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1,900 | a9245e38-6ddd-11ea-92dd-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-each-compound-ch2-with-molar-mass-42-08-g-mol | C3H6 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 7 7 12 13 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C3H6"}] | [{"type":"other","value":"The empirical formula of the compound is CH2."},{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{42.08 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of each compound CH2 with molar mass 42.08 g/mol?</h1> | null | C3H6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest, whole number ratio defining constituent atoms in a species. The <mathjax>#"molecular formula"#</mathjax> is a whole number multiple of the <mathjax>#"empirical formula"#</mathjax>, and of course the multiple MIGHT be ONE. </p>
<p>And thus....<mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax></p>
<p>And we take the relevant masses...</p>
<p><mathjax>#[(2xx12.011+4xx1.00794)*g*mol^-1]xxn-=underbrace(42.08*g*mol^-1)_"given molecular mass"#</mathjax></p>
<p>And so we solve for <mathjax>#n#</mathjax> by the quotient....</p>
<p><mathjax>#n=(42.08*g*mol^-1)/[1xx12.011+2xx1.00794*g*mol^-1]=3#</mathjax></p>
<p>And so our molecular formula is <mathjax>#3xxCH_2-=C_3H_6#</mathjax>...this is propylene, or cyclopropane...</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Don't we got <mathjax>#C_3H_6#</mathjax>...I,e, probably propylene...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest, whole number ratio defining constituent atoms in a species. The <mathjax>#"molecular formula"#</mathjax> is a whole number multiple of the <mathjax>#"empirical formula"#</mathjax>, and of course the multiple MIGHT be ONE. </p>
<p>And thus....<mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax></p>
<p>And we take the relevant masses...</p>
<p><mathjax>#[(2xx12.011+4xx1.00794)*g*mol^-1]xxn-=underbrace(42.08*g*mol^-1)_"given molecular mass"#</mathjax></p>
<p>And so we solve for <mathjax>#n#</mathjax> by the quotient....</p>
<p><mathjax>#n=(42.08*g*mol^-1)/[1xx12.011+2xx1.00794*g*mol^-1]=3#</mathjax></p>
<p>And so our molecular formula is <mathjax>#3xxCH_2-=C_3H_6#</mathjax>...this is propylene, or cyclopropane...</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of each compound CH2 with molar mass 42.08 g/mol?</h1>
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<div class="markdown"><p>Don't we got <mathjax>#C_3H_6#</mathjax>...I,e, probably propylene...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest, whole number ratio defining constituent atoms in a species. The <mathjax>#"molecular formula"#</mathjax> is a whole number multiple of the <mathjax>#"empirical formula"#</mathjax>, and of course the multiple MIGHT be ONE. </p>
<p>And thus....<mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax></p>
<p>And we take the relevant masses...</p>
<p><mathjax>#[(2xx12.011+4xx1.00794)*g*mol^-1]xxn-=underbrace(42.08*g*mol^-1)_"given molecular mass"#</mathjax></p>
<p>And so we solve for <mathjax>#n#</mathjax> by the quotient....</p>
<p><mathjax>#n=(42.08*g*mol^-1)/[1xx12.011+2xx1.00794*g*mol^-1]=3#</mathjax></p>
<p>And so our molecular formula is <mathjax>#3xxCH_2-=C_3H_6#</mathjax>...this is propylene, or cyclopropane...</p></div>
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</article> | What is the molecular formula of each compound CH2 with molar mass 42.08 g/mol? | null |
1,901 | aa1033f4-6ddd-11ea-8a6c-ccda262736ce | https://socratic.org/questions/how-many-moles-of-cobalt-are-in-6-97-10-23-atoms-of-cobalt | 1.16 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] cobalt [IN] moles"}] | [{"type":"physical unit","value":"1.16 moles"}] | [{"type":"physical unit","value":"Number [OF] cobalt atoms [=] \\pu{6.97 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles of cobalt are in #6.97*10^23# atoms of cobalt?</h1> | null | 1.16 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Identify the relationship between 1 mol Co and Avogadro's number of atoms.</p>
<p><mathjax>#"1 mol Co"#</mathjax><mathjax>#=#</mathjax><mathjax>#6.022xx10^23"atoms Co"#</mathjax></p>
<p>Divide the given atoms of cobalt by the number of atoms per mol.</p>
<p><mathjax>#6.97xx10^23cancel"atoms Co"xx("1 mol Co")/(6.022xx10^23cancel"atoms Co")="1.16 mol Co"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>There are <mathjax>#"1.16 mol Co"#</mathjax> in <mathjax>#6.97xx10^23#</mathjax> atoms of Co.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Identify the relationship between 1 mol Co and Avogadro's number of atoms.</p>
<p><mathjax>#"1 mol Co"#</mathjax><mathjax>#=#</mathjax><mathjax>#6.022xx10^23"atoms Co"#</mathjax></p>
<p>Divide the given atoms of cobalt by the number of atoms per mol.</p>
<p><mathjax>#6.97xx10^23cancel"atoms Co"xx("1 mol Co")/(6.022xx10^23cancel"atoms Co")="1.16 mol Co"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of cobalt are in #6.97*10^23# atoms of cobalt?</h1>
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<div class="markdown"><p>There are <mathjax>#"1.16 mol Co"#</mathjax> in <mathjax>#6.97xx10^23#</mathjax> atoms of Co.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Identify the relationship between 1 mol Co and Avogadro's number of atoms.</p>
<p><mathjax>#"1 mol Co"#</mathjax><mathjax>#=#</mathjax><mathjax>#6.022xx10^23"atoms Co"#</mathjax></p>
<p>Divide the given atoms of cobalt by the number of atoms per mol.</p>
<p><mathjax>#6.97xx10^23cancel"atoms Co"xx("1 mol Co")/(6.022xx10^23cancel"atoms Co")="1.16 mol Co"#</mathjax></p></div>
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</article> | How many moles of cobalt are in #6.97*10^23# atoms of cobalt? | null |
1,902 | ab0e26c0-6ddd-11ea-b8cd-ccda262736ce | https://socratic.org/questions/how-much-gallons-of-deionized-water-contains-2-000-000-hydrogen-atoms-please-sol | 7.90 × 10^(-21) gallons | start physical_unit 4 5 volume gallon qc_end physical_unit 8 9 7 7 number qc_end end | [{"type":"physical unit","value":"Volume [OF] deionized water [IN] gallons"}] | [{"type":"physical unit","value":"7.90 × 10^(-21) gallons"}] | [{"type":"physical unit","value":"Number [OF] hydrogen atoms [=] \\pu{2,000,000}"}] | <h1 class="questionTitle" itemprop="name">How much gallons of deionized water contains 2,000,000 hydrogen atoms?? Please solve. Thanks a lot. </h1> | null | 7.90 × 10^(-21) gallons | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And I use <mathjax>#"US gallon"#</mathjax> because even here in the UK we have moved away (thankfully) from the gallon as a unit of volume......(NB <mathjax>#"1 Imperial Gallon"-=4546*mL#</mathjax>). Maybe now that we are leaving the EU, there will be a move to reinstate pounds, shillings, and pence, and tons, and pounds, and gallons, and pints and gills, as the units of measure. </p>
<p>And if there are <mathjax>#2xx10^6*"hydrogen atoms"#</mathjax>, there are <mathjax>#1xx10^6#</mathjax> individual water molecules.....</p>
<p>And thus there are........</p>
<p><mathjax>#(1xx10^6*"water molecules")/(6.022xx10^23*"water molecules"*mol^-1)xx18.01*g*mol^-1=#</mathjax></p>
<p><mathjax>#3.00xx10^-17*g-=3.00xx10^-17*mL-=7.90xx10^-21*"US gallons."#</mathjax> </p>
<p>This measurement is almost meaningless, and it is really only an exercise. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Not that many gallons, and first we specify <mathjax>#"1 US gallon"-=3785.4*mL#</mathjax>...........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And I use <mathjax>#"US gallon"#</mathjax> because even here in the UK we have moved away (thankfully) from the gallon as a unit of volume......(NB <mathjax>#"1 Imperial Gallon"-=4546*mL#</mathjax>). Maybe now that we are leaving the EU, there will be a move to reinstate pounds, shillings, and pence, and tons, and pounds, and gallons, and pints and gills, as the units of measure. </p>
<p>And if there are <mathjax>#2xx10^6*"hydrogen atoms"#</mathjax>, there are <mathjax>#1xx10^6#</mathjax> individual water molecules.....</p>
<p>And thus there are........</p>
<p><mathjax>#(1xx10^6*"water molecules")/(6.022xx10^23*"water molecules"*mol^-1)xx18.01*g*mol^-1=#</mathjax></p>
<p><mathjax>#3.00xx10^-17*g-=3.00xx10^-17*mL-=7.90xx10^-21*"US gallons."#</mathjax> </p>
<p>This measurement is almost meaningless, and it is really only an exercise. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How much gallons of deionized water contains 2,000,000 hydrogen atoms?? Please solve. Thanks a lot. </h1>
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<span class="dateCreated" datetime="2017-09-16T11:53:05" itemprop="dateCreated">
Sep 16, 2017
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<div class="markdown"><p>Not that many gallons, and first we specify <mathjax>#"1 US gallon"-=3785.4*mL#</mathjax>...........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And I use <mathjax>#"US gallon"#</mathjax> because even here in the UK we have moved away (thankfully) from the gallon as a unit of volume......(NB <mathjax>#"1 Imperial Gallon"-=4546*mL#</mathjax>). Maybe now that we are leaving the EU, there will be a move to reinstate pounds, shillings, and pence, and tons, and pounds, and gallons, and pints and gills, as the units of measure. </p>
<p>And if there are <mathjax>#2xx10^6*"hydrogen atoms"#</mathjax>, there are <mathjax>#1xx10^6#</mathjax> individual water molecules.....</p>
<p>And thus there are........</p>
<p><mathjax>#(1xx10^6*"water molecules")/(6.022xx10^23*"water molecules"*mol^-1)xx18.01*g*mol^-1=#</mathjax></p>
<p><mathjax>#3.00xx10^-17*g-=3.00xx10^-17*mL-=7.90xx10^-21*"US gallons."#</mathjax> </p>
<p>This measurement is almost meaningless, and it is really only an exercise. </p></div>
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</article> | How much gallons of deionized water contains 2,000,000 hydrogen atoms?? Please solve. Thanks a lot. | null |
1,903 | aafaa834-6ddd-11ea-b3a5-ccda262736ce | https://socratic.org/questions/if-10-ml-of-a-9-g-l-stock-solution-of-ferric-chloride-are-added-to-a-hach-contai | 18 mg/L | start physical_unit 10 11 concentration mg/l qc_end physical_unit 8 8 1 2 volume qc_end physical_unit 23 23 20 21 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] ferric chloride [IN] mg/L"}] | [{"type":"physical unit","value":"18 mg/L"}] | [{"type":"physical unit","value":"Volume1 [OF] ferric chloride stock solution [=] \\pu{10 ml}"},{"type":"physical unit","value":"Concentration1 [OF] ferric chloride stock solution [=] \\pu{9 g/L}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{5 L}"}] | <h1 class="questionTitle" itemprop="name">If 10 ml of a 9 g/L stock solution of ferric chloride are added to a Hach container that has 5 L of water, what will be the concentration of ferric chloride in the water?</h1> | null | 18 mg/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the dilution equation to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#c_1#</mathjax> and <mathjax>#c_2#</mathjax> are the initial and final concentrations, and<br/>
<mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes.</p>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#c_1 = "9 g/L"; V_1 = "10 mL" color(white)(mmmll) = "0.010 L"#</mathjax><br/>
<mathjax>#c_2 = ?;color(white)(mm) V_2 = "5 L + 0.010 L" = "5.010 L"#</mathjax></p>
<p>∴ <mathjax>#c_2 = "9 g/L" × (0.010 color(red)(cancel(color(black)("L"))))/(5.010 color(red)(cancel(color(black)("L")))) = "0.018 g/L" = "18 mg/L"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The new concentration will be 18 mg/L.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the dilution equation to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#c_1#</mathjax> and <mathjax>#c_2#</mathjax> are the initial and final concentrations, and<br/>
<mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes.</p>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#c_1 = "9 g/L"; V_1 = "10 mL" color(white)(mmmll) = "0.010 L"#</mathjax><br/>
<mathjax>#c_2 = ?;color(white)(mm) V_2 = "5 L + 0.010 L" = "5.010 L"#</mathjax></p>
<p>∴ <mathjax>#c_2 = "9 g/L" × (0.010 color(red)(cancel(color(black)("L"))))/(5.010 color(red)(cancel(color(black)("L")))) = "0.018 g/L" = "18 mg/L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If 10 ml of a 9 g/L stock solution of ferric chloride are added to a Hach container that has 5 L of water, what will be the concentration of ferric chloride in the water?</h1>
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<div class="markdown"><p>The new concentration will be 18 mg/L.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the dilution equation to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#c_1#</mathjax> and <mathjax>#c_2#</mathjax> are the initial and final concentrations, and<br/>
<mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes.</p>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#c_1 = "9 g/L"; V_1 = "10 mL" color(white)(mmmll) = "0.010 L"#</mathjax><br/>
<mathjax>#c_2 = ?;color(white)(mm) V_2 = "5 L + 0.010 L" = "5.010 L"#</mathjax></p>
<p>∴ <mathjax>#c_2 = "9 g/L" × (0.010 color(red)(cancel(color(black)("L"))))/(5.010 color(red)(cancel(color(black)("L")))) = "0.018 g/L" = "18 mg/L"#</mathjax></p></div>
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</article> | If 10 ml of a 9 g/L stock solution of ferric chloride are added to a Hach container that has 5 L of water, what will be the concentration of ferric chloride in the water? | null |
1,904 | a86072f8-6ddd-11ea-be0d-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-nh3-h2so4-nh4-2so4 | 2 NH3 + H2SO4 -> (NH4)2SO4 | start chemical_equation qc_end chemical_equation 7 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 NH3 + H2SO4 -> (NH4)2SO4"}] | [{"type":"chemical equation","value":"NH3 + H2SO4 -> (NH4)2SO4"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation: NH3 + H2SO4 --> (NH4)2SO4?
</h1> | null | 2 NH3 + H2SO4 -> (NH4)2SO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance the below reaction you will need to multiply the <mathjax>#NH_3#</mathjax> by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)NH_3+H_2SO_4->(NH_4)_2SO_4#</mathjax></p>
<p>Then you will have:</p>
<p><mathjax>#2N#</mathjax> to the left and <mathjax>#2N#</mathjax> to the right.</p>
<p><mathjax>#8H=(2xxcolor(blue)(3)) " from " 2NH_(color(blue)(3)) +color(red)(2)" from " H_(color(red)(2))SO_4#</mathjax> to the left and <mathjax>#8H=2xxcolor(green)(4) " from " (NH_(color(green)(4)))_2SO_4#</mathjax> to the right.</p>
<p><mathjax>#SO_4#</mathjax> to the left and <mathjax>#SO_4#</mathjax> to the right.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Multiply the <mathjax>#NH_3#</mathjax> by <mathjax>#2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance the below reaction you will need to multiply the <mathjax>#NH_3#</mathjax> by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)NH_3+H_2SO_4->(NH_4)_2SO_4#</mathjax></p>
<p>Then you will have:</p>
<p><mathjax>#2N#</mathjax> to the left and <mathjax>#2N#</mathjax> to the right.</p>
<p><mathjax>#8H=(2xxcolor(blue)(3)) " from " 2NH_(color(blue)(3)) +color(red)(2)" from " H_(color(red)(2))SO_4#</mathjax> to the left and <mathjax>#8H=2xxcolor(green)(4) " from " (NH_(color(green)(4)))_2SO_4#</mathjax> to the right.</p>
<p><mathjax>#SO_4#</mathjax> to the left and <mathjax>#SO_4#</mathjax> to the right.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you balance the following equation: NH3 + H2SO4 --> (NH4)2SO4?
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Dr. Hayek
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Dec 21, 2015
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<div class="markdown"><p>Multiply the <mathjax>#NH_3#</mathjax> by <mathjax>#2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance the below reaction you will need to multiply the <mathjax>#NH_3#</mathjax> by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)NH_3+H_2SO_4->(NH_4)_2SO_4#</mathjax></p>
<p>Then you will have:</p>
<p><mathjax>#2N#</mathjax> to the left and <mathjax>#2N#</mathjax> to the right.</p>
<p><mathjax>#8H=(2xxcolor(blue)(3)) " from " 2NH_(color(blue)(3)) +color(red)(2)" from " H_(color(red)(2))SO_4#</mathjax> to the left and <mathjax>#8H=2xxcolor(green)(4) " from " (NH_(color(green)(4)))_2SO_4#</mathjax> to the right.</p>
<p><mathjax>#SO_4#</mathjax> to the left and <mathjax>#SO_4#</mathjax> to the right.</p></div>
</div>
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</article> | How would you balance the following equation: NH3 + H2SO4 --> (NH4)2SO4?
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1,905 | a8e138d4-6ddd-11ea-968c-ccda262736ce | https://socratic.org/questions/how-many-moles-of-hcl-are-required-to-completely-neutralize-0-30-moles-of-ca-oh- | 0.60 moles | start physical_unit 4 4 mole mol qc_end physical_unit 13 13 10 11 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] HCl [IN] moles"}] | [{"type":"physical unit","value":"0.60 moles"}] | [{"type":"physical unit","value":"Mole [OF] Ca(OH)2 [=] \\pu{0.30 moles}"},{"type":"other","value":"Completely neutralize."}] | <h1 class="questionTitle" itemprop="name">How many moles of HCl are required to completely neutralize 0.30 moles of #Ca(OH)_2#?</h1> | null | 0.60 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know?</p>
<p>Well, we interrogate the stoichiometric equation...</p>
<p><mathjax>#Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)#</mathjax></p>
<p>Two equiv of <mathjax>#HCl#</mathjax> are required per equiv <mathjax>#"calcium hydroxide"#</mathjax>..</p>
<p>What number of <mathjax>#HCl#</mathjax> molecules are required for equivalence?</p></div>
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<div class="markdown"><p>Why <mathjax>#0.60*mol#</mathjax>....</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know?</p>
<p>Well, we interrogate the stoichiometric equation...</p>
<p><mathjax>#Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)#</mathjax></p>
<p>Two equiv of <mathjax>#HCl#</mathjax> are required per equiv <mathjax>#"calcium hydroxide"#</mathjax>..</p>
<p>What number of <mathjax>#HCl#</mathjax> molecules are required for equivalence?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of HCl are required to completely neutralize 0.30 moles of #Ca(OH)_2#?</h1>
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<div class="markdown"><p>Why <mathjax>#0.60*mol#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how do we know?</p>
<p>Well, we interrogate the stoichiometric equation...</p>
<p><mathjax>#Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)#</mathjax></p>
<p>Two equiv of <mathjax>#HCl#</mathjax> are required per equiv <mathjax>#"calcium hydroxide"#</mathjax>..</p>
<p>What number of <mathjax>#HCl#</mathjax> molecules are required for equivalence?</p></div>
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</article> | How many moles of HCl are required to completely neutralize 0.30 moles of #Ca(OH)_2#? | null |
1,906 | ab91c258-6ddd-11ea-aa0d-ccda262736ce | https://socratic.org/questions/how-many-moles-of-gas-are-in-a-volume-of-48-6-l-at-stp | 2.17 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 10 11 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mole [OF] gas [IN] moles"}] | [{"type":"physical unit","value":"2.17 moles"}] | [{"type":"physical unit","value":"Volume [OF] gas [=] \\pu{48.6 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many moles of gas are in a volume of 48.6 L at STP?</h1> | null | 2.17 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP, so we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>.<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx48.6cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 2.17 mol#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>2.17 moles of gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP, so we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>.<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx48.6cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 2.17 mol#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of gas are in a volume of 48.6 L at STP?</h1>
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<div class="markdown"><p>2.17 moles of gas.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are at STP, so we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>.<br/>
<mathjax>#PxxV = nxxRxxT#</mathjax>. </p>
<ul>
<li>P represents pressure (could have units of atm, depending on the units of the universal gas constant)</li>
<li>V represents volume (must have units of liters)</li>
<li>n represents the number of moles</li>
<li>R is the universal gas constant (has units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax>)</li>
<li>T represents the temperature, which must be in Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the number of moles. Our known variables are P,V,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></p>
<p>Now we have to rearrange the equation to solve for n:</p>
<p><mathjax># n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (1cancel"atm"xx48.6cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#</mathjax><br/>
<mathjax>#n = 2.17 mol#</mathjax></p></div>
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</article> | How many moles of gas are in a volume of 48.6 L at STP? | null |
1,907 | ac2f7ced-6ddd-11ea-ab74-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-20-0-ml-of-nitrogen-gas | 0.02 grams | start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] nitrogen gas [IN] grams"}] | [{"type":"physical unit","value":"0.02 grams"}] | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [=] \\pu{20.0 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 20.0 mL of nitrogen gas?</h1> | null | 0.02 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we solve for <mathjax>#n#</mathjax> in the old Ideal Gas equation....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx20.0*mLxx10^-3*L*mL^-1)/(0.0821*(L*atm)/(K*mol)*298*K)#</mathjax></p>
<p>And I make this about <mathjax>#8xx10^-4*mol#</mathjax> with respect to dinitrogen...</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well we must specify a pressure of <mathjax>#1*atm#</mathjax> and a temperature of <mathjax>#298*K#</mathjax>...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we solve for <mathjax>#n#</mathjax> in the old Ideal Gas equation....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx20.0*mLxx10^-3*L*mL^-1)/(0.0821*(L*atm)/(K*mol)*298*K)#</mathjax></p>
<p>And I make this about <mathjax>#8xx10^-4*mol#</mathjax> with respect to dinitrogen...</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 20.0 mL of nitrogen gas?</h1>
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<div class="markdown"><p>Well we must specify a pressure of <mathjax>#1*atm#</mathjax> and a temperature of <mathjax>#298*K#</mathjax>...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And we solve for <mathjax>#n#</mathjax> in the old Ideal Gas equation....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx20.0*mLxx10^-3*L*mL^-1)/(0.0821*(L*atm)/(K*mol)*298*K)#</mathjax></p>
<p>And I make this about <mathjax>#8xx10^-4*mol#</mathjax> with respect to dinitrogen...</p></div>
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</article> | What is the mass of 20.0 mL of nitrogen gas? | null |
1,908 | ab4e7edc-6ddd-11ea-8e96-ccda262736ce | https://socratic.org/questions/what-would-be-the-final-temperature-if-3-31-10-3-joules-of-heat-were-added-to-18 | 64.60 ℃ | start physical_unit 19 19 temperature °c qc_end physical_unit 19 19 7 10 heat_energy qc_end physical_unit 19 19 21 22 temperature qc_end physical_unit 19 19 16 17 mass qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"64.60 ℃"}] | [{"type":"physical unit","value":"Added heat [OF] water [=] \\pu{3.31 × 10^3 joules}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{22.0 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{18.5 grams}"}] | <h1 class="questionTitle" itemprop="name">What would be the final temperature if #3.31 * 10^3# joules of heat were added to 18.5 grams of water at 22.0°C?</h1> | null | 64.60 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can try using the general relationship between heat <mathjax>#Q#</mathjax> (lost or gained) and change in temperature <mathjax>#DeltaT#</mathjax> of a mass <mathjax>#m#</mathjax> of a substance of specific heat <mathjax>#c#</mathjax> (in our case water: <mathjax>#c_("water")=4.2J/(g^@C)#</mathjax>):</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with our data:<br/>
<mathjax>#3.31xx10^3=18.5*4.2*(T_f-22)#</mathjax><br/>
rearranging we get:<br/>
<mathjax>#T_f=64.6^@C#</mathjax></p></div>
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<div class="markdown"><p>I found: <mathjax>#T_f=64.6^@C#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can try using the general relationship between heat <mathjax>#Q#</mathjax> (lost or gained) and change in temperature <mathjax>#DeltaT#</mathjax> of a mass <mathjax>#m#</mathjax> of a substance of specific heat <mathjax>#c#</mathjax> (in our case water: <mathjax>#c_("water")=4.2J/(g^@C)#</mathjax>):</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with our data:<br/>
<mathjax>#3.31xx10^3=18.5*4.2*(T_f-22)#</mathjax><br/>
rearranging we get:<br/>
<mathjax>#T_f=64.6^@C#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What would be the final temperature if #3.31 * 10^3# joules of heat were added to 18.5 grams of water at 22.0°C?</h1>
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<div class="markdown"><p>I found: <mathjax>#T_f=64.6^@C#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can try using the general relationship between heat <mathjax>#Q#</mathjax> (lost or gained) and change in temperature <mathjax>#DeltaT#</mathjax> of a mass <mathjax>#m#</mathjax> of a substance of specific heat <mathjax>#c#</mathjax> (in our case water: <mathjax>#c_("water")=4.2J/(g^@C)#</mathjax>):</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with our data:<br/>
<mathjax>#3.31xx10^3=18.5*4.2*(T_f-22)#</mathjax><br/>
rearranging we get:<br/>
<mathjax>#T_f=64.6^@C#</mathjax></p></div>
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</article> | What would be the final temperature if #3.31 * 10^3# joules of heat were added to 18.5 grams of water at 22.0°C? | null |
1,909 | aab3b2d4-6ddd-11ea-b134-ccda262736ce | https://socratic.org/questions/59a65e9a11ef6b7acf25f722 | 82.24% | start physical_unit 8 8 mass_percent none qc_end substance 10 10 qc_end end | [{"type":"physical unit","value":"Percentage composition by mass [OF] nitrogen"}] | [{"type":"physical unit","value":"82.24%"}] | [{"type":"substance name","value":"Ammonia"}] | <h1 class="questionTitle" itemprop="name">What is the percentage composition by mass of nitrogen in ammonia?</h1> | null | 82.24% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus taking molar masses......we get:</p>
<p><mathjax>#(14.01*g*mol^-1)/(14.01*g*mol^-1+3xx1.00794*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#~=80%#</mathjax></p></div>
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<div class="markdown"><p>Well it is simply <mathjax>#"Mass of nitrogen"/"Mass of nitrogen and hydrogen"xx100%#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus taking molar masses......we get:</p>
<p><mathjax>#(14.01*g*mol^-1)/(14.01*g*mol^-1+3xx1.00794*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#~=80%#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the percentage composition by mass of nitrogen in ammonia?</h1>
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<div class="markdown"><p>Well it is simply <mathjax>#"Mass of nitrogen"/"Mass of nitrogen and hydrogen"xx100%#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus taking molar masses......we get:</p>
<p><mathjax>#(14.01*g*mol^-1)/(14.01*g*mol^-1+3xx1.00794*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#~=80%#</mathjax></p></div>
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<div class="markdown"><p>The mass percent of <mathjax>#"N"#</mathjax> in <mathjax>#"NH"_3#</mathjax> is <mathjax>#82.244%#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Divide the mass of <mathjax>#"N"#</mathjax> by the mass of <mathjax>#"NH"_3#</mathjax>, then multiply by <mathjax>#100#</mathjax></p>
<p>The chemical formula for ammonia is <mathjax>#"NH"_3#</mathjax>. The formula indicates that in each mole of <mathjax>#"NH"_3#</mathjax>, there is one mole of <mathjax>#"N"#</mathjax> atoms.</p>
<p>The molar mass of ammonia is <mathjax>#"17.031 g/mol"#</mathjax>. The molar mass of <mathjax>#"N"#</mathjax> is <mathjax>#"14.007 g/mol"#</mathjax>.</p>
<p><mathjax>#"mass percent N"#</mathjax><mathjax>#=#</mathjax><mathjax>#(14.007color(red)cancel(color(black)("g/mol")))/(17.031color(red)cancel(color(black)("g/mol")))xx100="82.244%"#</mathjax></p></div>
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</article> | What is the percentage composition by mass of nitrogen in ammonia? | null |
1,910 | ac33019b-6ddd-11ea-be25-ccda262736ce | https://socratic.org/questions/if-a-sample-of-n-gas-was-put-into-a-flexible-container-at-an-pressure-of-25-5kpa | 705.35 ℃ | start physical_unit 34 35 temperature °c qc_end physical_unit 34 35 43 44 pressure qc_end physical_unit 34 35 51 52 volume qc_end physical_unit 2 5 26 27 volume qc_end physical_unit 2 5 16 17 pressure qc_end physical_unit 2 5 20 21 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] ℃"}] | [{"type":"physical unit","value":"705.35 ℃"}] | [{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{0.625 atm}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{625 mL}"},{"type":"physical unit","value":"Volume1 [OF] N2 gas sample [=] \\pu{500 mL}"},{"type":"physical unit","value":"Pressure1 [OF] N2 gas sample [=] \\pu{25.5 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] N2 gas sample [=] \\pu{35.9 ℃}"}] | <h1 class="questionTitle" itemprop="name">If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL? </h1> | null | 705.35 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First look at the things that need to be converted</p>
<p><mathjax>#35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"#</mathjax></p>
<p><mathjax>#"25.5 kPa = 0.2467 atm"#</mathjax></p>
<p><mathjax>#"500 mL = 0.500 L"#</mathjax></p>
<p><mathjax>#"625 mL = 0.625 L"#</mathjax> </p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>Plug in the variables</p>
<p><mathjax>#("0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L")/T_2#</mathjax></p>
<p><mathjax>#T_2 = 978.5K#</mathjax></p>
<p>or <mathjax>#T_2 = (978.5 - 273.15) ^@"C" = 705 ^@"C"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#705 ^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First look at the things that need to be converted</p>
<p><mathjax>#35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"#</mathjax></p>
<p><mathjax>#"25.5 kPa = 0.2467 atm"#</mathjax></p>
<p><mathjax>#"500 mL = 0.500 L"#</mathjax></p>
<p><mathjax>#"625 mL = 0.625 L"#</mathjax> </p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>Plug in the variables</p>
<p><mathjax>#("0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L")/T_2#</mathjax></p>
<p><mathjax>#T_2 = 978.5K#</mathjax></p>
<p>or <mathjax>#T_2 = (978.5 - 273.15) ^@"C" = 705 ^@"C"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL? </h1>
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Mar 5, 2017
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<div class="markdown"><p><mathjax>#705 ^@"C"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First look at the things that need to be converted</p>
<p><mathjax>#35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"#</mathjax></p>
<p><mathjax>#"25.5 kPa = 0.2467 atm"#</mathjax></p>
<p><mathjax>#"500 mL = 0.500 L"#</mathjax></p>
<p><mathjax>#"625 mL = 0.625 L"#</mathjax> </p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>Plug in the variables</p>
<p><mathjax>#("0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L")/T_2#</mathjax></p>
<p><mathjax>#T_2 = 978.5K#</mathjax></p>
<p>or <mathjax>#T_2 = (978.5 - 273.15) ^@"C" = 705 ^@"C"#</mathjax></p></div>
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</article> | If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL? | null |
1,911 | aaa2601e-6ddd-11ea-a584-ccda262736ce | https://socratic.org/questions/a-1-50-l-buffer-solution-is-0-250-m-hf-and-0-250-m-naf-what-is-the-ph-of-the-sol | 3.57 | start physical_unit 18 19 ph none qc_end physical_unit 3 4 1 2 volume qc_end physical_unit 8 8 6 7 molarity qc_end physical_unit 12 12 6 7 molarity qc_end physical_unit 27 28 24 25 mole qc_end physical_unit 8 8 37 39 ka qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"3.57"}] | [{"type":"physical unit","value":"Volume [OF] buffer solution [=] \\pu{1.50 L}"},{"type":"physical unit","value":"Molarity [OF] HF [=] \\pu{0.250 M}"},{"type":"physical unit","value":"Molarity [OF] NaF [=] \\pu{0.250 M}"},{"type":"physical unit","value":"Mole [OF] solid NaOH [=] \\pu{0.0500 moles}"},{"type":"physical unit","value":"Ka [OF] HF [=] \\pu{3.5 × 10^(-4)}"},{"type":"other","value":"Assume no volume change."}] | <h1 class="questionTitle" itemprop="name">A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change.
Ka for HF is 3.5x10^-4
</h1> | null | 3.57 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your buffer contains <em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, weak acid, and <em>sodium fluoride</em>, <mathjax>#"NaF"#</mathjax>, the salt of its conjugate base, the fluoride anion, <mathjax>#"F"^(-)#</mathjax>.</p>
<p>When the sodium hydroxide solution is added, assuming with no change in the total volume of the buffer, you can expect the weak acid and the <strong>strong base</strong> to <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> each other.</p>
<p>Moreover, this reaction will result in the production of more <em>conjugate base</em>. </p>
<p>So, the balanced chemical equation for the reaction between hydrofluoric acid and sodium hydroxide looks like this </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "NaOH"_text((aq]) -> "NaF"_text((aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrofluoric acid will react with <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide and produce <mathjax>#1#</mathjax> <strong>mole</strong> of sodium fluoride. </p>
<p>Use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrofluoric acid and the volume of the buffer to determine how many moles you have in solution</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HF) = "0.250 M" * "1.5 L" = "0.375 moles HF"#</mathjax></p>
</blockquote>
<p>Do the same for the conjugate base</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.250 M" * "1.5 L" = "0.375 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Now, you're adding <mathjax>#0.0500#</mathjax> moles of sodium hydroxide to the buffer. Since you have <strong>fewer moles</strong> of strong base than on weak acid, it follows that the sodium hydroxide will be <strong>completely consumed</strong>. </p>
<p>The number of moles of hydrofluoric acid will be changed to </p>
<blockquote>
<p><mathjax>#n_(HF) = "0.375 moles" = "0.0500 moles" = "0.325 moles"#</mathjax></p>
</blockquote>
<p>The number of moles of fluoride anions will <strong>increase</strong> by the same amount</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.375 moles" + "0.0500 moles" = "0.425 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Use the volume of the buffer to calculate the new molarities of the weak acid and its conjugate base</p>
<blockquote>
<p><mathjax>#["HF"] = "0.325 moles"/"1.5 L" = "0.21667 M"#</mathjax></p>
<p><mathjax>#["F"^(-)] = "0.425 moles"/"1.5 L" = "0.28333 M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Use the acid dissociation constant, <mathjax>#K_a#</mathjax>, to get the value of <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#pK_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>The pH of the solution will thus be </p>
<blockquote>
<p><mathjax>#"pH" = -log(K_a) + log( (["F"^(-)])/(["HF"]))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5 * 10^(-4)) + log( (0.28333color(red)(cancel(color(black)("M"))))/(0.21667color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = color(green)(3.57)#</mathjax></p>
</blockquote>
<p>Finally, does this result make sense?</p>
<p>Notice that your starting buffer had <strong>equal concentrations</strong> of weak acid and conjugate base. This means that the H-H equation can be reduced to</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log(1) = pK_a#</mathjax></p>
</blockquote>
<p>Initially, the pH of the solution was equal to <mathjax>#3.46#</mathjax>. </p>
<p>The buffer then <em>converts</em> the strong base to weak base, and so the concentration of conjugate base is <strong>bigger</strong> than that of the weak acid. This is why the pH of the buffer <strong>increases</strong> after the addition of the strong base. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 3.57#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your buffer contains <em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, weak acid, and <em>sodium fluoride</em>, <mathjax>#"NaF"#</mathjax>, the salt of its conjugate base, the fluoride anion, <mathjax>#"F"^(-)#</mathjax>.</p>
<p>When the sodium hydroxide solution is added, assuming with no change in the total volume of the buffer, you can expect the weak acid and the <strong>strong base</strong> to <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> each other.</p>
<p>Moreover, this reaction will result in the production of more <em>conjugate base</em>. </p>
<p>So, the balanced chemical equation for the reaction between hydrofluoric acid and sodium hydroxide looks like this </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "NaOH"_text((aq]) -> "NaF"_text((aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrofluoric acid will react with <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide and produce <mathjax>#1#</mathjax> <strong>mole</strong> of sodium fluoride. </p>
<p>Use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrofluoric acid and the volume of the buffer to determine how many moles you have in solution</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HF) = "0.250 M" * "1.5 L" = "0.375 moles HF"#</mathjax></p>
</blockquote>
<p>Do the same for the conjugate base</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.250 M" * "1.5 L" = "0.375 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Now, you're adding <mathjax>#0.0500#</mathjax> moles of sodium hydroxide to the buffer. Since you have <strong>fewer moles</strong> of strong base than on weak acid, it follows that the sodium hydroxide will be <strong>completely consumed</strong>. </p>
<p>The number of moles of hydrofluoric acid will be changed to </p>
<blockquote>
<p><mathjax>#n_(HF) = "0.375 moles" = "0.0500 moles" = "0.325 moles"#</mathjax></p>
</blockquote>
<p>The number of moles of fluoride anions will <strong>increase</strong> by the same amount</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.375 moles" + "0.0500 moles" = "0.425 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Use the volume of the buffer to calculate the new molarities of the weak acid and its conjugate base</p>
<blockquote>
<p><mathjax>#["HF"] = "0.325 moles"/"1.5 L" = "0.21667 M"#</mathjax></p>
<p><mathjax>#["F"^(-)] = "0.425 moles"/"1.5 L" = "0.28333 M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Use the acid dissociation constant, <mathjax>#K_a#</mathjax>, to get the value of <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#pK_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>The pH of the solution will thus be </p>
<blockquote>
<p><mathjax>#"pH" = -log(K_a) + log( (["F"^(-)])/(["HF"]))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5 * 10^(-4)) + log( (0.28333color(red)(cancel(color(black)("M"))))/(0.21667color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = color(green)(3.57)#</mathjax></p>
</blockquote>
<p>Finally, does this result make sense?</p>
<p>Notice that your starting buffer had <strong>equal concentrations</strong> of weak acid and conjugate base. This means that the H-H equation can be reduced to</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log(1) = pK_a#</mathjax></p>
</blockquote>
<p>Initially, the pH of the solution was equal to <mathjax>#3.46#</mathjax>. </p>
<p>The buffer then <em>converts</em> the strong base to weak base, and so the concentration of conjugate base is <strong>bigger</strong> than that of the weak acid. This is why the pH of the buffer <strong>increases</strong> after the addition of the strong base. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change.
Ka for HF is 3.5x10^-4
</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 3.57#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your buffer contains <em>hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, weak acid, and <em>sodium fluoride</em>, <mathjax>#"NaF"#</mathjax>, the salt of its conjugate base, the fluoride anion, <mathjax>#"F"^(-)#</mathjax>.</p>
<p>When the sodium hydroxide solution is added, assuming with no change in the total volume of the buffer, you can expect the weak acid and the <strong>strong base</strong> to <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> each other.</p>
<p>Moreover, this reaction will result in the production of more <em>conjugate base</em>. </p>
<p>So, the balanced chemical equation for the reaction between hydrofluoric acid and sodium hydroxide looks like this </p>
<blockquote>
<p><mathjax>#"HF"_text((aq]) + "NaOH"_text((aq]) -> "NaF"_text((aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that <mathjax>#1#</mathjax> <strong>mole</strong> of hydrofluoric acid will react with <mathjax>#1#</mathjax> <strong>mole</strong> of sodium hydroxide and produce <mathjax>#1#</mathjax> <strong>mole</strong> of sodium fluoride. </p>
<p>Use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrofluoric acid and the volume of the buffer to determine how many moles you have in solution</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(HF) = "0.250 M" * "1.5 L" = "0.375 moles HF"#</mathjax></p>
</blockquote>
<p>Do the same for the conjugate base</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.250 M" * "1.5 L" = "0.375 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Now, you're adding <mathjax>#0.0500#</mathjax> moles of sodium hydroxide to the buffer. Since you have <strong>fewer moles</strong> of strong base than on weak acid, it follows that the sodium hydroxide will be <strong>completely consumed</strong>. </p>
<p>The number of moles of hydrofluoric acid will be changed to </p>
<blockquote>
<p><mathjax>#n_(HF) = "0.375 moles" = "0.0500 moles" = "0.325 moles"#</mathjax></p>
</blockquote>
<p>The number of moles of fluoride anions will <strong>increase</strong> by the same amount</p>
<blockquote>
<p><mathjax>#n_(F^(-)) = "0.375 moles" + "0.0500 moles" = "0.425 moles F"^(-)#</mathjax></p>
</blockquote>
<p>Use the volume of the buffer to calculate the new molarities of the weak acid and its conjugate base</p>
<blockquote>
<p><mathjax>#["HF"] = "0.325 moles"/"1.5 L" = "0.21667 M"#</mathjax></p>
<p><mathjax>#["F"^(-)] = "0.425 moles"/"1.5 L" = "0.28333 M"#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>Henderson - Hasselbalch equation</strong> to find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>Use the acid dissociation constant, <mathjax>#K_a#</mathjax>, to get the value of <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#pK_a = - log(K_a)#</mathjax></p>
</blockquote>
<p>The pH of the solution will thus be </p>
<blockquote>
<p><mathjax>#"pH" = -log(K_a) + log( (["F"^(-)])/(["HF"]))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5 * 10^(-4)) + log( (0.28333color(red)(cancel(color(black)("M"))))/(0.21667color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pH" = color(green)(3.57)#</mathjax></p>
</blockquote>
<p>Finally, does this result make sense?</p>
<p>Notice that your starting buffer had <strong>equal concentrations</strong> of weak acid and conjugate base. This means that the H-H equation can be reduced to</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log(1) = pK_a#</mathjax></p>
</blockquote>
<p>Initially, the pH of the solution was equal to <mathjax>#3.46#</mathjax>. </p>
<p>The buffer then <em>converts</em> the strong base to weak base, and so the concentration of conjugate base is <strong>bigger</strong> than that of the weak acid. This is why the pH of the buffer <strong>increases</strong> after the addition of the strong base. </p></div>
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</article> | A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change.
Ka for HF is 3.5x10^-4
| null |
1,912 | abe47e14-6ddd-11ea-98bf-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-which-is-comprised-of-2-677-g-ba- | BaBr2 | start chemical_formula qc_end physical_unit 14 14 12 13 mass qc_end physical_unit 18 18 16 17 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"BaBr2"}] | [{"type":"physical unit","value":"Mass [OF] Ba [=] \\pu{2.677 g}"},{"type":"physical unit","value":"Mass [OF] Br [=] \\pu{3.115 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound which is comprised of 2.677 g Ba and 3.115 g Br?</h1> | null | BaBr2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Masses:<br/>
<mathjax>#Ba=2.677g#</mathjax><br/>
<mathjax>#Br=3.115g#</mathjax></p>
<p>Atomic Masses:</p>
<p><mathjax>#Ba=137.327g/(mol)#</mathjax><br/>
<mathjax>#Br=79.904g/(mol)#</mathjax></p>
<p>Looking for the values in mol:<br/>
<mathjax>#Ba=2.677cancel(gBa)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBa)/(137.327cancel(gBa)#</mathjax> = <mathjax>#0.01949molBa#</mathjax><br/>
<mathjax>#Br=3.115cancel(gBr)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBr)/(79.904cancel(gBr)#</mathjax>=<mathjax>#0.038983molBr#</mathjax></p>
<p>Therefore:<br/>
<mathjax>#Ba#</mathjax> = <mathjax>#(0.01949mol)/(0.01949mol)#</mathjax>=1<br/>
<mathjax>#Br#</mathjax>=<mathjax>#(0.03898mol)/(0.01949mol)#</mathjax>=2</p>
<h2>Empirical Formula<mathjax>#=BaBr_2#</mathjax></h2></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#BaBr_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Masses:<br/>
<mathjax>#Ba=2.677g#</mathjax><br/>
<mathjax>#Br=3.115g#</mathjax></p>
<p>Atomic Masses:</p>
<p><mathjax>#Ba=137.327g/(mol)#</mathjax><br/>
<mathjax>#Br=79.904g/(mol)#</mathjax></p>
<p>Looking for the values in mol:<br/>
<mathjax>#Ba=2.677cancel(gBa)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBa)/(137.327cancel(gBa)#</mathjax> = <mathjax>#0.01949molBa#</mathjax><br/>
<mathjax>#Br=3.115cancel(gBr)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBr)/(79.904cancel(gBr)#</mathjax>=<mathjax>#0.038983molBr#</mathjax></p>
<p>Therefore:<br/>
<mathjax>#Ba#</mathjax> = <mathjax>#(0.01949mol)/(0.01949mol)#</mathjax>=1<br/>
<mathjax>#Br#</mathjax>=<mathjax>#(0.03898mol)/(0.01949mol)#</mathjax>=2</p>
<h2>Empirical Formula<mathjax>#=BaBr_2#</mathjax></h2></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound which is comprised of 2.677 g Ba and 3.115 g Br?</h1>
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<div class="markdown"><p><mathjax>#BaBr_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Masses:<br/>
<mathjax>#Ba=2.677g#</mathjax><br/>
<mathjax>#Br=3.115g#</mathjax></p>
<p>Atomic Masses:</p>
<p><mathjax>#Ba=137.327g/(mol)#</mathjax><br/>
<mathjax>#Br=79.904g/(mol)#</mathjax></p>
<p>Looking for the values in mol:<br/>
<mathjax>#Ba=2.677cancel(gBa)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBa)/(137.327cancel(gBa)#</mathjax> = <mathjax>#0.01949molBa#</mathjax><br/>
<mathjax>#Br=3.115cancel(gBr)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#(1molBr)/(79.904cancel(gBr)#</mathjax>=<mathjax>#0.038983molBr#</mathjax></p>
<p>Therefore:<br/>
<mathjax>#Ba#</mathjax> = <mathjax>#(0.01949mol)/(0.01949mol)#</mathjax>=1<br/>
<mathjax>#Br#</mathjax>=<mathjax>#(0.03898mol)/(0.01949mol)#</mathjax>=2</p>
<h2>Empirical Formula<mathjax>#=BaBr_2#</mathjax></h2></div>
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</article> | What is the empirical formula for the compound which is comprised of 2.677 g Ba and 3.115 g Br? | null |
1,913 | a83ab452-6ddd-11ea-ba70-ccda262736ce | https://socratic.org/questions/calculate-the-ph-during-a-titration-when-9-54-ml-of-a-0-15m-hcl-solution-has-rea | 12.74 | start physical_unit 14 14 ph none qc_end physical_unit 13 14 7 8 volume qc_end physical_unit 13 14 11 12 molarity qc_end physical_unit 14 14 18 19 volume qc_end physical_unit 23 23 21 22 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] solution"}] | [{"type":"physical unit","value":"12.74"}] | [{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{9.54 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.15 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{22.88 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.14 M}"}] | <h1 class="questionTitle" itemprop="name">
Calculate the pH during a titration when 9.54 mL of a 0.15 M HCl solution has reacted with 22.88 mL of 0.14 M NaOH?</h1> | null | 12.74 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of hydrochloric acid and the number of moles of sodium hydroxide that took part in the reaction. </p>
<blockquote>
<p><mathjax>#9.54 color(red)(cancel(color(black)("mL solution"))) * "0.15 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.001431 moles HCl"#</mathjax></p>
<p><mathjax>#22.88 color(red)(cancel(color(black)("mL solution"))) * "0.14 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.003203 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid and sodium hydroxide neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>.</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This tells that in order to have a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong>, you need to mix <strong>equal numbers</strong> of moles of hydrochloric acid and of sodium hydroxide. </p>
<p>In your case, you have more moles of sodium hydroxide than of hydrochloric acid, so you can say that the acid will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed <em>before</em> all the moles of sodium hydroxide will get the chance to react. </p>
<p>This means that after the reaction takes places, your solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of hydrochloric acid and </p>
<blockquote>
<p><mathjax>#"0.003203 moles " - " 0.001431 moles" = "0.001772 moles NaOH"#</mathjax></p>
</blockquote>
<p>At this point, you should be able to say that the <mathjax>#"pH"#</mathjax> of the resulting solution <strong>must</strong> be <mathjax>#>7#</mathjax> because you're left with a solution that contains a <strong>strong base</strong>.</p>
<p>As you know, sodium hydroxide dissociates completely in aqueous solution to produce sodium cations and hydroxide anions. </p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that the resulting solution will contain <mathjax>#0.0017772#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>The <strong>total volume</strong> of the resulting solution will be</p>
<blockquote>
<p><mathjax>#"9.54 mL + 22.88 mL = 32.42 mL"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydroxide anions will be--don't forget to convert the volume of the solution to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.001772 moles"/(32.42 * 10^(-3) quad "L") = "0.05466 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Now, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>Since you know that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)])))#</mathjax></p>
</blockquote>
<p>you can say that</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(0.05466) = color(darkgreen)(ul(color(black)(12.74)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <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 concentrations of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 12.74#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of hydrochloric acid and the number of moles of sodium hydroxide that took part in the reaction. </p>
<blockquote>
<p><mathjax>#9.54 color(red)(cancel(color(black)("mL solution"))) * "0.15 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.001431 moles HCl"#</mathjax></p>
<p><mathjax>#22.88 color(red)(cancel(color(black)("mL solution"))) * "0.14 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.003203 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid and sodium hydroxide neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>.</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This tells that in order to have a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong>, you need to mix <strong>equal numbers</strong> of moles of hydrochloric acid and of sodium hydroxide. </p>
<p>In your case, you have more moles of sodium hydroxide than of hydrochloric acid, so you can say that the acid will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed <em>before</em> all the moles of sodium hydroxide will get the chance to react. </p>
<p>This means that after the reaction takes places, your solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of hydrochloric acid and </p>
<blockquote>
<p><mathjax>#"0.003203 moles " - " 0.001431 moles" = "0.001772 moles NaOH"#</mathjax></p>
</blockquote>
<p>At this point, you should be able to say that the <mathjax>#"pH"#</mathjax> of the resulting solution <strong>must</strong> be <mathjax>#>7#</mathjax> because you're left with a solution that contains a <strong>strong base</strong>.</p>
<p>As you know, sodium hydroxide dissociates completely in aqueous solution to produce sodium cations and hydroxide anions. </p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that the resulting solution will contain <mathjax>#0.0017772#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>The <strong>total volume</strong> of the resulting solution will be</p>
<blockquote>
<p><mathjax>#"9.54 mL + 22.88 mL = 32.42 mL"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydroxide anions will be--don't forget to convert the volume of the solution to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.001772 moles"/(32.42 * 10^(-3) quad "L") = "0.05466 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Now, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>Since you know that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)])))#</mathjax></p>
</blockquote>
<p>you can say that</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(0.05466) = color(darkgreen)(ul(color(black)(12.74)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <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 concentrations of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p></div>
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<h1 class="questionTitle" itemprop="name">
Calculate the pH during a titration when 9.54 mL of a 0.15 M HCl solution has reacted with 22.88 mL of 0.14 M NaOH?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 12.74#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of moles of hydrochloric acid and the number of moles of sodium hydroxide that took part in the reaction. </p>
<blockquote>
<p><mathjax>#9.54 color(red)(cancel(color(black)("mL solution"))) * "0.15 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.001431 moles HCl"#</mathjax></p>
<p><mathjax>#22.88 color(red)(cancel(color(black)("mL solution"))) * "0.14 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.003203 moles NaOH"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid and sodium hydroxide neutralize each other in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>.</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This tells that in order to have a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong>, you need to mix <strong>equal numbers</strong> of moles of hydrochloric acid and of sodium hydroxide. </p>
<p>In your case, you have more moles of sodium hydroxide than of hydrochloric acid, so you can say that the acid will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>, i.e. it will be completely consumed <em>before</em> all the moles of sodium hydroxide will get the chance to react. </p>
<p>This means that after the reaction takes places, your solution will contain <mathjax>#0#</mathjax> <strong>moles</strong> of hydrochloric acid and </p>
<blockquote>
<p><mathjax>#"0.003203 moles " - " 0.001431 moles" = "0.001772 moles NaOH"#</mathjax></p>
</blockquote>
<p>At this point, you should be able to say that the <mathjax>#"pH"#</mathjax> of the resulting solution <strong>must</strong> be <mathjax>#>7#</mathjax> because you're left with a solution that contains a <strong>strong base</strong>.</p>
<p>As you know, sodium hydroxide dissociates completely in aqueous solution to produce sodium cations and hydroxide anions. </p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that the resulting solution will contain <mathjax>#0.0017772#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>The <strong>total volume</strong> of the resulting solution will be</p>
<blockquote>
<p><mathjax>#"9.54 mL + 22.88 mL = 32.42 mL"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydroxide anions will be--don't forget to convert the volume of the solution to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = "0.001772 moles"/(32.42 * 10^(-3) quad "L") = "0.05466 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Now, an aqueous solution at <mathjax>#25^@"C"#</mathjax> has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH + pOH = 14")))#</mathjax></p>
</blockquote>
<p>Since you know that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)])))#</mathjax></p>
</blockquote>
<p>you can say that</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(0.05466) = color(darkgreen)(ul(color(black)(12.74)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <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 concentrations of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. </p></div>
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Calculate the pH during a titration when 9.54 mL of a 0.15 M HCl solution has reacted with 22.88 mL of 0.14 M NaOH? | null |
1,914 | ac2d4348-6ddd-11ea-9c1f-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-of-nano-3-84-99-g-mol-which-contains-11-8-gra | 1.49 M | start physical_unit 8 8 molarity mol/l qc_end physical_unit 8 8 9 10 molar_mass qc_end physical_unit 16 16 13 14 mass qc_end physical_unit 6 6 22 23 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaNO3 solution [IN] M"}] | [{"type":"physical unit","value":"1.49 M"}] | [{"type":"physical unit","value":"Molar mass [OF] NaNO3 [=] \\pu{84.99 g/mol}"},{"type":"physical unit","value":"Mass [OF] solute [=] \\pu{11.8 grams}"},{"type":"physical unit","value":"Volume [OF] NaNO3 solution [=] \\pu{93.9 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?</h1> | null | 1.49 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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<div class="markdown"><p>1.49 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?</h1>
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MM
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<div class="markdown"><p>1.49 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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anor277
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<span class="dateCreated" datetime="2017-04-12T05:20:15" itemprop="dateCreated">
Apr 12, 2017
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<div class="markdown"><p><mathjax>#"Molarity"=1.48*mol*L^-1...................#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>For this problem.......</p>
<p><mathjax>#"Molarity"=((11.8*cancelg)/(84.99*cancelg*mol^-1))/(93.9*cancel(mL)xx10^-3*L*cancel(mL^-1))#</mathjax></p>
<p><mathjax>#=??*mol*L^-1#</mathjax>...........</p></div>
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</article> | What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL? | null |
1,915 | a9899848-6ddd-11ea-a701-ccda262736ce | https://socratic.org/questions/the-hormone-thyroxine-is-secreted-by-the-thyroid-gland-and-has-the-formula-c-15h | 9796.68 milligrams | start physical_unit 18 18 mass mg qc_end physical_unit 13 13 23 24 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] iodine [IN] milligrams"}] | [{"type":"physical unit","value":"9796.68 milligrams"}] | [{"type":"physical unit","value":"Mass [OF] C15H17NO4I4 [=] \\pu{15.0 grams}"}] | <h1 class="questionTitle" itemprop="name">The hormone, thyroxine is secreted by the thyroid gland, and has the formula #C_15H_17NO_4I_4#. How many milligrams of iodine can be extracted from 15.0 grams of thyroxine?</h1> | null | 9796.68 milligrams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of thyroxine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15*g)/(776.86 *g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*mol#</mathjax></p>
<p>And thus <mathjax>#"mg of iodine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*molxx4xx126.9*g*mol^-1xx10^3*mg*g^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9801*mg#</mathjax>. </p>
<p>Iodine is a massive element, and thus provides most of the mass in a given sample. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Thyroxine has a formula of <mathjax>#C_15H_11I_4NO_4#</mathjax>, and a molar mass of <mathjax>#776.86*g*mol^-1#</mathjax>. Almost <mathjax>#10*g#</mathjax> iodine are present in the sample. How many <mathjax>#mg#</mathjax>?</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of thyroxine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15*g)/(776.86 *g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*mol#</mathjax></p>
<p>And thus <mathjax>#"mg of iodine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*molxx4xx126.9*g*mol^-1xx10^3*mg*g^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9801*mg#</mathjax>. </p>
<p>Iodine is a massive element, and thus provides most of the mass in a given sample. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The hormone, thyroxine is secreted by the thyroid gland, and has the formula #C_15H_17NO_4I_4#. How many milligrams of iodine can be extracted from 15.0 grams of thyroxine?</h1>
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<div class="markdown"><p>Thyroxine has a formula of <mathjax>#C_15H_11I_4NO_4#</mathjax>, and a molar mass of <mathjax>#776.86*g*mol^-1#</mathjax>. Almost <mathjax>#10*g#</mathjax> iodine are present in the sample. How many <mathjax>#mg#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of thyroxine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15*g)/(776.86 *g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*mol#</mathjax></p>
<p>And thus <mathjax>#"mg of iodine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0193*molxx4xx126.9*g*mol^-1xx10^3*mg*g^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9801*mg#</mathjax>. </p>
<p>Iodine is a massive element, and thus provides most of the mass in a given sample. </p></div>
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</article> | The hormone, thyroxine is secreted by the thyroid gland, and has the formula #C_15H_17NO_4I_4#. How many milligrams of iodine can be extracted from 15.0 grams of thyroxine? | null |
1,916 | ace508e6-6ddd-11ea-b9d5-ccda262736ce | https://socratic.org/questions/if-the-initial-temperature-of-an-ideal-gas-at-2-250-atm-is-62-00-c-what-final-te | 238.2 K | start physical_unit 6 7 temperature k qc_end physical_unit 6 7 12 13 temperature qc_end physical_unit 6 7 9 10 pressure qc_end physical_unit 6 7 25 26 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the ideal gas [IN] K"}] | [{"type":"physical unit","value":"238.2 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the ideal gas [=] \\pu{62.00 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the ideal gas [=] \\pu{2.250 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the ideal gas [=] \\pu{1.600 atm}"}] | <h1 class="questionTitle" itemprop="name">If the initial temperature of an ideal gas at 2.250 atm is 62.00 °C, what final temperature would cause the pressure to be reduced to 1.600 atm?</h1> | null | 238.2 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Gay-Lussac's law:</strong> For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.</p>
<p>Mathematically,</p>
<blockquote>
<p><mathjax>#P ∝ T#</mathjax></p>
</blockquote>
<p><mathjax>#P_f/P_i = T_f/T_i#</mathjax></p>
<p><mathjax>#T_f = T_i × P_f/P_i#</mathjax></p>
<p><mathjax>#T_f = "(62.00 + 273.15) Kelvin" × (1.600 cancel"atm")/(2.250 cancel"atm") = "238.2 Kelvin"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"238.2 Kelvin"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Gay-Lussac's law:</strong> For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.</p>
<p>Mathematically,</p>
<blockquote>
<p><mathjax>#P ∝ T#</mathjax></p>
</blockquote>
<p><mathjax>#P_f/P_i = T_f/T_i#</mathjax></p>
<p><mathjax>#T_f = T_i × P_f/P_i#</mathjax></p>
<p><mathjax>#T_f = "(62.00 + 273.15) Kelvin" × (1.600 cancel"atm")/(2.250 cancel"atm") = "238.2 Kelvin"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the initial temperature of an ideal gas at 2.250 atm is 62.00 °C, what final temperature would cause the pressure to be reduced to 1.600 atm?</h1>
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<div class="markdown"><p><mathjax>#"238.2 Kelvin"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Gay-Lussac's law:</strong> For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.</p>
<p>Mathematically,</p>
<blockquote>
<p><mathjax>#P ∝ T#</mathjax></p>
</blockquote>
<p><mathjax>#P_f/P_i = T_f/T_i#</mathjax></p>
<p><mathjax>#T_f = T_i × P_f/P_i#</mathjax></p>
<p><mathjax>#T_f = "(62.00 + 273.15) Kelvin" × (1.600 cancel"atm")/(2.250 cancel"atm") = "238.2 Kelvin"#</mathjax></p></div>
</div>
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</article> | If the initial temperature of an ideal gas at 2.250 atm is 62.00 °C, what final temperature would cause the pressure to be reduced to 1.600 atm? | null |
1,917 | ac12f938-6ddd-11ea-b0ed-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-molecules-in-1-00-mol-of-iodine-crystals | 6.02 × 10^23 | start physical_unit 5 5 number none qc_end physical_unit 10 11 7 8 mole qc_end end | [{"type":"physical unit","value":"Number [OF] molecules"}] | [{"type":"physical unit","value":"6.02 × 10^23"}] | [{"type":"physical unit","value":"Mole [OF] iodine crystals [=] \\pu{1.00 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the number of molecules in 1.00 mol of iodine crystals?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Is the answer 6.022 x 10^25 molec. of I2 ? or do I have double this answer, given that I2 is a diatomic molecule?? </p></div>
</h2>
</div>
</div> | 6.02 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants to make sure that you are familiar with the definition of <strong>one mole</strong> of a given substance.</p>
<p>In this case, you're dealing with a <em>covalent compound</em>, so you're looking for the number of molecules of iodine, <mathjax>#"I"_2#</mathjax>, present in exactly <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals. </p>
<p>Now, a <strong>mole</strong> is defined as very, very large collection of particles. In order to have <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>molecules</strong> of iodine.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole I"_2 = 6.022 * 10^(23)"molecules I"_2#</mathjax> <mathjax>#->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"1.00 moles I"_2 = 6.02 * 10^(23)"molecules I"_2#</mathjax></p>
</blockquote>
<p>The answer must be rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles of iodine crystals.</p>
<p>Notice that each iodine molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of iodine, <mathjax>#2 xx "I"#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of iodine molecules contains <mathjax>#2#</mathjax> <strong>moles</strong> of iodine atoms. </p>
<p>In <mathjax>#1.00#</mathjax> moles of iodine molecules you have <mathjax>#2#</mathjax> moles of iodine atoms, or</p>
<blockquote>
<p><mathjax>#overbrace(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2))))^(color(blue)(= "1.00 moles I"_2)) * "2 atoms of I"/(1color(red)(cancel(color(black)("molecule I"_2)))) = overbrace(1.20 * 10^(24)"atoms of I")^(color(purple)(="2.00 moles of I"))#</mathjax></p>
</blockquote>
<p>Once again, the answer must be rounded to three sig figs. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#6.02 * 10^(23)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants to make sure that you are familiar with the definition of <strong>one mole</strong> of a given substance.</p>
<p>In this case, you're dealing with a <em>covalent compound</em>, so you're looking for the number of molecules of iodine, <mathjax>#"I"_2#</mathjax>, present in exactly <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals. </p>
<p>Now, a <strong>mole</strong> is defined as very, very large collection of particles. In order to have <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>molecules</strong> of iodine.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole I"_2 = 6.022 * 10^(23)"molecules I"_2#</mathjax> <mathjax>#->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"1.00 moles I"_2 = 6.02 * 10^(23)"molecules I"_2#</mathjax></p>
</blockquote>
<p>The answer must be rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles of iodine crystals.</p>
<p>Notice that each iodine molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of iodine, <mathjax>#2 xx "I"#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of iodine molecules contains <mathjax>#2#</mathjax> <strong>moles</strong> of iodine atoms. </p>
<p>In <mathjax>#1.00#</mathjax> moles of iodine molecules you have <mathjax>#2#</mathjax> moles of iodine atoms, or</p>
<blockquote>
<p><mathjax>#overbrace(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2))))^(color(blue)(= "1.00 moles I"_2)) * "2 atoms of I"/(1color(red)(cancel(color(black)("molecule I"_2)))) = overbrace(1.20 * 10^(24)"atoms of I")^(color(purple)(="2.00 moles of I"))#</mathjax></p>
</blockquote>
<p>Once again, the answer must be rounded to three sig figs. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of molecules in 1.00 mol of iodine crystals?</h1>
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<div class="markdown"><p>Is the answer 6.022 x 10^25 molec. of I2 ? or do I have double this answer, given that I2 is a diatomic molecule?? </p></div>
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Stefan V.
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Jan 21, 2017
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<div class="markdown"><p><mathjax>#6.02 * 10^(23)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem wants to make sure that you are familiar with the definition of <strong>one mole</strong> of a given substance.</p>
<p>In this case, you're dealing with a <em>covalent compound</em>, so you're looking for the number of molecules of iodine, <mathjax>#"I"_2#</mathjax>, present in exactly <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals. </p>
<p>Now, a <strong>mole</strong> is defined as very, very large collection of particles. In order to have <mathjax>#1.00#</mathjax> <strong>moles</strong> of iodine crystals, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>molecules</strong> of iodine.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole I"_2 = 6.022 * 10^(23)"molecules I"_2#</mathjax> <mathjax>#->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"1.00 moles I"_2 = 6.02 * 10^(23)"molecules I"_2#</mathjax></p>
</blockquote>
<p>The answer must be rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles of iodine crystals.</p>
<p>Notice that each iodine molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of iodine, <mathjax>#2 xx "I"#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of iodine molecules contains <mathjax>#2#</mathjax> <strong>moles</strong> of iodine atoms. </p>
<p>In <mathjax>#1.00#</mathjax> moles of iodine molecules you have <mathjax>#2#</mathjax> moles of iodine atoms, or</p>
<blockquote>
<p><mathjax>#overbrace(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2))))^(color(blue)(= "1.00 moles I"_2)) * "2 atoms of I"/(1color(red)(cancel(color(black)("molecule I"_2)))) = overbrace(1.20 * 10^(24)"atoms of I")^(color(purple)(="2.00 moles of I"))#</mathjax></p>
</blockquote>
<p>Once again, the answer must be rounded to three sig figs. </p></div>
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</article> | What is the number of molecules in 1.00 mol of iodine crystals? |
Is the answer 6.022 x 10^25 molec. of I2 ? or do I have double this answer, given that I2 is a diatomic molecule??
|
1,918 | abc76758-6ddd-11ea-aaf9-ccda262736ce | https://socratic.org/questions/how-many-atoms-are-in-a-mole-of-barium-chloride-bacl-2 | 1.81 × 10^24 | start physical_unit 2 2 number none qc_end end | [{"type":"physical unit","value":"Number [OF] atoms"}] | [{"type":"physical unit","value":"1.81 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] BaCl2 [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How many atoms are in a mole of barium chloride, #BaCl_2#?</h1> | null | 1.81 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I asked how many atoms there were in a dozen formula units of <mathjax>#BaCl_2#</mathjax>, I think you would quickly answer 3 dozen: 12 barium atoms, and 24 chlorine atoms.</p>
<p>The question specified molar quantities, and <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is just another number, admittedly a very large number, i.e. <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>.</p>
<p>So in <mathjax>#1#</mathjax> mole of <mathjax>#BaCl_2#</mathjax>, there are <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#N_A#</mathjax> barium and chlorine atoms, where <mathjax>#N_A#</mathjax> is Avogadro's number, <mathjax>#6.022xx10^23#</mathjax>. </p>
<p>What is the actual mass of this quantity of barium chloride?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Clearly there are <mathjax>#3#</mathjax> <mathjax>#"moles"#</mathjax> of atoms in one mole of <mathjax>#BaCl_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I asked how many atoms there were in a dozen formula units of <mathjax>#BaCl_2#</mathjax>, I think you would quickly answer 3 dozen: 12 barium atoms, and 24 chlorine atoms.</p>
<p>The question specified molar quantities, and <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is just another number, admittedly a very large number, i.e. <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>.</p>
<p>So in <mathjax>#1#</mathjax> mole of <mathjax>#BaCl_2#</mathjax>, there are <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#N_A#</mathjax> barium and chlorine atoms, where <mathjax>#N_A#</mathjax> is Avogadro's number, <mathjax>#6.022xx10^23#</mathjax>. </p>
<p>What is the actual mass of this quantity of barium chloride?</p></div>
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<h1 class="questionTitle" itemprop="name">How many atoms are in a mole of barium chloride, #BaCl_2#?</h1>
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<div class="markdown"><p>Clearly there are <mathjax>#3#</mathjax> <mathjax>#"moles"#</mathjax> of atoms in one mole of <mathjax>#BaCl_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I asked how many atoms there were in a dozen formula units of <mathjax>#BaCl_2#</mathjax>, I think you would quickly answer 3 dozen: 12 barium atoms, and 24 chlorine atoms.</p>
<p>The question specified molar quantities, and <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is just another number, admittedly a very large number, i.e. <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.022xx10^23#</mathjax> <mathjax>#mol^-1#</mathjax>.</p>
<p>So in <mathjax>#1#</mathjax> mole of <mathjax>#BaCl_2#</mathjax>, there are <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#N_A#</mathjax> barium and chlorine atoms, where <mathjax>#N_A#</mathjax> is Avogadro's number, <mathjax>#6.022xx10^23#</mathjax>. </p>
<p>What is the actual mass of this quantity of barium chloride?</p></div>
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</article> | How many atoms are in a mole of barium chloride, #BaCl_2#? | null |
1,919 | ac583852-6ddd-11ea-bf6c-ccda262736ce | https://socratic.org/questions/a-chemist-performs-the-synthesis-of-sodium-chloride-from-its-elements-the-chemis | 2 moles | start physical_unit 23 23 mole mol qc_end physical_unit 6 6 15 16 mass qc_end chemical_equation 26 32 qc_end end | [{"type":"physical unit","value":"Mole [OF] chlorine [IN] moles"}] | [{"type":"physical unit","value":"2 moles"}] | [{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{46 grams}"},{"type":"chemical equation","value":"2 Na + Cl2 -> 2 NaCl"}] | <h1 class="questionTitle" itemprop="name">A chemist performs the synthesis of sodium chloride from its elements. The chemist begins with 46 grams of sodium. How many moles of chlorine are needed?
2 Na + Cl2 → 2 NaCl</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>I didn't understand that is it asking about Cl (chlorine) or Cl2 (Chlorine GAS)...</p></div>
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</div> | 2 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium "=(46.0*g)/(22.9*g*mol^-1)=2*mol#</mathjax>.</p>
<p>Given your equation, clearly we need the ONE mole of <mathjax>#Cl_2#</mathjax> gas. What is the mass of this quantity of chlorine?</p>
<p>It is fact that all of the elemental gases EXCEPT for the Noble Gases are diatomic: <mathjax>#N_2, O_2, F_2, Cl_2#</mathjax>. Of course, all the halogens, as the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, are diatomic, but not all are gaseous. </p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>We have <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> of sodium metal to start; clearly we need the 1 mole of <mathjax>#Cl_2(g)#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium "=(46.0*g)/(22.9*g*mol^-1)=2*mol#</mathjax>.</p>
<p>Given your equation, clearly we need the ONE mole of <mathjax>#Cl_2#</mathjax> gas. What is the mass of this quantity of chlorine?</p>
<p>It is fact that all of the elemental gases EXCEPT for the Noble Gases are diatomic: <mathjax>#N_2, O_2, F_2, Cl_2#</mathjax>. Of course, all the halogens, as the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, are diatomic, but not all are gaseous. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A chemist performs the synthesis of sodium chloride from its elements. The chemist begins with 46 grams of sodium. How many moles of chlorine are needed?
2 Na + Cl2 → 2 NaCl</h1>
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<div class="markdown"><p>I didn't understand that is it asking about Cl (chlorine) or Cl2 (Chlorine GAS)...</p></div>
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<div class="markdown"><p>We have <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> of sodium metal to start; clearly we need the 1 mole of <mathjax>#Cl_2(g)#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium "=(46.0*g)/(22.9*g*mol^-1)=2*mol#</mathjax>.</p>
<p>Given your equation, clearly we need the ONE mole of <mathjax>#Cl_2#</mathjax> gas. What is the mass of this quantity of chlorine?</p>
<p>It is fact that all of the elemental gases EXCEPT for the Noble Gases are diatomic: <mathjax>#N_2, O_2, F_2, Cl_2#</mathjax>. Of course, all the halogens, as the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, are diatomic, but not all are gaseous. </p></div>
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</article> | A chemist performs the synthesis of sodium chloride from its elements. The chemist begins with 46 grams of sodium. How many moles of chlorine are needed?
2 Na + Cl2 → 2 NaCl |
I didn't understand that is it asking about Cl (chlorine) or Cl2 (Chlorine GAS)...
|
1,920 | a8e4e0d8-6ddd-11ea-b235-ccda262736ce | https://socratic.org/questions/if-sulfur-dioxide-and-oxygen-can-be-made-into-sulfur-trioxide-what-is-the-revers | 2 SO3 -> 2 SO2 + O2 | start chemical_equation qc_end substance 1 2 qc_end substance 4 4 qc_end substance 9 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reverse reaction"}] | [{"type":"chemical equation","value":"2 SO3 -> 2 SO2 + O2"}] | [{"type":"substance name","value":"Sulfur dioxide"},{"type":"substance name","value":"Oxygen"},{"type":"substance name","value":"Sulfur trioxide"}] | <h1 class="questionTitle" itemprop="name">If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction? </h1> | null | 2 SO3 -> 2 SO2 + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Vast quantities of <mathjax>#"sulfur trioxide"#</mathjax> are utilized industrially for sulfuric acid synthesis:</p>
<p><mathjax>#SO_3 + H_2Orightleftharpoons H_2SO_4#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#SO_3(g) rightleftharpoonsSO_2(g) + 1/2O_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Vast quantities of <mathjax>#"sulfur trioxide"#</mathjax> are utilized industrially for sulfuric acid synthesis:</p>
<p><mathjax>#SO_3 + H_2Orightleftharpoons H_2SO_4#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction? </h1>
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<div class="markdown"><p><mathjax>#SO_3(g) rightleftharpoonsSO_2(g) + 1/2O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Vast quantities of <mathjax>#"sulfur trioxide"#</mathjax> are utilized industrially for sulfuric acid synthesis:</p>
<p><mathjax>#SO_3 + H_2Orightleftharpoons H_2SO_4#</mathjax></p></div>
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</article> | If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction? | null |
1,921 | a9f5c668-6ddd-11ea-8620-ccda262736ce | https://socratic.org/questions/given-the-specific-heat-of-lead-is-0-129-j-g-k-and-that-it-takes-93-4-j-of-energ | 40.91 kg | start physical_unit 22 24 mass kg qc_end physical_unit 5 5 7 10 specific_heat qc_end physical_unit 22 24 15 16 heat_energy qc_end physical_unit 22 24 26 27 temperature qc_end physical_unit 22 24 29 30 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] lead sample [IN] kg"}] | [{"type":"physical unit","value":"40.91 kg"}] | [{"type":"physical unit","value":"Specific heat [OF] lead [=] \\pu{0.129 J/(g * K)}"},{"type":"physical unit","value":"Taken energy [OF] lead sample [=] \\pu{93.4 J}"},{"type":"physical unit","value":"Temperature1 [OF] lead sample [=] \\pu{ 22.3 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] lead sample [=] \\pu{40.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">Given the specific heat of lead is 0.129 J/g#*#K and that it takes 93.4 J of energy to heat a sample of lead from 22.3°C to 40.°C, how do you find the mass of the lead.?</h1> | null | 40.91 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation you want is </p>
<p><mathjax>#E=mctheta#</mathjax></p>
<p>where <mathjax>#E#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity (SHC) and <mathjax>#theta#</mathjax> is the change in temperature (note: <em>change</em>). </p>
<p>In the question, we want to find mass, so we rearrange to make mass the subject:</p>
<p><mathjax>#E=mctheta -> m=E/(ctheta)#</mathjax></p>
<p>We know that <mathjax>#E=93.4J#</mathjax>, <mathjax>#c=0.129J/(gK)#</mathjax> and <mathjax>#theta=40-22.3=17.7K#</mathjax> </p>
<p>(We can easily convert between <mathjax>#K#</mathjax> and Celsius because they are on the same scale, but have different starting points. Celsius starts from water freezing at <mathjax>#0^oC#</mathjax>, while Kelvin starts at <mathjax>#0#</mathjax> thermal energy, which is about <mathjax>#-273.15^oC#</mathjax>. Calculating a change in Celsius gives the same change in Kelvin, because you can ignore the starting points of the scales.) </p>
<p>Now we have</p>
<p><mathjax>#m=93.4/(0.129xx17.7)=40.9057kg#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#40.9kg#</mathjax> lead </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation you want is </p>
<p><mathjax>#E=mctheta#</mathjax></p>
<p>where <mathjax>#E#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity (SHC) and <mathjax>#theta#</mathjax> is the change in temperature (note: <em>change</em>). </p>
<p>In the question, we want to find mass, so we rearrange to make mass the subject:</p>
<p><mathjax>#E=mctheta -> m=E/(ctheta)#</mathjax></p>
<p>We know that <mathjax>#E=93.4J#</mathjax>, <mathjax>#c=0.129J/(gK)#</mathjax> and <mathjax>#theta=40-22.3=17.7K#</mathjax> </p>
<p>(We can easily convert between <mathjax>#K#</mathjax> and Celsius because they are on the same scale, but have different starting points. Celsius starts from water freezing at <mathjax>#0^oC#</mathjax>, while Kelvin starts at <mathjax>#0#</mathjax> thermal energy, which is about <mathjax>#-273.15^oC#</mathjax>. Calculating a change in Celsius gives the same change in Kelvin, because you can ignore the starting points of the scales.) </p>
<p>Now we have</p>
<p><mathjax>#m=93.4/(0.129xx17.7)=40.9057kg#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Given the specific heat of lead is 0.129 J/g#*#K and that it takes 93.4 J of energy to heat a sample of lead from 22.3°C to 40.°C, how do you find the mass of the lead.?</h1>
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<div class="markdown"><p><mathjax>#40.9kg#</mathjax> lead </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation you want is </p>
<p><mathjax>#E=mctheta#</mathjax></p>
<p>where <mathjax>#E#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity (SHC) and <mathjax>#theta#</mathjax> is the change in temperature (note: <em>change</em>). </p>
<p>In the question, we want to find mass, so we rearrange to make mass the subject:</p>
<p><mathjax>#E=mctheta -> m=E/(ctheta)#</mathjax></p>
<p>We know that <mathjax>#E=93.4J#</mathjax>, <mathjax>#c=0.129J/(gK)#</mathjax> and <mathjax>#theta=40-22.3=17.7K#</mathjax> </p>
<p>(We can easily convert between <mathjax>#K#</mathjax> and Celsius because they are on the same scale, but have different starting points. Celsius starts from water freezing at <mathjax>#0^oC#</mathjax>, while Kelvin starts at <mathjax>#0#</mathjax> thermal energy, which is about <mathjax>#-273.15^oC#</mathjax>. Calculating a change in Celsius gives the same change in Kelvin, because you can ignore the starting points of the scales.) </p>
<p>Now we have</p>
<p><mathjax>#m=93.4/(0.129xx17.7)=40.9057kg#</mathjax></p></div>
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</article> | Given the specific heat of lead is 0.129 J/g#*#K and that it takes 93.4 J of energy to heat a sample of lead from 22.3°C to 40.°C, how do you find the mass of the lead.? | null |
1,922 | aa89371e-6ddd-11ea-9ac7-ccda262736ce | https://socratic.org/questions/how-would-you-balance-al-h2so4-al2-so4-3-h2 | 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2"}] | [{"type":"chemical equation","value":"Al + H2SO4 -> Al2(SO4)3 + H2"}] | <h1 class="questionTitle" itemprop="name">How would you balance Al + H2SO4 = Al2(SO4)3 + H2?
</h1> | null | 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>When balancing chemical equations I always start with oxygen and identify which compound has the most, which in our case is Al2(SO4)3. So because we have 4 O in SO4, and there are 3 of moles of SO4, so we multiple 4x3=12. So now, go to the opposite compound with O (H2SO4) and manipulate it to even out the O, in this case multiple by 3. Now we have 12 O on each side, 3 S on each side, and all we have to do here is multiply the lone Al on the reactants side by 2, and then because we multiplied H2SO4 by 3, we have 6 H on that side, so just multiple H2 on the other side by 3 as well and your done</p></div>
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<div class="markdown"><p>2Al + 3H2SO4 = Al2(SO4)3 + 3H2</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>When balancing chemical equations I always start with oxygen and identify which compound has the most, which in our case is Al2(SO4)3. So because we have 4 O in SO4, and there are 3 of moles of SO4, so we multiple 4x3=12. So now, go to the opposite compound with O (H2SO4) and manipulate it to even out the O, in this case multiple by 3. Now we have 12 O on each side, 3 S on each side, and all we have to do here is multiply the lone Al on the reactants side by 2, and then because we multiplied H2SO4 by 3, we have 6 H on that side, so just multiple H2 on the other side by 3 as well and your done</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance Al + H2SO4 = Al2(SO4)3 + H2?
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<div class="markdown"><p>2Al + 3H2SO4 = Al2(SO4)3 + 3H2</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>When balancing chemical equations I always start with oxygen and identify which compound has the most, which in our case is Al2(SO4)3. So because we have 4 O in SO4, and there are 3 of moles of SO4, so we multiple 4x3=12. So now, go to the opposite compound with O (H2SO4) and manipulate it to even out the O, in this case multiple by 3. Now we have 12 O on each side, 3 S on each side, and all we have to do here is multiply the lone Al on the reactants side by 2, and then because we multiplied H2SO4 by 3, we have 6 H on that side, so just multiple H2 on the other side by 3 as well and your done</p></div>
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</article> | How would you balance Al + H2SO4 = Al2(SO4)3 + H2?
| null |
1,923 | abb20c24-6ddd-11ea-b8a2-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-10-ml-0-1-m-nh3-solution-when-a-0-1-ml-0-1-m-naoh-solution-w | 11.30 | start physical_unit 11 11 ph none qc_end physical_unit 10 11 6 7 volume qc_end physical_unit 10 11 8 9 molarity qc_end physical_unit 18 19 14 15 volume qc_end physical_unit 18 19 8 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"11.30"}] | [{"type":"physical unit","value":"Volume [OF] NH3 solution [=] \\pu{10 mL}"},{"type":"physical unit","value":"Molarity [OF] NH3 solution [=] \\pu{0.1 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{0.1 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.1 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 10 mL, 0.1 M NH3 solution when a 0.1 mL, 0.1 M NaOH solution was added into it?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I tried using an ICE table, but do not come up with a reasonable answer. Thank you in advance!</p></div>
</h2>
</div>
</div> | 11.30 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To calculate the new concentration of <mathjax>#"NaOH"#</mathjax> in the ammonia solution, you can use the dilution formula</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#c_1 = "0.1 mol/L"; V_1 = "0.1 mL"#</mathjax><br/>
<mathjax>#c_2 = "?";color(white)(mmmml) V_2 = "(10 + 0.1) mL" = "10.1 mL"#</mathjax></p>
<p><mathjax>#c_2 = 0.1 "mol/L" × (0.10 color(red)(cancel(color(black)("mL"))))/(10.1 color(red)(cancel(color(black)("mL")))) = 9.9 ×10^"-4" "mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we can use an ICE table to solve the problem.</p>
<p><mathjax>#color(white)(mmmmmm)"NH"_3 + "H"_2"O" → "NH"_4^"+" color(white)(m)+ color(white)(m)"OH"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m)0.1color(white)(mmmmmmm) 0color(white)(mmml)9.9×10^"-4"#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(ll) "-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmmmll) "+"x #</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(l) "0.1 -" xcolor(white)(mmmmmm) x color(white)(mml)x + 9.9×10^"-4"#</mathjax></p>
<p><mathjax>#K_"b" = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#K_"b " = (x(x + 9.9×10^"-4"))/(0.1 - x) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#0.1/(1.8 × 10^"-5") = 5600 > 400#</mathjax></p>
<p>∴<mathjax># x ≪ 0.1#</mathjax></p>
<blockquote></blockquote>
<p>Then</p>
<p><mathjax>#(x(x + 9.9×10^"-4"))/0.1 = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#x(x + 9.9×10^"-4") = 0.1 × 1.8 × 10^"-5"= 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x = 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x - 1.8 × 10^"-6" = 0#</mathjax></p>
<p><mathjax>#x = 9.9 × 10^"-4"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (9.9 × 10^"-4" +x) color(white)(l)"mol/L" = (9.9 × 10^"-4" + 9.9 × 10^"-4") color(white)(l)"mol/L"#</mathjax><br/>
<mathjax>#= 2.0 × 10^"-3"color(white)(l) "mol/L"#</mathjax></p>
<p><mathjax>#"pOH" = "-"log["OH"^"-"] = "-"log(2.0 ×10^"-3") = 2.7#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pOH" = "14.00 - 2.7" = 11.3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <mathjax>#"pH = 11.3"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To calculate the new concentration of <mathjax>#"NaOH"#</mathjax> in the ammonia solution, you can use the dilution formula</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#c_1 = "0.1 mol/L"; V_1 = "0.1 mL"#</mathjax><br/>
<mathjax>#c_2 = "?";color(white)(mmmml) V_2 = "(10 + 0.1) mL" = "10.1 mL"#</mathjax></p>
<p><mathjax>#c_2 = 0.1 "mol/L" × (0.10 color(red)(cancel(color(black)("mL"))))/(10.1 color(red)(cancel(color(black)("mL")))) = 9.9 ×10^"-4" "mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we can use an ICE table to solve the problem.</p>
<p><mathjax>#color(white)(mmmmmm)"NH"_3 + "H"_2"O" → "NH"_4^"+" color(white)(m)+ color(white)(m)"OH"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m)0.1color(white)(mmmmmmm) 0color(white)(mmml)9.9×10^"-4"#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(ll) "-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmmmll) "+"x #</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(l) "0.1 -" xcolor(white)(mmmmmm) x color(white)(mml)x + 9.9×10^"-4"#</mathjax></p>
<p><mathjax>#K_"b" = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#K_"b " = (x(x + 9.9×10^"-4"))/(0.1 - x) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#0.1/(1.8 × 10^"-5") = 5600 > 400#</mathjax></p>
<p>∴<mathjax># x ≪ 0.1#</mathjax></p>
<blockquote></blockquote>
<p>Then</p>
<p><mathjax>#(x(x + 9.9×10^"-4"))/0.1 = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#x(x + 9.9×10^"-4") = 0.1 × 1.8 × 10^"-5"= 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x = 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x - 1.8 × 10^"-6" = 0#</mathjax></p>
<p><mathjax>#x = 9.9 × 10^"-4"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (9.9 × 10^"-4" +x) color(white)(l)"mol/L" = (9.9 × 10^"-4" + 9.9 × 10^"-4") color(white)(l)"mol/L"#</mathjax><br/>
<mathjax>#= 2.0 × 10^"-3"color(white)(l) "mol/L"#</mathjax></p>
<p><mathjax>#"pOH" = "-"log["OH"^"-"] = "-"log(2.0 ×10^"-3") = 2.7#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pOH" = "14.00 - 2.7" = 11.3#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 10 mL, 0.1 M NH3 solution when a 0.1 mL, 0.1 M NaOH solution was added into it?</h1>
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<div class="markdown"><p>I tried using an ICE table, but do not come up with a reasonable answer. Thank you in advance!</p></div>
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Ernest Z.
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<div class="markdown"><p>The <mathjax>#"pH = 11.3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To calculate the new concentration of <mathjax>#"NaOH"#</mathjax> in the ammonia solution, you can use the dilution formula</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c_2 = c_1 × V_1/V_2#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#c_1 = "0.1 mol/L"; V_1 = "0.1 mL"#</mathjax><br/>
<mathjax>#c_2 = "?";color(white)(mmmml) V_2 = "(10 + 0.1) mL" = "10.1 mL"#</mathjax></p>
<p><mathjax>#c_2 = 0.1 "mol/L" × (0.10 color(red)(cancel(color(black)("mL"))))/(10.1 color(red)(cancel(color(black)("mL")))) = 9.9 ×10^"-4" "mol/L"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we can use an ICE table to solve the problem.</p>
<p><mathjax>#color(white)(mmmmmm)"NH"_3 + "H"_2"O" → "NH"_4^"+" color(white)(m)+ color(white)(m)"OH"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m)0.1color(white)(mmmmmmm) 0color(white)(mmml)9.9×10^"-4"#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(ll) "-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmmmll) "+"x #</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(l) "0.1 -" xcolor(white)(mmmmmm) x color(white)(mml)x + 9.9×10^"-4"#</mathjax></p>
<p><mathjax>#K_"b" = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#K_"b " = (x(x + 9.9×10^"-4"))/(0.1 - x) = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#0.1/(1.8 × 10^"-5") = 5600 > 400#</mathjax></p>
<p>∴<mathjax># x ≪ 0.1#</mathjax></p>
<blockquote></blockquote>
<p>Then</p>
<p><mathjax>#(x(x + 9.9×10^"-4"))/0.1 = 1.8 × 10^"-5"#</mathjax></p>
<p><mathjax>#x(x + 9.9×10^"-4") = 0.1 × 1.8 × 10^"-5"= 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x = 1.8 × 10^"-6"#</mathjax></p>
<p><mathjax>#x^2 + 9.9×10^"-4"x - 1.8 × 10^"-6" = 0#</mathjax></p>
<p><mathjax>#x = 9.9 × 10^"-4"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (9.9 × 10^"-4" +x) color(white)(l)"mol/L" = (9.9 × 10^"-4" + 9.9 × 10^"-4") color(white)(l)"mol/L"#</mathjax><br/>
<mathjax>#= 2.0 × 10^"-3"color(white)(l) "mol/L"#</mathjax></p>
<p><mathjax>#"pOH" = "-"log["OH"^"-"] = "-"log(2.0 ×10^"-3") = 2.7#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pOH" = "14.00 - 2.7" = 11.3#</mathjax></p></div>
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</article> | What is the pH of a 10 mL, 0.1 M NH3 solution when a 0.1 mL, 0.1 M NaOH solution was added into it? |
I tried using an ICE table, but do not come up with a reasonable answer. Thank you in advance!
|
1,924 | a91fca02-6ddd-11ea-b479-ccda262736ce | https://socratic.org/questions/591b63ef7c01497a1a7d25be | 40.00 g | start physical_unit 3 4 mass g qc_end physical_unit 15 16 10 11 volume qc_end physical_unit 15 16 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium hydroxide [IN] g"}] | [{"type":"physical unit","value":"40.00 g"}] | [{"type":"physical unit","value":"Volume [OF] pamoic acid [=] \\pu{500 mL}"},{"type":"physical unit","value":"Molarity [OF] pamoic acid [=] \\pu{1 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide do I need to neutralize 500 mL of 1 mol/L pamoic acid?</h1> | null | 40.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Assuming that pamoic acid is that soluble, here's how you would do the calculation.</p>
<p>Pamoic acid is a dibasic acid with the formula <mathjax>#"C"_23"H"_16"O"_6#</mathjax>.</p>
<p>We can write the formula as <mathjax>#"H"_2"C"_23"H"_14"O"_6#</mathjax>.</p>
<p>Then the equation for the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is</p>
<p><mathjax>#M_text(r):color(white)(m)338.375color(white)(mmm)39.997#</mathjax><br/>
<mathjax>#color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"#</mathjax><br/>
<mathjax>#color(white)(mmmml)"H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"H"_2"A"#</mathjax></strong></p>
<p><mathjax>#"Moles of H"_2"A" = 0.500 color(red)(cancel(color(black)("L H"_2"A"))) × ("1 mol H"_2"A")/(1 color(red)(cancel(color(black)("L H"_2"A")))) = "0.50 mol H"_2"A"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Moles of NaOH" = 0.50 color(red)(cancel(color(black)("mol H"_2"A"))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"A")))) = "1.0 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Mass of NaOH" = 1.0 color(red)(cancel(color(black)("mol NaOH"))) × "39.997 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "40 g NaOH"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>You need 40 g of <mathjax>#"NaOH"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Assuming that pamoic acid is that soluble, here's how you would do the calculation.</p>
<p>Pamoic acid is a dibasic acid with the formula <mathjax>#"C"_23"H"_16"O"_6#</mathjax>.</p>
<p>We can write the formula as <mathjax>#"H"_2"C"_23"H"_14"O"_6#</mathjax>.</p>
<p>Then the equation for the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is</p>
<p><mathjax>#M_text(r):color(white)(m)338.375color(white)(mmm)39.997#</mathjax><br/>
<mathjax>#color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"#</mathjax><br/>
<mathjax>#color(white)(mmmml)"H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"H"_2"A"#</mathjax></strong></p>
<p><mathjax>#"Moles of H"_2"A" = 0.500 color(red)(cancel(color(black)("L H"_2"A"))) × ("1 mol H"_2"A")/(1 color(red)(cancel(color(black)("L H"_2"A")))) = "0.50 mol H"_2"A"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Moles of NaOH" = 0.50 color(red)(cancel(color(black)("mol H"_2"A"))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"A")))) = "1.0 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Mass of NaOH" = 1.0 color(red)(cancel(color(black)("mol NaOH"))) × "39.997 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "40 g NaOH"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of sodium hydroxide do I need to neutralize 500 mL of 1 mol/L pamoic acid?</h1>
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<div class="markdown"><p>You need 40 g of <mathjax>#"NaOH"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Assuming that pamoic acid is that soluble, here's how you would do the calculation.</p>
<p>Pamoic acid is a dibasic acid with the formula <mathjax>#"C"_23"H"_16"O"_6#</mathjax>.</p>
<p>We can write the formula as <mathjax>#"H"_2"C"_23"H"_14"O"_6#</mathjax>.</p>
<p>Then the equation for the <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> is</p>
<p><mathjax>#M_text(r):color(white)(m)338.375color(white)(mmm)39.997#</mathjax><br/>
<mathjax>#color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"#</mathjax><br/>
<mathjax>#color(white)(mmmml)"H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"H"_2"A"#</mathjax></strong></p>
<p><mathjax>#"Moles of H"_2"A" = 0.500 color(red)(cancel(color(black)("L H"_2"A"))) × ("1 mol H"_2"A")/(1 color(red)(cancel(color(black)("L H"_2"A")))) = "0.50 mol H"_2"A"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Moles of NaOH" = 0.50 color(red)(cancel(color(black)("mol H"_2"A"))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"A")))) = "1.0 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong></p>
<p><mathjax>#"Mass of NaOH" = 1.0 color(red)(cancel(color(black)("mol NaOH"))) × "39.997 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "40 g NaOH"#</mathjax></p></div>
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</article> | What mass of sodium hydroxide do I need to neutralize 500 mL of 1 mol/L pamoic acid? | null |
1,925 | abc8a0c0-6ddd-11ea-98a6-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-change-in-ph-when-3-00-ml-of-0-100-m-hcl-aq-is-added-to | -0.03 | start physical_unit 22 23 ph none qc_end physical_unit 23 23 9 10 volume qc_end physical_unit 14 14 12 13 molarity qc_end physical_unit 22 23 18 19 volume qc_end physical_unit 29 29 12 13 molarity qc_end physical_unit 34 34 12 13 molarity qc_end end | [{"type":"physical unit","value":"Change in pH [OF] buffer solution"}] | [{"type":"physical unit","value":"-0.03"}] | [{"type":"physical unit","value":"Volume [OF] HCl(aq) solution [=] \\pu{3.00 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl(aq) solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Volume [OF] buffer solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Molarity [OF] NH3(aq) solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Molarity [OF] NH4Cl(aq) solution [=] \\pu{0.100 M}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the change in pH when 3.00 mL of 0.100 M #"HCl"(aq)# is added to 100.0 mL of a buffer solution that is 0.100 M in #"NH"_3(aq)# and 0.100 M in #"NH"_4"Cl"(aq)# ?</h1> | null | -0.03 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're adding hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, to your buffer, so right from the start you should expect it <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> to <strong>decrease</strong>. </p>
<p>This implies that the change in pH will be <em>negative</em></p>
<blockquote>
<p><mathjax>#Delta_ "pH" = "pH"_ "final" - "pH"_ "initial" <0#</mathjax></p>
</blockquote>
<p>However, the fact that you're dealing with a buffer solution lets you know that this change will <em><strong>not</strong></em> be significant since the role of a buffer is to <strong>resist</strong> significant changes in pH that result from the addition of strong acid or strong bases. </p>
<p>So, hydrochloric acid will react with ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to produce the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, the ammonia's conjugate acid, and water. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The reaction consumes hydrochloric acid and ammonia in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, <strong>one mole</strong> of ammonium cations is produced. Keep this in mind. </p>
<p>Use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the ammonia solution and the <em>volume</em> of the buffer to calculate how many <strong>moles</strong> it contains </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Do the same for the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Calculate how many <strong>moles</strong> of hydrochloric acid are being added to the buffer</p>
<blockquote>
<p><mathjax>#n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.000300 moles HCl"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You know that hydrochloric acid and ammonia react in a <mathjax>#1:1#</mathjax> mole ratio, which means that the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_("HCl") = "0 moles HCl" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("NH"_3) = "0.0100 moles" - "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0097 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#="0.0103 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the buffer will be </p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL" + "3.00 mL" = "103.0 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of ammonia and ammonium cations will be </p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"#</mathjax></p>
<p><mathjax>#["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"#</mathjax></p>
</blockquote>
<p>Notice that the concentration of ammonium cations remained virtually <em>unchanged</em> because the <strong>increase</strong> in the number of moles was counteracted by the <strong>increase</strong> in volume. </p>
<p>Now, you can find the <em>change</em> in pH by using the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak base</strong> and its <em>conjugate acid</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = "p"K_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that the the initial buffer solution had <strong>equal concentrations</strong> of weak base and conjugate acid. This means that its pOH was equal to </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b#</mathjax></p>
</blockquote>
<p><strong>After</strong> the strong acid is added to the buffer, the pOH of the buffer will be </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b + 0.026#</mathjax></p>
</blockquote>
<p>You know that an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>For the <strong>initial solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "initial" = 14 - "p"K_b#</mathjax></p>
</blockquote>
<p>For the <strong>final solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "final" = 14 - ("p"K_b + 0.026)#</mathjax></p>
<p><mathjax>#"pH"_ "final" = 14 - "p"K_b - 0.026#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the change in pH is equal to </p>
<blockquote>
<p><mathjax>#Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))#</mathjax></p>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(Delta_"pH" = -0.026)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>As predicted, the change in pH is <em>negative</em> because the pH of the buffer <strong>decreased</strong> as a result of the addition of a strong acid. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Delta_"pH" = -0.026#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're adding hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, to your buffer, so right from the start you should expect it <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> to <strong>decrease</strong>. </p>
<p>This implies that the change in pH will be <em>negative</em></p>
<blockquote>
<p><mathjax>#Delta_ "pH" = "pH"_ "final" - "pH"_ "initial" <0#</mathjax></p>
</blockquote>
<p>However, the fact that you're dealing with a buffer solution lets you know that this change will <em><strong>not</strong></em> be significant since the role of a buffer is to <strong>resist</strong> significant changes in pH that result from the addition of strong acid or strong bases. </p>
<p>So, hydrochloric acid will react with ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to produce the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, the ammonia's conjugate acid, and water. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The reaction consumes hydrochloric acid and ammonia in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, <strong>one mole</strong> of ammonium cations is produced. Keep this in mind. </p>
<p>Use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the ammonia solution and the <em>volume</em> of the buffer to calculate how many <strong>moles</strong> it contains </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Do the same for the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Calculate how many <strong>moles</strong> of hydrochloric acid are being added to the buffer</p>
<blockquote>
<p><mathjax>#n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.000300 moles HCl"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You know that hydrochloric acid and ammonia react in a <mathjax>#1:1#</mathjax> mole ratio, which means that the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_("HCl") = "0 moles HCl" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("NH"_3) = "0.0100 moles" - "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0097 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#="0.0103 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the buffer will be </p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL" + "3.00 mL" = "103.0 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of ammonia and ammonium cations will be </p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"#</mathjax></p>
<p><mathjax>#["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"#</mathjax></p>
</blockquote>
<p>Notice that the concentration of ammonium cations remained virtually <em>unchanged</em> because the <strong>increase</strong> in the number of moles was counteracted by the <strong>increase</strong> in volume. </p>
<p>Now, you can find the <em>change</em> in pH by using the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak base</strong> and its <em>conjugate acid</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = "p"K_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that the the initial buffer solution had <strong>equal concentrations</strong> of weak base and conjugate acid. This means that its pOH was equal to </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b#</mathjax></p>
</blockquote>
<p><strong>After</strong> the strong acid is added to the buffer, the pOH of the buffer will be </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b + 0.026#</mathjax></p>
</blockquote>
<p>You know that an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>For the <strong>initial solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "initial" = 14 - "p"K_b#</mathjax></p>
</blockquote>
<p>For the <strong>final solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "final" = 14 - ("p"K_b + 0.026)#</mathjax></p>
<p><mathjax>#"pH"_ "final" = 14 - "p"K_b - 0.026#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the change in pH is equal to </p>
<blockquote>
<p><mathjax>#Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))#</mathjax></p>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(Delta_"pH" = -0.026)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>As predicted, the change in pH is <em>negative</em> because the pH of the buffer <strong>decreased</strong> as a result of the addition of a strong acid. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you calculate the change in pH when 3.00 mL of 0.100 M #"HCl"(aq)# is added to 100.0 mL of a buffer solution that is 0.100 M in #"NH"_3(aq)# and 0.100 M in #"NH"_4"Cl"(aq)# ?</h1>
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Stefan V.
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Aug 2, 2016
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<div class="markdown"><p><mathjax>#Delta_"pH" = -0.026#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're adding hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, to your buffer, so right from the start you should expect it <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> to <strong>decrease</strong>. </p>
<p>This implies that the change in pH will be <em>negative</em></p>
<blockquote>
<p><mathjax>#Delta_ "pH" = "pH"_ "final" - "pH"_ "initial" <0#</mathjax></p>
</blockquote>
<p>However, the fact that you're dealing with a buffer solution lets you know that this change will <em><strong>not</strong></em> be significant since the role of a buffer is to <strong>resist</strong> significant changes in pH that result from the addition of strong acid or strong bases. </p>
<p>So, hydrochloric acid will react with ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to produce the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, the ammonia's conjugate acid, and water. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>The reaction consumes hydrochloric acid and ammonia in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, <strong>one mole</strong> of ammonium cations is produced. Keep this in mind. </p>
<p>Use the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> of the ammonia solution and the <em>volume</em> of the buffer to calculate how many <strong>moles</strong> it contains </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Do the same for the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0100 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Calculate how many <strong>moles</strong> of hydrochloric acid are being added to the buffer</p>
<blockquote>
<p><mathjax>#n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.000300 moles HCl"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>You know that hydrochloric acid and ammonia react in a <mathjax>#1:1#</mathjax> mole ratio, which means that the resulting solution will contain </p>
<blockquote>
<p><mathjax>#n_("HCl") = "0 moles HCl" ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("NH"_3) = "0.0100 moles" - "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#= "0.0097 moles NH"_3#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#="0.0103 moles NH"_4^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The <strong>total volume</strong> of the buffer will be </p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL" + "3.00 mL" = "103.0 mL"#</mathjax></p>
</blockquote>
<p>The new concentrations of ammonia and ammonium cations will be </p>
<blockquote>
<p><mathjax>#["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"#</mathjax></p>
<p><mathjax>#["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"#</mathjax></p>
</blockquote>
<p>Notice that the concentration of ammonium cations remained virtually <em>unchanged</em> because the <strong>increase</strong> in the number of moles was counteracted by the <strong>increase</strong> in volume. </p>
<p>Now, you can find the <em>change</em> in pH by using the <strong>Henderson - Hasselbalch equation</strong>, which for a buffer that contains a <strong>weak base</strong> and its <em>conjugate acid</em> looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = "p"K_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that the the initial buffer solution had <strong>equal concentrations</strong> of weak base and conjugate acid. This means that its pOH was equal to </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b#</mathjax></p>
</blockquote>
<p><strong>After</strong> the strong acid is added to the buffer, the pOH of the buffer will be </p>
<blockquote>
<p><mathjax>#"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = "p"K_b + 0.026#</mathjax></p>
</blockquote>
<p>You know that an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>For the <strong>initial solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "initial" = 14 - "p"K_b#</mathjax></p>
</blockquote>
<p>For the <strong>final solution</strong>, you have</p>
<blockquote>
<p><mathjax>#"pH"_ "final" = 14 - ("p"K_b + 0.026)#</mathjax></p>
<p><mathjax>#"pH"_ "final" = 14 - "p"K_b - 0.026#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the change in pH is equal to </p>
<blockquote>
<p><mathjax>#Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))#</mathjax></p>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(Delta_"pH" = -0.026)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>As predicted, the change in pH is <em>negative</em> because the pH of the buffer <strong>decreased</strong> as a result of the addition of a strong acid. </p></div>
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</article> | How do you calculate the change in pH when 3.00 mL of 0.100 M #"HCl"(aq)# is added to 100.0 mL of a buffer solution that is 0.100 M in #"NH"_3(aq)# and 0.100 M in #"NH"_4"Cl"(aq)# ? | null |
1,926 | a99be5d8-6ddd-11ea-8036-ccda262736ce | https://socratic.org/questions/a-45-8-mg-sample-of-silicon-reacts-with-hydrogen-to-form-50-7-mg-of-the-compound | SiH3 | start chemical_formula qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 14 15 11 12 mass qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] silicon hydride [IN] empirical"}] | [{"type":"chemical equation","value":"SiH3"}] | [{"type":"physical unit","value":"Mass [OF] silicon sample [=] \\pu{45.8 mg}"},{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{50.7 mg}"},{"type":"substance name","value":"Hydrogen"}] | <h1 class="questionTitle" itemprop="name">A 45.8 mg sample of silicon reacts with hydrogen to form 50.7 mg of the compound. What is the empirical formula of the silicon hydride?</h1> | null | SiH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#color(white)(mll)"Si"color(white)(ll) +color(white)(ll) "H"_2 color(white)(l)→color(white)(l) ("Si, H")#</mathjax><br/>
<mathjax>#"45.8 mg"color(white)(m) xcolor(white)(l) "mg"color(white)(mm) "50.7 mg"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Mass of H" = xcolor(white)(l) "mg" = "mass of (Si, H) – mass of Si"#</mathjax></p>
<p><mathjax>#"= 50.7 mg – 45.8 mg = 4.9 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of Si " = 45.8color(red)(cancel(color(black)("mg Si"))) × "1 mmol Si"/(28.08 color(red)(cancel(color(black)("mg Si")))) = "1.631 mmol Si"#</mathjax></p>
<p><mathjax>#"Moles of H" = 4.9 color(red)(cancel(color(black)("mg H"))) × "1 mmol H"/(1.008 color(red)(cancel(color(black)("mg H")))) = "4.86 mmol H"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"1.631"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "mmol"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(m)"Si" color(white)(XXXm)45.8 color(white)(Xmm)"1.631"
color(white)(Xm)1color(white)(Xmmmm)1#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXll)4.9 color(white)(mmm)"4.86" color(white)(Xmll)2.98 color(white)(XXXl)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"SiH"_3#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#"SiH"_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#color(white)(mll)"Si"color(white)(ll) +color(white)(ll) "H"_2 color(white)(l)→color(white)(l) ("Si, H")#</mathjax><br/>
<mathjax>#"45.8 mg"color(white)(m) xcolor(white)(l) "mg"color(white)(mm) "50.7 mg"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Mass of H" = xcolor(white)(l) "mg" = "mass of (Si, H) – mass of Si"#</mathjax></p>
<p><mathjax>#"= 50.7 mg – 45.8 mg = 4.9 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of Si " = 45.8color(red)(cancel(color(black)("mg Si"))) × "1 mmol Si"/(28.08 color(red)(cancel(color(black)("mg Si")))) = "1.631 mmol Si"#</mathjax></p>
<p><mathjax>#"Moles of H" = 4.9 color(red)(cancel(color(black)("mg H"))) × "1 mmol H"/(1.008 color(red)(cancel(color(black)("mg H")))) = "4.86 mmol H"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"1.631"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "mmol"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(m)"Si" color(white)(XXXm)45.8 color(white)(Xmm)"1.631"
color(white)(Xm)1color(white)(Xmmmm)1#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXll)4.9 color(white)(mmm)"4.86" color(white)(Xmll)2.98 color(white)(XXXl)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"SiH"_3#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 45.8 mg sample of silicon reacts with hydrogen to form 50.7 mg of the compound. What is the empirical formula of the silicon hydride?</h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#"SiH"_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#color(white)(mll)"Si"color(white)(ll) +color(white)(ll) "H"_2 color(white)(l)→color(white)(l) ("Si, H")#</mathjax><br/>
<mathjax>#"45.8 mg"color(white)(m) xcolor(white)(l) "mg"color(white)(mm) "50.7 mg"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Mass of H" = xcolor(white)(l) "mg" = "mass of (Si, H) – mass of Si"#</mathjax></p>
<p><mathjax>#"= 50.7 mg – 45.8 mg = 4.9 mg"#</mathjax></p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of Si " = 45.8color(red)(cancel(color(black)("mg Si"))) × "1 mmol Si"/(28.08 color(red)(cancel(color(black)("mg Si")))) = "1.631 mmol Si"#</mathjax></p>
<p><mathjax>#"Moles of H" = 4.9 color(red)(cancel(color(black)("mg H"))) × "1 mmol H"/(1.008 color(red)(cancel(color(black)("mg H")))) = "4.86 mmol H"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"1.631"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/mg"color(white)(X) "mmol"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(m)"Si" color(white)(XXXm)45.8 color(white)(Xmm)"1.631"
color(white)(Xm)1color(white)(Xmmmm)1#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXll)4.9 color(white)(mmm)"4.86" color(white)(Xmll)2.98 color(white)(XXXl)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"SiH"_3#</mathjax>.</p></div>
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</article> | A 45.8 mg sample of silicon reacts with hydrogen to form 50.7 mg of the compound. What is the empirical formula of the silicon hydride? | null |
1,927 | aa391e7b-6ddd-11ea-a066-ccda262736ce | https://socratic.org/questions/if-the-initial-temperature-of-an-ideal-gas-at-2-250-atm-is-62-0-degrees-c-what-f | 260.71 K | start physical_unit 6 7 temperature k qc_end physical_unit 6 7 12 14 temperature qc_end physical_unit 6 7 9 10 pressure qc_end physical_unit 6 7 26 27 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the ideal gas [IN] K"}] | [{"type":"physical unit","value":"260.71 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the ideal gas [=] \\pu{62.0 degrees C}"},{"type":"physical unit","value":"Pressure1 [OF] the ideal gas [=] \\pu{2.250 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the ideal gas [=] \\pu{1.750 atm}"}] | <h1 class="questionTitle" itemprop="name">If the initial temperature of an ideal gas at 2.250 atm is 62.0 degrees C, what final temperature would cause the pressure to be reduced to 1.750 atm?</h1> | null | 260.71 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question is an example of Gay-Lussac's law relating temperature and pressure. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, the temperature also increases, and vice versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The temperature must be in Kelvins. To convert degrees Celsius to Kelvins, add <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize the data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#P_1="2.250 atm"#</mathjax></p>
<p><mathjax>#T_1="62.0"^@"C" + 273.15="335.2 K"#</mathjax></p>
<p><mathjax>#P_2="1.750 atm"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Insert the known data and solve.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.750"atm"xx335.2"K")/(2.250"atm")="260.7 K"#</mathjax></p>
<p>To convert Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#260.7-273.15="-12.45"^@"C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final temperature would be <mathjax>#"260.7 K"#</mathjax> or <mathjax>#-12.45^@"C"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question is an example of Gay-Lussac's law relating temperature and pressure. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, the temperature also increases, and vice versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The temperature must be in Kelvins. To convert degrees Celsius to Kelvins, add <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize the data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#P_1="2.250 atm"#</mathjax></p>
<p><mathjax>#T_1="62.0"^@"C" + 273.15="335.2 K"#</mathjax></p>
<p><mathjax>#P_2="1.750 atm"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Insert the known data and solve.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.750"atm"xx335.2"K")/(2.250"atm")="260.7 K"#</mathjax></p>
<p>To convert Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#260.7-273.15="-12.45"^@"C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the initial temperature of an ideal gas at 2.250 atm is 62.0 degrees C, what final temperature would cause the pressure to be reduced to 1.750 atm?</h1>
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<div class="markdown"><p>The final temperature would be <mathjax>#"260.7 K"#</mathjax> or <mathjax>#-12.45^@"C"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This question is an example of Gay-Lussac's law relating temperature and pressure. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, the temperature also increases, and vice versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The temperature must be in Kelvins. To convert degrees Celsius to Kelvins, add <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize the data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#P_1="2.250 atm"#</mathjax></p>
<p><mathjax>#T_1="62.0"^@"C" + 273.15="335.2 K"#</mathjax></p>
<p><mathjax>#P_2="1.750 atm"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#T_2#</mathjax>. Insert the known data and solve.</p>
<p><mathjax>#T_2=(P_2T_1)/(P_1)#</mathjax></p>
<p><mathjax>#T_2=(1.750"atm"xx335.2"K")/(2.250"atm")="260.7 K"#</mathjax></p>
<p>To convert Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#260.7-273.15="-12.45"^@"C"#</mathjax></p></div>
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</article> | If the initial temperature of an ideal gas at 2.250 atm is 62.0 degrees C, what final temperature would cause the pressure to be reduced to 1.750 atm? | null |
1,928 | aa0d7786-6ddd-11ea-ab09-ccda262736ce | https://socratic.org/questions/given-a-500-m-sample-of-h-2-at-2-00-atm-pressure-what-will-be-the-volume-when-th | 1055.6 mL | start physical_unit 4 6 volume ml qc_end physical_unit 4 6 2 3 volume qc_end physical_unit 4 6 8 9 pressure qc_end physical_unit 4 6 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] H2 sample [IN] mL"}] | [{"type":"physical unit","value":"1055.6 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] H2 sample [=] \\pu{500 mL}"},{"type":"physical unit","value":"Pressure1 [OF] H2 sample [=] \\pu{2.00 atm}"},{"type":"physical unit","value":"Pressure2 [OF] H2 sample [=] \\pu{720 torr}"}] | <h1 class="questionTitle" itemprop="name">Given a 500 m sample of H#_2# at 2.00 atm pressure. What will be the volume when the pressure is changed to 720. torr? </h1> | null | 1055.6 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(2*atmxx500*mL)/((720*"Torr")/(760*"Torr"*"atm"^-1))=1055.6*mL#</mathjax>.</p></div>
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<div class="markdown"><p>Well, <mathjax>#P_1V_1=P_2V_2#</mathjax>........so <mathjax>#V_2=1055.6*mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(2*atmxx500*mL)/((720*"Torr")/(760*"Torr"*"atm"^-1))=1055.6*mL#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Given a 500 m sample of H#_2# at 2.00 atm pressure. What will be the volume when the pressure is changed to 720. torr? </h1>
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<div class="markdown"><p>Well, <mathjax>#P_1V_1=P_2V_2#</mathjax>........so <mathjax>#V_2=1055.6*mL#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=(2*atmxx500*mL)/((720*"Torr")/(760*"Torr"*"atm"^-1))=1055.6*mL#</mathjax>.</p></div>
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</article> | Given a 500 m sample of H#_2# at 2.00 atm pressure. What will be the volume when the pressure is changed to 720. torr? | null |
1,929 | ad1165b6-6ddd-11ea-8060-ccda262736ce | https://socratic.org/questions/a-chemistry-teacher-adds-50-0-ml-of-1-50-m-h-2so-4-solution-to-200-ml-of-water-w | 0.30 M | start physical_unit 9 10 concentration mol/l qc_end physical_unit 9 10 4 5 volume qc_end physical_unit 9 10 7 8 concentration qc_end physical_unit 15 15 12 13 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] H2SO4 solution [IN] M"}] | [{"type":"physical unit","value":"0.30 M"}] | [{"type":"physical unit","value":"Volume1 [OF] H2SO4 solution [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Concentration1 [OF] H2SO4 solution [=] \\pu{1.50 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{200 mL}"}] | <h1 class="questionTitle" itemprop="name"> A chemistry teacher adds 50.0 mL of 1.50 M #H_2SO_4# solution to 200 mL of water. What is the concentration of the final solution?</h1> | null | 0.30 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of substance"/"Volume"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(50xx10^-3Lxx1.50*mol*L^-1)/((50+200)xx10^-3*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.30*mol*L^-1"with respect to "H_2SO_4#</mathjax>.</p>
<p>What is the concentration with respect to <mathjax>#H_3O^+#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#0.30*mol*L^-1#</mathjax> with respect to sulfuric acid. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of substance"/"Volume"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(50xx10^-3Lxx1.50*mol*L^-1)/((50+200)xx10^-3*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.30*mol*L^-1"with respect to "H_2SO_4#</mathjax>.</p>
<p>What is the concentration with respect to <mathjax>#H_3O^+#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name"> A chemistry teacher adds 50.0 mL of 1.50 M #H_2SO_4# solution to 200 mL of water. What is the concentration of the final solution?</h1>
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<div class="markdown"><p><mathjax>#0.30*mol*L^-1#</mathjax> with respect to sulfuric acid. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of substance"/"Volume"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(50xx10^-3Lxx1.50*mol*L^-1)/((50+200)xx10^-3*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.30*mol*L^-1"with respect to "H_2SO_4#</mathjax>.</p>
<p>What is the concentration with respect to <mathjax>#H_3O^+#</mathjax>?</p></div>
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</article> | A chemistry teacher adds 50.0 mL of 1.50 M #H_2SO_4# solution to 200 mL of water. What is the concentration of the final solution? | null |
1,930 | ac9cf4bb-6ddd-11ea-9ef1-ccda262736ce | https://socratic.org/questions/how-do-you-write-the-complete-equation-for-c-10h-20-l-o-2-g | C10H20(l) + 15 O2(g) -> 10 CO2(g) + 10 H2O(l) | start chemical_equation qc_end chemical_equation 8 8 qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the complete equation"}] | [{"type":"chemical equation","value":"C10H20(l) + 15 O2(g) -> 10 CO2(g) + 10 H2O(l)"}] | [{"type":"chemical equation","value":"C10H20(l)"},{"type":"chemical equation","value":"O2(g)"}] | <h1 class="questionTitle" itemprop="name">How do you write the complete equation for #C_10H_20(l) + O_2(g)#?</h1> | null | C10H20(l) + 15 O2(g) -> 10 CO2(g) + 10 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is figure out what the <strong>products</strong> of this reaction will be. </p>
<p>As you can see, you're dealing with the combustion of a <strong>hydrocarbon</strong>, which is a compound that only contains carbon and hydrogen. </p>
<p>The <strong>complete combustion</strong> of a hydrocarbon will <strong>always</strong> produce <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. This means that the <em>unbalanced chemical equation</em> for this presumed complete combustion will look like this</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>An important thing to realize here is that you have an <em>isolated</em> oxygen molecule on the reactants' side. This means that you can use <strong>fractional coefficients</strong> as a tool to get the balanced chemical equation. </p>
<p>So, focus on balancing the carbon and hydrogen atoms first. </p>
<p>The thing to remember here is that <strong>every atom</strong> that is found on the reactants' side <strong>must be accounted for</strong> on the products' side. </p>
<p>Notice that you have <mathjax>#10#</mathjax> carbon atoms on the reactants' side, but only <mathjax>#1#</mathjax> on the products' side, as part of the carbon dioxide molecule. </p>
<p>Multiply this molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now balance the hydrogen atoms. You know that you have <mathjax>#20#</mathjax> hydrogen atoms on the reactants' side, but only <mathjax>#2#</mathjax> on the products' side, as part of the water molecule.</p>
<p>Once again, multiply the water molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, focus on balancing the oxygen atoms. Notice that you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side as part of the oxygen molecule, but a total of </p>
<blockquote>
<p><mathjax>#color(red)(10) xx 2 + color(blue)(10) xx 1 = 30#</mathjax></p>
</blockquote>
<p>atoms of oxygen on the products' side. This means that you must multiply the oxygen molecule by <mathjax>#15#</mathjax> to get the <em>balanced chemical equation</em> for this reaction</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + color(green)(15)"O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_10"H"_text(20(l]) + 15"O"_text(2(g]) -> 10"CO"_text(2(g]) + 10"H"_2"O"_text((l])#</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 figure out what the <strong>products</strong> of this reaction will be. </p>
<p>As you can see, you're dealing with the combustion of a <strong>hydrocarbon</strong>, which is a compound that only contains carbon and hydrogen. </p>
<p>The <strong>complete combustion</strong> of a hydrocarbon will <strong>always</strong> produce <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. This means that the <em>unbalanced chemical equation</em> for this presumed complete combustion will look like this</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>An important thing to realize here is that you have an <em>isolated</em> oxygen molecule on the reactants' side. This means that you can use <strong>fractional coefficients</strong> as a tool to get the balanced chemical equation. </p>
<p>So, focus on balancing the carbon and hydrogen atoms first. </p>
<p>The thing to remember here is that <strong>every atom</strong> that is found on the reactants' side <strong>must be accounted for</strong> on the products' side. </p>
<p>Notice that you have <mathjax>#10#</mathjax> carbon atoms on the reactants' side, but only <mathjax>#1#</mathjax> on the products' side, as part of the carbon dioxide molecule. </p>
<p>Multiply this molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now balance the hydrogen atoms. You know that you have <mathjax>#20#</mathjax> hydrogen atoms on the reactants' side, but only <mathjax>#2#</mathjax> on the products' side, as part of the water molecule.</p>
<p>Once again, multiply the water molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, focus on balancing the oxygen atoms. Notice that you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side as part of the oxygen molecule, but a total of </p>
<blockquote>
<p><mathjax>#color(red)(10) xx 2 + color(blue)(10) xx 1 = 30#</mathjax></p>
</blockquote>
<p>atoms of oxygen on the products' side. This means that you must multiply the oxygen molecule by <mathjax>#15#</mathjax> to get the <em>balanced chemical equation</em> for this reaction</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + color(green)(15)"O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you write the complete equation for #C_10H_20(l) + O_2(g)#?</h1>
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<div class="answer" id="204428" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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Stefan V.
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<span class="dateCreated" datetime="2015-12-27T17:26:39" itemprop="dateCreated">
Dec 27, 2015
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<div class="markdown"><p><mathjax>#"C"_10"H"_text(20(l]) + 15"O"_text(2(g]) -> 10"CO"_text(2(g]) + 10"H"_2"O"_text((l])#</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 figure out what the <strong>products</strong> of this reaction will be. </p>
<p>As you can see, you're dealing with the combustion of a <strong>hydrocarbon</strong>, which is a compound that only contains carbon and hydrogen. </p>
<p>The <strong>complete combustion</strong> of a hydrocarbon will <strong>always</strong> produce <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. This means that the <em>unbalanced chemical equation</em> for this presumed complete combustion will look like this</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>An important thing to realize here is that you have an <em>isolated</em> oxygen molecule on the reactants' side. This means that you can use <strong>fractional coefficients</strong> as a tool to get the balanced chemical equation. </p>
<p>So, focus on balancing the carbon and hydrogen atoms first. </p>
<p>The thing to remember here is that <strong>every atom</strong> that is found on the reactants' side <strong>must be accounted for</strong> on the products' side. </p>
<p>Notice that you have <mathjax>#10#</mathjax> carbon atoms on the reactants' side, but only <mathjax>#1#</mathjax> on the products' side, as part of the carbon dioxide molecule. </p>
<p>Multiply this molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Now balance the hydrogen atoms. You know that you have <mathjax>#20#</mathjax> hydrogen atoms on the reactants' side, but only <mathjax>#2#</mathjax> on the products' side, as part of the water molecule.</p>
<p>Once again, multiply the water molecule by <mathjax>#10#</mathjax> to balance the carbon atoms</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + "O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Finally, focus on balancing the oxygen atoms. Notice that you have <mathjax>#2#</mathjax> oxygen atoms on the reactants' side as part of the oxygen molecule, but a total of </p>
<blockquote>
<p><mathjax>#color(red)(10) xx 2 + color(blue)(10) xx 1 = 30#</mathjax></p>
</blockquote>
<p>atoms of oxygen on the products' side. This means that you must multiply the oxygen molecule by <mathjax>#15#</mathjax> to get the <em>balanced chemical equation</em> for this reaction</p>
<blockquote>
<p><mathjax>#"C"_10"H"_text(20(l]) + color(green)(15)"O"_text(2(g]) -> color(red)(10)"CO"_text(2(g]) + color(blue)(10)"H"_2"O"_text((l])#</mathjax></p>
</blockquote></div>
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</article> | How do you write the complete equation for #C_10H_20(l) + O_2(g)#? | null |
1,931 | acd557c3-6ddd-11ea-8508-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-barium-phosphate | Ba3(PO4)2 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] barium phosphate [IN] default"}] | [{"type":"chemical equation","value":"Ba3(PO4)2"}] | [{"type":"substance name","value":"Barium phosphate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for barium phosphate?</h1> | null | Ba3(PO4)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The salt derives barium ion, <mathjax>#Ba^(2+)#</mathjax>, and phosphate ion, <mathjax>#PO_4^(3-)#</mathjax>. A neutral salt is formed from the formulation <mathjax>#Ba_3(PO_4)_2#</mathjax>.</p>
<p>Most of the time, we probably deal with the bisphosphate, <mathjax>#BaHPO_4#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Nominally <mathjax>#Ba_3(PO_4)_2#</mathjax>......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The salt derives barium ion, <mathjax>#Ba^(2+)#</mathjax>, and phosphate ion, <mathjax>#PO_4^(3-)#</mathjax>. A neutral salt is formed from the formulation <mathjax>#Ba_3(PO_4)_2#</mathjax>.</p>
<p>Most of the time, we probably deal with the bisphosphate, <mathjax>#BaHPO_4#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the formula for barium phosphate?</h1>
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anor277
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<div class="markdown"><p>Nominally <mathjax>#Ba_3(PO_4)_2#</mathjax>......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The salt derives barium ion, <mathjax>#Ba^(2+)#</mathjax>, and phosphate ion, <mathjax>#PO_4^(3-)#</mathjax>. A neutral salt is formed from the formulation <mathjax>#Ba_3(PO_4)_2#</mathjax>.</p>
<p>Most of the time, we probably deal with the bisphosphate, <mathjax>#BaHPO_4#</mathjax>.</p></div>
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</article> | What is the formula for barium phosphate? | null |
1,932 | ab9e0da2-6ddd-11ea-9cea-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-that-is-31-9-potassium-28-9-chlorin | KClO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"KClO3"}] | [{"type":"physical unit","value":"Percent [OF] potassium in the compound [=] \\pu{31.9%}"},{"type":"physical unit","value":"Percent [OF] chlorine in the compound [=] \\pu{28.9%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{39.2%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?</h1> | null | KClO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of stuff, there are 31.9 g potassium, 28.9 g of chlorine, and 39.2 g oxygen.</p>
<p>We divide these gram quantities by the respective atomic masses of each of the atoms to get the atomic ratio of constituents.</p>
<p><mathjax>#K#</mathjax>: <mathjax>#(31.9*g)/(39.1*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#Cl#</mathjax>: <mathjax>#(28.9*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#O#</mathjax>: <mathjax>#(39.2*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.45*mol#</mathjax>.</p>
<p>If we divide thru by the least molar quantity (that of potassium or oxygen), we get an empirical formula of <mathjax>#KClO_3#</mathjax>, potassium chlorate.</p>
<p>Note that because this is an ionic, and not a molecular species, we can quote the empirical as the actual formula.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#KClO_3#</mathjax>, potassium chlorate.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of stuff, there are 31.9 g potassium, 28.9 g of chlorine, and 39.2 g oxygen.</p>
<p>We divide these gram quantities by the respective atomic masses of each of the atoms to get the atomic ratio of constituents.</p>
<p><mathjax>#K#</mathjax>: <mathjax>#(31.9*g)/(39.1*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#Cl#</mathjax>: <mathjax>#(28.9*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#O#</mathjax>: <mathjax>#(39.2*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.45*mol#</mathjax>.</p>
<p>If we divide thru by the least molar quantity (that of potassium or oxygen), we get an empirical formula of <mathjax>#KClO_3#</mathjax>, potassium chlorate.</p>
<p>Note that because this is an ionic, and not a molecular species, we can quote the empirical as the actual formula.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?</h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#KClO_3#</mathjax>, potassium chlorate.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of stuff, there are 31.9 g potassium, 28.9 g of chlorine, and 39.2 g oxygen.</p>
<p>We divide these gram quantities by the respective atomic masses of each of the atoms to get the atomic ratio of constituents.</p>
<p><mathjax>#K#</mathjax>: <mathjax>#(31.9*g)/(39.1*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#Cl#</mathjax>: <mathjax>#(28.9*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.816*mol#</mathjax>.</p>
<p><mathjax>#O#</mathjax>: <mathjax>#(39.2*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.45*mol#</mathjax>.</p>
<p>If we divide thru by the least molar quantity (that of potassium or oxygen), we get an empirical formula of <mathjax>#KClO_3#</mathjax>, potassium chlorate.</p>
<p>Note that because this is an ionic, and not a molecular species, we can quote the empirical as the actual formula.</p></div>
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</article> | What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen? | null |
1,933 | acb55aee-6ddd-11ea-a33b-ccda262736ce | https://socratic.org/questions/mg-and-o-2-react-in-a-2-1-molar-ratio-2-moles-m-g-1-mole-o-2-if-a-reaction-used- | 49.4 g | start physical_unit 0 0 mass g qc_end physical_unit 0 2 6 6 molar_ratio qc_end physical_unit 0 0 9 10 mole qc_end physical_unit 2 2 13 14 mole qc_end physical_unit 2 2 20 21 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Mg [IN] g"}] | [{"type":"physical unit","value":"49.4 g"}] | [{"type":"physical unit","value":"Molar ratio [OF] Mg and O2 [=] \\pu{2:1}"},{"type":"physical unit","value":"Mole [OF] Mg [=] \\pu{2 moles}"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{1 mole}"},{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{32.5 g}"}] | <h1 class="questionTitle" itemprop="name">#Mg# and #O_2# react in a 2.1 molar ratio. 2 moles #M_g# = 1 mole #O_2#. If a reaction used 32.5 g of #O_2#, how many g of #Mg# reacted?</h1> | null | 49.4 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"O"_2#</mathjax>.</strong></p>
<p><mathjax>#32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Use the molar ratio to calculate the moles of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"#</mathjax></p>
<p>The reaction required <mathjax>#"49.4 g Mg"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"49.4 g of Mg"#</mathjax> reacted.</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"_2#</mathjax>.</strong></p>
<p><mathjax>#32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Use the molar ratio to calculate the moles of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"#</mathjax></p>
<p>The reaction required <mathjax>#"49.4 g Mg"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">#Mg# and #O_2# react in a 2.1 molar ratio. 2 moles #M_g# = 1 mole #O_2#. If a reaction used 32.5 g of #O_2#, how many g of #Mg# reacted?</h1>
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<div class="markdown"><p><mathjax>#"49.4 g of Mg"#</mathjax> reacted.</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"_2#</mathjax>.</strong></p>
<p><mathjax>#32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Use the molar ratio to calculate the moles of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the mass of <mathjax>#"Mg"#</mathjax>.</strong></p>
<p><mathjax>#2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"#</mathjax></p>
<p>The reaction required <mathjax>#"49.4 g Mg"#</mathjax>.</p></div>
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</article> | #Mg# and #O_2# react in a 2.1 molar ratio. 2 moles #M_g# = 1 mole #O_2#. If a reaction used 32.5 g of #O_2#, how many g of #Mg# reacted? | null |
1,934 | a979d34c-6ddd-11ea-95ef-ccda262736ce | https://socratic.org/questions/the-ph-of-a-solution-changes-fro-300-to-6-00-by-what-factor-does-the-h-3o-change | 1000 | start physical_unit 15 15 changed_factor none qc_end physical_unit 3 4 7 7 ph qc_end physical_unit 3 4 9 9 ph qc_end end | [{"type":"physical unit","value":"Changed factor [OF] [H3O+]"}] | [{"type":"physical unit","value":"1000"}] | [{"type":"physical unit","value":"pH1 [OF] a solution [=] \\pu{3.00}"},{"type":"physical unit","value":"pH2 [OF] a solution [=] \\pu{6.00}"}] | <h1 class="questionTitle" itemprop="name">The pH of a solution changes fro 300 to 6.00. By what factor does the #[H_3O^+]# change?</h1> | null | 1000 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>If <mathjax>#pH_1=3.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-3*mol*L^-1.#</mathjax></p>
<p>If <mathjax>#pH_2=6.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-6*mol*L^-1.#</mathjax></p>
<p><mathjax>#(pH_1)/(pH_2)=(10^-3*mol*L^-1)/(10^-6*mol*L^-1)=10^3#</mathjax></p>
<p>The latter concentration is less concentrated with respect to <mathjax>#H_3O^+#</mathjax> than the former by a factor of <mathjax>#1000#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>By a factor of <mathjax>#1000#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>If <mathjax>#pH_1=3.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-3*mol*L^-1.#</mathjax></p>
<p>If <mathjax>#pH_2=6.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-6*mol*L^-1.#</mathjax></p>
<p><mathjax>#(pH_1)/(pH_2)=(10^-3*mol*L^-1)/(10^-6*mol*L^-1)=10^3#</mathjax></p>
<p>The latter concentration is less concentrated with respect to <mathjax>#H_3O^+#</mathjax> than the former by a factor of <mathjax>#1000#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">The pH of a solution changes fro 300 to 6.00. By what factor does the #[H_3O^+]# change?</h1>
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<div class="markdown"><p>By a factor of <mathjax>#1000#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>If <mathjax>#pH_1=3.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-3*mol*L^-1.#</mathjax></p>
<p>If <mathjax>#pH_2=6.00#</mathjax>, then <mathjax>#[H_3O^+]=10^-6*mol*L^-1.#</mathjax></p>
<p><mathjax>#(pH_1)/(pH_2)=(10^-3*mol*L^-1)/(10^-6*mol*L^-1)=10^3#</mathjax></p>
<p>The latter concentration is less concentrated with respect to <mathjax>#H_3O^+#</mathjax> than the former by a factor of <mathjax>#1000#</mathjax>.</p></div>
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</article> | The pH of a solution changes fro 300 to 6.00. By what factor does the #[H_3O^+]# change? | null |
1,935 | ac36109e-6ddd-11ea-8b7d-ccda262736ce | https://socratic.org/questions/how-many-moles-of-air-are-in-a-6-06-l-tire-at-stp | 0.27 moles | start physical_unit 4 4 mole mol qc_end c_other STP qc_end physical_unit 4 4 8 9 volume qc_end end | [{"type":"physical unit","value":"Mole [OF] air [IN] moles"}] | [{"type":"physical unit","value":"0.27 moles"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Volume [OF] air [=] \\pu{6.06 L}"}] | <h1 class="questionTitle" itemprop="name">How many moles of air are in a 6.06-L tire at STP? </h1> | null | 0.27 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus at <mathjax>#"STP"#</mathjax>, <mathjax>#6.06*L#</mathjax> represents <mathjax>#(6.06*L)/(22.4*L*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax>, <mathjax>#~=1/4*mol#</mathjax>.</p></div>
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<div class="markdown"><p><a href="http://www.chemteam.info/GasLaw/MolarVolume.html" rel="nofollow">The molar volume at STP</a> is <mathjax>#22.4*L#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus at <mathjax>#"STP"#</mathjax>, <mathjax>#6.06*L#</mathjax> represents <mathjax>#(6.06*L)/(22.4*L*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax>, <mathjax>#~=1/4*mol#</mathjax>.</p></div>
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<div class="markdown"><p><a href="http://www.chemteam.info/GasLaw/MolarVolume.html" rel="nofollow">The molar volume at STP</a> is <mathjax>#22.4*L#</mathjax> </p></div>
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<div class="markdown"><p>And thus at <mathjax>#"STP"#</mathjax>, <mathjax>#6.06*L#</mathjax> represents <mathjax>#(6.06*L)/(22.4*L*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax>, <mathjax>#~=1/4*mol#</mathjax>.</p></div>
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</article> | How many moles of air are in a 6.06-L tire at STP? | null |
1,936 | a8e35a66-6ddd-11ea-aea6-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-300-ml-at-300-mm-hg-what-will-its-volume-be-if-the-pressur | 180 mL | start physical_unit 1 1 volume ml qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 11 pressure qc_end physical_unit 1 1 23 25 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas [IN] mL"}] | [{"type":"physical unit","value":"180 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{300 mL}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{300 mm Hg}"},{"type":"physical unit","value":"Pressure2 [OF] gas [=] \\pu{500 mm Hg}"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?</h1> | null | 180 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this question, we will issue that the change happens at constant temperature and that the gas behaves ideally.</p>
<p>Thus, we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>Since the change is occurring at constant temperature and number of mole, we can modify the ideal gas law as follows:</p>
<p><mathjax>#PV=k#</mathjax> where, <mathjax>#k=nRT#</mathjax>.</p>
<p>For position 1: we have <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax>.</p>
<p>For position 2: we will have <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax>.</p>
<p>Since <mathjax>#PV=k#</mathjax> we can write <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>Therefore, <mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#=>V_2=(300cancel(mmHg)xx300mL)/(500cancel(mmHg))=180mL#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_2=180mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this question, we will issue that the change happens at constant temperature and that the gas behaves ideally.</p>
<p>Thus, we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>Since the change is occurring at constant temperature and number of mole, we can modify the ideal gas law as follows:</p>
<p><mathjax>#PV=k#</mathjax> where, <mathjax>#k=nRT#</mathjax>.</p>
<p>For position 1: we have <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax>.</p>
<p>For position 2: we will have <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax>.</p>
<p>Since <mathjax>#PV=k#</mathjax> we can write <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>Therefore, <mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#=>V_2=(300cancel(mmHg)xx300mL)/(500cancel(mmHg))=180mL#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?</h1>
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<div class="markdown"><p><mathjax>#V_2=180mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this question, we will issue that the change happens at constant temperature and that the gas behaves ideally.</p>
<p>Thus, we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>Since the change is occurring at constant temperature and number of mole, we can modify the ideal gas law as follows:</p>
<p><mathjax>#PV=k#</mathjax> where, <mathjax>#k=nRT#</mathjax>.</p>
<p>For position 1: we have <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax>.</p>
<p>For position 2: we will have <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax>.</p>
<p>Since <mathjax>#PV=k#</mathjax> we can write <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>Therefore, <mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#=>V_2=(300cancel(mmHg)xx300mL)/(500cancel(mmHg))=180mL#</mathjax></p></div>
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</article> | A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg? | null |
1,937 | a9eb3f6a-6ddd-11ea-b1ac-ccda262736ce | https://socratic.org/questions/the-chemical-formula-for-rust-is-fe-2o-3-how-many-moles-of-fe-are-present-in-1-0 | 13 moles | start physical_unit 11 11 mole mol qc_end physical_unit 18 19 15 16 mass qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Mole [OF] Fe [IN] moles"}] | [{"type":"physical unit","value":"13 moles"}] | [{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{1.0 kg}"},{"type":"chemical equation","value":"Fe2O3"}] | <h1 class="questionTitle" itemprop="name">The chemical formula for rust is #Fe_2O_3#. How many moles of #Fe# are present in 1.0 kg of the compound?</h1> | null | 13 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 mol Fe"_2"O"_3"#</mathjax> contains <mathjax>#"2 mol Fe ions and 3 mol O ions"#</mathjax></p>
<p><mathjax>#"1.0 kg=1000 g"#</mathjax></p>
<p><mathjax>#"Molar mass Fe"_2"O"_3="159.6882 g/mol"#</mathjax><br/>
<a href="http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top</a></p>
<p>Determine the moles of <mathjax>#"Fe"_2"O"_3"#</mathjax> by dividing the given mass by its molar mass, then multiply times the mol ratio between <mathjax>#"Fe"#</mathjax> ions and <mathjax>#"Fe"_2"O"_3"#</mathjax> to get the moles of <mathjax>#"Fe"#</mathjax> ions in <mathjax>#"1000 g Fe"_2"O"_3"#</mathjax>.</p>
<p><mathjax>#1000cancel("g Fe"_2"O"_3)xx(1cancel("mol Fe"_2"O"_3))/(159.6882cancel("g Fe"_2"O"_3))xx(2"mol Fe")/(1cancel("mol Fe"_2"O"_3")#</mathjax><mathjax>#="13 mol Fe ions"#</mathjax><br/>
(rounded to two significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.0 kg Fe"_2"O"_3"#</mathjax> contains <mathjax>#"13 mol Fe ions"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 mol Fe"_2"O"_3"#</mathjax> contains <mathjax>#"2 mol Fe ions and 3 mol O ions"#</mathjax></p>
<p><mathjax>#"1.0 kg=1000 g"#</mathjax></p>
<p><mathjax>#"Molar mass Fe"_2"O"_3="159.6882 g/mol"#</mathjax><br/>
<a href="http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top</a></p>
<p>Determine the moles of <mathjax>#"Fe"_2"O"_3"#</mathjax> by dividing the given mass by its molar mass, then multiply times the mol ratio between <mathjax>#"Fe"#</mathjax> ions and <mathjax>#"Fe"_2"O"_3"#</mathjax> to get the moles of <mathjax>#"Fe"#</mathjax> ions in <mathjax>#"1000 g Fe"_2"O"_3"#</mathjax>.</p>
<p><mathjax>#1000cancel("g Fe"_2"O"_3)xx(1cancel("mol Fe"_2"O"_3))/(159.6882cancel("g Fe"_2"O"_3))xx(2"mol Fe")/(1cancel("mol Fe"_2"O"_3")#</mathjax><mathjax>#="13 mol Fe ions"#</mathjax><br/>
(rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">The chemical formula for rust is #Fe_2O_3#. How many moles of #Fe# are present in 1.0 kg of the compound?</h1>
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Jan 15, 2016
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<div class="markdown"><p><mathjax>#"1.0 kg Fe"_2"O"_3"#</mathjax> contains <mathjax>#"13 mol Fe ions"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 mol Fe"_2"O"_3"#</mathjax> contains <mathjax>#"2 mol Fe ions and 3 mol O ions"#</mathjax></p>
<p><mathjax>#"1.0 kg=1000 g"#</mathjax></p>
<p><mathjax>#"Molar mass Fe"_2"O"_3="159.6882 g/mol"#</mathjax><br/>
<a href="http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/14833#section=Top</a></p>
<p>Determine the moles of <mathjax>#"Fe"_2"O"_3"#</mathjax> by dividing the given mass by its molar mass, then multiply times the mol ratio between <mathjax>#"Fe"#</mathjax> ions and <mathjax>#"Fe"_2"O"_3"#</mathjax> to get the moles of <mathjax>#"Fe"#</mathjax> ions in <mathjax>#"1000 g Fe"_2"O"_3"#</mathjax>.</p>
<p><mathjax>#1000cancel("g Fe"_2"O"_3)xx(1cancel("mol Fe"_2"O"_3))/(159.6882cancel("g Fe"_2"O"_3))xx(2"mol Fe")/(1cancel("mol Fe"_2"O"_3")#</mathjax><mathjax>#="13 mol Fe ions"#</mathjax><br/>
(rounded to two significant figures)</p></div>
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</article> | The chemical formula for rust is #Fe_2O_3#. How many moles of #Fe# are present in 1.0 kg of the compound? | null |
1,938 | a992637e-6ddd-11ea-925f-ccda262736ce | https://socratic.org/questions/if-450-ml-of-6-0-m-kbr-solution-is-diluted-to-900-ml-what-is-the-concentration-o | 3 M | start physical_unit 6 7 concentration mol/l qc_end physical_unit 6 7 4 5 concentration qc_end physical_unit 6 7 1 2 volume qc_end physical_unit 6 7 11 12 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] KBr solution [IN] M"}] | [{"type":"physical unit","value":"3 M"}] | [{"type":"physical unit","value":"Concentration1 [OF] KBr solution [=] \\pu{6.0 M}"},{"type":"physical unit","value":"Volume1 [OF] KBr solution [=] \\pu{450 mL}"},{"type":"physical unit","value":"Volume2 [OF] KBr solution [=] \\pu{900 mL}"}] | <h1 class="questionTitle" itemprop="name">If 450 mL of 6.0 M KBr solution is diluted to 900 mL, what is the concentration of the KBr solution?</h1> | null | 3 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the dilution equation <mathjax>#M_1V_1=M_2V_2#</mathjax>, where <mathjax>#M#</mathjax> is <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> in mol/L, and <mathjax>#V#</mathjax> is volume in liters.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#M_1="6.0 M"="6.0 mol/L"#</mathjax><br/>
<mathjax>#V_1=450cancel"mL"xx(1"L")/(1000cancel"mL")="0.45 L"#</mathjax><br/>
<mathjax>#V_2=900cancel"mL"xx(1"L")/(1000cancel"mL")="0.9 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#M_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution equation to isolate <mathjax>#M_2#</mathjax> and solve.<br/>
<mathjax>#M_1V_1=M_2V_2#</mathjax></p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=(6.0"mol"/"L"xx0.45cancel"L")/((0.9cancel"L"))="3 mol/L"="3 M"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The concentration of the diluted KBr solution is 3 M.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the dilution equation <mathjax>#M_1V_1=M_2V_2#</mathjax>, where <mathjax>#M#</mathjax> is <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> in mol/L, and <mathjax>#V#</mathjax> is volume in liters.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#M_1="6.0 M"="6.0 mol/L"#</mathjax><br/>
<mathjax>#V_1=450cancel"mL"xx(1"L")/(1000cancel"mL")="0.45 L"#</mathjax><br/>
<mathjax>#V_2=900cancel"mL"xx(1"L")/(1000cancel"mL")="0.9 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#M_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution equation to isolate <mathjax>#M_2#</mathjax> and solve.<br/>
<mathjax>#M_1V_1=M_2V_2#</mathjax></p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=(6.0"mol"/"L"xx0.45cancel"L")/((0.9cancel"L"))="3 mol/L"="3 M"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If 450 mL of 6.0 M KBr solution is diluted to 900 mL, what is the concentration of the KBr solution?</h1>
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<div class="markdown"><p>The concentration of the diluted KBr solution is 3 M.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the dilution equation <mathjax>#M_1V_1=M_2V_2#</mathjax>, where <mathjax>#M#</mathjax> is <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> in mol/L, and <mathjax>#V#</mathjax> is volume in liters.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#M_1="6.0 M"="6.0 mol/L"#</mathjax><br/>
<mathjax>#V_1=450cancel"mL"xx(1"L")/(1000cancel"mL")="0.45 L"#</mathjax><br/>
<mathjax>#V_2=900cancel"mL"xx(1"L")/(1000cancel"mL")="0.9 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#M_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution equation to isolate <mathjax>#M_2#</mathjax> and solve.<br/>
<mathjax>#M_1V_1=M_2V_2#</mathjax></p>
<p><mathjax>#M_2=(M_1V_1)/(V_2)#</mathjax></p>
<p><mathjax>#M_2=(6.0"mol"/"L"xx0.45cancel"L")/((0.9cancel"L"))="3 mol/L"="3 M"#</mathjax></p></div>
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</article> | If 450 mL of 6.0 M KBr solution is diluted to 900 mL, what is the concentration of the KBr solution? | null |
1,939 | aad9e9e8-6ddd-11ea-9b8c-ccda262736ce | https://socratic.org/questions/how-would-you-balance-s8-f2-sf6 | S8 + 24 F2 -> 8 SF6 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"S8 + 24 F2 -> 8 SF6"}] | [{"type":"chemical equation","value":"S8 + F2 -> SF6"}] | <h1 class="questionTitle" itemprop="name">How would you balance __S8+__F2--> __SF6?</h1> | null | S8 + 24 F2 -> 8 SF6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing you need to do is to tally the atoms.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#SF_6#</mathjax> (unbalanced)</p>
<p>Based on the subscripts,</p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = 1; F = 6</p>
<p>Since you cannot balance an equation by changing subscripts, all you can do is insert the coefficients. Let's start with the easier atom to balance.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = 8; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = 48</p>
<p>Please notice that since <mathjax>#SF_6#</mathjax> is a substance, you need to multiply the coefficient, not only to S but also to F.</p>
<p>Now to balance the F.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#color (green) 24 F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = <strong>8</strong>; F = (2 x <mathjax>#color (green) 24#</mathjax>) = <strong>48</strong><br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = <strong>8</strong>; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = <strong>48</strong></p>
<p>The equation is now balanced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#S_8#</mathjax> + <mathjax>#24F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8SF_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing you need to do is to tally the atoms.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#SF_6#</mathjax> (unbalanced)</p>
<p>Based on the subscripts,</p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = 1; F = 6</p>
<p>Since you cannot balance an equation by changing subscripts, all you can do is insert the coefficients. Let's start with the easier atom to balance.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = 8; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = 48</p>
<p>Please notice that since <mathjax>#SF_6#</mathjax> is a substance, you need to multiply the coefficient, not only to S but also to F.</p>
<p>Now to balance the F.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#color (green) 24 F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = <strong>8</strong>; F = (2 x <mathjax>#color (green) 24#</mathjax>) = <strong>48</strong><br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = <strong>8</strong>; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = <strong>48</strong></p>
<p>The equation is now balanced.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How would you balance __S8+__F2--> __SF6?</h1>
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Nikka C.
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<span class="dateCreated" datetime="2015-10-27T13:02:04" itemprop="dateCreated">
Oct 27, 2015
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<div class="markdown"><p><mathjax>#S_8#</mathjax> + <mathjax>#24F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8SF_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing you need to do is to tally the atoms.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#SF_6#</mathjax> (unbalanced)</p>
<p>Based on the subscripts,</p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = 1; F = 6</p>
<p>Since you cannot balance an equation by changing subscripts, all you can do is insert the coefficients. Let's start with the easier atom to balance.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = 8; F = 2<br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = 8; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = 48</p>
<p>Please notice that since <mathjax>#SF_6#</mathjax> is a substance, you need to multiply the coefficient, not only to S but also to F.</p>
<p>Now to balance the F.</p>
<p><mathjax>#S_8#</mathjax> + <mathjax>#color (green) 24 F_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (blue) 8 SF_6#</mathjax></p>
<p>Left side: S = <strong>8</strong>; F = (2 x <mathjax>#color (green) 24#</mathjax>) = <strong>48</strong><br/>
Right side: S = (1 x <mathjax>#color (blue) 8#</mathjax>) = <strong>8</strong>; F = (6 x <mathjax>#color (blue) 8#</mathjax>) = <strong>48</strong></p>
<p>The equation is now balanced.</p></div>
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</article> | How would you balance __S8+__F2--> __SF6? | null |
1,940 | acbc6c6c-6ddd-11ea-bc5e-ccda262736ce | https://socratic.org/questions/how-many-moles-of-cl-ions-are-needed-to-completely-combine-with-0-25-moles-of-mg | 0.50 moles | start physical_unit 4 5 mole mol qc_end physical_unit 15 16 12 13 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] Cl- ions [IN] moles"}] | [{"type":"physical unit","value":"0.50 moles"}] | [{"type":"physical unit","value":"Mole [OF] Mg^2+ ions [=] \\pu{0.25 moles}"},{"type":"other","value":"Completely combine."}] | <h1 class="questionTitle" itemprop="name">How many moles of #Cl^-# ions are needed to completely combine with 0.25 moles of #Mg^(2+)# ions? </h1> | null | 0.50 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all these sorts of problems, reference to a balanced chemical equation is necessary as a first step:</p>
<p><mathjax>#Mg^(2+) + 2Cl^(-) rarr MgCl_2#</mathjax></p>
<p>This equation is balanced with respect to (i) mass, and (ii) charge, and is therefore a reasonable representation of chemical reality, of physical reality if you like. The equation clearly states that each equiv magnesium ion requires 2 equiv of chloride ion to form the salt. Since we started with a given NUMBER of <mathjax>#Mg^(2+)#</mathjax> ions (<mathjax>#0.25*mol#</mathjax>) it follows that we need TWICE this number of chloride ions for reaction. </p>
<p>Why do I refer to the <mathjax>#"mole"#</mathjax> as a number?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>To form <mathjax>#MgCl_2#</mathjax>? Clearly we need <mathjax>#0.50*mol#</mathjax> of chloride ion. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all these sorts of problems, reference to a balanced chemical equation is necessary as a first step:</p>
<p><mathjax>#Mg^(2+) + 2Cl^(-) rarr MgCl_2#</mathjax></p>
<p>This equation is balanced with respect to (i) mass, and (ii) charge, and is therefore a reasonable representation of chemical reality, of physical reality if you like. The equation clearly states that each equiv magnesium ion requires 2 equiv of chloride ion to form the salt. Since we started with a given NUMBER of <mathjax>#Mg^(2+)#</mathjax> ions (<mathjax>#0.25*mol#</mathjax>) it follows that we need TWICE this number of chloride ions for reaction. </p>
<p>Why do I refer to the <mathjax>#"mole"#</mathjax> as a number?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #Cl^-# ions are needed to completely combine with 0.25 moles of #Mg^(2+)# ions? </h1>
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<div class="markdown"><p>To form <mathjax>#MgCl_2#</mathjax>? Clearly we need <mathjax>#0.50*mol#</mathjax> of chloride ion. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all these sorts of problems, reference to a balanced chemical equation is necessary as a first step:</p>
<p><mathjax>#Mg^(2+) + 2Cl^(-) rarr MgCl_2#</mathjax></p>
<p>This equation is balanced with respect to (i) mass, and (ii) charge, and is therefore a reasonable representation of chemical reality, of physical reality if you like. The equation clearly states that each equiv magnesium ion requires 2 equiv of chloride ion to form the salt. Since we started with a given NUMBER of <mathjax>#Mg^(2+)#</mathjax> ions (<mathjax>#0.25*mol#</mathjax>) it follows that we need TWICE this number of chloride ions for reaction. </p>
<p>Why do I refer to the <mathjax>#"mole"#</mathjax> as a number?</p></div>
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</article> | How many moles of #Cl^-# ions are needed to completely combine with 0.25 moles of #Mg^(2+)# ions? | null |
1,941 | ac909662-6ddd-11ea-b3be-ccda262736ce | https://socratic.org/questions/at-1-01-atm-and-15-0-c-a-constant-pressure-tank-has-255-m-3-of-propane-what-is-t | 251 m^3 | start physical_unit 21 22 volume m^3 qc_end physical_unit 15 15 1 2 pressure qc_end physical_unit 15 15 4 5 temperature qc_end c_other OTHER qc_end physical_unit 15 15 12 13 volume qc_end physical_unit 21 22 24 25 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] m^3"}] | [{"type":"physical unit","value":"251 m^3"}] | [{"type":"physical unit","value":"Pressure1 [OF] propane [=] \\pu{1.01 atm}"},{"type":"physical unit","value":"Temperature1 [OF] propane [=] \\pu{15.0 ℃}"},{"type":"other","value":"Constant - pressure tank."},{"type":"physical unit","value":"Volume1 [OF] propane [=] \\pu{255 m^3}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{48.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">At 1.01 atm and 15.0°C a constant - pressure tank has 255 #m^3# of propane. What is the volume of the gas at 48.0°C?</h1> | null | 251 m^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>, with the equation <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax>. </p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="1.01 atm"#</mathjax><br/>
<mathjax>#V_1="225 m"^3"#</mathjax><br/>
<mathjax>#T_1="15.0"^"o""C"+273.15="288.2 K"#</mathjax><br/>
<mathjax>#P_2="1.01 atm"#</mathjax><br/>
<mathjax>#T_2="48.0"^"o""C"+273.15="321.2 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(1.01cancel"atm"xx225"m"^3xx321.2cancel"K")/(288.2cancel"K"xx1.01cancel"atm")="251 m"^3"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume at <mathjax>#"321.2 K"#</mathjax> <mathjax>#("48.0"^"o""C")#</mathjax> is <mathjax>#"251 m"^3"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>, with the equation <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax>. </p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="1.01 atm"#</mathjax><br/>
<mathjax>#V_1="225 m"^3"#</mathjax><br/>
<mathjax>#T_1="15.0"^"o""C"+273.15="288.2 K"#</mathjax><br/>
<mathjax>#P_2="1.01 atm"#</mathjax><br/>
<mathjax>#T_2="48.0"^"o""C"+273.15="321.2 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(1.01cancel"atm"xx225"m"^3xx321.2cancel"K")/(288.2cancel"K"xx1.01cancel"atm")="251 m"^3"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At 1.01 atm and 15.0°C a constant - pressure tank has 255 #m^3# of propane. What is the volume of the gas at 48.0°C?</h1>
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Dec 9, 2015
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<div class="markdown"><p>The volume at <mathjax>#"321.2 K"#</mathjax> <mathjax>#("48.0"^"o""C")#</mathjax> is <mathjax>#"251 m"^3"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>, with the equation <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax>. </p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="1.01 atm"#</mathjax><br/>
<mathjax>#V_1="225 m"^3"#</mathjax><br/>
<mathjax>#T_1="15.0"^"o""C"+273.15="288.2 K"#</mathjax><br/>
<mathjax>#P_2="1.01 atm"#</mathjax><br/>
<mathjax>#T_2="48.0"^"o""C"+273.15="321.2 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(1.01cancel"atm"xx225"m"^3xx321.2cancel"K")/(288.2cancel"K"xx1.01cancel"atm")="251 m"^3"#</mathjax></p></div>
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</article> | At 1.01 atm and 15.0°C a constant - pressure tank has 255 #m^3# of propane. What is the volume of the gas at 48.0°C? | null |
1,942 | a9d0581a-6ddd-11ea-bd56-ccda262736ce | https://socratic.org/questions/573f05ee7c01496ff8f34662 | CuSO4.5H2O | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] hydrate [IN] empirical"}] | [{"type":"chemical equation","value":"CuSO4.5H2O"}] | [{"type":"physical unit","value":"Percent [OF] Cu in the copper salt [=] \\pu{25.5%}"},{"type":"physical unit","value":"Percent [OF] S in the copper salt [=] \\pu{12.9%}"},{"type":"physical unit","value":"Percent [OF] O in the copper salt [=] \\pu{25.6%}"},{"type":"physical unit","value":"Percent [OF] water in the copper salt [=] \\pu{36.0%}"}] | <h1 class="questionTitle" itemprop="name">A copper salt contains #25.5%# #Cu#, #12.9%# #S#, #25.6%# #O#, and #36.0%# #"water"#...what is the empirical formula of the hydrate? </h1> | null | CuSO4.5H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We ASSUME that there are <mathjax>#100*g#</mathjax> of copper salt, and work out the molar quantities:</p>
<p><mathjax>#Cu:#</mathjax> <mathjax>#(25.5*g)/(63.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#Cu#</mathjax>.</p>
<p><mathjax>#S:#</mathjax> <mathjax>#(12.9*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#S#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(25.6*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.60*mol#</mathjax> <mathjax>#O#</mathjax></p>
<p><mathjax>#H_2O:#</mathjax> <mathjax>#(36.0*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.00*mol#</mathjax> <mathjax>#H_2O#</mathjax></p>
<p>To get the empirical formula, we divide thru by the LOWEST molar quantity, i.e. <mathjax>#0.401*mol#</mathjax>, to give:</p>
<p><mathjax>#CuSO_4*5H_2O#</mathjax>, <mathjax>#"copper sulfate pentahydrate"#</mathjax></p>
<p>I have treated the water here as I would normally treat an element; I am certainly free to do so inasmuch as we could dry the salt and measure the actual water loss. </p>
<p>What is the colour of this salt?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CuSO_4*5H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We ASSUME that there are <mathjax>#100*g#</mathjax> of copper salt, and work out the molar quantities:</p>
<p><mathjax>#Cu:#</mathjax> <mathjax>#(25.5*g)/(63.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#Cu#</mathjax>.</p>
<p><mathjax>#S:#</mathjax> <mathjax>#(12.9*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#S#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(25.6*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.60*mol#</mathjax> <mathjax>#O#</mathjax></p>
<p><mathjax>#H_2O:#</mathjax> <mathjax>#(36.0*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.00*mol#</mathjax> <mathjax>#H_2O#</mathjax></p>
<p>To get the empirical formula, we divide thru by the LOWEST molar quantity, i.e. <mathjax>#0.401*mol#</mathjax>, to give:</p>
<p><mathjax>#CuSO_4*5H_2O#</mathjax>, <mathjax>#"copper sulfate pentahydrate"#</mathjax></p>
<p>I have treated the water here as I would normally treat an element; I am certainly free to do so inasmuch as we could dry the salt and measure the actual water loss. </p>
<p>What is the colour of this salt?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A copper salt contains #25.5%# #Cu#, #12.9%# #S#, #25.6%# #O#, and #36.0%# #"water"#...what is the empirical formula of the hydrate? </h1>
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<div class="markdown"><p><mathjax>#CuSO_4*5H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We ASSUME that there are <mathjax>#100*g#</mathjax> of copper salt, and work out the molar quantities:</p>
<p><mathjax>#Cu:#</mathjax> <mathjax>#(25.5*g)/(63.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#Cu#</mathjax>.</p>
<p><mathjax>#S:#</mathjax> <mathjax>#(12.9*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.401*mol#</mathjax> <mathjax>#S#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(25.6*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.60*mol#</mathjax> <mathjax>#O#</mathjax></p>
<p><mathjax>#H_2O:#</mathjax> <mathjax>#(36.0*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.00*mol#</mathjax> <mathjax>#H_2O#</mathjax></p>
<p>To get the empirical formula, we divide thru by the LOWEST molar quantity, i.e. <mathjax>#0.401*mol#</mathjax>, to give:</p>
<p><mathjax>#CuSO_4*5H_2O#</mathjax>, <mathjax>#"copper sulfate pentahydrate"#</mathjax></p>
<p>I have treated the water here as I would normally treat an element; I am certainly free to do so inasmuch as we could dry the salt and measure the actual water loss. </p>
<p>What is the colour of this salt?</p></div>
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</article> | A copper salt contains #25.5%# #Cu#, #12.9%# #S#, #25.6%# #O#, and #36.0%# #"water"#...what is the empirical formula of the hydrate? | null |
1,943 | a94daa13-6ddd-11ea-95a7-ccda262736ce | https://socratic.org/questions/how-many-isomeric-alcohols-exist-with-the-formula-c-4h-10o | 7 | start physical_unit 2 3 number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Number [OF] isomeric alcohols"}] | [{"type":"physical unit","value":"7"}] | [{"type":"chemical equation","value":"C4H10O"}] | <h1 class="questionTitle" itemprop="name">How many isomeric alcohols exist with the formula #C_4H_10O#?</h1> | null | 7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The following is an image showing the seven <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a>.</p>
<p><img alt="http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Images/structure2.gif" src="https://useruploads.socratic.org/zfpxnJYISeiPeWD8FTxb_structure2.gif"/> </p></div>
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<div class="markdown"><p>There are seven isomers of the compound <mathjax>#"C"_4"H"_10"O"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The following is an image showing the seven <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a>.</p>
<p><img alt="http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Images/structure2.gif" src="https://useruploads.socratic.org/zfpxnJYISeiPeWD8FTxb_structure2.gif"/> </p></div>
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<h1 class="questionTitle" itemprop="name">How many isomeric alcohols exist with the formula #C_4H_10O#?</h1>
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<div class="markdown"><p>There are seven isomers of the compound <mathjax>#"C"_4"H"_10"O"#</mathjax>.</p></div>
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<div class="markdown"><p>The following is an image showing the seven <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a>.</p>
<p><img alt="http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Images/structure2.gif" src="https://useruploads.socratic.org/zfpxnJYISeiPeWD8FTxb_structure2.gif"/> </p></div>
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</article> | How many isomeric alcohols exist with the formula #C_4H_10O#? | null |
1,944 | ac5c1a07-6ddd-11ea-8e7b-ccda262736ce | https://socratic.org/questions/how-do-you-write-and-balance-the-equation-for-the-complete-combustion-of-acetyle | C2H2 + 5/2 O2 -> 2 CO2 + H2O | start chemical_equation qc_end chemical_equation 14 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the complete combustion"}] | [{"type":"chemical equation","value":"C2H2 + 5/2 O2 -> 2 CO2 + H2O"}] | [{"type":"chemical equation","value":"C2H2"}] | <h1 class="questionTitle" itemprop="name">How do you write and balance the equation for the complete combustion of acetylene, #C_2H_2#?</h1> | null | C2H2 + 5/2 O2 -> 2 CO2 + H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is the reaction stoichiometrically balanced? You can double the equation if you like to remove the half-integral coefficient; the arithmetic when determining equivalence is easier in the present form. </p>
<p>Do you think this reaction is endothermic or exothermic?</p></div>
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<div class="markdown"><p><mathjax>#HC-=CH(g) + 5/2O_2(g) rarr 2CO_2(g) + H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is the reaction stoichiometrically balanced? You can double the equation if you like to remove the half-integral coefficient; the arithmetic when determining equivalence is easier in the present form. </p>
<p>Do you think this reaction is endothermic or exothermic?</p></div>
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<h1 class="questionTitle" itemprop="name">How do you write and balance the equation for the complete combustion of acetylene, #C_2H_2#?</h1>
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<div class="markdown"><p><mathjax>#HC-=CH(g) + 5/2O_2(g) rarr 2CO_2(g) + H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is the reaction stoichiometrically balanced? You can double the equation if you like to remove the half-integral coefficient; the arithmetic when determining equivalence is easier in the present form. </p>
<p>Do you think this reaction is endothermic or exothermic?</p></div>
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</article> | How do you write and balance the equation for the complete combustion of acetylene, #C_2H_2#? | null |
1,945 | ab645392-6ddd-11ea-a94d-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-chemical-formula-of-a-compound-that-has-has-mass-percent-6 | C3H6O | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"C3H6O"}] | [{"type":"physical unit","value":"Mass percent [OF] carbon in the compound [=] \\pu{62.1%}"},{"type":"physical unit","value":"Mass percent [OF] hydrogen in the compound [=] \\pu{10.5%}"},{"type":"physical unit","value":"Mass percent [OF] oxygen in the compound [=] \\pu{27.4%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical chemical formula of a compound that has has mass percent 62.1% carbon, 10.5% hydrogen, and 27.4% oxygen?</h1> | null | C3H6O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to get a compound's <strong>empirical formula</strong>, you basically need to determine the <strong>smallest whole number ratio</strong> that exists between the number of moles of its <em>constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></em>. </p>
<p>This means that your goal here will be to convert the given <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> to <em>grams</em>, then convert the <em>grams</em> to <em>moles</em> by using the <strong>molar masses</strong> of the elements. </p>
<p>In order to make the calculations easier, pick a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li><mathjax>#"62.1 g " ->#</mathjax> <em><strong>carbon</strong></em></li>
<li><mathjax>#"10.5 g " ->#</mathjax> <em><strong>hydrogen</strong></em></li>
<li><mathjax>#"27.4 g " ->#</mathjax> <em><strong>oxygen</strong></em></li>
</ul>
</blockquote>
<p>Now, use the <strong>molar masses</strong> of the three elements to determine how many <em>moles</em> of each you have in this sample. </p>
<blockquote>
<p><mathjax>#"For C: " 62.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "5.1703 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 10.5color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "10.417 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 27.4color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.7126 moles O"#</mathjax></p>
</blockquote>
<p>To get the <strong>mole ratio</strong> that exists between the elements in the compound, divide all values by the <em>smallest</em> one. This will get you </p>
<blockquote>
<p><mathjax>#"For C: " (5.1703color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 3.019 ~~ 3#</mathjax></p>
<p><mathjax>#"For H: " (10.417color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 6.083 ~~ 6#</mathjax></p>
<p><mathjax>#"For O: " (1.7126color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#3:6:1#</mathjax> represents the <strong>smallest whole number ratio</strong> that can exist between the three elements, the empirical formula of the unknown compound will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"C"_3"H"_6"O"_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/lyWAGMEKzSY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_3"H"_6"O"_1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to get a compound's <strong>empirical formula</strong>, you basically need to determine the <strong>smallest whole number ratio</strong> that exists between the number of moles of its <em>constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></em>. </p>
<p>This means that your goal here will be to convert the given <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> to <em>grams</em>, then convert the <em>grams</em> to <em>moles</em> by using the <strong>molar masses</strong> of the elements. </p>
<p>In order to make the calculations easier, pick a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li><mathjax>#"62.1 g " ->#</mathjax> <em><strong>carbon</strong></em></li>
<li><mathjax>#"10.5 g " ->#</mathjax> <em><strong>hydrogen</strong></em></li>
<li><mathjax>#"27.4 g " ->#</mathjax> <em><strong>oxygen</strong></em></li>
</ul>
</blockquote>
<p>Now, use the <strong>molar masses</strong> of the three elements to determine how many <em>moles</em> of each you have in this sample. </p>
<blockquote>
<p><mathjax>#"For C: " 62.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "5.1703 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 10.5color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "10.417 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 27.4color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.7126 moles O"#</mathjax></p>
</blockquote>
<p>To get the <strong>mole ratio</strong> that exists between the elements in the compound, divide all values by the <em>smallest</em> one. This will get you </p>
<blockquote>
<p><mathjax>#"For C: " (5.1703color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 3.019 ~~ 3#</mathjax></p>
<p><mathjax>#"For H: " (10.417color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 6.083 ~~ 6#</mathjax></p>
<p><mathjax>#"For O: " (1.7126color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#3:6:1#</mathjax> represents the <strong>smallest whole number ratio</strong> that can exist between the three elements, the empirical formula of the unknown compound will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"C"_3"H"_6"O"_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/lyWAGMEKzSY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical chemical formula of a compound that has has mass percent 62.1% carbon, 10.5% hydrogen, and 27.4% oxygen?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-17T01:18:39" itemprop="dateCreated">
Mar 17, 2016
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<div class="markdown"><p><mathjax>#"C"_3"H"_6"O"_1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In order to get a compound's <strong>empirical formula</strong>, you basically need to determine the <strong>smallest whole number ratio</strong> that exists between the number of moles of its <em>constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a></em>. </p>
<p>This means that your goal here will be to convert the given <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> to <em>grams</em>, then convert the <em>grams</em> to <em>moles</em> by using the <strong>molar masses</strong> of the elements. </p>
<p>In order to make the calculations easier, pick a <mathjax>#"100-g"#</mathjax> sample of this unknown compound. According to the given <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a>, this sample will contain</p>
<blockquote>
<ul>
<li><mathjax>#"62.1 g " ->#</mathjax> <em><strong>carbon</strong></em></li>
<li><mathjax>#"10.5 g " ->#</mathjax> <em><strong>hydrogen</strong></em></li>
<li><mathjax>#"27.4 g " ->#</mathjax> <em><strong>oxygen</strong></em></li>
</ul>
</blockquote>
<p>Now, use the <strong>molar masses</strong> of the three elements to determine how many <em>moles</em> of each you have in this sample. </p>
<blockquote>
<p><mathjax>#"For C: " 62.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "5.1703 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 10.5color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "10.417 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 27.4color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.7126 moles O"#</mathjax></p>
</blockquote>
<p>To get the <strong>mole ratio</strong> that exists between the elements in the compound, divide all values by the <em>smallest</em> one. This will get you </p>
<blockquote>
<p><mathjax>#"For C: " (5.1703color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 3.019 ~~ 3#</mathjax></p>
<p><mathjax>#"For H: " (10.417color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 6.083 ~~ 6#</mathjax></p>
<p><mathjax>#"For O: " (1.7126color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#3:6:1#</mathjax> represents the <strong>smallest whole number ratio</strong> that can exist between the three elements, the empirical formula of the unknown compound will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"C"_3"H"_6"O"_1color(white)(a/a)|)))#</mathjax></p>
</blockquote>
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</article> | What is the empirical chemical formula of a compound that has has mass percent 62.1% carbon, 10.5% hydrogen, and 27.4% oxygen? | null |
1,946 | a98d9028-6ddd-11ea-ac58-ccda262736ce | https://socratic.org/questions/the-air-in-lungs-has-0-153-mol-of-gas-particles-at-302-k-and-101-3-of-pressure-w | 3.79 L | start physical_unit 0 1 volume l qc_end physical_unit 8 9 5 6 mole qc_end physical_unit 8 9 11 12 temperature qc_end physical_unit 8 9 14 15 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] the air [IN] L"}] | [{"type":"physical unit","value":"3.79 L"}] | [{"type":"physical unit","value":"Mole [OF] gas particles [=] \\pu{0.153 mol}"},{"type":"physical unit","value":"Temperature [OF] gas particles [=] \\pu{302 K}"},{"type":"physical unit","value":"Pressure [OF] gas particles [=] \\pu{101.3 kPa}"}] | <h1 class="questionTitle" itemprop="name">The air in lungs has 0.153 mol of gas particles at 302 K and 101.3 kPa of pressure. What is the volume of the air?</h1> | null | 3.79 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This can be solved applying equation of state for ideal gas, which i as follows</p>
<ul>
<li><mathjax>#PV=nRT#</mathjax></li>
</ul>
<p>Where for air in the lungs</p>
<ul>
<li>
<p><mathjax>#"Pressure" (P)= 101.3kPa = 1atm#</mathjax></p>
</li>
<li>
<p><mathjax>#"No. of moles"(n)=0.153#</mathjax></p>
</li>
<li>
<p><mathjax>#"Temperature"(T)=302K#</mathjax></p>
</li>
<li>
<p><mathjax>#"Universal gas constant " (R)= 0.082Latm K^-1mol^-1#</mathjax></p>
</li>
<li>
<p><mathjax>#"Volume of air" (V) =?#</mathjax></p>
</li>
</ul>
<p><mathjax>#:.V=(nRT)/P=(0.153xx0.082xx302)/1L=3.79L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>3.79L</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This can be solved applying equation of state for ideal gas, which i as follows</p>
<ul>
<li><mathjax>#PV=nRT#</mathjax></li>
</ul>
<p>Where for air in the lungs</p>
<ul>
<li>
<p><mathjax>#"Pressure" (P)= 101.3kPa = 1atm#</mathjax></p>
</li>
<li>
<p><mathjax>#"No. of moles"(n)=0.153#</mathjax></p>
</li>
<li>
<p><mathjax>#"Temperature"(T)=302K#</mathjax></p>
</li>
<li>
<p><mathjax>#"Universal gas constant " (R)= 0.082Latm K^-1mol^-1#</mathjax></p>
</li>
<li>
<p><mathjax>#"Volume of air" (V) =?#</mathjax></p>
</li>
</ul>
<p><mathjax>#:.V=(nRT)/P=(0.153xx0.082xx302)/1L=3.79L#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The air in lungs has 0.153 mol of gas particles at 302 K and 101.3 kPa of pressure. What is the volume of the air?</h1>
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<div class="markdown"><p>3.79L</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This can be solved applying equation of state for ideal gas, which i as follows</p>
<ul>
<li><mathjax>#PV=nRT#</mathjax></li>
</ul>
<p>Where for air in the lungs</p>
<ul>
<li>
<p><mathjax>#"Pressure" (P)= 101.3kPa = 1atm#</mathjax></p>
</li>
<li>
<p><mathjax>#"No. of moles"(n)=0.153#</mathjax></p>
</li>
<li>
<p><mathjax>#"Temperature"(T)=302K#</mathjax></p>
</li>
<li>
<p><mathjax>#"Universal gas constant " (R)= 0.082Latm K^-1mol^-1#</mathjax></p>
</li>
<li>
<p><mathjax>#"Volume of air" (V) =?#</mathjax></p>
</li>
</ul>
<p><mathjax>#:.V=(nRT)/P=(0.153xx0.082xx302)/1L=3.79L#</mathjax></p></div>
</div>
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</article> | The air in lungs has 0.153 mol of gas particles at 302 K and 101.3 kPa of pressure. What is the volume of the air? | null |
1,947 | a9b25e24-6ddd-11ea-b378-ccda262736ce | https://socratic.org/questions/what-is-the-percent-by-volume-of-isopropyl-alcohol-in-a-solution-made-by-dissolv | 4.39% | start physical_unit 7 11 volume_percent none qc_end physical_unit 7 8 15 16 volume qc_end physical_unit 24 24 21 22 volume qc_end end | [{"type":"physical unit","value":"Percent by volume [OF] isopropyl alcohol in a solution"}] | [{"type":"physical unit","value":"4.39%"}] | [{"type":"physical unit","value":"Volume [OF] isopropyl alcohol [=] \\pu{18.9 mL}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{412 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the percent by volume of isopropyl alcohol in a solution made by dissolving 18.9mL of isopropyl alcohol in 412mL of water?</h1> | null | 4.39% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We express a solution's <strong>volume by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong>, <mathjax>#"% v/v"#</mathjax>, by calculating how many milliliters of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present in <mathjax>#"100 mL"#</mathjax> of <em><strong>solution</strong></em>. </p>
<p>In essence, all you have to do to determine a solution's volume by volume percent concentration is figure out how many milliliters of solute you get in <mathjax>#"100 mL"#</mathjax> of solution, then tag along a <mathjax>#%#</mathjax> sign to <em>that value</em>. </p>
<p>In your case, you know that you get <mathjax>#"18.9 mL"#</mathjax> of isopropyl alcohol, your solute, in a <strong>total volume</strong> of </p>
<blockquote>
<p><mathjax>#V_"total" = V_"solute" + V_"solvent"#</mathjax></p>
<p><mathjax>#V_"total" = "18.9 mL" + "412 mL" = "430.9 mL"#</mathjax></p>
</blockquote>
<p>of <strong>solution</strong>. In order to find the solution's <mathjax>#"% v/v"#</mathjax> concentration, you must pick pick a <mathjax>#"100 mL"#</mathjax> of this solution and use its <strong>known composition</strong> to calculate how many milliliters of isopropyl alcohol it contains</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("mL solution"))) * overbrace("18.9 mL isopropyl alcohol"/(430.9color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the solution's composition")) = "4.39 mL isopropyl alcohol"#</mathjax></p>
</blockquote>
<p>Now all you have to do is replace the <mathjax>#"mL"#</mathjax> symbol with a <mathjax>#%#</mathjax> sign</p>
<blockquote>
<p><mathjax>#"% v/v" = "4.39 mL isopropyl alcohol"/"100 mL solution" = color(green)(|bar(ul(color(white)(a/a)color(black)(4.39%)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4.39%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We express a solution's <strong>volume by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong>, <mathjax>#"% v/v"#</mathjax>, by calculating how many milliliters of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present in <mathjax>#"100 mL"#</mathjax> of <em><strong>solution</strong></em>. </p>
<p>In essence, all you have to do to determine a solution's volume by volume percent concentration is figure out how many milliliters of solute you get in <mathjax>#"100 mL"#</mathjax> of solution, then tag along a <mathjax>#%#</mathjax> sign to <em>that value</em>. </p>
<p>In your case, you know that you get <mathjax>#"18.9 mL"#</mathjax> of isopropyl alcohol, your solute, in a <strong>total volume</strong> of </p>
<blockquote>
<p><mathjax>#V_"total" = V_"solute" + V_"solvent"#</mathjax></p>
<p><mathjax>#V_"total" = "18.9 mL" + "412 mL" = "430.9 mL"#</mathjax></p>
</blockquote>
<p>of <strong>solution</strong>. In order to find the solution's <mathjax>#"% v/v"#</mathjax> concentration, you must pick pick a <mathjax>#"100 mL"#</mathjax> of this solution and use its <strong>known composition</strong> to calculate how many milliliters of isopropyl alcohol it contains</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("mL solution"))) * overbrace("18.9 mL isopropyl alcohol"/(430.9color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the solution's composition")) = "4.39 mL isopropyl alcohol"#</mathjax></p>
</blockquote>
<p>Now all you have to do is replace the <mathjax>#"mL"#</mathjax> symbol with a <mathjax>#%#</mathjax> sign</p>
<blockquote>
<p><mathjax>#"% v/v" = "4.39 mL isopropyl alcohol"/"100 mL solution" = color(green)(|bar(ul(color(white)(a/a)color(black)(4.39%)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the percent by volume of isopropyl alcohol in a solution made by dissolving 18.9mL of isopropyl alcohol in 412mL of water?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-28T12:35:14" itemprop="dateCreated">
Jul 28, 2016
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<div class="markdown"><p><mathjax>#4.39%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We express a solution's <strong>volume by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong>, <mathjax>#"% v/v"#</mathjax>, by calculating how many milliliters of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present in <mathjax>#"100 mL"#</mathjax> of <em><strong>solution</strong></em>. </p>
<p>In essence, all you have to do to determine a solution's volume by volume percent concentration is figure out how many milliliters of solute you get in <mathjax>#"100 mL"#</mathjax> of solution, then tag along a <mathjax>#%#</mathjax> sign to <em>that value</em>. </p>
<p>In your case, you know that you get <mathjax>#"18.9 mL"#</mathjax> of isopropyl alcohol, your solute, in a <strong>total volume</strong> of </p>
<blockquote>
<p><mathjax>#V_"total" = V_"solute" + V_"solvent"#</mathjax></p>
<p><mathjax>#V_"total" = "18.9 mL" + "412 mL" = "430.9 mL"#</mathjax></p>
</blockquote>
<p>of <strong>solution</strong>. In order to find the solution's <mathjax>#"% v/v"#</mathjax> concentration, you must pick pick a <mathjax>#"100 mL"#</mathjax> of this solution and use its <strong>known composition</strong> to calculate how many milliliters of isopropyl alcohol it contains</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("mL solution"))) * overbrace("18.9 mL isopropyl alcohol"/(430.9color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the solution's composition")) = "4.39 mL isopropyl alcohol"#</mathjax></p>
</blockquote>
<p>Now all you have to do is replace the <mathjax>#"mL"#</mathjax> symbol with a <mathjax>#%#</mathjax> sign</p>
<blockquote>
<p><mathjax>#"% v/v" = "4.39 mL isopropyl alcohol"/"100 mL solution" = color(green)(|bar(ul(color(white)(a/a)color(black)(4.39%)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> | What is the percent by volume of isopropyl alcohol in a solution made by dissolving 18.9mL of isopropyl alcohol in 412mL of water? | null |
1,948 | ac3447b6-6ddd-11ea-b553-ccda262736ce | https://socratic.org/questions/59aef4b2b72cff12cd8b6f33 | 475 mL | start physical_unit 7 7 volume ml qc_end physical_unit 7 7 15 16 pressure qc_end physical_unit 7 7 18 19 pressure qc_end physical_unit 7 7 25 26 volume qc_end end | [{"type":"physical unit","value":"Volume1 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"475 mL"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{6.20 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{9.150 atm}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{322 mL}"}] | <h1 class="questionTitle" itemprop="name">What was the original volume of a gas that has changed from a pressure of 6.20 atm to 9.150 atm and a new volume of 322 mL?</h1> | null | 475 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve this problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)p_1V_1 = p_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_1 = V_2 × p_2/p_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = "6.20 atm"; color(white)(ll)V_1 = "?"#</mathjax><br/>
<mathjax>#p_2 ="9.150 atm"; V_2 = "322 mL"#</mathjax></p>
<p>∴ <mathjax>#V_1 = "322 mL" × (9.150 color(red)(cancel(color(black)("atm"))))/(6.20 color(red)(cancel(color(black)("atm")))) = "475 mL"#</mathjax></p></div>
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<div class="markdown"><blockquote></blockquote>
<p>The original volume was 475 mL.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve this problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)p_1V_1 = p_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_1 = V_2 × p_2/p_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = "6.20 atm"; color(white)(ll)V_1 = "?"#</mathjax><br/>
<mathjax>#p_2 ="9.150 atm"; V_2 = "322 mL"#</mathjax></p>
<p>∴ <mathjax>#V_1 = "322 mL" × (9.150 color(red)(cancel(color(black)("atm"))))/(6.20 color(red)(cancel(color(black)("atm")))) = "475 mL"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What was the original volume of a gas that has changed from a pressure of 6.20 atm to 9.150 atm and a new volume of 322 mL?</h1>
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<p>The original volume was 475 mL.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve this problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)p_1V_1 = p_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_1 = V_2 × p_2/p_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = "6.20 atm"; color(white)(ll)V_1 = "?"#</mathjax><br/>
<mathjax>#p_2 ="9.150 atm"; V_2 = "322 mL"#</mathjax></p>
<p>∴ <mathjax>#V_1 = "322 mL" × (9.150 color(red)(cancel(color(black)("atm"))))/(6.20 color(red)(cancel(color(black)("atm")))) = "475 mL"#</mathjax></p></div>
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</article> | What was the original volume of a gas that has changed from a pressure of 6.20 atm to 9.150 atm and a new volume of 322 mL? | null |
1,949 | ac63e23a-6ddd-11ea-ad45-ccda262736ce | https://socratic.org/questions/what-is-water-on-the-ph-scale | 7.00 | start physical_unit 2 2 ph none qc_end substance 2 2 qc_end end | [{"type":"physical unit","value":"pH [OF] water"}] | [{"type":"physical unit","value":"7.00"}] | [{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is water on the pH scale?</h1> | null | 7.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, is a measure of hydrogen ion concentration in water (I think there is an accent in there somewhere but nevermind). </p>
<p>Aqueous acidity is conceived to be a measure of hydrogen ion concentration according to the following equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + H_3O^+#</mathjax></p>
<p>And at <mathjax>#298K#</mathjax>, <mathjax>#K_w=[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>.</p>
<p>Taking <mathjax>#-log_10#</mathjax> of both sides, we get the defining relationship:</p>
<p><mathjax>#pH + pOH=-log_10(10^-14)=14#</mathjax>. </p>
<p>At neutrality under standard conditions, thus <mathjax>#pH=pOH=7#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>Distilled water has <mathjax>#pH=7#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, is a measure of hydrogen ion concentration in water (I think there is an accent in there somewhere but nevermind). </p>
<p>Aqueous acidity is conceived to be a measure of hydrogen ion concentration according to the following equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + H_3O^+#</mathjax></p>
<p>And at <mathjax>#298K#</mathjax>, <mathjax>#K_w=[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>.</p>
<p>Taking <mathjax>#-log_10#</mathjax> of both sides, we get the defining relationship:</p>
<p><mathjax>#pH + pOH=-log_10(10^-14)=14#</mathjax>. </p>
<p>At neutrality under standard conditions, thus <mathjax>#pH=pOH=7#</mathjax>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is water on the pH scale?</h1>
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<div class="markdown"><p>Distilled water has <mathjax>#pH=7#</mathjax>.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, is a measure of hydrogen ion concentration in water (I think there is an accent in there somewhere but nevermind). </p>
<p>Aqueous acidity is conceived to be a measure of hydrogen ion concentration according to the following equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + H_3O^+#</mathjax></p>
<p>And at <mathjax>#298K#</mathjax>, <mathjax>#K_w=[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>.</p>
<p>Taking <mathjax>#-log_10#</mathjax> of both sides, we get the defining relationship:</p>
<p><mathjax>#pH + pOH=-log_10(10^-14)=14#</mathjax>. </p>
<p>At neutrality under standard conditions, thus <mathjax>#pH=pOH=7#</mathjax>.</p></div>
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</article> | What is water on the pH scale? | null |
1,950 | ab27da82-6ddd-11ea-887d-ccda262736ce | https://socratic.org/questions/598b4cf8b72cff56605a9b65 | 1.70 | start physical_unit 6 6 ph none qc_end physical_unit 8 9 11 14 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] sulfuric acid solution"}] | [{"type":"physical unit","value":"1.70"}] | [{"type":"physical unit","value":"Concentration [OF] sulfuric acid solution [=] \\pu{2 × 10^(-2) mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# of a solution of #"sulfuric acid"# of #2xx10^-2*mol*L^-1# concentration?</h1> | null | 1.70 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>Now for <mathjax>#H_2SO_4#</mathjax> with a concentration of <mathjax>#10^-2*mol*L^-1#</mathjax>, clearly, <mathjax>#[H_3O^+]=2xx10^-2*mol*L^-1#</mathjax>.</p>
<p>That is to a first approx. <mathjax>#H_2SO_4#</mathjax> is considered to be a diacid in aqueous solution and gives 2 equiv of <mathjax>#H_3O^+#</mathjax>. (In the absence of a value for <mathjax>#pK_(a2)#</mathjax> I am perfectly justified in making this assumption!)</p>
<p>And so <mathjax>#pH=-log_10(2xx10^-2)=-log_10(0.02)=1.70#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pH=1.70...........#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>Now for <mathjax>#H_2SO_4#</mathjax> with a concentration of <mathjax>#10^-2*mol*L^-1#</mathjax>, clearly, <mathjax>#[H_3O^+]=2xx10^-2*mol*L^-1#</mathjax>.</p>
<p>That is to a first approx. <mathjax>#H_2SO_4#</mathjax> is considered to be a diacid in aqueous solution and gives 2 equiv of <mathjax>#H_3O^+#</mathjax>. (In the absence of a value for <mathjax>#pK_(a2)#</mathjax> I am perfectly justified in making this assumption!)</p>
<p>And so <mathjax>#pH=-log_10(2xx10^-2)=-log_10(0.02)=1.70#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# of a solution of #"sulfuric acid"# of #2xx10^-2*mol*L^-1# concentration?</h1>
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<div class="markdown"><p><mathjax>#pH=1.70...........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, <mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>Now for <mathjax>#H_2SO_4#</mathjax> with a concentration of <mathjax>#10^-2*mol*L^-1#</mathjax>, clearly, <mathjax>#[H_3O^+]=2xx10^-2*mol*L^-1#</mathjax>.</p>
<p>That is to a first approx. <mathjax>#H_2SO_4#</mathjax> is considered to be a diacid in aqueous solution and gives 2 equiv of <mathjax>#H_3O^+#</mathjax>. (In the absence of a value for <mathjax>#pK_(a2)#</mathjax> I am perfectly justified in making this assumption!)</p>
<p>And so <mathjax>#pH=-log_10(2xx10^-2)=-log_10(0.02)=1.70#</mathjax></p></div>
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</article> | What is the #pH# of a solution of #"sulfuric acid"# of #2xx10^-2*mol*L^-1# concentration? | null |
1,951 | a9b2fac3-6ddd-11ea-981c-ccda262736ce | https://socratic.org/questions/585d45ca11ef6b45e2fc7824 | 12.23 mL | start physical_unit 1 1 volume ml qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 20 21 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas [IN] mL"}] | [{"type":"physical unit","value":"12.23 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{12.7 mL}"},{"type":"physical unit","value":"Temperature1 [OF] gas [=] \\pu{295 K}"},{"type":"physical unit","value":"Temperature2 [OF] gas [=] \\pu{284 K}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies a volume of #12.7*mL# at #295*K# in a piston; what volume will it occupy at #284*K#?</h1> | null | 12.23 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#(12.7*mL)/(295*K)xx284*K=12.2*mL#</mathjax></p>
<p>AS expected, the volume is slightly reduced at slightly reduced temperature. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>We use old Charles' law.....<mathjax>#V_1/T_1=V_2/T_2#</mathjax>, which establishes the influence of absolute temperature on gaseous volume......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#(12.7*mL)/(295*K)xx284*K=12.2*mL#</mathjax></p>
<p>AS expected, the volume is slightly reduced at slightly reduced temperature. </p></div>
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<h1 class="questionTitle" itemprop="name">A gas occupies a volume of #12.7*mL# at #295*K# in a piston; what volume will it occupy at #284*K#?</h1>
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<div class="markdown"><p>We use old Charles' law.....<mathjax>#V_1/T_1=V_2/T_2#</mathjax>, which establishes the influence of absolute temperature on gaseous volume......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#(12.7*mL)/(295*K)xx284*K=12.2*mL#</mathjax></p>
<p>AS expected, the volume is slightly reduced at slightly reduced temperature. </p></div>
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</article> | A gas occupies a volume of #12.7*mL# at #295*K# in a piston; what volume will it occupy at #284*K#? | null |
1,952 | a96e467b-6ddd-11ea-95c8-ccda262736ce | https://socratic.org/questions/an-ion-has-79-protons-76-electrons-116-neutrons-what-is-the-correct-formula-of-t | Au^3+ | start chemical_formula qc_end physical_unit 4 4 3 3 number qc_end physical_unit 6 6 5 5 number qc_end physical_unit 8 8 7 7 number qc_end end | [{"type":"other","value":"Chemical Formula [OF] ion [IN] default"}] | [{"type":"chemical equation","value":"Au^3+"}] | [{"type":"physical unit","value":"Number [OF] protons [=] \\pu{79}"},{"type":"physical unit","value":"Number [OF] electrons [=] \\pu{76}"},{"type":"physical unit","value":"Number [OF] neutrons [=] \\pu{116}"}] | <h1 class="questionTitle" itemprop="name">An ion has 79 protons, 76 electrons, 116 neutrons. What is the correct formula of the ion?</h1> | null | Au^3+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, get the element from the proton number. 79 protons corresponds to gold, Au.</p>
<p>A gold atom has originally 79 electrons. (Must be the same number as proton to be electrically neutral) Since now it only has 76 electrons, it has lost 3 electrons. It means now it has a state of +3.</p></div>
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<div class="markdown"><p><mathjax>#"Au"^{3+}#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, get the element from the proton number. 79 protons corresponds to gold, Au.</p>
<p>A gold atom has originally 79 electrons. (Must be the same number as proton to be electrically neutral) Since now it only has 76 electrons, it has lost 3 electrons. It means now it has a state of +3.</p></div>
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<h1 class="questionTitle" itemprop="name">An ion has 79 protons, 76 electrons, 116 neutrons. What is the correct formula of the ion?</h1>
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Sep 10, 2016
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<div class="markdown"><p><mathjax>#"Au"^{3+}#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>First, get the element from the proton number. 79 protons corresponds to gold, Au.</p>
<p>A gold atom has originally 79 electrons. (Must be the same number as proton to be electrically neutral) Since now it only has 76 electrons, it has lost 3 electrons. It means now it has a state of +3.</p></div>
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</article> | An ion has 79 protons, 76 electrons, 116 neutrons. What is the correct formula of the ion? | null |
1,953 | ac78385c-6ddd-11ea-9df5-ccda262736ce | https://socratic.org/questions/the-specific-heat-of-liquid-bromine-is-0-226j-g-k-and-the-density-is-3-12g-ml-ho | 16.2 J | start physical_unit 4 5 heat_energy j qc_end physical_unit 4 5 7 10 specific_heat qc_end physical_unit 4 5 15 16 density qc_end physical_unit 4 5 28 29 volume qc_end physical_unit 4 5 33 34 temperature qc_end physical_unit 4 5 36 37 temperature qc_end end | [{"type":"physical unit","value":"Required heat [OF] liquid bromine [IN] J"}] | [{"type":"physical unit","value":"16.2 J"}] | [{"type":"physical unit","value":"specific heat [OF] liquid bromine [=] \\pu{0.226 J/(g * K)}"},{"type":"physical unit","value":"density [OF] liquid bromine [=] \\pu{3.12 g/mL}"},{"type":"physical unit","value":"Volume [OF] liquid bromine [=] \\pu{10.0 mL}"},{"type":"physical unit","value":"Temperature1 [OF] liquid bromine [=] \\pu{25.00 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] liquid bromine [=] \\pu{27.30 ℃}"}] | <h1 class="questionTitle" itemprop="name">The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C?</h1> | null | 16.2 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use bromine's <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to determine how many grams of bromine you have in that <mathjax>#"10.0-mL"#</mathjax> sample. </p>
<p>A density of <mathjax>#"3.12 g/mL"#</mathjax> tells you that you get a mass of <mathjax>#"3.12 g"#</mathjax> <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of bromine. This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"#</mathjax></p>
</blockquote>
<p>Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values and solve for <mathjax>#q#</mathjax> to get</p>
<blockquote>
<p><mathjax>#q = 31.2color(red)(cancel(color(black)("g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = color(green)("16.2 J")#</mathjax></p>
</blockquote>
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<iframe src="https://www.youtube.com/embed/llsBOI7BORI?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p><mathjax>#"16.2 J"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use bromine's <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to determine how many grams of bromine you have in that <mathjax>#"10.0-mL"#</mathjax> sample. </p>
<p>A density of <mathjax>#"3.12 g/mL"#</mathjax> tells you that you get a mass of <mathjax>#"3.12 g"#</mathjax> <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of bromine. This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"#</mathjax></p>
</blockquote>
<p>Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values and solve for <mathjax>#q#</mathjax> to get</p>
<blockquote>
<p><mathjax>#q = 31.2color(red)(cancel(color(black)("g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = color(green)("16.2 J")#</mathjax></p>
</blockquote>
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<iframe src="https://www.youtube.com/embed/llsBOI7BORI?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C?</h1>
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Nov 14, 2015
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<div class="markdown"><p><mathjax>#"16.2 J"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use bromine's <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to determine how many grams of bromine you have in that <mathjax>#"10.0-mL"#</mathjax> sample. </p>
<p>A density of <mathjax>#"3.12 g/mL"#</mathjax> tells you that you get a mass of <mathjax>#"3.12 g"#</mathjax> <strong>for every</strong> <mathjax>#"100 mL"#</mathjax> of bromine. This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"#</mathjax></p>
</blockquote>
<p>Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values and solve for <mathjax>#q#</mathjax> to get</p>
<blockquote>
<p><mathjax>#q = 31.2color(red)(cancel(color(black)("g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = color(green)("16.2 J")#</mathjax></p>
</blockquote>
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</article> | The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C? | null |
1,954 | abc19b92-6ddd-11ea-9c1d-ccda262736ce | https://socratic.org/questions/a-0-100-m-solution-of-an-acid-ha-has-a-ph-2-00-what-is-the-value-of-the-ionizati | 0.11 | start physical_unit 23 24 ka none qc_end physical_unit 7 7 1 2 molarity qc_end physical_unit 3 3 12 12 ph qc_end end | [{"type":"physical unit","value":"Ka [OF] this acid"}] | [{"type":"physical unit","value":"0.11"}] | [{"type":"physical unit","value":"Molarity [OF] HA solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"pH [OF] HA solution [=] \\pu{2.00}"}] | <h1 class="questionTitle" itemprop="name">A 0.100 M solution of an acid, HA, has a pH = 2.00. what is the value of the ionization constant, #K_a# for this acid?</h1> | null | 0.11 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(ml)0.100color(white)(mmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)+xcolor(white)(ml)+x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmll)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.00"#</mathjax></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) #</mathjax></p>
<p>∴ <mathjax>#K_"a" = x^2/(0.100-x) = 0.100^2/(0.100-0.010) = 0.0100/0.090 =0.11#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_"a" = 0.11#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(ml)0.100color(white)(mmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)+xcolor(white)(ml)+x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmll)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.00"#</mathjax></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) #</mathjax></p>
<p>∴ <mathjax>#K_"a" = x^2/(0.100-x) = 0.100^2/(0.100-0.010) = 0.0100/0.090 =0.11#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 0.100 M solution of an acid, HA, has a pH = 2.00. what is the value of the ionization constant, #K_a# for this acid?</h1>
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Ernest Z.
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<div class="markdown"><p><mathjax>#K_"a" = 0.11#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(ml)0.100color(white)(mmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)+xcolor(white)(ml)+x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmll)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.00"#</mathjax></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) #</mathjax></p>
<p>∴ <mathjax>#K_"a" = x^2/(0.100-x) = 0.100^2/(0.100-0.010) = 0.0100/0.090 =0.11#</mathjax></p></div>
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</article> | A 0.100 M solution of an acid, HA, has a pH = 2.00. what is the value of the ionization constant, #K_a# for this acid? | null |
1,955 | abcb44cb-6ddd-11ea-af41-ccda262736ce | https://socratic.org/questions/what-volume-of-0-181-m-na-3po-4-solution-is-necessary-to-completely-react-with-9 | 35.86 mL | start physical_unit 5 6 volume ml qc_end physical_unit 5 6 3 4 molarity qc_end physical_unit 6 6 13 14 volume qc_end physical_unit 18 18 16 17 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] Na3PO4 solution [IN] mL"}] | [{"type":"physical unit","value":"35.86 mL"}] | [{"type":"physical unit","value":"Molarity [OF] Na3PO4 solution [=] \\pu{0.181 M}"},{"type":"physical unit","value":"Volume [OF] CuCl2 solution [=] \\pu{91.0 mL}"},{"type":"physical unit","value":"Molarity [OF] CuCl2 solution [=] \\pu{0.107 M}"},{"type":"other","value":"Completely react."}] | <h1 class="questionTitle" itemprop="name">What volume of 0.181 M #Na_3PO_4# solution is necessary to completely react with 91.0 mL of 0.107 M #CuCl_2#?</h1> | null | 35.86 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the balanced equation:</p>
<p><mathjax>#3CuCl_(2(aq))+2Na_3PO_(4(aq))rarrCu_3(PO_4)_(2(s))+6NaCl(aq)#</mathjax></p>
<p>This tells us that 3 moles of <mathjax>#CuCl_(2)#</mathjax> react with 2 moles <mathjax>#Na_3PO_4#</mathjax>.</p>
<p><mathjax>#:.#</mathjax> 1 mole <mathjax>#CuCl_2#</mathjax> will react with 2/3 moles <mathjax>#Na_3PO_4#</mathjax></p>
<p>We know that concentration = moles/volume i.e:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p><mathjax>#:. nCuCl_2=0.107xx91.0/1000=9.737xx10^(-3)#</mathjax></p>
<p>I divided by 1000 to convert <mathjax>#"ml"#</mathjax> to <mathjax>#"L"#</mathjax></p>
<p><mathjax>#:.nNa_3PO_4=9.737xx10^(-3)xx2/3=6.491xx10^(-3)#</mathjax></p>
<p><mathjax>#v=n/c=(6.491xx10^(-3))/(0.181)=35.86xx10^(-3)"L"#</mathjax></p>
<p><mathjax>#:.v=35.86"ml"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#35.9"ml"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the balanced equation:</p>
<p><mathjax>#3CuCl_(2(aq))+2Na_3PO_(4(aq))rarrCu_3(PO_4)_(2(s))+6NaCl(aq)#</mathjax></p>
<p>This tells us that 3 moles of <mathjax>#CuCl_(2)#</mathjax> react with 2 moles <mathjax>#Na_3PO_4#</mathjax>.</p>
<p><mathjax>#:.#</mathjax> 1 mole <mathjax>#CuCl_2#</mathjax> will react with 2/3 moles <mathjax>#Na_3PO_4#</mathjax></p>
<p>We know that concentration = moles/volume i.e:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p><mathjax>#:. nCuCl_2=0.107xx91.0/1000=9.737xx10^(-3)#</mathjax></p>
<p>I divided by 1000 to convert <mathjax>#"ml"#</mathjax> to <mathjax>#"L"#</mathjax></p>
<p><mathjax>#:.nNa_3PO_4=9.737xx10^(-3)xx2/3=6.491xx10^(-3)#</mathjax></p>
<p><mathjax>#v=n/c=(6.491xx10^(-3))/(0.181)=35.86xx10^(-3)"L"#</mathjax></p>
<p><mathjax>#:.v=35.86"ml"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume of 0.181 M #Na_3PO_4# solution is necessary to completely react with 91.0 mL of 0.107 M #CuCl_2#?</h1>
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<div class="markdown"><p><mathjax>#35.9"ml"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Start with the balanced equation:</p>
<p><mathjax>#3CuCl_(2(aq))+2Na_3PO_(4(aq))rarrCu_3(PO_4)_(2(s))+6NaCl(aq)#</mathjax></p>
<p>This tells us that 3 moles of <mathjax>#CuCl_(2)#</mathjax> react with 2 moles <mathjax>#Na_3PO_4#</mathjax>.</p>
<p><mathjax>#:.#</mathjax> 1 mole <mathjax>#CuCl_2#</mathjax> will react with 2/3 moles <mathjax>#Na_3PO_4#</mathjax></p>
<p>We know that concentration = moles/volume i.e:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p><mathjax>#:. nCuCl_2=0.107xx91.0/1000=9.737xx10^(-3)#</mathjax></p>
<p>I divided by 1000 to convert <mathjax>#"ml"#</mathjax> to <mathjax>#"L"#</mathjax></p>
<p><mathjax>#:.nNa_3PO_4=9.737xx10^(-3)xx2/3=6.491xx10^(-3)#</mathjax></p>
<p><mathjax>#v=n/c=(6.491xx10^(-3))/(0.181)=35.86xx10^(-3)"L"#</mathjax></p>
<p><mathjax>#:.v=35.86"ml"#</mathjax></p></div>
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</article> | What volume of 0.181 M #Na_3PO_4# solution is necessary to completely react with 91.0 mL of 0.107 M #CuCl_2#? | null |
1,956 | ac2500be-6ddd-11ea-a4dd-ccda262736ce | https://socratic.org/questions/a-gas-sitting-in-a-5l-container-at-12-degrees-celcius-at-3atm-how-many-moles-do- | 0.64 moles | start physical_unit 1 1 mole mol qc_end physical_unit 1 1 5 6 volume qc_end physical_unit 1 1 9 11 temperature qc_end physical_unit 1 1 13 14 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] moles"}] | [{"type":"physical unit","value":"0.64 moles"}] | [{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{5 L}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{12 degrees celcius}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{3 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas sitting in a 5L container at 12 degrees celcius at 3atm, how many moles do you have?</h1> | null | 0.64 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We simply solve the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(3*atmxx5.0*L)/(0.0821*(L*atm)/(K*mol)*(12+273.15)*K)#</mathjax></p>
<p><mathjax>#=??*mol...#</mathjax></p></div>
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<div class="markdown"><p>A bit over half a mole....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We simply solve the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(3*atmxx5.0*L)/(0.0821*(L*atm)/(K*mol)*(12+273.15)*K)#</mathjax></p>
<p><mathjax>#=??*mol...#</mathjax></p></div>
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anor277
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<span class="dateCreated" datetime="2018-07-09T20:47:05" itemprop="dateCreated">
Jul 9, 2018
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<div class="markdown"><p>A bit over half a mole....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We simply solve the Ideal Gas equation...</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#=(3*atmxx5.0*L)/(0.0821*(L*atm)/(K*mol)*(12+273.15)*K)#</mathjax></p>
<p><mathjax>#=??*mol...#</mathjax></p></div>
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Harish Chandra Rajpoot
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Truong-Son N.
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<div class="markdown"><p><mathjax>#0.641435498778218\approx 0.641\ \text{moles}#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From ideal gas equation </p>
<blockquote>
<p><mathjax>#PV=nRT#</mathjax></p>
</blockquote>
<p>where:</p>
<ul>
<li>
<p><mathjax>#P#</mathjax> is absolute pressure of gas </p>
</li>
<li>
<p><mathjax>#V#</mathjax> is volume of gas</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is number of moles of gas</p>
</li>
<li>
<p><mathjax>#R=8.314\ \text{J/mole K}#</mathjax> is universal gas constant.</p>
</li>
<li>
<p><mathjax>#T#</mathjax> is absolute temperature of gas</p>
</li>
</ul>
<blockquote>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
</blockquote>
<p>Setting the values </p>
<ul>
<li>
<p><mathjax>#P=3\ "atm"=3 xx 101325\ "Pa"#</mathjax>, </p>
</li>
<li>
<p><mathjax>#V=5\ "L"=5 xx 10^{-3}\ "m"^3#</mathjax></p>
</li>
<li>
<p><mathjax>#R=8.314 \ \text{J/mole K}#</mathjax></p>
</li>
<li>
<p><mathjax>#T=12^0 "C"=12+273=285\ "K"#</mathjax></p>
</li>
</ul>
<p>we get, the number of moles </p>
<p><mathjax>#n=\frac{3\times 101325\times 5\times 10^{-3} }{8.314\times 285}#</mathjax></p>
<p><mathjax>#=0.641435498778218\ \text{moles}#</mathjax></p>
<p><mathjax>#\approx 0.641\ \text{moles}#</mathjax></p></div>
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Meave60
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<span class="dateCreated" datetime="2018-07-10T01:54:18" itemprop="dateCreated">
Jul 10, 2018
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<div class="markdown"><p>You have <mathjax>#"0.6 mol"#</mathjax> of gas.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where: </p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature in Kelvins.</p>
<p><strong>Known</strong></p>
<p><mathjax>#P="3 atm"#</mathjax></p>
<p><mathjax>#V="5 L"#</mathjax></p>
<p><mathjax>#R="0.08206 L atm K"^(-1) "mol"^(-1)#</mathjax></p>
<p><mathjax>#T="12"^@"C + 273.15"="285 K"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#n#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate moles. Plug in the known values and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(3color(red)cancel(color(black)("atm"))xx5color(red)cancel(color(black)("L")))/(0.08206color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx285color(red)cancel(color(black)("K")))="0.6 mol"#</mathjax> <br/>
(rounded to one significant figure)</p></div>
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</article> | A gas sitting in a 5L container at 12 degrees celcius at 3atm, how many moles do you have? | null |
1,957 | aaa9eb70-6ddd-11ea-950c-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-3-54-10-24-molecules-of-rubidium-oxide | 186.94 g | start physical_unit 10 11 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] rubidium oxide [IN] g"}] | [{"type":"physical unit","value":"186.94 g"}] | [{"type":"physical unit","value":"Number [OF] rubidium oxide molecules [=] \\pu{3.54 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #3.54*10^24# molecules of rubidium oxide?</h1> | null | 186.94 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that <mathjax>#6.02214xx10^24#</mathjax> formula units of <mathjax>#Rb_2O#</mathjax> have a mass of <mathjax>#186.94*g#</mathjax>, the given number of <mathjax>#Rb_2O#</mathjax> formula units have a mass of:</p>
<p><mathjax>#(3.54xx10^24" formula units of " Rb_2O)/(6.02214xx10^24" formula units "mol^-1*Rb_2O)#</mathjax> </p>
<p><mathjax>#=#</mathjax> </p>
<p><mathjax>#??" moles of rubidium oxide" #</mathjax> </p>
<p>So the mass <mathjax>#~=#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx186.94*g*mol^-1, "i.e. "93*g#</mathjax>.</p></div>
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<div class="markdown"><p>It is a fact that <mathjax>#N_A#</mathjax> <mathjax>#Rb_2O#</mathjax> particles have a mass of <mathjax>#186.94*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given that <mathjax>#6.02214xx10^24#</mathjax> formula units of <mathjax>#Rb_2O#</mathjax> have a mass of <mathjax>#186.94*g#</mathjax>, the given number of <mathjax>#Rb_2O#</mathjax> formula units have a mass of:</p>
<p><mathjax>#(3.54xx10^24" formula units of " Rb_2O)/(6.02214xx10^24" formula units "mol^-1*Rb_2O)#</mathjax> </p>
<p><mathjax>#=#</mathjax> </p>
<p><mathjax>#??" moles of rubidium oxide" #</mathjax> </p>
<p>So the mass <mathjax>#~=#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx186.94*g*mol^-1, "i.e. "93*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of #3.54*10^24# molecules of rubidium oxide?</h1>
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<div class="markdown"><p>It is a fact that <mathjax>#N_A#</mathjax> <mathjax>#Rb_2O#</mathjax> particles have a mass of <mathjax>#186.94*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given that <mathjax>#6.02214xx10^24#</mathjax> formula units of <mathjax>#Rb_2O#</mathjax> have a mass of <mathjax>#186.94*g#</mathjax>, the given number of <mathjax>#Rb_2O#</mathjax> formula units have a mass of:</p>
<p><mathjax>#(3.54xx10^24" formula units of " Rb_2O)/(6.02214xx10^24" formula units "mol^-1*Rb_2O)#</mathjax> </p>
<p><mathjax>#=#</mathjax> </p>
<p><mathjax>#??" moles of rubidium oxide" #</mathjax> </p>
<p>So the mass <mathjax>#~=#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#xx186.94*g*mol^-1, "i.e. "93*g#</mathjax>.</p></div>
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</article> | What is the mass of #3.54*10^24# molecules of rubidium oxide? | null |
1,958 | aa94cf5c-6ddd-11ea-a231-ccda262736ce | https://socratic.org/questions/each-macbook-ordered-this-winter-contains-6-87-10-23-lithium-atoms-in-each-batte | 1.14 moles | start physical_unit 9 10 mole mol qc_end physical_unit 9 10 6 8 number qc_end end | [{"type":"physical unit","value":"Mole [OF] lithium atoms [IN] moles"}] | [{"type":"physical unit","value":"1.14 moles"}] | [{"type":"physical unit","value":"Number [OF] lithium atoms [=] \\pu{6.87 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">Each Macbook ordered this winter contains #6.87 * 10^23# lithium atoms in each battery. How do you convert the amount of atoms in each battery to moles of lithium atoms?</h1> | null | 1.14 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of an element contains <mathjax>#6.022xx10^23#</mathjax> atoms of the element. <mathjax>#"1 mol Li"=6.022xx10^23"atoms Li"#</mathjax>. In order to convert atoms to moles, divide the given number of atoms by <mathjax>#6.022xx10^23"atoms/mol"#</mathjax>.</p>
<p><mathjax>#6.87xx10^23cancel"atoms Li"xx(1"mol Li")/(6.022xx10^23cancel"atoms Li")="1.14 mol Li"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#6.87xx10^23#</mathjax> atoms Li make up <mathjax>#"1.14 mol Li"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of an element contains <mathjax>#6.022xx10^23#</mathjax> atoms of the element. <mathjax>#"1 mol Li"=6.022xx10^23"atoms Li"#</mathjax>. In order to convert atoms to moles, divide the given number of atoms by <mathjax>#6.022xx10^23"atoms/mol"#</mathjax>.</p>
<p><mathjax>#6.87xx10^23cancel"atoms Li"xx(1"mol Li")/(6.022xx10^23cancel"atoms Li")="1.14 mol Li"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Each Macbook ordered this winter contains #6.87 * 10^23# lithium atoms in each battery. How do you convert the amount of atoms in each battery to moles of lithium atoms?</h1>
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<div class="markdown"><p><mathjax>#6.87xx10^23#</mathjax> atoms Li make up <mathjax>#"1.14 mol Li"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>One mole of an element contains <mathjax>#6.022xx10^23#</mathjax> atoms of the element. <mathjax>#"1 mol Li"=6.022xx10^23"atoms Li"#</mathjax>. In order to convert atoms to moles, divide the given number of atoms by <mathjax>#6.022xx10^23"atoms/mol"#</mathjax>.</p>
<p><mathjax>#6.87xx10^23cancel"atoms Li"xx(1"mol Li")/(6.022xx10^23cancel"atoms Li")="1.14 mol Li"#</mathjax></p></div>
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</article> | Each Macbook ordered this winter contains #6.87 * 10^23# lithium atoms in each battery. How do you convert the amount of atoms in each battery to moles of lithium atoms? | null |
1,959 | a8cff695-6ddd-11ea-a4d6-ccda262736ce | https://socratic.org/questions/a-soap-bubble-is-formed-using-a-mixture-of-detergent-and-water-the-surface-tensi | 3 Pa | start physical_unit 22 23 gauge_pressure pa qc_end physical_unit 16 17 19 20 surface_tension qc_end physical_unit 22 23 28 29 radius qc_end end | [{"type":"physical unit","value":"Gauge pressure [OF] the bubble [IN] Pa"}] | [{"type":"physical unit","value":"3 Pa"}] | [{"type":"physical unit","value":"Surface tension [OF] the mixture [=] \\pu{0.030 Nm^(−1)}"},{"type":"physical unit","value":"Radius [OF] the bubble [=] \\pu{2 cm}"},{"type":"physical unit","value":"Pressure [OF] atmosphere [=] \\pu{101.2 kPa}"}] | <h1 class="questionTitle" itemprop="name">A soap bubble is formed using a mixture of detergent and water. The surface tension of the mixture is #0.030 ##Nm^-1#. If the bubble has a radius of #2# #cm# and atmospheric pressure is #101.2# #kPa#, what is the gauge pressure inside the bubble?</h1> | null | 3 Pa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Laplace - Young equation</strong>, which allows you to calculate the <em>Laplace pressure</em>, i.e. the <em>pressure difference</em> that exists between the inside and the outside of a curved surface that acts as a boundary between a liquid region and a gas region</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma (1/R_1 + 1/R_2)color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#gamma#</mathjax> - the <em>surface tension</em> of the liquid<br/>
<mathjax>#R_1#</mathjax> ,<mathjax>#R_2#</mathjax> - the principal radii of curvature</p>
<p>The idea here is that a <em>difference in pressure</em> between the two sides of a <em>flat</em> liquid surface gives rise to a force that can only be balanced by the surface tension of the liquid if the surface of the liquid is <strong>curved</strong>. </p>
<p>The great thing about a <em>spherical shape</em> such as a bubble is that the two principal radii of curvature are <strong>equal</strong>. This implies tha the Laplace pressure will be equal to </p>
<blockquote>
<p><mathjax>#Deltap = gamma * (1/R + 1/R)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma * 2/Rcolor(white)(a/a)|))) ->#</mathjax> <em>the Laplace pressure for <strong>spherical shapes</strong></em></p>
</blockquote>
<p>Before plugging in your values to get the Laplace pressure for your soap bubble, make sure that you convert the radius of the bubble from <em>centimeters</em> to <em>meters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 m" = 10^2"cm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#Deltap = "0.030 N m"^(-1) * 2/(2 * 10^(-2)"m") = "3 N m"^(-2)#</mathjax></p>
</blockquote>
<p>As you know, <mathjax>#"N" * m"^(-2)#</mathjax> is equivalent to <em>Pascal</em>, <mathjax>#"Pa"#</mathjax>, which means that the Laplace pressure is equal to </p>
<blockquote>
<p><mathjax>#Deltap = "3 Pa"#</mathjax></p>
</blockquote>
<p>Now, <strong>gauge pressure</strong> essentially tells you the <strong>difference</strong> that exists between the absolute pressure and <em>ambient air pressure</em>. In other words, gauge pressure is zero-referenced against normal pressure, i.e. air pressure at sea level, given to you as <mathjax>#"101.2 kPa"#</mathjax>. </p>
<p>In this context, the Laplace pressure will actually be equivalent to the <strong>gauge pressure</strong>. In other words, the pressure <strong>inside</strong> the soap bubble will be equal to </p>
<blockquote>
<p><mathjax>#P_"inside bubble" = "101,200 Pa" + "3 Pa" = "101,203 Pa"#</mathjax></p>
</blockquote>
<p>The gauge pressure uses <mathjax>#"101,200 Pa"#</mathjax> as the <strong>zero reference</strong>, so you will get a gauge reading of </p>
<blockquote>
<p><mathjax>#P_"gauge" = Deltap = "3 Pa"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3 Pa"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Laplace - Young equation</strong>, which allows you to calculate the <em>Laplace pressure</em>, i.e. the <em>pressure difference</em> that exists between the inside and the outside of a curved surface that acts as a boundary between a liquid region and a gas region</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma (1/R_1 + 1/R_2)color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#gamma#</mathjax> - the <em>surface tension</em> of the liquid<br/>
<mathjax>#R_1#</mathjax> ,<mathjax>#R_2#</mathjax> - the principal radii of curvature</p>
<p>The idea here is that a <em>difference in pressure</em> between the two sides of a <em>flat</em> liquid surface gives rise to a force that can only be balanced by the surface tension of the liquid if the surface of the liquid is <strong>curved</strong>. </p>
<p>The great thing about a <em>spherical shape</em> such as a bubble is that the two principal radii of curvature are <strong>equal</strong>. This implies tha the Laplace pressure will be equal to </p>
<blockquote>
<p><mathjax>#Deltap = gamma * (1/R + 1/R)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma * 2/Rcolor(white)(a/a)|))) ->#</mathjax> <em>the Laplace pressure for <strong>spherical shapes</strong></em></p>
</blockquote>
<p>Before plugging in your values to get the Laplace pressure for your soap bubble, make sure that you convert the radius of the bubble from <em>centimeters</em> to <em>meters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 m" = 10^2"cm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#Deltap = "0.030 N m"^(-1) * 2/(2 * 10^(-2)"m") = "3 N m"^(-2)#</mathjax></p>
</blockquote>
<p>As you know, <mathjax>#"N" * m"^(-2)#</mathjax> is equivalent to <em>Pascal</em>, <mathjax>#"Pa"#</mathjax>, which means that the Laplace pressure is equal to </p>
<blockquote>
<p><mathjax>#Deltap = "3 Pa"#</mathjax></p>
</blockquote>
<p>Now, <strong>gauge pressure</strong> essentially tells you the <strong>difference</strong> that exists between the absolute pressure and <em>ambient air pressure</em>. In other words, gauge pressure is zero-referenced against normal pressure, i.e. air pressure at sea level, given to you as <mathjax>#"101.2 kPa"#</mathjax>. </p>
<p>In this context, the Laplace pressure will actually be equivalent to the <strong>gauge pressure</strong>. In other words, the pressure <strong>inside</strong> the soap bubble will be equal to </p>
<blockquote>
<p><mathjax>#P_"inside bubble" = "101,200 Pa" + "3 Pa" = "101,203 Pa"#</mathjax></p>
</blockquote>
<p>The gauge pressure uses <mathjax>#"101,200 Pa"#</mathjax> as the <strong>zero reference</strong>, so you will get a gauge reading of </p>
<blockquote>
<p><mathjax>#P_"gauge" = Deltap = "3 Pa"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A soap bubble is formed using a mixture of detergent and water. The surface tension of the mixture is #0.030 ##Nm^-1#. If the bubble has a radius of #2# #cm# and atmospheric pressure is #101.2# #kPa#, what is the gauge pressure inside the bubble?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"3 Pa"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Laplace - Young equation</strong>, which allows you to calculate the <em>Laplace pressure</em>, i.e. the <em>pressure difference</em> that exists between the inside and the outside of a curved surface that acts as a boundary between a liquid region and a gas region</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma (1/R_1 + 1/R_2)color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#gamma#</mathjax> - the <em>surface tension</em> of the liquid<br/>
<mathjax>#R_1#</mathjax> ,<mathjax>#R_2#</mathjax> - the principal radii of curvature</p>
<p>The idea here is that a <em>difference in pressure</em> between the two sides of a <em>flat</em> liquid surface gives rise to a force that can only be balanced by the surface tension of the liquid if the surface of the liquid is <strong>curved</strong>. </p>
<p>The great thing about a <em>spherical shape</em> such as a bubble is that the two principal radii of curvature are <strong>equal</strong>. This implies tha the Laplace pressure will be equal to </p>
<blockquote>
<p><mathjax>#Deltap = gamma * (1/R + 1/R)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)Deltap = gamma * 2/Rcolor(white)(a/a)|))) ->#</mathjax> <em>the Laplace pressure for <strong>spherical shapes</strong></em></p>
</blockquote>
<p>Before plugging in your values to get the Laplace pressure for your soap bubble, make sure that you convert the radius of the bubble from <em>centimeters</em> to <em>meters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 m" = 10^2"cm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#Deltap = "0.030 N m"^(-1) * 2/(2 * 10^(-2)"m") = "3 N m"^(-2)#</mathjax></p>
</blockquote>
<p>As you know, <mathjax>#"N" * m"^(-2)#</mathjax> is equivalent to <em>Pascal</em>, <mathjax>#"Pa"#</mathjax>, which means that the Laplace pressure is equal to </p>
<blockquote>
<p><mathjax>#Deltap = "3 Pa"#</mathjax></p>
</blockquote>
<p>Now, <strong>gauge pressure</strong> essentially tells you the <strong>difference</strong> that exists between the absolute pressure and <em>ambient air pressure</em>. In other words, gauge pressure is zero-referenced against normal pressure, i.e. air pressure at sea level, given to you as <mathjax>#"101.2 kPa"#</mathjax>. </p>
<p>In this context, the Laplace pressure will actually be equivalent to the <strong>gauge pressure</strong>. In other words, the pressure <strong>inside</strong> the soap bubble will be equal to </p>
<blockquote>
<p><mathjax>#P_"inside bubble" = "101,200 Pa" + "3 Pa" = "101,203 Pa"#</mathjax></p>
</blockquote>
<p>The gauge pressure uses <mathjax>#"101,200 Pa"#</mathjax> as the <strong>zero reference</strong>, so you will get a gauge reading of </p>
<blockquote>
<p><mathjax>#P_"gauge" = Deltap = "3 Pa"#</mathjax></p>
</blockquote></div>
</div>
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</article> | A soap bubble is formed using a mixture of detergent and water. The surface tension of the mixture is #0.030 ##Nm^-1#. If the bubble has a radius of #2# #cm# and atmospheric pressure is #101.2# #kPa#, what is the gauge pressure inside the bubble? | null |
1,960 | a9fb92a4-6ddd-11ea-938d-ccda262736ce | https://socratic.org/questions/the-ksp-for-baco3-is-5-1-10-9-how-many-grams-of-baco2-will-dissolve-in-1000-ml-o | 0.01 grams | start physical_unit 3 3 mass g qc_end physical_unit 3 3 5 7 equilibrium_constant_k qc_end physical_unit 19 19 16 17 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] BaCO3 [IN] grams"}] | [{"type":"physical unit","value":"0.01 grams"}] | [{"type":"physical unit","value":"Ksp [OF] BaCO3 [=] \\pu{5.1 × 10^(-9)}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{1000 ml}"}] | <h1 class="questionTitle" itemprop="name">The Ksp for BaCO3 is 5.1*10^-9. How many grams of BaCO2 will dissolve in 1000 ml of water?</h1> | null | 0.01 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Te idea here is that you need to use an <strong>ICE table</strong> to determine barium carbonate's <em>molar solubility</em>, then use the compound's molar mass to determine how many <em>grams</em> will dissolve in that much water. </p>
<p><em>Barium carbonate</em>, <mathjax>#"BaCO"_3#</mathjax>, is <strong>Insoluble</strong> in aqueous solution, which means that adding it to water will result in the formation of an <em>equilibrium reaction</em> between the undissolved solid and the dissolved ions. </p>
<p>If you take <mathjax>#s#</mathjax> to be the <strong>molar solubility</strong> of calcium carbonate, you can use an <strong>ICE table</strong> to write</p>
<blockquote>
<p><mathjax>#"BaCO"_text(3(s]) " "rightleftharpoons" " "Ba"_text((aq])^(2+) " "+" " "CO"_text(3(aq])^(2-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "- " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "- " " " " " " " " " "(+s)" " " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "- " " " " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a>, <mathjax>#K_(sp)#</mathjax>, will be equal to</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["CO"_3^(2-)]#</mathjax></p>
<p><mathjax>#K_(sp) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of barium carbonate will be equal to</p>
<blockquote>
<p><mathjax>#s = sqrt(K_(sp)) = sqrt(5.1 * 10^(-9)) = 7.14 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>To determine how many <em>grams</em> of barium carbonate you can dissolve in <mathjax>#"1.00 L"#</mathjax> of water, use barium carbonate's molar mass</p>
<blockquote>
<p><mathjax>#7.14 * 10^(-5)color(red)(cancel(color(black)("moles")))/"L" * "197.34 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.014 g/L")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have one sig figs for the volume of the water. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.014 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Te idea here is that you need to use an <strong>ICE table</strong> to determine barium carbonate's <em>molar solubility</em>, then use the compound's molar mass to determine how many <em>grams</em> will dissolve in that much water. </p>
<p><em>Barium carbonate</em>, <mathjax>#"BaCO"_3#</mathjax>, is <strong>Insoluble</strong> in aqueous solution, which means that adding it to water will result in the formation of an <em>equilibrium reaction</em> between the undissolved solid and the dissolved ions. </p>
<p>If you take <mathjax>#s#</mathjax> to be the <strong>molar solubility</strong> of calcium carbonate, you can use an <strong>ICE table</strong> to write</p>
<blockquote>
<p><mathjax>#"BaCO"_text(3(s]) " "rightleftharpoons" " "Ba"_text((aq])^(2+) " "+" " "CO"_text(3(aq])^(2-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "- " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "- " " " " " " " " " "(+s)" " " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "- " " " " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a>, <mathjax>#K_(sp)#</mathjax>, will be equal to</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["CO"_3^(2-)]#</mathjax></p>
<p><mathjax>#K_(sp) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of barium carbonate will be equal to</p>
<blockquote>
<p><mathjax>#s = sqrt(K_(sp)) = sqrt(5.1 * 10^(-9)) = 7.14 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>To determine how many <em>grams</em> of barium carbonate you can dissolve in <mathjax>#"1.00 L"#</mathjax> of water, use barium carbonate's molar mass</p>
<blockquote>
<p><mathjax>#7.14 * 10^(-5)color(red)(cancel(color(black)("moles")))/"L" * "197.34 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.014 g/L")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have one sig figs for the volume of the water. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The Ksp for BaCO3 is 5.1*10^-9. How many grams of BaCO2 will dissolve in 1000 ml of water?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-22T21:00:39" itemprop="dateCreated">
Nov 22, 2015
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<div class="markdown"><p><mathjax>#"0.014 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Te idea here is that you need to use an <strong>ICE table</strong> to determine barium carbonate's <em>molar solubility</em>, then use the compound's molar mass to determine how many <em>grams</em> will dissolve in that much water. </p>
<p><em>Barium carbonate</em>, <mathjax>#"BaCO"_3#</mathjax>, is <strong>Insoluble</strong> in aqueous solution, which means that adding it to water will result in the formation of an <em>equilibrium reaction</em> between the undissolved solid and the dissolved ions. </p>
<p>If you take <mathjax>#s#</mathjax> to be the <strong>molar solubility</strong> of calcium carbonate, you can use an <strong>ICE table</strong> to write</p>
<blockquote>
<p><mathjax>#"BaCO"_text(3(s]) " "rightleftharpoons" " "Ba"_text((aq])^(2+) " "+" " "CO"_text(3(aq])^(2-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "- " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "- " " " " " " " " " "(+s)" " " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "- " " " " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the <a href="https://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a>, <mathjax>#K_(sp)#</mathjax>, will be equal to</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["CO"_3^(2-)]#</mathjax></p>
<p><mathjax>#K_(sp) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of barium carbonate will be equal to</p>
<blockquote>
<p><mathjax>#s = sqrt(K_(sp)) = sqrt(5.1 * 10^(-9)) = 7.14 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>To determine how many <em>grams</em> of barium carbonate you can dissolve in <mathjax>#"1.00 L"#</mathjax> of water, use barium carbonate's molar mass</p>
<blockquote>
<p><mathjax>#7.14 * 10^(-5)color(red)(cancel(color(black)("moles")))/"L" * "197.34 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.014 g/L")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, despite the fact that you only have one sig figs for the volume of the water. </p></div>
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</article> | The Ksp for BaCO3 is 5.1*10^-9. How many grams of BaCO2 will dissolve in 1000 ml of water? | null |
1,961 | a9a58d2d-6ddd-11ea-8add-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-reaction-for-photosynthesis | 6 CO2 + 6 H2O + hv -> C6H12O6 + 6O2 | start chemical_equation qc_end c_other Photosynthesis qc_end end | [{"type":"other","value":"Chemical Equation [OF] photosynthesis"}] | [{"type":"chemical equation","value":"6 CO2 + 6 H2O + hv -> C6H12O6 + 6O2"}] | [{"type":"other","value":"Photosynthesis"}] | <h1 class="questionTitle" itemprop="name">What is the chemical reaction for photosynthesis? </h1> | null | 6 CO2 + 6 H2O + hv -> C6H12O6 + 6O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So is mass balanced; is charge balanced? You must learn to look at chemical reactions to see that they are indeed stoichiometrically balanced. If they are unbalanced, they cannot be accepted as a model for chemical reactivity. What does the <mathjax>#hnu#</mathjax> term mean in the reaction?</p></div>
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<div class="markdown"><p><mathjax>#6CO_2 + 6H_2O + hnurarr C_6H_12O_6 + 6O_2#</mathjax></p></div>
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<div class="markdown"><p>So is mass balanced; is charge balanced? You must learn to look at chemical reactions to see that they are indeed stoichiometrically balanced. If they are unbalanced, they cannot be accepted as a model for chemical reactivity. What does the <mathjax>#hnu#</mathjax> term mean in the reaction?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical reaction for photosynthesis? </h1>
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anor277
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<div class="markdown"><p><mathjax>#6CO_2 + 6H_2O + hnurarr C_6H_12O_6 + 6O_2#</mathjax></p></div>
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<div class="markdown"><p>So is mass balanced; is charge balanced? You must learn to look at chemical reactions to see that they are indeed stoichiometrically balanced. If they are unbalanced, they cannot be accepted as a model for chemical reactivity. What does the <mathjax>#hnu#</mathjax> term mean in the reaction?</p></div>
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</article> | What is the chemical reaction for photosynthesis? | null |
1,962 | a8d3404a-6ddd-11ea-9d79-ccda262736ce | https://socratic.org/questions/how-many-ml-of-6-0-m-hcl-solution-will-provide-164-g-of-hcl | 749.67 mL | start physical_unit 6 7 volume ml qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 6 6 10 11 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] HCl solution [IN] mL"}] | [{"type":"physical unit","value":"749.67 mL"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{6.0 M}"},{"type":"physical unit","value":"Mass [OF] HCl [=] \\pu{164 g}"}] | <h1 class="questionTitle" itemprop="name">How many mL of 6.0 M HCl solution will provide 164 g of HCl?</h1> | null | 749.67 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many <em>moles</em> you get in <mathjax>#"164 g"#</mathjax> of hydrochloric acid, then use the given solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> to figure out the volume that would contain that many moles. </p>
<p>So, hydrochloric acid has a molar mass of <mathjax>#"36.46 g/mol"#</mathjax>, which means that the <mathjax>#"164-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#164color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "4.498 moles HCl"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in this case is hydrochloric acid, divided by liters of solution</p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>A <mathjax>#"6.0-M"#</mathjax> hydrochloric acid solution will contain <mathjax>#6#</mathjax> moles of hydrochloric acid <strong>for every</strong> liter of solution. This means that <mathjax>#4.498#</mathjax> moles will come with a volume of </p>
<blockquote>
<p><mathjax>#c = n/V implies V = n/c#</mathjax></p>
<p><mathjax>#V = (4.498color(red)(cancel(color(black)("moles"))))/(6.0color(red)(cancel(color(black)("moles")))/"L") = "0.7497 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#V = color(green)("0.75 L")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/KCVRRl_lFGo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.75 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many <em>moles</em> you get in <mathjax>#"164 g"#</mathjax> of hydrochloric acid, then use the given solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> to figure out the volume that would contain that many moles. </p>
<p>So, hydrochloric acid has a molar mass of <mathjax>#"36.46 g/mol"#</mathjax>, which means that the <mathjax>#"164-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#164color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "4.498 moles HCl"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in this case is hydrochloric acid, divided by liters of solution</p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>A <mathjax>#"6.0-M"#</mathjax> hydrochloric acid solution will contain <mathjax>#6#</mathjax> moles of hydrochloric acid <strong>for every</strong> liter of solution. This means that <mathjax>#4.498#</mathjax> moles will come with a volume of </p>
<blockquote>
<p><mathjax>#c = n/V implies V = n/c#</mathjax></p>
<p><mathjax>#V = (4.498color(red)(cancel(color(black)("moles"))))/(6.0color(red)(cancel(color(black)("moles")))/"L") = "0.7497 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#V = color(green)("0.75 L")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/KCVRRl_lFGo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many mL of 6.0 M HCl solution will provide 164 g of HCl?</h1>
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Stefan V.
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Nov 18, 2015
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<div class="markdown"><p><mathjax>#"0.75 L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many <em>moles</em> you get in <mathjax>#"164 g"#</mathjax> of hydrochloric acid, then use the given solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> to figure out the volume that would contain that many moles. </p>
<p>So, hydrochloric acid has a molar mass of <mathjax>#"36.46 g/mol"#</mathjax>, which means that the <mathjax>#"164-g"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#164color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "4.498 moles HCl"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in this case is hydrochloric acid, divided by liters of solution</p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>A <mathjax>#"6.0-M"#</mathjax> hydrochloric acid solution will contain <mathjax>#6#</mathjax> moles of hydrochloric acid <strong>for every</strong> liter of solution. This means that <mathjax>#4.498#</mathjax> moles will come with a volume of </p>
<blockquote>
<p><mathjax>#c = n/V implies V = n/c#</mathjax></p>
<p><mathjax>#V = (4.498color(red)(cancel(color(black)("moles"))))/(6.0color(red)(cancel(color(black)("moles")))/"L") = "0.7497 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#V = color(green)("0.75 L")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/KCVRRl_lFGo?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How many mL of 6.0 M HCl solution will provide 164 g of HCl? | null |
1,963 | a9b807e7-6ddd-11ea-a6e0-ccda262736ce | https://socratic.org/questions/how-do-you-balance-ch3oh-o2-co2-h2o | 2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O"}] | [{"type":"chemical equation","value":"CH3OH + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance CH3OH + O2 -- CO2 + H2O? </h1> | null | 2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"CH"_3"OH"#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 C"#</mathjax> on the left, so we need <mathjax>#"1 C"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> because we have two oxygen-containing molecules without coefficients. ∴ Let's balance <mathjax>#"H"#</mathjax> instead.</p>
<p>We have <mathjax>#"4 H"#</mathjax> on the left, so we need <mathjax>#"4 H"#</mathjax> on the right. There are already <mathjax>#"2 H"#</mathjax> atoms on the right. We must put a 2 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax></strong>:</p>
<p>We have fixed <mathjax>#"4 O"#</mathjax> on the right and <mathjax>#"1 O"#</mathjax> on the left. We need <mathjax>#"3 O"#</mathjax> on the left.</p>
<p><strong>Uh, oh! Fractions!</strong></p>
<p>We start over, this time doubling all the coefficients.</p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Now we can balance <mathjax>#"O"#</mathjax> by putting a 3 in front of <mathjax>#"O"_2#</mathjax></p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let's check.</p>
<p><mathjax>#"Atom" color(white)(m)"lhs"color(white)(m)"rhs"#</mathjax><br/>
<mathjax>#color(white)(m)"C"color(white)(mml)2color(white)(mm)2#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mml)8color(white)(mm)8#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mml)8color(white)(mm)8#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The balanced equation is <mathjax>#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"CH"_3"OH"#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 C"#</mathjax> on the left, so we need <mathjax>#"1 C"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> because we have two oxygen-containing molecules without coefficients. ∴ Let's balance <mathjax>#"H"#</mathjax> instead.</p>
<p>We have <mathjax>#"4 H"#</mathjax> on the left, so we need <mathjax>#"4 H"#</mathjax> on the right. There are already <mathjax>#"2 H"#</mathjax> atoms on the right. We must put a 2 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax></strong>:</p>
<p>We have fixed <mathjax>#"4 O"#</mathjax> on the right and <mathjax>#"1 O"#</mathjax> on the left. We need <mathjax>#"3 O"#</mathjax> on the left.</p>
<p><strong>Uh, oh! Fractions!</strong></p>
<p>We start over, this time doubling all the coefficients.</p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Now we can balance <mathjax>#"O"#</mathjax> by putting a 3 in front of <mathjax>#"O"_2#</mathjax></p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let's check.</p>
<p><mathjax>#"Atom" color(white)(m)"lhs"color(white)(m)"rhs"#</mathjax><br/>
<mathjax>#color(white)(m)"C"color(white)(mml)2color(white)(mm)2#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mml)8color(white)(mm)8#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mml)8color(white)(mm)8#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance CH3OH + O2 -- CO2 + H2O? </h1>
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Ernest Z.
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Stefan V.
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Oct 30, 2016
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<div class="markdown"><p>The balanced equation is <mathjax>#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"CH"_3"OH"#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 C"#</mathjax> on the left, so we need <mathjax>#"1 C"#</mathjax> on the right. We put a 1 in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> because we have two oxygen-containing molecules without coefficients. ∴ Let's balance <mathjax>#"H"#</mathjax> instead.</p>
<p>We have <mathjax>#"4 H"#</mathjax> on the left, so we need <mathjax>#"4 H"#</mathjax> on the right. There are already <mathjax>#"2 H"#</mathjax> atoms on the right. We must put a 2 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"CH"_3"OH" + "O"_2 → color(blue)(1)"CO"_2 + color(orange)(2)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax></strong>:</p>
<p>We have fixed <mathjax>#"4 O"#</mathjax> on the right and <mathjax>#"1 O"#</mathjax> on the left. We need <mathjax>#"3 O"#</mathjax> on the left.</p>
<p><strong>Uh, oh! Fractions!</strong></p>
<p>We start over, this time doubling all the coefficients.</p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + "O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Now we can balance <mathjax>#"O"#</mathjax> by putting a 3 in front of <mathjax>#"O"_2#</mathjax></p>
<p><mathjax>#color(red)(2)"CH"_3"OH" + color(purple)(3)"O"_2 → color(blue)(2)"CO"_2 + color(orange)(4)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let's check.</p>
<p><mathjax>#"Atom" color(white)(m)"lhs"color(white)(m)"rhs"#</mathjax><br/>
<mathjax>#color(white)(m)"C"color(white)(mml)2color(white)(mm)2#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mml)8color(white)(mm)8#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mml)8color(white)(mm)8#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#2"CH"_3"OH" + 3"O"_2 → 2"CO"_2 + 4"H"_2"O"#</mathjax></p></div>
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Aditya Banerjee.
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<span class="dateCreated" datetime="2016-10-31T12:06:25" itemprop="dateCreated">
Oct 31, 2016
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<div class="markdown"><p><mathjax>#2CH_3OH#</mathjax> <mathjax>#+ 3O_2#</mathjax> <mathjax>#rarr 2CO_2 +4H_2O#</mathjax> . </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ul>
<li>First try to balance atoms other than O and H. So first balance the C atoms.<br/>
<mathjax>#2CH_3OH#</mathjax> <mathjax>#+ O_2#</mathjax> <mathjax>#rarr 2CO_2 +H_2O#</mathjax> . </li>
<li>Then balance the H atoms, as no. of H atoms is less than no. of O atoms.<br/>
<mathjax>#2CH_3OH#</mathjax> <mathjax>#+ O_2#</mathjax> <mathjax>#rarr 2CO_2 +4H_2O#</mathjax> .</li>
<li>Now, balance the remaining O atoms.<br/>
<mathjax>#2CH_3OH#</mathjax> <mathjax>#+3 O_2#</mathjax> <mathjax>#rarr 2CO_2 +4H_2O#</mathjax> .</li>
</ul>
<p>Now, check that the no. of all individual atoms in L.H.S and R.H.S are equal.</p></div>
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</article> | How do you balance CH3OH + O2 -- CO2 + H2O? | null |
1,964 | a8fa1288-6ddd-11ea-a306-ccda262736ce | https://socratic.org/questions/how-much-water-should-be-added-to-5-00-g-of-kci-to-prepare-a-0-500-m-solution | 133.4 cm^3 | start physical_unit 2 2 volume cm^3 qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] cm^3"}] | [{"type":"physical unit","value":"133.4 cm^3"}] | [{"type":"physical unit","value":"Mass [OF] KCl [=] \\pu{5.00 g}"},{"type":"physical unit","value":"Molarity [OF] KCl solution [=] \\pu{0.500 M}"}] | <h1 class="questionTitle" itemprop="name">How much water should be added to 5.00 g of KCI to prepare a 0.500 M solution?</h1> | null | 133.4 cm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by number of litres of solution.</p>
<p><mathjax>#M=n/V#</mathjax></p>
<p>Rearrange for V to give <mathjax>#V=n/M#</mathjax></p>
<p>Number of moles of KCl is the mass divided by the molar mass of KCl which is 75.5 g/mol, so (5/75.5) = 0.0667 moles.</p>
<p>Now plug in the numbers: <mathjax>#V=n/m = 0.0667/0.5 = 0.1334#</mathjax> litres</p>
<p>So you need to add 0.1334 litres of 133.4 <mathjax>#cm^3#</mathjax> of water.</p>
<p>(NB: symbol for molarity is upper case M. m is metres).</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>133.4 <mathjax>#cm^3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by number of litres of solution.</p>
<p><mathjax>#M=n/V#</mathjax></p>
<p>Rearrange for V to give <mathjax>#V=n/M#</mathjax></p>
<p>Number of moles of KCl is the mass divided by the molar mass of KCl which is 75.5 g/mol, so (5/75.5) = 0.0667 moles.</p>
<p>Now plug in the numbers: <mathjax>#V=n/m = 0.0667/0.5 = 0.1334#</mathjax> litres</p>
<p>So you need to add 0.1334 litres of 133.4 <mathjax>#cm^3#</mathjax> of water.</p>
<p>(NB: symbol for molarity is upper case M. m is metres).</p></div>
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<h1 class="questionTitle" itemprop="name">How much water should be added to 5.00 g of KCI to prepare a 0.500 M solution?</h1>
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<div class="markdown"><p>133.4 <mathjax>#cm^3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by number of litres of solution.</p>
<p><mathjax>#M=n/V#</mathjax></p>
<p>Rearrange for V to give <mathjax>#V=n/M#</mathjax></p>
<p>Number of moles of KCl is the mass divided by the molar mass of KCl which is 75.5 g/mol, so (5/75.5) = 0.0667 moles.</p>
<p>Now plug in the numbers: <mathjax>#V=n/m = 0.0667/0.5 = 0.1334#</mathjax> litres</p>
<p>So you need to add 0.1334 litres of 133.4 <mathjax>#cm^3#</mathjax> of water.</p>
<p>(NB: symbol for molarity is upper case M. m is metres).</p></div>
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</article> | How much water should be added to 5.00 g of KCI to prepare a 0.500 M solution? | null |
1,965 | ab3b378a-6ddd-11ea-a0b5-ccda262736ce | https://socratic.org/questions/how-many-moles-of-pb-is-in-9-3-10-15-atoms-of-pb | 1.5 × 10^(-8) moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] Pb [IN] moles"}] | [{"type":"physical unit","value":"1.5 × 10^(-8) moles"}] | [{"type":"physical unit","value":"Number [OF] Pb atoms [=] \\pu{9.3 × 10^15}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #Pb# is in #9.3 * 10^15# atoms of #Pb#?</h1> | null | 1.5 × 10^(-8) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of any element contains <mathjax>#6.022xx10^23"atoms"#</mathjax> of that element.</p>
<p>In order to determine how many moles are in a given number of atoms, divide the given number of atoms by <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#9.3xx10^15cancel"atoms Pb"xx(1"mol Pb")/(6.022xx10^23cancel"atoms Pb")=1.5xx10^(-8)"mol Pb"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#9.3xx10^(15)"atoms Pb"#</mathjax> make up <mathjax>#color(blue)(1.5xx10^(-8)"mol Pb")#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of any element contains <mathjax>#6.022xx10^23"atoms"#</mathjax> of that element.</p>
<p>In order to determine how many moles are in a given number of atoms, divide the given number of atoms by <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#9.3xx10^15cancel"atoms Pb"xx(1"mol Pb")/(6.022xx10^23cancel"atoms Pb")=1.5xx10^(-8)"mol Pb"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #Pb# is in #9.3 * 10^15# atoms of #Pb#?</h1>
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<div class="markdown"><p><mathjax>#9.3xx10^(15)"atoms Pb"#</mathjax> make up <mathjax>#color(blue)(1.5xx10^(-8)"mol Pb")#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of any element contains <mathjax>#6.022xx10^23"atoms"#</mathjax> of that element.</p>
<p>In order to determine how many moles are in a given number of atoms, divide the given number of atoms by <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#9.3xx10^15cancel"atoms Pb"xx(1"mol Pb")/(6.022xx10^23cancel"atoms Pb")=1.5xx10^(-8)"mol Pb"#</mathjax></p></div>
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</article> | How many moles of #Pb# is in #9.3 * 10^15# atoms of #Pb#? | null |
1,966 | aacae743-6ddd-11ea-a6b9-ccda262736ce | https://socratic.org/questions/how-much-ammonia-nh3-can-be-obtained-when-3-0g-h2-reacts-with-100-g-n2 | 17.00 g | start physical_unit 3 3 mass g qc_end physical_unit 10 10 8 9 mass qc_end physical_unit 15 15 13 14 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] NH3 [IN] g"}] | [{"type":"physical unit","value":"17.00 g"}] | [{"type":"physical unit","value":"Mass [OF] H2 [=] \\pu{3.0 g}"},{"type":"physical unit","value":"Mass [OF] N2 [=] \\pu{100 g}"}] | <h1 class="questionTitle" itemprop="name">How much ammonia (NH3) can be obtained when 3.0g H2 reacts with 100.g N2?</h1> | null | 17.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the stoichiometric equation above. Clearly, the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> is dihydrogen gas, as we have only <mathjax>#(3.0*cancel(g))/(2*cancel(g)*mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#N_2(g) + 3H_2(g) rarr 2NH_3(g)#</mathjax></p></div>
</div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the stoichiometric equation above. Clearly, the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> is dihydrogen gas, as we have only <mathjax>#(3.0*cancel(g))/(2*cancel(g)*mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How much ammonia (NH3) can be obtained when 3.0g H2 reacts with 100.g N2?</h1>
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anor277
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<span class="dateCreated" datetime="2015-11-29T18:03:52" itemprop="dateCreated">
Nov 29, 2015
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<div class="markdown"><p><mathjax>#N_2(g) + 3H_2(g) rarr 2NH_3(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the stoichiometric equation above. Clearly, the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> is dihydrogen gas, as we have only <mathjax>#(3.0*cancel(g))/(2*cancel(g)*mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax>.</p></div>
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</article> | How much ammonia (NH3) can be obtained when 3.0g H2 reacts with 100.g N2? | null |
1,967 | aab5fcbe-6ddd-11ea-8059-ccda262736ce | https://socratic.org/questions/how-many-grams-of-hcl-are-required-to-prepare-250-milliliters-of-a-0-158-m-solut | 1.44 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 4 4 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] HCl [IN] grams"}] | [{"type":"physical unit","value":"1.44 grams"}] | [{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{250 milliliters}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.158 M}"}] | <h1 class="questionTitle" itemprop="name"> How many grams of #HCl# are required to prepare 250 milliliters of a 0.158 M solution?</h1> | null | 1.44 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem gives you the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and volume of the target solution, so right from the start you know that you can use that information to determine the <strong>number of moles</strong> of hydrochloric acid that it must contain. </p>
<p>So, <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> measures a solution's concentration in terms of how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, you get <strong>per liter of solution</strong>. </p>
<p>A <mathjax>#"0.158 M"#</mathjax> hydrochloric acid solution contains <mathjax>#0.158#</mathjax> <strong>moles</strong> of hydrochloric acid <em>for very liter</em> of solution. Since you're dealing with <mathjax>#"250 mL"#</mathjax>, the equivalent of <mathjax>#1/4"th"#</mathjax> of a liter, your solution will contain </p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.158 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "0.0395 moles HCl"#</mathjax></p>
</blockquote>
<p>Now that you know how many <em>moles</em> of hydrochloric acid are needed to prepare your solution, use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.0395 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"1.44 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem gives you the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and volume of the target solution, so right from the start you know that you can use that information to determine the <strong>number of moles</strong> of hydrochloric acid that it must contain. </p>
<p>So, <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> measures a solution's concentration in terms of how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, you get <strong>per liter of solution</strong>. </p>
<p>A <mathjax>#"0.158 M"#</mathjax> hydrochloric acid solution contains <mathjax>#0.158#</mathjax> <strong>moles</strong> of hydrochloric acid <em>for very liter</em> of solution. Since you're dealing with <mathjax>#"250 mL"#</mathjax>, the equivalent of <mathjax>#1/4"th"#</mathjax> of a liter, your solution will contain </p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.158 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "0.0395 moles HCl"#</mathjax></p>
</blockquote>
<p>Now that you know how many <em>moles</em> of hydrochloric acid are needed to prepare your solution, use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.0395 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name"> How many grams of #HCl# are required to prepare 250 milliliters of a 0.158 M solution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"1.44 g"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem gives you the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and volume of the target solution, so right from the start you know that you can use that information to determine the <strong>number of moles</strong> of hydrochloric acid that it must contain. </p>
<p>So, <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> measures a solution's concentration in terms of how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is hydrochloric acid, you get <strong>per liter of solution</strong>. </p>
<p>A <mathjax>#"0.158 M"#</mathjax> hydrochloric acid solution contains <mathjax>#0.158#</mathjax> <strong>moles</strong> of hydrochloric acid <em>for very liter</em> of solution. Since you're dealing with <mathjax>#"250 mL"#</mathjax>, the equivalent of <mathjax>#1/4"th"#</mathjax> of a liter, your solution will contain </p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.158 moles HCl"/(1color(red)(cancel(color(black)("L")))) = "0.0395 moles HCl"#</mathjax></p>
</blockquote>
<p>Now that you know how many <em>moles</em> of hydrochloric acid are needed to prepare your solution, use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.0395 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | How many grams of #HCl# are required to prepare 250 milliliters of a 0.158 M solution? | null |
1,968 | ac7015f0-6ddd-11ea-a075-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-0-350-l-solution-of-nh-4no-3-containing-0-56-moles-of- | 1.60 M | start physical_unit 10 10 molarity mol/l qc_end physical_unit 8 8 6 7 volume qc_end physical_unit 15 15 12 13 mole qc_end end | [{"type":"physical unit","value":"Molarity [OF] NH4NO3 solution [IN] M"}] | [{"type":"physical unit","value":"1.60 M"}] | [{"type":"physical unit","value":"Volume [OF] NH4NO3 solution [=] \\pu{0.350 L}"},{"type":"physical unit","value":"Mole [OF] solute [=] \\pu{0.56 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a 0.350 L solution of #NH_4NO_3# containing 0.56 moles of solute?</h1> | null | 1.60 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molarity (M) means moles per liter.</p>
<p>For your question, divide the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <mathjax>#("NH"_4"NO"_3)#</mathjax> by the volume of the solution.</p>
<p><mathjax>#"M"=("mol"/"L")#</mathjax></p>
<p><mathjax>#"M"=(0.56"mol NH"_4"NO"_3)/(0.350"L")="1.6 mol/L"#</mathjax></p></div>
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<div class="markdown"><p>The <mathjax>#"NH"_4"NO"_3#</mathjax> solution has a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"1.6 mol/L"#</mathjax>, or <mathjax>#"1.6 M"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molarity (M) means moles per liter.</p>
<p>For your question, divide the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <mathjax>#("NH"_4"NO"_3)#</mathjax> by the volume of the solution.</p>
<p><mathjax>#"M"=("mol"/"L")#</mathjax></p>
<p><mathjax>#"M"=(0.56"mol NH"_4"NO"_3)/(0.350"L")="1.6 mol/L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a 0.350 L solution of #NH_4NO_3# containing 0.56 moles of solute?</h1>
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<div class="markdown"><p>The <mathjax>#"NH"_4"NO"_3#</mathjax> solution has a <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"1.6 mol/L"#</mathjax>, or <mathjax>#"1.6 M"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molarity (M) means moles per liter.</p>
<p>For your question, divide the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <mathjax>#("NH"_4"NO"_3)#</mathjax> by the volume of the solution.</p>
<p><mathjax>#"M"=("mol"/"L")#</mathjax></p>
<p><mathjax>#"M"=(0.56"mol NH"_4"NO"_3)/(0.350"L")="1.6 mol/L"#</mathjax></p></div>
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</article> | What is the molarity of a 0.350 L solution of #NH_4NO_3# containing 0.56 moles of solute? | null |
1,969 | ab9357d2-6ddd-11ea-b287-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-in-atmospheres-of-0-246-gram-of-hydrogen-gas-occupying-a-vo | 58.90 atmospheres | start physical_unit 10 11 pressure atm qc_end physical_unit 10 11 7 8 mass qc_end physical_unit 10 11 16 17 volume qc_end physical_unit 10 11 19 20 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] hydrogen gas [IN] atmospheres"}] | [{"type":"physical unit","value":"58.90 atmospheres"}] | [{"type":"physical unit","value":"Mass [OF] hydrogen gas [=] \\pu{0.246 gram}"},{"type":"physical unit","value":"Volume [OF] hydrogen gas [=] \\pu{0.0500 liters}"},{"type":"physical unit","value":"Temperature [OF] hydrogen gas [=] \\pu{21.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure in atmospheres of 0.246 gram of hydrogen gas occupying a volume of 0.0500 liters at 21.0°C?</h1> | null | 58.90 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You need the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to answer this question. </p>
<p><img alt="https://jahschem.wikispaces.com/ideal+gas+law" src="https://useruploads.socratic.org/fcYPEOQo2ztIEKfNMJsA_ideal.png%22%3B+size%3D%2231420"/> </p>
<p>First convert the mass of hydrogen gas <mathjax>#("H"_2")#</mathjax> to moles by dividing the given mass by its molar mass. </p>
<p>Molar mass <mathjax>#"H"_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#(1.008"g/mol"xx2=2.016"g/mol")#</mathjax>.</p>
<p>Moles <mathjax>#"H"_2"#</mathjax></p>
<p><mathjax>#(0.246cancel"g H"_2)/(2.016cancel"g/mol H"_2")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.122 mol H"_2"#</mathjax></p>
<p>Now you need to determine your known and unknown variables.</p>
<p><strong>Known</strong><br/>
<mathjax>#V="0.0500 L"#</mathjax><br/>
<mathjax>#n="0.122 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="21.0"^@"C"+"273.15=294.2K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the ideal gas equation to isolate <mathjax>#P#</mathjax> and solve.</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((0.122cancel"mol")xx(0.082057338 cancel"L" "atm" cancel"K"^(-1) cancel"mol"^(-1))xx(294.2cancel"K"))/(0.0500cancel"L")="58.9 atm"#</mathjax><br/>
rounded to three significant figures</p></div>
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<div class="markdown"><p>The pressure of <mathjax>#"H"_2"#</mathjax> gas will be 58.9 atm, rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to answer this question. </p>
<p><img alt="https://jahschem.wikispaces.com/ideal+gas+law" src="https://useruploads.socratic.org/fcYPEOQo2ztIEKfNMJsA_ideal.png%22%3B+size%3D%2231420"/> </p>
<p>First convert the mass of hydrogen gas <mathjax>#("H"_2")#</mathjax> to moles by dividing the given mass by its molar mass. </p>
<p>Molar mass <mathjax>#"H"_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#(1.008"g/mol"xx2=2.016"g/mol")#</mathjax>.</p>
<p>Moles <mathjax>#"H"_2"#</mathjax></p>
<p><mathjax>#(0.246cancel"g H"_2)/(2.016cancel"g/mol H"_2")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.122 mol H"_2"#</mathjax></p>
<p>Now you need to determine your known and unknown variables.</p>
<p><strong>Known</strong><br/>
<mathjax>#V="0.0500 L"#</mathjax><br/>
<mathjax>#n="0.122 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="21.0"^@"C"+"273.15=294.2K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the ideal gas equation to isolate <mathjax>#P#</mathjax> and solve.</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((0.122cancel"mol")xx(0.082057338 cancel"L" "atm" cancel"K"^(-1) cancel"mol"^(-1))xx(294.2cancel"K"))/(0.0500cancel"L")="58.9 atm"#</mathjax><br/>
rounded to three significant figures</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pressure in atmospheres of 0.246 gram of hydrogen gas occupying a volume of 0.0500 liters at 21.0°C?</h1>
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<div class="markdown"><p>The pressure of <mathjax>#"H"_2"#</mathjax> gas will be 58.9 atm, rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You need the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to answer this question. </p>
<p><img alt="https://jahschem.wikispaces.com/ideal+gas+law" src="https://useruploads.socratic.org/fcYPEOQo2ztIEKfNMJsA_ideal.png%22%3B+size%3D%2231420"/> </p>
<p>First convert the mass of hydrogen gas <mathjax>#("H"_2")#</mathjax> to moles by dividing the given mass by its molar mass. </p>
<p>Molar mass <mathjax>#"H"_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#(1.008"g/mol"xx2=2.016"g/mol")#</mathjax>.</p>
<p>Moles <mathjax>#"H"_2"#</mathjax></p>
<p><mathjax>#(0.246cancel"g H"_2)/(2.016cancel"g/mol H"_2")#</mathjax><mathjax>#=#</mathjax><mathjax>#"0.122 mol H"_2"#</mathjax></p>
<p>Now you need to determine your known and unknown variables.</p>
<p><strong>Known</strong><br/>
<mathjax>#V="0.0500 L"#</mathjax><br/>
<mathjax>#n="0.122 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="21.0"^@"C"+"273.15=294.2K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the ideal gas equation to isolate <mathjax>#P#</mathjax> and solve.</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((0.122cancel"mol")xx(0.082057338 cancel"L" "atm" cancel"K"^(-1) cancel"mol"^(-1))xx(294.2cancel"K"))/(0.0500cancel"L")="58.9 atm"#</mathjax><br/>
rounded to three significant figures</p></div>
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</article> | What is the pressure in atmospheres of 0.246 gram of hydrogen gas occupying a volume of 0.0500 liters at 21.0°C? | null |
1,970 | aad5fbc9-6ddd-11ea-9c18-ccda262736ce | https://socratic.org/questions/a-10-000g-sample-of-a-compound-of-nitrogen-and-oxygen-contains-3-686-g-of-nitrog | N2O3 | start chemical_formula qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 8 8 12 13 mass qc_end substance 10 10 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"N2O3"}] | [{"type":"physical unit","value":"Mass [OF] the compound sample [=] \\pu{10.000 g}"},{"type":"physical unit","value":"Mass [OF] nitrogen [=] \\pu{3.686 g}"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">A 10.000g sample of a compound of nitrogen and oxygen contains 3.686 g of nitrogen. What is the empirical formula? I'm unsure if it is N2O.</h1> | null | N2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that represents constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a species....</p>
<p>And given the composition, we simply take the molar quantities of each element ......</p>
<p><mathjax>#"Moles of nitrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.686*g)/(14.01*g*mol^-1)=0.2630*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.00*g-3.686*g)/(15.999*g*mol^-1)=0.3947*mol#</mathjax></p>
<p>And we divide thru by the smallest molar quantity to get......</p>
<p><mathjax>#NO_(1.50)#</mathjax>, but since we require whole numbers we get......</p>
<p><mathjax>#N_2O_3#</mathjax>.</p>
<p>This nitrogen oxide is probably not one we would commonly encounter.....<mathjax>#O=N-stackrel(+)N(=O)O^(-)#</mathjax> is a possibility, and also <mathjax>#"trinitramide"#</mathjax> <mathjax>#N_4O_6#</mathjax>, a potential rocket fuel. Neither is something you would want to play with..........</p>
<p>These are unstable nitrogen oxides that would only be found in specialist labs........and certainly not in the quantity quoted in the question. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Empirical formula"=N_2O_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that represents constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a species....</p>
<p>And given the composition, we simply take the molar quantities of each element ......</p>
<p><mathjax>#"Moles of nitrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.686*g)/(14.01*g*mol^-1)=0.2630*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.00*g-3.686*g)/(15.999*g*mol^-1)=0.3947*mol#</mathjax></p>
<p>And we divide thru by the smallest molar quantity to get......</p>
<p><mathjax>#NO_(1.50)#</mathjax>, but since we require whole numbers we get......</p>
<p><mathjax>#N_2O_3#</mathjax>.</p>
<p>This nitrogen oxide is probably not one we would commonly encounter.....<mathjax>#O=N-stackrel(+)N(=O)O^(-)#</mathjax> is a possibility, and also <mathjax>#"trinitramide"#</mathjax> <mathjax>#N_4O_6#</mathjax>, a potential rocket fuel. Neither is something you would want to play with..........</p>
<p>These are unstable nitrogen oxides that would only be found in specialist labs........and certainly not in the quantity quoted in the question. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 10.000g sample of a compound of nitrogen and oxygen contains 3.686 g of nitrogen. What is the empirical formula? I'm unsure if it is N2O.</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Empirical formula"=N_2O_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that represents constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a species....</p>
<p>And given the composition, we simply take the molar quantities of each element ......</p>
<p><mathjax>#"Moles of nitrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.686*g)/(14.01*g*mol^-1)=0.2630*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(10.00*g-3.686*g)/(15.999*g*mol^-1)=0.3947*mol#</mathjax></p>
<p>And we divide thru by the smallest molar quantity to get......</p>
<p><mathjax>#NO_(1.50)#</mathjax>, but since we require whole numbers we get......</p>
<p><mathjax>#N_2O_3#</mathjax>.</p>
<p>This nitrogen oxide is probably not one we would commonly encounter.....<mathjax>#O=N-stackrel(+)N(=O)O^(-)#</mathjax> is a possibility, and also <mathjax>#"trinitramide"#</mathjax> <mathjax>#N_4O_6#</mathjax>, a potential rocket fuel. Neither is something you would want to play with..........</p>
<p>These are unstable nitrogen oxides that would only be found in specialist labs........and certainly not in the quantity quoted in the question. </p></div>
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</article> | A 10.000g sample of a compound of nitrogen and oxygen contains 3.686 g of nitrogen. What is the empirical formula? I'm unsure if it is N2O. | null |
1,971 | a8d8cff0-6ddd-11ea-9d71-ccda262736ce | https://socratic.org/questions/56e75bb27c014905ff328fd5 | As2O3(s) + 2 NO3-(aq) + 2 H+(aq) + 2H2O(l) -> 2 H3AsO4(aq) + N2O3(aq) | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"As2O3(s) + 2 NO3-(aq) + 2 H+(aq) + 2H2O(l) -> 2 H3AsO4(aq) + N2O3(aq)"}] | [{"type":"chemical equation","value":"As2O3(s) + NO3-(aq) -> H3AsO4(aq) + N2O3(aq)"}] | <h1 class="questionTitle" itemprop="name">Balance this redox reaction in acidic media? #"As"_2"O"_3(s) + "NO"_3^(-)(aq) -> "H"_3"AsO"_4(aq) + "N"_2"O"_3(aq)#</h1> | null | As2O3(s) + 2 NO3-(aq) + 2 H+(aq) + 2H2O(l) -> 2 H3AsO4(aq) + N2O3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by assigning <strong><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to the atoms that are taking part in the reaction.</p>
<p>For simplicity, I won't add the <strong>state symbols</strong> for the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O"_3^(-)) -> stackrel(color(blue)(+1))("H")_3stackrel(color(blue)(+5))("As")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+3))("N")_2stackrel(color(blue)(-2))("O")_3#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of arsenic, <mathjax>#"As"#</mathjax>, changes from <mathjax>#color(blue)(+3)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+5)#</mathjax> on the products' side. This tells you that arsenic is being <strong>oxidized</strong>, since its oxidation number <em>increases</em>.</p>
<p>On the other hand, the oxidation number of nitrogen, <mathjax>#"N"#</mathjax>, changes from <mathjax>#color(blue)(+5)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+3)#</mathjax> on the products' side. This tells you that nitrogen is being <strong>reduced</strong>, since its oxidation number <em>decreases</em>. </p>
<p>Write and balance the oxidation and reduction half-reactions.</p>
<ul>
<li><em>the <strong>oxidation-half reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> "H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Balance the atoms of arsenic first </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of arsenic <strong>loses two electrons</strong>, which means that <strong>two</strong> atoms of arsenic will lose a total of <strong>four electrons</strong>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Since you're in <em>acidic solution</em>, you can balance the <strong>oxygen atoms</strong> by adding <em>water</em> to the side of the reaction that needs oxygen atoms, and the <strong>hydrogen atoms</strong> by adding <em>protons</em>, <mathjax>#"H"^(+)#</mathjax>, to the side that needs hydrogen atoms. </p>
<p>In this case, you need <strong>five</strong> extra atoms of oxygen on the reactants' side, so you'll have </p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Similarly, the half-reaction needs <strong>four</strong> extra atoms of hydrogen on the products' side, so add four protons in there, too.</p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)#</mathjax></p>
</blockquote>
<ul>
<li><em>the <strong>reduction half-reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the nitrogen atoms first</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of nitrogen <strong>gains</strong> two electrons, so <strong>two</strong> atoms of nitrogen will gain a total of <strong>four electrons</strong></p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the oxygen by adding <strong>three</strong> molecules of water on the products' side</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>Finally, balance the hydrogen by adding <strong>six</strong> protons on the reactants' side</p>
<blockquote>
<p><mathjax>#6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>In any redox reaction, the number of electrons <strong>gained</strong> in the reduction half-reaction <strong>must be equal</strong> to the number of electrons <strong>lost</strong> in the oxidation half-reaction. </p>
<p>Since <mathjax>#4"e"^(-)#</mathjax> are both <em>gained</em> and <em>lost</em> in the respective half-reactions, you can go ahead and <strong>add</strong> the two half-reactions to get the balanced <em>overall chemical equation</em></p>
<blockquote>
<p><mathjax>#{(5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)), (6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O") :}#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#color(red)(cancel(color(black)(3"H"_2"O"))) + 2"H"_2"O" + "As"_2"O"_3 + color(red)(cancel(color(black)(4"H"^(+)))) + 2"H"^(+) + 2"NO"_3^(-) + color(red)(cancel(color(black)(4"e"^(-)))) -> 2"H"_3"AsO"_4 + color(red)(cancel(color(black)(4"e"^(-)))) + color(red)(cancel(color(black)(4"H"^(+)))) + "N"_2"O"_3 + color(red)(cancel(color(black)(3"H"_2"O")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#"As"_2"O"_3 + 2"NO"_3^(-) + 2"H"^(+) + 2"H"_2"O" -> 2"H"_3"AsO"_4 + "N"_2"O"_3#</mathjax></p>
</blockquote>
<p>The balanced chemical equation - with <strong>state symbols</strong> included - will thus be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq]))|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by assigning <strong><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to the atoms that are taking part in the reaction.</p>
<p>For simplicity, I won't add the <strong>state symbols</strong> for the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O"_3^(-)) -> stackrel(color(blue)(+1))("H")_3stackrel(color(blue)(+5))("As")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+3))("N")_2stackrel(color(blue)(-2))("O")_3#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of arsenic, <mathjax>#"As"#</mathjax>, changes from <mathjax>#color(blue)(+3)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+5)#</mathjax> on the products' side. This tells you that arsenic is being <strong>oxidized</strong>, since its oxidation number <em>increases</em>.</p>
<p>On the other hand, the oxidation number of nitrogen, <mathjax>#"N"#</mathjax>, changes from <mathjax>#color(blue)(+5)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+3)#</mathjax> on the products' side. This tells you that nitrogen is being <strong>reduced</strong>, since its oxidation number <em>decreases</em>. </p>
<p>Write and balance the oxidation and reduction half-reactions.</p>
<ul>
<li><em>the <strong>oxidation-half reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> "H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Balance the atoms of arsenic first </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of arsenic <strong>loses two electrons</strong>, which means that <strong>two</strong> atoms of arsenic will lose a total of <strong>four electrons</strong>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Since you're in <em>acidic solution</em>, you can balance the <strong>oxygen atoms</strong> by adding <em>water</em> to the side of the reaction that needs oxygen atoms, and the <strong>hydrogen atoms</strong> by adding <em>protons</em>, <mathjax>#"H"^(+)#</mathjax>, to the side that needs hydrogen atoms. </p>
<p>In this case, you need <strong>five</strong> extra atoms of oxygen on the reactants' side, so you'll have </p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Similarly, the half-reaction needs <strong>four</strong> extra atoms of hydrogen on the products' side, so add four protons in there, too.</p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)#</mathjax></p>
</blockquote>
<ul>
<li><em>the <strong>reduction half-reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the nitrogen atoms first</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of nitrogen <strong>gains</strong> two electrons, so <strong>two</strong> atoms of nitrogen will gain a total of <strong>four electrons</strong></p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the oxygen by adding <strong>three</strong> molecules of water on the products' side</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>Finally, balance the hydrogen by adding <strong>six</strong> protons on the reactants' side</p>
<blockquote>
<p><mathjax>#6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>In any redox reaction, the number of electrons <strong>gained</strong> in the reduction half-reaction <strong>must be equal</strong> to the number of electrons <strong>lost</strong> in the oxidation half-reaction. </p>
<p>Since <mathjax>#4"e"^(-)#</mathjax> are both <em>gained</em> and <em>lost</em> in the respective half-reactions, you can go ahead and <strong>add</strong> the two half-reactions to get the balanced <em>overall chemical equation</em></p>
<blockquote>
<p><mathjax>#{(5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)), (6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O") :}#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#color(red)(cancel(color(black)(3"H"_2"O"))) + 2"H"_2"O" + "As"_2"O"_3 + color(red)(cancel(color(black)(4"H"^(+)))) + 2"H"^(+) + 2"NO"_3^(-) + color(red)(cancel(color(black)(4"e"^(-)))) -> 2"H"_3"AsO"_4 + color(red)(cancel(color(black)(4"e"^(-)))) + color(red)(cancel(color(black)(4"H"^(+)))) + "N"_2"O"_3 + color(red)(cancel(color(black)(3"H"_2"O")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#"As"_2"O"_3 + 2"NO"_3^(-) + 2"H"^(+) + 2"H"_2"O" -> 2"H"_3"AsO"_4 + "N"_2"O"_3#</mathjax></p>
</blockquote>
<p>The balanced chemical equation - with <strong>state symbols</strong> included - will thus be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq]))|)))#</mathjax></p>
</blockquote></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Balance this redox reaction in acidic media? #"As"_2"O"_3(s) + "NO"_3^(-)(aq) -> "H"_3"AsO"_4(aq) + "N"_2"O"_3(aq)#</h1>
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Truong-Son N.
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<div class="markdown"><p><mathjax>#"As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq])#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by assigning <strong><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></strong> to the atoms that are taking part in the reaction.</p>
<p>For simplicity, I won't add the <strong>state symbols</strong> for the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+5))("N")stackrel(color(blue)(-2))("O"_3^(-)) -> stackrel(color(blue)(+1))("H")_3stackrel(color(blue)(+5))("As")stackrel(color(blue)(-2))("O")_4 + stackrel(color(blue)(+3))("N")_2stackrel(color(blue)(-2))("O")_3#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of arsenic, <mathjax>#"As"#</mathjax>, changes from <mathjax>#color(blue)(+3)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+5)#</mathjax> on the products' side. This tells you that arsenic is being <strong>oxidized</strong>, since its oxidation number <em>increases</em>.</p>
<p>On the other hand, the oxidation number of nitrogen, <mathjax>#"N"#</mathjax>, changes from <mathjax>#color(blue)(+5)#</mathjax> on the reactants' side, to <mathjax>#color(blue)(+3)#</mathjax> on the products' side. This tells you that nitrogen is being <strong>reduced</strong>, since its oxidation number <em>decreases</em>. </p>
<p>Write and balance the oxidation and reduction half-reactions.</p>
<ul>
<li><em>the <strong>oxidation-half reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> "H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Balance the atoms of arsenic first </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of arsenic <strong>loses two electrons</strong>, which means that <strong>two</strong> atoms of arsenic will lose a total of <strong>four electrons</strong>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Since you're in <em>acidic solution</em>, you can balance the <strong>oxygen atoms</strong> by adding <em>water</em> to the side of the reaction that needs oxygen atoms, and the <strong>hydrogen atoms</strong> by adding <em>protons</em>, <mathjax>#"H"^(+)#</mathjax>, to the side that needs hydrogen atoms. </p>
<p>In this case, you need <strong>five</strong> extra atoms of oxygen on the reactants' side, so you'll have </p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-)#</mathjax></p>
</blockquote>
<p>Similarly, the half-reaction needs <strong>four</strong> extra atoms of hydrogen on the products' side, so add four protons in there, too.</p>
<blockquote>
<p><mathjax>#5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)#</mathjax></p>
</blockquote>
<ul>
<li><em>the <strong>reduction half-reaction</strong></em></li>
</ul>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the nitrogen atoms first</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Here <strong>each</strong> atom of nitrogen <strong>gains</strong> two electrons, so <strong>two</strong> atoms of nitrogen will gain a total of <strong>four electrons</strong></p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3#</mathjax></p>
</blockquote>
<p>Balance the oxygen by adding <strong>three</strong> molecules of water on the products' side</p>
<blockquote>
<p><mathjax>#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>Finally, balance the hydrogen by adding <strong>six</strong> protons on the reactants' side</p>
<blockquote>
<p><mathjax>#6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O"#</mathjax></p>
</blockquote>
<p>In any redox reaction, the number of electrons <strong>gained</strong> in the reduction half-reaction <strong>must be equal</strong> to the number of electrons <strong>lost</strong> in the oxidation half-reaction. </p>
<p>Since <mathjax>#4"e"^(-)#</mathjax> are both <em>gained</em> and <em>lost</em> in the respective half-reactions, you can go ahead and <strong>add</strong> the two half-reactions to get the balanced <em>overall chemical equation</em></p>
<blockquote>
<p><mathjax>#{(5"H"_2"O" + stackrel(color(blue)(+3))("As")_2"O"_3 -> 2"H"_3stackrel(color(blue)(+5))("As")"O"_4 + 4"e"^(-) + 4"H"^(+)), (6"H"^(+) + 2stackrel(color(blue)(+5))("N")"O"_3^(-) + 4"e"^(-) -> stackrel(color(blue)(+3))("N")_2"O"_3 + 3"H"_2"O") :}#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#color(red)(cancel(color(black)(3"H"_2"O"))) + 2"H"_2"O" + "As"_2"O"_3 + color(red)(cancel(color(black)(4"H"^(+)))) + 2"H"^(+) + 2"NO"_3^(-) + color(red)(cancel(color(black)(4"e"^(-)))) -> 2"H"_3"AsO"_4 + color(red)(cancel(color(black)(4"e"^(-)))) + color(red)(cancel(color(black)(4"H"^(+)))) + "N"_2"O"_3 + color(red)(cancel(color(black)(3"H"_2"O")))#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#"As"_2"O"_3 + 2"NO"_3^(-) + 2"H"^(+) + 2"H"_2"O" -> 2"H"_3"AsO"_4 + "N"_2"O"_3#</mathjax></p>
</blockquote>
<p>The balanced chemical equation - with <strong>state symbols</strong> included - will thus be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("As"_2"O"_text(3(s]) + 2"NO"_text(3(aq])^(-) + 2"H"_text((aq])^(+) + 2"H"_2"O"_text((l]) -> 2"H"_3"AsO"_text(4(aq]) + "N"_2"O"_text(3(aq]))|)))#</mathjax></p>
</blockquote></div>
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</article> | Balance this redox reaction in acidic media? #"As"_2"O"_3(s) + "NO"_3^(-)(aq) -> "H"_3"AsO"_4(aq) + "N"_2"O"_3(aq)# | null |
1,972 | abf3fe88-6ddd-11ea-9268-ccda262736ce | https://socratic.org/questions/when-1-57-mol-o-2-reacts-with-h-2-to-form-h-2o-how-many-moles-of-h-2-are-consume | 3.14 moles | start physical_unit 6 6 mole mol qc_end physical_unit 3 3 1 2 mole qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Mole [OF] H2 [IN] moles"}] | [{"type":"physical unit","value":"3.14 moles"}] | [{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{1.57 mol}"},{"type":"chemical equation","value":"H2O"}] | <h1 class="questionTitle" itemprop="name">When 1.57 mol #O_2# reacts with #H_2# to form #H_2O#, how many moles of #H_2# are consumed in the process?</h1> | null | 3.14 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax>, or I could simply double the equation,</p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax></p>
<p>Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are <mathjax>#1.57*mol#</mathjax> <mathjax>#O_2#</mathjax>, stoichiometric equivalence requires <mathjax>#2xx1.57*mol#</mathjax> dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Look at the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, the chemical equivalence, of the reaction: </p>
<p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax>, or I could simply double the equation,</p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax></p>
<p>Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are <mathjax>#1.57*mol#</mathjax> <mathjax>#O_2#</mathjax>, stoichiometric equivalence requires <mathjax>#2xx1.57*mol#</mathjax> dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?</p></div>
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<h1 class="questionTitle" itemprop="name">When 1.57 mol #O_2# reacts with #H_2# to form #H_2O#, how many moles of #H_2# are consumed in the process?</h1>
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<div class="markdown"><p>Look at the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, the chemical equivalence, of the reaction: </p>
<p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax>, or I could simply double the equation,</p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax></p>
<p>Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are <mathjax>#1.57*mol#</mathjax> <mathjax>#O_2#</mathjax>, stoichiometric equivalence requires <mathjax>#2xx1.57*mol#</mathjax> dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?</p></div>
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</article> | When 1.57 mol #O_2# reacts with #H_2# to form #H_2O#, how many moles of #H_2# are consumed in the process? | null |
1,973 | abfbecee-6ddd-11ea-b95c-ccda262736ce | https://socratic.org/questions/5911805711ef6b6d556c664f | 85.21 L | start physical_unit 11 12 volume l qc_end c_other OTHER qc_end chemical_equation 27 32 qc_end end | [{"type":"physical unit","value":"Volume [OF] dioxygen gas [IN] L"}] | [{"type":"physical unit","value":"85.21 L"}] | [{"type":"physical unit","value":"Number [OF] KClO3 formula units [=] \\pu{2.25 × 10^24}"},{"type":"other","value":"Under standard conditions."},{"type":"chemical equation","value":"KClO3(s) ->[\\delta] KCl(s) + 3/2 O2(g)"}] | <h1 class="questionTitle" itemprop="name">Given #2.25xx10^24# formula units of #KClO_3#, what volume of dioxygen gas under standard conditions would result on heating?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>Potassium chlorate undergoes the following thermal decomposition:</p>
<p><mathjax>#KClO_3(s) + Delta rarrKCl(s) + 3/2O_2(g)#</mathjax></p></div>
</h2>
</div>
</div> | 85.21 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2KClO_3(s) + Delta rarr 2KCl(s) + 3O_2#</mathjax></p>
<p><mathjax>#"Moles of potassium chlorate"=(2.25xx10^24*"formula units")/(6.022xx10^23*"formula units"*mol^-1)=3.74*mol#</mathjax></p>
<p>And thus, given stoichiometric reaction, we should get,</p>
<p><mathjax>#3.74*molxx3/2*mol#</mathjax> <mathjax>#"dioxygen gas"#</mathjax> evolved, i.e. <mathjax>#5.60*mol#</mathjax></p>
<p>And we use the Ideal Gas equation, <mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#=(5.60*molxx0.0821*(L*atm)/(K*mol)xx556*K)/(3.00*atm)=??L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>You have the stoichiometric equation........and I get a volume of over <mathjax>#"80 litres.........."#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2KClO_3(s) + Delta rarr 2KCl(s) + 3O_2#</mathjax></p>
<p><mathjax>#"Moles of potassium chlorate"=(2.25xx10^24*"formula units")/(6.022xx10^23*"formula units"*mol^-1)=3.74*mol#</mathjax></p>
<p>And thus, given stoichiometric reaction, we should get,</p>
<p><mathjax>#3.74*molxx3/2*mol#</mathjax> <mathjax>#"dioxygen gas"#</mathjax> evolved, i.e. <mathjax>#5.60*mol#</mathjax></p>
<p>And we use the Ideal Gas equation, <mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#=(5.60*molxx0.0821*(L*atm)/(K*mol)xx556*K)/(3.00*atm)=??L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Given #2.25xx10^24# formula units of #KClO_3#, what volume of dioxygen gas under standard conditions would result on heating?</h1>
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<div class="markdown"><p>Potassium chlorate undergoes the following thermal decomposition:</p>
<p><mathjax>#KClO_3(s) + Delta rarrKCl(s) + 3/2O_2(g)#</mathjax></p></div>
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anor277
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<div class="markdown"><p>You have the stoichiometric equation........and I get a volume of over <mathjax>#"80 litres.........."#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2KClO_3(s) + Delta rarr 2KCl(s) + 3O_2#</mathjax></p>
<p><mathjax>#"Moles of potassium chlorate"=(2.25xx10^24*"formula units")/(6.022xx10^23*"formula units"*mol^-1)=3.74*mol#</mathjax></p>
<p>And thus, given stoichiometric reaction, we should get,</p>
<p><mathjax>#3.74*molxx3/2*mol#</mathjax> <mathjax>#"dioxygen gas"#</mathjax> evolved, i.e. <mathjax>#5.60*mol#</mathjax></p>
<p>And we use the Ideal Gas equation, <mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#=(5.60*molxx0.0821*(L*atm)/(K*mol)xx556*K)/(3.00*atm)=??L#</mathjax></p></div>
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</article> | Given #2.25xx10^24# formula units of #KClO_3#, what volume of dioxygen gas under standard conditions would result on heating? |
Potassium chlorate undergoes the following thermal decomposition:
#KClO_3(s) + Delta rarrKCl(s) + 3/2O_2(g)#
|
1,974 | ac682594-6ddd-11ea-9d87-ccda262736ce | https://socratic.org/questions/at-627-deg-c-and-one-atmosphere-s0-3-is-partially-dissociated-into-s0-2-and-0-2- | 0.35 | start physical_unit 30 30 temperature none qc_end physical_unit 6 6 1 2 temperature qc_end physical_unit 6 6 4 5 pressure qc_end physical_unit 20 21 23 24 density qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Degree [OF] dissociation"}] | [{"type":"physical unit","value":"0.35"}] | [{"type":"physical unit","value":"Temperature [OF] SO3 [=] \\pu{627 ℃}"},{"type":"physical unit","value":"Pressure [OF] SO3 [=] \\pu{1 atm}"},{"type":"physical unit","value":"Desity [OF] equilibrium mixture [=] \\pu{0.925 g/L}"},{"type":"other","value":"SO3 is partially dissociated into SO2 and O2. "}] | <h1 class="questionTitle" itemprop="name">At #627^@"C"# and #"1 atm"#, #"SO"_3# is partially dissociated into #"SO"_2"# and #"O"_2# .
If the density of the equilibrium mixture is #"0.925 g/L"#, what is the degree of dissociation?
</h1> | null | 0.35 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to calculate the <strong>average molar mass</strong> of the mixture, as this will help you calculate how many <strong>moles</strong> of sulfur trioxide dissociated to produce sulfur dioxide and oxygen gas. </p>
<p>Your tool of choice will be this equation--I <em>will not</em> derive the equation here!</p>
<blockquote>
<p><mathjax>#M_M = (rho * R * T)/P#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#M_M#</mathjax> is the <strong>molar mass</strong> of an ideal gas</li>
<li><mathjax>#rho#</mathjax> is the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas at a <strong>pressure</strong> <mathjax>#P#</mathjax> and an <strong>absolute temperature</strong> <mathjax>#T#</mathjax></li>
<li><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.0821 quad ("atm" * "L")/("mol" * "K")#</mathjax></li>
</ul>
</blockquote>
<p>In your case, this equation will help you calculate the <strong>average molar mass</strong> of the mixture. Plug in your values to find--do not forget to convert the temperature of the mixture to <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_M = ("0.925 g" color(red)(cancel(color(black)("L"^(-1)))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (627 + 273.15) color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "68.36 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that this actually represents the <strong>weighted average</strong> of the molar masses of each individual gas. In other words, each gas will contribute to the average molar mass <strong>in proportion</strong> to its <strong>mole fraction</strong> in the mixture. </p>
<p>The balanced chemical equation that describes this equilibrium looks like this</p>
<blockquote>
<p><mathjax>#2"SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur trioxide <strong>that dissociate</strong>, the reaction produces <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur dioxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>To make the calculations easier, let's assume that you start with <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide. if you take <mathjax>#x#</mathjax> <strong>moles</strong> to be the number of moles of sulfur trioxide <strong>that dissociate</strong>, you can say that, at equilibrium, the reaction vessel will contain</p>
<blockquote>
<ul>
<li><mathjax>#(1 - x) quad "moles SO"_3#</mathjax></li>
<li><mathjax>#x quad "moles SO"_2#</mathjax></li>
<li><mathjax>#x/2 quad "moles O"_2#</mathjax></li>
</ul>
<blockquote>
<p>This is the case because the number of moles of oxygen gas produced will always be <strong>half</strong> the number of moles of sulfur trioxide <em>that dissociate</em>.</p>
</blockquote>
</blockquote>
<p>This means that. at equilibrium, the <strong>total number of moles</strong> of gas present in the reaction vessel is equal to </p>
<blockquote>
<p><mathjax>#(1 - color(red)(cancel(color(black)(x))) + color(red)(cancel(color(black)(x))) + x/2) quad "moles" = (1 + x/2) quad "moles"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sulfur trioxide will be</p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 3) = ((1-x) quad color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (1-x)/(1+x/2)#</mathjax></p>
</blockquote>
<p>Similarly, the <strong>mole fractions</strong> of sulfur dioxide and of oxygen gas will be </p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 2) = (x color(red)(cancel(color(black)("moles"))))/((1 + x/2)color(red)(cancel(color(black)("moles")))) = x/(1+x/2)#</mathjax></p>
<p><mathjax>#chi_ ("O"_ 2) = (x/2 color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (x/2)/(1+x/2)#</mathjax></p>
</blockquote>
<p>This means that the <strong>average molar mas</strong> of the mixture can be written as </p>
<blockquote>
<p><mathjax>#68.36 color(red)(cancel(color(black)("g mol"^(-1))))= overbrace( ((1-x)/(1+x/2)) * 80.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_3)) + #</mathjax> </p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace((x/(1+x/2)) * 64.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_2)) + #</mathjax></p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace( ((x/2)/(1+x/2)) * 32.0 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of O"_2))#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#68.36 * (1 + x/2) = (1-x) * 80.066 + x * 64.066 + x/2 * 32.0#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#136.72 + 68.36 * x = 160.132 - 160.132 * x + 128.132 * x + 32.0 * x#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#68.36 * x = 23.96 implies x= 23.96/68.36 = 0.35#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>degree of dissociation</strong> is equal to</p>
<blockquote>
<p><mathjax>#alpha = (0.35 color(red)(cancel(color(black)("moles"))))/(1color(red)(cancel(color(black)("mole")))) = 0.35#</mathjax></p>
<blockquote>
<p>This basically means that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide present in the mixture, <mathjax>#0.35#</mathjax> <strong>moles</strong> will dissociate to produce sulfur dioxide and oxygen gas. </p>
</blockquote>
</blockquote>
<p>I'll 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 have one significant figure for the pressure of the mixture. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#alpha = 0.35#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to calculate the <strong>average molar mass</strong> of the mixture, as this will help you calculate how many <strong>moles</strong> of sulfur trioxide dissociated to produce sulfur dioxide and oxygen gas. </p>
<p>Your tool of choice will be this equation--I <em>will not</em> derive the equation here!</p>
<blockquote>
<p><mathjax>#M_M = (rho * R * T)/P#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#M_M#</mathjax> is the <strong>molar mass</strong> of an ideal gas</li>
<li><mathjax>#rho#</mathjax> is the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas at a <strong>pressure</strong> <mathjax>#P#</mathjax> and an <strong>absolute temperature</strong> <mathjax>#T#</mathjax></li>
<li><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.0821 quad ("atm" * "L")/("mol" * "K")#</mathjax></li>
</ul>
</blockquote>
<p>In your case, this equation will help you calculate the <strong>average molar mass</strong> of the mixture. Plug in your values to find--do not forget to convert the temperature of the mixture to <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_M = ("0.925 g" color(red)(cancel(color(black)("L"^(-1)))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (627 + 273.15) color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "68.36 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that this actually represents the <strong>weighted average</strong> of the molar masses of each individual gas. In other words, each gas will contribute to the average molar mass <strong>in proportion</strong> to its <strong>mole fraction</strong> in the mixture. </p>
<p>The balanced chemical equation that describes this equilibrium looks like this</p>
<blockquote>
<p><mathjax>#2"SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur trioxide <strong>that dissociate</strong>, the reaction produces <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur dioxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>To make the calculations easier, let's assume that you start with <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide. if you take <mathjax>#x#</mathjax> <strong>moles</strong> to be the number of moles of sulfur trioxide <strong>that dissociate</strong>, you can say that, at equilibrium, the reaction vessel will contain</p>
<blockquote>
<ul>
<li><mathjax>#(1 - x) quad "moles SO"_3#</mathjax></li>
<li><mathjax>#x quad "moles SO"_2#</mathjax></li>
<li><mathjax>#x/2 quad "moles O"_2#</mathjax></li>
</ul>
<blockquote>
<p>This is the case because the number of moles of oxygen gas produced will always be <strong>half</strong> the number of moles of sulfur trioxide <em>that dissociate</em>.</p>
</blockquote>
</blockquote>
<p>This means that. at equilibrium, the <strong>total number of moles</strong> of gas present in the reaction vessel is equal to </p>
<blockquote>
<p><mathjax>#(1 - color(red)(cancel(color(black)(x))) + color(red)(cancel(color(black)(x))) + x/2) quad "moles" = (1 + x/2) quad "moles"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sulfur trioxide will be</p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 3) = ((1-x) quad color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (1-x)/(1+x/2)#</mathjax></p>
</blockquote>
<p>Similarly, the <strong>mole fractions</strong> of sulfur dioxide and of oxygen gas will be </p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 2) = (x color(red)(cancel(color(black)("moles"))))/((1 + x/2)color(red)(cancel(color(black)("moles")))) = x/(1+x/2)#</mathjax></p>
<p><mathjax>#chi_ ("O"_ 2) = (x/2 color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (x/2)/(1+x/2)#</mathjax></p>
</blockquote>
<p>This means that the <strong>average molar mas</strong> of the mixture can be written as </p>
<blockquote>
<p><mathjax>#68.36 color(red)(cancel(color(black)("g mol"^(-1))))= overbrace( ((1-x)/(1+x/2)) * 80.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_3)) + #</mathjax> </p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace((x/(1+x/2)) * 64.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_2)) + #</mathjax></p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace( ((x/2)/(1+x/2)) * 32.0 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of O"_2))#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#68.36 * (1 + x/2) = (1-x) * 80.066 + x * 64.066 + x/2 * 32.0#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#136.72 + 68.36 * x = 160.132 - 160.132 * x + 128.132 * x + 32.0 * x#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#68.36 * x = 23.96 implies x= 23.96/68.36 = 0.35#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>degree of dissociation</strong> is equal to</p>
<blockquote>
<p><mathjax>#alpha = (0.35 color(red)(cancel(color(black)("moles"))))/(1color(red)(cancel(color(black)("mole")))) = 0.35#</mathjax></p>
<blockquote>
<p>This basically means that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide present in the mixture, <mathjax>#0.35#</mathjax> <strong>moles</strong> will dissociate to produce sulfur dioxide and oxygen gas. </p>
</blockquote>
</blockquote>
<p>I'll 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 have one significant figure for the pressure of the mixture. </p></div>
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<h1 class="questionTitle" itemprop="name">At #627^@"C"# and #"1 atm"#, #"SO"_3# is partially dissociated into #"SO"_2"# and #"O"_2# .
If the density of the equilibrium mixture is #"0.925 g/L"#, what is the degree of dissociation?
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Stefan V.
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<span class="dateCreated" datetime="2018-03-12T00:56:35" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#alpha = 0.35#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to calculate the <strong>average molar mass</strong> of the mixture, as this will help you calculate how many <strong>moles</strong> of sulfur trioxide dissociated to produce sulfur dioxide and oxygen gas. </p>
<p>Your tool of choice will be this equation--I <em>will not</em> derive the equation here!</p>
<blockquote>
<p><mathjax>#M_M = (rho * R * T)/P#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#M_M#</mathjax> is the <strong>molar mass</strong> of an ideal gas</li>
<li><mathjax>#rho#</mathjax> is the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas at a <strong>pressure</strong> <mathjax>#P#</mathjax> and an <strong>absolute temperature</strong> <mathjax>#T#</mathjax></li>
<li><mathjax>#R#</mathjax> is the <strong>universal gas constant</strong>, equal to <mathjax>#0.0821 quad ("atm" * "L")/("mol" * "K")#</mathjax></li>
</ul>
</blockquote>
<p>In your case, this equation will help you calculate the <strong>average molar mass</strong> of the mixture. Plug in your values to find--do not forget to convert the temperature of the mixture to <em>Kelvin</em>!</p>
<blockquote>
<p><mathjax>#M_M = ("0.925 g" color(red)(cancel(color(black)("L"^(-1)))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (627 + 273.15) color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "68.36 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Now, it's important to realize that this actually represents the <strong>weighted average</strong> of the molar masses of each individual gas. In other words, each gas will contribute to the average molar mass <strong>in proportion</strong> to its <strong>mole fraction</strong> in the mixture. </p>
<p>The balanced chemical equation that describes this equilibrium looks like this</p>
<blockquote>
<p><mathjax>#2"SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur trioxide <strong>that dissociate</strong>, the reaction produces <mathjax>#2#</mathjax> <strong>moles</strong> of sulfur dioxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. </p>
<p>To make the calculations easier, let's assume that you start with <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide. if you take <mathjax>#x#</mathjax> <strong>moles</strong> to be the number of moles of sulfur trioxide <strong>that dissociate</strong>, you can say that, at equilibrium, the reaction vessel will contain</p>
<blockquote>
<ul>
<li><mathjax>#(1 - x) quad "moles SO"_3#</mathjax></li>
<li><mathjax>#x quad "moles SO"_2#</mathjax></li>
<li><mathjax>#x/2 quad "moles O"_2#</mathjax></li>
</ul>
<blockquote>
<p>This is the case because the number of moles of oxygen gas produced will always be <strong>half</strong> the number of moles of sulfur trioxide <em>that dissociate</em>.</p>
</blockquote>
</blockquote>
<p>This means that. at equilibrium, the <strong>total number of moles</strong> of gas present in the reaction vessel is equal to </p>
<blockquote>
<p><mathjax>#(1 - color(red)(cancel(color(black)(x))) + color(red)(cancel(color(black)(x))) + x/2) quad "moles" = (1 + x/2) quad "moles"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sulfur trioxide will be</p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 3) = ((1-x) quad color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (1-x)/(1+x/2)#</mathjax></p>
</blockquote>
<p>Similarly, the <strong>mole fractions</strong> of sulfur dioxide and of oxygen gas will be </p>
<blockquote>
<p><mathjax>#chi_ ("SO"_ 2) = (x color(red)(cancel(color(black)("moles"))))/((1 + x/2)color(red)(cancel(color(black)("moles")))) = x/(1+x/2)#</mathjax></p>
<p><mathjax>#chi_ ("O"_ 2) = (x/2 color(red)(cancel(color(black)("moles"))))/((1+x/2)color(red)(cancel(color(black)("moles")))) = (x/2)/(1+x/2)#</mathjax></p>
</blockquote>
<p>This means that the <strong>average molar mas</strong> of the mixture can be written as </p>
<blockquote>
<p><mathjax>#68.36 color(red)(cancel(color(black)("g mol"^(-1))))= overbrace( ((1-x)/(1+x/2)) * 80.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_3)) + #</mathjax> </p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace((x/(1+x/2)) * 64.066 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of SO"_2)) + #</mathjax></p>
<p><mathjax>#color(white)(68.36 color(white)(cancel(color(white)("g mol"^(-1)))) = ) + overbrace( ((x/2)/(1+x/2)) * 32.0 color(red)(cancel(color(black)("g mol"^(-1)))))^(color(blue)("the contribution of O"_2))#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#68.36 * (1 + x/2) = (1-x) * 80.066 + x * 64.066 + x/2 * 32.0#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#136.72 + 68.36 * x = 160.132 - 160.132 * x + 128.132 * x + 32.0 * x#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#68.36 * x = 23.96 implies x= 23.96/68.36 = 0.35#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>degree of dissociation</strong> is equal to</p>
<blockquote>
<p><mathjax>#alpha = (0.35 color(red)(cancel(color(black)("moles"))))/(1color(red)(cancel(color(black)("mole")))) = 0.35#</mathjax></p>
<blockquote>
<p>This basically means that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sulfur trioxide present in the mixture, <mathjax>#0.35#</mathjax> <strong>moles</strong> will dissociate to produce sulfur dioxide and oxygen gas. </p>
</blockquote>
</blockquote>
<p>I'll 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 have one significant figure for the pressure of the mixture. </p></div>
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</article> | At #627^@"C"# and #"1 atm"#, #"SO"_3# is partially dissociated into #"SO"_2"# and #"O"_2# .
If the density of the equilibrium mixture is #"0.925 g/L"#, what is the degree of dissociation?
| null |
1,975 | ad0eab9c-6ddd-11ea-afa7-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-0-50-mole-of-naoh-in-0-50-liter | 1.00 M | start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 6 6 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [IN] M"}] | [{"type":"physical unit","value":"1.00 M"}] | [{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{0.50 mole}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{0.50 liter}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 0.50 mole of #NaOH# in 0.50 liter of solution? </h1> | null | 1.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is determined by the equation <mathjax>#Molarity#</mathjax> <mathjax>#= (mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 0.50 NaOH#</mathjax><br/>
<mathjax>#L = 0.50 L#</mathjax></p>
<p><mathjax>#M = (0.50 mol)/(0.50 L)#</mathjax></p>
<p><mathjax>#M = 1.0M#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#M = (0.50 mol)/(0.50 L)#</mathjax></p>
<p><mathjax>#M = 1.0M#</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/solutions-and-their-behavior/molarity">Molarity</a> is determined by the equation <mathjax>#Molarity#</mathjax> <mathjax>#= (mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 0.50 NaOH#</mathjax><br/>
<mathjax>#L = 0.50 L#</mathjax></p>
<p><mathjax>#M = (0.50 mol)/(0.50 L)#</mathjax></p>
<p><mathjax>#M = 1.0M#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 0.50 mole of #NaOH# in 0.50 liter of solution? </h1>
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<div class="markdown"><p><mathjax>#M = (0.50 mol)/(0.50 L)#</mathjax></p>
<p><mathjax>#M = 1.0M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is determined by the equation <mathjax>#Molarity#</mathjax> <mathjax>#= (mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 0.50 NaOH#</mathjax><br/>
<mathjax>#L = 0.50 L#</mathjax></p>
<p><mathjax>#M = (0.50 mol)/(0.50 L)#</mathjax></p>
<p><mathjax>#M = 1.0M#</mathjax></p></div>
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</article> | What is the molarity of a solution that contains 0.50 mole of #NaOH# in 0.50 liter of solution? | null |
1,976 | acb62ac0-6ddd-11ea-9922-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-butane | C4H10 | start chemical_formula qc_end substance 6 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] butane [IN] empirical"}] | [{"type":"chemical equation","value":"C4H10"}] | [{"type":"substance name","value":"Butane"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for butane?
</h1> | null | C4H10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula for butane is <mathjax>#C_4H_10#</mathjax> because hydrocarbons have the formula <mathjax>#C_nH_(2n+2)#</mathjax></p>
<p>The empirical formula for butane, however, is <mathjax>#C_2H_5#</mathjax> </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#C_2H_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula for butane is <mathjax>#C_4H_10#</mathjax> because hydrocarbons have the formula <mathjax>#C_nH_(2n+2)#</mathjax></p>
<p>The empirical formula for butane, however, is <mathjax>#C_2H_5#</mathjax> </p></div>
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<div class="markdown"><p><mathjax>#C_2H_5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The formula for butane is <mathjax>#C_4H_10#</mathjax> because hydrocarbons have the formula <mathjax>#C_nH_(2n+2)#</mathjax></p>
<p>The empirical formula for butane, however, is <mathjax>#C_2H_5#</mathjax> </p></div>
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</article> | What is the empirical formula for butane?
| null |
1,977 | abd6de0d-6ddd-11ea-98e0-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>
</div>
</div>
<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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_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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<h4 class="answerHeader">Explanation:</h4>
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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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<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>
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<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 |
1,978 | aaaba6da-6ddd-11ea-a514-ccda262736ce | https://socratic.org/questions/56f385357c014959e950a928 | 42 grams | start physical_unit 4 5 mass g qc_end physical_unit 16 17 12 13 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] lithium metal [IN] grams"}] | [{"type":"physical unit","value":"42 grams"}] | [{"type":"physical unit","value":"Mass [OF] dinitrogen gas [=] \\pu{28 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of lithium metal are required to react with a #28*g# mass of dinitrogen gas?</h1> | null | 42 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium is one of the few metals that will react with nitrogen gas directly at room temperature. Lithium nitride is one VERY expensive way to produce ammonia gas:</p>
<p><mathjax>#Li_3N(s) + 3H_2O(l) rarr NH_3(aq) + 3LiOH(aq)#</mathjax></p>
<p>So, given the stoichimetry, 42 g of lithium metal are required to react with 28 g of dinitrogen gas. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#6Li(s) + N_2(g) rarr 2Li_3N(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium is one of the few metals that will react with nitrogen gas directly at room temperature. Lithium nitride is one VERY expensive way to produce ammonia gas:</p>
<p><mathjax>#Li_3N(s) + 3H_2O(l) rarr NH_3(aq) + 3LiOH(aq)#</mathjax></p>
<p>So, given the stoichimetry, 42 g of lithium metal are required to react with 28 g of dinitrogen gas. </p></div>
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<div class="markdown"><p><mathjax>#6Li(s) + N_2(g) rarr 2Li_3N(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Lithium is one of the few metals that will react with nitrogen gas directly at room temperature. Lithium nitride is one VERY expensive way to produce ammonia gas:</p>
<p><mathjax>#Li_3N(s) + 3H_2O(l) rarr NH_3(aq) + 3LiOH(aq)#</mathjax></p>
<p>So, given the stoichimetry, 42 g of lithium metal are required to react with 28 g of dinitrogen gas. </p></div>
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</article> | How many grams of lithium metal are required to react with a #28*g# mass of dinitrogen gas? | null |
1,979 | a9414b53-6ddd-11ea-989a-ccda262736ce | https://socratic.org/questions/a-substance-has-a-molecular-formula-of-c-8h-10n-4o-2-what-is-the-empirical-formu | C4H5N2O | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the substance [IN] empirical"}] | [{"type":"chemical equation","value":"C4H5N2O"}] | [{"type":"other","value":"A substance has a molecular formula of C8H10N4O2."}] | <h1 class="questionTitle" itemprop="name">A substance has a molecular formula of #C_8H_10N_4O_2#. What is the empirical formula?</h1> | null | C4H5N2O | <div class="answerDescription">
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<div class="markdown"><p>The empirical formula is the simplest whole number ratio that defines constituent atoms in a species. The molecular formula is always a multiple of the empirical formula. We divide the molecular formula by <mathjax>#2#</mathjax> to get the empirical formula.</p></div>
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<div class="markdown"><p><mathjax>#C_4H_5N_2O#</mathjax></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. The molecular formula is always a multiple of the empirical formula. We divide the molecular formula by <mathjax>#2#</mathjax> to get the empirical formula.</p></div>
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<h1 class="questionTitle" itemprop="name">A substance has a molecular formula of #C_8H_10N_4O_2#. What is the empirical formula?</h1>
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<div class="markdown"><p><mathjax>#C_4H_5N_2O#</mathjax></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. The molecular formula is always a multiple of the empirical formula. We divide the molecular formula by <mathjax>#2#</mathjax> to get the empirical formula.</p></div>
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</article> | A substance has a molecular formula of #C_8H_10N_4O_2#. What is the empirical formula? | null |
1,980 | abdd4d0c-6ddd-11ea-a1f2-ccda262736ce | https://socratic.org/questions/a-solution-contains-35-grams-of-kno-3-dissolved-in-100-grams-of-water-at-40-c-ho | 32 g | start physical_unit 6 6 mass g qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 1 1 14 15 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Added mass [OF] KNO3 [IN] g"}] | [{"type":"physical unit","value":"32 g"}] | [{"type":"physical unit","value":"Mass [OF] KNO3 [=] \\pu{35 grams}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 grams}"},{"type":"physical unit","value":"Temperature [OF] the solution [=] \\pu{40 ℃}"},{"type":"other","value":"Saturated solution."}] | <h1 class="questionTitle" itemprop="name">A solution contains 35 grams of #KNO_3# dissolved in 100 grams of water at 40°C. How much more #KNO_3# would have to be added to make it a saturated solution? </h1> | null | 32 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>solubility graph</strong> for potassium nitrate, <mathjax>#"KNO"_3#</mathjax>, which looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.mts.net/~alou/Chemistry%2011/Unit%204%20-%20Solutions%20Lessons" src="https://useruploads.socratic.org/cETxwwRQdOh9LKT6Ne1w_lesson%25203k.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, potassium nitrate has s solubility of about <mathjax>#"67 g / 100g H"_2"O"#</mathjax> at <mathjax>#40^@"C"#</mathjax>. This means that at <mathjax>#40^@"C"#</mathjax>, a <strong>saturated</strong> potassium nitrate solution will contain <mathjax>#"67 g"#</mathjax> of <em>dissolved</em> salt <strong>for every</strong> <mathjax>#"10 0g"#</mathjax> of water. </p>
<p>You know that at this temperature, your solution contains <mathjax>#"35 g"#</mathjax> of potassium nitrate in <mathjax>#"100 g"#</mathjax> of water. This solution will be <strong>unsaturated</strong> because it contains <em>less</em> potassium nitrate than the maximum amount that can be dissolved. </p>
<p>In order to make a <strong>saturated solution</strong>, you must get the total mass of potassium nitrate to <mathjax>#"67 g"#</mathjax>, which means that you must add</p>
<blockquote>
<p><mathjax>#"mass of KNO"_3 color(white)(a)"to be added" = "67 g" - "35 g" = color(green)(|bar(ul(color(white)(a/a)color(black)("32 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, dissolving another <mathjax>#"32 g"#</mathjax> of potassium nitrate in your solution at <mathjax>#40^@"C"#</mathjax> will get you a <strong>saturated</strong> solution. Any extra amount of potassium nitrate that you will add past this point will <em>remain undissolved</em>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"32 g KNO"_3#</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>solubility graph</strong> for potassium nitrate, <mathjax>#"KNO"_3#</mathjax>, which looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.mts.net/~alou/Chemistry%2011/Unit%204%20-%20Solutions%20Lessons" src="https://useruploads.socratic.org/cETxwwRQdOh9LKT6Ne1w_lesson%25203k.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, potassium nitrate has s solubility of about <mathjax>#"67 g / 100g H"_2"O"#</mathjax> at <mathjax>#40^@"C"#</mathjax>. This means that at <mathjax>#40^@"C"#</mathjax>, a <strong>saturated</strong> potassium nitrate solution will contain <mathjax>#"67 g"#</mathjax> of <em>dissolved</em> salt <strong>for every</strong> <mathjax>#"10 0g"#</mathjax> of water. </p>
<p>You know that at this temperature, your solution contains <mathjax>#"35 g"#</mathjax> of potassium nitrate in <mathjax>#"100 g"#</mathjax> of water. This solution will be <strong>unsaturated</strong> because it contains <em>less</em> potassium nitrate than the maximum amount that can be dissolved. </p>
<p>In order to make a <strong>saturated solution</strong>, you must get the total mass of potassium nitrate to <mathjax>#"67 g"#</mathjax>, which means that you must add</p>
<blockquote>
<p><mathjax>#"mass of KNO"_3 color(white)(a)"to be added" = "67 g" - "35 g" = color(green)(|bar(ul(color(white)(a/a)color(black)("32 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, dissolving another <mathjax>#"32 g"#</mathjax> of potassium nitrate in your solution at <mathjax>#40^@"C"#</mathjax> will get you a <strong>saturated</strong> solution. Any extra amount of potassium nitrate that you will add past this point will <em>remain undissolved</em>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A solution contains 35 grams of #KNO_3# dissolved in 100 grams of water at 40°C. How much more #KNO_3# would have to be added to make it a saturated solution? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"32 g KNO"_3#</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>solubility graph</strong> for potassium nitrate, <mathjax>#"KNO"_3#</mathjax>, which looks like this </p>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.mts.net/~alou/Chemistry%2011/Unit%204%20-%20Solutions%20Lessons" src="https://useruploads.socratic.org/cETxwwRQdOh9LKT6Ne1w_lesson%25203k.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, potassium nitrate has s solubility of about <mathjax>#"67 g / 100g H"_2"O"#</mathjax> at <mathjax>#40^@"C"#</mathjax>. This means that at <mathjax>#40^@"C"#</mathjax>, a <strong>saturated</strong> potassium nitrate solution will contain <mathjax>#"67 g"#</mathjax> of <em>dissolved</em> salt <strong>for every</strong> <mathjax>#"10 0g"#</mathjax> of water. </p>
<p>You know that at this temperature, your solution contains <mathjax>#"35 g"#</mathjax> of potassium nitrate in <mathjax>#"100 g"#</mathjax> of water. This solution will be <strong>unsaturated</strong> because it contains <em>less</em> potassium nitrate than the maximum amount that can be dissolved. </p>
<p>In order to make a <strong>saturated solution</strong>, you must get the total mass of potassium nitrate to <mathjax>#"67 g"#</mathjax>, which means that you must add</p>
<blockquote>
<p><mathjax>#"mass of KNO"_3 color(white)(a)"to be added" = "67 g" - "35 g" = color(green)(|bar(ul(color(white)(a/a)color(black)("32 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, dissolving another <mathjax>#"32 g"#</mathjax> of potassium nitrate in your solution at <mathjax>#40^@"C"#</mathjax> will get you a <strong>saturated</strong> solution. Any extra amount of potassium nitrate that you will add past this point will <em>remain undissolved</em>. </p></div>
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</article> | A solution contains 35 grams of #KNO_3# dissolved in 100 grams of water at 40°C. How much more #KNO_3# would have to be added to make it a saturated solution? | null |
1,981 | ad01ff14-6ddd-11ea-9bb1-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-equation-zn-h-2so-4-znso-4-h-2 | Zn + H2SO4 -> ZnSO4 + H2 | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"Zn + H2SO4 -> ZnSO4 + H2"}] | [{"type":"chemical equation","value":"Zn + H2SO4 -> ZnSO4 + H2"}] | <h1 class="questionTitle" itemprop="name">How do you balance this equation: #?Zn + ?H_2SO_4 -> ?ZnSO_4 + H_2#</h1> | null | Zn + H2SO4 -> ZnSO4 + H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"ZnSO"_4#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 Zn"#</mathjax> on the right, so we need <mathjax>#"1 Zn"#</mathjax> on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"S"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"1 S"#</mathjax> on the right. We need <mathjax>#"1 S"#</mathjax> on the left. Put a 1 in front of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>Done.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"2 H"#</mathjax> on the left, so we need <mathjax>#"2 H"#</mathjax> on the right. Put a 1 in front of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 + color(purple)(1)"H"_2#</mathjax></p>
<blockquote></blockquote>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<p>Let's check.</p>
<p><mathjax>#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#</mathjax><br/>
<mathjax>#color(white)(m)"Zn"color(white)(mml)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmll)2color(white)(mml)2#</mathjax><br/>
<mathjax>#color(white)(m)"S"color(white)(mmm)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmll)4color(white)(mml)4#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 →"ZnSO"_4 + "H"_2"SO"_4#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"ZnSO"_4#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 Zn"#</mathjax> on the right, so we need <mathjax>#"1 Zn"#</mathjax> on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"S"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"1 S"#</mathjax> on the right. We need <mathjax>#"1 S"#</mathjax> on the left. Put a 1 in front of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>Done.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"2 H"#</mathjax> on the left, so we need <mathjax>#"2 H"#</mathjax> on the right. Put a 1 in front of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 + color(purple)(1)"H"_2#</mathjax></p>
<blockquote></blockquote>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<p>Let's check.</p>
<p><mathjax>#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#</mathjax><br/>
<mathjax>#color(white)(m)"Zn"color(white)(mml)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmll)2color(white)(mml)2#</mathjax><br/>
<mathjax>#color(white)(m)"S"color(white)(mmm)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmll)4color(white)(mml)4#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 →"ZnSO"_4 + "H"_2"SO"_4#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance this equation: #?Zn + ?H_2SO_4 -> ?ZnSO_4 + H_2#</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2017-01-09T12:48:08" itemprop="dateCreated">
Jan 9, 2017
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<div class="markdown"><p><mathjax>#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#</mathjax></p>
<p>A method that often works is first to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax>, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"ZnSO"_4#</mathjax>. We put a 1 in front of it to remind ourselves that the number is now fixed.</p>
<blockquote></blockquote>
<p>We start with</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Zn"#</mathjax>:</strong></p>
<p>We have <mathjax>#"1 Zn"#</mathjax> on the right, so we need <mathjax>#"1 Zn"#</mathjax> on the left. We put a 1 in front of the <mathjax>#"Zn"#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"S"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"1 S"#</mathjax> on the right. We need <mathjax>#"1 S"#</mathjax> on the left. Put a 1 in front of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>Done.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"2 H"#</mathjax> on the left, so we need <mathjax>#"2 H"#</mathjax> on the right. Put a 1 in front of <mathjax>#"H"_2#</mathjax>.</p>
<p><mathjax>#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 + color(purple)(1)"H"_2#</mathjax></p>
<blockquote></blockquote>
<p>Every formula now has a coefficient. We should have a balanced equation.</p>
<p>Let's check.</p>
<p><mathjax>#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#</mathjax><br/>
<mathjax>#color(white)(m)"Zn"color(white)(mml)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmll)2color(white)(mml)2#</mathjax><br/>
<mathjax>#color(white)(m)"S"color(white)(mmm)1color(white)(mml)1#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmll)4color(white)(mml)4#</mathjax></p>
<blockquote></blockquote>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"Zn" + "H"_2"SO"_4 →"ZnSO"_4 + "H"_2"SO"_4#</mathjax></p></div>
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</article> | How do you balance this equation: #?Zn + ?H_2SO_4 -> ?ZnSO_4 + H_2# | null |
1,982 | ac103a4a-6ddd-11ea-8ffc-ccda262736ce | https://socratic.org/questions/a-sample-of-glass-that-has-a-mass-of-6-0-g-gives-off-12-j-of-heat-if-the-tempera | 0.50 J/(g * ℃) | start physical_unit 36 37 specific_heat j/(°c_·_g) qc_end physical_unit 1 3 9 10 mass qc_end physical_unit 21 22 25 26 temperature qc_end physical_unit 1 3 13 14 heat_energy qc_end end | [{"type":"physical unit","value":"Specific heat [OF] the glass [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.50 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] glass sample [=] \\pu{6.0 g}"},{"type":"physical unit","value":"Changed temperature [OF] the sample [=] \\pu{4.0 ℃}"},{"type":"physical unit","value":"Given off heat [OF] glass sample [=] \\pu{12 J}"}] | <h1 class="questionTitle" itemprop="name">A sample of glass that has a mass of 6.0 g gives off 12 J of heat. If the temperature of the sample changes by 4.0°C during this change, what is the specific heat of the glass? </h1> | null | 0.50 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To obtain the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of that glass, let's use the equation below:<br/>
<img alt="http://slideplayer.com/slide/4758097/" src="https://useruploads.socratic.org/l2tcEXnhQWC3dqQhAkJW_slide_1.jpg"/> </p>
<p>Based on the information you've provided, we know the following variables:</p>
<p><mathjax>#DeltaT#</mathjax> = <mathjax>#4^oC#</mathjax><br/>
<mathjax>#Q#</mathjax> = 12 J<br/>
<mathjax>#m#</mathjax> = <mathjax>#6.0g#</mathjax></p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#C#</mathjax>. This can be accomplished by dividing both sides by <mathjax>#m#</mathjax> and <mathjax>#Delta T#</mathjax> to get <mathjax>#C#</mathjax> by itself like this:</p>
<p><mathjax>#C = Q/(DeltaTxxm)#</mathjax></p>
<p>Now, we just plug in the known values:</p>
<p><mathjax>#C = (12J)/(4^oC"xx6.0g)#</mathjax></p>
<p><mathjax>#C = 0.50J/(gxx^oC) #</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.50 J/(g^oC) #</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To obtain the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of that glass, let's use the equation below:<br/>
<img alt="http://slideplayer.com/slide/4758097/" src="https://useruploads.socratic.org/l2tcEXnhQWC3dqQhAkJW_slide_1.jpg"/> </p>
<p>Based on the information you've provided, we know the following variables:</p>
<p><mathjax>#DeltaT#</mathjax> = <mathjax>#4^oC#</mathjax><br/>
<mathjax>#Q#</mathjax> = 12 J<br/>
<mathjax>#m#</mathjax> = <mathjax>#6.0g#</mathjax></p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#C#</mathjax>. This can be accomplished by dividing both sides by <mathjax>#m#</mathjax> and <mathjax>#Delta T#</mathjax> to get <mathjax>#C#</mathjax> by itself like this:</p>
<p><mathjax>#C = Q/(DeltaTxxm)#</mathjax></p>
<p>Now, we just plug in the known values:</p>
<p><mathjax>#C = (12J)/(4^oC"xx6.0g)#</mathjax></p>
<p><mathjax>#C = 0.50J/(gxx^oC) #</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of glass that has a mass of 6.0 g gives off 12 J of heat. If the temperature of the sample changes by 4.0°C during this change, what is the specific heat of the glass? </h1>
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<div class="markdown"><p><mathjax>#0.50 J/(g^oC) #</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To obtain the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of that glass, let's use the equation below:<br/>
<img alt="http://slideplayer.com/slide/4758097/" src="https://useruploads.socratic.org/l2tcEXnhQWC3dqQhAkJW_slide_1.jpg"/> </p>
<p>Based on the information you've provided, we know the following variables:</p>
<p><mathjax>#DeltaT#</mathjax> = <mathjax>#4^oC#</mathjax><br/>
<mathjax>#Q#</mathjax> = 12 J<br/>
<mathjax>#m#</mathjax> = <mathjax>#6.0g#</mathjax></p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#C#</mathjax>. This can be accomplished by dividing both sides by <mathjax>#m#</mathjax> and <mathjax>#Delta T#</mathjax> to get <mathjax>#C#</mathjax> by itself like this:</p>
<p><mathjax>#C = Q/(DeltaTxxm)#</mathjax></p>
<p>Now, we just plug in the known values:</p>
<p><mathjax>#C = (12J)/(4^oC"xx6.0g)#</mathjax></p>
<p><mathjax>#C = 0.50J/(gxx^oC) #</mathjax></p></div>
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</article> | A sample of glass that has a mass of 6.0 g gives off 12 J of heat. If the temperature of the sample changes by 4.0°C during this change, what is the specific heat of the glass? | null |
1,983 | a9a8cc3e-6ddd-11ea-988d-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-in-2-l-balloon-which-contains-5-moles-of-gas-at-a-temperatu | 51.29 atm | start physical_unit 13 13 pressure atm qc_end physical_unit 7 7 5 6 volume qc_end physical_unit 13 13 10 11 mole qc_end physical_unit 13 13 18 19 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] gas [IN] atm"}] | [{"type":"physical unit","value":"51.29 atm"}] | [{"type":"physical unit","value":"Volume [OF] balloon [=] \\pu{2 L}"},{"type":"physical unit","value":"Mole [OF] gas [=] \\pu{5 moles}"},{"type":"physical unit","value":"Temperature [OF] gas [=] \\pu{250 K}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure in 2 L balloon which contains 5 moles of gas at a temperature of 250 K?</h1> | null | 51.29 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>We can rearrange to solve specifically for pressure: <mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#n="moles"=5#</mathjax><br/>
<mathjax>#R="ideal gas constant"=0.8206#</mathjax><br/>
<mathjax>#T="temperature in Kelvin"=250#</mathjax><br/>
<mathjax>#V="volume in liters"=2#</mathjax></p>
<p>Plug in these values to find an answer of <mathjax>#51.3" atm"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#51.3" atm"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>We can rearrange to solve specifically for pressure: <mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#n="moles"=5#</mathjax><br/>
<mathjax>#R="ideal gas constant"=0.8206#</mathjax><br/>
<mathjax>#T="temperature in Kelvin"=250#</mathjax><br/>
<mathjax>#V="volume in liters"=2#</mathjax></p>
<p>Plug in these values to find an answer of <mathjax>#51.3" atm"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pressure in 2 L balloon which contains 5 moles of gas at a temperature of 250 K?</h1>
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<div class="markdown"><p><mathjax>#51.3" atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></p>
<p>We can rearrange to solve specifically for pressure: <mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#n="moles"=5#</mathjax><br/>
<mathjax>#R="ideal gas constant"=0.8206#</mathjax><br/>
<mathjax>#T="temperature in Kelvin"=250#</mathjax><br/>
<mathjax>#V="volume in liters"=2#</mathjax></p>
<p>Plug in these values to find an answer of <mathjax>#51.3" atm"#</mathjax>.</p></div>
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</article> | What is the pressure in 2 L balloon which contains 5 moles of gas at a temperature of 250 K? | null |
1,984 | acec77b0-6ddd-11ea-ac14-ccda262736ce | https://socratic.org/questions/what-is-the-mass-in-grams-of-12-10-23-molecules-of-co-2 | 87.58 grams | start physical_unit 12 12 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] CO2 [IN] grams"}] | [{"type":"physical unit","value":"87.58 grams"}] | [{"type":"physical unit","value":"Number [OF] CO2 molecules [=] \\pu{12 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">What is the mass in grams of #12 * 10^23# molecules of #CO_2#?</h1> | null | 87.58 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the number of moles of <mathjax>#"CO"_2#</mathjax>.</strong></p>
<p><mathjax>#12 × 10^23 color(red)(cancel(color(black)("molecules CO"_2))) × "1 mol CO"_2/(6.022 × 10^23 color(red)(cancel(color(black)("molecules CO"_2)))) = "1.99 mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"CO"_2#</mathjax></strong></p>
<p><mathjax>#1.99 color(red)(cancel(color(black)("mol CO"_2))) × "44.01 g CO"_2/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "88 g CO"_2#</mathjax></p></div>
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<div class="markdown"><p>The mass of <mathjax>#"CO"_2#</mathjax> is 88 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the number of moles of <mathjax>#"CO"_2#</mathjax>.</strong></p>
<p><mathjax>#12 × 10^23 color(red)(cancel(color(black)("molecules CO"_2))) × "1 mol CO"_2/(6.022 × 10^23 color(red)(cancel(color(black)("molecules CO"_2)))) = "1.99 mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"CO"_2#</mathjax></strong></p>
<p><mathjax>#1.99 color(red)(cancel(color(black)("mol CO"_2))) × "44.01 g CO"_2/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "88 g CO"_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass in grams of #12 * 10^23# molecules of #CO_2#?</h1>
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<div class="markdown"><p>The mass of <mathjax>#"CO"_2#</mathjax> is 88 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the number of moles of <mathjax>#"CO"_2#</mathjax>.</strong></p>
<p><mathjax>#12 × 10^23 color(red)(cancel(color(black)("molecules CO"_2))) × "1 mol CO"_2/(6.022 × 10^23 color(red)(cancel(color(black)("molecules CO"_2)))) = "1.99 mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"CO"_2#</mathjax></strong></p>
<p><mathjax>#1.99 color(red)(cancel(color(black)("mol CO"_2))) × "44.01 g CO"_2/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "88 g CO"_2#</mathjax></p></div>
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</article> | What is the mass in grams of #12 * 10^23# molecules of #CO_2#? | null |
1,985 | aaf65766-6ddd-11ea-951b-ccda262736ce | https://socratic.org/questions/2-76-g-of-k-2co-3-was-treated-by-a-series-of-reagents-so-as-to-convert-all-of-it | 1.16 g | start physical_unit 25 26 mass g qc_end physical_unit 3 3 0 1 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"weight [OF] the product [IN] g"}] | [{"type":"physical unit","value":"1.16 g"}] | [{"type":"physical unit","value":"Mass [OF] K2CO3 [=] \\pu{2.76 g}"},{"type":"other","value":"K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3[Fe(CN)6]2."}] | <h1 class="questionTitle" itemprop="name">2.76 #g# of #K_2CO_3# was treated by a series of reagents so as to convert all of its carbon to #K_2Zn_3[Fe(CN)_6]_2#. Calculate the weight of the product?</h1> | null | 1.16 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really have to do to solve this problem is figure out how much carbon you had in the <strong>2.76-g</strong> sample of potassium carbonate. </p>
<p>Once you do that, you use the fact that <em>all of the carbon</em> present in the potassium carbonate sample is now a part of the final product. </p>
<p>So, use carbon and potassium carbonate's molar masses to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of carbon in the compound</p>
<p><mathjax>#(12.011cancel("g/mol"))/(138.206cancel("g/mol")) * 100 = "8.69% C"#</mathjax></p>
<p>Potassium carbonate contains <strong>8.69 g</strong> of carbon for <em>every</em> 100 g of compound. This means that your sample contained </p>
<p><mathjax>#2.76cancel("g"K_2CO_3) * "8.69 g C"/(100cancel("g"K_2CO_3)) = "0.2398 g C"#</mathjax></p>
<p>All this carbon is now found in the final product. Once again, use molar masses to determine the carbon <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#K_2Zn_3[Fe(CN)_6]_2#</mathjax>.</p>
<p>Keep in mind, however, that you have a total of <strong>12</strong> carbon atoms in the final product!</p>
<p><mathjax>#(12 * 12.011cancel("g/mol"))/(698.235cancel("g/mol")) * 100 = "20.64% C"#</mathjax></p>
<p>Once again, this tells you that you have <strong>20.64 g</strong> of carbon for <em>every</em> 100 g of product. </p>
<p>Therefore,</p>
<p><mathjax>#0.2398cancel("g C") * "100 g product"/(20.64cancel("g C")) = color(green)("1.16 g product")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass of the product is <strong>1.16 g</strong>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really have to do to solve this problem is figure out how much carbon you had in the <strong>2.76-g</strong> sample of potassium carbonate. </p>
<p>Once you do that, you use the fact that <em>all of the carbon</em> present in the potassium carbonate sample is now a part of the final product. </p>
<p>So, use carbon and potassium carbonate's molar masses to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of carbon in the compound</p>
<p><mathjax>#(12.011cancel("g/mol"))/(138.206cancel("g/mol")) * 100 = "8.69% C"#</mathjax></p>
<p>Potassium carbonate contains <strong>8.69 g</strong> of carbon for <em>every</em> 100 g of compound. This means that your sample contained </p>
<p><mathjax>#2.76cancel("g"K_2CO_3) * "8.69 g C"/(100cancel("g"K_2CO_3)) = "0.2398 g C"#</mathjax></p>
<p>All this carbon is now found in the final product. Once again, use molar masses to determine the carbon <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#K_2Zn_3[Fe(CN)_6]_2#</mathjax>.</p>
<p>Keep in mind, however, that you have a total of <strong>12</strong> carbon atoms in the final product!</p>
<p><mathjax>#(12 * 12.011cancel("g/mol"))/(698.235cancel("g/mol")) * 100 = "20.64% C"#</mathjax></p>
<p>Once again, this tells you that you have <strong>20.64 g</strong> of carbon for <em>every</em> 100 g of product. </p>
<p>Therefore,</p>
<p><mathjax>#0.2398cancel("g C") * "100 g product"/(20.64cancel("g C")) = color(green)("1.16 g product")#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">2.76 #g# of #K_2CO_3# was treated by a series of reagents so as to convert all of its carbon to #K_2Zn_3[Fe(CN)_6]_2#. Calculate the weight of the product?</h1>
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<div class="markdown"><p>The mass of the product is <strong>1.16 g</strong>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really have to do to solve this problem is figure out how much carbon you had in the <strong>2.76-g</strong> sample of potassium carbonate. </p>
<p>Once you do that, you use the fact that <em>all of the carbon</em> present in the potassium carbonate sample is now a part of the final product. </p>
<p>So, use carbon and potassium carbonate's molar masses to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of carbon in the compound</p>
<p><mathjax>#(12.011cancel("g/mol"))/(138.206cancel("g/mol")) * 100 = "8.69% C"#</mathjax></p>
<p>Potassium carbonate contains <strong>8.69 g</strong> of carbon for <em>every</em> 100 g of compound. This means that your sample contained </p>
<p><mathjax>#2.76cancel("g"K_2CO_3) * "8.69 g C"/(100cancel("g"K_2CO_3)) = "0.2398 g C"#</mathjax></p>
<p>All this carbon is now found in the final product. Once again, use molar masses to determine the carbon <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#K_2Zn_3[Fe(CN)_6]_2#</mathjax>.</p>
<p>Keep in mind, however, that you have a total of <strong>12</strong> carbon atoms in the final product!</p>
<p><mathjax>#(12 * 12.011cancel("g/mol"))/(698.235cancel("g/mol")) * 100 = "20.64% C"#</mathjax></p>
<p>Once again, this tells you that you have <strong>20.64 g</strong> of carbon for <em>every</em> 100 g of product. </p>
<p>Therefore,</p>
<p><mathjax>#0.2398cancel("g C") * "100 g product"/(20.64cancel("g C")) = color(green)("1.16 g product")#</mathjax></p></div>
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</article> | 2.76 #g# of #K_2CO_3# was treated by a series of reagents so as to convert all of its carbon to #K_2Zn_3[Fe(CN)_6]_2#. Calculate the weight of the product? | null |
1,986 | aa1a6c99-6ddd-11ea-b43e-ccda262736ce | https://socratic.org/questions/5909740fb72cff7213453ce9 | 0.31 moles | start physical_unit 3 3 mole mol qc_end physical_unit 3 3 6 7 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] silicon [IN] moles"}] | [{"type":"physical unit","value":"0.31 moles"}] | [{"type":"physical unit","value":"Mass [OF] silicon [=] \\pu{10 g}"}] | <h1 class="questionTitle" itemprop="name">How moles of silicon in a #10*g# mass of same?</h1> | null | 0.31 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of stuff"="Mass"/"Molar Mass"#</mathjax></p>
<p>And for this problem,</p>
<p><mathjax>#"Moles of silicon"=(10*g)/(32.1*g*mol^-1)=??*mol??#</mathjax></p>
<p>From where did I get the molar mass of silicon?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>I take it you want the <mathjax>#"number of moles"#</mathjax> of element in a <mathjax>#10*g#</mathjax> mass....</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of stuff"="Mass"/"Molar Mass"#</mathjax></p>
<p>And for this problem,</p>
<p><mathjax>#"Moles of silicon"=(10*g)/(32.1*g*mol^-1)=??*mol??#</mathjax></p>
<p>From where did I get the molar mass of silicon?</p></div>
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<h1 class="questionTitle" itemprop="name">How moles of silicon in a #10*g# mass of same?</h1>
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anor277
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<div class="markdown"><p>I take it you want the <mathjax>#"number of moles"#</mathjax> of element in a <mathjax>#10*g#</mathjax> mass....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of stuff"="Mass"/"Molar Mass"#</mathjax></p>
<p>And for this problem,</p>
<p><mathjax>#"Moles of silicon"=(10*g)/(32.1*g*mol^-1)=??*mol??#</mathjax></p>
<p>From where did I get the molar mass of silicon?</p></div>
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</article> | How moles of silicon in a #10*g# mass of same? | null |
1,987 | acdf8eca-6ddd-11ea-9cb9-ccda262736ce | https://socratic.org/questions/58cf2b20b72cff11ef61c173 | 1.2 × 10^(-6) mL | start physical_unit 26 27 volume ml qc_end physical_unit 26 27 3 4 pressure qc_end physical_unit 26 27 6 9 volume qc_end physical_unit 26 27 16 17 pressure qc_end c_other constant_temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"1.2 × 10^(-6) mL"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{553 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2.6 × 10^(-6) mL}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{1.55 atm}"},{"type":"other","value":"ConstantTemperature"},{"type":"other","value":"The amount of the gas is held constant."}] | <h1 class="questionTitle" itemprop="name">A gas at #"553 mmHg"# occupies #2.6xx10^(-6)# #"mL"#. If the pressure is increased to #"1.55 atm"#, what is the new pressure? The amount of the gas and the temperature are held constant.</h1> | null | 1.2 × 10^(-6) mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas is indirectly proportional to the pressure, as long as temperature and amount are held constant. The equation to use:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax>,</p>
<p>where <mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressure; and <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volume.</p>
<p>The two pressures do not have the same units. One will have to be converted into the other. I'm going to convert mmHg to atm because atm is more common.</p>
<p><mathjax>#"1 atm=760.0 mmHg"#</mathjax></p>
<p><strong>Given</strong><br/>
<mathjax>#P_1=553color(red)(cancel(color(black)("mmHg")))xx(1"atm")/(760.0color(red)cancel(color(black)("mmHg")))="0.7276 atm"#</mathjax></p>
<p><mathjax>#V_1=2.6xx10^(-6)"mL"#</mathjax></p>
<p><mathjax>#P_2="1.55 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=((0.7276color(red)cancel(color(black)("atm")))xx(2.6xx10^(-6)"mL"))/(1.55color(red)cancel(color(black)("atm")))=1.2xx10^(-6)"mL"#</mathjax></p>
<p>(rounded to two significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V_2=1.2xx10^(-6)color(white)(.)"mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas is indirectly proportional to the pressure, as long as temperature and amount are held constant. The equation to use:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax>,</p>
<p>where <mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressure; and <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volume.</p>
<p>The two pressures do not have the same units. One will have to be converted into the other. I'm going to convert mmHg to atm because atm is more common.</p>
<p><mathjax>#"1 atm=760.0 mmHg"#</mathjax></p>
<p><strong>Given</strong><br/>
<mathjax>#P_1=553color(red)(cancel(color(black)("mmHg")))xx(1"atm")/(760.0color(red)cancel(color(black)("mmHg")))="0.7276 atm"#</mathjax></p>
<p><mathjax>#V_1=2.6xx10^(-6)"mL"#</mathjax></p>
<p><mathjax>#P_2="1.55 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=((0.7276color(red)cancel(color(black)("atm")))xx(2.6xx10^(-6)"mL"))/(1.55color(red)cancel(color(black)("atm")))=1.2xx10^(-6)"mL"#</mathjax></p>
<p>(rounded to two significant figures)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas at #"553 mmHg"# occupies #2.6xx10^(-6)# #"mL"#. If the pressure is increased to #"1.55 atm"#, what is the new pressure? The amount of the gas and the temperature are held constant.</h1>
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Meave60
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<span class="dateCreated" datetime="2017-03-25T04:33:14" itemprop="dateCreated">
Mar 25, 2017
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<div class="markdown"><p><mathjax>#V_2=1.2xx10^(-6)color(white)(.)"mL"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question involves <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas is indirectly proportional to the pressure, as long as temperature and amount are held constant. The equation to use:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax>,</p>
<p>where <mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressure; and <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volume.</p>
<p>The two pressures do not have the same units. One will have to be converted into the other. I'm going to convert mmHg to atm because atm is more common.</p>
<p><mathjax>#"1 atm=760.0 mmHg"#</mathjax></p>
<p><strong>Given</strong><br/>
<mathjax>#P_1=553color(red)(cancel(color(black)("mmHg")))xx(1"atm")/(760.0color(red)cancel(color(black)("mmHg")))="0.7276 atm"#</mathjax></p>
<p><mathjax>#V_1=2.6xx10^(-6)"mL"#</mathjax></p>
<p><mathjax>#P_2="1.55 atm"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=((0.7276color(red)cancel(color(black)("atm")))xx(2.6xx10^(-6)"mL"))/(1.55color(red)cancel(color(black)("atm")))=1.2xx10^(-6)"mL"#</mathjax></p>
<p>(rounded to two significant figures)</p></div>
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</article> | A gas at #"553 mmHg"# occupies #2.6xx10^(-6)# #"mL"#. If the pressure is increased to #"1.55 atm"#, what is the new pressure? The amount of the gas and the temperature are held constant. | null |
1,988 | a88c4264-6ddd-11ea-8441-ccda262736ce | https://socratic.org/questions/61-5l-of-oxygen-at-18-c-and-an-absolute-pressure-of-2-45-atm-is-compressed-to-48 | 3.43 atm | start physical_unit 3 3 pressure atm qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 3 3 12 13 pressure qc_end physical_unit 3 3 17 18 volume qc_end physical_unit 3 3 5 6 temperature qc_end physical_unit 3 3 29 30 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] oxygen gas [IN] atm"}] | [{"type":"physical unit","value":"3.43 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] oxygen gas [=] \\pu{61.5 L}"},{"type":"physical unit","value":"absolute pressure1 [OF] oxygen gas [=] \\pu{2.45 atm}"},{"type":"physical unit","value":"Volume2 [OF] oxygen gas [=] \\pu{ 48.8 L}"},{"type":"physical unit","value":"Temperature1 [OF] oxygen gas [=] \\pu{18 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] oxygen gas [=] \\pu{50 ℃}"}] | <h1 class="questionTitle" itemprop="name">61.5L of oxygen at 18°C and an absolute pressure of 2.45 atm is compressed to 48.8L and at the same time the temperature is raised to 50°C . what is the new pressure of the gas?</h1> | null | 3.43 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you're dealing with a clear-cut case of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> problem. </p>
<p>More specifically, you know that <em>pressure</em>, <em>temperature</em>, and <em>volume</em> are all changing from the initial state of the gas to the final state of the gas. </p>
<p>This of course in the context of the number of moles of gas being <strong>kept constant</strong>. </p>
<p>So let's say that you're not familiar with the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to describe the gas at the <strong>initial state</strong></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1#</mathjax></p>
</blockquote>
<p>and at the <strong>final state</strong> </p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2#</mathjax></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R))) * T_1)/(color(red)(cancel(color(black)(n * R))) * T_2)#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#color(blue)( (P_1V_1)/T_1 = (P_2V_2)/T_2) ->#</mathjax> <em>the combined gas law equation</em></p>
</blockquote>
<p>It is <strong>very important</strong> to remember that the temperature of the gas <strong>must</strong> be expressed in Kelvin! </p>
<p>You want to know the final pressure of the gas, so rearrange this equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#P_2 = (61.5color(red)(cancel(color(black)("L"))))/(48.8color(red)(cancel(color(black)("L")))) * ((273.15 + 50)color(red)(cancel(color(black)("K"))))/((273.15 + 18)color(red)(cancel(color(black)("K")))) * "2.45 atm"#</mathjax></p>
<p><mathjax>#P_2 = "3.4270 atm"#</mathjax></p>
</blockquote>
<p>Now, you <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the second temperature of the gas, but I'll leave it rounded off to two sig figs</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("3.4 atm")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/bftkRnTcFj8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.4 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you're dealing with a clear-cut case of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> problem. </p>
<p>More specifically, you know that <em>pressure</em>, <em>temperature</em>, and <em>volume</em> are all changing from the initial state of the gas to the final state of the gas. </p>
<p>This of course in the context of the number of moles of gas being <strong>kept constant</strong>. </p>
<p>So let's say that you're not familiar with the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to describe the gas at the <strong>initial state</strong></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1#</mathjax></p>
</blockquote>
<p>and at the <strong>final state</strong> </p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2#</mathjax></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R))) * T_1)/(color(red)(cancel(color(black)(n * R))) * T_2)#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#color(blue)( (P_1V_1)/T_1 = (P_2V_2)/T_2) ->#</mathjax> <em>the combined gas law equation</em></p>
</blockquote>
<p>It is <strong>very important</strong> to remember that the temperature of the gas <strong>must</strong> be expressed in Kelvin! </p>
<p>You want to know the final pressure of the gas, so rearrange this equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#P_2 = (61.5color(red)(cancel(color(black)("L"))))/(48.8color(red)(cancel(color(black)("L")))) * ((273.15 + 50)color(red)(cancel(color(black)("K"))))/((273.15 + 18)color(red)(cancel(color(black)("K")))) * "2.45 atm"#</mathjax></p>
<p><mathjax>#P_2 = "3.4270 atm"#</mathjax></p>
</blockquote>
<p>Now, you <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the second temperature of the gas, but I'll leave it rounded off to two sig figs</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("3.4 atm")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/bftkRnTcFj8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">61.5L of oxygen at 18°C and an absolute pressure of 2.45 atm is compressed to 48.8L and at the same time the temperature is raised to 50°C . what is the new pressure of the gas?</h1>
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Stefan V.
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Dec 10, 2015
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<div class="markdown"><p><mathjax>#"3.4 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that you're dealing with a clear-cut case of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> problem. </p>
<p>More specifically, you know that <em>pressure</em>, <em>temperature</em>, and <em>volume</em> are all changing from the initial state of the gas to the final state of the gas. </p>
<p>This of course in the context of the number of moles of gas being <strong>kept constant</strong>. </p>
<p>So let's say that you're not familiar with the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation. Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to describe the gas at the <strong>initial state</strong></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1#</mathjax></p>
</blockquote>
<p>and at the <strong>final state</strong> </p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2#</mathjax></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R))) * T_1)/(color(red)(cancel(color(black)(n * R))) * T_2)#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#color(blue)( (P_1V_1)/T_1 = (P_2V_2)/T_2) ->#</mathjax> <em>the combined gas law equation</em></p>
</blockquote>
<p>It is <strong>very important</strong> to remember that the temperature of the gas <strong>must</strong> be expressed in Kelvin! </p>
<p>You want to know the final pressure of the gas, so rearrange this equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#P_2 = (61.5color(red)(cancel(color(black)("L"))))/(48.8color(red)(cancel(color(black)("L")))) * ((273.15 + 50)color(red)(cancel(color(black)("K"))))/((273.15 + 18)color(red)(cancel(color(black)("K")))) * "2.45 atm"#</mathjax></p>
<p><mathjax>#P_2 = "3.4270 atm"#</mathjax></p>
</blockquote>
<p>Now, you <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, the number of sig figs you have for the second temperature of the gas, but I'll leave it rounded off to two sig figs</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("3.4 atm")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/bftkRnTcFj8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | 61.5L of oxygen at 18°C and an absolute pressure of 2.45 atm is compressed to 48.8L and at the same time the temperature is raised to 50°C . what is the new pressure of the gas? | null |
1,989 | ab44483e-6ddd-11ea-98f6-ccda262736ce | https://socratic.org/questions/given-the-following-reactions-the-enthalpy-of-reaction-for-2n2o-g-2no-g-n2-g-is- | +17.5 kJ | start physical_unit 7 7 enthalpy kj qc_end chemical_equation 18 27 qc_end chemical_equation 28 38 qc_end chemical_equation 9 15 qc_end end | [{"type":"physical unit","value":"Enthalpy [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"+17.5 kJ"}] | [{"type":"chemical equation","value":"N2(g) + O2(g) -> 2 NO(g) deltaH = +180.7 kJ"},{"type":"chemical equation","value":"2 N2O(g) -> O2(g) + 2 N2(g) deltaH = -163.2 kJ"},{"type":"chemical equation","value":"2 N2O(g-) -> 2 NO(g) + N2(g)"}] | <h1 class="questionTitle" itemprop="name">Given the following reactions the enthalpy of reaction for 2N2O(g) --> 2NO(g) +N2(g)
is __________ kJ?
N2 (g) + O2(g) --> 2NO(g) ΔH = +180.7 kJ
2N2O(g) --> O2 (g) + 2N2(g) ΔH = -163.2 kJ</h1> | null | +17.5 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all this kind of problems the first step is to balance the equations, in this case this is a very simple one, because if you add the two equations without any change you would obtain the asked one.</p>
<p>(1) <mathjax>#N_(2(g)) +O_(2(g)) rarr 2NO_((g)) Delta H_1 = +"180.7 kJ"#</mathjax> </p>
<p>(2) <mathjax>#2N_2O_((g)) rarr O_(2(g)) + 2N_(2(g)) Delta H_2 = -"163.7 kJ"#</mathjax> </p>
<p>(3) <mathjax>#2N_2O_((g)) rarr 2NO_((g)) + N_(2(g)) Delta H_2 = "? kJ"#</mathjax> </p>
<p>then <mathjax>#(1) + (2) = (3) #</mathjax></p>
<p>as <mathjax>#DeltaH_1 + DeltaH_2 = DeltaH_3 #</mathjax></p>
<p><mathjax>#:.DeltaH_3 = +17.5KJ#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>17.5 KJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all this kind of problems the first step is to balance the equations, in this case this is a very simple one, because if you add the two equations without any change you would obtain the asked one.</p>
<p>(1) <mathjax>#N_(2(g)) +O_(2(g)) rarr 2NO_((g)) Delta H_1 = +"180.7 kJ"#</mathjax> </p>
<p>(2) <mathjax>#2N_2O_((g)) rarr O_(2(g)) + 2N_(2(g)) Delta H_2 = -"163.7 kJ"#</mathjax> </p>
<p>(3) <mathjax>#2N_2O_((g)) rarr 2NO_((g)) + N_(2(g)) Delta H_2 = "? kJ"#</mathjax> </p>
<p>then <mathjax>#(1) + (2) = (3) #</mathjax></p>
<p>as <mathjax>#DeltaH_1 + DeltaH_2 = DeltaH_3 #</mathjax></p>
<p><mathjax>#:.DeltaH_3 = +17.5KJ#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Given the following reactions the enthalpy of reaction for 2N2O(g) --> 2NO(g) +N2(g)
is __________ kJ?
N2 (g) + O2(g) --> 2NO(g) ΔH = +180.7 kJ
2N2O(g) --> O2 (g) + 2N2(g) ΔH = -163.2 kJ</h1>
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<div class="markdown"><p>17.5 KJ</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In all this kind of problems the first step is to balance the equations, in this case this is a very simple one, because if you add the two equations without any change you would obtain the asked one.</p>
<p>(1) <mathjax>#N_(2(g)) +O_(2(g)) rarr 2NO_((g)) Delta H_1 = +"180.7 kJ"#</mathjax> </p>
<p>(2) <mathjax>#2N_2O_((g)) rarr O_(2(g)) + 2N_(2(g)) Delta H_2 = -"163.7 kJ"#</mathjax> </p>
<p>(3) <mathjax>#2N_2O_((g)) rarr 2NO_((g)) + N_(2(g)) Delta H_2 = "? kJ"#</mathjax> </p>
<p>then <mathjax>#(1) + (2) = (3) #</mathjax></p>
<p>as <mathjax>#DeltaH_1 + DeltaH_2 = DeltaH_3 #</mathjax></p>
<p><mathjax>#:.DeltaH_3 = +17.5KJ#</mathjax></p></div>
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</article> | Given the following reactions the enthalpy of reaction for 2N2O(g) --> 2NO(g) +N2(g)
is __________ kJ?
N2 (g) + O2(g) --> 2NO(g) ΔH = +180.7 kJ
2N2O(g) --> O2 (g) + 2N2(g) ΔH = -163.2 kJ | null |
1,990 | a8d56274-6ddd-11ea-ae2b-ccda262736ce | https://socratic.org/questions/in-the-reaction-2na-cl-2-2nacl-how-many-moles-of-cl-2-react-with-100-moles-of-so | 50.00 moles | start physical_unit 6 6 mole mol qc_end chemical_equation 3 9 qc_end physical_unit 20 20 17 18 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] Cl2 [IN] moles"}] | [{"type":"physical unit","value":"50.00 moles"}] | [{"type":"chemical equation","value":"2 Na + Cl2 -> 2 NaCl"},{"type":"physical unit","value":"Mole [OF] sodium [=] \\pu{100 moles}"}] | <h1 class="questionTitle" itemprop="name">In the reaction #2Na+Cl_2 -> 2NaCl#, how many moles of #Cl_2# react with 100 moles of sodium to form sodium chloride?</h1> | null | 50.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Remember that <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is a NUMERICAL (or extensive) quantity. I could say just as correctly that 2 dozen sodium atoms react with 1 dozen chlorine molecules to form 2 dozen sodium chloride entities.</p>
<p>You have started with 100 mol sodium metal. Clearly I need 50 mol of chlorine molecules for a stoichiometric reaction. Given these quantities, how many moles of sodium chloride will result? </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Look at the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction. One mole of chlorine gas reacts with 2 mol sodium metal to give 2 mol sodium chloride. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Remember that <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is a NUMERICAL (or extensive) quantity. I could say just as correctly that 2 dozen sodium atoms react with 1 dozen chlorine molecules to form 2 dozen sodium chloride entities.</p>
<p>You have started with 100 mol sodium metal. Clearly I need 50 mol of chlorine molecules for a stoichiometric reaction. Given these quantities, how many moles of sodium chloride will result? </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">In the reaction #2Na+Cl_2 -> 2NaCl#, how many moles of #Cl_2# react with 100 moles of sodium to form sodium chloride?</h1>
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<div class="markdown"><p>Look at the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction. One mole of chlorine gas reacts with 2 mol sodium metal to give 2 mol sodium chloride. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Remember that <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is a NUMERICAL (or extensive) quantity. I could say just as correctly that 2 dozen sodium atoms react with 1 dozen chlorine molecules to form 2 dozen sodium chloride entities.</p>
<p>You have started with 100 mol sodium metal. Clearly I need 50 mol of chlorine molecules for a stoichiometric reaction. Given these quantities, how many moles of sodium chloride will result? </p></div>
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</article> | In the reaction #2Na+Cl_2 -> 2NaCl#, how many moles of #Cl_2# react with 100 moles of sodium to form sodium chloride? | null |
1,991 | aca6b01e-6ddd-11ea-8bc8-ccda262736ce | https://socratic.org/questions/if-a-gas-sample-has-a-pressure-of-30-7-kpa-at-0-00-c-by-how-much-does-the-temper | -20.46 ℃ | start physical_unit 2 3 temperature °c qc_end physical_unit 2 3 8 9 pressure qc_end physical_unit 2 3 11 12 temperature qc_end physical_unit 2 3 27 28 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas sample [IN] ℃"}] | [{"type":"physical unit","value":"-20.46 ℃"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{30.7 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{0.00 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas sample [=] \\pu{28.4 kPa}"}] | <h1 class="questionTitle" itemprop="name">If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa?</h1> | null | -20.46 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question I assume that the volume is unchanged and therefore, we can use the following law:</p>
<p><mathjax>#P/T=k#</mathjax></p>
<p>Then, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>T_2=(T_1)/(P_1)xxP_2#</mathjax></p>
<p><mathjax>#=>P_2=("273.15 K")/(30.7cancel("kPa"))xx28.4cancel("kPa") = "252.69 K"#</mathjax></p>
<p>Therefore, the temperature should be lowered to: <mathjax>#252.69 - 273.15 = - 20.464^@ "C"#</mathjax></p>
<blockquote>
<p><mathjax>#=> color(blue)(T_2 = -20.5^@"C")#</mathjax> to 3 sig figs.</p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#T_2 = -20.5^@ "C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question I assume that the volume is unchanged and therefore, we can use the following law:</p>
<p><mathjax>#P/T=k#</mathjax></p>
<p>Then, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>T_2=(T_1)/(P_1)xxP_2#</mathjax></p>
<p><mathjax>#=>P_2=("273.15 K")/(30.7cancel("kPa"))xx28.4cancel("kPa") = "252.69 K"#</mathjax></p>
<p>Therefore, the temperature should be lowered to: <mathjax>#252.69 - 273.15 = - 20.464^@ "C"#</mathjax></p>
<blockquote>
<p><mathjax>#=> color(blue)(T_2 = -20.5^@"C")#</mathjax> to 3 sig figs.</p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa?</h1>
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Mar 16, 2016
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<div class="markdown"><p><mathjax>#T_2 = -20.5^@ "C"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question I assume that the volume is unchanged and therefore, we can use the following law:</p>
<p><mathjax>#P/T=k#</mathjax></p>
<p>Then, <mathjax>#(P_1)/(T_1)=(P_2)/(T_2)=>T_2=(T_1)/(P_1)xxP_2#</mathjax></p>
<p><mathjax>#=>P_2=("273.15 K")/(30.7cancel("kPa"))xx28.4cancel("kPa") = "252.69 K"#</mathjax></p>
<p>Therefore, the temperature should be lowered to: <mathjax>#252.69 - 273.15 = - 20.464^@ "C"#</mathjax></p>
<blockquote>
<p><mathjax>#=> color(blue)(T_2 = -20.5^@"C")#</mathjax> to 3 sig figs.</p>
</blockquote></div>
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</article> | If a gas sample has a pressure of 30.7 kPa at 0.00°C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa? | null |
1,992 | ad241274-6ddd-11ea-9da5-ccda262736ce | https://socratic.org/questions/a-compound-is-composed-of-85-64-carbon-and-14-36-hydrogen-the-compound-has-a-for | C3H6 | start chemical_formula qc_end physical_unit 10 11 17 18 molecular_weight qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C3H6"}] | [{"type":"physical unit","value":"Percent [OF] carbon in the compound [=] \\pu{85.64%}"},{"type":"physical unit","value":"Percent [OF] hydrogen in the compound [=] \\pu{14.36%}"},{"type":"physical unit","value":"Formula mass [OF] the compound [=] \\pu{42.08 grams/mol}"}] | <h1 class="questionTitle" itemprop="name">A compound is composed of 85.64% carbon and 14.36% hydrogen. The compound has a formula mass of 42.08 grams. What is the molecular formula?</h1> | null | C3H6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all of these problems, we assume <mathjax>#100*g#</mathjax> of unknown compound.</p>
<p>And thus, we determine the elemental composition by the given percentages.</p>
<p><mathjax>#"Moles of carbon"=(85.64*g)/(12.011*g*mol^-1)=7.13*mol#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"=(14.36*g)/(1.00794*g*mol^-1)=14.25*mol#</mathjax>.</p>
<p>Clearly, there are <mathjax>#2#</mathjax> <mathjax>#"moles"#</mathjax> of hydrogen per mol of carbon. And thus the empirical formula is <mathjax>#CH_(2)#</mathjax>.</p>
<p>And <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#nxx("empirical formula")#</mathjax></p>
<p>Thus, </p>
<p><mathjax>#42.08*g*mol^-1=nxx(12.011+2xx1.00794)*g*mol^-1#</mathjax></p>
<p>And thus <mathjax>#n=3#</mathjax>, and <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_3H_6#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"Molecular formula"= C_3H_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all of these problems, we assume <mathjax>#100*g#</mathjax> of unknown compound.</p>
<p>And thus, we determine the elemental composition by the given percentages.</p>
<p><mathjax>#"Moles of carbon"=(85.64*g)/(12.011*g*mol^-1)=7.13*mol#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"=(14.36*g)/(1.00794*g*mol^-1)=14.25*mol#</mathjax>.</p>
<p>Clearly, there are <mathjax>#2#</mathjax> <mathjax>#"moles"#</mathjax> of hydrogen per mol of carbon. And thus the empirical formula is <mathjax>#CH_(2)#</mathjax>.</p>
<p>And <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#nxx("empirical formula")#</mathjax></p>
<p>Thus, </p>
<p><mathjax>#42.08*g*mol^-1=nxx(12.011+2xx1.00794)*g*mol^-1#</mathjax></p>
<p>And thus <mathjax>#n=3#</mathjax>, and <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_3H_6#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound is composed of 85.64% carbon and 14.36% hydrogen. The compound has a formula mass of 42.08 grams. What is the molecular formula?</h1>
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<div class="markdown"><p><mathjax>#"Molecular formula"= C_3H_6#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all of these problems, we assume <mathjax>#100*g#</mathjax> of unknown compound.</p>
<p>And thus, we determine the elemental composition by the given percentages.</p>
<p><mathjax>#"Moles of carbon"=(85.64*g)/(12.011*g*mol^-1)=7.13*mol#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"=(14.36*g)/(1.00794*g*mol^-1)=14.25*mol#</mathjax>.</p>
<p>Clearly, there are <mathjax>#2#</mathjax> <mathjax>#"moles"#</mathjax> of hydrogen per mol of carbon. And thus the empirical formula is <mathjax>#CH_(2)#</mathjax>.</p>
<p>And <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#nxx("empirical formula")#</mathjax></p>
<p>Thus, </p>
<p><mathjax>#42.08*g*mol^-1=nxx(12.011+2xx1.00794)*g*mol^-1#</mathjax></p>
<p>And thus <mathjax>#n=3#</mathjax>, and <mathjax>#"molecular formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_3H_6#</mathjax>.</p></div>
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</article> | A compound is composed of 85.64% carbon and 14.36% hydrogen. The compound has a formula mass of 42.08 grams. What is the molecular formula? | null |
1,993 | aa495e3a-6ddd-11ea-9332-ccda262736ce | https://socratic.org/questions/a-sample-of-carbon-monoxide-occupies-2-44-l-at-295-0-k-and-771-torr-find-its-vol | 4.18 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 18 20 temperature qc_end physical_unit 1 4 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] carbon monoxide sample [IN] L"}] | [{"type":"physical unit","value":"4.18 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] carbon monoxide sample [=] \\pu{771 torr}"},{"type":"physical unit","value":"Volume1 [OF] carbon monoxide sample [=] \\pu{2.44 L}"},{"type":"physical unit","value":"Temperature1 [OF] carbon monoxide sample [=] \\pu{295.0 K}"},{"type":"physical unit","value":"Temperature2 [OF] carbon monoxide sample [=] \\pu{-47 degrees Celsius}"},{"type":"physical unit","value":"Pressure2 [OF] carbon monoxide sample [=] \\pu{350 torr}"}] | <h1 class="questionTitle" itemprop="name">A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?</h1> | null | 4.18 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>; <mathjax>#V_2~=4*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?</h1>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>; <mathjax>#V_2~=4*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#4.18"L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As there are a constant number of moles:<br/>
<mathjax>#("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2#</mathjax></p>
<p><mathjax>#-47^circC#</mathjax>=226^circK#</p>
<p><mathjax>#2.44"L"=0.00244"m"^3#</mathjax></p>
<p><mathjax>#771"torr"=102791.546"Pa"#</mathjax><br/>
<mathjax>#350"torr"=46662.8288"Pa"#</mathjax></p>
<p>If <mathjax>#("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2#</mathjax>, then <mathjax>#(102791.546*0.00244)/295=(46662.8288*"V"_2)/226#</mathjax>.</p>
<p><mathjax>#"V"_2=(102791.546*0.00244*226)/(295*46662.8288)=0.00411777473"m"^3~~0.00418"m"^3=4.18"L"#</mathjax>'</p></div>
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</article> | A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr? | null |
1,994 | ab695c6c-6ddd-11ea-baaa-ccda262736ce | https://socratic.org/questions/how-much-volume-would-63-9-g-of-carbon-monoxide-take-up | 32.60 L | start physical_unit 7 8 volume l qc_end physical_unit 7 8 4 5 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon monoxide [IN] L"}] | [{"type":"physical unit","value":"32.60 L"}] | [{"type":"physical unit","value":"Mass [OF] carbon monoxide [=] \\pu{63.9 g}"}] | <h1 class="questionTitle" itemprop="name">How much volume would 63.9 g of carbon monoxide take up? </h1> | null | 32.60 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p>The volume of <mathjax>#"63.9 g"#</mathjax> carbon dioxide at STP is <mathjax>#"32.6 L"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">How much volume would 63.9 g of carbon monoxide take up? </h1>
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<div class="markdown"><p>The volume of <mathjax>#"63.9 g"#</mathjax> carbon dioxide at STP is <mathjax>#"32.6 L"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p>51.1 liters at STP</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of Carbon Monoxide <mathjax>#CO#</mathjax> is 28 grams/mole</p>
<p>1 x C = 1 x 12 = 12 <br/>
1 x O = 1 x 16 = 16<br/>
Total = 28 grams/ mole </p>
<p>The number of moles is found by dividing the mass by the grams/mole </p>
<p><mathjax># 63.9/28 = 2.28 #</mathjax> moles.</p>
<p>At STP ( Standard Temperature Pressure) The molar volume is 22.4 liters/mole </p>
<p>to find the volume multiple the number of moles time the molar volume. </p>
<p># 2.28 xx 22.4 = 51.1 liters. </p></div>
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</article> | How much volume would 63.9 g of carbon monoxide take up? | null |
1,995 | ab910dfa-6ddd-11ea-9d95-ccda262736ce | https://socratic.org/questions/20-000-ml-aliquots-of-a-phosphoric-acid-solution-with-an-unknown-concentration-i | 0.38 mol/L | start physical_unit 5 6 concentration mol/l qc_end physical_unit 5 7 0 1 volume qc_end physical_unit 18 20 16 17 concentration qc_end physical_unit 25 27 29 30 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] phosphoric acid [IN] mol/L"}] | [{"type":"physical unit","value":"0.38 mol/L"}] | [{"type":"physical unit","value":"Volume [OF] phosphoric acid solution [=] \\pu{20.000 mL}"},{"type":"physical unit","value":"Concentration [OF] sodium hydroxide solution [=] \\pu{0.624 mol/L}"},{"type":"physical unit","value":"Average volume [OF] sodium hydroxide added [=] \\pu{24.01 mL}"}] | <h1 class="questionTitle" itemprop="name">20.000 mL aliquots of a phosphoric acid solution with an unknown concentration is titrated with some 0.624 mol/L sodium hydroxide solution.The average volume of the sodium hydroxide added was 24.01 mL. What is the concentration of the phosphoric acid?</h1> | null | 0.38 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the reaction............</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>Note that the phosphoric acid is A DIACID not a triacid in aqueous solution, and we use this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> accordingly. </p>
<p><mathjax>#"Concentration of acid"=(0.624*mol*L^-1xx24.01xx10^-3*L)/(2xx20.00xx10^-3*L)#</mathjax></p>
<p><mathjax>#=0.375*mol*L^-1#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#[H_3PO_4]=0.375*mol*L^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the reaction............</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>Note that the phosphoric acid is A DIACID not a triacid in aqueous solution, and we use this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> accordingly. </p>
<p><mathjax>#"Concentration of acid"=(0.624*mol*L^-1xx24.01xx10^-3*L)/(2xx20.00xx10^-3*L)#</mathjax></p>
<p><mathjax>#=0.375*mol*L^-1#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">20.000 mL aliquots of a phosphoric acid solution with an unknown concentration is titrated with some 0.624 mol/L sodium hydroxide solution.The average volume of the sodium hydroxide added was 24.01 mL. What is the concentration of the phosphoric acid?</h1>
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<div class="markdown"><p><mathjax>#[H_3PO_4]=0.375*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the reaction............</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>Note that the phosphoric acid is A DIACID not a triacid in aqueous solution, and we use this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> accordingly. </p>
<p><mathjax>#"Concentration of acid"=(0.624*mol*L^-1xx24.01xx10^-3*L)/(2xx20.00xx10^-3*L)#</mathjax></p>
<p><mathjax>#=0.375*mol*L^-1#</mathjax></p></div>
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</article> | 20.000 mL aliquots of a phosphoric acid solution with an unknown concentration is titrated with some 0.624 mol/L sodium hydroxide solution.The average volume of the sodium hydroxide added was 24.01 mL. What is the concentration of the phosphoric acid? | null |
1,996 | ad294068-6ddd-11ea-af05-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-copper-l-sulfide | Cu2S | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] copper (l) sulfide [IN] default"}] | [{"type":"chemical equation","value":"Cu2S"}] | [{"type":"substance name","value":"Copper (l) sulfide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for copper (l) sulfide?</h1> | null | Cu2S | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper metal can access two common oxidation states, i.e. <mathjax>#Cu^+#</mathjax> or <mathjax>#Cu^(2+)#</mathjax>: </p>
<p><mathjax>#underbrace(Cu_2S)_"copper(I) sulfide or cuprous sulfide"#</mathjax>;</p>
<p><mathjax>#underbrace(CuS)_"copper(II) sulfide or cupric sulfide"#</mathjax>.</p>
<p>As are most sulfide salts, both salts would be as soluble as bricks in aqueous solution, and are poorly characterizable... </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Copper(I) sulfide"-="cuprous sulfide"-=Cu_2S#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper metal can access two common oxidation states, i.e. <mathjax>#Cu^+#</mathjax> or <mathjax>#Cu^(2+)#</mathjax>: </p>
<p><mathjax>#underbrace(Cu_2S)_"copper(I) sulfide or cuprous sulfide"#</mathjax>;</p>
<p><mathjax>#underbrace(CuS)_"copper(II) sulfide or cupric sulfide"#</mathjax>.</p>
<p>As are most sulfide salts, both salts would be as soluble as bricks in aqueous solution, and are poorly characterizable... </p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for copper (l) sulfide?</h1>
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<div class="markdown"><p><mathjax>#"Copper(I) sulfide"-="cuprous sulfide"-=Cu_2S#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Copper metal can access two common oxidation states, i.e. <mathjax>#Cu^+#</mathjax> or <mathjax>#Cu^(2+)#</mathjax>: </p>
<p><mathjax>#underbrace(Cu_2S)_"copper(I) sulfide or cuprous sulfide"#</mathjax>;</p>
<p><mathjax>#underbrace(CuS)_"copper(II) sulfide or cupric sulfide"#</mathjax>.</p>
<p>As are most sulfide salts, both salts would be as soluble as bricks in aqueous solution, and are poorly characterizable... </p></div>
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</article> | What is the formula for copper (l) sulfide? | null |
1,997 | ab5b36fb-6ddd-11ea-813c-ccda262736ce | https://socratic.org/questions/how-do-you-write-an-equation-to-represent-azomethane-c-2h-6n-2-decomposes-to-for | C2H6N2(g) -> C2H6(g) + N2(g) | start chemical_equation qc_end chemical_equation 9 9 qc_end chemical_equation 14 14 qc_end substance 16 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C2H6N2(g) -> C2H6(g) + N2(g)"}] | [{"type":"chemical equation","value":"C2H6N2"},{"type":"chemical equation","value":"C2H6"},{"type":"substance name","value":"Nitrogen gas"}] | <h1 class="questionTitle" itemprop="name">How do you write an equation to represent "Azomethane (#C_2H_6N_2#) decomposes to form ethane (#C_2H_6#) and nitrogen gas"?</h1> | null | C2H6N2(g) -> C2H6(g) + N2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is make sure that the chemical equation that describes this <strong>decomposition reaction</strong> obeys the <strong><a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">Law of mass conservation</a></strong>. </p>
<p><em>Azomethane</em>, <mathjax>#"C"_2"H"_6"N"_2#</mathjax>, is said to decompose to form <em>ethane</em>, <mathjax>#"C"_2"H"_6#</mathjax>, and <em>nitrogen gas</em>. Let's assume for a second that you don't know the chemical formula of nitrogen gas. </p>
<p>You could write the chemical equation as </p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_(6(g)) + "nitrogen gas"#</mathjax></p>
</blockquote>
<p>Now, the <em>Law of <a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">mass conservation</a></em> states that <strong>all the atoms</strong> present on the reactants' side <strong>must also be present</strong> on the products' side. </p>
<p>Since matter cannot be <em>created</em> or <em>destroyed</em> in a chemical reaction, you know <strong>for a fact</strong> that each atom that took part in the reaction must now be a part of the products. </p>
<p>The reactants' side, which contains azomethane, will have </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
<li><em><strong>two atoms</strong> of nitrogen</em>, <mathjax>#2 xx "N"#</mathjax></li>
</ul>
</blockquote>
<p>This is exactly how many atoms you must end up with once the reaction is finished. As you can see, the reactants' side already contains </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
</ul>
</blockquote>
<p>This means that nitrogen gas <strong>must contain</strong> <mathjax>#2#</mathjax> <strong>atoms</strong> of nitrogen, since you know that two atoms of nitrogen are needed to balance out what's on the reactants' side. </p>
<p>Since you're dealing with <em>gaseous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></em>, the chemical formula for nitrogen gas will be <mathjax>#"N"_2#</mathjax>, i.e. the two nitrogen atoms will be bonded together. </p>
<p>Therefore, the chemical reaction that describes this decomposition reaction will look like this </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_ (6(g)) + "N"_(2(g)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_ (6(g)) + "N"_(2(g))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is make sure that the chemical equation that describes this <strong>decomposition reaction</strong> obeys the <strong><a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">Law of mass conservation</a></strong>. </p>
<p><em>Azomethane</em>, <mathjax>#"C"_2"H"_6"N"_2#</mathjax>, is said to decompose to form <em>ethane</em>, <mathjax>#"C"_2"H"_6#</mathjax>, and <em>nitrogen gas</em>. Let's assume for a second that you don't know the chemical formula of nitrogen gas. </p>
<p>You could write the chemical equation as </p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_(6(g)) + "nitrogen gas"#</mathjax></p>
</blockquote>
<p>Now, the <em>Law of <a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">mass conservation</a></em> states that <strong>all the atoms</strong> present on the reactants' side <strong>must also be present</strong> on the products' side. </p>
<p>Since matter cannot be <em>created</em> or <em>destroyed</em> in a chemical reaction, you know <strong>for a fact</strong> that each atom that took part in the reaction must now be a part of the products. </p>
<p>The reactants' side, which contains azomethane, will have </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
<li><em><strong>two atoms</strong> of nitrogen</em>, <mathjax>#2 xx "N"#</mathjax></li>
</ul>
</blockquote>
<p>This is exactly how many atoms you must end up with once the reaction is finished. As you can see, the reactants' side already contains </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
</ul>
</blockquote>
<p>This means that nitrogen gas <strong>must contain</strong> <mathjax>#2#</mathjax> <strong>atoms</strong> of nitrogen, since you know that two atoms of nitrogen are needed to balance out what's on the reactants' side. </p>
<p>Since you're dealing with <em>gaseous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></em>, the chemical formula for nitrogen gas will be <mathjax>#"N"_2#</mathjax>, i.e. the two nitrogen atoms will be bonded together. </p>
<p>Therefore, the chemical reaction that describes this decomposition reaction will look like this </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_ (6(g)) + "N"_(2(g)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">How do you write an equation to represent "Azomethane (#C_2H_6N_2#) decomposes to form ethane (#C_2H_6#) and nitrogen gas"?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-04-18T11:25:45" itemprop="dateCreated">
Apr 18, 2016
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<div class="markdown"><p><mathjax>#"C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_ (6(g)) + "N"_(2(g))#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is make sure that the chemical equation that describes this <strong>decomposition reaction</strong> obeys the <strong><a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">Law of mass conservation</a></strong>. </p>
<p><em>Azomethane</em>, <mathjax>#"C"_2"H"_6"N"_2#</mathjax>, is said to decompose to form <em>ethane</em>, <mathjax>#"C"_2"H"_6#</mathjax>, and <em>nitrogen gas</em>. Let's assume for a second that you don't know the chemical formula of nitrogen gas. </p>
<p>You could write the chemical equation as </p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_(6(g)) + "nitrogen gas"#</mathjax></p>
</blockquote>
<p>Now, the <em>Law of <a href="https://socratic.org/chemistry/a-closer-look-at-the-atom/mass-conservation">mass conservation</a></em> states that <strong>all the atoms</strong> present on the reactants' side <strong>must also be present</strong> on the products' side. </p>
<p>Since matter cannot be <em>created</em> or <em>destroyed</em> in a chemical reaction, you know <strong>for a fact</strong> that each atom that took part in the reaction must now be a part of the products. </p>
<p>The reactants' side, which contains azomethane, will have </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
<li><em><strong>two atoms</strong> of nitrogen</em>, <mathjax>#2 xx "N"#</mathjax></li>
</ul>
</blockquote>
<p>This is exactly how many atoms you must end up with once the reaction is finished. As you can see, the reactants' side already contains </p>
<blockquote>
<ul>
<li><em><strong>two atoms</strong> of carbon</em>, <mathjax>#2 xx "C"#</mathjax></li>
<li><em><strong>six atoms</strong> of hydrogen</em>, <mathjax>#6 xx "H"#</mathjax></li>
</ul>
</blockquote>
<p>This means that nitrogen gas <strong>must contain</strong> <mathjax>#2#</mathjax> <strong>atoms</strong> of nitrogen, since you know that two atoms of nitrogen are needed to balance out what's on the reactants' side. </p>
<p>Since you're dealing with <em>gaseous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></em>, the chemical formula for nitrogen gas will be <mathjax>#"N"_2#</mathjax>, i.e. the two nitrogen atoms will be bonded together. </p>
<p>Therefore, the chemical reaction that describes this decomposition reaction will look like this </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("C"_ 2"H"_ 6"N"_ (2(g)) -> "C"_ 2"H"_ (6(g)) + "N"_(2(g)))color(white)(a/a)|)))#</mathjax></p>
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</article> | How do you write an equation to represent "Azomethane (#C_2H_6N_2#) decomposes to form ethane (#C_2H_6#) and nitrogen gas"? | null |
1,998 | acb24936-6ddd-11ea-870f-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-of-iron-ill-sulfide | Fe2S3 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] iron(Ill) sulfide [IN] default"}] | [{"type":"chemical equation","value":"Fe2S3"}] | [{"type":"substance name","value":"Iron(Ill) sulfide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula of iron(Ill) sulfide?</h1> | null | Fe2S3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Iron%28III%29_sulfide" rel="nofollow">site.</a> </p>
<p>Sulfur typically has a <mathjax>#-2#</mathjax> oxidation state (it is a Group VI atom, and its redox chemistry follows that of oxygen). And <mathjax>#Fe^(3+)#</mathjax>, ferric ion, has a <mathjax>#+3#</mathjax> oxidation state. We cross multiply and get the given formula. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Fe_2S_3#</mathjax>; i.e. an analogue of ferric oxide, <mathjax>#Fe_2O_3#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Iron%28III%29_sulfide" rel="nofollow">site.</a> </p>
<p>Sulfur typically has a <mathjax>#-2#</mathjax> oxidation state (it is a Group VI atom, and its redox chemistry follows that of oxygen). And <mathjax>#Fe^(3+)#</mathjax>, ferric ion, has a <mathjax>#+3#</mathjax> oxidation state. We cross multiply and get the given formula. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula of iron(Ill) sulfide?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Fe_2S_3#</mathjax>; i.e. an analogue of ferric oxide, <mathjax>#Fe_2O_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>See this <a href="https://en.wikipedia.org/wiki/Iron%28III%29_sulfide" rel="nofollow">site.</a> </p>
<p>Sulfur typically has a <mathjax>#-2#</mathjax> oxidation state (it is a Group VI atom, and its redox chemistry follows that of oxygen). And <mathjax>#Fe^(3+)#</mathjax>, ferric ion, has a <mathjax>#+3#</mathjax> oxidation state. We cross multiply and get the given formula. </p></div>
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</article> | What is the chemical formula of iron(Ill) sulfide? | null |
1,999 | a8afbc52-6ddd-11ea-8a34-ccda262736ce | https://socratic.org/questions/a-liquid-has-a-molecular-weight-of-105-141g-mol-its-density-is-1-173g-ml-what-is | 11.16 M | start physical_unit 19 21 molarity mol/l qc_end physical_unit 0 1 7 8 molecular_weight qc_end physical_unit 0 1 12 13 density qc_end end | [{"type":"physical unit","value":"Molarity [OF] the pure liquid [IN] M"}] | [{"type":"physical unit","value":"11.16 M"}] | [{"type":"physical unit","value":"Molecular weight [OF] a liquid [=] \\pu{105.141 g/mol}"},{"type":"physical unit","value":"Density [OF] a liquid [=] \\pu{1.173 g/mL}"}] | <h1 class="questionTitle" itemprop="name"> A liquid has a molecular weight of 105.141g/mol. Its density is 1.173g/mL. What is the molarity of the pure liquid?</h1> | null | 11.16 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to pick a sample of this compound and try to find how many <strong>moles</strong> it contains <strong>per liter</strong>. </p>
<p>Although <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is usually defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> per <strong>liters</strong> of solution, you can extend this concept to <em>pure liquids</em>, provided that you figure out how many moles you'd get <strong>per liter</strong> of pure liquid. </p>
<p>As you know, a substance's <em>molar mass</em> tells you the mass of <strong>one mole</strong> of that substance. In this case, the liquid is said to have a molar mass of <mathjax>#"105.141 g/mol"#</mathjax>, which means that <strong>one mole</strong> will have a mass of <mathjax>#"105.141 g"#</mathjax>.</p>
<p>To make the calculations easier, let's pick a <mathjax>#"105.141-g"#</mathjax> sample of this liquid, the equivalent of <em>one mole</em>. You know from the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> that <strong>every milliliter</strong> of this liquid will have a mass of <mathjax>#"1.173 g"#</mathjax>.</p>
<p>This means that our sample will occupy a volume of </p>
<blockquote>
<p><mathjax>#105.141 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.173color(red)(cancel(color(black)("g")))) = "89.63427 mL"#</mathjax></p>
</blockquote>
<p>Convert this volume <strong>to liters</strong></p>
<blockquote>
<p><mathjax>#89.63427 color(red)(cancel(color(black)("mL"))) * "1 L"/(1000color(red)(cancel(color(black)("mL")))) = "0.08963427 L"#</mathjax></p>
</blockquote>
<p>Finally, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will be - remember that our sample contains exactly <strong>one mole</strong> of this liquid</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = "1 mole"/"0.08963427 L" = color(green)("11.16 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the density of the liquid. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"11.16 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to pick a sample of this compound and try to find how many <strong>moles</strong> it contains <strong>per liter</strong>. </p>
<p>Although <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is usually defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> per <strong>liters</strong> of solution, you can extend this concept to <em>pure liquids</em>, provided that you figure out how many moles you'd get <strong>per liter</strong> of pure liquid. </p>
<p>As you know, a substance's <em>molar mass</em> tells you the mass of <strong>one mole</strong> of that substance. In this case, the liquid is said to have a molar mass of <mathjax>#"105.141 g/mol"#</mathjax>, which means that <strong>one mole</strong> will have a mass of <mathjax>#"105.141 g"#</mathjax>.</p>
<p>To make the calculations easier, let's pick a <mathjax>#"105.141-g"#</mathjax> sample of this liquid, the equivalent of <em>one mole</em>. You know from the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> that <strong>every milliliter</strong> of this liquid will have a mass of <mathjax>#"1.173 g"#</mathjax>.</p>
<p>This means that our sample will occupy a volume of </p>
<blockquote>
<p><mathjax>#105.141 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.173color(red)(cancel(color(black)("g")))) = "89.63427 mL"#</mathjax></p>
</blockquote>
<p>Convert this volume <strong>to liters</strong></p>
<blockquote>
<p><mathjax>#89.63427 color(red)(cancel(color(black)("mL"))) * "1 L"/(1000color(red)(cancel(color(black)("mL")))) = "0.08963427 L"#</mathjax></p>
</blockquote>
<p>Finally, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will be - remember that our sample contains exactly <strong>one mole</strong> of this liquid</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = "1 mole"/"0.08963427 L" = color(green)("11.16 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the density of the liquid. </p></div>
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<h1 class="questionTitle" itemprop="name"> A liquid has a molecular weight of 105.141g/mol. Its density is 1.173g/mL. What is the molarity of the pure liquid?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-25T01:55:21" itemprop="dateCreated">
Dec 25, 2015
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<div class="markdown"><p><mathjax>#"11.16 M"#</mathjax></p></div>
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<div class="markdown"><p>Your strategy here will be to pick a sample of this compound and try to find how many <strong>moles</strong> it contains <strong>per liter</strong>. </p>
<p>Although <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is usually defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> per <strong>liters</strong> of solution, you can extend this concept to <em>pure liquids</em>, provided that you figure out how many moles you'd get <strong>per liter</strong> of pure liquid. </p>
<p>As you know, a substance's <em>molar mass</em> tells you the mass of <strong>one mole</strong> of that substance. In this case, the liquid is said to have a molar mass of <mathjax>#"105.141 g/mol"#</mathjax>, which means that <strong>one mole</strong> will have a mass of <mathjax>#"105.141 g"#</mathjax>.</p>
<p>To make the calculations easier, let's pick a <mathjax>#"105.141-g"#</mathjax> sample of this liquid, the equivalent of <em>one mole</em>. You know from the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> that <strong>every milliliter</strong> of this liquid will have a mass of <mathjax>#"1.173 g"#</mathjax>.</p>
<p>This means that our sample will occupy a volume of </p>
<blockquote>
<p><mathjax>#105.141 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.173color(red)(cancel(color(black)("g")))) = "89.63427 mL"#</mathjax></p>
</blockquote>
<p>Convert this volume <strong>to liters</strong></p>
<blockquote>
<p><mathjax>#89.63427 color(red)(cancel(color(black)("mL"))) * "1 L"/(1000color(red)(cancel(color(black)("mL")))) = "0.08963427 L"#</mathjax></p>
</blockquote>
<p>Finally, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will be - remember that our sample contains exactly <strong>one mole</strong> of this liquid</p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = "1 mole"/"0.08963427 L" = color(green)("11.16 M")#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the density of the liquid. </p></div>
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</article> | A liquid has a molecular weight of 105.141g/mol. Its density is 1.173g/mL. What is the molarity of the pure liquid? | null |
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