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1,600 | a837a703-6ddd-11ea-bc03-ccda262736ce | https://socratic.org/questions/a-container-is-filled-with-258-liters-of-nitrogen-gas-at-a-temperature-of-78-c-w | 403.54 liters | start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 volume qc_end physical_unit 8 9 14 15 temperature qc_end physical_unit 8 9 31 32 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen gas [IN] liters"}] | [{"type":"physical unit","value":"403.54 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] nitrogen gas [=] \\pu{258 liters}"},{"type":"physical unit","value":"Temperature1 [OF] nitrogen gas [=] \\pu{78 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] nitrogen gas [=] \\pu{122 ℃}"}] | <h1 class="questionTitle" itemprop="name">A container is filled with 258 liters of nitrogen gas at a temperature of 78°C. What would the volume (in liters) of the gas be if the temperature changed to 122°C?</h1> | null | 403.54 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant pressure <mathjax>#Vprop T#</mathjax>,</p>
<p>i.e. <mathjax>#V_1/T_1 =V_2/T_2#</mathjax>, absolute temperatures are used.</p>
<p>So <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#258*Lxx(395*K)/(351*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>As long as the temperature is in degrees Kelvin, I could use any units of volume, litres, cubic metres, pints, gallons, gills. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Use Charles' Law. I got under <mathjax>#300#</mathjax> <mathjax>#L#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant pressure <mathjax>#Vprop T#</mathjax>,</p>
<p>i.e. <mathjax>#V_1/T_1 =V_2/T_2#</mathjax>, absolute temperatures are used.</p>
<p>So <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#258*Lxx(395*K)/(351*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>As long as the temperature is in degrees Kelvin, I could use any units of volume, litres, cubic metres, pints, gallons, gills. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A container is filled with 258 liters of nitrogen gas at a temperature of 78°C. What would the volume (in liters) of the gas be if the temperature changed to 122°C?</h1>
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anor277
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Apr 22, 2016
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<div class="markdown"><p>Use Charles' Law. I got under <mathjax>#300#</mathjax> <mathjax>#L#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant pressure <mathjax>#Vprop T#</mathjax>,</p>
<p>i.e. <mathjax>#V_1/T_1 =V_2/T_2#</mathjax>, absolute temperatures are used.</p>
<p>So <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#258*Lxx(395*K)/(351*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>As long as the temperature is in degrees Kelvin, I could use any units of volume, litres, cubic metres, pints, gallons, gills. </p></div>
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</article> | A container is filled with 258 liters of nitrogen gas at a temperature of 78°C. What would the volume (in liters) of the gas be if the temperature changed to 122°C? | null |
1,601 | acd9a0e6-6ddd-11ea-ae5a-ccda262736ce | https://socratic.org/questions/how-would-you-determine-the-molecular-formula-from-the-following-empirical-formu | N3P3Cl6 | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the substance [IN] molecular"}] | [{"type":"chemical equation","value":"N3P3Cl6"}] | [{"type":"physical unit","value":"Molar mass [OF] the substance [=] \\pu{348 g/mol}"},{"type":"other","value":"Empirical formula of the substance is NPCl2."}] | <h1 class="questionTitle" itemprop="name">How would you determine the molecular formula from the following empirical formula and molar mass: NPCl2, 348 g/mol? </h1> | null | N3P3Cl6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> is <mathjax>#"NPCl"_2#</mathjax>. </p>
<p>As you know, a compound's empirical formula tells you the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up said compound is. </p>
<p>In other words, the empirical formula doesn't tell you <strong>how many moles</strong> of each element the compound's molecule contains, it only tells you what the basic <em>building block</em> of the molecule is. </p>
<p>The <strong>molar mass</strong>, on the other hand, tells you what the mass of <strong>one mole</strong> of that compound is. </p>
<p>This means that the molar mass takes into account the molar mass of <strong>every atom</strong> that's a part of the molecule. So, if you know what the <em>building block</em> looks like, and what the mass of a mole of the compound is, you can use the molar mass of the <em>building block</em> to determine how many of those you need. </p>
<blockquote>
<p><mathjax>#color(blue)("molecular formula" = "empirical formula" xx n)#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, add the molar mass of the atoms that are a part of the empirical formula - do not forget that you have <strong>2 atoms</strong> of chlorine in there!</p>
<blockquote>
<p><mathjax>#M_"M empirical formula" = M_"M N" + M_"M P" + 2 xx M_"M Cl"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "14.00674 g/mol" + "30.9736 g/mol" + 2 xx "35.4527 g/mol"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "115.886 g/mol"#</mathjax></p>
</blockquote>
<p>So, you now know that </p>
<blockquote>
<p><mathjax>#115.886 color(red)(cancel(color(black)("g/mol"))) * color(blue)(n) = 348 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#color(blue)(n)#</mathjax> will be </p>
<blockquote>
<p><mathjax>#color(blue)(n = 348/115.886 = 3.003 ~~ 3#</mathjax></p>
</blockquote>
<p>The <em>molecular formula</em> of the compound will thus be </p>
<blockquote>
<p><mathjax>#("NPCl"_2)_color(blue)(3) ->#</mathjax> <em>hexachlorophosphazene</em></p>
</blockquote>
<p><img alt="http://www.chemtube3d.com/gallery/structurepages/PNCl2_3.html" src="https://useruploads.socratic.org/qG3zMK3TnmpSRRiRnLcw_PNCl2_3.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#("NPCl"_2)_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> is <mathjax>#"NPCl"_2#</mathjax>. </p>
<p>As you know, a compound's empirical formula tells you the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up said compound is. </p>
<p>In other words, the empirical formula doesn't tell you <strong>how many moles</strong> of each element the compound's molecule contains, it only tells you what the basic <em>building block</em> of the molecule is. </p>
<p>The <strong>molar mass</strong>, on the other hand, tells you what the mass of <strong>one mole</strong> of that compound is. </p>
<p>This means that the molar mass takes into account the molar mass of <strong>every atom</strong> that's a part of the molecule. So, if you know what the <em>building block</em> looks like, and what the mass of a mole of the compound is, you can use the molar mass of the <em>building block</em> to determine how many of those you need. </p>
<blockquote>
<p><mathjax>#color(blue)("molecular formula" = "empirical formula" xx n)#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, add the molar mass of the atoms that are a part of the empirical formula - do not forget that you have <strong>2 atoms</strong> of chlorine in there!</p>
<blockquote>
<p><mathjax>#M_"M empirical formula" = M_"M N" + M_"M P" + 2 xx M_"M Cl"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "14.00674 g/mol" + "30.9736 g/mol" + 2 xx "35.4527 g/mol"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "115.886 g/mol"#</mathjax></p>
</blockquote>
<p>So, you now know that </p>
<blockquote>
<p><mathjax>#115.886 color(red)(cancel(color(black)("g/mol"))) * color(blue)(n) = 348 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#color(blue)(n)#</mathjax> will be </p>
<blockquote>
<p><mathjax>#color(blue)(n = 348/115.886 = 3.003 ~~ 3#</mathjax></p>
</blockquote>
<p>The <em>molecular formula</em> of the compound will thus be </p>
<blockquote>
<p><mathjax>#("NPCl"_2)_color(blue)(3) ->#</mathjax> <em>hexachlorophosphazene</em></p>
</blockquote>
<p><img alt="http://www.chemtube3d.com/gallery/structurepages/PNCl2_3.html" src="https://useruploads.socratic.org/qG3zMK3TnmpSRRiRnLcw_PNCl2_3.jpg"/> </p></div>
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<h1 class="questionTitle" itemprop="name">How would you determine the molecular formula from the following empirical formula and molar mass: NPCl2, 348 g/mol? </h1>
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Stefan V.
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Nov 26, 2015
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<div class="markdown"><p><mathjax>#("NPCl"_2)_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, you know that the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> is <mathjax>#"NPCl"_2#</mathjax>. </p>
<p>As you know, a compound's empirical formula tells you the <strong>smallest whole number ratio</strong> that exists between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up said compound is. </p>
<p>In other words, the empirical formula doesn't tell you <strong>how many moles</strong> of each element the compound's molecule contains, it only tells you what the basic <em>building block</em> of the molecule is. </p>
<p>The <strong>molar mass</strong>, on the other hand, tells you what the mass of <strong>one mole</strong> of that compound is. </p>
<p>This means that the molar mass takes into account the molar mass of <strong>every atom</strong> that's a part of the molecule. So, if you know what the <em>building block</em> looks like, and what the mass of a mole of the compound is, you can use the molar mass of the <em>building block</em> to determine how many of those you need. </p>
<blockquote>
<p><mathjax>#color(blue)("molecular formula" = "empirical formula" xx n)#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, add the molar mass of the atoms that are a part of the empirical formula - do not forget that you have <strong>2 atoms</strong> of chlorine in there!</p>
<blockquote>
<p><mathjax>#M_"M empirical formula" = M_"M N" + M_"M P" + 2 xx M_"M Cl"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "14.00674 g/mol" + "30.9736 g/mol" + 2 xx "35.4527 g/mol"#</mathjax></p>
<p><mathjax>#M_"M empirical formula" = "115.886 g/mol"#</mathjax></p>
</blockquote>
<p>So, you now know that </p>
<blockquote>
<p><mathjax>#115.886 color(red)(cancel(color(black)("g/mol"))) * color(blue)(n) = 348 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#color(blue)(n)#</mathjax> will be </p>
<blockquote>
<p><mathjax>#color(blue)(n = 348/115.886 = 3.003 ~~ 3#</mathjax></p>
</blockquote>
<p>The <em>molecular formula</em> of the compound will thus be </p>
<blockquote>
<p><mathjax>#("NPCl"_2)_color(blue)(3) ->#</mathjax> <em>hexachlorophosphazene</em></p>
</blockquote>
<p><img alt="http://www.chemtube3d.com/gallery/structurepages/PNCl2_3.html" src="https://useruploads.socratic.org/qG3zMK3TnmpSRRiRnLcw_PNCl2_3.jpg"/> </p></div>
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</article> | How would you determine the molecular formula from the following empirical formula and molar mass: NPCl2, 348 g/mol? | null |
1,602 | abef1bf1-6ddd-11ea-a7a5-ccda262736ce | https://socratic.org/questions/how-many-moles-of-kcl-are-produced-from-2-50-moles-of-k-and-excess-cl-2 | 2.50 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 8 9 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] KCl [IN] moles"}] | [{"type":"physical unit","value":"2.50 moles"}] | [{"type":"physical unit","value":"Mole [OF] K [=] \\pu{2.50 moles}"},{"type":"other","value":"Excess Cl2."}] | <h1 class="questionTitle" itemprop="name">How many moles of #KCl# are produced from 2.50 moles of #K# and excess #Cl_2#?</h1> | null | 2.50 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction is<br/>
<mathjax>#2K + Cl_2 = 2KCl#</mathjax><br/>
Thus the number of moles of potassium and potassium chloride is same</p></div>
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<div class="markdown"><p><mathjax>#2.5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction is<br/>
<mathjax>#2K + Cl_2 = 2KCl#</mathjax><br/>
Thus the number of moles of potassium and potassium chloride is same</p></div>
</div>
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<div class="markdown"><p><mathjax>#2.5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The reaction is<br/>
<mathjax>#2K + Cl_2 = 2KCl#</mathjax><br/>
Thus the number of moles of potassium and potassium chloride is same</p></div>
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</article> | How many moles of #KCl# are produced from 2.50 moles of #K# and excess #Cl_2#? | null |
1,603 | aaf3983f-6ddd-11ea-81d9-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-silver-acetate | CH3COOAg | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] silver acetate [IN] default"}] | [{"type":"chemical equation","value":"CH3COOAg"}] | [{"type":"substance name","value":"Silver acetate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for silver acetate?</h1> | null | CH3COOAg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Acetate salts are routinely soluble.</p></div>
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<div class="markdown"><p><mathjax>#H_3C-C(=O)O^(-)Ag^+#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Acetate salts are routinely soluble.</p></div>
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<div class="markdown"><p><mathjax>#H_3C-C(=O)O^(-)Ag^+#</mathjax>.</p></div>
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<div class="markdown"><p>Acetate salts are routinely soluble.</p></div>
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</article> | What is the formula for silver acetate? | null |
1,604 | a9fddca8-6ddd-11ea-82e1-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-that-is-made-of-1-85-moles-of-nit | N2O5 | start chemical_formula qc_end physical_unit 15 15 12 13 mole qc_end physical_unit 20 20 17 18 mole qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"N2O5"}] | [{"type":"physical unit","value":"Mole [OF] nitrogen [=] \\pu{1.85 moles}"},{"type":"physical unit","value":"Mole [OF] oxygen [=] \\pu{4.63 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound that is made of 1.85 moles of nitrogen and 4.63 moles of oxygen? </h1> | null | N2O5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Relative number of moles of nitrogen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.85*mol)/(1.85*mol)=1#</mathjax>.</p>
<p><mathjax>#"Relative number of moles of oxygen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(4.63*mol)/(1.85*mol)=2.5#</mathjax>.</p>
<p>Because, BY DEFINITION, <mathjax>#"the empiricial formula is the simplest"#</mathjax> <mathjax>#"whole number ratio that defines constituent atoms in a species,"#</mathjax> we double this ratio to get <mathjax>#N_2O_5#</mathjax>. Capisce?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Normalize the formula by dividing thru by the LOWEST molar quantity. We get <mathjax>#N_2O_5#</mathjax> finally. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Relative number of moles of nitrogen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.85*mol)/(1.85*mol)=1#</mathjax>.</p>
<p><mathjax>#"Relative number of moles of oxygen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(4.63*mol)/(1.85*mol)=2.5#</mathjax>.</p>
<p>Because, BY DEFINITION, <mathjax>#"the empiricial formula is the simplest"#</mathjax> <mathjax>#"whole number ratio that defines constituent atoms in a species,"#</mathjax> we double this ratio to get <mathjax>#N_2O_5#</mathjax>. Capisce?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound that is made of 1.85 moles of nitrogen and 4.63 moles of oxygen? </h1>
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<div class="markdown"><p>Normalize the formula by dividing thru by the LOWEST molar quantity. We get <mathjax>#N_2O_5#</mathjax> finally. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Relative number of moles of nitrogen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.85*mol)/(1.85*mol)=1#</mathjax>.</p>
<p><mathjax>#"Relative number of moles of oxygen:"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(4.63*mol)/(1.85*mol)=2.5#</mathjax>.</p>
<p>Because, BY DEFINITION, <mathjax>#"the empiricial formula is the simplest"#</mathjax> <mathjax>#"whole number ratio that defines constituent atoms in a species,"#</mathjax> we double this ratio to get <mathjax>#N_2O_5#</mathjax>. Capisce?</p></div>
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</article> | What is the empirical formula for the compound that is made of 1.85 moles of nitrogen and 4.63 moles of oxygen? | null |
1,605 | aada5f6e-6ddd-11ea-b689-ccda262736ce | https://socratic.org/questions/suppose-0-440-kg-of-ethane-are-burned-in-air-at-a-pressure-of-exactly-1-atm-and- | 690 L | start physical_unit 27 29 volume l qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 4 4 14 15 pressure qc_end physical_unit 4 4 20 21 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon dioxide gas [IN] L"}] | [{"type":"physical unit","value":"690 L"}] | [{"type":"physical unit","value":"Mass [OF] ethane [=] \\pu{0.440 kg}"},{"type":"physical unit","value":"Pressure [OF] ethane [=] \\pu{1 atm}"},{"type":"physical unit","value":"Temperature [OF] ethane [=] \\pu{13.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">Suppose 0.440 kg of ethane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. What is the volume of carbon dioxide gas that is produced?</h1> | null | 690 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> color(blue)(2)"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that every <mathjax>#1#</mathjax> <strong>mole</strong> of ethane that undergoes combustion produces <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon dioxide, so start by calculating the number of moles of ethane present in your sample. </p>
<p>To do that, use the <strong>molar mass</strong> of ethane</p>
<blockquote>
<p><mathjax>#0.440 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "14.63 moles C"_2"H"_6#</mathjax></p>
</blockquote>
<p>This means that the reaction will produce</p>
<blockquote>
<p><mathjax>#14.63 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(blue)(2)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_6)))) = "29.26 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that the carbon dioxide is collected at a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of</p>
<blockquote>
<p><mathjax>#13.0^@"C" = 13.0^@"C" + 273.15 = "286.15 K"#</mathjax></p>
</blockquote>
<p>At this point, you can calculate the volume of carbon dioxide produced by the reaction by using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (29.26 color(red)(cancel(color(black)("moles"))) * 0/0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 286.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(darkgreen)(ul(color(black)("690 L")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the pressure of the gas, which implies that the answer <em>should</em> be <mathjax>#"700 L"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"690 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> color(blue)(2)"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that every <mathjax>#1#</mathjax> <strong>mole</strong> of ethane that undergoes combustion produces <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon dioxide, so start by calculating the number of moles of ethane present in your sample. </p>
<p>To do that, use the <strong>molar mass</strong> of ethane</p>
<blockquote>
<p><mathjax>#0.440 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "14.63 moles C"_2"H"_6#</mathjax></p>
</blockquote>
<p>This means that the reaction will produce</p>
<blockquote>
<p><mathjax>#14.63 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(blue)(2)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_6)))) = "29.26 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that the carbon dioxide is collected at a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of</p>
<blockquote>
<p><mathjax>#13.0^@"C" = 13.0^@"C" + 273.15 = "286.15 K"#</mathjax></p>
</blockquote>
<p>At this point, you can calculate the volume of carbon dioxide produced by the reaction by using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (29.26 color(red)(cancel(color(black)("moles"))) * 0/0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 286.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(darkgreen)(ul(color(black)("690 L")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the pressure of the gas, which implies that the answer <em>should</em> be <mathjax>#"700 L"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Suppose 0.440 kg of ethane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. What is the volume of carbon dioxide gas that is produced?</h1>
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Stefan V.
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Jun 10, 2017
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<div class="markdown"><p><mathjax>#"690 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation that describes this combustion reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (6(g)) + 7/2"O"_ (2(g)) -> color(blue)(2)"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that every <mathjax>#1#</mathjax> <strong>mole</strong> of ethane that undergoes combustion produces <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon dioxide, so start by calculating the number of moles of ethane present in your sample. </p>
<p>To do that, use the <strong>molar mass</strong> of ethane</p>
<blockquote>
<p><mathjax>#0.440 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "14.63 moles C"_2"H"_6#</mathjax></p>
</blockquote>
<p>This means that the reaction will produce</p>
<blockquote>
<p><mathjax>#14.63 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(blue)(2)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_6)))) = "29.26 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that the carbon dioxide is collected at a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of</p>
<blockquote>
<p><mathjax>#13.0^@"C" = 13.0^@"C" + 273.15 = "286.15 K"#</mathjax></p>
</blockquote>
<p>At this point, you can calculate the volume of carbon dioxide produced by the reaction by using the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies V = (nRT)/P#</mathjax></p>
</blockquote>
<p>and plug in your values to find</p>
<blockquote>
<p><mathjax>#V = (29.26 color(red)(cancel(color(black)("moles"))) * 0/0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 286.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#V = color(darkgreen)(ul(color(black)("690 L")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the pressure of the gas, which implies that the answer <em>should</em> be <mathjax>#"700 L"#</mathjax>.</p></div>
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</article> | Suppose 0.440 kg of ethane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. What is the volume of carbon dioxide gas that is produced? | null |
1,606 | aad42710-6ddd-11ea-82e8-ccda262736ce | https://socratic.org/questions/at-25-c-3-m-of-a-gas-occupy-a-volume-of-4-0-l-at-1-9-kpa-if-the-volume-decreases | 3.8 kPa | start physical_unit 7 7 pressure kpa qc_end physical_unit 7 7 12 13 volume qc_end physical_unit 7 7 15 16 pressure qc_end physical_unit 7 7 1 2 temperature qc_end physical_unit 7 7 3 4 molarity qc_end physical_unit 7 7 22 23 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] kPa"}] | [{"type":"physical unit","value":"3.8 kPa"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{4.0 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.9 kPa}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Molarity [OF] the gas [=] \\pu{3 M}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{2.0 L}"}] | <h1 class="questionTitle" itemprop="name">At 25°C, 3 M of a gas occupy a volume of 4.0 L at 1.9 kPa. If the volume decreases to 2.0 L, what will the new pressure be?</h1> | null | 3.8 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that the <em>temperature</em> and <em>number of moles</em> of gas are <strong>kept constant</strong>, which should let you know that you can use the equation for <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to find the new pressure of the gas. </p>
<p><em><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></em> states that pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant. </p>
<blockquote>
<blockquote>
<p><img alt="http://www.daviddarling.info/encyclopedia/B/Boyles_law.html" src="https://useruploads.socratic.org/TSkNrUlwTHakASrFdvhD_Boyles_law.jpg"/> </p>
</blockquote>
</blockquote>
<p>In simple terms, when pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This happens because as you <em>increase</em> the volume of the gas, you essentially increase the space in which the molecules are moving. Since their <em>average kinetic speed</em> is <strong>constant</strong>, i.e. since temperature is constant, the molecules will hit the walls of the container <strong>less frequently</strong> than they did in the <em>smaller volume</em> <mathjax>#->#</mathjax> pressure will <strong>decrease</strong>. </p>
<p>In your case, the volume of the gas <em>decreased</em> to half of its original value, which means that you can expect the pressure to <strong>double</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of a gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange the equation to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to get</p>
<blockquote>
<p><mathjax>#P_2 = (4.0 color(red)(cancel(color(black)("L"))))/(2.0color(red)(cancel(color(black)("L")))) * "1.9 kPa" = color(green)(|bar(ul(color(white)(a/a)"3.8 kPa"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.8 kPa"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that the <em>temperature</em> and <em>number of moles</em> of gas are <strong>kept constant</strong>, which should let you know that you can use the equation for <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to find the new pressure of the gas. </p>
<p><em><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></em> states that pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant. </p>
<blockquote>
<blockquote>
<p><img alt="http://www.daviddarling.info/encyclopedia/B/Boyles_law.html" src="https://useruploads.socratic.org/TSkNrUlwTHakASrFdvhD_Boyles_law.jpg"/> </p>
</blockquote>
</blockquote>
<p>In simple terms, when pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This happens because as you <em>increase</em> the volume of the gas, you essentially increase the space in which the molecules are moving. Since their <em>average kinetic speed</em> is <strong>constant</strong>, i.e. since temperature is constant, the molecules will hit the walls of the container <strong>less frequently</strong> than they did in the <em>smaller volume</em> <mathjax>#->#</mathjax> pressure will <strong>decrease</strong>. </p>
<p>In your case, the volume of the gas <em>decreased</em> to half of its original value, which means that you can expect the pressure to <strong>double</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of a gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange the equation to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to get</p>
<blockquote>
<p><mathjax>#P_2 = (4.0 color(red)(cancel(color(black)("L"))))/(2.0color(red)(cancel(color(black)("L")))) * "1.9 kPa" = color(green)(|bar(ul(color(white)(a/a)"3.8 kPa"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">At 25°C, 3 M of a gas occupy a volume of 4.0 L at 1.9 kPa. If the volume decreases to 2.0 L, what will the new pressure be?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-08T13:48:52" itemprop="dateCreated">
Mar 8, 2016
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<div class="markdown"><p><mathjax>#"3.8 kPa"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to realize here is that the <em>temperature</em> and <em>number of moles</em> of gas are <strong>kept constant</strong>, which should let you know that you can use the equation for <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to find the new pressure of the gas. </p>
<p><em><a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></em> states that pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant. </p>
<blockquote>
<blockquote>
<p><img alt="http://www.daviddarling.info/encyclopedia/B/Boyles_law.html" src="https://useruploads.socratic.org/TSkNrUlwTHakASrFdvhD_Boyles_law.jpg"/> </p>
</blockquote>
</blockquote>
<p>In simple terms, when pressure <strong>increases</strong>, volume <strong>decreases</strong>, and when pressure <strong>decreases</strong>, volume <strong>increases</strong>. </p>
<p>This happens because as you <em>increase</em> the volume of the gas, you essentially increase the space in which the molecules are moving. Since their <em>average kinetic speed</em> is <strong>constant</strong>, i.e. since temperature is constant, the molecules will hit the walls of the container <strong>less frequently</strong> than they did in the <em>smaller volume</em> <mathjax>#->#</mathjax> pressure will <strong>decrease</strong>. </p>
<p>In your case, the volume of the gas <em>decreased</em> to half of its original value, which means that you can expect the pressure to <strong>double</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume of a gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume of the gas at a final state</p>
<p>Rearrange the equation to solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies P_2 = V_1/V_2 * P_1#</mathjax></p>
</blockquote>
<p>and plug in your values to get</p>
<blockquote>
<p><mathjax>#P_2 = (4.0 color(red)(cancel(color(black)("L"))))/(2.0color(red)(cancel(color(black)("L")))) * "1.9 kPa" = color(green)(|bar(ul(color(white)(a/a)"3.8 kPa"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | At 25°C, 3 M of a gas occupy a volume of 4.0 L at 1.9 kPa. If the volume decreases to 2.0 L, what will the new pressure be? | null |
1,607 | abf9a66d-6ddd-11ea-8b48-ccda262736ce | https://socratic.org/questions/59b94625b72cff7c56dce30d | 3.23 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 2 2 8 9 mole qc_end end | [{"type":"physical unit","value":"Number [OF] gold atoms"}] | [{"type":"physical unit","value":"3.23 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] gold [=] \\pu{5.37 mol}"}] | <h1 class="questionTitle" itemprop="name">How many gold atoms are there in a #5.37*mol# quantity of the metal?</h1> | null | 3.23 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How many? Well, <mathjax>#5.37*molxxN_A.....#</mathjax> where <mathjax>#N_A="Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we takes the product....</p>
<p><mathjax>#5.37*molxx6.022xx10^23*mol^-1=3.23xx10^24#</mathjax> individual gold atoms; pretty vast but not infinite.</p>
<p>The current price of gold is <mathjax>#$1,323.86*USD*oz^-1#</mathjax>. Given that <mathjax>#1*oz-=28.3*g#</mathjax>, how much is this molar quantity of gold worth? </p></div>
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<div class="markdown"><p>How many? Well, <mathjax>#5.37*molxxN_A.....#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How many? Well, <mathjax>#5.37*molxxN_A.....#</mathjax> where <mathjax>#N_A="Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we takes the product....</p>
<p><mathjax>#5.37*molxx6.022xx10^23*mol^-1=3.23xx10^24#</mathjax> individual gold atoms; pretty vast but not infinite.</p>
<p>The current price of gold is <mathjax>#$1,323.86*USD*oz^-1#</mathjax>. Given that <mathjax>#1*oz-=28.3*g#</mathjax>, how much is this molar quantity of gold worth? </p></div>
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<h1 class="questionTitle" itemprop="name">How many gold atoms are there in a #5.37*mol# quantity of the metal?</h1>
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<div class="markdown"><p>How many? Well, <mathjax>#5.37*molxxN_A.....#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>How many? Well, <mathjax>#5.37*molxxN_A.....#</mathjax> where <mathjax>#N_A="Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we takes the product....</p>
<p><mathjax>#5.37*molxx6.022xx10^23*mol^-1=3.23xx10^24#</mathjax> individual gold atoms; pretty vast but not infinite.</p>
<p>The current price of gold is <mathjax>#$1,323.86*USD*oz^-1#</mathjax>. Given that <mathjax>#1*oz-=28.3*g#</mathjax>, how much is this molar quantity of gold worth? </p></div>
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</article> | How many gold atoms are there in a #5.37*mol# quantity of the metal? | null |
1,608 | abda6a29-6ddd-11ea-82b4-ccda262736ce | https://socratic.org/questions/how-do-you-balance-na-c-l-rightarrow-n-a-c-l-2 | 2 NaCl -> 2 Na + Cl2 | 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 NaCl -> 2 Na + Cl2"}] | [{"type":"chemical equation","value":"NaCl -> Na + Cl2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Na C l \rightarrow N a + C l _ { 2}#?</h1> | null | 2 NaCl -> 2 Na + Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NaCl → Na + Cl_2#</mathjax></p>
<p>To balance this equation, we make the number of each element the same on both sides.</p>
<p>So Left Hand Side<br/>
Na= 1<br/>
Cl=1</p>
<p>And Right Hand Side<br/>
Na=1 <br/>
Cl=2</p>
<p>We balance the Cl first by placing a 2 in front of the NaCl<br/>
<mathjax>#2NaCl → Na + Cl_2#</mathjax></p>
<p>We now have<br/>
LHS<br/>
Na=2 Cl= 2</p>
<p>RHS<br/>
Na=1 Cl=2</p>
<p>So now we need to balance the Na by placing a 2 in front of it<br/>
<mathjax>#2NaCl → 2Na + Cl_2#</mathjax></p>
<p>LHS <br/>
Na=2 Cl=2</p>
<p>RHS<br/>
Na=2 Cl=2</p>
<p>Success!</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#2NaCl → 2Na + Cl_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NaCl → Na + Cl_2#</mathjax></p>
<p>To balance this equation, we make the number of each element the same on both sides.</p>
<p>So Left Hand Side<br/>
Na= 1<br/>
Cl=1</p>
<p>And Right Hand Side<br/>
Na=1 <br/>
Cl=2</p>
<p>We balance the Cl first by placing a 2 in front of the NaCl<br/>
<mathjax>#2NaCl → Na + Cl_2#</mathjax></p>
<p>We now have<br/>
LHS<br/>
Na=2 Cl= 2</p>
<p>RHS<br/>
Na=1 Cl=2</p>
<p>So now we need to balance the Na by placing a 2 in front of it<br/>
<mathjax>#2NaCl → 2Na + Cl_2#</mathjax></p>
<p>LHS <br/>
Na=2 Cl=2</p>
<p>RHS<br/>
Na=2 Cl=2</p>
<p>Success!</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Na C l \rightarrow N a + C l _ { 2}#?</h1>
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<div class="markdown"><p><mathjax>#2NaCl → 2Na + Cl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NaCl → Na + Cl_2#</mathjax></p>
<p>To balance this equation, we make the number of each element the same on both sides.</p>
<p>So Left Hand Side<br/>
Na= 1<br/>
Cl=1</p>
<p>And Right Hand Side<br/>
Na=1 <br/>
Cl=2</p>
<p>We balance the Cl first by placing a 2 in front of the NaCl<br/>
<mathjax>#2NaCl → Na + Cl_2#</mathjax></p>
<p>We now have<br/>
LHS<br/>
Na=2 Cl= 2</p>
<p>RHS<br/>
Na=1 Cl=2</p>
<p>So now we need to balance the Na by placing a 2 in front of it<br/>
<mathjax>#2NaCl → 2Na + Cl_2#</mathjax></p>
<p>LHS <br/>
Na=2 Cl=2</p>
<p>RHS<br/>
Na=2 Cl=2</p>
<p>Success!</p></div>
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</article> | How do you balance #Na C l \rightarrow N a + C l _ { 2}#? | null |
1,609 | a8982e50-6ddd-11ea-b9dc-ccda262736ce | https://socratic.org/questions/how-many-moles-of-oxygen-are-lost-in-the-following-problem | 0.2 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 24 25 mole qc_end physical_unit 30 31 39 40 pressure qc_end c_other OTHER qc_end physical_unit 30 31 57 58 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"losted mole3 [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"0.2 moles"}] | [{"type":"physical unit","value":"Volume1 [OF] oxygen in cylinder with a moveable piston [=] \\pu{12.7 L}"},{"type":"physical unit","value":"Mole1 [OF] oxygen [=] \\pu{3.5 mol}"},{"type":"physical unit","value":"pressure1 [OF] the gas [=] \\pu{5.83 atm}"},{"type":"other","value":"the cylinder develops a leak"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{12.1 L}"},{"type":"other","value":"at same pressure"}] | <h1 class="questionTitle" itemprop="name">How many moles of oxygen are lost in the following problem?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>A cylinder with a moveable piston records a volume of 12.7 L when 3.5 mol of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak, and the volume of the gas is now recorded to be 12.1 L at the same pressure. </p></div>
</h2>
</div>
</div> | 0.2 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyles Law to find the volume of Oxygen lost at standard pressure of 1 atm. </p>
<p><mathjax># P_1 xx V_1 = P_2 xx V_2#</mathjax> </p>
<p><mathjax>#P_1 = 5.83 #</mathjax><br/>
<mathjax>#V_1 = 0.6 #</mathjax> <br/>
<mathjax>#P_2 = 1.00#</mathjax><br/>
<mathjax>#V_2#</mathjax> = Volume of Oxygen lost. </p>
<p>Solving the problem to find the Volume of Oxygen lost </p>
<p><mathjax># 5.83 xx 0.6 = 1 xx V_2#</mathjax> </p>
<p>3.50 liters = <mathjax>#V_2#</mathjax> </p>
<p>22.4 liters = 1 mole at STP for an ideal gas. </p>
<p>Assuming Oxygen is close to an ideal gas </p>
<p><mathjax># 3.50/22.4#</mathjax> = moles Oxygen. </p>
<p>0.156 = moles of Oxygen. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>.16 moles of Oxygen assuming the temperature is at standard temperature. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyles Law to find the volume of Oxygen lost at standard pressure of 1 atm. </p>
<p><mathjax># P_1 xx V_1 = P_2 xx V_2#</mathjax> </p>
<p><mathjax>#P_1 = 5.83 #</mathjax><br/>
<mathjax>#V_1 = 0.6 #</mathjax> <br/>
<mathjax>#P_2 = 1.00#</mathjax><br/>
<mathjax>#V_2#</mathjax> = Volume of Oxygen lost. </p>
<p>Solving the problem to find the Volume of Oxygen lost </p>
<p><mathjax># 5.83 xx 0.6 = 1 xx V_2#</mathjax> </p>
<p>3.50 liters = <mathjax>#V_2#</mathjax> </p>
<p>22.4 liters = 1 mole at STP for an ideal gas. </p>
<p>Assuming Oxygen is close to an ideal gas </p>
<p><mathjax># 3.50/22.4#</mathjax> = moles Oxygen. </p>
<p>0.156 = moles of Oxygen. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of oxygen are lost in the following problem?</h1>
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<div class="markdown"><p>A cylinder with a moveable piston records a volume of 12.7 L when 3.5 mol of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak, and the volume of the gas is now recorded to be 12.1 L at the same pressure. </p></div>
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<div class="markdown"><p>.16 moles of Oxygen assuming the temperature is at standard temperature. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyles Law to find the volume of Oxygen lost at standard pressure of 1 atm. </p>
<p><mathjax># P_1 xx V_1 = P_2 xx V_2#</mathjax> </p>
<p><mathjax>#P_1 = 5.83 #</mathjax><br/>
<mathjax>#V_1 = 0.6 #</mathjax> <br/>
<mathjax>#P_2 = 1.00#</mathjax><br/>
<mathjax>#V_2#</mathjax> = Volume of Oxygen lost. </p>
<p>Solving the problem to find the Volume of Oxygen lost </p>
<p><mathjax># 5.83 xx 0.6 = 1 xx V_2#</mathjax> </p>
<p>3.50 liters = <mathjax>#V_2#</mathjax> </p>
<p>22.4 liters = 1 mole at STP for an ideal gas. </p>
<p>Assuming Oxygen is close to an ideal gas </p>
<p><mathjax># 3.50/22.4#</mathjax> = moles Oxygen. </p>
<p>0.156 = moles of Oxygen. </p></div>
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<div class="markdown"><p>The cylinder has lost 0.2 mol of oxygen.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The pressure and temperature are constant.</p>
<p>Only the volume and the number of moles are changing.</p>
<p>The appropriate relationship is therefore <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/avogadro-s-law">Avogadro's law</a></strong>: the volume of a gas is directly proportional to the number of moles, if all other variables are held constant.</p>
<p>In symbols,</p>
<p><mathjax>#V ∝ n#</mathjax></p>
<p><mathjax>#V = kn#</mathjax> or <mathjax>#V/n = k#</mathjax></p>
<p>This leads to the relation</p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) V_1/n_1 = V_2/n_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#V_1 = "12.7 L"; n_1 = "3.5 mol"#</mathjax><br/>
<mathjax>#V_2 = "12.1 L"; n_2 = "?"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#n_2 = n_1 × V_2/V_1 = "3.5 mol" × (12.1 color(red)(cancel(color(black)("L"))))/(12.7 color(red)(cancel(color(black)("L")))) = "3.33 mol"#</mathjax></p>
<p><mathjax>#Δn = n_2 - n_1 = "3.33 mol - 3.5 mol" = "-0.2 mol"#</mathjax> (1 significant figure)</p></div>
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</article> | How many moles of oxygen are lost in the following problem? |
A cylinder with a moveable piston records a volume of 12.7 L when 3.5 mol of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak, and the volume of the gas is now recorded to be 12.1 L at the same pressure.
|
1,610 | aa717522-6ddd-11ea-85a0-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-2-00-moles-of-chlorine-cl-2-at-stp-to-the-nearest-tenth-of | 44.8 liters | start physical_unit 9 9 volume l qc_end physical_unit 9 9 5 6 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] Cl2 [IN] liters"}] | [{"type":"physical unit","value":"44.8 liters"}] | [{"type":"physical unit","value":"Mole [OF] Cl2 [=] \\pu{2.00 moles}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 2.00 moles of chlorine (#Cl_2#) at STP, to the nearest tenth of a liter?</h1> | null | 44.8 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, the molar volume of all gases is <mathjax>#22.4dm^3//mol#</mathjax>.</p>
<p>Hence <mathjax>#2mol#</mathjax> of <mathjax>#Cl_2#</mathjax> at STP occupies a volume of <mathjax>#44.8dm^3#</mathjax>.</p>
<p>To convert this to litres, we use conversion factors :</p>
<p><mathjax>#1L=1000cm^3=1dm^3#</mathjax> </p>
<p>Hence <mathjax>#44.8dm^3=44.8L#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#44.8#</mathjax> litres</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, the molar volume of all gases is <mathjax>#22.4dm^3//mol#</mathjax>.</p>
<p>Hence <mathjax>#2mol#</mathjax> of <mathjax>#Cl_2#</mathjax> at STP occupies a volume of <mathjax>#44.8dm^3#</mathjax>.</p>
<p>To convert this to litres, we use conversion factors :</p>
<p><mathjax>#1L=1000cm^3=1dm^3#</mathjax> </p>
<p>Hence <mathjax>#44.8dm^3=44.8L#</mathjax> </p></div>
</div>
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Trevor Ryan.
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<span class="dateCreated" datetime="2016-01-28T22:03:47" itemprop="dateCreated">
Jan 28, 2016
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<div class="markdown"><p><mathjax>#44.8#</mathjax> litres</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, the molar volume of all gases is <mathjax>#22.4dm^3//mol#</mathjax>.</p>
<p>Hence <mathjax>#2mol#</mathjax> of <mathjax>#Cl_2#</mathjax> at STP occupies a volume of <mathjax>#44.8dm^3#</mathjax>.</p>
<p>To convert this to litres, we use conversion factors :</p>
<p><mathjax>#1L=1000cm^3=1dm^3#</mathjax> </p>
<p>Hence <mathjax>#44.8dm^3=44.8L#</mathjax> </p></div>
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</article> | What is the volume of 2.00 moles of chlorine (#Cl_2#) at STP, to the nearest tenth of a liter? | null |
1,611 | a893c042-6ddd-11ea-897a-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-pressure-in-atmospheres-of-1-00-mol-of-argon-in-a-500-l | 50 atmospheres | start physical_unit 12 12 pressure atm qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 17 17 15 16 volume qc_end physical_unit 12 12 19 20 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] argon [IN] atmospheres"}] | [{"type":"physical unit","value":"50 atmospheres"}] | [{"type":"physical unit","value":"Mole [OF] argon [=] \\pu{1.00 mol}"},{"type":"physical unit","value":"Volume [OF] container [=] \\pu{500 L}"},{"type":"physical unit","value":"Temperature [OF] argon [=] \\pu{29.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the pressure in atmospheres of 1.00 mol of argon in a .500-L container at 29.0°C?</h1> | null | 50 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation.........<mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#(1.00*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)*cancel(mol^-1)xx302*cancelK)/(0.500*cancelL)=??*atm#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#P~=50*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation.........<mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#(1.00*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)*cancel(mol^-1)xx302*cancelK)/(0.500*cancelL)=??*atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the pressure in atmospheres of 1.00 mol of argon in a .500-L container at 29.0°C?</h1>
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<div class="markdown"><p><mathjax>#P~=50*atm#</mathjax></p></div>
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<div class="markdown"><p>We use the Ideal Gas Equation.........<mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#(1.00*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)*cancel(mol^-1)xx302*cancelK)/(0.500*cancelL)=??*atm#</mathjax></p></div>
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</article> | How do you calculate the pressure in atmospheres of 1.00 mol of argon in a .500-L container at 29.0°C? | null |
1,612 | a83d7341-6ddd-11ea-b037-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-tin-iv-nitride | Sn3N4 | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] tin (IV) nitride [IN] default"}] | [{"type":"chemical equation","value":"Sn3N4"}] | [{"type":"substance name","value":"Tin (IV) nitride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for tin (IV) nitride?</h1> | null | Sn3N4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Stannic nitride is composed of stannic ion, <mathjax>#Sn^(4+)#</mathjax>, and nitride ion, <mathjax>#N^(3-)#</mathjax>. Cross multiplication of ions (<mathjax>#3xxSn^(4+)+4xxN^(-3)#</mathjax>) gives a neutral salt. </p></div>
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<div class="markdown"><p><mathjax>#"Stannic nitride, "Sn_3N_4#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Stannic nitride is composed of stannic ion, <mathjax>#Sn^(4+)#</mathjax>, and nitride ion, <mathjax>#N^(3-)#</mathjax>. Cross multiplication of ions (<mathjax>#3xxSn^(4+)+4xxN^(-3)#</mathjax>) gives a neutral salt. </p></div>
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<div class="markdown"><p><mathjax>#"Stannic nitride, "Sn_3N_4#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Stannic nitride is composed of stannic ion, <mathjax>#Sn^(4+)#</mathjax>, and nitride ion, <mathjax>#N^(3-)#</mathjax>. Cross multiplication of ions (<mathjax>#3xxSn^(4+)+4xxN^(-3)#</mathjax>) gives a neutral salt. </p></div>
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</article> | What is the formula for tin (IV) nitride? | null |
1,613 | ac42c124-6ddd-11ea-842c-ccda262736ce | https://socratic.org/questions/if-you-wish-to-warm-100-kg-of-water-by-20-degrees-c-for-your-bath-how-much-heat- | 8.37 × 10^6 J | start physical_unit 8 8 heat_energy j qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 8 8 10 12 temperature qc_end end | [{"type":"physical unit","value":"Required heat [OF] water [IN] J"}] | [{"type":"physical unit","value":"8.37 × 10^6 J"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 kg}"},{"type":"physical unit","value":"Raised temperature [OF] water [=] \\pu{20 degrees C}"}] | <h1 class="questionTitle" itemprop="name">If you wish to warm 100 kg of water by 20 degrees C for your bath how much heat is required?</h1> | null | 8.37 × 10^6 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat capacity of water is 1 calorie for 1 gram raised 1 degree C<br/>
or 4.186 joules of energy for 1 gram raised 1 degree C ( or <mathjax>#K^o#</mathjax>)</p>
<p>l Kg = 1000 grams so </p>
<p><mathjax># 100 Kg xx 1000 g /1 #</mathjax> = 100000 grams of water. </p>
<p>It takes 10 000 calories to raise 10 000 grams of water 1 degree</p>
<p>so 20 degrees x 10 000 grams = 2 000 000 calories. </p>
<p>The same type calculations can be done for heat in joules </p>
<p>10, 000 grams x 4.186 joules x 20 degrees. = 8372000 joules.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>2000000 heat calories. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat capacity of water is 1 calorie for 1 gram raised 1 degree C<br/>
or 4.186 joules of energy for 1 gram raised 1 degree C ( or <mathjax>#K^o#</mathjax>)</p>
<p>l Kg = 1000 grams so </p>
<p><mathjax># 100 Kg xx 1000 g /1 #</mathjax> = 100000 grams of water. </p>
<p>It takes 10 000 calories to raise 10 000 grams of water 1 degree</p>
<p>so 20 degrees x 10 000 grams = 2 000 000 calories. </p>
<p>The same type calculations can be done for heat in joules </p>
<p>10, 000 grams x 4.186 joules x 20 degrees. = 8372000 joules.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If you wish to warm 100 kg of water by 20 degrees C for your bath how much heat is required?</h1>
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David Drayer
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Aug 29, 2016
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<div class="markdown"><p>2000000 heat calories. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The heat capacity of water is 1 calorie for 1 gram raised 1 degree C<br/>
or 4.186 joules of energy for 1 gram raised 1 degree C ( or <mathjax>#K^o#</mathjax>)</p>
<p>l Kg = 1000 grams so </p>
<p><mathjax># 100 Kg xx 1000 g /1 #</mathjax> = 100000 grams of water. </p>
<p>It takes 10 000 calories to raise 10 000 grams of water 1 degree</p>
<p>so 20 degrees x 10 000 grams = 2 000 000 calories. </p>
<p>The same type calculations can be done for heat in joules </p>
<p>10, 000 grams x 4.186 joules x 20 degrees. = 8372000 joules.</p></div>
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Ernest Z.
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Sep 4, 2016
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<div class="markdown"><p>The heat required is 8.4 MJ.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>The energy required to heat an object is given by the formula</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) q = mcΔTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the energy required<br/>
<mathjax>#m#</mathjax> is the mass<br/>
<mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#ΔT#</mathjax> is the change in temperature</p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#m = "100 kg" = "100 000 g"#</mathjax><br/>
<mathjax>#c = "4.179 J·°C"^"-1"·"g"^"-1"#</mathjax><br/>
<mathjax>#ΔT = "20 °C"#</mathjax></p>
<p>∴ <mathjax>#q = "100 000" color(red)(cancel(color(black)("g"))) × "4.179 J"·color(red)(cancel(color(black)("°C"^"-1"·"g"^"-1"))) × 20 color(red)(cancel(color(black)("°C"))) = 8.4 ×10^6color(white)(l) "J" = "8.4 MJ"#</mathjax></p></div>
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</article> | If you wish to warm 100 kg of water by 20 degrees C for your bath how much heat is required? | null |
1,614 | abfb2aee-6ddd-11ea-b4af-ccda262736ce | https://socratic.org/questions/what-weight-of-potassium-permanganate-is-required-to-produce-300ml-of-solution-s | 1.5 g | start physical_unit 3 4 mass g qc_end physical_unit 12 12 9 10 volume qc_end physical_unit 12 12 15 16 volume qc_end physical_unit 12 12 22 23 volume qc_end end | [{"type":"physical unit","value":"Weight [OF] potassium permanganate [IN] g"}] | [{"type":"physical unit","value":"1.5 g"}] | [{"type":"physical unit","value":"Volume1 [OF] solution [=] \\pu{300 mL}"},{"type":"physical unit","value":"Volume2 [OF] solution [=] \\pu{5 mL}"},{"type":"physical unit","value":"Volume3 [OF] solution [=] \\pu{250 mL}"},{"type":"physical unit","value":"w/v [OF] potassium permanganate in solution [=] \\pu{0.01%}"}] | <h1 class="questionTitle" itemprop="name">What weight of potassium permanganate is required to produce 300mL of solution such that 5mL of this solution diluted to 250mL gives a 0.01% w/v solution?</h1> | null | 1.5 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to start from the diluted solution and work your way back to the original solution. </p>
<p>So, you know that a solution's <strong>mass by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every <mathjax>#"100 mL"#</mathjax> of the solution.</p>
<p>This means that your diluted solution contains <mathjax>#"0.01 g"#</mathjax> of potassium permanganate for every <mathjax>#"100 mL"#</mathjax> of the solution. Consequently, you can say that the <mathjax>#"250-mL"#</mathjax> sample of this solution contains</p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL solution"))) * "0.01 g KMnO"_4/(100color(red)(cancel(color(black)("mL solution")))) = "0.025 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that when you <strong>dilute a solution</strong>, you <strong>decrease</strong> its concentration by keeping the mass of solute <strong>constant</strong> and <strong>increasing</strong> its volume. </p>
<p>This means that when you dilute the <mathjax>#"5-mL"#</mathjax> sample to <mathjax>#"250 mL"#</mathjax>, the mass of potassium permanganate <strong>does not change</strong>. </p>
<p>You can thus say that the <mathjax>#"5-mL"#</mathjax> sample contained <mathjax>#"0.025 g"#</mathjax> of potassium permanganate, the amount of solute present in the diluted solution. </p>
<p>So, if you have <mathjax>#"0.025 g"#</mathjax> of solute in <mathjax>#"5 mL"#</mathjax> of the solution, how many grams would you have in <mathjax>#"300 mL"#</mathjax> of <em>the same solution</em>?</p>
<blockquote>
<p><mathjax>#300 color(red)(cancel(color(black)("mL solution"))) * "0.025 g KMnO"_4/(5color(red)(cancel(color(black)("mL solution")))) = "1.5 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Since you have one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong> for the volume of the initial solution and of the <mathjax>#"5-mL"#</mathjax> sample, you should report the answer as</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass KMnO"_4 = "2 g")))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2 g"#</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 start from the diluted solution and work your way back to the original solution. </p>
<p>So, you know that a solution's <strong>mass by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every <mathjax>#"100 mL"#</mathjax> of the solution.</p>
<p>This means that your diluted solution contains <mathjax>#"0.01 g"#</mathjax> of potassium permanganate for every <mathjax>#"100 mL"#</mathjax> of the solution. Consequently, you can say that the <mathjax>#"250-mL"#</mathjax> sample of this solution contains</p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL solution"))) * "0.01 g KMnO"_4/(100color(red)(cancel(color(black)("mL solution")))) = "0.025 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that when you <strong>dilute a solution</strong>, you <strong>decrease</strong> its concentration by keeping the mass of solute <strong>constant</strong> and <strong>increasing</strong> its volume. </p>
<p>This means that when you dilute the <mathjax>#"5-mL"#</mathjax> sample to <mathjax>#"250 mL"#</mathjax>, the mass of potassium permanganate <strong>does not change</strong>. </p>
<p>You can thus say that the <mathjax>#"5-mL"#</mathjax> sample contained <mathjax>#"0.025 g"#</mathjax> of potassium permanganate, the amount of solute present in the diluted solution. </p>
<p>So, if you have <mathjax>#"0.025 g"#</mathjax> of solute in <mathjax>#"5 mL"#</mathjax> of the solution, how many grams would you have in <mathjax>#"300 mL"#</mathjax> of <em>the same solution</em>?</p>
<blockquote>
<p><mathjax>#300 color(red)(cancel(color(black)("mL solution"))) * "0.025 g KMnO"_4/(5color(red)(cancel(color(black)("mL solution")))) = "1.5 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Since you have one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong> for the volume of the initial solution and of the <mathjax>#"5-mL"#</mathjax> sample, you should report the answer as</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass KMnO"_4 = "2 g")))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What weight of potassium permanganate is required to produce 300mL of solution such that 5mL of this solution diluted to 250mL gives a 0.01% w/v solution?</h1>
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Stefan V.
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Dec 11, 2017
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<div class="markdown"><p><mathjax>#"2 g"#</mathjax></p></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 start from the diluted solution and work your way back to the original solution. </p>
<p>So, you know that a solution's <strong>mass by volume <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a></strong> tells you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every <mathjax>#"100 mL"#</mathjax> of the solution.</p>
<p>This means that your diluted solution contains <mathjax>#"0.01 g"#</mathjax> of potassium permanganate for every <mathjax>#"100 mL"#</mathjax> of the solution. Consequently, you can say that the <mathjax>#"250-mL"#</mathjax> sample of this solution contains</p>
<blockquote>
<p><mathjax>#250 color(red)(cancel(color(black)("mL solution"))) * "0.01 g KMnO"_4/(100color(red)(cancel(color(black)("mL solution")))) = "0.025 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that when you <strong>dilute a solution</strong>, you <strong>decrease</strong> its concentration by keeping the mass of solute <strong>constant</strong> and <strong>increasing</strong> its volume. </p>
<p>This means that when you dilute the <mathjax>#"5-mL"#</mathjax> sample to <mathjax>#"250 mL"#</mathjax>, the mass of potassium permanganate <strong>does not change</strong>. </p>
<p>You can thus say that the <mathjax>#"5-mL"#</mathjax> sample contained <mathjax>#"0.025 g"#</mathjax> of potassium permanganate, the amount of solute present in the diluted solution. </p>
<p>So, if you have <mathjax>#"0.025 g"#</mathjax> of solute in <mathjax>#"5 mL"#</mathjax> of the solution, how many grams would you have in <mathjax>#"300 mL"#</mathjax> of <em>the same solution</em>?</p>
<blockquote>
<p><mathjax>#300 color(red)(cancel(color(black)("mL solution"))) * "0.025 g KMnO"_4/(5color(red)(cancel(color(black)("mL solution")))) = "1.5 g KMnO"_4#</mathjax></p>
</blockquote>
<p>Since you have one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong> for the volume of the initial solution and of the <mathjax>#"5-mL"#</mathjax> sample, you should report the answer as</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("mass KMnO"_4 = "2 g")))#</mathjax></p>
</blockquote></div>
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</article> | What weight of potassium permanganate is required to produce 300mL of solution such that 5mL of this solution diluted to 250mL gives a 0.01% w/v solution? | null |
1,615 | ac34ae98-6ddd-11ea-925e-ccda262736ce | https://socratic.org/questions/how-do-you-balance-feo-3-s-co-g-fe-l-co-2-g | FeO3(s) + 3 CO(g) -> Fe(l) + 3 CO2(g) | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"FeO3(s) + 3 CO(g) -> Fe(l) + 3 CO2(g)"}] | [{"type":"chemical equation","value":"FeO3(s) + CO(g) -> Fe(l) + CO2(g)"}] | <h1 class="questionTitle" itemprop="name">How do you balance #FeO_3(s) + CO(g) -> Fe(l) + CO_2(g)#?</h1> | null | FeO3(s) + 3 CO(g) -> Fe(l) + 3 CO2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The transition from carbon monoxide to carbon dioxide uses up one part of oxygen. So to use the three parts of oxygen in <mathjax>#FeO_3#</mathjax> combine with <mathjax>#3#</mathjax> parts carbon monoxide.</p></div>
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<div class="markdown"><p><mathjax>#FeO_3 + 3CO -> Fe + 3CO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The transition from carbon monoxide to carbon dioxide uses up one part of oxygen. So to use the three parts of oxygen in <mathjax>#FeO_3#</mathjax> combine with <mathjax>#3#</mathjax> parts carbon monoxide.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #FeO_3(s) + CO(g) -> Fe(l) + CO_2(g)#?</h1>
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<div class="markdown"><p><mathjax>#FeO_3 + 3CO -> Fe + 3CO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The transition from carbon monoxide to carbon dioxide uses up one part of oxygen. So to use the three parts of oxygen in <mathjax>#FeO_3#</mathjax> combine with <mathjax>#3#</mathjax> parts carbon monoxide.</p></div>
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</article> | How do you balance #FeO_3(s) + CO(g) -> Fe(l) + CO_2(g)#? | null |
1,616 | aca5a95e-6ddd-11ea-80f0-ccda262736ce | https://socratic.org/questions/how-many-moles-of-xenon-do-5-66-x-10-23-atoms-equal | 0.94 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] xenon [IN] moles"}] | [{"type":"physical unit","value":"0.94 moles"}] | [{"type":"physical unit","value":"Number [OF] xenon atoms [=] \\pu{5.66 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles of xenon do #5.66 xx 10^23# atoms equal? </h1> | null | 0.94 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Avogadro's number"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we take the quotient:</p>
<p><mathjax>#(5.66xx10^23*cancel"atoms")/(6.022xx10^23*cancel"atoms"*mol^-1)=??mol#</mathjax></p>
<p>This is dimensionally consistent, because <mathjax>#1/(mol^-1)=1/(1/(mol^-1))=mol#</mathjax> as required.........</p></div>
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<div class="markdown"><p>A bit more than <mathjax>#5/6#</mathjax> <mathjax>#"moles"#</mathjax>......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Avogadro's number"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we take the quotient:</p>
<p><mathjax>#(5.66xx10^23*cancel"atoms")/(6.022xx10^23*cancel"atoms"*mol^-1)=??mol#</mathjax></p>
<p>This is dimensionally consistent, because <mathjax>#1/(mol^-1)=1/(1/(mol^-1))=mol#</mathjax> as required.........</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of xenon do #5.66 xx 10^23# atoms equal? </h1>
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<div class="markdown"><p>A bit more than <mathjax>#5/6#</mathjax> <mathjax>#"moles"#</mathjax>......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Avogadro's number"#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And so we take the quotient:</p>
<p><mathjax>#(5.66xx10^23*cancel"atoms")/(6.022xx10^23*cancel"atoms"*mol^-1)=??mol#</mathjax></p>
<p>This is dimensionally consistent, because <mathjax>#1/(mol^-1)=1/(1/(mol^-1))=mol#</mathjax> as required.........</p></div>
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<div class="markdown"><p><mathjax>#5.66xx10^23"Xe atoms"#</mathjax> equals <mathjax>#"0.940 mol Xe atoms"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of anything, including xenon atoms is <mathjax>#6.022xx10^23#</mathjax>.</p>
<p>This gives us an equality from which we can derive two conversion factors.</p>
<p><mathjax>#1"mole Xe atoms"=6.022xx10^23color(white)(.)"atoms"#</mathjax></p>
<p>Now we can write two conversion factors.</p>
<p><mathjax>#(1"mol Xe atoms")/(6.022xx10^23color(white)(.)"Xe atoms")#</mathjax> and <mathjax>#(6.022xx10^23color(white)(.)"Xe atoms")/(1"mol Xe atoms")#</mathjax></p>
<p>Now mutliply the given number of atoms by the conversion factor that has moles in the numerator. This will result in the Xe atoms canceling.</p>
<p><mathjax>#5.66xx10^23cancel"Xe atoms"xx(1"mol Xe")/(6.022xx10^23cancel"Xe atoms")="0.940 mol Xe atoms"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | How many moles of xenon do #5.66 xx 10^23# atoms equal? | null |
1,617 | ab7115fb-6ddd-11ea-a30b-ccda262736ce | https://socratic.org/questions/591e828211ef6b6cb931cf50 | 7.51 L | start physical_unit 7 7 volume l qc_end physical_unit 16 17 12 13 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] SO2(g) [IN] L"}] | [{"type":"physical unit","value":"7.51 L"}] | [{"type":"physical unit","value":"Mass [OF] elemental sulfur [=] \\pu{10 g}"},{"type":"other","value":"Under conditions of NTP."}] | <h1 class="questionTitle" itemprop="name">Under conditions of #"NTP"#, what volume of #SO_2(g)# results from burning a #10*g# mass of elemental sulfur?</h1> | null | 7.51 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#S(s)+O_2(g) rarrSO_2(g)#</mathjax></p>
<p>Now depending on your definition of <mathjax>#"NTP"#</mathjax> (they seem to change with the wind across syllabuses) we can work out the volume. I use values of <mathjax>#"NTP"#</mathjax> to specify <mathjax>#P=1*atm#</mathjax>, and <mathjax>#T=293.15*K#</mathjax>.</p>
<p><mathjax>#"Moles of sulfur "-=" Moles of sulfur dioxide"#</mathjax></p>
<p><mathjax>#"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol#</mathjax>.</p>
<p>And thus there are <mathjax>#0.312*mol#</mathjax> <mathjax>#SO_2#</mathjax> gas........given conversion stoichiometric conversion.</p>
<p><mathjax>#V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)#</mathjax></p>
<p><mathjax>#=7.5*L#</mathjax></p></div>
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<div class="markdown"><p>We address the stoichiometric equation...........and I get a volume of under <mathjax>#10*L#</mathjax>...............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#S(s)+O_2(g) rarrSO_2(g)#</mathjax></p>
<p>Now depending on your definition of <mathjax>#"NTP"#</mathjax> (they seem to change with the wind across syllabuses) we can work out the volume. I use values of <mathjax>#"NTP"#</mathjax> to specify <mathjax>#P=1*atm#</mathjax>, and <mathjax>#T=293.15*K#</mathjax>.</p>
<p><mathjax>#"Moles of sulfur "-=" Moles of sulfur dioxide"#</mathjax></p>
<p><mathjax>#"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol#</mathjax>.</p>
<p>And thus there are <mathjax>#0.312*mol#</mathjax> <mathjax>#SO_2#</mathjax> gas........given conversion stoichiometric conversion.</p>
<p><mathjax>#V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)#</mathjax></p>
<p><mathjax>#=7.5*L#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Under conditions of #"NTP"#, what volume of #SO_2(g)# results from burning a #10*g# mass of elemental sulfur?</h1>
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<div class="markdown"><p>We address the stoichiometric equation...........and I get a volume of under <mathjax>#10*L#</mathjax>...............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#S(s)+O_2(g) rarrSO_2(g)#</mathjax></p>
<p>Now depending on your definition of <mathjax>#"NTP"#</mathjax> (they seem to change with the wind across syllabuses) we can work out the volume. I use values of <mathjax>#"NTP"#</mathjax> to specify <mathjax>#P=1*atm#</mathjax>, and <mathjax>#T=293.15*K#</mathjax>.</p>
<p><mathjax>#"Moles of sulfur "-=" Moles of sulfur dioxide"#</mathjax></p>
<p><mathjax>#"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol#</mathjax>.</p>
<p>And thus there are <mathjax>#0.312*mol#</mathjax> <mathjax>#SO_2#</mathjax> gas........given conversion stoichiometric conversion.</p>
<p><mathjax>#V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)#</mathjax></p>
<p><mathjax>#=7.5*L#</mathjax></p></div>
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</article> | Under conditions of #"NTP"#, what volume of #SO_2(g)# results from burning a #10*g# mass of elemental sulfur? | null |
1,618 | ac2771f4-6ddd-11ea-9e6d-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-that-has-46-grams-of-sodium-64-gr | Na2S2O3 | start chemical_formula qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 17 17 14 15 mass qc_end physical_unit 22 22 19 20 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Na2S2O3"}] | [{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{46 grams}"},{"type":"physical unit","value":"Mass [OF] sulfur [=] \\pu{64 grams}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{48 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen? </h1> | null | Na2S2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em>. </p>
<p>Grab one and look for the <strong>molar mass</strong> of sodium, <mathjax>#"Na"#</mathjax>, sulfur, <mathjax>#"S"#</mathjax>, and oxygen, <mathjax>#"O"#</mathjax>. Notice that you have</p>
<blockquote>
<p><mathjax>#M_("M Na") ~~ "23 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M S") ~~ "32 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M O") ~~ "16 g mol"^(-1)#</mathjax> </p>
</blockquote>
<p>This tells you that the sample given to you contains </p>
<blockquote>
<p><mathjax># 46 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23color(red)(cancel(color(black)("g")))) = "2 moles Na"#</mathjax></p>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32color(red)(cancel(color(black)("g")))) = "2 moles S"#</mathjax></p>
<p><mathjax>#48 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16color(red)(cancel(color(black)("g")))) = "3 moles O"#</mathjax></p>
</blockquote>
<p>Now, in order to write the <strong>empirical formula</strong> of the compound, you must find the <strong>smallest whole number ratio</strong> that exists between its constituent elements. </p>
<p>Since the <mathjax>#2:2:3#</mathjax> ratio is already in its smallest whole number form, you can say that the empirical formula of the compound is </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Na"_2"S"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#"Na"_2"S"_2"O"_3#</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 here will be the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em>. </p>
<p>Grab one and look for the <strong>molar mass</strong> of sodium, <mathjax>#"Na"#</mathjax>, sulfur, <mathjax>#"S"#</mathjax>, and oxygen, <mathjax>#"O"#</mathjax>. Notice that you have</p>
<blockquote>
<p><mathjax>#M_("M Na") ~~ "23 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M S") ~~ "32 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M O") ~~ "16 g mol"^(-1)#</mathjax> </p>
</blockquote>
<p>This tells you that the sample given to you contains </p>
<blockquote>
<p><mathjax># 46 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23color(red)(cancel(color(black)("g")))) = "2 moles Na"#</mathjax></p>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32color(red)(cancel(color(black)("g")))) = "2 moles S"#</mathjax></p>
<p><mathjax>#48 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16color(red)(cancel(color(black)("g")))) = "3 moles O"#</mathjax></p>
</blockquote>
<p>Now, in order to write the <strong>empirical formula</strong> of the compound, you must find the <strong>smallest whole number ratio</strong> that exists between its constituent elements. </p>
<p>Since the <mathjax>#2:2:3#</mathjax> ratio is already in its smallest whole number form, you can say that the empirical formula of the compound is </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Na"_2"S"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen? </h1>
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<div class="markdown"><p><mathjax>#"Na"_2"S"_2"O"_3#</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 here will be the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em>. </p>
<p>Grab one and look for the <strong>molar mass</strong> of sodium, <mathjax>#"Na"#</mathjax>, sulfur, <mathjax>#"S"#</mathjax>, and oxygen, <mathjax>#"O"#</mathjax>. Notice that you have</p>
<blockquote>
<p><mathjax>#M_("M Na") ~~ "23 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M S") ~~ "32 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M O") ~~ "16 g mol"^(-1)#</mathjax> </p>
</blockquote>
<p>This tells you that the sample given to you contains </p>
<blockquote>
<p><mathjax># 46 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23color(red)(cancel(color(black)("g")))) = "2 moles Na"#</mathjax></p>
<p><mathjax>#64 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32color(red)(cancel(color(black)("g")))) = "2 moles S"#</mathjax></p>
<p><mathjax>#48 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16color(red)(cancel(color(black)("g")))) = "3 moles O"#</mathjax></p>
</blockquote>
<p>Now, in order to write the <strong>empirical formula</strong> of the compound, you must find the <strong>smallest whole number ratio</strong> that exists between its constituent elements. </p>
<p>Since the <mathjax>#2:2:3#</mathjax> ratio is already in its smallest whole number form, you can say that the empirical formula of the compound is </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("Na"_2"S"_2"O"_3)color(white)(a/a)|)))#</mathjax></p>
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</article> | What is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen? | null |
1,619 | a8c2741a-6ddd-11ea-a3aa-ccda262736ce | https://socratic.org/questions/5923b3797c0149507530456d | 714 g | start physical_unit 3 3 mass g qc_end physical_unit 10 11 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] alumina [IN] g"}] | [{"type":"physical unit","value":"714 g"}] | [{"type":"physical unit","value":"Mole [OF] aluminum metal [=] \\pu{14 mol}"}] | <h1 class="questionTitle" itemprop="name">What mass of alumina will result if #14*mol# of aluminum metal are oxidized?</h1> | null | 714 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given the stoichiometric equation, if <mathjax>#14*mol#</mathjax> of metal are oxidized. then <mathjax>#7*mol#</mathjax> of <mathjax>#"alumina"#</mathjax> will be generated............</p>
<p><mathjax>#"Mass"=7*molxx101.96*g*mol^-1~=714*g#</mathjax>..........</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the stoichiometric equation, if <mathjax>#14*mol#</mathjax> of metal are oxidized. then <mathjax>#7*mol#</mathjax> of <mathjax>#"alumina"#</mathjax> will be generated............</p>
<p><mathjax>#"Mass"=7*molxx101.96*g*mol^-1~=714*g#</mathjax>..........</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of alumina will result if #14*mol# of aluminum metal are oxidized?</h1>
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<div class="markdown"><p><mathjax>#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given the stoichiometric equation, if <mathjax>#14*mol#</mathjax> of metal are oxidized. then <mathjax>#7*mol#</mathjax> of <mathjax>#"alumina"#</mathjax> will be generated............</p>
<p><mathjax>#"Mass"=7*molxx101.96*g*mol^-1~=714*g#</mathjax>..........</p></div>
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</article> | What mass of alumina will result if #14*mol# of aluminum metal are oxidized? | null |
1,620 | a90be606-6ddd-11ea-9450-ccda262736ce | https://socratic.org/questions/how-many-water-molecules-are-present-in-54-gram-of-water | 1.81 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 2 2 7 8 mass qc_end end | [{"type":"physical unit","value":"Number [OF] water molecules"}] | [{"type":"physical unit","value":"1.81 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{54 grams}"}] | <h1 class="questionTitle" itemprop="name">How many water molecules are present in 54 gram of water?
</h1> | null | 1.81 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the molar quantity of water in the usual way...</p>
<p><mathjax>#"Number of moles"="mass"/"molar mass"#</mathjax></p>
<p><mathjax>#-=(54*g)/(18.01*g*mol^-1)=3.00*mol#</mathjax></p>
<p>But the <mathjax>#mol#</mathjax> is like the <mathjax>#"dozen"#</mathjax>, or the <mathjax>#"gross"#</mathjax>, it specifies a NUMBER, here <mathjax>#N_A-=6.022xx10^23*mol#</mathjax>. Why do we use such an absurdly large number? Because <mathjax>#N_A*""^1H*"atoms"#</mathjax> have a MASS of <mathjax>#1*g#</mathjax> precisely...and thus <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is the link between the micro world of atoms, and molecules, to the macro world of grams and litres....</p>
<p>And so (finally!) we take the product...</p>
<p><mathjax>#"Number of water molecules"="molar quantity"xx"Avogadro's number."#</mathjax></p>
<p><mathjax>#=underbrace(3.00*cancel(mol)xx6.022xx10^23*cancel(mol^-1))_"a number as required.."=??*"water molecules"#</mathjax></p></div>
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<div class="markdown"><p>We get <mathjax>#1.81xx10^24#</mathjax> water molecules...</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the molar quantity of water in the usual way...</p>
<p><mathjax>#"Number of moles"="mass"/"molar mass"#</mathjax></p>
<p><mathjax>#-=(54*g)/(18.01*g*mol^-1)=3.00*mol#</mathjax></p>
<p>But the <mathjax>#mol#</mathjax> is like the <mathjax>#"dozen"#</mathjax>, or the <mathjax>#"gross"#</mathjax>, it specifies a NUMBER, here <mathjax>#N_A-=6.022xx10^23*mol#</mathjax>. Why do we use such an absurdly large number? Because <mathjax>#N_A*""^1H*"atoms"#</mathjax> have a MASS of <mathjax>#1*g#</mathjax> precisely...and thus <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is the link between the micro world of atoms, and molecules, to the macro world of grams and litres....</p>
<p>And so (finally!) we take the product...</p>
<p><mathjax>#"Number of water molecules"="molar quantity"xx"Avogadro's number."#</mathjax></p>
<p><mathjax>#=underbrace(3.00*cancel(mol)xx6.022xx10^23*cancel(mol^-1))_"a number as required.."=??*"water molecules"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many water molecules are present in 54 gram of water?
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<div class="markdown"><p>We get <mathjax>#1.81xx10^24#</mathjax> water molecules...</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assess the molar quantity of water in the usual way...</p>
<p><mathjax>#"Number of moles"="mass"/"molar mass"#</mathjax></p>
<p><mathjax>#-=(54*g)/(18.01*g*mol^-1)=3.00*mol#</mathjax></p>
<p>But the <mathjax>#mol#</mathjax> is like the <mathjax>#"dozen"#</mathjax>, or the <mathjax>#"gross"#</mathjax>, it specifies a NUMBER, here <mathjax>#N_A-=6.022xx10^23*mol#</mathjax>. Why do we use such an absurdly large number? Because <mathjax>#N_A*""^1H*"atoms"#</mathjax> have a MASS of <mathjax>#1*g#</mathjax> precisely...and thus <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> is the link between the micro world of atoms, and molecules, to the macro world of grams and litres....</p>
<p>And so (finally!) we take the product...</p>
<p><mathjax>#"Number of water molecules"="molar quantity"xx"Avogadro's number."#</mathjax></p>
<p><mathjax>#=underbrace(3.00*cancel(mol)xx6.022xx10^23*cancel(mol^-1))_"a number as required.."=??*"water molecules"#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-06-25T04:52:54" itemprop="dateCreated">
Jun 25, 2018
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<div class="markdown"><p><mathjax>#1.8 xx 10^24#</mathjax> water molecules.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are <mathjax>#6.022 xx 10^23#</mathjax> molecules in a mole.</p>
<p>There are <mathjax>#18.01528#</mathjax> grams of water per mole of water.</p>
<p>These can be figured out using the concept of moles and molecular weight. I have attached a video that provides a good explanation.</p>
<p><a href="https://www.youtube.com/watch?v=JC76NR8EtTQ" rel="nofollow">Explanation of moles and molecular weight</a></p>
<p>Now you have all the information needed to solve the problem. </p>
<p>You start out with 54 grams of water. You want to convert that into molecules, but how can you do that? You can't go directly from grams to molecules because there is no easy unit conversion. However, you can go from grams to moles using </p>
<p><mathjax>#18.01528#</mathjax> grams of water per mole of water.</p>
<p>From there, you can go from moles to molecules using <mathjax>#6.022 xx 10^23#</mathjax> molecules in a mole.</p>
<p>A T chart is useful for this kind of problem.</p>
<p><mathjax>#"54 g H"_2"O" xx ("1 mol H"_2"O")/("18.0152 g H"_2"O") xx(6.022 xx 10^23 \ "molecules")/("1 mol H"_2"O")#</mathjax></p>
<p><mathjax>#"54" cancel("g H"_2"O") xx (cancel("1 mol H"_2"O"))/(18.0152cancel ("g H"_2"O")) xx (6.022 xx 10^23 \ "molecules")/(cancel("1 mol H"_2"O"))#</mathjax></p>
<p>Notice when you apply this T chart, the units cancel out, leaving molecules in the final answer, which is the correct units.</p>
<p>Now actually multiply everything out</p>
<p><mathjax>#(54 xx 6.022xx10^23)/18.0152= 1.805 xx 10^24#</mathjax></p>
<p>You are left with <mathjax>#1.8 xx 10^24#</mathjax> water molecules (rounded to two sig figs).</p></div>
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</article> | How many water molecules are present in 54 gram of water?
| null |
1,621 | a86a3412-6ddd-11ea-a1f7-ccda262736ce | https://socratic.org/questions/how-many-moles-of-ch-4-contain-1-5-10-23-particles | 0.25 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] CH4 [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Number [OF] CH4 particles [=] \\pu{1.5 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #CH_4# contain #1.5 * 10^23# particles?</h1> | null | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The particles of <mathjax>#"CH"_4#</mathjax> are molecules. </p>
<p><mathjax>#1"mol CH"_4=6.022xx10^23"molecules CH"_4"#</mathjax></p>
<p>To determine the moles of <mathjax>#"CH"_4"#</mathjax>, divide the given number of molecules by <mathjax>#6.022xx10^23#</mathjax> molecules/mol.</p>
<p><mathjax>#1.5xx10^23cancel"molecules CH"_4xx(1"mol CH"_4)/(6.022xx10^23cancel"molecules CH"_4)="0.25 mol CH"_4"#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p><mathjax>#"0.25 mol CH"_4"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The particles of <mathjax>#"CH"_4#</mathjax> are molecules. </p>
<p><mathjax>#1"mol CH"_4=6.022xx10^23"molecules CH"_4"#</mathjax></p>
<p>To determine the moles of <mathjax>#"CH"_4"#</mathjax>, divide the given number of molecules by <mathjax>#6.022xx10^23#</mathjax> molecules/mol.</p>
<p><mathjax>#1.5xx10^23cancel"molecules CH"_4xx(1"mol CH"_4)/(6.022xx10^23cancel"molecules CH"_4)="0.25 mol CH"_4"#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #CH_4# contain #1.5 * 10^23# particles?</h1>
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<div class="markdown"><p><mathjax>#"0.25 mol CH"_4"#</mathjax></p></div>
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<div class="markdown"><p>The particles of <mathjax>#"CH"_4#</mathjax> are molecules. </p>
<p><mathjax>#1"mol CH"_4=6.022xx10^23"molecules CH"_4"#</mathjax></p>
<p>To determine the moles of <mathjax>#"CH"_4"#</mathjax>, divide the given number of molecules by <mathjax>#6.022xx10^23#</mathjax> molecules/mol.</p>
<p><mathjax>#1.5xx10^23cancel"molecules CH"_4xx(1"mol CH"_4)/(6.022xx10^23cancel"molecules CH"_4)="0.25 mol CH"_4"#</mathjax> (rounded to two significant figures)</p></div>
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</article> | How many moles of #CH_4# contain #1.5 * 10^23# particles? | null |
1,622 | ad07ef06-6ddd-11ea-92e0-ccda262736ce | https://socratic.org/questions/5961e72eb72cff16ecc4a307 | 31.58% | start physical_unit 7 7 mass_percent none qc_end physical_unit 15 15 10 11 molarity qc_end end | [{"type":"physical unit","value":"Mass percent [OF] NaOH in solution"}] | [{"type":"physical unit","value":"31.58%"}] | [{"type":"physical unit","value":"ρ [OF] NaOH solution [=] \\pu{1.33 g/mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{10.5 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the mass percent of a solution that is #10.5*mol*L^-1# with respect to #NaOH#, and for which #rho_"solution"=1.33*g*mL^-1#...?</h1> | null | 31.58% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that we have the molar concentration, and the solution density, we can fairly easily calculate the given quotient......</p>
<p>We work out the quotient on the basis of a <mathjax>#1*mL#</mathjax> volume, which has a mass of <mathjax>#1*mLxx1.33*g*mL^-1=1.33*g#</mathjax>.</p>
<p>For a <mathjax>#1*mL#</mathjax> volume, there are..........</p>
<p><mathjax>#10.5*mol*L^-1xx1xx10^-3*Lxx40.0*g*mol^-1=0.420*g#</mathjax></p>
<p>And thus <mathjax>#%"mass"=(0.420*g)/(1*mLxx1.33*g*mL^-1)xx100%=#</mathjax></p>
<p><mathjax>#=31.6%m/m#</mathjax>.</p>
<p>What safety precautions MUST BE observed when you make such a solution? </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We want <mathjax>#"Mass of caustic of soda"/"Mass of solution"xx100%#</mathjax>....</p>
<p>We get <mathjax>#"%NaOH(m/m)"-=32%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that we have the molar concentration, and the solution density, we can fairly easily calculate the given quotient......</p>
<p>We work out the quotient on the basis of a <mathjax>#1*mL#</mathjax> volume, which has a mass of <mathjax>#1*mLxx1.33*g*mL^-1=1.33*g#</mathjax>.</p>
<p>For a <mathjax>#1*mL#</mathjax> volume, there are..........</p>
<p><mathjax>#10.5*mol*L^-1xx1xx10^-3*Lxx40.0*g*mol^-1=0.420*g#</mathjax></p>
<p>And thus <mathjax>#%"mass"=(0.420*g)/(1*mLxx1.33*g*mL^-1)xx100%=#</mathjax></p>
<p><mathjax>#=31.6%m/m#</mathjax>.</p>
<p>What safety precautions MUST BE observed when you make such a solution? </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the mass percent of a solution that is #10.5*mol*L^-1# with respect to #NaOH#, and for which #rho_"solution"=1.33*g*mL^-1#...?</h1>
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anor277
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<div class="markdown"><p>We want <mathjax>#"Mass of caustic of soda"/"Mass of solution"xx100%#</mathjax>....</p>
<p>We get <mathjax>#"%NaOH(m/m)"-=32%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that we have the molar concentration, and the solution density, we can fairly easily calculate the given quotient......</p>
<p>We work out the quotient on the basis of a <mathjax>#1*mL#</mathjax> volume, which has a mass of <mathjax>#1*mLxx1.33*g*mL^-1=1.33*g#</mathjax>.</p>
<p>For a <mathjax>#1*mL#</mathjax> volume, there are..........</p>
<p><mathjax>#10.5*mol*L^-1xx1xx10^-3*Lxx40.0*g*mol^-1=0.420*g#</mathjax></p>
<p>And thus <mathjax>#%"mass"=(0.420*g)/(1*mLxx1.33*g*mL^-1)xx100%=#</mathjax></p>
<p><mathjax>#=31.6%m/m#</mathjax>.</p>
<p>What safety precautions MUST BE observed when you make such a solution? </p></div>
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</article> | What is the mass percent of a solution that is #10.5*mol*L^-1# with respect to #NaOH#, and for which #rho_"solution"=1.33*g*mL^-1#...? | null |
1,623 | a9cdc094-6ddd-11ea-adf2-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-capacity-of-substance-if-2-41times10-4-j-are-needed-to | 1.02 J/(g * ℃) | start physical_unit 7 7 specific_heat_capacity j/(°c_·_g) qc_end physical_unit 7 7 20 21 mass qc_end physical_unit 7 7 25 26 temperature qc_end physical_unit 7 7 28 29 temperature qc_end physical_unit 7 7 9 12 energy qc_end end | [{"type":"physical unit","value":"Specific heat capacity [OF] substance [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"1.02 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] substance [=] \\pu{105.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] substance [=] \\pu{25.00 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] substance [=] \\pu{250.0 ℃}"},{"type":"physical unit","value":"Needed energy [OF] substance [=] \\pu{2.41 × 10^4 J}"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat capacity of substance if #2.41times10^4# #J# are needed to change the temperature of #105.0# #g# of it from #25.00°C# to #250.0°C#? </h1> | null | 1.02 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use is <mathjax>#color(red)(Q=mcDeltat)#</mathjax>, where <mathjax>#Q#</mathjax> is the amount of energy gained or lost, <mathjax>#m#</mathjax> is mass in grams, <mathjax>#c#</mathjax> is specific heat capacity, and <mathjax>#color(blue)(Deltat#</mathjax> is the change in temperature, <mathjax>#color(blue)(t_f-t_i#</mathjax>.</p>
<p>To answer this question, rearrange the equation to isolate specific heat capacity. Substitute the known values into the equation and solve.</p>
<p><mathjax>#c=Q/(mDeltat)#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx(250.0^@"C"-25.0^@"C"))#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx225.0^@"C")#</mathjax></p>
<p><mathjax>#c=1.02"J"/("g"*^@"C")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the substance is <mathjax>#1.02"J"/("g"*^@"C")#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use is <mathjax>#color(red)(Q=mcDeltat)#</mathjax>, where <mathjax>#Q#</mathjax> is the amount of energy gained or lost, <mathjax>#m#</mathjax> is mass in grams, <mathjax>#c#</mathjax> is specific heat capacity, and <mathjax>#color(blue)(Deltat#</mathjax> is the change in temperature, <mathjax>#color(blue)(t_f-t_i#</mathjax>.</p>
<p>To answer this question, rearrange the equation to isolate specific heat capacity. Substitute the known values into the equation and solve.</p>
<p><mathjax>#c=Q/(mDeltat)#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx(250.0^@"C"-25.0^@"C"))#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx225.0^@"C")#</mathjax></p>
<p><mathjax>#c=1.02"J"/("g"*^@"C")#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the specific heat capacity of substance if #2.41times10^4# #J# are needed to change the temperature of #105.0# #g# of it from #25.00°C# to #250.0°C#? </h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the substance is <mathjax>#1.02"J"/("g"*^@"C")#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use is <mathjax>#color(red)(Q=mcDeltat)#</mathjax>, where <mathjax>#Q#</mathjax> is the amount of energy gained or lost, <mathjax>#m#</mathjax> is mass in grams, <mathjax>#c#</mathjax> is specific heat capacity, and <mathjax>#color(blue)(Deltat#</mathjax> is the change in temperature, <mathjax>#color(blue)(t_f-t_i#</mathjax>.</p>
<p>To answer this question, rearrange the equation to isolate specific heat capacity. Substitute the known values into the equation and solve.</p>
<p><mathjax>#c=Q/(mDeltat)#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx(250.0^@"C"-25.0^@"C"))#</mathjax></p>
<p><mathjax>#c=(2.41xx10^4"J")/(105.0"g"xx225.0^@"C")#</mathjax></p>
<p><mathjax>#c=1.02"J"/("g"*^@"C")#</mathjax></p></div>
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</article> | What is the specific heat capacity of substance if #2.41times10^4# #J# are needed to change the temperature of #105.0# #g# of it from #25.00°C# to #250.0°C#? | null |
1,624 | a9f689b7-6ddd-11ea-b8e6-ccda262736ce | https://socratic.org/questions/a-sample-of-carbon-dioxide-gas-at-a-pressure-of-1-07-atm-and-a-temperature-of-16 | 311.32 ℃ | start physical_unit 41 43 temperature °c qc_end physical_unit 1 5 16 17 temperature qc_end physical_unit 1 5 10 11 pressure qc_end physical_unit 1 5 22 23 volume qc_end physical_unit 41 43 36 37 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas sample [IN] ℃"}] | [{"type":"physical unit","value":"311.32 ℃"}] | [{"type":"physical unit","value":"Temperature1 [OF] carbon dioxide gas sample [=] \\pu{166 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] carbon dioxide gas sample [=] \\pu{1.07 atm}"},{"type":"physical unit","value":"Volume1 [OF] carbon dioxide gas sample [=] \\pu{686 mL}"},{"type":"physical unit","value":"Volume2 [OF] the gas sample [=] \\pu{913 mL}"},{"type":"other","value":"Constant pressure."}] | <h1 class="questionTitle" itemprop="name">A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C.</h1> | null | 311.32 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume and the temperature of a gas have a <strong>direct relationship</strong> when the pressure and the number of moles of gas are being <em>kept constant</em> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong>.</p>
<p>A <strong>very important</strong> thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>. In other words, you must always work with the <strong>absolute temperature</strong> of a gas. </p>
<p>So, start by converting the temperature of the gas to Kelvin by using</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#T = 166^@"C" + 273.15 = "439.15 K"#</mathjax></p>
</blockquote>
<p>The volume of the gas will <strong>increase</strong> as temperature <em>increases</em> and <strong>decrease</strong> as temperature <em>decreases</em>. Mathematically, this can be written as </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#T_1#</mathjax> represent the volume and the temperature of the gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#T_2#</mathjax> represent the volume and the temperature of the gas at a final state </li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (913 color(red)(cancel(color(black)("mL"))))/(686color(red)(cancel(color(black)("mL")))) * "439.15 K" = "584.47 K"#</mathjax></p>
</blockquote>
<p>Finally, convert this to degrees Celsius </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)(t[""^@"C"] = "584.47 K" - 273.15 = 311^@"C")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#311^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume and the temperature of a gas have a <strong>direct relationship</strong> when the pressure and the number of moles of gas are being <em>kept constant</em> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong>.</p>
<p>A <strong>very important</strong> thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>. In other words, you must always work with the <strong>absolute temperature</strong> of a gas. </p>
<p>So, start by converting the temperature of the gas to Kelvin by using</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#T = 166^@"C" + 273.15 = "439.15 K"#</mathjax></p>
</blockquote>
<p>The volume of the gas will <strong>increase</strong> as temperature <em>increases</em> and <strong>decrease</strong> as temperature <em>decreases</em>. Mathematically, this can be written as </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#T_1#</mathjax> represent the volume and the temperature of the gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#T_2#</mathjax> represent the volume and the temperature of the gas at a final state </li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (913 color(red)(cancel(color(black)("mL"))))/(686color(red)(cancel(color(black)("mL")))) * "439.15 K" = "584.47 K"#</mathjax></p>
</blockquote>
<p>Finally, convert this to degrees Celsius </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)(t[""^@"C"] = "584.47 K" - 273.15 = 311^@"C")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C.</h1>
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<div class="answer" id="401216" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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<a class="topContributorPic" href="/users/stefan-zdre"><img alt="" class="" src="https://profilepictures.socratic.org/LrguokJzR9yQlbiWbCvr_proba_1.jpg" title=""/></a>
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Stefan V.
</a></div>
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<span class="dateCreated" datetime="2017-04-04T00:29:46" itemprop="dateCreated">
Apr 4, 2017
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<div class="markdown"><p><mathjax>#311^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume and the temperature of a gas have a <strong>direct relationship</strong> when the pressure and the number of moles of gas are being <em>kept constant</em> <mathjax>#->#</mathjax> this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong>.</p>
<p>A <strong>very important</strong> thing to remember is that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>. In other words, you must always work with the <strong>absolute temperature</strong> of a gas. </p>
<p>So, start by converting the temperature of the gas to Kelvin by using</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#T = 166^@"C" + 273.15 = "439.15 K"#</mathjax></p>
</blockquote>
<p>The volume of the gas will <strong>increase</strong> as temperature <em>increases</em> and <strong>decrease</strong> as temperature <em>decreases</em>. Mathematically, this can be written as </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#V_1#</mathjax> and <mathjax>#T_1#</mathjax> represent the volume and the temperature of the gas at an initial state</li>
<li><mathjax>#V_2#</mathjax> and <mathjax>#T_2#</mathjax> represent the volume and the temperature of the gas at a final state </li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (913 color(red)(cancel(color(black)("mL"))))/(686color(red)(cancel(color(black)("mL")))) * "439.15 K" = "584.47 K"#</mathjax></p>
</blockquote>
<p>Finally, convert this to degrees Celsius </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)(t[""^@"C"] = "584.47 K" - 273.15 = 311^@"C")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C. | null |
1,625 | ab32414d-6ddd-11ea-b70e-ccda262736ce | https://socratic.org/questions/58c088d77c014933f9405c1c | NH3(aq) + HCl(aq) -> NH4Cl(aq) | start chemical_equation qc_end substance 11 11 qc_end substance 13 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"NH3(aq) + HCl(aq) -> NH4Cl(aq)"}] | [{"type":"substance name","value":"Ammonia"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation that describes the reaction between ammonia and hydrochloric acid?</h1> | null | NH3(aq) + HCl(aq) -> NH4Cl(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonia and hydrochloric acid will react to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, as described by the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to release hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and produce chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl" _((aq))^(-)#</mathjax></p>
</blockquote>
<p>Ammonium chloride is a <strong>soluble</strong> ionic compound, so it too will exist as ions in solution. </p>
<p>This means that the <strong>complete ionic equation</strong> that describes this reaction can be written like this</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + "Cl" _ ((aq))^(-) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>To get the <strong>net ionic equation</strong>, you must remove the <em>spectator ions</em>, i.e. the ions present on both sides of the equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-)))) -> "NH"_ (4(aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-))))#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonia and hydrochloric acid will react to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, as described by the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to release hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and produce chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl" _((aq))^(-)#</mathjax></p>
</blockquote>
<p>Ammonium chloride is a <strong>soluble</strong> ionic compound, so it too will exist as ions in solution. </p>
<p>This means that the <strong>complete ionic equation</strong> that describes this reaction can be written like this</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + "Cl" _ ((aq))^(-) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>To get the <strong>net ionic equation</strong>, you must remove the <em>spectator ions</em>, i.e. the ions present on both sides of the equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-)))) -> "NH"_ (4(aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-))))#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation that describes the reaction between ammonia and hydrochloric acid?</h1>
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Stefan V.
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Mar 9, 2017
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<div class="markdown"><p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonia and hydrochloric acid will react to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, as described by the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to release hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and produce chloride anions, <mathjax>#"Cl"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl" _((aq))^(-)#</mathjax></p>
</blockquote>
<p>Ammonium chloride is a <strong>soluble</strong> ionic compound, so it too will exist as ions in solution. </p>
<p>This means that the <strong>complete ionic equation</strong> that describes this reaction can be written like this</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + "Cl" _ ((aq))^(-) -> "NH"_ (4(aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>To get the <strong>net ionic equation</strong>, you must remove the <em>spectator ions</em>, i.e. the ions present on both sides of the equation</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-)))) -> "NH"_ (4(aq))^(+) + color(red)(cancel(color(black)("Cl" _ ((aq))^(-))))#</mathjax></p>
</blockquote>
<p>You will end up with</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote></div>
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</article> | What is the balanced chemical equation that describes the reaction between ammonia and hydrochloric acid? | null |
1,626 | ab9b1f74-6ddd-11ea-9ae8-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-composed-of-5-85-g-of-potassium-iodide-ki-dis | 0.28 mol/L | start physical_unit 14 14 molarity mol/l qc_end physical_unit 14 14 9 10 mass qc_end physical_unit 6 6 21 22 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] KI solution [IN] mol/L"}] | [{"type":"physical unit","value":"0.28 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] KI [=] \\pu{5.85 g}"},{"type":"physical unit","value":"Volume [OF] KI solution [=] \\pu{0.125 L}"},{"type":"other","value":"Enough water."}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution composed of 5.85 g of potassium iodide, #KI#, dissolved in enough water to make 0.125 L of solution?</h1> | null | 0.28 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> tells you how many <strong>moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium iodide, <mathjax>#"KI"#</mathjax>, are dissolved in exactly <strong>one liter</strong> of solution. </p>
<p>The first thing to do here is use the <strong>molar mass</strong> of potassium iodide to figure out how many moles you have in that sample </p>
<blockquote>
<p><mathjax>#5.85 color(red)(cancel(color(black)("g"))) * "1 mole KI"/(166.0color(red)(cancel(color(black)("g")))) = "0.03524 moles KI"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles of potassium iodide are dissolved in <mathjax>#"0.125 L"#</mathjax> of solution. Since you're interested in figuring out how many moles would be present in <mathjax>#"1 L"#</mathjax> of solution, all you have to do is <em>scale up</em> your current volume by a factor of </p>
<blockquote>
<p><mathjax>#(1 color(red)(cancel(color(black)("L"))))/(0.125color(red)(cancel(color(black)("L")))) = 8#</mathjax></p>
</blockquote>
<p>The volume goes up by a factor of <mathjax>#8#</mathjax>, which can only mean that the number of moles of solute must go up <strong>by the same factor</strong></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.03524 moles KI"/(0.125color(red)(cancel(color(black)("L solution")))) = "0.282 moles KI"#</mathjax></p>
</blockquote>
<p>Since you have <mathjax>#0.282#</mathjax> <strong>moles</strong> of potassium iodide in <mathjax>#"1 L"#</mathjax> of solution, it follows that the molarity will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.282 mol L"^(-1) = "0.282 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.282 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> tells you how many <strong>moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium iodide, <mathjax>#"KI"#</mathjax>, are dissolved in exactly <strong>one liter</strong> of solution. </p>
<p>The first thing to do here is use the <strong>molar mass</strong> of potassium iodide to figure out how many moles you have in that sample </p>
<blockquote>
<p><mathjax>#5.85 color(red)(cancel(color(black)("g"))) * "1 mole KI"/(166.0color(red)(cancel(color(black)("g")))) = "0.03524 moles KI"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles of potassium iodide are dissolved in <mathjax>#"0.125 L"#</mathjax> of solution. Since you're interested in figuring out how many moles would be present in <mathjax>#"1 L"#</mathjax> of solution, all you have to do is <em>scale up</em> your current volume by a factor of </p>
<blockquote>
<p><mathjax>#(1 color(red)(cancel(color(black)("L"))))/(0.125color(red)(cancel(color(black)("L")))) = 8#</mathjax></p>
</blockquote>
<p>The volume goes up by a factor of <mathjax>#8#</mathjax>, which can only mean that the number of moles of solute must go up <strong>by the same factor</strong></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.03524 moles KI"/(0.125color(red)(cancel(color(black)("L solution")))) = "0.282 moles KI"#</mathjax></p>
</blockquote>
<p>Since you have <mathjax>#0.282#</mathjax> <strong>moles</strong> of potassium iodide in <mathjax>#"1 L"#</mathjax> of solution, it follows that the molarity will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.282 mol L"^(-1) = "0.282 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution composed of 5.85 g of potassium iodide, #KI#, dissolved in enough water to make 0.125 L of solution?</h1>
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<div class="markdown"><p><mathjax>#"0.282 mol L"^(-1)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></strong> tells you how many <strong>moles</strong> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is potassium iodide, <mathjax>#"KI"#</mathjax>, are dissolved in exactly <strong>one liter</strong> of solution. </p>
<p>The first thing to do here is use the <strong>molar mass</strong> of potassium iodide to figure out how many moles you have in that sample </p>
<blockquote>
<p><mathjax>#5.85 color(red)(cancel(color(black)("g"))) * "1 mole KI"/(166.0color(red)(cancel(color(black)("g")))) = "0.03524 moles KI"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles of potassium iodide are dissolved in <mathjax>#"0.125 L"#</mathjax> of solution. Since you're interested in figuring out how many moles would be present in <mathjax>#"1 L"#</mathjax> of solution, all you have to do is <em>scale up</em> your current volume by a factor of </p>
<blockquote>
<p><mathjax>#(1 color(red)(cancel(color(black)("L"))))/(0.125color(red)(cancel(color(black)("L")))) = 8#</mathjax></p>
</blockquote>
<p>The volume goes up by a factor of <mathjax>#8#</mathjax>, which can only mean that the number of moles of solute must go up <strong>by the same factor</strong></p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.03524 moles KI"/(0.125color(red)(cancel(color(black)("L solution")))) = "0.282 moles KI"#</mathjax></p>
</blockquote>
<p>Since you have <mathjax>#0.282#</mathjax> <strong>moles</strong> of potassium iodide in <mathjax>#"1 L"#</mathjax> of solution, it follows that the molarity will be </p>
<blockquote>
<p><mathjax>#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.282 mol L"^(-1) = "0.282 M")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | What is the molarity of a solution composed of 5.85 g of potassium iodide, #KI#, dissolved in enough water to make 0.125 L of solution? | null |
1,627 | aa7cb2b0-6ddd-11ea-9959-ccda262736ce | https://socratic.org/questions/on-hot-days-you-may-have-noticed-that-potato-chip-bags-seem-to-inflate-even-thou | 270.68 mL | start physical_unit 54 55 volume ml qc_end physical_unit 54 55 25 26 volume qc_end physical_unit 54 55 32 33 temperature qc_end physical_unit 54 55 46 47 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the bag [IN] mL"}] | [{"type":"physical unit","value":"270.68 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the bag [=] \\pu{250 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the bag [=] \\pu{19 F}"},{"type":"physical unit","value":"Temperature2 [OF] the bag [=] \\pu{60 F}"}] | <h1 class="questionTitle" itemprop="name">On hot days you may have noticed that potato chip bags seem to inflate even though they have not been opened. If I have a 250 mL bag at a temperature of 19 F and I leave it in my car which has a temperature of 60 F what will the new voulume of the bag be?</h1> | null | 270.68 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mechanical strength of the material of the bag provides limits the expansion. Once it reaches the maximum volume, the pressure builds up in the bag. Assuming that the bag does not provide any resistance to the expansion of the gas within, and also assuming that the chips in the bag occupy negligible volume and ideal gas assumption is valid, you may use PV = nRT to solve the problem</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Cannot be determined</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mechanical strength of the material of the bag provides limits the expansion. Once it reaches the maximum volume, the pressure builds up in the bag. Assuming that the bag does not provide any resistance to the expansion of the gas within, and also assuming that the chips in the bag occupy negligible volume and ideal gas assumption is valid, you may use PV = nRT to solve the problem</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">On hot days you may have noticed that potato chip bags seem to inflate even though they have not been opened. If I have a 250 mL bag at a temperature of 19 F and I leave it in my car which has a temperature of 60 F what will the new voulume of the bag be?</h1>
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<div class="markdown"><p>Cannot be determined</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mechanical strength of the material of the bag provides limits the expansion. Once it reaches the maximum volume, the pressure builds up in the bag. Assuming that the bag does not provide any resistance to the expansion of the gas within, and also assuming that the chips in the bag occupy negligible volume and ideal gas assumption is valid, you may use PV = nRT to solve the problem</p></div>
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Rithin Mohan
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Apr 6, 2017
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<div class="markdown"><p>The new volume will be 271 ml. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The equation for Charles Law is <mathjax># V_1/T_1 = V_2/T_2#</mathjax> <br/>
Charles law is a direct relationship for volume and temperature as measured in degrees K or absolute temperature. </p>
<p>The temperatures are given in degree F not degree's K. </p>
<p>The temperatures must be changed first to degree's C and then degrees C must be changed to K (Kelvin). </p>
<p><mathjax>#C^o = ( F -32 ) xx 5/9#</mathjax> </p>
<p><mathjax>#T_1 C^o = (19-32) xx 5/9#</mathjax> </p>
<p><mathjax># T_1C^o = -7.2 #</mathjax></p>
<p><mathjax>#T_2C^o = ( 60 -32) xx 5/9#</mathjax> </p>
<p><mathjax># T_2C^o = 15.5 #</mathjax></p>
<p><mathjax># K = C^o + 273 #</mathjax></p>
<p><mathjax># T_1 K = -7.2 + 273 #</mathjax> </p>
<p><mathjax># T_1 K = 266 #</mathjax></p>
<p><mathjax># T_2K_ = 15.5 + 273 #</mathjax> </p>
<p><mathjax># T_2K = 288 #</mathjax> </p>
<p>These values for<mathjax># T_1 #</mathjax> and <mathjax># T_2#</mathjax> can be put into the equation for Charles Law </p>
<p><mathjax># V_1 = 250 ml
#</mathjax>V_2 = new volume. </p>
<p><mathjax># 250/266 = V_2/288#</mathjax> </p>
<p><mathjax># (250 xx 288)/266 = V_2#</mathjax> </p>
<p><mathjax># 271 ml = V_2#</mathjax> </p></div>
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</article> | On hot days you may have noticed that potato chip bags seem to inflate even though they have not been opened. If I have a 250 mL bag at a temperature of 19 F and I leave it in my car which has a temperature of 60 F what will the new voulume of the bag be? | null |
1,628 | ac2af880-6ddd-11ea-a29b-ccda262736ce | https://socratic.org/questions/what-volume-in-ml-of-220-m-ethanol-solution-contains-2-90-x-10-3-mol-ethanol | 13.2 mL | start physical_unit 7 8 volume ml qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 7 7 10 13 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] ethanol solution [IN] mL"}] | [{"type":"physical unit","value":"13.2 mL"}] | [{"type":"physical unit","value":"Molarity [OF] ethanol solution [=] \\pu{0.220 M}"},{"type":"physical unit","value":"Mole [OF] ethanol [=] \\pu{2.90 × 10^(-3) mol}"}] | <h1 class="questionTitle" itemprop="name">What volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol?</h1> | null | 13.2 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
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<div class="markdown"><p>The answer is 13.2 mL.</p></div>
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<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol?</h1>
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<div class="markdown"><p>The answer is 13.2 mL.</p></div>
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<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"13.2 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The idea here is that the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution can be converted to the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> </p>
</blockquote>
<p>of solution. In your case, a <mathjax>#"0.220-M"#</mathjax> solution of ethanol will contain <mathjax>#0.220#</mathjax> <strong>moles</strong> of ethanol for every <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. </p>
<p>This implies that <mathjax>#2.90 * 10^(-3)#</mathjax> <strong>moles</strong> of ethanol will be present in </p>
<blockquote>
<p><mathjax>#2.90 * color(blue)(cancel(color(black)(10^(-3))))color(red)(cancel(color(black)("moles ethanol"))) * overbrace((color(blue)(cancel(color(black)(10^3)))color(white)(.) "mL solution")/(0.220color(red)(cancel(color(black)("moles ethanol")))))^(color(blue)("= 0.220 M"))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)("13.2 mL solution")))#</mathjax></p>
</blockquote>
</blockquote>
</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 volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol? | null |
1,629 | ac11a166-6ddd-11ea-ac2e-ccda262736ce | https://socratic.org/questions/what-is-the-volume-in-the-liters-occupied-by-4-moles-of-so-2-at-6-atmospheres-an | 30 liters | start physical_unit 12 12 volume l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 12 12 14 15 pressure qc_end physical_unit 12 12 17 18 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] SO2 [IN] liters"}] | [{"type":"physical unit","value":"30 liters"}] | [{"type":"physical unit","value":"Mole [OF] SO2 [=] \\pu{4 moles}"},{"type":"physical unit","value":"Pressure [OF] SO2 [=] \\pu{6 atmospheres}"},{"type":"physical unit","value":"Temperature [OF] SO2 [=] \\pu{270 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the volume in the liters occupied by 4 moles of #SO_2# at 6 atmospheres and 270°C?</h1> | null | 30 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume.</p>
<blockquote></blockquote>
<p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n = "4 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(270 + 273.15) K" = "543.15 K"#</mathjax><br/>
<mathjax>#P = "6 atm"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V = (nRT)/P = (4 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"· color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 543.15 color(red)(cancel(color(black)("K"))))/(6 color(red)(cancel(color(black)("atm")))) = "30 L"#</mathjax><br/>
(1 significant figure)</p></div>
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<div class="markdown"><p>The volume is 30 L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume.</p>
<blockquote></blockquote>
<p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n = "4 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(270 + 273.15) K" = "543.15 K"#</mathjax><br/>
<mathjax>#P = "6 atm"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V = (nRT)/P = (4 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"· color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 543.15 color(red)(cancel(color(black)("K"))))/(6 color(red)(cancel(color(black)("atm")))) = "30 L"#</mathjax><br/>
(1 significant figure)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume in the liters occupied by 4 moles of #SO_2# at 6 atmospheres and 270°C?</h1>
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Dec 29, 2016
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<div class="markdown"><p>The volume is 30 L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the volume.</p>
<blockquote></blockquote>
<p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the Ideal Gas Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V = (nRT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#n = "4 mol"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(270 + 273.15) K" = "543.15 K"#</mathjax><br/>
<mathjax>#P = "6 atm"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V = (nRT)/P = (4 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"· color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 543.15 color(red)(cancel(color(black)("K"))))/(6 color(red)(cancel(color(black)("atm")))) = "30 L"#</mathjax><br/>
(1 significant figure)</p></div>
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</article> | What is the volume in the liters occupied by 4 moles of #SO_2# at 6 atmospheres and 270°C? | null |
1,630 | a8e138d3-6ddd-11ea-8e7e-ccda262736ce | https://socratic.org/questions/what-is-the-weight-of-4-30-mole-of-sodium | 98.86 g | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium [IN] g"}] | [{"type":"physical unit","value":"98.86 g"}] | [{"type":"physical unit","value":"Mole [OF] sodium [=] \\pu{4.30 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 4.30 mole of sodium? </h1> | null | 98.86 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>My Periodic Table tells me that it has a mass of <mathjax>#22.99*g*mol^-1#</mathjax>.</p>
<p>And for the mass associated with <mathjax>#4.30*mol#</mathjax> of sodium...we take the product...</p>
<p><mathjax>#4.30*molxx22.99*g*mol^-1~=99*g.#</mathjax></p></div>
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<div class="markdown"><p>Well, what is the mass of ONE MOLE of sodium metal?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>My Periodic Table tells me that it has a mass of <mathjax>#22.99*g*mol^-1#</mathjax>.</p>
<p>And for the mass associated with <mathjax>#4.30*mol#</mathjax> of sodium...we take the product...</p>
<p><mathjax>#4.30*molxx22.99*g*mol^-1~=99*g.#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 4.30 mole of sodium? </h1>
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<div class="markdown"><p>Well, what is the mass of ONE MOLE of sodium metal?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>My Periodic Table tells me that it has a mass of <mathjax>#22.99*g*mol^-1#</mathjax>.</p>
<p>And for the mass associated with <mathjax>#4.30*mol#</mathjax> of sodium...we take the product...</p>
<p><mathjax>#4.30*molxx22.99*g*mol^-1~=99*g.#</mathjax></p></div>
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<div class="markdown"><p>The mass of <mathjax>#"4.30 mol Na"#</mathjax> is <mathjax>#"98.9 g"#</mathjax>.</p></div>
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<div class="markdown"><p>Multiply the given moles of sodium by its molar mass <mathjax>#("22.990 g/mol")#</mathjax>.</p>
<p><mathjax>#4.30color(red)cancel(color(black)("mol Na"))xx(22.990"g Na")/(1color(red)cancel(color(black)("mol Na")))="98.9 g Na"#</mathjax></p></div>
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</article> | What is the mass of 4.30 mole of sodium? | null |
1,631 | a83de968-6ddd-11ea-b0ae-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-the-combination-of-cobalt-ii-and-chlorine | CoCl2 | start chemical_formula qc_end substance 9 10 qc_end substance 12 12 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the combination [IN] default"}] | [{"type":"chemical equation","value":"CoCl2"}] | [{"type":"substance name","value":"Cobalt (II)"},{"type":"substance name","value":"Chlorine"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for the combination of cobalt (II) and chlorine?</h1> | null | CoCl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"ous"#</mathjax> endings traditionally denoted the lower oxidation states. Anhydrous <mathjax>#CoCl_2#</mathjax> is a light blue powder. Its coordination complex with water is a beautiful purple colour. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Cobaltous chloride"#</mathjax> is <mathjax>#CoCl_2#</mathjax>, </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"ous"#</mathjax> endings traditionally denoted the lower oxidation states. Anhydrous <mathjax>#CoCl_2#</mathjax> is a light blue powder. Its coordination complex with water is a beautiful purple colour. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for the combination of cobalt (II) and chlorine?</h1>
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<div class="markdown"><p><mathjax>#"Cobaltous chloride"#</mathjax> is <mathjax>#CoCl_2#</mathjax>, </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <mathjax>#"ous"#</mathjax> endings traditionally denoted the lower oxidation states. Anhydrous <mathjax>#CoCl_2#</mathjax> is a light blue powder. Its coordination complex with water is a beautiful purple colour. </p></div>
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</article> | What is the chemical formula for the combination of cobalt (II) and chlorine? | null |
1,632 | aadc8208-6ddd-11ea-809c-ccda262736ce | https://socratic.org/questions/a-mixture-of-gases-with-a-pressure-of-800-0-hg-contains-60-nitrogen-and-40-oxyge | 0.42 atm | start physical_unit 15 15 partial_pressure atm qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] oxygen [IN] atm"}] | [{"type":"physical unit","value":"0.42 atm"}] | [{"type":"physical unit","value":"Pressure [OF] gases mixture [=] \\pu{800.0 mmHg}"},{"type":"physical unit","value":"Percent by volume [OF] nitrogen in gases mixture [=] \\pu{60%}"},{"type":"physical unit","value":"Percent by volume [OF] oxygen in gases mixture [=] \\pu{40%}"}] | <h1 class="questionTitle" itemprop="name">A mixture of gases with a pressure of 800.0 Hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture? </h1> | null | 0.42 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I am unhappy in quoting a pressure measurement of <a href="https://socratic.org/questions/how-many-milliliters-of-hydrogen-will-be-produced-by-the-reaction-of-23-699-g-zn">over 760 mm Hg</a>, as it is something you would try to avoid in the lab.......</p>
<p>We have <mathjax>#P_"Total"=1.052*atm#</mathjax>. But by definition, <mathjax>#P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm#</mathjax></p>
<p>..............and thus <mathjax>#P_"dioxygen"=0.421*atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We assume a pressure of <mathjax>#(800*mm*Hg)/(760*mm*Hg*atm^-1)=1.052*atm#</mathjax>. We finally get <mathjax>#P_"dioxygen"=0.421*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I am unhappy in quoting a pressure measurement of <a href="https://socratic.org/questions/how-many-milliliters-of-hydrogen-will-be-produced-by-the-reaction-of-23-699-g-zn">over 760 mm Hg</a>, as it is something you would try to avoid in the lab.......</p>
<p>We have <mathjax>#P_"Total"=1.052*atm#</mathjax>. But by definition, <mathjax>#P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm#</mathjax></p>
<p>..............and thus <mathjax>#P_"dioxygen"=0.421*atm#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A mixture of gases with a pressure of 800.0 Hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture? </h1>
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anor277
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<div class="markdown"><p>We assume a pressure of <mathjax>#(800*mm*Hg)/(760*mm*Hg*atm^-1)=1.052*atm#</mathjax>. We finally get <mathjax>#P_"dioxygen"=0.421*atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I am unhappy in quoting a pressure measurement of <a href="https://socratic.org/questions/how-many-milliliters-of-hydrogen-will-be-produced-by-the-reaction-of-23-699-g-zn">over 760 mm Hg</a>, as it is something you would try to avoid in the lab.......</p>
<p>We have <mathjax>#P_"Total"=1.052*atm#</mathjax>. But by definition, <mathjax>#P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm#</mathjax></p>
<p>..............and thus <mathjax>#P_"dioxygen"=0.421*atm#</mathjax></p></div>
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</article> | A mixture of gases with a pressure of 800.0 Hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture? | null |
1,633 | a8bccf68-6ddd-11ea-97fb-ccda262736ce | https://socratic.org/questions/write-the-net-ionic-equation-that-describes-the-reaction-that-occurs-when-a-solu | Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s) | start chemical_equation qc_end substance 15 16 qc_end substance 23 24 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)"}] | [{"type":"substance name","value":"Barium nitrate"},{"type":"substance name","value":"Sulfate ions"}] | <h1 class="questionTitle" itemprop="name">Write the net ionic equation that describes the reaction that occurs when a solution of barium nitrate is added to a flask containing sulfate ions?
</h1> | null | Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that barium nitrate is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>So when barium nitrate is dissolved in water, it exists as dissociated ions</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix this solution with a solution that contains sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of the solution will be produced. This insoluble solid is called <em>barium sulfate</em>, <mathjax>#"BaSO"_4#</mathjax>. </p>
<p>So you can say that when these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, you have</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Since the nitrate anions are present on both sides of the chemical equation, you can say that they are <strong>spectator ions</strong>.</p>
<p>This implies that you can eliminate them</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#</mathjax></p>
</blockquote>
<p>Barium sulfate is a <em>white solid</em> that will precipitate out of the solution.</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000IXJ9uV4OiRM" src="https://useruploads.socratic.org/pTtVPO35S864AT6Lbj2O_Fphoto-50110011B-2RM.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that barium nitrate is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>So when barium nitrate is dissolved in water, it exists as dissociated ions</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix this solution with a solution that contains sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of the solution will be produced. This insoluble solid is called <em>barium sulfate</em>, <mathjax>#"BaSO"_4#</mathjax>. </p>
<p>So you can say that when these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, you have</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Since the nitrate anions are present on both sides of the chemical equation, you can say that they are <strong>spectator ions</strong>.</p>
<p>This implies that you can eliminate them</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#</mathjax></p>
</blockquote>
<p>Barium sulfate is a <em>white solid</em> that will precipitate out of the solution.</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000IXJ9uV4OiRM" src="https://useruploads.socratic.org/pTtVPO35S864AT6Lbj2O_Fphoto-50110011B-2RM.jpg"/> </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Write the net ionic equation that describes the reaction that occurs when a solution of barium nitrate is added to a flask containing sulfate ions?
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<div class="markdown"><p><mathjax>#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that barium nitrate is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>So when barium nitrate is dissolved in water, it exists as dissociated ions</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix this solution with a solution that contains sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of the solution will be produced. This insoluble solid is called <em>barium sulfate</em>, <mathjax>#"BaSO"_4#</mathjax>. </p>
<p>So you can say that when these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, you have</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + 2"NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Since the nitrate anions are present on both sides of the chemical equation, you can say that they are <strong>spectator ions</strong>.</p>
<p>This implies that you can eliminate them</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#</mathjax></p>
</blockquote>
<p>Barium sulfate is a <em>white solid</em> that will precipitate out of the solution.</p>
<p><img alt="http://fphoto.photoshelter.com/image/I0000IXJ9uV4OiRM" src="https://useruploads.socratic.org/pTtVPO35S864AT6Lbj2O_Fphoto-50110011B-2RM.jpg"/> </p></div>
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</article> | Write the net ionic equation that describes the reaction that occurs when a solution of barium nitrate is added to a flask containing sulfate ions?
| null |
1,634 | a837d30a-6ddd-11ea-826f-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-7-mol-h-2o-2 | 238 g | start physical_unit 7 7 mass g qc_end physical_unit 7 7 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] H2O2 [IN] g"}] | [{"type":"physical unit","value":"238 g"}] | [{"type":"physical unit","value":"Mole [OF] H2O2 [=] \\pu{7 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 7 mol #H_2O_2#?</h1> | null | 238 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#7*(2*1+2*16) = 238#</mathjax></p></div>
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<div class="markdown"><p>238</p></div>
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<div class="markdown"><p><mathjax>#7*(2*1+2*16) = 238#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 7 mol #H_2O_2#?</h1>
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Mc'Coob
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<div class="markdown"><p>238</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#7*(2*1+2*16) = 238#</mathjax></p></div>
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</article> | What is the mass of 7 mol #H_2O_2#? | null |
1,635 | acad8ec0-6ddd-11ea-9aa3-ccda262736ce | https://socratic.org/questions/consider-the-reaction-2-no-g-o2-g-2-no2-g-at-430-c-an-equilibrium-mixture-consis | 1.5 × 10^5 | start physical_unit 24 25 kp none qc_end physical_unit 24 25 1 2 temperature qc_end physical_unit 11 11 8 9 mole qc_end physical_unit 15 15 12 13 mole qc_end physical_unit 20 20 17 18 mole qc_end chemical_equation 33 39 qc_end end | [{"type":"physical unit","value":"Kp [OF] the reaction"}] | [{"type":"physical unit","value":"1.5 × 10^5"}] | [{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{430 ℃}"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{0.020 moles}"},{"type":"physical unit","value":"Mole [OF] NO [=] \\pu{0.040 moles}"},{"type":"physical unit","value":"Mole [OF] NO2 [=] \\pu{0.96 moles}"},{"type":"physical unit","value":"Total pressure [OF] the gases [=] \\pu{0.20 atm}"},{"type":"chemical equation","value":"2 NO(g) + O2(g) <=> 2 NO2(g)"}] | <h1 class="questionTitle" itemprop="name">At #430^@"C"#, an equilibrium mixture consists of #0.020# moles of #"O"_2#, #0.040# moles of #"NO"#, and #0.96# moles of #"NO"_2#. Calculate #Kp# for the reaction, given the total pressure is #"0.20 atm"# ? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#2"NO"(g) + "O"_2(g) rightleftharpoons 2"NO"_2(g)#</mathjax>. </p>
<p>I know that the answer is <mathjax>#1.5 * 10^5#</mathjax> atm with significant figures, but I am not sure how to get this. Help is much appreciated! </p></div>
</h2>
</div>
</div> | 1.5 × 10^5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that</p>
<blockquote>
<p><mathjax>#2"NO"_ ((g)) + "O"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>At <mathjax>#430^@"C"#</mathjax>, the reaction vessel contains</p>
<blockquote>
<ul>
<li><mathjax>#"0.020 moles O"_2#</mathjax></li>
<li><mathjax>#"0.040 moles NO"#</mathjax></li>
<li><mathjax>#"0.96 moles NO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Even without doing any calculations, you should be able to predict that </p>
<blockquote>
<p><mathjax>#K_p " >> " 1#</mathjax></p>
</blockquote>
<p>because the equilibrium mixture contains significantly more <em>product</em> than reactants, which implies that the equilibrium lies <strong>to the right</strong>, i.e. the forward reaction is favored at this temperature. </p>
<p>Now, at equilibrium, the <strong>total pressure</strong> inside the reaction vessel is equal to <mathjax>#"0.20 atm"#</mathjax>.</p>
<p>As you know, the <strong>partial pressures</strong> of the three gases will depend on their respective <em><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></em>, as described by <strong>Dalton's Law of Partial Pressures</strong>. </p>
<p>For nitrogen oxide, you will have</p>
<blockquote>
<p><mathjax>#P_ "NO" = chi_ "NO" * P_"total"#</mathjax></p>
</blockquote>
<p>Here </p>
<blockquote>
<p><mathjax>#chi_ "NO" = (0.040 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) = 0.040/1.02#</mathjax></p>
</blockquote>
<p>represents the <strong>mole fraction</strong> of nitrogen oxide in the reaction vessel <em>at equilibrium</em>. </p>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#P_ "NO" = 0.040/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Use the same approach to calculate the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen gas and nitrogen dioxide. For oxygen gas, you will have</p>
<blockquote>
<p><mathjax>#P_ ("O"_ 2) = (0.020 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("O"_ 2)= 0.020/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Similarly, for nitrogen dioxide, you will have</p>
<blockquote>
<p><mathjax>#P_ ("NO"_ 2) = (0.96 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("NO"_ 2) = 0.96/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>The equilibrium constant for this reaction can be written using the partial pressures of the three gases</p>
<blockquote>
<p><mathjax>#K_p = (P_ ("NO"_ 2)^2)/( P_ "NO"^2 * P_ ("O"_ 2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = ( (0.96/1.02 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))))/( (0.040/1.20 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))) * 0.020/1.02 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 1.02)/(0.040^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 0.020 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * 1.02)/(0.040^2 * 0.020 * "0.20 atm") = color(darkgreen)(ul(color(black)(1.5 * 10^5color(white)(.)"atm"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.5 * 10^5"atm"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that</p>
<blockquote>
<p><mathjax>#2"NO"_ ((g)) + "O"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>At <mathjax>#430^@"C"#</mathjax>, the reaction vessel contains</p>
<blockquote>
<ul>
<li><mathjax>#"0.020 moles O"_2#</mathjax></li>
<li><mathjax>#"0.040 moles NO"#</mathjax></li>
<li><mathjax>#"0.96 moles NO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Even without doing any calculations, you should be able to predict that </p>
<blockquote>
<p><mathjax>#K_p " >> " 1#</mathjax></p>
</blockquote>
<p>because the equilibrium mixture contains significantly more <em>product</em> than reactants, which implies that the equilibrium lies <strong>to the right</strong>, i.e. the forward reaction is favored at this temperature. </p>
<p>Now, at equilibrium, the <strong>total pressure</strong> inside the reaction vessel is equal to <mathjax>#"0.20 atm"#</mathjax>.</p>
<p>As you know, the <strong>partial pressures</strong> of the three gases will depend on their respective <em><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></em>, as described by <strong>Dalton's Law of Partial Pressures</strong>. </p>
<p>For nitrogen oxide, you will have</p>
<blockquote>
<p><mathjax>#P_ "NO" = chi_ "NO" * P_"total"#</mathjax></p>
</blockquote>
<p>Here </p>
<blockquote>
<p><mathjax>#chi_ "NO" = (0.040 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) = 0.040/1.02#</mathjax></p>
</blockquote>
<p>represents the <strong>mole fraction</strong> of nitrogen oxide in the reaction vessel <em>at equilibrium</em>. </p>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#P_ "NO" = 0.040/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Use the same approach to calculate the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen gas and nitrogen dioxide. For oxygen gas, you will have</p>
<blockquote>
<p><mathjax>#P_ ("O"_ 2) = (0.020 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("O"_ 2)= 0.020/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Similarly, for nitrogen dioxide, you will have</p>
<blockquote>
<p><mathjax>#P_ ("NO"_ 2) = (0.96 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("NO"_ 2) = 0.96/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>The equilibrium constant for this reaction can be written using the partial pressures of the three gases</p>
<blockquote>
<p><mathjax>#K_p = (P_ ("NO"_ 2)^2)/( P_ "NO"^2 * P_ ("O"_ 2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = ( (0.96/1.02 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))))/( (0.040/1.20 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))) * 0.020/1.02 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 1.02)/(0.040^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 0.020 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * 1.02)/(0.040^2 * 0.020 * "0.20 atm") = color(darkgreen)(ul(color(black)(1.5 * 10^5color(white)(.)"atm"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">At #430^@"C"#, an equilibrium mixture consists of #0.020# moles of #"O"_2#, #0.040# moles of #"NO"#, and #0.96# moles of #"NO"_2#. Calculate #Kp# for the reaction, given the total pressure is #"0.20 atm"# ? </h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#2"NO"(g) + "O"_2(g) rightleftharpoons 2"NO"_2(g)#</mathjax>. </p>
<p>I know that the answer is <mathjax>#1.5 * 10^5#</mathjax> atm with significant figures, but I am not sure how to get this. Help is much appreciated! </p></div>
</h2>
</div>
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Stefan V.
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May 22, 2017
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<div class="markdown"><p><mathjax>#1.5 * 10^5"atm"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that</p>
<blockquote>
<p><mathjax>#2"NO"_ ((g)) + "O"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>At <mathjax>#430^@"C"#</mathjax>, the reaction vessel contains</p>
<blockquote>
<ul>
<li><mathjax>#"0.020 moles O"_2#</mathjax></li>
<li><mathjax>#"0.040 moles NO"#</mathjax></li>
<li><mathjax>#"0.96 moles NO"_2#</mathjax></li>
</ul>
</blockquote>
<p>Even without doing any calculations, you should be able to predict that </p>
<blockquote>
<p><mathjax>#K_p " >> " 1#</mathjax></p>
</blockquote>
<p>because the equilibrium mixture contains significantly more <em>product</em> than reactants, which implies that the equilibrium lies <strong>to the right</strong>, i.e. the forward reaction is favored at this temperature. </p>
<p>Now, at equilibrium, the <strong>total pressure</strong> inside the reaction vessel is equal to <mathjax>#"0.20 atm"#</mathjax>.</p>
<p>As you know, the <strong>partial pressures</strong> of the three gases will depend on their respective <em><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></em>, as described by <strong>Dalton's Law of Partial Pressures</strong>. </p>
<p>For nitrogen oxide, you will have</p>
<blockquote>
<p><mathjax>#P_ "NO" = chi_ "NO" * P_"total"#</mathjax></p>
</blockquote>
<p>Here </p>
<blockquote>
<p><mathjax>#chi_ "NO" = (0.040 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) = 0.040/1.02#</mathjax></p>
</blockquote>
<p>represents the <strong>mole fraction</strong> of nitrogen oxide in the reaction vessel <em>at equilibrium</em>. </p>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#P_ "NO" = 0.040/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Use the same approach to calculate the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of oxygen gas and nitrogen dioxide. For oxygen gas, you will have</p>
<blockquote>
<p><mathjax>#P_ ("O"_ 2) = (0.020 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("O"_ 2)= 0.020/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>Similarly, for nitrogen dioxide, you will have</p>
<blockquote>
<p><mathjax>#P_ ("NO"_ 2) = (0.96 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"#</mathjax></p>
<p><mathjax>#P_ ("NO"_ 2) = 0.96/1.02 * "0.20 atm"#</mathjax></p>
</blockquote>
<p>The equilibrium constant for this reaction can be written using the partial pressures of the three gases</p>
<blockquote>
<p><mathjax>#K_p = (P_ ("NO"_ 2)^2)/( P_ "NO"^2 * P_ ("O"_ 2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = ( (0.96/1.02 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))))/( (0.040/1.20 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))) * 0.020/1.02 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 1.02)/(0.040^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 0.020 * "0.20 atm")#</mathjax></p>
<p><mathjax>#K_p = (0.96^2 * 1.02)/(0.040^2 * 0.020 * "0.20 atm") = color(darkgreen)(ul(color(black)(1.5 * 10^5color(white)(.)"atm"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | At #430^@"C"#, an equilibrium mixture consists of #0.020# moles of #"O"_2#, #0.040# moles of #"NO"#, and #0.96# moles of #"NO"_2#. Calculate #Kp# for the reaction, given the total pressure is #"0.20 atm"# ? |
#2"NO"(g) + "O"_2(g) rightleftharpoons 2"NO"_2(g)#.
I know that the answer is #1.5 * 10^5# atm with significant figures, but I am not sure how to get this. Help is much appreciated!
|
1,636 | acf20158-6ddd-11ea-9f35-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-molar-solubility-of-mn-oh-2-ksp-1-6e-13-in-pure-wate | 3.42 × 10^(-5) | start physical_unit 8 8 molal_solubility none qc_end physical_unit 8 8 11 13 equilibrium_constant_k qc_end substance 15 16 qc_end end | [{"type":"physical unit","value":"Molal solubility [OF] Mn(OH)2"}] | [{"type":"physical unit","value":"3.42 × 10^(-5)"}] | [{"type":"physical unit","value":"Ksp [OF] Mn(OH)2 [=] \\pu{1.6 × 10^(-13)}"},{"type":"substance name","value":"Pure water"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in
pure water?
</h1> | null | 3.42 × 10^(-5) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K_(sp) = 1.6xx10^(-13)=[Mn^(2+)][OH^-]^2#</mathjax></p>
<p>If we call the solubility of manganese(II) hydroxide <mathjax>#S#</mathjax>, then by the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction:</p>
<p><mathjax>#K_(sp)=S(2S)^2=4S^3#</mathjax>.</p>
<p>So <mathjax>#S = root(3)((1.6 xx 10^(-13))/4)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Mn(OH)_2(s) rightleftharpoons Mn^(2+) + 2OH^-#</mathjax></p>
<p><mathjax>#K_(sp) = 1.6xx10^(-13)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K_(sp) = 1.6xx10^(-13)=[Mn^(2+)][OH^-]^2#</mathjax></p>
<p>If we call the solubility of manganese(II) hydroxide <mathjax>#S#</mathjax>, then by the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction:</p>
<p><mathjax>#K_(sp)=S(2S)^2=4S^3#</mathjax>.</p>
<p>So <mathjax>#S = root(3)((1.6 xx 10^(-13))/4)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in
pure water?
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<div class="markdown"><p><mathjax>#Mn(OH)_2(s) rightleftharpoons Mn^(2+) + 2OH^-#</mathjax></p>
<p><mathjax>#K_(sp) = 1.6xx10^(-13)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#K_(sp) = 1.6xx10^(-13)=[Mn^(2+)][OH^-]^2#</mathjax></p>
<p>If we call the solubility of manganese(II) hydroxide <mathjax>#S#</mathjax>, then by the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the reaction:</p>
<p><mathjax>#K_(sp)=S(2S)^2=4S^3#</mathjax>.</p>
<p>So <mathjax>#S = root(3)((1.6 xx 10^(-13))/4)#</mathjax></p></div>
</div>
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</article> | How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in
pure water?
| null |
1,637 | aa72f007-6ddd-11ea-b2a2-ccda262736ce | https://socratic.org/questions/375ml-of-a-0-455m-sodium-chloride-nacl-solution-is-diluted-with-1-88l-of-water-w | 0.08 M | start physical_unit 8 8 concentration mol/l qc_end physical_unit 9 9 0 1 volume qc_end physical_unit 8 8 4 5 concentration qc_end physical_unit 16 16 13 14 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] NaCl solution [IN] M"}] | [{"type":"physical unit","value":"0.08 M"}] | [{"type":"physical unit","value":"Volume [OF] NaCl solution [=] \\pu{375 mL}"},{"type":"physical unit","value":"Concentration1 [OF] NaCl solution [=] \\pu{0.455 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{1.88 L}"}] | <h1 class="questionTitle" itemprop="name">375mL of a 0.455M sodium chloride (#NaCl#) solution is diluted with 1.88L of water. What is the new concentration in molarity?</h1> | null | 0.08 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to keep in mind when you're dealing with a <strong>dilution</strong> is that the <em>number of moles of solute</em> <strong>must remain constant</strong>.</p>
<p>That means that the stock solution <strong>and</strong> the diluted solution will contain the <strong>same number of moles</strong> of solute. </p>
<p>So, your strategy here will be to calculate how many moles of solute, which in your case is sodium chloride, <mathjax>#"NaCl"#</mathjax>, are present in the stock solution. </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_(NaCl) = "0.455 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(375 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(green)("volume expressed in liters"))#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.1706 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of sodium chloride must be present in the <em>diluted solution</em>. Calculate the <strong>volume</strong> of the diluted solution first</p>
<blockquote>
<p><mathjax>#V_"diluted" = V_"stock" + V_"water"#</mathjax></p>
<p><mathjax>#V_"diluted" = 375 * 10^(-3)"L" + "1.88 L" = "2.255 L"#</mathjax></p>
</blockquote>
<p>This means that the concentration of the diluted solution will be </p>
<blockquote>
<p><mathjax>#c_"diluted" = "0.1706 moles"/"2.255 L" = color(green)(|bar(ul(color(white)(a/a)"0.0767 mol L"^(-1)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>#"0.0767 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to keep in mind when you're dealing with a <strong>dilution</strong> is that the <em>number of moles of solute</em> <strong>must remain constant</strong>.</p>
<p>That means that the stock solution <strong>and</strong> the diluted solution will contain the <strong>same number of moles</strong> of solute. </p>
<p>So, your strategy here will be to calculate how many moles of solute, which in your case is sodium chloride, <mathjax>#"NaCl"#</mathjax>, are present in the stock solution. </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_(NaCl) = "0.455 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(375 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(green)("volume expressed in liters"))#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.1706 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of sodium chloride must be present in the <em>diluted solution</em>. Calculate the <strong>volume</strong> of the diluted solution first</p>
<blockquote>
<p><mathjax>#V_"diluted" = V_"stock" + V_"water"#</mathjax></p>
<p><mathjax>#V_"diluted" = 375 * 10^(-3)"L" + "1.88 L" = "2.255 L"#</mathjax></p>
</blockquote>
<p>This means that the concentration of the diluted solution will be </p>
<blockquote>
<p><mathjax>#c_"diluted" = "0.1706 moles"/"2.255 L" = color(green)(|bar(ul(color(white)(a/a)"0.0767 mol L"^(-1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">375mL of a 0.455M sodium chloride (#NaCl#) solution is diluted with 1.88L of water. What is the new concentration in molarity?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.0767 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to keep in mind when you're dealing with a <strong>dilution</strong> is that the <em>number of moles of solute</em> <strong>must remain constant</strong>.</p>
<p>That means that the stock solution <strong>and</strong> the diluted solution will contain the <strong>same number of moles</strong> of solute. </p>
<p>So, your strategy here will be to calculate how many moles of solute, which in your case is sodium chloride, <mathjax>#"NaCl"#</mathjax>, are present in the stock solution. </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_(NaCl) = "0.455 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(375 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(green)("volume expressed in liters"))#</mathjax></p>
<p><mathjax>#n_(NaCl) = "0.1706 moles NaCl"#</mathjax></p>
</blockquote>
<p>This is exactly how many moles of sodium chloride must be present in the <em>diluted solution</em>. Calculate the <strong>volume</strong> of the diluted solution first</p>
<blockquote>
<p><mathjax>#V_"diluted" = V_"stock" + V_"water"#</mathjax></p>
<p><mathjax>#V_"diluted" = 375 * 10^(-3)"L" + "1.88 L" = "2.255 L"#</mathjax></p>
</blockquote>
<p>This means that the concentration of the diluted solution will be </p>
<blockquote>
<p><mathjax>#c_"diluted" = "0.1706 moles"/"2.255 L" = color(green)(|bar(ul(color(white)(a/a)"0.0767 mol L"^(-1)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> | 375mL of a 0.455M sodium chloride (#NaCl#) solution is diluted with 1.88L of water. What is the new concentration in molarity? | null |
1,638 | ac590d6e-6ddd-11ea-b159-ccda262736ce | https://socratic.org/questions/a-7-96-gram-sample-of-silver-reacts-with-oxygen-to-form-8-55-gram-of-the-metal-o | Ag2O | start chemical_formula qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 15 16 11 12 mass qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the oxide [IN] default"}] | [{"type":"chemical equation","value":"Ag2O"}] | [{"type":"physical unit","value":"Mass [OF] silver sample [=] \\pu{7.96 grams}"},{"type":"physical unit","value":"Mass [OF] metal oxide [=] \\pu{8.55 grams}"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide. What is the formula of the oxide?</h1> | null | Ag2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The empirical formula is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of <mathjax>#"Ag"#</mathjax> to <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"Mass of Ag = 7.96 g"#</mathjax></p>
<p><mathjax>#"Mass of silver oxide = mass of Ag + mass of O"#</mathjax></p>
<p><mathjax>#"8.55 g = 7.96 g + mass of O"#</mathjax></p>
<p><mathjax>#"Mass of O = (8.55 – 7.96) g = 0.59 g"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of Ag" = 7.96 color(red)(cancel(color(black)("g Ag"))) × "1 mol Ag"/(107.9color(red)(cancel(color(black)( "g Ag")))) = "0.073 77 mol Ag"#</mathjax></p>
<p><mathjax>#"Moles of O "= 0.59 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.0369 mol O"#</mathjax></p>
<p>To get this into an integer ratio, we divide both numbers by the smaller value.</p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(m)"Ag" color(white)(XXXm)7.96 color(white)(Xm)0.073 77
color(white)(Xll)2.00color(white)(mmm)2)#</mathjax><br/>
<mathjax>#color(white)(ml)"O" color(white)(XXXXl)0.59 color(white)(mm)"0.0369 color(white)(Xml)1 color(white)(mmmml)1#</mathjax></p>
<p>There are 2 mol of <mathjax>#"Ag"#</mathjax> for 1 mol of <mathjax>#"O"#</mathjax>.</p>
<p>The empirical formula of silver oxide is <mathjax>#"Ag"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Here is a video that illustrates how to determine an empirical formula.</p>
<p>
<iframe src="https://www.youtube.com/embed/r1OsFpz_-qY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#"Ag"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The empirical formula is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of <mathjax>#"Ag"#</mathjax> to <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"Mass of Ag = 7.96 g"#</mathjax></p>
<p><mathjax>#"Mass of silver oxide = mass of Ag + mass of O"#</mathjax></p>
<p><mathjax>#"8.55 g = 7.96 g + mass of O"#</mathjax></p>
<p><mathjax>#"Mass of O = (8.55 – 7.96) g = 0.59 g"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of Ag" = 7.96 color(red)(cancel(color(black)("g Ag"))) × "1 mol Ag"/(107.9color(red)(cancel(color(black)( "g Ag")))) = "0.073 77 mol Ag"#</mathjax></p>
<p><mathjax>#"Moles of O "= 0.59 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.0369 mol O"#</mathjax></p>
<p>To get this into an integer ratio, we divide both numbers by the smaller value.</p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(m)"Ag" color(white)(XXXm)7.96 color(white)(Xm)0.073 77
color(white)(Xll)2.00color(white)(mmm)2)#</mathjax><br/>
<mathjax>#color(white)(ml)"O" color(white)(XXXXl)0.59 color(white)(mm)"0.0369 color(white)(Xml)1 color(white)(mmmml)1#</mathjax></p>
<p>There are 2 mol of <mathjax>#"Ag"#</mathjax> for 1 mol of <mathjax>#"O"#</mathjax>.</p>
<p>The empirical formula of silver oxide is <mathjax>#"Ag"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Here is a video that illustrates how to determine an empirical formula.</p>
<p>
<iframe src="https://www.youtube.com/embed/r1OsFpz_-qY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide. What is the formula of the oxide?</h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#"Ag"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The empirical formula is the simplest whole-number ratio of atoms in a compound.</p>
<p>The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of <mathjax>#"Ag"#</mathjax> to <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"Mass of Ag = 7.96 g"#</mathjax></p>
<p><mathjax>#"Mass of silver oxide = mass of Ag + mass of O"#</mathjax></p>
<p><mathjax>#"8.55 g = 7.96 g + mass of O"#</mathjax></p>
<p><mathjax>#"Mass of O = (8.55 – 7.96) g = 0.59 g"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of Ag" = 7.96 color(red)(cancel(color(black)("g Ag"))) × "1 mol Ag"/(107.9color(red)(cancel(color(black)( "g Ag")))) = "0.073 77 mol Ag"#</mathjax></p>
<p><mathjax>#"Moles of O "= 0.59 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.0369 mol O"#</mathjax></p>
<p>To get this into an integer ratio, we divide both numbers by the smaller value.</p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-———)(color(white)(m)"Ag" color(white)(XXXm)7.96 color(white)(Xm)0.073 77
color(white)(Xll)2.00color(white)(mmm)2)#</mathjax><br/>
<mathjax>#color(white)(ml)"O" color(white)(XXXXl)0.59 color(white)(mm)"0.0369 color(white)(Xml)1 color(white)(mmmml)1#</mathjax></p>
<p>There are 2 mol of <mathjax>#"Ag"#</mathjax> for 1 mol of <mathjax>#"O"#</mathjax>.</p>
<p>The empirical formula of silver oxide is <mathjax>#"Ag"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Here is a video that illustrates how to determine an empirical formula.</p>
<p>
<iframe src="https://www.youtube.com/embed/r1OsFpz_-qY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide. What is the formula of the oxide? | null |
1,639 | a8d9098a-6ddd-11ea-8c92-ccda262736ce | https://socratic.org/questions/how-many-moles-of-methane-ch-4-are-there-in-80-grams-of-methane | 5 moles | start physical_unit 5 5 mole mol qc_end physical_unit 4 4 9 10 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] CH4 [IN] moles"}] | [{"type":"physical unit","value":"5 moles"}] | [{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{80 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles of methane #CH_4# are there in 80 grams of methane?</h1> | null | 5 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the number of moles, all we have to do is use the following relationship:</p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p>And the molar mass of methane may be read straight off your Periodic Table:</p>
<p><mathjax>#"Molar mass of methane, "CH_4=(12.011+4xx1.00794)*g*mol^-1#</mathjax></p>
<p><mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(80*cancelg)/(16.0*cancelg*mol^-1)=??mol#</mathjax>.</p>
<p>Note that this calculation is consistent dimensionally:</p>
<p><mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5*mol#</mathjax> of methane are present in this mass..........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the number of moles, all we have to do is use the following relationship:</p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p>And the molar mass of methane may be read straight off your Periodic Table:</p>
<p><mathjax>#"Molar mass of methane, "CH_4=(12.011+4xx1.00794)*g*mol^-1#</mathjax></p>
<p><mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(80*cancelg)/(16.0*cancelg*mol^-1)=??mol#</mathjax>.</p>
<p>Note that this calculation is consistent dimensionally:</p>
<p><mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of methane #CH_4# are there in 80 grams of methane?</h1>
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anor277
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<div class="markdown"><p><mathjax>#5*mol#</mathjax> of methane are present in this mass..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To find the number of moles, all we have to do is use the following relationship:</p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p>And the molar mass of methane may be read straight off your Periodic Table:</p>
<p><mathjax>#"Molar mass of methane, "CH_4=(12.011+4xx1.00794)*g*mol^-1#</mathjax></p>
<p><mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(80*cancelg)/(16.0*cancelg*mol^-1)=??mol#</mathjax>.</p>
<p>Note that this calculation is consistent dimensionally:</p>
<p><mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax></p></div>
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</article> | How many moles of methane #CH_4# are there in 80 grams of methane? | null |
1,640 | a8485a58-6ddd-11ea-8a9b-ccda262736ce | https://socratic.org/questions/a-25-0-g-sample-of-an-aluminum-mercury-amalgam-used-for-filling-cavities-is-7-00 | 23.3 g | start physical_unit 6 6 mass g qc_end physical_unit 6 7 1 2 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum [IN] g"}] | [{"type":"physical unit","value":"23.3 g"}] | [{"type":"physical unit","value":"Mass [OF] aluminum/mercury amalgam sample [=] \\pu{25.0 g}"},{"type":"physical unit","value":"Percentage by mass [OF] mercury in sample [=] \\pu{7.00%}"}] | <h1 class="questionTitle" itemprop="name">A 25.0 g sample of an aluminum/mercury amalgam used for filling cavities is 7.00% mercury by mass. What is the mass of aluminum in the 25.0 g sample? (Use correct numbers of significant figures)</h1> | null | 23.3 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that the sample is <mathjax>#7.00%#</mathjax> mercury <strong>by mass</strong>, which automatically implies that it is <mathjax>#93.00%#</mathjax> <em>aluminium</em> <strong>by mass</strong>. </p>
<p>That is the case because the amalgam contains two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, aluminium and mercury, and so their respective <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass <strong>must</strong> add up to give <mathjax>#100%#</mathjax>. </p>
<p>Now, a compound's <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> tells you how many grams of each of its constituent elements you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of said compound. </p>
<p>In your case, you know that you get <mathjax>#"93.00 g"#</mathjax> of aluminium and <mathjax>#"7.00 g"#</mathjax> of mercury <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of amalgam. This means that the given sample will contain</p>
<blockquote>
<p><mathjax>#25.0 color(red)(cancel(color(black)("g amalgam"))) * "93.00 g Al"/(100color(red)(cancel(color(black)("g amalgam")))) = "23.25 g Al"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("mass of Al " = " 23.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#"23.3 g Al"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that the sample is <mathjax>#7.00%#</mathjax> mercury <strong>by mass</strong>, which automatically implies that it is <mathjax>#93.00%#</mathjax> <em>aluminium</em> <strong>by mass</strong>. </p>
<p>That is the case because the amalgam contains two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, aluminium and mercury, and so their respective <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass <strong>must</strong> add up to give <mathjax>#100%#</mathjax>. </p>
<p>Now, a compound's <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> tells you how many grams of each of its constituent elements you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of said compound. </p>
<p>In your case, you know that you get <mathjax>#"93.00 g"#</mathjax> of aluminium and <mathjax>#"7.00 g"#</mathjax> of mercury <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of amalgam. This means that the given sample will contain</p>
<blockquote>
<p><mathjax>#25.0 color(red)(cancel(color(black)("g amalgam"))) * "93.00 g Al"/(100color(red)(cancel(color(black)("g amalgam")))) = "23.25 g Al"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("mass of Al " = " 23.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A 25.0 g sample of an aluminum/mercury amalgam used for filling cavities is 7.00% mercury by mass. What is the mass of aluminum in the 25.0 g sample? (Use correct numbers of significant figures)</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-10-20T23:56:31" itemprop="dateCreated">
Oct 20, 2016
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<div class="markdown"><p><mathjax>#"23.3 g Al"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that the sample is <mathjax>#7.00%#</mathjax> mercury <strong>by mass</strong>, which automatically implies that it is <mathjax>#93.00%#</mathjax> <em>aluminium</em> <strong>by mass</strong>. </p>
<p>That is the case because the amalgam contains two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, aluminium and mercury, and so their respective <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass <strong>must</strong> add up to give <mathjax>#100%#</mathjax>. </p>
<p>Now, a compound's <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> tells you how many grams of each of its constituent elements you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of said compound. </p>
<p>In your case, you know that you get <mathjax>#"93.00 g"#</mathjax> of aluminium and <mathjax>#"7.00 g"#</mathjax> of mercury <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of amalgam. This means that the given sample will contain</p>
<blockquote>
<p><mathjax>#25.0 color(red)(cancel(color(black)("g amalgam"))) * "93.00 g Al"/(100color(red)(cancel(color(black)("g amalgam")))) = "23.25 g Al"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("mass of Al " = " 23.3 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | A 25.0 g sample of an aluminum/mercury amalgam used for filling cavities is 7.00% mercury by mass. What is the mass of aluminum in the 25.0 g sample? (Use correct numbers of significant figures) | null |
1,641 | a8f7ef94-6ddd-11ea-8eda-ccda262736ce | https://socratic.org/questions/how-many-ml-of-a-0-124-moles-l-solution-of-potassium-permanganate-contain-0-722- | 36.8 mL | start physical_unit 7 7 volume ml qc_end physical_unit 9 10 5 6 molarity qc_end physical_unit 15 16 12 13 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] potassium permanganate solution [IN] mL"}] | [{"type":"physical unit","value":"36.8 mL"}] | [{"type":"physical unit","value":"Molarity [OF] potassium permanganate solution [=] \\pu{0.124 moles/L}"},{"type":"physical unit","value":"Mass [OF] the salt [=] \\pu{0.722 g}"}] | <h1 class="questionTitle" itemprop="name">How many mL of a 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt?</h1> | null | 36.8 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is a measure of how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>for every liter</strong> of a given solution. </p>
<p>In your case, a solution of sodium permanganate, <mathjax>#"KMnO"_4#</mathjax>, is said to have a <strong>molarity</strong> of <mathjax>#"0.124 mol L"^(-1)#</mathjax>. This tells you that <strong>every</strong> <mathjax>#"1 L"#</mathjax> of this solution will contain <mathjax>#0.124#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, one cool way of solving this problem would be to <strong>convert</strong> the molarity of the solution from <em>moles per liter</em> to <em>grams per milliliter</em>.</p>
<p>Potassium permanganate has a <strong>molar mass</strong> of <mathjax>#"158.034 g mol"^(-1)#</mathjax>, which means that <mathjax>#0.124#</mathjax> <strong>moles</strong> of potassium permanganate would be equivalent to</p>
<blockquote>
<p><mathjax>#0.124 color(red)(cancel(color(black)("moles KMnO"_4))) * "158.034 g"/(1color(red)(cancel(color(black)("mole KMnO"_4)))) = "19.60 g"#</mathjax></p>
</blockquote>
<p>You also know that you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 L" = 10^3"mL")))#</mathjax></p>
</blockquote>
<p>which means that the molarity of the solution is equivalent to a concentration of</p>
<blockquote>
<p><mathjax>#("0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution")#</mathjax></p>
</blockquote>
<p>Now you're ready to use this as a <strong>conversion factor</strong> to determine how many <em>milliliters</em> of solution would contain <mathjax>#"0.722 g"#</mathjax> of potassium permanganate</p>
<blockquote>
<p><mathjax>#0.722 color(red)(cancel(color(black)("g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"36.8 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is a measure of how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>for every liter</strong> of a given solution. </p>
<p>In your case, a solution of sodium permanganate, <mathjax>#"KMnO"_4#</mathjax>, is said to have a <strong>molarity</strong> of <mathjax>#"0.124 mol L"^(-1)#</mathjax>. This tells you that <strong>every</strong> <mathjax>#"1 L"#</mathjax> of this solution will contain <mathjax>#0.124#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, one cool way of solving this problem would be to <strong>convert</strong> the molarity of the solution from <em>moles per liter</em> to <em>grams per milliliter</em>.</p>
<p>Potassium permanganate has a <strong>molar mass</strong> of <mathjax>#"158.034 g mol"^(-1)#</mathjax>, which means that <mathjax>#0.124#</mathjax> <strong>moles</strong> of potassium permanganate would be equivalent to</p>
<blockquote>
<p><mathjax>#0.124 color(red)(cancel(color(black)("moles KMnO"_4))) * "158.034 g"/(1color(red)(cancel(color(black)("mole KMnO"_4)))) = "19.60 g"#</mathjax></p>
</blockquote>
<p>You also know that you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 L" = 10^3"mL")))#</mathjax></p>
</blockquote>
<p>which means that the molarity of the solution is equivalent to a concentration of</p>
<blockquote>
<p><mathjax>#("0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution")#</mathjax></p>
</blockquote>
<p>Now you're ready to use this as a <strong>conversion factor</strong> to determine how many <em>milliliters</em> of solution would contain <mathjax>#"0.722 g"#</mathjax> of potassium permanganate</p>
<blockquote>
<p><mathjax>#0.722 color(red)(cancel(color(black)("g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many mL of a 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt?</h1>
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Stefan V.
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Nov 21, 2016
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<div class="markdown"><p><mathjax>#"36.8 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is a measure of how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>for every liter</strong> of a given solution. </p>
<p>In your case, a solution of sodium permanganate, <mathjax>#"KMnO"_4#</mathjax>, is said to have a <strong>molarity</strong> of <mathjax>#"0.124 mol L"^(-1)#</mathjax>. This tells you that <strong>every</strong> <mathjax>#"1 L"#</mathjax> of this solution will contain <mathjax>#0.124#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, one cool way of solving this problem would be to <strong>convert</strong> the molarity of the solution from <em>moles per liter</em> to <em>grams per milliliter</em>.</p>
<p>Potassium permanganate has a <strong>molar mass</strong> of <mathjax>#"158.034 g mol"^(-1)#</mathjax>, which means that <mathjax>#0.124#</mathjax> <strong>moles</strong> of potassium permanganate would be equivalent to</p>
<blockquote>
<p><mathjax>#0.124 color(red)(cancel(color(black)("moles KMnO"_4))) * "158.034 g"/(1color(red)(cancel(color(black)("mole KMnO"_4)))) = "19.60 g"#</mathjax></p>
</blockquote>
<p>You also know that you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 L" = 10^3"mL")))#</mathjax></p>
</blockquote>
<p>which means that the molarity of the solution is equivalent to a concentration of</p>
<blockquote>
<p><mathjax>#("0.124 moles KMnO"_4)/("1 L solution") = ("19.60 g KMnO"_4)/(10^3"mL solution")#</mathjax></p>
</blockquote>
<p>Now you're ready to use this as a <strong>conversion factor</strong> to determine how many <em>milliliters</em> of solution would contain <mathjax>#"0.722 g"#</mathjax> of potassium permanganate</p>
<blockquote>
<p><mathjax>#0.722 color(red)(cancel(color(black)("g KMnO"_4))) * (10^3"mL solution")/(19.60color(red)(cancel(color(black)("g KMnO"_4)))) = color(darkgreen)(ul(color(black)("36.8 mL solution")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many mL of a 0.124 moles/L solution of potassium permanganate contain 0.722 g of the salt? | null |
1,642 | a8aaf0a8-6ddd-11ea-bdc1-ccda262736ce | https://socratic.org/questions/if-the-theoretical-yield-for-a-chemical-reaction-is-100-g-and-the-percentage-yie | 70.00 g | start physical_unit 6 7 actual_yield g qc_end physical_unit 5 7 9 10 theoretical_yield qc_end physical_unit 5 7 16 16 percent_yield qc_end end | [{"type":"physical unit","value":"Actual yield [OF] the chemical reaction [IN] g"}] | [{"type":"physical unit","value":"70.00 g"}] | [{"type":"physical unit","value":"Theoretical yield [OF] a chemical reaction [=] \\pu{100 g}"},{"type":"physical unit","value":"Percentage yield [OF] a chemical reaction [=] \\pu{70.0%}"}] | <h1 class="questionTitle" itemprop="name">If the theoretical yield for a chemical reaction is 100. g and the percentage yield is 70.0%, what is the actual yield?</h1> | null | 70.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>actual yield</strong> is a term used to describe the amount of product that is <em>actually</em> produced by a given chemical reaction. </p>
<p>By comparison, the <strong>theoretical yield</strong> is a term we use to denote the amount of product that <em>could</em> be produced by a given chemical reaction. </p>
<p>In other words, the theoretical yield corresponds to the case in which a reaction produces the <strong>maximum amount</strong> of a product given a certain amount of reactants. </p>
<p>Now, the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> is a measure of the difference between what we <em>could</em> get and what we <em>actually</em> get. Simply put, a reaction's percent yield tells us how much product we actually get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that we <em>could</em> theoretically get. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("% yield" = "what we actually get"/"what we could theoretically get" xx 100%)))#</mathjax></p>
</blockquote>
<p>In this case, you know that the theoretical yield for given reaction is equal to <mathjax>#"100. g"#</mathjax> and that its percent yield is equal to <mathjax>#70.0%#</mathjax>. </p>
<p>This means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that could theoretically be produced, we only get <mathjax>#"70.0 g"#</mathjax>. </p>
<p>Therefore, you can say that the actual yield is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("actual yield = 70.0 g")))#</mathjax></p>
</blockquote>
<p>Notice that you get the same result by using the equation</p>
<blockquote>
<p><mathjax>#70.0% = "actual yield"/"100. g" xx 100%#</mathjax></p>
</blockquote>
<p>Rearrange to find</p>
<blockquote>
<p><mathjax>#"actual yield" = (70.0 color(red)(cancel(color(black)(%))) * color(blue)(cancel(color(black)(100.))) "g")/(color(blue)(cancel(color(black)(100.)))color(red)(cancel(color(black)(%)))) = color(darkgreen)(ul(color(black)("70.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"70.0 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>actual yield</strong> is a term used to describe the amount of product that is <em>actually</em> produced by a given chemical reaction. </p>
<p>By comparison, the <strong>theoretical yield</strong> is a term we use to denote the amount of product that <em>could</em> be produced by a given chemical reaction. </p>
<p>In other words, the theoretical yield corresponds to the case in which a reaction produces the <strong>maximum amount</strong> of a product given a certain amount of reactants. </p>
<p>Now, the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> is a measure of the difference between what we <em>could</em> get and what we <em>actually</em> get. Simply put, a reaction's percent yield tells us how much product we actually get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that we <em>could</em> theoretically get. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("% yield" = "what we actually get"/"what we could theoretically get" xx 100%)))#</mathjax></p>
</blockquote>
<p>In this case, you know that the theoretical yield for given reaction is equal to <mathjax>#"100. g"#</mathjax> and that its percent yield is equal to <mathjax>#70.0%#</mathjax>. </p>
<p>This means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that could theoretically be produced, we only get <mathjax>#"70.0 g"#</mathjax>. </p>
<p>Therefore, you can say that the actual yield is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("actual yield = 70.0 g")))#</mathjax></p>
</blockquote>
<p>Notice that you get the same result by using the equation</p>
<blockquote>
<p><mathjax>#70.0% = "actual yield"/"100. g" xx 100%#</mathjax></p>
</blockquote>
<p>Rearrange to find</p>
<blockquote>
<p><mathjax>#"actual yield" = (70.0 color(red)(cancel(color(black)(%))) * color(blue)(cancel(color(black)(100.))) "g")/(color(blue)(cancel(color(black)(100.)))color(red)(cancel(color(black)(%)))) = color(darkgreen)(ul(color(black)("70.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">If the theoretical yield for a chemical reaction is 100. g and the percentage yield is 70.0%, what is the actual yield?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"70.0 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>actual yield</strong> is a term used to describe the amount of product that is <em>actually</em> produced by a given chemical reaction. </p>
<p>By comparison, the <strong>theoretical yield</strong> is a term we use to denote the amount of product that <em>could</em> be produced by a given chemical reaction. </p>
<p>In other words, the theoretical yield corresponds to the case in which a reaction produces the <strong>maximum amount</strong> of a product given a certain amount of reactants. </p>
<p>Now, the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong> is a measure of the difference between what we <em>could</em> get and what we <em>actually</em> get. Simply put, a reaction's percent yield tells us how much product we actually get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that we <em>could</em> theoretically get. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("% yield" = "what we actually get"/"what we could theoretically get" xx 100%)))#</mathjax></p>
</blockquote>
<p>In this case, you know that the theoretical yield for given reaction is equal to <mathjax>#"100. g"#</mathjax> and that its percent yield is equal to <mathjax>#70.0%#</mathjax>. </p>
<p>This means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of product that could theoretically be produced, we only get <mathjax>#"70.0 g"#</mathjax>. </p>
<p>Therefore, you can say that the actual yield is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("actual yield = 70.0 g")))#</mathjax></p>
</blockquote>
<p>Notice that you get the same result by using the equation</p>
<blockquote>
<p><mathjax>#70.0% = "actual yield"/"100. g" xx 100%#</mathjax></p>
</blockquote>
<p>Rearrange to find</p>
<blockquote>
<p><mathjax>#"actual yield" = (70.0 color(red)(cancel(color(black)(%))) * color(blue)(cancel(color(black)(100.))) "g")/(color(blue)(cancel(color(black)(100.)))color(red)(cancel(color(black)(%)))) = color(darkgreen)(ul(color(black)("70.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If the theoretical yield for a chemical reaction is 100. g and the percentage yield is 70.0%, what is the actual yield? | null |
1,643 | ab44b709-6ddd-11ea-a585-ccda262736ce | https://socratic.org/questions/what-is-a-pattern-in-which-an-element-is-dentical-throughout-known-as | 2 AlBr3 + 3 Cl2 -> 2 AlCl3 + 3 Br2 | start chemical_equation qc_end substance 5 6 qc_end substance 8 9 qc_end substance 11 12 qc_end substance 14 15 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 AlBr3 + 3 Cl2 -> 2 AlCl3 + 3 Br2"}] | [{"type":"substance name","value":"Aluminum bromide"},{"type":"substance name","value":"Chlorine gas"},{"type":"substance name","value":"Aluminum chloride"},{"type":"substance name","value":"Bromine gas"}] | <h1 class="questionTitle" itemprop="name">What is the equation for aluminum bromide plus chlorine gas yields aluminum chloride and bromine gas?</h1> | null | 2 AlBr3 + 3 Cl2 -> 2 AlCl3 + 3 Br2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Translate names to symbols<br/>
<mathjax>#AlBr_3 + Cl_2 -> AlCl_3 + Br_2#</mathjax><br/>
Balance Cl and Br by adding atoms<br/>
<mathjax>#AlBr_3 + 3/2Cl_2 -> AlCl_3 + 3/2Br_2#</mathjax><br/>
Multiply by 2 to get to the nearest whole number<br/>
<mathjax>#2AlBr_3 + 3Cl_2 -> 2AlCl_3 + 3Br_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2AlBr_3 + 3Cl_2 -> 2AlCl_3 + 3Br_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Translate names to symbols<br/>
<mathjax>#AlBr_3 + Cl_2 -> AlCl_3 + Br_2#</mathjax><br/>
Balance Cl and Br by adding atoms<br/>
<mathjax>#AlBr_3 + 3/2Cl_2 -> AlCl_3 + 3/2Br_2#</mathjax><br/>
Multiply by 2 to get to the nearest whole number<br/>
<mathjax>#2AlBr_3 + 3Cl_2 -> 2AlCl_3 + 3Br_2#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the equation for aluminum bromide plus chlorine gas yields aluminum chloride and bromine gas?</h1>
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<div class="markdown"><p><mathjax>#2AlBr_3 + 3Cl_2 -> 2AlCl_3 + 3Br_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Translate names to symbols<br/>
<mathjax>#AlBr_3 + Cl_2 -> AlCl_3 + Br_2#</mathjax><br/>
Balance Cl and Br by adding atoms<br/>
<mathjax>#AlBr_3 + 3/2Cl_2 -> AlCl_3 + 3/2Br_2#</mathjax><br/>
Multiply by 2 to get to the nearest whole number<br/>
<mathjax>#2AlBr_3 + 3Cl_2 -> 2AlCl_3 + 3Br_2#</mathjax></p></div>
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How can I balance this equation? ___ AlBr3 + ____ K2SO4 ---> ____ KBr + ____ Al2(SO4)3
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</article> | What is the equation for aluminum bromide plus chlorine gas yields aluminum chloride and bromine gas? | null |
1,644 | aa0cfd5a-6ddd-11ea-bbb7-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-equation-for-the-complete-combustion-of-c-4h-8 | 2 C4H8 + 13 O2 -> 8 CO2 + 10 H2O | start chemical_equation qc_end chemical_equation 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the complete combustion"}] | [{"type":"chemical equation","value":"2 C4H8 + 13 O2 -> 8 CO2 + 10 H2O"}] | [{"type":"chemical equation","value":"C4H8"}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced equation for the complete combustion of #C_4H_8#?</h1> | null | 2 C4H8 + 13 O2 -> 8 CO2 + 10 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>In a combustion reaction, <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> are converted to <mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>Start by writing the unbalanced equation.</p>
<p><mathjax>#"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Put a <mathjax>#1#</mathjax> in front of the most complicated formula, <mathjax>#"C"_4"H"_10#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1.</strong> Balance <mathjax>#"C"#</mathjax>.</p>
<p>Put a <mathjax>#4#</mathjax> in front of <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Balance <mathjax>#"H"#</mathjax>.</p>
<p>Put a <mathjax>#5#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + color(green)(5)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Balance <mathjax>#"O"#</mathjax>.</p>
<p>We have 13 <mathjax>#"O"#</mathjax> atoms on the right and 2 <mathjax>#"O"#</mathjax> atoms on the left.</p>
<p>We can't balance <mathjax>#"O"#</mathjax> without using fractions.</p>
<p>We start over, multiplying each coefficient by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)"C"_4"H"_10 + "O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Balance <mathjax>#"O"#</mathjax> atoms by putting <mathjax>#13#</mathjax> in front of <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax># color(red)(2)"C"_4"H"_10 + color(blue)(13)"O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient.</p>
<blockquote></blockquote>
<p><strong>Step 4.</strong> Check that atoms are balanced.</p>
<p><mathjax>#"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"#</mathjax><br/>
<mathjax>#color(white)(m)"C" color(white)(mmmmmll)8color(white)(mmmmmmml)8#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmmmm)20color(white)(mmmmmmm)20#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmmmm)26color(white)(mmmmmmm)26#</mathjax></p>
<p>All atoms are balanced.</p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>In a combustion reaction, <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> are converted to <mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>Start by writing the unbalanced equation.</p>
<p><mathjax>#"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Put a <mathjax>#1#</mathjax> in front of the most complicated formula, <mathjax>#"C"_4"H"_10#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1.</strong> Balance <mathjax>#"C"#</mathjax>.</p>
<p>Put a <mathjax>#4#</mathjax> in front of <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Balance <mathjax>#"H"#</mathjax>.</p>
<p>Put a <mathjax>#5#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + color(green)(5)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Balance <mathjax>#"O"#</mathjax>.</p>
<p>We have 13 <mathjax>#"O"#</mathjax> atoms on the right and 2 <mathjax>#"O"#</mathjax> atoms on the left.</p>
<p>We can't balance <mathjax>#"O"#</mathjax> without using fractions.</p>
<p>We start over, multiplying each coefficient by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)"C"_4"H"_10 + "O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Balance <mathjax>#"O"#</mathjax> atoms by putting <mathjax>#13#</mathjax> in front of <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax># color(red)(2)"C"_4"H"_10 + color(blue)(13)"O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient.</p>
<blockquote></blockquote>
<p><strong>Step 4.</strong> Check that atoms are balanced.</p>
<p><mathjax>#"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"#</mathjax><br/>
<mathjax>#color(white)(m)"C" color(white)(mmmmmll)8color(white)(mmmmmmml)8#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmmmm)20color(white)(mmmmmmm)20#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmmmm)26color(white)(mmmmmmm)26#</mathjax></p>
<p>All atoms are balanced.</p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you write a balanced equation for the complete combustion of #C_4H_8#?</h1>
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Ernest Z.
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<div class="markdown"><p><mathjax>#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>In a combustion reaction, <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> are converted to <mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>Start by writing the unbalanced equation.</p>
<p><mathjax>#"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<p>Put a <mathjax>#1#</mathjax> in front of the most complicated formula, <mathjax>#"C"_4"H"_10#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1.</strong> Balance <mathjax>#"C"#</mathjax>.</p>
<p>Put a <mathjax>#4#</mathjax> in front of <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Balance <mathjax>#"H"#</mathjax>.</p>
<p>Put a <mathjax>#5#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_4"H"_10 + "O"_2 → color(orange)(4)"CO"_2 + color(green)(5)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Balance <mathjax>#"O"#</mathjax>.</p>
<p>We have 13 <mathjax>#"O"#</mathjax> atoms on the right and 2 <mathjax>#"O"#</mathjax> atoms on the left.</p>
<p>We can't balance <mathjax>#"O"#</mathjax> without using fractions.</p>
<p>We start over, multiplying each coefficient by <mathjax>#2#</mathjax>.</p>
<p><mathjax>#color(red)(2)"C"_4"H"_10 + "O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Balance <mathjax>#"O"#</mathjax> atoms by putting <mathjax>#13#</mathjax> in front of <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax># color(red)(2)"C"_4"H"_10 + color(blue)(13)"O"_2 → color(orange)(8)"CO"_2 + color(green)(10)"H"_2"O"#</mathjax></p>
<p>Every formula now has a coefficient.</p>
<blockquote></blockquote>
<p><strong>Step 4.</strong> Check that atoms are balanced.</p>
<p><mathjax>#"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"#</mathjax><br/>
<mathjax>#color(white)(m)"C" color(white)(mmmmmll)8color(white)(mmmmmmml)8#</mathjax><br/>
<mathjax>#color(white)(m)"H"color(white)(mmmmm)20color(white)(mmmmmmm)20#</mathjax><br/>
<mathjax>#color(white)(m)"O"color(white)(mmmmm)26color(white)(mmmmmmm)26#</mathjax></p>
<p>All atoms are balanced.</p>
<blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#2"C"_4"H"_10 + 13"O"_2 → 8"CO"_2 + 10"H"_2"O"#</mathjax></p></div>
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</article> | How do you write a balanced equation for the complete combustion of #C_4H_8#? | null |
1,645 | ac3b2d78-6ddd-11ea-9e45-ccda262736ce | https://socratic.org/questions/hydrogen-sulfide-reacts-with-sulfur-dioxide-to-give-h2o-and-s-h2s-so2-h2o-s-soli-1 | 806.85 Celsius | start physical_unit 23 24 temperature °c qc_end chemical_equation 11 17 qc_end physical_unit 23 24 20 21 volume qc_end physical_unit 23 24 26 27 pressure qc_end physical_unit 4 4 29 30 mass qc_end end | [{"type":"physical unit","value":"Temperature [OF] H2S gas [IN] Celsius"}] | [{"type":"physical unit","value":"806.85 Celsius"}] | [{"type":"chemical equation","value":"H2S + SO2 -> H2O + S(solid)"},{"type":"physical unit","value":"Volume [OF] H2S gas [=] \\pu{6.0 L}"},{"type":"physical unit","value":"Pressure [OF] H2S gas [=] \\pu{750 torr}"},{"type":"physical unit","value":"Mass [OF] sulfur [=] \\pu{3.2 g}"}] | <h1 class="questionTitle" itemprop="name">Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,
H2S + SO2 = H2O + S(solid), unbalanced.
If 6.0 L of H2S gas at 750 torr produced 3.2 g of sulfur, what is the temperature in Celsius?</h1> | null | 806.85 Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are four steps in this problem:</p>
<ol>
<li>Write the balanced chemical equation.</li>
<li>Calculate the moles of sulfur.</li>
<li>Calculate the moles of H₂S.</li>
<li>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the temperature.</li>
</ol>
<p><strong>1.</strong> Write the balanced chemical equation</p>
<p>2H₂S + SO₂ → 2H₂O + 3S</p>
<p><strong>2.</strong> Calculate the moles of sulfur</p>
<p>Moles of S = 3.2 g S × <mathjax>#(1"mol S")/(32.06"g S")#</mathjax> = 0.0998 mol S</p>
<p><strong>3.</strong> Calculate the moles of H₂S</p>
<p>Moles of H₂S = 0.0998 mol S × <mathjax>#(2"mol H₂S")/(3"mol S")#</mathjax> = 0.0665 mol H₂S</p>
<p><strong>4.</strong> Calculate the temperature</p>
<p><mathjax>#P#</mathjax> = 750 Torr × <mathjax>#(1"atm")/(760"Torr")#</mathjax> = 0.9868 atm</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (0.9868"atm" × 6.0"L")/(0.0665"mol" × 0.082 06"L·atm·K⁻¹mol⁻¹")#</mathjax> = 1080 K</p>
<p><mathjax>#T#</mathjax> = (1080 – 273.15) °C = 810 °C (2 significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The temperature is 810 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are four steps in this problem:</p>
<ol>
<li>Write the balanced chemical equation.</li>
<li>Calculate the moles of sulfur.</li>
<li>Calculate the moles of H₂S.</li>
<li>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the temperature.</li>
</ol>
<p><strong>1.</strong> Write the balanced chemical equation</p>
<p>2H₂S + SO₂ → 2H₂O + 3S</p>
<p><strong>2.</strong> Calculate the moles of sulfur</p>
<p>Moles of S = 3.2 g S × <mathjax>#(1"mol S")/(32.06"g S")#</mathjax> = 0.0998 mol S</p>
<p><strong>3.</strong> Calculate the moles of H₂S</p>
<p>Moles of H₂S = 0.0998 mol S × <mathjax>#(2"mol H₂S")/(3"mol S")#</mathjax> = 0.0665 mol H₂S</p>
<p><strong>4.</strong> Calculate the temperature</p>
<p><mathjax>#P#</mathjax> = 750 Torr × <mathjax>#(1"atm")/(760"Torr")#</mathjax> = 0.9868 atm</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (0.9868"atm" × 6.0"L")/(0.0665"mol" × 0.082 06"L·atm·K⁻¹mol⁻¹")#</mathjax> = 1080 K</p>
<p><mathjax>#T#</mathjax> = (1080 – 273.15) °C = 810 °C (2 significant figures)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,
H2S + SO2 = H2O + S(solid), unbalanced.
If 6.0 L of H2S gas at 750 torr produced 3.2 g of sulfur, what is the temperature in Celsius?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2014-08-09T18:58:35" itemprop="dateCreated">
Aug 9, 2014
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<div class="markdown"><p>The temperature is 810 °C.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are four steps in this problem:</p>
<ol>
<li>Write the balanced chemical equation.</li>
<li>Calculate the moles of sulfur.</li>
<li>Calculate the moles of H₂S.</li>
<li>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to calculate the temperature.</li>
</ol>
<p><strong>1.</strong> Write the balanced chemical equation</p>
<p>2H₂S + SO₂ → 2H₂O + 3S</p>
<p><strong>2.</strong> Calculate the moles of sulfur</p>
<p>Moles of S = 3.2 g S × <mathjax>#(1"mol S")/(32.06"g S")#</mathjax> = 0.0998 mol S</p>
<p><strong>3.</strong> Calculate the moles of H₂S</p>
<p>Moles of H₂S = 0.0998 mol S × <mathjax>#(2"mol H₂S")/(3"mol S")#</mathjax> = 0.0665 mol H₂S</p>
<p><strong>4.</strong> Calculate the temperature</p>
<p><mathjax>#P#</mathjax> = 750 Torr × <mathjax>#(1"atm")/(760"Torr")#</mathjax> = 0.9868 atm</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (0.9868"atm" × 6.0"L")/(0.0665"mol" × 0.082 06"L·atm·K⁻¹mol⁻¹")#</mathjax> = 1080 K</p>
<p><mathjax>#T#</mathjax> = (1080 – 273.15) °C = 810 °C (2 significant figures)</p></div>
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</article> | Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,
H2S + SO2 = H2O + S(solid), unbalanced.
If 6.0 L of H2S gas at 750 torr produced 3.2 g of sulfur, what is the temperature in Celsius? | null |
1,646 | acda8524-6ddd-11ea-b1b4-ccda262736ce | https://socratic.org/questions/what-is-the-mass-in-grams-of-1-02-10-24-atoms-manganese | 9.0 grams | start physical_unit 11 11 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] manganese [IN] grams"}] | [{"type":"physical unit","value":"9.0 grams"}] | [{"type":"physical unit","value":"Number [OF] manganese atoms [=] \\pu{1.02 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">What is the mass in grams of #1.02 * 10^24# atoms manganese?</h1> | null | 9.0 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of manganese has a mass of <mathjax>#54.94*g#</mathjax>.<br/>
Thus, <mathjax>#"Mass"#</mathjax><br/>
<mathjax>#=(1.02xx10^24" atoms manganese")/(6.022xx10^24" atoms of manganese "mol^-1)xx54.94*g*mol^-1#</mathjax> </p>
<p><mathjax>#~=9.0*g#</mathjax> </p></div>
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<div class="markdown"><p>One mole of stuff represents <mathjax>#6.022140857(74)xx10^23#</mathjax> individual items of that stuff.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of manganese has a mass of <mathjax>#54.94*g#</mathjax>.<br/>
Thus, <mathjax>#"Mass"#</mathjax><br/>
<mathjax>#=(1.02xx10^24" atoms manganese")/(6.022xx10^24" atoms of manganese "mol^-1)xx54.94*g*mol^-1#</mathjax> </p>
<p><mathjax>#~=9.0*g#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass in grams of #1.02 * 10^24# atoms manganese?</h1>
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Jun 21, 2016
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<div class="markdown"><p>One mole of stuff represents <mathjax>#6.022140857(74)xx10^23#</mathjax> individual items of that stuff.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of manganese has a mass of <mathjax>#54.94*g#</mathjax>.<br/>
Thus, <mathjax>#"Mass"#</mathjax><br/>
<mathjax>#=(1.02xx10^24" atoms manganese")/(6.022xx10^24" atoms of manganese "mol^-1)xx54.94*g*mol^-1#</mathjax> </p>
<p><mathjax>#~=9.0*g#</mathjax> </p></div>
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</article> | What is the mass in grams of #1.02 * 10^24# atoms manganese? | null |
1,647 | ab6e1a52-6ddd-11ea-be37-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-with-oh-2-0-times-10-10-m | 4.30 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 10 13 [oh-] qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"4.30"}] | [{"type":"physical unit","value":"[OH-] [OF] the solution [=] \\pu{2.0 × 10^(-10) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution with #[OH^-] = 2.0 times 10^-10# #M#?</h1> | null | 4.30 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#2H_2Orightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>We know that in aqueous solution under standard conditions that:</p>
<p><mathjax>#[H_3O^+][HO^-]=K_w=10^-14#</mathjax></p>
<p>We take <mathjax>#log_10#</mathjax> OF BOTH SIDES:</p>
<p><mathjax>#log_10[H_3O^+] +log_10[HO^-]=log_10(10^-14)#</mathjax></p>
<p>But by definition, the logarithmic function is that power to which you raise the base to get the specific answer.........i.e. if <mathjax>#log_aB=c#</mathjax>, then <mathjax>#a^c=B#</mathjax>. </p>
<p>And thus <mathjax>#log_10(10^-14)=-14#</mathjax></p>
<p>And so <mathjax>#log_10[H_3O^+] +log_10[HO^-]=-14#</mathjax></p>
<p>OR <mathjax>#-log_10[H_3O^+] -log_10[HO^-]=14#</mathjax></p>
<p>But we define <mathjax>#pH=-log_10[H_3O^+]#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax>, and so...........</p>
<p><mathjax>#14=pH+pOH#</mathjax>. </p>
<p>And now finally, <mathjax>#pH=14-pOH#</mathjax> </p>
<p><mathjax>#14-(-log_10(2.0xx10^-10)=4.30#</mathjax></p></div>
</div>
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<div class="markdown"><p>We find <mathjax>#pH~=4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#2H_2Orightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>We know that in aqueous solution under standard conditions that:</p>
<p><mathjax>#[H_3O^+][HO^-]=K_w=10^-14#</mathjax></p>
<p>We take <mathjax>#log_10#</mathjax> OF BOTH SIDES:</p>
<p><mathjax>#log_10[H_3O^+] +log_10[HO^-]=log_10(10^-14)#</mathjax></p>
<p>But by definition, the logarithmic function is that power to which you raise the base to get the specific answer.........i.e. if <mathjax>#log_aB=c#</mathjax>, then <mathjax>#a^c=B#</mathjax>. </p>
<p>And thus <mathjax>#log_10(10^-14)=-14#</mathjax></p>
<p>And so <mathjax>#log_10[H_3O^+] +log_10[HO^-]=-14#</mathjax></p>
<p>OR <mathjax>#-log_10[H_3O^+] -log_10[HO^-]=14#</mathjax></p>
<p>But we define <mathjax>#pH=-log_10[H_3O^+]#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax>, and so...........</p>
<p><mathjax>#14=pH+pOH#</mathjax>. </p>
<p>And now finally, <mathjax>#pH=14-pOH#</mathjax> </p>
<p><mathjax>#14-(-log_10(2.0xx10^-10)=4.30#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a solution with #[OH^-] = 2.0 times 10^-10# #M#?</h1>
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<div class="markdown"><p>We find <mathjax>#pH~=4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#2H_2Orightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>We know that in aqueous solution under standard conditions that:</p>
<p><mathjax>#[H_3O^+][HO^-]=K_w=10^-14#</mathjax></p>
<p>We take <mathjax>#log_10#</mathjax> OF BOTH SIDES:</p>
<p><mathjax>#log_10[H_3O^+] +log_10[HO^-]=log_10(10^-14)#</mathjax></p>
<p>But by definition, the logarithmic function is that power to which you raise the base to get the specific answer.........i.e. if <mathjax>#log_aB=c#</mathjax>, then <mathjax>#a^c=B#</mathjax>. </p>
<p>And thus <mathjax>#log_10(10^-14)=-14#</mathjax></p>
<p>And so <mathjax>#log_10[H_3O^+] +log_10[HO^-]=-14#</mathjax></p>
<p>OR <mathjax>#-log_10[H_3O^+] -log_10[HO^-]=14#</mathjax></p>
<p>But we define <mathjax>#pH=-log_10[H_3O^+]#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax>, and so...........</p>
<p><mathjax>#14=pH+pOH#</mathjax>. </p>
<p>And now finally, <mathjax>#pH=14-pOH#</mathjax> </p>
<p><mathjax>#14-(-log_10(2.0xx10^-10)=4.30#</mathjax></p></div>
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<div class="markdown"><p>4.30</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Right off the bat, if you don't understand the general concepts of <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> and pOH, I suggest you watch this 5-minute <a href="https://www.youtube.com/watch?v=3dFRD_bI3Rg&t=1s" rel="nofollow">video</a> before proceeding to read this answer.</p>
<p>Recall that pH is an index for describing the concentration of hydronium ion. Mathematically, pH is described as:</p>
<p><mathjax>#pH = -log[H_3O^+]#</mathjax></p>
<p>A common mistake to make in this sort of problem is to move too quickly, and take <mathjax>#-log(2.0 * 10^-10)#</mathjax>, and get your pH as 9.70. This is wrong, because you're not given a hydronium ion concentration, but a <em>hydroxide</em> ion concentration. Therefore, what you've calculated is really something called the <strong>pOH</strong>, not the pH. </p>
<p>So how do you work this out? Basically, you'll need to know that the product of the concentration of hydroxide and hydronium ions in a solution will always equal <mathjax>#10^-14#</mathjax> at <mathjax>#24^oC#</mathjax> Mathematically speaking:</p>
<p><mathjax>#[H_3O^+][OH^-] = 10^-14#</mathjax></p>
<p>Taking the negative log of both sides of the equation yields:</p>
<p><mathjax>#pH + pOH = 14#</mathjax></p>
<p>Given this, there's 2 ways to do this problem. Method one is to use the 1st equation, and solve for <mathjax>#[H_3O^+]#</mathjax>. Then, you just take the negative log of that. </p>
<p>Method 2 (easier, in my opinion) is to solve for the pOH, and then subtract it from 14. </p>
<p>Either way, you should get an answer of <strong>4.30.</strong></p>
<p>Do a reality check: your concentration of hydroxide is very small. Therefore, your pOH value would be fairly large, meaning that your pH would be small. This aligns with what we got. </p>
<p>Hope that helps :) </p></div>
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</article> | What is the pH of a solution with #[OH^-] = 2.0 times 10^-10# #M#? | null |
1,648 | a840ff0b-6ddd-11ea-9355-ccda262736ce | https://socratic.org/questions/how-do-you-determine-the-mass-of-potassium-in-33-5-g-of-kcl | 17.57 g | start physical_unit 7 7 mass g qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium [IN] g"}] | [{"type":"physical unit","value":"17.57 g"}] | [{"type":"physical unit","value":"Mass [OF] KCl [=] \\pu{33.5 g}"}] | <h1 class="questionTitle" itemprop="name">How do you determine the mass of potassium in 33.5 g of KCl?</h1> | null | 17.57 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of KCl = <mathjax>#74.5513g*mol^-1#</mathjax><br/>
Moles of KCl in the given sample = <mathjax>#33.5/74.5513mol#</mathjax><br/>
We can easily see that <mathjax>#x#</mathjax> mol of KCl contains <mathjax>#x#</mathjax> mol of potassium atoms<br/>
Mass of potassium in given sample = moles of K <mathjax>#xx#</mathjax> molar mass of K =<mathjax>#33.5/74.5513 xx 39.0983#</mathjax> = <mathjax>#17.569 g #</mathjax></p></div>
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<div class="markdown"><p>Mass of potassium = <mathjax>#17.569 g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molar mass of KCl = <mathjax>#74.5513g*mol^-1#</mathjax><br/>
Moles of KCl in the given sample = <mathjax>#33.5/74.5513mol#</mathjax><br/>
We can easily see that <mathjax>#x#</mathjax> mol of KCl contains <mathjax>#x#</mathjax> mol of potassium atoms<br/>
Mass of potassium in given sample = moles of K <mathjax>#xx#</mathjax> molar mass of K =<mathjax>#33.5/74.5513 xx 39.0983#</mathjax> = <mathjax>#17.569 g #</mathjax></p></div>
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<div class="markdown"><p>Mass of potassium = <mathjax>#17.569 g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molar mass of KCl = <mathjax>#74.5513g*mol^-1#</mathjax><br/>
Moles of KCl in the given sample = <mathjax>#33.5/74.5513mol#</mathjax><br/>
We can easily see that <mathjax>#x#</mathjax> mol of KCl contains <mathjax>#x#</mathjax> mol of potassium atoms<br/>
Mass of potassium in given sample = moles of K <mathjax>#xx#</mathjax> molar mass of K =<mathjax>#33.5/74.5513 xx 39.0983#</mathjax> = <mathjax>#17.569 g #</mathjax></p></div>
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</article> | How do you determine the mass of potassium in 33.5 g of KCl? | null |
1,649 | ac38131c-6ddd-11ea-a26f-ccda262736ce | https://socratic.org/questions/if-2-54-l-of-hydrogen-reacts-with-3-13-l-of-oxygen-at-temperatures-above-100-deg | 2.54 L | start physical_unit 21 21 volume l qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 10 10 7 8 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] H2O(g) [IN] L"}] | [{"type":"physical unit","value":"2.54 L"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{2.54 L}"},{"type":"physical unit","value":"Volume [OF] oxygen [=] \\pu{3.13 L}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{100 degrees Celsius}"}] | <h1 class="questionTitle" itemprop="name">If 2.54 L of hydrogen reacts with 3.13 L of oxygen at temperatures above 100 degrees Celsius, how many L of H2O(g) will form?</h1> | null | 2.54 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with a balanced equation.</p>
<p><mathjax>#"2H"_2("g")" + O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_2"O"("g")#</mathjax></p>
<p>This is a <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> gas problem. Since the reactants and product are identified as liters (volume), we can use the relationships between the liters as we would do with moles.</p>
<p>This also a limiting reactant problem. You need to determine the volume of water that can be produced from the volumes given for each reactant gas. The gas that produces the least amount of water is the limiting reactant, and the amount of water that can be by the reaction is determined by the limiting reactant.</p>
<p><mathjax>#color(blue)("Determine the volume of water produced by hydrogen gas"#</mathjax>.</p>
<p>The ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#2:2#</mathjax>, which is equal to <mathjax>#1:1#</mathjax>.</p>
<p><mathjax>#2.54color(red)cancel(color(black)("L H"_2))xx(1"L H"_2"O")/(1color(red)cancel(color(black)("L H"_2)))="2.54 L H"_2"O"#</mathjax></p>
<p><mathjax>#color(blue)("Determine the volume of water produced by oxygen gas"#</mathjax>.</p>
<p>The ratio between the volume of <mathjax>#"O"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#1:2#</mathjax>.</p>
<p><mathjax>#3.13color(red)cancel(color(black)("L O"_2))xx(2"L H"_2"O")/(1color(red)cancel(color(black)("L O"_2)))="6.26 L H"_2"O"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"H"_2"#</mathjax>, which will produce <mathjax>#"2.54 L H"_2"O"#</mathjax>.</p></div>
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<div class="markdown"><p>Hydrogen gas is the limiting reactant, so the given reaction and volumes will produce 2.54 L of water.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with a balanced equation.</p>
<p><mathjax>#"2H"_2("g")" + O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_2"O"("g")#</mathjax></p>
<p>This is a <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> gas problem. Since the reactants and product are identified as liters (volume), we can use the relationships between the liters as we would do with moles.</p>
<p>This also a limiting reactant problem. You need to determine the volume of water that can be produced from the volumes given for each reactant gas. The gas that produces the least amount of water is the limiting reactant, and the amount of water that can be by the reaction is determined by the limiting reactant.</p>
<p><mathjax>#color(blue)("Determine the volume of water produced by hydrogen gas"#</mathjax>.</p>
<p>The ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#2:2#</mathjax>, which is equal to <mathjax>#1:1#</mathjax>.</p>
<p><mathjax>#2.54color(red)cancel(color(black)("L H"_2))xx(1"L H"_2"O")/(1color(red)cancel(color(black)("L H"_2)))="2.54 L H"_2"O"#</mathjax></p>
<p><mathjax>#color(blue)("Determine the volume of water produced by oxygen gas"#</mathjax>.</p>
<p>The ratio between the volume of <mathjax>#"O"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#1:2#</mathjax>.</p>
<p><mathjax>#3.13color(red)cancel(color(black)("L O"_2))xx(2"L H"_2"O")/(1color(red)cancel(color(black)("L O"_2)))="6.26 L H"_2"O"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"H"_2"#</mathjax>, which will produce <mathjax>#"2.54 L H"_2"O"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">If 2.54 L of hydrogen reacts with 3.13 L of oxygen at temperatures above 100 degrees Celsius, how many L of H2O(g) will form?</h1>
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<div class="markdown"><p>Hydrogen gas is the limiting reactant, so the given reaction and volumes will produce 2.54 L of water.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Start with a balanced equation.</p>
<p><mathjax>#"2H"_2("g")" + O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_2"O"("g")#</mathjax></p>
<p>This is a <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> gas problem. Since the reactants and product are identified as liters (volume), we can use the relationships between the liters as we would do with moles.</p>
<p>This also a limiting reactant problem. You need to determine the volume of water that can be produced from the volumes given for each reactant gas. The gas that produces the least amount of water is the limiting reactant, and the amount of water that can be by the reaction is determined by the limiting reactant.</p>
<p><mathjax>#color(blue)("Determine the volume of water produced by hydrogen gas"#</mathjax>.</p>
<p>The ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#2:2#</mathjax>, which is equal to <mathjax>#1:1#</mathjax>.</p>
<p><mathjax>#2.54color(red)cancel(color(black)("L H"_2))xx(1"L H"_2"O")/(1color(red)cancel(color(black)("L H"_2)))="2.54 L H"_2"O"#</mathjax></p>
<p><mathjax>#color(blue)("Determine the volume of water produced by oxygen gas"#</mathjax>.</p>
<p>The ratio between the volume of <mathjax>#"O"_2"#</mathjax> and <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#1:2#</mathjax>.</p>
<p><mathjax>#3.13color(red)cancel(color(black)("L O"_2))xx(2"L H"_2"O")/(1color(red)cancel(color(black)("L O"_2)))="6.26 L H"_2"O"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"H"_2"#</mathjax>, which will produce <mathjax>#"2.54 L H"_2"O"#</mathjax>.</p></div>
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<div class="markdown"><p>This problem involves determining the limiting reactant and using it to calculate the volume of <mathjax>#H_2O (g)#</mathjax> that forms, using the reaction</p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(g)#</mathjax></p>
<p>We're not given a specific temperature or pressure, so we'll (albeit impractically) ignore the fact that the temperature is above <mathjax>#100 ^oC#</mathjax> for now. </p>
<p>To calculate the limiting reactant, we'll convert the volumes of the reactant gases to moles, and we'll assume that one mole of a gas occupies <mathjax>#22.4L#</mathjax>, since we're assuming standard temperature and pressure conditions. We'll need to divide the calculated moles for <mathjax>#H_2#</mathjax> by <mathjax>#2#</mathjax> to compensate the fact that there is a coefficient of <mathjax>#2#</mathjax> in front of it in the chemical equation:</p>
<p><mathjax>#2.54LH_2((1molH_2)/(22.4LH_2)) = (0.113molH_2)/2 = 0.0567molH_2#</mathjax></p>
<p><mathjax>#3.13LO_2((1molO_2)/(22.4LO_2)) = 0.140molO_2#</mathjax></p>
<p>It is clear that <mathjax>#H_2#</mathjax> is the limiting reactant, as it is present in the smaller amount.</p>
<p>We'll use the calculated value of <mathjax>#0.113molH_2#</mathjax> to find the moles, and then the volume in liters of the <mathjax>#H_2O(g)#</mathjax> that will form if the reaction is complete:</p>
<p><mathjax>#0.113molH_2((2molH_2O)/(2 molH_2))((22.4LH_2O)/(1molH_2O)) = 2.54 L H_2O#</mathjax></p>
<p>You'll notice this is the same value as the volume of <mathjax>#H_2#</mathjax> originally present, because the coefficients in front of <mathjax>#H_2#</mathjax> and <mathjax>#H_2O#</mathjax> are equal.</p></div>
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</article> | If 2.54 L of hydrogen reacts with 3.13 L of oxygen at temperatures above 100 degrees Celsius, how many L of H2O(g) will form? | null |
1,650 | aa45423a-6ddd-11ea-8ea2-ccda262736ce | https://socratic.org/questions/how-do-you-determine-the-mass-of-5-20-moles-of-c-6h-12-gram-formula-mass-84-2-gr | 437.84 grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 15 15 molar_mass qc_end physical_unit 10 10 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] C6H12 [IN] grams"}] | [{"type":"physical unit","value":"437.84 grams"}] | [{"type":"physical unit","value":"Gram-formula mass [OF] C6H12 [=] \\pu{grams/mole}"},{"type":"physical unit","value":"Mole [OF] C6H12 [=] \\pu{5.20 moles}"}] | <h1 class="questionTitle" itemprop="name">How do you determine the mass of 5.20 moles of #C_6H_12# (gram-formula mass = 84.2 grams/mole)?</h1> | null | 437.84 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Divide the given moles by the molar mass of the compound. I prefer to do this by multiplying by the inverse of the molar mass, since it is a fraction. I believe it's easier to see what happens with the units.</p>
<p><mathjax>#5.20color(red)cancel(color(black)("mol C"_6"H"_12))xx(84.2"g C"_6"H"_12)/(1color(red)cancel(color(black)("mol C"_6"H"_12)))="438 g C"_6"H"_12"#</mathjax> (rounded to two significant figures)</p>
<p>The mass of <mathjax>#"5.20 mol C"_6"H"_12"#</mathjax> is <mathjax>#"438 g C"_6"H"_12"#</mathjax>.</p></div>
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<div class="markdown"><p>The mass of <mathjax>#"5.20 mol C"_6"H"_12"#</mathjax> is <mathjax>#"438 g C"_6"H"_12"#</mathjax>.</p>
<p>Refer to the explanation for the process.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Divide the given moles by the molar mass of the compound. I prefer to do this by multiplying by the inverse of the molar mass, since it is a fraction. I believe it's easier to see what happens with the units.</p>
<p><mathjax>#5.20color(red)cancel(color(black)("mol C"_6"H"_12))xx(84.2"g C"_6"H"_12)/(1color(red)cancel(color(black)("mol C"_6"H"_12)))="438 g C"_6"H"_12"#</mathjax> (rounded to two significant figures)</p>
<p>The mass of <mathjax>#"5.20 mol C"_6"H"_12"#</mathjax> is <mathjax>#"438 g C"_6"H"_12"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you determine the mass of 5.20 moles of #C_6H_12# (gram-formula mass = 84.2 grams/mole)?</h1>
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<div class="markdown"><p>The mass of <mathjax>#"5.20 mol C"_6"H"_12"#</mathjax> is <mathjax>#"438 g C"_6"H"_12"#</mathjax>.</p>
<p>Refer to the explanation for the process.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Divide the given moles by the molar mass of the compound. I prefer to do this by multiplying by the inverse of the molar mass, since it is a fraction. I believe it's easier to see what happens with the units.</p>
<p><mathjax>#5.20color(red)cancel(color(black)("mol C"_6"H"_12))xx(84.2"g C"_6"H"_12)/(1color(red)cancel(color(black)("mol C"_6"H"_12)))="438 g C"_6"H"_12"#</mathjax> (rounded to two significant figures)</p>
<p>The mass of <mathjax>#"5.20 mol C"_6"H"_12"#</mathjax> is <mathjax>#"438 g C"_6"H"_12"#</mathjax>.</p></div>
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</article> | How do you determine the mass of 5.20 moles of #C_6H_12# (gram-formula mass = 84.2 grams/mole)? | null |
1,651 | ac6461fa-6ddd-11ea-ad7f-ccda262736ce | https://socratic.org/questions/what-mass-of-potassium-hydroxide-is-required-to-react-completely-with-2-70-g-of- | 3.09 grams | start physical_unit 3 4 mass g qc_end physical_unit 14 15 11 12 mass qc_end c_other OTHER qc_end substance 18 19 qc_end substance 21 21 qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"3.09 grams"}] | [{"type":"physical unit","value":"Mass [OF] sulfuric acid [=] \\pu{2.70 g}"},{"type":"other","value":"React completely."},{"type":"substance name","value":"Potassium sulfate"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What mass of potassium hydroxide is required to react completely with 2.70 g of sulfuric acid to produce potassium sulfate and water?</h1> | null | 3.09 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This problem is a bit unrealistic, as I could not really measure such a mass of sulfuric acid. Realistically, I could take a volume of sulfuric acid of a given concentration, which had an equivalent molar quantity to <mathjax>#(2.70*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0275*mol#</mathjax>.</p>
<p>Given the equation above, i.e. sulfuric acid is a diacid that requires 2 equiv of base, we require <mathjax>#2xx0.0275*molxx56.11*g*mol^-1=3.09*g#</mathjax></p>
<p>If I had a <mathjax>#1.00*mol*L^-1#</mathjax> solution of sulfuric acid, what volume would I require for equivalence? Such question would be a lot more reasonable. </p>
<p>As you no doubt realize, the ability to write a stoichiometric equation is absolutely vital to answer the question. </p></div>
</div>
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<div class="markdown"><p><mathjax>#2KOH + H_2SO_4 rarr K_2SO_4 + 2H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This problem is a bit unrealistic, as I could not really measure such a mass of sulfuric acid. Realistically, I could take a volume of sulfuric acid of a given concentration, which had an equivalent molar quantity to <mathjax>#(2.70*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0275*mol#</mathjax>.</p>
<p>Given the equation above, i.e. sulfuric acid is a diacid that requires 2 equiv of base, we require <mathjax>#2xx0.0275*molxx56.11*g*mol^-1=3.09*g#</mathjax></p>
<p>If I had a <mathjax>#1.00*mol*L^-1#</mathjax> solution of sulfuric acid, what volume would I require for equivalence? Such question would be a lot more reasonable. </p>
<p>As you no doubt realize, the ability to write a stoichiometric equation is absolutely vital to answer the question. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of potassium hydroxide is required to react completely with 2.70 g of sulfuric acid to produce potassium sulfate and water?</h1>
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<div class="markdown"><p><mathjax>#2KOH + H_2SO_4 rarr K_2SO_4 + 2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This problem is a bit unrealistic, as I could not really measure such a mass of sulfuric acid. Realistically, I could take a volume of sulfuric acid of a given concentration, which had an equivalent molar quantity to <mathjax>#(2.70*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0275*mol#</mathjax>.</p>
<p>Given the equation above, i.e. sulfuric acid is a diacid that requires 2 equiv of base, we require <mathjax>#2xx0.0275*molxx56.11*g*mol^-1=3.09*g#</mathjax></p>
<p>If I had a <mathjax>#1.00*mol*L^-1#</mathjax> solution of sulfuric acid, what volume would I require for equivalence? Such question would be a lot more reasonable. </p>
<p>As you no doubt realize, the ability to write a stoichiometric equation is absolutely vital to answer the question. </p></div>
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</article> | What mass of potassium hydroxide is required to react completely with 2.70 g of sulfuric acid to produce potassium sulfate and water? | null |
1,652 | a8c5cefa-6ddd-11ea-a28f-ccda262736ce | https://socratic.org/questions/how-do-you-balance-k-mgbr-kbr-mg | 2 K + MgBr2 -> 2 KBr + Mg | 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 K + MgBr2 -> 2 KBr + Mg"}] | [{"type":"chemical equation","value":"K + MgBr2 -> KBr + Mg"}] | <h1 class="questionTitle" itemprop="name">How do you balance #K+MgBr -> KBr + Mg#?</h1> | null | 2 K + MgBr2 -> 2 KBr + Mg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>K(s) = 0 charge. <br/>
Mg = +2 charge.<br/>
Br = Br-1 charge. <br/>
Mg(s) = 0 charge. </p>
<p><mathjax># MgBr = Mg^+2Br_2^-1#</mathjax>( balanced compound)<br/>
<mathjax># KBr = K^+1Br^-1#</mathjax> ( balanced compound)</p>
<p><mathjax># K(s_ + MgBr_2 = KBr + Mg(s)#</mathjax> There are two Br in the reactants but only 1 Br in the products so the KBr must be multiplied by 2 </p>
<p><mathjax># K(s) + MgBr_2 = 2 KBr + Mg(s)#</mathjax> Now there are two K in the product but only 1 K in the reactants so the K must be multiplied by 2</p>
<p><mathjax># 2 K(s) + MgBr_2 = 2 KBr + Mg(s) #</mathjax> </p>
<p>This is the balanced chemical reaction <br/>
2K in the reactants 2 K in the products. <br/>
1 Mg in the reactants 1 Mg in the products<br/>
2 Br in the reactants 2 Br in the products. </p>
<p>This is an example of a single replacement reaction. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>First it necessary to balance all of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. Then make sure that there is the same number of atoms on each side of the equation. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>K(s) = 0 charge. <br/>
Mg = +2 charge.<br/>
Br = Br-1 charge. <br/>
Mg(s) = 0 charge. </p>
<p><mathjax># MgBr = Mg^+2Br_2^-1#</mathjax>( balanced compound)<br/>
<mathjax># KBr = K^+1Br^-1#</mathjax> ( balanced compound)</p>
<p><mathjax># K(s_ + MgBr_2 = KBr + Mg(s)#</mathjax> There are two Br in the reactants but only 1 Br in the products so the KBr must be multiplied by 2 </p>
<p><mathjax># K(s) + MgBr_2 = 2 KBr + Mg(s)#</mathjax> Now there are two K in the product but only 1 K in the reactants so the K must be multiplied by 2</p>
<p><mathjax># 2 K(s) + MgBr_2 = 2 KBr + Mg(s) #</mathjax> </p>
<p>This is the balanced chemical reaction <br/>
2K in the reactants 2 K in the products. <br/>
1 Mg in the reactants 1 Mg in the products<br/>
2 Br in the reactants 2 Br in the products. </p>
<p>This is an example of a single replacement reaction. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #K+MgBr -> KBr + Mg#?</h1>
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<div class="markdown"><p>First it necessary to balance all of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. Then make sure that there is the same number of atoms on each side of the equation. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>K(s) = 0 charge. <br/>
Mg = +2 charge.<br/>
Br = Br-1 charge. <br/>
Mg(s) = 0 charge. </p>
<p><mathjax># MgBr = Mg^+2Br_2^-1#</mathjax>( balanced compound)<br/>
<mathjax># KBr = K^+1Br^-1#</mathjax> ( balanced compound)</p>
<p><mathjax># K(s_ + MgBr_2 = KBr + Mg(s)#</mathjax> There are two Br in the reactants but only 1 Br in the products so the KBr must be multiplied by 2 </p>
<p><mathjax># K(s) + MgBr_2 = 2 KBr + Mg(s)#</mathjax> Now there are two K in the product but only 1 K in the reactants so the K must be multiplied by 2</p>
<p><mathjax># 2 K(s) + MgBr_2 = 2 KBr + Mg(s) #</mathjax> </p>
<p>This is the balanced chemical reaction <br/>
2K in the reactants 2 K in the products. <br/>
1 Mg in the reactants 1 Mg in the products<br/>
2 Br in the reactants 2 Br in the products. </p>
<p>This is an example of a single replacement reaction. </p></div>
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<div class="markdown"><p>Refer to the explanation.<br/>
The balanced equation is <mathjax>#"2K" + "MgBr"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2KBr" + "Mg"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to make sure the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have the correct formulas. <mathjax>#"MgBr"#</mathjax> is not the correct formula for magnesium bromide. The correct formula is <mathjax>#"MgBr"_2"#</mathjax>, because the magnesium ion has a charge of <mathjax>#2^+#</mathjax> and each bromide ion has a charge of <mathjax>#1^-#</mathjax>. So two bromide ions are needed to bond with magnesium in order to make the compound neutral.</p>
<p>Now, we need to balance the equation. A balanced equation has the same number of atoms of each element on both sides. We do that by changing the amount of the reactants and products as needed by adding coefficients (numbers in front) of the formulas. Each coefficient is multiplied by the subscripts for each element. We never change a formula, so never try to balance by changing subscripts. No coefficient or subscript is understood to be <mathjax>#1#</mathjax>.</p>
<p><mathjax>#color(purple)("K" + "MgBr"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)("KBr" + "Mg"#</mathjax></p>
<p>There are two bromide ions on the left side and one on the right. So we need to place a coefficient of <mathjax>#2#</mathjax> in front of <mathjax>#"KBr"#</mathjax></p>
<p><mathjax>#color(purple)("K" + "MgBr"_2")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(magenta)(2)color(purple)("KBr" + "Mg")#</mathjax></p>
<p>Now we have two potassium ions on the right side, but only one on the left side. So we need to place a coefficient of <mathjax>#2#</mathjax> in front of <mathjax>#"K"#</mathjax>.</p>
<p><mathjax>#color(magenta)(2)color(purple)("K" + "MgBr"_2")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(magenta)(2)color(purple)("KBr" + "Mg")#</mathjax></p>
<p>Now we need to verify that the equation has the same number of atoms of each element on each side.</p>
<p><strong>Left:</strong><br/>
<mathjax>#"2 K"#</mathjax><br/>
<mathjax>#"1 Mg"#</mathjax><br/>
<mathjax>#"2 Br"#</mathjax></p>
<p><strong>Right:</strong><br/>
<mathjax>#"2 K"#</mathjax><br/>
<mathjax>#"1 Mg"#</mathjax><br/>
<mathjax>#"2 Br"#</mathjax></p>
<p>So both sides have the same number of atoms of each element, and the equation is balanced.</p></div>
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</article> | How do you balance #K+MgBr -> KBr + Mg#? | null |
1,653 | a9cac47a-6ddd-11ea-9d29-ccda262736ce | https://socratic.org/questions/consider-the-following-unbalanced-half-reaction-cro-2-oh-cro-4-2-h-2o-3e-if-the- | 2 | start physical_unit 28 30 coefficient none qc_end chemical_equation 5 16 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] the water molecule"}] | [{"type":"physical unit","value":"2"}] | [{"type":"chemical equation","value":"CrO2 + ? OH- -> CrO4^2- + ? H2O + 3 e-"}] | <h1 class="questionTitle" itemprop="name">Consider the following unbalanced half-reaction: #CrO_2 + ?OH -> CrO_4^(-2) + ?H_2O + 3e^-#. If the equation were balanced, what would be the coefficient of the water molecule? </h1> | null | 2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation should be as follows<br/>
<mathjax>#CrO_2 + 4OH ^"-" -> CrO_4^(-2) + 2H_2O + 2e^-#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CrO_2 + 4OH ^"-" -> CrO_4^(-2) + 2H_2O + 2e^-#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation should be as follows<br/>
<mathjax>#CrO_2 + 4OH ^"-" -> CrO_4^(-2) + 2H_2O + 2e^-#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Consider the following unbalanced half-reaction: #CrO_2 + ?OH -> CrO_4^(-2) + ?H_2O + 3e^-#. If the equation were balanced, what would be the coefficient of the water molecule? </h1>
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P dilip_k
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Jul 1, 2016
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<div class="markdown"><p><mathjax>#CrO_2 + 4OH ^"-" -> CrO_4^(-2) + 2H_2O + 2e^-#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation should be as follows<br/>
<mathjax>#CrO_2 + 4OH ^"-" -> CrO_4^(-2) + 2H_2O + 2e^-#</mathjax></p></div>
</div>
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</article> | Consider the following unbalanced half-reaction: #CrO_2 + ?OH -> CrO_4^(-2) + ?H_2O + 3e^-#. If the equation were balanced, what would be the coefficient of the water molecule? | null |
1,654 | aac6a81d-6ddd-11ea-83fa-ccda262736ce | https://socratic.org/questions/a-laboratory-procedure-calls-for-making-453-3-ml-of-a-2-62m-salt-solution-if-the | 258.98 g | start physical_unit 19 20 mass g qc_end physical_unit 19 20 22 23 molar_mass qc_end physical_unit 12 13 6 7 volume qc_end physical_unit 12 13 10 11 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] the salt [IN] g"}] | [{"type":"physical unit","value":"258.98 g"}] | [{"type":"physical unit","value":"molar mass [OF] the salt [=] \\pu{218 g/mol}"},{"type":"physical unit","value":"Volume [OF] salt solution [=] \\pu{453.3 ml}"},{"type":"physical unit","value":"Molarity [OF] salt solution [=] \\pu{2.62 M}"}] | <h1 class="questionTitle" itemprop="name">A laboratory procedure calls for making 453.3 ml of a 2.62M salt solution. If the molar mass of the salt is 218 g/mol what mass is required?</h1> | null | 258.98 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<blockquote>
<ul>
<li><em>use the <strong>molarity</strong> and <strong>volume</strong> of the solution to determine how many <strong>moles of solute</strong> it must contain</em></li>
<li><em>use the <strong>molar mass</strong> of the salt to calculate how many grams would contain that many moles</em></li>
</ul>
</blockquote>
<p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you have in <em><strong>one liter of solution</strong></em>. </p>
<p>In your case, the solution must have a molarity of <mathjax>#"2.62 mol L"^(-1)#</mathjax>, which means that every liter of solution will contain <mathjax>#2.62#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, your sample is said to have a volume of <mathjax>#"453.3 mL"#</mathjax>. Use its molarity to figure out how many moles of solute it contains -- <strong>do not</strong> forget to convert the volume to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#453.3 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("2.62 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("known molarity")) = "1.188 moles solute"#</mathjax></p>
</blockquote>
<p>As you know, a compound's <strong>molar mass</strong> tells you the mass of <strong>one mole</strong> of said compound. In your case, the salt has a molar mass of <mathjax>#"218 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> has a mass of <mathjax>#"218 g"#</mathjax>.</p>
<p>This means that the mass of <mathjax>#1.188#</mathjax> <strong>moles</strong> will be </p>
<blockquote>
<p><mathjax>#1.188 color(red)(cancel(color(black)("moles"))) * "218 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("259 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"259 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<blockquote>
<ul>
<li><em>use the <strong>molarity</strong> and <strong>volume</strong> of the solution to determine how many <strong>moles of solute</strong> it must contain</em></li>
<li><em>use the <strong>molar mass</strong> of the salt to calculate how many grams would contain that many moles</em></li>
</ul>
</blockquote>
<p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you have in <em><strong>one liter of solution</strong></em>. </p>
<p>In your case, the solution must have a molarity of <mathjax>#"2.62 mol L"^(-1)#</mathjax>, which means that every liter of solution will contain <mathjax>#2.62#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, your sample is said to have a volume of <mathjax>#"453.3 mL"#</mathjax>. Use its molarity to figure out how many moles of solute it contains -- <strong>do not</strong> forget to convert the volume to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#453.3 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("2.62 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("known molarity")) = "1.188 moles solute"#</mathjax></p>
</blockquote>
<p>As you know, a compound's <strong>molar mass</strong> tells you the mass of <strong>one mole</strong> of said compound. In your case, the salt has a molar mass of <mathjax>#"218 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> has a mass of <mathjax>#"218 g"#</mathjax>.</p>
<p>This means that the mass of <mathjax>#1.188#</mathjax> <strong>moles</strong> will be </p>
<blockquote>
<p><mathjax>#1.188 color(red)(cancel(color(black)("moles"))) * "218 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("259 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A laboratory procedure calls for making 453.3 ml of a 2.62M salt solution. If the molar mass of the salt is 218 g/mol what mass is required?</h1>
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Stefan V.
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Jul 8, 2016
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<div class="markdown"><p><mathjax>#"259 g"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<blockquote>
<ul>
<li><em>use the <strong>molarity</strong> and <strong>volume</strong> of the solution to determine how many <strong>moles of solute</strong> it must contain</em></li>
<li><em>use the <strong>molar mass</strong> of the salt to calculate how many grams would contain that many moles</em></li>
</ul>
</blockquote>
<p>A solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you have in <em><strong>one liter of solution</strong></em>. </p>
<p>In your case, the solution must have a molarity of <mathjax>#"2.62 mol L"^(-1)#</mathjax>, which means that every liter of solution will contain <mathjax>#2.62#</mathjax> <strong>moles</strong> of solute. </p>
<p>Now, your sample is said to have a volume of <mathjax>#"453.3 mL"#</mathjax>. Use its molarity to figure out how many moles of solute it contains -- <strong>do not</strong> forget to convert the volume to <em>liters</em>! </p>
<blockquote>
<p><mathjax>#453.3 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("2.62 moles solute"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("known molarity")) = "1.188 moles solute"#</mathjax></p>
</blockquote>
<p>As you know, a compound's <strong>molar mass</strong> tells you the mass of <strong>one mole</strong> of said compound. In your case, the salt has a molar mass of <mathjax>#"218 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> has a mass of <mathjax>#"218 g"#</mathjax>.</p>
<p>This means that the mass of <mathjax>#1.188#</mathjax> <strong>moles</strong> will be </p>
<blockquote>
<p><mathjax>#1.188 color(red)(cancel(color(black)("moles"))) * "218 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("259 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | A laboratory procedure calls for making 453.3 ml of a 2.62M salt solution. If the molar mass of the salt is 218 g/mol what mass is required? | null |
1,655 | a8fd6246-6ddd-11ea-b102-ccda262736ce | https://socratic.org/questions/the-ionization-energy-of-aluminum-is-577-6kj-mol-what-is-the-longest-wavelength- | 2.07 × 10^(-7) m | start physical_unit 14 14 longest_wavelength m qc_end physical_unit 4 4 6 7 ionization_energy qc_end end | [{"type":"physical unit","value":"Longest wavelength [OF] light [IN] m"}] | [{"type":"physical unit","value":"2.07 × 10^(-7) m"}] | [{"type":"physical unit","value":"Ionization energy [OF] aluminum [=] \\pu{577.6 kJ/mol}"},{"type":"physical unit","value":"Number [OF] photon [=] \\pu{1}"},{"type":"physical unit","value":"Number [OF] aluminum atom [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">The ionization energy of aluminum is 577.6kJ/mol. What is the longest wavelength of light to which a single photon could ionize an aluminum atom?</h1> | null | 2.07 × 10^(-7) m | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to calculate the <em>ionization energy</em> for <strong>one atom</strong> of aluminium, <mathjax>#"Al"#</mathjax>, by suing the fact that you know the energy needed to ionize <strong>one mole</strong> of atoms of aluminium. </p>
<p>As you know, <strong>one mole</strong> of any element contains exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element, as given by <em>Avogadro's number</em>. </p>
<p>You can thus say that the energy needed to ionize <strong>one atom</strong>of aluminium is equal to </p>
<blockquote>
<p><mathjax>#577.6 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole Al"))))/(6.022 * 10^(23)"atoms of Al") = 9.5915 * 10^(-22)"kJ / atom"#</mathjax></p>
</blockquote>
<p>Now, the energy of a photon is <strong>directly proportional</strong> to its <em>frequency</em> as given by the <strong>Planck - Einstein equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)E = h * nu color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#E#</mathjax> - the energy of the photon <br/>
<mathjax>#nu#</mathjax> - its <em>frequency</em><br/>
<mathjax>#h#</mathjax> - <strong><a href="https://socratic.org/chemistry/the-bohr-model-of-the-atom/calculations-with-plancks-constant-and-frequency">Planck's constant</a></strong>, equal to <mathjax>#6.626 * 10^(-34)"J s"#</mathjax></p>
<p>Use the above equation to calculate the frequency of a photon that carries enough energy to ionize a single atom of aluminium</p>
<blockquote>
<p><mathjax>#E = h * nu implies nu = E/h#</mathjax></p>
</blockquote>
<p>Notice that Planck's constant is expressed in <em>joules second</em>, <mathjax>#"J s"#</mathjax>, which means that you must convert the ionization energy from <em>kilojoules per atom</em> to <em>joules</em> per atom</p>
<blockquote>
<p><mathjax>#E = (9.5915 * 10^(-22) * 10^(3)color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J")))"s") = 1.4476 * 10^(15)"s"^(-1)#</mathjax></p>
</blockquote>
<p>Now, the frequency of a photon has an <strong>inverse relationship</strong> with its wavelength as given by the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)lamda * nu = c color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#lamda#</mathjax> - the <em>wavelength</em> of the photon<br/>
<mathjax>#c#</mathjax> - the <em>speed of light</em> in a vacuum, usually given as <mathjax>#3 * 10^8"m s"^(-1)#</mathjax></p>
<p>Rearrange this equation to solve for <mathjax>#lamda#</mathjax></p>
<blockquote>
<p><mathjax>#lamda * nu = c implies lamda = c/ (nu)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#lamda = (3 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))) )/(1.4476 * 10^(15)color(red)(cancel(color(black)("s"^(-1))))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.072 * 10^(-7)"m")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.072 * 10^(-7)"m"#</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 to calculate the <em>ionization energy</em> for <strong>one atom</strong> of aluminium, <mathjax>#"Al"#</mathjax>, by suing the fact that you know the energy needed to ionize <strong>one mole</strong> of atoms of aluminium. </p>
<p>As you know, <strong>one mole</strong> of any element contains exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element, as given by <em>Avogadro's number</em>. </p>
<p>You can thus say that the energy needed to ionize <strong>one atom</strong>of aluminium is equal to </p>
<blockquote>
<p><mathjax>#577.6 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole Al"))))/(6.022 * 10^(23)"atoms of Al") = 9.5915 * 10^(-22)"kJ / atom"#</mathjax></p>
</blockquote>
<p>Now, the energy of a photon is <strong>directly proportional</strong> to its <em>frequency</em> as given by the <strong>Planck - Einstein equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)E = h * nu color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#E#</mathjax> - the energy of the photon <br/>
<mathjax>#nu#</mathjax> - its <em>frequency</em><br/>
<mathjax>#h#</mathjax> - <strong><a href="https://socratic.org/chemistry/the-bohr-model-of-the-atom/calculations-with-plancks-constant-and-frequency">Planck's constant</a></strong>, equal to <mathjax>#6.626 * 10^(-34)"J s"#</mathjax></p>
<p>Use the above equation to calculate the frequency of a photon that carries enough energy to ionize a single atom of aluminium</p>
<blockquote>
<p><mathjax>#E = h * nu implies nu = E/h#</mathjax></p>
</blockquote>
<p>Notice that Planck's constant is expressed in <em>joules second</em>, <mathjax>#"J s"#</mathjax>, which means that you must convert the ionization energy from <em>kilojoules per atom</em> to <em>joules</em> per atom</p>
<blockquote>
<p><mathjax>#E = (9.5915 * 10^(-22) * 10^(3)color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J")))"s") = 1.4476 * 10^(15)"s"^(-1)#</mathjax></p>
</blockquote>
<p>Now, the frequency of a photon has an <strong>inverse relationship</strong> with its wavelength as given by the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)lamda * nu = c color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#lamda#</mathjax> - the <em>wavelength</em> of the photon<br/>
<mathjax>#c#</mathjax> - the <em>speed of light</em> in a vacuum, usually given as <mathjax>#3 * 10^8"m s"^(-1)#</mathjax></p>
<p>Rearrange this equation to solve for <mathjax>#lamda#</mathjax></p>
<blockquote>
<p><mathjax>#lamda * nu = c implies lamda = c/ (nu)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#lamda = (3 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))) )/(1.4476 * 10^(15)color(red)(cancel(color(black)("s"^(-1))))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.072 * 10^(-7)"m")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The ionization energy of aluminum is 577.6kJ/mol. What is the longest wavelength of light to which a single photon could ionize an aluminum atom?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-08-18T00:58:52" itemprop="dateCreated">
Aug 18, 2016
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<div class="markdown"><p><mathjax>#2.072 * 10^(-7)"m"#</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 to calculate the <em>ionization energy</em> for <strong>one atom</strong> of aluminium, <mathjax>#"Al"#</mathjax>, by suing the fact that you know the energy needed to ionize <strong>one mole</strong> of atoms of aluminium. </p>
<p>As you know, <strong>one mole</strong> of any element contains exactly <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element, as given by <em>Avogadro's number</em>. </p>
<p>You can thus say that the energy needed to ionize <strong>one atom</strong>of aluminium is equal to </p>
<blockquote>
<p><mathjax>#577.6 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole Al"))))/(6.022 * 10^(23)"atoms of Al") = 9.5915 * 10^(-22)"kJ / atom"#</mathjax></p>
</blockquote>
<p>Now, the energy of a photon is <strong>directly proportional</strong> to its <em>frequency</em> as given by the <strong>Planck - Einstein equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)E = h * nu color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#E#</mathjax> - the energy of the photon <br/>
<mathjax>#nu#</mathjax> - its <em>frequency</em><br/>
<mathjax>#h#</mathjax> - <strong><a href="https://socratic.org/chemistry/the-bohr-model-of-the-atom/calculations-with-plancks-constant-and-frequency">Planck's constant</a></strong>, equal to <mathjax>#6.626 * 10^(-34)"J s"#</mathjax></p>
<p>Use the above equation to calculate the frequency of a photon that carries enough energy to ionize a single atom of aluminium</p>
<blockquote>
<p><mathjax>#E = h * nu implies nu = E/h#</mathjax></p>
</blockquote>
<p>Notice that Planck's constant is expressed in <em>joules second</em>, <mathjax>#"J s"#</mathjax>, which means that you must convert the ionization energy from <em>kilojoules per atom</em> to <em>joules</em> per atom</p>
<blockquote>
<p><mathjax>#E = (9.5915 * 10^(-22) * 10^(3)color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J")))"s") = 1.4476 * 10^(15)"s"^(-1)#</mathjax></p>
</blockquote>
<p>Now, the frequency of a photon has an <strong>inverse relationship</strong> with its wavelength as given by the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)lamda * nu = c color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#lamda#</mathjax> - the <em>wavelength</em> of the photon<br/>
<mathjax>#c#</mathjax> - the <em>speed of light</em> in a vacuum, usually given as <mathjax>#3 * 10^8"m s"^(-1)#</mathjax></p>
<p>Rearrange this equation to solve for <mathjax>#lamda#</mathjax></p>
<blockquote>
<p><mathjax>#lamda * nu = c implies lamda = c/ (nu)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#lamda = (3 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))) )/(1.4476 * 10^(15)color(red)(cancel(color(black)("s"^(-1))))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.072 * 10^(-7)"m")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to four <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | The ionization energy of aluminum is 577.6kJ/mol. What is the longest wavelength of light to which a single photon could ionize an aluminum atom? | null |
1,656 | a8ac2974-6ddd-11ea-ac22-ccda262736ce | https://socratic.org/questions/573dba207c01497c6022862a | -1 | start physical_unit 6 6 ph none qc_end physical_unit 8 9 11 12 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] sulfuric acid solution"}] | [{"type":"physical unit","value":"-1"}] | [{"type":"physical unit","value":"Concentration [OF] sulfuric acid solution [=] \\pu{5.98 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# of a solution of sulfuric acid of #5.98*mol*L^-1# concentration? </h1> | null | -1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[2xx5.98]#</mathjax> <mathjax>#~=-1#</mathjax>.</p>
<p>Why did I multiply the given concentration by 2?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[2xx5.98]#</mathjax> <mathjax>#~=-1#</mathjax>.</p>
<p>Why did I multiply the given concentration by 2?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# of a solution of sulfuric acid of #5.98*mol*L^-1# concentration? </h1>
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anor277
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Jun 22, 2016
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<div class="markdown"><p><mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=-log_(10)[2xx5.98]#</mathjax> <mathjax>#~=-1#</mathjax>.</p>
<p>Why did I multiply the given concentration by 2?</p></div>
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</article> | What is the #pH# of a solution of sulfuric acid of #5.98*mol*L^-1# concentration? | null |
1,657 | ace5db5d-6ddd-11ea-b500-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-for-this-reaction-k-3po-4-ca-no-3-2-ca-3-po-4-2-kn | 2 K3PO4 + 3 Ca(NO3)2 -> Ca3(PO4)2 + 6 KNO3 | start chemical_equation qc_end chemical_equation 8 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"2 K3PO4 + 3 Ca(NO3)2 -> Ca3(PO4)2 + 6 KNO3"}] | [{"type":"chemical equation","value":"K3PO4 + Ca(NO3)2 -> Ca3(PO4)2 + KNO3"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation for this reaction: #?K_3PO_4 + ?Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#?</h1> | null | 2 K3PO4 + 3 Ca(NO3)2 -> Ca3(PO4)2 + 6 KNO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So there's not really a whole lot of concepts to explain here. However, one handy thing to keep in mind here is to <strong>treat <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a> as individual atoms</strong>. This will save you a load of time, and makes visualizing the scenario much less trickier. </p>
<p>Since there's really no one correct mathematical way to do this problem, I'm just going to briefly walk you through my thought process as I did it:</p>
<p>My first idea was to try and get the same number of calciums on both sides. To do this, I added a 3 in front of <mathjax>#Ca(NO_3)_2#</mathjax> This took care of the calciums, but left me with <mathjax>#6 (NO_3)#</mathjax>s as opposed to 1 on the other: </p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#</mathjax>?</p>
<p>To even this out, I added a 6 in front of the <mathjax>#KNO_3#</mathjax>. Now, however, I had 6 Potassiums on one side as opposed to 3 on the other:</p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + 6KNO_3#</mathjax>?</p>
<p>To even this out, I added a 2 in front of the <mathjax>#K_3PO_4#</mathjax>, which took care of the Potassiums, and also evened out the number of <mathjax>#(PO_4)#</mathjax>s on both sides. This left me with my final answer:</p>
<p><mathjax>#2K_3PO_4 + 3Ca(NO_3)_2 -> Ca_3(PO_4)_2 + 6KNO_3#</mathjax></p>
<p>As you can see, there's no one right way to solve these problems. Rather, it's just about fixing one thing, and seeing where that leads you. </p>
<p>Hope that helped :) </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2K_3PO_4 + 3Ca(NO_3)_2 -> Ca_3(PO_4)_2 + 6KNO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So there's not really a whole lot of concepts to explain here. However, one handy thing to keep in mind here is to <strong>treat <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a> as individual atoms</strong>. This will save you a load of time, and makes visualizing the scenario much less trickier. </p>
<p>Since there's really no one correct mathematical way to do this problem, I'm just going to briefly walk you through my thought process as I did it:</p>
<p>My first idea was to try and get the same number of calciums on both sides. To do this, I added a 3 in front of <mathjax>#Ca(NO_3)_2#</mathjax> This took care of the calciums, but left me with <mathjax>#6 (NO_3)#</mathjax>s as opposed to 1 on the other: </p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#</mathjax>?</p>
<p>To even this out, I added a 6 in front of the <mathjax>#KNO_3#</mathjax>. Now, however, I had 6 Potassiums on one side as opposed to 3 on the other:</p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + 6KNO_3#</mathjax>?</p>
<p>To even this out, I added a 2 in front of the <mathjax>#K_3PO_4#</mathjax>, which took care of the Potassiums, and also evened out the number of <mathjax>#(PO_4)#</mathjax>s on both sides. This left me with my final answer:</p>
<p><mathjax>#2K_3PO_4 + 3Ca(NO_3)_2 -> Ca_3(PO_4)_2 + 6KNO_3#</mathjax></p>
<p>As you can see, there's no one right way to solve these problems. Rather, it's just about fixing one thing, and seeing where that leads you. </p>
<p>Hope that helped :) </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation for this reaction: #?K_3PO_4 + ?Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#?</h1>
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Darshan Senthil
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<span class="dateCreated" datetime="2016-12-10T02:17:02" itemprop="dateCreated">
Dec 10, 2016
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<div class="markdown"><p><mathjax>#2K_3PO_4 + 3Ca(NO_3)_2 -> Ca_3(PO_4)_2 + 6KNO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So there's not really a whole lot of concepts to explain here. However, one handy thing to keep in mind here is to <strong>treat <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a> as individual atoms</strong>. This will save you a load of time, and makes visualizing the scenario much less trickier. </p>
<p>Since there's really no one correct mathematical way to do this problem, I'm just going to briefly walk you through my thought process as I did it:</p>
<p>My first idea was to try and get the same number of calciums on both sides. To do this, I added a 3 in front of <mathjax>#Ca(NO_3)_2#</mathjax> This took care of the calciums, but left me with <mathjax>#6 (NO_3)#</mathjax>s as opposed to 1 on the other: </p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#</mathjax>?</p>
<p>To even this out, I added a 6 in front of the <mathjax>#KNO_3#</mathjax>. Now, however, I had 6 Potassiums on one side as opposed to 3 on the other:</p>
<p><mathjax>#?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + 6KNO_3#</mathjax>?</p>
<p>To even this out, I added a 2 in front of the <mathjax>#K_3PO_4#</mathjax>, which took care of the Potassiums, and also evened out the number of <mathjax>#(PO_4)#</mathjax>s on both sides. This left me with my final answer:</p>
<p><mathjax>#2K_3PO_4 + 3Ca(NO_3)_2 -> Ca_3(PO_4)_2 + 6KNO_3#</mathjax></p>
<p>As you can see, there's no one right way to solve these problems. Rather, it's just about fixing one thing, and seeing where that leads you. </p>
<p>Hope that helped :) </p></div>
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</article> | What is the balanced equation for this reaction: #?K_3PO_4 + ?Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3#? | null |
1,658 | a8d95730-6ddd-11ea-88fd-ccda262736ce | https://socratic.org/questions/how-much-calcium-hydroxide-ca-oh-2-in-grams-is-needed-much-produce-1-5-l-of-a-0- | 27.75 grams | start physical_unit 4 4 mass g qc_end physical_unit 17 17 11 12 volume qc_end physical_unit 4 4 15 16 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] Ca(OH)2 [IN] grams"}] | [{"type":"physical unit","value":"27.75 grams"}] | [{"type":"physical unit","value":"Volume [OF] Ca(OH)2 solution [=] \\pu{1.5 L}"},{"type":"physical unit","value":"Molarity [OF] Ca(OH)2 solution [=] \\pu{0.25 M}"}] | <h1 class="questionTitle" itemprop="name">How much calcium hydroxide (#Ca(OH)_2#), in grams, is needed much produce 1.5 L of a 0.25M solution?</h1> | null | 27.75 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as = moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / litres of solution</p>
<p>to produce 1.5 L of 0.25 M solution , <br/>
moles of calcium hydroxide needed are = <mathjax># 0.25*1.5 #</mathjax><br/>
= 0.375 mol<br/>
now amount of calcium hydroxide in grams needed can be calculated by multiplying moles by molar mass : <mathjax># 0.375*74 #</mathjax><br/>
= 27.75 gm</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>27.75 gm</p></div>
</div>
</div>
<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 defined as = moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / litres of solution</p>
<p>to produce 1.5 L of 0.25 M solution , <br/>
moles of calcium hydroxide needed are = <mathjax># 0.25*1.5 #</mathjax><br/>
= 0.375 mol<br/>
now amount of calcium hydroxide in grams needed can be calculated by multiplying moles by molar mass : <mathjax># 0.375*74 #</mathjax><br/>
= 27.75 gm</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How much calcium hydroxide (#Ca(OH)_2#), in grams, is needed much produce 1.5 L of a 0.25M solution?</h1>
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<div class="markdown"><p>27.75 gm</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 defined as = moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / litres of solution</p>
<p>to produce 1.5 L of 0.25 M solution , <br/>
moles of calcium hydroxide needed are = <mathjax># 0.25*1.5 #</mathjax><br/>
= 0.375 mol<br/>
now amount of calcium hydroxide in grams needed can be calculated by multiplying moles by molar mass : <mathjax># 0.375*74 #</mathjax><br/>
= 27.75 gm</p></div>
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</article> | How much calcium hydroxide (#Ca(OH)_2#), in grams, is needed much produce 1.5 L of a 0.25M solution? | null |
1,659 | a875d108-6ddd-11ea-b06d-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-standard-enthalpy-change-for-the-reaction-2a-b-2c-2d | +365 kJ | start physical_unit 9 10 standard_enthalpy_change kj qc_end physical_unit 12 12 24 26 standard_enthalpy qc_end physical_unit 14 14 29 31 standard_enthalpy qc_end physical_unit 17 17 34 35 standard_enthalpy qc_end physical_unit 20 20 38 40 standard_enthalpy qc_end chemical_equation 11 20 qc_end end | [{"type":"physical unit","value":"Standard enthalpy change [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"+365 kJ"}] | [{"type":"physical unit","value":"standard enthalpy [OF] A [=] \\pu{- 269 kJ/mol}"},{"type":"physical unit","value":"standard enthalpy [OF] B [=] \\pu{- 411 kJ/mol}"},{"type":"physical unit","value":"standard enthalpy [OF] C [=] \\pu{189 kJ/mol}"},{"type":"physical unit","value":"standard enthalpy [OF] D [=] \\pu{- 481 kJ/mol}"},{"type":"chemical equation","value":"2 A + B = 2 C + 2 D"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the standard enthalpy change for the reaction 2A + B = 2C + 2D, if A = -269
B = -411
C = 189
D = -481?</h1> | null | +365 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the reaction</p>
<blockquote>
<p><mathjax>#2"A" + "B" -> 2"C" + 2"D"#</mathjax></p>
</blockquote>
<p>Now, I assume that the values given to you represent the respective <em>standard enthalpies of formation</em> for the species that take part in the reaction. </p>
<p>Standard enthalpies of formation are measured when <strong>one mole</strong> of each of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their pure state. </p>
<p>As a result, it is expressed in <em>kilojoules per mole</em>.</p>
<p>This is a very important thing to keep in mind - standard enthalpies of formation are <strong>always</strong> given in <mathjax>#"kJ/mol"#</mathjax>!</p>
<p>In your case, you would have</p>
<blockquote>
<ul>
<li><mathjax>#"A" -> -"269 kJ/mol"#</mathjax></li>
<li><mathjax>#"B" -> -"411 kJ/mol"#</mathjax></li>
<li><mathjax>#"C" -> +"189 kJ/mol"#</mathjax></li>
<li><mathjax>#"D" -> -"481 kJ/mol"#</mathjax></li>
</ul>
</blockquote>
<p>You can calculate the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction using standard enthalpies of formation by taking into account the fact that enthalpy changes are <em>additive</em> - this is known as <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>. </p>
<p>In other words, you can express the enthlapy change of reaction by using each individual reaction that corresponds to the standard enthalpy change of formation for the products and for the reactants.</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx DeltaH_"f products"^@) - sum(m xx DeltaH_"f reactants"^@))#</mathjax></p>
</blockquote>
<p>Here <mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> represent the stoichiometric coefficients of the compounds in the reaction. </p>
<p>So, considering that this reaction consumes <strong>2 moles</strong> of <mathjax>#"A"#</mathjax> and <strong>one mole</strong> of <mathjax>#"B"#</mathjax>, and produces <strong>2 moles</strong> of <mathjax>#"C"#</mathjax> and <strong>2 moles</strong> of <mathjax>#"D"#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (+189"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-481"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-269"kJ"/color(red)(cancel(color(black)("mole")))) + 1color(red)(cancel(color(black)("mole"))) xx (-411"kJ"/color(red)(cancel(color(black)("mole"))))]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"584 kJ" - (-"949 kJ") = color(green)(+"365 kJ")#</mathjax></p>
</blockquote>
<p>Check out this video on how standard enthalpies of formation are used to find the enthalpy change of formation for a given reaction</p>
<p>
<iframe src="https://www.youtube.com/embed/1W6jMOJunQM?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>#DeltaH_"rxn" = +"365 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the reaction</p>
<blockquote>
<p><mathjax>#2"A" + "B" -> 2"C" + 2"D"#</mathjax></p>
</blockquote>
<p>Now, I assume that the values given to you represent the respective <em>standard enthalpies of formation</em> for the species that take part in the reaction. </p>
<p>Standard enthalpies of formation are measured when <strong>one mole</strong> of each of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their pure state. </p>
<p>As a result, it is expressed in <em>kilojoules per mole</em>.</p>
<p>This is a very important thing to keep in mind - standard enthalpies of formation are <strong>always</strong> given in <mathjax>#"kJ/mol"#</mathjax>!</p>
<p>In your case, you would have</p>
<blockquote>
<ul>
<li><mathjax>#"A" -> -"269 kJ/mol"#</mathjax></li>
<li><mathjax>#"B" -> -"411 kJ/mol"#</mathjax></li>
<li><mathjax>#"C" -> +"189 kJ/mol"#</mathjax></li>
<li><mathjax>#"D" -> -"481 kJ/mol"#</mathjax></li>
</ul>
</blockquote>
<p>You can calculate the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction using standard enthalpies of formation by taking into account the fact that enthalpy changes are <em>additive</em> - this is known as <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>. </p>
<p>In other words, you can express the enthlapy change of reaction by using each individual reaction that corresponds to the standard enthalpy change of formation for the products and for the reactants.</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx DeltaH_"f products"^@) - sum(m xx DeltaH_"f reactants"^@))#</mathjax></p>
</blockquote>
<p>Here <mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> represent the stoichiometric coefficients of the compounds in the reaction. </p>
<p>So, considering that this reaction consumes <strong>2 moles</strong> of <mathjax>#"A"#</mathjax> and <strong>one mole</strong> of <mathjax>#"B"#</mathjax>, and produces <strong>2 moles</strong> of <mathjax>#"C"#</mathjax> and <strong>2 moles</strong> of <mathjax>#"D"#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (+189"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-481"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-269"kJ"/color(red)(cancel(color(black)("mole")))) + 1color(red)(cancel(color(black)("mole"))) xx (-411"kJ"/color(red)(cancel(color(black)("mole"))))]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"584 kJ" - (-"949 kJ") = color(green)(+"365 kJ")#</mathjax></p>
</blockquote>
<p>Check out this video on how standard enthalpies of formation are used to find the enthalpy change of formation for a given reaction</p>
<p>
<iframe src="https://www.youtube.com/embed/1W6jMOJunQM?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How would you calculate the standard enthalpy change for the reaction 2A + B = 2C + 2D, if A = -269
B = -411
C = 189
D = -481?</h1>
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Stefan V.
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Oct 26, 2015
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<div class="markdown"><p><mathjax>#DeltaH_"rxn" = +"365 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the reaction</p>
<blockquote>
<p><mathjax>#2"A" + "B" -> 2"C" + 2"D"#</mathjax></p>
</blockquote>
<p>Now, I assume that the values given to you represent the respective <em>standard enthalpies of formation</em> for the species that take part in the reaction. </p>
<p>Standard enthalpies of formation are measured when <strong>one mole</strong> of each of these <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their pure state. </p>
<p>As a result, it is expressed in <em>kilojoules per mole</em>.</p>
<p>This is a very important thing to keep in mind - standard enthalpies of formation are <strong>always</strong> given in <mathjax>#"kJ/mol"#</mathjax>!</p>
<p>In your case, you would have</p>
<blockquote>
<ul>
<li><mathjax>#"A" -> -"269 kJ/mol"#</mathjax></li>
<li><mathjax>#"B" -> -"411 kJ/mol"#</mathjax></li>
<li><mathjax>#"C" -> +"189 kJ/mol"#</mathjax></li>
<li><mathjax>#"D" -> -"481 kJ/mol"#</mathjax></li>
</ul>
</blockquote>
<p>You can calculate the standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction using standard enthalpies of formation by taking into account the fact that enthalpy changes are <em>additive</em> - this is known as <strong><a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a></strong>. </p>
<p>In other words, you can express the enthlapy change of reaction by using each individual reaction that corresponds to the standard enthalpy change of formation for the products and for the reactants.</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx DeltaH_"f products"^@) - sum(m xx DeltaH_"f reactants"^@))#</mathjax></p>
</blockquote>
<p>Here <mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> represent the stoichiometric coefficients of the compounds in the reaction. </p>
<p>So, considering that this reaction consumes <strong>2 moles</strong> of <mathjax>#"A"#</mathjax> and <strong>one mole</strong> of <mathjax>#"B"#</mathjax>, and produces <strong>2 moles</strong> of <mathjax>#"C"#</mathjax> and <strong>2 moles</strong> of <mathjax>#"D"#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (+189"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-481"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-269"kJ"/color(red)(cancel(color(black)("mole")))) + 1color(red)(cancel(color(black)("mole"))) xx (-411"kJ"/color(red)(cancel(color(black)("mole"))))]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"584 kJ" - (-"949 kJ") = color(green)(+"365 kJ")#</mathjax></p>
</blockquote>
<p>Check out this video on how standard enthalpies of formation are used to find the enthalpy change of formation for a given reaction</p>
<p>
<iframe src="https://www.youtube.com/embed/1W6jMOJunQM?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | How would you calculate the standard enthalpy change for the reaction 2A + B = 2C + 2D, if A = -269
B = -411
C = 189
D = -481? | null |
1,660 | ab5451b8-6ddd-11ea-8ff6-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-which-contains-0-0134-g-of-iron-0-0 | FeSO3 | start chemical_formula qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 17 17 14 15 mass qc_end physical_unit 22 22 19 20 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] empirical"}] | [{"type":"chemical equation","value":"FeSO3"}] | [{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{0.0134 g}"},{"type":"physical unit","value":"Mass [OF] sulfur [=] \\pu{0.00769 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{0.0115 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?</h1> | null | FeSO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Convert to % per 100 wt and follow the scheme... </p>
<p>Given <mathjax>#0.0134 g Fe + 0.00769 g S + 0.0115 g O = 0.03259 g#</mathjax></p>
<p>=> <mathjax>#%Fe = ((0.0134g)/(0.03259g))100% = 41.1169% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%S = ((0.00769g)/(0.03259g))100% = 23.5962% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%O = ((0.0115g)/(0.03259g))100% = 35.2669% per 100wt#</mathjax></p>
<p>%/100wt => g/100wt => moles => normalize (divide by smallest moles) => Empirical Ratio => Empirical Formula</p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/XSPoRdRLTyaQpY9MNcSQ_Empirical%20Formula%20FeSO3_0002.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#FeSO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Convert to % per 100 wt and follow the scheme... </p>
<p>Given <mathjax>#0.0134 g Fe + 0.00769 g S + 0.0115 g O = 0.03259 g#</mathjax></p>
<p>=> <mathjax>#%Fe = ((0.0134g)/(0.03259g))100% = 41.1169% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%S = ((0.00769g)/(0.03259g))100% = 23.5962% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%O = ((0.0115g)/(0.03259g))100% = 35.2669% per 100wt#</mathjax></p>
<p>%/100wt => g/100wt => moles => normalize (divide by smallest moles) => Empirical Ratio => Empirical Formula</p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/XSPoRdRLTyaQpY9MNcSQ_Empirical%20Formula%20FeSO3_0002.jpg"/> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?</h1>
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Doc048
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<span class="dateCreated" datetime="2017-07-02T02:10:14" itemprop="dateCreated">
Jul 2, 2017
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<div class="markdown"><p><mathjax>#FeSO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Convert to % per 100 wt and follow the scheme... </p>
<p>Given <mathjax>#0.0134 g Fe + 0.00769 g S + 0.0115 g O = 0.03259 g#</mathjax></p>
<p>=> <mathjax>#%Fe = ((0.0134g)/(0.03259g))100% = 41.1169% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%S = ((0.00769g)/(0.03259g))100% = 23.5962% per 100 wt#</mathjax></p>
<p>=> <mathjax>#%O = ((0.0115g)/(0.03259g))100% = 35.2669% per 100wt#</mathjax></p>
<p>%/100wt => g/100wt => moles => normalize (divide by smallest moles) => Empirical Ratio => Empirical Formula</p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/XSPoRdRLTyaQpY9MNcSQ_Empirical%20Formula%20FeSO3_0002.jpg"/> </p></div>
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Ernest Z.
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<span class="dateCreated" datetime="2017-07-02T13:46:35" itemprop="dateCreated">
Jul 2, 2017
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<div class="markdown"><p>The empirical formula is <mathjax>#"FeSO"_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The <strong>empirical formula</strong> is the simplest whole number molar ratio of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>We must convert the masses of <mathjax>#"Fe"#</mathjax>, <mathjax>#"S"#</mathjax>, and <mathjax>#"O"#</mathjax> to moles and then find the ratio.</p>
<blockquote></blockquote>
<p><mathjax>#"Moles of Fe" = 0.0134 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = 2.400 × 10^"-4"color(white)(l) "mol Fe"#</mathjax></p>
<p><mathjax>#"Moles of S" = "0.007 69" color(red)(cancel(color(black)("g S"))) × "1 mol S"/(32.06 color(red)(cancel(color(black)("g S")))) = 2.399 × 10^"-4"color(white)(l)"mol S"#</mathjax></p>
<p><mathjax>#"Moles of O" = 0.0115 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = 7.188 × 10^"-4"color(white)(l) "mol O"#</mathjax></p>
<blockquote></blockquote>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#bb("Element"color(white)(Ag) "Mass/g"color(white)(Xm) "Moles"color(white)(Xmm) "Ratio"color(white)(m)color(white)(l)"Integers")#</mathjax><br/>
<mathjax>#color(white)(m)"Fe" color(white)(XXXml)0.0134 color(white)(Xml)2.400 × 10^"-4" color(white)(Xll)1.000color(white)(mmmll)1#</mathjax><br/>
<mathjax>#color(white)(m)"S" color(white)(XXXXm)0.00769 color(white)(mll)2.399 × 10^"-4" color(white)(Xll)1 color(white)(mmmmmll)1#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXXll)0.0115 color(white)(mml)7.188 × 10^"-4" color(white)(Xll)2.996 color(white)(mmmll)3#</mathjax></p>
<p>The molar ratios are <mathjax>#"Fe:S:O = 1:1:3"#</mathjax>.</p>
<p>The empirical formula is <mathjax>#"FeSO"_3#</mathjax>.</p>
<blockquote></blockquote>
<p>Here is a video that illustrates how to determine an empirical formula.</p>
<p>
<iframe src="https://www.youtube.com/embed/r1OsFpz_-qY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen? | null |
1,661 | a9daeb00-6ddd-11ea-8702-ccda262736ce | https://socratic.org/questions/564e9387581e2a7cd735d6d0 | Ba(NO3)2(aq) + Na2CO3(aq) -> BaCO3(s) + 2 NaNO3(aq) | start chemical_equation qc_end substance 2 3 qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Ba(NO3)2(aq) + Na2CO3(aq) -> BaCO3(s) + 2 NaNO3(aq)"}] | [{"type":"substance name","value":"Barium nitrate"},{"type":"substance name","value":"Sodium carbonate"}] | <h1 class="questionTitle" itemprop="name">How do barium nitrate and sodium carbonate react in aqueous solution?</h1> | null | Ba(NO3)2(aq) + Na2CO3(aq) -> BaCO3(s) + 2 NaNO3(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All nitrates are soluble (in aqueous solution). All carbonates are INSOLUBLE (except for those of the alkali metals). Barium carbonate is certainly insoluble. </p>
<p>The net ionic equation for the above:</p>
<p><mathjax>#Ba^(2+) + CO_3^(2-) rarr BaCO_3(s)darr#</mathjax></p>
<p>As for any equation, both mass and charge are conserved. What does this mean with respect to the 2nd equation?</p></div>
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<div class="markdown"><p><mathjax>#Ba(NO_3)_2(aq) + Na_2CO_3(aq) rarr BaCO_3(s)darr + 2Na(NO_3)(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All nitrates are soluble (in aqueous solution). All carbonates are INSOLUBLE (except for those of the alkali metals). Barium carbonate is certainly insoluble. </p>
<p>The net ionic equation for the above:</p>
<p><mathjax>#Ba^(2+) + CO_3^(2-) rarr BaCO_3(s)darr#</mathjax></p>
<p>As for any equation, both mass and charge are conserved. What does this mean with respect to the 2nd equation?</p></div>
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<h1 class="questionTitle" itemprop="name">How do barium nitrate and sodium carbonate react in aqueous solution?</h1>
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<div class="markdown"><p><mathjax>#Ba(NO_3)_2(aq) + Na_2CO_3(aq) rarr BaCO_3(s)darr + 2Na(NO_3)(aq)#</mathjax></p></div>
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<div class="markdown"><p>All nitrates are soluble (in aqueous solution). All carbonates are INSOLUBLE (except for those of the alkali metals). Barium carbonate is certainly insoluble. </p>
<p>The net ionic equation for the above:</p>
<p><mathjax>#Ba^(2+) + CO_3^(2-) rarr BaCO_3(s)darr#</mathjax></p>
<p>As for any equation, both mass and charge are conserved. What does this mean with respect to the 2nd equation?</p></div>
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</article> | How do barium nitrate and sodium carbonate react in aqueous solution? | null |
1,662 | a8e3cf7a-6ddd-11ea-9d96-ccda262736ce | https://socratic.org/questions/ten-grams-of-a-gas-occupies-12-5-liters-at-a-pressure-of-42-0-cm-hg-what-is-the- | 7 liters | start physical_unit 4 4 volume l qc_end physical_unit 4 4 6 7 volume qc_end physical_unit 4 4 12 14 pressure qc_end physical_unit 4 4 25 27 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"7 liters"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{10 grams}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{12.5 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{42.0 cm Hg}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{75.0 cm Hg}"}] | <h1 class="questionTitle" itemprop="name"> Ten grams of a gas occupies 12.5 liters at a pressure of 42.0 cm Hg. What is the volume when the pressure has increased to 75.0 cm Hg?</h1> | null | 7 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is convert the <mathjax>#cm*Hg#</mathjax> into a pressure reading, and calculate <mathjax>#V_2#</mathjax>.</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax>; of course, given the relationship the units of pressure cancel out.</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.5*Lxx420*mm*Hg)/(750*mm*Hg)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#7*L#</mathjax>.</p>
<p>Note that here we use a unit of length to measure the pressure. That is <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. Ordinarily, you would be expected to calculate the pressure with authentic units of <mathjax>#kPa#</mathjax> or <mathjax>#"atmospheres"#</mathjax>. Here, we can afford to be a little slack. </p></div>
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<div class="markdown"><p>At constant temperature, and constant amount of gas, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is convert the <mathjax>#cm*Hg#</mathjax> into a pressure reading, and calculate <mathjax>#V_2#</mathjax>.</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax>; of course, given the relationship the units of pressure cancel out.</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.5*Lxx420*mm*Hg)/(750*mm*Hg)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#7*L#</mathjax>.</p>
<p>Note that here we use a unit of length to measure the pressure. That is <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. Ordinarily, you would be expected to calculate the pressure with authentic units of <mathjax>#kPa#</mathjax> or <mathjax>#"atmospheres"#</mathjax>. Here, we can afford to be a little slack. </p></div>
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<h1 class="questionTitle" itemprop="name"> Ten grams of a gas occupies 12.5 liters at a pressure of 42.0 cm Hg. What is the volume when the pressure has increased to 75.0 cm Hg?</h1>
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<div class="markdown"><p>At constant temperature, and constant amount of gas, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is convert the <mathjax>#cm*Hg#</mathjax> into a pressure reading, and calculate <mathjax>#V_2#</mathjax>.</p>
<p><mathjax>#1*atm-=760*mm*Hg#</mathjax>; of course, given the relationship the units of pressure cancel out.</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.5*Lxx420*mm*Hg)/(750*mm*Hg)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#7*L#</mathjax>.</p>
<p>Note that here we use a unit of length to measure the pressure. That is <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. Ordinarily, you would be expected to calculate the pressure with authentic units of <mathjax>#kPa#</mathjax> or <mathjax>#"atmospheres"#</mathjax>. Here, we can afford to be a little slack. </p></div>
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</article> | Ten grams of a gas occupies 12.5 liters at a pressure of 42.0 cm Hg. What is the volume when the pressure has increased to 75.0 cm Hg? | null |
1,663 | ab6d309e-6ddd-11ea-b503-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-perfluoropropane-if-the-compound-contains-81-f | C3F8 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] perfluoropropane [IN] empirical"}] | [{"type":"chemical equation","value":"C3F8"}] | [{"type":"physical unit","value":"Percent by mass [OF] fluorine in the compound [=] \\pu{81%}"},{"type":"physical unit","value":"Percent by mass [OF] carbon in the compound [=] \\pu{19%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for perfluoropropane if the compound contains 81% fluorine and 19% carbon by mass? </h1> | null | C3F8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always with the problems it is useful to assume an <mathjax>#100*g#</mathjax> mass of compound, and then interrogate the atomic composition...</p>
<p><mathjax>#"Moles of fluorine"=(81*g)/(19.0*g*mol^-1)=4.26*mol#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"=(19*g)/(12.011*g*mol^-1)=1.59*mol#</mathjax>.</p>
<p>We then divide thru by the LOWEST molar quantity to get a trial empirical formula of ....<mathjax>#C_((1.59*mol)/(1.59*mol))F_((4.26*mol)/(1.59*mol))=CF_2.68#</mathjax>...</p>
<p>But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get...<mathjax>#C_3F_8#</mathjax>...<mathjax>#"perfluoropropane"#</mathjax>...this is the old <mathjax>#"Freon 218"#</mathjax>...and you must have twisted some analyst's arm to make the determination. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_3F_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always with the problems it is useful to assume an <mathjax>#100*g#</mathjax> mass of compound, and then interrogate the atomic composition...</p>
<p><mathjax>#"Moles of fluorine"=(81*g)/(19.0*g*mol^-1)=4.26*mol#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"=(19*g)/(12.011*g*mol^-1)=1.59*mol#</mathjax>.</p>
<p>We then divide thru by the LOWEST molar quantity to get a trial empirical formula of ....<mathjax>#C_((1.59*mol)/(1.59*mol))F_((4.26*mol)/(1.59*mol))=CF_2.68#</mathjax>...</p>
<p>But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get...<mathjax>#C_3F_8#</mathjax>...<mathjax>#"perfluoropropane"#</mathjax>...this is the old <mathjax>#"Freon 218"#</mathjax>...and you must have twisted some analyst's arm to make the determination. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for perfluoropropane if the compound contains 81% fluorine and 19% carbon by mass? </h1>
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anor277
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<span class="dateCreated" datetime="2018-03-29T04:06:29" itemprop="dateCreated">
Mar 29, 2018
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<div class="markdown"><p><mathjax>#C_3F_8#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always with the problems it is useful to assume an <mathjax>#100*g#</mathjax> mass of compound, and then interrogate the atomic composition...</p>
<p><mathjax>#"Moles of fluorine"=(81*g)/(19.0*g*mol^-1)=4.26*mol#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"=(19*g)/(12.011*g*mol^-1)=1.59*mol#</mathjax>.</p>
<p>We then divide thru by the LOWEST molar quantity to get a trial empirical formula of ....<mathjax>#C_((1.59*mol)/(1.59*mol))F_((4.26*mol)/(1.59*mol))=CF_2.68#</mathjax>...</p>
<p>But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get...<mathjax>#C_3F_8#</mathjax>...<mathjax>#"perfluoropropane"#</mathjax>...this is the old <mathjax>#"Freon 218"#</mathjax>...and you must have twisted some analyst's arm to make the determination. </p></div>
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</article> | What is the empirical formula for perfluoropropane if the compound contains 81% fluorine and 19% carbon by mass? | null |
1,664 | aa7a1a94-6ddd-11ea-92a2-ccda262736ce | https://socratic.org/questions/what-is-the-boiling-point-of-an-ionic-solution-containing-29-7-g-na-2so-4-and-84 | 103.81 ℃ | start physical_unit 7 8 boiling_point_temperature °c qc_end physical_unit 12 12 10 11 mass qc_end physical_unit 16 16 14 15 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Boiling point [OF] ionic solution [IN] ℃"}] | [{"type":"physical unit","value":"103.81 ℃"}] | [{"type":"physical unit","value":"Mass [OF] Na2SO4 [=] \\pu{29.7 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{84.4 g}"},{"type":"other","value":"100% ionization."}] | <h1 class="questionTitle" itemprop="name">What is the boiling point of an ionic solution containing 29.7 g #Na_2SO_4# and 84.4 g water, assuming 100% ionization?</h1> | null | 103.81 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every formula unit</strong> of sodium sulfate produces <strong>three ions</strong> in solution, tow sodium cations and one sulfate anion. This means that the solution's <strong>van't Hoff factor</strong> will be equal to <mathjax>#3#</mathjax>. </p>
<p>The equation for <em>boiling-point elevation</em> looks like this</p>
<p><mathjax>#DeltaT_b = i * K_b * b#</mathjax>, where</p>
<p><mathjax>#i#</mathjax> - the van't Hoff factor;<br/>
<mathjax>#K_b#</mathjax> - the <strong>ebullioscopic constant</strong>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution. <br/>
<mathjax>#DeltaT_b#</mathjax> - the poiling point elevation - defined as <mathjax>#T_"b" - T_"b"^@#</mathjax></p>
<p>The <em>ebullioscopic constant</em> of water is equal to <mathjax>#0.512^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>Now, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution is defined as the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium sulfate, divided by the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, expressed in <strong>kilograms</strong>. </p>
<p>Use sodium sulfate's molar mass to determine how many moles you have in that <mathjax>#"29.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#29.7color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.2091 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution will thus be </p>
<blockquote>
<p><mathjax>#color(blue)(b = n_"solute"/m_"solvent")#</mathjax></p>
<p><mathjax>#b = "0.2091 moles"/(84.4 * 10^(-3)"kg") = "2.477 moles kg"^(-1) = "2.477 molal"#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#DeltaT_b#</mathjax>, the boiling-point elevation of the solution</p>
<blockquote>
<p><mathjax>#DeltaT_b = 3 * 0.512^@"C"color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 2.477color(red)(cancel(color(black)("moles")))color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 3.805^@"C"#</mathjax></p>
</blockquote>
<p>The boiling point of the solution will thus be </p>
<blockquote>
<p><mathjax>#DeltaT_"b" = T_"b" - T_"b"^@" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_"b"#</mathjax> - the boiling point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"b"^@#</mathjax> - the boiling point of the solution</p>
<blockquote>
<p><mathjax>#T_"b" = DeltaT_"b" + T_"b"^@#</mathjax></p>
<p><mathjax>#T_"b" = 3.805^@"C" + 100^@"C" = 103.805^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#T_"b" = color(green)(104^@"C")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#104^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every formula unit</strong> of sodium sulfate produces <strong>three ions</strong> in solution, tow sodium cations and one sulfate anion. This means that the solution's <strong>van't Hoff factor</strong> will be equal to <mathjax>#3#</mathjax>. </p>
<p>The equation for <em>boiling-point elevation</em> looks like this</p>
<p><mathjax>#DeltaT_b = i * K_b * b#</mathjax>, where</p>
<p><mathjax>#i#</mathjax> - the van't Hoff factor;<br/>
<mathjax>#K_b#</mathjax> - the <strong>ebullioscopic constant</strong>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution. <br/>
<mathjax>#DeltaT_b#</mathjax> - the poiling point elevation - defined as <mathjax>#T_"b" - T_"b"^@#</mathjax></p>
<p>The <em>ebullioscopic constant</em> of water is equal to <mathjax>#0.512^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>Now, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution is defined as the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium sulfate, divided by the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, expressed in <strong>kilograms</strong>. </p>
<p>Use sodium sulfate's molar mass to determine how many moles you have in that <mathjax>#"29.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#29.7color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.2091 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution will thus be </p>
<blockquote>
<p><mathjax>#color(blue)(b = n_"solute"/m_"solvent")#</mathjax></p>
<p><mathjax>#b = "0.2091 moles"/(84.4 * 10^(-3)"kg") = "2.477 moles kg"^(-1) = "2.477 molal"#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#DeltaT_b#</mathjax>, the boiling-point elevation of the solution</p>
<blockquote>
<p><mathjax>#DeltaT_b = 3 * 0.512^@"C"color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 2.477color(red)(cancel(color(black)("moles")))color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 3.805^@"C"#</mathjax></p>
</blockquote>
<p>The boiling point of the solution will thus be </p>
<blockquote>
<p><mathjax>#DeltaT_"b" = T_"b" - T_"b"^@" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_"b"#</mathjax> - the boiling point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"b"^@#</mathjax> - the boiling point of the solution</p>
<blockquote>
<p><mathjax>#T_"b" = DeltaT_"b" + T_"b"^@#</mathjax></p>
<p><mathjax>#T_"b" = 3.805^@"C" + 100^@"C" = 103.805^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#T_"b" = color(green)(104^@"C")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the boiling point of an ionic solution containing 29.7 g #Na_2SO_4# and 84.4 g water, assuming 100% ionization?</h1>
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<div>
<div class="markdown"><p><mathjax>#104^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium sulfate</em>, <mathjax>#"Na"_2"SO"_4#</mathjax>, will dissociate in aqueous solution to form sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and sulfate anions, <mathjax>#"SO"_4^(2-)#</mathjax></p>
<blockquote>
<p><mathjax>#"Na"_2"SO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every formula unit</strong> of sodium sulfate produces <strong>three ions</strong> in solution, tow sodium cations and one sulfate anion. This means that the solution's <strong>van't Hoff factor</strong> will be equal to <mathjax>#3#</mathjax>. </p>
<p>The equation for <em>boiling-point elevation</em> looks like this</p>
<p><mathjax>#DeltaT_b = i * K_b * b#</mathjax>, where</p>
<p><mathjax>#i#</mathjax> - the van't Hoff factor;<br/>
<mathjax>#K_b#</mathjax> - the <strong>ebullioscopic constant</strong>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution. <br/>
<mathjax>#DeltaT_b#</mathjax> - the poiling point elevation - defined as <mathjax>#T_"b" - T_"b"^@#</mathjax></p>
<p>The <em>ebullioscopic constant</em> of water is equal to <mathjax>#0.512^@"C kg mol"^(-1)#</mathjax></p>
<p><a href="http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf" rel="nofollow" target="_blank">http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf</a></p>
<p>Now, the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution is defined as the number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium sulfate, divided by the mass of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, which in your case is water, expressed in <strong>kilograms</strong>. </p>
<p>Use sodium sulfate's molar mass to determine how many moles you have in that <mathjax>#"29.7-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#29.7color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.2091 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution will thus be </p>
<blockquote>
<p><mathjax>#color(blue)(b = n_"solute"/m_"solvent")#</mathjax></p>
<p><mathjax>#b = "0.2091 moles"/(84.4 * 10^(-3)"kg") = "2.477 moles kg"^(-1) = "2.477 molal"#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#DeltaT_b#</mathjax>, the boiling-point elevation of the solution</p>
<blockquote>
<p><mathjax>#DeltaT_b = 3 * 0.512^@"C"color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 2.477color(red)(cancel(color(black)("moles")))color(red)(cancel(color(black)("kg"^(-1))))#</mathjax></p>
<p><mathjax>#DeltaT_b = 3.805^@"C"#</mathjax></p>
</blockquote>
<p>The boiling point of the solution will thus be </p>
<blockquote>
<p><mathjax>#DeltaT_"b" = T_"b" - T_"b"^@" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#T_"b"#</mathjax> - the boiling point of the <strong>pure solvent</strong><br/>
<mathjax>#T_"b"^@#</mathjax> - the boiling point of the solution</p>
<blockquote>
<p><mathjax>#T_"b" = DeltaT_"b" + T_"b"^@#</mathjax></p>
<p><mathjax>#T_"b" = 3.805^@"C" + 100^@"C" = 103.805^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#T_"b" = color(green)(104^@"C")#</mathjax></p>
</blockquote></div>
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</article> | What is the boiling point of an ionic solution containing 29.7 g #Na_2SO_4# and 84.4 g water, assuming 100% ionization? | null |
1,665 | ac94db98-6ddd-11ea-a2f7-ccda262736ce | https://socratic.org/questions/592c5281b72cff02377e34c3 | 1.11 g | start physical_unit 3 4 mass g qc_end physical_unit 12 13 10 11 mass qc_end physical_unit 21 21 18 19 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] dihydrogen gas [IN] g"}] | [{"type":"physical unit","value":"1.11 g"}] | [{"type":"physical unit","value":"Mass [OF] sodium metal [=] \\pu{20 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{10 g}"}] | <h1 class="questionTitle" itemprop="name">What mass of dihydrogen gas results from the oxidation of #20*g# sodium metal by a mass of #10*g# of water?</h1> | null | 1.11 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We represent the reaction by the following stoichiometric equation:</p>
<p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p>
<p>You can double this if you like. I think the arithmetic is easier if you include the <mathjax>#1/2#</mathjax> coefficient.......</p>
<p><mathjax>#"Moles of sodium"=(20*g)/(22.99*g*mol^-1)=0.870*mol#</mathjax></p>
<p><mathjax>#"Moles of water"=(10*g)/(18.01*g*mol^-1)=0.555*mol#</mathjax>.</p>
<p>Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only <mathjax>#0.555*mol#</mathjax> of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.</p>
<p>And thus we get <mathjax>#(0.555*mol)/2#</mathjax> dihydrogen gas evolved, i.e. <mathjax>#(0.555*mol)/2xx2.02*g*mol^-1=1.11*g#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well first we write a stoichiometric equation, and get a bit over <mathjax>#1*g#</mathjax> of <mathjax>#H_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We represent the reaction by the following stoichiometric equation:</p>
<p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p>
<p>You can double this if you like. I think the arithmetic is easier if you include the <mathjax>#1/2#</mathjax> coefficient.......</p>
<p><mathjax>#"Moles of sodium"=(20*g)/(22.99*g*mol^-1)=0.870*mol#</mathjax></p>
<p><mathjax>#"Moles of water"=(10*g)/(18.01*g*mol^-1)=0.555*mol#</mathjax>.</p>
<p>Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only <mathjax>#0.555*mol#</mathjax> of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.</p>
<p>And thus we get <mathjax>#(0.555*mol)/2#</mathjax> dihydrogen gas evolved, i.e. <mathjax>#(0.555*mol)/2xx2.02*g*mol^-1=1.11*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of dihydrogen gas results from the oxidation of #20*g# sodium metal by a mass of #10*g# of water?</h1>
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anor277
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<div class="markdown"><p>Well first we write a stoichiometric equation, and get a bit over <mathjax>#1*g#</mathjax> of <mathjax>#H_2(g)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We represent the reaction by the following stoichiometric equation:</p>
<p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p>
<p>You can double this if you like. I think the arithmetic is easier if you include the <mathjax>#1/2#</mathjax> coefficient.......</p>
<p><mathjax>#"Moles of sodium"=(20*g)/(22.99*g*mol^-1)=0.870*mol#</mathjax></p>
<p><mathjax>#"Moles of water"=(10*g)/(18.01*g*mol^-1)=0.555*mol#</mathjax>.</p>
<p>Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only <mathjax>#0.555*mol#</mathjax> of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.</p>
<p>And thus we get <mathjax>#(0.555*mol)/2#</mathjax> dihydrogen gas evolved, i.e. <mathjax>#(0.555*mol)/2xx2.02*g*mol^-1=1.11*g#</mathjax></p></div>
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</article> | What mass of dihydrogen gas results from the oxidation of #20*g# sodium metal by a mass of #10*g# of water? | null |
1,666 | a8f3d0ca-6ddd-11ea-bffd-ccda262736ce | https://socratic.org/questions/how-much-energy-is-absorbed-as-heat-by-21-g-of-gold-when-it-is-heated-from-25-c- | 27.09 J | start physical_unit 11 11 heat_energy j qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 11 11 17 18 temperature qc_end physical_unit 11 11 20 21 temperature qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] gold [IN] J"}] | [{"type":"physical unit","value":"27.09 J"}] | [{"type":"physical unit","value":"Mass [OF] gold [=] \\pu{21 g}"},{"type":"physical unit","value":"Temperature1 [OF] gold [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] gold [=] \\pu{35 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much energy is absorbed as heat by 21 g of gold when it is heated from 25°C to 35°C?</h1> | null | 27.09 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat absorbed by the system could be calculated by:</p>
<p><mathjax>#q=mxxsxxDeltaT#</mathjax></p>
<p>where, <br/>
<mathjax>#q#</mathjax> is the heat, <br/>
<mathjax>#m#</mathjax> is the mass of the gold sample, <br/>
<mathjax>#s=0.129 J/(g*^@C)#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of gold<br/>
and <mathjax>#DeltaT=T_f-T_i=35^@-25^@=10^@C#</mathjax> is the change on temperature of the sample.</p>
<p><mathjax>#=>q=21cancel(g)xx0.129 J/(cancel(g)*cancel(""^@C))xx10cancel(""^@C)=27.09 J#</mathjax></p>
<p>Here is a video that explains more about this topic:<br/>
<strong>Thermochemistry | <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> and <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">Calorimetry</a>. </strong><br/>
<iframe src="https://www.youtube.com/embed/RuBReH04pyE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#q=27.09 J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat absorbed by the system could be calculated by:</p>
<p><mathjax>#q=mxxsxxDeltaT#</mathjax></p>
<p>where, <br/>
<mathjax>#q#</mathjax> is the heat, <br/>
<mathjax>#m#</mathjax> is the mass of the gold sample, <br/>
<mathjax>#s=0.129 J/(g*^@C)#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of gold<br/>
and <mathjax>#DeltaT=T_f-T_i=35^@-25^@=10^@C#</mathjax> is the change on temperature of the sample.</p>
<p><mathjax>#=>q=21cancel(g)xx0.129 J/(cancel(g)*cancel(""^@C))xx10cancel(""^@C)=27.09 J#</mathjax></p>
<p>Here is a video that explains more about this topic:<br/>
<strong>Thermochemistry | <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> and <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">Calorimetry</a>. </strong><br/>
<iframe src="https://www.youtube.com/embed/RuBReH04pyE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">How much energy is absorbed as heat by 21 g of gold when it is heated from 25°C to 35°C?</h1>
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Dr. Hayek
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<span class="dateCreated" datetime="2016-01-03T17:29:21" itemprop="dateCreated">
Jan 3, 2016
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<div class="markdown"><p><mathjax>#q=27.09 J#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat absorbed by the system could be calculated by:</p>
<p><mathjax>#q=mxxsxxDeltaT#</mathjax></p>
<p>where, <br/>
<mathjax>#q#</mathjax> is the heat, <br/>
<mathjax>#m#</mathjax> is the mass of the gold sample, <br/>
<mathjax>#s=0.129 J/(g*^@C)#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of gold<br/>
and <mathjax>#DeltaT=T_f-T_i=35^@-25^@=10^@C#</mathjax> is the change on temperature of the sample.</p>
<p><mathjax>#=>q=21cancel(g)xx0.129 J/(cancel(g)*cancel(""^@C))xx10cancel(""^@C)=27.09 J#</mathjax></p>
<p>Here is a video that explains more about this topic:<br/>
<strong>Thermochemistry | <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> and <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">Calorimetry</a>. </strong><br/>
<iframe src="https://www.youtube.com/embed/RuBReH04pyE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How much energy is absorbed as heat by 21 g of gold when it is heated from 25°C to 35°C? | null |
1,667 | a8d22b47-6ddd-11ea-978b-ccda262736ce | https://socratic.org/questions/how-many-grams-of-nacl-are-in-225-ml-of-a-0-75-m-nacl-solution-the-molecular-wei | 9.9 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 14 11 12 molarity qc_end physical_unit 13 14 7 8 volume qc_end physical_unit 4 4 21 22 molecular_weight qc_end end | [{"type":"physical unit","value":"Mass [OF] NaCl [IN] grams"}] | [{"type":"physical unit","value":"9.9 grams"}] | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [=] \\pu{0.75 M}"},{"type":"physical unit","value":"Volume [OF] NaCl solution [=] \\pu{225 mL}"},{"type":"physical unit","value":"Molecular weight [OF] NaCl [=] \\pu{58.44 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of NaCl are in 225 mL of a 0.75 M NaCl solution? (The molecular weight of NaCl is 58.44 g.)</h1> | null | 9.9 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing you need to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution to determine how many <strong>moles</strong> of sodium chloride you have in your sample. </p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium chloride, present in <mathjax>#"1 L"#</mathjax> of solution. This means that a <mathjax>#"0.75 M"#</mathjax> sodium chloride solution will contain <mathjax>#0.75#</mathjax> <strong>moles</strong> of sodium chloride <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution. </p>
<blockquote>
<p><mathjax>#"0.75 M: " "0.75 moles NaCl " stackrel( color(white)(a) color(blue)("for every") color(white)(aaa) )(rarr) "1 L of solution"#</mathjax></p>
</blockquote>
<p>You can thus use the molarity of the solution as a conversion factor to find the number of moles of sodium chloride present in your sample</p>
<blockquote>
<p><mathjax>#225 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.75 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.16875 moles NaCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of sodium chloride</p>
<blockquote>
<p><mathjax>#0.16875 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("9.9 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"9.9 g NaCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing you need to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution to determine how many <strong>moles</strong> of sodium chloride you have in your sample. </p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium chloride, present in <mathjax>#"1 L"#</mathjax> of solution. This means that a <mathjax>#"0.75 M"#</mathjax> sodium chloride solution will contain <mathjax>#0.75#</mathjax> <strong>moles</strong> of sodium chloride <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution. </p>
<blockquote>
<p><mathjax>#"0.75 M: " "0.75 moles NaCl " stackrel( color(white)(a) color(blue)("for every") color(white)(aaa) )(rarr) "1 L of solution"#</mathjax></p>
</blockquote>
<p>You can thus use the molarity of the solution as a conversion factor to find the number of moles of sodium chloride present in your sample</p>
<blockquote>
<p><mathjax>#225 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.75 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.16875 moles NaCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of sodium chloride</p>
<blockquote>
<p><mathjax>#0.16875 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("9.9 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of NaCl are in 225 mL of a 0.75 M NaCl solution? (The molecular weight of NaCl is 58.44 g.)</h1>
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<div class="markdown"><p><mathjax>#"9.9 g NaCl"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing you need to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution to determine how many <strong>moles</strong> of sodium chloride you have in your sample. </p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium chloride, present in <mathjax>#"1 L"#</mathjax> of solution. This means that a <mathjax>#"0.75 M"#</mathjax> sodium chloride solution will contain <mathjax>#0.75#</mathjax> <strong>moles</strong> of sodium chloride <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution. </p>
<blockquote>
<p><mathjax>#"0.75 M: " "0.75 moles NaCl " stackrel( color(white)(a) color(blue)("for every") color(white)(aaa) )(rarr) "1 L of solution"#</mathjax></p>
</blockquote>
<p>You can thus use the molarity of the solution as a conversion factor to find the number of moles of sodium chloride present in your sample</p>
<blockquote>
<p><mathjax>#225 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.75 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))) = "0.16875 moles NaCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of sodium chloride</p>
<blockquote>
<p><mathjax>#0.16875 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("9.9 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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</article> | How many grams of NaCl are in 225 mL of a 0.75 M NaCl solution? (The molecular weight of NaCl is 58.44 g.) | null |
1,668 | ac27995d-6ddd-11ea-9911-ccda262736ce | https://socratic.org/questions/what-is-the-k-sp-for-silver-chloride | 1.76 × 10^(-10) | start physical_unit 2 3 equilibrium_constant_k none qc_end physical_unit 2 3 6 7 molar_mass qc_end physical_unit 14 14 11 12 volume qc_end physical_unit 2 3 15 18 mass qc_end end | [{"type":"physical unit","value":"Ksp [OF] silver chloride"}] | [{"type":"physical unit","value":"1.76 × 10^(-10)"}] | [{"type":"physical unit","value":"MM [OF] silver chloride [=] \\pu{143.3 g/mol}"},{"type":"physical unit","value":"Volume [OF] H2O [=] \\pu{100 mL}"},{"type":"physical unit","value":"Mass [OF] silver chloride [=] \\pu{1.9 × 10^(-4) grams}"}] | <h1 class="questionTitle" itemprop="name">What is the #K_(sp)# for silver chloride? </h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>When solid silver chloride (MM=143.3) is added to 100 mL of <mathjax>#H_2O#</mathjax>, <mathjax>#1.9 xx 10^-4#</mathjax> grams dissolves. What is the <mathjax>#K_(sp)#</mathjax> for silver chloride? </p>
<p>The answer should be <mathjax>#1.8 xx 10^-10#</mathjax>, but I'm not quite sure how, could someone please explain? Thanks </p></div>
</h2>
</div>
</div> | 1.76 × 10^(-10) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For which <mathjax>#K_"sp"=[Ag^+][Cl^-]=??#</mathjax></p>
<p>We are given that <mathjax>#[AgCl]=((1.9xx10^-4*g)/(143.32*g*mol^-1))/(0.100*L)=1.33xx10^-5*mol*L^-1#</mathjax>...</p>
<p>But by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>...</p>
<p><mathjax>#[Ag^+]=[Cl^-]=1.33xx10^-5*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_"sp"=[1.33xx10^-5]^2=1.76xx10^-10#</mathjax>...the which is close enuff to your answer....</p>
<p>Happy?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>WE address the equilibrium...<mathjax>#Ag^+ +Cl^(-) rightleftharpoonsAgCl(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For which <mathjax>#K_"sp"=[Ag^+][Cl^-]=??#</mathjax></p>
<p>We are given that <mathjax>#[AgCl]=((1.9xx10^-4*g)/(143.32*g*mol^-1))/(0.100*L)=1.33xx10^-5*mol*L^-1#</mathjax>...</p>
<p>But by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>...</p>
<p><mathjax>#[Ag^+]=[Cl^-]=1.33xx10^-5*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_"sp"=[1.33xx10^-5]^2=1.76xx10^-10#</mathjax>...the which is close enuff to your answer....</p>
<p>Happy?</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the #K_(sp)# for silver chloride? </h1>
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<div class="markdown"><p>When solid silver chloride (MM=143.3) is added to 100 mL of <mathjax>#H_2O#</mathjax>, <mathjax>#1.9 xx 10^-4#</mathjax> grams dissolves. What is the <mathjax>#K_(sp)#</mathjax> for silver chloride? </p>
<p>The answer should be <mathjax>#1.8 xx 10^-10#</mathjax>, but I'm not quite sure how, could someone please explain? Thanks </p></div>
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anor277
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Feb 24, 2018
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<div class="markdown"><p>WE address the equilibrium...<mathjax>#Ag^+ +Cl^(-) rightleftharpoonsAgCl(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For which <mathjax>#K_"sp"=[Ag^+][Cl^-]=??#</mathjax></p>
<p>We are given that <mathjax>#[AgCl]=((1.9xx10^-4*g)/(143.32*g*mol^-1))/(0.100*L)=1.33xx10^-5*mol*L^-1#</mathjax>...</p>
<p>But by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>...</p>
<p><mathjax>#[Ag^+]=[Cl^-]=1.33xx10^-5*mol*L^-1#</mathjax></p>
<p>And thus <mathjax>#K_"sp"=[1.33xx10^-5]^2=1.76xx10^-10#</mathjax>...the which is close enuff to your answer....</p>
<p>Happy?</p></div>
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<div class="markdown"><p><mathjax>#1.8 * 10^(-10)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong>solubility product constant</strong> for silver chloride, you need to determine the <strong>equilibrium concentrations</strong> of the dissolved ions.</p>
<p>As you know, silver chloride is considered <em>insoluble</em> in water, and so when you dissolve this salt, an equilibrium is established in aqueous solution between the undissolved solid and the dissolved ions.</p>
<blockquote>
<p><mathjax>#"AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, you are told that when you dissolve silver chloride in water, only <mathjax>#1.9 * 10^(-4)#</mathjax> <mathjax>#"g"#</mathjax> per <mathjax>#"100 mL"#</mathjax> of water will <strong>dissociate</strong> into ions to give a <strong>saturated solution</strong> of silver chloride. </p>
<p>To convert this mass to <em>moles</em>, use the <strong>molar mass</strong> of silver chloride.</p>
<blockquote>
<p><mathjax>#1.9 * 10^(-4) color(red)(cancel(color(black)("g"))) * "1 mole AgCl"/(143.3color(red)(cancel(color(black)("g")))) = 1.326 * 10^(-6) quad "moles AgCl"#</mathjax></p>
</blockquote>
<p>So, you know for a fact that at the temperature at which you're performing your experiment, only <mathjax>#1.326 * 10^(-6)#</mathjax> <strong>moles</strong> of silver chloride for every <mathjax>#"100 mL"#</mathjax> of water <strong>will dissociate</strong> to produce ions.</p>
<p>Since every mole of silver chloride <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of silver(I) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of chloride anions, you can say that, at equilibrium, your solution will contain <mathjax>#1.326 * 10^(-6)#</mathjax> <strong>moles</strong> of silver(I) cations and <mathjax>#1.326 * 10^(-6)#</mathjax> <strong>moles</strong> of chloride anons <strong>per</strong> <mathjax>#"100 mL"#</mathjax> of water. </p>
<p>To calculate the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the ions, scale up the volume to <mathjax>#"1 " = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of water, which, for all intents and purposes, will be <em>equal</em> to the volume of the solution. </p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * (1.326 * 10^(-6) quad "moles Ag"^(+))/(100color(red)(cancel(color(black)("mL solution")))) = 1.326 * 10^(-5) quad "moles Ag"^(+)#</mathjax></p>
</blockquote>
<p>The calculation is the same for the chloride anions, which means that, at equilibrium, your solution contains</p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = 1.326 * 10^(-5) quad "M"#</mathjax></p>
</blockquote>
<p>By definition, the solubility product constant for silver chloride is equal to</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the value <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_(sp) = (1.326 8 10^(-5)) * (1.326 * 10^(-5)) = 1.758 * 10^(-10)#</mathjax></p>
</blockquote>
<p>Based on the values you have for the volume of water, the answer should be rounded to one significant figure, but if you go by the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the mass of silver chloride, you can round the answer to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)(K_(sp) = 1.8 * 10^(-10))))#</mathjax></p>
</blockquote></div>
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</article> | What is the #K_(sp)# for silver chloride? |
When solid silver chloride (MM=143.3) is added to 100 mL of #H_2O#, #1.9 xx 10^-4# grams dissolves. What is the #K_(sp)# for silver chloride?
The answer should be #1.8 xx 10^-10#, but I'm not quite sure how, could someone please explain? Thanks
|
1,669 | acbead0c-6ddd-11ea-b5a3-ccda262736ce | https://socratic.org/questions/to-what-temperature-will-a-45-0-g-piece-of-glass-raise-if-it-absorbs-6875-joules | 323 ℃ | start physical_unit 9 9 temperature °c qc_end physical_unit 9 9 5 6 mass qc_end physical_unit 9 9 14 15 heat_energy qc_end physical_unit 9 9 24 27 specific_heat_capacity qc_end physical_unit 9 9 35 36 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] glass [IN] ℃"}] | [{"type":"physical unit","value":"323 ℃"}] | [{"type":"physical unit","value":"Mass [OF] glass [=] \\pu{45.0 g}"},{"type":"physical unit","value":"Absorbed heat [OF] glass [=] \\pu{6875 joules}"},{"type":"physical unit","value":"Specific heat capacity [OF] glass [=] \\pu{0.50 J/(g * ℃)}"},{"type":"physical unit","value":"Temperature1 [OF] glass [=] \\pu{18.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">To what temperature will a 45.0 g piece of glass rise if it absorbs 6875 joules of heat and its specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 18.0°C.</h1> | null | 323 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat absorbed <mathjax>#Q#</mathjax> is given by:</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with your daata we get:</p>
<p><mathjax>#6875=45*0.5*(T_f-18)#</mathjax></p>
<p>rearranging:</p>
<p><mathjax>#T_f=323^@C#</mathjax></p></div>
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<div class="markdown"><p>I got: <mathjax>#323^@C#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat absorbed <mathjax>#Q#</mathjax> is given by:</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with your daata we get:</p>
<p><mathjax>#6875=45*0.5*(T_f-18)#</mathjax></p>
<p>rearranging:</p>
<p><mathjax>#T_f=323^@C#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">To what temperature will a 45.0 g piece of glass rise if it absorbs 6875 joules of heat and its specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 18.0°C.</h1>
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<div class="markdown"><p>I got: <mathjax>#323^@C#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat absorbed <mathjax>#Q#</mathjax> is given by:</p>
<p><mathjax>#Q=mcDeltaT#</mathjax></p>
<p>with your daata we get:</p>
<p><mathjax>#6875=45*0.5*(T_f-18)#</mathjax></p>
<p>rearranging:</p>
<p><mathjax>#T_f=323^@C#</mathjax></p></div>
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</article> | To what temperature will a 45.0 g piece of glass rise if it absorbs 6875 joules of heat and its specific heat capacity is 0.50 J/g°C? The initial temperature of the glass is 18.0°C. | null |
1,670 | ace61a07-6ddd-11ea-a309-ccda262736ce | https://socratic.org/questions/how-many-li-atoms-are-in-5-8-moles-of-li | 3.49 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 2 2 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] Li atoms"}] | [{"type":"physical unit","value":"3.49 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] Li [=] \\pu{5.8 moles}"}] | <h1 class="questionTitle" itemprop="name">How many Li atoms are in 5.8 moles of Li?</h1> | null | 3.49 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#5.8xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number, " 6.022xx10^23*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many Li atoms are in 5.8 moles of Li?</h1>
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<div class="markdown"><p><mathjax>#5.8xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number, " 6.022xx10^23*mol^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#3.49*10^24#</mathjax> atoms Li</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Consider:<img alt="enter image source here" src="https://useruploads.socratic.org/bZS7TxCUTuWGviVQXz3W_Atoms%20Li.png"/> </p></div>
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</article> | How many Li atoms are in 5.8 moles of Li? | null |
1,671 | a8e9c2fa-6ddd-11ea-8c56-ccda262736ce | https://socratic.org/questions/5773a81c11ef6b20975dd6cf | 175 milliliters | start physical_unit 2 2 volume ml qc_end physical_unit 2 2 24 25 mass qc_end physical_unit 2 2 5 5 density qc_end end | [{"type":"physical unit","value":"Volume [OF] the solution [IN] milliliters"}] | [{"type":"physical unit","value":"175 milliliters"}] | [{"type":"physical unit","value":"Mass [OF] the solution [=] \\pu{70 grams}"},{"type":"physical unit","value":"Mass to volume ratio [OF] the solution [=] \\pu{40%}"}] | <h1 class="questionTitle" itemprop="name">If a solution has a #40%# mass (in grams)to volume (in milliliters) ratio; what will be the volume corresponding to a mass of 70 grams?</h1> | null | 175 milliliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#40%#</mathjax> mass/volume solution can be thought of as the ratio<br/>
<mathjax>#color(white)("XXX")(40 " mass units")/(100 " volume units")#</mathjax></p>
<p>If the <mathjax>#"mass units"#</mathjax> are <mathjax>#"grams"#</mathjax><br/>
and the <mathjax>#"volume units"#</mathjax> are <mathjax>#"milliliters"#</mathjax></p>
<p>then we are looking for a number, <mathjax>#x#</mathjax>, of liters such that<br/>
<mathjax>#color(white)("XXX")(40 " grams")/(100" milliliters")= (70" grams")/(x" milliliters")#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x" milliliters"=(70 cancel(" grams"))/(40 cancel(" grams"))xx100" milliliters"#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x " milliliters" = 175 " milliliters"#</mathjax></p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">Solute</a> volume <mathjax># = color(blue)(175 " milliliters")#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#40%#</mathjax> mass/volume solution can be thought of as the ratio<br/>
<mathjax>#color(white)("XXX")(40 " mass units")/(100 " volume units")#</mathjax></p>
<p>If the <mathjax>#"mass units"#</mathjax> are <mathjax>#"grams"#</mathjax><br/>
and the <mathjax>#"volume units"#</mathjax> are <mathjax>#"milliliters"#</mathjax></p>
<p>then we are looking for a number, <mathjax>#x#</mathjax>, of liters such that<br/>
<mathjax>#color(white)("XXX")(40 " grams")/(100" milliliters")= (70" grams")/(x" milliliters")#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x" milliliters"=(70 cancel(" grams"))/(40 cancel(" grams"))xx100" milliliters"#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x " milliliters" = 175 " milliliters"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If a solution has a #40%# mass (in grams)to volume (in milliliters) ratio; what will be the volume corresponding to a mass of 70 grams?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">Solute</a> volume <mathjax># = color(blue)(175 " milliliters")#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#40%#</mathjax> mass/volume solution can be thought of as the ratio<br/>
<mathjax>#color(white)("XXX")(40 " mass units")/(100 " volume units")#</mathjax></p>
<p>If the <mathjax>#"mass units"#</mathjax> are <mathjax>#"grams"#</mathjax><br/>
and the <mathjax>#"volume units"#</mathjax> are <mathjax>#"milliliters"#</mathjax></p>
<p>then we are looking for a number, <mathjax>#x#</mathjax>, of liters such that<br/>
<mathjax>#color(white)("XXX")(40 " grams")/(100" milliliters")= (70" grams")/(x" milliliters")#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x" milliliters"=(70 cancel(" grams"))/(40 cancel(" grams"))xx100" milliliters"#</mathjax></p>
<p><mathjax>#color(white)("XXX")rarr x " milliliters" = 175 " milliliters"#</mathjax></p></div>
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</article> | If a solution has a #40%# mass (in grams)to volume (in milliliters) ratio; what will be the volume corresponding to a mass of 70 grams? | null |
1,672 | a9fc07cb-6ddd-11ea-9b0a-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-mercury-l-nitrate | Hg(NO3)2 | start chemical_formula qc_end substance 6 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] mercury (l) nitrate [IN] default"}] | [{"type":"chemical equation","value":"Hg(NO3)2"}] | [{"type":"substance name","value":"Mercury (l) nitrate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for mercury (l) nitrate? </h1> | null | Hg(NO3)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mercurous ion is <mathjax>#Hg_2^(2+)#</mathjax>. Clearly the oxidation state of each mercury centre is <mathjax>#+I#</mathjax>. I have never used this <mathjax>#"mercurous nitrate"#</mathjax> reagent, but given that we combine a subvalent metal, and a potent oxidizing agent in the nitrate ion, I don't think it would be too stable. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For <mathjax>#"mercurous nitrate"#</mathjax>? <mathjax>#Hg_2(NO_3)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mercurous ion is <mathjax>#Hg_2^(2+)#</mathjax>. Clearly the oxidation state of each mercury centre is <mathjax>#+I#</mathjax>. I have never used this <mathjax>#"mercurous nitrate"#</mathjax> reagent, but given that we combine a subvalent metal, and a potent oxidizing agent in the nitrate ion, I don't think it would be too stable. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for mercury (l) nitrate? </h1>
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anor277
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<div class="markdown"><p>For <mathjax>#"mercurous nitrate"#</mathjax>? <mathjax>#Hg_2(NO_3)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mercurous ion is <mathjax>#Hg_2^(2+)#</mathjax>. Clearly the oxidation state of each mercury centre is <mathjax>#+I#</mathjax>. I have never used this <mathjax>#"mercurous nitrate"#</mathjax> reagent, but given that we combine a subvalent metal, and a potent oxidizing agent in the nitrate ion, I don't think it would be too stable. </p></div>
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</article> | What is the chemical formula for mercury (l) nitrate? | null |
1,673 | ab27413a-6ddd-11ea-b5c8-ccda262736ce | https://socratic.org/questions/the-ratio-ofthe-volume-of-milk-to-that-of-milk-coffee-in-a-cup-of-cappuccino-is- | 525.00 mL | start physical_unit 26 27 volume ml qc_end physical_unit 6 6 34 35 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] the coffee [IN] mL"}] | [{"type":"physical unit","value":"525.00 mL"}] | [{"type":"physical unit","value":"Ratio of the volume [OF] milk to milk coffee [=] \\pu{2:5}"},{"type":"physical unit","value":"Volume [OF] milk [=] \\pu{150 mL}"}] | <h1 class="questionTitle" itemprop="name">The ratio ofthe volume of milk to that of milk coffee in a cup of cappuccino is 2 : 5. How would you find the volume of the coffee in a cup of cappuccino if 150 mL of milk is used?</h1> | null | 525.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>(I'm assuming that the "milk coffee" is different from the "coffee" itself.)</p>
<p><mathjax>#"milk"/"milk coffee" = 2/5#</mathjax></p>
<p>Let <mathjax>#x#</mathjax> be the volume of the milk coffee.</p>
<blockquote>
<p><mathjax>#2/5=(150"mL")/x#</mathjax></p>
<p><mathjax>#2x=5*150#</mathjax></p>
<p><mathjax>#2x=750#</mathjax></p>
<p><mathjax>#x=750/2#</mathjax></p>
<p><mathjax>#x=375#</mathjax></p>
</blockquote>
<p>So, there is <mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> of milk coffee in the cup.</p>
<p>Add this to the volume of milk:</p>
<p><mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> milk coffee + <mathjax>#150#</mathjax> <mathjax>#"mL"#</mathjax> milk <mathjax>#= 525#</mathjax> <mathjax>#"mL"#</mathjax> coffee</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#525#</mathjax> <mathjax>#"mL"#</mathjax> coffee</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>(I'm assuming that the "milk coffee" is different from the "coffee" itself.)</p>
<p><mathjax>#"milk"/"milk coffee" = 2/5#</mathjax></p>
<p>Let <mathjax>#x#</mathjax> be the volume of the milk coffee.</p>
<blockquote>
<p><mathjax>#2/5=(150"mL")/x#</mathjax></p>
<p><mathjax>#2x=5*150#</mathjax></p>
<p><mathjax>#2x=750#</mathjax></p>
<p><mathjax>#x=750/2#</mathjax></p>
<p><mathjax>#x=375#</mathjax></p>
</blockquote>
<p>So, there is <mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> of milk coffee in the cup.</p>
<p>Add this to the volume of milk:</p>
<p><mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> milk coffee + <mathjax>#150#</mathjax> <mathjax>#"mL"#</mathjax> milk <mathjax>#= 525#</mathjax> <mathjax>#"mL"#</mathjax> coffee</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The ratio ofthe volume of milk to that of milk coffee in a cup of cappuccino is 2 : 5. How would you find the volume of the coffee in a cup of cappuccino if 150 mL of milk is used?</h1>
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<div class="markdown"><p><mathjax>#525#</mathjax> <mathjax>#"mL"#</mathjax> coffee</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>(I'm assuming that the "milk coffee" is different from the "coffee" itself.)</p>
<p><mathjax>#"milk"/"milk coffee" = 2/5#</mathjax></p>
<p>Let <mathjax>#x#</mathjax> be the volume of the milk coffee.</p>
<blockquote>
<p><mathjax>#2/5=(150"mL")/x#</mathjax></p>
<p><mathjax>#2x=5*150#</mathjax></p>
<p><mathjax>#2x=750#</mathjax></p>
<p><mathjax>#x=750/2#</mathjax></p>
<p><mathjax>#x=375#</mathjax></p>
</blockquote>
<p>So, there is <mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> of milk coffee in the cup.</p>
<p>Add this to the volume of milk:</p>
<p><mathjax>#375#</mathjax> <mathjax>#"mL"#</mathjax> milk coffee + <mathjax>#150#</mathjax> <mathjax>#"mL"#</mathjax> milk <mathjax>#= 525#</mathjax> <mathjax>#"mL"#</mathjax> coffee</p></div>
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</article> | The ratio ofthe volume of milk to that of milk coffee in a cup of cappuccino is 2 : 5. How would you find the volume of the coffee in a cup of cappuccino if 150 mL of milk is used? | null |
1,674 | aa0cb35f-6ddd-11ea-8ab8-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-empirical-formula-of-acetic-acid-ch3cooh | CH2O | start chemical_formula qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Formula [OF] CH3COOH [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O"}] | [{"type":"chemical equation","value":"CH3COOH"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the empirical formula of acetic acid (CH3COOH)? </h1> | null | CH2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_3"COOH"#</mathjax> can be written as <mathjax>#"C"_2"H"_4"O"_2"#</mathjax>. </p>
<p>An empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound.</p>
<p>The subscripts in the formula <mathjax>#"C"_2"H"_4"O"_2"#</mathjax> can be reduced to the simplest whole number ratio by dividing the subscripts in the molecular formula by <mathjax>#2#</mathjax>.</p>
<p>The empirical formula of acetic acid is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The empirical formula for acetic acid is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_3"COOH"#</mathjax> can be written as <mathjax>#"C"_2"H"_4"O"_2"#</mathjax>. </p>
<p>An empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound.</p>
<p>The subscripts in the formula <mathjax>#"C"_2"H"_4"O"_2"#</mathjax> can be reduced to the simplest whole number ratio by dividing the subscripts in the molecular formula by <mathjax>#2#</mathjax>.</p>
<p>The empirical formula of acetic acid is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you calculate the empirical formula of acetic acid (CH3COOH)? </h1>
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<div class="markdown"><p>The empirical formula for acetic acid is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_3"COOH"#</mathjax> can be written as <mathjax>#"C"_2"H"_4"O"_2"#</mathjax>. </p>
<p>An empirical formula represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in a compound.</p>
<p>The subscripts in the formula <mathjax>#"C"_2"H"_4"O"_2"#</mathjax> can be reduced to the simplest whole number ratio by dividing the subscripts in the molecular formula by <mathjax>#2#</mathjax>.</p>
<p>The empirical formula of acetic acid is <mathjax>#"CH"_2"O"#</mathjax>.</p></div>
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</article> | How would you calculate the empirical formula of acetic acid (CH3COOH)? | null |
1,675 | ab3ae918-6ddd-11ea-8dea-ccda262736ce | https://socratic.org/questions/how-many-moles-of-sodium-hypochlorite-should-be-added-to-1l-of-0-05m-hypochlorou | 2.13 moles | start physical_unit 4 5 mole mol qc_end physical_unit 21 21 10 11 volume qc_end physical_unit 15 16 13 14 molarity qc_end physical_unit 20 21 24 24 ph qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium hypochlorite solution [IN] moles"}] | [{"type":"physical unit","value":"2.13 moles"}] | [{"type":"physical unit","value":"Volume [OF] hypochlorous acid solution [=] \\pu{1 L}"},{"type":"physical unit","value":"Molarity [OF] hypochlorous acid solution [=] \\pu{0.05 M}"},{"type":"physical unit","value":"pH [OF] buffer solution [=] \\pu{9.15}"},{"type":"other","value":"Assume no changes of volume occurs when NaOCl salt is added."}] | <h1 class="questionTitle" itemprop="name">How many moles of sodium hypochlorite should be added to 1L of 0.05M hypochlorous acid to form a buffer solution with pH 9.15ssume no changes of volume occurs when NaOCl salt is added?</h1> | null | 2.13 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first - you need to know the <em>acid dissociation constant</em> for hypochlorous acid, <mathjax>#"HClO"#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#K_a = 3.0 * 10^(-8)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, look at the information provided to you and try to get a feel for what you're dealing with. </p>
<p>You know that <em>sodium hypochlorite</em>, <mathjax>#"NaClO"#</mathjax>, is the salt of hypochlorous acid's <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>, which is the hypochlorite anion, <mathjax>#"ClO"^(-)#</mathjax>. </p>
<p>Your solution will thus contain hypochlorous acid, a <strong>weak acid</strong>, and its conjugate base, which means that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a>. </p>
<p>Before moving forward, calculate the acid's <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#color(blue)(pK_a = - log(K_a))#</mathjax></p>
</blockquote>
<p>In your csae, </p>
<blockquote>
<p><mathjax>#pK_a = - log(3.0 * 10^(-8)) = 7.52#</mathjax></p>
</blockquote>
<p>Notice the the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the target solution is <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>, which means that it must contain <strong>more conjugate base</strong> than weak acid. </p>
<p>But how <em>much more</em> conjugate base? </p>
<p>To figure that out, you can use the <strong>Henderson - Hasselbalch equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>You know that your solution has </p>
<blockquote>
<p><mathjax>#["HClO"] = "0.05 M"#</mathjax></p>
</blockquote>
<p>Which means that you have everything you need to find <mathjax>#["ClO"^(-)]#</mathjax> by using the H - H equation. </p>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#9.15 = 7.52 + log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
<p><mathjax>#1.63 = log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["ClO"^(-)])/(["HClO"])) = 10^1.63#</mathjax></p>
<p><mathjax># (["ClO"^(-)])/(["HClO"]) = 42.66#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 42.66 * ["HClO"]#</mathjax></p>
<p><mathjax>#["ClO"^(-)] = 42.66 * "0.05 M" = "2.133 M"#</mathjax></p>
</blockquote>
<p>Since you are told the the volume of the solution <strong>does not change</strong> upon the addition of the salt, you can use its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume to determine how many <strong>moles</strong> of salt must be added. </p>
<p>Keep in mind, sodium hypochlorite dissociates in a <mathjax>#1:1#</mathjax> mole ratio with the hypochlorite anion, so </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = ["NaClO"]#</mathjax></p>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaClO) = "2.133 M" * "1 L" = "2.133 moles ClO"^(-)#</mathjax></p>
</blockquote>
<p>You <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, but I will leave the answer rounded to two sig figs</p>
<blockquote>
<p><mathjax>#n_(NaClO) = color(green)("2.1 moles")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.1 moles NaClO"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first - you need to know the <em>acid dissociation constant</em> for hypochlorous acid, <mathjax>#"HClO"#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#K_a = 3.0 * 10^(-8)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, look at the information provided to you and try to get a feel for what you're dealing with. </p>
<p>You know that <em>sodium hypochlorite</em>, <mathjax>#"NaClO"#</mathjax>, is the salt of hypochlorous acid's <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>, which is the hypochlorite anion, <mathjax>#"ClO"^(-)#</mathjax>. </p>
<p>Your solution will thus contain hypochlorous acid, a <strong>weak acid</strong>, and its conjugate base, which means that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a>. </p>
<p>Before moving forward, calculate the acid's <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#color(blue)(pK_a = - log(K_a))#</mathjax></p>
</blockquote>
<p>In your csae, </p>
<blockquote>
<p><mathjax>#pK_a = - log(3.0 * 10^(-8)) = 7.52#</mathjax></p>
</blockquote>
<p>Notice the the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the target solution is <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>, which means that it must contain <strong>more conjugate base</strong> than weak acid. </p>
<p>But how <em>much more</em> conjugate base? </p>
<p>To figure that out, you can use the <strong>Henderson - Hasselbalch equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>You know that your solution has </p>
<blockquote>
<p><mathjax>#["HClO"] = "0.05 M"#</mathjax></p>
</blockquote>
<p>Which means that you have everything you need to find <mathjax>#["ClO"^(-)]#</mathjax> by using the H - H equation. </p>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#9.15 = 7.52 + log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
<p><mathjax>#1.63 = log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["ClO"^(-)])/(["HClO"])) = 10^1.63#</mathjax></p>
<p><mathjax># (["ClO"^(-)])/(["HClO"]) = 42.66#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 42.66 * ["HClO"]#</mathjax></p>
<p><mathjax>#["ClO"^(-)] = 42.66 * "0.05 M" = "2.133 M"#</mathjax></p>
</blockquote>
<p>Since you are told the the volume of the solution <strong>does not change</strong> upon the addition of the salt, you can use its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume to determine how many <strong>moles</strong> of salt must be added. </p>
<p>Keep in mind, sodium hypochlorite dissociates in a <mathjax>#1:1#</mathjax> mole ratio with the hypochlorite anion, so </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = ["NaClO"]#</mathjax></p>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaClO) = "2.133 M" * "1 L" = "2.133 moles ClO"^(-)#</mathjax></p>
</blockquote>
<p>You <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, but I will leave the answer rounded to two sig figs</p>
<blockquote>
<p><mathjax>#n_(NaClO) = color(green)("2.1 moles")#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of sodium hypochlorite should be added to 1L of 0.05M hypochlorous acid to form a buffer solution with pH 9.15ssume no changes of volume occurs when NaOCl salt is added?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"2.1 moles NaClO"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first - you need to know the <em>acid dissociation constant</em> for hypochlorous acid, <mathjax>#"HClO"#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#K_a = 3.0 * 10^(-8)#</mathjax></p>
</blockquote>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, look at the information provided to you and try to get a feel for what you're dealing with. </p>
<p>You know that <em>sodium hypochlorite</em>, <mathjax>#"NaClO"#</mathjax>, is the salt of hypochlorous acid's <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a>, which is the hypochlorite anion, <mathjax>#"ClO"^(-)#</mathjax>. </p>
<p>Your solution will thus contain hypochlorous acid, a <strong>weak acid</strong>, and its conjugate base, which means that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a>. </p>
<p>Before moving forward, calculate the acid's <mathjax>#pK_a#</mathjax></p>
<blockquote>
<p><mathjax>#color(blue)(pK_a = - log(K_a))#</mathjax></p>
</blockquote>
<p>In your csae, </p>
<blockquote>
<p><mathjax>#pK_a = - log(3.0 * 10^(-8)) = 7.52#</mathjax></p>
</blockquote>
<p>Notice the the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the target solution is <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>, which means that it must contain <strong>more conjugate base</strong> than weak acid. </p>
<p>But how <em>much more</em> conjugate base? </p>
<p>To figure that out, you can use the <strong>Henderson - Hasselbalch equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#</mathjax></p>
</blockquote>
<p>You know that your solution has </p>
<blockquote>
<p><mathjax>#["HClO"] = "0.05 M"#</mathjax></p>
</blockquote>
<p>Which means that you have everything you need to find <mathjax>#["ClO"^(-)]#</mathjax> by using the H - H equation. </p>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#9.15 = 7.52 + log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
<p><mathjax>#1.63 = log( (["ClO"^(-)])/(["HClO"]))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#10^log( (["ClO"^(-)])/(["HClO"])) = 10^1.63#</mathjax></p>
<p><mathjax># (["ClO"^(-)])/(["HClO"]) = 42.66#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = 42.66 * ["HClO"]#</mathjax></p>
<p><mathjax>#["ClO"^(-)] = 42.66 * "0.05 M" = "2.133 M"#</mathjax></p>
</blockquote>
<p>Since you are told the the volume of the solution <strong>does not change</strong> upon the addition of the salt, you can use its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume to determine how many <strong>moles</strong> of salt must be added. </p>
<p>Keep in mind, sodium hypochlorite dissociates in a <mathjax>#1:1#</mathjax> mole ratio with the hypochlorite anion, so </p>
<blockquote>
<p><mathjax>#["ClO"^(-)] = ["NaClO"]#</mathjax></p>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(NaClO) = "2.133 M" * "1 L" = "2.133 moles ClO"^(-)#</mathjax></p>
</blockquote>
<p>You <em>should</em> round this off to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, but I will leave the answer rounded to two sig figs</p>
<blockquote>
<p><mathjax>#n_(NaClO) = color(green)("2.1 moles")#</mathjax></p>
</blockquote></div>
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</article> | How many moles of sodium hypochlorite should be added to 1L of 0.05M hypochlorous acid to form a buffer solution with pH 9.15ssume no changes of volume occurs when NaOCl salt is added? | null |
1,676 | aa1515da-6ddd-11ea-a21b-ccda262736ce | https://socratic.org/questions/5757afea11ef6b1b49bca702 | 5.42 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 10 11 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"5.42 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] calcium carbonate [=] \\pu{300 g}"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms in a #300*g# mass of calcium carbonate?</h1> | null | 5.42 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate, "(300*g)/(100.09*g*mol^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3*mol#</mathjax>.</p>
<p>Given the elemental makeup, in such a quantity there are <mathjax>#3#</mathjax> moles of calcium, <mathjax>#3#</mathjax> moles of carbon, and <mathjax>#9#</mathjax> moles of oxygen.</p>
<p>But what is a mole? It is simply a LARGE number that relates to a given mass of stuff. In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of stuff there are <mathjax>#6.022xx10^23#</mathjax> individual particles of that stuff.</p>
<p>So given 3 moles of calcium carbonate, there are 3 moles of calcium, 3 moles ofcarbon, and 9 moles of oxygen atoms.</p>
<p>So <mathjax>#"number of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9*molxxN_A#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*mol^-1xx9*mol" oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"oxygen atoms"??#</mathjax> </p>
<p>What is the mass of this quantity of oxygen atoms, and how do you know?</p>
<p>See <a href="https://socratic.org/questions/precious-metals-are-commonly-measured-in-troy-ounces-a-troy-ounce-is-equivalent-">here</a> for another examples of this process. </p></div>
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<div class="markdown"><p>We work out (i) the molar quantity of <mathjax>#CaCO_3#</mathjax>, and (ii) the constituent quantity of oxygen.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate, "(300*g)/(100.09*g*mol^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3*mol#</mathjax>.</p>
<p>Given the elemental makeup, in such a quantity there are <mathjax>#3#</mathjax> moles of calcium, <mathjax>#3#</mathjax> moles of carbon, and <mathjax>#9#</mathjax> moles of oxygen.</p>
<p>But what is a mole? It is simply a LARGE number that relates to a given mass of stuff. In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of stuff there are <mathjax>#6.022xx10^23#</mathjax> individual particles of that stuff.</p>
<p>So given 3 moles of calcium carbonate, there are 3 moles of calcium, 3 moles ofcarbon, and 9 moles of oxygen atoms.</p>
<p>So <mathjax>#"number of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9*molxxN_A#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*mol^-1xx9*mol" oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"oxygen atoms"??#</mathjax> </p>
<p>What is the mass of this quantity of oxygen atoms, and how do you know?</p>
<p>See <a href="https://socratic.org/questions/precious-metals-are-commonly-measured-in-troy-ounces-a-troy-ounce-is-equivalent-">here</a> for another examples of this process. </p></div>
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<h1 class="questionTitle" itemprop="name">How many oxygen atoms in a #300*g# mass of calcium carbonate?</h1>
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<div class="markdown"><p>We work out (i) the molar quantity of <mathjax>#CaCO_3#</mathjax>, and (ii) the constituent quantity of oxygen.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate, "(300*g)/(100.09*g*mol^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3*mol#</mathjax>.</p>
<p>Given the elemental makeup, in such a quantity there are <mathjax>#3#</mathjax> moles of calcium, <mathjax>#3#</mathjax> moles of carbon, and <mathjax>#9#</mathjax> moles of oxygen.</p>
<p>But what is a mole? It is simply a LARGE number that relates to a given mass of stuff. In <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of stuff there are <mathjax>#6.022xx10^23#</mathjax> individual particles of that stuff.</p>
<p>So given 3 moles of calcium carbonate, there are 3 moles of calcium, 3 moles ofcarbon, and 9 moles of oxygen atoms.</p>
<p>So <mathjax>#"number of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9*molxxN_A#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#6.022xx10^23*mol^-1xx9*mol" oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"oxygen atoms"??#</mathjax> </p>
<p>What is the mass of this quantity of oxygen atoms, and how do you know?</p>
<p>See <a href="https://socratic.org/questions/precious-metals-are-commonly-measured-in-troy-ounces-a-troy-ounce-is-equivalent-">here</a> for another examples of this process. </p></div>
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</article> | How many oxygen atoms in a #300*g# mass of calcium carbonate? | null |
1,677 | aaf129a6-6ddd-11ea-a082-ccda262736ce | https://socratic.org/questions/a-0-500-l-solution-of-6-m-hcl-has-to-be-made-how-much-12-m-hci-is-needed | 0.25 L | start physical_unit 3 3 volume l qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 7 7 14 15 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] HCl solution [IN] L"}] | [{"type":"physical unit","value":"0.25 L"}] | [{"type":"physical unit","value":"Volume1 [OF] HCl solution [=] \\pu{0.500 L}"},{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{6 M}"},{"type":"physical unit","value":"Molarity2 [OF] HCl solution [=] \\pu{12 M}"}] | <h1 class="questionTitle" itemprop="name">A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?</h1> | null | 0.25 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax> </p>
<p>And thus <mathjax>#"moles of solute"=6*mol*L^-1xx0.500*L=3*mol#</mathjax> required initially.</p>
<p>Since this comes from a volume of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#HCl#</mathjax>.....we get</p>
<p><mathjax>#(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L#</mathjax></p>
<p>The concentrated hydrochloric acid used in the lab (<mathjax>#"36%w/w"#</mathjax>) is <mathjax>#10.6*mol*L^-1#</mathjax> so this is another poor question. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use the relationship, <mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax> TWICE to get a volume of ..............<mathjax>#0.25*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax> </p>
<p>And thus <mathjax>#"moles of solute"=6*mol*L^-1xx0.500*L=3*mol#</mathjax> required initially.</p>
<p>Since this comes from a volume of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#HCl#</mathjax>.....we get</p>
<p><mathjax>#(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L#</mathjax></p>
<p>The concentrated hydrochloric acid used in the lab (<mathjax>#"36%w/w"#</mathjax>) is <mathjax>#10.6*mol*L^-1#</mathjax> so this is another poor question. </p></div>
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<h1 class="questionTitle" itemprop="name">A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?</h1>
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anor277
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<div class="markdown"><p>We use the relationship, <mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax> TWICE to get a volume of ..............<mathjax>#0.25*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax> </p>
<p>And thus <mathjax>#"moles of solute"=6*mol*L^-1xx0.500*L=3*mol#</mathjax> required initially.</p>
<p>Since this comes from a volume of <mathjax>#12*mol*L^-1#</mathjax> <mathjax>#HCl#</mathjax>.....we get</p>
<p><mathjax>#(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L#</mathjax></p>
<p>The concentrated hydrochloric acid used in the lab (<mathjax>#"36%w/w"#</mathjax>) is <mathjax>#10.6*mol*L^-1#</mathjax> so this is another poor question. </p></div>
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</article> | A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed? | null |
1,678 | abd0f3ba-6ddd-11ea-aee4-ccda262736ce | https://socratic.org/questions/the-density-of-a-gas-at-s-t-p-is-1-750-g-dm-3-3-how-do-you-calculate-the-relativ | 39.74 g/mol | start physical_unit 20 21 relative_molecular_weight g/mol qc_end c_other STP qc_end physical_unit 20 21 8 10 density qc_end end | [{"type":"physical unit","value":"Relative molecular mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"39.74 g/mol"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Density [OF] the gas [=] \\pu{1.750 g dm^(-3)}"}] | <h1 class="questionTitle" itemprop="name">The density of a gas at s.t.p is 1.750 g #dm^(-3)#3. How do you calculate the relative molecular mass of the gas?</h1> | null | 39.74 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (ρRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p><mathjax>#ρ = "1.750 g/dm"^3#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax><br/>
<mathjax>#P = "1 bar"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("1.750 g"·color(red)(cancel(color(black)("dm"^"-3"))) × "0.083 14" color(red)(cancel(color(black)("bar·dm"^3"·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "39.74 g/mol"#</mathjax></p>
<p>∴ <mathjax>#M_r = 39.74#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The relative molecular mass of the gas is 39.74.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (ρRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p><mathjax>#ρ = "1.750 g/dm"^3#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax><br/>
<mathjax>#P = "1 bar"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("1.750 g"·color(red)(cancel(color(black)("dm"^"-3"))) × "0.083 14" color(red)(cancel(color(black)("bar·dm"^3"·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "39.74 g/mol"#</mathjax></p>
<p>∴ <mathjax>#M_r = 39.74#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The density of a gas at s.t.p is 1.750 g #dm^(-3)#3. How do you calculate the relative molecular mass of the gas?</h1>
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<div class="markdown"><p>The relative molecular mass of the gas is 39.74.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>,</p>
<p>we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (m/V)(RT)/P#</mathjax></p>
</blockquote>
</blockquote>
<p>or</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)M = (ρRT)/Pcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p><mathjax>#ρ = "1.750 g/dm"^3#</mathjax><br/>
<mathjax>#R = "0.083 14 bar·dm"^3"·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "273.15 K"#</mathjax><br/>
<mathjax>#P = "1 bar"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("1.750 g"·color(red)(cancel(color(black)("dm"^"-3"))) × "0.083 14" color(red)(cancel(color(black)("bar·dm"^3"·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "39.74 g/mol"#</mathjax></p>
<p>∴ <mathjax>#M_r = 39.74#</mathjax></p></div>
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</article> | The density of a gas at s.t.p is 1.750 g #dm^(-3)#3. How do you calculate the relative molecular mass of the gas? | null |
1,679 | ab6fb6a2-6ddd-11ea-b280-ccda262736ce | https://socratic.org/questions/20g-of-nacl-were-added-to-an-8-180g-of-nacl-solution-what-is-the-mass-percent-of | 17.20% | start physical_unit 20 23 mass_percent none qc_end physical_unit 20 23 8 8 mass_percent qc_end physical_unit 3 3 0 1 mass qc_end end | [{"type":"physical unit","value":"Mass percent2 [OF] NaCl in the solution"}] | [{"type":"physical unit","value":"17.20%"}] | [{"type":"physical unit","value":"Mass percent1 [OF] NaCl in the solution [=] \\pu{8%}"},{"type":"physical unit","value":"Mass1 [OF] NaCl solution [=] \\pu{180g}"},{"type":"physical unit","value":"Mass2 [OF] NaCl [=] \\pu{20 g}"}] | <h1 class="questionTitle" itemprop="name">20g of NaCl were added to an 8% 180g of NaCl solution. What is the mass percent of NaCl in the solution now?</h1> | null | 17.20% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>It seems we should have a three-part strategy:</p>
<ol>
<li>Calculate the mass of <mathjax>#"NaCl"#</mathjax> in the original solution.</li>
<li>Calculate the new masses of <mathjax>#"NaCl"#</mathjax> and solution after adding <mathjax>#"NaCl"#</mathjax>.</li>
<li>Calculate the percent of <mathjax>#"NaCl"#</mathjax> in the new solution.</li>
</ol>
<blockquote></blockquote>
<p><strong>1. Mass of <mathjax>#bb"NaCl"#</mathjax> in original solution</strong></p>
<p>The formula for <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Mass %" = "mass of component"/"mass of mixture" × "100 %"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to</p>
<p><mathjax>#"Mass of component" = ("mass of mixture" × 100 %)/"mass %"#</mathjax></p>
<p>∴ <mathjax>#"Mass of NaCl" = ("180 g" × 8 color(red)(cancel(color(black)(%))))/(100 color(red)(cancel(color(black)(%)))) = "14.4 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. New masses</strong></p>
<p><mathjax>#"Mass of NaCl" = "14.4 g + 20 g" = "34.4 g"#</mathjax></p>
<p><mathjax>#"Mass of solution" = "180 g + 20 g" = "200 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. New percent composition</strong></p>
<p><mathjax>#"Mass %" = "mass of NaCl"/"mass of solution" × "100 %" = (34.4 color(red)(cancel(color(black)("g"))))/(200 color(red)(cancel(color(black)("g")))) × 100 % = 17 %#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new solution is <mathjax>#17 % ("w/w")color(white)(l) "NaCl"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>It seems we should have a three-part strategy:</p>
<ol>
<li>Calculate the mass of <mathjax>#"NaCl"#</mathjax> in the original solution.</li>
<li>Calculate the new masses of <mathjax>#"NaCl"#</mathjax> and solution after adding <mathjax>#"NaCl"#</mathjax>.</li>
<li>Calculate the percent of <mathjax>#"NaCl"#</mathjax> in the new solution.</li>
</ol>
<blockquote></blockquote>
<p><strong>1. Mass of <mathjax>#bb"NaCl"#</mathjax> in original solution</strong></p>
<p>The formula for <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Mass %" = "mass of component"/"mass of mixture" × "100 %"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to</p>
<p><mathjax>#"Mass of component" = ("mass of mixture" × 100 %)/"mass %"#</mathjax></p>
<p>∴ <mathjax>#"Mass of NaCl" = ("180 g" × 8 color(red)(cancel(color(black)(%))))/(100 color(red)(cancel(color(black)(%)))) = "14.4 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. New masses</strong></p>
<p><mathjax>#"Mass of NaCl" = "14.4 g + 20 g" = "34.4 g"#</mathjax></p>
<p><mathjax>#"Mass of solution" = "180 g + 20 g" = "200 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. New percent composition</strong></p>
<p><mathjax>#"Mass %" = "mass of NaCl"/"mass of solution" × "100 %" = (34.4 color(red)(cancel(color(black)("g"))))/(200 color(red)(cancel(color(black)("g")))) × 100 % = 17 %#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">20g of NaCl were added to an 8% 180g of NaCl solution. What is the mass percent of NaCl in the solution now?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-06-19T23:57:24" itemprop="dateCreated">
Jun 19, 2016
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<div class="markdown"><p>The new solution is <mathjax>#17 % ("w/w")color(white)(l) "NaCl"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>It seems we should have a three-part strategy:</p>
<ol>
<li>Calculate the mass of <mathjax>#"NaCl"#</mathjax> in the original solution.</li>
<li>Calculate the new masses of <mathjax>#"NaCl"#</mathjax> and solution after adding <mathjax>#"NaCl"#</mathjax>.</li>
<li>Calculate the percent of <mathjax>#"NaCl"#</mathjax> in the new solution.</li>
</ol>
<blockquote></blockquote>
<p><strong>1. Mass of <mathjax>#bb"NaCl"#</mathjax> in original solution</strong></p>
<p>The formula for <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Mass %" = "mass of component"/"mass of mixture" × "100 %"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to</p>
<p><mathjax>#"Mass of component" = ("mass of mixture" × 100 %)/"mass %"#</mathjax></p>
<p>∴ <mathjax>#"Mass of NaCl" = ("180 g" × 8 color(red)(cancel(color(black)(%))))/(100 color(red)(cancel(color(black)(%)))) = "14.4 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. New masses</strong></p>
<p><mathjax>#"Mass of NaCl" = "14.4 g + 20 g" = "34.4 g"#</mathjax></p>
<p><mathjax>#"Mass of solution" = "180 g + 20 g" = "200 g"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. New percent composition</strong></p>
<p><mathjax>#"Mass %" = "mass of NaCl"/"mass of solution" × "100 %" = (34.4 color(red)(cancel(color(black)("g"))))/(200 color(red)(cancel(color(black)("g")))) × 100 % = 17 %#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-20T00:04:03" itemprop="dateCreated">
Jun 20, 2016
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<div>
<div class="markdown"><p><mathjax>#17%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that adding the <mathjax>#"20 g"#</mathjax> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, will <strong>increase</strong> the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</em> of the solution, so right from the start you should expect the concentration of the target solution to be <strong>higher</strong> than <mathjax>#8%#</mathjax>.</p>
<p>A solution's <strong>percent concentration by mass</strong>, <mathjax>#"% m/m"#</mathjax>, essentially tells you how many grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case is sodium chloride, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>Initially, your solution has a mass of <mathjax>#"180 g"#</mathjax> and a percent concentration by mass equal to <mathjax>#8%#</mathjax>. This implies that the initial solution contains </p>
<blockquote>
<p><mathjax>#180 color(red)(cancel(color(black)("g solution"))) * overbrace("8 g NaCl"/(100color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 8% m/m NaCl")) = "14.4 g NaCl"#</mathjax></p>
</blockquote>
<p>The <strong>target solution</strong> will contain a total of </p>
<blockquote>
<p><mathjax>#m_"NaCl" = "14.4 g" + "20 g" = "34.4 g NaCl"#</mathjax></p>
</blockquote>
<p>The <strong>total mass of the solution</strong> will now be </p>
<blockquote>
<p><mathjax>#m_"solution" = "180 g" + "20 g" = "200 g"#</mathjax></p>
</blockquote>
<p>Since <mathjax>#"200 g"#</mathjax> of solution contain <mathjax>#"34.4 g"#</mathjax> of sodium chloride, it follows that <mathjax>#"100 g"#</mathjax> will contain</p>
<blockquote>
<p><mathjax>#100color(red)(cancel(color(black)("g solution"))) * "34.4 g NaCl"/(200color(red)(cancel(color(black)("g solution")))) = "17.2 g NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, the target solution's percent concentration by mass is</p>
<blockquote>
<p><mathjax>#"% m/m" = color(green)(|bar(ul(color(white)(a/a)color(black)("17% NaCl")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>As predicted, the percent concentration of the solution <strong>increased</strong> upon the addition of more solute. </p></div>
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</article> | 20g of NaCl were added to an 8% 180g of NaCl solution. What is the mass percent of NaCl in the solution now? | null |
1,680 | a8b1d340-6ddd-11ea-8c2d-ccda262736ce | https://socratic.org/questions/a-gas-at-155-kpa-and-25-c-has-an-initial-volume-of-1-00-l-the-pressure-of-the-ga-1 | 0.34 L | start physical_unit 18 19 volume l qc_end physical_unit 18 19 3 4 pressure qc_end physical_unit 18 19 6 7 temperature qc_end physical_unit 18 19 13 14 volume qc_end physical_unit 18 19 22 23 pressure qc_end physical_unit 18 19 30 31 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.34 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{155 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 °C}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{605 kPa}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{125 °C}"}] | <h1 class="questionTitle" itemprop="name">A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?</h1> | null | 0.34 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V_2=0.342*L#</mathjax>...the gas is compressed....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?</h1>
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<div class="markdown"><p><mathjax>#V_2=0.342*L#</mathjax>...the gas is compressed....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old <mathjax>#"combined gas law..."#</mathjax></p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And if we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>...the which product CLEARLY has the units of <mathjax>#"volume"#</mathjax>. And we plug in the values....</p>
<p><mathjax>#V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L#</mathjax></p>
<p>Pressure wins here over temperature....</p></div>
</div>
</div>
</div>
</div>
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</article> | A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume? | null |
1,681 | a8a7489e-6ddd-11ea-ae89-ccda262736ce | https://socratic.org/questions/how-many-moles-of-glucose-are-required-to-form-122-g-of-ethanol | 1.32 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] glucose [IN] moles"}] | [{"type":"physical unit","value":"1.32 moles"}] | [{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{122 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of glucose are required to form 122 g of ethanol? </h1> | null | 1.32 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molecular mass of ethanol, <mathjax>#46.07#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>We require, <mathjax>#(122.0*g)/(46.07*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.65#</mathjax> <mathjax>#mol#</mathjax> ethanol.</p>
<p>Given that 1 mol of glucose gives 2 moles of ethyl alcohol upon fermentation, we need to ferment at least <mathjax>#1.33#</mathjax> <mathjax>#mol#</mathjax> of sugar. What is this amount in grams? </p></div>
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<div class="markdown"><p><mathjax>#C_6H_12O_6(aq) rarr 2C_2H_5OH(aq) + 2CO_2(g)#</mathjax></p></div>
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<div class="markdown"><p>Molecular mass of ethanol, <mathjax>#46.07#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>We require, <mathjax>#(122.0*g)/(46.07*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.65#</mathjax> <mathjax>#mol#</mathjax> ethanol.</p>
<p>Given that 1 mol of glucose gives 2 moles of ethyl alcohol upon fermentation, we need to ferment at least <mathjax>#1.33#</mathjax> <mathjax>#mol#</mathjax> of sugar. What is this amount in grams? </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of glucose are required to form 122 g of ethanol? </h1>
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<div class="markdown"><p><mathjax>#C_6H_12O_6(aq) rarr 2C_2H_5OH(aq) + 2CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molecular mass of ethanol, <mathjax>#46.07#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>We require, <mathjax>#(122.0*g)/(46.07*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.65#</mathjax> <mathjax>#mol#</mathjax> ethanol.</p>
<p>Given that 1 mol of glucose gives 2 moles of ethyl alcohol upon fermentation, we need to ferment at least <mathjax>#1.33#</mathjax> <mathjax>#mol#</mathjax> of sugar. What is this amount in grams? </p></div>
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</article> | How many moles of glucose are required to form 122 g of ethanol? | null |
1,682 | ab023eaa-6ddd-11ea-81ae-ccda262736ce | https://socratic.org/questions/a-silver-block-initially-at-58-5-c-is-submerged-into-100-0-of-water-at-24-8-c-in | 77.56 g | start physical_unit 1 2 mass g qc_end physical_unit 1 2 5 6 temperature qc_end physical_unit 26 26 32 33 temperature qc_end physical_unit 13 13 15 16 temperature qc_end physical_unit 13 13 10 11 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] silver block [IN] g"}] | [{"type":"physical unit","value":"77.56 g"}] | [{"type":"physical unit","value":"Temperature1 [OF] silver block [=] \\pu{58.5 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] mixture [=] \\pu{26.2 ℃}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{24.8 ℃}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{100.0 mL}"}] | <h1 class="questionTitle" itemprop="name"> A silver block, initially at 58.5°C, is submerged into 100.0 of water at 24.8°C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.2 °C. What is the mass of the silver block?</h1> | null | 77.56 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Given the water's temperature, its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> at that particular temperature can be found through interpolation; i.e., ( <em>Density of water at various temperature are given in the attached table</em> ). <a href="http://www.engineeringcivil.com/relative-density-of-water.html" rel="nofollow" target="_blank">http://www.engineeringcivil.com/relative-density-of-water.html</a><br/>
<mathjax>#rho_(water) "@"24.8^oC=0.9971244" g/ml"#</mathjax></li>
<li>Through the density formula, mass of the water in the container can be computed as shown below:<br/>
<mathjax>#"mass"(m)=rhoxx"volume(V)"#</mathjax><br/>
<mathjax>#m=(0.9971244g)/cancel(ml)xx100.0cancel(ml)#</mathjax><br/>
<mathjax>#m=99.71g#</mathjax></li>
<li>Now, find the heat absorbed during the submersion of the silver. Knowing that the thermal equilibrium is at <mathjax>#26.2^oC#</mathjax>, the <mathjax>#Q_("gained")#</mathjax> by the water is computed as follows: <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_("gained")=m_wCp_wDeltaT#</mathjax> <br/>
<mathjax>#where#</mathjax>:<br/>
<mathjax>#DeltaT=T_f-Ti#</mathjax>=change in temperature(final and initial)<br/>
<mathjax>#Q_("gained")=mCp_w(T_f-T_i)#</mathjax><br/>
<mathjax>#Q_("gained")=(99.71g)((4.186J)/(g*^oC))(26.2-24.8)*^oC#</mathjax><br/>
<mathjax>#Q_("gained")=583.753J~~584J#</mathjax></li>
<li>Take note that <mathjax>#Q_("lost")=Q_("gained")#</mathjax>; so that, this relationship can be used to find the mass of the silver being cooled in the water. The Cp of the silver is given in the attached table. Rearrange the formula as needed. <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_(lost)=mCpDeltaT#</mathjax><br/>
Note: <mathjax>#Q_("lost")=-Q#</mathjax><br/>
<mathjax>#m=(-Q)/(CpDeltaT)#</mathjax>; where: <mathjax>#DeltaT=T_f-T_i#</mathjax><br/>
<mathjax>#m=(-Q)/((Cp)(T_f-T_i))#</mathjax><br/>
<mathjax>#m=(-584cancel(J))/(((0.233cancel(J))/(g*cancel(^oC)))(26.2-58.5)*cancel(^oC))#</mathjax><br/>
<mathjax>#m=(-584g)/(-7.53)#</mathjax><br/>
<mathjax>#m=77.6g#</mathjax></li>
</ol></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#=77.6g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Given the water's temperature, its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> at that particular temperature can be found through interpolation; i.e., ( <em>Density of water at various temperature are given in the attached table</em> ). <a href="http://www.engineeringcivil.com/relative-density-of-water.html" rel="nofollow" target="_blank">http://www.engineeringcivil.com/relative-density-of-water.html</a><br/>
<mathjax>#rho_(water) "@"24.8^oC=0.9971244" g/ml"#</mathjax></li>
<li>Through the density formula, mass of the water in the container can be computed as shown below:<br/>
<mathjax>#"mass"(m)=rhoxx"volume(V)"#</mathjax><br/>
<mathjax>#m=(0.9971244g)/cancel(ml)xx100.0cancel(ml)#</mathjax><br/>
<mathjax>#m=99.71g#</mathjax></li>
<li>Now, find the heat absorbed during the submersion of the silver. Knowing that the thermal equilibrium is at <mathjax>#26.2^oC#</mathjax>, the <mathjax>#Q_("gained")#</mathjax> by the water is computed as follows: <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_("gained")=m_wCp_wDeltaT#</mathjax> <br/>
<mathjax>#where#</mathjax>:<br/>
<mathjax>#DeltaT=T_f-Ti#</mathjax>=change in temperature(final and initial)<br/>
<mathjax>#Q_("gained")=mCp_w(T_f-T_i)#</mathjax><br/>
<mathjax>#Q_("gained")=(99.71g)((4.186J)/(g*^oC))(26.2-24.8)*^oC#</mathjax><br/>
<mathjax>#Q_("gained")=583.753J~~584J#</mathjax></li>
<li>Take note that <mathjax>#Q_("lost")=Q_("gained")#</mathjax>; so that, this relationship can be used to find the mass of the silver being cooled in the water. The Cp of the silver is given in the attached table. Rearrange the formula as needed. <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_(lost)=mCpDeltaT#</mathjax><br/>
Note: <mathjax>#Q_("lost")=-Q#</mathjax><br/>
<mathjax>#m=(-Q)/(CpDeltaT)#</mathjax>; where: <mathjax>#DeltaT=T_f-T_i#</mathjax><br/>
<mathjax>#m=(-Q)/((Cp)(T_f-T_i))#</mathjax><br/>
<mathjax>#m=(-584cancel(J))/(((0.233cancel(J))/(g*cancel(^oC)))(26.2-58.5)*cancel(^oC))#</mathjax><br/>
<mathjax>#m=(-584g)/(-7.53)#</mathjax><br/>
<mathjax>#m=77.6g#</mathjax></li>
</ol></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> A silver block, initially at 58.5°C, is submerged into 100.0 of water at 24.8°C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.2 °C. What is the mass of the silver block?</h1>
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<div class="markdown"><p><mathjax>#=77.6g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>Given the water's temperature, its <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> at that particular temperature can be found through interpolation; i.e., ( <em>Density of water at various temperature are given in the attached table</em> ). <a href="http://www.engineeringcivil.com/relative-density-of-water.html" rel="nofollow" target="_blank">http://www.engineeringcivil.com/relative-density-of-water.html</a><br/>
<mathjax>#rho_(water) "@"24.8^oC=0.9971244" g/ml"#</mathjax></li>
<li>Through the density formula, mass of the water in the container can be computed as shown below:<br/>
<mathjax>#"mass"(m)=rhoxx"volume(V)"#</mathjax><br/>
<mathjax>#m=(0.9971244g)/cancel(ml)xx100.0cancel(ml)#</mathjax><br/>
<mathjax>#m=99.71g#</mathjax></li>
<li>Now, find the heat absorbed during the submersion of the silver. Knowing that the thermal equilibrium is at <mathjax>#26.2^oC#</mathjax>, the <mathjax>#Q_("gained")#</mathjax> by the water is computed as follows: <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_("gained")=m_wCp_wDeltaT#</mathjax> <br/>
<mathjax>#where#</mathjax>:<br/>
<mathjax>#DeltaT=T_f-Ti#</mathjax>=change in temperature(final and initial)<br/>
<mathjax>#Q_("gained")=mCp_w(T_f-T_i)#</mathjax><br/>
<mathjax>#Q_("gained")=(99.71g)((4.186J)/(g*^oC))(26.2-24.8)*^oC#</mathjax><br/>
<mathjax>#Q_("gained")=583.753J~~584J#</mathjax></li>
<li>Take note that <mathjax>#Q_("lost")=Q_("gained")#</mathjax>; so that, this relationship can be used to find the mass of the silver being cooled in the water. The Cp of the silver is given in the attached table. Rearrange the formula as needed. <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html</a><br/>
<mathjax>#Q_(lost)=mCpDeltaT#</mathjax><br/>
Note: <mathjax>#Q_("lost")=-Q#</mathjax><br/>
<mathjax>#m=(-Q)/(CpDeltaT)#</mathjax>; where: <mathjax>#DeltaT=T_f-T_i#</mathjax><br/>
<mathjax>#m=(-Q)/((Cp)(T_f-T_i))#</mathjax><br/>
<mathjax>#m=(-584cancel(J))/(((0.233cancel(J))/(g*cancel(^oC)))(26.2-58.5)*cancel(^oC))#</mathjax><br/>
<mathjax>#m=(-584g)/(-7.53)#</mathjax><br/>
<mathjax>#m=77.6g#</mathjax></li>
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</article> | A silver block, initially at 58.5°C, is submerged into 100.0 of water at 24.8°C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.2 °C. What is the mass of the silver block? | null |
1,683 | ac907418-6ddd-11ea-88ce-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-maximum-number-of-grams-of-cacl2-that-could-be-produ | 111.00 grams | start physical_unit 10 10 mass g qc_end physical_unit 19 19 16 17 mass qc_end chemical_equation 24 32 qc_end end | [{"type":"physical unit","value":"Mass [OF] CaCl2 [IN] grams"}] | [{"type":"physical unit","value":"111.00 grams"}] | [{"type":"physical unit","value":"Mass [OF] Ca(OH)2 [=] \\pu{74 g}"},{"type":"chemical equation","value":"2 HCl + Ca(OH)2 -> 2 H2O + CaCl2"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the maximum number of grams of CaCl2 that could be produced from 74g of Ca(OH)2 in the following reaction:
2HCl + Ca(OH)2 --> 2H2O + CaCl2?</h1> | null | 111.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the number of moles of calcium hydroxide in the 74 grams :<br/>
<mathjax>#n=m/(M_r)=74/(40+32+2)=1mol#</mathjax></p>
<p>Therefore, providing sufficient hydrochloric acid is added, the maximum amount of calcium calcium chloride that can be formed is also 1 mole (in order to satisfy <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in the balanced chemical reaction).</p>
<p>But 1 mole of calcium chloride corresponds to a mass of </p>
<p><mathjax>#m=nxxM_r=1xx(40+2xx35,5)=111g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#111g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the number of moles of calcium hydroxide in the 74 grams :<br/>
<mathjax>#n=m/(M_r)=74/(40+32+2)=1mol#</mathjax></p>
<p>Therefore, providing sufficient hydrochloric acid is added, the maximum amount of calcium calcium chloride that can be formed is also 1 mole (in order to satisfy <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in the balanced chemical reaction).</p>
<p>But 1 mole of calcium chloride corresponds to a mass of </p>
<p><mathjax>#m=nxxM_r=1xx(40+2xx35,5)=111g#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How would you calculate the maximum number of grams of CaCl2 that could be produced from 74g of Ca(OH)2 in the following reaction:
2HCl + Ca(OH)2 --> 2H2O + CaCl2?</h1>
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Trevor Ryan.
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<div class="markdown"><p><mathjax>#111g#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First calculate the number of moles of calcium hydroxide in the 74 grams :<br/>
<mathjax>#n=m/(M_r)=74/(40+32+2)=1mol#</mathjax></p>
<p>Therefore, providing sufficient hydrochloric acid is added, the maximum amount of calcium calcium chloride that can be formed is also 1 mole (in order to satisfy <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in the balanced chemical reaction).</p>
<p>But 1 mole of calcium chloride corresponds to a mass of </p>
<p><mathjax>#m=nxxM_r=1xx(40+2xx35,5)=111g#</mathjax></p></div>
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</article> | How would you calculate the maximum number of grams of CaCl2 that could be produced from 74g of Ca(OH)2 in the following reaction:
2HCl + Ca(OH)2 --> 2H2O + CaCl2? | null |
1,684 | aa4e70f4-6ddd-11ea-855b-ccda262736ce | https://socratic.org/questions/569ed6747c0149480b33b917 | KClO3(s) ->[\Delta] KClO(s) + O2(g)^ | start chemical_equation qc_end substance 8 9 qc_end substance 12 13 qc_end substance 15 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] thermal decomposition"}] | [{"type":"chemical equation","value":"KClO3(s) ->[\\Delta] KClO(s) + O2(g)^"}] | [{"type":"substance name","value":"Potassium chlorate"},{"type":"substance name","value":"Potassium hypochlorite"},{"type":"substance name","value":"Dioxygen gas"}] | <h1 class="questionTitle" itemprop="name">How do we represent the thermal decomposition of potassium chlorate to give potassium hypochlorite and dioxygen gas?</h1> | null | KClO3(s) ->[\Delta] KClO(s) + O2(g)^ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With suitable catalysis, <mathjax>#KClO#</mathjax> can be further reduced to <mathjax>#KCl#</mathjax>, and more oxygen given off. It is thus a convenient source of oxygen gas that you don't have to store in a bottle. Potassium chlorate is present in many fireworks, and some very zippy experiments can be done in the lab. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#KClO_3(s) + Delta rarr KClO(s) + O_2(g)uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With suitable catalysis, <mathjax>#KClO#</mathjax> can be further reduced to <mathjax>#KCl#</mathjax>, and more oxygen given off. It is thus a convenient source of oxygen gas that you don't have to store in a bottle. Potassium chlorate is present in many fireworks, and some very zippy experiments can be done in the lab. </p></div>
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<h1 class="questionTitle" itemprop="name">How do we represent the thermal decomposition of potassium chlorate to give potassium hypochlorite and dioxygen gas?</h1>
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anor277
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<div class="markdown"><p><mathjax>#KClO_3(s) + Delta rarr KClO(s) + O_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With suitable catalysis, <mathjax>#KClO#</mathjax> can be further reduced to <mathjax>#KCl#</mathjax>, and more oxygen given off. It is thus a convenient source of oxygen gas that you don't have to store in a bottle. Potassium chlorate is present in many fireworks, and some very zippy experiments can be done in the lab. </p></div>
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</article> | How do we represent the thermal decomposition of potassium chlorate to give potassium hypochlorite and dioxygen gas? | null |
1,685 | ac4115fe-6ddd-11ea-9026-ccda262736ce | https://socratic.org/questions/how-many-grams-are-contained-in-2-00-mol-of-neon | 40.40 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] neon [IN] grams"}] | [{"type":"physical unit","value":"40.40 grams"}] | [{"type":"physical unit","value":"Mole [OF] neon [=] \\pu{2.00 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams are contained in 2.00 mol of neon?</h1> | null | 40.40 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"2.00 mol Ne"#</mathjax> has a mass of <mathjax>#"40.4 g"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams are contained in 2.00 mol of neon?</h1>
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<div class="markdown"><p><mathjax>#"2.00 mol Ne"#</mathjax> has a mass of <mathjax>#"40.4 g"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Multiply the given moles times the molar mass of neon, <mathjax>#"20.180 g/mol"#</mathjax> (periodic table).</p>
<p><mathjax>#2.00cancel"mol Ne"xx(20.180"g Ne")/(1cancel"mol Ne")="40.4 g Ne"#</mathjax></p></div>
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<div class="markdown"><p>2 moles <mathjax>#xx#</mathjax> 20.2 g/mol = ??</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume that:<br/>
n - number of moles<br/>
m - mass of substance<br/>
M - molar mass<br/>
<mathjax>#n = m -: M#</mathjax></p>
<p>2.00 moles (n) of Ne has been provided for you.</p>
<p>Now you have to find the molar mass (M) of Ne. If you look into your periodic table, the molar mass of Neon is 20.2 g/mol.</p>
<p>Final step:<br/>
flip-over the formula and it becomes <mathjax>#m = M xx n#</mathjax>.<br/>
The mass of Neon is: [m = 20.2 g/mol <mathjax>#xx#</mathjax> 2 moles]<br/>
= 40.4 grams.</p>
<p>Therefore, 40.4 grams are contained in 2 moles of Neon.</p></div>
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</article> | How many grams are contained in 2.00 mol of neon? | null |
1,686 | ab6b6012-6ddd-11ea-9c30-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-s-in-s-2o-3-2 | +2 | start physical_unit 6 6 oxidation_state none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] S"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"S2O3^2-"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of S in #S_2O_3^-2#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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<div class="markdown"><p><mathjax>#+2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of S in #S_2O_3^-2#?</h1>
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<div class="markdown"><p><mathjax>#+2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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<div class="markdown"><p>Formally we gots <mathjax>#stackrel(VI+)S#</mathjax> and <mathjax>#stackrel(-II)S#</mathjax>...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#S_"average oxidation state"=(VI+(-II))/2=+II#</mathjax>.</p>
<p><mathjax>#"Thiosulfate ion"#</mathjax> is an interesting customer in terms of sulfur oxidation state. If we look at sulfate, <mathjax>#SO_4^(2-)#</mathjax>, CLEARLY we got <mathjax>#S^(VI+)#</mathjax> and <mathjax>#4xxO^(-II)#</mathjax>..and as usual, the weighted sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a>, <mathjax>#6-8=-2#</mathjax>, i.e. the charge on the ion.</p>
<p>In <mathjax>#"thiosulfate"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide. And thus the central sulfur is <mathjax>#+VI#</mathjax>, and the terminal sulfur is <mathjax>#S^(-II)#</mathjax>, i.e. its oxidation state is PRECISELY the same as its Group 16 congener, oxygen. Of course, this is a formalism, but so is the whole concept of oxidation state and oxidation number.</p>
<p>Claro? </p></div>
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<div class="markdown"><p><mathjax>#+2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We got the thiosulfate ion <mathjax>#S_2O_3^(2-)#</mathjax>.</p>
<p>Since oxygen is more electronegative than sulfur, then oxygen will have its usual <mathjax>#-2#</mathjax> oxidation state. There are three oxygen atoms, and so the total charge of the oxygens is <mathjax>#-2*3=-6#</mathjax>.</p>
<p>Let <mathjax>#x#</mathjax> be the total charge of two sulfur atoms.</p>
<p>We got:</p>
<p><mathjax>#x-6=-2#</mathjax></p>
<p><mathjax>#x=-2+6#</mathjax></p>
<p><mathjax>#x=4#</mathjax></p>
<p>So, the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of the sulfur atoms is <mathjax>#+4#</mathjax>. If we consider both sulfur atoms to have the same oxidation number, then each sulfur will have an oxidation number of <mathjax>#(+4)/2=+2#</mathjax>.</p></div>
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</article> | What is the oxidation state of S in #S_2O_3^-2#? | null |
1,687 | ac639424-6ddd-11ea-991c-ccda262736ce | https://socratic.org/questions/58dfbb8b7c01497dff5e99d5 | 3 SO2 + K2Cr2O7 + H2SO4 -> K2SO4 + Cr2(SO4)3 + H2O | start chemical_equation qc_end chemical_equation 7 7 qc_end chemical_equation 11 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"3 SO2 + K2Cr2O7 + H2SO4 -> K2SO4 + Cr2(SO4)3 + H2O"}] | [{"type":"chemical equation","value":"SO2"},{"type":"chemical equation","value":"K2Cr2O7"}] | <h1 class="questionTitle" itemprop="name">What is the reaction between sulfur dioxide, #"SO"_2#, and potassium dichromate, #"K"_2 "Cr"_2"O"_7# ?</h1> | null | 3 SO2 + K2Cr2O7 + H2SO4 -> K2SO4 + Cr2(SO4)3 + H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#3"SO"_2#</mathjax> + <mathjax>#"K"_2"Cr"_2"O"_7#</mathjax> + <mathjax>#"H"_2"SO"_4#</mathjax> <mathjax>#->#</mathjax> <mathjax>#"K"_2"SO"_4#</mathjax> + <mathjax>#"Cr"_2("SO"_4)_3#</mathjax> + <mathjax>#"H"_2"O"#</mathjax></p></div>
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<div class="markdown"><p>In presence of <mathjax>#"H"_2"SO"_4#</mathjax>,potassium dicromate reacts with <mathjax>#"SO"_2#</mathjax> as following.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#3"SO"_2#</mathjax> + <mathjax>#"K"_2"Cr"_2"O"_7#</mathjax> + <mathjax>#"H"_2"SO"_4#</mathjax> <mathjax>#->#</mathjax> <mathjax>#"K"_2"SO"_4#</mathjax> + <mathjax>#"Cr"_2("SO"_4)_3#</mathjax> + <mathjax>#"H"_2"O"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the reaction between sulfur dioxide, #"SO"_2#, and potassium dichromate, #"K"_2 "Cr"_2"O"_7# ?</h1>
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Sunayan S
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<div class="markdown"><p>In presence of <mathjax>#"H"_2"SO"_4#</mathjax>,potassium dicromate reacts with <mathjax>#"SO"_2#</mathjax> as following.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#3"SO"_2#</mathjax> + <mathjax>#"K"_2"Cr"_2"O"_7#</mathjax> + <mathjax>#"H"_2"SO"_4#</mathjax> <mathjax>#->#</mathjax> <mathjax>#"K"_2"SO"_4#</mathjax> + <mathjax>#"Cr"_2("SO"_4)_3#</mathjax> + <mathjax>#"H"_2"O"#</mathjax></p></div>
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anor277
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<div class="markdown"><p>Well, we would assume that <mathjax>#"sulfur dioxide"#</mathjax> is oxidized to <mathjax>#"sulfur trioxide"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I think this is done industrially by <mathjax>#V_2O_5#</mathjax>.</p>
<p>But sulfur is oxidized to from <mathjax>#S(+IV)#</mathjax> to <mathjax>#S(+VI):#</mathjax></p>
<p><mathjax>#SO_2(g) +H_2O(l) rarr SO_3(g) +2H^(+) + 2e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And dichromate is reduced from <mathjax>#Cr(+VI)#</mathjax> to <mathjax>#Cr(+III):#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+) +6e^(-)rarr 2Cr^(3+) +7H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#3xx(i) + (ii)#</mathjax> to eliminate the electrons:</p>
<p><mathjax>#Cr_2O_7^(2-) +8H^(+) +3SO_2(g) rarr 2Cr^(3+) +4H_2O(l)+3SO_3(g)#</mathjax></p>
<p>Are mass and charge balanced? This is your problem not mine. </p></div>
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</article> | What is the reaction between sulfur dioxide, #"SO"_2#, and potassium dichromate, #"K"_2 "Cr"_2"O"_7# ? | null |
1,688 | a9260be4-6ddd-11ea-895b-ccda262736ce | https://socratic.org/questions/how-do-you-balance-agno-3-mgcl-2-agcl-mg-no-3-2 | 2 AgNO3 + MgCl2 -> 2 AgCl + Mg(NO3)2 | 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 AgNO3 + MgCl2 -> 2 AgCl + Mg(NO3)2"}] | [{"type":"chemical equation","value":"AgNO3 + MgCl2 -> AgCl + Mg(NO3)2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2#?</h1> | null | 2 AgNO3 + MgCl2 -> 2 AgCl + Mg(NO3)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>On the RHS, we have two nitrate ions. Therefore, we need <mathjax>#2#</mathjax> mols of <mathjax>#"AgNO"_3#</mathjax> to balance them. We now have two mols of <mathjax>#"Ag"^+#</mathjax> on the LHS but only one mol on the RHS. Thus we also need <mathjax>#2#</mathjax> mols <mathjax>#"AgC"l#</mathjax>. </p></div>
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<div class="markdown"><p><mathjax>#2"AgNO"_3 +"MgC"l_2 → 2"AgC"l + "Mg(NO"_3")"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>On the RHS, we have two nitrate ions. Therefore, we need <mathjax>#2#</mathjax> mols of <mathjax>#"AgNO"_3#</mathjax> to balance them. We now have two mols of <mathjax>#"Ag"^+#</mathjax> on the LHS but only one mol on the RHS. Thus we also need <mathjax>#2#</mathjax> mols <mathjax>#"AgC"l#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2#?</h1>
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Monzur R.
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<div class="markdown"><p><mathjax>#2"AgNO"_3 +"MgC"l_2 → 2"AgC"l + "Mg(NO"_3")"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>On the RHS, we have two nitrate ions. Therefore, we need <mathjax>#2#</mathjax> mols of <mathjax>#"AgNO"_3#</mathjax> to balance them. We now have two mols of <mathjax>#"Ag"^+#</mathjax> on the LHS but only one mol on the RHS. Thus we also need <mathjax>#2#</mathjax> mols <mathjax>#"AgC"l#</mathjax>. </p></div>
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</article> | How do you balance #AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2#? | null |
1,689 | aa268910-6ddd-11ea-bfc7-ccda262736ce | https://socratic.org/questions/calculate-the-standard-enthalpy-change-in-kj-for-the-production-of-12-2-g-of-h2o | -380.59 kJ | start physical_unit 17 17 standard_enthalpy_change kj qc_end physical_unit 14 14 11 12 mass qc_end chemical_equation 21 21 qc_end chemical_equation 22 22 qc_end chemical_equation 27 27 qc_end end | [{"type":"physical unit","value":"Standard enthalpy change [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"-380.59 kJ"}] | [{"type":"physical unit","value":"Mass [OF] H2O(l) [=] \\pu{12.2 g}"},{"type":"chemical equation","value":"H2S(g)"},{"type":"chemical equation","value":"O2(g)"},{"type":"chemical equation","value":"SO2(g)"}] | <h1 class="questionTitle" itemprop="name">Calculate the standard enthalpy change in kJ for the production of 12.2 g of H2O(l) from a reaction with the reactants: H2S(g), O2(g) and the products: H2O(l), SO2(g)?</h1> | null | -380.59 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is write a balanced chemical equation for this reaction</p>
<blockquote>
<p><mathjax>#2"H"_2"S"_text((g]) + 3"O"_text(2(g]) -> 2"SO"_text(2(g]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>SInce the <em>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"^@#</mathjax>, for this reaction was not provided, you're going to have to calculate it using <em>standard enthalpies of formation</em></p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"For H"_2"S: " DeltaH_text(f)^@ = -"20.63 kJ/mol"#</mathjax></p>
<p><mathjax>#"For O"_2 : " "DeltaH_text(f)^@ = "0 kJ/mol"#</mathjax></p>
<p><mathjax>#"For SO"_2: " "DeltaH_text(f)^@ = - "296.84 kJ/mol"#</mathjax></p>
<p><mathjax>#"For H"_2"O"_text((l]): DeltaH_text(f)^@ = -"285.8 kJ/mol"#</mathjax></p>
</blockquote>
<p>The standard enthalpy change of reaction can be calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx Delta_"f products") - sum(m xx DeltaH_"reactants"))" "#</mathjax> , where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the number of moles of each product and reactant, respectively</p>
<p>In your case, you would have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (-296.84"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-285.8"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-20.63"kJ"/color(red)(cancel(color(black)("mole")))) + "3 moles" * 0]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"1165.28 kJ" - (-"41.26 kJ")#</mathjax></p>
<p><mathjax>#DeltaH_"rxn" = -"1124.02 kJ"#</mathjax></p>
</blockquote>
<p>Now, this is the enthalpy change of reaction when <strong>2 moles</strong> of hydrogen sulfide react with <strong>3 moles</strong> of oxygen gas to produce <strong>2 moles</strong> of sulfur dioxide and <strong>2 moles</strong> of water. </p>
<p>In your case, the mass of water produced by the reaction will help you determine how many moles of each species were actually involved in the reaction. </p>
<p>Use water's molar mass to get</p>
<blockquote>
<p><mathjax>#12.2color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.6772 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>If the reaction will release <mathjax>#"1124.02 kJ"#</mathjax> of heat when <strong>2 moles</strong> of water are produced, it follows that you will get</p>
<blockquote>
<p><mathjax>#0.6772color(red)(cancel(color(black)("moles H"_2"O"))) * (-"1124.02 kJ")/(2color(red)(cancel(color(black)("moles H"_2"O")))) = color(green)(-"381 kJ")#</mathjax></p>
</blockquote>
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<div class="markdown"><p><mathjax>#DeltaH_"rxn" = -"381 kJ"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is write a balanced chemical equation for this reaction</p>
<blockquote>
<p><mathjax>#2"H"_2"S"_text((g]) + 3"O"_text(2(g]) -> 2"SO"_text(2(g]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>SInce the <em>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"^@#</mathjax>, for this reaction was not provided, you're going to have to calculate it using <em>standard enthalpies of formation</em></p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"For H"_2"S: " DeltaH_text(f)^@ = -"20.63 kJ/mol"#</mathjax></p>
<p><mathjax>#"For O"_2 : " "DeltaH_text(f)^@ = "0 kJ/mol"#</mathjax></p>
<p><mathjax>#"For SO"_2: " "DeltaH_text(f)^@ = - "296.84 kJ/mol"#</mathjax></p>
<p><mathjax>#"For H"_2"O"_text((l]): DeltaH_text(f)^@ = -"285.8 kJ/mol"#</mathjax></p>
</blockquote>
<p>The standard enthalpy change of reaction can be calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx Delta_"f products") - sum(m xx DeltaH_"reactants"))" "#</mathjax> , where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the number of moles of each product and reactant, respectively</p>
<p>In your case, you would have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (-296.84"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-285.8"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-20.63"kJ"/color(red)(cancel(color(black)("mole")))) + "3 moles" * 0]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"1165.28 kJ" - (-"41.26 kJ")#</mathjax></p>
<p><mathjax>#DeltaH_"rxn" = -"1124.02 kJ"#</mathjax></p>
</blockquote>
<p>Now, this is the enthalpy change of reaction when <strong>2 moles</strong> of hydrogen sulfide react with <strong>3 moles</strong> of oxygen gas to produce <strong>2 moles</strong> of sulfur dioxide and <strong>2 moles</strong> of water. </p>
<p>In your case, the mass of water produced by the reaction will help you determine how many moles of each species were actually involved in the reaction. </p>
<p>Use water's molar mass to get</p>
<blockquote>
<p><mathjax>#12.2color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.6772 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>If the reaction will release <mathjax>#"1124.02 kJ"#</mathjax> of heat when <strong>2 moles</strong> of water are produced, it follows that you will get</p>
<blockquote>
<p><mathjax>#0.6772color(red)(cancel(color(black)("moles H"_2"O"))) * (-"1124.02 kJ")/(2color(red)(cancel(color(black)("moles H"_2"O")))) = color(green)(-"381 kJ")#</mathjax></p>
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<h1 class="questionTitle" itemprop="name">Calculate the standard enthalpy change in kJ for the production of 12.2 g of H2O(l) from a reaction with the reactants: H2S(g), O2(g) and the products: H2O(l), SO2(g)?</h1>
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<div class="markdown"><p><mathjax>#DeltaH_"rxn" = -"381 kJ"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The first thing to do here is write a balanced chemical equation for this reaction</p>
<blockquote>
<p><mathjax>#2"H"_2"S"_text((g]) + 3"O"_text(2(g]) -> 2"SO"_text(2(g]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>SInce the <em>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"^@#</mathjax>, for this reaction was not provided, you're going to have to calculate it using <em>standard enthalpies of formation</em></p>
<p><a href="https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29</a></p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"For H"_2"S: " DeltaH_text(f)^@ = -"20.63 kJ/mol"#</mathjax></p>
<p><mathjax>#"For O"_2 : " "DeltaH_text(f)^@ = "0 kJ/mol"#</mathjax></p>
<p><mathjax>#"For SO"_2: " "DeltaH_text(f)^@ = - "296.84 kJ/mol"#</mathjax></p>
<p><mathjax>#"For H"_2"O"_text((l]): DeltaH_text(f)^@ = -"285.8 kJ/mol"#</mathjax></p>
</blockquote>
<p>The standard enthalpy change of reaction can be calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(DeltaH_"rxn"^@ = sum(n xx Delta_"f products") - sum(m xx DeltaH_"reactants"))" "#</mathjax> , where</p>
</blockquote>
<p><mathjax>#n#</mathjax>, <mathjax>#m#</mathjax> - the number of moles of each product and reactant, respectively</p>
<p>In your case, you would have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (-296.84"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-285.8"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-20.63"kJ"/color(red)(cancel(color(black)("mole")))) + "3 moles" * 0]#</mathjax></p>
<p><mathjax>#DeltaH_"rxn"^@ = -"1165.28 kJ" - (-"41.26 kJ")#</mathjax></p>
<p><mathjax>#DeltaH_"rxn" = -"1124.02 kJ"#</mathjax></p>
</blockquote>
<p>Now, this is the enthalpy change of reaction when <strong>2 moles</strong> of hydrogen sulfide react with <strong>3 moles</strong> of oxygen gas to produce <strong>2 moles</strong> of sulfur dioxide and <strong>2 moles</strong> of water. </p>
<p>In your case, the mass of water produced by the reaction will help you determine how many moles of each species were actually involved in the reaction. </p>
<p>Use water's molar mass to get</p>
<blockquote>
<p><mathjax>#12.2color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.6772 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>If the reaction will release <mathjax>#"1124.02 kJ"#</mathjax> of heat when <strong>2 moles</strong> of water are produced, it follows that you will get</p>
<blockquote>
<p><mathjax>#0.6772color(red)(cancel(color(black)("moles H"_2"O"))) * (-"1124.02 kJ")/(2color(red)(cancel(color(black)("moles H"_2"O")))) = color(green)(-"381 kJ")#</mathjax></p>
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</article> | Calculate the standard enthalpy change in kJ for the production of 12.2 g of H2O(l) from a reaction with the reactants: H2S(g), O2(g) and the products: H2O(l), SO2(g)? | null |
1,690 | a86cf7b6-6ddd-11ea-9a96-ccda262736ce | https://socratic.org/questions/581165ac7c014952676d91d4 | 4.91 L | start physical_unit 23 24 volume l qc_end physical_unit 23 24 6 7 volume qc_end c_other STP qc_end c_other SATP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"4.91 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{4.5 L}"},{"type":"other","value":"STP"},{"type":"other","value":"SATP"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water?</h1> | null | 4.91 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH~=13#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water?</h1>
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<div class="markdown"><p><mathjax>#pH~=13#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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</article> | What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water? | null |
1,691 | ab01a1ef-6ddd-11ea-b15a-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-equation-for-the-combination-of-magnesium-and-oxygen | 2 Mg(s) + O2(g) -> 2 MgO | start chemical_equation qc_end substance 11 11 qc_end substance 13 13 qc_end chemical_equation 18 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced equation"}] | [{"type":"chemical equation","value":"2 Mg(s) + O2(g) -> 2 MgO"}] | [{"type":"substance name","value":"Magnesium"},{"type":"substance name","value":"Oxygen"},{"type":"chemical equation","value":"MgO"}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced equation for the combination of magnesium and oxygen to form magnesium oxide, #MgO#?</h1> | null | 2 Mg(s) + O2(g) -> 2 MgO | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we need to identify the type of bond and reaction. This is an ionic bond so atoms are giving and accepting electrons. This is a synthesis reaction. </p>
<p>Magnesium has a charge of <mathjax>#2+#</mathjax>, while oxygen has a charge of <mathjax>#2-#</mathjax>. The burning of magnesium metal reacts with oxygen found in the air (oxygen gas) to form magnesium oxide. </p>
<blockquote>
<p><mathjax>#2 Mg (s) + O_"2" (g) -> 2 MgO#</mathjax></p>
</blockquote>
<p>Oxygen is a diatomic molecule and is almost always reacting with <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> as a gas, <mathjax>#O_"2"#</mathjax>.</p>
<p>Here is a short video showing the reaction of magnesium and oxygen. The white ash and smoke formed in the reaction is the magnesium oxide.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>Hope this helps :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2 Mg (s) + O_"2" (g) -> 2 MgO#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we need to identify the type of bond and reaction. This is an ionic bond so atoms are giving and accepting electrons. This is a synthesis reaction. </p>
<p>Magnesium has a charge of <mathjax>#2+#</mathjax>, while oxygen has a charge of <mathjax>#2-#</mathjax>. The burning of magnesium metal reacts with oxygen found in the air (oxygen gas) to form magnesium oxide. </p>
<blockquote>
<p><mathjax>#2 Mg (s) + O_"2" (g) -> 2 MgO#</mathjax></p>
</blockquote>
<p>Oxygen is a diatomic molecule and is almost always reacting with <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> as a gas, <mathjax>#O_"2"#</mathjax>.</p>
<p>Here is a short video showing the reaction of magnesium and oxygen. The white ash and smoke formed in the reaction is the magnesium oxide.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>Hope this helps :)</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you write a balanced equation for the combination of magnesium and oxygen to form magnesium oxide, #MgO#?</h1>
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Nov 25, 2016
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<div class="markdown"><p><mathjax>#2 Mg (s) + O_"2" (g) -> 2 MgO#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we need to identify the type of bond and reaction. This is an ionic bond so atoms are giving and accepting electrons. This is a synthesis reaction. </p>
<p>Magnesium has a charge of <mathjax>#2+#</mathjax>, while oxygen has a charge of <mathjax>#2-#</mathjax>. The burning of magnesium metal reacts with oxygen found in the air (oxygen gas) to form magnesium oxide. </p>
<blockquote>
<p><mathjax>#2 Mg (s) + O_"2" (g) -> 2 MgO#</mathjax></p>
</blockquote>
<p>Oxygen is a diatomic molecule and is almost always reacting with <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> as a gas, <mathjax>#O_"2"#</mathjax>.</p>
<p>Here is a short video showing the reaction of magnesium and oxygen. The white ash and smoke formed in the reaction is the magnesium oxide.</p>
<p>
<iframe src="https://www.youtube.com/embed/X_NyWhG_uWc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>Hope this helps :)</p></div>
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</article> | How do you write a balanced equation for the combination of magnesium and oxygen to form magnesium oxide, #MgO#? | null |
1,692 | a8fe2c91-6ddd-11ea-9fcc-ccda262736ce | https://socratic.org/questions/how-do-you-convert-4-00-moles-of-n-2-to-liters | 100 liters | start physical_unit 7 7 volume l qc_end physical_unit 7 7 4 5 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] N2 [IN] liters"}] | [{"type":"chemical equation","value":"100 liters"}] | [{"type":"physical unit","value":"Mole [OF] N2 [=] \\pu{4.00 moles}"}] | <h1 class="questionTitle" itemprop="name">How do you convert 4.00 moles of #N_2# to liters?</h1> | null | 100 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is multiply the number of moles by the molar volume:</p>
<p><mathjax>#4*molxx25.4#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#mol^-1#</mathjax> or <mathjax>#4*molxx25.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#mol^-1#</mathjax> (<mathjax>#1L=1*dm^3#</mathjax>) <mathjax>#=#</mathjax> <mathjax>#??" litres"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>First we assume a temperature and pressure; the molar volume at <mathjax>#RT#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> is <mathjax>#25.4#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#mol^-1#</mathjax>. We get a volume of slightly over <mathjax>#100#</mathjax> <mathjax>#L#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is multiply the number of moles by the molar volume:</p>
<p><mathjax>#4*molxx25.4#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#mol^-1#</mathjax> or <mathjax>#4*molxx25.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#mol^-1#</mathjax> (<mathjax>#1L=1*dm^3#</mathjax>) <mathjax>#=#</mathjax> <mathjax>#??" litres"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you convert 4.00 moles of #N_2# to liters?</h1>
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<div class="markdown"><p>First we assume a temperature and pressure; the molar volume at <mathjax>#RT#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> is <mathjax>#25.4#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#mol^-1#</mathjax>. We get a volume of slightly over <mathjax>#100#</mathjax> <mathjax>#L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we have to do is multiply the number of moles by the molar volume:</p>
<p><mathjax>#4*molxx25.4#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#mol^-1#</mathjax> or <mathjax>#4*molxx25.4#</mathjax> <mathjax>#L#</mathjax> <mathjax>#mol^-1#</mathjax> (<mathjax>#1L=1*dm^3#</mathjax>) <mathjax>#=#</mathjax> <mathjax>#??" litres"#</mathjax></p></div>
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</article> | How do you convert 4.00 moles of #N_2# to liters? | null |
1,693 | a94b601e-6ddd-11ea-a06d-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-the-compound-sulfur-ii-chloride | SCl2 | start chemical_formula qc_end substance 7 9 qc_end end | [{"type":"other","value":"Chemical Formula [OF] sulfur (II) chloride [IN] default"}] | [{"type":"chemical equation","value":"SCl2"}] | [{"type":"substance name","value":"Sulfur (II) chloride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for the compound sulfur (II) chloride? </h1> | null | SCl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I have never used this, but it sounds pretty nasty. It is cherry-red, volatile liquid..........possibly there are unopened bottles in your inorganic store that are older than you (or me!). </p></div>
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<div class="markdown"><p>It is simply <mathjax>#SCl_2#</mathjax>........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I have never used this, but it sounds pretty nasty. It is cherry-red, volatile liquid..........possibly there are unopened bottles in your inorganic store that are older than you (or me!). </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the formula for the compound sulfur (II) chloride? </h1>
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<div class="markdown"><p>It is simply <mathjax>#SCl_2#</mathjax>........</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I have never used this, but it sounds pretty nasty. It is cherry-red, volatile liquid..........possibly there are unopened bottles in your inorganic store that are older than you (or me!). </p></div>
</div>
</div>
</div>
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</article> | What is the formula for the compound sulfur (II) chloride? | null |
1,694 | ab023ea8-6ddd-11ea-8ec2-ccda262736ce | https://socratic.org/questions/if-20-00-moles-of-calcium-reacted-how-many-grams-of-calcium-oxide-are-produced | 1121.54 grams | start physical_unit 10 11 mass g qc_end physical_unit 4 4 1 2 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] calcium oxide [IN] grams"}] | [{"type":"physical unit","value":"1121.54 grams"}] | [{"type":"physical unit","value":"Mole [OF] calcium [=] \\pu{20.00 moles}"}] | <h1 class="questionTitle" itemprop="name">If 20.00 moles of calcium reacted, how many grams of calcium oxide are produced?</h1> | null | 1121.54 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to create the skeletal equation for the reaction. Since calcium oxide (CaO) is produced, oxygen (<mathjax>#O_2)#</mathjax> is required. </p>
<p>Skeletal equation: <mathjax>#Ca + O_2 -> CaO#</mathjax></p>
<p><strong>Balance the equation</strong> based on the law of conservation of mass. Since there are two oxygen atoms on the left, a <mathjax>#2#</mathjax> coefficient is required: <br/>
<mathjax>#" "Ca + O_2 -> 2CaO#</mathjax></p>
<p>Now you need to have <mathjax>#2#</mathjax> calcium molecules on the left to balance:</p>
<p><mathjax>#" "2Ca + O_2 -> 2CaO#</mathjax></p>
<p>This equation is balanced. The equation says that for every two moles of calcium, there are two moles of calcium oxide produced.</p>
<p>The molar mass of CaO <mathjax>#= 40.078 + 15.999 = (56.077" g CaO")/(1 "mol CaO") #</mathjax></p>
<p><mathjax>#20.00 " mol Ca" xx (2" mol CaO")/(2" mol Ca") xx (56.077" g CaO")/(1 " mol CaO") #</mathjax></p>
<p><mathjax>#= 1121.54 " g CaO"#</mathjax></p>
<p>If you take into consideration <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (SF), the smallest number of SF is 4. This would mean your answer should be <mathjax>#1122 " g CaO"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#1122" g CaO"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to create the skeletal equation for the reaction. Since calcium oxide (CaO) is produced, oxygen (<mathjax>#O_2)#</mathjax> is required. </p>
<p>Skeletal equation: <mathjax>#Ca + O_2 -> CaO#</mathjax></p>
<p><strong>Balance the equation</strong> based on the law of conservation of mass. Since there are two oxygen atoms on the left, a <mathjax>#2#</mathjax> coefficient is required: <br/>
<mathjax>#" "Ca + O_2 -> 2CaO#</mathjax></p>
<p>Now you need to have <mathjax>#2#</mathjax> calcium molecules on the left to balance:</p>
<p><mathjax>#" "2Ca + O_2 -> 2CaO#</mathjax></p>
<p>This equation is balanced. The equation says that for every two moles of calcium, there are two moles of calcium oxide produced.</p>
<p>The molar mass of CaO <mathjax>#= 40.078 + 15.999 = (56.077" g CaO")/(1 "mol CaO") #</mathjax></p>
<p><mathjax>#20.00 " mol Ca" xx (2" mol CaO")/(2" mol Ca") xx (56.077" g CaO")/(1 " mol CaO") #</mathjax></p>
<p><mathjax>#= 1121.54 " g CaO"#</mathjax></p>
<p>If you take into consideration <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (SF), the smallest number of SF is 4. This would mean your answer should be <mathjax>#1122 " g CaO"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If 20.00 moles of calcium reacted, how many grams of calcium oxide are produced?</h1>
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<div class="markdown"><p><mathjax>#1122" g CaO"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First you need to create the skeletal equation for the reaction. Since calcium oxide (CaO) is produced, oxygen (<mathjax>#O_2)#</mathjax> is required. </p>
<p>Skeletal equation: <mathjax>#Ca + O_2 -> CaO#</mathjax></p>
<p><strong>Balance the equation</strong> based on the law of conservation of mass. Since there are two oxygen atoms on the left, a <mathjax>#2#</mathjax> coefficient is required: <br/>
<mathjax>#" "Ca + O_2 -> 2CaO#</mathjax></p>
<p>Now you need to have <mathjax>#2#</mathjax> calcium molecules on the left to balance:</p>
<p><mathjax>#" "2Ca + O_2 -> 2CaO#</mathjax></p>
<p>This equation is balanced. The equation says that for every two moles of calcium, there are two moles of calcium oxide produced.</p>
<p>The molar mass of CaO <mathjax>#= 40.078 + 15.999 = (56.077" g CaO")/(1 "mol CaO") #</mathjax></p>
<p><mathjax>#20.00 " mol Ca" xx (2" mol CaO")/(2" mol Ca") xx (56.077" g CaO")/(1 " mol CaO") #</mathjax></p>
<p><mathjax>#= 1121.54 " g CaO"#</mathjax></p>
<p>If you take into consideration <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (SF), the smallest number of SF is 4. This would mean your answer should be <mathjax>#1122 " g CaO"#</mathjax></p></div>
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</article> | If 20.00 moles of calcium reacted, how many grams of calcium oxide are produced? | null |
1,695 | aca9eba8-6ddd-11ea-b623-ccda262736ce | https://socratic.org/questions/when-sodium-is-placed-in-water-sodium-hydroxide-and-hydrogen-are-formed-how-do-y | Na(s) + H2O(l) -> NaOH(aq) + 1/2 H2(g) | start chemical_equation qc_end substance 1 1 qc_end substance 5 5 qc_end substance 6 7 qc_end substance 9 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"Na(s) + H2O(l) -> NaOH(aq) + 1/2 H2(g)"}] | [{"type":"substance name","value":"Sodium"},{"type":"substance name","value":"Water"},{"type":"substance name","value":"Sodium hydroxide"},{"type":"substance name","value":"Hydrogen"}] | <h1 class="questionTitle" itemprop="name">When sodium is placed in water, sodium hydroxide and hydrogen are formed. How do you write the balanced equation for this reaction?</h1> | null | Na(s) + H2O(l) -> NaOH(aq) + 1/2 H2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Are mass and charge balanced? They should be. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Are mass and charge balanced? They should be. </p></div>
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<h1 class="questionTitle" itemprop="name">When sodium is placed in water, sodium hydroxide and hydrogen are formed. How do you write the balanced equation for this reaction?</h1>
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<div class="markdown"><p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Are mass and charge balanced? They should be. </p></div>
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</article> | When sodium is placed in water, sodium hydroxide and hydrogen are formed. How do you write the balanced equation for this reaction? | null |
1,696 | a921511c-6ddd-11ea-8326-ccda262736ce | https://socratic.org/questions/what-is-the-molar-solubility-of-ca-3-po-4-2-in-0-0015-m-ca-no-3-2-if-k-sp-for-ca | 3.92 × 10^(-13) M | start physical_unit 6 6 molar_solubility mol/l qc_end physical_unit 10 10 8 9 molarity qc_end physical_unit 6 6 16 18 equilibrium_constant_k qc_end end | [{"type":"physical unit","value":"Molar solubility [OF] Ca3(PO4)2 [IN] M"}] | [{"type":"physical unit","value":"3.92 × 10^(-13) M"}] | [{"type":"physical unit","value":"Molarity [OF] Ca(NO3)2 [=] \\pu{0.0015 M}"},{"type":"physical unit","value":"Ksp [OF] Ca3(PO4)2 [=] \\pu{1.0 × 10^(-18)}"}] | <h1 class="questionTitle" itemprop="name">What is the molar solubility of #Ca_3(PO_4)_2# in 0.0015 M #Ca(NO_3)_2# if #K_(sp)# for #Ca_3(PO_4)_2# is #1.0xx10^(-18)#?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I'm aware that's probably not the accurate <mathjax>#K_(sp)#</mathjax> for said compound, but is it possible to work out a similarly structured process with that given "constant" ?</p></div>
</h2>
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</div> | 3.92 × 10^(-13) M | <div class="answerDescription">
<div>
<div class="markdown"><p><mathjax>#s = 3.92 xx 10^(-13) "M"#</mathjax>.</p>
<hr/>
<p>Using <a href="http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf" rel="nofollow">the <mathjax>#K_(sp)#</mathjax> of calcium phosphate</a>, which is <mathjax>#2.07 xx 10^(-33)#</mathjax>, we first note that calcium nitrate introduces the <mathjax>#"Ca"^(2+)#</mathjax> common ion. </p>
<p>However, since calcium nitrate is a <strong>strong electrolyte</strong>, AND it contains <strong>only one</strong> <mathjax>#"Ca"^(2+)#</mathjax> <strong>per</strong> formula unit, it introduces <mathjax>#100%#</mathjax> of its concentration as <mathjax>#"Ca"^(2+)#</mathjax> and <mathjax>#200%#</mathjax> of its concentration as <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>Hence, that becomes the <strong>initial concentration</strong> of <mathjax>#"Ca"^(2+)#</mathjax> (whereas the nitrate just sits around), instead of zero:</p>
<blockquote>
<blockquote>
<p><mathjax>#"Ca"_3("PO"_4)_2(s) rightleftharpoons color(red)(3)"Ca"^(2+)(aq) + color(red)(2)"PO"_4^(3-)(aq)#</mathjax></p>
</blockquote>
<p><mathjax>#"I"" "-" "" "" "" "" "" "0.0015" "" "" "0#</mathjax><br/>
<mathjax>#"C"" "-" "" "" "" "" "+color(red)(3)s" "" "" "+color(red)(2)s#</mathjax><br/>
<mathjax>#"E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" "color(red)(2)s#</mathjax></p>
<p>where <mathjax>#s#</mathjax> is the <strong>molar solubility</strong> of calcium phosphate. Each product is forming in water, so they get <mathjax>#+#</mathjax> in the ICE table.</p>
</blockquote>
<p><em>Remember that the coefficients go into the change in concentration, as well as the exponents.</em></p>
<blockquote>
<p><mathjax>#K_(sp) = 2.07 xx 10^(-33) = ["Ca"^(2+)]^color(red)(3)["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#= (0.0015 + 3color(red)(s))^color(red)(3)(color(red)(2)s)^color(red)(2)#</mathjax></p>
</blockquote>
<p>Now, since <mathjax>#K_(sp)#</mathjax> is so small (even the one given in the problem), <mathjax>#3s#</mathjax> <mathjax>#"<<"#</mathjax> <mathjax>#0.0015#</mathjax>, so:</p>
<blockquote>
<p><mathjax>#2.07 xx 10^(-33) = (0.0015)^3(2s)^2#</mathjax></p>
<p><mathjax>#= 3.38 xx 10^(-9) cdot 4s^2#</mathjax></p>
<p><mathjax>#= 1.35 xx 10^(-8)s^2#</mathjax></p>
</blockquote>
<p>Now you can solve for the <strong>molar solubility</strong> of calcium phosphate without working with a fifth order polynomial.</p>
<blockquote>
<p><mathjax>#color(blue)(s) = sqrt((2.07 xx 10^(-33))/(1.35 xx 10^(-8)))#</mathjax></p>
<p><mathjax>#= color(blue)(3.92 xx 10^(-13) "M")#</mathjax></p>
</blockquote>
<p>What would then be the molar solubility of <mathjax>#"Ca"^(2+)#</mathjax> in terms of <mathjax>#s#</mathjax>? What about <mathjax>#"PO"_4^(3-)#</mathjax> in terms of <mathjax>#s#</mathjax>?</p>
<hr/>
<p>On the other hand, without <mathjax>#"Ca"("NO"_3)_2#</mathjax> in solution,</p>
<blockquote>
<p><mathjax>#K_(sp) = (3s)^3(2s)^2 = 108s^5 = 2.07 xx 10^(-33)#</mathjax></p>
</blockquote>
<p>And this molar solubility in <em>pure water</em> is then:</p>
<blockquote>
<p><mathjax>#s = (K_(sp)/108)^(1//5) = 1.14 xx 10^(-7) "M"#</mathjax></p>
</blockquote>
<p>This is about <mathjax>#1000000#</mathjax> times less soluble in <mathjax>#"0.0015 M"#</mathjax> calcium nitrate than in pure water, so the calcium common ion <strong>suppresses</strong> the solubility of calcium phosphate. </p>
<p>That is the <strong>common ion effect</strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerDescription">
<div>
<div class="markdown"><p><mathjax>#s = 3.92 xx 10^(-13) "M"#</mathjax>.</p>
<hr/>
<p>Using <a href="http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf" rel="nofollow">the <mathjax>#K_(sp)#</mathjax> of calcium phosphate</a>, which is <mathjax>#2.07 xx 10^(-33)#</mathjax>, we first note that calcium nitrate introduces the <mathjax>#"Ca"^(2+)#</mathjax> common ion. </p>
<p>However, since calcium nitrate is a <strong>strong electrolyte</strong>, AND it contains <strong>only one</strong> <mathjax>#"Ca"^(2+)#</mathjax> <strong>per</strong> formula unit, it introduces <mathjax>#100%#</mathjax> of its concentration as <mathjax>#"Ca"^(2+)#</mathjax> and <mathjax>#200%#</mathjax> of its concentration as <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>Hence, that becomes the <strong>initial concentration</strong> of <mathjax>#"Ca"^(2+)#</mathjax> (whereas the nitrate just sits around), instead of zero:</p>
<blockquote>
<blockquote>
<p><mathjax>#"Ca"_3("PO"_4)_2(s) rightleftharpoons color(red)(3)"Ca"^(2+)(aq) + color(red)(2)"PO"_4^(3-)(aq)#</mathjax></p>
</blockquote>
<p><mathjax>#"I"" "-" "" "" "" "" "" "0.0015" "" "" "0#</mathjax><br/>
<mathjax>#"C"" "-" "" "" "" "" "+color(red)(3)s" "" "" "+color(red)(2)s#</mathjax><br/>
<mathjax>#"E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" "color(red)(2)s#</mathjax></p>
<p>where <mathjax>#s#</mathjax> is the <strong>molar solubility</strong> of calcium phosphate. Each product is forming in water, so they get <mathjax>#+#</mathjax> in the ICE table.</p>
</blockquote>
<p><em>Remember that the coefficients go into the change in concentration, as well as the exponents.</em></p>
<blockquote>
<p><mathjax>#K_(sp) = 2.07 xx 10^(-33) = ["Ca"^(2+)]^color(red)(3)["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#= (0.0015 + 3color(red)(s))^color(red)(3)(color(red)(2)s)^color(red)(2)#</mathjax></p>
</blockquote>
<p>Now, since <mathjax>#K_(sp)#</mathjax> is so small (even the one given in the problem), <mathjax>#3s#</mathjax> <mathjax>#"<<"#</mathjax> <mathjax>#0.0015#</mathjax>, so:</p>
<blockquote>
<p><mathjax>#2.07 xx 10^(-33) = (0.0015)^3(2s)^2#</mathjax></p>
<p><mathjax>#= 3.38 xx 10^(-9) cdot 4s^2#</mathjax></p>
<p><mathjax>#= 1.35 xx 10^(-8)s^2#</mathjax></p>
</blockquote>
<p>Now you can solve for the <strong>molar solubility</strong> of calcium phosphate without working with a fifth order polynomial.</p>
<blockquote>
<p><mathjax>#color(blue)(s) = sqrt((2.07 xx 10^(-33))/(1.35 xx 10^(-8)))#</mathjax></p>
<p><mathjax>#= color(blue)(3.92 xx 10^(-13) "M")#</mathjax></p>
</blockquote>
<p>What would then be the molar solubility of <mathjax>#"Ca"^(2+)#</mathjax> in terms of <mathjax>#s#</mathjax>? What about <mathjax>#"PO"_4^(3-)#</mathjax> in terms of <mathjax>#s#</mathjax>?</p>
<hr/>
<p>On the other hand, without <mathjax>#"Ca"("NO"_3)_2#</mathjax> in solution,</p>
<blockquote>
<p><mathjax>#K_(sp) = (3s)^3(2s)^2 = 108s^5 = 2.07 xx 10^(-33)#</mathjax></p>
</blockquote>
<p>And this molar solubility in <em>pure water</em> is then:</p>
<blockquote>
<p><mathjax>#s = (K_(sp)/108)^(1//5) = 1.14 xx 10^(-7) "M"#</mathjax></p>
</blockquote>
<p>This is about <mathjax>#1000000#</mathjax> times less soluble in <mathjax>#"0.0015 M"#</mathjax> calcium nitrate than in pure water, so the calcium common ion <strong>suppresses</strong> the solubility of calcium phosphate. </p>
<p>That is the <strong>common ion effect</strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the molar solubility of #Ca_3(PO_4)_2# in 0.0015 M #Ca(NO_3)_2# if #K_(sp)# for #Ca_3(PO_4)_2# is #1.0xx10^(-18)#?</h1>
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<div class="markdown"><p>I'm aware that's probably not the accurate <mathjax>#K_(sp)#</mathjax> for said compound, but is it possible to work out a similarly structured process with that given "constant" ?</p></div>
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Truong-Son N.
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<span class="dateCreated" datetime="2018-06-30T23:55:04" itemprop="dateCreated">
Jun 30, 2018
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<div class="markdown"><p><mathjax>#s = 3.92 xx 10^(-13) "M"#</mathjax>.</p>
<hr/>
<p>Using <a href="http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf" rel="nofollow">the <mathjax>#K_(sp)#</mathjax> of calcium phosphate</a>, which is <mathjax>#2.07 xx 10^(-33)#</mathjax>, we first note that calcium nitrate introduces the <mathjax>#"Ca"^(2+)#</mathjax> common ion. </p>
<p>However, since calcium nitrate is a <strong>strong electrolyte</strong>, AND it contains <strong>only one</strong> <mathjax>#"Ca"^(2+)#</mathjax> <strong>per</strong> formula unit, it introduces <mathjax>#100%#</mathjax> of its concentration as <mathjax>#"Ca"^(2+)#</mathjax> and <mathjax>#200%#</mathjax> of its concentration as <mathjax>#"NO"_3^(-)#</mathjax>.</p>
<p>Hence, that becomes the <strong>initial concentration</strong> of <mathjax>#"Ca"^(2+)#</mathjax> (whereas the nitrate just sits around), instead of zero:</p>
<blockquote>
<blockquote>
<p><mathjax>#"Ca"_3("PO"_4)_2(s) rightleftharpoons color(red)(3)"Ca"^(2+)(aq) + color(red)(2)"PO"_4^(3-)(aq)#</mathjax></p>
</blockquote>
<p><mathjax>#"I"" "-" "" "" "" "" "" "0.0015" "" "" "0#</mathjax><br/>
<mathjax>#"C"" "-" "" "" "" "" "+color(red)(3)s" "" "" "+color(red)(2)s#</mathjax><br/>
<mathjax>#"E"" "-" "" "" "" "" "0.0015+color(red)(3)s" "" "color(red)(2)s#</mathjax></p>
<p>where <mathjax>#s#</mathjax> is the <strong>molar solubility</strong> of calcium phosphate. Each product is forming in water, so they get <mathjax>#+#</mathjax> in the ICE table.</p>
</blockquote>
<p><em>Remember that the coefficients go into the change in concentration, as well as the exponents.</em></p>
<blockquote>
<p><mathjax>#K_(sp) = 2.07 xx 10^(-33) = ["Ca"^(2+)]^color(red)(3)["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#= (0.0015 + 3color(red)(s))^color(red)(3)(color(red)(2)s)^color(red)(2)#</mathjax></p>
</blockquote>
<p>Now, since <mathjax>#K_(sp)#</mathjax> is so small (even the one given in the problem), <mathjax>#3s#</mathjax> <mathjax>#"<<"#</mathjax> <mathjax>#0.0015#</mathjax>, so:</p>
<blockquote>
<p><mathjax>#2.07 xx 10^(-33) = (0.0015)^3(2s)^2#</mathjax></p>
<p><mathjax>#= 3.38 xx 10^(-9) cdot 4s^2#</mathjax></p>
<p><mathjax>#= 1.35 xx 10^(-8)s^2#</mathjax></p>
</blockquote>
<p>Now you can solve for the <strong>molar solubility</strong> of calcium phosphate without working with a fifth order polynomial.</p>
<blockquote>
<p><mathjax>#color(blue)(s) = sqrt((2.07 xx 10^(-33))/(1.35 xx 10^(-8)))#</mathjax></p>
<p><mathjax>#= color(blue)(3.92 xx 10^(-13) "M")#</mathjax></p>
</blockquote>
<p>What would then be the molar solubility of <mathjax>#"Ca"^(2+)#</mathjax> in terms of <mathjax>#s#</mathjax>? What about <mathjax>#"PO"_4^(3-)#</mathjax> in terms of <mathjax>#s#</mathjax>?</p>
<hr/>
<p>On the other hand, without <mathjax>#"Ca"("NO"_3)_2#</mathjax> in solution,</p>
<blockquote>
<p><mathjax>#K_(sp) = (3s)^3(2s)^2 = 108s^5 = 2.07 xx 10^(-33)#</mathjax></p>
</blockquote>
<p>And this molar solubility in <em>pure water</em> is then:</p>
<blockquote>
<p><mathjax>#s = (K_(sp)/108)^(1//5) = 1.14 xx 10^(-7) "M"#</mathjax></p>
</blockquote>
<p>This is about <mathjax>#1000000#</mathjax> times less soluble in <mathjax>#"0.0015 M"#</mathjax> calcium nitrate than in pure water, so the calcium common ion <strong>suppresses</strong> the solubility of calcium phosphate. </p>
<p>That is the <strong>common ion effect</strong>.</p></div>
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seol
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<div class="markdown"><p>The molar solubility of <mathjax>#Ca_3(PO_4)2#</mathjax> is <mathjax>#8.6xx10^-6#</mathjax> M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ICE table:<br/>
<mathjax>#" " " | " " Ca_3PO_4(s)\rightleftharpoons3Ca^(2+)(aq)+2PO_4^(3-)(aq)#</mathjax><br/>
<mathjax>#\text(I) " " | " [solid] " " " " " " " 0.0015 " " " " " 0#</mathjax><br/>
<mathjax>#\text(C) " | " -s " " " " " " " " " " +3s " " " " +2s#</mathjax><br/>
<mathjax>#\text(E) " | [solid]-s " " " 0.0015+3s " " " " 2s#</mathjax></p>
<p><mathjax>#K_(sp)=1.0xx10^(-18)\rArr\color(red)(108x^5)??#</mathjax> (I don't know what this red part means, really)<br/>
<mathjax>#" " " "\rArr[0.0015+3x]^3[2x]^2#</mathjax><br/>
<mathjax>#" " " "\cong(0.0015)^3*4x^2#</mathjax><br/>
<mathjax>#\thereforex=\sqrt((1.0xx10^(-18))/(0.0015^3*4))\approx8.6xx10^-6#</mathjax> M</p>
<p>And as <mathjax>#s\hArrx=[Ca_3(PO_4)_2]#</mathjax>, so the molar solubility of <mathjax>#Ca_3(PO_4)2#</mathjax> is <mathjax>#8.6xx10^-6#</mathjax> M</p></div>
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</article> | What is the molar solubility of #Ca_3(PO_4)_2# in 0.0015 M #Ca(NO_3)_2# if #K_(sp)# for #Ca_3(PO_4)_2# is #1.0xx10^(-18)#? |
I'm aware that's probably not the accurate #K_(sp)# for said compound, but is it possible to work out a similarly structured process with that given "constant" ?
|
1,697 | ab1b6dc9-6ddd-11ea-b5ce-ccda262736ce | https://socratic.org/questions/15g-of-copper-ii-chloride-reacts-with-20g-of-sodium-nitrate-calculate-the-mass-o | 12.15 g | start physical_unit 16 17 mass g qc_end physical_unit 3 4 0 1 mass qc_end physical_unit 10 11 7 8 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium chloride [IN] g"}] | [{"type":"physical unit","value":"12.15 g"}] | [{"type":"physical unit","value":"Mass [OF] copper(ii) chloride [=] \\pu{15 g}"},{"type":"physical unit","value":"Mass [OF] sodium nitrate [=] \\pu{20 g}"}] | <h1 class="questionTitle" itemprop="name">15g of copper(ii) chloride reacts with 20g of sodium nitrate. Calculate the mass of sodium chloride formed?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>I calculated the answer to be around 11.7g. Can someone help verify this?</p></div>
</h2>
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</div> | 12.15 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I got elemental weights as (Na=22.99 grams; Cu=63.55 grams; Cl=31.45 grams; N=14.01 grams; O=16 grams). The answer (Mass of sodium chloride after reaction) is 12.15 grams </p></div>
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<div class="markdown"><p>12.15 grams</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I got elemental weights as (Na=22.99 grams; Cu=63.55 grams; Cl=31.45 grams; N=14.01 grams; O=16 grams). The answer (Mass of sodium chloride after reaction) is 12.15 grams </p></div>
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<h1 class="questionTitle" itemprop="name">15g of copper(ii) chloride reacts with 20g of sodium nitrate. Calculate the mass of sodium chloride formed?</h1>
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<div class="markdown"><p>I calculated the answer to be around 11.7g. Can someone help verify this?</p></div>
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<div class="markdown"><p>12.15 grams</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I got elemental weights as (Na=22.99 grams; Cu=63.55 grams; Cl=31.45 grams; N=14.01 grams; O=16 grams). The answer (Mass of sodium chloride after reaction) is 12.15 grams </p></div>
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Sarita Rana
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<div class="markdown"><p><mathjax>#NaCl#</mathjax> formed = <mathjax>#13.03 g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#CuCl_2#</mathjax> + <mathjax>#2NaNO_3#</mathjax> ------------> <mathjax>#2NaCl#</mathjax> + <mathjax>#Cu(NO_3)_2#</mathjax></p>
<p>no. of moles of <mathjax>#CuCl_2#</mathjax> = <mathjax>#15g#</mathjax>/<mathjax>#134.45#</mathjax> g.mol -1 = <mathjax>#0.1115#</mathjax> moles ..............(1)</p>
<p>no. of moles of <mathjax>#NaNO_3#</mathjax> = <mathjax>#20g#</mathjax>/<mathjax>#84.99#</mathjax> g.mol -1 = <mathjax>#0.2353#</mathjax> moles........(2)</p>
<p>According to equation <br/>
1 mole of <mathjax>#CuCl_2#</mathjax> produces <mathjax>#NaCl#</mathjax> = <mathjax>#2#</mathjax> moles<br/>
so, <mathjax>#0.1115#</mathjax> moles of <mathjax>#CuCl_2#</mathjax> will produce <mathjax>#NaCl#</mathjax> = <mathjax>#2#</mathjax> x <mathjax>#0.1115#</mathjax> moles = <mathjax>#0.223#</mathjax> moles ........(3)</p>
<p>2 mole of <mathjax>#NaNO_3#</mathjax> produces <mathjax>#NaCl#</mathjax> = <mathjax>#2#</mathjax> moles<br/>
so, <mathjax>#0.2353#</mathjax> moles of <mathjax>#NaNO_3#</mathjax> will produce <mathjax>#NaCl#</mathjax> = <mathjax>#2#</mathjax> x <mathjax>#0.2353#</mathjax> moles...........(4)</p>
<p><mathjax>#0.1115#</mathjax> moles of <mathjax>#CuCl_2#</mathjax> produces lesser number of moles of <mathjax>#NaCl#</mathjax>. hence <mathjax>#CuCl_2#</mathjax> is limiting reagent..................(5)</p>
<p>Therefore mass of <mathjax>#NaCl#</mathjax> produced = <mathjax>#0.223#</mathjax> moles.......(6)<br/>
Molar mass of <mathjax>#NaCl#</mathjax> = <mathjax>#58.44#</mathjax> g/mol<br/>
mass of <mathjax>#NaCl#</mathjax> = <mathjax>#0.223#</mathjax> moles x <mathjax>#58.44#</mathjax> g/mol = <mathjax># **13.03** g #</mathjax></p></div>
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</article> | 15g of copper(ii) chloride reacts with 20g of sodium nitrate. Calculate the mass of sodium chloride formed? |
I calculated the answer to be around 11.7g. Can someone help verify this?
|
1,698 | a8a6ac01-6ddd-11ea-adea-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-lead-iv-oxide | PbO2 | start chemical_formula qc_end substance 6 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] lead (IV) oxide [IN] default"}] | [{"type":"chemical equation","value":"PbO2"}] | [{"type":"substance name","value":"Lead (IV) oxide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for lead (IV) oxide?</h1> | null | PbO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is also sometimes called <em>lead dioxide</em>.</p>
<p>The (IV) indicates that lead is in the <mathjax>#4+#</mathjax> oxidation state in this compound. Since oxygen almost always has an oxidation state of <mathjax>#2-#</mathjax>, we use simple mathematics to yield a net oxidation state of <mathjax>#0#</mathjax>:</p>
<p><mathjax>#overbrace((4))^"Pb" + overbrace((x)(-2))^"O" = 0#</mathjax></p>
<p><mathjax>#x = color(red)(2#</mathjax></p>
<p>There is thus one lead atom and two oxygen atoms per compound of lead(IV) oxide.</p>
<p><mathjax>#"PbO"_(color(red)(2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"PbO"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is also sometimes called <em>lead dioxide</em>.</p>
<p>The (IV) indicates that lead is in the <mathjax>#4+#</mathjax> oxidation state in this compound. Since oxygen almost always has an oxidation state of <mathjax>#2-#</mathjax>, we use simple mathematics to yield a net oxidation state of <mathjax>#0#</mathjax>:</p>
<p><mathjax>#overbrace((4))^"Pb" + overbrace((x)(-2))^"O" = 0#</mathjax></p>
<p><mathjax>#x = color(red)(2#</mathjax></p>
<p>There is thus one lead atom and two oxygen atoms per compound of lead(IV) oxide.</p>
<p><mathjax>#"PbO"_(color(red)(2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for lead (IV) oxide?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#"PbO"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is also sometimes called <em>lead dioxide</em>.</p>
<p>The (IV) indicates that lead is in the <mathjax>#4+#</mathjax> oxidation state in this compound. Since oxygen almost always has an oxidation state of <mathjax>#2-#</mathjax>, we use simple mathematics to yield a net oxidation state of <mathjax>#0#</mathjax>:</p>
<p><mathjax>#overbrace((4))^"Pb" + overbrace((x)(-2))^"O" = 0#</mathjax></p>
<p><mathjax>#x = color(red)(2#</mathjax></p>
<p>There is thus one lead atom and two oxygen atoms per compound of lead(IV) oxide.</p>
<p><mathjax>#"PbO"_(color(red)(2#</mathjax></p></div>
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</article> | What is the chemical formula for lead (IV) oxide? | null |
1,699 | aaac6a24-6ddd-11ea-ae69-ccda262736ce | https://socratic.org/questions/12-00-moles-of-naclo3-will-produce-how-many-grams-of-o2-2-naclo3-2-nacl-3-o2 | 576.00 grams | start physical_unit 10 10 mass g qc_end physical_unit 3 3 0 1 mole qc_end chemical_equation 10 17 qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 [IN] grams"}] | [{"type":"physical unit","value":"576.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] NaClO3 [=] \\pu{12.00 moles}"},{"type":"chemical equation","value":"2 NaClO3 -> 2 NaCl + 3 O2"}] | <h1 class="questionTitle" itemprop="name">12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2
</h1> | null | 576.00 grams | <div class="answerDescription">
<div>
<div class="markdown"><p>How many grams of <mathjax>#O_2#</mathjax> are produced if 12.00 moles of <mathjax>#NaClO_3#</mathjax>?</p>
<p>We begin with a balanced chemical equation that is provided in the question.</p>
<p><mathjax>#2NaClO_3 -> 2NaCl + 3O_2#</mathjax></p>
<p>Next we determine what we have and what we want.<br/>
We have 12.00 moles of <mathjax>#NaClO_3#</mathjax> and we want grams of <mathjax>#O_2#</mathjax>.</p>
<p>We set up a roadmap to solve the problem</p>
<p><mathjax>#mol NaClO_3 -> mol O_2 -> grams O_2#</mathjax></p>
<p>We need the molar mass (<mathjax>#gfm#</mathjax>) of <mathjax>#O_2#</mathjax>.<br/>
<mathjax>#O_2#</mathjax> = 32.0 g/mol</p>
<p>We need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#NaClO_3#</mathjax>:<mathjax>#O_2#</mathjax> 2:3. This comes from the coefficients of the balanced chemical equation.</p>
<p>Now we set up conversion factors following the roadmap from above.<br/>
Unit we want in the numerator, unit to cancel in the denominator.</p>
<p><mathjax>#12.00 mol NaClO_3 x (3 mol O_2)/(2 mol NaClO_3) x(32.0. g O_2)/(1 mol O_2) =#</mathjax> </p>
<p>Multiply the numerators, and divide the denominators.</p>
<p>The outcome is <br/>
<mathjax>#576.0 g O_2#</mathjax> <br/>
will be produced from 12.00 moles <mathjax>#NaClO_3#</mathjax></p>
<p>I hope this was helpful.<br/>
SMARTER TEACHER</p>
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<div class="markdown"><p>How many grams of <mathjax>#O_2#</mathjax> are produced if 12.00 moles of <mathjax>#NaClO_3#</mathjax>?</p>
<p>We begin with a balanced chemical equation that is provided in the question.</p>
<p><mathjax>#2NaClO_3 -> 2NaCl + 3O_2#</mathjax></p>
<p>Next we determine what we have and what we want.<br/>
We have 12.00 moles of <mathjax>#NaClO_3#</mathjax> and we want grams of <mathjax>#O_2#</mathjax>.</p>
<p>We set up a roadmap to solve the problem</p>
<p><mathjax>#mol NaClO_3 -> mol O_2 -> grams O_2#</mathjax></p>
<p>We need the molar mass (<mathjax>#gfm#</mathjax>) of <mathjax>#O_2#</mathjax>.<br/>
<mathjax>#O_2#</mathjax> = 32.0 g/mol</p>
<p>We need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#NaClO_3#</mathjax>:<mathjax>#O_2#</mathjax> 2:3. This comes from the coefficients of the balanced chemical equation.</p>
<p>Now we set up conversion factors following the roadmap from above.<br/>
Unit we want in the numerator, unit to cancel in the denominator.</p>
<p><mathjax>#12.00 mol NaClO_3 x (3 mol O_2)/(2 mol NaClO_3) x(32.0. g O_2)/(1 mol O_2) =#</mathjax> </p>
<p>Multiply the numerators, and divide the denominators.</p>
<p>The outcome is <br/>
<mathjax>#576.0 g O_2#</mathjax> <br/>
will be produced from 12.00 moles <mathjax>#NaClO_3#</mathjax></p>
<p>I hope this was helpful.<br/>
SMARTER TEACHER</p>
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<h1 class="questionTitle" itemprop="name">12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2
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BRIAN M.
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May 14, 2014
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<div class="markdown"><p>How many grams of <mathjax>#O_2#</mathjax> are produced if 12.00 moles of <mathjax>#NaClO_3#</mathjax>?</p>
<p>We begin with a balanced chemical equation that is provided in the question.</p>
<p><mathjax>#2NaClO_3 -> 2NaCl + 3O_2#</mathjax></p>
<p>Next we determine what we have and what we want.<br/>
We have 12.00 moles of <mathjax>#NaClO_3#</mathjax> and we want grams of <mathjax>#O_2#</mathjax>.</p>
<p>We set up a roadmap to solve the problem</p>
<p><mathjax>#mol NaClO_3 -> mol O_2 -> grams O_2#</mathjax></p>
<p>We need the molar mass (<mathjax>#gfm#</mathjax>) of <mathjax>#O_2#</mathjax>.<br/>
<mathjax>#O_2#</mathjax> = 32.0 g/mol</p>
<p>We need <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#NaClO_3#</mathjax>:<mathjax>#O_2#</mathjax> 2:3. This comes from the coefficients of the balanced chemical equation.</p>
<p>Now we set up conversion factors following the roadmap from above.<br/>
Unit we want in the numerator, unit to cancel in the denominator.</p>
<p><mathjax>#12.00 mol NaClO_3 x (3 mol O_2)/(2 mol NaClO_3) x(32.0. g O_2)/(1 mol O_2) =#</mathjax> </p>
<p>Multiply the numerators, and divide the denominators.</p>
<p>The outcome is <br/>
<mathjax>#576.0 g O_2#</mathjax> <br/>
will be produced from 12.00 moles <mathjax>#NaClO_3#</mathjax></p>
<p>I hope this was helpful.<br/>
SMARTER TEACHER</p>
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<span class="dateCreated" datetime="2016-04-25T05:24:15" itemprop="dateCreated">
Apr 25, 2016
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<div class="markdown"><p>The number of moles for NaClO3 = 12 moles<br/>
The number of moles for O2 = 18 moles<br/>
Molar mass of O2 = 32 g/mol<br/>
Final Answer (mass of O2) = PLEASE SEE BELOW!!!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If 2 moles of NaClO3 gives you 3 moles of O2, <br/>
it's because <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between NaClO3 : O2 is 2 : 3.<br/>
Then 12 moles of NaClO3 must give you [12 ÷ 2 then<mathjax>#xx#</mathjax>3] = 18 moles of O2.<br/>
You have found the number of moles (n) for O2. </p>
<p>Note that number of moles (n) = mass of substance (m)<mathjax>#-:#</mathjax>molar mass (M). <br/>
<mathjax>#n = m -: M#</mathjax> </p>
<p>The next step is to find out what is the molar mass (M) of O2.<br/>
If you look into your periodic table, the atomic weight is equivalent to the molar mass.<br/>
The molar mass of one O is = 16 g/mol.<br/>
Since there are two O present in the reaction (refer to equation), the molar mass of O2 is [16 g/mol<mathjax>#xx#</mathjax>2] = 32 g/mol<br/>
You've now found the molar mass (M).</p>
<p>The final step is to find out the mass (m) of O2. Since you found the number of moles (n) and molar mass (M), you can now find out the mass of substance.<br/>
Looking back at the formula <mathjax>#n = m -: M#</mathjax>, to find m, we need to make the formula look: m = n<mathjax>#xx#</mathjax>M.<br/>
The mass of O2 is [18 moles <mathjax>#xx#</mathjax>32 g/mol] = 576 grams.</p>
<p>Overall, 12 moles of NaClO3 will produce 576 grams of O2.</p></div>
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</article> | 12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2
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