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1,400 | aa44f224-6ddd-11ea-869c-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-burning-wood | C6H10O5(s) + 6 O2(g) -> 6 CO2(g) + 5 H2O(l) | start chemical_equation qc_end substance 7 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] burning wood"}] | [{"type":"chemical equation","value":"C6H10O5(s) + 6 O2(g) -> 6 CO2(g) + 5 H2O(l)"}] | [{"type":"substance name","value":"Wood"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for burning wood?</h1> | null | C6H10O5(s) + 6 O2(g) -> 6 CO2(g) + 5 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we got thru the usual rigmarole...balance the carbons as carbon dioxide......and then the hydrogens as water, AND then balance the oxygens...</p>
<p><mathjax>#C_6H_10O_5(s)+6O_2(g) rarr 6CO_2(g) + 5H_2O(l)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>For simplicity, we deal with the formula of cellulose, whose formula is more or less that of wood...., i.e. <mathjax>#C_6H_10O_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we got thru the usual rigmarole...balance the carbons as carbon dioxide......and then the hydrogens as water, AND then balance the oxygens...</p>
<p><mathjax>#C_6H_10O_5(s)+6O_2(g) rarr 6CO_2(g) + 5H_2O(l)#</mathjax></p></div>
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Jun 27, 2018
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<div class="markdown"><p>For simplicity, we deal with the formula of cellulose, whose formula is more or less that of wood...., i.e. <mathjax>#C_6H_10O_5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so we got thru the usual rigmarole...balance the carbons as carbon dioxide......and then the hydrogens as water, AND then balance the oxygens...</p>
<p><mathjax>#C_6H_10O_5(s)+6O_2(g) rarr 6CO_2(g) + 5H_2O(l)#</mathjax></p></div>
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</article> | What is the chemical equation for burning wood? | null |
1,401 | a84fa210-6ddd-11ea-b07f-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-the-container-needed-to-store-0-8-moles-of-argon-gas-at-5- | 6.19 L | start physical_unit 13 14 volume l qc_end physical_unit 13 14 10 11 mole qc_end physical_unit 13 14 19 20 temperature qc_end physical_unit 13 14 16 17 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] argon gas [IN] L"}] | [{"type":"physical unit","value":"6.19 L"}] | [{"type":"physical unit","value":"Mole [OF] argon gas [=] \\pu{0.8 moles}"},{"type":"physical unit","value":"Temperature [OF] argon gas [=] \\pu{227 ℃}"},{"type":"physical unit","value":"Pressure [OF] argon gas [=] \\pu{5.3 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?</h1> | null | 6.19 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V=(0.8*molxx0.0821*L*atm*K^-1mol^(-1)*500*K)/(5.3*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#V=(nRT)/P#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V=(0.8*molxx0.0821*L*atm*K^-1mol^(-1)*500*K)/(5.3*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V=(nRT)/P#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V=(0.8*molxx0.0821*L*atm*K^-1mol^(-1)*500*K)/(5.3*atm)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#=6.19L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By Ideal gas equation we know,</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>Where</p>
<ul>
<li>Pressure <mathjax>#P=5.3atm#</mathjax></li>
<li>Volume <mathjax>#V=?L#</mathjax></li>
<li>Universal gas constant <mathjax>#R=0.082LatmK^-1mol^-1#</mathjax></li>
<li>Number moles of gas <mathjax>#n =0.8mol#</mathjax></li>
<li>The temperature of the gas <mathjax>#T=227+273=500K#</mathjax> </li>
</ul>
<p>Putting inthe gas equation we get</p>
<p><mathjax>#5.3V=0.8xx0.082xx500#</mathjax></p>
<p><mathjax>#V=(0.8xx0.082xx500)/5.3L=6.19L#</mathjax></p></div>
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</article> | What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C? | null |
1,402 | ab09951c-6ddd-11ea-913e-ccda262736ce | https://socratic.org/questions/a-hydrate-containing-aluminium-sulphate-has-the-formula-al-2-so-4-3-x-h-2-o-and- | 8 | start physical_unit 21 21 value none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Value [OF] x"}] | [{"type":"physical unit","value":"8"}] | [{"type":"chemical equation","value":"Al2(SO4)3.xH2O"},{"type":"physical unit","value":"Percent by mass [OF] aluminium in aluminium sulphate [=] \\pu{11.11%}"}] | <h1 class="questionTitle" itemprop="name">A hydrate containing aluminium sulphate has the formula #"Al"_2 ("SO"_4)_3 * x"H"_2 "O"# and it contains 11.11% of aluminium by mass. Calculate the value of #x# in the hydrate formula.
?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>If possible, could you type out your thought process that went into solving it as well. Thank you</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out exactly how many <em>moles</em> of water of hydration are present <strong>for every mole</strong> of hydrate.</p>
<p>As you can see, the chemical formula of the hydrate tells you that you get <mathjax>#x#</mathjax> <strong>moles</strong> of water of hydration for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrate. Keep this in mind.</p>
<p>You know that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of aluminium in this unknown aluminium sulfate hydrate is <mathjax>#11.11%#</mathjax>, which means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of hydrate you get <mathjax>#"11.11 g"#</mathjax> of aluminium. </p>
<p>If you pick a <mathjax>#"100-g"#</mathjax> sample of this hydrate, you can calculate the number of <em>moles</em> of aluminium it contains by using the <strong>molar mass</strong> of this element. </p>
<blockquote>
<p><mathjax>#11.11 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.4118 moles Al"#</mathjax></p>
</blockquote>
<p>Now, the chemical formula of the <em>anhydrous</em> salt tells you that <strong>every mole</strong> of aluminium sulfate contains <mathjax>#2#</mathjax> <strong>moles</strong> of aluminium. </p>
<blockquote>
<p><mathjax>#"Al"_ color(red)(2)("SO"_4)_3 -> "1 mole Al"_color(red)(2)("SO"_4)_3color(white)(.)"contains"color(white)(.)color(red)(2)color(white)(.)"moles Al"#</mathjax></p>
</blockquote>
<p>This means that the sample contains</p>
<blockquote>
<p><mathjax>#0.4118 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2("SO"_4)_3)/(2color(red)(cancel(color(black)("moles Al")))) = "0.2059 moles Al"_2("SO"_4)_3#</mathjax></p>
</blockquote>
<p>You can determine the <em>mass</em> of the anhydrous salt present in the hydrate by using the <strong>molar mass</strong> of aluminium sulfate</p>
<blockquote>
<p><mathjax>#0.2059 color(red)(cancel(color(black)("moles Al"_2("SO"_4)_3))) * "342. 15 g"/(1color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = "70.45 g"#</mathjax></p>
</blockquote>
<p>This implies that the sample contains</p>
<blockquote>
<p><mathjax>#overbrace("100 g")^(color(blue)("mass of hydrate")) - overbrace("70.45 g")^(color(blue)("mass of Al"_2("SO"_4)_3)) = overbrace("29.55 g")^(color(blue)("mass of water"))#</mathjax></p>
</blockquote>
<p>Convert this to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#29.55 color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("moles H"_2"O")))) = "1.640 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, if you know that you get <mathjax>#1.640#</mathjax> <strong>moles</strong> of water for <mathjax>#0.2059#</mathjax> <strong>moles</strong> of anhydrous aluminium sulfate, you can say that <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium sulfate will get</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3))) * ("1.640 moles H"_2"O")/(0.2059color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = 7.965 ~~ "8 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous aluminium sulfate you get <mathjax>#8#</mathjax> <strong>moles</strong> of water, so </p>
<blockquote>
<p><mathjax>#x = 8#</mathjax></p>
</blockquote>
<p>and the unknown hydrate is <em>aluminium sulfate octahydrate</em>, <mathjax>#"Al"_2("SO"_4)_3 * 8"H"_2"O"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#x = 8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out exactly how many <em>moles</em> of water of hydration are present <strong>for every mole</strong> of hydrate.</p>
<p>As you can see, the chemical formula of the hydrate tells you that you get <mathjax>#x#</mathjax> <strong>moles</strong> of water of hydration for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrate. Keep this in mind.</p>
<p>You know that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of aluminium in this unknown aluminium sulfate hydrate is <mathjax>#11.11%#</mathjax>, which means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of hydrate you get <mathjax>#"11.11 g"#</mathjax> of aluminium. </p>
<p>If you pick a <mathjax>#"100-g"#</mathjax> sample of this hydrate, you can calculate the number of <em>moles</em> of aluminium it contains by using the <strong>molar mass</strong> of this element. </p>
<blockquote>
<p><mathjax>#11.11 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.4118 moles Al"#</mathjax></p>
</blockquote>
<p>Now, the chemical formula of the <em>anhydrous</em> salt tells you that <strong>every mole</strong> of aluminium sulfate contains <mathjax>#2#</mathjax> <strong>moles</strong> of aluminium. </p>
<blockquote>
<p><mathjax>#"Al"_ color(red)(2)("SO"_4)_3 -> "1 mole Al"_color(red)(2)("SO"_4)_3color(white)(.)"contains"color(white)(.)color(red)(2)color(white)(.)"moles Al"#</mathjax></p>
</blockquote>
<p>This means that the sample contains</p>
<blockquote>
<p><mathjax>#0.4118 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2("SO"_4)_3)/(2color(red)(cancel(color(black)("moles Al")))) = "0.2059 moles Al"_2("SO"_4)_3#</mathjax></p>
</blockquote>
<p>You can determine the <em>mass</em> of the anhydrous salt present in the hydrate by using the <strong>molar mass</strong> of aluminium sulfate</p>
<blockquote>
<p><mathjax>#0.2059 color(red)(cancel(color(black)("moles Al"_2("SO"_4)_3))) * "342. 15 g"/(1color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = "70.45 g"#</mathjax></p>
</blockquote>
<p>This implies that the sample contains</p>
<blockquote>
<p><mathjax>#overbrace("100 g")^(color(blue)("mass of hydrate")) - overbrace("70.45 g")^(color(blue)("mass of Al"_2("SO"_4)_3)) = overbrace("29.55 g")^(color(blue)("mass of water"))#</mathjax></p>
</blockquote>
<p>Convert this to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#29.55 color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("moles H"_2"O")))) = "1.640 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, if you know that you get <mathjax>#1.640#</mathjax> <strong>moles</strong> of water for <mathjax>#0.2059#</mathjax> <strong>moles</strong> of anhydrous aluminium sulfate, you can say that <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium sulfate will get</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3))) * ("1.640 moles H"_2"O")/(0.2059color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = 7.965 ~~ "8 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous aluminium sulfate you get <mathjax>#8#</mathjax> <strong>moles</strong> of water, so </p>
<blockquote>
<p><mathjax>#x = 8#</mathjax></p>
</blockquote>
<p>and the unknown hydrate is <em>aluminium sulfate octahydrate</em>, <mathjax>#"Al"_2("SO"_4)_3 * 8"H"_2"O"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A hydrate containing aluminium sulphate has the formula #"Al"_2 ("SO"_4)_3 * x"H"_2 "O"# and it contains 11.11% of aluminium by mass. Calculate the value of #x# in the hydrate formula.
?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-09-15T14:56:49" itemprop="dateCreated">
Sep 15, 2017
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<div class="markdown"><p><mathjax>#x = 8#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out exactly how many <em>moles</em> of water of hydration are present <strong>for every mole</strong> of hydrate.</p>
<p>As you can see, the chemical formula of the hydrate tells you that you get <mathjax>#x#</mathjax> <strong>moles</strong> of water of hydration for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrate. Keep this in mind.</p>
<p>You know that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of aluminium in this unknown aluminium sulfate hydrate is <mathjax>#11.11%#</mathjax>, which means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of hydrate you get <mathjax>#"11.11 g"#</mathjax> of aluminium. </p>
<p>If you pick a <mathjax>#"100-g"#</mathjax> sample of this hydrate, you can calculate the number of <em>moles</em> of aluminium it contains by using the <strong>molar mass</strong> of this element. </p>
<blockquote>
<p><mathjax>#11.11 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.4118 moles Al"#</mathjax></p>
</blockquote>
<p>Now, the chemical formula of the <em>anhydrous</em> salt tells you that <strong>every mole</strong> of aluminium sulfate contains <mathjax>#2#</mathjax> <strong>moles</strong> of aluminium. </p>
<blockquote>
<p><mathjax>#"Al"_ color(red)(2)("SO"_4)_3 -> "1 mole Al"_color(red)(2)("SO"_4)_3color(white)(.)"contains"color(white)(.)color(red)(2)color(white)(.)"moles Al"#</mathjax></p>
</blockquote>
<p>This means that the sample contains</p>
<blockquote>
<p><mathjax>#0.4118 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2("SO"_4)_3)/(2color(red)(cancel(color(black)("moles Al")))) = "0.2059 moles Al"_2("SO"_4)_3#</mathjax></p>
</blockquote>
<p>You can determine the <em>mass</em> of the anhydrous salt present in the hydrate by using the <strong>molar mass</strong> of aluminium sulfate</p>
<blockquote>
<p><mathjax>#0.2059 color(red)(cancel(color(black)("moles Al"_2("SO"_4)_3))) * "342. 15 g"/(1color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = "70.45 g"#</mathjax></p>
</blockquote>
<p>This implies that the sample contains</p>
<blockquote>
<p><mathjax>#overbrace("100 g")^(color(blue)("mass of hydrate")) - overbrace("70.45 g")^(color(blue)("mass of Al"_2("SO"_4)_3)) = overbrace("29.55 g")^(color(blue)("mass of water"))#</mathjax></p>
</blockquote>
<p>Convert this to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#29.55 color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("moles H"_2"O")))) = "1.640 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, if you know that you get <mathjax>#1.640#</mathjax> <strong>moles</strong> of water for <mathjax>#0.2059#</mathjax> <strong>moles</strong> of anhydrous aluminium sulfate, you can say that <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium sulfate will get</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3))) * ("1.640 moles H"_2"O")/(0.2059color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = 7.965 ~~ "8 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous aluminium sulfate you get <mathjax>#8#</mathjax> <strong>moles</strong> of water, so </p>
<blockquote>
<p><mathjax>#x = 8#</mathjax></p>
</blockquote>
<p>and the unknown hydrate is <em>aluminium sulfate octahydrate</em>, <mathjax>#"Al"_2("SO"_4)_3 * 8"H"_2"O"#</mathjax>.</p></div>
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</article> | A hydrate containing aluminium sulphate has the formula #"Al"_2 ("SO"_4)_3 * x"H"_2 "O"# and it contains 11.11% of aluminium by mass. Calculate the value of #x# in the hydrate formula.
? |
If possible, could you type out your thought process that went into solving it as well. Thank you
|
1,403 | aacbaa98-6ddd-11ea-99cb-ccda262736ce | https://socratic.org/questions/given-the-following-what-is-the-specific-heat-capacity-of-ice-if-the-specific-he | 2.04 J/(g * degrees C) | start physical_unit 10 10 specific_heat_capacity j/(g_·_°c) qc_end physical_unit 17 18 20 24 specific_heat_capacity qc_end physical_unit 18 18 43 44 mass qc_end physical_unit 10 10 43 44 mass qc_end physical_unit 18 18 67 69 temperature qc_end physical_unit 18 18 71 73 temperature qc_end physical_unit 10 10 81 83 temperature qc_end physical_unit 10 10 85 87 temperature qc_end end | [{"type":"physical unit","value":"specific heat capacity [OF] ice [IN] J/(g * degrees C)"}] | [{"type":"physical unit","value":"2.04 J/(g * degrees C)"}] | [{"type":"physical unit","value":"specific heat capacity [OF] liquid water [=] \\pu{4.2 J/(g * degrees C)}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 g}"},{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{100 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{22 degrees C}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{24 degrees C}"},{"type":"physical unit","value":"Temperature1 [OF] ice [=] \\pu{-7 degrees C}"},{"type":"physical unit","value":"Temperature2 [OF] ice [=] \\pu{-3 degrees C}"}] | <h1 class="questionTitle" itemprop="name">Given the following what is the specific heat capacity of ice if the specific heat capactiy of liquid water is 4.2 J/g/degrees C?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Some students did an investigation to determine the specific heat capacity of ice. They did this by heating 100g of water and 100 g of ice in separate beakers simultaneously on the same hot plate. The temperature of water changed from 22 degrees C to 24 degrees C while the temperature of ice changed from -7 degrees C to -3 degrees C. </p></div>
</h2>
</div>
</div> | 2.04 J/(g * degrees C) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>for your experiment Qwater = Qice<br/>
Qwater =<mathjax># m cp(water) (T_f-T_i) = 100 g xx4,2 J/(g°C) xx2 °C = 840 J#</mathjax><br/>
Qice =<mathjax># m cp (T_f-T_i) = 100 g xx cp(ice) J/(g°C) xx 4 °C = 840 J#</mathjax><br/>
from that you obtain cp(ice)<mathjax># = 840J/(300 g°C)=2,1 J/(g°C) #</mathjax><br/>
the experiment is not very accurate because the temperature variation is too little, you don't calculate the cp of the becher...<br/>
the real value is about 2,04 J/(g°C) </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>2,1 J/(g°C)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>for your experiment Qwater = Qice<br/>
Qwater =<mathjax># m cp(water) (T_f-T_i) = 100 g xx4,2 J/(g°C) xx2 °C = 840 J#</mathjax><br/>
Qice =<mathjax># m cp (T_f-T_i) = 100 g xx cp(ice) J/(g°C) xx 4 °C = 840 J#</mathjax><br/>
from that you obtain cp(ice)<mathjax># = 840J/(300 g°C)=2,1 J/(g°C) #</mathjax><br/>
the experiment is not very accurate because the temperature variation is too little, you don't calculate the cp of the becher...<br/>
the real value is about 2,04 J/(g°C) </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Given the following what is the specific heat capacity of ice if the specific heat capactiy of liquid water is 4.2 J/g/degrees C?</h1>
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<div class="markdown"><p>Some students did an investigation to determine the specific heat capacity of ice. They did this by heating 100g of water and 100 g of ice in separate beakers simultaneously on the same hot plate. The temperature of water changed from 22 degrees C to 24 degrees C while the temperature of ice changed from -7 degrees C to -3 degrees C. </p></div>
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<div class="markdown"><p>2,1 J/(g°C)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>for your experiment Qwater = Qice<br/>
Qwater =<mathjax># m cp(water) (T_f-T_i) = 100 g xx4,2 J/(g°C) xx2 °C = 840 J#</mathjax><br/>
Qice =<mathjax># m cp (T_f-T_i) = 100 g xx cp(ice) J/(g°C) xx 4 °C = 840 J#</mathjax><br/>
from that you obtain cp(ice)<mathjax># = 840J/(300 g°C)=2,1 J/(g°C) #</mathjax><br/>
the experiment is not very accurate because the temperature variation is too little, you don't calculate the cp of the becher...<br/>
the real value is about 2,04 J/(g°C) </p></div>
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</article> | Given the following what is the specific heat capacity of ice if the specific heat capactiy of liquid water is 4.2 J/g/degrees C? |
Some students did an investigation to determine the specific heat capacity of ice. They did this by heating 100g of water and 100 g of ice in separate beakers simultaneously on the same hot plate. The temperature of water changed from 22 degrees C to 24 degrees C while the temperature of ice changed from -7 degrees C to -3 degrees C.
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1,404 | ac35490c-6ddd-11ea-9d52-ccda262736ce | https://socratic.org/questions/two-20-0-g-ice-cubes-at-11-0-c-are-placed-into-265-g-of-water-at-25-0-c-assuming | 17.27 ℃ | start physical_unit 14 14 temperature °c qc_end physical_unit 3 4 1 2 mass qc_end physical_unit 3 4 6 7 temperature qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 14 14 16 17 temperature qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"17.27 ℃"}] | [{"type":"physical unit","value":"Mass [OF] ice cubes [=] \\pu{20.0 g}"},{"type":"physical unit","value":"Temperature [OF] ice cubes [=] \\pu{-11.0 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{265 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{25.0 ℃}"},{"type":"other","value":"Assuming no energy is transferred to or from the surroundings."},{"type":"other","value":"After all the ice melts."}] | <h1 class="questionTitle" itemprop="name">Two 20.0-g ice cubes at -11.0 °C are placed into 265 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, how do you calculate the final temperature of the water after all the ice melts?</h1> | null | 17.27 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let <mathjax>#T_f#</mathjax> be the final temperature of mixture. The ice absorbs heat during two process</p>
<p><mathjax>#\text{Ice}\to\text{cooling from}-11^\circ C \ \text{to }0^\circ C \to \text{melting at }0^\circ C\to \text{cooling from }0^\circ C \ \text{to}\ T_f #</mathjax></p>
<p>As there is no loss of heat to or from the surrounding hence by conservation of energy,</p>
<p><mathjax>#\text{heat absorbed by ice of 20 gm}=\text{heat rejected by water of 265 gm} #</mathjax></p>
<p><mathjax>#M_{\text{ice}}(C_{\text{ice}}\Delta T+LH_{\text{ice}}+C_{\text{water}}\Delta T)=M_{\text{water}}C_{\text{water}}\Delta T#</mathjax></p>
<p><mathjax>#20(2.108(0-(-11))+333.55+4.187(T_f-0))=265(4.187)(25-T_f)#</mathjax></p>
<p><mathjax>#7134.76+83.74T_f=27738.875-1109.555T_f#</mathjax></p>
<p><mathjax>#1193.295T_f=20604.115#</mathjax></p>
<p><mathjax>#T_f=\frac{20604.115}{1193.295}#</mathjax></p>
<p><mathjax>#=17.266^\circ C#</mathjax></p>
<p>Hence, the final temperature of mixture (ice +water) will be <mathjax>#T_f=17.266^\circ C#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#17.266^\circ C#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let <mathjax>#T_f#</mathjax> be the final temperature of mixture. The ice absorbs heat during two process</p>
<p><mathjax>#\text{Ice}\to\text{cooling from}-11^\circ C \ \text{to }0^\circ C \to \text{melting at }0^\circ C\to \text{cooling from }0^\circ C \ \text{to}\ T_f #</mathjax></p>
<p>As there is no loss of heat to or from the surrounding hence by conservation of energy,</p>
<p><mathjax>#\text{heat absorbed by ice of 20 gm}=\text{heat rejected by water of 265 gm} #</mathjax></p>
<p><mathjax>#M_{\text{ice}}(C_{\text{ice}}\Delta T+LH_{\text{ice}}+C_{\text{water}}\Delta T)=M_{\text{water}}C_{\text{water}}\Delta T#</mathjax></p>
<p><mathjax>#20(2.108(0-(-11))+333.55+4.187(T_f-0))=265(4.187)(25-T_f)#</mathjax></p>
<p><mathjax>#7134.76+83.74T_f=27738.875-1109.555T_f#</mathjax></p>
<p><mathjax>#1193.295T_f=20604.115#</mathjax></p>
<p><mathjax>#T_f=\frac{20604.115}{1193.295}#</mathjax></p>
<p><mathjax>#=17.266^\circ C#</mathjax></p>
<p>Hence, the final temperature of mixture (ice +water) will be <mathjax>#T_f=17.266^\circ C#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Two 20.0-g ice cubes at -11.0 °C are placed into 265 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, how do you calculate the final temperature of the water after all the ice melts?</h1>
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Harish Chandra Rajpoot
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<div class="markdown"><p><mathjax>#17.266^\circ C#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Let <mathjax>#T_f#</mathjax> be the final temperature of mixture. The ice absorbs heat during two process</p>
<p><mathjax>#\text{Ice}\to\text{cooling from}-11^\circ C \ \text{to }0^\circ C \to \text{melting at }0^\circ C\to \text{cooling from }0^\circ C \ \text{to}\ T_f #</mathjax></p>
<p>As there is no loss of heat to or from the surrounding hence by conservation of energy,</p>
<p><mathjax>#\text{heat absorbed by ice of 20 gm}=\text{heat rejected by water of 265 gm} #</mathjax></p>
<p><mathjax>#M_{\text{ice}}(C_{\text{ice}}\Delta T+LH_{\text{ice}}+C_{\text{water}}\Delta T)=M_{\text{water}}C_{\text{water}}\Delta T#</mathjax></p>
<p><mathjax>#20(2.108(0-(-11))+333.55+4.187(T_f-0))=265(4.187)(25-T_f)#</mathjax></p>
<p><mathjax>#7134.76+83.74T_f=27738.875-1109.555T_f#</mathjax></p>
<p><mathjax>#1193.295T_f=20604.115#</mathjax></p>
<p><mathjax>#T_f=\frac{20604.115}{1193.295}#</mathjax></p>
<p><mathjax>#=17.266^\circ C#</mathjax></p>
<p>Hence, the final temperature of mixture (ice +water) will be <mathjax>#T_f=17.266^\circ C#</mathjax></p></div>
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</article> | Two 20.0-g ice cubes at -11.0 °C are placed into 265 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, how do you calculate the final temperature of the water after all the ice melts? | null |
1,405 | ac2d1c1a-6ddd-11ea-b1ec-ccda262736ce | https://socratic.org/questions/the-standard-heat-of-formation-of-fe2o3-s-is-824-2kj-mol-calculate-heat-change-f | -1648.40 kJ | start physical_unit 14 15 heat_energy kj qc_end physical_unit 6 6 8 9 standard_enthalpy_formation qc_end chemical_equation 16 23 qc_end end | [{"type":"physical unit","value":"Heat change [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"-1648.40 kJ"}] | [{"type":"physical unit","value":"Standard heat of formation [OF] Fe2O3(s) [=] \\pu{824.2 kJ/mol}"},{"type":"chemical equation","value":"4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)"}] | <h1 class="questionTitle" itemprop="name">The standard heat of formation of Fe2O3(s) is 824.2kJ/mol. Calculate heat change for the reaction
4Fe(s)+3O2(g)=2Fe2O3(s)?</h1> | null | -1648.40 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Standard Heat of Formation (<mathjax>#DeltaH_(f)^o#</mathjax>) is given in terms of Kj per 1 mole. By definition <mathjax>#DeltaH_f^o#</mathjax> is the heat liberated or gained on formation of one mole of substance from basic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their standard states. The Standard <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> of Formation for elements in their basic standard states is equal to 'zero' Kj/mole. The form/state can be identified in Thermodynamics Tables in the appendix of most college level general chemistry text books. </p>
<p>For <mathjax>#Fe_2O_3#</mathjax> the Standard State Equation would be represented as follows ...</p>
<p><mathjax>#2Fe^o(s)#</mathjax> + <mathjax>#3/2O_2(g)#</mathjax> => <mathjax>#Fe_2O_3(s)#</mathjax>; <mathjax>#DeltaH_f^o#</mathjax>= <mathjax>#-
824.2 "Kj"/"mole"#</mathjax></p>
<p>For the reaction in the post it is showing 2 moles of <mathjax>#Fe_2O_3#</mathjax>. Therefore, the <mathjax>#Delta H_(Rxn)^o#</mathjax> = <mathjax>#2xxDeltaH_f^o#</mathjax> = <mathjax>#2cancel("moles")(-824.2 "Kj"/cancel("mole"))#</mathjax> = <mathjax>#-1648.4 Kj#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaH_(Rxn)^o = -1648.4 Kj#</mathjax> <br/>
Note, the <mathjax>#Delta H_"f"^o(Fe_2O_3(s))#</mathjax> = <mathjax>#- 824.2 "Kj"/"mole"#</mathjax> (exothermic)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Standard Heat of Formation (<mathjax>#DeltaH_(f)^o#</mathjax>) is given in terms of Kj per 1 mole. By definition <mathjax>#DeltaH_f^o#</mathjax> is the heat liberated or gained on formation of one mole of substance from basic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their standard states. The Standard <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> of Formation for elements in their basic standard states is equal to 'zero' Kj/mole. The form/state can be identified in Thermodynamics Tables in the appendix of most college level general chemistry text books. </p>
<p>For <mathjax>#Fe_2O_3#</mathjax> the Standard State Equation would be represented as follows ...</p>
<p><mathjax>#2Fe^o(s)#</mathjax> + <mathjax>#3/2O_2(g)#</mathjax> => <mathjax>#Fe_2O_3(s)#</mathjax>; <mathjax>#DeltaH_f^o#</mathjax>= <mathjax>#-
824.2 "Kj"/"mole"#</mathjax></p>
<p>For the reaction in the post it is showing 2 moles of <mathjax>#Fe_2O_3#</mathjax>. Therefore, the <mathjax>#Delta H_(Rxn)^o#</mathjax> = <mathjax>#2xxDeltaH_f^o#</mathjax> = <mathjax>#2cancel("moles")(-824.2 "Kj"/cancel("mole"))#</mathjax> = <mathjax>#-1648.4 Kj#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The standard heat of formation of Fe2O3(s) is 824.2kJ/mol. Calculate heat change for the reaction
4Fe(s)+3O2(g)=2Fe2O3(s)?</h1>
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<div class="markdown"><p><mathjax>#DeltaH_(Rxn)^o = -1648.4 Kj#</mathjax> <br/>
Note, the <mathjax>#Delta H_"f"^o(Fe_2O_3(s))#</mathjax> = <mathjax>#- 824.2 "Kj"/"mole"#</mathjax> (exothermic)</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Standard Heat of Formation (<mathjax>#DeltaH_(f)^o#</mathjax>) is given in terms of Kj per 1 mole. By definition <mathjax>#DeltaH_f^o#</mathjax> is the heat liberated or gained on formation of one mole of substance from basic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their standard states. The Standard <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> of Formation for elements in their basic standard states is equal to 'zero' Kj/mole. The form/state can be identified in Thermodynamics Tables in the appendix of most college level general chemistry text books. </p>
<p>For <mathjax>#Fe_2O_3#</mathjax> the Standard State Equation would be represented as follows ...</p>
<p><mathjax>#2Fe^o(s)#</mathjax> + <mathjax>#3/2O_2(g)#</mathjax> => <mathjax>#Fe_2O_3(s)#</mathjax>; <mathjax>#DeltaH_f^o#</mathjax>= <mathjax>#-
824.2 "Kj"/"mole"#</mathjax></p>
<p>For the reaction in the post it is showing 2 moles of <mathjax>#Fe_2O_3#</mathjax>. Therefore, the <mathjax>#Delta H_(Rxn)^o#</mathjax> = <mathjax>#2xxDeltaH_f^o#</mathjax> = <mathjax>#2cancel("moles")(-824.2 "Kj"/cancel("mole"))#</mathjax> = <mathjax>#-1648.4 Kj#</mathjax></p></div>
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</article> | The standard heat of formation of Fe2O3(s) is 824.2kJ/mol. Calculate heat change for the reaction
4Fe(s)+3O2(g)=2Fe2O3(s)? | null |
1,406 | abd6de0c-6ddd-11ea-846b-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-at-a-pressure-of-121-59-kpa-a-volume-of-31-l-and-a-temperature-o | 1.3 moles | start physical_unit 1 3 mole mol qc_end physical_unit 1 3 8 9 pressure qc_end physical_unit 1 3 13 14 volume qc_end physical_unit 1 3 19 20 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] gas sample [IN] moles"}] | [{"type":"physical unit","value":"1.3 moles"}] | [{"type":"physical unit","value":"Pressure [OF] gas sample [=] \\pu{121.59 kPa}"},{"type":"physical unit","value":"Volume [OF] gas sample [=] \\pu{31 L}"},{"type":"physical unit","value":"Temperature [OF] gas sample [=] \\pu{360 K}"}] | <h1 class="questionTitle" itemprop="name">A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? </h1> | null | 1.3 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> as shown below.</p>
<p><img alt="https://arenahanna.wordpress.com/gas-laws/" src="https://useruploads.socratic.org/8AFau7ntT82iIWxsldJ6_ideal_gas.png"/> </p>
<p><mathjax>#P="121.59 kPa"#</mathjax><br/>
<mathjax>#V="31 L"#</mathjax><br/>
<mathjax>#T="360 K"#</mathjax><br/>
<mathjax>#R=8.3144598 "L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>, plug in the known values, and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(121.59cancel"kPa"xx31cancel"L")/(8.3144598cancel"L kPa K"^(-1) "mol"^(-1)xx360cancel"K")="1.3 mol"#</mathjax> (rounded to two significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The sample of gas contains 1.3 moles of gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> as shown below.</p>
<p><img alt="https://arenahanna.wordpress.com/gas-laws/" src="https://useruploads.socratic.org/8AFau7ntT82iIWxsldJ6_ideal_gas.png"/> </p>
<p><mathjax>#P="121.59 kPa"#</mathjax><br/>
<mathjax>#V="31 L"#</mathjax><br/>
<mathjax>#T="360 K"#</mathjax><br/>
<mathjax>#R=8.3144598 "L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>, plug in the known values, and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(121.59cancel"kPa"xx31cancel"L")/(8.3144598cancel"L kPa K"^(-1) "mol"^(-1)xx360cancel"K")="1.3 mol"#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? </h1>
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<div class="markdown"><p>The sample of gas contains 1.3 moles of gas.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> as shown below.</p>
<p><img alt="https://arenahanna.wordpress.com/gas-laws/" src="https://useruploads.socratic.org/8AFau7ntT82iIWxsldJ6_ideal_gas.png"/> </p>
<p><mathjax>#P="121.59 kPa"#</mathjax><br/>
<mathjax>#V="31 L"#</mathjax><br/>
<mathjax>#T="360 K"#</mathjax><br/>
<mathjax>#R=8.3144598 "L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p>Rearrange the equation to isolate <mathjax>#n#</mathjax>, plug in the known values, and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(121.59cancel"kPa"xx31cancel"L")/(8.3144598cancel"L kPa K"^(-1) "mol"^(-1)xx360cancel"K")="1.3 mol"#</mathjax> (rounded to two significant figures)</p></div>
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</article> | A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? | null |
1,407 | a871d590-6ddd-11ea-ae25-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-alcohol-present-in-620-ml-of-a-40-0-viv-solution-of-alcoho | 248.00 mL | start physical_unit 5 5 volume ml qc_end physical_unit 14 14 8 9 volume qc_end end | [{"type":"physical unit","value":"volume [OF] alcohol [IN] mL"}] | [{"type":"physical unit","value":"248.00 mL"}] | [{"type":"physical unit","value":"volume [OF] solution [=] \\pu{620 mL}"},{"type":"physical unit","value":"(viv) percent [OF] alcohol in solution [=] \\pu{40.0%}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of alcohol present in 620 mL of a 40.0% (viv) solution of alcohol?</h1> | null | 248.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, we have a solution that is <mathjax>#40%#</mathjax> <mathjax>#v/v#</mathjax>.</p>
<p>That is <mathjax>#"Volume of alcohol"/"Volume of solution"xx100%=40%#</mathjax>.</p>
<p>And <mathjax>#620*mLxx40%=248*mL#</mathjax>. </p>
<p>Note that I have not assumed what the alcohol is, and what the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is - we have assumed also (fairly reasonably) that the volumes are additive. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We get a volume of <mathjax>#248*mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, we have a solution that is <mathjax>#40%#</mathjax> <mathjax>#v/v#</mathjax>.</p>
<p>That is <mathjax>#"Volume of alcohol"/"Volume of solution"xx100%=40%#</mathjax>.</p>
<p>And <mathjax>#620*mLxx40%=248*mL#</mathjax>. </p>
<p>Note that I have not assumed what the alcohol is, and what the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is - we have assumed also (fairly reasonably) that the volumes are additive. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the volume of alcohol present in 620 mL of a 40.0% (viv) solution of alcohol?</h1>
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anor277
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<div class="markdown"><p>We get a volume of <mathjax>#248*mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, we have a solution that is <mathjax>#40%#</mathjax> <mathjax>#v/v#</mathjax>.</p>
<p>That is <mathjax>#"Volume of alcohol"/"Volume of solution"xx100%=40%#</mathjax>.</p>
<p>And <mathjax>#620*mLxx40%=248*mL#</mathjax>. </p>
<p>Note that I have not assumed what the alcohol is, and what the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is - we have assumed also (fairly reasonably) that the volumes are additive. </p></div>
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</article> | What is the volume of alcohol present in 620 mL of a 40.0% (viv) solution of alcohol? | null |
1,408 | aaaf6d78-6ddd-11ea-8f27-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-phosphorus-in-po4-3 | +5 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] phosphorus"}] | [{"type":"physical unit","value":"+5"}] | [{"type":"chemical equation","value":"PO3^4-"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in #"PO"_4^(3-)#?
</h1> | null | +5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with an ion, the <em>phosphate anion</em> to be precise, which means that the oxidation states of <strong>each atom</strong> must add up to give the <strong>overall charge</strong> of the ion. </p>
<p>So, start by writing down what you know. You know that</p>
<ul>
<li><em>the ion has a</em> <mathjax>#"(3-)"#</mathjax> <em>overall charge</em></li>
<li><em>the ion is made up of one phosphous atom and four oxygen atoms</em></li>
<li><em>oxygen's oxidation state in most <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is</em> <mathjax>#-2#</mathjax>.</li>
</ul>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#overbrace(1 xx ON_"P")^(color(blue)("one phosphorus atom")) + underbrace(4 xx ON_"oxygen")_(color(red)("four oxygen atoms")) = -3#</mathjax></p>
</blockquote>
<p>Solve the equation for <mathjax>#ON_"P"#</mathjax> to get</p>
<blockquote>
<p><mathjax>#ON_"P" + 4 xx (-2) = -3#</mathjax></p>
<p><mathjax>#ON_"P" = -3 + 8 = color(green)(+5)#</mathjax></p>
</blockquote>
<p>So, phosphorus has a <mathjax>#+5#</mathjax> <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> in the phosphate anion, <mathjax>#"PO"_4^(3-)#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with an ion, the <em>phosphate anion</em> to be precise, which means that the oxidation states of <strong>each atom</strong> must add up to give the <strong>overall charge</strong> of the ion. </p>
<p>So, start by writing down what you know. You know that</p>
<ul>
<li><em>the ion has a</em> <mathjax>#"(3-)"#</mathjax> <em>overall charge</em></li>
<li><em>the ion is made up of one phosphous atom and four oxygen atoms</em></li>
<li><em>oxygen's oxidation state in most <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is</em> <mathjax>#-2#</mathjax>.</li>
</ul>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#overbrace(1 xx ON_"P")^(color(blue)("one phosphorus atom")) + underbrace(4 xx ON_"oxygen")_(color(red)("four oxygen atoms")) = -3#</mathjax></p>
</blockquote>
<p>Solve the equation for <mathjax>#ON_"P"#</mathjax> to get</p>
<blockquote>
<p><mathjax>#ON_"P" + 4 xx (-2) = -3#</mathjax></p>
<p><mathjax>#ON_"P" = -3 + 8 = color(green)(+5)#</mathjax></p>
</blockquote>
<p>So, phosphorus has a <mathjax>#+5#</mathjax> <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> in the phosphate anion, <mathjax>#"PO"_4^(3-)#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in #"PO"_4^(3-)#?
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Stefan V.
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<span class="dateCreated" datetime="2015-10-23T10:27:02" itemprop="dateCreated">
Oct 23, 2015
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<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with an ion, the <em>phosphate anion</em> to be precise, which means that the oxidation states of <strong>each atom</strong> must add up to give the <strong>overall charge</strong> of the ion. </p>
<p>So, start by writing down what you know. You know that</p>
<ul>
<li><em>the ion has a</em> <mathjax>#"(3-)"#</mathjax> <em>overall charge</em></li>
<li><em>the ion is made up of one phosphous atom and four oxygen atoms</em></li>
<li><em>oxygen's oxidation state in most <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is</em> <mathjax>#-2#</mathjax>.</li>
</ul>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#overbrace(1 xx ON_"P")^(color(blue)("one phosphorus atom")) + underbrace(4 xx ON_"oxygen")_(color(red)("four oxygen atoms")) = -3#</mathjax></p>
</blockquote>
<p>Solve the equation for <mathjax>#ON_"P"#</mathjax> to get</p>
<blockquote>
<p><mathjax>#ON_"P" + 4 xx (-2) = -3#</mathjax></p>
<p><mathjax>#ON_"P" = -3 + 8 = color(green)(+5)#</mathjax></p>
</blockquote>
<p>So, phosphorus has a <mathjax>#+5#</mathjax> <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> in the phosphate anion, <mathjax>#"PO"_4^(3-)#</mathjax>.</p></div>
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</article> | What is the oxidation number of phosphorus in #"PO"_4^(3-)#?
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1,409 | ac60fc00-6ddd-11ea-b38f-ccda262736ce | https://socratic.org/questions/591912d87c01496e5f856445 | TeO3^2- + 4 I- + 6 H+ -> Te + 2 I2 + 3 H2O | start chemical_equation qc_end chemical_equation 8 8 qc_end substance 10 11 qc_end substance 16 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reduction"}] | [{"type":"chemical equation","value":"TeO3^2- + 4 I- + 6 H+ -> Te + 2 I2 + 3 H2O"}] | [{"type":"chemical equation","value":"TeO3^2-"},{"type":"substance name","value":"Tellurium metal"},{"type":"substance name","value":"Iodide ion"}] | <h1 class="questionTitle" itemprop="name">Can you represent the reduction of tellurite ion, #TeO_3^(2-)#, to tellurium metal by the oxidation of iodide ion?</h1> | null | TeO3^2- + 4 I- + 6 H+ -> Te + 2 I2 + 3 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#I^(-) rarr1/2I_2(s) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Tellurite ion is reduced to tellurium metal...........</p>
<p><mathjax>#TeO_3^(2-) +6H^+ + 4e^(-) rarr stackrel(0)Te + 3H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Mass and charge are balanced in each instance. And so we take <mathjax>#4xx(i)+(ii):#</mathjax></p>
<p><mathjax>#TeO_3^(2-) +4I^(-) +6H^(+)rarr Te + 2I_2 + 3H_2O #</mathjax></p>
<p>Which is balanced with respect to mass and charge. Whether it represents an actual chemical reaction, I don't know...........</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Iodide anion is oxidized to iodine...........</p>
<p><mathjax>#I^(-) rarr1/2I_2(s) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#I^(-) rarr1/2I_2(s) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Tellurite ion is reduced to tellurium metal...........</p>
<p><mathjax>#TeO_3^(2-) +6H^+ + 4e^(-) rarr stackrel(0)Te + 3H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Mass and charge are balanced in each instance. And so we take <mathjax>#4xx(i)+(ii):#</mathjax></p>
<p><mathjax>#TeO_3^(2-) +4I^(-) +6H^(+)rarr Te + 2I_2 + 3H_2O #</mathjax></p>
<p>Which is balanced with respect to mass and charge. Whether it represents an actual chemical reaction, I don't know...........</p></div>
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<h1 class="questionTitle" itemprop="name">Can you represent the reduction of tellurite ion, #TeO_3^(2-)#, to tellurium metal by the oxidation of iodide ion?</h1>
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<div class="markdown"><p>Iodide anion is oxidized to iodine...........</p>
<p><mathjax>#I^(-) rarr1/2I_2(s) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#I^(-) rarr1/2I_2(s) + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Tellurite ion is reduced to tellurium metal...........</p>
<p><mathjax>#TeO_3^(2-) +6H^+ + 4e^(-) rarr stackrel(0)Te + 3H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Mass and charge are balanced in each instance. And so we take <mathjax>#4xx(i)+(ii):#</mathjax></p>
<p><mathjax>#TeO_3^(2-) +4I^(-) +6H^(+)rarr Te + 2I_2 + 3H_2O #</mathjax></p>
<p>Which is balanced with respect to mass and charge. Whether it represents an actual chemical reaction, I don't know...........</p></div>
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</article> | Can you represent the reduction of tellurite ion, #TeO_3^(2-)#, to tellurium metal by the oxidation of iodide ion? | null |
1,410 | abe1499a-6ddd-11ea-a6a6-ccda262736ce | https://socratic.org/questions/58fc3962b72cff794a377eb6 | 2.54 × 10^(-5) | start physical_unit 4 4 ka none qc_end end | [{"type":"physical unit","value":"Ka [OF] HOCl"}] | [{"type":"physical unit","value":"2.54 × 10^(-5)"}] | [{"type":"physical unit","value":"[HOCl] [OF] the solution [=] \\pu{0.08 mol/dm^3}"},{"type":"physical unit","value":"pH [OF] HOCl solution [=] \\pu{2.85}"}] | <h1 class="questionTitle" itemprop="name">What is #K_a# for #HOCl#, given that initially #[HOCl]=0.08*mol*dm^3#, and #pH=2.85#?</h1> | null | 2.54 × 10^(-5) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"acid dissociation constant,"#</mathjax> <mathjax>#K_a#</mathjax>, measures the particular acid dissociation............</p>
<p><mathjax>#HOCl(aq) + H_2O rightleftharpoonsH_3O^(+) + ""^(-)OCl#</mathjax></p>
<p>Where <mathjax>#K_a=([H_3O^+][""^(-)OCl])/([HOCl])#</mathjax></p>
<p>We are given that the initial concentration was <mathjax>#0.08*mol*dm^-3#</mathjax>, and that <mathjax>#pH=2.85#</mathjax>. But we know that <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>But, by <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> <mathjax>#[""^(-)OCl]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>And, thus......................</p>
<p><mathjax>#[HOCl]=(0.08-10^(-2.85))*mol*L^-1=...............#</mathjax></p>
<p><mathjax>#=0.0786*mol*L^-1#</mathjax>.</p>
<p>And thus we can calculate <mathjax>#K_a#</mathjax> sans approximation, i.e.</p>
<p><mathjax>#K_a=((10^(-2.85))^2)/(0.0786)=2.54xx10^-5#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#K_a=([H_3O^+][""^(-)OCl])/([HOCl])=2.54xx10^-5#</mathjax>.........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"acid dissociation constant,"#</mathjax> <mathjax>#K_a#</mathjax>, measures the particular acid dissociation............</p>
<p><mathjax>#HOCl(aq) + H_2O rightleftharpoonsH_3O^(+) + ""^(-)OCl#</mathjax></p>
<p>Where <mathjax>#K_a=([H_3O^+][""^(-)OCl])/([HOCl])#</mathjax></p>
<p>We are given that the initial concentration was <mathjax>#0.08*mol*dm^-3#</mathjax>, and that <mathjax>#pH=2.85#</mathjax>. But we know that <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>But, by <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> <mathjax>#[""^(-)OCl]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>And, thus......................</p>
<p><mathjax>#[HOCl]=(0.08-10^(-2.85))*mol*L^-1=...............#</mathjax></p>
<p><mathjax>#=0.0786*mol*L^-1#</mathjax>.</p>
<p>And thus we can calculate <mathjax>#K_a#</mathjax> sans approximation, i.e.</p>
<p><mathjax>#K_a=((10^(-2.85))^2)/(0.0786)=2.54xx10^-5#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is #K_a# for #HOCl#, given that initially #[HOCl]=0.08*mol*dm^3#, and #pH=2.85#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#K_a=([H_3O^+][""^(-)OCl])/([HOCl])=2.54xx10^-5#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <mathjax>#"acid dissociation constant,"#</mathjax> <mathjax>#K_a#</mathjax>, measures the particular acid dissociation............</p>
<p><mathjax>#HOCl(aq) + H_2O rightleftharpoonsH_3O^(+) + ""^(-)OCl#</mathjax></p>
<p>Where <mathjax>#K_a=([H_3O^+][""^(-)OCl])/([HOCl])#</mathjax></p>
<p>We are given that the initial concentration was <mathjax>#0.08*mol*dm^-3#</mathjax>, and that <mathjax>#pH=2.85#</mathjax>. But we know that <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax>.</p>
<p>And thus <mathjax>#[H_3O^+]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>But, by <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> <mathjax>#[""^(-)OCl]=10^(-2.85)*mol*L^-1#</mathjax>.</p>
<p>And, thus......................</p>
<p><mathjax>#[HOCl]=(0.08-10^(-2.85))*mol*L^-1=...............#</mathjax></p>
<p><mathjax>#=0.0786*mol*L^-1#</mathjax>.</p>
<p>And thus we can calculate <mathjax>#K_a#</mathjax> sans approximation, i.e.</p>
<p><mathjax>#K_a=((10^(-2.85))^2)/(0.0786)=2.54xx10^-5#</mathjax>.</p></div>
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</article> | What is #K_a# for #HOCl#, given that initially #[HOCl]=0.08*mol*dm^3#, and #pH=2.85#? | null |
1,411 | a8389136-6ddd-11ea-bbe9-ccda262736ce | https://socratic.org/questions/59316f18b72cff625456d606 | C3H8 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 10 11 7 8 volume qc_end physical_unit 20 22 24 25 volume qc_end end | [{"type":"other","value":"Chemical Formula [OF] the alkane [IN] default"}] | [{"type":"chemical equation","value":"C3H8"}] | [{"type":"other","value":"The combustion of a gaseous alkane produces carbon dioxide plus water vapour."},{"type":"physical unit","value":"Volume [OF] carbon dioxide [=] \\pu{7 L}"},{"type":"physical unit","value":"Original volume [OF] alkane plus oxygen [=] \\pu{6 L}"}] | <h1 class="questionTitle" itemprop="name">The combustion of a gaseous alkane produces 7 L of carbon dioxide plus water vapour. If the original volume of alkane plus oxygen was 6 L, what is the formula of the alkane?</h1> | null | C3H8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction used is a normal total oxidation of the hydrocarbon by oxygen:</p>
<p><mathjax>#C_nH_"2n+2"+[n+((n+1)/2)]O_2 -> nCo_2 + [n+1]H_2O#</mathjax></p>
<p>The coefficient for the O_2 is coming from the addition of the ones for the products.</p>
<p>Using the ideal gas formula PV=nRT, where everything is constant but V and n, we can relate both states, before and after the reaction.</p>
<p><mathjax>#V_1/V_2 = n_1/ n_2 = 600/700 = 6/7#</mathjax></p>
<p>where <mathjax>#n_1#</mathjax> and <mathjax>#n_2#</mathjax> are the total number of moles, for the first side of the equation that means the oxygen and the hydrocarbon, while in the second the carbon dioxide and the water.</p>
<p>Then the first side</p>
<p><mathjax>#n_1 = 1 + [n+((n+1)/2)] = 1 +((3n+1)/2) = (3n+3)/2 = 6#</mathjax></p>
<p>solving n</p>
<p><mathjax>#(3n+3)/2 = 6#</mathjax></p>
<p><mathjax>#3n+3 = 12#</mathjax> factorizing 3</p>
<p><mathjax>#n+1 = 4#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p>
<p>Then the formula is <mathjax>#C_3H_8#</mathjax></p>
<p>Now, the same result should be obtained from the other side, lets try it out.</p>
<p><mathjax>#n_2 = n+(n+1) = 2n+1 = 7#</mathjax></p>
<p><mathjax>#2n+1 = 7#</mathjax></p>
<p><mathjax>#n = (7-1)/2#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#C_3H_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction used is a normal total oxidation of the hydrocarbon by oxygen:</p>
<p><mathjax>#C_nH_"2n+2"+[n+((n+1)/2)]O_2 -> nCo_2 + [n+1]H_2O#</mathjax></p>
<p>The coefficient for the O_2 is coming from the addition of the ones for the products.</p>
<p>Using the ideal gas formula PV=nRT, where everything is constant but V and n, we can relate both states, before and after the reaction.</p>
<p><mathjax>#V_1/V_2 = n_1/ n_2 = 600/700 = 6/7#</mathjax></p>
<p>where <mathjax>#n_1#</mathjax> and <mathjax>#n_2#</mathjax> are the total number of moles, for the first side of the equation that means the oxygen and the hydrocarbon, while in the second the carbon dioxide and the water.</p>
<p>Then the first side</p>
<p><mathjax>#n_1 = 1 + [n+((n+1)/2)] = 1 +((3n+1)/2) = (3n+3)/2 = 6#</mathjax></p>
<p>solving n</p>
<p><mathjax>#(3n+3)/2 = 6#</mathjax></p>
<p><mathjax>#3n+3 = 12#</mathjax> factorizing 3</p>
<p><mathjax>#n+1 = 4#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p>
<p>Then the formula is <mathjax>#C_3H_8#</mathjax></p>
<p>Now, the same result should be obtained from the other side, lets try it out.</p>
<p><mathjax>#n_2 = n+(n+1) = 2n+1 = 7#</mathjax></p>
<p><mathjax>#2n+1 = 7#</mathjax></p>
<p><mathjax>#n = (7-1)/2#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The combustion of a gaseous alkane produces 7 L of carbon dioxide plus water vapour. If the original volume of alkane plus oxygen was 6 L, what is the formula of the alkane?</h1>
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<div class="markdown"><p><mathjax>#C_3H_8#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction used is a normal total oxidation of the hydrocarbon by oxygen:</p>
<p><mathjax>#C_nH_"2n+2"+[n+((n+1)/2)]O_2 -> nCo_2 + [n+1]H_2O#</mathjax></p>
<p>The coefficient for the O_2 is coming from the addition of the ones for the products.</p>
<p>Using the ideal gas formula PV=nRT, where everything is constant but V and n, we can relate both states, before and after the reaction.</p>
<p><mathjax>#V_1/V_2 = n_1/ n_2 = 600/700 = 6/7#</mathjax></p>
<p>where <mathjax>#n_1#</mathjax> and <mathjax>#n_2#</mathjax> are the total number of moles, for the first side of the equation that means the oxygen and the hydrocarbon, while in the second the carbon dioxide and the water.</p>
<p>Then the first side</p>
<p><mathjax>#n_1 = 1 + [n+((n+1)/2)] = 1 +((3n+1)/2) = (3n+3)/2 = 6#</mathjax></p>
<p>solving n</p>
<p><mathjax>#(3n+3)/2 = 6#</mathjax></p>
<p><mathjax>#3n+3 = 12#</mathjax> factorizing 3</p>
<p><mathjax>#n+1 = 4#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p>
<p>Then the formula is <mathjax>#C_3H_8#</mathjax></p>
<p>Now, the same result should be obtained from the other side, lets try it out.</p>
<p><mathjax>#n_2 = n+(n+1) = 2n+1 = 7#</mathjax></p>
<p><mathjax>#2n+1 = 7#</mathjax></p>
<p><mathjax>#n = (7-1)/2#</mathjax></p>
<p><mathjax>#n = 3#</mathjax></p></div>
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</article> | The combustion of a gaseous alkane produces 7 L of carbon dioxide plus water vapour. If the original volume of alkane plus oxygen was 6 L, what is the formula of the alkane? | null |
1,412 | aba958c8-6ddd-11ea-9bc3-ccda262736ce | https://socratic.org/questions/what-mass-in-grams-of-sodium-hydroxide-is-produced-if-20-0-g-of-sodium-metal-rea | 34.78 grams | start physical_unit 5 6 mass g qc_end physical_unit 13 14 10 11 mass qc_end c_other OTHER qc_end chemical_equation 24 33 qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"34.78 grams"}] | [{"type":"physical unit","value":"Mass [OF] sodium metal [=] \\pu{20.0 g}"},{"type":"other","value":"Excess water."},{"type":"chemical equation","value":"2 Na(s) + 2 H2O(l) -> 2 NaOH(aq) + H2(g)"}] | <h1 class="questionTitle" itemprop="name">What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation #2Na(s) 2H_2O(l)->2NaOH(aq)+H_2(g)#?</h1> | null | 34.78 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of gas occupies <mathjax>#24.5#</mathjax> <mathjax>#L#</mathjax> at <mathjax>#298#</mathjax> <mathjax>#K#</mathjax>, and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>. What volume of dihydrogen would be evolved under the above conditions?</p></div>
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<div class="markdown"><p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)#</mathjax></p>
<p>Moles of NaOH <mathjax>#=#</mathjax> <mathjax>#(20.0*g)/(22.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of gas occupies <mathjax>#24.5#</mathjax> <mathjax>#L#</mathjax> at <mathjax>#298#</mathjax> <mathjax>#K#</mathjax>, and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>. What volume of dihydrogen would be evolved under the above conditions?</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation #2Na(s) 2H_2O(l)->2NaOH(aq)+H_2(g)#?</h1>
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<div class="markdown"><p><mathjax>#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)#</mathjax></p>
<p>Moles of NaOH <mathjax>#=#</mathjax> <mathjax>#(20.0*g)/(22.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of gas occupies <mathjax>#24.5#</mathjax> <mathjax>#L#</mathjax> at <mathjax>#298#</mathjax> <mathjax>#K#</mathjax>, and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>. What volume of dihydrogen would be evolved under the above conditions?</p></div>
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</article> | What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation #2Na(s) 2H_2O(l)->2NaOH(aq)+H_2(g)#? | null |
1,413 | ad11b363-6ddd-11ea-81b4-ccda262736ce | https://socratic.org/questions/a-gas-occupies-a-volume-of-2-4-l-at-14-1-kpa-what-volume-will-the-gas-occupy-at- | 0.40 L | start physical_unit 14 15 volume l qc_end physical_unit 14 15 6 7 volume qc_end physical_unit 14 15 9 10 pressure qc_end physical_unit 14 15 18 19 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.40 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2.4 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{14.1 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{84.6 kPa}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa? </h1> | null | 0.40 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.1*kPaxx2.4*L)/(84.6*kPa)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p>
<p>The volume has reduced reasonably, because the pressure has increased almost tenfold. </p></div>
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<div class="markdown"><p>Under <mathjax>#1/2*L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.1*kPaxx2.4*L)/(84.6*kPa)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p>
<p>The volume has reduced reasonably, because the pressure has increased almost tenfold. </p></div>
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<h1 class="questionTitle" itemprop="name">A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa? </h1>
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<div class="markdown"><p>Under <mathjax>#1/2*L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(14.1*kPaxx2.4*L)/(84.6*kPa)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*L#</mathjax></p>
<p>The volume has reduced reasonably, because the pressure has increased almost tenfold. </p></div>
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</article> | A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa? | null |
1,414 | ace05115-6ddd-11ea-9f08-ccda262736ce | https://socratic.org/questions/583048eb7c014916cedd9d5a | 0.05 molar | start physical_unit 13 13 mole mol/l qc_end physical_unit 5 6 1 2 mole qc_end substance 16 16 qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium [IN] molar"}] | [{"type":"physical unit","value":"0.05 molar"}] | [{"type":"physical unit","value":"Mole [OF] hydrogen gas [=] \\pu{0.025 mol}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">An #0.025*mol# quantity of hydrogen gas was collected from the reaction of sodium metal with water. What molar quantity of sodium was used?</h1> | null | 0.05 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For the reaction,</p>
<p><mathjax>#Na(s) + H_2O(l) rarr 1/2H_2(g)uarr + NaOH(aq)#</mathjax>, </p>
<p>we were given a specified molar quantity of reactant.</p>
<p>For the mass we simply mulitply the number of moles by the molar mass, as given in the intro. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#0.05*molxx22.99*g*mol^-1#</mathjax> of sodium metal were reacted. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For the reaction,</p>
<p><mathjax>#Na(s) + H_2O(l) rarr 1/2H_2(g)uarr + NaOH(aq)#</mathjax>, </p>
<p>we were given a specified molar quantity of reactant.</p>
<p>For the mass we simply mulitply the number of moles by the molar mass, as given in the intro. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">An #0.025*mol# quantity of hydrogen gas was collected from the reaction of sodium metal with water. What molar quantity of sodium was used?</h1>
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<div class="markdown"><p><mathjax>#0.05*molxx22.99*g*mol^-1#</mathjax> of sodium metal were reacted. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For the reaction,</p>
<p><mathjax>#Na(s) + H_2O(l) rarr 1/2H_2(g)uarr + NaOH(aq)#</mathjax>, </p>
<p>we were given a specified molar quantity of reactant.</p>
<p>For the mass we simply mulitply the number of moles by the molar mass, as given in the intro. </p></div>
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</article> | An #0.025*mol# quantity of hydrogen gas was collected from the reaction of sodium metal with water. What molar quantity of sodium was used? | null |
1,415 | a8d44f7b-6ddd-11ea-9c35-ccda262736ce | https://socratic.org/questions/5792cab47c01494109296ee1 | 232.3 L | start physical_unit 9 10 volume l qc_end physical_unit 9 10 6 7 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [IN] L"}] | [{"type":"physical unit","value":"232.3 L"}] | [{"type":"physical unit","value":"Mole [OF] nitrogen gas [=] \\pu{10.23 moles}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume occupied by #"10.23"# moles of nitrogen gas at STP?</h1> | null | 232.3 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>STP conditions</strong> are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. The thing to remember about gases kept under STP conditions is that <strong>one mole</strong> of any ideal gas will <em>always</em> occupy <mathjax>#"22.71 L"#</mathjax>. </p>
<p>In other words, if you know how many <em>moles</em> of gas you have under STP conditions, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong>, which is equal to <mathjax>#"22.71 L mol"^(-1)#</mathjax>, as a <strong>conversion factor</strong> to help you find the volume occupied by the sample. </p>
<p>In your case, the sample contains <mathjax>#10.23#</mathjax> <strong>moles</strong> of nitrogen gas kept under STP conditions. This means that the volume of the sample will be </p>
<blockquote>
<p><mathjax>#10.23 color(red)(cancel(color(black)("moles N"_2))) * overbrace("22.71 L"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)color(black)("232.3 L")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>
<p><strong>SIDE NOTE</strong> <em>More often than not, the definition of <strong>STP conditions</strong> will be given to you as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, the <strong>molar volume of a gas</strong> is equal to</em> <mathjax>#"22.423 L mol"^(-1)#</mathjax>. <em>If this is the definition given to you, simply redo the calculation using</em> <mathjax>#"22.423 L"#</mathjax> <em>instead of</em> <mathjax>#"22.71 L"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"232.3 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>STP conditions</strong> are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. The thing to remember about gases kept under STP conditions is that <strong>one mole</strong> of any ideal gas will <em>always</em> occupy <mathjax>#"22.71 L"#</mathjax>. </p>
<p>In other words, if you know how many <em>moles</em> of gas you have under STP conditions, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong>, which is equal to <mathjax>#"22.71 L mol"^(-1)#</mathjax>, as a <strong>conversion factor</strong> to help you find the volume occupied by the sample. </p>
<p>In your case, the sample contains <mathjax>#10.23#</mathjax> <strong>moles</strong> of nitrogen gas kept under STP conditions. This means that the volume of the sample will be </p>
<blockquote>
<p><mathjax>#10.23 color(red)(cancel(color(black)("moles N"_2))) * overbrace("22.71 L"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)color(black)("232.3 L")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>
<p><strong>SIDE NOTE</strong> <em>More often than not, the definition of <strong>STP conditions</strong> will be given to you as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, the <strong>molar volume of a gas</strong> is equal to</em> <mathjax>#"22.423 L mol"^(-1)#</mathjax>. <em>If this is the definition given to you, simply redo the calculation using</em> <mathjax>#"22.423 L"#</mathjax> <em>instead of</em> <mathjax>#"22.71 L"#</mathjax>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the volume occupied by #"10.23"# moles of nitrogen gas at STP?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-24T00:06:46" itemprop="dateCreated">
Jul 24, 2016
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<div class="markdown"><p><mathjax>#"232.3 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>STP conditions</strong> are <em><strong>currently</strong></em> defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. The thing to remember about gases kept under STP conditions is that <strong>one mole</strong> of any ideal gas will <em>always</em> occupy <mathjax>#"22.71 L"#</mathjax>. </p>
<p>In other words, if you know how many <em>moles</em> of gas you have under STP conditions, you can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a></strong>, which is equal to <mathjax>#"22.71 L mol"^(-1)#</mathjax>, as a <strong>conversion factor</strong> to help you find the volume occupied by the sample. </p>
<p>In your case, the sample contains <mathjax>#10.23#</mathjax> <strong>moles</strong> of nitrogen gas kept under STP conditions. This means that the volume of the sample will be </p>
<blockquote>
<p><mathjax>#10.23 color(red)(cancel(color(black)("moles N"_2))) * overbrace("22.71 L"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)color(black)("232.3 L")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>
<p><strong>SIDE NOTE</strong> <em>More often than not, the definition of <strong>STP conditions</strong> will be given to you as a pressure of</em> <mathjax>#"1 atm"#</mathjax> <em>and a temperature of</em> <mathjax>#0^@"C"#</mathjax>. </p>
<p><em>Under these conditions, the <strong>molar volume of a gas</strong> is equal to</em> <mathjax>#"22.423 L mol"^(-1)#</mathjax>. <em>If this is the definition given to you, simply redo the calculation using</em> <mathjax>#"22.423 L"#</mathjax> <em>instead of</em> <mathjax>#"22.71 L"#</mathjax>. </p></div>
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<a href="https://socratic.org/answers/291317" itemprop="url">Answer link</a>
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</article> | What is the volume occupied by #"10.23"# moles of nitrogen gas at STP? | null |
1,416 | abc783f8-6ddd-11ea-99fb-ccda262736ce | https://socratic.org/questions/what-is-the-mole-ratio-of-cl-2-to-br-2-in-the-reaction-cl-2-2nabr-2nacl-br-2 | 1:1 | start physical_unit 6 8 mole_fraction none qc_end chemical_equation 12 20 qc_end end | [{"type":"physical unit","value":"Mole ratio [OF] Cl2 to Br2"}] | [{"type":"physical unit","value":"1:1"}] | [{"type":"chemical equation","value":"Cl2 + 2 NaBr -> 2 NaCl + Br2"}] | <h1 class="questionTitle" itemprop="name">What is the mole ratio of #Cl_2# to #Br_2# in the reaction #Cl_2+2NaBr->2NaCl+Br_2#?</h1> | null | 1:1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is no coefficient in front of the <mathjax>#"Cl"_2"#</mathjax> or the <mathjax>#"Br"_2"#</mathjax>, which is understood to be 1 mole.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio of <mathjax>#"Cl"_2#</mathjax> and <mathjax>#"Br"_2#</mathjax> is <mathjax>#color(blue)("1 mol Cl"_2")#</mathjax><mathjax>#:#</mathjax><mathjax>#color(blue)("1 mol Br"_2")#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is no coefficient in front of the <mathjax>#"Cl"_2"#</mathjax> or the <mathjax>#"Br"_2"#</mathjax>, which is understood to be 1 mole.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the mole ratio of #Cl_2# to #Br_2# in the reaction #Cl_2+2NaBr->2NaCl+Br_2#?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio of <mathjax>#"Cl"_2#</mathjax> and <mathjax>#"Br"_2#</mathjax> is <mathjax>#color(blue)("1 mol Cl"_2")#</mathjax><mathjax>#:#</mathjax><mathjax>#color(blue)("1 mol Br"_2")#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>There is no coefficient in front of the <mathjax>#"Cl"_2"#</mathjax> or the <mathjax>#"Br"_2"#</mathjax>, which is understood to be 1 mole.</p></div>
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</article> | What is the mole ratio of #Cl_2# to #Br_2# in the reaction #Cl_2+2NaBr->2NaCl+Br_2#? | null |
1,417 | abfef812-6ddd-11ea-94c9-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-of-0-250-mol-of-air-contained-in-a-balloon-having-a-volume- | 4.22 atm | start physical_unit 8 8 pressure atm qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 8 8 17 18 volume qc_end physical_unit 8 8 20 21 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] air [IN] atm"}] | [{"type":"physical unit","value":"4.22 atm"}] | [{"type":"physical unit","value":"Mole [OF] air [=] \\pu{0.250 mol}"},{"type":"physical unit","value":"Volume [OF] air [=] \\pu{1.50 L}"},{"type":"physical unit","value":"Temperature [OF] air [=] \\pu{35.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure of 0.250 mol of air contained in a balloon having a volume of 1.50 L at 35.0 °C? </h1> | null | 4.22 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=(0.250*molxx0.0821*(L*atm)/(K*mol)xx308.1*K)/(1.50*L)#</mathjax></p>
<p><mathjax>#=??atm..........#</mathjax></p></div>
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<div class="markdown"><p>Well, if we assume ideal behaviour...........I get a pressure of under <mathjax>#5*atm#</mathjax>. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=(0.250*molxx0.0821*(L*atm)/(K*mol)xx308.1*K)/(1.50*L)#</mathjax></p>
<p><mathjax>#=??atm..........#</mathjax></p></div>
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<div class="markdown"><p>Well, if we assume ideal behaviour...........I get a pressure of under <mathjax>#5*atm#</mathjax>. </p></div>
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<div class="markdown"><p><mathjax>#P=(nRT)/V=(0.250*molxx0.0821*(L*atm)/(K*mol)xx308.1*K)/(1.50*L)#</mathjax></p>
<p><mathjax>#=??atm..........#</mathjax></p></div>
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</article> | What is the pressure of 0.250 mol of air contained in a balloon having a volume of 1.50 L at 35.0 °C? | null |
1,418 | a9739d51-6ddd-11ea-9050-ccda262736ce | https://socratic.org/questions/what-is-the-total-number-of-moles-of-water-needed-to-make-2-5-moles-of-c-6h-12o- | 15.00 moles | start physical_unit 8 8 mole mol qc_end physical_unit 15 15 12 13 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] water [IN] moles"}] | [{"type":"physical unit","value":"15.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] C6H12O6 [=] \\pu{2.5 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the total number of moles of water needed to make 2.5 moles of #C_6H_12O_6#?</h1> | null | 15.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole carbohydrate contains 6 moles water<br/>
so 2.5 moles contain 6x2.5=15moles</p></div>
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<div class="markdown"><p>as per the formula of given carbohydrate the answer is 15 moles</p></div>
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<div class="markdown"><p>1 mole carbohydrate contains 6 moles water<br/>
so 2.5 moles contain 6x2.5=15moles</p></div>
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<div class="markdown"><p>as per the formula of given carbohydrate the answer is 15 moles</p></div>
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<div class="markdown"><p>1 mole carbohydrate contains 6 moles water<br/>
so 2.5 moles contain 6x2.5=15moles</p></div>
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</article> | What is the total number of moles of water needed to make 2.5 moles of #C_6H_12O_6#? | null |
1,419 | acd52db0-6ddd-11ea-a4dc-ccda262736ce | https://socratic.org/questions/if-the-cr-3-concentration-in-a-saturation-solution-of-cr-oh-3-is-4-0x10-6-m-how- | 6.91 × 10^(-21) | start physical_unit 9 9 equilibrium_constant_k none qc_end physical_unit 2 2 11 14 concentration qc_end end | [{"type":"physical unit","value":"Ksp [OF] Cr(OH)3"}] | [{"type":"physical unit","value":"6.91 × 10^(-21)"}] | [{"type":"physical unit","value":"Concentration [OF] Cr^3+ [=] \\pu{4.0 × 10^(-6) M}"}] | <h1 class="questionTitle" itemprop="name">If the #Cr_3^+# concentration in a saturation solution of #Cr(OH)_3# is #4.0xx10^-6 M#. How do you calculate the Ksp?</h1> | null | 6.91 × 10^(-21) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Chromium(III) hydroxide</em>, <mathjax>#"Cr"("OH")_3#</mathjax>, is made up of chromium(III) cations, <mathjax>#"Cr"^(3+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Chromium(III) hydroxide is considered to be <strong>insoluble</strong> in water, which means that when the salt dissolves, only very, very small amounts will actually dissociate to produce ions in solution.</p>
<blockquote>
<p><mathjax>#"Cr"("OH")_ (color(red)(3)(s)) rightleftharpoons "Cr"_ ((aq))^(3+) + color(red)(3)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of chromium(III) hydroxide that <em>dissociates</em> you get <mathjax>#1#</mathjax> <strong>mole</strong> of chromium cations and <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>This means that at equilibrium, an aqueous solution of chromium(III) hydroxide will contain <strong>three times as many moles</strong> of hydroxide anions than of chromium cations. </p>
<p>Therefore, for any concentration of chromium(III) cations you have in solution, you will also have </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx ["Cr"^(3+)]#</mathjax></p>
</blockquote>
<p>In your case, you know that </p>
<blockquote>
<p><mathjax>#["Cr"^(3+)] = 4.0 * 10^(-6)"M"#</mathjax></p>
</blockquote>
<p>which means that this <em>saturated solution</em> contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx 4.0 * 10^(-6)"M" = 1.2 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em> for this dissociation equilibrium is equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Cr"^(3+)] * ["OH"^(-)]^color(red)(3)#</mathjax></p>
</blockquote>
<p>Plug in your values to find </p>
<blockquote>
<p><mathjax>#K_(sp) = 4.0 * 10^(-6)"M" * (1.2 * 10^(-5)"M")^color(red)(3)#</mathjax></p>
<p><mathjax>#K_(sp) = 6.9 * 10^(-21)"M"^4#</mathjax></p>
</blockquote>
<p>You'll usually see the <mathjax>#K_(sp)#</mathjax> expressed <em>without added units</em>, so you can say that you have </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.9 * 10^(-21))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_(sp) = 6.9 * 10^(-21)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Chromium(III) hydroxide</em>, <mathjax>#"Cr"("OH")_3#</mathjax>, is made up of chromium(III) cations, <mathjax>#"Cr"^(3+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Chromium(III) hydroxide is considered to be <strong>insoluble</strong> in water, which means that when the salt dissolves, only very, very small amounts will actually dissociate to produce ions in solution.</p>
<blockquote>
<p><mathjax>#"Cr"("OH")_ (color(red)(3)(s)) rightleftharpoons "Cr"_ ((aq))^(3+) + color(red)(3)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of chromium(III) hydroxide that <em>dissociates</em> you get <mathjax>#1#</mathjax> <strong>mole</strong> of chromium cations and <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>This means that at equilibrium, an aqueous solution of chromium(III) hydroxide will contain <strong>three times as many moles</strong> of hydroxide anions than of chromium cations. </p>
<p>Therefore, for any concentration of chromium(III) cations you have in solution, you will also have </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx ["Cr"^(3+)]#</mathjax></p>
</blockquote>
<p>In your case, you know that </p>
<blockquote>
<p><mathjax>#["Cr"^(3+)] = 4.0 * 10^(-6)"M"#</mathjax></p>
</blockquote>
<p>which means that this <em>saturated solution</em> contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx 4.0 * 10^(-6)"M" = 1.2 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em> for this dissociation equilibrium is equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Cr"^(3+)] * ["OH"^(-)]^color(red)(3)#</mathjax></p>
</blockquote>
<p>Plug in your values to find </p>
<blockquote>
<p><mathjax>#K_(sp) = 4.0 * 10^(-6)"M" * (1.2 * 10^(-5)"M")^color(red)(3)#</mathjax></p>
<p><mathjax>#K_(sp) = 6.9 * 10^(-21)"M"^4#</mathjax></p>
</blockquote>
<p>You'll usually see the <mathjax>#K_(sp)#</mathjax> expressed <em>without added units</em>, so you can say that you have </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.9 * 10^(-21))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the #Cr_3^+# concentration in a saturation solution of #Cr(OH)_3# is #4.0xx10^-6 M#. How do you calculate the Ksp?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#K_(sp) = 6.9 * 10^(-21)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Chromium(III) hydroxide</em>, <mathjax>#"Cr"("OH")_3#</mathjax>, is made up of chromium(III) cations, <mathjax>#"Cr"^(3+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>Chromium(III) hydroxide is considered to be <strong>insoluble</strong> in water, which means that when the salt dissolves, only very, very small amounts will actually dissociate to produce ions in solution.</p>
<blockquote>
<p><mathjax>#"Cr"("OH")_ (color(red)(3)(s)) rightleftharpoons "Cr"_ ((aq))^(3+) + color(red)(3)"OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of chromium(III) hydroxide that <em>dissociates</em> you get <mathjax>#1#</mathjax> <strong>mole</strong> of chromium cations and <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of hydroxide anions. </p>
<p>This means that at equilibrium, an aqueous solution of chromium(III) hydroxide will contain <strong>three times as many moles</strong> of hydroxide anions than of chromium cations. </p>
<p>Therefore, for any concentration of chromium(III) cations you have in solution, you will also have </p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx ["Cr"^(3+)]#</mathjax></p>
</blockquote>
<p>In your case, you know that </p>
<blockquote>
<p><mathjax>#["Cr"^(3+)] = 4.0 * 10^(-6)"M"#</mathjax></p>
</blockquote>
<p>which means that this <em>saturated solution</em> contains</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = color(red)(3) xx 4.0 * 10^(-6)"M" = 1.2 * 10^(-5)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em> for this dissociation equilibrium is equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Cr"^(3+)] * ["OH"^(-)]^color(red)(3)#</mathjax></p>
</blockquote>
<p>Plug in your values to find </p>
<blockquote>
<p><mathjax>#K_(sp) = 4.0 * 10^(-6)"M" * (1.2 * 10^(-5)"M")^color(red)(3)#</mathjax></p>
<p><mathjax>#K_(sp) = 6.9 * 10^(-21)"M"^4#</mathjax></p>
</blockquote>
<p>You'll usually see the <mathjax>#K_(sp)#</mathjax> expressed <em>without added units</em>, so you can say that you have </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.9 * 10^(-21))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If the #Cr_3^+# concentration in a saturation solution of #Cr(OH)_3# is #4.0xx10^-6 M#. How do you calculate the Ksp? | null |
1,420 | abd557e8-6ddd-11ea-ab78-ccda262736ce | https://socratic.org/questions/how-would-you-determine-the-molecular-formula-of-a-substance-that-contains-92-3- | C8H8 | start chemical_formula qc_end physical_unit 9 9 31 32 density qc_end c_other STP qc_end end | [{"type":"other","value":"Chemical Formula [OF] the substance [IN] molecular"}] | [{"type":"chemical equation","value":"C8H8"}] | [{"type":"physical unit","value":"Percent by mass [OF] C in the substance [=] \\pu{92.3%}"},{"type":"physical unit","value":"Percent by mass [OF] H in the substance [=] \\pu{7.7%}"},{"type":"physical unit","value":"Density [OF] the substance [=] \\pu{0.00454 g/ml}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How would you determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00454g/ml at stp conditions?</h1> | null | C8H8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<ul>
<li><em>use a sample of this compound to help you determine its <strong>empirical formula</strong></em></li>
<li><em>use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, and the known STP conditions for pressure and temperature to find th gas' <strong>molar mass</strong></em></li>
<li><em>use the empirical formula and the gas' molar mass to find its <strong>molecular formula</strong></em></li>
</ul>
<p>So, you know that you're dealing with a <strong>hydrocarbon</strong> that contains <mathjax>#92.3%#</mathjax> carbon and <mathjax>#7.7%#</mathjax> hydrogen by mass. To make the calculations easier, select a <mathjax>#"100-g"#</mathjax> sample of this compound. </p>
<p>You know from the aforementioned <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> that this sample will contain <mathjax>#"92.3 g"#</mathjax> of carbon and <mathjax>#"7.7 g"#</mathjax> of hydrogen. </p>
<p>Use the molar masses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to figure out how many <em>moles</em> of each you'd get in this sample</p>
<blockquote>
<p><mathjax>#92.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "7.6846 moles C"#</mathjax></p>
<p><mathjax>#7.7 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "7.6393 moles H"#</mathjax></p>
</blockquote>
<p>Divide both values by the <em>smallest one</em> to get the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the compound</p>
<blockquote>
<p><mathjax>#"For C: " (7.6846 color(red)(cancel(color(black)("moles"))))/(7.6393color(red)(cancel(color(black)("moles")))) = 1.006 ~~1#</mathjax></p>
<p><mathjax>#"For H: " (0.76393color(red)(cancel(color(black)("moles"))))/(0.76393color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>The <strong>empirical formula</strong> for this hydrocarbon is </p>
<blockquote>
<p><mathjax>#"C"_1"H"_1 implies "CH"#</mathjax></p>
</blockquote>
<p>Now, <strong>STP conditions</strong> are characterized by a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Keep this in mind. </p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>can be rewritten using the definition of <mathjax>#n#</mathjax>, the number of moles of gas. As you know, the number of moles if equal to the ratio between the <strong>mass</strong> and the <strong>molar mass</strong> of a compound</p>
<blockquote>
<p><mathjax>#n = m/M_M#</mathjax></p>
</blockquote>
<p>Plug this into the ideal gas law equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get <mathjax>#M_M#</mathjax> isolated on one side of the equation</p>
<blockquote>
<p><mathjax>#M_M = overbrace(m/V)^(color(blue)(=rho)) * (RT)/P#</mathjax></p>
</blockquote>
<p>Since density is defined as mass per unit of volume, you can say that </p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Plug in the STP pressure and temperature and solve for <mathjax>#M_M#</mathjax> - <strong>do not</strong> forget to convert the pressure from <em>kPa</em> to <em>atm</em> and the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>! </p>
<p>Also, convert the density of the gas from <em>grams per milliliter</em> to <em>grams per liter</em></p>
<blockquote>
<p><mathjax>#0.00454 "g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "4.54 g/L"#</mathjax></p>
<p><mathjax>#M_M = 4.54"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "103.2 g/mol"#</mathjax></p>
</blockquote>
<p>Now, the <strong>molecular formula</strong> will always be a multiple of the <em>empirical formula</em>. The molar mass of the empirical formula is </p>
<blockquote>
<p><mathjax>#1 xx "12./011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#13.019 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 103.2 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This will get you</p>
<blockquote>
<p><mathjax>#color(blue)(n) = 103.2/13.019 = 7.93 ~~ 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon will be </p>
<blockquote>
<p><mathjax>#("CH")_color(blue)(8) implies color(green)("C"_8"H"_8)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_8"H"_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<ul>
<li><em>use a sample of this compound to help you determine its <strong>empirical formula</strong></em></li>
<li><em>use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, and the known STP conditions for pressure and temperature to find th gas' <strong>molar mass</strong></em></li>
<li><em>use the empirical formula and the gas' molar mass to find its <strong>molecular formula</strong></em></li>
</ul>
<p>So, you know that you're dealing with a <strong>hydrocarbon</strong> that contains <mathjax>#92.3%#</mathjax> carbon and <mathjax>#7.7%#</mathjax> hydrogen by mass. To make the calculations easier, select a <mathjax>#"100-g"#</mathjax> sample of this compound. </p>
<p>You know from the aforementioned <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> that this sample will contain <mathjax>#"92.3 g"#</mathjax> of carbon and <mathjax>#"7.7 g"#</mathjax> of hydrogen. </p>
<p>Use the molar masses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to figure out how many <em>moles</em> of each you'd get in this sample</p>
<blockquote>
<p><mathjax>#92.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "7.6846 moles C"#</mathjax></p>
<p><mathjax>#7.7 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "7.6393 moles H"#</mathjax></p>
</blockquote>
<p>Divide both values by the <em>smallest one</em> to get the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the compound</p>
<blockquote>
<p><mathjax>#"For C: " (7.6846 color(red)(cancel(color(black)("moles"))))/(7.6393color(red)(cancel(color(black)("moles")))) = 1.006 ~~1#</mathjax></p>
<p><mathjax>#"For H: " (0.76393color(red)(cancel(color(black)("moles"))))/(0.76393color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>The <strong>empirical formula</strong> for this hydrocarbon is </p>
<blockquote>
<p><mathjax>#"C"_1"H"_1 implies "CH"#</mathjax></p>
</blockquote>
<p>Now, <strong>STP conditions</strong> are characterized by a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Keep this in mind. </p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>can be rewritten using the definition of <mathjax>#n#</mathjax>, the number of moles of gas. As you know, the number of moles if equal to the ratio between the <strong>mass</strong> and the <strong>molar mass</strong> of a compound</p>
<blockquote>
<p><mathjax>#n = m/M_M#</mathjax></p>
</blockquote>
<p>Plug this into the ideal gas law equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get <mathjax>#M_M#</mathjax> isolated on one side of the equation</p>
<blockquote>
<p><mathjax>#M_M = overbrace(m/V)^(color(blue)(=rho)) * (RT)/P#</mathjax></p>
</blockquote>
<p>Since density is defined as mass per unit of volume, you can say that </p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Plug in the STP pressure and temperature and solve for <mathjax>#M_M#</mathjax> - <strong>do not</strong> forget to convert the pressure from <em>kPa</em> to <em>atm</em> and the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>! </p>
<p>Also, convert the density of the gas from <em>grams per milliliter</em> to <em>grams per liter</em></p>
<blockquote>
<p><mathjax>#0.00454 "g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "4.54 g/L"#</mathjax></p>
<p><mathjax>#M_M = 4.54"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "103.2 g/mol"#</mathjax></p>
</blockquote>
<p>Now, the <strong>molecular formula</strong> will always be a multiple of the <em>empirical formula</em>. The molar mass of the empirical formula is </p>
<blockquote>
<p><mathjax>#1 xx "12./011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#13.019 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 103.2 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This will get you</p>
<blockquote>
<p><mathjax>#color(blue)(n) = 103.2/13.019 = 7.93 ~~ 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon will be </p>
<blockquote>
<p><mathjax>#("CH")_color(blue)(8) implies color(green)("C"_8"H"_8)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00454g/ml at stp conditions?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-23T23:06:25" itemprop="dateCreated">
Dec 23, 2015
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<div class="answerContainer clearfix">
<div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_8"H"_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to</p>
<ul>
<li><em>use a sample of this compound to help you determine its <strong>empirical formula</strong></em></li>
<li><em>use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, the gas' <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, and the known STP conditions for pressure and temperature to find th gas' <strong>molar mass</strong></em></li>
<li><em>use the empirical formula and the gas' molar mass to find its <strong>molecular formula</strong></em></li>
</ul>
<p>So, you know that you're dealing with a <strong>hydrocarbon</strong> that contains <mathjax>#92.3%#</mathjax> carbon and <mathjax>#7.7%#</mathjax> hydrogen by mass. To make the calculations easier, select a <mathjax>#"100-g"#</mathjax> sample of this compound. </p>
<p>You know from the aforementioned <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> that this sample will contain <mathjax>#"92.3 g"#</mathjax> of carbon and <mathjax>#"7.7 g"#</mathjax> of hydrogen. </p>
<p>Use the molar masses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to figure out how many <em>moles</em> of each you'd get in this sample</p>
<blockquote>
<p><mathjax>#92.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "7.6846 moles C"#</mathjax></p>
<p><mathjax>#7.7 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "7.6393 moles H"#</mathjax></p>
</blockquote>
<p>Divide both values by the <em>smallest one</em> to get the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the compound</p>
<blockquote>
<p><mathjax>#"For C: " (7.6846 color(red)(cancel(color(black)("moles"))))/(7.6393color(red)(cancel(color(black)("moles")))) = 1.006 ~~1#</mathjax></p>
<p><mathjax>#"For H: " (0.76393color(red)(cancel(color(black)("moles"))))/(0.76393color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>The <strong>empirical formula</strong> for this hydrocarbon is </p>
<blockquote>
<p><mathjax>#"C"_1"H"_1 implies "CH"#</mathjax></p>
</blockquote>
<p>Now, <strong>STP conditions</strong> are characterized by a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. Keep this in mind. </p>
<p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation </p>
<blockquote>
<p><mathjax>#color(blue)(PV = nRT)#</mathjax></p>
</blockquote>
<p>can be rewritten using the definition of <mathjax>#n#</mathjax>, the number of moles of gas. As you know, the number of moles if equal to the ratio between the <strong>mass</strong> and the <strong>molar mass</strong> of a compound</p>
<blockquote>
<p><mathjax>#n = m/M_M#</mathjax></p>
</blockquote>
<p>Plug this into the ideal gas law equation to get</p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get <mathjax>#M_M#</mathjax> isolated on one side of the equation</p>
<blockquote>
<p><mathjax>#M_M = overbrace(m/V)^(color(blue)(=rho)) * (RT)/P#</mathjax></p>
</blockquote>
<p>Since density is defined as mass per unit of volume, you can say that </p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Plug in the STP pressure and temperature and solve for <mathjax>#M_M#</mathjax> - <strong>do not</strong> forget to convert the pressure from <em>kPa</em> to <em>atm</em> and the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>! </p>
<p>Also, convert the density of the gas from <em>grams per milliliter</em> to <em>grams per liter</em></p>
<blockquote>
<p><mathjax>#0.00454 "g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "4.54 g/L"#</mathjax></p>
<p><mathjax>#M_M = 4.54"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = "103.2 g/mol"#</mathjax></p>
</blockquote>
<p>Now, the <strong>molecular formula</strong> will always be a multiple of the <em>empirical formula</em>. The molar mass of the empirical formula is </p>
<blockquote>
<p><mathjax>#1 xx "12./011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#13.019 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 103.2 color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>This will get you</p>
<blockquote>
<p><mathjax>#color(blue)(n) = 103.2/13.019 = 7.93 ~~ 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon will be </p>
<blockquote>
<p><mathjax>#("CH")_color(blue)(8) implies color(green)("C"_8"H"_8)#</mathjax></p>
</blockquote></div>
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</article> | How would you determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00454g/ml at stp conditions? | null |
1,421 | ac073826-6ddd-11ea-9e50-ccda262736ce | https://socratic.org/questions/what-is-the-volume-in-liters-of-a-3-00-mole-sample-of-carbon-dioxide-at-stp | 67.2 liters | start physical_unit 10 13 volume l qc_end physical_unit 10 13 8 9 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon dioxide sample [IN] liters"}] | [{"type":"physical unit","value":"67.2 liters"}] | [{"type":"physical unit","value":"Mole [OF] carbon dioxide sample [=] \\pu{3.00 mole}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume, in liters, of a 3.00 mole sample of carbon dioxide at STP? </h1> | null | 67.2 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, we treat carbon dioxide an an ideal gas:</p>
<p><mathjax>#3#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4#</mathjax> <mathjax>#L*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p>One mole of ideal gas occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> volume at STP.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, we treat carbon dioxide an an ideal gas:</p>
<p><mathjax>#3#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4#</mathjax> <mathjax>#L*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p>One mole of ideal gas occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax> volume at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, we treat carbon dioxide an an ideal gas:</p>
<p><mathjax>#3#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4#</mathjax> <mathjax>#L*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</article> | What is the volume, in liters, of a 3.00 mole sample of carbon dioxide at STP? | null |
1,422 | a92e2c36-6ddd-11ea-94a6-ccda262736ce | https://socratic.org/questions/how-many-sodium-atoms-are-present-in-7na-3po-4 | 1.26 × 10^25 | start physical_unit 2 3 number none qc_end end | [{"type":"physical unit","value":"Number [OF] sodium atoms"}] | [{"type":"physical unit","value":"1.26 × 10^25"}] | [{"type":"physical unit","value":"Mole [OF] Na3PO4 [=] \\pu{7 mol}"}] | <h1 class="questionTitle" itemprop="name">How many sodium atoms are present in #7Na_3PO_4#?</h1> | null | 1.26 × 10^25 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I can count <mathjax>#3#</mathjax> in the one formula unit. In one mole of compound, there are also <mathjax>#3*mol#</mathjax> of sodium.</p>
<p>Now you have quoted <mathjax>#7*Na_3PO_3#</mathjax>, i.e. <mathjax>#7*"formula units"#</mathjax> OR <mathjax>#7*"moles"#</mathjax> of sodium phosphate. And thus <mathjax>#3xx7=??#</mathjax> <mathjax>#"how many sodium atoms or how many moles of sodium atoms?"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, how many sodium atoms in one formula unit of <mathjax>#Na_3PO_4#</mathjax>?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I can count <mathjax>#3#</mathjax> in the one formula unit. In one mole of compound, there are also <mathjax>#3*mol#</mathjax> of sodium.</p>
<p>Now you have quoted <mathjax>#7*Na_3PO_3#</mathjax>, i.e. <mathjax>#7*"formula units"#</mathjax> OR <mathjax>#7*"moles"#</mathjax> of sodium phosphate. And thus <mathjax>#3xx7=??#</mathjax> <mathjax>#"how many sodium atoms or how many moles of sodium atoms?"#</mathjax></p></div>
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<div class="markdown"><p>Well, how many sodium atoms in one formula unit of <mathjax>#Na_3PO_4#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I can count <mathjax>#3#</mathjax> in the one formula unit. In one mole of compound, there are also <mathjax>#3*mol#</mathjax> of sodium.</p>
<p>Now you have quoted <mathjax>#7*Na_3PO_3#</mathjax>, i.e. <mathjax>#7*"formula units"#</mathjax> OR <mathjax>#7*"moles"#</mathjax> of sodium phosphate. And thus <mathjax>#3xx7=??#</mathjax> <mathjax>#"how many sodium atoms or how many moles of sodium atoms?"#</mathjax></p></div>
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</article> | How many sodium atoms are present in #7Na_3PO_4#? | null |
1,423 | a95b9f9e-6ddd-11ea-874e-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-3-01-10-23-formula-units-of-cacl-2 | 55.48 grams | start physical_unit 11 11 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] CaCl2 [IN] grams"}] | [{"type":"physical unit","value":"55.48 grams"}] | [{"type":"physical unit","value":"Number [OF] CaCl2 formula units [=] \\pu{3.01 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in #3.01*10^23# formula units of #CaCl_2#?</h1> | null | 55.48 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you need to do here is use <strong>Avogadro's number</strong> to calculate how many <em>moles</em> of calcium chloride, <mathjax>#"CaCl"_2#</mathjax>, are equivalent to that many formula units of this salt. </p>
<p>Once you know how many <em>moles</em> you have, you can use calcium chloride's <strong>molar mass</strong> to convert those to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which essentially acts as the definition of a <strong>mole</strong>, tells you that <strong>one mole</strong> of any ionic compound contains <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of that compound. </p>
<p>As you can see, the sample given to you contains approximately <strong>half a mole</strong> of calcium chloride, since</p>
<blockquote>
<p><mathjax>#3.01 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2))) * "1 mole CaCl"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = "0.4998 moles CaCl"_2#</mathjax></p>
</blockquote>
<p>Now, calcium chloride has a molar mass of approximately <mathjax>#"111 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of calcium chloride has a mass of <mathjax>#"111 g"#</mathjax>. </p>
<p>Your sample will thus be equivalent to </p>
<blockquote>
<p><mathjax>#0.4998 color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("55.5 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>#"55.5 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you need to do here is use <strong>Avogadro's number</strong> to calculate how many <em>moles</em> of calcium chloride, <mathjax>#"CaCl"_2#</mathjax>, are equivalent to that many formula units of this salt. </p>
<p>Once you know how many <em>moles</em> you have, you can use calcium chloride's <strong>molar mass</strong> to convert those to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which essentially acts as the definition of a <strong>mole</strong>, tells you that <strong>one mole</strong> of any ionic compound contains <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of that compound. </p>
<p>As you can see, the sample given to you contains approximately <strong>half a mole</strong> of calcium chloride, since</p>
<blockquote>
<p><mathjax>#3.01 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2))) * "1 mole CaCl"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = "0.4998 moles CaCl"_2#</mathjax></p>
</blockquote>
<p>Now, calcium chloride has a molar mass of approximately <mathjax>#"111 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of calcium chloride has a mass of <mathjax>#"111 g"#</mathjax>. </p>
<p>Your sample will thus be equivalent to </p>
<blockquote>
<p><mathjax>#0.4998 color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("55.5 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams are in #3.01*10^23# formula units of #CaCl_2#?</h1>
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Stefan V.
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Aug 21, 2016
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<div class="markdown"><p><mathjax>#"55.5 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you need to do here is use <strong>Avogadro's number</strong> to calculate how many <em>moles</em> of calcium chloride, <mathjax>#"CaCl"_2#</mathjax>, are equivalent to that many formula units of this salt. </p>
<p>Once you know how many <em>moles</em> you have, you can use calcium chloride's <strong>molar mass</strong> to convert those to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which essentially acts as the definition of a <strong>mole</strong>, tells you that <strong>one mole</strong> of any ionic compound contains <mathjax>#6.022 * 10^(23)#</mathjax> <strong>formula units</strong> of that compound. </p>
<p>As you can see, the sample given to you contains approximately <strong>half a mole</strong> of calcium chloride, since</p>
<blockquote>
<p><mathjax>#3.01 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2))) * "1 mole CaCl"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = "0.4998 moles CaCl"_2#</mathjax></p>
</blockquote>
<p>Now, calcium chloride has a molar mass of approximately <mathjax>#"111 g mol"^(-1)#</mathjax>, which means that <strong>every mole</strong> of calcium chloride has a mass of <mathjax>#"111 g"#</mathjax>. </p>
<p>Your sample will thus be equivalent to </p>
<blockquote>
<p><mathjax>#0.4998 color(red)(cancel(color(black)("moles CaCl"_2))) * "111 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("55.5 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many grams are in #3.01*10^23# formula units of #CaCl_2#? | null |
1,424 | a8934b6c-6ddd-11ea-a17b-ccda262736ce | https://socratic.org/questions/a-11-04g-sample-of-a-hydrocarbon-produces-34-71g-of-co2-and-14-20g-h2o-what-is-e | CH2 | start chemical_formula qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 15 15 13 14 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] hydrocarbon [IN] empirical"}] | [{"type":"chemical equation","value":"CH2"}] | [{"type":"physical unit","value":"Mass [OF] hydrocarbon sample [=] \\pu{11.04 g}"},{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{34.71 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{14.20 g}"}] | <h1 class="questionTitle" itemprop="name">A 11.04g sample of a hydrocarbon produces 34.71g of CO2 and 14.20g H2O. What is empirical formula of hydrocarbon?</h1> | null | CH2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to realize that you're dealing with a <em>hydrocarbon</em>, that is, a compound that <strong>contains only</strong> carbon and hydrogen. </p>
<p>Notice that the products of this combustion reaction are carbon dioxide, <mathjax>#"CO"_2#</mathjax>, and water, <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>This tells you that <strong>all the carbon</strong> that was initially a part of the hydrocarbon will now be part of the carbon dioxide. Likewise, <strong>all the hydrogen</strong> that was initially a part of the hydrocarbon is now a part of the water.</p>
<p>This means that you can use the number of moles of water and carbon dioxide, respectively, to determine how many moles of carbon and of hydrogen were originally present in the hdyrocarbon. </p>
<p>So, for water you have</p>
<blockquote>
<p><mathjax>#14.20color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.78823 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>and for carbon dioxide</p>
<blockquote>
<p><mathjax>#34.71color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.78868 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that <em>every mole</em> of water contains <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen, which means that the reaction produced </p>
<blockquote>
<p><mathjax>#0.78823color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "1.5765 moles H"#</mathjax></p>
</blockquote>
<p>SInce <em>every mole</em> of carbon dioxide contains <strong>1 mole</strong> of carbon and <strong>2 moles</strong> of oxygen, it follows that the reaction also produced</p>
<blockquote>
<p><mathjax>#0.78868color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.78868 moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between carbon and hydrogen in the hydrocarbon, divide these values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For C: " (0.78868color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.5765color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1.999 ~~ 2#</mathjax></p>
</blockquote>
<p>The empirical formula of the hydrocarbon will thus be </p>
<blockquote>
<p><mathjax>#"C"_1"H"_2 implies "CH"_2#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/EtS0oPmPzTs?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>#"CH"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to realize that you're dealing with a <em>hydrocarbon</em>, that is, a compound that <strong>contains only</strong> carbon and hydrogen. </p>
<p>Notice that the products of this combustion reaction are carbon dioxide, <mathjax>#"CO"_2#</mathjax>, and water, <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>This tells you that <strong>all the carbon</strong> that was initially a part of the hydrocarbon will now be part of the carbon dioxide. Likewise, <strong>all the hydrogen</strong> that was initially a part of the hydrocarbon is now a part of the water.</p>
<p>This means that you can use the number of moles of water and carbon dioxide, respectively, to determine how many moles of carbon and of hydrogen were originally present in the hdyrocarbon. </p>
<p>So, for water you have</p>
<blockquote>
<p><mathjax>#14.20color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.78823 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>and for carbon dioxide</p>
<blockquote>
<p><mathjax>#34.71color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.78868 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that <em>every mole</em> of water contains <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen, which means that the reaction produced </p>
<blockquote>
<p><mathjax>#0.78823color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "1.5765 moles H"#</mathjax></p>
</blockquote>
<p>SInce <em>every mole</em> of carbon dioxide contains <strong>1 mole</strong> of carbon and <strong>2 moles</strong> of oxygen, it follows that the reaction also produced</p>
<blockquote>
<p><mathjax>#0.78868color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.78868 moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between carbon and hydrogen in the hydrocarbon, divide these values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For C: " (0.78868color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.5765color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1.999 ~~ 2#</mathjax></p>
</blockquote>
<p>The empirical formula of the hydrocarbon will thus be </p>
<blockquote>
<p><mathjax>#"C"_1"H"_2 implies "CH"_2#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/EtS0oPmPzTs?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">A 11.04g sample of a hydrocarbon produces 34.71g of CO2 and 14.20g H2O. What is empirical formula of hydrocarbon?</h1>
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Stefan V.
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Oct 31, 2015
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<div class="markdown"><p><mathjax>#"CH"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key here is to realize that you're dealing with a <em>hydrocarbon</em>, that is, a compound that <strong>contains only</strong> carbon and hydrogen. </p>
<p>Notice that the products of this combustion reaction are carbon dioxide, <mathjax>#"CO"_2#</mathjax>, and water, <mathjax>#"H"_2"O"#</mathjax>.</p>
<p>This tells you that <strong>all the carbon</strong> that was initially a part of the hydrocarbon will now be part of the carbon dioxide. Likewise, <strong>all the hydrogen</strong> that was initially a part of the hydrocarbon is now a part of the water.</p>
<p>This means that you can use the number of moles of water and carbon dioxide, respectively, to determine how many moles of carbon and of hydrogen were originally present in the hdyrocarbon. </p>
<p>So, for water you have</p>
<blockquote>
<p><mathjax>#14.20color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.78823 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>and for carbon dioxide</p>
<blockquote>
<p><mathjax>#34.71color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.78868 moles CO"_2#</mathjax></p>
</blockquote>
<p>Now, you know that <em>every mole</em> of water contains <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen, which means that the reaction produced </p>
<blockquote>
<p><mathjax>#0.78823color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "1.5765 moles H"#</mathjax></p>
</blockquote>
<p>SInce <em>every mole</em> of carbon dioxide contains <strong>1 mole</strong> of carbon and <strong>2 moles</strong> of oxygen, it follows that the reaction also produced</p>
<blockquote>
<p><mathjax>#0.78868color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.78868 moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that exists between carbon and hydrogen in the hydrocarbon, divide these values by the <em>smallest one</em> </p>
<blockquote>
<p><mathjax>#"For C: " (0.78868color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For H: " (1.5765color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1.999 ~~ 2#</mathjax></p>
</blockquote>
<p>The empirical formula of the hydrocarbon will thus be </p>
<blockquote>
<p><mathjax>#"C"_1"H"_2 implies "CH"_2#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/EtS0oPmPzTs?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | A 11.04g sample of a hydrocarbon produces 34.71g of CO2 and 14.20g H2O. What is empirical formula of hydrocarbon? | null |
1,425 | ab44909c-6ddd-11ea-82e2-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-chemical-equation-for-methane-ch-4-reacting-with-oxy | CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g) | start chemical_equation qc_end chemical_equation 10 10 qc_end substance 13 14 qc_end substance 17 17 qc_end substance 19 20 qc_end end | [{"type":"other","value":"Chemical Equation [OF] balanced chemical equation"}] | [{"type":"chemical equation","value":"CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)"}] | [{"type":"chemical equation","value":"CH4"},{"type":"substance name","value":"Oxygen gas"},{"type":"substance name","value":"Water"},{"type":"substance name","value":"Carbon Dioxide"}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced chemical equation for methane, #CH_4# reacting with oxygen gas to produce water and Carbon Dioxide?</h1> | null | CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this balanced? How do you know? How does energy transfer in this reaction?</p></div>
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<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is this balanced? How do you know? How does energy transfer in this reaction?</p></div>
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<h1 class="questionTitle" itemprop="name">How do you write a balanced chemical equation for methane, #CH_4# reacting with oxygen gas to produce water and Carbon Dioxide?</h1>
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<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) +2H_2O(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is this balanced? How do you know? How does energy transfer in this reaction?</p></div>
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</article> | How do you write a balanced chemical equation for methane, #CH_4# reacting with oxygen gas to produce water and Carbon Dioxide? | null |
1,426 | aa5f39b4-6ddd-11ea-bcd8-ccda262736ce | https://socratic.org/questions/5647929611ef6b5c1ccd6dcf | 1.67 atm | start physical_unit 8 8 pressure atm qc_end physical_unit 8 8 4 5 volume qc_end physical_unit 8 8 10 11 pressure qc_end physical_unit 8 8 16 17 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"1.67 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{5.00 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1 atm}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{3.00 L}"},{"type":"other","value":"Constant temperature."}] | <h1 class="questionTitle" itemprop="name">Under constant temperature, a #5.00*L# volume of gas at #1*atm# pressure is compressed to #3.00*L#. What is the new pressure?</h1> | null | 1.67 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5.00L)/(3.00L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5/3atm)#</mathjax>.</p>
<p>Here I have used non-standard units. I am free to do so as long as I do so consistently. (Note that a pressure measurement of 760 mm Hg is equivalent to 1 atmosphere).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_1V_1 = P_2V_2#</mathjax> (Boyle's law, pressure is inversely proportional to volume at constant temperature.) </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5.00L)/(3.00L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5/3atm)#</mathjax>.</p>
<p>Here I have used non-standard units. I am free to do so as long as I do so consistently. (Note that a pressure measurement of 760 mm Hg is equivalent to 1 atmosphere).</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Under constant temperature, a #5.00*L# volume of gas at #1*atm# pressure is compressed to #3.00*L#. What is the new pressure?</h1>
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<div class="markdown"><p><mathjax>#P_1V_1 = P_2V_2#</mathjax> (Boyle's law, pressure is inversely proportional to volume at constant temperature.) </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5.00L)/(3.00L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5/3atm)#</mathjax>.</p>
<p>Here I have used non-standard units. I am free to do so as long as I do so consistently. (Note that a pressure measurement of 760 mm Hg is equivalent to 1 atmosphere).</p></div>
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</article> | Under constant temperature, a #5.00*L# volume of gas at #1*atm# pressure is compressed to #3.00*L#. What is the new pressure? | null |
1,427 | a8e9c2f8-6ddd-11ea-8ffc-ccda262736ce | https://socratic.org/questions/how-do-i-calculate-the-ph-of-the-buffer-solution-formed-when-10-0-cm-3-of-0-80-m | 4.43 | start physical_unit 7 9 ph none qc_end physical_unit 17 18 12 13 volume qc_end physical_unit 17 18 15 16 molarity qc_end physical_unit 27 28 22 23 volume qc_end physical_unit 27 28 25 26 molarity qc_end physical_unit 27 28 31 34 ka qc_end end | [{"type":"physical unit","value":"pH [OF] the buffer solution"}] | [{"type":"physical unit","value":"4.43"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide [=] \\pu{10.0 cm^3}"},{"type":"physical unit","value":"Molarity [OF] sodium hydroxide [=] \\pu{0.80 mol/dm^3}"},{"type":"physical unit","value":"Volume [OF] ethanoic acid [=] \\pu{50.0 cm^3}"},{"type":"physical unit","value":"Molarity [OF] ethanoic acid [=] \\pu{0.50 mol/dm^3}"},{"type":"physical unit","value":"Ka [OF] ethanoic acid [=] \\pu{1.74 × 10^(−5) mol/dm^3}"}] | <h1 class="questionTitle" itemprop="name">How do I calculate the #"pH"# of the buffer solution formed when #10.0# #"cm"^3# of #0.80# #"mol"# #"dm"^-3# sodium hydroxide is mixed with #50.0# #"cm"^3# of #0.50# #"mol"# #"dm"^-3# ethanoic acid (#"K"_"a"=1.74xx10^-5# #"mol"# #"dm"^-3#)?</h1> | null | 4.43 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The first thing you must recognize is that there will be an acid-base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction.</p>
<p>Your first task is to calculate the concentrations of the species present at the end of the reaction.</p>
<p>We can summarize the calculations in an ICE table.</p>
<p><mathjax>#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"#</mathjax><br/>
<mathjax>#"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"#</mathjax><br/>
<mathjax>#"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080#</mathjax></p>
<p><mathjax>#"Moles HA" = 0.0500 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA"))) × "0.50 mol HA"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA")))) = "0.025 mol HA"#</mathjax></p>
<p><mathjax>#"Moles OH"^"-" = 0.0100 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-"))) × "0.80 mol A"^"-"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-")))) = "0.0080 mol A"^"-"#</mathjax></p>
<p>So, at the end of the reaction, we have a solution containing 0.017 mol of <mathjax>#"HA"#</mathjax> and 0.0080 mol of <mathjax>#"A"^"-"#</mathjax>.</p>
<p>A mixture of a weak acid and its salt is a <strong>buffer</strong>.</p>
<blockquote></blockquote>
<p>We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.</p>
<p>∴ <mathjax>#"pH" = "-log"(1.74 × 10^"-5") + log((0.0080 color(red)(cancel(color(black)("mol"))))/(0.017 color(red)(cancel(color(black)("mol"))))) = "4.76 - 0.33" = 4.43#</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/acids-and-bases/the-ph-concept">pH</a> of the buffer is 4.43.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The first thing you must recognize is that there will be an acid-base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction.</p>
<p>Your first task is to calculate the concentrations of the species present at the end of the reaction.</p>
<p>We can summarize the calculations in an ICE table.</p>
<p><mathjax>#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"#</mathjax><br/>
<mathjax>#"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"#</mathjax><br/>
<mathjax>#"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080#</mathjax></p>
<p><mathjax>#"Moles HA" = 0.0500 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA"))) × "0.50 mol HA"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA")))) = "0.025 mol HA"#</mathjax></p>
<p><mathjax>#"Moles OH"^"-" = 0.0100 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-"))) × "0.80 mol A"^"-"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-")))) = "0.0080 mol A"^"-"#</mathjax></p>
<p>So, at the end of the reaction, we have a solution containing 0.017 mol of <mathjax>#"HA"#</mathjax> and 0.0080 mol of <mathjax>#"A"^"-"#</mathjax>.</p>
<p>A mixture of a weak acid and its salt is a <strong>buffer</strong>.</p>
<blockquote></blockquote>
<p>We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.</p>
<p>∴ <mathjax>#"pH" = "-log"(1.74 × 10^"-5") + log((0.0080 color(red)(cancel(color(black)("mol"))))/(0.017 color(red)(cancel(color(black)("mol"))))) = "4.76 - 0.33" = 4.43#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do I calculate the #"pH"# of the buffer solution formed when #10.0# #"cm"^3# of #0.80# #"mol"# #"dm"^-3# sodium hydroxide is mixed with #50.0# #"cm"^3# of #0.50# #"mol"# #"dm"^-3# ethanoic acid (#"K"_"a"=1.74xx10^-5# #"mol"# #"dm"^-3#)?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the buffer is 4.43.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The first thing you must recognize is that there will be an acid-base <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction.</p>
<p>Your first task is to calculate the concentrations of the species present at the end of the reaction.</p>
<p>We can summarize the calculations in an ICE table.</p>
<p><mathjax>#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"#</mathjax><br/>
<mathjax>#"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"#</mathjax><br/>
<mathjax>#"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080#</mathjax></p>
<p><mathjax>#"Moles HA" = 0.0500 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA"))) × "0.50 mol HA"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA")))) = "0.025 mol HA"#</mathjax></p>
<p><mathjax>#"Moles OH"^"-" = 0.0100 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-"))) × "0.80 mol A"^"-"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-")))) = "0.0080 mol A"^"-"#</mathjax></p>
<p>So, at the end of the reaction, we have a solution containing 0.017 mol of <mathjax>#"HA"#</mathjax> and 0.0080 mol of <mathjax>#"A"^"-"#</mathjax>.</p>
<p>A mixture of a weak acid and its salt is a <strong>buffer</strong>.</p>
<blockquote></blockquote>
<p>We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.</p>
<p>∴ <mathjax>#"pH" = "-log"(1.74 × 10^"-5") + log((0.0080 color(red)(cancel(color(black)("mol"))))/(0.017 color(red)(cancel(color(black)("mol"))))) = "4.76 - 0.33" = 4.43#</mathjax></p></div>
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</article> | How do I calculate the #"pH"# of the buffer solution formed when #10.0# #"cm"^3# of #0.80# #"mol"# #"dm"^-3# sodium hydroxide is mixed with #50.0# #"cm"^3# of #0.50# #"mol"# #"dm"^-3# ethanoic acid (#"K"_"a"=1.74xx10^-5# #"mol"# #"dm"^-3#)? | null |
1,428 | aca9492c-6ddd-11ea-b0c9-ccda262736ce | https://socratic.org/questions/how-many-ml-of-a-0-100-m-solution-are-required-have-1-32-x-10-2-moles-of-solute | 1.32 × 10^6 mL | start physical_unit 7 7 volume ml qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 16 16 11 14 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] solution [IN] mL"}] | [{"type":"physical unit","value":"1.32 × 10^6 mL"}] | [{"type":"physical unit","value":"Molarity [OF] solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Mole [OF] solute [=] \\pu{1.32 × 10^2 moles}"}] | <h1 class="questionTitle" itemprop="name">How many mL of a 0.100 M solution are required have #1.32 x 10^2# moles of solute? </h1> | null | 1.32 × 10^6 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>And, typically, <mathjax>#"concentration"#</mathjax> has units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Here, clearly, <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(1.32xx10^2*mol)/(0.100*mol*L^-1)=1320*L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Rather a large volume, over <mathjax>#1*m^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>And, typically, <mathjax>#"concentration"#</mathjax> has units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Here, clearly, <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(1.32xx10^2*mol)/(0.100*mol*L^-1)=1320*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many mL of a 0.100 M solution are required have #1.32 x 10^2# moles of solute? </h1>
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anor277
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<div class="markdown"><p>Rather a large volume, over <mathjax>#1*m^3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>And, typically, <mathjax>#"concentration"#</mathjax> has units of <mathjax>#mol*L^-1#</mathjax>.</p>
<p>Here, clearly, <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(1.32xx10^2*mol)/(0.100*mol*L^-1)=1320*L#</mathjax></p></div>
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</article> | How many mL of a 0.100 M solution are required have #1.32 x 10^2# moles of solute? | null |
1,429 | a99c8220-6ddd-11ea-b29f-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-hydronium-ions-in-a-solution-with-a-ph-3-9 | 10^(-3.9) mol/L | start physical_unit 5 6 concentration mol/l qc_end physical_unit 9 9 14 14 ph qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydronium ions [IN] mol/L"}] | [{"type":"physical unit","value":"10^(-3.9) mol/L"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{3.9}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of hydronium ions in a solution with a pH = 3.9?</h1> | null | 10^(-3.9) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definiion, <br/>
<mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>, and <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>In water, under standard conditions <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Both <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> measure the extent of the equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax>, the dissociation constant of water, </p>
<p><mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax></p>
<p>The <mathjax>#p#</mathjax> means take the logarithm to the base 10, and mulitply this by <mathjax>#-1#</mathjax>.</p>
<p>And thus <mathjax>#pK_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)10^(-14)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14#</mathjax> <mathjax>#=#</mathjax> <mathjax>#pH+pOH#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.9#</mathjax>, then <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-3.9)*mol*L^-1#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definiion, <br/>
<mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>, and <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>In water, under standard conditions <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Both <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> measure the extent of the equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax>, the dissociation constant of water, </p>
<p><mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax></p>
<p>The <mathjax>#p#</mathjax> means take the logarithm to the base 10, and mulitply this by <mathjax>#-1#</mathjax>.</p>
<p>And thus <mathjax>#pK_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)10^(-14)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14#</mathjax> <mathjax>#=#</mathjax> <mathjax>#pH+pOH#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of hydronium ions in a solution with a pH = 3.9?</h1>
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anor277
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<div class="markdown"><p>If <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.9#</mathjax>, then <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-3.9)*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definiion, <br/>
<mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>, and <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>In water, under standard conditions <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Both <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> measure the extent of the equilibrium:</p>
<p><mathjax>#2H_2O rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax>, the dissociation constant of water, </p>
<p><mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#[H_3O^+][HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax></p>
<p>The <mathjax>#p#</mathjax> means take the logarithm to the base 10, and mulitply this by <mathjax>#-1#</mathjax>.</p>
<p>And thus <mathjax>#pK_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)10^(-14)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#14#</mathjax> <mathjax>#=#</mathjax> <mathjax>#pH+pOH#</mathjax>.</p></div>
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</article> | What is the concentration of hydronium ions in a solution with a pH = 3.9? | null |
1,430 | ab23360c-6ddd-11ea-bfdd-ccda262736ce | https://socratic.org/questions/59ad9ae5b72cff4c215f2807 | 18.60 g | start physical_unit 3 4 mass g qc_end physical_unit 19 20 16 17 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] calcium phosphate [IN] g"}] | [{"type":"physical unit","value":"18.60 g"}] | [{"type":"physical unit","value":"Mass [OF] sodium phosphate [=] \\pu{14.2 g}"},{"type":"other","value":"Calcium phosphate contains the same number of ions as sodium phosphate."}] | <h1 class="questionTitle" itemprop="name">What mass of calcium phosphate will I need to contain the same number of ions as 14.2 g of sodium phosphate?</h1> | null | 18.60 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 14.2 color(red)(cancel(color(black)("g Na"_2"SO"_4))) × ( "1 mol Na"_2"SO"_4)/(142.04 color(red)(cancel(color(black)("g Na"_2"SO"_4)))) = "0.099 97 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ions</strong></p>
<p><mathjax>#"Na"_2"SO"_4 → "2Na"^"+" + "SO"_4^"2-"#</mathjax></p>
<p>So, 1 mol of <mathjax>#"Na"_2"SO"_4#</mathjax> produces 3 mol of ions.</p>
<p><mathjax>#"Moles of ions" = "0.099 97" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × "3 mol ions"/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "0.2999 mol ions"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Ca"_3("PO"_4)_2 → "3Ca"^"2+" + "2PO"_4^"3-"#</mathjax></p>
<p>∴ 1 mol of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax> produces 5 mol of ions.</p>
<p><mathjax>#" Moles of Ca"_3("PO"_4)_2 = 0.2999 color(red)(cancel(color(black)("mol ions"))) × (1 "mol Ca"_3("PO"_4)_2)/(5 color(red)(cancel(color(black)("mol ions")))) = "0.059 98 mol Ca"_3("PO"_4)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"_3("PO"_4)_2 = "0.059 98" color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("310.18 g Ca"_3("PO"_4)_2)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "18.6 g Ca"_3("PO"_4)_2#</mathjax></p></div>
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<p>You will need 18.6 g of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 14.2 color(red)(cancel(color(black)("g Na"_2"SO"_4))) × ( "1 mol Na"_2"SO"_4)/(142.04 color(red)(cancel(color(black)("g Na"_2"SO"_4)))) = "0.099 97 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ions</strong></p>
<p><mathjax>#"Na"_2"SO"_4 → "2Na"^"+" + "SO"_4^"2-"#</mathjax></p>
<p>So, 1 mol of <mathjax>#"Na"_2"SO"_4#</mathjax> produces 3 mol of ions.</p>
<p><mathjax>#"Moles of ions" = "0.099 97" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × "3 mol ions"/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "0.2999 mol ions"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Ca"_3("PO"_4)_2 → "3Ca"^"2+" + "2PO"_4^"3-"#</mathjax></p>
<p>∴ 1 mol of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax> produces 5 mol of ions.</p>
<p><mathjax>#" Moles of Ca"_3("PO"_4)_2 = 0.2999 color(red)(cancel(color(black)("mol ions"))) × (1 "mol Ca"_3("PO"_4)_2)/(5 color(red)(cancel(color(black)("mol ions")))) = "0.059 98 mol Ca"_3("PO"_4)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"_3("PO"_4)_2 = "0.059 98" color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("310.18 g Ca"_3("PO"_4)_2)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "18.6 g Ca"_3("PO"_4)_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of calcium phosphate will I need to contain the same number of ions as 14.2 g of sodium phosphate?</h1>
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<p>You will need 18.6 g of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 14.2 color(red)(cancel(color(black)("g Na"_2"SO"_4))) × ( "1 mol Na"_2"SO"_4)/(142.04 color(red)(cancel(color(black)("g Na"_2"SO"_4)))) = "0.099 97 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ions</strong></p>
<p><mathjax>#"Na"_2"SO"_4 → "2Na"^"+" + "SO"_4^"2-"#</mathjax></p>
<p>So, 1 mol of <mathjax>#"Na"_2"SO"_4#</mathjax> produces 3 mol of ions.</p>
<p><mathjax>#"Moles of ions" = "0.099 97" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × "3 mol ions"/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "0.2999 mol ions"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Ca"_3("PO"_4)_2 → "3Ca"^"2+" + "2PO"_4^"3-"#</mathjax></p>
<p>∴ 1 mol of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax> produces 5 mol of ions.</p>
<p><mathjax>#" Moles of Ca"_3("PO"_4)_2 = 0.2999 color(red)(cancel(color(black)("mol ions"))) × (1 "mol Ca"_3("PO"_4)_2)/(5 color(red)(cancel(color(black)("mol ions")))) = "0.059 98 mol Ca"_3("PO"_4)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of <mathjax>#"Ca"_3("PO"_4)_2#</mathjax></strong></p>
<p><mathjax>#"Mass of Ca"_3("PO"_4)_2 = "0.059 98" color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("310.18 g Ca"_3("PO"_4)_2)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "18.6 g Ca"_3("PO"_4)_2#</mathjax></p></div>
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</article> | What mass of calcium phosphate will I need to contain the same number of ions as 14.2 g of sodium phosphate? | null |
1,431 | ac42e577-6ddd-11ea-bdf1-ccda262736ce | https://socratic.org/questions/58ec78de7c014941196c5b37 | 5.00 × 10^(-3) L | start physical_unit 17 18 volume l qc_end physical_unit 18 18 1 2 volume qc_end physical_unit 18 18 5 6 concentration qc_end physical_unit 17 18 20 21 concentration qc_end end | [{"type":"physical unit","value":"Volume2 [OF] mother solution [IN] L"}] | [{"type":"physical unit","value":"5.00 × 10^(-3) L"}] | [{"type":"physical unit","value":"Volume1 [OF] solution [=] \\pu{0.25 L}"},{"type":"physical unit","value":"Concentration1 [OF] solution [=] \\pu{0.010 mol/L}"},{"type":"physical unit","value":"Concentration2 [OF] mother solution [=] \\pu{0.50 mol/L}"}] | <h1 class="questionTitle" itemprop="name">A #0.25*L# volume of #0.010*mol*L^-1# concentration was obtained by diluting a WHAT volume of a mother solution of #0.50*mol*L^-1# concentrations?</h1> | null | 5.00 × 10^(-3) L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#"concentration"xx"volume"#</mathjax> has units of <mathjax>#"moles per litre"xx"litre"#</mathjax>, i.e. clearly an answer in <mathjax>#"moles"#</mathjax>................</p>
<p>And thus, since we dilute a given molar quantity of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>...........</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax></p>
<p>We solve for <mathjax>#V_1#</mathjax>, and thus we solve..........</p>
<p><mathjax>#V_1=(C_2V_2)/C_1=(0.010*cancel(mol)*cancel(L^-1)xx0.25*cancelL)/(0.50*mol*L^-1)=0.0050*L-=0.005(1/(1/L))=5.0xx10^-3L=5.0*mL#</mathjax>. </p></div>
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<div class="markdown"><p>We need a volume of.........<mathjax>#5.0*mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The product <mathjax>#"concentration"xx"volume"#</mathjax> has units of <mathjax>#"moles per litre"xx"litre"#</mathjax>, i.e. clearly an answer in <mathjax>#"moles"#</mathjax>................</p>
<p>And thus, since we dilute a given molar quantity of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>...........</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax></p>
<p>We solve for <mathjax>#V_1#</mathjax>, and thus we solve..........</p>
<p><mathjax>#V_1=(C_2V_2)/C_1=(0.010*cancel(mol)*cancel(L^-1)xx0.25*cancelL)/(0.50*mol*L^-1)=0.0050*L-=0.005(1/(1/L))=5.0xx10^-3L=5.0*mL#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">A #0.25*L# volume of #0.010*mol*L^-1# concentration was obtained by diluting a WHAT volume of a mother solution of #0.50*mol*L^-1# concentrations?</h1>
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<div class="markdown"><p>We need a volume of.........<mathjax>#5.0*mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#"concentration"xx"volume"#</mathjax> has units of <mathjax>#"moles per litre"xx"litre"#</mathjax>, i.e. clearly an answer in <mathjax>#"moles"#</mathjax>................</p>
<p>And thus, since we dilute a given molar quantity of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>...........</p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax></p>
<p>We solve for <mathjax>#V_1#</mathjax>, and thus we solve..........</p>
<p><mathjax>#V_1=(C_2V_2)/C_1=(0.010*cancel(mol)*cancel(L^-1)xx0.25*cancelL)/(0.50*mol*L^-1)=0.0050*L-=0.005(1/(1/L))=5.0xx10^-3L=5.0*mL#</mathjax>. </p></div>
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</article> | A #0.25*L# volume of #0.010*mol*L^-1# concentration was obtained by diluting a WHAT volume of a mother solution of #0.50*mol*L^-1# concentrations? | null |
1,432 | ac5e1662-6ddd-11ea-835e-ccda262736ce | https://socratic.org/questions/a-rigid-plastic-container-holds-1-00-l-methane-gas-at-650-torr-pressure-when-the | 0.92 atm | start physical_unit 7 8 pressure atm qc_end physical_unit 7 8 10 11 pressure qc_end physical_unit 7 8 5 6 volume qc_end physical_unit 7 8 17 18 temperature qc_end physical_unit 7 8 33 34 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] methane gas [IN] atm"}] | [{"type":"physical unit","value":"0.92 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] methane gas [=] \\pu{650 torr}"},{"type":"physical unit","value":"Volume1 [OF] methane gas [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Temperature1 [OF] methane gas [=] \\pu{20.9 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] methane gas [=] \\pu{44.1 ℃}"},{"type":"other","value":"A rigid plastic container holds methane gas."}] | <h1 class="questionTitle" itemprop="name">A rigid plastic container holds 1.00 L methane gas at 650 torr pressure when the temperature is 20.9°C. How much more pressure will the gas exert if the temperature is raised to 44.1°C? </h1> | null | 0.92 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, old <mathjax>#"Charles' Law"#</mathjax> holds that <mathjax>#P_1/T_1=P_2/T_2#</mathjax> given constant volume, and constant quantity of gas, the which is the scenario we have here...we use <mathjax>#"absolute temperature"#</mathjax>....we ALSO know (or should know) that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high...i.e. <mathjax>#1*atm-=760*"mm Hg"#</mathjax></p>
<p><mathjax>#P_2=P_1/T_1xxT_2=((650*mm*Hg)/(760*mm*Hg*atm^-1))/(294.05*K)xx317.25*K=0.923*atm#</mathjax></p>
<p>...approx. <mathjax>#701.3*mm*Hg#</mathjax>...</p></div>
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<div class="markdown"><p><mathjax>#P_2=701.3*mm*Hg#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, old <mathjax>#"Charles' Law"#</mathjax> holds that <mathjax>#P_1/T_1=P_2/T_2#</mathjax> given constant volume, and constant quantity of gas, the which is the scenario we have here...we use <mathjax>#"absolute temperature"#</mathjax>....we ALSO know (or should know) that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high...i.e. <mathjax>#1*atm-=760*"mm Hg"#</mathjax></p>
<p><mathjax>#P_2=P_1/T_1xxT_2=((650*mm*Hg)/(760*mm*Hg*atm^-1))/(294.05*K)xx317.25*K=0.923*atm#</mathjax></p>
<p>...approx. <mathjax>#701.3*mm*Hg#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">A rigid plastic container holds 1.00 L methane gas at 650 torr pressure when the temperature is 20.9°C. How much more pressure will the gas exert if the temperature is raised to 44.1°C? </h1>
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<div class="markdown"><p><mathjax>#P_2=701.3*mm*Hg#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Well, old <mathjax>#"Charles' Law"#</mathjax> holds that <mathjax>#P_1/T_1=P_2/T_2#</mathjax> given constant volume, and constant quantity of gas, the which is the scenario we have here...we use <mathjax>#"absolute temperature"#</mathjax>....we ALSO know (or should know) that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high...i.e. <mathjax>#1*atm-=760*"mm Hg"#</mathjax></p>
<p><mathjax>#P_2=P_1/T_1xxT_2=((650*mm*Hg)/(760*mm*Hg*atm^-1))/(294.05*K)xx317.25*K=0.923*atm#</mathjax></p>
<p>...approx. <mathjax>#701.3*mm*Hg#</mathjax>...</p></div>
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</article> | A rigid plastic container holds 1.00 L methane gas at 650 torr pressure when the temperature is 20.9°C. How much more pressure will the gas exert if the temperature is raised to 44.1°C? | null |
1,433 | aa53ef2e-6ddd-11ea-9641-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-the-mn-in-kmno-4 | +7 | start physical_unit 7 7 oxidation_number none qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] Mn"}] | [{"type":"physical unit","value":"+7"}] | [{"type":"chemical equation","value":"KMnO4"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of the #Mn# in #KMnO_4#?</h1> | null | +7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As a rule, except for peroxides and superoxides, oxygen is always -2. so if we have 4 oxygens, that is a total of -8.</p>
<p>As a rule alkali metals (group 1 except H) are always in the +1 oxidation state. There is one group 1 metal, K. So that's +1</p>
<p>In order to balance out the total charge to give a net of zero, we have to balance <mathjax>#+1 -8 = -7#</mathjax> charges. Therefore the Mn has to be +7.</p>
<p>To solve this sort of question, you need to know the common oxidation states of group 1, group 2, group 17, and period 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Those don't change, except in rare circumstances. Everything else you can get by subtraction of the known oxidation states. </p>
<p>For example, <mathjax>#SiO_2#</mathjax> the silicon is +4.</p></div>
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<div class="markdown"><p>+7</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As a rule, except for peroxides and superoxides, oxygen is always -2. so if we have 4 oxygens, that is a total of -8.</p>
<p>As a rule alkali metals (group 1 except H) are always in the +1 oxidation state. There is one group 1 metal, K. So that's +1</p>
<p>In order to balance out the total charge to give a net of zero, we have to balance <mathjax>#+1 -8 = -7#</mathjax> charges. Therefore the Mn has to be +7.</p>
<p>To solve this sort of question, you need to know the common oxidation states of group 1, group 2, group 17, and period 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Those don't change, except in rare circumstances. Everything else you can get by subtraction of the known oxidation states. </p>
<p>For example, <mathjax>#SiO_2#</mathjax> the silicon is +4.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of the #Mn# in #KMnO_4#?</h1>
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<div class="markdown"><p>+7</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As a rule, except for peroxides and superoxides, oxygen is always -2. so if we have 4 oxygens, that is a total of -8.</p>
<p>As a rule alkali metals (group 1 except H) are always in the +1 oxidation state. There is one group 1 metal, K. So that's +1</p>
<p>In order to balance out the total charge to give a net of zero, we have to balance <mathjax>#+1 -8 = -7#</mathjax> charges. Therefore the Mn has to be +7.</p>
<p>To solve this sort of question, you need to know the common oxidation states of group 1, group 2, group 17, and period 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Those don't change, except in rare circumstances. Everything else you can get by subtraction of the known oxidation states. </p>
<p>For example, <mathjax>#SiO_2#</mathjax> the silicon is +4.</p></div>
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<div class="markdown"><p>We have <mathjax>#Mn(+VII)#</mathjax>............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Permanganate ion is a potent oxidizing agent, and it is (typically) conceived to accept 5 electrons upon reduction:</p>
<p><mathjax>#MnO_4^(-) +8H^+ + 5e^(-) rarr Mn^(2+) +4H_2O#</mathjax></p>
<p>This reaction is self-indicating in that permanganate is brightly coloured (deep-red), whereas <mathjax>#Mn^(2+)#</mathjax> is almost colourless. Are charge and mass balanced here?</p>
<p>As to how you calculate the oxidation number, the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> is equal to the charge of the ions, here <mathjax>#-1#</mathjax>. Now the oxidation number of oxygen is usually <mathjax>#-II#</mathjax>, and it is here.</p>
<p>Thus <mathjax>#Mn_"oxidation number" +4xx(-II)=-1#</mathjax> </p>
<p>And thus <mathjax>#Mn_"oxidation number"=+VII#</mathjax>.</p></div>
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</article> | What is the oxidation number of the #Mn# in #KMnO_4#? | null |
1,434 | abb2abd8-6ddd-11ea-8c2d-ccda262736ce | https://socratic.org/questions/how-many-kilograms-of-o-2-gas-are-needed-to-give-an-oxygen-pressure-of-3-48-atm- | 4.60 kilograms | start physical_unit 4 5 mass kg qc_end physical_unit 4 5 14 15 pressure qc_end physical_unit 4 5 17 18 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 gas [IN] kilograms"}] | [{"type":"physical unit","value":"4.60 kilograms"}] | [{"type":"physical unit","value":"Pressure [OF] O2 gas [=] \\pu{3.48 atm}"},{"type":"physical unit","value":"Temperature [OF] O2 gas [=] \\pu{22 ℃}"}] | <h1 class="questionTitle" itemprop="name">How many kilograms of #O_2# gas are needed to give an oxygen pressure of 3.48 atm at 22°C?</h1> | null | 4.60 kilograms | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation, with <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And thus <mathjax>#n=(PV)/(RT)=(3.48*cancel(atm)xx1*cancel(m^3)xx10^3*cancelL*cancel(m^-3))/(0.0821*cancel(L*atm)*cancel(K^-1)*mol^-1xx295*cancelK)#</mathjax></p>
<p><mathjax>#=143.7*mol#</mathjax> (note that the answer has units <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax> <mathjax>#"as required"#</mathjax></p>
<p>And we convert this molar quantity into a mass:</p>
<p><mathjax>#143.7*cancel(mol)xx32.00*cancel(g*mol^-1)xx10^-3*kg*cancel(g^-1)#</mathjax></p>
<p><mathjax>#=4.60*kg.#</mathjax></p>
<p>Please review my arithmetic. There are no money-back guarantees.</p>
<p>Note that <mathjax>#1*m^3#</mathjax> is a very large volume, and is equal to <mathjax>#1000*L#</mathjax>. Chemists tend to deal with <mathjax>#"litres"#</mathjax> and <mathjax>#mL#</mathjax> and <mathjax>#cm^3#</mathjax> (<mathjax>#1*mL-=1*cm^3-=10^-3*L-=10^-6*m^3#</mathjax>). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa. </p></div>
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<div class="markdown"><p>Just to retire this question, let us specify a volume of <mathjax>#1*m^3#</mathjax>; and I get a mass of dioxygen gas of under <mathjax>#5*kg#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation, with <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And thus <mathjax>#n=(PV)/(RT)=(3.48*cancel(atm)xx1*cancel(m^3)xx10^3*cancelL*cancel(m^-3))/(0.0821*cancel(L*atm)*cancel(K^-1)*mol^-1xx295*cancelK)#</mathjax></p>
<p><mathjax>#=143.7*mol#</mathjax> (note that the answer has units <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax> <mathjax>#"as required"#</mathjax></p>
<p>And we convert this molar quantity into a mass:</p>
<p><mathjax>#143.7*cancel(mol)xx32.00*cancel(g*mol^-1)xx10^-3*kg*cancel(g^-1)#</mathjax></p>
<p><mathjax>#=4.60*kg.#</mathjax></p>
<p>Please review my arithmetic. There are no money-back guarantees.</p>
<p>Note that <mathjax>#1*m^3#</mathjax> is a very large volume, and is equal to <mathjax>#1000*L#</mathjax>. Chemists tend to deal with <mathjax>#"litres"#</mathjax> and <mathjax>#mL#</mathjax> and <mathjax>#cm^3#</mathjax> (<mathjax>#1*mL-=1*cm^3-=10^-3*L-=10^-6*m^3#</mathjax>). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa. </p></div>
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<h1 class="questionTitle" itemprop="name">How many kilograms of #O_2# gas are needed to give an oxygen pressure of 3.48 atm at 22°C?</h1>
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<div class="markdown"><p>Just to retire this question, let us specify a volume of <mathjax>#1*m^3#</mathjax>; and I get a mass of dioxygen gas of under <mathjax>#5*kg#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation, with <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And thus <mathjax>#n=(PV)/(RT)=(3.48*cancel(atm)xx1*cancel(m^3)xx10^3*cancelL*cancel(m^-3))/(0.0821*cancel(L*atm)*cancel(K^-1)*mol^-1xx295*cancelK)#</mathjax></p>
<p><mathjax>#=143.7*mol#</mathjax> (note that the answer has units <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax> <mathjax>#"as required"#</mathjax></p>
<p>And we convert this molar quantity into a mass:</p>
<p><mathjax>#143.7*cancel(mol)xx32.00*cancel(g*mol^-1)xx10^-3*kg*cancel(g^-1)#</mathjax></p>
<p><mathjax>#=4.60*kg.#</mathjax></p>
<p>Please review my arithmetic. There are no money-back guarantees.</p>
<p>Note that <mathjax>#1*m^3#</mathjax> is a very large volume, and is equal to <mathjax>#1000*L#</mathjax>. Chemists tend to deal with <mathjax>#"litres"#</mathjax> and <mathjax>#mL#</mathjax> and <mathjax>#cm^3#</mathjax> (<mathjax>#1*mL-=1*cm^3-=10^-3*L-=10^-6*m^3#</mathjax>). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa. </p></div>
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</article> | How many kilograms of #O_2# gas are needed to give an oxygen pressure of 3.48 atm at 22°C? | null |
1,435 | a96bfc06-6ddd-11ea-946d-ccda262736ce | https://socratic.org/questions/9-suppose-two-gases-in-a-container-have-a-total-pressure-of-1-20-atm-what-is-the | 0.45 atm | start physical_unit 18 19 partial_pressure atm qc_end physical_unit 25 26 28 29 partial_pressure qc_end physical_unit 1 2 11 12 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] gas B [IN] atm"}] | [{"type":"physical unit","value":"0.45 atm"}] | [{"type":"physical unit","value":"Partial pressure [OF] gas A [=] \\pu{0.75 atm}"},{"type":"physical unit","value":"Total pressure [OF] two gases [=] \\pu{1.20 atm}"}] | <h1 class="questionTitle" itemprop="name">9. Suppose two gases in a container have a total pressure of 1.20 atm. What is the pressure of gas B if the partial pressure of gas A is 0.75 atm? </h1> | null | 0.45 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> holds that in a gaseous mixture, the pressure exerted by a component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressure.</p>
<p>i.e. <mathjax>#P_"Total"=P_1+P_2+P_3.........#</mathjax></p>
<p>And here <mathjax>#P_"Total"=1.20*atm=P_A+P_B=0.75*atm+P_B#</mathjax>,</p>
<p>i.e. <mathjax>#P_B=1.20*atm-0.75*atm=??*atm.#</mathjax></p>
<p>Note that we can use this law to access the number of moles of each gas, given that <mathjax>#P=(nRT)/V#</mathjax>,</p>
<p>i.e. <mathjax>#P_"Total"=(RT)/V{n_1+n_2+...........n_n}#</mathjax>, where <mathjax>#n_i#</mathjax> is the number of moles of gas <mathjax>#i#</mathjax>. </p>
<p>And we can show that:</p>
<p><mathjax>#P_i/P_"Total"=n_i/n_"Total"#</mathjax>, </p>
<p>i.e. partial pressure is equivalent to mole fraction. </p></div>
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<div class="markdown"><p><mathjax>#P_B=0.45*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> holds that in a gaseous mixture, the pressure exerted by a component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressure.</p>
<p>i.e. <mathjax>#P_"Total"=P_1+P_2+P_3.........#</mathjax></p>
<p>And here <mathjax>#P_"Total"=1.20*atm=P_A+P_B=0.75*atm+P_B#</mathjax>,</p>
<p>i.e. <mathjax>#P_B=1.20*atm-0.75*atm=??*atm.#</mathjax></p>
<p>Note that we can use this law to access the number of moles of each gas, given that <mathjax>#P=(nRT)/V#</mathjax>,</p>
<p>i.e. <mathjax>#P_"Total"=(RT)/V{n_1+n_2+...........n_n}#</mathjax>, where <mathjax>#n_i#</mathjax> is the number of moles of gas <mathjax>#i#</mathjax>. </p>
<p>And we can show that:</p>
<p><mathjax>#P_i/P_"Total"=n_i/n_"Total"#</mathjax>, </p>
<p>i.e. partial pressure is equivalent to mole fraction. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">9. Suppose two gases in a container have a total pressure of 1.20 atm. What is the pressure of gas B if the partial pressure of gas A is 0.75 atm? </h1>
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<div class="markdown"><p><mathjax>#P_B=0.45*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's Law of <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> holds that in a gaseous mixture, the pressure exerted by a component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressure.</p>
<p>i.e. <mathjax>#P_"Total"=P_1+P_2+P_3.........#</mathjax></p>
<p>And here <mathjax>#P_"Total"=1.20*atm=P_A+P_B=0.75*atm+P_B#</mathjax>,</p>
<p>i.e. <mathjax>#P_B=1.20*atm-0.75*atm=??*atm.#</mathjax></p>
<p>Note that we can use this law to access the number of moles of each gas, given that <mathjax>#P=(nRT)/V#</mathjax>,</p>
<p>i.e. <mathjax>#P_"Total"=(RT)/V{n_1+n_2+...........n_n}#</mathjax>, where <mathjax>#n_i#</mathjax> is the number of moles of gas <mathjax>#i#</mathjax>. </p>
<p>And we can show that:</p>
<p><mathjax>#P_i/P_"Total"=n_i/n_"Total"#</mathjax>, </p>
<p>i.e. partial pressure is equivalent to mole fraction. </p></div>
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</article> | 9. Suppose two gases in a container have a total pressure of 1.20 atm. What is the pressure of gas B if the partial pressure of gas A is 0.75 atm? | null |
1,436 | ab5589e2-6ddd-11ea-bbbd-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-net-ionic-equation-for-h-3po-4-aq-ca-oh-2-aq-ca-3-po-4-2-aq | H+(aq) + OH-(aq) -> H2O(l) | start chemical_equation qc_end chemical_equation 8 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] net ionic equation"}] | [{"type":"chemical equation","value":"H+(aq) + OH-(aq) -> H2O(l)"}] | [{"type":"chemical equation","value":"H3PO4(aq) + Ca(OH)2(aq) -> Ca3(PO4)2(aq) + H2O(l)"}] | <h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for #H_3PO_4(aq) + Ca(OH)_2(aq) -> Ca_3(PO_4)_2(aq) + H_2O(l)#?</h1> | null | H+(aq) + OH-(aq) -> H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Firstly, balance your equation:</p>
<p><mathjax>#2"H"_3"PO"_4(aq) + 3"Ca(OH)"_2(aq) -> "Ca"_3"(PO"_4")"_2(aq) + 6"H"_2"O"#</mathjax></p>
<p>Then, split up your aqueous substances into their respective ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 2"PO"_4^(3-)(aq) + 3"Ca"^(2+)(aq) + 6"OH"^(-)(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq) + 6"H"_2"O"(l)#</mathjax></p>
<p>Then we cancel ions that appear on both sides of the equation: these are <strong>spectator ions</strong> that do not participate in the reaction.</p>
<p><mathjax>#6"H"^(+)(aq) + cancel(2"PO"_4^(3-)(aq)) + cancel(3"Ca"^(2+)(aq)) + 6"OH"^(-)(aq) -> cancel(3"Ca"^(2+)(aq)) + cancel(2"PO"_4^(3-)(aq)) + 6"H"_2"O"(l)#</mathjax></p>
<p>And write out the new net ionic equation that excludes these ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)#</mathjax></p>
<p>And by cancelling down coefficients we get:</p>
<p><mathjax>#"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)#</mathjax></p>
<p>Which is the balanced net ionic equation for any <strong>neutralisation reaction</strong> : the type of reaction we were dealing with in the first place.</p></div>
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<div class="markdown"><p><mathjax>#6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)#</mathjax></p>
<p><strong>OR</strong></p>
<p><mathjax>#"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Firstly, balance your equation:</p>
<p><mathjax>#2"H"_3"PO"_4(aq) + 3"Ca(OH)"_2(aq) -> "Ca"_3"(PO"_4")"_2(aq) + 6"H"_2"O"#</mathjax></p>
<p>Then, split up your aqueous substances into their respective ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 2"PO"_4^(3-)(aq) + 3"Ca"^(2+)(aq) + 6"OH"^(-)(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq) + 6"H"_2"O"(l)#</mathjax></p>
<p>Then we cancel ions that appear on both sides of the equation: these are <strong>spectator ions</strong> that do not participate in the reaction.</p>
<p><mathjax>#6"H"^(+)(aq) + cancel(2"PO"_4^(3-)(aq)) + cancel(3"Ca"^(2+)(aq)) + 6"OH"^(-)(aq) -> cancel(3"Ca"^(2+)(aq)) + cancel(2"PO"_4^(3-)(aq)) + 6"H"_2"O"(l)#</mathjax></p>
<p>And write out the new net ionic equation that excludes these ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)#</mathjax></p>
<p>And by cancelling down coefficients we get:</p>
<p><mathjax>#"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)#</mathjax></p>
<p>Which is the balanced net ionic equation for any <strong>neutralisation reaction</strong> : the type of reaction we were dealing with in the first place.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced net ionic equation for #H_3PO_4(aq) + Ca(OH)_2(aq) -> Ca_3(PO_4)_2(aq) + H_2O(l)#?</h1>
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<div class="markdown"><p><mathjax>#6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)#</mathjax></p>
<p><strong>OR</strong></p>
<p><mathjax>#"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Firstly, balance your equation:</p>
<p><mathjax>#2"H"_3"PO"_4(aq) + 3"Ca(OH)"_2(aq) -> "Ca"_3"(PO"_4")"_2(aq) + 6"H"_2"O"#</mathjax></p>
<p>Then, split up your aqueous substances into their respective ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 2"PO"_4^(3-)(aq) + 3"Ca"^(2+)(aq) + 6"OH"^(-)(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq) + 6"H"_2"O"(l)#</mathjax></p>
<p>Then we cancel ions that appear on both sides of the equation: these are <strong>spectator ions</strong> that do not participate in the reaction.</p>
<p><mathjax>#6"H"^(+)(aq) + cancel(2"PO"_4^(3-)(aq)) + cancel(3"Ca"^(2+)(aq)) + 6"OH"^(-)(aq) -> cancel(3"Ca"^(2+)(aq)) + cancel(2"PO"_4^(3-)(aq)) + 6"H"_2"O"(l)#</mathjax></p>
<p>And write out the new net ionic equation that excludes these ions:</p>
<p><mathjax>#6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)#</mathjax></p>
<p>And by cancelling down coefficients we get:</p>
<p><mathjax>#"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)#</mathjax></p>
<p>Which is the balanced net ionic equation for any <strong>neutralisation reaction</strong> : the type of reaction we were dealing with in the first place.</p></div>
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</article> | What is the balanced net ionic equation for #H_3PO_4(aq) + Ca(OH)_2(aq) -> Ca_3(PO_4)_2(aq) + H_2O(l)#? | null |
1,437 | ab4371f6-6ddd-11ea-9703-ccda262736ce | https://socratic.org/questions/for-an-ideal-gas-how-do-you-calculate-the-pressure-in-atm-if-8-25-x-10-2-mol-occ | 19.00 atm | start physical_unit 2 3 pressure atm qc_end physical_unit 2 3 13 16 mole qc_end physical_unit 2 3 18 19 volume qc_end physical_unit 2 3 21 22 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] the ideal gas [IN] atm"}] | [{"type":"physical unit","value":"19.00 atm"}] | [{"type":"physical unit","value":"Mole [OF] the ideal gas [=] \\pu{8.25 × 10^(-2) mol}"},{"type":"physical unit","value":"Volume [OF] the ideal gas [=] \\pu{174 mL}"},{"type":"physical unit","value":"Temperature [OF] the ideal gas [=] \\pu{215 ℃}"}] | <h1 class="questionTitle" itemprop="name">For an ideal gas, how do you calculate the pressure in atm if #8.25 x 10^-2# mol occupies 174 mL at 215°C? </h1> | null | 19.00 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have been quoted the moles, <mathjax>#n#</mathjax>, the temperature, <mathjax>#488*K#</mathjax> on the absolute scale, the volume, <mathjax>#0.174*L#</mathjax>. All we need to do is choose an appropriate gas constant. <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>.</p>
<p>Thus, <mathjax>#P=(8.25xx10^-2*molxx0.0821*(L*atm)/(K*mol)xx488*K)/(0.174*L)#</mathjax> for an answer in atmospheres. </p>
<p>Thx to Shelby who pointed out an error on my part!</p></div>
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<div class="markdown"><p><mathjax>#P=(nRT)/V=??*atm#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have been quoted the moles, <mathjax>#n#</mathjax>, the temperature, <mathjax>#488*K#</mathjax> on the absolute scale, the volume, <mathjax>#0.174*L#</mathjax>. All we need to do is choose an appropriate gas constant. <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>.</p>
<p>Thus, <mathjax>#P=(8.25xx10^-2*molxx0.0821*(L*atm)/(K*mol)xx488*K)/(0.174*L)#</mathjax> for an answer in atmospheres. </p>
<p>Thx to Shelby who pointed out an error on my part!</p></div>
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<h1 class="questionTitle" itemprop="name">For an ideal gas, how do you calculate the pressure in atm if #8.25 x 10^-2# mol occupies 174 mL at 215°C? </h1>
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<div class="markdown"><p><mathjax>#P=(nRT)/V=??*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have been quoted the moles, <mathjax>#n#</mathjax>, the temperature, <mathjax>#488*K#</mathjax> on the absolute scale, the volume, <mathjax>#0.174*L#</mathjax>. All we need to do is choose an appropriate gas constant. <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax>.</p>
<p>Thus, <mathjax>#P=(8.25xx10^-2*molxx0.0821*(L*atm)/(K*mol)xx488*K)/(0.174*L)#</mathjax> for an answer in atmospheres. </p>
<p>Thx to Shelby who pointed out an error on my part!</p></div>
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</article> | For an ideal gas, how do you calculate the pressure in atm if #8.25 x 10^-2# mol occupies 174 mL at 215°C? | null |
1,438 | a8389134-6ddd-11ea-ba5d-ccda262736ce | https://socratic.org/questions/a-sample-of-carbon-monoxide-occupies-2-44-l-at-295-0-k-and-771-torr-find-its-vol | 4.18 L | start physical_unit 1 4 volume l qc_end physical_unit 1 4 12 13 pressure qc_end physical_unit 1 4 6 7 volume qc_end physical_unit 1 4 9 10 temperature qc_end physical_unit 1 4 18 20 temperature qc_end physical_unit 1 4 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] carbon monoxide sample [IN] L"}] | [{"type":"physical unit","value":"4.18 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] carbon monoxide sample [=] \\pu{771 torr}"},{"type":"physical unit","value":"Volume1 [OF] carbon monoxide sample [=] \\pu{2.44 L}"},{"type":"physical unit","value":"Temperature1 [OF] carbon monoxide sample [=] \\pu{295.0 K}"},{"type":"physical unit","value":"Temperature2 [OF] carbon monoxide sample [=] \\pu{-47 degrees Celsius}"},{"type":"physical unit","value":"Pressure2 [OF] carbon monoxide sample [=] \\pu{350 torr}"}] | <h1 class="questionTitle" itemprop="name">A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?</h1> | null | 4.18 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>; <mathjax>#V_2~=4*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr?</h1>
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anor277
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Jul 5, 2017
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>; <mathjax>#V_2~=4*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that a mercury column is used as a highly visual measure of pressure. One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high.</p>
<p>i.e. <mathjax>#1*atm-=760*mm*Hg#</mathjax></p>
<p>....And thus we can use the length measurement to represent pressure...We also realize that we must use <mathjax>#"Absolute temperature"#</mathjax>, where <mathjax>#"Absolute Temperature"-=""^@C+273.15*K#</mathjax>.</p>
<p>So we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax>, and clearly the RHS gives an answer with units of volume.......</p>
<p>And so........</p>
<p><mathjax>#(771*mm*Hgxx2.44*L)/(295*K)xx(226.1*K)/(350*mm*Hg)#</mathjax>....</p>
<p><mathjax>#~=4*L#</mathjax></p>
<p>Of course, I could have reduced each pressure measurement to units of <mathjax>#atm#</mathjax>, but the increase in volume is consistent with a decrease in PRESSURE, and the moderate decrease in temperature. </p>
<p>Do you see how the units cancel in the expression to give an answer in <mathjax>#"litres"#</mathjax>, i.e. as required for a <mathjax>#"volume"#</mathjax>?</p></div>
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IDKwhatName
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<span class="dateCreated" datetime="2017-07-05T04:46:07" itemprop="dateCreated">
Jul 5, 2017
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<div class="markdown"><p><mathjax>#4.18"L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As there are a constant number of moles:<br/>
<mathjax>#("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2#</mathjax></p>
<p><mathjax>#-47^circC#</mathjax>=226^circK#</p>
<p><mathjax>#2.44"L"=0.00244"m"^3#</mathjax></p>
<p><mathjax>#771"torr"=102791.546"Pa"#</mathjax><br/>
<mathjax>#350"torr"=46662.8288"Pa"#</mathjax></p>
<p>If <mathjax>#("P"_1"V"_1)/"T"_1=("P"_2"V"_2)/"T"_2#</mathjax>, then <mathjax>#(102791.546*0.00244)/295=(46662.8288*"V"_2)/226#</mathjax>.</p>
<p><mathjax>#"V"_2=(102791.546*0.00244*226)/(295*46662.8288)=0.00411777473"m"^3~~0.00418"m"^3=4.18"L"#</mathjax>'</p></div>
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</article> | A sample of carbon monoxide occupies 2.44 L at 295.0 K and 771 torr. Find its volume at -47 degrees Celsius and 350 torr? | null |
1,439 | abc851d8-6ddd-11ea-90c2-ccda262736ce | https://socratic.org/questions/the-molar-heat-of-fusion-of-iron-is-14-9-kj-mol-what-is-the-energy-in-kj-needed- | 1.20 kJ | start physical_unit 6 6 energy kj qc_end physical_unit 6 6 8 9 molar_heat qc_end physical_unit 6 6 19 20 mass qc_end end | [{"type":"physical unit","value":"Needed energy [OF] iron [IN] kJ"}] | [{"type":"physical unit","value":"1.20 kJ"}] | [{"type":"physical unit","value":"Molar heat of fusion [OF] iron [=] \\pu{14.9 kJ/mol}"},{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{4.50 g}"}] | <h1 class="questionTitle" itemprop="name">The molar heat of fusion of iron is #14.9# kJ/mol. What is the energy (in kJ) needed to melt #4.50g# of iron?</h1> | null | 1.20 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To determine the energy that is needed to melt the iron, we have to calculate the amount of mol <mathjax>#Fe#</mathjax> (iron) we have. This can be done with the molar mass of iron, which is <mathjax>#55.85 u#</mathjax> (look up in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>!)</p>
<p>Now we use the formula:<br/>
<mathjax>#mol=("mass" color(white)(a) color(blue)((gram)) )/("molar mass"color(white)(a)color(blue)(((gram)/(mol)))#</mathjax></p>
<p>We fill the numbers in and we obtain:<br/>
<mathjax>#color(white)(a) color(red)("mol Fe")=(4.5color(white)(a)color(red)(cancel(color(blue)(gram))))/(55.85color(white)(a)color(red)(cancel(color(blue)(gram)))/color(blue)(mol))=0.081#</mathjax></p>
<p>We use the molar heat of iron (<mathjax>#14.9#</mathjax> kJ/mol) to calculate the amount of energy that is needed to melt it. </p>
<p>We make a table:</p>
<p><mathjax>#color(red)("mol Fe") color(white)(aaaaa)color(red)("Energy needed (kJ)"#</mathjax><br/>
<mathjax>#color(white)(aa)1color(white)(aaaaaa):color(white)(aaaaaaa)14.9#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaa)cancel(color(black)(\))#</mathjax><br/>
<mathjax>#color(white)(aa)0.081color(white)(aaa):color(white)(aaaaaaa)?#</mathjax></p>
<p>To calculate the energy needed we use the ratio (the cross) and calculate:<br/>
<mathjax>#(0.081xx14.9)/1=1.20#</mathjax> kJ </p>
<p>Therefore to melt down the 4.50 gram of iron we need 1.20 kJ energy. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The energy needed is <mathjax>#1.20#</mathjax> kJ for <mathjax>#4.50#</mathjax> gram iron.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To determine the energy that is needed to melt the iron, we have to calculate the amount of mol <mathjax>#Fe#</mathjax> (iron) we have. This can be done with the molar mass of iron, which is <mathjax>#55.85 u#</mathjax> (look up in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>!)</p>
<p>Now we use the formula:<br/>
<mathjax>#mol=("mass" color(white)(a) color(blue)((gram)) )/("molar mass"color(white)(a)color(blue)(((gram)/(mol)))#</mathjax></p>
<p>We fill the numbers in and we obtain:<br/>
<mathjax>#color(white)(a) color(red)("mol Fe")=(4.5color(white)(a)color(red)(cancel(color(blue)(gram))))/(55.85color(white)(a)color(red)(cancel(color(blue)(gram)))/color(blue)(mol))=0.081#</mathjax></p>
<p>We use the molar heat of iron (<mathjax>#14.9#</mathjax> kJ/mol) to calculate the amount of energy that is needed to melt it. </p>
<p>We make a table:</p>
<p><mathjax>#color(red)("mol Fe") color(white)(aaaaa)color(red)("Energy needed (kJ)"#</mathjax><br/>
<mathjax>#color(white)(aa)1color(white)(aaaaaa):color(white)(aaaaaaa)14.9#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaa)cancel(color(black)(\))#</mathjax><br/>
<mathjax>#color(white)(aa)0.081color(white)(aaa):color(white)(aaaaaaa)?#</mathjax></p>
<p>To calculate the energy needed we use the ratio (the cross) and calculate:<br/>
<mathjax>#(0.081xx14.9)/1=1.20#</mathjax> kJ </p>
<p>Therefore to melt down the 4.50 gram of iron we need 1.20 kJ energy. </p></div>
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<h1 class="questionTitle" itemprop="name">The molar heat of fusion of iron is #14.9# kJ/mol. What is the energy (in kJ) needed to melt #4.50g# of iron?</h1>
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Martin M.
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<div class="markdown"><p>The energy needed is <mathjax>#1.20#</mathjax> kJ for <mathjax>#4.50#</mathjax> gram iron.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To determine the energy that is needed to melt the iron, we have to calculate the amount of mol <mathjax>#Fe#</mathjax> (iron) we have. This can be done with the molar mass of iron, which is <mathjax>#55.85 u#</mathjax> (look up in <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>!)</p>
<p>Now we use the formula:<br/>
<mathjax>#mol=("mass" color(white)(a) color(blue)((gram)) )/("molar mass"color(white)(a)color(blue)(((gram)/(mol)))#</mathjax></p>
<p>We fill the numbers in and we obtain:<br/>
<mathjax>#color(white)(a) color(red)("mol Fe")=(4.5color(white)(a)color(red)(cancel(color(blue)(gram))))/(55.85color(white)(a)color(red)(cancel(color(blue)(gram)))/color(blue)(mol))=0.081#</mathjax></p>
<p>We use the molar heat of iron (<mathjax>#14.9#</mathjax> kJ/mol) to calculate the amount of energy that is needed to melt it. </p>
<p>We make a table:</p>
<p><mathjax>#color(red)("mol Fe") color(white)(aaaaa)color(red)("Energy needed (kJ)"#</mathjax><br/>
<mathjax>#color(white)(aa)1color(white)(aaaaaa):color(white)(aaaaaaa)14.9#</mathjax><br/>
<mathjax>#color(white)(aaaaaaaaa)cancel(color(black)(\))#</mathjax><br/>
<mathjax>#color(white)(aa)0.081color(white)(aaa):color(white)(aaaaaaa)?#</mathjax></p>
<p>To calculate the energy needed we use the ratio (the cross) and calculate:<br/>
<mathjax>#(0.081xx14.9)/1=1.20#</mathjax> kJ </p>
<p>Therefore to melt down the 4.50 gram of iron we need 1.20 kJ energy. </p></div>
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</article> | The molar heat of fusion of iron is #14.9# kJ/mol. What is the energy (in kJ) needed to melt #4.50g# of iron? | null |
1,440 | a9f996d4-6ddd-11ea-9ab2-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-created-by-mixing-3-7-moles-of-substance-in-2 | 1.76 M | start physical_unit 6 6 molarity mol/l qc_end physical_unit 13 13 10 11 mole qc_end physical_unit 6 6 15 16 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] solution [IN] M"}] | [{"type":"physical unit","value":"1.76 M"}] | [{"type":"physical unit","value":"Mole [OF] substance [=] \\pu{3.7 moles}"},{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{2.1 liters}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution created by mixing 3.7 moles of substance in 2.1 liters of solution?</h1> | null | 1.76 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is the ratio of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution.</p>
<p><mathjax>#M=(mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 3.7mol#</mathjax><br/>
<mathjax>#L=2.1 L#</mathjax></p>
<p><mathjax>#?M = (3.7 mol)/ (2.1 L)#</mathjax></p>
<p><mathjax>#M = 1.76M#</mathjax></p>
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<div class="markdown"><p><mathjax>#M = 1.76M#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is the ratio of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution.</p>
<p><mathjax>#M=(mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 3.7mol#</mathjax><br/>
<mathjax>#L=2.1 L#</mathjax></p>
<p><mathjax>#?M = (3.7 mol)/ (2.1 L)#</mathjax></p>
<p><mathjax>#M = 1.76M#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/loL46MQt1vQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution created by mixing 3.7 moles of substance in 2.1 liters of solution?</h1>
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<div class="markdown"><p><mathjax>#M = 1.76M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is the ratio of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution.</p>
<p><mathjax>#M=(mol)/L#</mathjax></p>
<p><mathjax>#M = ?#</mathjax><br/>
<mathjax>#mol = 3.7mol#</mathjax><br/>
<mathjax>#L=2.1 L#</mathjax></p>
<p><mathjax>#?M = (3.7 mol)/ (2.1 L)#</mathjax></p>
<p><mathjax>#M = 1.76M#</mathjax></p>
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</article> | What is the molarity of a solution created by mixing 3.7 moles of substance in 2.1 liters of solution? | null |
1,441 | a8bafaf4-6ddd-11ea-9e5f-ccda262736ce | https://socratic.org/questions/57f3e13d11ef6b4d111a0d9f | BaF2 | start chemical_formula qc_end chemical_equation 9 9 qc_end chemical_equation 11 11 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"BaF2"}] | [{"type":"chemical equation","value":"Ba"},{"type":"chemical equation","value":"F"}] | <h1 class="questionTitle" itemprop="name">What is the formula of the compound formed by Ba and F?</h1> | null | BaF2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Ba"#</mathjax> is the chemical symbol for the metal barium, and <mathjax>#"F"#</mathjax> is the chemical symbol for the nonmetal fluorine. The name of this compound is barium fluoride. Usually <a href="https://socratic.org/chemistry/the-periodic-table/metals-and-nonmetals">metals and nonmetals</a> form <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> when chemically combined. </p>
<p>Since this is an ionic compound, its chemical formula is called a formula unit because there are no discrete molecules. A formula unit represents the lowest whole number ratio of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>Barium is a group 2 element. Group 2 elements form cations with a <mathjax>#2^+#</mathjax> charge. The formula for the barium ion is <mathjax>#"Ba"^(2+)"#</mathjax>. Fluorine is a group 17 element. Group 17 elements form anions with a <mathjax>#1^-#</mathjax> charge. The formula for the fluoride ion is <mathjax>#"F"^(-)"#</mathjax>.</p>
<p>Ionic compounds are neutral, so the charges of the positive and negative ions must equal zero when added. So for every <mathjax>#"Ba"^(2+)"#</mathjax> ion, two <mathjax>#"F"^(-)"#</mathjax> ions are needed. So the formula unit for barium and fluorine compound is <mathjax>#"BaF"_2"#</mathjax>.</p>
<p><mathjax>#"Ba" + "2F"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#["Ba"]^(2+) + 2["F"]^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"BaF"_2"#</mathjax></p>
<p>You can also use the "crisscross method" to determine the formula for an ionic compound as shown below. </p>
<p>
<iframe src="https://www.youtube.com/embed/d9Xpt5Xh_D4?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The formula unit is <mathjax>#"BaF"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Ba"#</mathjax> is the chemical symbol for the metal barium, and <mathjax>#"F"#</mathjax> is the chemical symbol for the nonmetal fluorine. The name of this compound is barium fluoride. Usually <a href="https://socratic.org/chemistry/the-periodic-table/metals-and-nonmetals">metals and nonmetals</a> form <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> when chemically combined. </p>
<p>Since this is an ionic compound, its chemical formula is called a formula unit because there are no discrete molecules. A formula unit represents the lowest whole number ratio of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>Barium is a group 2 element. Group 2 elements form cations with a <mathjax>#2^+#</mathjax> charge. The formula for the barium ion is <mathjax>#"Ba"^(2+)"#</mathjax>. Fluorine is a group 17 element. Group 17 elements form anions with a <mathjax>#1^-#</mathjax> charge. The formula for the fluoride ion is <mathjax>#"F"^(-)"#</mathjax>.</p>
<p>Ionic compounds are neutral, so the charges of the positive and negative ions must equal zero when added. So for every <mathjax>#"Ba"^(2+)"#</mathjax> ion, two <mathjax>#"F"^(-)"#</mathjax> ions are needed. So the formula unit for barium and fluorine compound is <mathjax>#"BaF"_2"#</mathjax>.</p>
<p><mathjax>#"Ba" + "2F"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#["Ba"]^(2+) + 2["F"]^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"BaF"_2"#</mathjax></p>
<p>You can also use the "crisscross method" to determine the formula for an ionic compound as shown below. </p>
<p>
<iframe src="https://www.youtube.com/embed/d9Xpt5Xh_D4?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula of the compound formed by Ba and F?</h1>
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<div class="markdown"><p>The formula unit is <mathjax>#"BaF"_2"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Ba"#</mathjax> is the chemical symbol for the metal barium, and <mathjax>#"F"#</mathjax> is the chemical symbol for the nonmetal fluorine. The name of this compound is barium fluoride. Usually <a href="https://socratic.org/chemistry/the-periodic-table/metals-and-nonmetals">metals and nonmetals</a> form <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> when chemically combined. </p>
<p>Since this is an ionic compound, its chemical formula is called a formula unit because there are no discrete molecules. A formula unit represents the lowest whole number ratio of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>Barium is a group 2 element. Group 2 elements form cations with a <mathjax>#2^+#</mathjax> charge. The formula for the barium ion is <mathjax>#"Ba"^(2+)"#</mathjax>. Fluorine is a group 17 element. Group 17 elements form anions with a <mathjax>#1^-#</mathjax> charge. The formula for the fluoride ion is <mathjax>#"F"^(-)"#</mathjax>.</p>
<p>Ionic compounds are neutral, so the charges of the positive and negative ions must equal zero when added. So for every <mathjax>#"Ba"^(2+)"#</mathjax> ion, two <mathjax>#"F"^(-)"#</mathjax> ions are needed. So the formula unit for barium and fluorine compound is <mathjax>#"BaF"_2"#</mathjax>.</p>
<p><mathjax>#"Ba" + "2F"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#["Ba"]^(2+) + 2["F"]^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"BaF"_2"#</mathjax></p>
<p>You can also use the "crisscross method" to determine the formula for an ionic compound as shown below. </p>
<p>
<iframe src="https://www.youtube.com/embed/d9Xpt5Xh_D4?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | What is the formula of the compound formed by Ba and F? | null |
1,442 | ad18ce51-6ddd-11ea-92b4-ccda262736ce | https://socratic.org/questions/how-do-you-balance-the-following-chemical-equation-c-cl-4-c-cl-2 | CCl4 -> C + 2 Cl2 | start chemical_equation qc_end chemical_equation 8 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the following chemical equation"}] | [{"type":"chemical equation","value":"CCl4 -> C + 2 Cl2"}] | [{"type":"chemical equation","value":"CCl4 -> C + Cl2"}] | <h1 class="questionTitle" itemprop="name">How do you balance the following chemical equation? #C Cl_4 -> C + Cl_2#</h1> | null | CCl4 -> C + 2 Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>I usually just play around with the numbers until they come out even...</p>
<p>Hope I helped. </p></div>
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<div class="markdown"><p><mathjax>#2CCCl_"4" rarr 2C + 4Cl_"2"#</mathjax></p></div>
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<div class="markdown"><p>I usually just play around with the numbers until they come out even...</p>
<p>Hope I helped. </p></div>
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<div class="markdown"><p><mathjax>#2CCCl_"4" rarr 2C + 4Cl_"2"#</mathjax></p></div>
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<p>Hope I helped. </p></div>
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</article> | How do you balance the following chemical equation? #C Cl_4 -> C + Cl_2# | null |
1,443 | a85d0210-6ddd-11ea-b8da-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-which-contains-67-1-zinc-and-the-re | ZnO2 | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"ZnO2"}] | [{"type":"physical unit","value":"Percentage [OF] zinc in the compound [=] \\pu{67.1%}"},{"type":"other","value":"The compound contains zinc and the rest is oxygen."}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen? </h1> | null | ZnO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With all these problems, we assume that there are <mathjax>#100*g#</mathjax> of unknown compound, and then we proceed to calculate the empirical formula:</p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(67.1*g)/(65.4*g*mol^-1)=1.03*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.9*g)/(15.999*g*mol^-1)=2.06*mol#</mathjax></p>
<p>Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of <mathjax>#ZnO_2#</mathjax>. The examiner has been nasty in that the expectation would have been <mathjax>#ZnO#</mathjax>. However, zinc does form a peroxo species composed of <mathjax>#Zn^(2+)#</mathjax> and <mathjax>#O_2^(2-)#</mathjax> ions.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#ZnO_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With all these problems, we assume that there are <mathjax>#100*g#</mathjax> of unknown compound, and then we proceed to calculate the empirical formula:</p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(67.1*g)/(65.4*g*mol^-1)=1.03*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.9*g)/(15.999*g*mol^-1)=2.06*mol#</mathjax></p>
<p>Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of <mathjax>#ZnO_2#</mathjax>. The examiner has been nasty in that the expectation would have been <mathjax>#ZnO#</mathjax>. However, zinc does form a peroxo species composed of <mathjax>#Zn^(2+)#</mathjax> and <mathjax>#O_2^(2-)#</mathjax> ions.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen? </h1>
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<div class="markdown"><p><mathjax>#ZnO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With all these problems, we assume that there are <mathjax>#100*g#</mathjax> of unknown compound, and then we proceed to calculate the empirical formula:</p>
<p><mathjax>#"Moles of zinc"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(67.1*g)/(65.4*g*mol^-1)=1.03*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.9*g)/(15.999*g*mol^-1)=2.06*mol#</mathjax></p>
<p>Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of <mathjax>#ZnO_2#</mathjax>. The examiner has been nasty in that the expectation would have been <mathjax>#ZnO#</mathjax>. However, zinc does form a peroxo species composed of <mathjax>#Zn^(2+)#</mathjax> and <mathjax>#O_2^(2-)#</mathjax> ions.</p></div>
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</article> | What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen? | null |
1,444 | acc78b6e-6ddd-11ea-822f-ccda262736ce | https://socratic.org/questions/594177777c0149794032cc00 | 2.12 × 10^(-3) molar | start physical_unit 2 2 mole mol/l qc_end physical_unit 9 9 5 6 volume qc_end physical_unit 9 9 11 12 pressure qc_end physical_unit 9 9 14 15 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] zinc [IN] molar"}] | [{"type":"physical unit","value":"2.12 × 10^(-3) molar"}] | [{"type":"physical unit","value":"Volume [OF] dihydrogen [=] \\pu{47.5 mL}"},{"type":"physical unit","value":"Pressure [OF] dihydrogen [=] \\pu{1 atm}"},{"type":"physical unit","value":"Temperature [OF] dihydrogen [=] \\pu{273.15 K}"}] | <h1 class="questionTitle" itemprop="name">Oxidation of zinc produced a #47.5*mL# volume of dihydrogen at #1*atm# and #273.15*K#. What is the molar quantity of zinc that reacted?</h1> | null | 2.12 × 10^(-3) molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the redox reaction...........</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#</mathjax></p>
<p>We use the old Ideal gas equation to access the molar quantity....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx47.5*mLxx10^-3*L*mL^-1)/(0.0821*L*atm*K^-1*mol^-1xx273.15*K)#</mathjax></p>
<p><mathjax>#=2.12xx10^-3*mol#</mathjax>.</p>
<p>I am reluctant to use <mathjax>#"SATP"#</mathjax> because the values seem to change across curricula. You will have to adapt this question to whatever values of <mathjax>#"SATP"#</mathjax> apply. As a chemist, I tend to like measurement of pressure in <mathjax>#"atmospheres"#</mathjax>, and measurement of volume in <mathjax>#"millilitres"#</mathjax>. As a student you have to be adaptable. </p></div>
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<div class="markdown"><p>I will do this for <mathjax>#1*atm#</mathjax>, and <mathjax>#273.15*K#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the redox reaction...........</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#</mathjax></p>
<p>We use the old Ideal gas equation to access the molar quantity....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx47.5*mLxx10^-3*L*mL^-1)/(0.0821*L*atm*K^-1*mol^-1xx273.15*K)#</mathjax></p>
<p><mathjax>#=2.12xx10^-3*mol#</mathjax>.</p>
<p>I am reluctant to use <mathjax>#"SATP"#</mathjax> because the values seem to change across curricula. You will have to adapt this question to whatever values of <mathjax>#"SATP"#</mathjax> apply. As a chemist, I tend to like measurement of pressure in <mathjax>#"atmospheres"#</mathjax>, and measurement of volume in <mathjax>#"millilitres"#</mathjax>. As a student you have to be adaptable. </p></div>
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<h1 class="questionTitle" itemprop="name">Oxidation of zinc produced a #47.5*mL# volume of dihydrogen at #1*atm# and #273.15*K#. What is the molar quantity of zinc that reacted?</h1>
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<div class="markdown"><p>I will do this for <mathjax>#1*atm#</mathjax>, and <mathjax>#273.15*K#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We interrogate the redox reaction...........</p>
<p><mathjax>#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)#</mathjax></p>
<p>We use the old Ideal gas equation to access the molar quantity....</p>
<p><mathjax>#n=(PV)/(RT)=(1*atmxx47.5*mLxx10^-3*L*mL^-1)/(0.0821*L*atm*K^-1*mol^-1xx273.15*K)#</mathjax></p>
<p><mathjax>#=2.12xx10^-3*mol#</mathjax>.</p>
<p>I am reluctant to use <mathjax>#"SATP"#</mathjax> because the values seem to change across curricula. You will have to adapt this question to whatever values of <mathjax>#"SATP"#</mathjax> apply. As a chemist, I tend to like measurement of pressure in <mathjax>#"atmospheres"#</mathjax>, and measurement of volume in <mathjax>#"millilitres"#</mathjax>. As a student you have to be adaptable. </p></div>
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</article> | Oxidation of zinc produced a #47.5*mL# volume of dihydrogen at #1*atm# and #273.15*K#. What is the molar quantity of zinc that reacted? | null |
1,445 | acafd7a4-6ddd-11ea-ac3e-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-the-sulfur-atom-in-li-2so-4 | +6 | start physical_unit 7 8 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] sulfur atom"}] | [{"type":"physical unit","value":"+6"}] | [{"type":"chemical equation","value":"Li2SO4"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of the sulfur atom in #Li_2SO_4#?</h1> | null | +6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur in sulfate expresses it maximum oxidation state, its Group Number, <mathjax>#+VI#</mathjax>.</p>
<p>How? Well, as you know the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> MUST equal the charge on the ion..........</p>
<p>And so <mathjax>#S_"oxidation number"+4xxO_"oxidation number"=-2#</mathjax></p>
<p>Now oxygen generally assumes an oxidation number of <mathjax>#-II#</mathjax> in its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, and certainly it does so here.........</p>
<p><mathjax>#S_"oxidation number"+4xx(-2)=-2#</mathjax></p>
<p>Add <mathjax>#8#</mathjax> to both sides of the equation..........</p>
<p><mathjax>#S_"oxidation number"+cancel8+ cancel(4xx(-2))=-2+8#</mathjax></p>
<p><mathjax>#S_"oxidation number"=+6=+VI#</mathjax>..........</p>
<p>What about <mathjax>#"thiosulfate"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>? What are the oxidation numbers of sulfur here?</p></div>
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<div class="markdown"><p>We has <mathjax>#S^(VI+)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur in sulfate expresses it maximum oxidation state, its Group Number, <mathjax>#+VI#</mathjax>.</p>
<p>How? Well, as you know the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> MUST equal the charge on the ion..........</p>
<p>And so <mathjax>#S_"oxidation number"+4xxO_"oxidation number"=-2#</mathjax></p>
<p>Now oxygen generally assumes an oxidation number of <mathjax>#-II#</mathjax> in its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, and certainly it does so here.........</p>
<p><mathjax>#S_"oxidation number"+4xx(-2)=-2#</mathjax></p>
<p>Add <mathjax>#8#</mathjax> to both sides of the equation..........</p>
<p><mathjax>#S_"oxidation number"+cancel8+ cancel(4xx(-2))=-2+8#</mathjax></p>
<p><mathjax>#S_"oxidation number"=+6=+VI#</mathjax>..........</p>
<p>What about <mathjax>#"thiosulfate"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>? What are the oxidation numbers of sulfur here?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of the sulfur atom in #Li_2SO_4#?</h1>
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anor277
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<div class="markdown"><p>We has <mathjax>#S^(VI+)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur in sulfate expresses it maximum oxidation state, its Group Number, <mathjax>#+VI#</mathjax>.</p>
<p>How? Well, as you know the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> MUST equal the charge on the ion..........</p>
<p>And so <mathjax>#S_"oxidation number"+4xxO_"oxidation number"=-2#</mathjax></p>
<p>Now oxygen generally assumes an oxidation number of <mathjax>#-II#</mathjax> in its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, and certainly it does so here.........</p>
<p><mathjax>#S_"oxidation number"+4xx(-2)=-2#</mathjax></p>
<p>Add <mathjax>#8#</mathjax> to both sides of the equation..........</p>
<p><mathjax>#S_"oxidation number"+cancel8+ cancel(4xx(-2))=-2+8#</mathjax></p>
<p><mathjax>#S_"oxidation number"=+6=+VI#</mathjax>..........</p>
<p>What about <mathjax>#"thiosulfate"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>? What are the oxidation numbers of sulfur here?</p></div>
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<div class="markdown"><p><mathjax>#+6#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Charge of <mathjax>#Li_2SO_4#</mathjax> is balanced</p>
<p>The charge of <mathjax>#Li#</mathjax> is <mathjax>#+1#</mathjax> (because it is a <mathjax>#1^(st)#</mathjax> group element)</p>
<p>There are <mathjax>#2#</mathjax> <mathjax>#Li#</mathjax> atoms therefore the charge of <mathjax>#Li_2#</mathjax> is <mathjax>#+2#</mathjax> and <mathjax>#SO_4#</mathjax> must have a total charge of <mathjax>#-2#</mathjax> (to balance the charge)</p>
<p>Oxygen "in this case (because it is not a peroxide)" has a charge of <mathjax>#-2#</mathjax> (if it was a peroxide "<mathjax>#(O_2)^(2-)#</mathjax>" it would have had a charge of <mathjax>#-1#</mathjax>). Because there is <mathjax>#4#</mathjax> oxygen atoms, the total charge of oxygen will be <mathjax>#4*(-2)=color(red)(-8)#</mathjax></p>
<p>Sulfur must have a charge that will make( <mathjax>#color(red)(-8)+#</mathjax> its charge)<mathjax>#=-2#</mathjax> ("charge of <mathjax>#SO_4#</mathjax>")</p>
<p>Therefore the charge of sulfur <mathjax>#=-2+8=color(blue)(+6)#</mathjax></p></div>
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</article> | What is the oxidation number of the sulfur atom in #Li_2SO_4#? | null |
1,446 | aa890f81-6ddd-11ea-9dcb-ccda262736ce | https://socratic.org/questions/58ebe336b72cff3f95658746 | 1.67 × 10^(-3) mol/L | start physical_unit 5 6 concentration mol/l qc_end physical_unit 11 11 15 16 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] ammonium ion [IN] mol/L"}] | [{"type":"physical unit","value":"1.67 × 10^(-3) mol/L"}] | [{"type":"physical unit","value":"Concentration [OF] NH3(aq) solution [=] \\pu{0.157 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of ammonium ion in a solution of #NH_3(aq)# that is nominally #0.157*mol*L^-1#?</h1> | null | 1.67 × 10^(-3) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH#</mathjax>,</p>
<p>for which <mathjax>#K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#</mathjax></p>
<p>Initially, <mathjax>#[NH_3]=0.157*mol*L^-1#</mathjax>, and we assume some ammonia accepts a proton from water to give stoichiometric <mathjax>#NH_4^+#</mathjax> and <mathjax>#HO^-#</mathjax>. If we let the amount of protonation be <mathjax>#x#</mathjax>, then we write:</p>
<p><mathjax>#K_b=1.8xx10^-5=(x^2)/(0.157-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax>, which we could solve exactly, but we make that approximation that <mathjax>#0.157-x~=0.157#</mathjax>, and thus........</p>
<p><mathjax>#x_1~=sqrt(1.8xx10^-5xx0.157)=1.68xx10^-3#</mathjax></p>
<p>This is indeed small compared to <mathjax>#0.157#</mathjax>, but we could plug it in again, now that we have an approx. for <mathjax>#x#</mathjax>.</p>
<p><mathjax>#x_2=1.67xx10^-3#</mathjax>, and since the values have converged, I am willing to accept this value for our answer. </p>
<p>Now <mathjax>#x=[NH_4^+]=[HO^-]#</mathjax> by our definition. What is the <mathjax>#pH#</mathjax> of this solution?</p>
<p>I hope this agrees with your model answer............If you want something re-explained or rephrased, ax, and someone will help you. </p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#[NH_4^+]=1.67xx10^-3*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH#</mathjax>,</p>
<p>for which <mathjax>#K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#</mathjax></p>
<p>Initially, <mathjax>#[NH_3]=0.157*mol*L^-1#</mathjax>, and we assume some ammonia accepts a proton from water to give stoichiometric <mathjax>#NH_4^+#</mathjax> and <mathjax>#HO^-#</mathjax>. If we let the amount of protonation be <mathjax>#x#</mathjax>, then we write:</p>
<p><mathjax>#K_b=1.8xx10^-5=(x^2)/(0.157-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax>, which we could solve exactly, but we make that approximation that <mathjax>#0.157-x~=0.157#</mathjax>, and thus........</p>
<p><mathjax>#x_1~=sqrt(1.8xx10^-5xx0.157)=1.68xx10^-3#</mathjax></p>
<p>This is indeed small compared to <mathjax>#0.157#</mathjax>, but we could plug it in again, now that we have an approx. for <mathjax>#x#</mathjax>.</p>
<p><mathjax>#x_2=1.67xx10^-3#</mathjax>, and since the values have converged, I am willing to accept this value for our answer. </p>
<p>Now <mathjax>#x=[NH_4^+]=[HO^-]#</mathjax> by our definition. What is the <mathjax>#pH#</mathjax> of this solution?</p>
<p>I hope this agrees with your model answer............If you want something re-explained or rephrased, ax, and someone will help you. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of ammonium ion in a solution of #NH_3(aq)# that is nominally #0.157*mol*L^-1#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[NH_4^+]=1.67xx10^-3*mol*L^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We interrogate the equilibrium:</p>
<p><mathjax>#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH#</mathjax>,</p>
<p>for which <mathjax>#K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])#</mathjax></p>
<p>Initially, <mathjax>#[NH_3]=0.157*mol*L^-1#</mathjax>, and we assume some ammonia accepts a proton from water to give stoichiometric <mathjax>#NH_4^+#</mathjax> and <mathjax>#HO^-#</mathjax>. If we let the amount of protonation be <mathjax>#x#</mathjax>, then we write:</p>
<p><mathjax>#K_b=1.8xx10^-5=(x^2)/(0.157-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax>, which we could solve exactly, but we make that approximation that <mathjax>#0.157-x~=0.157#</mathjax>, and thus........</p>
<p><mathjax>#x_1~=sqrt(1.8xx10^-5xx0.157)=1.68xx10^-3#</mathjax></p>
<p>This is indeed small compared to <mathjax>#0.157#</mathjax>, but we could plug it in again, now that we have an approx. for <mathjax>#x#</mathjax>.</p>
<p><mathjax>#x_2=1.67xx10^-3#</mathjax>, and since the values have converged, I am willing to accept this value for our answer. </p>
<p>Now <mathjax>#x=[NH_4^+]=[HO^-]#</mathjax> by our definition. What is the <mathjax>#pH#</mathjax> of this solution?</p>
<p>I hope this agrees with your model answer............If you want something re-explained or rephrased, ax, and someone will help you. </p></div>
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</article> | What is the concentration of ammonium ion in a solution of #NH_3(aq)# that is nominally #0.157*mol*L^-1#? | null |
1,447 | aa154226-6ddd-11ea-8f1a-ccda262736ce | https://socratic.org/questions/if-the-initial-pressure-of-a-system-was-1-00-atm-and-the-volume-was-halved-and-t | 6.00 atm | start physical_unit 6 6 pressure atm qc_end physical_unit 6 6 8 9 pressure qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the system [IN] atm"}] | [{"type":"physical unit","value":"6.00 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the system [=] \\pu{1.00 atm}"},{"type":"other","value":"The volume was halved and the temperature in Kelvin was tripled."}] | <h1 class="questionTitle" itemprop="name">If the initial pressure of a system was 1.00 atm and the volume was halved and the temperature in Kelvin was tripled, what is the final pressure?</h1> | null | 6.00 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even without doing any calculation, you can say that the pressure of the sample will <strong>increase</strong> by factor of <mathjax>#6#</mathjax>. </p>
<p>Here's why that is the case. </p>
<p>The idea here is to determine if the <em>change in volume</em> and the <em>change in temperature</em> will <strong>counteract each other</strong> or not. </p>
<p>The way to do that is to keep one <em>constant</em> and observe what happens to the pressure when the second changes, and vice versa. </p>
<p>So, let's say that temperature <strong>is kept constant</strong> and volume is halved. As you know, pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<blockquote>
<p><mathjax>#color(blue)(P * V = "constant")#</mathjax></p>
</blockquote>
<p>This means that a <strong>reduction</strong> in volume by a factor of <mathjax>#2#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#2#</mathjax>. </p>
<p>Now let's say that volume <strong>is <a href="http://socratic.org/chemistry/chemical-equilibrium/kep">kep</a> constant</strong> and temperature is tripled. As you know, pressure and temperature have a <strong>direct relationship</strong> when volume and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<blockquote>
<p><mathjax>#color(blue)(P/T = "constant")#</mathjax></p>
</blockquote>
<p>This means that an <strong>increase</strong> in temperature by a factor of <mathjax>#3#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#3#</mathjax>. </p>
<p>This tells you that <em>decreasing the volume</em> <strong>and</strong> <em>increasing the temperature</em> will lead to an overall <strong>increase</strong> in pressure by a factor of <mathjax>#2 xx 3 = 6#</mathjax>. </p>
<p>You can prove this by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax> and use </p>
<blockquote>
<p><mathjax>#V_2 = V_1/2 ->#</mathjax> <em>the volume is <strong>halved</strong></em></p>
<p><mathjax>#T_2 = 3 * T_1 ->#</mathjax> <em>the temperature is <strong>tripled</strong></em></p>
</blockquote>
<p>You will get</p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = color(red)(cancel(color(black)(V_1))) * 2/(color(red)(cancel(color(black)(V_1)))) * (3 * color(red)(cancel(color(black)(T_1))))/(color(red)(cancel(color(black)(T_1)))) * P_1#</mathjax></p>
<p><mathjax>#P_2 = 6 * P_1#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#P_2 = 6 * "1.00 atm" = color(green)("6.00 atm")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_2 = "6.00 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even without doing any calculation, you can say that the pressure of the sample will <strong>increase</strong> by factor of <mathjax>#6#</mathjax>. </p>
<p>Here's why that is the case. </p>
<p>The idea here is to determine if the <em>change in volume</em> and the <em>change in temperature</em> will <strong>counteract each other</strong> or not. </p>
<p>The way to do that is to keep one <em>constant</em> and observe what happens to the pressure when the second changes, and vice versa. </p>
<p>So, let's say that temperature <strong>is kept constant</strong> and volume is halved. As you know, pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<blockquote>
<p><mathjax>#color(blue)(P * V = "constant")#</mathjax></p>
</blockquote>
<p>This means that a <strong>reduction</strong> in volume by a factor of <mathjax>#2#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#2#</mathjax>. </p>
<p>Now let's say that volume <strong>is <a href="http://socratic.org/chemistry/chemical-equilibrium/kep">kep</a> constant</strong> and temperature is tripled. As you know, pressure and temperature have a <strong>direct relationship</strong> when volume and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<blockquote>
<p><mathjax>#color(blue)(P/T = "constant")#</mathjax></p>
</blockquote>
<p>This means that an <strong>increase</strong> in temperature by a factor of <mathjax>#3#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#3#</mathjax>. </p>
<p>This tells you that <em>decreasing the volume</em> <strong>and</strong> <em>increasing the temperature</em> will lead to an overall <strong>increase</strong> in pressure by a factor of <mathjax>#2 xx 3 = 6#</mathjax>. </p>
<p>You can prove this by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax> and use </p>
<blockquote>
<p><mathjax>#V_2 = V_1/2 ->#</mathjax> <em>the volume is <strong>halved</strong></em></p>
<p><mathjax>#T_2 = 3 * T_1 ->#</mathjax> <em>the temperature is <strong>tripled</strong></em></p>
</blockquote>
<p>You will get</p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = color(red)(cancel(color(black)(V_1))) * 2/(color(red)(cancel(color(black)(V_1)))) * (3 * color(red)(cancel(color(black)(T_1))))/(color(red)(cancel(color(black)(T_1)))) * P_1#</mathjax></p>
<p><mathjax>#P_2 = 6 * P_1#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#P_2 = 6 * "1.00 atm" = color(green)("6.00 atm")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the initial pressure of a system was 1.00 atm and the volume was halved and the temperature in Kelvin was tripled, what is the final pressure?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#P_2 = "6.00 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even without doing any calculation, you can say that the pressure of the sample will <strong>increase</strong> by factor of <mathjax>#6#</mathjax>. </p>
<p>Here's why that is the case. </p>
<p>The idea here is to determine if the <em>change in volume</em> and the <em>change in temperature</em> will <strong>counteract each other</strong> or not. </p>
<p>The way to do that is to keep one <em>constant</em> and observe what happens to the pressure when the second changes, and vice versa. </p>
<p>So, let's say that temperature <strong>is kept constant</strong> and volume is halved. As you know, pressure and volume have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<blockquote>
<p><mathjax>#color(blue)(P * V = "constant")#</mathjax></p>
</blockquote>
<p>This means that a <strong>reduction</strong> in volume by a factor of <mathjax>#2#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#2#</mathjax>. </p>
<p>Now let's say that volume <strong>is <a href="http://socratic.org/chemistry/chemical-equilibrium/kep">kep</a> constant</strong> and temperature is tripled. As you know, pressure and temperature have a <strong>direct relationship</strong> when volume and number of moles of gas are kept constant - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<blockquote>
<p><mathjax>#color(blue)(P/T = "constant")#</mathjax></p>
</blockquote>
<p>This means that an <strong>increase</strong> in temperature by a factor of <mathjax>#3#</mathjax> will trigger an <strong>increase</strong> in pressure by the same factor of <mathjax>#3#</mathjax>. </p>
<p>This tells you that <em>decreasing the volume</em> <strong>and</strong> <em>increasing the temperature</em> will lead to an overall <strong>increase</strong> in pressure by a factor of <mathjax>#2 xx 3 = 6#</mathjax>. </p>
<p>You can prove this by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#P_2#</mathjax> and use </p>
<blockquote>
<p><mathjax>#V_2 = V_1/2 ->#</mathjax> <em>the volume is <strong>halved</strong></em></p>
<p><mathjax>#T_2 = 3 * T_1 ->#</mathjax> <em>the temperature is <strong>tripled</strong></em></p>
</blockquote>
<p>You will get</p>
<blockquote>
<p><mathjax>#P_2 = V_1/V_2 * T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = color(red)(cancel(color(black)(V_1))) * 2/(color(red)(cancel(color(black)(V_1)))) * (3 * color(red)(cancel(color(black)(T_1))))/(color(red)(cancel(color(black)(T_1)))) * P_1#</mathjax></p>
<p><mathjax>#P_2 = 6 * P_1#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#P_2 = 6 * "1.00 atm" = color(green)("6.00 atm")#</mathjax></p>
</blockquote></div>
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</article> | If the initial pressure of a system was 1.00 atm and the volume was halved and the temperature in Kelvin was tripled, what is the final pressure? | null |
1,448 | ab67e458-6ddd-11ea-85cb-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-7-5x10-3-m3-h-solution | 2.13 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] H+ solution"}] | [{"type":"physical unit","value":"2.13"}] | [{"type":"physical unit","value":"Molarity [OF] H+ solution [=] \\pu{7.5 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #7.5x10^-3 M3 #H^+# solution?</h1> | null | 2.13 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-(-2.13)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.13#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=2.13#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-(-2.13)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.13#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=2.13#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-(-2.13)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.13#</mathjax></p></div>
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</article> | What is the pH of a #7.5x10^-3 M3 #H^+# solution? | null |
1,449 | ab2828ac-6ddd-11ea-a624-ccda262736ce | https://socratic.org/questions/57fdad5fb72cff4ce364f698 | 0.28 grams | start physical_unit 3 4 mass g qc_end physical_unit 14 14 10 11 volume qc_end physical_unit 4 4 18 19 concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] magnesium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"0.28 grams"}] | [{"type":"physical unit","value":"Volume [OF] [Mg(OH)2] [=] \\pu{1 L}"},{"type":"physical unit","value":"Concentration [OF] hydroxide [=] \\pu{0.01 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What mass of #"magnesium hydroxide"# is required to give a #1*L# volume of #[Mg(OH)_2]# whose concentration is #0.01*mol*L^-1# with respect to the hydroxide?</h1> | null | 0.28 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.01*mol)/(1.0*L)#</mathjax>.</p>
<p>And thus we need to dissolve <mathjax>#0.01*molxx58.32*g*mol^-1xx1/2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.28*g#</mathjax> in a <mathjax>#1*L#</mathjax> volume. Why did I include the <mathjax>#1/2#</mathjax>?</p>
<p>However, <mathjax>#K_"sp",Mg(OH)_2=5.61×10^(−12)#</mathjax> at <mathjax>#298K#</mathjax>, which gives a solubility of <mathjax>#6.4xx10^-3*g*L^-1#</mathjax>. The question was thus not well-proposed. </p>
<p>Magnesium hydroxide is thus too insoluble to provide such a concentration. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For a <mathjax>#1*L#</mathjax> volume of <mathjax>#0.01*mol*L^-1#</mathjax> <mathjax>#Mg(OH)_2#</mathjax>, <mathjax>#0.28*g#</mathjax> salt are required. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.01*mol)/(1.0*L)#</mathjax>.</p>
<p>And thus we need to dissolve <mathjax>#0.01*molxx58.32*g*mol^-1xx1/2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.28*g#</mathjax> in a <mathjax>#1*L#</mathjax> volume. Why did I include the <mathjax>#1/2#</mathjax>?</p>
<p>However, <mathjax>#K_"sp",Mg(OH)_2=5.61×10^(−12)#</mathjax> at <mathjax>#298K#</mathjax>, which gives a solubility of <mathjax>#6.4xx10^-3*g*L^-1#</mathjax>. The question was thus not well-proposed. </p>
<p>Magnesium hydroxide is thus too insoluble to provide such a concentration. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of #"magnesium hydroxide"# is required to give a #1*L# volume of #[Mg(OH)_2]# whose concentration is #0.01*mol*L^-1# with respect to the hydroxide?</h1>
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<div class="markdown"><p>For a <mathjax>#1*L#</mathjax> volume of <mathjax>#0.01*mol*L^-1#</mathjax> <mathjax>#Mg(OH)_2#</mathjax>, <mathjax>#0.28*g#</mathjax> salt are required. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.01*mol)/(1.0*L)#</mathjax>.</p>
<p>And thus we need to dissolve <mathjax>#0.01*molxx58.32*g*mol^-1xx1/2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.28*g#</mathjax> in a <mathjax>#1*L#</mathjax> volume. Why did I include the <mathjax>#1/2#</mathjax>?</p>
<p>However, <mathjax>#K_"sp",Mg(OH)_2=5.61×10^(−12)#</mathjax> at <mathjax>#298K#</mathjax>, which gives a solubility of <mathjax>#6.4xx10^-3*g*L^-1#</mathjax>. The question was thus not well-proposed. </p>
<p>Magnesium hydroxide is thus too insoluble to provide such a concentration. </p></div>
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</article> | What mass of #"magnesium hydroxide"# is required to give a #1*L# volume of #[Mg(OH)_2]# whose concentration is #0.01*mol*L^-1# with respect to the hydroxide? | null |
1,450 | a9577010-6ddd-11ea-89ff-ccda262736ce | https://socratic.org/questions/how-would-you-find-the-net-ionic-equation-of-hcl-zn-h2-zncl2 | Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(gk) | start chemical_equation qc_end physical_unit 37 38 33 34 molarity qc_end chemical_equation 9 15 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(gk)"}] | [{"type":"physical unit","value":"Molarity [OF] hydrochloric acid solution [=] \\pu{1.0 M}"},{"type":"physical unit","value":"Temperature [OF] hydrochloric acid solution [=] \\pu{25 ℃}"},{"type":"chemical equation","value":"HCl + Zn -> H2 + ZnCl2"}] | <h1 class="questionTitle" itemprop="name">How would you find the net ionic equation of #"HCl" + "Zn" -> "H"_2 + "ZnCl"_2# ?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>This information was given along with the question:</p>
<p>A piece of zinc metal is placed in a 1.0 M solution of hydrochloric acid at 25 C°.</p></div>
</h2>
</div>
</div> | Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(gk) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, make sure that you have a <em>balanced</em> equation to work with. To balance the equation given to you, multiply the hydrochloric acid by <mathjax>#2#</mathjax></p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>You know that hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to produce hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and chloride anions</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now ,zinc chloride, <mathjax>#"ZnCl"_2#</mathjax>, is <strong>soluble</strong> in aqueous solution, which means that it too will exist as <em>ions</em> </p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Now, the <strong>net ionic equation</strong> is obtained by removing the <em>spectator ions</em>, i.e. the ions that are present on both sides of the equation. </p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>The net ionic equation that describes this single replacement reaction will thus be</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g)))))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, make sure that you have a <em>balanced</em> equation to work with. To balance the equation given to you, multiply the hydrochloric acid by <mathjax>#2#</mathjax></p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>You know that hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to produce hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and chloride anions</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now ,zinc chloride, <mathjax>#"ZnCl"_2#</mathjax>, is <strong>soluble</strong> in aqueous solution, which means that it too will exist as <em>ions</em> </p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Now, the <strong>net ionic equation</strong> is obtained by removing the <em>spectator ions</em>, i.e. the ions that are present on both sides of the equation. </p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>The net ionic equation that describes this single replacement reaction will thus be</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g)))))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you find the net ionic equation of #"HCl" + "Zn" -> "H"_2 + "ZnCl"_2# ?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>This information was given along with the question:</p>
<p>A piece of zinc metal is placed in a 1.0 M solution of hydrochloric acid at 25 C°.</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-12-20T00:03:07" itemprop="dateCreated">
Dec 20, 2016
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<div class="markdown"><p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, make sure that you have a <em>balanced</em> equation to work with. To balance the equation given to you, multiply the hydrochloric acid by <mathjax>#2#</mathjax></p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>You know that hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates completely in aqueous solution to produce hydrogen ions, <mathjax>#"H"^(+)#</mathjax>, and chloride anions</p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now ,zinc chloride, <mathjax>#"ZnCl"_2#</mathjax>, is <strong>soluble</strong> in aqueous solution, which means that it too will exist as <em>ions</em> </p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that you can write</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Now, the <strong>net ionic equation</strong> is obtained by removing the <em>spectator ions</em>, i.e. the ions that are present on both sides of the equation. </p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>The net ionic equation that describes this single replacement reaction will thus be</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g)))))#</mathjax></p>
</blockquote></div>
</div>
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</article> | How would you find the net ionic equation of #"HCl" + "Zn" -> "H"_2 + "ZnCl"_2# ? |
This information was given along with the question:
A piece of zinc metal is placed in a 1.0 M solution of hydrochloric acid at 25 C°.
|
1,451 | abde5e10-6ddd-11ea-a810-ccda262736ce | https://socratic.org/questions/how-much-heat-does-it-take-to-raise-the-temperature-of-9-0-kg-of-wood-from-12-c- | 1285.2 kJ | start physical_unit 14 14 heat_energy kj qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 14 14 19 20 temperature qc_end physical_unit 14 14 27 30 specific_heat qc_end end | [{"type":"physical unit","value":"Taken heat [OF] wood [IN] kJ"}] | [{"type":"physical unit","value":"1285.2 kJ"}] | [{"type":"physical unit","value":"Mass [OF] wood [=] \\pu{9.0 kg}"},{"type":"physical unit","value":"Temperature1 [OF] wood [=] \\pu{12 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] wood [=] \\pu{97 ℃}"},{"type":"physical unit","value":"Specific heat [OF] wood [=] \\pu{1.68 kJ/(kg * ℃)}"}] | <h1 class="questionTitle" itemprop="name">How much heat does it take to raise the temperature of 9.0 kg of wood from 12°C to 97*C? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat of wood is 1.68 kJ/kg•°C. </p></div>
</h2>
</div>
</div> | 1285.2 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Heat energy = mass x <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity x temperature difference</p>
<p>Heat energy = 9.0 x 1.68 x (97-12)</p>
<p>Heat energy = 9.0 x 1.68 x 85</p>
<p>Heat energy = 1285.2kJ</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>1285.2kJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Heat energy = mass x <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity x temperature difference</p>
<p>Heat energy = 9.0 x 1.68 x (97-12)</p>
<p>Heat energy = 9.0 x 1.68 x 85</p>
<p>Heat energy = 1285.2kJ</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much heat does it take to raise the temperature of 9.0 kg of wood from 12°C to 97*C? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat of wood is 1.68 kJ/kg•°C. </p></div>
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<div class="markdown"><p>1285.2kJ</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Heat energy = mass x <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity x temperature difference</p>
<p>Heat energy = 9.0 x 1.68 x (97-12)</p>
<p>Heat energy = 9.0 x 1.68 x 85</p>
<p>Heat energy = 1285.2kJ</p></div>
</div>
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</article> | How much heat does it take to raise the temperature of 9.0 kg of wood from 12°C to 97*C? |
The specific heat of wood is 1.68 kJ/kg•°C.
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1,452 | ac11e7e9-6ddd-11ea-9ff9-ccda262736ce | https://socratic.org/questions/when-the-equation-c-2h-4-o-2-co-2-h-2o-is-balanced-using-smallest-whole-numbers- | 3 | start physical_unit 5 5 coefficient none qc_end chemical_equation 3 9 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Coefficient [OF] O2"}] | [{"type":"physical unit","value":"3"}] | [{"type":"chemical equation","value":"C2H4 + O2 -> CO2 + H2O"},{"type":"other","value":"Use smallest whole numbers."}] | <h1 class="questionTitle" itemprop="name">When the equation #C_2H_4+O_2 -> CO_2+H_2O# is balanced using smallest whole numbers, what is the coefficient of the #O_2#?</h1> | null | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer is <mathjax>#3#</mathjax>. Is this balanced? For every reactant particle, is there a corresponding product particle? There are <mathjax>#2C#</mathjax> reactant particles, and <mathjax>#2C#</mathjax> product particles. You have to count up the number of remaining hydrogen and oxygen reactant particles, and see if these correspond to the number of hydrogen and oxygen product particles.</p>
<p>If these numbers do not agree, you know that the reaction is unbalanced; therefore, it is not consistent with physical reality. </p></div>
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<div class="markdown"><p><mathjax>#C_2H_4(g) + 3O_2(g) rarr 2CO_2(g) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer is <mathjax>#3#</mathjax>. Is this balanced? For every reactant particle, is there a corresponding product particle? There are <mathjax>#2C#</mathjax> reactant particles, and <mathjax>#2C#</mathjax> product particles. You have to count up the number of remaining hydrogen and oxygen reactant particles, and see if these correspond to the number of hydrogen and oxygen product particles.</p>
<p>If these numbers do not agree, you know that the reaction is unbalanced; therefore, it is not consistent with physical reality. </p></div>
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<h1 class="questionTitle" itemprop="name">When the equation #C_2H_4+O_2 -> CO_2+H_2O# is balanced using smallest whole numbers, what is the coefficient of the #O_2#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#C_2H_4(g) + 3O_2(g) rarr 2CO_2(g) + 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>The answer is <mathjax>#3#</mathjax>. Is this balanced? For every reactant particle, is there a corresponding product particle? There are <mathjax>#2C#</mathjax> reactant particles, and <mathjax>#2C#</mathjax> product particles. You have to count up the number of remaining hydrogen and oxygen reactant particles, and see if these correspond to the number of hydrogen and oxygen product particles.</p>
<p>If these numbers do not agree, you know that the reaction is unbalanced; therefore, it is not consistent with physical reality. </p></div>
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</article> | When the equation #C_2H_4+O_2 -> CO_2+H_2O# is balanced using smallest whole numbers, what is the coefficient of the #O_2#? | null |
1,453 | aa5f1298-6ddd-11ea-bb8d-ccda262736ce | https://socratic.org/questions/how-many-grams-of-potassium-chloride-are-produced-if-25-0g-of-potassium-chlorate | 15.25 grams | start physical_unit 4 5 mass g qc_end physical_unit 12 13 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium chloride [IN] grams"}] | [{"type":"physical unit","value":"15.25 grams"}] | [{"type":"physical unit","value":"Mass [OF] potassium chlorate [=] \\pu{25.0 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of potassium chloride are produced if 25.0g of potassium chlorate decompose?</h1> | null | 15.25 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want. </p>
<p>change 25 grams of potassium chlorate to moles. </p>
<p>calculate the gram molecular mass of potassium chlorate. </p>
<p>Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.</p>
<p>1 K x 39 grams/mole <br/>
+1 Cl x 35.4 grams/ mole<br/>
+3 O x 16 grams/ mole<br/>
= 122.4 grams / mole Potassium Chlorate</p>
<p><mathjax>#25 /122.4#</mathjax> = moles. </p>
<p>2.05 moles of Potassium Chlorate. </p>
<p>There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride. </p>
<p>2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride. </p>
<p>Find the gram molecular mass of Potassium Chloride. </p>
<p>1 K x 39 = 39<br/>
+1 Cl x 35.4 = 35.4<br/>
= 74.4 grams / mole.</p>
<p>2.05 moles x 74.4 grams/ mole = 15.2 grams </p></div>
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<div class="markdown"><p>approximately 15.1 grams. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want. </p>
<p>change 25 grams of potassium chlorate to moles. </p>
<p>calculate the gram molecular mass of potassium chlorate. </p>
<p>Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.</p>
<p>1 K x 39 grams/mole <br/>
+1 Cl x 35.4 grams/ mole<br/>
+3 O x 16 grams/ mole<br/>
= 122.4 grams / mole Potassium Chlorate</p>
<p><mathjax>#25 /122.4#</mathjax> = moles. </p>
<p>2.05 moles of Potassium Chlorate. </p>
<p>There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride. </p>
<p>2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride. </p>
<p>Find the gram molecular mass of Potassium Chloride. </p>
<p>1 K x 39 = 39<br/>
+1 Cl x 35.4 = 35.4<br/>
= 74.4 grams / mole.</p>
<p>2.05 moles x 74.4 grams/ mole = 15.2 grams </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of potassium chloride are produced if 25.0g of potassium chlorate decompose?</h1>
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<div class="markdown"><p>approximately 15.1 grams. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want. </p>
<p>change 25 grams of potassium chlorate to moles. </p>
<p>calculate the gram molecular mass of potassium chlorate. </p>
<p>Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.</p>
<p>1 K x 39 grams/mole <br/>
+1 Cl x 35.4 grams/ mole<br/>
+3 O x 16 grams/ mole<br/>
= 122.4 grams / mole Potassium Chlorate</p>
<p><mathjax>#25 /122.4#</mathjax> = moles. </p>
<p>2.05 moles of Potassium Chlorate. </p>
<p>There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride. </p>
<p>2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride. </p>
<p>Find the gram molecular mass of Potassium Chloride. </p>
<p>1 K x 39 = 39<br/>
+1 Cl x 35.4 = 35.4<br/>
= 74.4 grams / mole.</p>
<p>2.05 moles x 74.4 grams/ mole = 15.2 grams </p></div>
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Ernest Z.
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Aug 16, 2016
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<div class="markdown"><p>The reaction will produce 15.3 g of <mathjax>#"KCl"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong>.</p>
<p><mathjax>#"2KClO"_3 → "2KCl" + "3O"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"KClO"_3#</mathjax></strong>.</p>
<p><mathjax>#"Moles of KClO"_3 = 25.0 color(red)(cancel(color(black)("g KClO"_3))) × ("1 mol KClO"_3)/(122.55 color(red)(cancel(color(black)("g KClO"_3)))) = "0.2046 mol KClO"_3"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"KCl""#</mathjax></strong>.</p>
<p><mathjax>#"Moles of KCl" = 0.2046 color(red)(cancel(color(black)("mol KClO"_3))) × ("2 mol KCl")/(2 color(red)(cancel(color(black)("mol KClO"_3)))) = "0.2046 mol KCl"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"KCl"#</mathjax></strong>.</p>
<p><mathjax>#"Mass of KCl" = 0.2046 color(red)(cancel(color(black)("mol KCl"))) × ("74.55 g KCl")/(1 color(red)(cancel(color(black)("mol KCl")))) = "15.3 g KCl"#</mathjax></p></div>
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</article> | How many grams of potassium chloride are produced if 25.0g of potassium chlorate decompose? | null |
1,454 | ac8f1022-6ddd-11ea-a00d-ccda262736ce | https://socratic.org/questions/molar-mass-of-substance-a-in-material-eu-a-3-is-238g-mol-0-476g-of-substance-a-w | C16H14O4 | start chemical_formula qc_end physical_unit 3 4 9 10 molar_mass qc_end physical_unit 23 23 20 21 mass qc_end physical_unit 28 28 25 26 mass qc_end physical_unit 3 4 11 12 mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] substance A [IN] default"}] | [{"type":"chemical equation","value":"C16H14O4"}] | [{"type":"physical unit","value":"Molar mass [OF] substance A [=] \\pu{238 g/mol}"},{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{1.408 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{0.252 g}"},{"type":"physical unit","value":"Mass [OF] substance A [=] \\pu{0.476 g}"},{"type":"other","value":"Substance A is made up of carbon, hydrogen and oxygen. "}] | <h1 class="questionTitle" itemprop="name">Molar mass of substance A in material Eu(A)3 is 238g/mol. 0.476g of substance A was burned and produced 1.408g of CO2 and 0.252g of H2O. Substance A is made up of carbon, hydrogen and oxygen. What is the chemical formula for substance A?</h1> | null | C16H14O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The way I see it, the part about how</p>
<blockquote>
<blockquote>
<p><em>The molar mass of substance</em> <mathjax>#"A"#</mathjax> <em>in material</em> <mathjax>#"Eu"("A")_3#</mathjax> <em>is</em> <mathjax>#"238 g mol"^(-1)#</mathjax> </p>
</blockquote>
</blockquote>
<p>does not make sense because the <strong>molar mass</strong> of a substance, which represents the mass of <strong>one mole</strong> of said substance, is <strong>constant</strong>. It could be that the problem meant something else entirely, but I'm not sure what that something else could be. </p>
<p>So my instinct would be to take the info given by the problem and say that the molar mass of substance <mathjax>#"A"#</mathjax> is <mathjax>#"238 g mol"^(-1)#</mathjax>. </p>
<p>Even if that is <em><strong>not</strong></em> the correct interpretation, you can use these techniques to find the molecular formula of substance <mathjax>#"A"#</mathjax> in any given context. </p>
<p>Now, the idea here is that <strong>all the carbon</strong> that was initially present in the sample is now present in the carbon dioxide, <mathjax>#"CO"_2#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#1.408 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole CO"_2))))/(44.01color(red)(cancel(color(black)("g")))) * (1color(red)(cancel(color(black)("mole C"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.38426 g C"#</mathjax></p>
</blockquote>
<p>Likewise, <strong>all the hydrogen</strong> that was initially present in the sample is now present in the water, <mathjax>#"H"_2"O"#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of hydrogen, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#0.252 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/(18.015color(red)(cancel(color(black)("g")))) * (color(blue)(2)color(white)(a)"moles H")/(1color(red)(cancel(color(black)("mole H"_2"O")))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.028120 g H"#</mathjax></p>
</blockquote>
<p>This means that the sample of substance <mathjax>#"A"#</mathjax> also contained </p>
<blockquote>
<p><mathjax>#m_"A" = m_"C" + m_"O" + m_"H"#</mathjax></p>
<p><mathjax>#m_"O" = "0.476 g" - ("0.38426 g" + "0.028120 g")#</mathjax></p>
<p><mathjax>#m_"O" = "0.063620 g O"#</mathjax></p>
</blockquote>
<p>At this point, you must figure out how many <strong>moles</strong> of each element you have in that sample. This will allow you to calculate the <em>empirical formula</em> of substance <mathjax>#"A"#</mathjax>.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " 0.38426 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.031992 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 0.028120 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.027898 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 0.063620 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "0.0039764 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exist between the three <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound, divide all values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For C: " (0.031992 color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 8.045 ~~ 8#</mathjax></p>
<p><mathjax>#"For H: " (0.027898color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 7.016 ~~ 7#</mathjax></p>
<p><mathjax>#"For O: " (0.0039764color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#8:7:1#</mathjax> is the <strong>smallest whole number ratio</strong> that can exist between the three elements, it follows that the empirical formula for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#"C"_8"H"_7"O"_1 ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote>
<p>Now, if you take <mathjax>#"238 g mol"^(-1)#</mathjax> to be the <strong>molar mass</strong> of substance <mathjax>#"A"#</mathjax>, you can say that you must have </p>
<blockquote>
<p><mathjax>#(8 xx 12.011 color(red)(cancel(color(black)("g mol"^(-1)))) + 7 xx 1.00794color(red)(cancel(color(black)("g mol"^(-1)))) + 1 xx 15.9994color(red)(cancel(color(black)("g mol"^(-1))))) xx color(blue)(n) = 238color(red)(cancel(color(black)("g mol"^(-1))))#</mathjax></p>
<p><mathjax>#119.43 * color(blue)(n) = 238 implies color(blue)(n) = 238/119.143 = 1.9976 ~~ 2#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#("C"_8"H"_7"O"_1)_color(blue)(2) = color(green)(|bar(ul(color(white)(a/a)color(black)("C"_16"H"_14"O"_2)color(white)(a/a)|))) ->#</mathjax> <em><strong>molecular formula</strong></em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_16"H"_14"O"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The way I see it, the part about how</p>
<blockquote>
<blockquote>
<p><em>The molar mass of substance</em> <mathjax>#"A"#</mathjax> <em>in material</em> <mathjax>#"Eu"("A")_3#</mathjax> <em>is</em> <mathjax>#"238 g mol"^(-1)#</mathjax> </p>
</blockquote>
</blockquote>
<p>does not make sense because the <strong>molar mass</strong> of a substance, which represents the mass of <strong>one mole</strong> of said substance, is <strong>constant</strong>. It could be that the problem meant something else entirely, but I'm not sure what that something else could be. </p>
<p>So my instinct would be to take the info given by the problem and say that the molar mass of substance <mathjax>#"A"#</mathjax> is <mathjax>#"238 g mol"^(-1)#</mathjax>. </p>
<p>Even if that is <em><strong>not</strong></em> the correct interpretation, you can use these techniques to find the molecular formula of substance <mathjax>#"A"#</mathjax> in any given context. </p>
<p>Now, the idea here is that <strong>all the carbon</strong> that was initially present in the sample is now present in the carbon dioxide, <mathjax>#"CO"_2#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#1.408 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole CO"_2))))/(44.01color(red)(cancel(color(black)("g")))) * (1color(red)(cancel(color(black)("mole C"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.38426 g C"#</mathjax></p>
</blockquote>
<p>Likewise, <strong>all the hydrogen</strong> that was initially present in the sample is now present in the water, <mathjax>#"H"_2"O"#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of hydrogen, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#0.252 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/(18.015color(red)(cancel(color(black)("g")))) * (color(blue)(2)color(white)(a)"moles H")/(1color(red)(cancel(color(black)("mole H"_2"O")))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.028120 g H"#</mathjax></p>
</blockquote>
<p>This means that the sample of substance <mathjax>#"A"#</mathjax> also contained </p>
<blockquote>
<p><mathjax>#m_"A" = m_"C" + m_"O" + m_"H"#</mathjax></p>
<p><mathjax>#m_"O" = "0.476 g" - ("0.38426 g" + "0.028120 g")#</mathjax></p>
<p><mathjax>#m_"O" = "0.063620 g O"#</mathjax></p>
</blockquote>
<p>At this point, you must figure out how many <strong>moles</strong> of each element you have in that sample. This will allow you to calculate the <em>empirical formula</em> of substance <mathjax>#"A"#</mathjax>.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " 0.38426 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.031992 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 0.028120 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.027898 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 0.063620 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "0.0039764 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exist between the three <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound, divide all values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For C: " (0.031992 color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 8.045 ~~ 8#</mathjax></p>
<p><mathjax>#"For H: " (0.027898color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 7.016 ~~ 7#</mathjax></p>
<p><mathjax>#"For O: " (0.0039764color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#8:7:1#</mathjax> is the <strong>smallest whole number ratio</strong> that can exist between the three elements, it follows that the empirical formula for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#"C"_8"H"_7"O"_1 ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote>
<p>Now, if you take <mathjax>#"238 g mol"^(-1)#</mathjax> to be the <strong>molar mass</strong> of substance <mathjax>#"A"#</mathjax>, you can say that you must have </p>
<blockquote>
<p><mathjax>#(8 xx 12.011 color(red)(cancel(color(black)("g mol"^(-1)))) + 7 xx 1.00794color(red)(cancel(color(black)("g mol"^(-1)))) + 1 xx 15.9994color(red)(cancel(color(black)("g mol"^(-1))))) xx color(blue)(n) = 238color(red)(cancel(color(black)("g mol"^(-1))))#</mathjax></p>
<p><mathjax>#119.43 * color(blue)(n) = 238 implies color(blue)(n) = 238/119.143 = 1.9976 ~~ 2#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#("C"_8"H"_7"O"_1)_color(blue)(2) = color(green)(|bar(ul(color(white)(a/a)color(black)("C"_16"H"_14"O"_2)color(white)(a/a)|))) ->#</mathjax> <em><strong>molecular formula</strong></em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Molar mass of substance A in material Eu(A)3 is 238g/mol. 0.476g of substance A was burned and produced 1.408g of CO2 and 0.252g of H2O. Substance A is made up of carbon, hydrogen and oxygen. What is the chemical formula for substance A?</h1>
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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.
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<span class="dateCreated" datetime="2016-06-21T14:07:50" itemprop="dateCreated">
Jun 21, 2016
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<div class="markdown"><p><mathjax>#"C"_16"H"_14"O"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The way I see it, the part about how</p>
<blockquote>
<blockquote>
<p><em>The molar mass of substance</em> <mathjax>#"A"#</mathjax> <em>in material</em> <mathjax>#"Eu"("A")_3#</mathjax> <em>is</em> <mathjax>#"238 g mol"^(-1)#</mathjax> </p>
</blockquote>
</blockquote>
<p>does not make sense because the <strong>molar mass</strong> of a substance, which represents the mass of <strong>one mole</strong> of said substance, is <strong>constant</strong>. It could be that the problem meant something else entirely, but I'm not sure what that something else could be. </p>
<p>So my instinct would be to take the info given by the problem and say that the molar mass of substance <mathjax>#"A"#</mathjax> is <mathjax>#"238 g mol"^(-1)#</mathjax>. </p>
<p>Even if that is <em><strong>not</strong></em> the correct interpretation, you can use these techniques to find the molecular formula of substance <mathjax>#"A"#</mathjax> in any given context. </p>
<p>Now, the idea here is that <strong>all the carbon</strong> that was initially present in the sample is now present in the carbon dioxide, <mathjax>#"CO"_2#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#1.408 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole CO"_2))))/(44.01color(red)(cancel(color(black)("g")))) * (1color(red)(cancel(color(black)("mole C"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.38426 g C"#</mathjax></p>
</blockquote>
<p>Likewise, <strong>all the hydrogen</strong> that was initially present in the sample is now present in the water, <mathjax>#"H"_2"O"#</mathjax>. Since <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of hydrogen, you can say that the sample contained</p>
<blockquote>
<p><mathjax>#0.252 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/(18.015color(red)(cancel(color(black)("g")))) * (color(blue)(2)color(white)(a)"moles H")/(1color(red)(cancel(color(black)("mole H"_2"O")))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.028120 g H"#</mathjax></p>
</blockquote>
<p>This means that the sample of substance <mathjax>#"A"#</mathjax> also contained </p>
<blockquote>
<p><mathjax>#m_"A" = m_"C" + m_"O" + m_"H"#</mathjax></p>
<p><mathjax>#m_"O" = "0.476 g" - ("0.38426 g" + "0.028120 g")#</mathjax></p>
<p><mathjax>#m_"O" = "0.063620 g O"#</mathjax></p>
</blockquote>
<p>At this point, you must figure out how many <strong>moles</strong> of each element you have in that sample. This will allow you to calculate the <em>empirical formula</em> of substance <mathjax>#"A"#</mathjax>.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For C: " 0.38426 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.031992 moles C"#</mathjax></p>
<p><mathjax>#"For H: " 0.028120 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.027898 moles H"#</mathjax></p>
<p><mathjax>#"For O: " 0.063620 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "0.0039764 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exist between the three <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound, divide all values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For C: " (0.031992 color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 8.045 ~~ 8#</mathjax></p>
<p><mathjax>#"For H: " (0.027898color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 7.016 ~~ 7#</mathjax></p>
<p><mathjax>#"For O: " (0.0039764color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since <mathjax>#8:7:1#</mathjax> is the <strong>smallest whole number ratio</strong> that can exist between the three elements, it follows that the empirical formula for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#"C"_8"H"_7"O"_1 ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote>
<p>Now, if you take <mathjax>#"238 g mol"^(-1)#</mathjax> to be the <strong>molar mass</strong> of substance <mathjax>#"A"#</mathjax>, you can say that you must have </p>
<blockquote>
<p><mathjax>#(8 xx 12.011 color(red)(cancel(color(black)("g mol"^(-1)))) + 7 xx 1.00794color(red)(cancel(color(black)("g mol"^(-1)))) + 1 xx 15.9994color(red)(cancel(color(black)("g mol"^(-1))))) xx color(blue)(n) = 238color(red)(cancel(color(black)("g mol"^(-1))))#</mathjax></p>
<p><mathjax>#119.43 * color(blue)(n) = 238 implies color(blue)(n) = 238/119.143 = 1.9976 ~~ 2#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for substance <mathjax>#"A"#</mathjax> is </p>
<blockquote>
<p><mathjax>#("C"_8"H"_7"O"_1)_color(blue)(2) = color(green)(|bar(ul(color(white)(a/a)color(black)("C"_16"H"_14"O"_2)color(white)(a/a)|))) ->#</mathjax> <em><strong>molecular formula</strong></em></p>
</blockquote></div>
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</article> | Molar mass of substance A in material Eu(A)3 is 238g/mol. 0.476g of substance A was burned and produced 1.408g of CO2 and 0.252g of H2O. Substance A is made up of carbon, hydrogen and oxygen. What is the chemical formula for substance A? | null |
1,455 | a8e4124a-6ddd-11ea-848d-ccda262736ce | https://socratic.org/questions/58eb4553b72cff1036e0cbf9 | PbO2(s) + 4 HCl(aq) -> PbCl2(s) + 2 H2O(l) + Cl2(g) | start chemical_equation qc_end substance 1 2 qc_end chemical_equation 4 4 qc_end substance 6 7 qc_end substance 9 10 qc_end substance 12 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"PbO2(s) + 4 HCl(aq) -> PbCl2(s) + 2 H2O(l) + Cl2(g)"}] | [{"type":"substance name","value":"Hydrochloric acid"},{"type":"chemical equation","value":"PbCl4"},{"type":"substance name","value":"Lead metal"},{"type":"substance name","value":"Oxidize chloride"},{"type":"substance name","value":"Chlorine gas"}] | <h1 class="questionTitle" itemprop="name">How hydrochloric acid reduce #PbCl_4# to lead metal, and oxidize chloride to chlorine gas?</h1> | null | PbO2(s) + 4 HCl(aq) -> PbCl2(s) + 2 H2O(l) + Cl2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P(+IV)#</mathjax> is reduced to <mathjax>#Pb(+II)#</mathjax>:</p>
<p><mathjax>#Pb^(4+) + 2e^(-) rarr Pb^(2+)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2(g) + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And thus <mathjax>#(i) + 2xx(ii):#</mathjax></p>
<p><mathjax>#Pb^(4+) + 2Cl^(-) + 2e^(-) rarr Pb^(2+) +Cl_2(g) +2e^-#</mathjax></p>
<p>Which is balanced with respect to mass and charge. </p>
<p>We can write a more conventional stoichiometric equation, by substituting the ions for the parent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.......</p>
<p><mathjax>#PbO_2 + 4HCl rarr PbCl_2(s)darr +Cl_2(g)uarr +2H_2O#</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#PbO_2(s) + 4HCl(aq) rarr PbCl_2(s) +2H_2O(l) + Cl_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P(+IV)#</mathjax> is reduced to <mathjax>#Pb(+II)#</mathjax>:</p>
<p><mathjax>#Pb^(4+) + 2e^(-) rarr Pb^(2+)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2(g) + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And thus <mathjax>#(i) + 2xx(ii):#</mathjax></p>
<p><mathjax>#Pb^(4+) + 2Cl^(-) + 2e^(-) rarr Pb^(2+) +Cl_2(g) +2e^-#</mathjax></p>
<p>Which is balanced with respect to mass and charge. </p>
<p>We can write a more conventional stoichiometric equation, by substituting the ions for the parent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.......</p>
<p><mathjax>#PbO_2 + 4HCl rarr PbCl_2(s)darr +Cl_2(g)uarr +2H_2O#</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How hydrochloric acid reduce #PbCl_4# to lead metal, and oxidize chloride to chlorine gas?</h1>
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anor277
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<div class="markdown"><p><mathjax>#PbO_2(s) + 4HCl(aq) rarr PbCl_2(s) +2H_2O(l) + Cl_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P(+IV)#</mathjax> is reduced to <mathjax>#Pb(+II)#</mathjax>:</p>
<p><mathjax>#Pb^(4+) + 2e^(-) rarr Pb^(2+)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2(g) + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And thus <mathjax>#(i) + 2xx(ii):#</mathjax></p>
<p><mathjax>#Pb^(4+) + 2Cl^(-) + 2e^(-) rarr Pb^(2+) +Cl_2(g) +2e^-#</mathjax></p>
<p>Which is balanced with respect to mass and charge. </p>
<p>We can write a more conventional stoichiometric equation, by substituting the ions for the parent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.......</p>
<p><mathjax>#PbO_2 + 4HCl rarr PbCl_2(s)darr +Cl_2(g)uarr +2H_2O#</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
</div>
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</article> | How hydrochloric acid reduce #PbCl_4# to lead metal, and oxidize chloride to chlorine gas? | null |
1,456 | ac5e645d-6ddd-11ea-b250-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-dinitrogen-tetroxide | N2O4 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] dinitrogen tetroxide [IN] default"}] | [{"type":"chemical equation","value":"N2O4"}] | [{"type":"substance name","value":"Dinitrogen tetroxide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for dinitrogen tetroxide? </h1> | null | N2O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molecular formula</strong> of <strong>dinitrogen tetroxide</strong>.</p>
<blockquote>
<blockquote></blockquote>
</blockquote>
<p>This is a compound of <strong>nitrogen</strong> and <strong>oxygen</strong> (two <em>nonmetals</em>), so we speculate that it is a <strong>covalent compound</strong> (the prefixes also indicate it is a covalent compound).</p>
<blockquote></blockquote>
<p>Nomenclature of chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is a fundamental aspect of chemistry, and for <strong>binary (two-element) <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compounds</a></strong> like dinitrogen tetroxide, what we do is examine the <strong>prefixes</strong>:</p>
<blockquote>
<p><img alt="http://scottscience.weebly.com" src="https://useruploads.socratic.org/FRpDT2zQyqnYhT6WuAcQ_804914350.gif"/> </p>
</blockquote>
<p>The prefix in front of nitrogen is <strong>di-</strong>, so according to the above image, we know that there are <mathjax>#color(red)("two nitrogen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>The prefix in front of the oxide (oxygen) species is <strong>tetra-</strong> (the "a" is left off if oxygen is the second element), so there are <mathjax>#color(blue)("four oxygen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>We put the numbers of each element as <strong>subscripts</strong> of the element's symbol, so the formula is</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "color(red)("N"_2)color(blue)("O"_4)" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"N"_2"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molecular formula</strong> of <strong>dinitrogen tetroxide</strong>.</p>
<blockquote>
<blockquote></blockquote>
</blockquote>
<p>This is a compound of <strong>nitrogen</strong> and <strong>oxygen</strong> (two <em>nonmetals</em>), so we speculate that it is a <strong>covalent compound</strong> (the prefixes also indicate it is a covalent compound).</p>
<blockquote></blockquote>
<p>Nomenclature of chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is a fundamental aspect of chemistry, and for <strong>binary (two-element) <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compounds</a></strong> like dinitrogen tetroxide, what we do is examine the <strong>prefixes</strong>:</p>
<blockquote>
<p><img alt="http://scottscience.weebly.com" src="https://useruploads.socratic.org/FRpDT2zQyqnYhT6WuAcQ_804914350.gif"/> </p>
</blockquote>
<p>The prefix in front of nitrogen is <strong>di-</strong>, so according to the above image, we know that there are <mathjax>#color(red)("two nitrogen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>The prefix in front of the oxide (oxygen) species is <strong>tetra-</strong> (the "a" is left off if oxygen is the second element), so there are <mathjax>#color(blue)("four oxygen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>We put the numbers of each element as <strong>subscripts</strong> of the element's symbol, so the formula is</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "color(red)("N"_2)color(blue)("O"_4)" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the formula for dinitrogen tetroxide? </h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-08-07T17:34:42" itemprop="dateCreated">
Aug 7, 2017
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<div class="markdown"><p><mathjax>#"N"_2"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molecular formula</strong> of <strong>dinitrogen tetroxide</strong>.</p>
<blockquote>
<blockquote></blockquote>
</blockquote>
<p>This is a compound of <strong>nitrogen</strong> and <strong>oxygen</strong> (two <em>nonmetals</em>), so we speculate that it is a <strong>covalent compound</strong> (the prefixes also indicate it is a covalent compound).</p>
<blockquote></blockquote>
<p>Nomenclature of chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> is a fundamental aspect of chemistry, and for <strong>binary (two-element) <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compounds</a></strong> like dinitrogen tetroxide, what we do is examine the <strong>prefixes</strong>:</p>
<blockquote>
<p><img alt="http://scottscience.weebly.com" src="https://useruploads.socratic.org/FRpDT2zQyqnYhT6WuAcQ_804914350.gif"/> </p>
</blockquote>
<p>The prefix in front of nitrogen is <strong>di-</strong>, so according to the above image, we know that there are <mathjax>#color(red)("two nitrogen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>The prefix in front of the oxide (oxygen) species is <strong>tetra-</strong> (the "a" is left off if oxygen is the second element), so there are <mathjax>#color(blue)("four oxygen atoms"#</mathjax> in the compound.</p>
<blockquote></blockquote>
<p>We put the numbers of each element as <strong>subscripts</strong> of the element's symbol, so the formula is</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "color(red)("N"_2)color(blue)("O"_4)" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div>
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</article> | What is the formula for dinitrogen tetroxide? | null |
1,457 | ab235d00-6ddd-11ea-93a8-ccda262736ce | https://socratic.org/questions/how-many-grams-of-potassium-nitrate-kno-3-are-formed-when-102-grams-of-nitric-ac | 163.47 grams | start physical_unit 6 6 mass g qc_end physical_unit 15 15 10 11 mass qc_end chemical_equation 20 20 qc_end end | [{"type":"physical unit","value":"Mass [OF] KNO3 [IN] grams"}] | [{"type":"physical unit","value":"163.47 grams"}] | [{"type":"physical unit","value":"Mass [OF] HNO3 [=] \\pu{102 grams}"},{"type":"chemical equation","value":"KOH"}] | <h1 class="questionTitle" itemprop="name">How many grams of potassium nitrate, #KNO_3#. are formed when 102 grams of nitric acid, #HNO_3#, react with potassium hydroxide #KOH#? </h1> | null | 163.47 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by creating the chemical equation for this reaction:</p>
<p><mathjax>#HNO_3(aq) + KOH(aq) rightleftharpoons KNO_3(aq) + H_2O(l)#</mathjax></p>
<p>This is a <em><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> reaction, in which an acid (<mathjax>#HNO_3#</mathjax>) and a base (<mathjax>#KOH#</mathjax>) react to form a salt and (most often liquid) water. </p>
<p>We'll go about this problem by first using the molar mass of <mathjax>#HNO_3#</mathjax> to calculate the number of moles present, then use the stoichiometric relationships (the coefficients, which are all <mathjax>#1#</mathjax> in this case), to calculate the moles of <mathjax>#KNO_3#</mathjax> that form, and lastly use its molar mass to find how many grams were formed.</p>
<p>Using dimensional analysis, the whole procedure is</p>
<p><mathjax>#102cancel(gHNO_3)((1cancel(molHNO_3))/(63.02cancel(gHNO_3)))((1cancel(molKNO_3))/(1cancel(molHNO_3)))((101.11gKNO_3)/(1cancel(molKNO_3))) = color(red)(164gKNO_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#164gKNO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by creating the chemical equation for this reaction:</p>
<p><mathjax>#HNO_3(aq) + KOH(aq) rightleftharpoons KNO_3(aq) + H_2O(l)#</mathjax></p>
<p>This is a <em><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> reaction, in which an acid (<mathjax>#HNO_3#</mathjax>) and a base (<mathjax>#KOH#</mathjax>) react to form a salt and (most often liquid) water. </p>
<p>We'll go about this problem by first using the molar mass of <mathjax>#HNO_3#</mathjax> to calculate the number of moles present, then use the stoichiometric relationships (the coefficients, which are all <mathjax>#1#</mathjax> in this case), to calculate the moles of <mathjax>#KNO_3#</mathjax> that form, and lastly use its molar mass to find how many grams were formed.</p>
<p>Using dimensional analysis, the whole procedure is</p>
<p><mathjax>#102cancel(gHNO_3)((1cancel(molHNO_3))/(63.02cancel(gHNO_3)))((1cancel(molKNO_3))/(1cancel(molHNO_3)))((101.11gKNO_3)/(1cancel(molKNO_3))) = color(red)(164gKNO_3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of potassium nitrate, #KNO_3#. are formed when 102 grams of nitric acid, #HNO_3#, react with potassium hydroxide #KOH#? </h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-05-22T00:19:20" itemprop="dateCreated">
May 22, 2017
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<div class="markdown"><p><mathjax>#164gKNO_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's start by creating the chemical equation for this reaction:</p>
<p><mathjax>#HNO_3(aq) + KOH(aq) rightleftharpoons KNO_3(aq) + H_2O(l)#</mathjax></p>
<p>This is a <em><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em> reaction, in which an acid (<mathjax>#HNO_3#</mathjax>) and a base (<mathjax>#KOH#</mathjax>) react to form a salt and (most often liquid) water. </p>
<p>We'll go about this problem by first using the molar mass of <mathjax>#HNO_3#</mathjax> to calculate the number of moles present, then use the stoichiometric relationships (the coefficients, which are all <mathjax>#1#</mathjax> in this case), to calculate the moles of <mathjax>#KNO_3#</mathjax> that form, and lastly use its molar mass to find how many grams were formed.</p>
<p>Using dimensional analysis, the whole procedure is</p>
<p><mathjax>#102cancel(gHNO_3)((1cancel(molHNO_3))/(63.02cancel(gHNO_3)))((1cancel(molKNO_3))/(1cancel(molHNO_3)))((101.11gKNO_3)/(1cancel(molKNO_3))) = color(red)(164gKNO_3#</mathjax></p></div>
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Meave60
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<div class="markdown"><p>There would be <mathjax>#"164 g KNO"_3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"HNO"_3("aq") + "KOH(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"KNO"_3("aq") + "H"_2"O(l)"#</mathjax></p>
<p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction, which is a type of double replacement reaction. <mathjax>#"KNO"_3"#</mathjax> is in aqueous solution, which means the <mathjax>#"K"^+"#</mathjax> and <mathjax>#"NO"_3"^(-)#</mathjax> ions are dissociated and there is no discrete compound of <mathjax>#"KNO"_3"#</mathjax> as a product. However, if the reaction went to completion, and the only products were potassium nitrate and water, you could evaporate the water to isolate the potassium nitrate, and you should get the calculated mass <mathjax>#+-#</mathjax>. Regardless, the purpose of this question is to help you learn how to do <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> problems, so I will go ahead and answer as if <mathjax>#"KNO"_3"#</mathjax> were a discrete product, which it would be if the water were evaporated.</p>
<p><strong>The process goes like this:</strong></p>
<p><mathjax>#color(red)("given mass HNO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)("moles HNO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)("moles KNO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(red)("mass KNO"_3"#</mathjax> </p>
<p><strong>We will need the MOLAR MASSES of <mathjax>#"HNO"_3"#</mathjax> and <mathjax>#"KNO"_3"#</mathjax>.</strong> <br/>
Multiply the molar mass of each element by its subscript and add. The molar mass is the element's atomic weight on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><mathjax>#"HNO"_3:#</mathjax><mathjax>#(1xx1.008"g/mol H")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")="63.012 g/mol HNO"_3"#</mathjax></p>
<p><mathjax>#"KNO"_3:#</mathjax><mathjax>#(1xx39.0983"g/mol K")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")="101.102 g/mol KNO"_3"#</mathjax></p>
<p><mathjax>#color(blue)("Given Mass of Nitric Acid to Moles"#</mathjax><br/>
<strong>Multiply the given mass of <mathjax>#"HNO"_3"#</mathjax> by the inverse of its molar mass.</strong></p>
<p><mathjax>#102color(red)cancel(color(black)("g HNO"_3))xx(1"mol HNO"_3)/(63.012color(red)cancel(color(black)("g HNO"_3)))="1.6187 mol HNO"_3"#</mathjax> </p>
<p><mathjax>#color(blue)("Moles Nitric Acid to Moles Potassium Nitrate"#</mathjax><br/>
<strong>Multiply the moles <mathjax>#"HNO"_3#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"HNO"_3#</mathjax> and <mathjax>#"KNO"_3#</mathjax> so that mol <mathjax>#"HNO"_3#</mathjax> cancels.</strong></p>
<p><mathjax>#1.6187color(red)cancel(color(black)("mol HNO"_3))xx(1"mol KNO"_3)/(1color(red)cancel(color(black)("mol HNO"_3)))="1.6187 mol KNO"_3"#</mathjax></p>
<p><mathjax>#color(blue)("Moles Potassium Nitrate to Mass Potassium Nitrate"#</mathjax><br/>
<strong>Multiply mol <mathjax>#"KNO"_3"#</mathjax> by its molar mass.</strong></p>
<p><mathjax>#1.6187color(red)cancel(color(black)("mol KNO"_3))xx(101.102"g KNO"_3)/(1color(red)cancel(color(black)("mol KNO"_3)))="164 g KNO"_3#</mathjax> (rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to 102 g)</p></div>
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</article> | How many grams of potassium nitrate, #KNO_3#. are formed when 102 grams of nitric acid, #HNO_3#, react with potassium hydroxide #KOH#? | null |
1,458 | a99c821e-6ddd-11ea-be96-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-h3po4-mg-oh-2-mg3-po4-2-h20-1 | 2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O"}] | [{"type":"chemical equation","value":"H3PO4 + Mg(OH)2 -> Mg3(PO4)2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation:
H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?</h1> | null | 2 H3PO4 + 3 Mg(OH)2 -> Mg3(PO4)2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"H"_3"PO"_4" + Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Balance the magnesium first. There are three magnesium ions on the right side and only one on the left side. Add a coefficient of <mathjax>#3#</mathjax> in front of <mathjax>#"Mg(OH)"_2"#</mathjax>..</p>
<p><mathjax>#"H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"PO"_4"#</mathjax>.</p>
<p>There are two <mathjax>#"PO"_4"#</mathjax> ions on the right side and one on the left. Place a coefficient of <mathjax>#2#</mathjax> in front of <mathjax>#"H"_3"PO"_4"#</mathjax>. </p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"H"#</mathjax>. There are twelve hydrogens on the left side and two on the right. Place a coefficient of <mathjax>#6#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax> on the right side.</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p>
<p>Balance the <mathjax>#"O"#</mathjax>. There are fourteen oxygen atoms on both sides of the equation, so it is already balanced.</p>
<p>Balanced Equation</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"H"_3"PO"_4" + Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Balance the magnesium first. There are three magnesium ions on the right side and only one on the left side. Add a coefficient of <mathjax>#3#</mathjax> in front of <mathjax>#"Mg(OH)"_2"#</mathjax>..</p>
<p><mathjax>#"H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"PO"_4"#</mathjax>.</p>
<p>There are two <mathjax>#"PO"_4"#</mathjax> ions on the right side and one on the left. Place a coefficient of <mathjax>#2#</mathjax> in front of <mathjax>#"H"_3"PO"_4"#</mathjax>. </p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"H"#</mathjax>. There are twelve hydrogens on the left side and two on the right. Place a coefficient of <mathjax>#6#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax> on the right side.</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p>
<p>Balance the <mathjax>#"O"#</mathjax>. There are fourteen oxygen atoms on both sides of the equation, so it is already balanced.</p>
<p>Balanced Equation</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you balance the following equation:
H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?</h1>
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<div class="markdown"><p>Balanced Equation</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"H"_3"PO"_4" + Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Balance the magnesium first. There are three magnesium ions on the right side and only one on the left side. Add a coefficient of <mathjax>#3#</mathjax> in front of <mathjax>#"Mg(OH)"_2"#</mathjax>..</p>
<p><mathjax>#"H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"PO"_4"#</mathjax>.</p>
<p>There are two <mathjax>#"PO"_4"#</mathjax> ions on the right side and one on the left. Place a coefficient of <mathjax>#2#</mathjax> in front of <mathjax>#"H"_3"PO"_4"#</mathjax>. </p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "H"_2"O"#</mathjax></p>
<p>Next balance the <mathjax>#"H"#</mathjax>. There are twelve hydrogens on the left side and two on the right. Place a coefficient of <mathjax>#6#</mathjax> in front of <mathjax>#"H"_2"O"#</mathjax> on the right side.</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p>
<p>Balance the <mathjax>#"O"#</mathjax>. There are fourteen oxygen atoms on both sides of the equation, so it is already balanced.</p>
<p>Balanced Equation</p>
<p><mathjax>#"2H"_3"PO"_4" + 3Mg(OH)"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Mg"_3"(PO"_4)_2 + "6H"_2"O"#</mathjax></p></div>
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</article> | How would you balance the following equation:
H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20? | null |
1,459 | a95c67ca-6ddd-11ea-98b8-ccda262736ce | https://socratic.org/questions/c3h7oh-is-propanol-suppose-you-have-85-9-g-sample-of-propanol-how-many-carbon-at | 2.58 × 10^24 | start physical_unit 13 14 number none qc_end physical_unit 0 0 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] carbon atoms"}] | [{"type":"physical unit","value":"2.58 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] C3H7OH sample [=] \\pu{85.9 g}"}] | <h1 class="questionTitle" itemprop="name">C3H7OH is propanol. Suppose you have 85.9 g sample of propanol. How many carbon atoms are in the sample?</h1> | null | 2.58 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of propanol is <mathjax>#C_3 H_8 O#</mathjax>.<br/>
First, we should calculate its molecular mass. I prefer to work with rounded numbers because that's how I was taught and frankly it makes life much easier. You should have memorised that the molar masses of carbon, hydrogen and oxygen are, respectively, <mathjax>#12g/(mol)#</mathjax>, <mathjax>#1g/(mol)#</mathjax> and <mathjax>#16g/(mol)#</mathjax>.<br/>
So let's add all that up:</p>
<blockquote>
<p><mathjax>#M = (3*12)+(8*1)+(1*16)#</mathjax><br/>
<mathjax>#M = 60g/(mol)#</mathjax></p>
</blockquote>
<p>Okay, but that doesn't tell us much by itself. <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a> make up substances in different proportions, because they have different masses, so how much of propanol is actually carbon, in terms of percentage? Remember, <mathjax>#36g#</mathjax> of that is carbon:</p>
<blockquote>
<p><mathjax>#{:(60g -> 100%), (36g -> x%) :}} implies x = 60%#</mathjax></p>
</blockquote>
<p>So now we know that any given sample of propanol is sixty percent carbon. Now we can use the information that we were given: <mathjax>#85.9g#</mathjax>. If that's the total mass of propanol, sixty percent of that must be carbon:</p>
<blockquote>
<p><mathjax>#{: (85.9g -> 100%), (ycolor(white)(x)g -> 60%) :}} implies y = 51.54g#</mathjax></p>
</blockquote>
<p>Now, let us return to the concept of a mole. A mole is a quantity of <mathjax>#6*10^23#</mathjax> of something. You might be used to using a more accurate value, but for now let us use that simplified number. Molar mass is the mass of a mole of something; so <mathjax>#6*10^23#</mathjax> atoms of carbon have a total mass of <mathjax>#12g#</mathjax>. With that in mind, we can now work out how many atoms of carbon have a total mass of <mathjax>#51.54g#</mathjax>:</p>
<p><mathjax>#{:(6*10^23 "atoms" -> 12g), (zcolor(white)(x) "atoms" -> 51.54g) :}} implies z = 2.58*10^24 "atoms"#</mathjax></p>
<p>My answer might be different from somebody else's because, as I said, I prefer to work with simpler values. I hope, however, that the logic is clear! Cheers.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>I got: <mathjax>#2.58 * 10^24#</mathjax> carbon atoms.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of propanol is <mathjax>#C_3 H_8 O#</mathjax>.<br/>
First, we should calculate its molecular mass. I prefer to work with rounded numbers because that's how I was taught and frankly it makes life much easier. You should have memorised that the molar masses of carbon, hydrogen and oxygen are, respectively, <mathjax>#12g/(mol)#</mathjax>, <mathjax>#1g/(mol)#</mathjax> and <mathjax>#16g/(mol)#</mathjax>.<br/>
So let's add all that up:</p>
<blockquote>
<p><mathjax>#M = (3*12)+(8*1)+(1*16)#</mathjax><br/>
<mathjax>#M = 60g/(mol)#</mathjax></p>
</blockquote>
<p>Okay, but that doesn't tell us much by itself. <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a> make up substances in different proportions, because they have different masses, so how much of propanol is actually carbon, in terms of percentage? Remember, <mathjax>#36g#</mathjax> of that is carbon:</p>
<blockquote>
<p><mathjax>#{:(60g -> 100%), (36g -> x%) :}} implies x = 60%#</mathjax></p>
</blockquote>
<p>So now we know that any given sample of propanol is sixty percent carbon. Now we can use the information that we were given: <mathjax>#85.9g#</mathjax>. If that's the total mass of propanol, sixty percent of that must be carbon:</p>
<blockquote>
<p><mathjax>#{: (85.9g -> 100%), (ycolor(white)(x)g -> 60%) :}} implies y = 51.54g#</mathjax></p>
</blockquote>
<p>Now, let us return to the concept of a mole. A mole is a quantity of <mathjax>#6*10^23#</mathjax> of something. You might be used to using a more accurate value, but for now let us use that simplified number. Molar mass is the mass of a mole of something; so <mathjax>#6*10^23#</mathjax> atoms of carbon have a total mass of <mathjax>#12g#</mathjax>. With that in mind, we can now work out how many atoms of carbon have a total mass of <mathjax>#51.54g#</mathjax>:</p>
<p><mathjax>#{:(6*10^23 "atoms" -> 12g), (zcolor(white)(x) "atoms" -> 51.54g) :}} implies z = 2.58*10^24 "atoms"#</mathjax></p>
<p>My answer might be different from somebody else's because, as I said, I prefer to work with simpler values. I hope, however, that the logic is clear! Cheers.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">C3H7OH is propanol. Suppose you have 85.9 g sample of propanol. How many carbon atoms are in the sample?</h1>
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Marilia E.K.
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Stefan V.
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<div class="markdown"><p>I got: <mathjax>#2.58 * 10^24#</mathjax> carbon atoms.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular formula of propanol is <mathjax>#C_3 H_8 O#</mathjax>.<br/>
First, we should calculate its molecular mass. I prefer to work with rounded numbers because that's how I was taught and frankly it makes life much easier. You should have memorised that the molar masses of carbon, hydrogen and oxygen are, respectively, <mathjax>#12g/(mol)#</mathjax>, <mathjax>#1g/(mol)#</mathjax> and <mathjax>#16g/(mol)#</mathjax>.<br/>
So let's add all that up:</p>
<blockquote>
<p><mathjax>#M = (3*12)+(8*1)+(1*16)#</mathjax><br/>
<mathjax>#M = 60g/(mol)#</mathjax></p>
</blockquote>
<p>Okay, but that doesn't tell us much by itself. <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a> make up substances in different proportions, because they have different masses, so how much of propanol is actually carbon, in terms of percentage? Remember, <mathjax>#36g#</mathjax> of that is carbon:</p>
<blockquote>
<p><mathjax>#{:(60g -> 100%), (36g -> x%) :}} implies x = 60%#</mathjax></p>
</blockquote>
<p>So now we know that any given sample of propanol is sixty percent carbon. Now we can use the information that we were given: <mathjax>#85.9g#</mathjax>. If that's the total mass of propanol, sixty percent of that must be carbon:</p>
<blockquote>
<p><mathjax>#{: (85.9g -> 100%), (ycolor(white)(x)g -> 60%) :}} implies y = 51.54g#</mathjax></p>
</blockquote>
<p>Now, let us return to the concept of a mole. A mole is a quantity of <mathjax>#6*10^23#</mathjax> of something. You might be used to using a more accurate value, but for now let us use that simplified number. Molar mass is the mass of a mole of something; so <mathjax>#6*10^23#</mathjax> atoms of carbon have a total mass of <mathjax>#12g#</mathjax>. With that in mind, we can now work out how many atoms of carbon have a total mass of <mathjax>#51.54g#</mathjax>:</p>
<p><mathjax>#{:(6*10^23 "atoms" -> 12g), (zcolor(white)(x) "atoms" -> 51.54g) :}} implies z = 2.58*10^24 "atoms"#</mathjax></p>
<p>My answer might be different from somebody else's because, as I said, I prefer to work with simpler values. I hope, however, that the logic is clear! Cheers.</p></div>
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Oct 14, 2015
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<div class="markdown"><p>The number of carbon atoms in an 85.9 g sample of propanol is <mathjax>#2.58xx10^24"atoms C"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Propanol: <mathjax>#"C"_3"H"_7"OH"#</mathjax><mathjax>#=#</mathjax><mathjax>#"C"_3"H"_8"O"#</mathjax> </p>
<p>The molar masses of carbon, hydrogen, and oxygen are:<br/>
<mathjax>#"C"="12.0107 g/mol"#</mathjax><br/>
<mathjax>#"H"="1.00794 g/mol"#</mathjax><br/>
<mathjax>#"O"="15.999 g/mol"#</mathjax></p>
<p><mathjax>#"Molar mass of propanol"=(3xx12.0107)+(8xx1.00794)+(1xx15.999)="60.095 g/mol"#</mathjax>. </p>
<p><strong>Identify the Molar Masses of Propanol and Carbon</strong></p>
<p>Molar mass of propanol: <mathjax>#"60.095 g/mol"#</mathjax><br/>
Molar mass of carbon: <mathjax>#(3xx12.0107)="36.032 g/mol"#</mathjax></p>
<p><strong>Determine the Percentage of Carbon</strong></p>
<p><mathjax>#"Percentage of C in propanol"=("molar mass of C")/("molar mass of propanol")xx100=(36.032"g/mol C")/(60.095"g/mol propanol")xx100="59.958% C"#</mathjax></p>
<p><strong>Determine the Mass of Carbon in 85.9 g of Propanol</strong></p>
<p>In order to determine the mass of carbon present in 85.9 g propanol, we need to multiply the mass of propanol times the percentage of carbon as a decimal (e.g. <mathjax>#(25%)/(100)=0.25#</mathjax>).</p>
<p><mathjax>#"85.9 g propanol"xx(0.59958 "C")="51.504 g C"#</mathjax></p>
<p>There are <mathjax>#"51.504 g C"#</mathjax> in <mathjax>#"85.9 g propanol"#</mathjax></p>
<p><strong>Determining the Number of Carbon Atoms</strong> </p>
<p>Use the mass of carbon in the compound and carbon's molar mass to convert to moles of carbon. Then multiply the moles of carbon times <mathjax>#6.022xx10^23 "atoms/mol"#</mathjax>.</p>
<p><mathjax>#51.504cancel"g C"xx(1cancel"mol C")/(12.0107cancel"g C")xx(6.022xx10^23"atoms C")/(1cancel"mol C")=2.58xx10^24"atoms C"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</article> | C3H7OH is propanol. Suppose you have 85.9 g sample of propanol. How many carbon atoms are in the sample? | null |
1,460 | acf4bfe2-6ddd-11ea-85e1-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-5-0-grams-of-co-2-at-stp | 2.58 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 5 6 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 [IN] L"}] | [{"type":"physical unit","value":"2.58 L"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{5.0 grams}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 5.0 grams of #CO_2# at STP?</h1> | null | 2.58 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(blue)("STP"#</mathjax></p>
<p>STP is currently <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax>, which are equal, though the Kelvin temperature scale is used for <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a>; and pressure is <mathjax>#"10"^5color(white)(.)"Pascals (Pa)"#</mathjax>, but most people use <mathjax>#"100 kPa"#</mathjax>, which is equal to <mathjax>#10^5color(white)(.)"Pa"#</mathjax>.</p>
<p>You will use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to answer this question. Its formula is:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is a gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><mathjax>#color(blue)("Determine moles"#</mathjax><br/>
You may have noticed that the equation requires moles <mathjax>#(n)#</mathjax>, but you have been given the mass of <mathjax>#"CO"_2"#</mathjax>. To determine moles, you <strong>multiply the given mass by the inverse of the molar mass</strong> of <mathjax>#"CO"_2"#</mathjax>, which is <mathjax>#"44.009 g/mol"#</mathjax>.</p>
<p><mathjax>#5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"#</mathjax></p>
<p><mathjax>#color(blue)("Organize your data"#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#n="0.1136 mol"#</mathjax><br/>
<mathjax>#R="8.3145 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V#</mathjax></p>
<p><mathjax>#color(blue)("Solve for volume using the ideal gas law."#</mathjax><br/>
Rearrange the formula to isolate <mathjax>#V#</mathjax>. Insert your data into the equation and solve.</p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to <mathjax>#"5.0 g"#</mathjax></p></div>
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<div class="markdown"><p>The volume of <mathjax>#"5.0 g CO"_2"#</mathjax> is <mathjax>#"2.6 L CO"_2"#</mathjax> at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(blue)("STP"#</mathjax></p>
<p>STP is currently <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax>, which are equal, though the Kelvin temperature scale is used for <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a>; and pressure is <mathjax>#"10"^5color(white)(.)"Pascals (Pa)"#</mathjax>, but most people use <mathjax>#"100 kPa"#</mathjax>, which is equal to <mathjax>#10^5color(white)(.)"Pa"#</mathjax>.</p>
<p>You will use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to answer this question. Its formula is:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is a gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><mathjax>#color(blue)("Determine moles"#</mathjax><br/>
You may have noticed that the equation requires moles <mathjax>#(n)#</mathjax>, but you have been given the mass of <mathjax>#"CO"_2"#</mathjax>. To determine moles, you <strong>multiply the given mass by the inverse of the molar mass</strong> of <mathjax>#"CO"_2"#</mathjax>, which is <mathjax>#"44.009 g/mol"#</mathjax>.</p>
<p><mathjax>#5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"#</mathjax></p>
<p><mathjax>#color(blue)("Organize your data"#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#n="0.1136 mol"#</mathjax><br/>
<mathjax>#R="8.3145 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V#</mathjax></p>
<p><mathjax>#color(blue)("Solve for volume using the ideal gas law."#</mathjax><br/>
Rearrange the formula to isolate <mathjax>#V#</mathjax>. Insert your data into the equation and solve.</p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to <mathjax>#"5.0 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 5.0 grams of #CO_2# at STP?</h1>
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<div class="markdown"><p>The volume of <mathjax>#"5.0 g CO"_2"#</mathjax> is <mathjax>#"2.6 L CO"_2"#</mathjax> at STP.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color(blue)("STP"#</mathjax></p>
<p>STP is currently <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax>, which are equal, though the Kelvin temperature scale is used for <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a>; and pressure is <mathjax>#"10"^5color(white)(.)"Pascals (Pa)"#</mathjax>, but most people use <mathjax>#"100 kPa"#</mathjax>, which is equal to <mathjax>#10^5color(white)(.)"Pa"#</mathjax>.</p>
<p>You will use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> to answer this question. Its formula is:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is a gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p><mathjax>#color(blue)("Determine moles"#</mathjax><br/>
You may have noticed that the equation requires moles <mathjax>#(n)#</mathjax>, but you have been given the mass of <mathjax>#"CO"_2"#</mathjax>. To determine moles, you <strong>multiply the given mass by the inverse of the molar mass</strong> of <mathjax>#"CO"_2"#</mathjax>, which is <mathjax>#"44.009 g/mol"#</mathjax>.</p>
<p><mathjax>#5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"#</mathjax></p>
<p><mathjax>#color(blue)("Organize your data"#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#n="0.1136 mol"#</mathjax><br/>
<mathjax>#R="8.3145 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#V#</mathjax></p>
<p><mathjax>#color(blue)("Solve for volume using the ideal gas law."#</mathjax><br/>
Rearrange the formula to isolate <mathjax>#V#</mathjax>. Insert your data into the equation and solve.</p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> due to <mathjax>#"5.0 g"#</mathjax></p></div>
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<div class="markdown"><p>I got 2.55 Liters</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of any gas at STP = 22.4 Liters<br/>
<mathjax>#5 g CO_2(g) = (5 g)/(44(g/"mole")) = 0.114 "mole" CO_2(g) #</mathjax><br/>
Volume of 0.114 mole <mathjax>#CO_2(g)#</mathjax> = (0.114 mole)(22.4 L/mole) = 2.55 Liters <mathjax>#CO_2#</mathjax>(g) at STP</p></div>
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</article> | What is the volume of 5.0 grams of #CO_2# at STP? | null |
1,461 | ac9f173e-6ddd-11ea-b54a-ccda262736ce | https://socratic.org/questions/how-milliliters-of-a-9-0-m-h-2so-solution-are-needed-to-make-0-35-l-of-a-3-5-m-s | 136.11 milliliters | start physical_unit 6 7 volume ml qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 6 7 12 13 volume qc_end physical_unit 6 7 16 17 molarity qc_end end | [{"type":"physical unit","value":"Volume1 [OF] H2SO4 solution [IN] milliliters"}] | [{"type":"physical unit","value":"136.11 milliliters"}] | [{"type":"physical unit","value":"Molarity1 [OF] H2SO4 solution [=] \\pu{9.0 M}"},{"type":"physical unit","value":"Volume2 [OF] H2SO4 solution [=] \\pu{0.35 L}"},{"type":"physical unit","value":"Molarity2 [OF] H2SO4 solution [=] \\pu{3.5 M}"}] | <h1 class="questionTitle" itemprop="name">How milliliters of a 9.0 M #H_2SO_# solution are needed to make 0.35 L of a 3.5 M solution? </h1> | null | 136.11 milliliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The unit for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) is <mathjax>#"moles of solute"/"liters of solution"="mol/L"#</mathjax>.</p>
<p>When diluting a solution, the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> remains constant, but the volume of the solution increases.</p>
<p>The formula for the dilution of a solution is given below:</p>
<p><mathjax>#"M"_1"V"_1"=M"_2"V"_2"#</mathjax>, </p>
<p>where <mathjax>#"M"#</mathjax> is molarity and <mathjax>#"V"#</mathjax> is volume of the solution in liters (L).</p>
<p><strong>Known</strong><br/>
<mathjax>#"M"_1="9.0 M"="9.0 mol/L"#</mathjax><br/>
<mathjax>#"M"_2="3.5 M"="3.5 mol/L"#</mathjax><br/>
<mathjax>#"V"_2="0.35 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#"V"_1"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution formula to isolate <mathjax>#"V"_1"#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#"V"_1=("M"_2"V"_2)/("M"_1)#</mathjax></p>
<p><mathjax>#"V"_1=(3.5cancel"M"xx"0.35 L")/(9.0cancel"M")="0.14 L"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><mathjax>#"V"_1=0.14 cancel"L"xx(1000 "mL")/(1 cancel"L")="140 mL"#</mathjax> rounded to two significant figures</p></div>
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<div class="markdown"><p>You will need <mathjax>#"140 mL"#</mathjax> of <mathjax>#"9.0 M H"_2"SO"_4"#</mathjax> to make <mathjax>#"350 mL"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The unit for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) is <mathjax>#"moles of solute"/"liters of solution"="mol/L"#</mathjax>.</p>
<p>When diluting a solution, the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> remains constant, but the volume of the solution increases.</p>
<p>The formula for the dilution of a solution is given below:</p>
<p><mathjax>#"M"_1"V"_1"=M"_2"V"_2"#</mathjax>, </p>
<p>where <mathjax>#"M"#</mathjax> is molarity and <mathjax>#"V"#</mathjax> is volume of the solution in liters (L).</p>
<p><strong>Known</strong><br/>
<mathjax>#"M"_1="9.0 M"="9.0 mol/L"#</mathjax><br/>
<mathjax>#"M"_2="3.5 M"="3.5 mol/L"#</mathjax><br/>
<mathjax>#"V"_2="0.35 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#"V"_1"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution formula to isolate <mathjax>#"V"_1"#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#"V"_1=("M"_2"V"_2)/("M"_1)#</mathjax></p>
<p><mathjax>#"V"_1=(3.5cancel"M"xx"0.35 L")/(9.0cancel"M")="0.14 L"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><mathjax>#"V"_1=0.14 cancel"L"xx(1000 "mL")/(1 cancel"L")="140 mL"#</mathjax> rounded to two significant figures</p></div>
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<h1 class="questionTitle" itemprop="name">How milliliters of a 9.0 M #H_2SO_# solution are needed to make 0.35 L of a 3.5 M solution? </h1>
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<div class="markdown"><p>You will need <mathjax>#"140 mL"#</mathjax> of <mathjax>#"9.0 M H"_2"SO"_4"#</mathjax> to make <mathjax>#"350 mL"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The unit for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> (M) is <mathjax>#"moles of solute"/"liters of solution"="mol/L"#</mathjax>.</p>
<p>When diluting a solution, the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> remains constant, but the volume of the solution increases.</p>
<p>The formula for the dilution of a solution is given below:</p>
<p><mathjax>#"M"_1"V"_1"=M"_2"V"_2"#</mathjax>, </p>
<p>where <mathjax>#"M"#</mathjax> is molarity and <mathjax>#"V"#</mathjax> is volume of the solution in liters (L).</p>
<p><strong>Known</strong><br/>
<mathjax>#"M"_1="9.0 M"="9.0 mol/L"#</mathjax><br/>
<mathjax>#"M"_2="3.5 M"="3.5 mol/L"#</mathjax><br/>
<mathjax>#"V"_2="0.35 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#"V"_1"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the dilution formula to isolate <mathjax>#"V"_1"#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#"V"_1=("M"_2"V"_2)/("M"_1)#</mathjax></p>
<p><mathjax>#"V"_1=(3.5cancel"M"xx"0.35 L")/(9.0cancel"M")="0.14 L"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><mathjax>#"V"_1=0.14 cancel"L"xx(1000 "mL")/(1 cancel"L")="140 mL"#</mathjax> rounded to two significant figures</p></div>
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</article> | How milliliters of a 9.0 M #H_2SO_# solution are needed to make 0.35 L of a 3.5 M solution? | null |
1,462 | a83e36cc-6ddd-11ea-934c-ccda262736ce | https://socratic.org/questions/58228fddb72cff56e22c1141 | Na2CO3(aq) + 2 HNO3(aq) -> 2 NaNO3(aq) + CO2(g)↑ + H2O(l) | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Na2CO3(aq) + 2 HNO3(aq) -> 2 NaNO3(aq) + CO2(g)↑ + H2O(l)"}] | [{"type":"substance name","value":"Sodium carbonate"},{"type":"substance name","value":"Nitric acid"}] | <h1 class="questionTitle" itemprop="name">How would sodium carbonate react with nitric acid?</h1> | null | Na2CO3(aq) + 2 HNO3(aq) -> 2 NaNO3(aq) + CO2(g)↑ + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid base reaction:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr 3H_2O(l) + CO_2(g)uarr#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is simply an acid base reaction:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr 3H_2O(l) + CO_2(g)uarr#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How would sodium carbonate react with nitric acid?</h1>
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<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is simply an acid base reaction:</p>
<p><mathjax>#CO_3^(2-) + 2H_3O^(+) rarr 3H_2O(l) + CO_2(g)uarr#</mathjax></p></div>
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</article> | How would sodium carbonate react with nitric acid? | null |
1,463 | a98ad090-6ddd-11ea-b025-ccda262736ce | https://socratic.org/questions/581915bf11ef6b15c0150418 | Sr(OH)2(s) + 2 HNO3(aq) -> Sr(NO3)2(aq) + 2 H2O(l) | start chemical_equation qc_end substance 5 6 qc_end substance 8 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Sr(OH)2(s) + 2 HNO3(aq) -> Sr(NO3)2(aq) + 2 H2O(l)"}] | [{"type":"substance name","value":"Strontium hydroxide"},{"type":"substance name","value":"Nitric acid"}] | <h1 class="questionTitle" itemprop="name">How is the reaction of strontium hydroxide with nitric acid represented?</h1> | null | Sr(OH)2(s) + 2 HNO3(aq) -> Sr(NO3)2(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a typical acid base reaction. Two equiv of acid react with two equiv of base to give a salt plus two equiv of water.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Sr(OH)_2(s) + 2HNO_3(aq) rarr Sr(NO_3)_2(aq) + 2H_2O(l)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a typical acid base reaction. Two equiv of acid react with two equiv of base to give a salt plus two equiv of water.</p></div>
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<h1 class="questionTitle" itemprop="name">How is the reaction of strontium hydroxide with nitric acid represented?</h1>
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<div class="markdown"><p><mathjax>#Sr(OH)_2(s) + 2HNO_3(aq) rarr Sr(NO_3)_2(aq) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a typical acid base reaction. Two equiv of acid react with two equiv of base to give a salt plus two equiv of water.</p></div>
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</article> | How is the reaction of strontium hydroxide with nitric acid represented? | null |
1,464 | aa5158e8-6ddd-11ea-9ef4-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-carbon-in-co2 | +4 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] carbon"}] | [{"type":"physical unit","value":"+4"}] | [{"type":"chemical equation","value":"CO2"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in CO2?
</h1> | null | +4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since oxygen is in group 6 on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, its valency is <mathjax>#2-#</mathjax> and hence so is its oxidation number.<br/>
But there are 2 oxygen atoms for each carbon atom so the total negative charge due the oxygen is <mathjax>#4-#</mathjax><br/>
But since a molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero in the free form, it implies that the oxidation number of the carbon atom must then be <mathjax>#+4#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4+#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since oxygen is in group 6 on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, its valency is <mathjax>#2-#</mathjax> and hence so is its oxidation number.<br/>
But there are 2 oxygen atoms for each carbon atom so the total negative charge due the oxygen is <mathjax>#4-#</mathjax><br/>
But since a molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero in the free form, it implies that the oxidation number of the carbon atom must then be <mathjax>#+4#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in CO2?
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Trevor Ryan.
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<div class="markdown"><p><mathjax>#4+#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since oxygen is in group 6 on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, its valency is <mathjax>#2-#</mathjax> and hence so is its oxidation number.<br/>
But there are 2 oxygen atoms for each carbon atom so the total negative charge due the oxygen is <mathjax>#4-#</mathjax><br/>
But since a molecule must have a <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero in the free form, it implies that the oxidation number of the carbon atom must then be <mathjax>#+4#</mathjax></p></div>
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anor277
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Oct 25, 2015
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<div class="markdown"><p>Two oxygens and one carbon; carbon clearly has <mathjax>#IV^+#</mathjax> oxidation state.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>What is the oxidation state of carbon in methane, ethane, and carbon monoxide?</p></div>
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</article> | What is the oxidation number of carbon in CO2?
| null |
1,465 | ab68e723-6ddd-11ea-82c4-ccda262736ce | https://socratic.org/questions/what-pressure-in-atm-will-0-44-moles-of-co-2-exert-in-a-2-6-l-container-at-25-c | 4.14 atm | start physical_unit 8 8 pressure atm qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 8 8 12 13 volume qc_end physical_unit 8 8 16 17 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] CO2 [IN] atm"}] | [{"type":"physical unit","value":"4.14 atm"}] | [{"type":"physical unit","value":"Mole [OF] CO2 [=] \\pu{0.44 moles}"},{"type":"physical unit","value":"Volume [OF] CO2 [=] \\pu{2.6 L}"},{"type":"physical unit","value":"Temperature [OF] CO2 [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">What pressure (in atm) will 0.44 moles of #CO_2# exert in a 2.6 L container at 25°C?</h1> | null | 4.14 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(0.44*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)cancel(mol^-1)xx298*cancelK)/(2.6*cancelL)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? atm#</mathjax>.</p>
<p>The <mathjax>#"Gas constant"#</mathjax>, <mathjax>#R#</mathjax>, most generally used by chemists is <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. Sometimes, we use <mathjax>#dm^3#</mathjax>, i.e. <mathjax>#1*dm^3=(1xx10^-1*m)^3=10^-3*m^3#</mathjax>, i.e. <mathjax>#1L=1*dm^3=10^-3m^3#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=??*atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(0.44*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)cancel(mol^-1)xx298*cancelK)/(2.6*cancelL)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? atm#</mathjax>.</p>
<p>The <mathjax>#"Gas constant"#</mathjax>, <mathjax>#R#</mathjax>, most generally used by chemists is <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. Sometimes, we use <mathjax>#dm^3#</mathjax>, i.e. <mathjax>#1*dm^3=(1xx10^-1*m)^3=10^-3*m^3#</mathjax>, i.e. <mathjax>#1L=1*dm^3=10^-3m^3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What pressure (in atm) will 0.44 moles of #CO_2# exert in a 2.6 L container at 25°C?</h1>
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<div class="markdown"><p><mathjax>#P=(nRT)/V=??*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#P=(0.44*cancel(mol)xx0.0821*cancelL*atm*cancel(K^-1)cancel(mol^-1)xx298*cancelK)/(2.6*cancelL)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? atm#</mathjax>.</p>
<p>The <mathjax>#"Gas constant"#</mathjax>, <mathjax>#R#</mathjax>, most generally used by chemists is <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. Sometimes, we use <mathjax>#dm^3#</mathjax>, i.e. <mathjax>#1*dm^3=(1xx10^-1*m)^3=10^-3*m^3#</mathjax>, i.e. <mathjax>#1L=1*dm^3=10^-3m^3#</mathjax>.</p></div>
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</article> | What pressure (in atm) will 0.44 moles of #CO_2# exert in a 2.6 L container at 25°C? | null |
1,466 | a8c5f8b6-6ddd-11ea-aecd-ccda262736ce | https://socratic.org/questions/a-ask-contains-a-gas-mixture-of-methane-hydrogen-and-nitrogen-with-partial-press | 3.3 atm | start physical_unit 29 30 total_pressure atm qc_end physical_unit 7 7 15 16 partial_pressure qc_end physical_unit 8 8 17 18 partial_pressure qc_end physical_unit 10 10 20 21 partial_pressure qc_end end | [{"type":"physical unit","value":"Total pressure [OF] the mixture [IN] atm"}] | [{"type":"physical unit","value":"3.3 atm"}] | [{"type":"physical unit","value":"Partial pressure [OF] methane [=] \\pu{1 atm}"},{"type":"physical unit","value":"Partial pressure [OF] hydrogen [=] \\pu{1.2 atm}"},{"type":"physical unit","value":"Partial pressure [OF] nitrogen [=] \\pu{1.1 atm}"}] | <h1 class="questionTitle" itemprop="name">A flask contains a gas mixture of methane, hydrogen. and nitrogen with partial pressures of 1 atm. 1.2 atm, and 1.1 aim, respectively. What is the total pressure of the mixture?</h1> | null | 3.3 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use <strong>Dalton's Law of Partial Pressures</strong> to express the <em>total pressure</em> of the gaseous mixture as the <em>sum</em> of the partial pressures of the individual components of the mixture. </p>
<blockquote>
<p><mathjax>#color(blue)(P_"total" = sum_i (P_"partial i")) " "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_"partial i"#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of component <mathjax>#i#</mathjax><br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p><img alt="https://www.alcaweb.org/arch.php/resource/view/86033" src="https://useruploads.socratic.org/LgX9xditSZe7Twyn1hFo_Blh7W8x3T3.jpg"/> </p>
<p>In your case, you know that the mixture contains three gases, methane, hydrogen, and nitrogen, each with its own partial pressure</p>
<blockquote>
<ul>
<li><mathjax>#"CH"_4 -> "1.0 atm"#</mathjax></li>
<li><mathjax>#"H"_2 -> "1.2 atm"#</mathjax></li>
<li><mathjax>#"N"_2 -> "1.1 atm"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the total pressure of the mixture will be </p>
<blockquote>
<p><mathjax>#P_"total" = P_(CH_4) + P_(H_2) + P_(N_2)#</mathjax></p>
<p><mathjax>#P_"total" = "1.0 atm" + "1.2 atm" + "1.1 atm" = color(green)("3.3 atm")#</mathjax></p>
</blockquote>
<p>The underlying principle that's at work here is the fact that at <em>constant temperature and volume</em>, the pressure of a gas is <strong>proportional</strong> ot the number of moles of gas.</p>
<p>
<iframe src="https://www.youtube.com/embed/RqffPYOoxd8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.3 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use <strong>Dalton's Law of Partial Pressures</strong> to express the <em>total pressure</em> of the gaseous mixture as the <em>sum</em> of the partial pressures of the individual components of the mixture. </p>
<blockquote>
<p><mathjax>#color(blue)(P_"total" = sum_i (P_"partial i")) " "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_"partial i"#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of component <mathjax>#i#</mathjax><br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p><img alt="https://www.alcaweb.org/arch.php/resource/view/86033" src="https://useruploads.socratic.org/LgX9xditSZe7Twyn1hFo_Blh7W8x3T3.jpg"/> </p>
<p>In your case, you know that the mixture contains three gases, methane, hydrogen, and nitrogen, each with its own partial pressure</p>
<blockquote>
<ul>
<li><mathjax>#"CH"_4 -> "1.0 atm"#</mathjax></li>
<li><mathjax>#"H"_2 -> "1.2 atm"#</mathjax></li>
<li><mathjax>#"N"_2 -> "1.1 atm"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the total pressure of the mixture will be </p>
<blockquote>
<p><mathjax>#P_"total" = P_(CH_4) + P_(H_2) + P_(N_2)#</mathjax></p>
<p><mathjax>#P_"total" = "1.0 atm" + "1.2 atm" + "1.1 atm" = color(green)("3.3 atm")#</mathjax></p>
</blockquote>
<p>The underlying principle that's at work here is the fact that at <em>constant temperature and volume</em>, the pressure of a gas is <strong>proportional</strong> ot the number of moles of gas.</p>
<p>
<iframe src="https://www.youtube.com/embed/RqffPYOoxd8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A flask contains a gas mixture of methane, hydrogen. and nitrogen with partial pressures of 1 atm. 1.2 atm, and 1.1 aim, respectively. What is the total pressure of the mixture?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-10-30T12:47:23" itemprop="dateCreated">
Oct 30, 2015
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<div class="markdown"><p><mathjax>#"3.3 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use <strong>Dalton's Law of Partial Pressures</strong> to express the <em>total pressure</em> of the gaseous mixture as the <em>sum</em> of the partial pressures of the individual components of the mixture. </p>
<blockquote>
<p><mathjax>#color(blue)(P_"total" = sum_i (P_"partial i")) " "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_"partial i"#</mathjax> - the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of component <mathjax>#i#</mathjax><br/>
<mathjax>#P_"total"#</mathjax> - the total pressure of the mixture</p>
<p><img alt="https://www.alcaweb.org/arch.php/resource/view/86033" src="https://useruploads.socratic.org/LgX9xditSZe7Twyn1hFo_Blh7W8x3T3.jpg"/> </p>
<p>In your case, you know that the mixture contains three gases, methane, hydrogen, and nitrogen, each with its own partial pressure</p>
<blockquote>
<ul>
<li><mathjax>#"CH"_4 -> "1.0 atm"#</mathjax></li>
<li><mathjax>#"H"_2 -> "1.2 atm"#</mathjax></li>
<li><mathjax>#"N"_2 -> "1.1 atm"#</mathjax></li>
</ul>
</blockquote>
<p>This means that the total pressure of the mixture will be </p>
<blockquote>
<p><mathjax>#P_"total" = P_(CH_4) + P_(H_2) + P_(N_2)#</mathjax></p>
<p><mathjax>#P_"total" = "1.0 atm" + "1.2 atm" + "1.1 atm" = color(green)("3.3 atm")#</mathjax></p>
</blockquote>
<p>The underlying principle that's at work here is the fact that at <em>constant temperature and volume</em>, the pressure of a gas is <strong>proportional</strong> ot the number of moles of gas.</p>
<p>
<iframe src="https://www.youtube.com/embed/RqffPYOoxd8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | A flask contains a gas mixture of methane, hydrogen. and nitrogen with partial pressures of 1 atm. 1.2 atm, and 1.1 aim, respectively. What is the total pressure of the mixture? | null |
1,467 | aac2afee-6ddd-11ea-bd12-ccda262736ce | https://socratic.org/questions/a-solution-with-a-volume-of-0-25-liters-contains-20-grams-of-hydrogen-fluoride-h | 4.00 M | start physical_unit 20 21 molarity mol/l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 14 14 9 10 mass qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"4.00 M"}] | [{"type":"physical unit","value":"Volume [OF] HF solution [=] \\pu{0.25 liters}"},{"type":"physical unit","value":"Mass [OF] HF [=] \\pu{20 grams}"}] | <h1 class="questionTitle" itemprop="name">A solution with a volume of 0.25 liters contains 20 grams of hydrogen fluoride, HF. What is the molarity of the solution? </h1> | null | 4.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>, you must determine how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>To determine the number of moles of hydrofluoric acid present in your sample, use the compound's <strong>molar mass</strong>. Hydrofluoric acid has a molar mass of <mathjax>#"20.01 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of this compound has a mass of <mathjax>#"20.01 g"#</mathjax>.</p>
<p>Since your sample has a mass of <mathjax>#"20 g"#</mathjax>, you can say that it contains</p>
<blockquote>
<p><mathjax>#20 color(red)(cancel(color(black)("g"))) * "1 mole HF"/(20.01color(red)(cancel(color(black)("g")))) = "0.9995 moles HF"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles are being dissolved in <mathjax>#"0.25 L"#</mathjax>, so you can say that <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.9995 moles HF"/(0.25color(red)(cancel(color(black)("L solution")))) = "3.998 moles HF"#</mathjax></p>
</blockquote>
<p>Rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the molarity of the solution will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("molarity HF solution" = "4 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"4 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>, you must determine how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>To determine the number of moles of hydrofluoric acid present in your sample, use the compound's <strong>molar mass</strong>. Hydrofluoric acid has a molar mass of <mathjax>#"20.01 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of this compound has a mass of <mathjax>#"20.01 g"#</mathjax>.</p>
<p>Since your sample has a mass of <mathjax>#"20 g"#</mathjax>, you can say that it contains</p>
<blockquote>
<p><mathjax>#20 color(red)(cancel(color(black)("g"))) * "1 mole HF"/(20.01color(red)(cancel(color(black)("g")))) = "0.9995 moles HF"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles are being dissolved in <mathjax>#"0.25 L"#</mathjax>, so you can say that <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.9995 moles HF"/(0.25color(red)(cancel(color(black)("L solution")))) = "3.998 moles HF"#</mathjax></p>
</blockquote>
<p>Rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the molarity of the solution will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("molarity HF solution" = "4 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A solution with a volume of 0.25 liters contains 20 grams of hydrogen fluoride, HF. What is the molarity of the solution? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-10-22T00:11:51" itemprop="dateCreated">
Oct 22, 2016
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<div class="markdown"><p><mathjax>#"4 mol L"^(-1)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>, you must determine how many <em>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> you get <strong>per liter</strong> of solution. </p>
<p>To determine the number of moles of hydrofluoric acid present in your sample, use the compound's <strong>molar mass</strong>. Hydrofluoric acid has a molar mass of <mathjax>#"20.01 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of this compound has a mass of <mathjax>#"20.01 g"#</mathjax>.</p>
<p>Since your sample has a mass of <mathjax>#"20 g"#</mathjax>, you can say that it contains</p>
<blockquote>
<p><mathjax>#20 color(red)(cancel(color(black)("g"))) * "1 mole HF"/(20.01color(red)(cancel(color(black)("g")))) = "0.9995 moles HF"#</mathjax></p>
</blockquote>
<p>Now, you know that this many moles are being dissolved in <mathjax>#"0.25 L"#</mathjax>, so you can say that <mathjax>#"1 L"#</mathjax> of this solution will contain</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * "0.9995 moles HF"/(0.25color(red)(cancel(color(black)("L solution")))) = "3.998 moles HF"#</mathjax></p>
</blockquote>
<p>Rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the molarity of the solution will be</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("molarity HF solution" = "4 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | A solution with a volume of 0.25 liters contains 20 grams of hydrogen fluoride, HF. What is the molarity of the solution? | null |
1,468 | a9701b18-6ddd-11ea-a7e1-ccda262736ce | https://socratic.org/questions/if-15-moles-of-water-was-made-in-this-reaction-how-many-grams-of-carbon-dioxide- | 660.15 grams | start physical_unit 14 15 mass g qc_end physical_unit 4 4 1 2 mole qc_end chemical_equation 18 22 qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] grams"}] | [{"type":"physical unit","value":"660.15 grams"}] | [{"type":"physical unit","value":"Mole [OF] water [=] \\pu{15 moles}"},{"type":"chemical equation","value":"H2CO3 -> H2O + CO2"}] | <h1 class="questionTitle" itemprop="name">If 15 moles of water was made in this reaction, how many grams of carbon dioxide were made?
#H_2CO_3rarrH_2O+CO_2#</h1> | null | 660.15 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the equation <mathjax>#H_2CO_3rarrH_2O+CO_2#</mathjax>.</p>
<p>The moles of water and carbon dioxide are in the ratio <mathjax>#1:1#</mathjax>.</p>
<p>So <mathjax>#15#</mathjax> moles of water and <mathjax>#15#</mathjax> moles of carbon dioxide are produced.</p>
<p>The formula for moles is <mathjax>#n=m/M#</mathjax>, where <mathjax>#m#</mathjax> is the mass of the substance and <mathjax>#M#</mathjax> its molar mass.</p>
<p>Rearranging the equation we get <mathjax>#m=nM#</mathjax>.</p>
<p>We have <mathjax>#15"mol"#</mathjax> of carbon dioxide and the molar mass of the gas is <mathjax>#44.01"g"/"mol"#</mathjax>.</p>
<p>Inputting:</p>
<p><mathjax>#m=15*44.01#</mathjax></p>
<p><mathjax>#m=660.15"g"#</mathjax> of carbon dioxide was produced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#660.15"g"#</mathjax> of carbon dioxide was produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the equation <mathjax>#H_2CO_3rarrH_2O+CO_2#</mathjax>.</p>
<p>The moles of water and carbon dioxide are in the ratio <mathjax>#1:1#</mathjax>.</p>
<p>So <mathjax>#15#</mathjax> moles of water and <mathjax>#15#</mathjax> moles of carbon dioxide are produced.</p>
<p>The formula for moles is <mathjax>#n=m/M#</mathjax>, where <mathjax>#m#</mathjax> is the mass of the substance and <mathjax>#M#</mathjax> its molar mass.</p>
<p>Rearranging the equation we get <mathjax>#m=nM#</mathjax>.</p>
<p>We have <mathjax>#15"mol"#</mathjax> of carbon dioxide and the molar mass of the gas is <mathjax>#44.01"g"/"mol"#</mathjax>.</p>
<p>Inputting:</p>
<p><mathjax>#m=15*44.01#</mathjax></p>
<p><mathjax>#m=660.15"g"#</mathjax> of carbon dioxide was produced.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If 15 moles of water was made in this reaction, how many grams of carbon dioxide were made?
#H_2CO_3rarrH_2O+CO_2#</h1>
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Surya K.
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Feb 12, 2018
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<div class="markdown"><p><mathjax>#660.15"g"#</mathjax> of carbon dioxide was produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have the equation <mathjax>#H_2CO_3rarrH_2O+CO_2#</mathjax>.</p>
<p>The moles of water and carbon dioxide are in the ratio <mathjax>#1:1#</mathjax>.</p>
<p>So <mathjax>#15#</mathjax> moles of water and <mathjax>#15#</mathjax> moles of carbon dioxide are produced.</p>
<p>The formula for moles is <mathjax>#n=m/M#</mathjax>, where <mathjax>#m#</mathjax> is the mass of the substance and <mathjax>#M#</mathjax> its molar mass.</p>
<p>Rearranging the equation we get <mathjax>#m=nM#</mathjax>.</p>
<p>We have <mathjax>#15"mol"#</mathjax> of carbon dioxide and the molar mass of the gas is <mathjax>#44.01"g"/"mol"#</mathjax>.</p>
<p>Inputting:</p>
<p><mathjax>#m=15*44.01#</mathjax></p>
<p><mathjax>#m=660.15"g"#</mathjax> of carbon dioxide was produced.</p></div>
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</article> | If 15 moles of water was made in this reaction, how many grams of carbon dioxide were made?
#H_2CO_3rarrH_2O+CO_2# | null |
1,469 | a887db8c-6ddd-11ea-a20f-ccda262736ce | https://socratic.org/questions/58f75e0311ef6b4cf4c17815 | ZnCO3(s) ＋ 2 HNO3(aq) -> Zn(NO3)2(aq) ＋ CO2(g) ＋ H2O(l) | start chemical_equation qc_end substance 11 12 qc_end substance 14 15 qc_end end | [{"type":"other","value":"Chemical Equation [OF] a stoichiometric equation"}] | [{"type":"chemical equation","value":"ZnCO3(s) ＋ 2 HNO3(aq) -> Zn(NO3)2(aq) ＋ CO2(g) ＋ H2O(l)"}] | [{"type":"substance name","value":"zinc carbonate"},{"type":"substance name","value":"nitric acid"}] | <h1 class="questionTitle" itemprop="name">How do you write a stoichiometric equation for the reaction between zinc carbonate and nitric acid?</h1> | null | ZnCO3(s) ＋ 2 HNO3(aq) -> Zn(NO3)2(aq) ＋ CO2(g) ＋ H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this balanced? What would you likely see in the reaction? Zinc carbonate is a bit insoluble. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#ZnCO_3(s) + 2HNO_3(aq) rarr Zn(NO_3)_2(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is this balanced? What would you likely see in the reaction? Zinc carbonate is a bit insoluble. </p></div>
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<div class="markdown"><p><mathjax>#ZnCO_3(s) + 2HNO_3(aq) rarr Zn(NO_3)_2(aq) + CO_2(g)uarr + H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is this balanced? What would you likely see in the reaction? Zinc carbonate is a bit insoluble. </p></div>
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</article> | How do you write a stoichiometric equation for the reaction between zinc carbonate and nitric acid? | null |
1,470 | a8df89a8-6ddd-11ea-905c-ccda262736ce | https://socratic.org/questions/hydrogen-is-collected-over-water-at-0-975-atm-and-28-c-what-is-the-partial-press | 0.94 atm | start physical_unit 16 16 partial_pressure atm qc_end substance 4 4 qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] H2 [IN] atm"}] | [{"type":"physical unit","value":"0.94 atm"}] | [{"type":"physical unit","value":"Temperature [OF] gas [=] \\pu{28 ℃}"},{"type":"physical unit","value":"Pressure [OF] gas [=] \\pu{0.975 atm}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">Hydrogen is collected over water at 0.975 atm and 28°C. What is the partial pressure of #H_2#?</h1> | null | 0.94 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"measured"=P_"dihydrogen"+P_"SVP"#</mathjax></p>
<p><mathjax>#P_"dihydrogen"=P_"measured"-P_"SVP"#</mathjax></p>
<p><a href="https://en.wikipedia.org/wiki/Vapour_pressure_of_water" rel="nofollow">This site gives</a> <mathjax>#P_"SVP"=28*mm*Hg#</mathjax> at <mathjax>#25#</mathjax> <mathjax>#""^@C#</mathjax>. These data SHOULD have been supplied with the question. </p>
<p>And so..............................</p>
<p><mathjax>#P_"dihydrogen"=0.975*atm-(28*mm*Hg)/(760*mm*Hg*atm^-1)=0.938*atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, we need the <mathjax>#"saturated vapour pressure"#</mathjax>, <mathjax>#P_"SVP"#</mathjax>, of water at <mathjax>#301.15*K#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"measured"=P_"dihydrogen"+P_"SVP"#</mathjax></p>
<p><mathjax>#P_"dihydrogen"=P_"measured"-P_"SVP"#</mathjax></p>
<p><a href="https://en.wikipedia.org/wiki/Vapour_pressure_of_water" rel="nofollow">This site gives</a> <mathjax>#P_"SVP"=28*mm*Hg#</mathjax> at <mathjax>#25#</mathjax> <mathjax>#""^@C#</mathjax>. These data SHOULD have been supplied with the question. </p>
<p>And so..............................</p>
<p><mathjax>#P_"dihydrogen"=0.975*atm-(28*mm*Hg)/(760*mm*Hg*atm^-1)=0.938*atm#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Hydrogen is collected over water at 0.975 atm and 28°C. What is the partial pressure of #H_2#?</h1>
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<div class="markdown"><p>Well, we need the <mathjax>#"saturated vapour pressure"#</mathjax>, <mathjax>#P_"SVP"#</mathjax>, of water at <mathjax>#301.15*K#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"measured"=P_"dihydrogen"+P_"SVP"#</mathjax></p>
<p><mathjax>#P_"dihydrogen"=P_"measured"-P_"SVP"#</mathjax></p>
<p><a href="https://en.wikipedia.org/wiki/Vapour_pressure_of_water" rel="nofollow">This site gives</a> <mathjax>#P_"SVP"=28*mm*Hg#</mathjax> at <mathjax>#25#</mathjax> <mathjax>#""^@C#</mathjax>. These data SHOULD have been supplied with the question. </p>
<p>And so..............................</p>
<p><mathjax>#P_"dihydrogen"=0.975*atm-(28*mm*Hg)/(760*mm*Hg*atm^-1)=0.938*atm#</mathjax></p></div>
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</article> | Hydrogen is collected over water at 0.975 atm and 28°C. What is the partial pressure of #H_2#? | null |
1,471 | a8ef8b10-6ddd-11ea-bb2c-ccda262736ce | https://socratic.org/questions/590cc01011ef6b197a2f8c4c | 2.7 mol/L | start physical_unit 18 18 molarity mol/l qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 6 6 11 12 molecular_weight qc_end physical_unit 18 18 20 21 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] mol/L"}] | [{"type":"physical unit","value":"2.7 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] the solute [=] \\pu{66.5 g}"},{"type":"physical unit","value":"Formula mass [OF] the solute [=] \\pu{80.05 g/mol}"},{"type":"physical unit","value":"Volume [OF] the solution [=] \\pu{315 mL}"}] | <h1 class="questionTitle" itemprop="name">A #66.5*g# mass of a solute whose formula mass is #80.05*g*mol^-1# is used to prepare a solution of #315*mL# volume. What is its molarity?</h1> | null | 2.7 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want the quotient, <mathjax>#"moles of stuff (mol)"/"volume of solution (L)"#</mathjax> to give the <mathjax>#"molarity"#</mathjax>.</p>
<p>And thus <mathjax>#((66.5*g)/(80.05*g*mol^-1))/(315*mLxx10^-3*L*mL^-1)~=2.7*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Molarity"~=2.7*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We want the quotient, <mathjax>#"moles of stuff (mol)"/"volume of solution (L)"#</mathjax> to give the <mathjax>#"molarity"#</mathjax>.</p>
<p>And thus <mathjax>#((66.5*g)/(80.05*g*mol^-1))/(315*mLxx10^-3*L*mL^-1)~=2.7*mol*L^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A #66.5*g# mass of a solute whose formula mass is #80.05*g*mol^-1# is used to prepare a solution of #315*mL# volume. What is its molarity?</h1>
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<div class="markdown"><p><mathjax>#"Molarity"~=2.7*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p>We want the quotient, <mathjax>#"moles of stuff (mol)"/"volume of solution (L)"#</mathjax> to give the <mathjax>#"molarity"#</mathjax>.</p>
<p>And thus <mathjax>#((66.5*g)/(80.05*g*mol^-1))/(315*mLxx10^-3*L*mL^-1)~=2.7*mol*L^-1#</mathjax></p></div>
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</article> | A #66.5*g# mass of a solute whose formula mass is #80.05*g*mol^-1# is used to prepare a solution of #315*mL# volume. What is its molarity? | null |
1,472 | ac06725e-6ddd-11ea-9e49-ccda262736ce | https://socratic.org/questions/if-you-have-200-grams-of-na-2o-how-many-moles-of-na-2o-do-you-have | 3.23 moles | start physical_unit 6 6 mole mol qc_end physical_unit 6 6 3 4 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] Na2O [IN] moles"}] | [{"type":"physical unit","value":"3.23 moles"}] | [{"type":"physical unit","value":"Mass [OF] Na2O [=] \\pu{200 grams}"}] | <h1 class="questionTitle" itemprop="name">If you have 200 grams of #Na_2O#, how many moles of #Na_2O# do you have?</h1> | null | 3.23 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles "= " Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(200*g)/(61.98*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??" moles"#</mathjax></p>
<p>The signal advantage of doing calculations this way, i.e. including units in the calculation is that it adds an extra level of validation. I wanted an answer in <mathjax>#"moles"#</mathjax>. The calculation as given me an answer in <mathjax>#"moles"#</mathjax>, i.e. <mathjax>#(200*cancelg)/(61.98*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?mol#</mathjax>. Why? Because <mathjax>#1/x^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/(1/x)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#x#</mathjax></p>
<p>This is an example of dimensional analysis. You can get very good at it and introduce another level of checking your answser. </p></div>
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<div class="markdown"><p>You have <mathjax>#3#</mathjax> and a bit moles of sodium oxide.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles "= " Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(200*g)/(61.98*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??" moles"#</mathjax></p>
<p>The signal advantage of doing calculations this way, i.e. including units in the calculation is that it adds an extra level of validation. I wanted an answer in <mathjax>#"moles"#</mathjax>. The calculation as given me an answer in <mathjax>#"moles"#</mathjax>, i.e. <mathjax>#(200*cancelg)/(61.98*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?mol#</mathjax>. Why? Because <mathjax>#1/x^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/(1/x)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#x#</mathjax></p>
<p>This is an example of dimensional analysis. You can get very good at it and introduce another level of checking your answser. </p></div>
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<h1 class="questionTitle" itemprop="name">If you have 200 grams of #Na_2O#, how many moles of #Na_2O# do you have?</h1>
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<div class="markdown"><p>You have <mathjax>#3#</mathjax> and a bit moles of sodium oxide.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles "= " Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(200*g)/(61.98*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??" moles"#</mathjax></p>
<p>The signal advantage of doing calculations this way, i.e. including units in the calculation is that it adds an extra level of validation. I wanted an answer in <mathjax>#"moles"#</mathjax>. The calculation as given me an answer in <mathjax>#"moles"#</mathjax>, i.e. <mathjax>#(200*cancelg)/(61.98*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?mol#</mathjax>. Why? Because <mathjax>#1/x^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/(1/x)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#x#</mathjax></p>
<p>This is an example of dimensional analysis. You can get very good at it and introduce another level of checking your answser. </p></div>
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</article> | If you have 200 grams of #Na_2O#, how many moles of #Na_2O# do you have? | null |
1,473 | a9ca2833-6ddd-11ea-a283-ccda262736ce | https://socratic.org/questions/how-would-you-balance-ko2-s-h2o-l-koh-aq-o2-g-h2o2-aq | 2 KO2(s) + 2 H2O(l) -> 2 KOH(aq) + O2(g) + H2O2(aq) | start chemical_equation qc_end chemical_equation 4 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 KO2(s) + 2 H2O(l) -> 2 KOH(aq) + O2(g) + H2O2(aq)"}] | [{"type":"chemical equation","value":"KO2(s) + H2O(l) -> KOH(aq) + O2(g) + H2O2(aq)"}] | <h1 class="questionTitle" itemprop="name">How would you balance: KO2(s) + H2O(l) --? KOH(aq) + O2(g) +H2O2(aq)?</h1> | null | 2 KO2(s) + 2 H2O(l) -> 2 KOH(aq) + O2(g) + H2O2(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First tally all the atoms involved in the reaction. You can do this by looking at the subscripts.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> (unbalanced)</p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 (do not add this up yet)<br/>
H = 2</p>
<p><em>right side:</em><br/>
K = 1<br/>
O = 1 + 2 + 2 (do not add this up yet)<br/>
H = 1 + 2 (do not add this up yet)</p>
<p>Always remember that in balancing equations, you are not allowed to change the subscript, but you are allowed to put coefficients before the chemical formulas.</p>
<p>Let's start to balance the easiest atom, in this case the <mathjax>#H#</mathjax> atom. Since there are 2 <mathjax>#H#</mathjax> atoms on the left and 3 <mathjax>#H#</mathjax> atoms on the right (one coming from <mathjax>#KOH#</mathjax> and the other two coming from <mathjax>#H_2O_2#</mathjax>), you need to find a factor that you can multiply in either one of the substances to produce a desirable number to balance. </p>
<p>Since <mathjax>#KOH#</mathjax> is the simplest of the three substances with <mathjax>#H#</mathjax> atoms, we'll start the balancing there.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (red) 2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 <br/>
H = 2</p>
<p><em>right side:</em><br/>
K = (1 x <mathjax>#color (red) 2#</mathjax>)<br/>
O = (1 x <mathjax>#color (red) 2#</mathjax>) + 2 + 2 <br/>
H = (1 x <mathjax>#color (red) 2#</mathjax>) + 2</p>
<p>Notice that since the <mathjax>#H#</mathjax> atom is bonded to one <mathjax>#K#</mathjax> atom and one <mathjax>#O#</mathjax> atom, we also need to apply the coefficients to these elements. Now you have a total of 4 <mathjax>#H#</mathjax> atoms on the right, so to balance the <mathjax>#H#</mathjax> atoms on the left,</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#color (blue) 2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + (1 x <mathjax>#color (blue) 2#</mathjax>) <br/>
H = 2 x <mathjax>#color (blue) 2#</mathjax> = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Also, you have two <mathjax>#K#</mathjax> atoms on the right but only one <mathjax>#K#</mathjax> atom on the left. Thus,</p>
<p><mathjax>#color (green) 2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x <mathjax>#color (green) 2#</mathjax> = <strong>2</strong><br/>
O = (2 x <mathjax>#color (green) 2#</mathjax>) + (1 x 2) <br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Now the only thing left to balance are the <mathjax>#O#</mathjax> atoms. But then again if you get the sum of <mathjax>#O#</mathjax> atoms on both sides of the equation,</p>
<p><mathjax>#2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (2 x 2) + (1 x 2) = <strong>6</strong><br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 = <strong>6</strong><br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>The equation is already balanced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Provided that the given reactants and products are correct, the balanced equation is <mathjax>#2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First tally all the atoms involved in the reaction. You can do this by looking at the subscripts.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> (unbalanced)</p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 (do not add this up yet)<br/>
H = 2</p>
<p><em>right side:</em><br/>
K = 1<br/>
O = 1 + 2 + 2 (do not add this up yet)<br/>
H = 1 + 2 (do not add this up yet)</p>
<p>Always remember that in balancing equations, you are not allowed to change the subscript, but you are allowed to put coefficients before the chemical formulas.</p>
<p>Let's start to balance the easiest atom, in this case the <mathjax>#H#</mathjax> atom. Since there are 2 <mathjax>#H#</mathjax> atoms on the left and 3 <mathjax>#H#</mathjax> atoms on the right (one coming from <mathjax>#KOH#</mathjax> and the other two coming from <mathjax>#H_2O_2#</mathjax>), you need to find a factor that you can multiply in either one of the substances to produce a desirable number to balance. </p>
<p>Since <mathjax>#KOH#</mathjax> is the simplest of the three substances with <mathjax>#H#</mathjax> atoms, we'll start the balancing there.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (red) 2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 <br/>
H = 2</p>
<p><em>right side:</em><br/>
K = (1 x <mathjax>#color (red) 2#</mathjax>)<br/>
O = (1 x <mathjax>#color (red) 2#</mathjax>) + 2 + 2 <br/>
H = (1 x <mathjax>#color (red) 2#</mathjax>) + 2</p>
<p>Notice that since the <mathjax>#H#</mathjax> atom is bonded to one <mathjax>#K#</mathjax> atom and one <mathjax>#O#</mathjax> atom, we also need to apply the coefficients to these elements. Now you have a total of 4 <mathjax>#H#</mathjax> atoms on the right, so to balance the <mathjax>#H#</mathjax> atoms on the left,</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#color (blue) 2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + (1 x <mathjax>#color (blue) 2#</mathjax>) <br/>
H = 2 x <mathjax>#color (blue) 2#</mathjax> = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Also, you have two <mathjax>#K#</mathjax> atoms on the right but only one <mathjax>#K#</mathjax> atom on the left. Thus,</p>
<p><mathjax>#color (green) 2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x <mathjax>#color (green) 2#</mathjax> = <strong>2</strong><br/>
O = (2 x <mathjax>#color (green) 2#</mathjax>) + (1 x 2) <br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Now the only thing left to balance are the <mathjax>#O#</mathjax> atoms. But then again if you get the sum of <mathjax>#O#</mathjax> atoms on both sides of the equation,</p>
<p><mathjax>#2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (2 x 2) + (1 x 2) = <strong>6</strong><br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 = <strong>6</strong><br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>The equation is already balanced.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you balance: KO2(s) + H2O(l) --? KOH(aq) + O2(g) +H2O2(aq)?</h1>
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<a class="topContributorPic" href="/users/nikka-c"><img alt="" class="" src="https://profilepictures.socratic.org/AQ8VOCyTM1J7U0TvIWhw_nkahara.jpg" title=""/></a>
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<div class="answerInfoTop clearfix"><a class="topContributorLink" href="/users/nikka-c">
Nikka C.
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Adam G
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<span class="dateCreated" datetime="2015-11-13T01:52:48" itemprop="dateCreated">
Nov 13, 2015
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<div class="markdown"><p>Provided that the given reactants and products are correct, the balanced equation is <mathjax>#2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First tally all the atoms involved in the reaction. You can do this by looking at the subscripts.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> (unbalanced)</p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 (do not add this up yet)<br/>
H = 2</p>
<p><em>right side:</em><br/>
K = 1<br/>
O = 1 + 2 + 2 (do not add this up yet)<br/>
H = 1 + 2 (do not add this up yet)</p>
<p>Always remember that in balancing equations, you are not allowed to change the subscript, but you are allowed to put coefficients before the chemical formulas.</p>
<p>Let's start to balance the easiest atom, in this case the <mathjax>#H#</mathjax> atom. Since there are 2 <mathjax>#H#</mathjax> atoms on the left and 3 <mathjax>#H#</mathjax> atoms on the right (one coming from <mathjax>#KOH#</mathjax> and the other two coming from <mathjax>#H_2O_2#</mathjax>), you need to find a factor that you can multiply in either one of the substances to produce a desirable number to balance. </p>
<p>Since <mathjax>#KOH#</mathjax> is the simplest of the three substances with <mathjax>#H#</mathjax> atoms, we'll start the balancing there.</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color (red) 2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + 1 <br/>
H = 2</p>
<p><em>right side:</em><br/>
K = (1 x <mathjax>#color (red) 2#</mathjax>)<br/>
O = (1 x <mathjax>#color (red) 2#</mathjax>) + 2 + 2 <br/>
H = (1 x <mathjax>#color (red) 2#</mathjax>) + 2</p>
<p>Notice that since the <mathjax>#H#</mathjax> atom is bonded to one <mathjax>#K#</mathjax> atom and one <mathjax>#O#</mathjax> atom, we also need to apply the coefficients to these elements. Now you have a total of 4 <mathjax>#H#</mathjax> atoms on the right, so to balance the <mathjax>#H#</mathjax> atoms on the left,</p>
<p><mathjax>#KO_2#</mathjax> + <mathjax>#color (blue) 2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1<br/>
O = 2 + (1 x <mathjax>#color (blue) 2#</mathjax>) <br/>
H = 2 x <mathjax>#color (blue) 2#</mathjax> = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Also, you have two <mathjax>#K#</mathjax> atoms on the right but only one <mathjax>#K#</mathjax> atom on the left. Thus,</p>
<p><mathjax>#color (green) 2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x <mathjax>#color (green) 2#</mathjax> = <strong>2</strong><br/>
O = (2 x <mathjax>#color (green) 2#</mathjax>) + (1 x 2) <br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 <br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>Now the only thing left to balance are the <mathjax>#O#</mathjax> atoms. But then again if you get the sum of <mathjax>#O#</mathjax> atoms on both sides of the equation,</p>
<p><mathjax>#2KO_2#</mathjax> + <mathjax>#2H_2O#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2KOH#</mathjax> + <mathjax>#O_2#</mathjax> + <mathjax>#H_2O_2#</mathjax> </p>
<p><em>left side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (2 x 2) + (1 x 2) = <strong>6</strong><br/>
H = 2 x 2 = <strong>4</strong></p>
<p><em>right side:</em><br/>
K = 1 x 2 = <strong>2</strong><br/>
O = (1 x 2) + 2 + 2 = <strong>6</strong><br/>
H = (1 x 2) + 2 = <strong>4</strong></p>
<p>The equation is already balanced.</p></div>
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</article> | How would you balance: KO2(s) + H2O(l) --? KOH(aq) + O2(g) +H2O2(aq)? | null |
1,474 | ad07ef0b-6ddd-11ea-8f33-ccda262736ce | https://socratic.org/questions/what-is-the-poh-of-a-6-5-10-12-m-oh-solution | 11.2 | start physical_unit 10 11 poh none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pOH [OF] OH- solution"}] | [{"type":"physical unit","value":"11.2"}] | [{"type":"physical unit","value":"Molarity [OF] OH- solution [=] \\pu{6.5 × 10^(-12) M}"}] | <h1 class="questionTitle" itemprop="name">What is the #pOH# of a #6.5*10^-12# #M# #OH^-# solution?</h1> | null | 11.2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#-log_10{6.5xx10^-12}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#11.2#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution? You should not need a calculator. </p></div>
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<div class="markdown"><p><mathjax>#pOH~=11.#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#-log_10{6.5xx10^-12}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#11.2#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution? You should not need a calculator. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pOH# of a #6.5*10^-12# #M# #OH^-# solution?</h1>
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<div class="markdown"><p><mathjax>#pOH~=11.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#-log_10{6.5xx10^-12}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#11.2#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution? You should not need a calculator. </p></div>
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</article> | What is the #pOH# of a #6.5*10^-12# #M# #OH^-# solution? | null |
1,475 | a8d4ea9a-6ddd-11ea-acd4-ccda262736ce | https://socratic.org/questions/combustion-of-a-0-9827-g-sample-of-a-compound-containing-only-carbon-hydrogen-an | C4H11O2 | start chemical_formula qc_end physical_unit 5 8 3 4 mass qc_end physical_unit 19 19 16 17 mass qc_end physical_unit 24 24 21 22 mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"C4H11O2"}] | [{"type":"physical unit","value":"Mass [OF] a compound sample [=] \\pu{0.9827 g}"},{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{1.900 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{1.070 g}"},{"type":"other","value":"A sample of a compound contains only carbon, hydrogen, and oxygen."}] | <h1 class="questionTitle" itemprop="name">Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of #CO_2# and 1.070 g of #H_2O#. What is the empirical formula of the compound? </h1> | null | C4H11O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here is the equation with masses:</p>
<p><mathjax>#"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2#</mathjax></p>
<p>Use 32.00 g/mol for the molar mass of <mathjax>#"O"_2#</mathjax><br/>
Use 18.03 g/mol for the molar mass of <mathjax>#"H"_2"O"#</mathjax><br/>
Use 44.01 g/mol for the molar mass of <mathjax>#"CO"_2#</mathjax></p>
<p>We do not know the molar mass of the hydrocarbon so we allow <mathjax>#a, b#</mathjax>, and <mathjax>#c#</mathjax> to be any positive real number.</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900
g")/("44.01 g/mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Perform the division:</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Matching coefficients, we get:</p>
<p><mathjax>#a = 0.04317#</mathjax><br/>
<mathjax>#b = 2 xx 0.05935 = 0.1187#</mathjax> <br/>
<mathjax>#c = 0.05935 + 2 xx 0.04317 - 2 xx 0.06210 = 0.02149#</mathjax></p>
<blockquote></blockquote>
<p>Divide every number by 0.02149:</p>
<p><mathjax>#a/0.02149 = 2.009#</mathjax><br/>
<mathjax>#b/0.02149 = 5.523#</mathjax> <br/>
<mathjax>#c/0.02149 = 1#</mathjax></p>
<blockquote></blockquote>
<p>Multiply every number by 2:</p>
<p><mathjax>#2.009 xx 2 = 4.018#</mathjax><br/>
<mathjax>#5.523 xx 2 = 11.05#</mathjax><br/>
<mathjax>#1 xx 2 = 2#</mathjax></p>
<p>Round off each number to the nearest integer.</p>
<p><mathjax>#4.018 ≈ color(white)(l)4#</mathjax><br/>
<mathjax>#11.05 ≈ 11#</mathjax><br/>
<mathjax>#color(white)(ll)2color(white)(ml) =color(white)(ll) 2#</mathjax></p>
<blockquote></blockquote>
<p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here is the equation with masses:</p>
<p><mathjax>#"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2#</mathjax></p>
<p>Use 32.00 g/mol for the molar mass of <mathjax>#"O"_2#</mathjax><br/>
Use 18.03 g/mol for the molar mass of <mathjax>#"H"_2"O"#</mathjax><br/>
Use 44.01 g/mol for the molar mass of <mathjax>#"CO"_2#</mathjax></p>
<p>We do not know the molar mass of the hydrocarbon so we allow <mathjax>#a, b#</mathjax>, and <mathjax>#c#</mathjax> to be any positive real number.</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900
g")/("44.01 g/mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Perform the division:</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Matching coefficients, we get:</p>
<p><mathjax>#a = 0.04317#</mathjax><br/>
<mathjax>#b = 2 xx 0.05935 = 0.1187#</mathjax> <br/>
<mathjax>#c = 0.05935 + 2 xx 0.04317 - 2 xx 0.06210 = 0.02149#</mathjax></p>
<blockquote></blockquote>
<p>Divide every number by 0.02149:</p>
<p><mathjax>#a/0.02149 = 2.009#</mathjax><br/>
<mathjax>#b/0.02149 = 5.523#</mathjax> <br/>
<mathjax>#c/0.02149 = 1#</mathjax></p>
<blockquote></blockquote>
<p>Multiply every number by 2:</p>
<p><mathjax>#2.009 xx 2 = 4.018#</mathjax><br/>
<mathjax>#5.523 xx 2 = 11.05#</mathjax><br/>
<mathjax>#1 xx 2 = 2#</mathjax></p>
<p>Round off each number to the nearest integer.</p>
<p><mathjax>#4.018 ≈ color(white)(l)4#</mathjax><br/>
<mathjax>#11.05 ≈ 11#</mathjax><br/>
<mathjax>#color(white)(ll)2color(white)(ml) =color(white)(ll) 2#</mathjax></p>
<blockquote></blockquote>
<p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of #CO_2# and 1.070 g of #H_2O#. What is the empirical formula of the compound? </h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here is the equation with masses:</p>
<p><mathjax>#"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2#</mathjax></p>
<p>Use 32.00 g/mol for the molar mass of <mathjax>#"O"_2#</mathjax><br/>
Use 18.03 g/mol for the molar mass of <mathjax>#"H"_2"O"#</mathjax><br/>
Use 44.01 g/mol for the molar mass of <mathjax>#"CO"_2#</mathjax></p>
<p>We do not know the molar mass of the hydrocarbon so we allow <mathjax>#a, b#</mathjax>, and <mathjax>#c#</mathjax> to be any positive real number.</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900
g")/("44.01 g/mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Perform the division:</p>
<p><mathjax>#"C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Matching coefficients, we get:</p>
<p><mathjax>#a = 0.04317#</mathjax><br/>
<mathjax>#b = 2 xx 0.05935 = 0.1187#</mathjax> <br/>
<mathjax>#c = 0.05935 + 2 xx 0.04317 - 2 xx 0.06210 = 0.02149#</mathjax></p>
<blockquote></blockquote>
<p>Divide every number by 0.02149:</p>
<p><mathjax>#a/0.02149 = 2.009#</mathjax><br/>
<mathjax>#b/0.02149 = 5.523#</mathjax> <br/>
<mathjax>#c/0.02149 = 1#</mathjax></p>
<blockquote></blockquote>
<p>Multiply every number by 2:</p>
<p><mathjax>#2.009 xx 2 = 4.018#</mathjax><br/>
<mathjax>#5.523 xx 2 = 11.05#</mathjax><br/>
<mathjax>#1 xx 2 = 2#</mathjax></p>
<p>Round off each number to the nearest integer.</p>
<p><mathjax>#4.018 ≈ color(white)(l)4#</mathjax><br/>
<mathjax>#11.05 ≈ 11#</mathjax><br/>
<mathjax>#color(white)(ll)2color(white)(ml) =color(white)(ll) 2#</mathjax></p>
<blockquote></blockquote>
<p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
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<div class="markdown"><p>The empirical formula is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>First, we calculate the masses of <mathjax>#"C"#</mathjax> and <mathjax>#"H"#</mathjax> from the masses of their oxides (<mathjax>#"CO"_2#</mathjax> and <mathjax>#"H"_2"O"#</mathjax>).</p>
<p><mathjax>#"Mass of C" = 1.900 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.5185 g C"#</mathjax></p>
<p><mathjax>#"Mass of H" = 1.070 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1197 g H"#</mathjax></p>
<p><mathjax>#"Mass of C + Mass of H" = "0.5185 g + 0.1197 g" = "0.6382 g"#</mathjax></p>
<p>This is less than the mass of the sample.</p>
<p>The missing mass must be caused by <mathjax>#"O"#</mathjax>.</p>
<p><mathjax>#"Mass of O = 0.9827 g - 0.6382 g = 0.3445 g"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we must convert these masses to moles and find their ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#bb("Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(mm) "Ratio"color(white)(ml)×2color(white)(mm)"Integers")#</mathjax><br/>
<mathjax>#color(white)(ml)"C" color(white)(XXXm)0.5185 color(white)(mll)0.04317color(white)(Xmlll)2.005color(white)(m)4.010color(white)(Xmmml)4#</mathjax><br/>
<mathjax>#color(white)(ml)"H" color(white)(XXXm)0.1197 color(white)(mll)0.1188 color(white)(mmml)5.518 color(white)(m)11.04color(white)(Xmmm)11#</mathjax><br/>
<mathjax>#color(white)(ml)"O" color(white)(XXXm)0.3445 color(white)(mll)0.02153 color(white)(mmm)1 color(white)(mmml)2color(white)(Xmmmmm)2#</mathjax></p>
<p>The <strong>empirical formula</strong> is <mathjax>#"C"_4"H"_11"O"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Note</strong>: This is an impossible empirical formula.</p>
<p>The molecular formula must have an even number of <mathjax>#"H"#</mathjax> atoms, e.g. <mathjax>#"C"_8"H"_22"O"_4#</mathjax>.</p>
<p>A compound with 8 <mathjax>#"C"#</mathjax> atoms can contain no more than 18 <mathjax>#"H"#</mathjax> atoms.</p></div>
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</article> | Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of #CO_2# and 1.070 g of #H_2O#. What is the empirical formula of the compound? | null |
1,476 | ab069874-6ddd-11ea-ae48-ccda262736ce | https://socratic.org/questions/a-0-2-mol-sample-of-n2-co2-with-a-pressure-of-840-torr-is-exposed-to-solid-cao-w | 0.12 moles | start physical_unit 7 7 mole mol qc_end physical_unit 3 3 1 2 mole qc_end physical_unit 3 3 12 13 pressure qc_end c_other OTHER qc_end physical_unit 3 3 41 42 pressure qc_end chemical_equation 18 18 qc_end chemical_equation 27 27 qc_end end | [{"type":"physical unit","value":"Mole [OF] CO2 [IN] moles"}] | [{"type":"physical unit","value":"0.12 moles"}] | [{"type":"physical unit","value":"Mole [OF] sample [=] \\pu{0.2 mol}"},{"type":"physical unit","value":"Pressure1 [OF] sample [=] \\pu{840 torr}"},{"type":"other","value":"Assume the reaction goes to completion complete pressure."},{"type":"physical unit","value":"Pressure2 [OF] sample [=] \\pu{320 torr}"},{"type":"chemical equation","value":"CaO"},{"type":"chemical equation","value":"CaCO3"}] | <h1 class="questionTitle" itemprop="name">A 0.2 mol sample of N2 , CO2 with a pressure of 840 torr is exposed to solid CaO which react with theCO2 to form solidCaCO3 assume the reaction goes to completion complete pressure the gas had dropped to 320 how many moles of CO2 in origin mixtur?
how</h1> | null | 0.12 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure in a gas is the sum of the partial pressures of the components, that, at the same temperature and constant volume, are proportional to the molar fractions of the components themselves.</p>
<p>The mixture of <mathjax>#"CO"_2#</mathjax> and <mathjax>#"N"_2#</mathjax> gases, after absorption of <mathjax>#"CO"_2#</mathjax> by <mathjax>#"CaO"#</mathjax>, drops by <mathjax>#"520 torr"#</mathjax>, which corresponds to the proportion of <mathjax>#"CO"_2#</mathjax> over the total gas.</p>
<p><mathjax>#520/820=x/0.2 #</mathjax></p>
<p>So </p>
<p><mathjax>#x = "0.124 moles"#</mathjax> </p>
<p>of <mathjax>#"CO"_2#</mathjax> were present in the original mixture.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.124 moles"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure in a gas is the sum of the partial pressures of the components, that, at the same temperature and constant volume, are proportional to the molar fractions of the components themselves.</p>
<p>The mixture of <mathjax>#"CO"_2#</mathjax> and <mathjax>#"N"_2#</mathjax> gases, after absorption of <mathjax>#"CO"_2#</mathjax> by <mathjax>#"CaO"#</mathjax>, drops by <mathjax>#"520 torr"#</mathjax>, which corresponds to the proportion of <mathjax>#"CO"_2#</mathjax> over the total gas.</p>
<p><mathjax>#520/820=x/0.2 #</mathjax></p>
<p>So </p>
<p><mathjax>#x = "0.124 moles"#</mathjax> </p>
<p>of <mathjax>#"CO"_2#</mathjax> were present in the original mixture.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 0.2 mol sample of N2 , CO2 with a pressure of 840 torr is exposed to solid CaO which react with theCO2 to form solidCaCO3 assume the reaction goes to completion complete pressure the gas had dropped to 320 how many moles of CO2 in origin mixtur?
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<div class="markdown"><p><mathjax>#"0.124 moles"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total pressure in a gas is the sum of the partial pressures of the components, that, at the same temperature and constant volume, are proportional to the molar fractions of the components themselves.</p>
<p>The mixture of <mathjax>#"CO"_2#</mathjax> and <mathjax>#"N"_2#</mathjax> gases, after absorption of <mathjax>#"CO"_2#</mathjax> by <mathjax>#"CaO"#</mathjax>, drops by <mathjax>#"520 torr"#</mathjax>, which corresponds to the proportion of <mathjax>#"CO"_2#</mathjax> over the total gas.</p>
<p><mathjax>#520/820=x/0.2 #</mathjax></p>
<p>So </p>
<p><mathjax>#x = "0.124 moles"#</mathjax> </p>
<p>of <mathjax>#"CO"_2#</mathjax> were present in the original mixture.</p></div>
</div>
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</article> | A 0.2 mol sample of N2 , CO2 with a pressure of 840 torr is exposed to solid CaO which react with theCO2 to form solidCaCO3 assume the reaction goes to completion complete pressure the gas had dropped to 320 how many moles of CO2 in origin mixtur?
how | null |
1,477 | aaabd8f4-6ddd-11ea-b5ec-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-sodium-chloride-solution-that-contains-3-moles-of-nacl | 0.75 M | start physical_unit 5 8 molarity mol/l qc_end physical_unit 14 14 11 12 mole qc_end physical_unit 8 8 16 17 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] a sodium chloride solution [IN] M"}] | [{"type":"physical unit","value":"0.75 M"}] | [{"type":"physical unit","value":"Mole [OF] NaCl [=] \\pu{3 moles}"},{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{4 liters}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a sodium chloride solution that contains 3 moles of #NaCl# in 4 liters of solution?</h1> | null | 0.75 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we need to do is to set up this quotient. You have been given the amount of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in moles, and the volume of solution.</p>
<p>So, <mathjax>#"Molarity" = (3*mol)/(4*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1?#</mathjax></p>
<p>In <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of this solution, what is the mass of the <mathjax>#NaCl#</mathjax> solute?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Molarity" = ("Moles of solute")/("Volume of SOLUTION")#</mathjax>. </p>
<p>And we get an answer with units, <mathjax>#mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we need to do is to set up this quotient. You have been given the amount of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in moles, and the volume of solution.</p>
<p>So, <mathjax>#"Molarity" = (3*mol)/(4*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1?#</mathjax></p>
<p>In <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of this solution, what is the mass of the <mathjax>#NaCl#</mathjax> solute?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a sodium chloride solution that contains 3 moles of #NaCl# in 4 liters of solution?</h1>
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Feb 14, 2016
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<div class="markdown"><p><mathjax>#"Molarity" = ("Moles of solute")/("Volume of SOLUTION")#</mathjax>. </p>
<p>And we get an answer with units, <mathjax>#mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So all we need to do is to set up this quotient. You have been given the amount of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in moles, and the volume of solution.</p>
<p>So, <mathjax>#"Molarity" = (3*mol)/(4*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#mol*L^-1?#</mathjax></p>
<p>In <mathjax>#1#</mathjax> <mathjax>#L#</mathjax> of this solution, what is the mass of the <mathjax>#NaCl#</mathjax> solute?</p></div>
</div>
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<a href="https://socratic.org/answers/226271" itemprop="url">Answer link</a>
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</article> | What is the molarity of a sodium chloride solution that contains 3 moles of #NaCl# in 4 liters of solution? | null |
1,478 | a99cf8f9-6ddd-11ea-9707-ccda262736ce | https://socratic.org/questions/the-pressure-of-a-gas-in-a-container-is-152-mm-hg-what-is-this-equivalent-to | 0.20 atm | start physical_unit 3 4 pressure atm qc_end physical_unit 3 4 9 11 pressure qc_end end | [{"type":"physical unit","value":"Preesure2 [OF] a gas [IN] atm"}] | [{"type":"physical unit","value":"0.20 atm"}] | [{"type":"physical unit","value":"Preesure1 [OF] a gas [=] \\pu{152 mm Hg}"}] | <h1 class="questionTitle" itemprop="name">The pressure of a gas in a container is 152 mm Hg. What is this equivalent to?</h1> | null | 0.20 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the pressure measurement is:</p>
<p><mathjax>#(152*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200*atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>It is a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> pressure will support a column of mercury <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> high. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the pressure measurement is:</p>
<p><mathjax>#(152*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200*atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The pressure of a gas in a container is 152 mm Hg. What is this equivalent to?</h1>
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<div class="markdown"><p>It is a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> pressure will support a column of mercury <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> high. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So the pressure measurement is:</p>
<p><mathjax>#(152*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200*atm#</mathjax></p></div>
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</article> | The pressure of a gas in a container is 152 mm Hg. What is this equivalent to? | null |
1,479 | a9db1859-6ddd-11ea-a8aa-ccda262736ce | https://socratic.org/questions/how-do-you-use-hess-s-law-to-calculate-enthalpy-for-this-reaction | +311.3 kJ | start physical_unit 53 54 deltah kj qc_end c_other OTHER qc_end chemical_equation 14 19 qc_end physical_unit 53 54 22 23 deltah qc_end chemical_equation 24 30 qc_end physical_unit 53 54 33 34 deltah qc_end chemical_equation 35 41 qc_end physical_unit 53 54 44 45 deltah qc_end chemical_equation 55 60 qc_end end | [{"type":"physical unit","value":"DeltaH4 [OF] the reaction [IN] kJ"}] | [{"type":"physical unit","value":"+311.3 kJ"}] | [{"type":"other","value":"Use Hess's Law."},{"type":"chemical equation","value":"N2(g) + O2(g) -> 2 NO(g)"},{"type":"physical unit","value":"DeltaH1 [OF] the reaction [=] \\pu{+180.7 kJ}"},{"type":"chemical equation","value":"2 NO(g) + O2(g) -> 2 NO2(g)"},{"type":"physical unit","value":"DeltaH2 [OF] the reaction [=] \\pu{-113.1 kJ}"},{"type":"chemical equation","value":"2 N2O(g) -> 2 N2(g) + O2(g)"},{"type":"physical unit","value":"DeltaH3 [OF] the reaction [=] \\pu{-163.2 kJ}"},{"type":"chemical equation","value":"N2O(g) + NO2(g) -> 3 NO(g)"}] | <h1 class="questionTitle" itemprop="name">How do you use Hess's Law to calculate enthalpy for this reaction?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Given Data:<br/>
<mathjax>#"N"_2(g) + "O"_2(g) -> 2"NO"(g)#</mathjax> </p>
<blockquote>
<p><mathjax>#DeltaH = +"180.7 kJ"#</mathjax></p>
</blockquote>
<p><mathjax>#2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaH = -"113.1 kJ"#</mathjax></p>
</blockquote>
<p><mathjax>#2"N"_2"O"(g) -> 2"N"_2(g) + "O"_2(g)#</mathjax> </p>
<blockquote>
<p><mathjax>#DeltaH = -"163.2 kJ"#</mathjax></p>
</blockquote>
<p>Use Hess' Law to calculate <mathjax>#DeltaH#</mathjax> for the reaction:</p>
<p><mathjax>#"N"_2"O"(g) + "NO"_2(g) -> 3"NO"(g)#</mathjax></p></div>
</h2>
</div>
</div> | +311.3 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>PC of cake</p>
<p><mathjax>#cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> </p>
<p><mathjax>#2NO_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p><mathjax>#2N_2 O#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#cancel"2"N_2#</mathjax> + <mathjax>#O_2#</mathjax> <br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding the 3 equations:</p>
<p><mathjax>#2NO_2#</mathjax> + <mathjax>#2N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#4 NO#</mathjax> + <mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p>Adding one more reaction<br/>
<mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#2 NO#</mathjax><br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding previous 2 equations and dividing by 2</p>
<p>Finally, </p>
<p><mathjax>#NO_2#</mathjax> + <mathjax>#N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#3 NO#</mathjax> </p>
<p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> is the sum of the all reactions:</p>
<p><mathjax>#"=>#</mathjax>180.7 113.1 + (- 163.2) + 180.7 = 311.3 kJ</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#ΔH = 311.3#</mathjax> <mathjax>#"kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>PC of cake</p>
<p><mathjax>#cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> </p>
<p><mathjax>#2NO_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p><mathjax>#2N_2 O#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#cancel"2"N_2#</mathjax> + <mathjax>#O_2#</mathjax> <br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding the 3 equations:</p>
<p><mathjax>#2NO_2#</mathjax> + <mathjax>#2N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#4 NO#</mathjax> + <mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p>Adding one more reaction<br/>
<mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#2 NO#</mathjax><br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding previous 2 equations and dividing by 2</p>
<p>Finally, </p>
<p><mathjax>#NO_2#</mathjax> + <mathjax>#N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#3 NO#</mathjax> </p>
<p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> is the sum of the all reactions:</p>
<p><mathjax>#"=>#</mathjax>180.7 113.1 + (- 163.2) + 180.7 = 311.3 kJ</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you use Hess's Law to calculate enthalpy for this reaction?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Given Data:<br/>
<mathjax>#"N"_2(g) + "O"_2(g) -> 2"NO"(g)#</mathjax> </p>
<blockquote>
<p><mathjax>#DeltaH = +"180.7 kJ"#</mathjax></p>
</blockquote>
<p><mathjax>#2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaH = -"113.1 kJ"#</mathjax></p>
</blockquote>
<p><mathjax>#2"N"_2"O"(g) -> 2"N"_2(g) + "O"_2(g)#</mathjax> </p>
<blockquote>
<p><mathjax>#DeltaH = -"163.2 kJ"#</mathjax></p>
</blockquote>
<p>Use Hess' Law to calculate <mathjax>#DeltaH#</mathjax> for the reaction:</p>
<p><mathjax>#"N"_2"O"(g) + "NO"_2(g) -> 3"NO"(g)#</mathjax></p></div>
</h2>
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<div class="markdown"><p><mathjax>#ΔH = 311.3#</mathjax> <mathjax>#"kJ"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>PC of cake</p>
<p><mathjax>#cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> </p>
<p><mathjax>#2NO_2#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#2NO#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p><mathjax>#2N_2 O#</mathjax> <mathjax>#" ----->"#</mathjax> <mathjax>#cancel"2"N_2#</mathjax> + <mathjax>#O_2#</mathjax> <br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding the 3 equations:</p>
<p><mathjax>#2NO_2#</mathjax> + <mathjax>#2N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#4 NO#</mathjax> + <mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> </p>
<p>Adding one more reaction<br/>
<mathjax># cancel"N"_2#</mathjax> + <mathjax>#cancel"O"_2#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#2 NO#</mathjax><br/>
<mathjax>#"---------------------------------"#</mathjax><br/>
Adding previous 2 equations and dividing by 2</p>
<p>Finally, </p>
<p><mathjax>#NO_2#</mathjax> + <mathjax>#N_2 O#</mathjax> <mathjax>#" -----> "#</mathjax><mathjax>#3 NO#</mathjax> </p>
<p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">Enthalpy</a> is the sum of the all reactions:</p>
<p><mathjax>#"=>#</mathjax>180.7 113.1 + (- 163.2) + 180.7 = 311.3 kJ</p></div>
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</article> | How do you use Hess's Law to calculate enthalpy for this reaction? |
Given Data:
#"N"_2(g) + "O"_2(g) -> 2"NO"(g)#
#DeltaH = +"180.7 kJ"#
#2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)#
#DeltaH = -"113.1 kJ"#
#2"N"_2"O"(g) -> 2"N"_2(g) + "O"_2(g)#
#DeltaH = -"163.2 kJ"#
Use Hess' Law to calculate #DeltaH# for the reaction:
#"N"_2"O"(g) + "NO"_2(g) -> 3"NO"(g)#
|
1,480 | ac2771f5-6ddd-11ea-9dff-ccda262736ce | https://socratic.org/questions/a-5-00-liter-balloon-of-gas-at-25-c-is-cooled-to-0-c-what-is-the-new-volume-lite | 4.5 liters | start physical_unit 21 22 volume l qc_end physical_unit 21 22 1 2 volume qc_end physical_unit 21 22 7 8 temperature qc_end physical_unit 21 22 12 13 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] liters"}] | [{"type":"physical unit","value":"4.5 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{5.00 liters}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the balloon [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 5.00 liter balloon of gas at 25 C is cooled to 0 C. What is the new volume (liters) of the balloon?</h1> | null | 4.5 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T#</mathjax> is specified to be <mathjax>#"Absolute Temperature"#</mathjax>:</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#= (5.00*L)/(298*K)xx273*K~=4.5*L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use <mathjax>#"Charles' Law"#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T#</mathjax> is specified to be <mathjax>#"Absolute Temperature"#</mathjax>:</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#= (5.00*L)/(298*K)xx273*K~=4.5*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A 5.00 liter balloon of gas at 25 C is cooled to 0 C. What is the new volume (liters) of the balloon?</h1>
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anor277
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Dec 13, 2016
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<div class="markdown"><p>We use <mathjax>#"Charles' Law"#</mathjax>, <mathjax>#V_1/T_1=V_2/T_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#T#</mathjax> is specified to be <mathjax>#"Absolute Temperature"#</mathjax>:</p>
<p><mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#= (5.00*L)/(298*K)xx273*K~=4.5*L#</mathjax></p></div>
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</article> | A 5.00 liter balloon of gas at 25 C is cooled to 0 C. What is the new volume (liters) of the balloon? | null |
1,481 | acaddbf7-6ddd-11ea-8883-ccda262736ce | https://socratic.org/questions/what-is-the-formula-of-the-ionic-compound-formed-by-the-elements-lithium-and-oxy | Li2O | start chemical_formula qc_end substance 12 12 qc_end substance 14 14 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the ionic compound [IN] default"}] | [{"type":"chemical equation","value":"Li2O"}] | [{"type":"substance name","value":"Lithium"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">What is the formula of the ionic compound formed by the elements lithium and oxygen? </h1> | null | Li2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium, a Group 1 metal, commonly (exclusively?) forms a <mathjax>#Li^+#</mathjax> ion. Oxygen a Group 16 non-metal commonly forms an <mathjax>#O^(2-)#</mathjax> ion. Their binary compound is of course neutral and has a formulation of <mathjax>#Li_2O#</mathjax>. How does <mathjax>#Li_2O#</mathjax> react with water?</p></div>
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<div class="markdown"><p><mathjax>#Li_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium, a Group 1 metal, commonly (exclusively?) forms a <mathjax>#Li^+#</mathjax> ion. Oxygen a Group 16 non-metal commonly forms an <mathjax>#O^(2-)#</mathjax> ion. Their binary compound is of course neutral and has a formulation of <mathjax>#Li_2O#</mathjax>. How does <mathjax>#Li_2O#</mathjax> react with water?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula of the ionic compound formed by the elements lithium and oxygen? </h1>
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<div class="markdown"><p><mathjax>#Li_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Lithium, a Group 1 metal, commonly (exclusively?) forms a <mathjax>#Li^+#</mathjax> ion. Oxygen a Group 16 non-metal commonly forms an <mathjax>#O^(2-)#</mathjax> ion. Their binary compound is of course neutral and has a formulation of <mathjax>#Li_2O#</mathjax>. How does <mathjax>#Li_2O#</mathjax> react with water?</p></div>
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</article> | What is the formula of the ionic compound formed by the elements lithium and oxygen? | null |
1,482 | a94b601f-6ddd-11ea-855f-ccda262736ce | https://socratic.org/questions/how-many-grams-of-calcium-are-in-7-0-moles | 280.56 grams | start physical_unit 4 4 mass g qc_end physical_unit 4 4 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] calcium [IN] grams"}] | [{"type":"physical unit","value":"280.56 grams"}] | [{"type":"physical unit","value":"Mole [OF] calcium [=] \\pu{7.0 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of calcium are in 7.0 moles?</h1> | null | 280.56 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#40.08*g*cancel(mol^-1)xx7.0*cancel(mol)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#280*g#</mathjax>.</p></div>
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<div class="markdown"><p>Well, there are <mathjax>#40.08*g#</mathjax> calcium in <mathjax>#1*mol#</mathjax> of metal.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#40.08*g*cancel(mol^-1)xx7.0*cancel(mol)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#280*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of calcium are in 7.0 moles?</h1>
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<div class="markdown"><p>Well, there are <mathjax>#40.08*g#</mathjax> calcium in <mathjax>#1*mol#</mathjax> of metal.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And <mathjax>#40.08*g*cancel(mol^-1)xx7.0*cancel(mol)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#280*g#</mathjax>.</p></div>
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</article> | How many grams of calcium are in 7.0 moles? | null |
1,483 | a8ddc6b5-6ddd-11ea-b57a-ccda262736ce | https://socratic.org/questions/a-gas-has-a-solubility-in-water-at-0-c-of-3-5-g-l-at-a-pressure-of-0-8-atm-what- | 2.0 atm | start physical_unit 32 34 pressure atm qc_end physical_unit 0 1 8 9 temperature qc_end physical_unit 0 1 11 12 solubility qc_end physical_unit 0 1 17 18 pressure qc_end physical_unit 32 34 29 30 solubility qc_end physical_unit 32 34 8 9 temperature qc_end substance 6 6 qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the same gas [IN] atm"}] | [{"type":"physical unit","value":"2.0 atm"}] | [{"type":"physical unit","value":"Temperature1 [OF] a gas [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Solubility1 [OF] a gas [=] \\pu{3.5 g/L}"},{"type":"physical unit","value":"Pressure1 [OF] a gas [=] \\pu{0.80 atm}"},{"type":"physical unit","value":"Solubility2 [OF] the same gas [=] \\pu{8.9 g/L}"},{"type":"physical unit","value":"Temperature2 [OF] the same gas [=] \\pu{0 ℃}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">A gas has a solubility in water at 0°C of 3.5 g/L at a pressure of 0.80 atm. What pressure is needed to produce an aqueous solution containing 8.9 g/L of the same gas at 0°C?</h1> | null | 2.0 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with <em>pressures</em> and <em>gas solubility</em>, which should automatically tell you that this is an example of a <strong>Henry's Law</strong> problem. </p>
<p>As you know, <strong>Henry's Law</strong> states that the <em>solubility</em> of a gas in a liquid is <strong>directly proportional</strong> to its <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></strong> of the gas above the liquid. </p>
<blockquote>
<p><mathjax>#color(blue)("solubility " prop " partial pressure")#</mathjax></p>
</blockquote>
<p>This means that <em>increasing</em> the pressure of the gas above the liquid will <strong>Increase</strong> its solubility. </p>
<p><img alt="http://ch302.cm.utexas.edu/physEQ/solutions/selector.php?name=henrys-law" src="https://useruploads.socratic.org/21RMycJFTmuFG2Xb6dL7_C12F14.gif"/> </p>
<p>Likewise, <em>decreasing</em> the pressure of the gas above the liquid will <strong>decrease</strong> its solubility. </p>
<p>Mathematically, you can express Henry's Law like this</p>
<blockquote>
<p><mathjax>#color(blue)(S_1/P_1 = S_2/P_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#S_1#</mathjax>, <mathjax>#P_1#</mathjax> - the solubility of the gas at a pressure <mathjax>#P_1#</mathjax><br/>
<mathjax>#S_2#</mathjax>, <mathjax>#P_2#</mathjax> - the solubility of the gas a pressure <mathjax>#P_2#</mathjax></p>
<p>You'll often be dealing with the <strong>molar solubility</strong> of the gas, measured in <em>moles per liter</em>, but for this problem you can use the solubility in <em>grams per liter</em>. </p>
<p>So, the solubility of the gas <strong>increased</strong> from <mathjax>#"3.5 g L"^(-1)#</mathjax> to <mathjax>#"8.9 g L"^(-1)#</mathjax>, which can only mean that the pressure <strong>must have increased</strong> as well. </p>
<p>Plug your values into the above equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = S_2/S_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = (8.9 color(red)(cancel(color(black)("g L"^(-1)))))/(3.5color(red)(cancel(color(black)("g L"^(-1))))) * "0.80 atm" = color(green)("2.0 atm")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.0 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with <em>pressures</em> and <em>gas solubility</em>, which should automatically tell you that this is an example of a <strong>Henry's Law</strong> problem. </p>
<p>As you know, <strong>Henry's Law</strong> states that the <em>solubility</em> of a gas in a liquid is <strong>directly proportional</strong> to its <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></strong> of the gas above the liquid. </p>
<blockquote>
<p><mathjax>#color(blue)("solubility " prop " partial pressure")#</mathjax></p>
</blockquote>
<p>This means that <em>increasing</em> the pressure of the gas above the liquid will <strong>Increase</strong> its solubility. </p>
<p><img alt="http://ch302.cm.utexas.edu/physEQ/solutions/selector.php?name=henrys-law" src="https://useruploads.socratic.org/21RMycJFTmuFG2Xb6dL7_C12F14.gif"/> </p>
<p>Likewise, <em>decreasing</em> the pressure of the gas above the liquid will <strong>decrease</strong> its solubility. </p>
<p>Mathematically, you can express Henry's Law like this</p>
<blockquote>
<p><mathjax>#color(blue)(S_1/P_1 = S_2/P_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#S_1#</mathjax>, <mathjax>#P_1#</mathjax> - the solubility of the gas at a pressure <mathjax>#P_1#</mathjax><br/>
<mathjax>#S_2#</mathjax>, <mathjax>#P_2#</mathjax> - the solubility of the gas a pressure <mathjax>#P_2#</mathjax></p>
<p>You'll often be dealing with the <strong>molar solubility</strong> of the gas, measured in <em>moles per liter</em>, but for this problem you can use the solubility in <em>grams per liter</em>. </p>
<p>So, the solubility of the gas <strong>increased</strong> from <mathjax>#"3.5 g L"^(-1)#</mathjax> to <mathjax>#"8.9 g L"^(-1)#</mathjax>, which can only mean that the pressure <strong>must have increased</strong> as well. </p>
<p>Plug your values into the above equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = S_2/S_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = (8.9 color(red)(cancel(color(black)("g L"^(-1)))))/(3.5color(red)(cancel(color(black)("g L"^(-1))))) * "0.80 atm" = color(green)("2.0 atm")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas has a solubility in water at 0°C of 3.5 g/L at a pressure of 0.80 atm. What pressure is needed to produce an aqueous solution containing 8.9 g/L of the same gas at 0°C?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-02-13T22:10:30" itemprop="dateCreated">
Feb 13, 2016
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<div class="markdown"><p><mathjax>#"2.0 atm"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with <em>pressures</em> and <em>gas solubility</em>, which should automatically tell you that this is an example of a <strong>Henry's Law</strong> problem. </p>
<p>As you know, <strong>Henry's Law</strong> states that the <em>solubility</em> of a gas in a liquid is <strong>directly proportional</strong> to its <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></strong> of the gas above the liquid. </p>
<blockquote>
<p><mathjax>#color(blue)("solubility " prop " partial pressure")#</mathjax></p>
</blockquote>
<p>This means that <em>increasing</em> the pressure of the gas above the liquid will <strong>Increase</strong> its solubility. </p>
<p><img alt="http://ch302.cm.utexas.edu/physEQ/solutions/selector.php?name=henrys-law" src="https://useruploads.socratic.org/21RMycJFTmuFG2Xb6dL7_C12F14.gif"/> </p>
<p>Likewise, <em>decreasing</em> the pressure of the gas above the liquid will <strong>decrease</strong> its solubility. </p>
<p>Mathematically, you can express Henry's Law like this</p>
<blockquote>
<p><mathjax>#color(blue)(S_1/P_1 = S_2/P_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#S_1#</mathjax>, <mathjax>#P_1#</mathjax> - the solubility of the gas at a pressure <mathjax>#P_1#</mathjax><br/>
<mathjax>#S_2#</mathjax>, <mathjax>#P_2#</mathjax> - the solubility of the gas a pressure <mathjax>#P_2#</mathjax></p>
<p>You'll often be dealing with the <strong>molar solubility</strong> of the gas, measured in <em>moles per liter</em>, but for this problem you can use the solubility in <em>grams per liter</em>. </p>
<p>So, the solubility of the gas <strong>increased</strong> from <mathjax>#"3.5 g L"^(-1)#</mathjax> to <mathjax>#"8.9 g L"^(-1)#</mathjax>, which can only mean that the pressure <strong>must have increased</strong> as well. </p>
<p>Plug your values into the above equation and solve for <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_2 = S_2/S_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = (8.9 color(red)(cancel(color(black)("g L"^(-1)))))/(3.5color(red)(cancel(color(black)("g L"^(-1))))) * "0.80 atm" = color(green)("2.0 atm")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | A gas has a solubility in water at 0°C of 3.5 g/L at a pressure of 0.80 atm. What pressure is needed to produce an aqueous solution containing 8.9 g/L of the same gas at 0°C? | null |
1,484 | abe2f73e-6ddd-11ea-ae9b-ccda262736ce | https://socratic.org/questions/58bc7a667c0149140a02230e | 56 L | start physical_unit 6 9 volume l qc_end physical_unit 6 9 4 5 mass qc_end c_other STP qc_end physical_unit 9 9 22 23 molar_volume qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrogen gas sample [IN] L"}] | [{"type":"physical unit","value":"56 L"}] | [{"type":"physical unit","value":"Mass [OF] hydrogen gas sample [=] \\pu{5.0 g}"},{"type":"other","value":"STP"},{"type":"physical unit","value":"Molar volume [OF] the gas [=] \\pu{22.4 L/mol}"}] | <h1 class="questionTitle" itemprop="name">What volume does a #"5.0-g"# sample of hydrogen gas occupy at STP?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The molar volume of a gas at STP <mathjax>#= "22.4 L mol"^(-1)#</mathjax></p></div>
</h2>
</div>
</div> | 56 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP</strong></p>
<blockquote>
<p><mathjax>#V_"n STP" = "22.4 L mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, you know that under STP conditions for pressure and temperature, <mathjax>#1#</mathjax> <strong>mole</strong> of any ideal gas will occupy <mathjax>#"22.4 L"#</mathjax>. </p>
<p>You can use the molar volume of a gas at STP as a <em>conversion factor</em> to help you determine the volume a sample of gas will occupy when kept under STP conditions. </p>
<p>In your case, you need to find <em>volume</em>, so rearrange the molar volume of a gas as </p>
<blockquote>
<p><mathjax>#"22.4 L"/"1 mole H"_2 ->#</mathjax> <em>this takes you from <strong>moles</strong> of</em> <mathjax>#"H"_2#</mathjax> <em>to <strong>liters</strong></em></p>
</blockquote>
<p>Now, in order to find the number of moles present in <mathjax>#"5.0 g"#</mathjax> of hydrogen gas, you can use the gas' <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "2.48 moles H"_2#</mathjax></p>
</blockquote>
<p>This sample of hydrogen gas will thus occupy</p>
<blockquote>
<p><mathjax>#2.48 color(red)(cancel(color(black)("moles H"_2))) * "22.4 L"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(darkgreen)(ul(color(black)("56 L")))#</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 mass of hydrogen gas. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"56 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP</strong></p>
<blockquote>
<p><mathjax>#V_"n STP" = "22.4 L mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, you know that under STP conditions for pressure and temperature, <mathjax>#1#</mathjax> <strong>mole</strong> of any ideal gas will occupy <mathjax>#"22.4 L"#</mathjax>. </p>
<p>You can use the molar volume of a gas at STP as a <em>conversion factor</em> to help you determine the volume a sample of gas will occupy when kept under STP conditions. </p>
<p>In your case, you need to find <em>volume</em>, so rearrange the molar volume of a gas as </p>
<blockquote>
<p><mathjax>#"22.4 L"/"1 mole H"_2 ->#</mathjax> <em>this takes you from <strong>moles</strong> of</em> <mathjax>#"H"_2#</mathjax> <em>to <strong>liters</strong></em></p>
</blockquote>
<p>Now, in order to find the number of moles present in <mathjax>#"5.0 g"#</mathjax> of hydrogen gas, you can use the gas' <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "2.48 moles H"_2#</mathjax></p>
</blockquote>
<p>This sample of hydrogen gas will thus occupy</p>
<blockquote>
<p><mathjax>#2.48 color(red)(cancel(color(black)("moles H"_2))) * "22.4 L"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(darkgreen)(ul(color(black)("56 L")))#</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 mass of hydrogen gas. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume does a #"5.0-g"# sample of hydrogen gas occupy at STP?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The molar volume of a gas at STP <mathjax>#= "22.4 L mol"^(-1)#</mathjax></p></div>
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Stefan V.
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Mar 5, 2017
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<div class="markdown"><p><mathjax>#"56 L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP</strong></p>
<blockquote>
<p><mathjax>#V_"n STP" = "22.4 L mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, you know that under STP conditions for pressure and temperature, <mathjax>#1#</mathjax> <strong>mole</strong> of any ideal gas will occupy <mathjax>#"22.4 L"#</mathjax>. </p>
<p>You can use the molar volume of a gas at STP as a <em>conversion factor</em> to help you determine the volume a sample of gas will occupy when kept under STP conditions. </p>
<p>In your case, you need to find <em>volume</em>, so rearrange the molar volume of a gas as </p>
<blockquote>
<p><mathjax>#"22.4 L"/"1 mole H"_2 ->#</mathjax> <em>this takes you from <strong>moles</strong> of</em> <mathjax>#"H"_2#</mathjax> <em>to <strong>liters</strong></em></p>
</blockquote>
<p>Now, in order to find the number of moles present in <mathjax>#"5.0 g"#</mathjax> of hydrogen gas, you can use the gas' <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "2.48 moles H"_2#</mathjax></p>
</blockquote>
<p>This sample of hydrogen gas will thus occupy</p>
<blockquote>
<p><mathjax>#2.48 color(red)(cancel(color(black)("moles H"_2))) * "22.4 L"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(darkgreen)(ul(color(black)("56 L")))#</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 mass of hydrogen gas. </p></div>
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</article> | What volume does a #"5.0-g"# sample of hydrogen gas occupy at STP? |
The molar volume of a gas at STP #= "22.4 L mol"^(-1)#
|
1,485 | ac2639a5-6ddd-11ea-afc6-ccda262736ce | https://socratic.org/questions/how-many-moles-is-5-3-g-of-kno3 | 0.05 moles | start physical_unit 9 9 mole mol qc_end physical_unit 9 9 6 7 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] KNO3 [IN] moles"}] | [{"type":"physical unit","value":"0.05 moles"}] | [{"type":"physical unit","value":"Mass [OF] KNO3 [=] \\pu{5.3 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in #"5.3 g"# of #"KNO"_3# ?</h1> | null | 0.05 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to first find the molar mass of <mathjax>#"KNO"_3#</mathjax>. You do this by adding the individual molar mass of each element, these can be found on a periodic table or online.</p>
<p><mathjax>#M_K = "39.0983 g/mol"#</mathjax><br/>
<mathjax>#M_N= "14.0067 g/mol"#</mathjax><br/>
<mathjax>#M_O = "15.999 g/mol"#</mathjax></p>
<p>The molar mass is the sum of all of the elements times the number of moles of each element.</p>
<p><mathjax>#M_ ("KNO"_3)= M_K + M_N + 3M_O#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3)= ("39.0983 g/mol") + ("14.0067 g/mol") + 3("15.999 g/mol")#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3) = "101.102 g/mol"#</mathjax></p>
<p>Use</p>
<p><mathjax>#n=m/M#</mathjax></p>
<p>where:</p>
<blockquote>
<ul>
<li><mathjax>#n=#</mathjax>number of moles</li>
<li><mathjax>#m=#</mathjax>mass</li>
<li><mathjax>#M=#</mathjax>molar mass</li>
</ul>
</blockquote>
<p>So</p>
<p><mathjax>#n="5.3g"/"101.102 g/mol"#</mathjax></p>
<p><mathjax># n = "0.052 moles"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>It is <mathjax>#0.052#</mathjax> moles. You need to use the molar masses to convert from grams to moles. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to first find the molar mass of <mathjax>#"KNO"_3#</mathjax>. You do this by adding the individual molar mass of each element, these can be found on a periodic table or online.</p>
<p><mathjax>#M_K = "39.0983 g/mol"#</mathjax><br/>
<mathjax>#M_N= "14.0067 g/mol"#</mathjax><br/>
<mathjax>#M_O = "15.999 g/mol"#</mathjax></p>
<p>The molar mass is the sum of all of the elements times the number of moles of each element.</p>
<p><mathjax>#M_ ("KNO"_3)= M_K + M_N + 3M_O#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3)= ("39.0983 g/mol") + ("14.0067 g/mol") + 3("15.999 g/mol")#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3) = "101.102 g/mol"#</mathjax></p>
<p>Use</p>
<p><mathjax>#n=m/M#</mathjax></p>
<p>where:</p>
<blockquote>
<ul>
<li><mathjax>#n=#</mathjax>number of moles</li>
<li><mathjax>#m=#</mathjax>mass</li>
<li><mathjax>#M=#</mathjax>molar mass</li>
</ul>
</blockquote>
<p>So</p>
<p><mathjax>#n="5.3g"/"101.102 g/mol"#</mathjax></p>
<p><mathjax># n = "0.052 moles"#</mathjax></p></div>
</div>
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<div class="markdown"><p>It is <mathjax>#0.052#</mathjax> moles. You need to use the molar masses to convert from grams to moles. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to first find the molar mass of <mathjax>#"KNO"_3#</mathjax>. You do this by adding the individual molar mass of each element, these can be found on a periodic table or online.</p>
<p><mathjax>#M_K = "39.0983 g/mol"#</mathjax><br/>
<mathjax>#M_N= "14.0067 g/mol"#</mathjax><br/>
<mathjax>#M_O = "15.999 g/mol"#</mathjax></p>
<p>The molar mass is the sum of all of the elements times the number of moles of each element.</p>
<p><mathjax>#M_ ("KNO"_3)= M_K + M_N + 3M_O#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3)= ("39.0983 g/mol") + ("14.0067 g/mol") + 3("15.999 g/mol")#</mathjax></p>
<p><mathjax>#M_ ("KNO"_3) = "101.102 g/mol"#</mathjax></p>
<p>Use</p>
<p><mathjax>#n=m/M#</mathjax></p>
<p>where:</p>
<blockquote>
<ul>
<li><mathjax>#n=#</mathjax>number of moles</li>
<li><mathjax>#m=#</mathjax>mass</li>
<li><mathjax>#M=#</mathjax>molar mass</li>
</ul>
</blockquote>
<p>So</p>
<p><mathjax>#n="5.3g"/"101.102 g/mol"#</mathjax></p>
<p><mathjax># n = "0.052 moles"#</mathjax></p></div>
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</article> | How many moles are there in #"5.3 g"# of #"KNO"_3# ? | null |
1,486 | aa0d9a0c-6ddd-11ea-9d4b-ccda262736ce | https://socratic.org/questions/54e55277581e2a76545ebe11 | 2039 mL | start physical_unit 19 20 volume ml qc_end physical_unit 19 20 6 7 volume qc_end physical_unit 19 20 9 10 temperature qc_end physical_unit 19 20 12 13 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"2039 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{720.0 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{3.00 atm}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of #"720.0 mL"# at #"20.0"^@"C"# and #"3.00 atm"#. What would the volume of the gas be at STP? </h1> | null | 2039 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>STP means standard temperature and pressure. The current values for STP are <mathjax>#0^("o")"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"10"^5#</mathjax> <mathjax>#"Pascals"#</mathjax> <mathjax>#("Pa")#</mathjax>, usually given as <mathjax>#"100 kPa"#</mathjax> to make it easier to work with. For the gas laws, the Celsius temperature must be converted to Kelvins by adding <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p>This question can be answered using the equation for the combined gas law:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressures, <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes, and <mathjax>#T_1#</mathjax> and <mathjax>#T_2#</mathjax> are the initial and final temperatures in Kelvins.</p>
<p><strong>Known/Given:</strong></p>
<p><mathjax>#P_1 = 3.00 color(red)cancel(color(black)("atm"))xx(101.325"kPa")/(1color(red)cancel(color(black)("atm")))="303.975 kPa"#</mathjax></p>
<p><mathjax>#V_1 = 720.0 "mL"#</mathjax></p>
<p><mathjax>#T_1 = 20.0^("o")"C" + 273.15 = 293.2"K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><mathjax>#T_2 = 273.15 "K"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#V_2#</mathjax></p>
<p><strong>Solution:</strong> Rearrange the combined gas law so that <mathjax>#V_2#</mathjax> is isolated, then solve for <mathjax>#V_2#</mathjax></p>
<p><mathjax>#V_2 =(P_1V_1T_2)/(T_1P_2) = (303.975color(red)cancel(color(black)("kPa"))*720.0 "mL"*273.15 color(red)cancel(color(black)("K")))/(293.2 color(red)cancel(color(black)("K"))*100color(red)cancel(color(black)("kPa"))) = "2039 mL"#</mathjax> (rounded to four significant figures)</p>
<p><strong>Note:</strong></p>
<p>If your teacher is still using <mathjax>#"1 atm"#</mathjax> for standard pressure, substitute <mathjax>#"3.00 atm"#</mathjax> for <mathjax>#"303.975 kPa"#</mathjax>, and <mathjax>#"1 atm"#</mathjax> for <mathjax>#"100 kPa"#</mathjax>. The final volume will be <mathjax>#"2010 mL"#</mathjax> rounded to three significant figures. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final volume will be <mathjax>#"2039 mL"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>STP means standard temperature and pressure. The current values for STP are <mathjax>#0^("o")"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"10"^5#</mathjax> <mathjax>#"Pascals"#</mathjax> <mathjax>#("Pa")#</mathjax>, usually given as <mathjax>#"100 kPa"#</mathjax> to make it easier to work with. For the gas laws, the Celsius temperature must be converted to Kelvins by adding <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p>This question can be answered using the equation for the combined gas law:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressures, <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes, and <mathjax>#T_1#</mathjax> and <mathjax>#T_2#</mathjax> are the initial and final temperatures in Kelvins.</p>
<p><strong>Known/Given:</strong></p>
<p><mathjax>#P_1 = 3.00 color(red)cancel(color(black)("atm"))xx(101.325"kPa")/(1color(red)cancel(color(black)("atm")))="303.975 kPa"#</mathjax></p>
<p><mathjax>#V_1 = 720.0 "mL"#</mathjax></p>
<p><mathjax>#T_1 = 20.0^("o")"C" + 273.15 = 293.2"K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><mathjax>#T_2 = 273.15 "K"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#V_2#</mathjax></p>
<p><strong>Solution:</strong> Rearrange the combined gas law so that <mathjax>#V_2#</mathjax> is isolated, then solve for <mathjax>#V_2#</mathjax></p>
<p><mathjax>#V_2 =(P_1V_1T_2)/(T_1P_2) = (303.975color(red)cancel(color(black)("kPa"))*720.0 "mL"*273.15 color(red)cancel(color(black)("K")))/(293.2 color(red)cancel(color(black)("K"))*100color(red)cancel(color(black)("kPa"))) = "2039 mL"#</mathjax> (rounded to four significant figures)</p>
<p><strong>Note:</strong></p>
<p>If your teacher is still using <mathjax>#"1 atm"#</mathjax> for standard pressure, substitute <mathjax>#"3.00 atm"#</mathjax> for <mathjax>#"303.975 kPa"#</mathjax>, and <mathjax>#"1 atm"#</mathjax> for <mathjax>#"100 kPa"#</mathjax>. The final volume will be <mathjax>#"2010 mL"#</mathjax> rounded to three significant figures. </p></div>
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<span class="dateCreated" datetime="2015-02-19T04:38:18" itemprop="dateCreated">
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<div class="markdown"><p>The final volume will be <mathjax>#"2039 mL"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>STP means standard temperature and pressure. The current values for STP are <mathjax>#0^("o")"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"10"^5#</mathjax> <mathjax>#"Pascals"#</mathjax> <mathjax>#("Pa")#</mathjax>, usually given as <mathjax>#"100 kPa"#</mathjax> to make it easier to work with. For the gas laws, the Celsius temperature must be converted to Kelvins by adding <mathjax>#273.15#</mathjax> to the Celsius temperature.</p>
<p>This question can be answered using the equation for the combined gas law:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P_1#</mathjax> and <mathjax>#P_2#</mathjax> are the initial and final pressures, <mathjax>#V_1#</mathjax> and <mathjax>#V_2#</mathjax> are the initial and final volumes, and <mathjax>#T_1#</mathjax> and <mathjax>#T_2#</mathjax> are the initial and final temperatures in Kelvins.</p>
<p><strong>Known/Given:</strong></p>
<p><mathjax>#P_1 = 3.00 color(red)cancel(color(black)("atm"))xx(101.325"kPa")/(1color(red)cancel(color(black)("atm")))="303.975 kPa"#</mathjax></p>
<p><mathjax>#V_1 = 720.0 "mL"#</mathjax></p>
<p><mathjax>#T_1 = 20.0^("o")"C" + 273.15 = 293.2"K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><mathjax>#T_2 = 273.15 "K"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#V_2#</mathjax></p>
<p><strong>Solution:</strong> Rearrange the combined gas law so that <mathjax>#V_2#</mathjax> is isolated, then solve for <mathjax>#V_2#</mathjax></p>
<p><mathjax>#V_2 =(P_1V_1T_2)/(T_1P_2) = (303.975color(red)cancel(color(black)("kPa"))*720.0 "mL"*273.15 color(red)cancel(color(black)("K")))/(293.2 color(red)cancel(color(black)("K"))*100color(red)cancel(color(black)("kPa"))) = "2039 mL"#</mathjax> (rounded to four significant figures)</p>
<p><strong>Note:</strong></p>
<p>If your teacher is still using <mathjax>#"1 atm"#</mathjax> for standard pressure, substitute <mathjax>#"3.00 atm"#</mathjax> for <mathjax>#"303.975 kPa"#</mathjax>, and <mathjax>#"1 atm"#</mathjax> for <mathjax>#"100 kPa"#</mathjax>. The final volume will be <mathjax>#"2010 mL"#</mathjax> rounded to three significant figures. </p></div>
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</article> | A gas has a volume of #"720.0 mL"# at #"20.0"^@"C"# and #"3.00 atm"#. What would the volume of the gas be at STP? | null |
1,487 | aade5c9a-6ddd-11ea-a869-ccda262736ce | https://socratic.org/questions/hydrogen-sulfide-reacts-with-sulfur-dioxide-to-give-h2o-and-s-h2s-so2-h2o-s-soli | 92.85 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":"92.85 Celsius"}] | [{"type":"chemical equation","value":"H2S + SO2 -> H2O + S(solid)"},{"type":"physical unit","value":"Volume [OF] H2S gas [=] \\pu{3.0 L}"},{"type":"physical unit","value":"Pressure [OF] H2S gas [=] \\pu{760 torr}"},{"type":"physical unit","value":"Mass [OF] sulfur [=] \\pu{4.8 g}"}] | <h1 class="questionTitle" itemprop="name">Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,
H2S + SO2 = H2O + S(solid), unbalanced.
If 3.0 L of H2S gas at 760 torr produced 4.8 g of sulfur, what is the the temperature in Celsius?</h1> | null | 92.85 Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You must convert</p>
<p><mathjax>#"grams of S" → "moles of S" → "moles of H"_2"S"#</mathjax></p>
<p>and then use the <strong>Ideal Gas Law</strong> to find the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>1.</strong> Write the balanced equation.</p>
<p>The balanced equation is</p>
<p><mathjax>#"2H"_2"S" + "SO"_2 → "2H"_2"O" + "3S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"H"_2"S"#</mathjax>.</p>
<p><mathjax>#"moles of H"_2"S" = 4.8 color(red)(cancel(color(black)("g S"))) × (1color(red)(cancel(color(black)("mol S"))))/(32.06color(red)(cancel(color(black)("g S")))) × ("2 mol H"_2"S")/(3color(red)(cancel(color(black)("mol S")))) = "0.0998 mol H"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Use the Ideal Gas Law to calculate the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = 760 color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760color(red)(cancel(color(black)("Torr")))) =" 1.00 atm"#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (1.00 color(red)(cancel(color(black)("atm"))) × 3.0color(red)(cancel(color(black)("L"))))/(0.0998color(red)(cancel(color(black)("mol"))) × "0.082 06"color(red)(cancel(color(black)("L·atm")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) = "366 K"#</mathjax></p>
<p><mathjax>#T = "(366 – 273.15) K" = "90 °C"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because the Kelvin temperature has only 2 significant digits. If you need more precision, you will have to recalculate.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The temperature of the gas is 90 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You must convert</p>
<p><mathjax>#"grams of S" → "moles of S" → "moles of H"_2"S"#</mathjax></p>
<p>and then use the <strong>Ideal Gas Law</strong> to find the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>1.</strong> Write the balanced equation.</p>
<p>The balanced equation is</p>
<p><mathjax>#"2H"_2"S" + "SO"_2 → "2H"_2"O" + "3S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"H"_2"S"#</mathjax>.</p>
<p><mathjax>#"moles of H"_2"S" = 4.8 color(red)(cancel(color(black)("g S"))) × (1color(red)(cancel(color(black)("mol S"))))/(32.06color(red)(cancel(color(black)("g S")))) × ("2 mol H"_2"S")/(3color(red)(cancel(color(black)("mol S")))) = "0.0998 mol H"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Use the Ideal Gas Law to calculate the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = 760 color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760color(red)(cancel(color(black)("Torr")))) =" 1.00 atm"#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (1.00 color(red)(cancel(color(black)("atm"))) × 3.0color(red)(cancel(color(black)("L"))))/(0.0998color(red)(cancel(color(black)("mol"))) × "0.082 06"color(red)(cancel(color(black)("L·atm")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) = "366 K"#</mathjax></p>
<p><mathjax>#T = "(366 – 273.15) K" = "90 °C"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because the Kelvin temperature has only 2 significant digits. If you need more precision, you will have to recalculate.</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 3.0 L of H2S gas at 760 torr produced 4.8 g of sulfur, what is the the temperature in Celsius?</h1>
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<div class="markdown"><p>The temperature of the gas is 90 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You must convert</p>
<p><mathjax>#"grams of S" → "moles of S" → "moles of H"_2"S"#</mathjax></p>
<p>and then use the <strong>Ideal Gas Law</strong> to find the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>1.</strong> Write the balanced equation.</p>
<p>The balanced equation is</p>
<p><mathjax>#"2H"_2"S" + "SO"_2 → "2H"_2"O" + "3S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2.</strong> Calculate the moles of <mathjax>#"H"_2"S"#</mathjax>.</p>
<p><mathjax>#"moles of H"_2"S" = 4.8 color(red)(cancel(color(black)("g S"))) × (1color(red)(cancel(color(black)("mol S"))))/(32.06color(red)(cancel(color(black)("g S")))) × ("2 mol H"_2"S")/(3color(red)(cancel(color(black)("mol S")))) = "0.0998 mol H"_2"S"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3.</strong> Use the Ideal Gas Law to calculate the temperature of the <mathjax>#"H"_2"S"#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = 760 color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760color(red)(cancel(color(black)("Torr")))) =" 1.00 atm"#</mathjax></p>
<p><mathjax>#T = (PV)/(nR) = (1.00 color(red)(cancel(color(black)("atm"))) × 3.0color(red)(cancel(color(black)("L"))))/(0.0998color(red)(cancel(color(black)("mol"))) × "0.082 06"color(red)(cancel(color(black)("L·atm")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) = "366 K"#</mathjax></p>
<p><mathjax>#T = "(366 – 273.15) K" = "90 °C"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because the Kelvin temperature has only 2 significant digits. If you need more precision, you will have to recalculate.</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 3.0 L of H2S gas at 760 torr produced 4.8 g of sulfur, what is the the temperature in Celsius? | null |
1,488 | a8d49c83-6ddd-11ea-b447-ccda262736ce | https://socratic.org/questions/how-much-nacl-can-be-dissolved-in-750-g-of-water-at-25-c | 270 g | start physical_unit 2 2 mass g qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 12 13 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] NaCl [IN] g"}] | [{"type":"physical unit","value":"270 g"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{750 g}"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much NaCl can be dissolved in 750 g of water at 25°C? </h1> | null | 270 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solubility of sodium chloride in water at 25 degrees C is 360 g per litre of water. </p>
<p>750 g of water is 750 ml, or 3/4 of a litre, so you should be able to dissolve around 270 g of NaCl at this temperature.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>About 270 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solubility of sodium chloride in water at 25 degrees C is 360 g per litre of water. </p>
<p>750 g of water is 750 ml, or 3/4 of a litre, so you should be able to dissolve around 270 g of NaCl at this temperature.</p></div>
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<h1 class="questionTitle" itemprop="name">How much NaCl can be dissolved in 750 g of water at 25°C? </h1>
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<div class="markdown"><p>About 270 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solubility of sodium chloride in water at 25 degrees C is 360 g per litre of water. </p>
<p>750 g of water is 750 ml, or 3/4 of a litre, so you should be able to dissolve around 270 g of NaCl at this temperature.</p></div>
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</article> | How much NaCl can be dissolved in 750 g of water at 25°C? | null |
1,489 | ac8ec10c-6ddd-11ea-a930-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-occupies-a-volume-of-272-ml-at-a-pressure-of-999-atm-and-a-temp- | 110.97 g/mol | start physical_unit 34 35 molecular_weight g/mol qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 14 15 pressure qc_end physical_unit 1 3 20 21 temperature qc_end physical_unit 1 3 26 27 mass qc_end end | [{"type":"physical unit","value":"Molecular weight [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"110.97 g/mol"}] | [{"type":"physical unit","value":"Volume [OF] gas sample [=] \\pu{272 mL}"},{"type":"physical unit","value":"Pressure [OF] gas sample [=] \\pu{0.999 atm}"},{"type":"physical unit","value":"Temperature [OF] gas sample [=] \\pu{357 K}"},{"type":"physical unit","value":"WeighT [OF] gas sample [=] \\pu{1.03 g}"}] | <h1 class="questionTitle" itemprop="name">A sample of gas occupies a volume of 272 mL at a pressure of .999 atm and a temp of 357K. If the sample weighs 1.03 g, what is the molecular weight of the gas?</h1> | null | 110.97 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that a gas occupies a volume of <mathjax>#"227 mL"#</mathjax> at those specific conditions for pressure and temperature. Moreover, you know that this sample has a mass of <mathjax>#"1.03 g"#</mathjax>. </p>
<p>Now, the <strong>molar mass</strong> of a substance tells you exactly what the mass of <em>one mole</em> of that substance is. This means that in order to be able to find the molar mass of the gas, you need to know exactly how many moles of gas you have in that sample. </p>
<p>To do that, use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#PV = nRT" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax> - keep in mind that the volume must be expressed in <em>liters</em>!</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (0.999color(red)(cancel(color(black)("atm"))) * 272 * 10^(-3)color(red)(cancel(color(black)("mL"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 357color(red)(cancel(color(black)("K")))) = "0.009282 moles"#</mathjax></p>
</blockquote>
<p>This means that the molar mass of the gas will be </p>
<blockquote>
<p><mathjax>#M_"M" = m/n#</mathjax></p>
<p><mathjax>#M_"M" = "1.03 g"/"0.009282 moles" = "110.97 g/mol"#</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>#M_"M" = color(green)("111 g/mol")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"111 g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that a gas occupies a volume of <mathjax>#"227 mL"#</mathjax> at those specific conditions for pressure and temperature. Moreover, you know that this sample has a mass of <mathjax>#"1.03 g"#</mathjax>. </p>
<p>Now, the <strong>molar mass</strong> of a substance tells you exactly what the mass of <em>one mole</em> of that substance is. This means that in order to be able to find the molar mass of the gas, you need to know exactly how many moles of gas you have in that sample. </p>
<p>To do that, use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#PV = nRT" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax> - keep in mind that the volume must be expressed in <em>liters</em>!</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (0.999color(red)(cancel(color(black)("atm"))) * 272 * 10^(-3)color(red)(cancel(color(black)("mL"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 357color(red)(cancel(color(black)("K")))) = "0.009282 moles"#</mathjax></p>
</blockquote>
<p>This means that the molar mass of the gas will be </p>
<blockquote>
<p><mathjax>#M_"M" = m/n#</mathjax></p>
<p><mathjax>#M_"M" = "1.03 g"/"0.009282 moles" = "110.97 g/mol"#</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>#M_"M" = color(green)("111 g/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of gas occupies a volume of 272 mL at a pressure of .999 atm and a temp of 357K. If the sample weighs 1.03 g, what is the molecular weight of the gas?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"111 g/mol"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that a gas occupies a volume of <mathjax>#"227 mL"#</mathjax> at those specific conditions for pressure and temperature. Moreover, you know that this sample has a mass of <mathjax>#"1.03 g"#</mathjax>. </p>
<p>Now, the <strong>molar mass</strong> of a substance tells you exactly what the mass of <em>one mole</em> of that substance is. This means that in order to be able to find the molar mass of the gas, you need to know exactly how many moles of gas you have in that sample. </p>
<p>To do that, use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</p>
<blockquote>
<p><mathjax>#PV = nRT" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.082("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the temperature of the gas, expressed in Kelvin</p>
<p>Plug in your values and solve for <mathjax>#n#</mathjax> - keep in mind that the volume must be expressed in <em>liters</em>!</p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
<p><mathjax>#n = (0.999color(red)(cancel(color(black)("atm"))) * 272 * 10^(-3)color(red)(cancel(color(black)("mL"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 357color(red)(cancel(color(black)("K")))) = "0.009282 moles"#</mathjax></p>
</blockquote>
<p>This means that the molar mass of the gas will be </p>
<blockquote>
<p><mathjax>#M_"M" = m/n#</mathjax></p>
<p><mathjax>#M_"M" = "1.03 g"/"0.009282 moles" = "110.97 g/mol"#</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>#M_"M" = color(green)("111 g/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div>
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</article> | A sample of gas occupies a volume of 272 mL at a pressure of .999 atm and a temp of 357K. If the sample weighs 1.03 g, what is the molecular weight of the gas? | null |
1,490 | a957e806-6ddd-11ea-a2c8-ccda262736ce | https://socratic.org/questions/how-many-joules-of-heat-energy-would-be-required-to-raise-the-temperature-of-150 | 17145.00 joules | start physical_unit 17 17 heat_energy j qc_end physical_unit 17 17 14 15 mass qc_end physical_unit 17 17 19 20 temperature qc_end physical_unit 17 17 22 23 temperature qc_end physical_unit 17 17 31 34 specific_heat_capacity qc_end end | [{"type":"physical unit","value":"Required heat energy [OF] aluminum [IN] joules"}] | [{"type":"physical unit","value":"17145.00 joules"}] | [{"type":"physical unit","value":"Mass [OF] aluminum [=] \\pu{150.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] aluminum [=] \\pu{23 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] aluminum [=] \\pu{150 ℃}"},{"type":"physical unit","value":"Specific heat capacity [OF] aluminum [=] \\pu{0.90 J/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">How many joules of heat energy would be required to raise the temperature of 150.0 g of aluminum from 23°C to 150°C? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat capacity of aluminum is .90 <mathjax>#J#</mathjax><mathjax>#/g*°C#</mathjax>. </p></div>
</h2>
</div>
</div> | 17145.00 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the following equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained / lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>The problem provides you with all the information that you need in order to determine the amount of heat, <mathjax>#q#</mathjax>, needed. </p>
<p>In your case, the change in temperature will be </p>
<blockquote>
<p><mathjax>#DeltaT = 150^@"C" - 23^@"C" = 127^@"C"#</mathjax></p>
</blockquote>
<p>All you have to do now is plug in your values into the above equation and solve for <mathjax>#q#</mathjax></p>
<blockquote>
<p><mathjax>#q = 150.0 color(red)(cancel(color(black)("g"))) * 0.90"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 127color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "17,145 J"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("17,000 J")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"17,000 J"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the following equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained / lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>The problem provides you with all the information that you need in order to determine the amount of heat, <mathjax>#q#</mathjax>, needed. </p>
<p>In your case, the change in temperature will be </p>
<blockquote>
<p><mathjax>#DeltaT = 150^@"C" - 23^@"C" = 127^@"C"#</mathjax></p>
</blockquote>
<p>All you have to do now is plug in your values into the above equation and solve for <mathjax>#q#</mathjax></p>
<blockquote>
<p><mathjax>#q = 150.0 color(red)(cancel(color(black)("g"))) * 0.90"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 127color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "17,145 J"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("17,000 J")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many joules of heat energy would be required to raise the temperature of 150.0 g of aluminum from 23°C to 150°C? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat capacity of aluminum is .90 <mathjax>#J#</mathjax><mathjax>#/g*°C#</mathjax>. </p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-05-29T22:37:54" itemprop="dateCreated">
May 29, 2016
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<div class="markdown"><p><mathjax>#"17,000 J"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the following equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained / lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>The problem provides you with all the information that you need in order to determine the amount of heat, <mathjax>#q#</mathjax>, needed. </p>
<p>In your case, the change in temperature will be </p>
<blockquote>
<p><mathjax>#DeltaT = 150^@"C" - 23^@"C" = 127^@"C"#</mathjax></p>
</blockquote>
<p>All you have to do now is plug in your values into the above equation and solve for <mathjax>#q#</mathjax></p>
<blockquote>
<p><mathjax>#q = 150.0 color(red)(cancel(color(black)("g"))) * 0.90"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 127color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "17,145 J"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#"amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("17,000 J")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | How many joules of heat energy would be required to raise the temperature of 150.0 g of aluminum from 23°C to 150°C? |
The specific heat capacity of aluminum is .90 #J##/g*°C#.
|
1,491 | a9149300-6ddd-11ea-bb72-ccda262736ce | https://socratic.org/questions/calculate-the-enthalpy-of-formation-of-acetic-acid-if-its-anthalpy-of-combustion | -491.8 kJ/mol | start physical_unit 6 7 enthalpy_of_formation kj/mol qc_end physical_unit 14 14 31 32 enthalpy_of_formation qc_end physical_unit 16 16 34 35 enthalpy_of_formation qc_end end | [{"type":"physical unit","value":"Enthalpy of formation [OF] acetic acid [IN] kJ/mol"}] | [{"type":"physical unit","value":"-491.8 kJ/mol"}] | [{"type":"physical unit","value":"Enthalpy of combustion [OF] acetic acid to CO2 and H2O [=] \\pu{-867 kJ/mol}"},{"type":"physical unit","value":"Enthalpies of formation [OF] CO2 [=] \\pu{-393.5 kJ/mol}"},{"type":"physical unit","value":"Enthalpies of formation [OF] H2O [=] \\pu{-285.9 kJ/mol}"}] | <h1 class="questionTitle" itemprop="name">Calculate the enthalpy of formation of acetic acid if its anthalpy of combustion to CO2 and H2O is - 867 kJ/mol and the enthalpies of formation of CO2 and H2O are respectively - 393.5 and -285.9 kJ / mol ? </h1> | null | -491.8 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Combustion reaction for above question is </p>
<p><mathjax>#C_2H_4O_2 + 2O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#DeltaH_f(combustion) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] - DeltaH_(C_2H_4O_2)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] -DeltaH_(combustion)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 867+(2times-679.4) = -491.8(kJ)/(mol)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>-491.8kJ/mol</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Combustion reaction for above question is </p>
<p><mathjax>#C_2H_4O_2 + 2O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#DeltaH_f(combustion) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] - DeltaH_(C_2H_4O_2)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] -DeltaH_(combustion)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 867+(2times-679.4) = -491.8(kJ)/(mol)#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Calculate the enthalpy of formation of acetic acid if its anthalpy of combustion to CO2 and H2O is - 867 kJ/mol and the enthalpies of formation of CO2 and H2O are respectively - 393.5 and -285.9 kJ / mol ? </h1>
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<div class="markdown"><p>-491.8kJ/mol</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Combustion reaction for above question is </p>
<p><mathjax>#C_2H_4O_2 + 2O_2 -> 2CO_2 + 2H_2O#</mathjax></p>
<p><mathjax>#DeltaH_f(combustion) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] - DeltaH_(C_2H_4O_2)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 2[DeltaH_(CO_2) + DeltaH_(H_2O)] -DeltaH_(combustion)#</mathjax></p>
<p><mathjax>#DeltaH_(C_2H_4O_2) = 867+(2times-679.4) = -491.8(kJ)/(mol)#</mathjax></p></div>
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</article> | Calculate the enthalpy of formation of acetic acid if its anthalpy of combustion to CO2 and H2O is - 867 kJ/mol and the enthalpies of formation of CO2 and H2O are respectively - 393.5 and -285.9 kJ / mol ? | null |
1,492 | a8db04db-6ddd-11ea-90fd-ccda262736ce | https://socratic.org/questions/580dff9fb72cff64785a11b8 | +3 | start physical_unit 6 6 oxidation_state none qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] carbon"}] | [{"type":"physical unit","value":"+3"}] | [{"type":"chemical equation","value":"C2O3"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of carbon in cyclic #C_2O_3#?</h1> | null | +3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As an hypothetical entity, each carbon atom has an oxidation state of <mathjax>#+III#</mathjax>.</p>
<p>Here, I assumed (reasonably) that the oxidation state of oxygen was <mathjax>#-II#</mathjax>.</p>
<p>Learned later that this is the cyclic anhydride of oxalic acid, i.e. <mathjax>#HO_2C-CO_2H#</mathjax>. It has a fleeting existence, and is a gleam in a spectroscopist's eye.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>As far as I know, this molecule is unknown.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As an hypothetical entity, each carbon atom has an oxidation state of <mathjax>#+III#</mathjax>.</p>
<p>Here, I assumed (reasonably) that the oxidation state of oxygen was <mathjax>#-II#</mathjax>.</p>
<p>Learned later that this is the cyclic anhydride of oxalic acid, i.e. <mathjax>#HO_2C-CO_2H#</mathjax>. It has a fleeting existence, and is a gleam in a spectroscopist's eye.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of carbon in cyclic #C_2O_3#?</h1>
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anor277
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Oct 29, 2016
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<div class="markdown"><p>As far as I know, this molecule is unknown.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As an hypothetical entity, each carbon atom has an oxidation state of <mathjax>#+III#</mathjax>.</p>
<p>Here, I assumed (reasonably) that the oxidation state of oxygen was <mathjax>#-II#</mathjax>.</p>
<p>Learned later that this is the cyclic anhydride of oxalic acid, i.e. <mathjax>#HO_2C-CO_2H#</mathjax>. It has a fleeting existence, and is a gleam in a spectroscopist's eye.</p></div>
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</article> | What is the oxidation state of carbon in cyclic #C_2O_3#? | null |
1,493 | a8fe0007-6ddd-11ea-bc3a-ccda262736ce | https://socratic.org/questions/55c6f88f581e2a77013a3012 | 2 KMnO4 + 5 H2O2 + 3 H2SO4 -> 2 MnSO4 + K2SO4 + 5 O2 + 8 H2O | start chemical_equation qc_end chemical_equation 7 19 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the redox reaction"}] | [{"type":"chemical equation","value":"2 KMnO4 + 5 H2O2 + 3 H2SO4 -> 2 MnSO4 + K2SO4 + 5 O2 + 8 H2O"}] | [{"type":"chemical equation","value":"KMnO4 + H2O2 + H2SO4 -> MnSO4 + K2SO4 + O2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance the redox reaction? #"KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O"# ?</h1> | null | 2 KMnO4 + 5 H2O2 + 3 H2SO4 -> 2 MnSO4 + K2SO4 + 5 O2 + 8 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>Before you start balancing this equation, it's important to recognize that you're in <em>acidic medium</em> because the reaction take place in the presence of sulfuric acid. </p>
<p>This means that <em>any protons</em> that will remain unbalanced on the reactants' side will come from the sulfuric acid. </p>
<p>Moreover, not all the species are relevant to the reaction. More precisely, the potassium cation and the sulfate anion are spectator ions. The <em>net ionic equation</em> actually looks like this </p>
<p><mathjax>#MnO_4^(-) + H_2O_2 + 2H^(+) -> Mn^(2+) + O_2 + H_2O#</mathjax></p>
<p>So, start by assigning <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to all the atoms that take part in the reaction</p>
<p><mathjax>#stackrel(color(blue)(+1))(K) stackrel(color(blue)(+7))Mn stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(+2))Mn stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))K_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(0))O_2 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-2))O#</mathjax></p>
<p>Notice that the oxidation number of mangenese changes from <mathjax>#"+7"#</mathjax> on the reactants' side to <mathjax>#"+2"#</mathjax> on the products' side, which means that mangenese is being <strong>reduced</strong>. </p>
<p>On the other hand, the oxidation number of oxygen in <em>hydrogen peroxide</em>, <mathjax>#H_2O_2#</mathjax>, changes from <mathjax>#"-1"#</mathjax> on the reactants' side to <mathjax>#"0"#</mathjax> in oxygen gas, <mathjax>#O_2#</mathjax>, on the products' side.</p>
<p>This means that oxygen is being <strong>oxidized</strong>. The two half-reactions will look like this </p>
<ul>
<li><em>reduction half-reaction</em></li>
</ul>
<p><mathjax>#stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+)#</mathjax></p>
<p>Here, each manganese atom gains <strong>5 electrons</strong>.</p>
<p>Since you're in acidic medium, you can balance the oxygen atoms by adding water to the side that needs oxygen and the hydrogen atoms by adding protons, <mathjax>#H^(+)#</mathjax>, to the side that needs hydrogen. </p>
<p>The reduction half-reaction will thus be </p>
<p><mathjax>#8H^(+) + stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+) + 4H_2O#</mathjax></p>
<ul>
<li><em>oxidation half-reaction</em></li>
</ul>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-)#</mathjax></p>
<p>Here, each oxygen <em>atom</em> loses one electron, so two oxygen atoms will lose <strong>2 electrons</strong>. </p>
<p>Once again, balance the oxygen and hydrogen accordingly</p>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-) + 2H^(+)#</mathjax></p>
<p>Now, in a <a href="http://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a>, the number of electrons <em>gained</em> in the <strong>reduction</strong> half-reaction must be <strong>equal to</strong> the number of electrons <em>lost</em> in the <strong>oxidation</strong> half-reaction. </p>
<p>This means that you need to multiply the reduction half-reaction by <strong>2</strong> and the oxidation half-reaction by <strong>5</strong> to get a total of <mathjax>#10e^(-)#</mathjax> transferred in the reaction. </p>
<p><mathjax>#{(16H^(+) + 2MnO_4^(-) + 10e^(-) -> 2Mn^(2+) + 8H_2O), (5H_2O_2 -> 5O_2 + 10e^(-) + 10H^(+)) :}#</mathjax></p>
<p>Add these two half-reactions to get </p>
<p><mathjax>#16H^(+) + 2MnO_4^(-) + color(red)(cancel(color(black)(10e^(-)))) + 5H_2O_2 -> 2Mn^(2+) + color(red)(cancel(color(black)(10e^(-)))) + 5O_2 + 10H^(+) + 8H_2O#</mathjax></p>
<p>Finally, reduce the number of protons to get </p>
<p><mathjax>#6H^(+) + 2MnO_4^(-) + 5H_2O_2 ->2Mn^(2+) + 5O_2 + 8H_2O#</mathjax></p>
<p>Now, all those protons are coming from the sulfuric acid, which means that the balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> will be - add the spectator ions, too</p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + color(red)(3)H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>This is actually the half-reaction method, not the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> method. The oxidation numbers method does not use water and protons (for reactions that take place in acidic medium) to balance hydrogen and oxygen.</em></p>
<p><em>To balance this equation using the oxidation number method, figure out the number of electrons transferred, which is of course equal to 10, multiply the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> so that these electrons are balanced, then balance the rest of the atoms by inspection</em>.</p>
<p><em>So, you know that you need to multiply the reduction half-reaction by</em> <mathjax>#2#</mathjax> <em>and the oxidation half-reaction by</em> <mathjax>#5#</mathjax>. <em>This means that you have</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Now balance the equation by balancing the number of sulfate anions present on both sides of the equation. Notice that you have one sulfate anion on the reactants' side and three on the products side, so multiply sulfuric acid by</em> <mathjax>#3#</mathjax> <em>to get</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Finally, balance hydrogen and oxygen by multiplying the water by</em> <mathjax>#8#</mathjax></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p>And there you have it. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>Before you start balancing this equation, it's important to recognize that you're in <em>acidic medium</em> because the reaction take place in the presence of sulfuric acid. </p>
<p>This means that <em>any protons</em> that will remain unbalanced on the reactants' side will come from the sulfuric acid. </p>
<p>Moreover, not all the species are relevant to the reaction. More precisely, the potassium cation and the sulfate anion are spectator ions. The <em>net ionic equation</em> actually looks like this </p>
<p><mathjax>#MnO_4^(-) + H_2O_2 + 2H^(+) -> Mn^(2+) + O_2 + H_2O#</mathjax></p>
<p>So, start by assigning <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to all the atoms that take part in the reaction</p>
<p><mathjax>#stackrel(color(blue)(+1))(K) stackrel(color(blue)(+7))Mn stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(+2))Mn stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))K_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(0))O_2 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-2))O#</mathjax></p>
<p>Notice that the oxidation number of mangenese changes from <mathjax>#"+7"#</mathjax> on the reactants' side to <mathjax>#"+2"#</mathjax> on the products' side, which means that mangenese is being <strong>reduced</strong>. </p>
<p>On the other hand, the oxidation number of oxygen in <em>hydrogen peroxide</em>, <mathjax>#H_2O_2#</mathjax>, changes from <mathjax>#"-1"#</mathjax> on the reactants' side to <mathjax>#"0"#</mathjax> in oxygen gas, <mathjax>#O_2#</mathjax>, on the products' side.</p>
<p>This means that oxygen is being <strong>oxidized</strong>. The two half-reactions will look like this </p>
<ul>
<li><em>reduction half-reaction</em></li>
</ul>
<p><mathjax>#stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+)#</mathjax></p>
<p>Here, each manganese atom gains <strong>5 electrons</strong>.</p>
<p>Since you're in acidic medium, you can balance the oxygen atoms by adding water to the side that needs oxygen and the hydrogen atoms by adding protons, <mathjax>#H^(+)#</mathjax>, to the side that needs hydrogen. </p>
<p>The reduction half-reaction will thus be </p>
<p><mathjax>#8H^(+) + stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+) + 4H_2O#</mathjax></p>
<ul>
<li><em>oxidation half-reaction</em></li>
</ul>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-)#</mathjax></p>
<p>Here, each oxygen <em>atom</em> loses one electron, so two oxygen atoms will lose <strong>2 electrons</strong>. </p>
<p>Once again, balance the oxygen and hydrogen accordingly</p>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-) + 2H^(+)#</mathjax></p>
<p>Now, in a <a href="http://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a>, the number of electrons <em>gained</em> in the <strong>reduction</strong> half-reaction must be <strong>equal to</strong> the number of electrons <em>lost</em> in the <strong>oxidation</strong> half-reaction. </p>
<p>This means that you need to multiply the reduction half-reaction by <strong>2</strong> and the oxidation half-reaction by <strong>5</strong> to get a total of <mathjax>#10e^(-)#</mathjax> transferred in the reaction. </p>
<p><mathjax>#{(16H^(+) + 2MnO_4^(-) + 10e^(-) -> 2Mn^(2+) + 8H_2O), (5H_2O_2 -> 5O_2 + 10e^(-) + 10H^(+)) :}#</mathjax></p>
<p>Add these two half-reactions to get </p>
<p><mathjax>#16H^(+) + 2MnO_4^(-) + color(red)(cancel(color(black)(10e^(-)))) + 5H_2O_2 -> 2Mn^(2+) + color(red)(cancel(color(black)(10e^(-)))) + 5O_2 + 10H^(+) + 8H_2O#</mathjax></p>
<p>Finally, reduce the number of protons to get </p>
<p><mathjax>#6H^(+) + 2MnO_4^(-) + 5H_2O_2 ->2Mn^(2+) + 5O_2 + 8H_2O#</mathjax></p>
<p>Now, all those protons are coming from the sulfuric acid, which means that the balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> will be - add the spectator ions, too</p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + color(red)(3)H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>This is actually the half-reaction method, not the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> method. The oxidation numbers method does not use water and protons (for reactions that take place in acidic medium) to balance hydrogen and oxygen.</em></p>
<p><em>To balance this equation using the oxidation number method, figure out the number of electrons transferred, which is of course equal to 10, multiply the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> so that these electrons are balanced, then balance the rest of the atoms by inspection</em>.</p>
<p><em>So, you know that you need to multiply the reduction half-reaction by</em> <mathjax>#2#</mathjax> <em>and the oxidation half-reaction by</em> <mathjax>#5#</mathjax>. <em>This means that you have</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Now balance the equation by balancing the number of sulfate anions present on both sides of the equation. Notice that you have one sulfate anion on the reactants' side and three on the products side, so multiply sulfuric acid by</em> <mathjax>#3#</mathjax> <em>to get</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Finally, balance hydrogen and oxygen by multiplying the water by</em> <mathjax>#8#</mathjax></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p>And there you have it. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance the redox reaction? #"KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O"# ?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>Before you start balancing this equation, it's important to recognize that you're in <em>acidic medium</em> because the reaction take place in the presence of sulfuric acid. </p>
<p>This means that <em>any protons</em> that will remain unbalanced on the reactants' side will come from the sulfuric acid. </p>
<p>Moreover, not all the species are relevant to the reaction. More precisely, the potassium cation and the sulfate anion are spectator ions. The <em>net ionic equation</em> actually looks like this </p>
<p><mathjax>#MnO_4^(-) + H_2O_2 + 2H^(+) -> Mn^(2+) + O_2 + H_2O#</mathjax></p>
<p>So, start by assigning <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to all the atoms that take part in the reaction</p>
<p><mathjax>#stackrel(color(blue)(+1))(K) stackrel(color(blue)(+7))Mn stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(+2))Mn stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(+1))K_2 stackrel(color(blue)(+6))S stackrel(color(blue)(-2))O_4 + stackrel(color(blue)(0))O_2 + stackrel(color(blue)(+1))H_2 stackrel(color(blue)(-2))O#</mathjax></p>
<p>Notice that the oxidation number of mangenese changes from <mathjax>#"+7"#</mathjax> on the reactants' side to <mathjax>#"+2"#</mathjax> on the products' side, which means that mangenese is being <strong>reduced</strong>. </p>
<p>On the other hand, the oxidation number of oxygen in <em>hydrogen peroxide</em>, <mathjax>#H_2O_2#</mathjax>, changes from <mathjax>#"-1"#</mathjax> on the reactants' side to <mathjax>#"0"#</mathjax> in oxygen gas, <mathjax>#O_2#</mathjax>, on the products' side.</p>
<p>This means that oxygen is being <strong>oxidized</strong>. The two half-reactions will look like this </p>
<ul>
<li><em>reduction half-reaction</em></li>
</ul>
<p><mathjax>#stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+)#</mathjax></p>
<p>Here, each manganese atom gains <strong>5 electrons</strong>.</p>
<p>Since you're in acidic medium, you can balance the oxygen atoms by adding water to the side that needs oxygen and the hydrogen atoms by adding protons, <mathjax>#H^(+)#</mathjax>, to the side that needs hydrogen. </p>
<p>The reduction half-reaction will thus be </p>
<p><mathjax>#8H^(+) + stackrel(color(blue)(+7))MnO_4^(-) + 5e^(-) -> stackrel(color(blue)(+2))Mn^(2+) + 4H_2O#</mathjax></p>
<ul>
<li><em>oxidation half-reaction</em></li>
</ul>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-)#</mathjax></p>
<p>Here, each oxygen <em>atom</em> loses one electron, so two oxygen atoms will lose <strong>2 electrons</strong>. </p>
<p>Once again, balance the oxygen and hydrogen accordingly</p>
<p><mathjax>#H_2stackrel(color(blue)(-1))O_2 -> stackrel(color(blue)(0))O_2 + 2e^(-) + 2H^(+)#</mathjax></p>
<p>Now, in a <a href="http://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a>, the number of electrons <em>gained</em> in the <strong>reduction</strong> half-reaction must be <strong>equal to</strong> the number of electrons <em>lost</em> in the <strong>oxidation</strong> half-reaction. </p>
<p>This means that you need to multiply the reduction half-reaction by <strong>2</strong> and the oxidation half-reaction by <strong>5</strong> to get a total of <mathjax>#10e^(-)#</mathjax> transferred in the reaction. </p>
<p><mathjax>#{(16H^(+) + 2MnO_4^(-) + 10e^(-) -> 2Mn^(2+) + 8H_2O), (5H_2O_2 -> 5O_2 + 10e^(-) + 10H^(+)) :}#</mathjax></p>
<p>Add these two half-reactions to get </p>
<p><mathjax>#16H^(+) + 2MnO_4^(-) + color(red)(cancel(color(black)(10e^(-)))) + 5H_2O_2 -> 2Mn^(2+) + color(red)(cancel(color(black)(10e^(-)))) + 5O_2 + 10H^(+) + 8H_2O#</mathjax></p>
<p>Finally, reduce the number of protons to get </p>
<p><mathjax>#6H^(+) + 2MnO_4^(-) + 5H_2O_2 ->2Mn^(2+) + 5O_2 + 8H_2O#</mathjax></p>
<p>Now, all those protons are coming from the sulfuric acid, which means that the balanced <a href="http://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> will be - add the spectator ions, too</p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + color(red)(3)H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>This is actually the half-reaction method, not the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> method. The oxidation numbers method does not use water and protons (for reactions that take place in acidic medium) to balance hydrogen and oxygen.</em></p>
<p><em>To balance this equation using the oxidation number method, figure out the number of electrons transferred, which is of course equal to 10, multiply the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> so that these electrons are balanced, then balance the rest of the atoms by inspection</em>.</p>
<p><em>So, you know that you need to multiply the reduction half-reaction by</em> <mathjax>#2#</mathjax> <em>and the oxidation half-reaction by</em> <mathjax>#5#</mathjax>. <em>This means that you have</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Now balance the equation by balancing the number of sulfate anions present on both sides of the equation. Notice that you have one sulfate anion on the reactants' side and three on the products side, so multiply sulfuric acid by</em> <mathjax>#3#</mathjax> <em>to get</em></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + H_2O#</mathjax></p>
<p><em>Finally, balance hydrogen and oxygen by multiplying the water by</em> <mathjax>#8#</mathjax></p>
<p><mathjax>#2KMnO_4 + 5H_2O_2 + 3H_2SO_4 -> 2MnSO_4 + K_2SO_4 + 5O_2 + 8H_2O#</mathjax></p>
<p>And there you have it. </p></div>
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</article> | How do you balance the redox reaction? #"KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O"# ? | null |
1,494 | abe6529e-6ddd-11ea-9167-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-3-mol-of-kbr | 357 grams | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] KBr [IN] grams"}] | [{"type":"physical unit","value":"357 grams"}] | [{"type":"physical unit","value":"Mole [OF] KBr [=] \\pu{3 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in 3 mol of #KBr#?</h1> | null | 357 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to use the <strong>molar mass</strong> of potassium bromide, <mathjax>#"KBr"#</mathjax>, as a <em>conversion factor</em> to help you find the mass of <strong>three moles</strong> of this compound. </p>
<p>So, a compound's <strong>molar mass</strong> essentially tells you the mass of <strong>one mole</strong> of said compound. Now, let's assume that you only have a periodic table to work with here. </p>
<p>Potassium bromide is an ionic compound that is made up of potassium cations, <mathjax>#"K"^(+)#</mathjax>, and bromide anions, <mathjax>#"Br"^(-)#</mathjax>. Essentially, one <strong>formula unit</strong> of potassium bromide contains a potassium atom and a bromine atom. </p>
<p>Use <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> to find the <strong>molar masses</strong> of these two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You will find</p>
<blockquote>
<p><mathjax>#"For K: " M_M = "39.0963 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br: " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements</p>
<blockquote>
<p><mathjax>#M_"M KBr" = "39.0963 g mol"^(-1) + "79.904 g mol"^(-1) ~~ "119 g mol"^(-)#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> of potassium bromide has a mas of <mathjax>#"119 g"#</mathjax>m it follows that <strong>three moles</strong> will have a mass of </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("moles KBr"))) * overbrace("119 g"/(1color(red)(cancel(color(black)("mole KBr")))))^(color(purple)("molar mass of KBr")) = "357 g"#</mathjax></p>
</blockquote>
<p>You should round this off to one <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#"mass of 3 moles of KBr" = color(green)(|bar(ul(color(white)(a/a)"360 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"360 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 use the <strong>molar mass</strong> of potassium bromide, <mathjax>#"KBr"#</mathjax>, as a <em>conversion factor</em> to help you find the mass of <strong>three moles</strong> of this compound. </p>
<p>So, a compound's <strong>molar mass</strong> essentially tells you the mass of <strong>one mole</strong> of said compound. Now, let's assume that you only have a periodic table to work with here. </p>
<p>Potassium bromide is an ionic compound that is made up of potassium cations, <mathjax>#"K"^(+)#</mathjax>, and bromide anions, <mathjax>#"Br"^(-)#</mathjax>. Essentially, one <strong>formula unit</strong> of potassium bromide contains a potassium atom and a bromine atom. </p>
<p>Use <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> to find the <strong>molar masses</strong> of these two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You will find</p>
<blockquote>
<p><mathjax>#"For K: " M_M = "39.0963 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br: " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements</p>
<blockquote>
<p><mathjax>#M_"M KBr" = "39.0963 g mol"^(-1) + "79.904 g mol"^(-1) ~~ "119 g mol"^(-)#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> of potassium bromide has a mas of <mathjax>#"119 g"#</mathjax>m it follows that <strong>three moles</strong> will have a mass of </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("moles KBr"))) * overbrace("119 g"/(1color(red)(cancel(color(black)("mole KBr")))))^(color(purple)("molar mass of KBr")) = "357 g"#</mathjax></p>
</blockquote>
<p>You should round this off to one <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#"mass of 3 moles of KBr" = color(green)(|bar(ul(color(white)(a/a)"360 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams are in 3 mol of #KBr#?</h1>
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Stefan V.
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Mar 24, 2016
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<div class="markdown"><p><mathjax>#"360 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 use the <strong>molar mass</strong> of potassium bromide, <mathjax>#"KBr"#</mathjax>, as a <em>conversion factor</em> to help you find the mass of <strong>three moles</strong> of this compound. </p>
<p>So, a compound's <strong>molar mass</strong> essentially tells you the mass of <strong>one mole</strong> of said compound. Now, let's assume that you only have a periodic table to work with here. </p>
<p>Potassium bromide is an ionic compound that is made up of potassium cations, <mathjax>#"K"^(+)#</mathjax>, and bromide anions, <mathjax>#"Br"^(-)#</mathjax>. Essentially, one <strong>formula unit</strong> of potassium bromide contains a potassium atom and a bromine atom. </p>
<p>Use <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> to find the <strong>molar masses</strong> of these two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. You will find</p>
<blockquote>
<p><mathjax>#"For K: " M_M = "39.0963 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br: " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements</p>
<blockquote>
<p><mathjax>#M_"M KBr" = "39.0963 g mol"^(-1) + "79.904 g mol"^(-1) ~~ "119 g mol"^(-)#</mathjax></p>
</blockquote>
<p>So, if <strong>one mole</strong> of potassium bromide has a mas of <mathjax>#"119 g"#</mathjax>m it follows that <strong>three moles</strong> will have a mass of </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("moles KBr"))) * overbrace("119 g"/(1color(red)(cancel(color(black)("mole KBr")))))^(color(purple)("molar mass of KBr")) = "357 g"#</mathjax></p>
</blockquote>
<p>You should round this off to one <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs</p>
<blockquote>
<p><mathjax>#"mass of 3 moles of KBr" = color(green)(|bar(ul(color(white)(a/a)"360 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | How many grams are in 3 mol of #KBr#? | null |
1,495 | ad0ef3e8-6ddd-11ea-9094-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-1-m-acid-solution | 1.00 | start physical_unit 8 9 ph none qc_end physical_unit 8 9 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] acid solution"}] | [{"type":"physical unit","value":"1.00"}] | [{"type":"physical unit","value":"Molarity [OF] acid solution [=] \\pu{0.1 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.1 M acid solution? </h1> | null | 1.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.</p>
<p><mathjax>#HX(aq)+H_2O(l)rightleftharpoonsX^(-) + H_3O^(+)#</mathjax> </p>
<p>Why do we use <mathjax>#pH#</mathjax>? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).</p>
<p>When I write <mathjax>#log_ab=c#</mathjax>, I explicitly say that <mathjax>#a^c=b#</mathjax>. Common logarithmic bases are <mathjax>#10#</mathjax> and <mathjax>#e#</mathjax> (I think you cover this in A level mathematics these days). <mathjax>#pH#</mathjax> and <mathjax>#pK_a#</mathjax> scales use logarithms to the base <mathjax>#10#</mathjax>.</p>
<p>Anyway given what I have said <mathjax>#log_(10)100=2#</mathjax>, <mathjax>#log_(10)1000=3#</mathjax>, and <mathjax>#log_(10)1=0#</mathjax>, and <mathjax>#log_(10)0.1=log_(10)10^-1=-1#</mathjax>. If you can grasp this, you will find the <mathjax>#pH#</mathjax> concept (literally <mathjax>#"pouvoir hydrogene, power of hydrogen"#</mathjax>) fairly straightforward. </p>
<p>With weaker acids, say acetic acid, <mathjax>#H_3C-C(=O)OH#</mathjax>, the acid-base equilibrium lies to the left. The equilibrium constant, <mathjax>#K_a#</mathjax>, can be measured and often reported as <mathjax>#pK_a#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_a#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For a strong acid, <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[0.1]=-log_10(10^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.</p>
<p><mathjax>#HX(aq)+H_2O(l)rightleftharpoonsX^(-) + H_3O^(+)#</mathjax> </p>
<p>Why do we use <mathjax>#pH#</mathjax>? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).</p>
<p>When I write <mathjax>#log_ab=c#</mathjax>, I explicitly say that <mathjax>#a^c=b#</mathjax>. Common logarithmic bases are <mathjax>#10#</mathjax> and <mathjax>#e#</mathjax> (I think you cover this in A level mathematics these days). <mathjax>#pH#</mathjax> and <mathjax>#pK_a#</mathjax> scales use logarithms to the base <mathjax>#10#</mathjax>.</p>
<p>Anyway given what I have said <mathjax>#log_(10)100=2#</mathjax>, <mathjax>#log_(10)1000=3#</mathjax>, and <mathjax>#log_(10)1=0#</mathjax>, and <mathjax>#log_(10)0.1=log_(10)10^-1=-1#</mathjax>. If you can grasp this, you will find the <mathjax>#pH#</mathjax> concept (literally <mathjax>#"pouvoir hydrogene, power of hydrogen"#</mathjax>) fairly straightforward. </p>
<p>With weaker acids, say acetic acid, <mathjax>#H_3C-C(=O)OH#</mathjax>, the acid-base equilibrium lies to the left. The equilibrium constant, <mathjax>#K_a#</mathjax>, can be measured and often reported as <mathjax>#pK_a#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_a#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.1 M acid solution? </h1>
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anor277
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<div class="markdown"><p>For a strong acid, <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[0.1]=-log_10(10^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.</p>
<p><mathjax>#HX(aq)+H_2O(l)rightleftharpoonsX^(-) + H_3O^(+)#</mathjax> </p>
<p>Why do we use <mathjax>#pH#</mathjax>? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).</p>
<p>When I write <mathjax>#log_ab=c#</mathjax>, I explicitly say that <mathjax>#a^c=b#</mathjax>. Common logarithmic bases are <mathjax>#10#</mathjax> and <mathjax>#e#</mathjax> (I think you cover this in A level mathematics these days). <mathjax>#pH#</mathjax> and <mathjax>#pK_a#</mathjax> scales use logarithms to the base <mathjax>#10#</mathjax>.</p>
<p>Anyway given what I have said <mathjax>#log_(10)100=2#</mathjax>, <mathjax>#log_(10)1000=3#</mathjax>, and <mathjax>#log_(10)1=0#</mathjax>, and <mathjax>#log_(10)0.1=log_(10)10^-1=-1#</mathjax>. If you can grasp this, you will find the <mathjax>#pH#</mathjax> concept (literally <mathjax>#"pouvoir hydrogene, power of hydrogen"#</mathjax>) fairly straightforward. </p>
<p>With weaker acids, say acetic acid, <mathjax>#H_3C-C(=O)OH#</mathjax>, the acid-base equilibrium lies to the left. The equilibrium constant, <mathjax>#K_a#</mathjax>, can be measured and often reported as <mathjax>#pK_a#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10K_a#</mathjax>.</p></div>
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</article> | What is the pH of a 0.1 M acid solution? | null |
1,496 | ac61e834-6ddd-11ea-aec8-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-equation-for-the-following-reaction-hydrochloric-acid-reac | HCl + NaOH -> NaCl + H2O + heat | start chemical_equation qc_end substance 9 10 qc_end substance 13 14 qc_end substance 18 18 qc_end substance 19 19 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"HCl + NaOH -> NaCl + H2O + heat"}] | [{"type":"substance name","value":"Hydrochloric acid"},{"type":"substance name","value":"Sodium hydroxide"},{"type":"substance name","value":"Salt"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the molecular equation for the following reaction: Hydrochloric acid reacts with sodium hydroxide to produce a salt, water and heat?</h1> | null | HCl + NaOH -> NaCl + H2O + heat | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction. In a neutralization reaction, a strong acid and a strong base react to produce a salt and water.</p>
<p>The general equation for a neutralization is:</p>
<p><img alt="http://study.com/academy/lesson/neutralization-reaction-definition-equation-examples.html#transcriptHeader" src="https://useruploads.socratic.org/Joxzp8WoTtOLvQ1hrVfT_screen_shot_2014-09-05_at_3.46.10_pm.png"/> </p></div>
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<div class="markdown"><p>The molecular equation for the reaction between HCl and NaOH is the following:</p>
<p><mathjax>#"HCl + NaOH"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaCl + H"_2"O + heat"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction. In a neutralization reaction, a strong acid and a strong base react to produce a salt and water.</p>
<p>The general equation for a neutralization is:</p>
<p><img alt="http://study.com/academy/lesson/neutralization-reaction-definition-equation-examples.html#transcriptHeader" src="https://useruploads.socratic.org/Joxzp8WoTtOLvQ1hrVfT_screen_shot_2014-09-05_at_3.46.10_pm.png"/> </p></div>
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<div class="markdown"><p>The molecular equation for the reaction between HCl and NaOH is the following:</p>
<p><mathjax>#"HCl + NaOH"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaCl + H"_2"O + heat"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction. In a neutralization reaction, a strong acid and a strong base react to produce a salt and water.</p>
<p>The general equation for a neutralization is:</p>
<p><img alt="http://study.com/academy/lesson/neutralization-reaction-definition-equation-examples.html#transcriptHeader" src="https://useruploads.socratic.org/Joxzp8WoTtOLvQ1hrVfT_screen_shot_2014-09-05_at_3.46.10_pm.png"/> </p></div>
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</article> | What is the molecular equation for the following reaction: Hydrochloric acid reacts with sodium hydroxide to produce a salt, water and heat? | null |
1,497 | a8b9e98c-6ddd-11ea-92c0-ccda262736ce | https://socratic.org/questions/equilibrium-questions-1 | 0.01 | start physical_unit 6 7 kp none qc_end chemical_equation 34 37 qc_end physical_unit 16 16 23 24 partial_pressure qc_end physical_unit 18 18 26 27 partial_pressure qc_end end | [{"type":"physical unit","value":"Kp [OF] the reaction"}] | [{"type":"physical unit","value":"0.01"}] | [{"type":"chemical equation","value":"N2O4(g) <=> 2 NO2(g)"},{"type":"physical unit","value":"Partial pressure [OF] N2O4 [=] \\pu{0.6 atm}"},{"type":"physical unit","value":"Partial pressure [OF] NO2 [=] \\pu{0.07 atm}"}] | <h1 class="questionTitle" itemprop="name">Under a particular set of conditions the reaction comes to equilibrium where the partial pressures of #N_2O_4# and #NO_2# are found to be #"0.6 atm"# and #"0.07"# atm respectively. Determine #K_p#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is:<br/>
<mathjax>#N_2O_4(g) rightleftharpoons 2NO_2(g)#</mathjax></p></div>
</h2>
</div>
</div> | 0.01 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You are working with the following equilibrium reaction</p>
<blockquote>
<p><mathjax>#"N"_ 2"O"_ (4(g)) rightleftharpoons color(red)(2)"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>For starters, the fact that, at equilibrium, the reaction vessel contains significantly more dinitrogen tetroxide than nitrogen dioxide--you can tell that this is the case because the equilibrium <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of the reactant is <strong>higher</strong> than the equilibrium partial pressure of the product--tells you that you have</p>
<blockquote>
<p><mathjax>#K_p <1#</mathjax></p>
</blockquote>
<p>This means that at the temperature at which the reaction takes place, the <strong>reverse reaction</strong> is favored, which implies that the equilibrium will lie <strong>to the left</strong>. </p>
<p>In other words, if you start with dinitrogen tetroxide and no nitrogen dioxide, you should expect to find that most of the dinitrogen tetroxide will <strong>not react</strong> to produce nitrogen dioxide.</p>
<p>Similarly, if you start with nitrogen dioxide and no dinitrogen tetroxide, you should expect to find that most of the nitrogen dioxide will <strong>react</strong> to produce dinitrogen tetroxide. </p>
<p>The <strong>equilibrium constant</strong> that governs this reaction can be expressed using the <strong>equilibrium partial pressures</strong> of the two gases.</p>
<blockquote>
<p><mathjax>#K_ p = (("NO"_ 2)^color(red)(2))/(("N"_ 2"O"_4))#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the expression of the equilibrium constant <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_p = (0.07)^color(red)(2)/(0.6) = 0.008#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_p = 0.008#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You are working with the following equilibrium reaction</p>
<blockquote>
<p><mathjax>#"N"_ 2"O"_ (4(g)) rightleftharpoons color(red)(2)"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>For starters, the fact that, at equilibrium, the reaction vessel contains significantly more dinitrogen tetroxide than nitrogen dioxide--you can tell that this is the case because the equilibrium <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of the reactant is <strong>higher</strong> than the equilibrium partial pressure of the product--tells you that you have</p>
<blockquote>
<p><mathjax>#K_p <1#</mathjax></p>
</blockquote>
<p>This means that at the temperature at which the reaction takes place, the <strong>reverse reaction</strong> is favored, which implies that the equilibrium will lie <strong>to the left</strong>. </p>
<p>In other words, if you start with dinitrogen tetroxide and no nitrogen dioxide, you should expect to find that most of the dinitrogen tetroxide will <strong>not react</strong> to produce nitrogen dioxide.</p>
<p>Similarly, if you start with nitrogen dioxide and no dinitrogen tetroxide, you should expect to find that most of the nitrogen dioxide will <strong>react</strong> to produce dinitrogen tetroxide. </p>
<p>The <strong>equilibrium constant</strong> that governs this reaction can be expressed using the <strong>equilibrium partial pressures</strong> of the two gases.</p>
<blockquote>
<p><mathjax>#K_ p = (("NO"_ 2)^color(red)(2))/(("N"_ 2"O"_4))#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the expression of the equilibrium constant <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_p = (0.07)^color(red)(2)/(0.6) = 0.008#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Under a particular set of conditions the reaction comes to equilibrium where the partial pressures of #N_2O_4# and #NO_2# are found to be #"0.6 atm"# and #"0.07"# atm respectively. Determine #K_p#?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is:<br/>
<mathjax>#N_2O_4(g) rightleftharpoons 2NO_2(g)#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-01-23T00:40:50" itemprop="dateCreated">
Jan 23, 2018
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<div class="markdown"><p><mathjax>#K_p = 0.008#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You are working with the following equilibrium reaction</p>
<blockquote>
<p><mathjax>#"N"_ 2"O"_ (4(g)) rightleftharpoons color(red)(2)"NO"_ (2(g))#</mathjax></p>
</blockquote>
<p>For starters, the fact that, at equilibrium, the reaction vessel contains significantly more dinitrogen tetroxide than nitrogen dioxide--you can tell that this is the case because the equilibrium <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of the reactant is <strong>higher</strong> than the equilibrium partial pressure of the product--tells you that you have</p>
<blockquote>
<p><mathjax>#K_p <1#</mathjax></p>
</blockquote>
<p>This means that at the temperature at which the reaction takes place, the <strong>reverse reaction</strong> is favored, which implies that the equilibrium will lie <strong>to the left</strong>. </p>
<p>In other words, if you start with dinitrogen tetroxide and no nitrogen dioxide, you should expect to find that most of the dinitrogen tetroxide will <strong>not react</strong> to produce nitrogen dioxide.</p>
<p>Similarly, if you start with nitrogen dioxide and no dinitrogen tetroxide, you should expect to find that most of the nitrogen dioxide will <strong>react</strong> to produce dinitrogen tetroxide. </p>
<p>The <strong>equilibrium constant</strong> that governs this reaction can be expressed using the <strong>equilibrium partial pressures</strong> of the two gases.</p>
<blockquote>
<p><mathjax>#K_ p = (("NO"_ 2)^color(red)(2))/(("N"_ 2"O"_4))#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the expression of the equilibrium constant <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_p = (0.07)^color(red)(2)/(0.6) = 0.008#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p></div>
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</article> | Under a particular set of conditions the reaction comes to equilibrium where the partial pressures of #N_2O_4# and #NO_2# are found to be #"0.6 atm"# and #"0.07"# atm respectively. Determine #K_p#? |
The reaction is:
#N_2O_4(g) rightleftharpoons 2NO_2(g)#
|
1,498 | aa72f006-6ddd-11ea-aaf4-ccda262736ce | https://socratic.org/questions/a-compound-contains-40-calcium-12-carbon-and-48-oxygen-by-mass-what-is-the-empir | CaCO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CaCO3"}] | [{"type":"physical unit","value":"Percent by mass [OF] calcium in compound [=] \\pu{40%}"},{"type":"physical unit","value":"Percent by mass [OF] carbon in compound [=] \\pu{12%}"},{"type":"physical unit","value":"Percent by mass [OF] oxygen in compound [=] \\pu{48%}"}] | <h1 class="questionTitle" itemprop="name">A compound contains 40% calcium, 12% carbon, and 48% oxygen by mass. What is the empirical formula of this compound?</h1> | null | CaCO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Although you have not been provided with direct masses, you can use the percentage composition values in place of masses.</p>
<p>First, calculate the number of moles of each element:</p>
<p><mathjax>#["mol"("Ca") = 40/40 = 1]#</mathjax><mathjax>#["mol"("C") = 12/12 = 1]#</mathjax><mathjax>#["mol"("0") = 48/16 = 3]#</mathjax></p>
<p>Your next step <em>would</em> be to divide by the smallest value, but since this is <mathjax>#1#</mathjax> this would have no effect on the ratios already present. Thus, the empirical formula of the compound would be:</p>
<p><mathjax>#" Ca ":" C ":" O "#</mathjax><br/>
<mathjax>#" "1" ":" "1" ":" "3#</mathjax></p>
<p><mathjax>#"CaCO"_3#</mathjax></p>
<p>This is the formula for the <strong>ionic compound</strong> <a href="https://en.wikipedia.org/wiki/Calcium_carbonate" rel="nofollow">calcium carbonate</a>, confirming that our result is indeed in the magnitude of a plausible answer. In fact, given the fact that calcium carbonate is ionic, it is also true that this empirical formula equates to the unit formula of the compound.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>Empirical formula</strong>: <mathjax>#"CaCO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Although you have not been provided with direct masses, you can use the percentage composition values in place of masses.</p>
<p>First, calculate the number of moles of each element:</p>
<p><mathjax>#["mol"("Ca") = 40/40 = 1]#</mathjax><mathjax>#["mol"("C") = 12/12 = 1]#</mathjax><mathjax>#["mol"("0") = 48/16 = 3]#</mathjax></p>
<p>Your next step <em>would</em> be to divide by the smallest value, but since this is <mathjax>#1#</mathjax> this would have no effect on the ratios already present. Thus, the empirical formula of the compound would be:</p>
<p><mathjax>#" Ca ":" C ":" O "#</mathjax><br/>
<mathjax>#" "1" ":" "1" ":" "3#</mathjax></p>
<p><mathjax>#"CaCO"_3#</mathjax></p>
<p>This is the formula for the <strong>ionic compound</strong> <a href="https://en.wikipedia.org/wiki/Calcium_carbonate" rel="nofollow">calcium carbonate</a>, confirming that our result is indeed in the magnitude of a plausible answer. In fact, given the fact that calcium carbonate is ionic, it is also true that this empirical formula equates to the unit formula of the compound.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound contains 40% calcium, 12% carbon, and 48% oxygen by mass. What is the empirical formula of this compound?</h1>
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Owen Bell
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<div class="markdown"><p><strong>Empirical formula</strong>: <mathjax>#"CaCO"_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Although you have not been provided with direct masses, you can use the percentage composition values in place of masses.</p>
<p>First, calculate the number of moles of each element:</p>
<p><mathjax>#["mol"("Ca") = 40/40 = 1]#</mathjax><mathjax>#["mol"("C") = 12/12 = 1]#</mathjax><mathjax>#["mol"("0") = 48/16 = 3]#</mathjax></p>
<p>Your next step <em>would</em> be to divide by the smallest value, but since this is <mathjax>#1#</mathjax> this would have no effect on the ratios already present. Thus, the empirical formula of the compound would be:</p>
<p><mathjax>#" Ca ":" C ":" O "#</mathjax><br/>
<mathjax>#" "1" ":" "1" ":" "3#</mathjax></p>
<p><mathjax>#"CaCO"_3#</mathjax></p>
<p>This is the formula for the <strong>ionic compound</strong> <a href="https://en.wikipedia.org/wiki/Calcium_carbonate" rel="nofollow">calcium carbonate</a>, confirming that our result is indeed in the magnitude of a plausible answer. In fact, given the fact that calcium carbonate is ionic, it is also true that this empirical formula equates to the unit formula of the compound.</p></div>
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</article> | A compound contains 40% calcium, 12% carbon, and 48% oxygen by mass. What is the empirical formula of this compound? | null |
1,499 | a91f0c74-6ddd-11ea-9fef-ccda262736ce | https://socratic.org/questions/how-many-hydrogen-atoms-would-need-to-combine-with-one-atom-of-nitrogen-so-that- | 3 | start physical_unit 2 3 number none qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] hydrogen atoms"}] | [{"type":"physical unit","value":"3"}] | [{"type":"other","value":"Hydrogen atoms combine with nitrogen so that all of them would be stable."},{"type":"physical unit","value":"Number [OF] nitrogen atom [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How many hydrogen atoms would need to combine with one atom of nitrogen so that all of them would be stable? </h1> | null | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Nitrogen has 5 valance electrons so it would need 3 more valance electrons to be stable. Since hydrogen has 1 valance electron, you would need 3 hydrogen atoms.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>3 hydrogen atoms</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Nitrogen has 5 valance electrons so it would need 3 more valance electrons to be stable. Since hydrogen has 1 valance electron, you would need 3 hydrogen atoms.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many hydrogen atoms would need to combine with one atom of nitrogen so that all of them would be stable? </h1>
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Feb 18, 2017
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<div class="markdown"><p>3 hydrogen atoms</p></div>
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<div class="markdown"><p>Nitrogen has 5 valance electrons so it would need 3 more valance electrons to be stable. Since hydrogen has 1 valance electron, you would need 3 hydrogen atoms.</p></div>
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</article> | How many hydrogen atoms would need to combine with one atom of nitrogen so that all of them would be stable? | null |
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