.5 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #O_2# are needed to react with #84.5# g of #NH_3#?</h1> | null | 198 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced equation</p>
<p><mathjax>#"4NH"_3("g") + "5O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NO(g) + 6H"_2"O(g)"#</mathjax></p>
<p>There are three steps required to answer this question.</p>
<p><mathjax>#color(red)1#</mathjax>. Convert mass <mathjax>#"NH"_3#</mathjax> to mol <mathjax>#"NH"_3"#</mathjax> by dividing the given mass by its molar mass <mathjax>#("17.031 g/mol")#</mathjax>. Do this by multiplying by the reciprocal of the molar mass.</p>
<p><mathjax>#color(blue)2#</mathjax>. Determine <mathjax>#"mol O"_2#</mathjax> by multiplying <mathjax>#"mol NH"_3"#</mathjax> by the mol ratio between <mathjax>#"NH"_3#</mathjax> and <mathjax>#"O"_2#</mathjax> in the balanced equation, with <mathjax>#"mol O"_2#</mathjax> in the numerator.</p>
<p><mathjax>#color(green)3#</mathjax>. Determine mass <mathjax>#"O"_2#</mathjax> by multiplying <mathjax>#"mol O"_2"#</mathjax> by its molar mass <mathjax>#("31.998 g/mol")#</mathjax>.</p>
<p><mathjax>#color(red)(84.5)color(black)cancel(color(red)("g NH"_3))xx(color(red)1color(black)cancel(color(red)("mol NH"_3)))/(color(red)17.031color(black)cancel(color(red)("g NH"_3)))xx(color(blue)5color(black)cancel(color(blue)("mol O"_2)))/(color(blue)4color(black)cancel(color(blue)("mol NH"_3)))xx(color(green)31.998color(black)cancel(color(green)("g O"_2)))/(color(green)1color(black)cancel(color(green)("mol O"_2)))=color(green)("198 g O"_2"#</mathjax> (rounded to three significant figures)</p>
<p><mathjax>#"198 g O"_2"#</mathjax> are needed to react with <mathjax>#"84.5 g NH"_3"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"198 g O"_2"#</mathjax> are needed to react with <mathjax>#"84.5 g NH"_3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balanced equation</p>
<p><mathjax>#"4NH"_3("g") + "5O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NO(g) + 6H"_2"O(g)"#</mathjax></p>
<p>There are three steps required to answer this question.</p>
<p><mathjax>#color(red)1#</mathjax>. Convert mass <mathjax>#"NH"_3#</mathjax> to mol <mathjax>#"NH"_3"#</mathjax> by dividing the given mass by its molar mass <mathjax>#("17.031 g/mol")#</mathjax>. Do this by multiplying by the reciprocal of the molar mass.</p>
<p><mathjax>#color(blue)2#</mathjax>. Determine <mathjax>#"mol O"_2#</mathjax> by multiplying <mathjax>#"mol NH"_3"#</mathjax> by the mol ratio between <mathjax>#"NH"_3#</mathjax> and <mathjax>#"O"_2#</mathjax> in the balanced equation, with <mathjax>#"mol O"_2#</mathjax> in the numerator.</p>
<p><mathjax>#color(green)3#</mathjax>. Determine mass <mathjax>#"O"_2#</mathjax> by multiplying <mathjax>#"mol O"_2"#</mathjax> by its molar mass <mathjax>#("31.998 g/mol")#</mathjax>.</p>
<p><mathjax>#color(red)(84.5)color(black)cancel(color(red)("g NH"_3))xx(color(red)1color(black)cancel(color(red)("mol NH"_3)))/(color(red)17.031color(black)cancel(color(red)("g NH"_3)))xx(color(blue)5color(black)cancel(color(blue)("mol O"_2)))/(color(blue)4color(black)cancel(color(blue)("mol NH"_3)))xx(color(green)31.998color(black)cancel(color(green)("g O"_2)))/(color(green)1color(black)cancel(color(green)("mol O"_2)))=color(green)("198 g O"_2"#</mathjax> (rounded to three significant figures)</p>
<p><mathjax>#"198 g O"_2"#</mathjax> are needed to react with <mathjax>#"84.5 g NH"_3"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #O_2# are needed to react with #84.5# g of #NH_3#?</h1>
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<div class="markdown"><p><mathjax>#"198 g O"_2"#</mathjax> are needed to react with <mathjax>#"84.5 g NH"_3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Balanced equation</p>
<p><mathjax>#"4NH"_3("g") + "5O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NO(g) + 6H"_2"O(g)"#</mathjax></p>
<p>There are three steps required to answer this question.</p>
<p><mathjax>#color(red)1#</mathjax>. Convert mass <mathjax>#"NH"_3#</mathjax> to mol <mathjax>#"NH"_3"#</mathjax> by dividing the given mass by its molar mass <mathjax>#("17.031 g/mol")#</mathjax>. Do this by multiplying by the reciprocal of the molar mass.</p>
<p><mathjax>#color(blue)2#</mathjax>. Determine <mathjax>#"mol O"_2#</mathjax> by multiplying <mathjax>#"mol NH"_3"#</mathjax> by the mol ratio between <mathjax>#"NH"_3#</mathjax> and <mathjax>#"O"_2#</mathjax> in the balanced equation, with <mathjax>#"mol O"_2#</mathjax> in the numerator.</p>
<p><mathjax>#color(green)3#</mathjax>. Determine mass <mathjax>#"O"_2#</mathjax> by multiplying <mathjax>#"mol O"_2"#</mathjax> by its molar mass <mathjax>#("31.998 g/mol")#</mathjax>.</p>
<p><mathjax>#color(red)(84.5)color(black)cancel(color(red)("g NH"_3))xx(color(red)1color(black)cancel(color(red)("mol NH"_3)))/(color(red)17.031color(black)cancel(color(red)("g NH"_3)))xx(color(blue)5color(black)cancel(color(blue)("mol O"_2)))/(color(blue)4color(black)cancel(color(blue)("mol NH"_3)))xx(color(green)31.998color(black)cancel(color(green)("g O"_2)))/(color(green)1color(black)cancel(color(green)("mol O"_2)))=color(green)("198 g O"_2"#</mathjax> (rounded to three significant figures)</p>
<p><mathjax>#"198 g O"_2"#</mathjax> are needed to react with <mathjax>#"84.5 g NH"_3"#</mathjax>.</p></div>
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</article> | How many grams of #O_2# are needed to react with #84.5# g of #NH_3#? | null |
2,577 | ac2cf506-6ddd-11ea-814f-ccda262736ce | https://socratic.org/questions/a-sample-of-calcium-nitrate-ca-no-3-2-with-a-formula-weight-of-164-g-mol-has-5-0 | 2.27 kilograms | start physical_unit 5 5 mass kg qc_end physical_unit 5 5 11 12 formula_weight qc_end end | [{"type":"physical unit","value":"Mass [OF] Ca(NO3)2 [IN] kilograms"}] | [{"type":"physical unit","value":"2.27 kilograms"}] | [{"type":"physical unit","value":"Formula weight [OF] Ca(NO3)2 [=] \\pu{164 g/mol}"},{"type":"physical unit","value":"Number [OF] oxygen atoms [=] \\pu{5.00 × 10^25}"}] | <h1 class="questionTitle" itemprop="name">A sample of calcium nitrate, #Ca(NO_3)_2#, with a formula weight of 164 g/mol, has #5.00 x 10^25# atoms of oxygen. How many kilograms of #Ca(NO_3)_2# are present?</h1> | null | 2.27 kilograms | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"O"#</mathjax> atoms</strong>.</p>
<p><mathjax>#"Moles of O" = 5.00 × 10^25 color(red)(cancel(color(black)("atoms O"))) × "1 mol O"/(6.022 × 10^23color(red)(cancel(color(black)("atoms O")))) = "83.03 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong>.</p>
<p>The formula tells us that 1 mol of <mathjax>#"Ca"("NO"_3)_2#</mathjax> contains 6 mol of <mathjax>#"O"#</mathjax> atoms.</p>
<p>∴ <mathjax>#"Moles of Ca"("NO"_3)_2 = 83.03color(red)(cancel(color(black)( "mol O"))) × ("1 mol Ca"("NO"_3)_2)/(6 color(red)(cancel(color(black)("mol O")))) = "13.84 mol Ca"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"("NO"_3)_2 = 13.84 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2))) × ("164 g Ca"("NO"_3)_2)/(1 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "2270 g Ca"("NO"_3)_2 = "2.27 kg Ca"("NO"_3)_2#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>There are 2.27 kg of <mathjax>#"Ca"("NO"_3)_2#</mathjax> present.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"O"#</mathjax> atoms</strong>.</p>
<p><mathjax>#"Moles of O" = 5.00 × 10^25 color(red)(cancel(color(black)("atoms O"))) × "1 mol O"/(6.022 × 10^23color(red)(cancel(color(black)("atoms O")))) = "83.03 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong>.</p>
<p>The formula tells us that 1 mol of <mathjax>#"Ca"("NO"_3)_2#</mathjax> contains 6 mol of <mathjax>#"O"#</mathjax> atoms.</p>
<p>∴ <mathjax>#"Moles of Ca"("NO"_3)_2 = 83.03color(red)(cancel(color(black)( "mol O"))) × ("1 mol Ca"("NO"_3)_2)/(6 color(red)(cancel(color(black)("mol O")))) = "13.84 mol Ca"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"("NO"_3)_2 = 13.84 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2))) × ("164 g Ca"("NO"_3)_2)/(1 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "2270 g Ca"("NO"_3)_2 = "2.27 kg Ca"("NO"_3)_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A sample of calcium nitrate, #Ca(NO_3)_2#, with a formula weight of 164 g/mol, has #5.00 x 10^25# atoms of oxygen. How many kilograms of #Ca(NO_3)_2# are present?</h1>
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Ernest Z.
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<div class="markdown"><p>There are 2.27 kg of <mathjax>#"Ca"("NO"_3)_2#</mathjax> present.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"O"#</mathjax> atoms</strong>.</p>
<p><mathjax>#"Moles of O" = 5.00 × 10^25 color(red)(cancel(color(black)("atoms O"))) × "1 mol O"/(6.022 × 10^23color(red)(cancel(color(black)("atoms O")))) = "83.03 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong>.</p>
<p>The formula tells us that 1 mol of <mathjax>#"Ca"("NO"_3)_2#</mathjax> contains 6 mol of <mathjax>#"O"#</mathjax> atoms.</p>
<p>∴ <mathjax>#"Moles of Ca"("NO"_3)_2 = 83.03color(red)(cancel(color(black)( "mol O"))) × ("1 mol Ca"("NO"_3)_2)/(6 color(red)(cancel(color(black)("mol O")))) = "13.84 mol Ca"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Ca"("NO"_3)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"("NO"_3)_2 = 13.84 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2))) × ("164 g Ca"("NO"_3)_2)/(1 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "2270 g Ca"("NO"_3)_2 = "2.27 kg Ca"("NO"_3)_2#</mathjax></p></div>
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</article> | A sample of calcium nitrate, #Ca(NO_3)_2#, with a formula weight of 164 g/mol, has #5.00 x 10^25# atoms of oxygen. How many kilograms of #Ca(NO_3)_2# are present? | null |
2,578 | a8517614-6ddd-11ea-81b5-ccda262736ce | https://socratic.org/questions/gas-a-effuses-0-68-times-as-fast-as-gas-b-if-the-molar-mass-of-gas-b-is-17-g-wha-1 | 37 g | start physical_unit 0 1 mass g qc_end physical_unit 8 9 18 19 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Gas A [IN] g"}] | [{"type":"physical unit","value":"37 g"}] | [{"type":"physical unit","value":"Time [OF] Gas A effused faster than Gas B [=] \\pu{0.68}"},{"type":"physical unit","value":"Molar mass [OF] Gas B [=] \\pu{17 g}"}] | <h1 class="questionTitle" itemprop="name">Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?</h1> | null | 37 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use <em>Graham's law of effusion</em> , which relates the relative rates of effusion of gases in terms of their molar masses:</p>
<p><mathjax>#(r_1)/(r_2) = sqrt((MM_2)/(MM_1)#</mathjax></p>
<p>Where <br/>
<mathjax>#(r_1)/(r_2)#</mathjax> is the ratio of effusion of gas <mathjax>#1#</mathjax> to gas <mathjax>#2#</mathjax>, and<br/>
<mathjax>#MM_(1, 2)#</mathjax> is the molar mass of the gases <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax>, respectively.</p>
<p>We're given that the ratio of effusion of gas A to gas B is <mathjax>#0.68#</mathjax>, and that <mathjax>#MM_B#</mathjax> is <mathjax>#17"g"/"mol"#</mathjax>, so</p>
<p><mathjax>#0.68 = (sqrt17)/sqrt(MM_A)#</mathjax></p>
<p><mathjax>#MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"#</mathjax></p>
<p>The mass of the gas is thus <mathjax>#37"g"#</mathjax>.</p></div>
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<div>
<div class="markdown"><p><mathjax>#37"g"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use <em>Graham's law of effusion</em> , which relates the relative rates of effusion of gases in terms of their molar masses:</p>
<p><mathjax>#(r_1)/(r_2) = sqrt((MM_2)/(MM_1)#</mathjax></p>
<p>Where <br/>
<mathjax>#(r_1)/(r_2)#</mathjax> is the ratio of effusion of gas <mathjax>#1#</mathjax> to gas <mathjax>#2#</mathjax>, and<br/>
<mathjax>#MM_(1, 2)#</mathjax> is the molar mass of the gases <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax>, respectively.</p>
<p>We're given that the ratio of effusion of gas A to gas B is <mathjax>#0.68#</mathjax>, and that <mathjax>#MM_B#</mathjax> is <mathjax>#17"g"/"mol"#</mathjax>, so</p>
<p><mathjax>#0.68 = (sqrt17)/sqrt(MM_A)#</mathjax></p>
<p><mathjax>#MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"#</mathjax></p>
<p>The mass of the gas is thus <mathjax>#37"g"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#37"g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use <em>Graham's law of effusion</em> , which relates the relative rates of effusion of gases in terms of their molar masses:</p>
<p><mathjax>#(r_1)/(r_2) = sqrt((MM_2)/(MM_1)#</mathjax></p>
<p>Where <br/>
<mathjax>#(r_1)/(r_2)#</mathjax> is the ratio of effusion of gas <mathjax>#1#</mathjax> to gas <mathjax>#2#</mathjax>, and<br/>
<mathjax>#MM_(1, 2)#</mathjax> is the molar mass of the gases <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax>, respectively.</p>
<p>We're given that the ratio of effusion of gas A to gas B is <mathjax>#0.68#</mathjax>, and that <mathjax>#MM_B#</mathjax> is <mathjax>#17"g"/"mol"#</mathjax>, so</p>
<p><mathjax>#0.68 = (sqrt17)/sqrt(MM_A)#</mathjax></p>
<p><mathjax>#MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"#</mathjax></p>
<p>The mass of the gas is thus <mathjax>#37"g"#</mathjax>.</p></div>
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</article> | Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A? | null |
2,579 | aced4ee4-6ddd-11ea-b9cf-ccda262736ce | https://socratic.org/questions/how-many-liters-at-stp-does-738-moles-of-a-gas-occupy | 16545.96 liters | start physical_unit 10 10 volume l qc_end physical_unit 10 10 6 7 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"16545.96 liters"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{738 moles}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many liters at STP does 738 moles of a gas occupy? </h1> | null | 16545.96 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume at STP is <mathjax>#22.42L/(mol)#</mathjax>.</p>
<p>For <mathjax>#738 mol#</mathjax> the gas will occupy then:</p>
<p><mathjax>#V=738cancel(mol)xx(22.42L)/(1cancel(mol))=16545.96L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#V=16545.96L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume at STP is <mathjax>#22.42L/(mol)#</mathjax>.</p>
<p>For <mathjax>#738 mol#</mathjax> the gas will occupy then:</p>
<p><mathjax>#V=738cancel(mol)xx(22.42L)/(1cancel(mol))=16545.96L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many liters at STP does 738 moles of a gas occupy? </h1>
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Dr. Hayek
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<div class="markdown"><p><mathjax>#V=16545.96L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume at STP is <mathjax>#22.42L/(mol)#</mathjax>.</p>
<p>For <mathjax>#738 mol#</mathjax> the gas will occupy then:</p>
<p><mathjax>#V=738cancel(mol)xx(22.42L)/(1cancel(mol))=16545.96L#</mathjax></p></div>
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</article> | How many liters at STP does 738 moles of a gas occupy? | null |
2,580 | a86e7aae-6ddd-11ea-84cb-ccda262736ce | https://socratic.org/questions/how-many-moles-of-carbon-dioxide-is-formed-when-4-0-moles-of-carbonic-acid-decom | 4.0 moles | start physical_unit 4 5 mole mol qc_end physical_unit 12 13 9 10 mole qc_end c_other decomposes qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon dioxide [IN] moles"}] | [{"type":"physical unit","value":"4.0 moles"}] | [{"type":"physical unit","value":"Mole [OF] carbonic acid [=] \\pu{4.0 moles}"},{"type":"other","value":"decomposes"},{"type":"other","value":"Water is the other product."}] | <h1 class="questionTitle" itemprop="name">How many moles of carbon dioxide is formed when 4.0 moles of carbonic acid decomposes (water is the other product)?</h1> | null | 4.0 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical equation for this reaction is</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" rightleftharpoons "CO"_2"(g)" + "H"_2"O(l)"#</mathjax></p>
<p>You'll notice that all the reagents in this reaction have equal amounts of relative moles, so if <mathjax>#4.0#</mathjax> moles of <mathjax>#H_2CO_3#</mathjax> decomposes completely, there will be <mathjax>#4.0#</mathjax> moles of each water and carbon dioxide produced.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#4.0#</mathjax> mol <mathjax>#"CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical equation for this reaction is</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" rightleftharpoons "CO"_2"(g)" + "H"_2"O(l)"#</mathjax></p>
<p>You'll notice that all the reagents in this reaction have equal amounts of relative moles, so if <mathjax>#4.0#</mathjax> moles of <mathjax>#H_2CO_3#</mathjax> decomposes completely, there will be <mathjax>#4.0#</mathjax> moles of each water and carbon dioxide produced.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of carbon dioxide is formed when 4.0 moles of carbonic acid decomposes (water is the other product)?</h1>
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Nathan L.
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May 23, 2017
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<div class="markdown"><p><mathjax>#4.0#</mathjax> mol <mathjax>#"CO"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical equation for this reaction is</p>
<p><mathjax>#"H"_2"CO"_3"(aq)" rightleftharpoons "CO"_2"(g)" + "H"_2"O(l)"#</mathjax></p>
<p>You'll notice that all the reagents in this reaction have equal amounts of relative moles, so if <mathjax>#4.0#</mathjax> moles of <mathjax>#H_2CO_3#</mathjax> decomposes completely, there will be <mathjax>#4.0#</mathjax> moles of each water and carbon dioxide produced.</p></div>
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SCooke
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May 23, 2017
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<div class="markdown"><p>Four (4).</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Carbonic acid is a chemical compound with the chemical formula <mathjax>#H_2CO_3#</mathjax>. <br/>
Thus, each mole of carbonic acid can only produce one mole of carbon dioxide because there is only one mole of carbon per mole of carbonic acid.</p></div>
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</article> | How many moles of carbon dioxide is formed when 4.0 moles of carbonic acid decomposes (water is the other product)? | null |
2,581 | abccf3e7-6ddd-11ea-8e4b-ccda262736ce | https://socratic.org/questions/how-many-water-molecules-are-in-15-mol-of-h2o | 9.03 × 10^22 | start physical_unit 2 3 number none qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] water molecules"}] | [{"type":"physical unit","value":"9.03 × 10^22"}] | [{"type":"physical unit","value":"Mole [OF] H2O [=] \\pu{0.15 moles}"}] | <h1 class="questionTitle" itemprop="name">How many water molecules are in #0.15# moles of #"H"_2"O"#?</h1> | null | 9.03 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1#</mathjax> mole has <mathjax>#6.022 xx 10^23#</mathjax> molecules.</p>
<p>Thus, <mathjax>#0.15#</mathjax> or <mathjax>#3//20#</mathjax> moles has </p>
<p><mathjax>#(6.022xx3xx10^23)//20 = 9.0xx 10^22#</mathjax> molecules of water </p></div>
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<div class="markdown"><p><mathjax>#9.0 xx 10^22#</mathjax> molecules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1#</mathjax> mole has <mathjax>#6.022 xx 10^23#</mathjax> molecules.</p>
<p>Thus, <mathjax>#0.15#</mathjax> or <mathjax>#3//20#</mathjax> moles has </p>
<p><mathjax>#(6.022xx3xx10^23)//20 = 9.0xx 10^22#</mathjax> molecules of water </p></div>
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<h1 class="questionTitle" itemprop="name">How many water molecules are in #0.15# moles of #"H"_2"O"#?</h1>
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<div class="markdown"><p><mathjax>#9.0 xx 10^22#</mathjax> molecules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#1#</mathjax> mole has <mathjax>#6.022 xx 10^23#</mathjax> molecules.</p>
<p>Thus, <mathjax>#0.15#</mathjax> or <mathjax>#3//20#</mathjax> moles has </p>
<p><mathjax>#(6.022xx3xx10^23)//20 = 9.0xx 10^22#</mathjax> molecules of water </p></div>
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</article> | How many water molecules are in #0.15# moles of #"H"_2"O"#? | null |
2,582 | a8c42154-6ddd-11ea-9c26-ccda262736ce | https://socratic.org/questions/what-is-the-volume-in-liters-of-6-75-10-24-molecules-of-ammonia-gas-at-stp | 254.55 liters | start physical_unit 12 13 volume l qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] ammonia gas [IN] liters"}] | [{"type":"physical unit","value":"254.55 liters"}] | [{"type":"physical unit","value":"Number [OF] ammonia gas molecules [=] \\pu{6.75 × 10^24}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume in liters of #6.75*10^24# molecules of ammonia gas at STP?</h1> | null | 254.55 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To go from molecules to volume, we need to convert to <strong>moles</strong>. Recall that one mole of a substance has <mathjax>#6.022*10^23#</mathjax> molecules (Avogadro's number). </p>
<p>Moles of ammonia gas:</p>
<p><mathjax>#(6.75*10^24)/(6.022*10^23)=11.2089...#</mathjax>moles</p>
<p>At STP, the <strong>molar volume</strong> of a gas is <mathjax>#22.71Lmol^-1#</mathjax>. This means that per mole of a gas, there are 22.71 litres of the gas. </p>
<p>Volume of ammonia gas:</p>
<p><mathjax>#11.2089... cancel(mol)*(22.71L)/(cancel(mol))=254.55L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#255L#</mathjax> (3 s.f.)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To go from molecules to volume, we need to convert to <strong>moles</strong>. Recall that one mole of a substance has <mathjax>#6.022*10^23#</mathjax> molecules (Avogadro's number). </p>
<p>Moles of ammonia gas:</p>
<p><mathjax>#(6.75*10^24)/(6.022*10^23)=11.2089...#</mathjax>moles</p>
<p>At STP, the <strong>molar volume</strong> of a gas is <mathjax>#22.71Lmol^-1#</mathjax>. This means that per mole of a gas, there are 22.71 litres of the gas. </p>
<p>Volume of ammonia gas:</p>
<p><mathjax>#11.2089... cancel(mol)*(22.71L)/(cancel(mol))=254.55L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume in liters of #6.75*10^24# molecules of ammonia gas at STP?</h1>
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<div class="markdown"><p><mathjax>#255L#</mathjax> (3 s.f.)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To go from molecules to volume, we need to convert to <strong>moles</strong>. Recall that one mole of a substance has <mathjax>#6.022*10^23#</mathjax> molecules (Avogadro's number). </p>
<p>Moles of ammonia gas:</p>
<p><mathjax>#(6.75*10^24)/(6.022*10^23)=11.2089...#</mathjax>moles</p>
<p>At STP, the <strong>molar volume</strong> of a gas is <mathjax>#22.71Lmol^-1#</mathjax>. This means that per mole of a gas, there are 22.71 litres of the gas. </p>
<p>Volume of ammonia gas:</p>
<p><mathjax>#11.2089... cancel(mol)*(22.71L)/(cancel(mol))=254.55L#</mathjax></p></div>
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</article> | What is the volume in liters of #6.75*10^24# molecules of ammonia gas at STP? | null |
2,583 | ac352201-6ddd-11ea-abd6-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-of-c6h6-o2-co2-h2o | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O"}] | [{"type":"chemical equation","value":"C6H6 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1> | null | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1>
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Alan P.
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<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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Nikka C.
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Feb 9, 2017
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<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p></div>
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Nikka C.
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<span class="dateCreated" datetime="2017-02-09T02:34:35" itemprop="dateCreated">
Feb 9, 2017
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<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p>
<p>If you don't like fractions, you can always multiply the whole chemical equation by the denominator.</p>
<p>[<mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O] x 2</p>
<p><mathjax>#2#</mathjax> <mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(12)#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(6)#</mathjax> <mathjax>#H_2#</mathjax>O (also acceptable)</p></div>
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</article> | What is the balanced equation of C6H6+O2 = CO2+H2O? | null |
2,584 | ace66851-6ddd-11ea-a28f-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-s8-o2-so3-2 | S8 + 12 O2 -> 8 SO3 | start chemical_equation qc_end chemical_equation 7 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the following equation"}] | [{"type":"chemical equation","value":"S8 + 12 O2 -> 8 SO3"}] | [{"type":"chemical equation","value":"S8 + O2 -> SO3"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation:
S8 + O2 --> SO3?</h1> | null | S8 + 12 O2 -> 8 SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To start balancing this reaction, we can first multiply <mathjax>#S#</mathjax> by <mathjax>#8#</mathjax> which will make the number of oxygen atoms in the product <mathjax>#24#</mathjax>, then we multiply <mathjax>#O#</mathjax> by <mathjax>#12#</mathjax>. </p>
<p>The balanced reaction is then:</p>
<p><mathjax>#S_8+color(red)(12)O_2->color(blue)(8)SO_3#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#S_8+color(red)(12)O_2->color(blue)(8)SO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To start balancing this reaction, we can first multiply <mathjax>#S#</mathjax> by <mathjax>#8#</mathjax> which will make the number of oxygen atoms in the product <mathjax>#24#</mathjax>, then we multiply <mathjax>#O#</mathjax> by <mathjax>#12#</mathjax>. </p>
<p>The balanced reaction is then:</p>
<p><mathjax>#S_8+color(red)(12)O_2->color(blue)(8)SO_3#</mathjax></p></div>
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S8 + O2 --> SO3?</h1>
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<div class="markdown"><p><mathjax>#S_8+color(red)(12)O_2->color(blue)(8)SO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To start balancing this reaction, we can first multiply <mathjax>#S#</mathjax> by <mathjax>#8#</mathjax> which will make the number of oxygen atoms in the product <mathjax>#24#</mathjax>, then we multiply <mathjax>#O#</mathjax> by <mathjax>#12#</mathjax>. </p>
<p>The balanced reaction is then:</p>
<p><mathjax>#S_8+color(red)(12)O_2->color(blue)(8)SO_3#</mathjax></p></div>
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</article> | How would you balance the following equation:
S8 + O2 --> SO3? | null |
2,585 | ab3ee0ee-6ddd-11ea-81b9-ccda262736ce | https://socratic.org/questions/56757fd17c014977b0f256da | Fe2O3(s) + 3 CO(g) -> 2 Fe(l) + 3 CO2(g) | start chemical_equation qc_end chemical_equation 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the essential redox reaction"}] | [{"type":"chemical equation","value":"Fe2O3(s) + 3 CO(g) -> 2 Fe(l) + 3 CO2(g)"}] | [{"type":"chemical equation","value":"Fe2O3"}] | <h1 class="questionTitle" itemprop="name">In the production of iron what is the essential redox reaction that #Fe_2O_3# undergoes?</h1> | null | Fe2O3(s) + 3 CO(g) -> 2 Fe(l) + 3 CO2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above is certainly a redox reaction, though it might not be the one you want. Can you write individual redox equations? Carbon is oxidized, and iron is reduced; from what state to what final state? </p></div>
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<div class="markdown"><p><mathjax>#Fe_2O_3(s) + 3CO(g) rarr 2Fe(l) + 3CO_2(g)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The reaction above is certainly a redox reaction, though it might not be the one you want. Can you write individual redox equations? Carbon is oxidized, and iron is reduced; from what state to what final state? </p></div>
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<h1 class="questionTitle" itemprop="name">In the production of iron what is the essential redox reaction that #Fe_2O_3# undergoes?</h1>
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<div class="markdown"><p><mathjax>#Fe_2O_3(s) + 3CO(g) rarr 2Fe(l) + 3CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The reaction above is certainly a redox reaction, though it might not be the one you want. Can you write individual redox equations? Carbon is oxidized, and iron is reduced; from what state to what final state? </p></div>
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</article> | In the production of iron what is the essential redox reaction that #Fe_2O_3# undergoes? | null |
2,586 | a9482862-6ddd-11ea-8eda-ccda262736ce | https://socratic.org/questions/if-the-mole-fraction-of-na-2s-in-a-solution-is-0-125-what-is-the-percent-by-mass | 31.67% | start physical_unit 5 8 mass_percent none qc_end physical_unit 5 8 10 10 mole_fraction qc_end end | [{"type":"physical unit","value":"Percent by mass [OF] NaCl in a solution"}] | [{"type":"physical unit","value":"31.67%"}] | [{"type":"physical unit","value":"Mole fraction [OF] NaCl in a solution [=] \\pu{0.125}"}] | <h1 class="questionTitle" itemprop="name">If the mole fraction of #"NaCl"# in a solution is #0.125#, what is the percent by mass of #"NaCl"#?</h1> | null | 31.67% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to pick a sample of this solution and use <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sodium chloride to find the total mass of the sample and the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>Assuming that your solution contains sodium chloride as the solute and water as the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, you can say that</p>
<blockquote>
<p><mathjax>#chi_ ("NaCl") + chi_ ("H"_ 2"O") = 1" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#chi_ ("NaCl")#</mathjax> is the <strong>mole fraction</strong> of sodium chloride</li>
<li><mathjax>#chi_ ("H"_ 2"O")#</mathjax> is the <strong>mole faction</strong> of water</li>
</ul>
</blockquote>
<p>Now, let's assume that you pick a sample of this solution that contains <strong>exactly</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of particles, i.e. that the <strong>total number of moles</strong> present in the sample is equal to <mathjax>#1#</mathjax>. </p>
<p>By definition, the mole fraction of sodium chloride is equal to the number of moles of sodium chloride divided by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_( "NaCl") = ("moles of NaCl")/"total moles"#</mathjax></p>
</blockquote>
<p>So in this case, you'd have</p>
<blockquote>
<p><mathjax>#0.125 = ("moles of NaCl")/"1 mole"#</mathjax></p>
</blockquote>
<p>which implies that the sample contains <mathjax>#0.125#</mathjax> <strong>moles</strong> of sodium chloride. Following the same logic or using equation <mathjax>#color(darkorange)("(*)")#</mathjax>, you can determine that this sample contains </p>
<blockquote>
<p><mathjax>#"1 mole" \ - \ "0.125 moles" = "0.875 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert the number of moles to grams.</p>
<blockquote>
<p><mathjax>#0.125 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "7.3050 g"#</mathjax></p>
<p><mathjax>#0.875 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "15.7631 g"#</mathjax></p>
</blockquote>
<p>You can now say that the total mass of the sample is</p>
<blockquote>
<p><mathjax>#"7.3050 g" \ + \ "15.7631 g" = "23.0681 g"#</mathjax></p>
</blockquote>
<p>Finally, to find the percent by mass of sodium chloride, you need to determine the mass of sodium chloride present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>Your sample contains <mathjax>#"7.3050 g"#</mathjax> of sodium chloride in <mathjax>#"23.0681 g"#</mathjax> of the solution, which means that you have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * ("7.3050 g NaCl")/(23.0681color(red)(cancel(color(black)("g solution")))) = "31.7 g NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the percent by mass of sodium chloride is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% NaCl" = 31.7%)))#</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 mole fraction of sodium chloride.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"31.7% NaCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to pick a sample of this solution and use <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sodium chloride to find the total mass of the sample and the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>Assuming that your solution contains sodium chloride as the solute and water as the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, you can say that</p>
<blockquote>
<p><mathjax>#chi_ ("NaCl") + chi_ ("H"_ 2"O") = 1" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#chi_ ("NaCl")#</mathjax> is the <strong>mole fraction</strong> of sodium chloride</li>
<li><mathjax>#chi_ ("H"_ 2"O")#</mathjax> is the <strong>mole faction</strong> of water</li>
</ul>
</blockquote>
<p>Now, let's assume that you pick a sample of this solution that contains <strong>exactly</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of particles, i.e. that the <strong>total number of moles</strong> present in the sample is equal to <mathjax>#1#</mathjax>. </p>
<p>By definition, the mole fraction of sodium chloride is equal to the number of moles of sodium chloride divided by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_( "NaCl") = ("moles of NaCl")/"total moles"#</mathjax></p>
</blockquote>
<p>So in this case, you'd have</p>
<blockquote>
<p><mathjax>#0.125 = ("moles of NaCl")/"1 mole"#</mathjax></p>
</blockquote>
<p>which implies that the sample contains <mathjax>#0.125#</mathjax> <strong>moles</strong> of sodium chloride. Following the same logic or using equation <mathjax>#color(darkorange)("(*)")#</mathjax>, you can determine that this sample contains </p>
<blockquote>
<p><mathjax>#"1 mole" \ - \ "0.125 moles" = "0.875 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert the number of moles to grams.</p>
<blockquote>
<p><mathjax>#0.125 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "7.3050 g"#</mathjax></p>
<p><mathjax>#0.875 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "15.7631 g"#</mathjax></p>
</blockquote>
<p>You can now say that the total mass of the sample is</p>
<blockquote>
<p><mathjax>#"7.3050 g" \ + \ "15.7631 g" = "23.0681 g"#</mathjax></p>
</blockquote>
<p>Finally, to find the percent by mass of sodium chloride, you need to determine the mass of sodium chloride present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>Your sample contains <mathjax>#"7.3050 g"#</mathjax> of sodium chloride in <mathjax>#"23.0681 g"#</mathjax> of the solution, which means that you have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * ("7.3050 g NaCl")/(23.0681color(red)(cancel(color(black)("g solution")))) = "31.7 g NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the percent by mass of sodium chloride is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% NaCl" = 31.7%)))#</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 mole fraction of sodium chloride.</p></div>
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<h1 class="questionTitle" itemprop="name">If the mole fraction of #"NaCl"# in a solution is #0.125#, what is the percent by mass of #"NaCl"#?</h1>
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Stefan V.
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Jul 14, 2018
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<div class="markdown"><p><mathjax>#"31.7% NaCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to pick a sample of this solution and use <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sodium chloride to find the total mass of the sample and the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>Assuming that your solution contains sodium chloride as the solute and water as the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, you can say that</p>
<blockquote>
<p><mathjax>#chi_ ("NaCl") + chi_ ("H"_ 2"O") = 1" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#chi_ ("NaCl")#</mathjax> is the <strong>mole fraction</strong> of sodium chloride</li>
<li><mathjax>#chi_ ("H"_ 2"O")#</mathjax> is the <strong>mole faction</strong> of water</li>
</ul>
</blockquote>
<p>Now, let's assume that you pick a sample of this solution that contains <strong>exactly</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of particles, i.e. that the <strong>total number of moles</strong> present in the sample is equal to <mathjax>#1#</mathjax>. </p>
<p>By definition, the mole fraction of sodium chloride is equal to the number of moles of sodium chloride divided by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_( "NaCl") = ("moles of NaCl")/"total moles"#</mathjax></p>
</blockquote>
<p>So in this case, you'd have</p>
<blockquote>
<p><mathjax>#0.125 = ("moles of NaCl")/"1 mole"#</mathjax></p>
</blockquote>
<p>which implies that the sample contains <mathjax>#0.125#</mathjax> <strong>moles</strong> of sodium chloride. Following the same logic or using equation <mathjax>#color(darkorange)("(*)")#</mathjax>, you can determine that this sample contains </p>
<blockquote>
<p><mathjax>#"1 mole" \ - \ "0.125 moles" = "0.875 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert the number of moles to grams.</p>
<blockquote>
<p><mathjax>#0.125 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "7.3050 g"#</mathjax></p>
<p><mathjax>#0.875 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "15.7631 g"#</mathjax></p>
</blockquote>
<p>You can now say that the total mass of the sample is</p>
<blockquote>
<p><mathjax>#"7.3050 g" \ + \ "15.7631 g" = "23.0681 g"#</mathjax></p>
</blockquote>
<p>Finally, to find the percent by mass of sodium chloride, you need to determine the mass of sodium chloride present <strong>for every</strong> <mathjax>#"100. g"#</mathjax> of this solution. </p>
<p>Your sample contains <mathjax>#"7.3050 g"#</mathjax> of sodium chloride in <mathjax>#"23.0681 g"#</mathjax> of the solution, which means that you have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * ("7.3050 g NaCl")/(23.0681color(red)(cancel(color(black)("g solution")))) = "31.7 g NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the percent by mass of sodium chloride is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% NaCl" = 31.7%)))#</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 mole fraction of sodium chloride.</p></div>
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</article> | If the mole fraction of #"NaCl"# in a solution is #0.125#, what is the percent by mass of #"NaCl"#? | null |
2,587 | ab26c97e-6ddd-11ea-bff4-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-hydroxide-ions-in-pure-water-at-30-0-c-if-k-w-at-th | 1.21 × 10^(-7) M | start physical_unit 5 6 concentration mol/l qc_end physical_unit 8 9 11 12 temperature qc_end physical_unit 8 9 19 21 kw qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydroxide ions [IN] M"}] | [{"type":"physical unit","value":"1.21 × 10^(-7) M"}] | [{"type":"physical unit","value":"Temperature [OF] pure water [=] \\pu{30.0 ℃}"},{"type":"physical unit","value":"Kw [OF] pure water [=] \\pu{1.47 × 10^(-14)}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of hydroxide ions in pure water at 30.0°C, if #K_w# at this temperature is #1.47*10^-14#?</h1> | null | 1.21 × 10^(-7) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#</mathjax>,</p>
<p>where the ion product, <mathjax>#K_w=1.47xx10^-14#</mathjax>, at <mathjax>#303*K#</mathjax>. Note that this is slightly higher than <mathjax>#K_w=10^-14#</mathjax> at <mathjax>#298*K#</mathjax>, and given that this is a bond breaking reaction this is perhaps reasonable.</p>
<p>From the equation, <mathjax>#[H_3O^+][HO^-]=1.47xx10^-14#</mathjax>.</p>
<p>Given that this is a neutral solution, <mathjax>#[H_3O^+]=[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#[HO^-]^2=1.47xx10^-14#</mathjax></p>
<p>And <mathjax>#[HO^-]=sqrt(1.47xx10^-14)=1.21xx10^-7*mol*L^-1#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We interrogate the equilibrium: <mathjax>#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#</mathjax>.</p>
<p><mathjax>#[HO^-]=1.21xx10^-7*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#</mathjax>,</p>
<p>where the ion product, <mathjax>#K_w=1.47xx10^-14#</mathjax>, at <mathjax>#303*K#</mathjax>. Note that this is slightly higher than <mathjax>#K_w=10^-14#</mathjax> at <mathjax>#298*K#</mathjax>, and given that this is a bond breaking reaction this is perhaps reasonable.</p>
<p>From the equation, <mathjax>#[H_3O^+][HO^-]=1.47xx10^-14#</mathjax>.</p>
<p>Given that this is a neutral solution, <mathjax>#[H_3O^+]=[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#[HO^-]^2=1.47xx10^-14#</mathjax></p>
<p>And <mathjax>#[HO^-]=sqrt(1.47xx10^-14)=1.21xx10^-7*mol*L^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of hydroxide ions in pure water at 30.0°C, if #K_w# at this temperature is #1.47*10^-14#?</h1>
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<div class="markdown"><p>We interrogate the equilibrium: <mathjax>#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#</mathjax>.</p>
<p><mathjax>#[HO^-]=1.21xx10^-7*mol*L^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#</mathjax>,</p>
<p>where the ion product, <mathjax>#K_w=1.47xx10^-14#</mathjax>, at <mathjax>#303*K#</mathjax>. Note that this is slightly higher than <mathjax>#K_w=10^-14#</mathjax> at <mathjax>#298*K#</mathjax>, and given that this is a bond breaking reaction this is perhaps reasonable.</p>
<p>From the equation, <mathjax>#[H_3O^+][HO^-]=1.47xx10^-14#</mathjax>.</p>
<p>Given that this is a neutral solution, <mathjax>#[H_3O^+]=[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#[HO^-]^2=1.47xx10^-14#</mathjax></p>
<p>And <mathjax>#[HO^-]=sqrt(1.47xx10^-14)=1.21xx10^-7*mol*L^-1#</mathjax></p></div>
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</article> | What is the concentration of hydroxide ions in pure water at 30.0°C, if #K_w# at this temperature is #1.47*10^-14#? | null |
2,588 | a840ff0a-6ddd-11ea-b588-ccda262736ce | https://socratic.org/questions/how-many-sulfur-atoms-are-in-0-42-mol-s | 2.53 × 10^23 | start physical_unit 2 3 number none qc_end physical_unit 8 8 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] Sulfur atoms"}] | [{"type":"physical unit","value":"2.53 × 10^23"}] | [{"type":"physical unit","value":"Mole [OF] S [=] \\pu{0.42 mol}"}] | <h1 class="questionTitle" itemprop="name">How many Sulfur atoms are in 0.42 mol S?</h1> | null | 2.53 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a number, admittedly a very large number: <mathjax>#N_A = 6.022 xx 10^(23)#</mathjax>. So, in fact I have <mathjax>#0.42 xx N_A#</mathjax> sulfur atoms.</p>
<p>To expand, Avogadro's number, <mathjax>#N_A#</mathjax>, is simply the link between the micro world of atoms and molecules, with the macro world of grams and kilograms. If I have a mass of <mathjax>#32.06#</mathjax> <mathjax>#g#</mathjax> of sulfur, which I can easily weigh out on a bench, I know, to a very good approximation, that I have <mathjax>#6.022 xx 10^(23)#</mathjax> sulfur atoms (these may be individual atoms or they may be part of <mathjax>#S_8#</mathjax> rings - the number of sulfur atoms is determined directly!)</p></div>
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<div class="markdown"><p>If I have half a dozen sulfur atoms; clearly I have 6 sulfur atoms. If I have 0.42 mol of sulfur there are <mathjax>#0.42 xx N_A#</mathjax> sulfur atoms, where <mathjax>#N_A =#</mathjax> Avogadro's number.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a number, admittedly a very large number: <mathjax>#N_A = 6.022 xx 10^(23)#</mathjax>. So, in fact I have <mathjax>#0.42 xx N_A#</mathjax> sulfur atoms.</p>
<p>To expand, Avogadro's number, <mathjax>#N_A#</mathjax>, is simply the link between the micro world of atoms and molecules, with the macro world of grams and kilograms. If I have a mass of <mathjax>#32.06#</mathjax> <mathjax>#g#</mathjax> of sulfur, which I can easily weigh out on a bench, I know, to a very good approximation, that I have <mathjax>#6.022 xx 10^(23)#</mathjax> sulfur atoms (these may be individual atoms or they may be part of <mathjax>#S_8#</mathjax> rings - the number of sulfur atoms is determined directly!)</p></div>
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<h1 class="questionTitle" itemprop="name">How many Sulfur atoms are in 0.42 mol S?</h1>
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<div class="markdown"><p>If I have half a dozen sulfur atoms; clearly I have 6 sulfur atoms. If I have 0.42 mol of sulfur there are <mathjax>#0.42 xx N_A#</mathjax> sulfur atoms, where <mathjax>#N_A =#</mathjax> Avogadro's number.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply a number, admittedly a very large number: <mathjax>#N_A = 6.022 xx 10^(23)#</mathjax>. So, in fact I have <mathjax>#0.42 xx N_A#</mathjax> sulfur atoms.</p>
<p>To expand, Avogadro's number, <mathjax>#N_A#</mathjax>, is simply the link between the micro world of atoms and molecules, with the macro world of grams and kilograms. If I have a mass of <mathjax>#32.06#</mathjax> <mathjax>#g#</mathjax> of sulfur, which I can easily weigh out on a bench, I know, to a very good approximation, that I have <mathjax>#6.022 xx 10^(23)#</mathjax> sulfur atoms (these may be individual atoms or they may be part of <mathjax>#S_8#</mathjax> rings - the number of sulfur atoms is determined directly!)</p></div>
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</article> | How many Sulfur atoms are in 0.42 mol S? | null |
2,589 | a8d4791a-6ddd-11ea-afab-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-h2o2 | HO | start chemical_formula qc_end chemical_equation 6 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] H2O2 [IN] empirical"}] | [{"type":"chemical equation","value":"HO"}] | [{"type":"chemical equation","value":"H2O2"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of H2O2?</h1> | null | HO | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the above definition the empirical formula of hydrogen peroxide is simply <mathjax>#OH#</mathjax>. Note that the molecular formula is ALWAYS a multiple of the empirical formula. That is <mathjax>#(EF)_n = MF#</mathjax>; where <mathjax>#EF, MF#</mathjax> are empirical and molecular formulae respectively and <mathjax>#n#</mathjax> is a whole number (what does <mathjax>#n#</mathjax> equal here?).</p>
<p>What are the empirical formulae of benzene, ethylene, and acetylene; the molecular formulae are <mathjax>#C_6H_6#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax>, and <mathjax>#HC-=HC#</mathjax>, respectively. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The empirical formula is the simplest whole number ratio that defines constituent atoms in a species.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the above definition the empirical formula of hydrogen peroxide is simply <mathjax>#OH#</mathjax>. Note that the molecular formula is ALWAYS a multiple of the empirical formula. That is <mathjax>#(EF)_n = MF#</mathjax>; where <mathjax>#EF, MF#</mathjax> are empirical and molecular formulae respectively and <mathjax>#n#</mathjax> is a whole number (what does <mathjax>#n#</mathjax> equal here?).</p>
<p>What are the empirical formulae of benzene, ethylene, and acetylene; the molecular formulae are <mathjax>#C_6H_6#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax>, and <mathjax>#HC-=HC#</mathjax>, respectively. </p></div>
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<div class="markdown"><p>The empirical formula is the simplest whole number ratio that defines constituent atoms in a species.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the above definition the empirical formula of hydrogen peroxide is simply <mathjax>#OH#</mathjax>. Note that the molecular formula is ALWAYS a multiple of the empirical formula. That is <mathjax>#(EF)_n = MF#</mathjax>; where <mathjax>#EF, MF#</mathjax> are empirical and molecular formulae respectively and <mathjax>#n#</mathjax> is a whole number (what does <mathjax>#n#</mathjax> equal here?).</p>
<p>What are the empirical formulae of benzene, ethylene, and acetylene; the molecular formulae are <mathjax>#C_6H_6#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax>, and <mathjax>#HC-=HC#</mathjax>, respectively. </p></div>
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<div class="markdown"><p>The empirical formula for the molecular formula <mathjax>#"H"_2"O"_2"#</mathjax> is <mathjax>#"HO"#</mathjax>.</p></div>
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<div class="markdown"><p>The empirical formula of a compound is the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound. This ratio becomes the subscripts for the empirical formula.</p>
<p>The molecular formula <mathjax>#"H"_2"O"_2"#</mathjax> is a multiple of the empirical formula. We can know this because the ratio of <mathjax>#"2H":2"O"#</mathjax> is not the lowest whole number ratio. By dividing the subscripts by <mathjax>#2#</mathjax>, we can determine the lowest whole number ratio, which is <mathjax>#"1H":1"O"#</mathjax>. Therefore, the empirical formula for <mathjax>#"H"_2"O"_2"#</mathjax> is <mathjax>#"HO"#</mathjax>.</p>
<p><img alt="http://study.com/academy/lesson/what-is-a-chemical-formula-definition-types-examples.html" src="https://useruploads.socratic.org/MIx5XCwKRdWNpUDVDH4C_empiricalformula2.png"/> </p>
<p>Sometimes the molecular and empirical formulas are the same.</p>
<p><img alt="http://study.com/academy/lesson/what-is-a-chemical-formula-definition-types-examples.html" src="https://useruploads.socratic.org/dCPwOEriQ0SGb6tp356s_empiricalformula3.png"/> </p></div>
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</article> | What is the empirical formula of H2O2? | null |
2,590 | a8f66ddf-6ddd-11ea-84cd-ccda262736ce | https://socratic.org/questions/592bc51a11ef6b2839d3beaf | 10 L | start physical_unit 3 3 volume l qc_end physical_unit 17 17 9 10 mass qc_end physical_unit 17 17 19 21 equilibrium_constant_k qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] L"}] | [{"type":"physical unit","value":"10 L"}] | [{"type":"physical unit","value":"Mass [OF] CaSO4 [=] \\pu{6.7 g}"},{"type":"physical unit","value":"Ksp [OF] CaSO4 [=] \\pu{2.4 × 10^(-5)}"}] | <h1 class="questionTitle" itemprop="name">What volume of water is required to dissolve a #6.7*g# mass of #"calcium sulfate"# if #K_(sp)*CaSO_4=2.4xx10^-5#?</h1> | null | 10 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the equilibrium reaction.......</p>
<p><mathjax>#CaSO_4(s) rightleftharpoonsCa^(2+) + SO_4^(2-)#</mathjax>,</p>
<p>Where <mathjax>#K_"eq"=K_"sp"=[Ca^(2+)][SO_4^(2-)]#</mathjax>.</p>
<p>And if we put <mathjax>#[Ca^(2+)]=[SO_4^(2-)]=x#</mathjax>..............</p>
<p>.......then <mathjax>#x=sqrt(K_"sp")=sqrt(2.4xx10^-5)=0.005*mol*L^-1#</mathjax></p>
<p>And so <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(??*L)=0.005*mol*L^-1#</mathjax></p>
<p>i.e. <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(0.005*mol*L^-1)~=10*L#</mathjax></p></div>
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<div class="markdown"><p>A volume of <mathjax>#10*L#</mathjax> is required........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the equilibrium reaction.......</p>
<p><mathjax>#CaSO_4(s) rightleftharpoonsCa^(2+) + SO_4^(2-)#</mathjax>,</p>
<p>Where <mathjax>#K_"eq"=K_"sp"=[Ca^(2+)][SO_4^(2-)]#</mathjax>.</p>
<p>And if we put <mathjax>#[Ca^(2+)]=[SO_4^(2-)]=x#</mathjax>..............</p>
<p>.......then <mathjax>#x=sqrt(K_"sp")=sqrt(2.4xx10^-5)=0.005*mol*L^-1#</mathjax></p>
<p>And so <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(??*L)=0.005*mol*L^-1#</mathjax></p>
<p>i.e. <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(0.005*mol*L^-1)~=10*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume of water is required to dissolve a #6.7*g# mass of #"calcium sulfate"# if #K_(sp)*CaSO_4=2.4xx10^-5#?</h1>
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<div class="markdown"><p>A volume of <mathjax>#10*L#</mathjax> is required........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assess the equilibrium reaction.......</p>
<p><mathjax>#CaSO_4(s) rightleftharpoonsCa^(2+) + SO_4^(2-)#</mathjax>,</p>
<p>Where <mathjax>#K_"eq"=K_"sp"=[Ca^(2+)][SO_4^(2-)]#</mathjax>.</p>
<p>And if we put <mathjax>#[Ca^(2+)]=[SO_4^(2-)]=x#</mathjax>..............</p>
<p>.......then <mathjax>#x=sqrt(K_"sp")=sqrt(2.4xx10^-5)=0.005*mol*L^-1#</mathjax></p>
<p>And so <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(??*L)=0.005*mol*L^-1#</mathjax></p>
<p>i.e. <mathjax>#((6.7*g)/(136.14*g*mol^-1))/(0.005*mol*L^-1)~=10*L#</mathjax></p></div>
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</article> | What volume of water is required to dissolve a #6.7*g# mass of #"calcium sulfate"# if #K_(sp)*CaSO_4=2.4xx10^-5#? | null |
2,591 | aaabf522-6ddd-11ea-8836-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-that-is-0-50-m-in-sodium-acetate-and-0-75-m-in-acet | 4.57 | start physical_unit 6 6 ph none qc_end physical_unit 12 13 9 10 molarity qc_end physical_unit 18 19 15 16 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"4.57"}] | [{"type":"physical unit","value":"Molarity [OF] sodium acetate solution [=] \\pu{0.50 M}"},{"type":"physical unit","value":"Molarity [OF] acetic acid solution [=] \\pu{0.75 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid? </h1> | null | 4.57 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>this is an acid tampon solution.<br/>
<mathjax>#pH = pKa + log ((Cs)/(Ca))#</mathjax> = 4,75 + log(0,5/0,75) =4,75 +log 0,66 = 4,57 #<br/>
pKa is the logarithm (changed of sign) of the dissociation constant of acetic acid; Cs the concentration of the salt (sodium acetate) ; Ca the concentration of the acid (acetic acid). As you can see <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> doesn't change with concetration, and change very little changing the amount of salt and of acid</p></div>
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<div class="markdown"><p>4,57</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>this is an acid tampon solution.<br/>
<mathjax>#pH = pKa + log ((Cs)/(Ca))#</mathjax> = 4,75 + log(0,5/0,75) =4,75 +log 0,66 = 4,57 #<br/>
pKa is the logarithm (changed of sign) of the dissociation constant of acetic acid; Cs the concentration of the salt (sodium acetate) ; Ca the concentration of the acid (acetic acid). As you can see <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> doesn't change with concetration, and change very little changing the amount of salt and of acid</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid? </h1>
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<div class="markdown"><p>4,57</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>this is an acid tampon solution.<br/>
<mathjax>#pH = pKa + log ((Cs)/(Ca))#</mathjax> = 4,75 + log(0,5/0,75) =4,75 +log 0,66 = 4,57 #<br/>
pKa is the logarithm (changed of sign) of the dissociation constant of acetic acid; Cs the concentration of the salt (sodium acetate) ; Ca the concentration of the acid (acetic acid). As you can see <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> doesn't change with concetration, and change very little changing the amount of salt and of acid</p></div>
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</article> | What is the pH of a solution that is 0.50 M in sodium acetate and 0.75 M in acetic acid? | null |
2,592 | a9b919a2-6ddd-11ea-a26c-ccda262736ce | https://socratic.org/questions/aluminum-reacts-with-aqueous-hbr-to-produce-hydrogen-gas-and-aluminum-bromide-ho | 432.28 mL | start physical_unit 7 8 volume ml qc_end c_other OTHER qc_end physical_unit 7 8 24 25 temperature qc_end physical_unit 0 0 27 28 mass qc_end physical_unit 7 8 21 22 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrogen gas [IN] mL"}] | [{"type":"physical unit","value":"432.28 mL"}] | [{"type":"other","value":"Excess of HBr ."},{"type":"physical unit","value":"Temperature [OF] hydrogen gas [=] \\pu{20.00 ℃}"},{"type":"physical unit","value":"Mass [OF] aluminum [=] \\pu{0.498 g}"},{"type":"physical unit","value":"Pressure [OF] hydrogen gas [=] \\pu{1.03 atm}"}] | <h1 class="questionTitle" itemprop="name">Aluminum reacts with aqueous #HBr# to produce hydrogen gas and aluminum bromide. How many mL of hydrogen gas are produced at 1.03 atm and 20.00 #"^o#C when 0.498 g of aluminum react with an excess of #HBr#?</h1> | null | 432.28 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Each equiv of metal reduces <mathjax>#3#</mathjax> equiv of acid, and <mathjax>#3/2#</mathjax> equiv of dihydrogen gas are evolved.</p>
<p><mathjax>#"Moles of aluminum"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.498*g)/(26.98*g*mol^-1)=0.0185*mol#</mathjax>.</p>
<p>Now, given the conditions, clearly the metal is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. We have the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, so a molar quantity of <mathjax>#3/2xx0.0185*mol#</mathjax> <mathjax>#"dihydrogen gas"#</mathjax> result.</p>
<p>And now it is an Ideal Gas calculation (mind you if we collect the gas over water, we should include the <mathjax>#"saturated vapour pressure"#</mathjax>, to account for the water vapour collected with the dihydrogen. Since there are no details in the initial conditions, I can ignore these data.)</p>
<p><mathjax>#V=(nRT)/P=#</mathjax><br/>
<mathjax>#(0.0185*molxx0.0821*L*atm*K^-1*mol^-1xx293.15*K)/(1.03*atm)=0.431*L=??*mL#</mathjax>.</p>
<p>This is experiment that could be performed in a high school lab with litre graduated cylinders, with which to collect the gas. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>First you need a stoichiometric equation:</p>
<p><mathjax>#Al(s) + 3HBr rarr AlBr_3(aq) + 3/2H_2(g)#</mathjax></p>
<p>Finally we get under a <mathjax>#1/2*L#</mathjax> volume of dihydrogen. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Each equiv of metal reduces <mathjax>#3#</mathjax> equiv of acid, and <mathjax>#3/2#</mathjax> equiv of dihydrogen gas are evolved.</p>
<p><mathjax>#"Moles of aluminum"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.498*g)/(26.98*g*mol^-1)=0.0185*mol#</mathjax>.</p>
<p>Now, given the conditions, clearly the metal is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. We have the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, so a molar quantity of <mathjax>#3/2xx0.0185*mol#</mathjax> <mathjax>#"dihydrogen gas"#</mathjax> result.</p>
<p>And now it is an Ideal Gas calculation (mind you if we collect the gas over water, we should include the <mathjax>#"saturated vapour pressure"#</mathjax>, to account for the water vapour collected with the dihydrogen. Since there are no details in the initial conditions, I can ignore these data.)</p>
<p><mathjax>#V=(nRT)/P=#</mathjax><br/>
<mathjax>#(0.0185*molxx0.0821*L*atm*K^-1*mol^-1xx293.15*K)/(1.03*atm)=0.431*L=??*mL#</mathjax>.</p>
<p>This is experiment that could be performed in a high school lab with litre graduated cylinders, with which to collect the gas. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Aluminum reacts with aqueous #HBr# to produce hydrogen gas and aluminum bromide. How many mL of hydrogen gas are produced at 1.03 atm and 20.00 #"^o#C when 0.498 g of aluminum react with an excess of #HBr#?</h1>
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<div class="markdown"><p>First you need a stoichiometric equation:</p>
<p><mathjax>#Al(s) + 3HBr rarr AlBr_3(aq) + 3/2H_2(g)#</mathjax></p>
<p>Finally we get under a <mathjax>#1/2*L#</mathjax> volume of dihydrogen. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Each equiv of metal reduces <mathjax>#3#</mathjax> equiv of acid, and <mathjax>#3/2#</mathjax> equiv of dihydrogen gas are evolved.</p>
<p><mathjax>#"Moles of aluminum"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.498*g)/(26.98*g*mol^-1)=0.0185*mol#</mathjax>.</p>
<p>Now, given the conditions, clearly the metal is the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>. We have the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, so a molar quantity of <mathjax>#3/2xx0.0185*mol#</mathjax> <mathjax>#"dihydrogen gas"#</mathjax> result.</p>
<p>And now it is an Ideal Gas calculation (mind you if we collect the gas over water, we should include the <mathjax>#"saturated vapour pressure"#</mathjax>, to account for the water vapour collected with the dihydrogen. Since there are no details in the initial conditions, I can ignore these data.)</p>
<p><mathjax>#V=(nRT)/P=#</mathjax><br/>
<mathjax>#(0.0185*molxx0.0821*L*atm*K^-1*mol^-1xx293.15*K)/(1.03*atm)=0.431*L=??*mL#</mathjax>.</p>
<p>This is experiment that could be performed in a high school lab with litre graduated cylinders, with which to collect the gas. </p></div>
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</article> | Aluminum reacts with aqueous #HBr# to produce hydrogen gas and aluminum bromide. How many mL of hydrogen gas are produced at 1.03 atm and 20.00 #"^o#C when 0.498 g of aluminum react with an excess of #HBr#? | null |
2,593 | ac5f2728-6ddd-11ea-9bf9-ccda262736ce | https://socratic.org/questions/5954a708b72cff186eace8a2 | 79.05 grams | start physical_unit 13 14 mass g qc_end physical_unit 6 6 2 3 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] grams"}] | [{"type":"physical unit","value":"79.05 grams"}] | [{"type":"physical unit","value":"Mass [OF] octane [=] \\pu{25.6 g}"},{"type":"other","value":"Completely combuste."}] | <h1 class="questionTitle" itemprop="name">If a #25.6*g# mass of #"octane"# is completely combusted, what mass of carbon dioxide will result?</h1> | null | 79.05 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometrically balanced equation.....</p>
<p><mathjax>#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It must be if we purport to represent chemical reality.</p>
<p>And (ii) we need equivalent quantities of octane.....</p>
<p><mathjax>#"Moles of octane"=(25.6*g)/(114.23*g*mol^-1)=0.224*mol#</mathjax></p>
<p>And given the equation, CLEARLY we KNOW that 8 equiv of carbon dioxide result per equiv of octane.....</p>
<p>And thus a molar quantity of <mathjax>#8xx0.224*mol#</mathjax> <mathjax>#CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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<div class="markdown"><p>Approx...<mathjax>#1.8*mol..........#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometrically balanced equation.....</p>
<p><mathjax>#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It must be if we purport to represent chemical reality.</p>
<p>And (ii) we need equivalent quantities of octane.....</p>
<p><mathjax>#"Moles of octane"=(25.6*g)/(114.23*g*mol^-1)=0.224*mol#</mathjax></p>
<p>And given the equation, CLEARLY we KNOW that 8 equiv of carbon dioxide result per equiv of octane.....</p>
<p>And thus a molar quantity of <mathjax>#8xx0.224*mol#</mathjax> <mathjax>#CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">If a #25.6*g# mass of #"octane"# is completely combusted, what mass of carbon dioxide will result?</h1>
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<div class="markdown"><p>Approx...<mathjax>#1.8*mol..........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need (i) a stoichiometrically balanced equation.....</p>
<p><mathjax>#C_8H_18(l) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It must be if we purport to represent chemical reality.</p>
<p>And (ii) we need equivalent quantities of octane.....</p>
<p><mathjax>#"Moles of octane"=(25.6*g)/(114.23*g*mol^-1)=0.224*mol#</mathjax></p>
<p>And given the equation, CLEARLY we KNOW that 8 equiv of carbon dioxide result per equiv of octane.....</p>
<p>And thus a molar quantity of <mathjax>#8xx0.224*mol#</mathjax> <mathjax>#CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax>.</p></div>
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<div class="markdown"><p>79 grams of carbondioxide</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#C_8H_18 + 12.5 O_2 -> 8CO_2 + 9H_2O#</mathjax></p>
<p>This is the chemical reaction. </p>
<p>1 mole of octane mass is 114 grams.</p>
<p>It means if you have 114 grams of octane, you will have 352 grams of <mathjax>#CO_2#</mathjax> after the reaction.</p>
<p>However you have only 25.6 grams of octane. You can get how much carbondioxide after the reaction.</p>
<p><mathjax>#=(25.6times352)/114#</mathjax></p>
<p><mathjax>#=79#</mathjax> grams</p></div>
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</article> | If a #25.6*g# mass of #"octane"# is completely combusted, what mass of carbon dioxide will result? | null |
2,594 | acaa5b4a-6ddd-11ea-becb-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-prepared-by-combining-125-ml-of-0-251-molar-h | 0.09 M | start physical_unit 6 6 molarity mol/l qc_end physical_unit 15 15 13 14 molarity qc_end physical_unit 6 6 10 11 volume qc_end physical_unit 20 21 17 18 volume qc_end end | [{"type":"physical unit","value":"Molarity2 [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"0.09 M"}] | [{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{0.251 molar}"},{"type":"physical unit","value":"Volume1 [OF] HCl solution [=] \\pu{125 mL}"},{"type":"physical unit","value":"Volume [OF] pure water [=] \\pu{250 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by combining 125 mL of 0.251 molar HCl with 250 mL of pure water?</h1> | null | 0.09 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're mixing <mathjax>#"125 mL"#</mathjax> of a <mathjax>#"0.251-M"#</mathjax> hydrochloric acid solution with <mathjax>#"250 mL"#</mathjax> of <em>pure water</em>. </p>
<p>In essence, you're <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">diluting</a> the initial solution by keeping the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, <strong>constant</strong> and increasing the amount of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>This means that you can expect the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the second solution to be <strong>smaller</strong> than that of the initial solution, which would be characteristic of a <em>diluted</em> solution. </p>
<p>Use the volume and the molarity of the initial solution to determine how many moles of hydrochloric acid you get in that volume</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_"HCl" = "0.251 M" * 125 * 10^(-3)"L" = "0.031375 moles HCl"#</mathjax></p>
</blockquote>
<p>The <strong>total volume</strong> of the final solution will be equal to </p>
<blockquote>
<p><mathjax>#V_"total" = "125 mL" + "250 mL" = "375 mL"#</mathjax></p>
</blockquote>
<p>The molarity of the second solution will thus be</p>
<blockquote>
<p><mathjax>#c = "0.031375 moles"/(375 * 10^(-3)"L") = color(green)("0.0874 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the target solution.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.084 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're mixing <mathjax>#"125 mL"#</mathjax> of a <mathjax>#"0.251-M"#</mathjax> hydrochloric acid solution with <mathjax>#"250 mL"#</mathjax> of <em>pure water</em>. </p>
<p>In essence, you're <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">diluting</a> the initial solution by keeping the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, <strong>constant</strong> and increasing the amount of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>This means that you can expect the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the second solution to be <strong>smaller</strong> than that of the initial solution, which would be characteristic of a <em>diluted</em> solution. </p>
<p>Use the volume and the molarity of the initial solution to determine how many moles of hydrochloric acid you get in that volume</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_"HCl" = "0.251 M" * 125 * 10^(-3)"L" = "0.031375 moles HCl"#</mathjax></p>
</blockquote>
<p>The <strong>total volume</strong> of the final solution will be equal to </p>
<blockquote>
<p><mathjax>#V_"total" = "125 mL" + "250 mL" = "375 mL"#</mathjax></p>
</blockquote>
<p>The molarity of the second solution will thus be</p>
<blockquote>
<p><mathjax>#c = "0.031375 moles"/(375 * 10^(-3)"L") = color(green)("0.0874 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the target solution.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution prepared by combining 125 mL of 0.251 molar HCl with 250 mL of pure water?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-09T23:29:33" itemprop="dateCreated">
Nov 9, 2015
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<div class="markdown"><p><mathjax>#"0.084 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that you're mixing <mathjax>#"125 mL"#</mathjax> of a <mathjax>#"0.251-M"#</mathjax> hydrochloric acid solution with <mathjax>#"250 mL"#</mathjax> of <em>pure water</em>. </p>
<p>In essence, you're <a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">diluting</a> the initial solution by keeping the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, <strong>constant</strong> and increasing the amount of <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>This means that you can expect the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the second solution to be <strong>smaller</strong> than that of the initial solution, which would be characteristic of a <em>diluted</em> solution. </p>
<p>Use the volume and the molarity of the initial solution to determine how many moles of hydrochloric acid you get in that volume</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_"HCl" = "0.251 M" * 125 * 10^(-3)"L" = "0.031375 moles HCl"#</mathjax></p>
</blockquote>
<p>The <strong>total volume</strong> of the final solution will be equal to </p>
<blockquote>
<p><mathjax>#V_"total" = "125 mL" + "250 mL" = "375 mL"#</mathjax></p>
</blockquote>
<p>The molarity of the second solution will thus be</p>
<blockquote>
<p><mathjax>#c = "0.031375 moles"/(375 * 10^(-3)"L") = color(green)("0.0874 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the volume of the target solution.</p></div>
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</article> | What is the molarity of a solution prepared by combining 125 mL of 0.251 molar HCl with 250 mL of pure water? | null |
2,595 | a9cc72a8-6ddd-11ea-8225-ccda262736ce | https://socratic.org/questions/57f2b64811ef6b1c778503b6 | 0.75 mol/L | start physical_unit 3 3 concentration mol/l qc_end physical_unit 9 9 11 12 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] IONIC [IN] mol/L"}] | [{"type":"physical unit","value":"0.75 mol/L"}] | [{"type":"physical unit","value":"Concentration [OF] MgCl2(aq) [=] \\pu{0.25 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is total IONIC concentration in a solution of #MgCl_2(aq)# at #0.25*mol*L^-1# concentration?</h1> | null | 0.75 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqeous solution, magnesium chloride ionizes according to the following equation:</p>
<p><mathjax>#MgCl_2(s) rarr Mg^(2+) + 2Cl^-#</mathjax> </p>
<p>Thus three equiv of ion result for each equiv of magnesium chloride: <mathjax>#1xxMg^(2+) + 2xxCl^-#</mathjax></p>
<p>So if the concentation of <mathjax>#MgCl_2(aq)=0.25*mol*L^-1#</mathjax>, then <mathjax>#[Mg^(2+)]=0.25*mol*L^-1#</mathjax>, and <mathjax>#[Cl^-]=0.50*mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The total concentration of ions from the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is <mathjax>#0.75*mol*L^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqeous solution, magnesium chloride ionizes according to the following equation:</p>
<p><mathjax>#MgCl_2(s) rarr Mg^(2+) + 2Cl^-#</mathjax> </p>
<p>Thus three equiv of ion result for each equiv of magnesium chloride: <mathjax>#1xxMg^(2+) + 2xxCl^-#</mathjax></p>
<p>So if the concentation of <mathjax>#MgCl_2(aq)=0.25*mol*L^-1#</mathjax>, then <mathjax>#[Mg^(2+)]=0.25*mol*L^-1#</mathjax>, and <mathjax>#[Cl^-]=0.50*mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is total IONIC concentration in a solution of #MgCl_2(aq)# at #0.25*mol*L^-1# concentration?</h1>
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anor277
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<div class="markdown"><p>The total concentration of ions from the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is <mathjax>#0.75*mol*L^-1#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqeous solution, magnesium chloride ionizes according to the following equation:</p>
<p><mathjax>#MgCl_2(s) rarr Mg^(2+) + 2Cl^-#</mathjax> </p>
<p>Thus three equiv of ion result for each equiv of magnesium chloride: <mathjax>#1xxMg^(2+) + 2xxCl^-#</mathjax></p>
<p>So if the concentation of <mathjax>#MgCl_2(aq)=0.25*mol*L^-1#</mathjax>, then <mathjax>#[Mg^(2+)]=0.25*mol*L^-1#</mathjax>, and <mathjax>#[Cl^-]=0.50*mol*L^-1#</mathjax>.</p>
<p>Capisce?</p></div>
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</article> | What is total IONIC concentration in a solution of #MgCl_2(aq)# at #0.25*mol*L^-1# concentration? | null |
2,596 | ac8be34c-6ddd-11ea-a2c0-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-of-750-0-ml-of-0-500-m-hno-2-to-which-25-00-grams-o | 3.39 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 8 9 volume qc_end physical_unit 13 13 11 12 molarity qc_end physical_unit 19 20 16 17 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pH [OF] HNO2 solution"}] | [{"type":"physical unit","value":"3.39"}] | [{"type":"physical unit","value":"Volume [OF] HNO2 solution [=] \\pu{750.0 mL}"},{"type":"physical unit","value":"Molarity [OF] HNO2 solution [=] \\pu{0.500 M}"},{"type":"physical unit","value":"Mass [OF] sodium nitrite [=] \\pu{25.00 grams}"},{"type":"other","value":"Assume no volume change."}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution of 750.0 mL of 0.500 M #HNO_2# to which 25.00 grams of sodium nitrite has been added (assume no volume change)?</h1> | null | 3.39 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <strong>buffer solution</strong> that contains nitrous acid, <mathjax>#"HNO"_2#</mathjax>, a <strong>weak acid</strong>, and the nitrite anion, <mathjax>#"NO"_2^(-)#</mathjax>, its <strong>conjugate base</strong>, delivered to the solution by the soluble salt sodium nitrite.</p>
<p>The <mathjax>#"pH"#</mathjax> of a weak acid/conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>and <mathjax>#K_a#</mathjax> is the <strong>acid dissociation constant</strong> of the weak acid. </p>
<p>Nitrous acid has an acid dissociation constant equal to</p>
<blockquote>
<p><mathjax>#K_a = 4.0 * 10^(-4)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>which means that you can write the Henderson - Hasselbalch equation as</p>
<blockquote>
<p><mathjax>#"pH" = -log(4.0 * 10^(-4)) + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
<p><mathjax>#"pH" = 3.40 + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
</blockquote>
<p>Now, the problem wants you to assume that no volume change occurs when you add the salt, so you can say that</p>
<blockquote>
<p><mathjax>#["HNO"_2] = "0.500 M"#</mathjax></p>
</blockquote>
<p>In order to find the concentration of the nitrite anion, use the <strong>molar mass</strong> of <em>sodium nitrite</em> to calculate the number of moles of salt dissolved in the solution.</p>
<blockquote>
<p><mathjax>#25.00 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_2/(68.9953color(red)(cancel(color(black)("g")))) = "0.36234 moles NaNO"_2#</mathjax></p>
</blockquote>
<p>Sine sodium nitrite dissociates in a <mathjax>#1:1#</mathjax> mole ratio to produce nitrite anions</p>
<blockquote>
<p><mathjax>#"NaNO"_ (2(aq)) -> "Na"_ ((aq))^(+) + "NO"_ (2(aq))^(-)#</mathjax></p>
</blockquote>
<p>you can say that the solution will contain <mathjax>#0.36234#</mathjax> <strong>moles</strong> of nitrite anions in <mathjax>#"750.0 mL"#</mathjax> of solution. </p>
<p>This implies that the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the nitrite anions, which is calculated by dividing the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> by the total volume of the solution <strong>in liters</strong>, will be</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.36234 moles"/(750.0 * 10^(-3)color(white)(.)"L") = "0.36234 moles"/0.750 * 1/"1 L"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.48312 moles"/"1 L" = "0.48312 M"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You can now plug this into the Henderson - Hasselbalch equation to find the <mathjax>#"pH"#</mathjax> of the solution</p>
<blockquote>
<p><mathjax>#"pH" = 3.40 + log( (0.48312 color(red)(cancel(color(black)("M"))))/(0.500color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 3.385)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of sig figs you have for the molarity of the nitrous acid. </p>
<p>Finally, does the result make sense?</p>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] > ["NO"_2]#</mathjax></p>
</blockquote>
<p>This tells you that the solution contains more weak acid than conjugate base, which implies that the <mathjax>#"pH"#</mathjax> of the solution will be <strong>lower</strong> than the <mathjax>#"p"K_a#</mathjax> of the weak acid.</p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 3.385 < "p"K_a = 3.40" " ->#</mathjax> makes sense. </p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 3.385#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <strong>buffer solution</strong> that contains nitrous acid, <mathjax>#"HNO"_2#</mathjax>, a <strong>weak acid</strong>, and the nitrite anion, <mathjax>#"NO"_2^(-)#</mathjax>, its <strong>conjugate base</strong>, delivered to the solution by the soluble salt sodium nitrite.</p>
<p>The <mathjax>#"pH"#</mathjax> of a weak acid/conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>and <mathjax>#K_a#</mathjax> is the <strong>acid dissociation constant</strong> of the weak acid. </p>
<p>Nitrous acid has an acid dissociation constant equal to</p>
<blockquote>
<p><mathjax>#K_a = 4.0 * 10^(-4)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>which means that you can write the Henderson - Hasselbalch equation as</p>
<blockquote>
<p><mathjax>#"pH" = -log(4.0 * 10^(-4)) + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
<p><mathjax>#"pH" = 3.40 + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
</blockquote>
<p>Now, the problem wants you to assume that no volume change occurs when you add the salt, so you can say that</p>
<blockquote>
<p><mathjax>#["HNO"_2] = "0.500 M"#</mathjax></p>
</blockquote>
<p>In order to find the concentration of the nitrite anion, use the <strong>molar mass</strong> of <em>sodium nitrite</em> to calculate the number of moles of salt dissolved in the solution.</p>
<blockquote>
<p><mathjax>#25.00 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_2/(68.9953color(red)(cancel(color(black)("g")))) = "0.36234 moles NaNO"_2#</mathjax></p>
</blockquote>
<p>Sine sodium nitrite dissociates in a <mathjax>#1:1#</mathjax> mole ratio to produce nitrite anions</p>
<blockquote>
<p><mathjax>#"NaNO"_ (2(aq)) -> "Na"_ ((aq))^(+) + "NO"_ (2(aq))^(-)#</mathjax></p>
</blockquote>
<p>you can say that the solution will contain <mathjax>#0.36234#</mathjax> <strong>moles</strong> of nitrite anions in <mathjax>#"750.0 mL"#</mathjax> of solution. </p>
<p>This implies that the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the nitrite anions, which is calculated by dividing the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> by the total volume of the solution <strong>in liters</strong>, will be</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.36234 moles"/(750.0 * 10^(-3)color(white)(.)"L") = "0.36234 moles"/0.750 * 1/"1 L"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.48312 moles"/"1 L" = "0.48312 M"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You can now plug this into the Henderson - Hasselbalch equation to find the <mathjax>#"pH"#</mathjax> of the solution</p>
<blockquote>
<p><mathjax>#"pH" = 3.40 + log( (0.48312 color(red)(cancel(color(black)("M"))))/(0.500color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 3.385)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of sig figs you have for the molarity of the nitrous acid. </p>
<p>Finally, does the result make sense?</p>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] > ["NO"_2]#</mathjax></p>
</blockquote>
<p>This tells you that the solution contains more weak acid than conjugate base, which implies that the <mathjax>#"pH"#</mathjax> of the solution will be <strong>lower</strong> than the <mathjax>#"p"K_a#</mathjax> of the weak acid.</p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 3.385 < "p"K_a = 3.40" " ->#</mathjax> makes sense. </p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the pH of a solution of 750.0 mL of 0.500 M #HNO_2# to which 25.00 grams of sodium nitrite has been added (assume no volume change)?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-06-16T00:46:00" itemprop="dateCreated">
Jun 16, 2017
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<div class="markdown"><p><mathjax>#"pH" = 3.385#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <strong>buffer solution</strong> that contains nitrous acid, <mathjax>#"HNO"_2#</mathjax>, a <strong>weak acid</strong>, and the nitrite anion, <mathjax>#"NO"_2^(-)#</mathjax>, its <strong>conjugate base</strong>, delivered to the solution by the soluble salt sodium nitrite.</p>
<p>The <mathjax>#"pH"#</mathjax> of a weak acid/conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#"p"K_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>and <mathjax>#K_a#</mathjax> is the <strong>acid dissociation constant</strong> of the weak acid. </p>
<p>Nitrous acid has an acid dissociation constant equal to</p>
<blockquote>
<p><mathjax>#K_a = 4.0 * 10^(-4)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>which means that you can write the Henderson - Hasselbalch equation as</p>
<blockquote>
<p><mathjax>#"pH" = -log(4.0 * 10^(-4)) + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
<p><mathjax>#"pH" = 3.40 + log( (["NO"_2^(-)])/(["HNO"_2]))#</mathjax></p>
</blockquote>
<p>Now, the problem wants you to assume that no volume change occurs when you add the salt, so you can say that</p>
<blockquote>
<p><mathjax>#["HNO"_2] = "0.500 M"#</mathjax></p>
</blockquote>
<p>In order to find the concentration of the nitrite anion, use the <strong>molar mass</strong> of <em>sodium nitrite</em> to calculate the number of moles of salt dissolved in the solution.</p>
<blockquote>
<p><mathjax>#25.00 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_2/(68.9953color(red)(cancel(color(black)("g")))) = "0.36234 moles NaNO"_2#</mathjax></p>
</blockquote>
<p>Sine sodium nitrite dissociates in a <mathjax>#1:1#</mathjax> mole ratio to produce nitrite anions</p>
<blockquote>
<p><mathjax>#"NaNO"_ (2(aq)) -> "Na"_ ((aq))^(+) + "NO"_ (2(aq))^(-)#</mathjax></p>
</blockquote>
<p>you can say that the solution will contain <mathjax>#0.36234#</mathjax> <strong>moles</strong> of nitrite anions in <mathjax>#"750.0 mL"#</mathjax> of solution. </p>
<p>This implies that the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the nitrite anions, which is calculated by dividing the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> by the total volume of the solution <strong>in liters</strong>, will be</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.36234 moles"/(750.0 * 10^(-3)color(white)(.)"L") = "0.36234 moles"/0.750 * 1/"1 L"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.48312 moles"/"1 L" = "0.48312 M"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You can now plug this into the Henderson - Hasselbalch equation to find the <mathjax>#"pH"#</mathjax> of the solution</p>
<blockquote>
<p><mathjax>#"pH" = 3.40 + log( (0.48312 color(red)(cancel(color(black)("M"))))/(0.500color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 3.385)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong>decimal places</strong>, the number of sig figs you have for the molarity of the nitrous acid. </p>
<p>Finally, does the result make sense?</p>
<p>Notice that you have</p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] > ["NO"_2]#</mathjax></p>
</blockquote>
<p>This tells you that the solution contains more weak acid than conjugate base, which implies that the <mathjax>#"pH"#</mathjax> of the solution will be <strong>lower</strong> than the <mathjax>#"p"K_a#</mathjax> of the weak acid.</p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"pH" = 3.385 < "p"K_a = 3.40" " ->#</mathjax> makes sense. </p>
</blockquote></div>
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</article> | What is the pH of a solution of 750.0 mL of 0.500 M #HNO_2# to which 25.00 grams of sodium nitrite has been added (assume no volume change)? | null |
2,597 | ab91f7b4-6ddd-11ea-90c3-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-that-has-a-molecular-mass-of-42-and- | C3H6 | start chemical_formula qc_end physical_unit 7 7 14 15 molecular_weight qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C3H6"}] | [{"type":"physical unit","value":"Molecular mass [OF] the compound [=] \\pu{42 g/mol}"},{"type":"other","value":"The compound has an empirical formula of CH2."}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound that has a molecular mass of 42 and an empirical formula of #CH_2#?</h1> | null | C3H6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula mass of <mathjax>#"CH"_2"#</mathjax> is 14 g/mol.</p>
<p>Divide the molecular mass by the empirical formula mass.</p>
<p><mathjax>#42/14=3#</mathjax></p>
<p>Multiply the subscripts of the empirical formula times three to get the molecular formula.</p>
<p>The molecular formula is <mathjax>#"C"_3"H"_6"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molecular formula is <mathjax>#"C"_3"H"_6"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula mass of <mathjax>#"CH"_2"#</mathjax> is 14 g/mol.</p>
<p>Divide the molecular mass by the empirical formula mass.</p>
<p><mathjax>#42/14=3#</mathjax></p>
<p>Multiply the subscripts of the empirical formula times three to get the molecular formula.</p>
<p>The molecular formula is <mathjax>#"C"_3"H"_6"#</mathjax>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound that has a molecular mass of 42 and an empirical formula of #CH_2#?</h1>
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<div class="markdown"><p>The molecular formula is <mathjax>#"C"_3"H"_6"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The empirical formula mass of <mathjax>#"CH"_2"#</mathjax> is 14 g/mol.</p>
<p>Divide the molecular mass by the empirical formula mass.</p>
<p><mathjax>#42/14=3#</mathjax></p>
<p>Multiply the subscripts of the empirical formula times three to get the molecular formula.</p>
<p>The molecular formula is <mathjax>#"C"_3"H"_6"#</mathjax>.</p></div>
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</article> | What is the molecular formula of a compound that has a molecular mass of 42 and an empirical formula of #CH_2#? | null |
2,598 | a9b807e6-6ddd-11ea-8c6e-ccda262736ce | https://socratic.org/questions/if-you-start-with-7-0-moles-of-c3h8-propane-and-7-0-moles-of-o2-is-the-percent-y-1 | 95.24% | start physical_unit 29 29 percent_yield none qc_end physical_unit 38 38 4 5 mole qc_end physical_unit 44 44 4 5 mole qc_end physical_unit 23 24 20 21 mole qc_end chemical_equation 58 67 qc_end end | [{"type":"physical unit","value":"Percent yield [OF] reaction"}] | [{"type":"physical unit","value":"95.24%"}] | [{"type":"physical unit","value":"Mole [OF] C3H8 [=] \\pu{7.0 moles}"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{7.0 moles}"},{"type":"physical unit","value":"Mole [OF] carbon dioxide [=] \\pu{4.0 moles}"},{"type":"chemical equation","value":"C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)"}] | <h1 class="questionTitle" itemprop="name">If you start with #7.0# moles of propane and #7.0# moles of oxygen gas what is the percent yield if #4.0# moles of carbon dioxide are produced?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, <strong>_</strong> is the percent yield if 4.0 moles of carbon dioxide is produced. <br/>
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)</p></div>
</h2>
</div>
</div> | 95.24% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <strong>theoretical yield</strong> of the reaction.</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>You know that the reaction consumes <mathjax>#5#</mathjax> <strong>moles</strong> of oxygen gas for every <mathjax>#1#</mathjax> <strong>mole</strong> of propane that takes part in the reaction. </p>
<p>This means that, in order for the reaction to consume <mathjax>#7#</mathjax> <strong>moles</strong> of propane, it must also consume</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * "5 moles O"_2/(1color(red)(cancel(color(black)("moles C"_3"H"_8)))) = "35 moles O"_2#</mathjax></p>
</blockquote>
<p>Since you have only <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas, you can say that oxygen gas will be the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed before all the moles of propane will get the chance to react. </p>
<p>So the reaction will consume <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas and</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_3"H"_8)/(5color(red)(cancel(color(black)("moles O"_2)))) = "1.4 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>In this case, you can say that the reaction can <strong>theoretically</strong> produce, i.e. what you would get at <mathjax>#100%#</mathjax> yield</p>
<blockquote>
<p><mathjax>#7 color(red)(cancel(color(black)("moles O"_2))) * "3 moles CO"_2/(5color(red)(cancel(color(black)("moles O"_2)))) = "4.2 moles CO"_2#</mathjax></p>
</blockquote>
<p>However, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#4.1#</mathjax> <strong>moles</strong> of carbon dioxide, so you can say that the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of the reaction was</p>
<blockquote>
<p><mathjax>#"% yield" = (4.0 color(red)(cancel(color(black)("moles CO"_2))))/(4.2color(red)(cancel(color(black)("moles CO"_2)))) * 100% = color(darkgreen)(ul(color(black)(95%)))#</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>#95%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <strong>theoretical yield</strong> of the reaction.</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>You know that the reaction consumes <mathjax>#5#</mathjax> <strong>moles</strong> of oxygen gas for every <mathjax>#1#</mathjax> <strong>mole</strong> of propane that takes part in the reaction. </p>
<p>This means that, in order for the reaction to consume <mathjax>#7#</mathjax> <strong>moles</strong> of propane, it must also consume</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * "5 moles O"_2/(1color(red)(cancel(color(black)("moles C"_3"H"_8)))) = "35 moles O"_2#</mathjax></p>
</blockquote>
<p>Since you have only <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas, you can say that oxygen gas will be the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed before all the moles of propane will get the chance to react. </p>
<p>So the reaction will consume <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas and</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_3"H"_8)/(5color(red)(cancel(color(black)("moles O"_2)))) = "1.4 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>In this case, you can say that the reaction can <strong>theoretically</strong> produce, i.e. what you would get at <mathjax>#100%#</mathjax> yield</p>
<blockquote>
<p><mathjax>#7 color(red)(cancel(color(black)("moles O"_2))) * "3 moles CO"_2/(5color(red)(cancel(color(black)("moles O"_2)))) = "4.2 moles CO"_2#</mathjax></p>
</blockquote>
<p>However, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#4.1#</mathjax> <strong>moles</strong> of carbon dioxide, so you can say that the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of the reaction was</p>
<blockquote>
<p><mathjax>#"% yield" = (4.0 color(red)(cancel(color(black)("moles CO"_2))))/(4.2color(red)(cancel(color(black)("moles CO"_2)))) * 100% = color(darkgreen)(ul(color(black)(95%)))#</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">If you start with #7.0# moles of propane and #7.0# moles of oxygen gas what is the percent yield if #4.0# moles of carbon dioxide are produced?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, <strong>_</strong> is the percent yield if 4.0 moles of carbon dioxide is produced. <br/>
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)</p></div>
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Feb 10, 2018
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<div class="markdown"><p><mathjax>#95%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <strong>theoretical yield</strong> of the reaction.</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> 3"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>You know that the reaction consumes <mathjax>#5#</mathjax> <strong>moles</strong> of oxygen gas for every <mathjax>#1#</mathjax> <strong>mole</strong> of propane that takes part in the reaction. </p>
<p>This means that, in order for the reaction to consume <mathjax>#7#</mathjax> <strong>moles</strong> of propane, it must also consume</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * "5 moles O"_2/(1color(red)(cancel(color(black)("moles C"_3"H"_8)))) = "35 moles O"_2#</mathjax></p>
</blockquote>
<p>Since you have only <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas, you can say that oxygen gas will be the <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed before all the moles of propane will get the chance to react. </p>
<p>So the reaction will consume <mathjax>#7#</mathjax> <strong>moles</strong> of oxygen gas and</p>
<blockquote>
<p><mathjax>#7.0 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_3"H"_8)/(5color(red)(cancel(color(black)("moles O"_2)))) = "1.4 moles C"_3"H"_8#</mathjax></p>
</blockquote>
<p>In this case, you can say that the reaction can <strong>theoretically</strong> produce, i.e. what you would get at <mathjax>#100%#</mathjax> yield</p>
<blockquote>
<p><mathjax>#7 color(red)(cancel(color(black)("moles O"_2))) * "3 moles CO"_2/(5color(red)(cancel(color(black)("moles O"_2)))) = "4.2 moles CO"_2#</mathjax></p>
</blockquote>
<p>However, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#4.1#</mathjax> <strong>moles</strong> of carbon dioxide, so you can say that the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> of the reaction was</p>
<blockquote>
<p><mathjax>#"% yield" = (4.0 color(red)(cancel(color(black)("moles CO"_2))))/(4.2color(red)(cancel(color(black)("moles CO"_2)))) * 100% = color(darkgreen)(ul(color(black)(95%)))#</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> | If you start with #7.0# moles of propane and #7.0# moles of oxygen gas what is the percent yield if #4.0# moles of carbon dioxide are produced? |
Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, _ is the percent yield if 4.0 moles of carbon dioxide is produced.
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
|
2,599 | a8668ac2-6ddd-11ea-83d2-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-85-m-nh-4cl-solution | 4.66 | start physical_unit 8 9 ph none qc_end physical_unit 8 9 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NH4Cl solution"}] | [{"type":"physical unit","value":"4.66"}] | [{"type":"physical unit","value":"Molarity [OF] NH4Cl solution [=] \\pu{0.85 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.85 M #NH_4Cl# solution?</h1> | null | 4.66 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonium chloride is a soluble salt that ionizes completely when dissolved in water.</p>
<p><mathjax>#" "NH_4Cl -> NH_4^++ Cl^-#</mathjax></p>
<p><mathjax>#Cl^(-)#</mathjax>is a spectator ion that does not affect the pH of the medium.</p>
<p>On the other hand, <mathjax>#NH_4^+#</mathjax>is the conjugate species of the weak base<mathjax>#\ NH_3\ #</mathjax> therefore, it is a weak acid stronger than water. It dissociates in water according to the following equation:</p>
<p><mathjax># " "NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+#</mathjax></p>
<p><mathjax>#(NH_4^+//NH_3)#</mathjax> are conjugate acid base pairs, and for conjugate acid base pairs:</p>
<p><mathjax>#" "K_axxK_b=K_w#</mathjax></p>
<p>Knowing the<mathjax>#\ K_b\ #</mathjax> value for <mathjax>#\ NH_3 (1.8xx10^-5)#</mathjax>, the <mathjax>#\ K_a\ #</mathjax> value for <mathjax>#NH_4^+ #</mathjax> is determined.</p>
<p><mathjax>#K_a =K_w/K_b#</mathjax></p>
<p><mathjax>#K_a =(1.00xx10^-14)/( 1.80xx10^-5)#</mathjax></p>
<p><mathjax>#K_a =5.56xx10^-10#</mathjax></p>
<p>Use the ICE table to figure out the concentration of each species present at equilibrium.</p>
<p><mathjax># " "ul(NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+)#</mathjax><br/>
<mathjax># \ "I "0.85 M" " - " " - #</mathjax><br/>
<mathjax>#"C " -x" " +x" "+x#</mathjax><br/>
<mathjax>#"E " ul( 0.85-x" "x" " x" "#</mathjax></p>
<p><mathjax>#K_a= ([NH_3]xx[ H_3O^+] )/([NH_4^+])#</mathjax></p>
<p><mathjax>#K_a= (x.x)/((0.85-x))=5.56xx10^-10#</mathjax></p>
<p><mathjax>#(0.85-x)~= 0.85#</mathjax> since <mathjax>#x#</mathjax> is too small compared to 0.85</p>
<p><mathjax>#(x^2)/(0.85)=5.56xx10^-10#</mathjax></p>
<p><mathjax>#x= sqrt(0.85xx5.56xx10^-10)#</mathjax></p>
<p><mathjax>#x~=2.2xx10^-5\ M#</mathjax></p>
<p><mathjax>#x=[H_3O^+]#</mathjax></p>
<p><mathjax>#pH=-log\ (2.2xx10^-5)#</mathjax></p>
<p><mathjax>#pH = 4.66#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH = 4.66#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonium chloride is a soluble salt that ionizes completely when dissolved in water.</p>
<p><mathjax>#" "NH_4Cl -> NH_4^++ Cl^-#</mathjax></p>
<p><mathjax>#Cl^(-)#</mathjax>is a spectator ion that does not affect the pH of the medium.</p>
<p>On the other hand, <mathjax>#NH_4^+#</mathjax>is the conjugate species of the weak base<mathjax>#\ NH_3\ #</mathjax> therefore, it is a weak acid stronger than water. It dissociates in water according to the following equation:</p>
<p><mathjax># " "NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+#</mathjax></p>
<p><mathjax>#(NH_4^+//NH_3)#</mathjax> are conjugate acid base pairs, and for conjugate acid base pairs:</p>
<p><mathjax>#" "K_axxK_b=K_w#</mathjax></p>
<p>Knowing the<mathjax>#\ K_b\ #</mathjax> value for <mathjax>#\ NH_3 (1.8xx10^-5)#</mathjax>, the <mathjax>#\ K_a\ #</mathjax> value for <mathjax>#NH_4^+ #</mathjax> is determined.</p>
<p><mathjax>#K_a =K_w/K_b#</mathjax></p>
<p><mathjax>#K_a =(1.00xx10^-14)/( 1.80xx10^-5)#</mathjax></p>
<p><mathjax>#K_a =5.56xx10^-10#</mathjax></p>
<p>Use the ICE table to figure out the concentration of each species present at equilibrium.</p>
<p><mathjax># " "ul(NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+)#</mathjax><br/>
<mathjax># \ "I "0.85 M" " - " " - #</mathjax><br/>
<mathjax>#"C " -x" " +x" "+x#</mathjax><br/>
<mathjax>#"E " ul( 0.85-x" "x" " x" "#</mathjax></p>
<p><mathjax>#K_a= ([NH_3]xx[ H_3O^+] )/([NH_4^+])#</mathjax></p>
<p><mathjax>#K_a= (x.x)/((0.85-x))=5.56xx10^-10#</mathjax></p>
<p><mathjax>#(0.85-x)~= 0.85#</mathjax> since <mathjax>#x#</mathjax> is too small compared to 0.85</p>
<p><mathjax>#(x^2)/(0.85)=5.56xx10^-10#</mathjax></p>
<p><mathjax>#x= sqrt(0.85xx5.56xx10^-10)#</mathjax></p>
<p><mathjax>#x~=2.2xx10^-5\ M#</mathjax></p>
<p><mathjax>#x=[H_3O^+]#</mathjax></p>
<p><mathjax>#pH=-log\ (2.2xx10^-5)#</mathjax></p>
<p><mathjax>#pH = 4.66#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.85 M #NH_4Cl# solution?</h1>
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<div class="markdown"><p><mathjax>#pH = 4.66#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonium chloride is a soluble salt that ionizes completely when dissolved in water.</p>
<p><mathjax>#" "NH_4Cl -> NH_4^++ Cl^-#</mathjax></p>
<p><mathjax>#Cl^(-)#</mathjax>is a spectator ion that does not affect the pH of the medium.</p>
<p>On the other hand, <mathjax>#NH_4^+#</mathjax>is the conjugate species of the weak base<mathjax>#\ NH_3\ #</mathjax> therefore, it is a weak acid stronger than water. It dissociates in water according to the following equation:</p>
<p><mathjax># " "NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+#</mathjax></p>
<p><mathjax>#(NH_4^+//NH_3)#</mathjax> are conjugate acid base pairs, and for conjugate acid base pairs:</p>
<p><mathjax>#" "K_axxK_b=K_w#</mathjax></p>
<p>Knowing the<mathjax>#\ K_b\ #</mathjax> value for <mathjax>#\ NH_3 (1.8xx10^-5)#</mathjax>, the <mathjax>#\ K_a\ #</mathjax> value for <mathjax>#NH_4^+ #</mathjax> is determined.</p>
<p><mathjax>#K_a =K_w/K_b#</mathjax></p>
<p><mathjax>#K_a =(1.00xx10^-14)/( 1.80xx10^-5)#</mathjax></p>
<p><mathjax>#K_a =5.56xx10^-10#</mathjax></p>
<p>Use the ICE table to figure out the concentration of each species present at equilibrium.</p>
<p><mathjax># " "ul(NH_4^+ + H_2O rightleftharpoons NH_3+ H_3O^+)#</mathjax><br/>
<mathjax># \ "I "0.85 M" " - " " - #</mathjax><br/>
<mathjax>#"C " -x" " +x" "+x#</mathjax><br/>
<mathjax>#"E " ul( 0.85-x" "x" " x" "#</mathjax></p>
<p><mathjax>#K_a= ([NH_3]xx[ H_3O^+] )/([NH_4^+])#</mathjax></p>
<p><mathjax>#K_a= (x.x)/((0.85-x))=5.56xx10^-10#</mathjax></p>
<p><mathjax>#(0.85-x)~= 0.85#</mathjax> since <mathjax>#x#</mathjax> is too small compared to 0.85</p>
<p><mathjax>#(x^2)/(0.85)=5.56xx10^-10#</mathjax></p>
<p><mathjax>#x= sqrt(0.85xx5.56xx10^-10)#</mathjax></p>
<p><mathjax>#x~=2.2xx10^-5\ M#</mathjax></p>
<p><mathjax>#x=[H_3O^+]#</mathjax></p>
<p><mathjax>#pH=-log\ (2.2xx10^-5)#</mathjax></p>
<p><mathjax>#pH = 4.66#</mathjax></p></div>
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