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2,800 | ab690e30-6ddd-11ea-bb25-ccda262736ce | https://socratic.org/questions/synthetic-diamonds-can-be-manufactured-at-pressures-of-6-00-times-10-4-atm-if-we | 0.03 mL | start physical_unit 37 38 volume ml qc_end physical_unit 0 1 8 11 pressure qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 18 18 20 21 pressure qc_end physical_unit 18 18 8 11 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] that gas [IN] mL"}] | [{"type":"physical unit","value":"0.03 mL"}] | [{"type":"physical unit","value":"Pressure [OF] synthetic diamonds [=] \\pu{6.00 × 10^4 atm}"},{"type":"physical unit","value":"Volume [OF] gas [=] \\pu{2.00 liters}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Pressure2 [OF] gas [=] \\pu{6.00 × 10^4 atm}"}] | <h1 class="questionTitle" itemprop="name">Synthetic diamonds can be manufactured at pressures of #6.00 times 10^4# atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure #6.00 times 10^4#
atm, what would the volume of that gas be?</h1> | null | 0.03 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#"Pressure "xx" Volume "=" Constant"#</mathjax></p>
<p>The temperature being constant</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=2L#</mathjax></p>
<p>The initial pressure <mathjax>#P_1=1atm#</mathjax></p>
<p>The final pressure is <mathjax>#P_2=6*10^4atm#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=1/(6*10^4)*2=1/3*10^-4L=1/30mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume is <mathjax>#=1/30mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#"Pressure "xx" Volume "=" Constant"#</mathjax></p>
<p>The temperature being constant</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=2L#</mathjax></p>
<p>The initial pressure <mathjax>#P_1=1atm#</mathjax></p>
<p>The final pressure is <mathjax>#P_2=6*10^4atm#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=1/(6*10^4)*2=1/3*10^-4L=1/30mL#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Synthetic diamonds can be manufactured at pressures of #6.00 times 10^4# atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure #6.00 times 10^4#
atm, what would the volume of that gas be?</h1>
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Narad T.
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<span class="dateCreated" datetime="2018-07-22T06:39:01" itemprop="dateCreated">
Jul 22, 2018
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<div class="markdown"><p>The volume is <mathjax>#=1/30mL#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#"Pressure "xx" Volume "=" Constant"#</mathjax></p>
<p>The temperature being constant</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>The initial volume is <mathjax>#V_1=2L#</mathjax></p>
<p>The initial pressure <mathjax>#P_1=1atm#</mathjax></p>
<p>The final pressure is <mathjax>#P_2=6*10^4atm#</mathjax></p>
<p>The final volume is</p>
<p><mathjax>#V_2=1/(6*10^4)*2=1/3*10^-4L=1/30mL#</mathjax></p></div>
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anor277
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Jul 22, 2018
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<div class="markdown"><p>Rather small, a fraction of a millilitre….</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> insists that <mathjax>#P_1V_1=P_2V_2#</mathjax>...</p>
<p>And so …………….</p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(2.00*Lxx1.00*atm)/(6.00xx10^4*atm)=3.33xx10^-5*L~=0.03*cm^3#</mathjax></p>
<p>And note our assumptions...we ASSUME (perhaps wrongly) that the gas would not undergo a phase change. </p></div>
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Harish Chandra Rajpoot
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<span class="dateCreated" datetime="2018-07-22T09:21:11" itemprop="dateCreated">
Jul 22, 2018
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<div class="markdown"><p><mathjax>#3.33\times 10^{-5}\ \text{ltrs#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming gas is <strong>ideal undergoing a constant temperature process</strong> then </p>
<p>According to <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>, at constant temperature <mathjax>#T#</mathjax>, the pressure <mathjax>#P#</mathjax> of a constant mass of an ideal gas is inversely proportional to its volume <mathjax>#V#</mathjax>. Mathematically given as</p>
<p><mathjax>#P\prop 1/V#</mathjax></p>
<p><mathjax>#PV=\text{constant#</mathjax> </p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p>As per given data, initial pressure <mathjax>#P_1=1\ text{atm#</mathjax> , initial volume <mathjax>#V_1=2\ text{ltrs#</mathjax> & final pressure <mathjax>#P_2=6\times 10^4\ text{atm#</mathjax>. Substituting these values in above equation, we get</p>
<p><mathjax>#1\times 2=6\times 10^4\times V_2#</mathjax></p>
<p><mathjax>#V_2=1/3\times 10^{-4}#</mathjax></p>
<p><mathjax>#=3.33\times 10^{-5}\ \text{ltrs}#</mathjax> </p>
<p>The final volume of gas becomes <mathjax>#V_2=3.33\times 10^{-5}\ \text{ltrs#</mathjax></p></div>
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</article> | Synthetic diamonds can be manufactured at pressures of #6.00 times 10^4# atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure #6.00 times 10^4#
atm, what would the volume of that gas be? | null |
2,801 | ad1d622f-6ddd-11ea-8c1e-ccda262736ce | https://socratic.org/questions/what-is-the-ionization-energy-of-a-hydrogen-atom-that-is-in-the-n-6-excited-stat | 36.46 kJ/mol | start physical_unit 7 8 ionization_energy kj/mol qc_end physical_unit 7 8 15 15 state_n qc_end end | [{"type":"physical unit","value":"Ionization energy [OF] hydrogen atom [IN] kJ/mol"}] | [{"type":"physical unit","value":"36.46 kJ/mol"}] | [{"type":"physical unit","value":"n [OF] hydrogen atom [=] \\pu{6}"}] | <h1 class="questionTitle" itemprop="name">What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? </h1> | null | 36.46 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Rydberg Expression is given by:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-1/n_2^2]#</mathjax></p>
<p><mathjax>#lambda#</mathjax> is the wavelength of the emission line</p>
<p><mathjax>#n_1#</mathjax> is the principle quantum number of the lower energy level</p>
<p><mathjax>#n_2#</mathjax> is the principle quantum number of the higher energy level</p>
<p><mathjax>#R#</mathjax> is the Rydberg Constant <mathjax>#1.097xx10^7" ""m^(-1)#</mathjax></p>
<p>The energy levels converge and coalesce:</p>
<p><img alt="www.stmary.ws" src="http://www.stmary.ws/HighSchool/Physics/home/notes/modPhysics/bohrs_3.gif"/></p>
<p>At higher and higher values of <mathjax>#n_2#</mathjax> the term <mathjax>#1/n_2^2#</mathjax> tends to zero. Effectively <mathjax>#n_2=oo#</mathjax> and the electron has left the atom, forming a hydrogen ion.</p>
<p>The Rydberg Expression refers to emission where an electron falls from a higher energy level to a lower one, emitting a photon.</p>
<p>In this case we can use it to find the energy required to move an electron from <mathjax>#n=6#</mathjax> to <mathjax>#n=oo#</mathjax>.</p>
<p>The expression now becomes:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-0]#</mathjax></p>
<p><mathjax>#:.1/lambda=R/n_1^2#</mathjax></p>
<p>Since <mathjax>#n_1=6#</mathjax> this becomes:</p>
<p><mathjax>#1/lambda=R/36#</mathjax></p>
<p><mathjax>#:.lambda=36/R=36/(1.097xx10^7)=32.816xx10^-7" ""m"#</mathjax></p>
<p>To convert this into energy we use the Planck expression:</p>
<p><mathjax>#E=hf=(hc)/lambda#</mathjax></p>
<p><mathjax>#:.E=(6.626xx10^(-34)xx3xx10^(8))/(32.816xx10^(-7))=6.0575xx10^(-20)"J"#</mathjax></p>
<p>You'll notice from the graphic that the energy of the <mathjax>#n=6#</mathjax> electron is <mathjax>#-0.38"eV"#</mathjax>. </p>
<p>This means the energy to remove it will be <mathjax>#+0.38"eV"#</mathjax>.</p>
<p>To convert this to Joules you multiply by the electronic charge <mathjax># =0.38xx1.6xx10^-19=6.08xx10^(-20)" ""J"#</mathjax></p>
<p>As you can see my calculated value is very close to this.</p>
<p>This is the energy required to ionise a single atom. To get the energy required to ionise a mole of atoms we multiply by the Avogadro Constant:</p>
<p><mathjax>#E=6.0575xx10^-20xx6.02xx10^23" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.461xx10^3" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.46" ""kJ/mol"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#36.46" ""kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Rydberg Expression is given by:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-1/n_2^2]#</mathjax></p>
<p><mathjax>#lambda#</mathjax> is the wavelength of the emission line</p>
<p><mathjax>#n_1#</mathjax> is the principle quantum number of the lower energy level</p>
<p><mathjax>#n_2#</mathjax> is the principle quantum number of the higher energy level</p>
<p><mathjax>#R#</mathjax> is the Rydberg Constant <mathjax>#1.097xx10^7" ""m^(-1)#</mathjax></p>
<p>The energy levels converge and coalesce:</p>
<p><img alt="www.stmary.ws" src="http://www.stmary.ws/HighSchool/Physics/home/notes/modPhysics/bohrs_3.gif"/></p>
<p>At higher and higher values of <mathjax>#n_2#</mathjax> the term <mathjax>#1/n_2^2#</mathjax> tends to zero. Effectively <mathjax>#n_2=oo#</mathjax> and the electron has left the atom, forming a hydrogen ion.</p>
<p>The Rydberg Expression refers to emission where an electron falls from a higher energy level to a lower one, emitting a photon.</p>
<p>In this case we can use it to find the energy required to move an electron from <mathjax>#n=6#</mathjax> to <mathjax>#n=oo#</mathjax>.</p>
<p>The expression now becomes:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-0]#</mathjax></p>
<p><mathjax>#:.1/lambda=R/n_1^2#</mathjax></p>
<p>Since <mathjax>#n_1=6#</mathjax> this becomes:</p>
<p><mathjax>#1/lambda=R/36#</mathjax></p>
<p><mathjax>#:.lambda=36/R=36/(1.097xx10^7)=32.816xx10^-7" ""m"#</mathjax></p>
<p>To convert this into energy we use the Planck expression:</p>
<p><mathjax>#E=hf=(hc)/lambda#</mathjax></p>
<p><mathjax>#:.E=(6.626xx10^(-34)xx3xx10^(8))/(32.816xx10^(-7))=6.0575xx10^(-20)"J"#</mathjax></p>
<p>You'll notice from the graphic that the energy of the <mathjax>#n=6#</mathjax> electron is <mathjax>#-0.38"eV"#</mathjax>. </p>
<p>This means the energy to remove it will be <mathjax>#+0.38"eV"#</mathjax>.</p>
<p>To convert this to Joules you multiply by the electronic charge <mathjax># =0.38xx1.6xx10^-19=6.08xx10^(-20)" ""J"#</mathjax></p>
<p>As you can see my calculated value is very close to this.</p>
<p>This is the energy required to ionise a single atom. To get the energy required to ionise a mole of atoms we multiply by the Avogadro Constant:</p>
<p><mathjax>#E=6.0575xx10^-20xx6.02xx10^23" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.461xx10^3" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.46" ""kJ/mol"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? </h1>
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Michael
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<span class="dateCreated" datetime="2016-06-04T17:00:45" itemprop="dateCreated">
Jun 4, 2016
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<div class="markdown"><p><mathjax>#36.46" ""kJ/mol"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Rydberg Expression is given by:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-1/n_2^2]#</mathjax></p>
<p><mathjax>#lambda#</mathjax> is the wavelength of the emission line</p>
<p><mathjax>#n_1#</mathjax> is the principle quantum number of the lower energy level</p>
<p><mathjax>#n_2#</mathjax> is the principle quantum number of the higher energy level</p>
<p><mathjax>#R#</mathjax> is the Rydberg Constant <mathjax>#1.097xx10^7" ""m^(-1)#</mathjax></p>
<p>The energy levels converge and coalesce:</p>
<p><img alt="www.stmary.ws" src="http://www.stmary.ws/HighSchool/Physics/home/notes/modPhysics/bohrs_3.gif"/></p>
<p>At higher and higher values of <mathjax>#n_2#</mathjax> the term <mathjax>#1/n_2^2#</mathjax> tends to zero. Effectively <mathjax>#n_2=oo#</mathjax> and the electron has left the atom, forming a hydrogen ion.</p>
<p>The Rydberg Expression refers to emission where an electron falls from a higher energy level to a lower one, emitting a photon.</p>
<p>In this case we can use it to find the energy required to move an electron from <mathjax>#n=6#</mathjax> to <mathjax>#n=oo#</mathjax>.</p>
<p>The expression now becomes:</p>
<p><mathjax>#1/lambda=R[1/n_1^2-0]#</mathjax></p>
<p><mathjax>#:.1/lambda=R/n_1^2#</mathjax></p>
<p>Since <mathjax>#n_1=6#</mathjax> this becomes:</p>
<p><mathjax>#1/lambda=R/36#</mathjax></p>
<p><mathjax>#:.lambda=36/R=36/(1.097xx10^7)=32.816xx10^-7" ""m"#</mathjax></p>
<p>To convert this into energy we use the Planck expression:</p>
<p><mathjax>#E=hf=(hc)/lambda#</mathjax></p>
<p><mathjax>#:.E=(6.626xx10^(-34)xx3xx10^(8))/(32.816xx10^(-7))=6.0575xx10^(-20)"J"#</mathjax></p>
<p>You'll notice from the graphic that the energy of the <mathjax>#n=6#</mathjax> electron is <mathjax>#-0.38"eV"#</mathjax>. </p>
<p>This means the energy to remove it will be <mathjax>#+0.38"eV"#</mathjax>.</p>
<p>To convert this to Joules you multiply by the electronic charge <mathjax># =0.38xx1.6xx10^-19=6.08xx10^(-20)" ""J"#</mathjax></p>
<p>As you can see my calculated value is very close to this.</p>
<p>This is the energy required to ionise a single atom. To get the energy required to ionise a mole of atoms we multiply by the Avogadro Constant:</p>
<p><mathjax>#E=6.0575xx10^-20xx6.02xx10^23" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.461xx10^3" ""J/mol"#</mathjax></p>
<p><mathjax>#E=36.46" ""kJ/mol"#</mathjax></p></div>
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</article> | What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? | null |
2,802 | ab0a6916-6ddd-11ea-b774-ccda262736ce | https://socratic.org/questions/what-volume-of-nitrogen-monoxide-is-produced-when-146-liters-of-ammonia-react-ac | 146 L | start physical_unit 3 4 volume l qc_end chemical_equation 34 41 qc_end physical_unit 11 11 8 9 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] nitrogen monoxide [IN] L"}] | [{"type":"physical unit","value":"146 L"}] | [{"type":"chemical equation","value":"ammonia(g) + oxygen(g) -> nitrogen monoxide(g) + water(g)"},{"type":"physical unit","value":"Volume [OF] ammonia [=] \\pu{146 L}"},{"type":"other","value":"All gases are at the same temperature and pressure."}] | <h1 class="questionTitle" itemprop="name">What volume of nitrogen monoxide is produced when #"146 L"# of ammonia react according to the following reaction? (All gases are at the same temperature and pressure.)
</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>Please explain I need help with this!</p>
<p><mathjax>#"ammonia"(g) + "oxygen"(g) -> "nitrogen monoxide"(g) + "water"(g)#</mathjax></p></div>
</h2>
</div>
</div> | 146 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by converting the word equation to a balanced chemical equation. </p>
<p>You know that ammonia, <mathjax>#"NH"_3#</mathjax>, reacts with oxygen gas, <mathjax>#"O"_2#</mathjax>, to produce nitric oxide, <mathjax>#"NO"#</mathjax>, which is just another name for <em>nitrogen monoxide</em>, and water vapor, so you can write a balanced chemical equation that looks like this:</p>
<blockquote>
<p><mathjax>#4"NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Now, notice that in order for the reaction to produce <mathjax>#4#</mathjax> <strong>moles</strong> of nitric oxide, it must consume <mathjax>#4#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>This means that if you have enough moles of oxygen gas available, the number of moles of nitric oxide produced by the reaction will be <strong>equal</strong> to the number of moles of ammonia consumed. </p>
<p>Since all gases are kept under the same conditions for pressure and temperature, you can express this <em>mole ratio</em> in terms of the <strong>volumes</strong> occupied by each gas. </p>
<p>In other words, when the reaction consumes <mathjax>#4#</mathjax> <strong>liters</strong> of ammonia ir produces <mathjax>#4#</mathjax> <strong>liters</strong> of nitric oxide, provided, of course, that you have enough oxygen gas available, i.e. <mathjax>#5#</mathjax> <strong>liters</strong>. </p>
<p>So you can say that when oxygen gas is in excess, as you have here, when the reaction consumes <mathjax>#"146 L"#</mathjax> of ammonia and, it produces</p>
<blockquote>
<p><mathjax>#146 color(red)(cancel(color(black)("L NH"_3))) * "4 L NO"/(4color(red)(cancel(color(black)("L NH"_3)))) = color(darkgreen)(ul(color(black)("146 L NO")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"146 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by converting the word equation to a balanced chemical equation. </p>
<p>You know that ammonia, <mathjax>#"NH"_3#</mathjax>, reacts with oxygen gas, <mathjax>#"O"_2#</mathjax>, to produce nitric oxide, <mathjax>#"NO"#</mathjax>, which is just another name for <em>nitrogen monoxide</em>, and water vapor, so you can write a balanced chemical equation that looks like this:</p>
<blockquote>
<p><mathjax>#4"NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Now, notice that in order for the reaction to produce <mathjax>#4#</mathjax> <strong>moles</strong> of nitric oxide, it must consume <mathjax>#4#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>This means that if you have enough moles of oxygen gas available, the number of moles of nitric oxide produced by the reaction will be <strong>equal</strong> to the number of moles of ammonia consumed. </p>
<p>Since all gases are kept under the same conditions for pressure and temperature, you can express this <em>mole ratio</em> in terms of the <strong>volumes</strong> occupied by each gas. </p>
<p>In other words, when the reaction consumes <mathjax>#4#</mathjax> <strong>liters</strong> of ammonia ir produces <mathjax>#4#</mathjax> <strong>liters</strong> of nitric oxide, provided, of course, that you have enough oxygen gas available, i.e. <mathjax>#5#</mathjax> <strong>liters</strong>. </p>
<p>So you can say that when oxygen gas is in excess, as you have here, when the reaction consumes <mathjax>#"146 L"#</mathjax> of ammonia and, it produces</p>
<blockquote>
<p><mathjax>#146 color(red)(cancel(color(black)("L NH"_3))) * "4 L NO"/(4color(red)(cancel(color(black)("L NH"_3)))) = color(darkgreen)(ul(color(black)("146 L NO")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of nitrogen monoxide is produced when #"146 L"# of ammonia react according to the following reaction? (All gases are at the same temperature and pressure.)
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<div class="markdown"><p>Please explain I need help with this!</p>
<p><mathjax>#"ammonia"(g) + "oxygen"(g) -> "nitrogen monoxide"(g) + "water"(g)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"146 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by converting the word equation to a balanced chemical equation. </p>
<p>You know that ammonia, <mathjax>#"NH"_3#</mathjax>, reacts with oxygen gas, <mathjax>#"O"_2#</mathjax>, to produce nitric oxide, <mathjax>#"NO"#</mathjax>, which is just another name for <em>nitrogen monoxide</em>, and water vapor, so you can write a balanced chemical equation that looks like this:</p>
<blockquote>
<p><mathjax>#4"NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Now, notice that in order for the reaction to produce <mathjax>#4#</mathjax> <strong>moles</strong> of nitric oxide, it must consume <mathjax>#4#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>This means that if you have enough moles of oxygen gas available, the number of moles of nitric oxide produced by the reaction will be <strong>equal</strong> to the number of moles of ammonia consumed. </p>
<p>Since all gases are kept under the same conditions for pressure and temperature, you can express this <em>mole ratio</em> in terms of the <strong>volumes</strong> occupied by each gas. </p>
<p>In other words, when the reaction consumes <mathjax>#4#</mathjax> <strong>liters</strong> of ammonia ir produces <mathjax>#4#</mathjax> <strong>liters</strong> of nitric oxide, provided, of course, that you have enough oxygen gas available, i.e. <mathjax>#5#</mathjax> <strong>liters</strong>. </p>
<p>So you can say that when oxygen gas is in excess, as you have here, when the reaction consumes <mathjax>#"146 L"#</mathjax> of ammonia and, it produces</p>
<blockquote>
<p><mathjax>#146 color(red)(cancel(color(black)("L NH"_3))) * "4 L NO"/(4color(red)(cancel(color(black)("L NH"_3)))) = color(darkgreen)(ul(color(black)("146 L NO")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What volume of nitrogen monoxide is produced when #"146 L"# of ammonia react according to the following reaction? (All gases are at the same temperature and pressure.)
|
Please explain I need help with this!
#"ammonia"(g) + "oxygen"(g) -> "nitrogen monoxide"(g) + "water"(g)#
|
2,803 | a8c903fb-6ddd-11ea-abbc-ccda262736ce | https://socratic.org/questions/57eee2657c014936a9a9a534 | 0.39 moles | start physical_unit 8 8 mole mol qc_end physical_unit 3 3 1 2 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon [IN] moles"}] | [{"type":"physical unit","value":"0.39 moles"}] | [{"type":"physical unit","value":"Mole [OF] CO2(g) [=] \\pu{0.39 mol}"}] | <h1 class="questionTitle" itemprop="name">In #0.39*mol# #CO_2(g)#, how many moles of carbon are present?</h1> | null | 0.39 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In every <mathjax>#"Avogadro's number, "N_A#</mathjax> of carbon dioxide, there are <mathjax>#N_A#</mathjax> carbon atoms, and <mathjax>#2xxN_A#</mathjax> oxygen atoms. And <mathjax>#N_A#</mathjax> carbon dioxide molecules have a mass of approx. <mathjax>#44*g#</mathjax>.</p>
<p>What is (i), the mass of the carbon in such a molar quantity; and (ii), the mass of the carbon dioxide in such a molar quantity?</p>
<p>If I asked, <mathjax>#"how many oxygen atoms in a dozen"#</mathjax> <mathjax>#CO_2#</mathjax> <mathjax>#"molecules?"#</mathjax>, would you be able to answer confidently? Because this is what the question in effect asks. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.39#</mathjax> <mathjax>#"moles C"#</mathjax></p></div>
</div>
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<div class="markdown"><p>In every <mathjax>#"Avogadro's number, "N_A#</mathjax> of carbon dioxide, there are <mathjax>#N_A#</mathjax> carbon atoms, and <mathjax>#2xxN_A#</mathjax> oxygen atoms. And <mathjax>#N_A#</mathjax> carbon dioxide molecules have a mass of approx. <mathjax>#44*g#</mathjax>.</p>
<p>What is (i), the mass of the carbon in such a molar quantity; and (ii), the mass of the carbon dioxide in such a molar quantity?</p>
<p>If I asked, <mathjax>#"how many oxygen atoms in a dozen"#</mathjax> <mathjax>#CO_2#</mathjax> <mathjax>#"molecules?"#</mathjax>, would you be able to answer confidently? Because this is what the question in effect asks. </p></div>
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<h1 class="questionTitle" itemprop="name">In #0.39*mol# #CO_2(g)#, how many moles of carbon are present?</h1>
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<div class="markdown"><p><mathjax>#0.39#</mathjax> <mathjax>#"moles C"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In every <mathjax>#"Avogadro's number, "N_A#</mathjax> of carbon dioxide, there are <mathjax>#N_A#</mathjax> carbon atoms, and <mathjax>#2xxN_A#</mathjax> oxygen atoms. And <mathjax>#N_A#</mathjax> carbon dioxide molecules have a mass of approx. <mathjax>#44*g#</mathjax>.</p>
<p>What is (i), the mass of the carbon in such a molar quantity; and (ii), the mass of the carbon dioxide in such a molar quantity?</p>
<p>If I asked, <mathjax>#"how many oxygen atoms in a dozen"#</mathjax> <mathjax>#CO_2#</mathjax> <mathjax>#"molecules?"#</mathjax>, would you be able to answer confidently? Because this is what the question in effect asks. </p></div>
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</article> | In #0.39*mol# #CO_2(g)#, how many moles of carbon are present? | null |
2,804 | aceaadb9-6ddd-11ea-bad9-ccda262736ce | https://socratic.org/questions/58d7ff6c11ef6b3bb4d4a49c | 96.00 g | start physical_unit 4 4 mass g qc_end physical_unit 14 14 10 11 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] SO4^2- [IN] g"}] | [{"type":"physical unit","value":"96.00 g"}] | [{"type":"physical unit","value":"Mass [OF] dioxygen [=] \\pu{64 g}"}] | <h1 class="questionTitle" itemprop="name">How much sulfate dianion, #SO_4^(2-)# could be made from a #64*g# mass of dioxygen?</h1> | null | 96.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#64*g#</mathjax> of oxygen atoms represents <mathjax>#(64*g)/(16.0*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"4 moles of oxygen atoms."#</mathjax></p>
<p>And thus, with mole of sulfur atoms, i.e. <mathjax>#32*g#</mathjax>, we could make <mathjax>#96*g#</mathjax> sulfate dianion, i.e. <mathjax>#SO_4^(2-)#</mathjax>. </p>
<p>All we have done here is to take equivalent masses of sulfur (<mathjax>#32.06*g*mol#</mathjax>) and oxygen, (<mathjax>#15.999*g*mol^-1#</mathjax>).</p></div>
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<div class="markdown"><p>You mean <mathjax>#"sulfate ion"#</mathjax>, <mathjax>#SO_4^(2-)#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#64*g#</mathjax> of oxygen atoms represents <mathjax>#(64*g)/(16.0*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"4 moles of oxygen atoms."#</mathjax></p>
<p>And thus, with mole of sulfur atoms, i.e. <mathjax>#32*g#</mathjax>, we could make <mathjax>#96*g#</mathjax> sulfate dianion, i.e. <mathjax>#SO_4^(2-)#</mathjax>. </p>
<p>All we have done here is to take equivalent masses of sulfur (<mathjax>#32.06*g*mol#</mathjax>) and oxygen, (<mathjax>#15.999*g*mol^-1#</mathjax>).</p></div>
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<h1 class="questionTitle" itemprop="name">How much sulfate dianion, #SO_4^(2-)# could be made from a #64*g# mass of dioxygen?</h1>
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<div class="markdown"><p>You mean <mathjax>#"sulfate ion"#</mathjax>, <mathjax>#SO_4^(2-)#</mathjax>.........</p></div>
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<div class="markdown"><p><mathjax>#64*g#</mathjax> of oxygen atoms represents <mathjax>#(64*g)/(16.0*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"4 moles of oxygen atoms."#</mathjax></p>
<p>And thus, with mole of sulfur atoms, i.e. <mathjax>#32*g#</mathjax>, we could make <mathjax>#96*g#</mathjax> sulfate dianion, i.e. <mathjax>#SO_4^(2-)#</mathjax>. </p>
<p>All we have done here is to take equivalent masses of sulfur (<mathjax>#32.06*g*mol#</mathjax>) and oxygen, (<mathjax>#15.999*g*mol^-1#</mathjax>).</p></div>
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</article> | How much sulfate dianion, #SO_4^(2-)# could be made from a #64*g# mass of dioxygen? | null |
2,805 | aa8ff3c3-6ddd-11ea-9984-ccda262736ce | https://socratic.org/questions/given-a-solution-with-a-ph-of-2-8-what-is-the-h-of-the-solution | 10^(-2.8) M | start physical_unit 11 11 molarity mol/l qc_end physical_unit 13 14 7 7 ph qc_end end | [{"type":"physical unit","value":"Molarity [OF] H+ [IN] M"}] | [{"type":"physical unit","value":"10^(-2.8) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{2.8}"}] | <h1 class="questionTitle" itemprop="name">Given a solution with a pH of 2.8, what is the #H^+# of the solution? </h1> | null | 10^(-2.8) M | <div class="answerDescription">
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<div class="markdown"><p><mathjax>#pH=-log_(10)[H^+]#</mathjax></p>
<p>Thus <mathjax>#[H^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-pH)*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#H^+#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-2.8)*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_(10)[H^+]#</mathjax></p>
<p>Thus <mathjax>#[H^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-pH)*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#H^+#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-2.8)*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=-log_(10)[H^+]#</mathjax></p>
<p>Thus <mathjax>#[H^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-pH)*mol*L^-1#</mathjax></p></div>
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</article> | Given a solution with a pH of 2.8, what is the #H^+# of the solution? | null |
2,806 | ac9ef04c-6ddd-11ea-b2dd-ccda262736ce | https://socratic.org/questions/5647904e581e2a5ad1730b66 | 0.63 L | start physical_unit 3 4 volume l qc_end physical_unit 3 4 1 2 volume qc_end physical_unit 3 4 14 15 pressure qc_end physical_unit 3 4 20 21 temperature qc_end physical_unit 3 4 33 34 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] air bubble [IN] L"}] | [{"type":"physical unit","value":"0.63 L"}] | [{"type":"physical unit","value":"Volume1 [OF] air bubble [=] \\pu{0.25 L}"},{"type":"physical unit","value":"Pressure1 [OF] air bubble [=] \\pu{2.4 atm}"},{"type":"physical unit","value":"Temperature1 [OF] air bubble [=] \\pu{288 K}"},{"type":"physical unit","value":"Temperature2 [OF] air bubble [=] \\pu{300 K}"},{"type":"other","value":"Express at the surface."}] | <h1 class="questionTitle" itemprop="name">An #0.25*L# air bubble is released by a diver at a pressure of #2.4*atm#, and a temperature of #288*K#. What volume will it express at the surface with #T=300*K#?</h1> | null | 0.63 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So <mathjax>#V_2 = (P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.4*cancel(atm)xx0.250*Lxx300*cancelK)/(288*cancelKxx1.0*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>It is fairly clearly that the volume of the bubble WILL expand, both on the basis of the change in pressure, and on the basis of the change in temperature. </p>
<p>Note that the golden rule of scuba diving as opposed to free diving is <mathjax>#"NEVER HOLD YOUR BREATH"#</mathjax>. Why should holding your breath be an issue for a scuba diver?</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>; we use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> and absolute temperature. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So <mathjax>#V_2 = (P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.4*cancel(atm)xx0.250*Lxx300*cancelK)/(288*cancelKxx1.0*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>It is fairly clearly that the volume of the bubble WILL expand, both on the basis of the change in pressure, and on the basis of the change in temperature. </p>
<p>Note that the golden rule of scuba diving as opposed to free diving is <mathjax>#"NEVER HOLD YOUR BREATH"#</mathjax>. Why should holding your breath be an issue for a scuba diver?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An #0.25*L# air bubble is released by a diver at a pressure of #2.4*atm#, and a temperature of #288*K#. What volume will it express at the surface with #T=300*K#?</h1>
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anor277
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<span class="dateCreated" datetime="2016-01-18T11:00:53" itemprop="dateCreated">
Jan 18, 2016
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax>; we use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> and absolute temperature. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So <mathjax>#V_2 = (P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.4*cancel(atm)xx0.250*Lxx300*cancelK)/(288*cancelKxx1.0*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p>
<p>It is fairly clearly that the volume of the bubble WILL expand, both on the basis of the change in pressure, and on the basis of the change in temperature. </p>
<p>Note that the golden rule of scuba diving as opposed to free diving is <mathjax>#"NEVER HOLD YOUR BREATH"#</mathjax>. Why should holding your breath be an issue for a scuba diver?</p></div>
</div>
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</article> | An #0.25*L# air bubble is released by a diver at a pressure of #2.4*atm#, and a temperature of #288*K#. What volume will it express at the surface with #T=300*K#? | null |
2,807 | abc98a10-6ddd-11ea-ab0b-ccda262736ce | https://socratic.org/questions/a-chemist-dissolves-0-01-mol-of-an-acid-hcl-in-enough-water-to-make-a-350-ml-sol | 1.54 | start physical_unit 25 27 ph none qc_end physical_unit 8 8 3 4 mole qc_end physical_unit 25 27 15 16 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pH [OF] the acid solution"}] | [{"type":"physical unit","value":"1.54"}] | [{"type":"physical unit","value":"Mole [OF] HCl [=] \\pu{0.01 mol}"},{"type":"physical unit","value":"Volume [OF] the acid solution [=] \\pu{350 mL}"},{"type":"other","value":"Enough water."}] | <h1 class="questionTitle" itemprop="name">A chemist dissolves 0.01 mol of an acid (HCl) in enough water to make a 350 mL solution of acid. What is the pH of the acid solution?</h1> | null | 1.54 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you need to calculate the molar concentration of the acid by figuring out how many moles of hydrochloric acid would be present in <mathjax>#"1 L" = 10^3 quad "mL"#</mathjax> of this solution. </p>
<p>To do that, use the fact that you dissolved <mathjax>#0.01#</mathjax> <strong>moles</strong> of hydrochloric acid to make <mathjax>#"350 mL"#</mathjax> of the solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01 moles HCl"/(350color(red)(cancel(color(black)("mL solution")))) = "0.02857 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that the molar concentration of the solution is equal to</p>
<blockquote>
<p><mathjax>#["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
<blockquote>
<p>This implies that <mathjax>#10^3 quad "mL" = "1 L"#</mathjax> of this solution contains <mathjax>#0.02857#</mathjax> <strong>moles</strong> of hydrochloric acid.</p>
</blockquote>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which implies that it dissociates completely in aqueous solution to produce hydronium cations and chloride anions. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that all the moles of hydrochloric acid that you add to the solution will dissociate to produce hydronium cations.</p>
<p>Therefore, you can say that your solution has</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution is calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_ 3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.02857) = color(darkgreen)(ul(color(black)(1.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the number of moles of hydrochloric acid present in the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 1.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you need to calculate the molar concentration of the acid by figuring out how many moles of hydrochloric acid would be present in <mathjax>#"1 L" = 10^3 quad "mL"#</mathjax> of this solution. </p>
<p>To do that, use the fact that you dissolved <mathjax>#0.01#</mathjax> <strong>moles</strong> of hydrochloric acid to make <mathjax>#"350 mL"#</mathjax> of the solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01 moles HCl"/(350color(red)(cancel(color(black)("mL solution")))) = "0.02857 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that the molar concentration of the solution is equal to</p>
<blockquote>
<p><mathjax>#["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
<blockquote>
<p>This implies that <mathjax>#10^3 quad "mL" = "1 L"#</mathjax> of this solution contains <mathjax>#0.02857#</mathjax> <strong>moles</strong> of hydrochloric acid.</p>
</blockquote>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which implies that it dissociates completely in aqueous solution to produce hydronium cations and chloride anions. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that all the moles of hydrochloric acid that you add to the solution will dissociate to produce hydronium cations.</p>
<p>Therefore, you can say that your solution has</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution is calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_ 3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.02857) = color(darkgreen)(ul(color(black)(1.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the number of moles of hydrochloric acid present in the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A chemist dissolves 0.01 mol of an acid (HCl) in enough water to make a 350 mL solution of acid. What is the pH of the acid solution?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-03-07T15:29:28" itemprop="dateCreated">
Mar 7, 2018
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<div class="markdown"><p><mathjax>#"pH" = 1.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you need to calculate the molar concentration of the acid by figuring out how many moles of hydrochloric acid would be present in <mathjax>#"1 L" = 10^3 quad "mL"#</mathjax> of this solution. </p>
<p>To do that, use the fact that you dissolved <mathjax>#0.01#</mathjax> <strong>moles</strong> of hydrochloric acid to make <mathjax>#"350 mL"#</mathjax> of the solution.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01 moles HCl"/(350color(red)(cancel(color(black)("mL solution")))) = "0.02857 moles HCl"#</mathjax></p>
</blockquote>
<p>This means that the molar concentration of the solution is equal to</p>
<blockquote>
<p><mathjax>#["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
<blockquote>
<p>This implies that <mathjax>#10^3 quad "mL" = "1 L"#</mathjax> of this solution contains <mathjax>#0.02857#</mathjax> <strong>moles</strong> of hydrochloric acid.</p>
</blockquote>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which implies that it dissociates completely in aqueous solution to produce hydronium cations and chloride anions. </p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This means that all the moles of hydrochloric acid that you add to the solution will dissociate to produce hydronium cations.</p>
<p>Therefore, you can say that your solution has</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["HCl"] = "0.02857 mol L"^(-1)#</mathjax></p>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of the solution is calculated by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_ 3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.02857) = color(darkgreen)(ul(color(black)(1.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the number of moles of hydrochloric acid present in the solution. </p></div>
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</article> | A chemist dissolves 0.01 mol of an acid (HCl) in enough water to make a 350 mL solution of acid. What is the pH of the acid solution? | null |
2,808 | a8bca830-6ddd-11ea-bd3b-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-0-02-moles-of-berylium-iode-bel2 | 5 grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] BeI2 [IN] grams"}] | [{"type":"physical unit","value":"5 grams"}] | [{"type":"physical unit","value":"Mole [OF] BeI2 [=] \\pu{0.02 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in 0.02 moles of beryllium iodide, #"Bel"_2# ?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I don't understand physical science, please help me!</p></div>
</h2>
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</div> | 5 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you need to do here is to use a <em>conversion factor</em> that can take you from grams of beryllium iodide to moles of beryllium iodide.</p>
<p>As you know, the <strong>molar mass</strong> of a compound tells you the mass of exactly <mathjax>#1#</mathjax> <strong>mole</strong> of that compound. Beryllium iodide has a molar mass of <mathjax>#"262.821 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of beryllium iodide has a mass of <mathjax>#"262.821 g"#</mathjax>.</p>
<p>So you can say that you have</p>
<blockquote>
<p><mathjax>#"262.821 g mol"^(-1) = "262.821 g"/"1 mole BeI"_2#</mathjax></p>
</blockquote>
<p>To find the mass of <mathjax>#0.02#</mathjax> <strong>moles</strong> of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.</p>
<blockquote>
<p><mathjax>#0.02 color(red)(cancel(color(black)("moles BeI"_2))) * "262.821 g"/(1color(red)(cancel(color(black)("mole BeI"_2)))) = color(darkgreen)(ul(color(black)("5 g")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles present in your sample. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"5 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you need to do here is to use a <em>conversion factor</em> that can take you from grams of beryllium iodide to moles of beryllium iodide.</p>
<p>As you know, the <strong>molar mass</strong> of a compound tells you the mass of exactly <mathjax>#1#</mathjax> <strong>mole</strong> of that compound. Beryllium iodide has a molar mass of <mathjax>#"262.821 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of beryllium iodide has a mass of <mathjax>#"262.821 g"#</mathjax>.</p>
<p>So you can say that you have</p>
<blockquote>
<p><mathjax>#"262.821 g mol"^(-1) = "262.821 g"/"1 mole BeI"_2#</mathjax></p>
</blockquote>
<p>To find the mass of <mathjax>#0.02#</mathjax> <strong>moles</strong> of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.</p>
<blockquote>
<p><mathjax>#0.02 color(red)(cancel(color(black)("moles BeI"_2))) * "262.821 g"/(1color(red)(cancel(color(black)("mole BeI"_2)))) = color(darkgreen)(ul(color(black)("5 g")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles present in your sample. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams are in 0.02 moles of beryllium iodide, #"Bel"_2# ?</h1>
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Stefan V.
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Feb 13, 2018
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<div class="markdown"><p><mathjax>#"5 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you need to do here is to use a <em>conversion factor</em> that can take you from grams of beryllium iodide to moles of beryllium iodide.</p>
<p>As you know, the <strong>molar mass</strong> of a compound tells you the mass of exactly <mathjax>#1#</mathjax> <strong>mole</strong> of that compound. Beryllium iodide has a molar mass of <mathjax>#"262.821 g mol"^(-1)#</mathjax>, which means that <mathjax>#1#</mathjax> <strong>mole</strong> of beryllium iodide has a mass of <mathjax>#"262.821 g"#</mathjax>.</p>
<p>So you can say that you have</p>
<blockquote>
<p><mathjax>#"262.821 g mol"^(-1) = "262.821 g"/"1 mole BeI"_2#</mathjax></p>
</blockquote>
<p>To find the mass of <mathjax>#0.02#</mathjax> <strong>moles</strong> of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.</p>
<blockquote>
<p><mathjax>#0.02 color(red)(cancel(color(black)("moles BeI"_2))) * "262.821 g"/(1color(red)(cancel(color(black)("mole BeI"_2)))) = color(darkgreen)(ul(color(black)("5 g")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the number of moles present in your sample. </p></div>
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</article> | How many grams are in 0.02 moles of beryllium iodide, #"Bel"_2# ? |
I don't understand physical science, please help me!
|
2,809 | aa391e7a-6ddd-11ea-8ad1-ccda262736ce | https://socratic.org/questions/earth-s-oceans-have-an-average-depth-of-3800-m-a-total-area-of-3-63-x-10-8-km-2- | 8.00 × 10^9 grams | start physical_unit 23 23 mass g qc_end physical_unit 0 1 7 8 depth qc_end physical_unit 0 1 13 16 area qc_end physical_unit 22 23 25 28 concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] gold [IN] grams"}] | [{"type":"physical unit","value":"8.00 × 10^9 grams"}] | [{"type":"physical unit","value":"Depth [OF] earth's oceans [=] \\pu{3800 m}"},{"type":"physical unit","value":"Total area [OF] earth's oceans [=] \\pu{3.63 × 10^8 km^2}"},{"type":"physical unit","value":"Average concentration [OF] dissolved gold [=] \\pu{5.8 × 10^(-9) g/L}"}] | <h1 class="questionTitle" itemprop="name">Earth's oceans have an average depth of 3800 m, a total area of #3.63 x 10^8 km^2#, and an average concentration of dissolved gold of #5.8 x 10^-9# #g#/#L#. How many grams of gold are in the oceans? </h1> | null | 8.00 × 10^9 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find the volume of the ocean in <mathjax>#m^3#</mathjax>, and then multiply this volume by the average concentration in <mathjax>#g*L^-1#</mathjax> knowing that there are <mathjax>#1000 *L#</mathjax> in a <mathjax>#m^3#</mathjax>.</p>
<p><mathjax>#"Volume of the oceans"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.63xx10^11*m^2xx3800*m#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*m^3#</mathjax>.</p>
<p><mathjax>#"Mass of gold"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume "xx" concentration"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*cancel(m^3)xx5.8xx10^-9*g*cancel(L^-1)xx1000*cancelL*cancel(m^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000000#</mathjax> <mathjax>#g#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000*kg#</mathjax></p>
<p>Anyway, go over my figures carefully. There is a lot of arithmetic here. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Over 8 million kilograms; i.e. <mathjax>#8xx10^9*g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find the volume of the ocean in <mathjax>#m^3#</mathjax>, and then multiply this volume by the average concentration in <mathjax>#g*L^-1#</mathjax> knowing that there are <mathjax>#1000 *L#</mathjax> in a <mathjax>#m^3#</mathjax>.</p>
<p><mathjax>#"Volume of the oceans"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.63xx10^11*m^2xx3800*m#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*m^3#</mathjax>.</p>
<p><mathjax>#"Mass of gold"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume "xx" concentration"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*cancel(m^3)xx5.8xx10^-9*g*cancel(L^-1)xx1000*cancelL*cancel(m^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000000#</mathjax> <mathjax>#g#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000*kg#</mathjax></p>
<p>Anyway, go over my figures carefully. There is a lot of arithmetic here. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Earth's oceans have an average depth of 3800 m, a total area of #3.63 x 10^8 km^2#, and an average concentration of dissolved gold of #5.8 x 10^-9# #g#/#L#. How many grams of gold are in the oceans? </h1>
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<div class="markdown"><p>Over 8 million kilograms; i.e. <mathjax>#8xx10^9*g#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find the volume of the ocean in <mathjax>#m^3#</mathjax>, and then multiply this volume by the average concentration in <mathjax>#g*L^-1#</mathjax> knowing that there are <mathjax>#1000 *L#</mathjax> in a <mathjax>#m^3#</mathjax>.</p>
<p><mathjax>#"Volume of the oceans"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.63xx10^11*m^2xx3800*m#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*m^3#</mathjax>.</p>
<p><mathjax>#"Mass of gold"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume "xx" concentration"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.38xx10^15*cancel(m^3)xx5.8xx10^-9*g*cancel(L^-1)xx1000*cancelL*cancel(m^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000000#</mathjax> <mathjax>#g#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#8004000*kg#</mathjax></p>
<p>Anyway, go over my figures carefully. There is a lot of arithmetic here. </p></div>
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</article> | Earth's oceans have an average depth of 3800 m, a total area of #3.63 x 10^8 km^2#, and an average concentration of dissolved gold of #5.8 x 10^-9# #g#/#L#. How many grams of gold are in the oceans? | null |
2,810 | a9b7c576-6ddd-11ea-97d8-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-composed-of-3-25-hydrogen-h-19-36-ca | CH2O3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O3"}] | [{"type":"physical unit","value":"Percent by mass [OF] H in the compound [=] \\pu{3.25%}"},{"type":"physical unit","value":"Percent by mass [OF] C in the compound [=] \\pu{19.36%}"},{"type":"physical unit","value":"Percent by mass [OF] O in the compound [=] \\pu{77.39%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? </h1> | null | CH2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? </h1>
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anor277
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May 7, 2016
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<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of unknown compound, and work out the elemental proportions in terms of moles:</p>
<p><mathjax>#H:#</mathjax> <mathjax>#(3.25*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.22*mol;#</mathjax></p>
<p><mathjax>#C:#</mathjax> <mathjax>#(19.36*g)/(12.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.61*mol;#</mathjax></p>
<p><mathjax>#O:#</mathjax> <mathjax>#(77.39*g)/(15.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4.84*mol;#</mathjax></p>
<p>We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:</p>
<p><mathjax>#CH_2O_3#</mathjax></p></div>
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Madi
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May 7, 2016
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<div class="markdown"><p><mathjax>#CH_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume that there is 100g of the substance, so there would be </p>
<p><mathjax>#3.25g\ H#</mathjax> <mathjax>#19.36g\ C#</mathjax> and <mathjax>#77.39g\ O#</mathjax></p>
<p>Divide each mass by the molar mass of their respective element.</p>
<p><mathjax>#(3.25g)/1 H#</mathjax> <mathjax>#(19.36g)/12 C#</mathjax> <mathjax>#(77.39g) /16 O#</mathjax></p>
<p><mathjax>#=3.25 H#</mathjax> <mathjax>#=1.61 C#</mathjax> <mathjax>#=4.84 O#</mathjax></p>
<p>Divide by the smallest number. In this case, <mathjax>#1.61#</mathjax></p>
<p><mathjax>#=2 H#</mathjax> <mathjax>#=1 C#</mathjax> <mathjax>#=3O#</mathjax></p>
<p>So the empirical formula is <mathjax>#CH_2O_3#</mathjax></p></div>
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</article> | What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass? | null |
2,811 | aa5c2cf0-6ddd-11ea-b829-ccda262736ce | https://socratic.org/questions/a-60-0-ml-volume-of-a-gas-was-found-to-have-a-pressure-of-2-0-atm-at-25-c-what-w | 240 mL | start physical_unit 6 6 volume ml qc_end physical_unit 6 6 1 2 volume qc_end physical_unit 6 6 14 15 pressure qc_end physical_unit 6 6 17 18 temperature qc_end physical_unit 6 6 31 32 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"240 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{60.0 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{2.0 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{0.50 atm}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name"> A 60.0 mL volume of a gas was found to have a pressure of 2.0 atm at 25°C. What will be the new volume if the pressure is changed to 0.50 atm at constant temperature?</h1> | null | 240 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of a gas kept at constant temperature varies inversely with the pressure. In other words, if the volume increases, the pressure decreases, and vice versa. The equation to use is <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#P_1="2.0 atm"#</mathjax><br/>
<mathjax>#V_1="60.0 mL"#</mathjax><br/>
<mathjax>#P_2="0.50 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2=???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=((2.0"atm"xx60.0"mL"))/(0.5"atm")="240 mL"#</mathjax> </p></div>
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<div class="markdown"><p>The second volume will be 240 mL.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of a gas kept at constant temperature varies inversely with the pressure. In other words, if the volume increases, the pressure decreases, and vice versa. The equation to use is <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#P_1="2.0 atm"#</mathjax><br/>
<mathjax>#V_1="60.0 mL"#</mathjax><br/>
<mathjax>#P_2="0.50 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2=???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=((2.0"atm"xx60.0"mL"))/(0.5"atm")="240 mL"#</mathjax> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name"> A 60.0 mL volume of a gas was found to have a pressure of 2.0 atm at 25°C. What will be the new volume if the pressure is changed to 0.50 atm at constant temperature?</h1>
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<div class="markdown"><p>The second volume will be 240 mL.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of a gas kept at constant temperature varies inversely with the pressure. In other words, if the volume increases, the pressure decreases, and vice versa. The equation to use is <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#P_1="2.0 atm"#</mathjax><br/>
<mathjax>#V_1="60.0 mL"#</mathjax><br/>
<mathjax>#P_2="0.50 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2=???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the given values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=((2.0"atm"xx60.0"mL"))/(0.5"atm")="240 mL"#</mathjax> </p></div>
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</article> | A 60.0 mL volume of a gas was found to have a pressure of 2.0 atm at 25°C. What will be the new volume if the pressure is changed to 0.50 atm at constant temperature? | null |
2,812 | a905c792-6ddd-11ea-b587-ccda262736ce | https://socratic.org/questions/what-is-the-molality-of-a-solution-in-which-15-g-of-i-2-is-dissolved-in-500-0-g- | 0.24 mol/kg | start physical_unit 6 6 molality mol/kg qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 19 19 16 17 mass qc_end end | [{"type":"physical unit","value":"Molality [OF] the solution [IN] mol/kg"}] | [{"type":"physical unit","value":"0.24 mol/kg"}] | [{"type":"physical unit","value":"Mass [OF] I2 [=] \\pu{15 g}"},{"type":"physical unit","value":"Mass [OF] alcohol [=] \\pu{500.0 g}"}] | <h1 class="questionTitle" itemprop="name">What is the molality of a solution in which 15 g of #I_2# is dissolved in 500.0 g of alcohol?</h1> | null | 0.24 mol/kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of moles of iodine is <mathjax>#n=15/126.9=0.118 mol es #</mathjax></p>
<p>The mass of alcohol is <mathjax>#=500/1000=0.5kg#</mathjax></p>
<p>The molality is</p>
<p><mathjax>#p=n_(solute)/m_(solvent)=0.118/0.5=0.236mo l e s kg^-1#</mathjax></p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> is <mathjax>#=0.236mo l e s kg^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of moles of iodine is <mathjax>#n=15/126.9=0.118 mol es #</mathjax></p>
<p>The mass of alcohol is <mathjax>#=500/1000=0.5kg#</mathjax></p>
<p>The molality is</p>
<p><mathjax>#p=n_(solute)/m_(solvent)=0.118/0.5=0.236mo l e s kg^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molality of a solution in which 15 g of #I_2# is dissolved in 500.0 g of alcohol?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> is <mathjax>#=0.236mo l e s kg^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The number of moles of iodine is <mathjax>#n=15/126.9=0.118 mol es #</mathjax></p>
<p>The mass of alcohol is <mathjax>#=500/1000=0.5kg#</mathjax></p>
<p>The molality is</p>
<p><mathjax>#p=n_(solute)/m_(solvent)=0.118/0.5=0.236mo l e s kg^-1#</mathjax></p></div>
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</article> | What is the molality of a solution in which 15 g of #I_2# is dissolved in 500.0 g of alcohol? | null |
2,813 | a8a7e3aa-6ddd-11ea-ab2d-ccda262736ce | https://socratic.org/questions/how-do-you-balance-si-oh-4-nabr-sibr-4-naoh | Si(OH)4 + 4 NaBr -> SiBr4 + 4 NaOH | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Si(OH)4 + 4 NaBr -> SiBr4 + 4 NaOH"}] | [{"type":"chemical equation","value":"Si(OH)4 + NaBr -> SiBr4 + NaOH"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Si(OH)_4 + NaBr -> SiBr_4 + NaOH#?</h1> | null | Si(OH)4 + 4 NaBr -> SiBr4 + 4 NaOH | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Unbalanced Equation</strong></p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+"NaOH"#</mathjax></p>
<p>The word equation for this reaction is, "Silicon tetrahydroxide plus sodium bromide produce silicon tetrabromide plus sodium hydroxide." </p>
<p>When balancing a chemical equation, we compare the numbers of atoms (or moles) of each element on both sides of the equation. If the number of each element is not the same, we can add coefficients in front of each formula.</p>
<p><strong>Balancing the Equation</strong></p>
<p>Think of <mathjax>#"OH"#</mathjax>, as a single species. There are 4 <mathjax>#"OH"#</mathjax> on the left and 1 on the right. To balance them, place a coefficient of <mathjax>#color(blue)4#</mathjax> in front of NaOH.</p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>The <mathjax>#"OH"#</mathjax> is now balanced with 4 on each side.</p>
<p>There are now 4 <mathjax>#"Na"#</mathjax> atoms on the right side and 1 on the left. To balance them, place a coefficient of <mathjax>#color(red)4#</mathjax> in front of NaBr.</p>
<p><mathjax>#"Si(OH)"_4+color(red)4"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>There are now 4 sodium atoms on both sides, and there are also now 4 bromine atoms, <mathjax>#"Br"#</mathjax>, on both sides. </p>
<p>The equation is now completely balanced, having the same number of atoms (or moles) of each element on both sides: 1 Si, 4 O, 4 H, 4 Na, 4 Br.</p>
<p>Notice that I did not change the subscripts. I only changed the quantity of the species involved in the reaction by adding coefficients.</p></div>
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<div class="markdown"><p><mathjax>#"Si(OH)"_4+color(red)4"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>The word equation is now, "One mole silicon tetrahydroxide plus four moles sodium bromide produce one mole silicon tetrabromide plus four moles sodium hydroxide."</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Unbalanced Equation</strong></p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+"NaOH"#</mathjax></p>
<p>The word equation for this reaction is, "Silicon tetrahydroxide plus sodium bromide produce silicon tetrabromide plus sodium hydroxide." </p>
<p>When balancing a chemical equation, we compare the numbers of atoms (or moles) of each element on both sides of the equation. If the number of each element is not the same, we can add coefficients in front of each formula.</p>
<p><strong>Balancing the Equation</strong></p>
<p>Think of <mathjax>#"OH"#</mathjax>, as a single species. There are 4 <mathjax>#"OH"#</mathjax> on the left and 1 on the right. To balance them, place a coefficient of <mathjax>#color(blue)4#</mathjax> in front of NaOH.</p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>The <mathjax>#"OH"#</mathjax> is now balanced with 4 on each side.</p>
<p>There are now 4 <mathjax>#"Na"#</mathjax> atoms on the right side and 1 on the left. To balance them, place a coefficient of <mathjax>#color(red)4#</mathjax> in front of NaBr.</p>
<p><mathjax>#"Si(OH)"_4+color(red)4"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>There are now 4 sodium atoms on both sides, and there are also now 4 bromine atoms, <mathjax>#"Br"#</mathjax>, on both sides. </p>
<p>The equation is now completely balanced, having the same number of atoms (or moles) of each element on both sides: 1 Si, 4 O, 4 H, 4 Na, 4 Br.</p>
<p>Notice that I did not change the subscripts. I only changed the quantity of the species involved in the reaction by adding coefficients.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Si(OH)_4 + NaBr -> SiBr_4 + NaOH#?</h1>
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<div class="markdown"><p><mathjax>#"Si(OH)"_4+color(red)4"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>The word equation is now, "One mole silicon tetrahydroxide plus four moles sodium bromide produce one mole silicon tetrabromide plus four moles sodium hydroxide."</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Unbalanced Equation</strong></p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+"NaOH"#</mathjax></p>
<p>The word equation for this reaction is, "Silicon tetrahydroxide plus sodium bromide produce silicon tetrabromide plus sodium hydroxide." </p>
<p>When balancing a chemical equation, we compare the numbers of atoms (or moles) of each element on both sides of the equation. If the number of each element is not the same, we can add coefficients in front of each formula.</p>
<p><strong>Balancing the Equation</strong></p>
<p>Think of <mathjax>#"OH"#</mathjax>, as a single species. There are 4 <mathjax>#"OH"#</mathjax> on the left and 1 on the right. To balance them, place a coefficient of <mathjax>#color(blue)4#</mathjax> in front of NaOH.</p>
<p><mathjax>#"Si(OH)"_4+"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>The <mathjax>#"OH"#</mathjax> is now balanced with 4 on each side.</p>
<p>There are now 4 <mathjax>#"Na"#</mathjax> atoms on the right side and 1 on the left. To balance them, place a coefficient of <mathjax>#color(red)4#</mathjax> in front of NaBr.</p>
<p><mathjax>#"Si(OH)"_4+color(red)4"NaBr"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SiBr"_4+color(blue)(4)"NaOH"#</mathjax></p>
<p>There are now 4 sodium atoms on both sides, and there are also now 4 bromine atoms, <mathjax>#"Br"#</mathjax>, on both sides. </p>
<p>The equation is now completely balanced, having the same number of atoms (or moles) of each element on both sides: 1 Si, 4 O, 4 H, 4 Na, 4 Br.</p>
<p>Notice that I did not change the subscripts. I only changed the quantity of the species involved in the reaction by adding coefficients.</p></div>
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</article> | How do you balance #Si(OH)_4 + NaBr -> SiBr_4 + NaOH#? | null |
2,814 | ac26172e-6ddd-11ea-9866-ccda262736ce | https://socratic.org/questions/how-much-heat-is-lost-when-a-640-g-piece-of-copper-cools-from-375-c-to-26-c | 86.49 kJ | start physical_unit 11 11 heat_energy kj qc_end physical_unit 11 11 7 8 mass qc_end physical_unit 11 11 14 15 temperature qc_end physical_unit 11 11 17 18 temperature qc_end end | [{"type":"physical unit","value":"Lost heat [OF] copper [IN] kJ"}] | [{"type":"physical unit","value":"86.49 kJ"}] | [{"type":"physical unit","value":"Mass [OF] copper [=] \\pu{640 g}"},{"type":"physical unit","value":"Temperature1 [OF] copper [=] \\pu{375 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] copper [=] \\pu{26 ℃}"}] | <h1 class="questionTitle" itemprop="name"> How much heat is lost when a 640 g piece of copper cools from 375°C to 26°C?</h1> | null | 86.49 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#E=mcdeltatheta#</mathjax></p>
<p><mathjax>#m#</mathjax>: mass (kg)<br/>
<mathjax>#deltatheta#</mathjax>: change in temperature <mathjax>#(°C)#</mathjax><br/>
<mathjax>#c#</mathjax>: <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity - the energy needed to heat <mathjax>#1#</mathjax>kg of a substance by <mathjax>#1°C#</mathjax>.</p>
<p>specific heat capacity of copper <mathjax>#=385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C#</mathjax></p>
<p>the energy needed to heat <mathjax>#1#</mathjax>kg of copper by <mathjax>#1°C#</mathjax> is lost if <mathjax>#1#</mathjax>kg of copper cools by <mathjax>#1°C#</mathjax>.</p>
<p>therefore:</p>
<p><mathjax>#E = mcdeltatheta#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * (375-26)°C#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * 351°C#</mathjax></p>
<p><mathjax>#86486.4J#</mathjax> or <mathjax>#86.486kJ#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#86486.4J#</mathjax> or <mathjax>#86.486kJ#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#E=mcdeltatheta#</mathjax></p>
<p><mathjax>#m#</mathjax>: mass (kg)<br/>
<mathjax>#deltatheta#</mathjax>: change in temperature <mathjax>#(°C)#</mathjax><br/>
<mathjax>#c#</mathjax>: <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity - the energy needed to heat <mathjax>#1#</mathjax>kg of a substance by <mathjax>#1°C#</mathjax>.</p>
<p>specific heat capacity of copper <mathjax>#=385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C#</mathjax></p>
<p>the energy needed to heat <mathjax>#1#</mathjax>kg of copper by <mathjax>#1°C#</mathjax> is lost if <mathjax>#1#</mathjax>kg of copper cools by <mathjax>#1°C#</mathjax>.</p>
<p>therefore:</p>
<p><mathjax>#E = mcdeltatheta#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * (375-26)°C#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * 351°C#</mathjax></p>
<p><mathjax>#86486.4J#</mathjax> or <mathjax>#86.486kJ#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name"> How much heat is lost when a 640 g piece of copper cools from 375°C to 26°C?</h1>
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LM
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<div class="markdown"><p><mathjax>#86486.4J#</mathjax> or <mathjax>#86.486kJ#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#E=mcdeltatheta#</mathjax></p>
<p><mathjax>#m#</mathjax>: mass (kg)<br/>
<mathjax>#deltatheta#</mathjax>: change in temperature <mathjax>#(°C)#</mathjax><br/>
<mathjax>#c#</mathjax>: <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity - the energy needed to heat <mathjax>#1#</mathjax>kg of a substance by <mathjax>#1°C#</mathjax>.</p>
<p>specific heat capacity of copper <mathjax>#=385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C#</mathjax></p>
<p>the energy needed to heat <mathjax>#1#</mathjax>kg of copper by <mathjax>#1°C#</mathjax> is lost if <mathjax>#1#</mathjax>kg of copper cools by <mathjax>#1°C#</mathjax>.</p>
<p>therefore:</p>
<p><mathjax>#E = mcdeltatheta#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * (375-26)°C#</mathjax></p>
<p><mathjax># = 0.640kg * 385J#</mathjax>/<mathjax>#kg#</mathjax>/<mathjax>#°C * 351°C#</mathjax></p>
<p><mathjax>#86486.4J#</mathjax> or <mathjax>#86.486kJ#</mathjax></p></div>
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</article> | How much heat is lost when a 640 g piece of copper cools from 375°C to 26°C? | null |
2,815 | ab1146f6-6ddd-11ea-8ebd-ccda262736ce | https://socratic.org/questions/the-rate-constant-k-for-a-particular-reaction-is-1-3x10-4-m-1-s-1-at-100-celsius | 64.21 kJ | start physical_unit 35 36 activation_barrier kj qc_end physical_unit 6 7 9 14 equilibrium_constant_k qc_end physical_unit 6 7 19 24 equilibrium_constant_k qc_end physical_unit 6 7 16 17 temperature qc_end physical_unit 6 7 26 27 temperature qc_end end | [{"type":"physical unit","value":"Ea [OF] this reaction [IN] kJ"}] | [{"type":"physical unit","value":"64.21 kJ"}] | [{"type":"physical unit","value":"k1 [OF] particular reaction [=] \\pu{1.3 × 10^(-4) (M * s)^(-1)}"},{"type":"physical unit","value":"k2 [OF] particular reaction [=] \\pu{1.5 × 10^(-3) (M * s)^(-1)}"},{"type":"physical unit","value":"Temperature1 [OF] particular reaction [=] \\pu{100 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] particular reaction [=] \\pu{150 ℃}"}] | <h1 class="questionTitle" itemprop="name">The rate constant, #k#, for a particular reaction is #1.3xx10^(-4)"M"^(-1)"s"^(-1)# at #100^@"C"# and #1.5xx10^(-3)"M"^(-1)"s"^(-1)# at #150^@"C"#. What is the #E_a# (in kJ) for this reaction?
</h1> | null | 64.21 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the absolute temperature at which the reaction takes place</p>
<p>As you can see, the Arrhenius equation establishes a relationship between the <em>rate constant</em> and the <strong>absolute temperature</strong> at which the reaction takes place. </p>
<p>In other words, this equation allows you to determine how a <strong>change in temperature</strong> affects the <strong><a href="http://socratic.org/chemistry/chemical-kinetics/rate-of-reactions">rate of the reaction</a></strong>.</p>
<p>Now, convert the two temperatures from <em>degrees Celsius</em> to <em>Kelvin</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#T_1 = 100^@"C" + 273.15 = "373.15 K"#</mathjax></p>
<p><mathjax>#T_2 = 150^@"C" + 273.15 = "423.15 K"#</mathjax></p>
</blockquote>
<p>So, you know that at <mathjax>#T_1#</mathjax>, the rate constant for your reaction takes the form </p>
<blockquote>
<p><mathjax>#k_1 = A * "exp"( -E_a/(R * T_1))#</mathjax></p>
</blockquote>
<p>Likewise, at <mathjax>#T_2#</mathjax> the rate constant takes the form</p>
<blockquote>
<p><mathjax>#k_2 = A * "exp"( -E_a/(R * T_2))#</mathjax></p>
</blockquote>
<p>Divide these two equations to get rid of the <em>pre-exponential factor</em>, <mathjax>#A#</mathjax></p>
<blockquote>
<p><mathjax>#k_1/k_2 = (color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_1)))/(color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_2)))#</mathjax></p>
</blockquote>
<p>This will be equivalent to - using the properties of exponents</p>
<blockquote>
<p><mathjax>#k_1/k_2 = "exp"[E_a/R * (1/T_2 - 1/T_1)]#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation to get</p>
<blockquote>
<p><mathjax>#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#E_a#</mathjax>, the activation energy of the reaction</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(E_a = R * ln(k_1/k_2)/(1/T_2 - 1/T_1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Finally, plug in your values to get </p>
<blockquote>
<p><mathjax>#E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))#</mathjax></p>
<p><mathjax>#E_a = "64,212.3 J mol"^(-1)#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em>, the answer will be </p>
<blockquote>
<p><mathjax>#E_a = color(green)(|bar(ul(color(white)(a/a)"64 kJ mol"^(-1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://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>#"64 kJ mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the absolute temperature at which the reaction takes place</p>
<p>As you can see, the Arrhenius equation establishes a relationship between the <em>rate constant</em> and the <strong>absolute temperature</strong> at which the reaction takes place. </p>
<p>In other words, this equation allows you to determine how a <strong>change in temperature</strong> affects the <strong><a href="http://socratic.org/chemistry/chemical-kinetics/rate-of-reactions">rate of the reaction</a></strong>.</p>
<p>Now, convert the two temperatures from <em>degrees Celsius</em> to <em>Kelvin</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#T_1 = 100^@"C" + 273.15 = "373.15 K"#</mathjax></p>
<p><mathjax>#T_2 = 150^@"C" + 273.15 = "423.15 K"#</mathjax></p>
</blockquote>
<p>So, you know that at <mathjax>#T_1#</mathjax>, the rate constant for your reaction takes the form </p>
<blockquote>
<p><mathjax>#k_1 = A * "exp"( -E_a/(R * T_1))#</mathjax></p>
</blockquote>
<p>Likewise, at <mathjax>#T_2#</mathjax> the rate constant takes the form</p>
<blockquote>
<p><mathjax>#k_2 = A * "exp"( -E_a/(R * T_2))#</mathjax></p>
</blockquote>
<p>Divide these two equations to get rid of the <em>pre-exponential factor</em>, <mathjax>#A#</mathjax></p>
<blockquote>
<p><mathjax>#k_1/k_2 = (color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_1)))/(color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_2)))#</mathjax></p>
</blockquote>
<p>This will be equivalent to - using the properties of exponents</p>
<blockquote>
<p><mathjax>#k_1/k_2 = "exp"[E_a/R * (1/T_2 - 1/T_1)]#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation to get</p>
<blockquote>
<p><mathjax>#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#E_a#</mathjax>, the activation energy of the reaction</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(E_a = R * ln(k_1/k_2)/(1/T_2 - 1/T_1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Finally, plug in your values to get </p>
<blockquote>
<p><mathjax>#E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))#</mathjax></p>
<p><mathjax>#E_a = "64,212.3 J mol"^(-1)#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em>, the answer will be </p>
<blockquote>
<p><mathjax>#E_a = color(green)(|bar(ul(color(white)(a/a)"64 kJ mol"^(-1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The rate constant, #k#, for a particular reaction is #1.3xx10^(-4)"M"^(-1)"s"^(-1)# at #100^@"C"# and #1.5xx10^(-3)"M"^(-1)"s"^(-1)# at #150^@"C"#. What is the #E_a# (in kJ) for this reaction?
</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"64 kJ mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the absolute temperature at which the reaction takes place</p>
<p>As you can see, the Arrhenius equation establishes a relationship between the <em>rate constant</em> and the <strong>absolute temperature</strong> at which the reaction takes place. </p>
<p>In other words, this equation allows you to determine how a <strong>change in temperature</strong> affects the <strong><a href="http://socratic.org/chemistry/chemical-kinetics/rate-of-reactions">rate of the reaction</a></strong>.</p>
<p>Now, convert the two temperatures from <em>degrees Celsius</em> to <em>Kelvin</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#T_1 = 100^@"C" + 273.15 = "373.15 K"#</mathjax></p>
<p><mathjax>#T_2 = 150^@"C" + 273.15 = "423.15 K"#</mathjax></p>
</blockquote>
<p>So, you know that at <mathjax>#T_1#</mathjax>, the rate constant for your reaction takes the form </p>
<blockquote>
<p><mathjax>#k_1 = A * "exp"( -E_a/(R * T_1))#</mathjax></p>
</blockquote>
<p>Likewise, at <mathjax>#T_2#</mathjax> the rate constant takes the form</p>
<blockquote>
<p><mathjax>#k_2 = A * "exp"( -E_a/(R * T_2))#</mathjax></p>
</blockquote>
<p>Divide these two equations to get rid of the <em>pre-exponential factor</em>, <mathjax>#A#</mathjax></p>
<blockquote>
<p><mathjax>#k_1/k_2 = (color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_1)))/(color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_2)))#</mathjax></p>
</blockquote>
<p>This will be equivalent to - using the properties of exponents</p>
<blockquote>
<p><mathjax>#k_1/k_2 = "exp"[E_a/R * (1/T_2 - 1/T_1)]#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation to get</p>
<blockquote>
<p><mathjax>#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#E_a#</mathjax>, the activation energy of the reaction</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(E_a = R * ln(k_1/k_2)/(1/T_2 - 1/T_1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Finally, plug in your values to get </p>
<blockquote>
<p><mathjax>#E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))#</mathjax></p>
<p><mathjax>#E_a = "64,212.3 J mol"^(-1)#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em>, the answer will be </p>
<blockquote>
<p><mathjax>#E_a = color(green)(|bar(ul(color(white)(a/a)"64 kJ mol"^(-1)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | The rate constant, #k#, for a particular reaction is #1.3xx10^(-4)"M"^(-1)"s"^(-1)# at #100^@"C"# and #1.5xx10^(-3)"M"^(-1)"s"^(-1)# at #150^@"C"#. What is the #E_a# (in kJ) for this reaction?
| null |
2,816 | a943beec-6ddd-11ea-bbe5-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-for-the-incomplete-combustion-of-heptene-c-7h-14 | C7H14(l) + 9 O2(g) -> 5 CO2(g) + CO(g) + C(s) + 7 H2O(l) | start chemical_equation qc_end chemical_equation 11 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the incomplete combustion"}] | [{"type":"chemical equation","value":"C7H14(l) + 9 O2(g) -> 5 CO2(g) + CO(g) + C(s) + 7 H2O(l)"}] | [{"type":"chemical equation","value":"C7H14"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation for the incomplete combustion of heptene, #C_7H_14#?</h1> | null | C7H14(l) + 9 O2(g) -> 5 CO2(g) + CO(g) + C(s) + 7 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the given equation <a href="https://socratic.org/questions/what-is-a-balanced-chemical-equation-how-would-you-use-the-law-of-conservation-o">stoichiometrically</a> balanced? </p>
<p>And how did I know that 1 mole each of carbon monoxide and carbon (as soot) would result? Of course, I don't KNOW that, but I can nevertheless represent the possibility: i.e. some of the carbon is incompletely combusted. I could also represent the thermodynamic properties of the equation. </p>
<p>Of course complete combustion demands complete carbon oxidation to give:</p>
<p><mathjax>#C_7H_14(l) + 21/2O_2(g) rarr 7CO_2(g)+ 7H_2O(l)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#C_7H_14(l) + 9O_2(g) rarr 5CO_2(g) + CO(g) + C(s)+ 7H_2O(l)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the given equation <a href="https://socratic.org/questions/what-is-a-balanced-chemical-equation-how-would-you-use-the-law-of-conservation-o">stoichiometrically</a> balanced? </p>
<p>And how did I know that 1 mole each of carbon monoxide and carbon (as soot) would result? Of course, I don't KNOW that, but I can nevertheless represent the possibility: i.e. some of the carbon is incompletely combusted. I could also represent the thermodynamic properties of the equation. </p>
<p>Of course complete combustion demands complete carbon oxidation to give:</p>
<p><mathjax>#C_7H_14(l) + 21/2O_2(g) rarr 7CO_2(g)+ 7H_2O(l)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation for the incomplete combustion of heptene, #C_7H_14#?</h1>
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<div class="markdown"><p><mathjax>#C_7H_14(l) + 9O_2(g) rarr 5CO_2(g) + CO(g) + C(s)+ 7H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the given equation <a href="https://socratic.org/questions/what-is-a-balanced-chemical-equation-how-would-you-use-the-law-of-conservation-o">stoichiometrically</a> balanced? </p>
<p>And how did I know that 1 mole each of carbon monoxide and carbon (as soot) would result? Of course, I don't KNOW that, but I can nevertheless represent the possibility: i.e. some of the carbon is incompletely combusted. I could also represent the thermodynamic properties of the equation. </p>
<p>Of course complete combustion demands complete carbon oxidation to give:</p>
<p><mathjax>#C_7H_14(l) + 21/2O_2(g) rarr 7CO_2(g)+ 7H_2O(l)#</mathjax></p></div>
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</article> | What is the balanced equation for the incomplete combustion of heptene, #C_7H_14#? | null |
2,817 | ac312a47-6ddd-11ea-84a7-ccda262736ce | https://socratic.org/questions/59dc0d147c01491acef4e92d | 0.50 molar | start physical_unit 9 9 mole mol/l qc_end physical_unit 9 10 6 8 number qc_end end | [{"type":"physical unit","value":"Mole [OF] magnesium [IN] molar"}] | [{"type":"physical unit","value":"0.50 molar"}] | [{"type":"physical unit","value":"Number [OF] magnesium atoms [=] \\pu{3.011 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">What molar quantity is represented by #3.011xx10^23*"magnesium atoms"#?</h1> | null | 0.50 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this question is PRECISELY the same as the one you asked EXCEPT that we use <mathjax>#"the dozen"#</mathjax> as our metric instead of the <mathjax>#"Avocado number"#</mathjax>, i.e. <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax></p>
<p>And so we take the product...</p>
<p><mathjax>#6.022xx10^23*mol^-1xx0.5*mol-=3.011xx10^23*"magnesium atoms"#</mathjax>. You know the mass of this quantity....</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>If I have a half a dozen magnesium atoms, how many atoms would that be?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this question is PRECISELY the same as the one you asked EXCEPT that we use <mathjax>#"the dozen"#</mathjax> as our metric instead of the <mathjax>#"Avocado number"#</mathjax>, i.e. <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax></p>
<p>And so we take the product...</p>
<p><mathjax>#6.022xx10^23*mol^-1xx0.5*mol-=3.011xx10^23*"magnesium atoms"#</mathjax>. You know the mass of this quantity....</p></div>
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<h1 class="questionTitle" itemprop="name">What molar quantity is represented by #3.011xx10^23*"magnesium atoms"#?</h1>
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<div class="markdown"><p>If I have a half a dozen magnesium atoms, how many atoms would that be?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And this question is PRECISELY the same as the one you asked EXCEPT that we use <mathjax>#"the dozen"#</mathjax> as our metric instead of the <mathjax>#"Avocado number"#</mathjax>, i.e. <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax></p>
<p>And so we take the product...</p>
<p><mathjax>#6.022xx10^23*mol^-1xx0.5*mol-=3.011xx10^23*"magnesium atoms"#</mathjax>. You know the mass of this quantity....</p></div>
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yagmur
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Stefan V.
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<div class="markdown"><p><mathjax>#3.01 xx10^23#</mathjax> <mathjax>#"atoms"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#6.02xx10^23#</mathjax> (Avogadro's constant) indicates the number of atoms in a mole of any element, so basically:</p>
<p><mathjax>#"1 mole of Mg " -> " " 6.02 xx10^23#</mathjax> <mathjax>#"atoms"#</mathjax><br/>
<mathjax>#"0.5 moles of Mg " -> " " "? atoms"#</mathjax></p>
<p>So</p>
<p><mathjax>#"0.5 moles" xx ( 6.02xx10^23 "atoms")/"1 mole"#</mathjax> </p>
<p>which equals to half of Avogadro's constant. </p></div>
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</article> | What molar quantity is represented by #3.011xx10^23*"magnesium atoms"#? | null |
2,818 | a9db59a6-6ddd-11ea-96f5-ccda262736ce | https://socratic.org/questions/how-many-moles-of-glucose-c-6h-12o-6-are-in-300-g-of-glucose | 1.67 moles | start physical_unit 5 5 mole mol qc_end physical_unit 5 5 8 9 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] C6H12O6 [IN] moles"}] | [{"type":"physical unit","value":"1.67 moles"}] | [{"type":"physical unit","value":"Mass [OF] C6H12O6 [=] \\pu{300 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of glucose, #C_6H_12O_6#, are in 300 g of glucose?</h1> | null | 1.67 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First determine the molar mass (mass of one mole) of glucose, which is <mathjax>#"180.15588 g/mol"#</mathjax>. <a href="http://pubchem.ncbi.nlm.nih.gov/compound/5793" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/5793</a></p>
<p>Next divide the given mass of glucose by its molar mass.</p>
<p><mathjax>#300"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)="2 mol C"_6"H"_12"O"_6"#</mathjax> (<mathjax>#1.7#</mathjax> rounded to one significant figure)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>There are 2 moles of <mathjax>#"C"_6"H"_12"O"_6"#</mathjax> in 300 g <mathjax>#"C"_6"H"_12"O"_6"#</mathjax>, rounded to one significant figure.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First determine the molar mass (mass of one mole) of glucose, which is <mathjax>#"180.15588 g/mol"#</mathjax>. <a href="http://pubchem.ncbi.nlm.nih.gov/compound/5793" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/5793</a></p>
<p>Next divide the given mass of glucose by its molar mass.</p>
<p><mathjax>#300"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)="2 mol C"_6"H"_12"O"_6"#</mathjax> (<mathjax>#1.7#</mathjax> rounded to one significant figure)</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of glucose, #C_6H_12O_6#, are in 300 g of glucose?</h1>
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<div class="markdown"><p>There are 2 moles of <mathjax>#"C"_6"H"_12"O"_6"#</mathjax> in 300 g <mathjax>#"C"_6"H"_12"O"_6"#</mathjax>, rounded to one significant figure.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>First determine the molar mass (mass of one mole) of glucose, which is <mathjax>#"180.15588 g/mol"#</mathjax>. <a href="http://pubchem.ncbi.nlm.nih.gov/compound/5793" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/5793</a></p>
<p>Next divide the given mass of glucose by its molar mass.</p>
<p><mathjax>#300"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)="2 mol C"_6"H"_12"O"_6"#</mathjax> (<mathjax>#1.7#</mathjax> rounded to one significant figure)</p></div>
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</article> | How many moles of glucose, #C_6H_12O_6#, are in 300 g of glucose? | null |
2,819 | a8e4e0da-6ddd-11ea-898c-ccda262736ce | https://socratic.org/questions/how-many-grams-of-calcium-carbonate-caco-3-contain-48-grams-of-oxygen-atoms | 100.09 grams | start physical_unit 6 6 mass g qc_end physical_unit 11 12 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] CaCO3 [IN] grams"}] | [{"type":"physical unit","value":"100.09 grams"}] | [{"type":"physical unit","value":"Mass [OF] oxygen atoms [=] \\pu{48 grams}"}] | <h1 class="questionTitle" itemprop="name">How many grams of calcium carbonate, #CaCO_3#, contain 48 grams of oxygen atoms?</h1> | null | 100.09 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of calcium carbonate, there are <mathjax>#40.08*g#</mathjax> calcium metal, <mathjax>#12.01*g#</mathjax> of carbon, and <mathjax>#3xx15.999*g=48*g#</mathjax> of oxygen. From where did I get these masses? Did I know them off the top of my head?</p></div>
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<div class="markdown"><p>A <mathjax>#100.09*g#</mathjax> mass of calcium carbonate contains <mathjax>#48*g#</mathjax> of oxygen atoms. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of calcium carbonate, there are <mathjax>#40.08*g#</mathjax> calcium metal, <mathjax>#12.01*g#</mathjax> of carbon, and <mathjax>#3xx15.999*g=48*g#</mathjax> of oxygen. From where did I get these masses? Did I know them off the top of my head?</p></div>
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<div class="markdown"><p>A <mathjax>#100.09*g#</mathjax> mass of calcium carbonate contains <mathjax>#48*g#</mathjax> of oxygen atoms. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In one mole of calcium carbonate, there are <mathjax>#40.08*g#</mathjax> calcium metal, <mathjax>#12.01*g#</mathjax> of carbon, and <mathjax>#3xx15.999*g=48*g#</mathjax> of oxygen. From where did I get these masses? Did I know them off the top of my head?</p></div>
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</article> | How many grams of calcium carbonate, #CaCO_3#, contain 48 grams of oxygen atoms? | null |
2,820 | ab7155ba-6ddd-11ea-b998-ccda262736ce | https://socratic.org/questions/how-much-of-a-5-4-m-sucrose-solution-should-you-use-to-make-81-0-ml-of-a-0-030-m | 0.45 mL | start physical_unit 6 7 volume ml qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 6 7 13 14 volume qc_end physical_unit 6 7 17 18 molarity qc_end end | [{"type":"physical unit","value":"Volume1 [OF] sucrose solution [IN] mL"}] | [{"type":"physical unit","value":"0.45 mL"}] | [{"type":"physical unit","value":"Molarity1 [OF] sucrose solution [=] \\pu{5.4 M}"},{"type":"physical unit","value":"Volume2 [OF] sucrose solution [=] \\pu{81.0 mL}"},{"type":"physical unit","value":"Molarity2 [OF] sucrose solution [=] \\pu{0.030 M}"}] | <h1 class="questionTitle" itemprop="name">How much of a 5.4 M sucrose solution should you use to make 81.0 mL of a 0.030 M solution? </h1> | null | 0.45 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration = no. moles/volume</p>
<p>Or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p>The number of moles of sucrose <mathjax>#n#</mathjax> is the same in both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> so we can write:</p>
<p><mathjax>#5.4xxv=0.03xx81#</mathjax></p>
<p><mathjax>#:.v=(0.03xx81)/(5.4)#</mathjax></p>
<p><mathjax>#v=0.45"ml"#</mathjax></p></div>
</div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.45"ml"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration = no. moles/volume</p>
<p>Or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p>The number of moles of sucrose <mathjax>#n#</mathjax> is the same in both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> so we can write:</p>
<p><mathjax>#5.4xxv=0.03xx81#</mathjax></p>
<p><mathjax>#:.v=(0.03xx81)/(5.4)#</mathjax></p>
<p><mathjax>#v=0.45"ml"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How much of a 5.4 M sucrose solution should you use to make 81.0 mL of a 0.030 M solution? </h1>
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<a class="topContributorPic" href="/users/michael-2"><img alt="" class="" src="https://lh3.googleusercontent.com/-gCv6FQRhls0/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7oOJS05Ylqn3KuDSW0LfnbOk7FezQ/mo/photo.jpg?sz=50" title=""/></a>
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Michael
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<span class="dateCreated" datetime="2016-04-15T17:10:08" itemprop="dateCreated">
Apr 15, 2016
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<div class="markdown"><p><mathjax>#0.45"ml"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration = no. moles/volume</p>
<p>Or:</p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#:.n=cxxv#</mathjax></p>
<p>The number of moles of sucrose <mathjax>#n#</mathjax> is the same in both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> so we can write:</p>
<p><mathjax>#5.4xxv=0.03xx81#</mathjax></p>
<p><mathjax>#:.v=(0.03xx81)/(5.4)#</mathjax></p>
<p><mathjax>#v=0.45"ml"#</mathjax></p></div>
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</article> | How much of a 5.4 M sucrose solution should you use to make 81.0 mL of a 0.030 M solution? | null |
2,821 | aa8d08be-6ddd-11ea-96a8-ccda262736ce | https://socratic.org/questions/is-the-following-chemical-equation-balanced-2al-30-2-2al-2o-3 | 4 Al + 3 O2 -> 2 Al2O3 | start chemical_equation qc_end chemical_equation 6 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"4 Al + 3 O2 -> 2 Al2O3"}] | [{"type":"chemical equation","value":"2 Al + 3 O2 -> 2 Al2O3"}] | <h1 class="questionTitle" itemprop="name">Is the following chemical equation balanced? #2Al + 30_2 -> 2Al_2O_3#?</h1> | null | 4 Al + 3 O2 -> 2 Al2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you would immediately recognize that you had been short-changed. </p>
<p>Now looking at the equation,</p>
<p><mathjax>#2Al + 3O_2 rarr 2Al_2O_3#</mathjax>, </p>
<p>on each side there are 6 oxygen atoms, but on the left hand side there are 2 aluminum atoms, and on the RHS, there are 4 aluminium atoms. And this the equation is UNBALANCED, and thus it CANNOT be accepted as a representation of chemical reality.</p>
<p>The stoichiometrically balanced equation has GARBAGE OUT EQUIVALENT TO GARBAGE IN..........i.e.</p>
<p><mathjax>#4Al+3O_2 rarr 2Al_2O_3#</mathjax></p>
<p>On the other hand, if you were presented with <mathjax>#2+4=10#</mathjax>, would you twig that the balance is shot?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If you come into my shop and you receive change of <mathjax>#£2-50#</mathjax> from a <mathjax>#£10-00#</mathjax> note on a <mathjax>#£6-50#</mathjax> purchase, have you been short-changed?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you would immediately recognize that you had been short-changed. </p>
<p>Now looking at the equation,</p>
<p><mathjax>#2Al + 3O_2 rarr 2Al_2O_3#</mathjax>, </p>
<p>on each side there are 6 oxygen atoms, but on the left hand side there are 2 aluminum atoms, and on the RHS, there are 4 aluminium atoms. And this the equation is UNBALANCED, and thus it CANNOT be accepted as a representation of chemical reality.</p>
<p>The stoichiometrically balanced equation has GARBAGE OUT EQUIVALENT TO GARBAGE IN..........i.e.</p>
<p><mathjax>#4Al+3O_2 rarr 2Al_2O_3#</mathjax></p>
<p>On the other hand, if you were presented with <mathjax>#2+4=10#</mathjax>, would you twig that the balance is shot?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Is the following chemical equation balanced? #2Al + 30_2 -> 2Al_2O_3#?</h1>
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anor277
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Mar 29, 2017
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<div class="markdown"><p>If you come into my shop and you receive change of <mathjax>#£2-50#</mathjax> from a <mathjax>#£10-00#</mathjax> note on a <mathjax>#£6-50#</mathjax> purchase, have you been short-changed?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I think you would immediately recognize that you had been short-changed. </p>
<p>Now looking at the equation,</p>
<p><mathjax>#2Al + 3O_2 rarr 2Al_2O_3#</mathjax>, </p>
<p>on each side there are 6 oxygen atoms, but on the left hand side there are 2 aluminum atoms, and on the RHS, there are 4 aluminium atoms. And this the equation is UNBALANCED, and thus it CANNOT be accepted as a representation of chemical reality.</p>
<p>The stoichiometrically balanced equation has GARBAGE OUT EQUIVALENT TO GARBAGE IN..........i.e.</p>
<p><mathjax>#4Al+3O_2 rarr 2Al_2O_3#</mathjax></p>
<p>On the other hand, if you were presented with <mathjax>#2+4=10#</mathjax>, would you twig that the balance is shot?</p></div>
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</article> | Is the following chemical equation balanced? #2Al + 30_2 -> 2Al_2O_3#? | null |
2,822 | a9fb6f98-6ddd-11ea-9f6d-ccda262736ce | https://socratic.org/questions/if-it-takes-33-2-ml-of-0-1-m-naoh-to-neutralize-40-ml-of-hcl-then-what-is-the-mo | 0.08 M | start physical_unit 14 14 molarity mol/l qc_end physical_unit 8 8 6 7 molarity qc_end end | [{"type":"physical unit","value":"Molarity [OF] HCl solution [IN] M"}] | [{"type":"physical unit","value":"0.08 M"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{33.2 ml}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.1 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{40 ml}"}] | <h1 class="questionTitle" itemprop="name">If it takes 33.2 ml of 0.1 M NaOH to neutralize 40 ml of HCl, then what is the molarity of the hcl?</h1> | null | 0.08 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the molar concentration (molarity) of the <mathjax>#"HCl"#</mathjax> solution, given some titration measurements. </p>
<p>Let's first write the chemical equation for this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction:</p>
<p><mathjax>#"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)#</mathjax></p>
<p>Since we're given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <mathjax>#"NaOH"#</mathjax> solution used, we can calculate the moles be using the molarity equation:</p>
<p><mathjax>#"mol solute" = ("molarity")("L soln")#</mathjax></p>
<p><mathjax>#= (0.1"mol"/(cancel("L")))(0.0332cancel("L")) = color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol NaOH"#</mathjax></p>
<p>Since all the coefficients in the chemical equation are <mathjax>#1#</mathjax>, the relative number of moles of <mathjax>#"HCl"#</mathjax> used is also <mathjax>#color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol"#</mathjax>. </p>
<p>Finally, let's use the molarity equation again to find the molarity of the <mathjax>#"HCl"#</mathjax> solution (given the volume of <mathjax>#"NaOH soln"#</mathjax> is <mathjax>#40#</mathjax> <mathjax>#"mL"#</mathjax>, which must be in <strong>liters</strong> when using molarity equations):</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p><mathjax>#= (color(red)(0.00332)color(white)(l)color(red)("mol HCl"))/(0.040color(white)(l)"L") = color(blue)(0.083M#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.083M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the molar concentration (molarity) of the <mathjax>#"HCl"#</mathjax> solution, given some titration measurements. </p>
<p>Let's first write the chemical equation for this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction:</p>
<p><mathjax>#"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)#</mathjax></p>
<p>Since we're given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <mathjax>#"NaOH"#</mathjax> solution used, we can calculate the moles be using the molarity equation:</p>
<p><mathjax>#"mol solute" = ("molarity")("L soln")#</mathjax></p>
<p><mathjax>#= (0.1"mol"/(cancel("L")))(0.0332cancel("L")) = color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol NaOH"#</mathjax></p>
<p>Since all the coefficients in the chemical equation are <mathjax>#1#</mathjax>, the relative number of moles of <mathjax>#"HCl"#</mathjax> used is also <mathjax>#color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol"#</mathjax>. </p>
<p>Finally, let's use the molarity equation again to find the molarity of the <mathjax>#"HCl"#</mathjax> solution (given the volume of <mathjax>#"NaOH soln"#</mathjax> is <mathjax>#40#</mathjax> <mathjax>#"mL"#</mathjax>, which must be in <strong>liters</strong> when using molarity equations):</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p><mathjax>#= (color(red)(0.00332)color(white)(l)color(red)("mol HCl"))/(0.040color(white)(l)"L") = color(blue)(0.083M#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If it takes 33.2 ml of 0.1 M NaOH to neutralize 40 ml of HCl, then what is the molarity of the hcl?</h1>
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Nathan L.
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Jul 3, 2017
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<div class="markdown"><p><mathjax>#0.083M#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the molar concentration (molarity) of the <mathjax>#"HCl"#</mathjax> solution, given some titration measurements. </p>
<p>Let's first write the chemical equation for this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction:</p>
<p><mathjax>#"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)#</mathjax></p>
<p>Since we're given the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <mathjax>#"NaOH"#</mathjax> solution used, we can calculate the moles be using the molarity equation:</p>
<p><mathjax>#"mol solute" = ("molarity")("L soln")#</mathjax></p>
<p><mathjax>#= (0.1"mol"/(cancel("L")))(0.0332cancel("L")) = color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol NaOH"#</mathjax></p>
<p>Since all the coefficients in the chemical equation are <mathjax>#1#</mathjax>, the relative number of moles of <mathjax>#"HCl"#</mathjax> used is also <mathjax>#color(red)(0.00332#</mathjax> <mathjax>#color(red)("mol"#</mathjax>. </p>
<p>Finally, let's use the molarity equation again to find the molarity of the <mathjax>#"HCl"#</mathjax> solution (given the volume of <mathjax>#"NaOH soln"#</mathjax> is <mathjax>#40#</mathjax> <mathjax>#"mL"#</mathjax>, which must be in <strong>liters</strong> when using molarity equations):</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p><mathjax>#= (color(red)(0.00332)color(white)(l)color(red)("mol HCl"))/(0.040color(white)(l)"L") = color(blue)(0.083M#</mathjax></p></div>
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</article> | If it takes 33.2 ml of 0.1 M NaOH to neutralize 40 ml of HCl, then what is the molarity of the hcl? | null |
2,823 | aab5fcbf-6ddd-11ea-8c62-ccda262736ce | https://socratic.org/questions/how-many-grams-of-solute-are-needed-to-prepare-5-00-l-of-a-o-250-m-solution-of-c | 410.23 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 17 17 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"410.23 grams"}] | [{"type":"physical unit","value":"Volume [OF] Ca(NO3)2 solution [=] \\pu{5.00 L}"},{"type":"physical unit","value":"Molarity [OF] Ca(NO3)2 solution [=] \\pu{0.250 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of solute are needed to prepare 5.00 L of a o.250 M solution of #Ca(NO_3)_2#?</h1> | null | 410.23 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus..........</p>
<p><mathjax>#"moles of solute"="concentration"xx"volume"#</mathjax></p>
<p><mathjax>#=0.250*mol*cancel(L^-1)xx5.00*cancelL=2.50*mol#</mathjax></p>
<p><mathjax>#"Mass of solute"="moles"xx"molar mass"#</mathjax></p>
<p><mathjax>#=2.50*molxx164.09*g*mol^-1=??*g#</mathjax></p></div>
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<div class="markdown"><p>Over <mathjax>#400*g#</mathjax> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus..........</p>
<p><mathjax>#"moles of solute"="concentration"xx"volume"#</mathjax></p>
<p><mathjax>#=0.250*mol*cancel(L^-1)xx5.00*cancelL=2.50*mol#</mathjax></p>
<p><mathjax>#"Mass of solute"="moles"xx"molar mass"#</mathjax></p>
<p><mathjax>#=2.50*molxx164.09*g*mol^-1=??*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of solute are needed to prepare 5.00 L of a o.250 M solution of #Ca(NO_3)_2#?</h1>
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<div class="markdown"><p>Over <mathjax>#400*g#</mathjax> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present..........</p></div>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus..........</p>
<p><mathjax>#"moles of solute"="concentration"xx"volume"#</mathjax></p>
<p><mathjax>#=0.250*mol*cancel(L^-1)xx5.00*cancelL=2.50*mol#</mathjax></p>
<p><mathjax>#"Mass of solute"="moles"xx"molar mass"#</mathjax></p>
<p><mathjax>#=2.50*molxx164.09*g*mol^-1=??*g#</mathjax></p></div>
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</article> | How many grams of solute are needed to prepare 5.00 L of a o.250 M solution of #Ca(NO_3)_2#? | null |
2,824 | ad07ef09-6ddd-11ea-ab7d-ccda262736ce | https://socratic.org/questions/write-the-net-ionic-equation-including-the-phases | H3O+(aq) + OH-(aq) -> 2 H2O(l) | start chemical_equation qc_end chemical_equation 13 19 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"H3O+(aq) + OH-(aq) -> 2 H2O(l)"}] | [{"type":"chemical equation","value":"HClO4(aq) + NaOH(aq) -> H2O(l) + NaClO4(aq)"}] | <h1 class="questionTitle" itemprop="name">Write the net ionic equation including the phases?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Based off of this reaction - </p>
<p><mathjax>#"HClO"_4(aq) + "NaOH" (aq) -> "H"_2"O"(l) + "NaClO"_4 (aq)#</mathjax></p></div>
</h2>
</div>
</div> | H3O+(aq) + OH-(aq) -> 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Perchloric acid is a <strong>strong acid</strong> and sodium hydroxide is a <strong>strong base</strong>, so right from the start, you should be able to say that the net ionic equation that describes this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction is</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Perchloric acid will ionize completely in aqueous solution to produce hydronium cations and perchlorate anions.</p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))-> "H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide will also ionize completely in aqueous solution to produce sodium cations and hydroxide anions</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, you will end up with </p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l))) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this--keep in mind that sodium perchlorate is <strong>soluble</strong> in aqueous solution!</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + "Na"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Eliminate the <em>spectator ions</em> </p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong></p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Perchloric acid is a <strong>strong acid</strong> and sodium hydroxide is a <strong>strong base</strong>, so right from the start, you should be able to say that the net ionic equation that describes this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction is</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Perchloric acid will ionize completely in aqueous solution to produce hydronium cations and perchlorate anions.</p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))-> "H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide will also ionize completely in aqueous solution to produce sodium cations and hydroxide anions</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, you will end up with </p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l))) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this--keep in mind that sodium perchlorate is <strong>soluble</strong> in aqueous solution!</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + "Na"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Eliminate the <em>spectator ions</em> </p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong></p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Write the net ionic equation including the phases?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Based off of this reaction - </p>
<p><mathjax>#"HClO"_4(aq) + "NaOH" (aq) -> "H"_2"O"(l) + "NaClO"_4 (aq)#</mathjax></p></div>
</h2>
</div>
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Stefan V.
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Nov 28, 2017
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<div class="markdown"><p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Perchloric acid is a <strong>strong acid</strong> and sodium hydroxide is a <strong>strong base</strong>, so right from the start, you should be able to say that the net ionic equation that describes this <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction is</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Perchloric acid will ionize completely in aqueous solution to produce hydronium cations and perchlorate anions.</p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))-> "H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide will also ionize completely in aqueous solution to produce sodium cations and hydroxide anions</p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When you mix these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>, you will end up with </p>
<blockquote>
<p><mathjax>#"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l))) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong> looks like this--keep in mind that sodium perchlorate is <strong>soluble</strong> in aqueous solution!</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + "Na"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)#</mathjax></p>
</blockquote>
<p>Eliminate the <em>spectator ions</em> </p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-))))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong></p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote></div>
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</article> | Write the net ionic equation including the phases? |
Based off of this reaction -
#"HClO"_4(aq) + "NaOH" (aq) -> "H"_2"O"(l) + "NaClO"_4 (aq)#
|
2,825 | a8386a2d-6ddd-11ea-bd59-ccda262736ce | https://socratic.org/questions/593e16417c014979398dee86 | - 2406 kJ | start physical_unit 22 23 enthalpy kj qc_end physical_unit 22 23 19 20 mole qc_end chemical_equation 0 6 qc_end physical_unit 7 7 9 10 deltah qc_end end | [{"type":"physical unit","value":"Enthalpy [OF] magnesium metal [IN] kJ"}] | [{"type":"physical unit","value":"- 2406 kJ"}] | [{"type":"physical unit","value":"Mole [OF] magnesium metal [=] \\pu{4 moles}"},{"type":"chemical equation","value":"2 Mg(s) + O2(g) -> 2 MgO(s)"},{"type":"physical unit","value":"DeltaH [OF] rxn [=] \\pu{-1203 kJ/mol}"}] | <h1 class="questionTitle" itemprop="name">Given....#2Mg(s) + O_2(g) rarr 2MgO(s)# #; DeltaH_"rxn"=-1203*kJ*mol^-1#
What enthalpy is associated with the combustion of FOUR moles of magnesium metal?</h1> | null | - 2406 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2Mg(s) + O_2(g) rarr 2MgO(s)#</mathjax> <mathjax># DeltaH_"rxn"=-1203*kJ*mol^-1#</mathjax></p>
<p>Enthalpy terms are written per mole of reaction as written. We combust 4 moles of magnesium metal, and should therefore generate <mathjax>#2xx-1203*kJ=-2406*kJ#</mathjax>.</p>
<p>Note that given these data we could write <mathjax>#DeltaH_f^@(MgO)=-601.5*kJ#</mathjax>. Why?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaH=-2406*kJ...........#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2Mg(s) + O_2(g) rarr 2MgO(s)#</mathjax> <mathjax># DeltaH_"rxn"=-1203*kJ*mol^-1#</mathjax></p>
<p>Enthalpy terms are written per mole of reaction as written. We combust 4 moles of magnesium metal, and should therefore generate <mathjax>#2xx-1203*kJ=-2406*kJ#</mathjax>.</p>
<p>Note that given these data we could write <mathjax>#DeltaH_f^@(MgO)=-601.5*kJ#</mathjax>. Why?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Given....#2Mg(s) + O_2(g) rarr 2MgO(s)# #; DeltaH_"rxn"=-1203*kJ*mol^-1#
What enthalpy is associated with the combustion of FOUR moles of magnesium metal?</h1>
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anor277
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<div class="markdown"><p><mathjax>#DeltaH=-2406*kJ...........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2Mg(s) + O_2(g) rarr 2MgO(s)#</mathjax> <mathjax># DeltaH_"rxn"=-1203*kJ*mol^-1#</mathjax></p>
<p>Enthalpy terms are written per mole of reaction as written. We combust 4 moles of magnesium metal, and should therefore generate <mathjax>#2xx-1203*kJ=-2406*kJ#</mathjax>.</p>
<p>Note that given these data we could write <mathjax>#DeltaH_f^@(MgO)=-601.5*kJ#</mathjax>. Why?</p></div>
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</article> | Given....#2Mg(s) + O_2(g) rarr 2MgO(s)# #; DeltaH_"rxn"=-1203*kJ*mol^-1#
What enthalpy is associated with the combustion of FOUR moles of magnesium metal? | null |
2,826 | abf47358-6ddd-11ea-9325-ccda262736ce | https://socratic.org/questions/2-30-g-of-ethanol-were-mixed-with-aqueous-acidified-potassium-dichromate-vi-the- | 2.10 g | start physical_unit 13 13 mass g qc_end physical_unit 3 3 0 1 mass qc_end substance 7 10 qc_end c_other OTHER qc_end physical_unit 13 13 27 27 percent_yield qc_end end | [{"type":"physical unit","value":"Mass [OF] product [IN] g"}] | [{"type":"physical unit","value":"2.10 g"}] | [{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{2.30 g}"},{"type":"substance name","value":"Aqueous acidified potassium dichromate(VI)"},{"type":"other","value":"The desired product was collected by immediate distillation under gentle warming."},{"type":"physical unit","value":"Yield [OF] product [=] \\pu{70.0%}"}] | <h1 class="questionTitle" itemprop="name">2.30 g of ethanol were mixed with aqueous acidified potassium dichromate(VI). The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%. What mass of product was collected?</h1> | null | 2.10 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We know that we will need molar masses and moles, so let's gather all the information in one place</p>
<blockquote></blockquote>
<p><strong>Step 1. Write the chemical equation</strong></p>
<p>The product of the reaction is acetic acid, (<mathjax>#"CH"_3"COOH"#</mathjax>).</p>
<p><mathjax>#M_text(r):color(white)(mmmm)46.07color(white)(mmmmmmmmmmmmmmm)60.05#</mathjax><br/>
<mathjax>#color(white)(mmmm)"3CH"_3"CH"_2"OH" + "2Cr"_2"O"_7^"2-" + "16H"^"+" → "3CH"_3"COOH" + "4Cr"^"3+" + "11H"_2"O"#</mathjax><br/>
<mathjax>#"Mass/g":color(white)(mml)2.30#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.049 92"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ethanol</strong></p>
<p><mathjax>#"Moles of ethanol" = 2.30 color(red)(cancel(color(black)("g ethanol"))) × "1 mol ethanol"/(46.07 color(red)(cancel(color(black)("g ethanol")))) = "0.049 92 mol ethanol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of acetic acid</strong></p>
<p><mathjax>#"Moles of acetic acid" = "0.049 92" color(red)(cancel(color(black)("g ethanol"))) × "1 mol acetic acid"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "0.049 92 mol acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the theoretical yield of acetic acid</strong></p>
<p><mathjax>#"Theor. yield" = "0.049 92" color(red)(cancel(color(black)("mol acetic acid"))) × "60.05 g acetic acid"/(1 color(red)(cancel(color(black)("mol acetic acid")))) ="2.998 g acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the actual yield</strong></p>
<p><mathjax>#"Actual yield" = 2.998 color(red)(cancel(color(black)("g theor."))) × "70.0 g"/{100 color(red)(cancel(color(black)("g theor.")))) = "2.10 g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You obtained 2.10 g of product.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We know that we will need molar masses and moles, so let's gather all the information in one place</p>
<blockquote></blockquote>
<p><strong>Step 1. Write the chemical equation</strong></p>
<p>The product of the reaction is acetic acid, (<mathjax>#"CH"_3"COOH"#</mathjax>).</p>
<p><mathjax>#M_text(r):color(white)(mmmm)46.07color(white)(mmmmmmmmmmmmmmm)60.05#</mathjax><br/>
<mathjax>#color(white)(mmmm)"3CH"_3"CH"_2"OH" + "2Cr"_2"O"_7^"2-" + "16H"^"+" → "3CH"_3"COOH" + "4Cr"^"3+" + "11H"_2"O"#</mathjax><br/>
<mathjax>#"Mass/g":color(white)(mml)2.30#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.049 92"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ethanol</strong></p>
<p><mathjax>#"Moles of ethanol" = 2.30 color(red)(cancel(color(black)("g ethanol"))) × "1 mol ethanol"/(46.07 color(red)(cancel(color(black)("g ethanol")))) = "0.049 92 mol ethanol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of acetic acid</strong></p>
<p><mathjax>#"Moles of acetic acid" = "0.049 92" color(red)(cancel(color(black)("g ethanol"))) × "1 mol acetic acid"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "0.049 92 mol acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the theoretical yield of acetic acid</strong></p>
<p><mathjax>#"Theor. yield" = "0.049 92" color(red)(cancel(color(black)("mol acetic acid"))) × "60.05 g acetic acid"/(1 color(red)(cancel(color(black)("mol acetic acid")))) ="2.998 g acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the actual yield</strong></p>
<p><mathjax>#"Actual yield" = 2.998 color(red)(cancel(color(black)("g theor."))) × "70.0 g"/{100 color(red)(cancel(color(black)("g theor.")))) = "2.10 g"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">2.30 g of ethanol were mixed with aqueous acidified potassium dichromate(VI). The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%. What mass of product was collected?</h1>
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<div class="markdown"><p>You obtained 2.10 g of product.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We know that we will need molar masses and moles, so let's gather all the information in one place</p>
<blockquote></blockquote>
<p><strong>Step 1. Write the chemical equation</strong></p>
<p>The product of the reaction is acetic acid, (<mathjax>#"CH"_3"COOH"#</mathjax>).</p>
<p><mathjax>#M_text(r):color(white)(mmmm)46.07color(white)(mmmmmmmmmmmmmmm)60.05#</mathjax><br/>
<mathjax>#color(white)(mmmm)"3CH"_3"CH"_2"OH" + "2Cr"_2"O"_7^"2-" + "16H"^"+" → "3CH"_3"COOH" + "4Cr"^"3+" + "11H"_2"O"#</mathjax><br/>
<mathjax>#"Mass/g":color(white)(mml)2.30#</mathjax><br/>
<mathjax>#"Moles:"color(white)(mmml)"0.049 92"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of ethanol</strong></p>
<p><mathjax>#"Moles of ethanol" = 2.30 color(red)(cancel(color(black)("g ethanol"))) × "1 mol ethanol"/(46.07 color(red)(cancel(color(black)("g ethanol")))) = "0.049 92 mol ethanol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of acetic acid</strong></p>
<p><mathjax>#"Moles of acetic acid" = "0.049 92" color(red)(cancel(color(black)("g ethanol"))) × "1 mol acetic acid"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "0.049 92 mol acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the theoretical yield of acetic acid</strong></p>
<p><mathjax>#"Theor. yield" = "0.049 92" color(red)(cancel(color(black)("mol acetic acid"))) × "60.05 g acetic acid"/(1 color(red)(cancel(color(black)("mol acetic acid")))) ="2.998 g acetic acid"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the actual yield</strong></p>
<p><mathjax>#"Actual yield" = 2.998 color(red)(cancel(color(black)("g theor."))) × "70.0 g"/{100 color(red)(cancel(color(black)("g theor.")))) = "2.10 g"#</mathjax></p></div>
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</article> | 2.30 g of ethanol were mixed with aqueous acidified potassium dichromate(VI). The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%. What mass of product was collected? | null |
2,827 | aa7315c2-6ddd-11ea-aed5-ccda262736ce | https://socratic.org/questions/what-is-the-percent-by-mass-of-kbr-if-58-2-grams-are-dissolved-in-344-grams-of-w | 14.47% | start physical_unit 7 7 mass_percent none qc_end physical_unit 7 7 9 10 mass qc_end physical_unit 17 17 14 15 mass qc_end end | [{"type":"physical unit","value":"Percent by mass [OF] KBr in solution"}] | [{"type":"physical unit","value":"14.47%"}] | [{"type":"physical unit","value":"Mass [OF] KBr [=] \\pu{58.2 grams}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{344 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the percent by mass of KBr if 58.2 grams are dissolved in 344 grams of water?</h1> | null | 14.47% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many grams of potassium bromide will be present for every <mathjax>#"100. g"#</mathjax> of this solution, which is what the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> represents. </p>
<p>If you dissolve <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"344 g"#</mathjax> of water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#"58.2 g + 344 g = 402.2 g"#</mathjax> </p>
</blockquote>
<p>So this solution contains <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"402.2 g"#</mathjax> of the solution, which means that in <mathjax>#"100. g"#</mathjax> of this solution you will have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * "58.2 g KBr"/(402.2 color(red)(cancel(color(black)("g solution")))) = "14.5 g KBr"#</mathjax></p>
</blockquote>
<p>You can thus say that the solution's <strong>percent concentration by mass</strong> is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% m/m = 14.5% KBr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your figures. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#14.5%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many grams of potassium bromide will be present for every <mathjax>#"100. g"#</mathjax> of this solution, which is what the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> represents. </p>
<p>If you dissolve <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"344 g"#</mathjax> of water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#"58.2 g + 344 g = 402.2 g"#</mathjax> </p>
</blockquote>
<p>So this solution contains <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"402.2 g"#</mathjax> of the solution, which means that in <mathjax>#"100. g"#</mathjax> of this solution you will have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * "58.2 g KBr"/(402.2 color(red)(cancel(color(black)("g solution")))) = "14.5 g KBr"#</mathjax></p>
</blockquote>
<p>You can thus say that the solution's <strong>percent concentration by mass</strong> is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% m/m = 14.5% KBr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your figures. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the percent by mass of KBr if 58.2 grams are dissolved in 344 grams of water?</h1>
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Stefan V.
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Apr 20, 2018
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<div class="markdown"><p><mathjax>#14.5%#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to determine how many grams of potassium bromide will be present for every <mathjax>#"100. g"#</mathjax> of this solution, which is what the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong> represents. </p>
<p>If you dissolve <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"344 g"#</mathjax> of water, the <strong>total mass</strong> of the solution will be </p>
<blockquote>
<p><mathjax>#"58.2 g + 344 g = 402.2 g"#</mathjax> </p>
</blockquote>
<p>So this solution contains <mathjax>#"58.2 g"#</mathjax> of potassium bromide in <mathjax>#"402.2 g"#</mathjax> of the solution, which means that in <mathjax>#"100. g"#</mathjax> of this solution you will have</p>
<blockquote>
<p><mathjax>#100. color(red)(cancel(color(black)("g solution"))) * "58.2 g KBr"/(402.2 color(red)(cancel(color(black)("g solution")))) = "14.5 g KBr"#</mathjax></p>
</blockquote>
<p>You can thus say that the solution's <strong>percent concentration by mass</strong> is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% m/m = 14.5% KBr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for your figures. </p></div>
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Meave60
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<div class="markdown"><p>The percent by mass of <mathjax>#"KBr"#</mathjax> is <mathjax>#"14.4%"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"%(m/m)"=("mass of solute")/("mass of solution")xx100#</mathjax></p>
<p>The mass of the solution is the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>.</p>
<p>The given mass of the solute <mathjax>#"KBr"#</mathjax> is <mathjax>#"58.2 g"#</mathjax>.</p>
<p>The mass of the solution is <mathjax>#"58.2 g KBr + 344 g H"_2"O"="404.2 g"#</mathjax></p>
<p>Plug in the known values and solve.</p>
<p><mathjax>#"%(m/m)"=(58.2"g KBr")/(404.2"g solution")xx100=14.4%#</mathjax></p></div>
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</article> | What is the percent by mass of KBr if 58.2 grams are dissolved in 344 grams of water? | null |
2,828 | aac98883-6ddd-11ea-aac4-ccda262736ce | https://socratic.org/questions/how-do-you-complete-the-equation-mg-oh-2-s-2hcl-aq-aq-2h-2o-l | Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l) | start chemical_equation qc_end chemical_equation 6 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l)"}] | [{"type":"chemical equation","value":"Mg(OH)2(s) + 2 HCl(aq) -> _(aq) + 2 H2O(l)"}] | <h1 class="questionTitle" itemprop="name">How do you complete the equation #Mg(OH)_2(s) + 2HCl(aq) -> __(aq) + 2H_2O(l)#?</h1> | null | Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid-base reaction, which we may represent by the balanced net ionic reaction:</p>
<p><mathjax>#Mg(OH)_2(s) + 2H^(+) rarr Mg^(2+) + 2H_2O(l)#</mathjax></p>
<p>Charge is conserved, as well as mass........What does this mean?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Mg(OH)_2(s) + 2HCl(aq)rarrMgCl_2(aq) + H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is simply an acid-base reaction, which we may represent by the balanced net ionic reaction:</p>
<p><mathjax>#Mg(OH)_2(s) + 2H^(+) rarr Mg^(2+) + 2H_2O(l)#</mathjax></p>
<p>Charge is conserved, as well as mass........What does this mean?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you complete the equation #Mg(OH)_2(s) + 2HCl(aq) -> __(aq) + 2H_2O(l)#?</h1>
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<div class="markdown"><p><mathjax>#Mg(OH)_2(s) + 2HCl(aq)rarrMgCl_2(aq) + 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, which we may represent by the balanced net ionic reaction:</p>
<p><mathjax>#Mg(OH)_2(s) + 2H^(+) rarr Mg^(2+) + 2H_2O(l)#</mathjax></p>
<p>Charge is conserved, as well as mass........What does this mean?</p></div>
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</article> | How do you complete the equation #Mg(OH)_2(s) + 2HCl(aq) -> __(aq) + 2H_2O(l)#? | null |
2,829 | a8fe4df4-6ddd-11ea-996d-ccda262736ce | https://socratic.org/questions/what-is-the-molality-of-a-solution-prepared-by-dissolving-150-0-g-c-6h-12o-6-in- | 1.39 M | start physical_unit 5 6 molality mol/l qc_end physical_unit 12 12 10 11 mass qc_end physical_unit 16 16 14 15 mass qc_end end | [{"type":"physical unit","value":"Molality [OF] a solution [IN] M"}] | [{"type":"physical unit","value":"1.39 M"}] | [{"type":"physical unit","value":"Mass [OF] C6H12O6 [=] \\pu{150.0 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{600.0 g}"}] | <h1 class="questionTitle" itemprop="name">What is the molality of a solution prepared by dissolving 150.0 g #C_6H_12O_6# in 600.0 g #H_2O#?</h1> | null | 1.39 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molality <mathjax>#("m")#</mathjax> is a way of indicating the concentration of a solute dissolved in a solution. It is calculated by dividing the moles of solute by the mass of the solvent in kg.</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p>You have been given the masses of both the solute and the the solvent in grams. You will need to convert the mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> to moles using its molar mass, and you will need to convert the mass of water from grams to kilograms.</p>
<p>Molar mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> is <mathjax>#"180.15588 g/mol"#</mathjax>.<br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top</a></p>
<p>Determine moles of <mathjax>#"C"_6"H"_12"O"_6"#</mathjax>:</p>
<p><mathjax>#"mol solute"=150.0"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)=color(blue)("0.8326 mol C"_6"H"_12"O"_6")#</mathjax></p>
<p>Convert mass of solvent from g to kg.</p>
<p><mathjax>#600.0"g H"_2"O"xx(1"kg H"_2"O")/(1000"g H"_2"O")=color(red)("0.6000 kg H"_2"O")#</mathjax></p>
<p>Calculate molality:</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p><mathjax>#"molality"=(color(blue)(0.8326"mol"))/(color(red)(0.6000"kg"))=color(purple)("1.388 mol/kg=1.388 m")#</mathjax></p></div>
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<div class="markdown"><p>The molality of the solution is 1.388 m.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molality <mathjax>#("m")#</mathjax> is a way of indicating the concentration of a solute dissolved in a solution. It is calculated by dividing the moles of solute by the mass of the solvent in kg.</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p>You have been given the masses of both the solute and the the solvent in grams. You will need to convert the mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> to moles using its molar mass, and you will need to convert the mass of water from grams to kilograms.</p>
<p>Molar mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> is <mathjax>#"180.15588 g/mol"#</mathjax>.<br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top</a></p>
<p>Determine moles of <mathjax>#"C"_6"H"_12"O"_6"#</mathjax>:</p>
<p><mathjax>#"mol solute"=150.0"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)=color(blue)("0.8326 mol C"_6"H"_12"O"_6")#</mathjax></p>
<p>Convert mass of solvent from g to kg.</p>
<p><mathjax>#600.0"g H"_2"O"xx(1"kg H"_2"O")/(1000"g H"_2"O")=color(red)("0.6000 kg H"_2"O")#</mathjax></p>
<p>Calculate molality:</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p><mathjax>#"molality"=(color(blue)(0.8326"mol"))/(color(red)(0.6000"kg"))=color(purple)("1.388 mol/kg=1.388 m")#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molality of a solution prepared by dissolving 150.0 g #C_6H_12O_6# in 600.0 g #H_2O#?</h1>
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<div class="markdown"><p>The molality of the solution is 1.388 m.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molality <mathjax>#("m")#</mathjax> is a way of indicating the concentration of a solute dissolved in a solution. It is calculated by dividing the moles of solute by the mass of the solvent in kg.</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p>You have been given the masses of both the solute and the the solvent in grams. You will need to convert the mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> to moles using its molar mass, and you will need to convert the mass of water from grams to kilograms.</p>
<p>Molar mass of <mathjax>#"C"_6"H"_12"O"_6#</mathjax> is <mathjax>#"180.15588 g/mol"#</mathjax>.<br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/5793#section=Top</a></p>
<p>Determine moles of <mathjax>#"C"_6"H"_12"O"_6"#</mathjax>:</p>
<p><mathjax>#"mol solute"=150.0"g C"_6"H"_12"O"_6xx(1"mol C"_6"H"_12"O"_6)/(180.15588"g C"_6"H"_12"O"_6)=color(blue)("0.8326 mol C"_6"H"_12"O"_6")#</mathjax></p>
<p>Convert mass of solvent from g to kg.</p>
<p><mathjax>#600.0"g H"_2"O"xx(1"kg H"_2"O")/(1000"g H"_2"O")=color(red)("0.6000 kg H"_2"O")#</mathjax></p>
<p>Calculate molality:</p>
<p><mathjax>#"molality"=("moles solute")/("kg solvent")#</mathjax></p>
<p><mathjax>#"molality"=(color(blue)(0.8326"mol"))/(color(red)(0.6000"kg"))=color(purple)("1.388 mol/kg=1.388 m")#</mathjax></p></div>
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</article> | What is the molality of a solution prepared by dissolving 150.0 g #C_6H_12O_6# in 600.0 g #H_2O#? | null |
2,830 | a845405e-6ddd-11ea-9798-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-containing-38-8-g-carbon-16-2-g-hyd | NCH5 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] empirical"}] | [{"type":"chemical equation","value":"NCH5"}] | [{"type":"physical unit","value":"mass [OF] carbon in compound [=] \\pu{38.8 g}"},{"type":"physical unit","value":"mass [OF] hydrogen in compound [=] \\pu{16.2 g}"},{"type":"physical unit","value":"mass [OF] nitrogen in compound [=] \\pu{45.1 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound containing 38.8 g carbon, 16.2 g hydrogen and 45.1 g nitrogen? </h1> | null | NCH5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the empirical formula change the mass to moles Then find a ratio between the moles. </p>
<p>For Carbon <mathjax>#38.8 /12 = 3.233#</mathjax> moles </p>
<p>For Hydrogen <mathjax>#16.2/1 = 16.2#</mathjax> moles </p>
<p>For Nitrogen <mathjax>#45.1/14 = 3.22#</mathjax> moles. </p>
<p>The ratio for Carbon and Nitrogen is <mathjax>#1:1#</mathjax> </p>
<p><mathjax># 3.233/3.22 = 1.00 #</mathjax></p>
<p>The ratio of Hydrogen to both Carbon and Nitrogen is 5:1 </p>
<p><mathjax># 16.2/ 3.22 = 5.03 #</mathjax> </p>
<p>So the ratio is 1 C to 1 N to 5 H. </p>
<p><mathjax># H_2N C H_3#</mathjax></p>
<p>Nitrogen usually forms three bonds one to Carbon and 2 to Hydrogen atoms. Carbon usually forms four bonds one to Nitrogen and 3 to Hydrogen atoms. </p></div>
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<div class="markdown"><p><mathjax># H_2NCH_3#</mathjax> or <mathjax># NCH_5#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the empirical formula change the mass to moles Then find a ratio between the moles. </p>
<p>For Carbon <mathjax>#38.8 /12 = 3.233#</mathjax> moles </p>
<p>For Hydrogen <mathjax>#16.2/1 = 16.2#</mathjax> moles </p>
<p>For Nitrogen <mathjax>#45.1/14 = 3.22#</mathjax> moles. </p>
<p>The ratio for Carbon and Nitrogen is <mathjax>#1:1#</mathjax> </p>
<p><mathjax># 3.233/3.22 = 1.00 #</mathjax></p>
<p>The ratio of Hydrogen to both Carbon and Nitrogen is 5:1 </p>
<p><mathjax># 16.2/ 3.22 = 5.03 #</mathjax> </p>
<p>So the ratio is 1 C to 1 N to 5 H. </p>
<p><mathjax># H_2N C H_3#</mathjax></p>
<p>Nitrogen usually forms three bonds one to Carbon and 2 to Hydrogen atoms. Carbon usually forms four bonds one to Nitrogen and 3 to Hydrogen atoms. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound containing 38.8 g carbon, 16.2 g hydrogen and 45.1 g nitrogen? </h1>
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<div class="markdown"><p><mathjax># H_2NCH_3#</mathjax> or <mathjax># NCH_5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To find the empirical formula change the mass to moles Then find a ratio between the moles. </p>
<p>For Carbon <mathjax>#38.8 /12 = 3.233#</mathjax> moles </p>
<p>For Hydrogen <mathjax>#16.2/1 = 16.2#</mathjax> moles </p>
<p>For Nitrogen <mathjax>#45.1/14 = 3.22#</mathjax> moles. </p>
<p>The ratio for Carbon and Nitrogen is <mathjax>#1:1#</mathjax> </p>
<p><mathjax># 3.233/3.22 = 1.00 #</mathjax></p>
<p>The ratio of Hydrogen to both Carbon and Nitrogen is 5:1 </p>
<p><mathjax># 16.2/ 3.22 = 5.03 #</mathjax> </p>
<p>So the ratio is 1 C to 1 N to 5 H. </p>
<p><mathjax># H_2N C H_3#</mathjax></p>
<p>Nitrogen usually forms three bonds one to Carbon and 2 to Hydrogen atoms. Carbon usually forms four bonds one to Nitrogen and 3 to Hydrogen atoms. </p></div>
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<span class="dateCreated" datetime="2018-05-26T02:39:31" itemprop="dateCreated">
May 26, 2018
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<div class="markdown"><p>The empirical formula is <mathjax>#"CH"_5"N"#</mathjax>.</p>
<p>This may be methylamine, <mathjax>#"CH"_3"NH"_2#</mathjax>, derived from ammonia, <mathjax>#"NH"_3"#</mathjax>, in which one <mathjax>#"H"#</mathjax> atom is replaced by a methyl group <mathjax>#"-CH"_3"#</mathjax>. However, we can't know this for certain unless we are given the molecular mass.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The empirical formula of a compound represents the simplest whole number ratio of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>We must first determine the moles of each element in the compound by dividing the given mass by the molar mass.</p>
<p>Since the masses of all of the elements equal <mathjax>#"100.1 g"#</mathjax>, we can use the masses as given.</p>
<p><mathjax>#"mol C":#</mathjax> <mathjax>#(38.8"g C")/(12.011"g"/"mol")="3.23 mol"#</mathjax></p>
<p><mathjax>#"mol H":#</mathjax> <mathjax>#(16.2"g H")/(1.008"g"/"mol")="16.1 mol"#</mathjax></p>
<p><mathjax>#"mol N":#</mathjax> <mathjax>#(45.1"g N")/(14.007"g"/"mol")="3.22 mol"#</mathjax></p>
<p>To determine the subscripts of the empirical formula, we find the lowest whole number ratio for each element by dividing the moles of each element by the least number of moles, in this case <mathjax>#"3.22 mol"#</mathjax>. </p>
<p><mathjax>#"C":#</mathjax> <mathjax>#(3.23)/(3.22)=1.00#</mathjax></p>
<p><mathjax>#"H":#</mathjax> <mathjax>#(16.1)/(3.22)=5.00#</mathjax></p>
<p><mathjax>#"N":#</mathjax> <mathjax>#(3.22)/(3.22)=1.00#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_5"N"#</mathjax>.</p>
<p><strong>Note:</strong> If the mol ratios are not all whole numbers, you will need to multiply each mole ratio by the same factor that will make all mol ratios whole numbers.</p>
<p>This may be methylamine, <mathjax>#"CH"_3"NH"_2#</mathjax>, derived from ammonia, <mathjax>#"NH"_3"#</mathjax>, in which one <mathjax>#"H"#</mathjax> atom is replaced by a methyl group <mathjax>#"-CH"_3"#</mathjax>. However, we can't know this for certain unless we are given the molecular mass.</p>
<p><img alt="https://en.wikipedia.org/wiki/Methylamine" src="https://useruploads.socratic.org/54q1ZVq8TCmAE4RJ2V9p_100px-Methylamine.svg.png"/> </p></div>
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</article> | What is the empirical formula for a compound containing 38.8 g carbon, 16.2 g hydrogen and 45.1 g nitrogen? | null |
2,831 | a85c1a68-6ddd-11ea-b108-ccda262736ce | https://socratic.org/questions/a-farmer-has-100-gallons-of-70-pure-disinfectant-he-wishes-to-mix-it-with-the-di | 33.3 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 3 4 volume qc_end physical_unit 8 8 6 6 percent qc_end physical_unit 8 8 19 19 percent qc_end physical_unit 8 8 23 23 percent qc_end end | [{"type":"physical unit","value":"Volume2 [OF] disinfectant [IN] L"}] | [{"type":"physical unit","value":"33.3 L"}] | [{"type":"physical unit","value":"Volume1 [OF] disinfectant [=] \\pu{100 gallons}"},{"type":"physical unit","value":"Percent1 [OF] disinfectant in solution [=] \\pu{70%}"},{"type":"physical unit","value":"Percent2 [OF] disinfectant in solution [=] \\pu{90%}"},{"type":"physical unit","value":"Percent3 [OF] disinfectant in solution [=] \\pu{75%}"}] | <h1 class="questionTitle" itemprop="name">A farmer has 100 gallons of 70% pure disinfectant. He wishes to mix it with the disinfectant which is 90% pure to obtain 75% pure disinfectant. How much of the 90% disinfectant must he use?</h1> | null | 33.3 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Amount of substance = concentration x volume</p>
<p>If <mathjax>#V#</mathjax> is the volume of the 90% disinfectant then we can set up this equation:</p>
<p><mathjax>#(100xx70)+(90xxV)=75(100+V)#</mathjax></p>
<p><mathjax>#:.7000+90V=7500+75V#</mathjax></p>
<p><mathjax>#:.90V-75V=7500-7000#</mathjax></p>
<p><mathjax>#:.15V=500#</mathjax></p>
<p><mathjax>#:.V=500/15=33.33"L"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#33.3"L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Amount of substance = concentration x volume</p>
<p>If <mathjax>#V#</mathjax> is the volume of the 90% disinfectant then we can set up this equation:</p>
<p><mathjax>#(100xx70)+(90xxV)=75(100+V)#</mathjax></p>
<p><mathjax>#:.7000+90V=7500+75V#</mathjax></p>
<p><mathjax>#:.90V-75V=7500-7000#</mathjax></p>
<p><mathjax>#:.15V=500#</mathjax></p>
<p><mathjax>#:.V=500/15=33.33"L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A farmer has 100 gallons of 70% pure disinfectant. He wishes to mix it with the disinfectant which is 90% pure to obtain 75% pure disinfectant. How much of the 90% disinfectant must he use?</h1>
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<div class="markdown"><p><mathjax>#33.3"L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Amount of substance = concentration x volume</p>
<p>If <mathjax>#V#</mathjax> is the volume of the 90% disinfectant then we can set up this equation:</p>
<p><mathjax>#(100xx70)+(90xxV)=75(100+V)#</mathjax></p>
<p><mathjax>#:.7000+90V=7500+75V#</mathjax></p>
<p><mathjax>#:.90V-75V=7500-7000#</mathjax></p>
<p><mathjax>#:.15V=500#</mathjax></p>
<p><mathjax>#:.V=500/15=33.33"L"#</mathjax></p></div>
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</article> | A farmer has 100 gallons of 70% pure disinfectant. He wishes to mix it with the disinfectant which is 90% pure to obtain 75% pure disinfectant. How much of the 90% disinfectant must he use? | null |
2,832 | a9af5192-6ddd-11ea-9c51-ccda262736ce | https://socratic.org/questions/solid-arsenic-reacts-with-oxygen-gas-to-synthesize-solid-arsenic-trioxide-how-ma | 2.26 moles | start physical_unit 9 10 mole mol qc_end physical_unit 4 5 21 22 mole qc_end substance 0 1 qc_end end | [{"type":"physical unit","value":"Mole [OF] arsenic trioxide [IN] moles"}] | [{"type":"physical unit","value":"2.26 moles"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{3.4 moles}"},{"type":"substance name","value":"Solid arsenic"}] | <h1 class="questionTitle" itemprop="name">Solid arsenic reacts with Oxygen gas to synthesize solid arsenic trioxide. How many moles of arsenic trioxide will be produced from 3.4 moles of oxygen gas? </h1> | null | 2.26 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#4As#</mathjax> + <mathjax># 3O_2#</mathjax> -----> <mathjax>#2As_2O_3#</mathjax></p>
<p>According to equation<br/>
<mathjax>#3#</mathjax> moles of oxygen gas is required to form arsenic trioxide = <mathjax>#2#</mathjax> moles<br/>
<mathjax>#3.4#</mathjax> moles of O2 will be required to form Arsenic trioxide = (<mathjax>#2#</mathjax> x <mathjax>#3.4#</mathjax>) / <mathjax>#3#</mathjax><br/>
= <mathjax>#2.26#</mathjax> moles</p></div>
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<div class="markdown"><p><mathjax>#2As_2O_3#</mathjax> = <mathjax>#2.26#</mathjax> moles</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#4As#</mathjax> + <mathjax># 3O_2#</mathjax> -----> <mathjax>#2As_2O_3#</mathjax></p>
<p>According to equation<br/>
<mathjax>#3#</mathjax> moles of oxygen gas is required to form arsenic trioxide = <mathjax>#2#</mathjax> moles<br/>
<mathjax>#3.4#</mathjax> moles of O2 will be required to form Arsenic trioxide = (<mathjax>#2#</mathjax> x <mathjax>#3.4#</mathjax>) / <mathjax>#3#</mathjax><br/>
= <mathjax>#2.26#</mathjax> moles</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Solid arsenic reacts with Oxygen gas to synthesize solid arsenic trioxide. How many moles of arsenic trioxide will be produced from 3.4 moles of oxygen gas? </h1>
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Sarita Rana
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Aug 19, 2017
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<div class="markdown"><p><mathjax>#2As_2O_3#</mathjax> = <mathjax>#2.26#</mathjax> moles</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#4As#</mathjax> + <mathjax># 3O_2#</mathjax> -----> <mathjax>#2As_2O_3#</mathjax></p>
<p>According to equation<br/>
<mathjax>#3#</mathjax> moles of oxygen gas is required to form arsenic trioxide = <mathjax>#2#</mathjax> moles<br/>
<mathjax>#3.4#</mathjax> moles of O2 will be required to form Arsenic trioxide = (<mathjax>#2#</mathjax> x <mathjax>#3.4#</mathjax>) / <mathjax>#3#</mathjax><br/>
= <mathjax>#2.26#</mathjax> moles</p></div>
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</article> | Solid arsenic reacts with Oxygen gas to synthesize solid arsenic trioxide. How many moles of arsenic trioxide will be produced from 3.4 moles of oxygen gas? | null |
2,833 | aa7e1294-6ddd-11ea-b36b-ccda262736ce | https://socratic.org/questions/when-1942-j-of-heat-energy-is-added-to-47-4-g-of-ethanol-c-2h-6o-the-temperature | 112.90 J/(mol * ℃) | start physical_unit 13 13 molar_heat j/(mol_·_°c) qc_end physical_unit 13 13 1 2 heat_energy qc_end physical_unit 13 13 9 10 mass qc_end physical_unit 13 13 18 19 temperature qc_end end | [{"type":"physical unit","value":"Molar heat capacity [OF] C2H6O [IN] J/(mol * ℃)"}] | [{"type":"physical unit","value":"112.90 J/(mol * ℃)"}] | [{"type":"physical unit","value":"Added heat energy [OF] C2H6O [=] \\pu{1942 J}"},{"type":"physical unit","value":"Mass [OF] C2H6O [=] \\pu{47.4 g}"},{"type":"physical unit","value":"Increased temperature [OF] C2H6O [=] \\pu{16.7 ℃}"}] | <h1 class="questionTitle" itemprop="name">When 1942 J of heat energy is added to 47.4 g of ethanol, #C_2H_6O#, the temperature increases by 16.7 °C. What is the molar heat capacity of #C_2H_6O#?</h1> | null | 112.90 J/(mol * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ethanol has MM=46,06 g/mol, so you have <mathjax>#n= (47,4g)/ (46,6 g/(mol)) = 1,030 mol #</mathjax> of ethanol. Since Q = n Cm (Tf-Ti) , the molar heat capacity of ethanol is<br/>
<mathjax># Cm = Q/ (n(Tf-Ti)) = (1942 J)/ (1,030 mol 16,7 °C) = 112 J/(mol °C)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#112 J/(mol °C)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ethanol has MM=46,06 g/mol, so you have <mathjax>#n= (47,4g)/ (46,6 g/(mol)) = 1,030 mol #</mathjax> of ethanol. Since Q = n Cm (Tf-Ti) , the molar heat capacity of ethanol is<br/>
<mathjax># Cm = Q/ (n(Tf-Ti)) = (1942 J)/ (1,030 mol 16,7 °C) = 112 J/(mol °C)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When 1942 J of heat energy is added to 47.4 g of ethanol, #C_2H_6O#, the temperature increases by 16.7 °C. What is the molar heat capacity of #C_2H_6O#?</h1>
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<div class="markdown"><p><mathjax>#112 J/(mol °C)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>ethanol has MM=46,06 g/mol, so you have <mathjax>#n= (47,4g)/ (46,6 g/(mol)) = 1,030 mol #</mathjax> of ethanol. Since Q = n Cm (Tf-Ti) , the molar heat capacity of ethanol is<br/>
<mathjax># Cm = Q/ (n(Tf-Ti)) = (1942 J)/ (1,030 mol 16,7 °C) = 112 J/(mol °C)#</mathjax></p></div>
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</article> | When 1942 J of heat energy is added to 47.4 g of ethanol, #C_2H_6O#, the temperature increases by 16.7 °C. What is the molar heat capacity of #C_2H_6O#? | null |
2,834 | a8f6bd06-6ddd-11ea-b428-ccda262736ce | https://socratic.org/questions/5997244911ef6b367b8c251f | 1.20 x 10^22 | start physical_unit 2 2 number none qc_end physical_unit 7 10 14 15 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"1.20 x 10^22"}] | [{"type":"physical unit","value":"Volume [OF] dioxygen gas sample [=] \\pu{0.448 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many electrons are found in a sample of dioxygen gas contained in a #"0.448 L"# container at STP?</h1> | null | 1.20 x 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The no of electrons present in448 ml of O2 at STP <br/>
At STP conditions 1 mole of O2 gas = <mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons</p>
<p><mathjax>#1#</mathjax> mole = <mathjax>#22.4#</mathjax> L<br/>
So, <mathjax>#22.4#</mathjax> L of <mathjax>#O_2#</mathjax> = (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)<br/>
<mathjax>#0.448#</mathjax> L of <mathjax>#O_2 #</mathjax>= (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)x (<mathjax>#0.448#</mathjax>)]/<mathjax>#22.4#</mathjax>L<br/>
= <mathjax>#1.20#</mathjax> x <mathjax>#10^22#</mathjax> electrons</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#1.20#</mathjax> x <mathjax>#10^22#</mathjax> electrons</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The no of electrons present in448 ml of O2 at STP <br/>
At STP conditions 1 mole of O2 gas = <mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons</p>
<p><mathjax>#1#</mathjax> mole = <mathjax>#22.4#</mathjax> L<br/>
So, <mathjax>#22.4#</mathjax> L of <mathjax>#O_2#</mathjax> = (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)<br/>
<mathjax>#0.448#</mathjax> L of <mathjax>#O_2 #</mathjax>= (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)x (<mathjax>#0.448#</mathjax>)]/<mathjax>#22.4#</mathjax>L<br/>
= <mathjax>#1.20#</mathjax> x <mathjax>#10^22#</mathjax> electrons</p></div>
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<h1 class="questionTitle" itemprop="name">How many electrons are found in a sample of dioxygen gas contained in a #"0.448 L"# container at STP?</h1>
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<div class="markdown"><p><mathjax>#1.20#</mathjax> x <mathjax>#10^22#</mathjax> electrons</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The no of electrons present in448 ml of O2 at STP <br/>
At STP conditions 1 mole of O2 gas = <mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons</p>
<p><mathjax>#1#</mathjax> mole = <mathjax>#22.4#</mathjax> L<br/>
So, <mathjax>#22.4#</mathjax> L of <mathjax>#O_2#</mathjax> = (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)<br/>
<mathjax>#0.448#</mathjax> L of <mathjax>#O_2 #</mathjax>= (<mathjax>#6.023#</mathjax> x <mathjax>#10^23#</mathjax> electrons)x (<mathjax>#0.448#</mathjax>)]/<mathjax>#22.4#</mathjax>L<br/>
= <mathjax>#1.20#</mathjax> x <mathjax>#10^22#</mathjax> electrons</p></div>
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Sunayan S
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<div class="markdown"><p>See.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#22400ml#</mathjax> <mathjax>#O_2#</mathjax> contains <mathjax>#16×N_A#</mathjax> electrons<br/>
<mathjax>#1ml#</mathjax> <mathjax>#O_2#</mathjax> contains <mathjax>#(16×N_A)/22400#</mathjax>electrons<br/>
Hence,<mathjax>#448ml#</mathjax> contains <mathjax>#(16×N_A×448)/22400#</mathjax> electrons=<mathjax>#0.32×N_A#</mathjax><br/>
Here, <mathjax>#N_A#</mathjax> is Avogadro number.</p></div>
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<div class="markdown"><p>About <mathjax>#1.93 xx 10^23#</mathjax> electrons.</p>
<hr/>
<p>Assuming your definition of STP is <mathjax>#0^@ "C"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the molar volume of <mathjax>#"O"_2#</mathjax>, assuming it is an ideal gas, is based on the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<blockquote>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>for pressure <mathjax>#P#</mathjax>, volume <mathjax>#V#</mathjax>, mols <mathjax>#n#</mathjax>, temperature <mathjax>#T#</mathjax>, and universal gas constant <mathjax>#R#</mathjax>.</p>
</blockquote>
<p>Rearranging, we get...</p>
<blockquote>
<p><mathjax>#V/n = (RT)/P#</mathjax></p>
<p><mathjax>#= (("0.082057 L"cdot"atm/mol"cdot"K")("273.15 K"))/"1 atm"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#"22.414 L/mol"#</mathjax></p>
</blockquote>
<p>So, with this molar volume, we compare ratios:</p>
<blockquote>
<p><mathjax>#"0.448 L"/("x mols") = "22.414 L"/"mol"#</mathjax></p>
<p><mathjax>#=> n_("O"_2) = "0.0200 mols"#</mathjax></p>
</blockquote>
<p>Each <mathjax>#"O"#</mathjax> atom contains <mathjax>#8#</mathjax> electrons, so each <mathjax>#"O"_2#</mathjax> molecule contains... well, <mathjax>#16#</mathjax>.</p>
<blockquote>
<p><mathjax>#"0.0200 mols O"_2 harr "0.320 mols e"^(-)#</mathjax></p>
</blockquote>
<p>And thus, the number of electrons is given by</p>
<blockquote>
<p><mathjax>#color(blue)("Number of electrons") = "0.320 mols e"^(-) xx 6.0221413 xx 10^(23) "mol"^(-1)#</mathjax></p>
<p><mathjax>#= ulcolor(blue)(1.93 xx 10^(23)color(white)(.)"e"^(-))#</mathjax></p>
</blockquote></div>
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</article> | How many electrons are found in a sample of dioxygen gas contained in a #"0.448 L"# container at STP? | null |
2,835 | ab27b3ca-6ddd-11ea-8ad2-ccda262736ce | https://socratic.org/questions/chlorine-reacts-with-hot-sodium-hydroxide-solution-a-sample-of-a-compound-formed | NaClO3 | start chemical_formula qc_end physical_unit 23 24 20 21 mass qc_end physical_unit 28 29 25 26 mass qc_end physical_unit 34 35 31 32 mass qc_end substance 0 0 qc_end substance 4 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"NaClO3"}] | [{"type":"physical unit","value":"Mass [OF] sodium atoms [=] \\pu{10.8 g}"},{"type":"physical unit","value":"Mass [OF] chlorine atoms [=] \\pu{16.7 g}"},{"type":"physical unit","value":"Mass [OF] oxygen atoms [=] \\pu{22.5 g}"},{"type":"substance name","value":"Chlorine"},{"type":"substance name","value":"Sodium hydroxide solution"}] | <h1 class="questionTitle" itemprop="name">Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed?</h1> | null | NaClO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol#</mathjax></p>
<p>We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of <mathjax>#NaClO_3#</mathjax>, i.e. <mathjax>#"sodium chlorate"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#NaClO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol#</mathjax></p>
<p>We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of <mathjax>#NaClO_3#</mathjax>, i.e. <mathjax>#"sodium chlorate"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed?</h1>
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anor277
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<div class="markdown"><p><mathjax>#NaClO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol#</mathjax></p>
<p>We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of <mathjax>#NaClO_3#</mathjax>, i.e. <mathjax>#"sodium chlorate"#</mathjax>.</p></div>
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<div class="markdown"><p>The empirical formula is <mathjax>#"NaClO"_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The empirical formula for a compound represents the lowest whole number ratio of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. In order to determine the empirical formula of a compound, you must find the number of moles of each element. Once you know the moles, you divide the moles of each element by the smallest number of moles. This will give you the empirical formula, unless you do not have whole numbers. If you don't have all whole numbers, you need to multiply by a number that will result in all whole numbers.</p>
<p><strong>Moles of Each Element</strong></p>
<p>Divide the given mass of each element by its molar mass (atomic weight on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol) by multiplying by its inverse.</p>
<p><mathjax>#"Na":#</mathjax><mathjax>#10.8color(red)cancel(color(black)("g Na"))xx(1"mol Na")/(22.990color(red)cancel(color(black)("g Na")))="0.470 mol Na"#</mathjax></p>
<p><mathjax>#"Cl":#</mathjax><mathjax>#16.7color(red)cancel(color(black)("g Cl"))xx(1"mol Cl")/(35.453color(red)cancel(color(black)("g Cl")))="0.471 mol Cl"#</mathjax></p>
<p><mathjax>#"O":#</mathjax><mathjax>#22.5color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="1.41 mol O"#</mathjax></p>
<p><strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">Mole Ratios</a> (Subscripts)</strong></p>
<p>Divide the moles of each element by the least number of moles.</p>
<p><mathjax>#"Na":#</mathjax><mathjax>#(0.470color(red)cancel(color(black)("mol")))/(0.470color(red)cancel(color(black)("mol")))=1.00#</mathjax></p>
<p><mathjax>#"Cl":#</mathjax><mathjax>#(0.471color(red)cancel(color(black)("mol")))/(0.470color(red)cancel(color(black)("mol")))=1.00#</mathjax></p>
<p><mathjax>#"O":#</mathjax><mathjax>#(1.41color(red)cancel(color(black)("mol")))/(0.470color(red)cancel(color(black)("mol")))=3.00#</mathjax></p>
<p>Since the mole ratios are all whole numbers, you do not need to do anything else except write the formula.</p>
<p>The empirical formula is <mathjax>#"NaClO"_3#</mathjax>, which is the ionic compound sodium chlorate, composed of the <mathjax>#"Na"^(2+)#</mathjax> sodium cation and the <mathjax>#"ClO"_3"^(1-)#</mathjax> chlorate anion.</p></div>
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</article> | Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed? | null |
2,836 | ac3d28cc-6ddd-11ea-8f2f-ccda262736ce | https://socratic.org/questions/niacin-c-6nh-5o-2-reacts-with-o-2-to-produce-co-2-h-2o-and-no-2-how-would-you-wr | 4 C6NH5O2 + 29 O2 -> 24 CO2 + 10 H2O + 4 NO2 | start chemical_equation qc_end chemical_equation 1 1 qc_end chemical_equation 4 4 qc_end chemical_equation 7 7 qc_end chemical_equation 8 8 qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"4 C6NH5O2 + 29 O2 -> 24 CO2 + 10 H2O + 4 NO2"}] | [{"type":"chemical equation","value":"C6NH5O2"},{"type":"chemical equation","value":"O2"},{"type":"chemical equation","value":"CO2"},{"type":"chemical equation","value":"H2O"},{"type":"chemical equation","value":"NO2"}] | <h1 class="questionTitle" itemprop="name">Niacin (#C_6NH_5O_2#) reacts with #O_2# to produce #CO_2#, #H_2O#, and #NO_2#. How would you write and balance the equation for this reaction?</h1> | null | 4 C6NH5O2 + 29 O2 -> 24 CO2 + 10 H2O + 4 NO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>After adjusting the no of atoms of 6C,1N and 5 H in RHS with LHS ,we get total <mathjax>#(14+5/2) #</mathjax> O at RHS.<br/>
To adjust the no of O atom in <br/>
LHS we can write <mathjax>#(6+5/4) #</mathjax>before <mathjax>#O_2#</mathjax> in LHS and the balanced equation becomes.</p>
<p><mathjax>#C_6NH_5O_2+(6+5/4)O_2->6CO_2+5/2H_2O+NO_2#</mathjax></p>
<p>Finally multiplying by 4 on both sides to remove fraction,we get the balanced equation as follows.</p>
<p><mathjax>#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>After adjusting the no of atoms of 6C,1N and 5 H in RHS with LHS ,we get total <mathjax>#(14+5/2) #</mathjax> O at RHS.<br/>
To adjust the no of O atom in <br/>
LHS we can write <mathjax>#(6+5/4) #</mathjax>before <mathjax>#O_2#</mathjax> in LHS and the balanced equation becomes.</p>
<p><mathjax>#C_6NH_5O_2+(6+5/4)O_2->6CO_2+5/2H_2O+NO_2#</mathjax></p>
<p>Finally multiplying by 4 on both sides to remove fraction,we get the balanced equation as follows.</p>
<p><mathjax>#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Niacin (#C_6NH_5O_2#) reacts with #O_2# to produce #CO_2#, #H_2O#, and #NO_2#. How would you write and balance the equation for this reaction?</h1>
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<div class="markdown"><p><mathjax>#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>After adjusting the no of atoms of 6C,1N and 5 H in RHS with LHS ,we get total <mathjax>#(14+5/2) #</mathjax> O at RHS.<br/>
To adjust the no of O atom in <br/>
LHS we can write <mathjax>#(6+5/4) #</mathjax>before <mathjax>#O_2#</mathjax> in LHS and the balanced equation becomes.</p>
<p><mathjax>#C_6NH_5O_2+(6+5/4)O_2->6CO_2+5/2H_2O+NO_2#</mathjax></p>
<p>Finally multiplying by 4 on both sides to remove fraction,we get the balanced equation as follows.</p>
<p><mathjax>#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#</mathjax></p></div>
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</article> | Niacin (#C_6NH_5O_2#) reacts with #O_2# to produce #CO_2#, #H_2O#, and #NO_2#. How would you write and balance the equation for this reaction? | null |
2,837 | aa8676c0-6ddd-11ea-a995-ccda262736ce | https://socratic.org/questions/what-is-the-equilibrium-constant-at-2000-c-for-the-reaction-n2-g-o2-g-2no-g-is-4 | 3.62 mol/L | start physical_unit 5 5 equilibrium_concentration mol/l qc_end physical_unit 11 11 13 13 equilibrium_concentration qc_end physical_unit 17 17 19 20 equilibrium_concentration qc_end physical_unit 28 29 37 39 equilibrium_constant_k qc_end physical_unit 28 29 25 26 temperature qc_end chemical_equation 30 35 qc_end end | [{"type":"physical unit","value":"Equilibrium concentration [OF] NO [IN] mol/L"}] | [{"type":"physical unit","value":"3.62 mol/L"}] | [{"type":"physical unit","value":"Equilibrium concentration [OF] N2 [=] \\pu{0.036mol/L}"},{"type":"physical unit","value":"Equilibrium concentration [OF] O2 [=] \\pu{0.0089 mol/L}"},{"type":"physical unit","value":"Equilibrium constant [OF] the reaction [=] \\pu{4.1 × 10^4}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{2000 ℃}"},{"type":"chemical equation","value":"N2(g) + O2(g) -> 2 NO(g)"}] | <h1 class="questionTitle" itemprop="name">Calculate the equilibrium concentration of #"NO"# if the equilibrium concentration of #"N"_2# is #"0.036mol L"^(-1)# and that of #"O"_2# is #"0.0089 mol L"^(-1)# ?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What is the equilibrium constant at <mathjax>#2000^@"C"#</mathjax> for the reaction </p>
<p><mathjax>#"N"_2(g) + "O"_2(g)-> 2"NO"(g)#</mathjax> is <mathjax>#4.1*10^4#</mathjax>.</p></div>
</h2>
</div>
</div> | 3.62 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that at <mathjax>#2000^@"C"#</mathjax>, the reaction</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + "O"_ (2(g)) -> 2"NO"_ ((g))#</mathjax></p>
</blockquote>
<p>has an <strong>equilibrium constant</strong> equal to</p>
<blockquote>
<p><mathjax>#K_c= 4.1 * 10^4#</mathjax></p>
</blockquote>
<p>Now, the equilibrium constant is calculated by taking the <strong>equilibrium concentrations</strong> of the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#K_c = (["NO"]^2)/(["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>At equilibrium, you have</p>
<blockquote>
<ul>
<li><mathjax>#["N"_2] = "0.036 mol L"^(-1)#</mathjax></li>
<li><mathjax>#["O"_2] = "0.0089 mol L"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Rearrange the equation to isolate the equilibrium concentration of nitric oxide.</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt( K_c * ["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt(4.1 * 10^4 * "0.036 mol L"^(-1) * "0.0089 mol L"^(-1))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(["NO"] = "3.6 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>Notice that the equilibrium concentration of nitric oxide is significantly higher than the equilibrium concentrations of the nitrogen gas and the oxygen gas.</p>
<p>This is the case because you have</p>
<blockquote>
<p><mathjax>#K_c ">>" 1#</mathjax></p>
</blockquote>
<p>which implies that the equilibrium <strong>lies to the right</strong>, i.e. at this temperature, the forward reaction is favored. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3.6 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that at <mathjax>#2000^@"C"#</mathjax>, the reaction</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + "O"_ (2(g)) -> 2"NO"_ ((g))#</mathjax></p>
</blockquote>
<p>has an <strong>equilibrium constant</strong> equal to</p>
<blockquote>
<p><mathjax>#K_c= 4.1 * 10^4#</mathjax></p>
</blockquote>
<p>Now, the equilibrium constant is calculated by taking the <strong>equilibrium concentrations</strong> of the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#K_c = (["NO"]^2)/(["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>At equilibrium, you have</p>
<blockquote>
<ul>
<li><mathjax>#["N"_2] = "0.036 mol L"^(-1)#</mathjax></li>
<li><mathjax>#["O"_2] = "0.0089 mol L"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Rearrange the equation to isolate the equilibrium concentration of nitric oxide.</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt( K_c * ["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt(4.1 * 10^4 * "0.036 mol L"^(-1) * "0.0089 mol L"^(-1))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(["NO"] = "3.6 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>Notice that the equilibrium concentration of nitric oxide is significantly higher than the equilibrium concentrations of the nitrogen gas and the oxygen gas.</p>
<p>This is the case because you have</p>
<blockquote>
<p><mathjax>#K_c ">>" 1#</mathjax></p>
</blockquote>
<p>which implies that the equilibrium <strong>lies to the right</strong>, i.e. at this temperature, the forward reaction is favored. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Calculate the equilibrium concentration of #"NO"# if the equilibrium concentration of #"N"_2# is #"0.036mol L"^(-1)# and that of #"O"_2# is #"0.0089 mol L"^(-1)# ?</h1>
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<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What is the equilibrium constant at <mathjax>#2000^@"C"#</mathjax> for the reaction </p>
<p><mathjax>#"N"_2(g) + "O"_2(g)-> 2"NO"(g)#</mathjax> is <mathjax>#4.1*10^4#</mathjax>.</p></div>
</h2>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-12T00:06:36" itemprop="dateCreated">
Feb 12, 2018
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<div class="markdown"><p><mathjax>#"3.6 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that at <mathjax>#2000^@"C"#</mathjax>, the reaction</p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + "O"_ (2(g)) -> 2"NO"_ ((g))#</mathjax></p>
</blockquote>
<p>has an <strong>equilibrium constant</strong> equal to</p>
<blockquote>
<p><mathjax>#K_c= 4.1 * 10^4#</mathjax></p>
</blockquote>
<p>Now, the equilibrium constant is calculated by taking the <strong>equilibrium concentrations</strong> of the chemical species involved in the reaction. </p>
<blockquote>
<p><mathjax>#K_c = (["NO"]^2)/(["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>At equilibrium, you have</p>
<blockquote>
<ul>
<li><mathjax>#["N"_2] = "0.036 mol L"^(-1)#</mathjax></li>
<li><mathjax>#["O"_2] = "0.0089 mol L"^(-1)#</mathjax></li>
</ul>
</blockquote>
<p>Rearrange the equation to isolate the equilibrium concentration of nitric oxide.</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt( K_c * ["N"_2] * ["O"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#["NO"] = sqrt(4.1 * 10^4 * "0.036 mol L"^(-1) * "0.0089 mol L"^(-1))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(["NO"] = "3.6 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>Notice that the equilibrium concentration of nitric oxide is significantly higher than the equilibrium concentrations of the nitrogen gas and the oxygen gas.</p>
<p>This is the case because you have</p>
<blockquote>
<p><mathjax>#K_c ">>" 1#</mathjax></p>
</blockquote>
<p>which implies that the equilibrium <strong>lies to the right</strong>, i.e. at this temperature, the forward reaction is favored. </p></div>
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</article> | Calculate the equilibrium concentration of #"NO"# if the equilibrium concentration of #"N"_2# is #"0.036mol L"^(-1)# and that of #"O"_2# is #"0.0089 mol L"^(-1)# ? |
What is the equilibrium constant at #2000^@"C"# for the reaction
#"N"_2(g) + "O"_2(g)-> 2"NO"(g)# is #4.1*10^4#.
|
2,838 | ac8b66be-6ddd-11ea-8473-ccda262736ce | https://socratic.org/questions/a-sample-of-sulfur-hexafluoride-gas-occupies-8-62-l-at-221-degrees-celsius-assum | -123.57 Celsius | start physical_unit 1 5 temperature °c qc_end physical_unit 1 5 7 8 volume qc_end physical_unit 1 5 10 12 temperature qc_end physical_unit 1 5 30 31 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] sulfur hexafluoride gas sample [IN] Celsius"}] | [{"type":"physical unit","value":"-123.57 Celsius"}] | [{"type":"physical unit","value":"Volume1 [OF] sulfur hexafluoride gas sample [=] \\pu{8.62 L}"},{"type":"physical unit","value":"Temperature1 [OF] sulfur hexafluoride gas sample [=] \\pu{221 degrees Celsius}"},{"type":"physical unit","value":"Volume2 [OF] sulfur hexafluoride gas sample [=] \\pu{2.61 L}"},{"type":"other","value":"Assuming that the pressure remains constant."}] | <h1 class="questionTitle" itemprop="name">A sample of sulfur hexafluoride gas occupies 8.62 L at 221 degrees Celsius. Assuming that the pressure remains constant, what temperature (in Celsius) is needed to reduce the volume to 2.61 L?</h1> | null | -123.57 Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your variables, <mathjax>#"L"#</mathjax> and <mathjax>#""^@"C"#</mathjax> represent volume and temperature. The gas law that involves the relationship between the volume of a gas and its temperature is <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>. <a href="http://chemistry.bd.psu.edu/jircitano/gases.html" rel="nofollow" target="_blank">http://chemistry.bd.psu.edu/jircitano/gases.html</a></p>
<p>Charles' law states that the volume of a given amount of gas held at constant pressure is directly proportional to its Kelvin temperature. This means that as the temperature increases, the volume also increases, and vice-versa. The equation used to solve gas problems involving Charles' law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, </p>
<p>where the temperatures must be in Kelvins. Add <mathjax>#273.15#</mathjax> to the Celsius temperature to get the Kelvin temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="8.62 L"#</mathjax></p>
<p><mathjax>#T_1="221"^@"C" + 273.15="494 K"#</mathjax></p>
<p><mathjax>#V_2="2.61 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation given above to isolate <mathjax>#"T_2#</mathjax>. Insert your data into the new equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(2.61color(red)cancel(color(black)("L"))xx494"K")/(8.62color(red)cancel(color(black)("L")))="150. K"#</mathjax></p>
<p>To convert from Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#"150. K"-273.15"="-123^@"C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The temperature needed to reduce the volume from <mathjax>#"8.62 L"#</mathjax> to <mathjax>#"2.61 L"#</mathjax> is <mathjax>#-123^@"C"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your variables, <mathjax>#"L"#</mathjax> and <mathjax>#""^@"C"#</mathjax> represent volume and temperature. The gas law that involves the relationship between the volume of a gas and its temperature is <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>. <a href="http://chemistry.bd.psu.edu/jircitano/gases.html" rel="nofollow" target="_blank">http://chemistry.bd.psu.edu/jircitano/gases.html</a></p>
<p>Charles' law states that the volume of a given amount of gas held at constant pressure is directly proportional to its Kelvin temperature. This means that as the temperature increases, the volume also increases, and vice-versa. The equation used to solve gas problems involving Charles' law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, </p>
<p>where the temperatures must be in Kelvins. Add <mathjax>#273.15#</mathjax> to the Celsius temperature to get the Kelvin temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="8.62 L"#</mathjax></p>
<p><mathjax>#T_1="221"^@"C" + 273.15="494 K"#</mathjax></p>
<p><mathjax>#V_2="2.61 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation given above to isolate <mathjax>#"T_2#</mathjax>. Insert your data into the new equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(2.61color(red)cancel(color(black)("L"))xx494"K")/(8.62color(red)cancel(color(black)("L")))="150. K"#</mathjax></p>
<p>To convert from Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#"150. K"-273.15"="-123^@"C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of sulfur hexafluoride gas occupies 8.62 L at 221 degrees Celsius. Assuming that the pressure remains constant, what temperature (in Celsius) is needed to reduce the volume to 2.61 L?</h1>
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<div class="markdown"><p>The temperature needed to reduce the volume from <mathjax>#"8.62 L"#</mathjax> to <mathjax>#"2.61 L"#</mathjax> is <mathjax>#-123^@"C"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your variables, <mathjax>#"L"#</mathjax> and <mathjax>#""^@"C"#</mathjax> represent volume and temperature. The gas law that involves the relationship between the volume of a gas and its temperature is <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>. <a href="http://chemistry.bd.psu.edu/jircitano/gases.html" rel="nofollow" target="_blank">http://chemistry.bd.psu.edu/jircitano/gases.html</a></p>
<p>Charles' law states that the volume of a given amount of gas held at constant pressure is directly proportional to its Kelvin temperature. This means that as the temperature increases, the volume also increases, and vice-versa. The equation used to solve gas problems involving Charles' law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, </p>
<p>where the temperatures must be in Kelvins. Add <mathjax>#273.15#</mathjax> to the Celsius temperature to get the Kelvin temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Known</strong></p>
<p><mathjax>#V_1="8.62 L"#</mathjax></p>
<p><mathjax>#T_1="221"^@"C" + 273.15="494 K"#</mathjax></p>
<p><mathjax>#V_2="2.61 L"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#T_2=?#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation given above to isolate <mathjax>#"T_2#</mathjax>. Insert your data into the new equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(2.61color(red)cancel(color(black)("L"))xx494"K")/(8.62color(red)cancel(color(black)("L")))="150. K"#</mathjax></p>
<p>To convert from Kelvins to degrees Celsius, subtract <mathjax>#273.15#</mathjax> from the Kelvin temperature.</p>
<p><mathjax>#"150. K"-273.15"="-123^@"C"#</mathjax></p></div>
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</article> | A sample of sulfur hexafluoride gas occupies 8.62 L at 221 degrees Celsius. Assuming that the pressure remains constant, what temperature (in Celsius) is needed to reduce the volume to 2.61 L? | null |
2,839 | aa98edb4-6ddd-11ea-8e67-ccda262736ce | https://socratic.org/questions/what-is-the-amount-of-energy-needed-to-heat-55-0-g-of-from-45-c-to-67-c | 5.83 kJ | start physical_unit 12 12 energy kj qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 14 15 temperature qc_end physical_unit 12 12 17 18 temperature qc_end end | [{"type":"physical unit","value":"Needed energy [OF] water [IN] kJ"}] | [{"type":"physical unit","value":"5.83 kJ"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{55.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{45 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{67 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the amount of energy needed to heat 55.0 g of water from 45°C to 67°C? </h1> | null | 5.83 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation, </p>
<p><mathjax>#Q=mcdeltaT#</mathjax></p>
<p>where Q is heat flow (energy), m is mass, c is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capcity and <mathjax>#deltaT#</mathjax> is the change in tempreture (<mathjax>#T2-T1#</mathjax>)</p>
<p>The specific heat capacity of water is <mathjax>#(4.816J/(g*C))#</mathjax></p>
<p><mathjax>#Q=(55.0 g)*(4.816J/(g*C))*(67 C -45C)#</mathjax></p>
<p>Solve and the answer is, </p>
<p>5,827 J or 5.827 kJ</p>
<p>To 2 significant digits,</p>
<p>5.8 kJ</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>5.8 kJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation, </p>
<p><mathjax>#Q=mcdeltaT#</mathjax></p>
<p>where Q is heat flow (energy), m is mass, c is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capcity and <mathjax>#deltaT#</mathjax> is the change in tempreture (<mathjax>#T2-T1#</mathjax>)</p>
<p>The specific heat capacity of water is <mathjax>#(4.816J/(g*C))#</mathjax></p>
<p><mathjax>#Q=(55.0 g)*(4.816J/(g*C))*(67 C -45C)#</mathjax></p>
<p>Solve and the answer is, </p>
<p>5,827 J or 5.827 kJ</p>
<p>To 2 significant digits,</p>
<p>5.8 kJ</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the amount of energy needed to heat 55.0 g of water from 45°C to 67°C? </h1>
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<div class="markdown"><p>5.8 kJ</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation, </p>
<p><mathjax>#Q=mcdeltaT#</mathjax></p>
<p>where Q is heat flow (energy), m is mass, c is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capcity and <mathjax>#deltaT#</mathjax> is the change in tempreture (<mathjax>#T2-T1#</mathjax>)</p>
<p>The specific heat capacity of water is <mathjax>#(4.816J/(g*C))#</mathjax></p>
<p><mathjax>#Q=(55.0 g)*(4.816J/(g*C))*(67 C -45C)#</mathjax></p>
<p>Solve and the answer is, </p>
<p>5,827 J or 5.827 kJ</p>
<p>To 2 significant digits,</p>
<p>5.8 kJ</p></div>
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</article> | What is the amount of energy needed to heat 55.0 g of water from 45°C to 67°C? | null |
2,840 | a9b1ed22-6ddd-11ea-9f42-ccda262736ce | https://socratic.org/questions/580052b17c01495a8ef9cf2a | ClO3- + 3 SO2(g) + 6 OH- -> 3 SO4^2- + Cl- + 3 H2O | start chemical_equation qc_end chemical_equation 10 10 qc_end chemical_equation 12 12 qc_end chemical_equation 15 15 qc_end chemical_equation 17 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the redox reaction"}] | [{"type":"chemical equation","value":"ClO3- + 3 SO2(g) + 6 OH- -> 3 SO4^2- + Cl- + 3 H2O"}] | [{"type":"chemical equation","value":"ClO3-"},{"type":"chemical equation","value":"SO2(g)"},{"type":"chemical equation","value":"Cl-"},{"type":"chemical equation","value":"SO4^2-"}] | <h1 class="questionTitle" itemprop="name">How do you represent the redox reaction of chlorate ion, #ClO_3^(-)# with #SO_2(g)# to give #Cl^(-)# and #SO_4^(2-)#?</h1> | null | ClO3- + 3 SO2(g) + 6 OH- -> 3 SO4^2- + Cl- + 3 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chlorate <mathjax>#(Cl(V))#</mathjax> is reduced to chloride <mathjax>#(Cl^(-)(-I))#</mathjax>:</p>
<p><mathjax>#ClO_3^(-)+6H^+ +6e^(-) rarr Cl^(-) + 3H_2O.#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And sulfur dioxide, <mathjax>#(S(IV))#</mathjax>, is oxidized to sulfate, <mathjax>#(S(VI))#</mathjax>:</p>
<p><mathjax>#SO_2(aq) +2H_2O(l) rarr SO_4^(2-) + 2e^(-) + 4H^(+).#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced in each equation, so we may cross mulitply to remove the electrons:</p>
<p><mathjax>#(i)+3xx(ii)#</mathjax></p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +3H_2O(aq) rarr 3SO_4^(2-) + Cl^(-)+ 6H^+#</mathjax></p>
<p>Are charge and mass balanced here? Do not trust my arithmetic! </p>
<p>On other words, chlorate oxidizes sulfurous acid to sulfate. </p>
<p>Ah, but you specified that the reaction was performed in base. We have the process in acid. What do we do? Easy! Simply add <mathjax>#6xxOH^-#</mathjax> to each side of the equation <br/>
(i.e. <mathjax>#HO^(-)+H^(+) rarr H_2O#</mathjax>):</p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O#</mathjax></p>
<p>Charge and mass are balanced, the which is absolutely required. If you are unclear of the steps I took and why, ask, and someone will help you. </p></div>
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<div class="markdown"><p><mathjax>#ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chlorate <mathjax>#(Cl(V))#</mathjax> is reduced to chloride <mathjax>#(Cl^(-)(-I))#</mathjax>:</p>
<p><mathjax>#ClO_3^(-)+6H^+ +6e^(-) rarr Cl^(-) + 3H_2O.#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And sulfur dioxide, <mathjax>#(S(IV))#</mathjax>, is oxidized to sulfate, <mathjax>#(S(VI))#</mathjax>:</p>
<p><mathjax>#SO_2(aq) +2H_2O(l) rarr SO_4^(2-) + 2e^(-) + 4H^(+).#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced in each equation, so we may cross mulitply to remove the electrons:</p>
<p><mathjax>#(i)+3xx(ii)#</mathjax></p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +3H_2O(aq) rarr 3SO_4^(2-) + Cl^(-)+ 6H^+#</mathjax></p>
<p>Are charge and mass balanced here? Do not trust my arithmetic! </p>
<p>On other words, chlorate oxidizes sulfurous acid to sulfate. </p>
<p>Ah, but you specified that the reaction was performed in base. We have the process in acid. What do we do? Easy! Simply add <mathjax>#6xxOH^-#</mathjax> to each side of the equation <br/>
(i.e. <mathjax>#HO^(-)+H^(+) rarr H_2O#</mathjax>):</p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O#</mathjax></p>
<p>Charge and mass are balanced, the which is absolutely required. If you are unclear of the steps I took and why, ask, and someone will help you. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you represent the redox reaction of chlorate ion, #ClO_3^(-)# with #SO_2(g)# to give #Cl^(-)# and #SO_4^(2-)#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chlorate <mathjax>#(Cl(V))#</mathjax> is reduced to chloride <mathjax>#(Cl^(-)(-I))#</mathjax>:</p>
<p><mathjax>#ClO_3^(-)+6H^+ +6e^(-) rarr Cl^(-) + 3H_2O.#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And sulfur dioxide, <mathjax>#(S(IV))#</mathjax>, is oxidized to sulfate, <mathjax>#(S(VI))#</mathjax>:</p>
<p><mathjax>#SO_2(aq) +2H_2O(l) rarr SO_4^(2-) + 2e^(-) + 4H^(+).#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced in each equation, so we may cross mulitply to remove the electrons:</p>
<p><mathjax>#(i)+3xx(ii)#</mathjax></p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +3H_2O(aq) rarr 3SO_4^(2-) + Cl^(-)+ 6H^+#</mathjax></p>
<p>Are charge and mass balanced here? Do not trust my arithmetic! </p>
<p>On other words, chlorate oxidizes sulfurous acid to sulfate. </p>
<p>Ah, but you specified that the reaction was performed in base. We have the process in acid. What do we do? Easy! Simply add <mathjax>#6xxOH^-#</mathjax> to each side of the equation <br/>
(i.e. <mathjax>#HO^(-)+H^(+) rarr H_2O#</mathjax>):</p>
<p><mathjax>#ClO_3^(-)+3SO_2(g) +6HO^(-) rarr 3SO_4^(2-) + Cl^(-) +3H_2O#</mathjax></p>
<p>Charge and mass are balanced, the which is absolutely required. If you are unclear of the steps I took and why, ask, and someone will help you. </p></div>
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</article> | How do you represent the redox reaction of chlorate ion, #ClO_3^(-)# with #SO_2(g)# to give #Cl^(-)# and #SO_4^(2-)#? | null |
2,841 | ab2d595e-6ddd-11ea-a116-ccda262736ce | https://socratic.org/questions/when-2211-j-of-heat-energy-is-added-to-47-7-g-of-ethanol-c-2h-6o-the-temperature | 113.03 J/(mol * ℃) | start physical_unit 13 13 molar_heat j/(mol_·_°c) qc_end physical_unit 13 13 1 2 heat_energy qc_end physical_unit 13 13 9 10 mass qc_end physical_unit 13 13 18 19 temperature qc_end end | [{"type":"physical unit","value":"Molar heat capacity [OF] C2H6O [IN] J/(mol * ℃)"}] | [{"type":"physical unit","value":"113.03 J/(mol * ℃)"}] | [{"type":"physical unit","value":"Added heat energy [OF] C2H6O [=] \\pu{2211 J}"},{"type":"physical unit","value":"Mass [OF] C2H6O [=] \\pu{47.7 g}"},{"type":"physical unit","value":"Increased temperature [OF] C2H6O [=] \\pu{18.9 ℃}"}] | <h1 class="questionTitle" itemprop="name">When 2211 J of heat energy is added to 47.7 g of ethanol, #C_2H_6O#, the temperature increases by 18.9°C. What is the molar heat capacity of #C_2H_6O#?</h1> | null | 113.03 J/(mol * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>A typical <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">calorimetry</a> formula for calculating the molar heat capacity of a substance is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul( q = ncΔT)|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li><mathjax>#q#</mathjax> = amount of heat transferred</li>
<li><mathjax>#n#</mathjax> = number of moles</li>
<li><mathjax>#c#</mathjax> = molar heat capacity</li>
<li><mathjax>#ΔT#</mathjax> = change in temperature</li>
</ul>
<p>We can rearrange this formula to</p>
<p><mathjax>#c = q/(nΔT)#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#q = "2211 J"#</mathjax></p>
<p><mathjax>#n = 47.7 color(red)(cancel(color(black)("g C"_2"H"_6"O"))) × ("1 mol C"_2"H"_6"O")/(46.07 color(red)(cancel(color(black)("g C"_2"H"_6"O")))) = "1.035 mol C"_2"H"_6"O"#</mathjax></p>
<p><mathjax>#ΔT = "19.9 °C"#</mathjax></p>
<blockquote></blockquote>
<p>Hence,</p>
<p><mathjax>#q = "2211 J"/("1.035 mol" × "19.9 °C") = "107 J·°C"^"-1""mol"^"-1"#</mathjax></p>
<p>The molar heat capacity of ethanol is <mathjax>#"107 J·°C"^"-1""mol"^"-1"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molar heat capacity of ethanol is <mathjax>#"107 J·°C"^"-1""mol"^"-1"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>A typical <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">calorimetry</a> formula for calculating the molar heat capacity of a substance is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul( q = ncΔT)|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li><mathjax>#q#</mathjax> = amount of heat transferred</li>
<li><mathjax>#n#</mathjax> = number of moles</li>
<li><mathjax>#c#</mathjax> = molar heat capacity</li>
<li><mathjax>#ΔT#</mathjax> = change in temperature</li>
</ul>
<p>We can rearrange this formula to</p>
<p><mathjax>#c = q/(nΔT)#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#q = "2211 J"#</mathjax></p>
<p><mathjax>#n = 47.7 color(red)(cancel(color(black)("g C"_2"H"_6"O"))) × ("1 mol C"_2"H"_6"O")/(46.07 color(red)(cancel(color(black)("g C"_2"H"_6"O")))) = "1.035 mol C"_2"H"_6"O"#</mathjax></p>
<p><mathjax>#ΔT = "19.9 °C"#</mathjax></p>
<blockquote></blockquote>
<p>Hence,</p>
<p><mathjax>#q = "2211 J"/("1.035 mol" × "19.9 °C") = "107 J·°C"^"-1""mol"^"-1"#</mathjax></p>
<p>The molar heat capacity of ethanol is <mathjax>#"107 J·°C"^"-1""mol"^"-1"#</mathjax>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">When 2211 J of heat energy is added to 47.7 g of ethanol, #C_2H_6O#, the temperature increases by 18.9°C. What is the molar heat capacity of #C_2H_6O#?</h1>
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Ernest Z.
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Mar 14, 2016
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<div class="markdown"><p>The molar heat capacity of ethanol is <mathjax>#"107 J·°C"^"-1""mol"^"-1"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>A typical <a href="http://socratic.org/chemistry/thermochemistry/calorimetry">calorimetry</a> formula for calculating the molar heat capacity of a substance is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul( q = ncΔT)|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li><mathjax>#q#</mathjax> = amount of heat transferred</li>
<li><mathjax>#n#</mathjax> = number of moles</li>
<li><mathjax>#c#</mathjax> = molar heat capacity</li>
<li><mathjax>#ΔT#</mathjax> = change in temperature</li>
</ul>
<p>We can rearrange this formula to</p>
<p><mathjax>#c = q/(nΔT)#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#q = "2211 J"#</mathjax></p>
<p><mathjax>#n = 47.7 color(red)(cancel(color(black)("g C"_2"H"_6"O"))) × ("1 mol C"_2"H"_6"O")/(46.07 color(red)(cancel(color(black)("g C"_2"H"_6"O")))) = "1.035 mol C"_2"H"_6"O"#</mathjax></p>
<p><mathjax>#ΔT = "19.9 °C"#</mathjax></p>
<blockquote></blockquote>
<p>Hence,</p>
<p><mathjax>#q = "2211 J"/("1.035 mol" × "19.9 °C") = "107 J·°C"^"-1""mol"^"-1"#</mathjax></p>
<p>The molar heat capacity of ethanol is <mathjax>#"107 J·°C"^"-1""mol"^"-1"#</mathjax>.</p></div>
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</article> | When 2211 J of heat energy is added to 47.7 g of ethanol, #C_2H_6O#, the temperature increases by 18.9°C. What is the molar heat capacity of #C_2H_6O#? | null |
2,842 | a86367ec-6ddd-11ea-8022-ccda262736ce | https://socratic.org/questions/a-closed-food-jar-has-a-fixed-volume-at-stp-what-would-the-new-pressure-be-at-45 | 318 kPa | start physical_unit 1 3 pressure kpa qc_end physical_unit 1 3 17 18 temperature qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pressure2 [OF] closed food jar [IN] kPa"}] | [{"type":"physical unit","value":"318 kPa"}] | [{"type":"physical unit","value":"Temperature2 [OF] closed food jar [=] \\pu{45 ℃}"},{"type":"other","value":"STP"},{"type":"other","value":"A closed food jar has a fixed volume."}] | <h1 class="questionTitle" itemprop="name">A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C?</h1> | null | 318 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall that Gay-Lussac's Law states that <strong>pressure is directly proportional to temperature</strong> , as long as the volume and number of moles of the gas remain constant. </p>
<p>Gay-Lussac's Law can be expressed mathematically with the formula,</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1=P_2/T_2color(white)(a/a)|)))#</mathjax></p>
<p>where:<br/>
<mathjax>#P_1=#</mathjax>initial pressure<br/>
<mathjax>#T_1=#</mathjax>inital temperature (Kelvin)<br/>
<mathjax>#P_2=#</mathjax>final pressure<br/>
<mathjax>#T_2=#</mathjax>final temperature (Kelvin)</p>
</blockquote>
<p><strong>Determining the Final Pressure</strong><br/>
<mathjax>#1#</mathjax>. Start by determining the values for each variable in the formula.</p>
<blockquote>
<p>STP conditions (initial):<br/>
initial pressure <mathjax>#color(orange)((P_1))#</mathjax>: <mathjax>#color(orange)(101.325kPa)#</mathjax><br/>
initial temperature <mathjax>#color(purple)((T_1))#</mathjax>: <mathjax>#color(purple)(273.15K)#</mathjax></p>
<p>New conditions (final)<br/>
final pressure <mathjax>#color(teal)((P_2))#</mathjax>: <mathjax>#color(teal)(P_2)#</mathjax><br/>
final temperature <mathjax>#color(magenta)((T_2))#</mathjax>: <mathjax>#45^@C+273.15=color(magenta)(318.15K)#</mathjax></p>
</blockquote>
<p><mathjax>#2#</mathjax>. Rearrange Gay-Lussac's formula in terms of <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(orange)(P_1)/color(purple)(T_1)=color(teal)(P_2)/color(magenta)(T_2)#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(T_2))((color(orange)(P_1))/(color(purple)(T_1)))#</mathjax></p>
</blockquote>
<p><mathjax>#3#</mathjax>. Using these values, substitute them into the rearranged formula.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15K))(color(orange)(101.325kPa)/color(purple)(273.15K))#</mathjax></p>
</blockquote>
<p><mathjax>#4#</mathjax>. Solve for <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15color(red)cancelcolor(magenta)K))(color(orange)(101.325kPa)/color(purple)(273.15color(red)cancelcolor(purple)K))#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=118.0177512kPa#</mathjax></p>
<p><mathjax>#color(teal)(P_2)~~color(green)(|bar(ul(color(white)(a/a)318kPacolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#:.#</mathjax>, the final pressure is <mathjax>#318kPa#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#318kPa#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall that Gay-Lussac's Law states that <strong>pressure is directly proportional to temperature</strong> , as long as the volume and number of moles of the gas remain constant. </p>
<p>Gay-Lussac's Law can be expressed mathematically with the formula,</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1=P_2/T_2color(white)(a/a)|)))#</mathjax></p>
<p>where:<br/>
<mathjax>#P_1=#</mathjax>initial pressure<br/>
<mathjax>#T_1=#</mathjax>inital temperature (Kelvin)<br/>
<mathjax>#P_2=#</mathjax>final pressure<br/>
<mathjax>#T_2=#</mathjax>final temperature (Kelvin)</p>
</blockquote>
<p><strong>Determining the Final Pressure</strong><br/>
<mathjax>#1#</mathjax>. Start by determining the values for each variable in the formula.</p>
<blockquote>
<p>STP conditions (initial):<br/>
initial pressure <mathjax>#color(orange)((P_1))#</mathjax>: <mathjax>#color(orange)(101.325kPa)#</mathjax><br/>
initial temperature <mathjax>#color(purple)((T_1))#</mathjax>: <mathjax>#color(purple)(273.15K)#</mathjax></p>
<p>New conditions (final)<br/>
final pressure <mathjax>#color(teal)((P_2))#</mathjax>: <mathjax>#color(teal)(P_2)#</mathjax><br/>
final temperature <mathjax>#color(magenta)((T_2))#</mathjax>: <mathjax>#45^@C+273.15=color(magenta)(318.15K)#</mathjax></p>
</blockquote>
<p><mathjax>#2#</mathjax>. Rearrange Gay-Lussac's formula in terms of <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(orange)(P_1)/color(purple)(T_1)=color(teal)(P_2)/color(magenta)(T_2)#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(T_2))((color(orange)(P_1))/(color(purple)(T_1)))#</mathjax></p>
</blockquote>
<p><mathjax>#3#</mathjax>. Using these values, substitute them into the rearranged formula.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15K))(color(orange)(101.325kPa)/color(purple)(273.15K))#</mathjax></p>
</blockquote>
<p><mathjax>#4#</mathjax>. Solve for <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15color(red)cancelcolor(magenta)K))(color(orange)(101.325kPa)/color(purple)(273.15color(red)cancelcolor(purple)K))#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=118.0177512kPa#</mathjax></p>
<p><mathjax>#color(teal)(P_2)~~color(green)(|bar(ul(color(white)(a/a)318kPacolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#:.#</mathjax>, the final pressure is <mathjax>#318kPa#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C?</h1>
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Johnson Z.
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Mar 15, 2016
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<div class="markdown"><p><mathjax>#318kPa#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall that Gay-Lussac's Law states that <strong>pressure is directly proportional to temperature</strong> , as long as the volume and number of moles of the gas remain constant. </p>
<p>Gay-Lussac's Law can be expressed mathematically with the formula,</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)P_1/T_1=P_2/T_2color(white)(a/a)|)))#</mathjax></p>
<p>where:<br/>
<mathjax>#P_1=#</mathjax>initial pressure<br/>
<mathjax>#T_1=#</mathjax>inital temperature (Kelvin)<br/>
<mathjax>#P_2=#</mathjax>final pressure<br/>
<mathjax>#T_2=#</mathjax>final temperature (Kelvin)</p>
</blockquote>
<p><strong>Determining the Final Pressure</strong><br/>
<mathjax>#1#</mathjax>. Start by determining the values for each variable in the formula.</p>
<blockquote>
<p>STP conditions (initial):<br/>
initial pressure <mathjax>#color(orange)((P_1))#</mathjax>: <mathjax>#color(orange)(101.325kPa)#</mathjax><br/>
initial temperature <mathjax>#color(purple)((T_1))#</mathjax>: <mathjax>#color(purple)(273.15K)#</mathjax></p>
<p>New conditions (final)<br/>
final pressure <mathjax>#color(teal)((P_2))#</mathjax>: <mathjax>#color(teal)(P_2)#</mathjax><br/>
final temperature <mathjax>#color(magenta)((T_2))#</mathjax>: <mathjax>#45^@C+273.15=color(magenta)(318.15K)#</mathjax></p>
</blockquote>
<p><mathjax>#2#</mathjax>. Rearrange Gay-Lussac's formula in terms of <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(orange)(P_1)/color(purple)(T_1)=color(teal)(P_2)/color(magenta)(T_2)#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(T_2))((color(orange)(P_1))/(color(purple)(T_1)))#</mathjax></p>
</blockquote>
<p><mathjax>#3#</mathjax>. Using these values, substitute them into the rearranged formula.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15K))(color(orange)(101.325kPa)/color(purple)(273.15K))#</mathjax></p>
</blockquote>
<p><mathjax>#4#</mathjax>. Solve for <mathjax>#color(teal)(P_2)#</mathjax>.</p>
<blockquote>
<p><mathjax>#color(teal)(P_2)=(color(magenta)(318.15color(red)cancelcolor(magenta)K))(color(orange)(101.325kPa)/color(purple)(273.15color(red)cancelcolor(purple)K))#</mathjax></p>
<p><mathjax>#color(teal)(P_2)=118.0177512kPa#</mathjax></p>
<p><mathjax>#color(teal)(P_2)~~color(green)(|bar(ul(color(white)(a/a)318kPacolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><mathjax>#:.#</mathjax>, the final pressure is <mathjax>#318kPa#</mathjax>.</p></div>
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</article> | A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C? | null |
2,843 | abdd9af0-6ddd-11ea-9b3f-ccda262736ce | https://socratic.org/questions/how-would-you-balance-this-chemical-equation-2sb-i-2-2sbi-3 | 2 Sb + 3 I2 -> 2 SbI3 | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this chemical equation"}] | [{"type":"chemical equation","value":"2 Sb + 3 I2 -> 2 SbI3"}] | [{"type":"chemical equation","value":"2 Sb + I2 -> 2 SbI3"}] | <h1 class="questionTitle" itemprop="name">How would you balance this chemical equation: #2Sb + I_2 -> 2SbI_3#?</h1> | null | 2 Sb + 3 I2 -> 2 SbI3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>There are six Iodines in the products so there must be six iodines in the reactants. </p>
<p>Iodine comes as diatomic gas molecule. ( as well as a grey crystal). </p>
<p><mathjax># 3 xx I_2 = 6 I^-1#</mathjax> therefore it requires 3 <mathjax>#I_2#</mathjax> molecules to balance the equation </p></div>
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<div class="markdown"><p><mathjax># 2 Sb + 3 I_2 === 2 SbI_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are six Iodines in the products so there must be six iodines in the reactants. </p>
<p>Iodine comes as diatomic gas molecule. ( as well as a grey crystal). </p>
<p><mathjax># 3 xx I_2 = 6 I^-1#</mathjax> therefore it requires 3 <mathjax>#I_2#</mathjax> molecules to balance the equation </p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance this chemical equation: #2Sb + I_2 -> 2SbI_3#?</h1>
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<div class="markdown"><p><mathjax># 2 Sb + 3 I_2 === 2 SbI_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>There are six Iodines in the products so there must be six iodines in the reactants. </p>
<p>Iodine comes as diatomic gas molecule. ( as well as a grey crystal). </p>
<p><mathjax># 3 xx I_2 = 6 I^-1#</mathjax> therefore it requires 3 <mathjax>#I_2#</mathjax> molecules to balance the equation </p></div>
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Ernest Z.
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<div class="markdown"><p>You follow a systematic procedure to balance the equation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"2Sb" + "I"_2 → "2SbI"_3#</mathjax></p>
<p>A method that often works is to balance everything other than <mathjax>#"O"#</mathjax> and <mathjax>#"H"#</mathjax> first, then balance <mathjax>#"O"#</mathjax>, and finally balance <mathjax>#"H"#</mathjax>.</p>
<p>Another useful procedure is to start with what looks like the most complicated formula.</p>
<p>The most complicated formula looks like <mathjax>#"SbI"_3#</mathjax>. It already has a 2 in front of it.</p>
<p><mathjax>#"2Sb" + "I"_2 → color(red)(2)"SbI"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"Sb"#</mathjax>:</strong></p>
<p>There are <mathjax>#"2 Sb"#</mathjax> on the right, and there are <mathjax>#"2 Sb"#</mathjax> on the right. <mathjax>#"Sb"#</mathjax> is balanced.</p>
<p><mathjax>#color(blue)(2)"Sb" + "I"_2 → color(red)(2)"SbI"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"I"#</mathjax>:</strong></p>
<p>We have fixed <mathjax>#"6 I"#</mathjax> on the right, so we need <mathjax>#"6 I"#</mathjax> on the left. Put a 3 in front of <mathjax>#"I"_2#</mathjax>.</p>
<p><mathjax>#color(blue)(2)"Sb" + color(orange)(3)"I"_2 → color(red)(2)"SbI"_3#</mathjax></p>
<blockquote></blockquote>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<p>Let’s check:</p>
<p><mathjax>#bb("Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side")#</mathjax><br/>
<mathjax>#color(white)(m)"Sb"color(white)(mmmmm)2color(white)(mmmmmmmm)2#</mathjax><br/>
<mathjax>#color(white)(m)"I"color(white)(mmmmmm)6color(white)(mmmmmmmm)6#</mathjax></p>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#2"Sb" + 3"I"_2 → 2"SbI"_3#</mathjax></p></div>
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</article> | How would you balance this chemical equation: #2Sb + I_2 -> 2SbI_3#? | null |
2,844 | abff1f5c-6ddd-11ea-9f3b-ccda262736ce | https://socratic.org/questions/when-14-0-g-of-calcium-metal-is-reacted-with-water-5-00-g-of-calcium-hydroxide-i | 19.3% | start physical_unit 27 28 percent_yield none qc_end physical_unit 4 5 1 2 mass qc_end physical_unit 13 14 10 11 mass qc_end chemical_equation 29 36 qc_end end | [{"type":"physical unit","value":"Percent yield [OF] the reaction"}] | [{"type":"physical unit","value":"19.3%"}] | [{"type":"physical unit","value":"Mass [OF] calcium metal [=] \\pu{14.0 g}"},{"type":"physical unit","value":"Mass [OF] calcium hydroxide [=] \\pu{5.00 g}"},{"type":"chemical equation","value":"Ca(s) + 2 H2O(l) -> Ca(OH)2(aq) + H2(g)"}] | <h1 class="questionTitle" itemprop="name">When 14.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction?
</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(aq) + "H"_2(g)#</mathjax></p>
<p>I'm more so worried about the execution rather than the answer so if you do not include the answer it is fine! Thanks so much :D </p></div>
</h2>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> of the reaction, you need to know two things</p>
<blockquote>
<ul>
<li>how many grams of calcium hydroxide would <em>theoretically</em> be produced by the reaction of <mathjax>#"14.0 g"#</mathjax> of calcium metal with excess water <mathjax>#->#</mathjax> this is the <strong>theoretical yield</strong> of the reaction</li>
<li>how many grams of calcium hydroxide are <em>actually</em> produced by this reaction <mathjax>#->#</mathjax> this is the <strong>actual yield</strong> of the reaction</li>
</ul>
</blockquote>
<p>The <strong>percent yield</strong> is calculated using the equation</p>
<blockquote>
<p><mathjax>#"% yield" = "actual yield"/"theoretical yield" * 100%#</mathjax></p>
</blockquote>
<p>Now, you know that the reaction produced <mathjax>#"5.00 g"#</mathjax> of calcium hydroxide, so you can say that</p>
<blockquote>
<p><mathjax>#"actual yield = 5.00 g"#</mathjax></p>
</blockquote>
<p>To find the <em>theoretical yield</em> of the reaction, use the fact that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of calcium hydroxide for every <mathjax>#1#</mathjax> <strong>mole</strong> of calcium metal that reacts.</p>
<blockquote>
<p><mathjax>#overbrace("Ca"_ ((s)))^(color(blue)("1 mole reacts")) + 2"H"_ 2"O"_ ((l)) -> overbrace("Ca"("OH")_ (2(aq)))^(color(blue)("1 mole is produced")) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Convert the mass of calcium to <em>moles</em> by using the <strong>molar mass</strong> of the metal.</p>
<blockquote>
<p><mathjax>#14.0 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.3493 moles Ca"#</mathjax></p>
</blockquote>
<p>Since you're looking for the <strong>theoretical yield</strong> of the reaction, i.e. what you would expect to get at <mathjax>#100%#</mathjax> yield, you can use the <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> that exists between calcium and calcium hydroxide (no information was provided about the mass of water that takes part in the reaction, so you can presume that water is <em>in excess</em>) to say that the reaction would theoretically produce</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"))) * ("1 mole Ca"("OH")_2)/(1color(red)(cancel(color(black)("mole Ca")))) = "0.3493 moles Ca"("OH")_2#</mathjax></p>
</blockquote>
<p>Convert the number of moles of calcium hydroxide to grams by using the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "74.093 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "25.88 g"#</mathjax></p>
</blockquote>
<p>This means that at <mathjax>#100%#</mathjax> yield, the <em>complete</em> reaction of <mathjax>#"14.0 g"#</mathjax> of calcium with enough water would produce <mathjax>#"25.88 g"#</mathjax> of calcium hydroxide. </p>
<p>Since the reaction produced <mathjax>#"5.00 g"#</mathjax>, you can say that the percent yield was equal to</p>
<blockquote>
<p><mathjax>#"% yield" = (5.00 color(red)(cancel(color(black)("g"))))/(25.88color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(19.3%)))#</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>#19.3%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> of the reaction, you need to know two things</p>
<blockquote>
<ul>
<li>how many grams of calcium hydroxide would <em>theoretically</em> be produced by the reaction of <mathjax>#"14.0 g"#</mathjax> of calcium metal with excess water <mathjax>#->#</mathjax> this is the <strong>theoretical yield</strong> of the reaction</li>
<li>how many grams of calcium hydroxide are <em>actually</em> produced by this reaction <mathjax>#->#</mathjax> this is the <strong>actual yield</strong> of the reaction</li>
</ul>
</blockquote>
<p>The <strong>percent yield</strong> is calculated using the equation</p>
<blockquote>
<p><mathjax>#"% yield" = "actual yield"/"theoretical yield" * 100%#</mathjax></p>
</blockquote>
<p>Now, you know that the reaction produced <mathjax>#"5.00 g"#</mathjax> of calcium hydroxide, so you can say that</p>
<blockquote>
<p><mathjax>#"actual yield = 5.00 g"#</mathjax></p>
</blockquote>
<p>To find the <em>theoretical yield</em> of the reaction, use the fact that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of calcium hydroxide for every <mathjax>#1#</mathjax> <strong>mole</strong> of calcium metal that reacts.</p>
<blockquote>
<p><mathjax>#overbrace("Ca"_ ((s)))^(color(blue)("1 mole reacts")) + 2"H"_ 2"O"_ ((l)) -> overbrace("Ca"("OH")_ (2(aq)))^(color(blue)("1 mole is produced")) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Convert the mass of calcium to <em>moles</em> by using the <strong>molar mass</strong> of the metal.</p>
<blockquote>
<p><mathjax>#14.0 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.3493 moles Ca"#</mathjax></p>
</blockquote>
<p>Since you're looking for the <strong>theoretical yield</strong> of the reaction, i.e. what you would expect to get at <mathjax>#100%#</mathjax> yield, you can use the <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> that exists between calcium and calcium hydroxide (no information was provided about the mass of water that takes part in the reaction, so you can presume that water is <em>in excess</em>) to say that the reaction would theoretically produce</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"))) * ("1 mole Ca"("OH")_2)/(1color(red)(cancel(color(black)("mole Ca")))) = "0.3493 moles Ca"("OH")_2#</mathjax></p>
</blockquote>
<p>Convert the number of moles of calcium hydroxide to grams by using the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "74.093 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "25.88 g"#</mathjax></p>
</blockquote>
<p>This means that at <mathjax>#100%#</mathjax> yield, the <em>complete</em> reaction of <mathjax>#"14.0 g"#</mathjax> of calcium with enough water would produce <mathjax>#"25.88 g"#</mathjax> of calcium hydroxide. </p>
<p>Since the reaction produced <mathjax>#"5.00 g"#</mathjax>, you can say that the percent yield was equal to</p>
<blockquote>
<p><mathjax>#"% yield" = (5.00 color(red)(cancel(color(black)("g"))))/(25.88color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(19.3%)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When 14.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction?
</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(aq) + "H"_2(g)#</mathjax></p>
<p>I'm more so worried about the execution rather than the answer so if you do not include the answer it is fine! Thanks so much :D </p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-10T23:02:31" itemprop="dateCreated">
Feb 10, 2018
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<div class="markdown"><p><mathjax>#19.3%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a> of the reaction, you need to know two things</p>
<blockquote>
<ul>
<li>how many grams of calcium hydroxide would <em>theoretically</em> be produced by the reaction of <mathjax>#"14.0 g"#</mathjax> of calcium metal with excess water <mathjax>#->#</mathjax> this is the <strong>theoretical yield</strong> of the reaction</li>
<li>how many grams of calcium hydroxide are <em>actually</em> produced by this reaction <mathjax>#->#</mathjax> this is the <strong>actual yield</strong> of the reaction</li>
</ul>
</blockquote>
<p>The <strong>percent yield</strong> is calculated using the equation</p>
<blockquote>
<p><mathjax>#"% yield" = "actual yield"/"theoretical yield" * 100%#</mathjax></p>
</blockquote>
<p>Now, you know that the reaction produced <mathjax>#"5.00 g"#</mathjax> of calcium hydroxide, so you can say that</p>
<blockquote>
<p><mathjax>#"actual yield = 5.00 g"#</mathjax></p>
</blockquote>
<p>To find the <em>theoretical yield</em> of the reaction, use the fact that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of calcium hydroxide for every <mathjax>#1#</mathjax> <strong>mole</strong> of calcium metal that reacts.</p>
<blockquote>
<p><mathjax>#overbrace("Ca"_ ((s)))^(color(blue)("1 mole reacts")) + 2"H"_ 2"O"_ ((l)) -> overbrace("Ca"("OH")_ (2(aq)))^(color(blue)("1 mole is produced")) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Convert the mass of calcium to <em>moles</em> by using the <strong>molar mass</strong> of the metal.</p>
<blockquote>
<p><mathjax>#14.0 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.3493 moles Ca"#</mathjax></p>
</blockquote>
<p>Since you're looking for the <strong>theoretical yield</strong> of the reaction, i.e. what you would expect to get at <mathjax>#100%#</mathjax> yield, you can use the <mathjax>#1:1#</mathjax> <strong>mole ratio</strong> that exists between calcium and calcium hydroxide (no information was provided about the mass of water that takes part in the reaction, so you can presume that water is <em>in excess</em>) to say that the reaction would theoretically produce</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"))) * ("1 mole Ca"("OH")_2)/(1color(red)(cancel(color(black)("mole Ca")))) = "0.3493 moles Ca"("OH")_2#</mathjax></p>
</blockquote>
<p>Convert the number of moles of calcium hydroxide to grams by using the <strong>molar mass</strong> of the compound.</p>
<blockquote>
<p><mathjax>#0.3493 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "74.093 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "25.88 g"#</mathjax></p>
</blockquote>
<p>This means that at <mathjax>#100%#</mathjax> yield, the <em>complete</em> reaction of <mathjax>#"14.0 g"#</mathjax> of calcium with enough water would produce <mathjax>#"25.88 g"#</mathjax> of calcium hydroxide. </p>
<p>Since the reaction produced <mathjax>#"5.00 g"#</mathjax>, you can say that the percent yield was equal to</p>
<blockquote>
<p><mathjax>#"% yield" = (5.00 color(red)(cancel(color(black)("g"))))/(25.88color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(19.3%)))#</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> | When 14.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction?
|
#"Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(aq) + "H"_2(g)#
I'm more so worried about the execution rather than the answer so if you do not include the answer it is fine! Thanks so much :D
|
2,845 | a8fa39d8-6ddd-11ea-988a-ccda262736ce | https://socratic.org/questions/how-much-heat-is-needed-to-raise-the-temperature-of-5-58-kilograms-of-iron-from- | 2.46 × 10^9 J | start physical_unit 13 13 heat_energy j qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 13 13 15 16 temperature qc_end physical_unit 13 13 18 19 temperature qc_end physical_unit 13 13 26 29 specific_heat qc_end end | [{"type":"physical unit","value":"Needed heat [OF] iron [IN] J"}] | [{"type":"physical unit","value":"2.46 × 10^9 J"}] | [{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{5.58 kilograms}"},{"type":"physical unit","value":"Temperature1 [OF] iron [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] iron [=] \\pu{1000.0 ℃}"},{"type":"physical unit","value":"Specific heat [OF] iron [=] \\pu{450,000 J/(kg * ℃)}"}] | <h1 class="questionTitle" itemprop="name">How much heat is needed to raise the temperature of 5.58 kilograms of iron from 20.0°C to 1000.0°C?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat of iron is 450,000 J(kg * °C).</p></div>
</h2>
</div>
</div> | 2.46 × 10^9 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:<br/>
<mathjax>#mass(m)=5.58kg#</mathjax><br/>
<mathjax>#DeltaT#</mathjax>=<mathjax>#(T_2#</mathjax>-<mathjax>#T_1)#</mathjax>=<mathjax>#(1000^oC-20^oC)=980^oC#</mathjax><br/>
<mathjax>#Cp= (450,000 J)/(kg.^oC)#</mathjax></p>
<p>Required:<br/>
<mathjax>#Heat (Q)= ? J#</mathjax></p>
<p>Solution:</p>
<p>From the Formula:</p>
<p><mathjax>#Q=mCpDeltaT#</mathjax><br/>
<mathjax>#Q=5.58cancel(kg)#</mathjax> <mathjax># xx#</mathjax> <mathjax>#(450,000J)/(cancel(kg.^oC)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#980.0cancel(^oC)#</mathjax><br/>
<mathjax>#Q=2.46#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^9J#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.46xx 10^9 J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:<br/>
<mathjax>#mass(m)=5.58kg#</mathjax><br/>
<mathjax>#DeltaT#</mathjax>=<mathjax>#(T_2#</mathjax>-<mathjax>#T_1)#</mathjax>=<mathjax>#(1000^oC-20^oC)=980^oC#</mathjax><br/>
<mathjax>#Cp= (450,000 J)/(kg.^oC)#</mathjax></p>
<p>Required:<br/>
<mathjax>#Heat (Q)= ? J#</mathjax></p>
<p>Solution:</p>
<p>From the Formula:</p>
<p><mathjax>#Q=mCpDeltaT#</mathjax><br/>
<mathjax>#Q=5.58cancel(kg)#</mathjax> <mathjax># xx#</mathjax> <mathjax>#(450,000J)/(cancel(kg.^oC)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#980.0cancel(^oC)#</mathjax><br/>
<mathjax>#Q=2.46#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^9J#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much heat is needed to raise the temperature of 5.58 kilograms of iron from 20.0°C to 1000.0°C?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The specific heat of iron is 450,000 J(kg * °C).</p></div>
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Jarni Renz
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Feb 27, 2016
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<div class="markdown"><p><mathjax>#2.46xx 10^9 J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:<br/>
<mathjax>#mass(m)=5.58kg#</mathjax><br/>
<mathjax>#DeltaT#</mathjax>=<mathjax>#(T_2#</mathjax>-<mathjax>#T_1)#</mathjax>=<mathjax>#(1000^oC-20^oC)=980^oC#</mathjax><br/>
<mathjax>#Cp= (450,000 J)/(kg.^oC)#</mathjax></p>
<p>Required:<br/>
<mathjax>#Heat (Q)= ? J#</mathjax></p>
<p>Solution:</p>
<p>From the Formula:</p>
<p><mathjax>#Q=mCpDeltaT#</mathjax><br/>
<mathjax>#Q=5.58cancel(kg)#</mathjax> <mathjax># xx#</mathjax> <mathjax>#(450,000J)/(cancel(kg.^oC)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#980.0cancel(^oC)#</mathjax><br/>
<mathjax>#Q=2.46#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^9J#</mathjax></p></div>
</div>
</div>
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</article> | How much heat is needed to raise the temperature of 5.58 kilograms of iron from 20.0°C to 1000.0°C? |
The specific heat of iron is 450,000 J(kg * °C).
|
2,846 | a8c050f5-6ddd-11ea-afd1-ccda262736ce | https://socratic.org/questions/how-many-moles-of-sodium-peroxide-are-needed-to-produce-1-00-mol-sodium-hydroxid | 0.50 moles | start physical_unit 4 5 mole mol qc_end physical_unit 12 13 10 11 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium peroxide [IN] moles"}] | [{"type":"physical unit","value":"0.50 moles"}] | [{"type":"physical unit","value":"Mole [OF] sodium hydroxide [=] \\pu{1.00 mol}"}] | <h1 class="questionTitle" itemprop="name">How many moles of sodium peroxide are needed to produce 1.00 mol sodium hydroxide? </h1> | null | 0.50 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"2Na"_2"O"_2 + "2H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NaOH + O"_2"#</mathjax></p>
<p><strong>Mole ratio between <mathjax>#"Na"_2"O"_2"#</mathjax> and <mathjax>#"NaOH"#</mathjax>.</strong></p>
<p><mathjax>#(2"mol Na"_2"O"_2)/(4"mol NaOH")#</mathjax> and <mathjax>#(4"mol NaOH")/(2"mol Na"_2"O"_2)#</mathjax></p>
<p><strong>Multiply given mol <mathjax>#"NaOH"#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio with <mathjax>#"Na"_2"O"_2"#</mathjax> in the numerator.</strong></p>
<p><mathjax>#1.00color(red)cancel(color(black)("mol NaOH"))xx(2"mol Na"_2"O"_2)/(4color(red)cancel(color(black)("mol NaOH")))="0.500 mol Na"_2"O"_2"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.500 mol Na"_2"O"_2"#</mathjax> will be needed to react with excess water to produce <mathjax>#"1.00 mol NaOH"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"2Na"_2"O"_2 + "2H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NaOH + O"_2"#</mathjax></p>
<p><strong>Mole ratio between <mathjax>#"Na"_2"O"_2"#</mathjax> and <mathjax>#"NaOH"#</mathjax>.</strong></p>
<p><mathjax>#(2"mol Na"_2"O"_2)/(4"mol NaOH")#</mathjax> and <mathjax>#(4"mol NaOH")/(2"mol Na"_2"O"_2)#</mathjax></p>
<p><strong>Multiply given mol <mathjax>#"NaOH"#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio with <mathjax>#"Na"_2"O"_2"#</mathjax> in the numerator.</strong></p>
<p><mathjax>#1.00color(red)cancel(color(black)("mol NaOH"))xx(2"mol Na"_2"O"_2)/(4color(red)cancel(color(black)("mol NaOH")))="0.500 mol Na"_2"O"_2"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of sodium peroxide are needed to produce 1.00 mol sodium hydroxide? </h1>
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<div class="markdown"><p><mathjax>#"0.500 mol Na"_2"O"_2"#</mathjax> will be needed to react with excess water to produce <mathjax>#"1.00 mol NaOH"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"2Na"_2"O"_2 + "2H"_2"O"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4NaOH + O"_2"#</mathjax></p>
<p><strong>Mole ratio between <mathjax>#"Na"_2"O"_2"#</mathjax> and <mathjax>#"NaOH"#</mathjax>.</strong></p>
<p><mathjax>#(2"mol Na"_2"O"_2)/(4"mol NaOH")#</mathjax> and <mathjax>#(4"mol NaOH")/(2"mol Na"_2"O"_2)#</mathjax></p>
<p><strong>Multiply given mol <mathjax>#"NaOH"#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio with <mathjax>#"Na"_2"O"_2"#</mathjax> in the numerator.</strong></p>
<p><mathjax>#1.00color(red)cancel(color(black)("mol NaOH"))xx(2"mol Na"_2"O"_2)/(4color(red)cancel(color(black)("mol NaOH")))="0.500 mol Na"_2"O"_2"#</mathjax></p></div>
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</article> | How many moles of sodium peroxide are needed to produce 1.00 mol sodium hydroxide? | null |
2,847 | aad203ed-6ddd-11ea-a69b-ccda262736ce | https://socratic.org/questions/what-volume-in-l-of-a-3-95-m-potassium-chloride-kcl-solution-would-be-needed-to- | 0.23 L | start physical_unit 11 11 volume l qc_end physical_unit 10 10 6 7 molarity qc_end physical_unit 10 10 21 22 molarity qc_end physical_unit 11 11 17 18 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] KCl solution [IN] L"}] | [{"type":"physical unit","value":"0.23 L"}] | [{"type":"physical unit","value":"Molarity2 [OF] KCl solution [=] \\pu{3.95 M}"},{"type":"physical unit","value":"Molarity1 [OF] KCl solution [=] \\pu{2.76 M}"},{"type":"physical unit","value":"Volume1 [OF] KCl solution [=] \\pu{325 mL }"}] | <h1 class="questionTitle" itemprop="name">What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution?</h1> | null | 0.23 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One possible strategy to use here is to find the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> first, then use it to find the volume of the stock solution that is needed in order to make your target solution. </p>
<p>The <em>dilution factor</em> essentially tells you <strong>how many times more concentrated</strong> the <em>stock solution</em> was compared with the target solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_"stock"/c_"target"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.95 color(red)(cancel(color(black)("M"))))/(2.76color(red)(cancel(color(black)("M")))) = 1.43#</mathjax></p>
</blockquote>
<p>Now, because a dilution implies that the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong>, the dilution factor also tells you the ratio that exists between the volume of the <strong>target solution</strong>, i.e. the <em>diluted solution</em>, and the volume of the <strong>stock solution</strong>, i. e. the <em>concentrated solution</em>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"target"/V_"stock"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the volume of the stock solution needed to make the target solution would be </p>
<blockquote>
<p><mathjax>#V_"stock" = V_"target"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "325 mL"/1.43 = "227 mL"#</mathjax></p>
</blockquote>
<p>Expressed in <em>liters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>So, in order to make <mathjax>#"325 mL"#</mathjax> of a <mathjax>#"2.76 M"#</mathjax> potassium chloride solution you must take <mathjax>#"227 mL"#</mathjax> of a <mathjax>#"3.95 M"#</mathjax> solution and add enough water to get the total volume to <mathjax>#"325 mL"#</mathjax>.</p>
<p>Notice that the concentration of the solution <strong>decreases</strong> by the same ratio by which the volume <strong>increases</strong>. This is what the dilution factor is useful for. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.227 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One possible strategy to use here is to find the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> first, then use it to find the volume of the stock solution that is needed in order to make your target solution. </p>
<p>The <em>dilution factor</em> essentially tells you <strong>how many times more concentrated</strong> the <em>stock solution</em> was compared with the target solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_"stock"/c_"target"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.95 color(red)(cancel(color(black)("M"))))/(2.76color(red)(cancel(color(black)("M")))) = 1.43#</mathjax></p>
</blockquote>
<p>Now, because a dilution implies that the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong>, the dilution factor also tells you the ratio that exists between the volume of the <strong>target solution</strong>, i.e. the <em>diluted solution</em>, and the volume of the <strong>stock solution</strong>, i. e. the <em>concentrated solution</em>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"target"/V_"stock"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the volume of the stock solution needed to make the target solution would be </p>
<blockquote>
<p><mathjax>#V_"stock" = V_"target"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "325 mL"/1.43 = "227 mL"#</mathjax></p>
</blockquote>
<p>Expressed in <em>liters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>So, in order to make <mathjax>#"325 mL"#</mathjax> of a <mathjax>#"2.76 M"#</mathjax> potassium chloride solution you must take <mathjax>#"227 mL"#</mathjax> of a <mathjax>#"3.95 M"#</mathjax> solution and add enough water to get the total volume to <mathjax>#"325 mL"#</mathjax>.</p>
<p>Notice that the concentration of the solution <strong>decreases</strong> by the same ratio by which the volume <strong>increases</strong>. This is what the dilution factor is useful for. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.227 L"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One possible strategy to use here is to find the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> first, then use it to find the volume of the stock solution that is needed in order to make your target solution. </p>
<p>The <em>dilution factor</em> essentially tells you <strong>how many times more concentrated</strong> the <em>stock solution</em> was compared with the target solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_"stock"/c_"target"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is equal to </p>
<blockquote>
<p><mathjax>#"D.F." = (3.95 color(red)(cancel(color(black)("M"))))/(2.76color(red)(cancel(color(black)("M")))) = 1.43#</mathjax></p>
</blockquote>
<p>Now, because a dilution implies that the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>remains constant</strong>, the dilution factor also tells you the ratio that exists between the volume of the <strong>target solution</strong>, i.e. the <em>diluted solution</em>, and the volume of the <strong>stock solution</strong>, i. e. the <em>concentrated solution</em>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"target"/V_"stock"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the volume of the stock solution needed to make the target solution would be </p>
<blockquote>
<p><mathjax>#V_"stock" = V_"target"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "325 mL"/1.43 = "227 mL"#</mathjax></p>
</blockquote>
<p>Expressed in <em>liters</em>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></em></p>
</blockquote>
<p>So, in order to make <mathjax>#"325 mL"#</mathjax> of a <mathjax>#"2.76 M"#</mathjax> potassium chloride solution you must take <mathjax>#"227 mL"#</mathjax> of a <mathjax>#"3.95 M"#</mathjax> solution and add enough water to get the total volume to <mathjax>#"325 mL"#</mathjax>.</p>
<p>Notice that the concentration of the solution <strong>decreases</strong> by the same ratio by which the volume <strong>increases</strong>. This is what the dilution factor is useful for. </p></div>
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</article> | What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution? | null |
2,848 | aae651c9-6ddd-11ea-b8e4-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-025-m-hcl-solution | 1.60 | 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] HCl solution"}] | [{"type":"physical unit","value":"1.60"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.025 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a .025 M #HCl# solution?</h1> | null | 1.60 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrochloric acid is a strong acid that completely dissociates in water. </p>
<p>For a 0.025 M HCl solution the concentration of the hydrogen ions, <mathjax>#["H"^+]#</mathjax>, is 0.025 moles/liter (i.e. 0.025 M).</p>
<p>pH is defined as <mathjax>#-log ["H"^+]#</mathjax>.</p>
<p>Substituting 0.025 into the equation for pH we find that the 0.025 M HCl solution has a pH of -log(0.025) = 1.60. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The pH is 1.60.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrochloric acid is a strong acid that completely dissociates in water. </p>
<p>For a 0.025 M HCl solution the concentration of the hydrogen ions, <mathjax>#["H"^+]#</mathjax>, is 0.025 moles/liter (i.e. 0.025 M).</p>
<p>pH is defined as <mathjax>#-log ["H"^+]#</mathjax>.</p>
<p>Substituting 0.025 into the equation for pH we find that the 0.025 M HCl solution has a pH of -log(0.025) = 1.60. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a .025 M #HCl# solution?</h1>
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<div class="markdown"><p>The pH is 1.60.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrochloric acid is a strong acid that completely dissociates in water. </p>
<p>For a 0.025 M HCl solution the concentration of the hydrogen ions, <mathjax>#["H"^+]#</mathjax>, is 0.025 moles/liter (i.e. 0.025 M).</p>
<p>pH is defined as <mathjax>#-log ["H"^+]#</mathjax>.</p>
<p>Substituting 0.025 into the equation for pH we find that the 0.025 M HCl solution has a pH of -log(0.025) = 1.60. </p></div>
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</article> | What is the pH of a .025 M #HCl# solution? | null |
2,849 | abe47e15-6ddd-11ea-a34f-ccda262736ce | https://socratic.org/questions/how-much-energy-is-needed-to-vaporize-0-22-kg-of-water | 496 kJ | start physical_unit 10 10 energy kj qc_end physical_unit 10 10 7 8 mass qc_end c_other vaporize qc_end end | [{"type":"physical unit","value":"Needed energy [OF] water [IN] kJ"}] | [{"type":"physical unit","value":"496 kJ"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{0.22 kg}"},{"type":"other","value":"vaporize"}] | <h1 class="questionTitle" itemprop="name">How much energy is needed to vaporize 0.22 kg of water? </h1> | null | 496 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume that the water is ready to be vaporized, i.e. it's already at the boiling point, and that we just need to put in enough energy to vaporize it. We need to know the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization for water (also called the latent heat of vaporization). This can be found from <a href="https://en.wikipedia.org/wiki/Enthalpy_of_vaporization" rel="nofollow">standard tables</a> as:</p>
<p><mathjax>#Delta H_(vap,water) = 2257 kJ//kg#</mathjax></p>
<p>Now we just need to use this like a conversion factor to find the energy we are looking for knowing the mass of the water we are working with:</p>
<p><mathjax>#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume that the water is ready to be vaporized, i.e. it's already at the boiling point, and that we just need to put in enough energy to vaporize it. We need to know the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization for water (also called the latent heat of vaporization). This can be found from <a href="https://en.wikipedia.org/wiki/Enthalpy_of_vaporization" rel="nofollow">standard tables</a> as:</p>
<p><mathjax>#Delta H_(vap,water) = 2257 kJ//kg#</mathjax></p>
<p>Now we just need to use this like a conversion factor to find the energy we are looking for knowing the mass of the water we are working with:</p>
<p><mathjax>#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How much energy is needed to vaporize 0.22 kg of water? </h1>
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<div class="markdown"><p><mathjax>#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume that the water is ready to be vaporized, i.e. it's already at the boiling point, and that we just need to put in enough energy to vaporize it. We need to know the <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization for water (also called the latent heat of vaporization). This can be found from <a href="https://en.wikipedia.org/wiki/Enthalpy_of_vaporization" rel="nofollow">standard tables</a> as:</p>
<p><mathjax>#Delta H_(vap,water) = 2257 kJ//kg#</mathjax></p>
<p>Now we just need to use this like a conversion factor to find the energy we are looking for knowing the mass of the water we are working with:</p>
<p><mathjax>#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#</mathjax></p></div>
</div>
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</article> | How much energy is needed to vaporize 0.22 kg of water? | null |
2,850 | a9b151d5-6ddd-11ea-81ed-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-sodium-carbonate-na2co3 | 105.99 g/mol | start physical_unit 8 8 molar_mass g/mol qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Molar mass [OF] Na2CO3 [IN] g/mol"}] | [{"type":"physical unit","value":"105.99 g/mol"}] | [{"type":"chemical equation","value":"Na2CO3"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of sodium carbonate, (Na2CO3)?</h1> | null | 105.99 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the molar mass of a compound, multiply the subscript of each element from the formula times the molar mass of the element. The molar mass of an element is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><strong>Molar mass of <mathjax>#"Na"_2"CO"_3"#</mathjax></strong></p>
<p><mathjax>#(2xx22.98976928"g/mol Na")+(1xx12.0107"g/mol C")+(3xx15.999"g/mol O")="105.987 g/mol Na"_2"CO"_3"#</mathjax></p></div>
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<div class="markdown"><p>The molar mass of sodium carbonate is <mathjax>#"105.987 g/mol"#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the molar mass of a compound, multiply the subscript of each element from the formula times the molar mass of the element. The molar mass of an element is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><strong>Molar mass of <mathjax>#"Na"_2"CO"_3"#</mathjax></strong></p>
<p><mathjax>#(2xx22.98976928"g/mol Na")+(1xx12.0107"g/mol C")+(3xx15.999"g/mol O")="105.987 g/mol Na"_2"CO"_3"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar mass of sodium carbonate, (Na2CO3)?</h1>
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<div class="markdown"><p>The molar mass of sodium carbonate is <mathjax>#"105.987 g/mol"#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To calculate the molar mass of a compound, multiply the subscript of each element from the formula times the molar mass of the element. The molar mass of an element is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p>
<p><strong>Molar mass of <mathjax>#"Na"_2"CO"_3"#</mathjax></strong></p>
<p><mathjax>#(2xx22.98976928"g/mol Na")+(1xx12.0107"g/mol C")+(3xx15.999"g/mol O")="105.987 g/mol Na"_2"CO"_3"#</mathjax></p></div>
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</article> | What is the molar mass of sodium carbonate, (Na2CO3)? | null |
2,851 | acf50e0e-6ddd-11ea-9b36-ccda262736ce | https://socratic.org/questions/what-is-the-mole-fraction-of-na2so4-in-a-solution-which-is-11-5-by-weight-na2so4 | 0.02 | start physical_unit 6 7 mole_fraction none qc_end physical_unit 6 7 12 12 mass_percent qc_end end | [{"type":"physical unit","value":"Mole fraction [OF] Na2SO4 in solution"}] | [{"type":"physical unit","value":"0.02"}] | [{"type":"physical unit","value":"Percent by weight [OF] Na2SO4 in solution [=] \\pu{11.5%}"}] | <h1 class="questionTitle" itemprop="name">What is the mole fraction of #"Na"_2"SO"_4# in a solution which is #11.5%# by weight #"Na"_2"SO"_4# ?</h1> | null | 0.02 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass. </p>
<p>Your solution is said to have a percent concentration by mass of <mathjax>#11.5%#</mathjax>, which means that for every <mathjax>#"100.0 g"#</mathjax> of this solution, you get <mathjax>#"11.5 g"#</mathjax> of sodium sulfate, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>To make the calculations easier, let's pick a sample of this solution that has a mass of exactly <mathjax>#"100.0 g"#</mathjax>. Since you know how many grams of sodium sulfate it contains, you can calculate the mass of water present in the sample. </p>
<blockquote>
<p><mathjax>#overbrace("mass of water")^(color(blue)("mass solvent")) = overbrace("100.0 g")^(color(blue)("mass solution")) - overbrace("11.5 g")^(color(blue)("mass solute"))#</mathjax></p>
<p><mathjax>#"mass of water = 88.5 g"#</mathjax></p>
</blockquote>
<p>Now, use the <strong>molar mass</strong> of water to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#88.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.9126 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Do the same for sodium sulfate, i.e. use its <strong>molar mass</strong> to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#11.5 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.08096 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>To find the <strong>mole fraction</strong> of sodium sulfate, <mathjax>#chi_ ("Na"_ 2"SO"_ 4)#</mathjax>, you must divide the number of moles of sodium sulfate by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_ ("Na"_ 2"SO"_ 4) = (0.08096 color(red)(cancel(color(black)("moles"))))/((4.9126 + 0.08096)color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)(0.0162)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.0162#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass. </p>
<p>Your solution is said to have a percent concentration by mass of <mathjax>#11.5%#</mathjax>, which means that for every <mathjax>#"100.0 g"#</mathjax> of this solution, you get <mathjax>#"11.5 g"#</mathjax> of sodium sulfate, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>To make the calculations easier, let's pick a sample of this solution that has a mass of exactly <mathjax>#"100.0 g"#</mathjax>. Since you know how many grams of sodium sulfate it contains, you can calculate the mass of water present in the sample. </p>
<blockquote>
<p><mathjax>#overbrace("mass of water")^(color(blue)("mass solvent")) = overbrace("100.0 g")^(color(blue)("mass solution")) - overbrace("11.5 g")^(color(blue)("mass solute"))#</mathjax></p>
<p><mathjax>#"mass of water = 88.5 g"#</mathjax></p>
</blockquote>
<p>Now, use the <strong>molar mass</strong> of water to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#88.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.9126 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Do the same for sodium sulfate, i.e. use its <strong>molar mass</strong> to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#11.5 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.08096 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>To find the <strong>mole fraction</strong> of sodium sulfate, <mathjax>#chi_ ("Na"_ 2"SO"_ 4)#</mathjax>, you must divide the number of moles of sodium sulfate by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_ ("Na"_ 2"SO"_ 4) = (0.08096 color(red)(cancel(color(black)("moles"))))/((4.9126 + 0.08096)color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)(0.0162)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the mole fraction of #"Na"_2"SO"_4# in a solution which is #11.5%# by weight #"Na"_2"SO"_4# ?</h1>
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<div class="markdown"><p><mathjax>#0.0162#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass. </p>
<p>Your solution is said to have a percent concentration by mass of <mathjax>#11.5%#</mathjax>, which means that for every <mathjax>#"100.0 g"#</mathjax> of this solution, you get <mathjax>#"11.5 g"#</mathjax> of sodium sulfate, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>To make the calculations easier, let's pick a sample of this solution that has a mass of exactly <mathjax>#"100.0 g"#</mathjax>. Since you know how many grams of sodium sulfate it contains, you can calculate the mass of water present in the sample. </p>
<blockquote>
<p><mathjax>#overbrace("mass of water")^(color(blue)("mass solvent")) = overbrace("100.0 g")^(color(blue)("mass solution")) - overbrace("11.5 g")^(color(blue)("mass solute"))#</mathjax></p>
<p><mathjax>#"mass of water = 88.5 g"#</mathjax></p>
</blockquote>
<p>Now, use the <strong>molar mass</strong> of water to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#88.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.9126 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>Do the same for sodium sulfate, i.e. use its <strong>molar mass</strong> to convert the mass to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#11.5 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.08096 moles Na"_2"SO"_4#</mathjax></p>
</blockquote>
<p>To find the <strong>mole fraction</strong> of sodium sulfate, <mathjax>#chi_ ("Na"_ 2"SO"_ 4)#</mathjax>, you must divide the number of moles of sodium sulfate by the <strong>total number of moles</strong> present in the solution. </p>
<blockquote>
<p><mathjax>#chi_ ("Na"_ 2"SO"_ 4) = (0.08096 color(red)(cancel(color(black)("moles"))))/((4.9126 + 0.08096)color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)(0.0162)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What is the mole fraction of #"Na"_2"SO"_4# in a solution which is #11.5%# by weight #"Na"_2"SO"_4# ? | null |
2,852 | aa7d27d1-6ddd-11ea-bffd-ccda262736ce | https://socratic.org/questions/if-you-have-a-gas-with-an-initial-pressure-of-5-atm-and-it-occupies-a-volume-of- | 3.85 atm | start physical_unit 34 35 pressure atm qc_end physical_unit 34 35 18 19 volume qc_end physical_unit 34 35 10 11 pressure qc_end physical_unit 34 35 26 27 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"3.85 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{10 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{5 atm}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{13 L}"}] | <h1 class="questionTitle" itemprop="name">If you have a gas with an initial pressure of 5 atm and it occupies a volume of 10L. If you increase the volume to 13L, what will the new pressure of the gas be?</h1> | null | 3.85 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyle's Law</p>
<p><mathjax>#"pressure(1)" xx "volume(1)" = "pressure(2)" xx "volume(2)"#</mathjax></p>
<p>where </p>
<blockquote>
<ul>
<li><mathjax>#"pressure(1) = 5 atm"#</mathjax> </li>
<li><mathjax>#"volume(1) = 10 L"#</mathjax></li>
<li><mathjax>#"pressure(2)" = "13 atm"#</mathjax></li>
</ul>
</blockquote>
<p>Therefore </p>
<p><mathjax>#"5 atm" xx "10 L" ="13 atm" xx "volume(2)"#</mathjax></p>
<p><mathjax>#"volume(2)" = ("5 atm" xx "10 L")/"13 atm" = "3.85 L" = "4 L" ->#</mathjax> one sigfig</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"4 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyle's Law</p>
<p><mathjax>#"pressure(1)" xx "volume(1)" = "pressure(2)" xx "volume(2)"#</mathjax></p>
<p>where </p>
<blockquote>
<ul>
<li><mathjax>#"pressure(1) = 5 atm"#</mathjax> </li>
<li><mathjax>#"volume(1) = 10 L"#</mathjax></li>
<li><mathjax>#"pressure(2)" = "13 atm"#</mathjax></li>
</ul>
</blockquote>
<p>Therefore </p>
<p><mathjax>#"5 atm" xx "10 L" ="13 atm" xx "volume(2)"#</mathjax></p>
<p><mathjax>#"volume(2)" = ("5 atm" xx "10 L")/"13 atm" = "3.85 L" = "4 L" ->#</mathjax> one sigfig</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If you have a gas with an initial pressure of 5 atm and it occupies a volume of 10L. If you increase the volume to 13L, what will the new pressure of the gas be?</h1>
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Gilbert
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Stefan V.
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<span class="dateCreated" datetime="2018-02-26T04:45:35" itemprop="dateCreated">
Feb 26, 2018
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<div>
<div class="markdown"><p><mathjax>#"4 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use Boyle's Law</p>
<p><mathjax>#"pressure(1)" xx "volume(1)" = "pressure(2)" xx "volume(2)"#</mathjax></p>
<p>where </p>
<blockquote>
<ul>
<li><mathjax>#"pressure(1) = 5 atm"#</mathjax> </li>
<li><mathjax>#"volume(1) = 10 L"#</mathjax></li>
<li><mathjax>#"pressure(2)" = "13 atm"#</mathjax></li>
</ul>
</blockquote>
<p>Therefore </p>
<p><mathjax>#"5 atm" xx "10 L" ="13 atm" xx "volume(2)"#</mathjax></p>
<p><mathjax>#"volume(2)" = ("5 atm" xx "10 L")/"13 atm" = "3.85 L" = "4 L" ->#</mathjax> one sigfig</p></div>
</div>
</div>
</div>
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</article> | If you have a gas with an initial pressure of 5 atm and it occupies a volume of 10L. If you increase the volume to 13L, what will the new pressure of the gas be? | null |
2,853 | a9414b52-6ddd-11ea-83f3-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-the-compound-formed-between-chromium-vi-and-iod | CrI6 | start chemical_formula qc_end substance 10 10 qc_end substance 12 12 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"CrI6"}] | [{"type":"substance name","value":"Chromium(VI)"},{"type":"substance name","value":"Iodine"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for the compound formed between chromium(VI) and iodine? </h1> | null | CrI6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chromium VI has an oxidation number value of +6. <br/>
Iodide an a negative ion normally has an oxidation number value of -1 </p>
<p><mathjax># 1 xx +6 + 6 ( -1) = 0#</mathjax> </p>
<p>so it takes 6 Iodide ions to balance the 1 Chromium ion of +6 </p>
<p>The formula then is </p>
<p><mathjax># CrI_6#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax># CrI_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chromium VI has an oxidation number value of +6. <br/>
Iodide an a negative ion normally has an oxidation number value of -1 </p>
<p><mathjax># 1 xx +6 + 6 ( -1) = 0#</mathjax> </p>
<p>so it takes 6 Iodide ions to balance the 1 Chromium ion of +6 </p>
<p>The formula then is </p>
<p><mathjax># CrI_6#</mathjax> </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the chemical formula for the compound formed between chromium(VI) and iodine? </h1>
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Hriman
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Mar 20, 2018
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<div class="markdown"><p><mathjax># CrI_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Chromium VI has an oxidation number value of +6. <br/>
Iodide an a negative ion normally has an oxidation number value of -1 </p>
<p><mathjax># 1 xx +6 + 6 ( -1) = 0#</mathjax> </p>
<p>so it takes 6 Iodide ions to balance the 1 Chromium ion of +6 </p>
<p>The formula then is </p>
<p><mathjax># CrI_6#</mathjax> </p></div>
</div>
</div>
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</article> | What is the chemical formula for the compound formed between chromium(VI) and iodine? | null |
2,854 | ab2f0675-6ddd-11ea-a167-ccda262736ce | https://socratic.org/questions/what-mass-of-ethanol-c-2h-5oh-l-must-be-burned-to-supply-500-0-kj-of-heat | 16.95 g | start physical_unit 4 4 mass g qc_end physical_unit 4 4 10 11 heat_energy qc_end end | [{"type":"physical unit","value":"Mass [OF] C2H5OH(l) [IN] g"}] | [{"type":"physical unit","value":"16.95 g"}] | [{"type":"physical unit","value":"Supplied heat [OF] C2H5OH(l) [=] \\pu{500.0 kJ}"}] | <h1 class="questionTitle" itemprop="name">What mass of ethanol #C_2H_5OH(l)# must be burned to supply 500.0 kJ of heat?</h1> | null | 16.95 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of combustion of ethanol is widely quoted in the public domain and is 1360 kJ/mol.</p>
<p>Therefore in order to supply 500 kJ of heat you would need to burn (500/1360) = 0.368 moles of ethanol.</p>
<p>The molecular weight of ethanol is 46.07 g/mol, so the mass required would be 0.368 x 46.07 = 16.95 g.</p>
<p>This is based on 100% ethanol - in practise ethanol will have a certain water content, and this of course does not burn or generate heat. If you have a water content of, say, 2%, then the mass of ethanol required would need to be slightly increased to compensate for this.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Based on 100% ethanol (zero water content) 16.95 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of combustion of ethanol is widely quoted in the public domain and is 1360 kJ/mol.</p>
<p>Therefore in order to supply 500 kJ of heat you would need to burn (500/1360) = 0.368 moles of ethanol.</p>
<p>The molecular weight of ethanol is 46.07 g/mol, so the mass required would be 0.368 x 46.07 = 16.95 g.</p>
<p>This is based on 100% ethanol - in practise ethanol will have a certain water content, and this of course does not burn or generate heat. If you have a water content of, say, 2%, then the mass of ethanol required would need to be slightly increased to compensate for this.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass of ethanol #C_2H_5OH(l)# must be burned to supply 500.0 kJ of heat?</h1>
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<div class="markdown"><p>Based on 100% ethanol (zero water content) 16.95 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of combustion of ethanol is widely quoted in the public domain and is 1360 kJ/mol.</p>
<p>Therefore in order to supply 500 kJ of heat you would need to burn (500/1360) = 0.368 moles of ethanol.</p>
<p>The molecular weight of ethanol is 46.07 g/mol, so the mass required would be 0.368 x 46.07 = 16.95 g.</p>
<p>This is based on 100% ethanol - in practise ethanol will have a certain water content, and this of course does not burn or generate heat. If you have a water content of, say, 2%, then the mass of ethanol required would need to be slightly increased to compensate for this.</p></div>
</div>
</div>
</div>
</div>
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</article> | What mass of ethanol #C_2H_5OH(l)# must be burned to supply 500.0 kJ of heat? | null |
2,855 | aaec6cae-6ddd-11ea-a612-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-oxygen-atoms-in-al-2-so-4-3 | 12 | start physical_unit 5 6 number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"12"}] | [{"type":"chemical equation","value":"Al2(SO4)3"}] | <h1 class="questionTitle" itemprop="name">What is the number of oxygen atoms in #Al_2(SO_4)_3#?</h1> | null | 12 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The subscript for the sulfate ion <mathjax>#("SO"_4"^(2-))#</mathjax> is <mathjax>#3#</mathjax>. The subscript for the oxygen atom is <mathjax>#4#</mathjax>. Multiply them together to determine the number of oxygen atoms in one formula unit of aluminum sulfate. <mathjax>#(3xx4=12)#</mathjax></p></div>
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<div class="markdown"><p>There are 12 oxygen atoms in one formula unit of <mathjax>#"Al(SO"_4)_3"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The subscript for the sulfate ion <mathjax>#("SO"_4"^(2-))#</mathjax> is <mathjax>#3#</mathjax>. The subscript for the oxygen atom is <mathjax>#4#</mathjax>. Multiply them together to determine the number of oxygen atoms in one formula unit of aluminum sulfate. <mathjax>#(3xx4=12)#</mathjax></p></div>
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<div class="markdown"><p>There are 12 oxygen atoms in one formula unit of <mathjax>#"Al(SO"_4)_3"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The subscript for the sulfate ion <mathjax>#("SO"_4"^(2-))#</mathjax> is <mathjax>#3#</mathjax>. The subscript for the oxygen atom is <mathjax>#4#</mathjax>. Multiply them together to determine the number of oxygen atoms in one formula unit of aluminum sulfate. <mathjax>#(3xx4=12)#</mathjax></p></div>
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</article> | What is the number of oxygen atoms in #Al_2(SO_4)_3#? | null |
2,856 | abddc29a-6ddd-11ea-b491-ccda262736ce | https://socratic.org/questions/how-many-grams-would-you-weigh-out-on-an-electronic-balance-if-a-reaction-requir | 103.5 grams | start physical_unit 18 19 mass g qc_end physical_unit 18 19 15 16 mole qc_end end | [{"type":"physical unit","value":"Weight [OF] lithium oxide [IN] grams"}] | [{"type":"physical unit","value":"103.5 grams"}] | [{"type":"physical unit","value":"Mole [OF] lithium oxide [=] \\pu{4.5 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams would you weigh out on an electronic balance if a reaction requires 4.5 moles of lithium oxide? </h1> | null | 103.5 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> given ln <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> is actually not just grams. It's grams per mole.</p>
<p>This means that 1 mole of Li is 7grams and 1 mole of O is 16grams.</p>
<p>Together, they have a combined molecular weight of:<br/>
7+16= 23 grams per mole (or g/mol as you'll often see it written)</p>
<p>So, now that you know there are 23 grams in one mole of lithium oxide, you can simply multiply it by the number of moles you need:</p>
<p>4.5<mathjax>#*#</mathjax>23 = 103.5 grams</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>103.5 grams</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> given ln <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> is actually not just grams. It's grams per mole.</p>
<p>This means that 1 mole of Li is 7grams and 1 mole of O is 16grams.</p>
<p>Together, they have a combined molecular weight of:<br/>
7+16= 23 grams per mole (or g/mol as you'll often see it written)</p>
<p>So, now that you know there are 23 grams in one mole of lithium oxide, you can simply multiply it by the number of moles you need:</p>
<p>4.5<mathjax>#*#</mathjax>23 = 103.5 grams</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams would you weigh out on an electronic balance if a reaction requires 4.5 moles of lithium oxide? </h1>
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<div class="markdown"><p>103.5 grams</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> given ln <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> is actually not just grams. It's grams per mole.</p>
<p>This means that 1 mole of Li is 7grams and 1 mole of O is 16grams.</p>
<p>Together, they have a combined molecular weight of:<br/>
7+16= 23 grams per mole (or g/mol as you'll often see it written)</p>
<p>So, now that you know there are 23 grams in one mole of lithium oxide, you can simply multiply it by the number of moles you need:</p>
<p>4.5<mathjax>#*#</mathjax>23 = 103.5 grams</p></div>
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</article> | How many grams would you weigh out on an electronic balance if a reaction requires 4.5 moles of lithium oxide? | null |
2,857 | ac8ed599-6ddd-11ea-80f6-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-350-ml-at-45-c-if-the-volume-changes-to-400-ml-what-is-the | 28.13 ℃ | start physical_unit 1 1 temperature °c qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 16 17 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] ℃"}] | [{"type":"physical unit","value":"28.13 ℃"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{350 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{45 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{400 mL}"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 350 mL at 45°C. If the volume changes to 400 mL, what is the new temperature?</h1> | null | 28.13 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The correct equation to use with these kinds of questions is the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>.</p>
<p><mathjax>#(P_1*V_1)/T_1 = (P_2*V_2)/T_2#</mathjax></p>
<p>This equation, though, is a little unnecessary at times because of the possibility of extra constants.</p>
<p>In this case you have a pressure constant which enables us to shorten your longer equation down to <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> because pressure remains the same and doesn't affect the calculation. </p>
<p><mathjax>#(V_1)/T_1 = (V_2)/T_2#</mathjax></p>
<p>The next step is to isolate the variable that you want, which is, in this case, <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (V_1T_1)/V_2#</mathjax></p>
<p>Now it is time to replace your known variables with numbers</p>
<p><mathjax>#T_2 = (350mL*45°C)/(400mL)#</mathjax></p>
<p>And calculate...</p>
<p><mathjax>#T_2 = 28.125°C#</mathjax></p>
<p>Significant digits...</p>
<p><mathjax>#T_2 = 28°C#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>28°C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The correct equation to use with these kinds of questions is the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>.</p>
<p><mathjax>#(P_1*V_1)/T_1 = (P_2*V_2)/T_2#</mathjax></p>
<p>This equation, though, is a little unnecessary at times because of the possibility of extra constants.</p>
<p>In this case you have a pressure constant which enables us to shorten your longer equation down to <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> because pressure remains the same and doesn't affect the calculation. </p>
<p><mathjax>#(V_1)/T_1 = (V_2)/T_2#</mathjax></p>
<p>The next step is to isolate the variable that you want, which is, in this case, <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (V_1T_1)/V_2#</mathjax></p>
<p>Now it is time to replace your known variables with numbers</p>
<p><mathjax>#T_2 = (350mL*45°C)/(400mL)#</mathjax></p>
<p>And calculate...</p>
<p><mathjax>#T_2 = 28.125°C#</mathjax></p>
<p>Significant digits...</p>
<p><mathjax>#T_2 = 28°C#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 350 mL at 45°C. If the volume changes to 400 mL, what is the new temperature?</h1>
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<div class="markdown"><p>28°C.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The correct equation to use with these kinds of questions is the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>.</p>
<p><mathjax>#(P_1*V_1)/T_1 = (P_2*V_2)/T_2#</mathjax></p>
<p>This equation, though, is a little unnecessary at times because of the possibility of extra constants.</p>
<p>In this case you have a pressure constant which enables us to shorten your longer equation down to <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> because pressure remains the same and doesn't affect the calculation. </p>
<p><mathjax>#(V_1)/T_1 = (V_2)/T_2#</mathjax></p>
<p>The next step is to isolate the variable that you want, which is, in this case, <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#T_2 = (V_1T_1)/V_2#</mathjax></p>
<p>Now it is time to replace your known variables with numbers</p>
<p><mathjax>#T_2 = (350mL*45°C)/(400mL)#</mathjax></p>
<p>And calculate...</p>
<p><mathjax>#T_2 = 28.125°C#</mathjax></p>
<p>Significant digits...</p>
<p><mathjax>#T_2 = 28°C#</mathjax></p></div>
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</article> | A gas has a volume of 350 mL at 45°C. If the volume changes to 400 mL, what is the new temperature? | null |
2,858 | a9a3fbf7-6ddd-11ea-b57d-ccda262736ce | https://socratic.org/questions/how-many-joules-are-required-to-heat-15-2-g-of-water-at-35-c-to-70-c | 2223.76 joules | start physical_unit 10 10 heat_energy j qc_end physical_unit 10 10 12 13 temperature qc_end physical_unit 10 10 15 16 temperature qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Required heat [OF] water [IN] joules"}] | [{"type":"physical unit","value":"2223.76 joules"}] | [{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{35.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{70.0 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{15.2 g}"}] | <h1 class="questionTitle" itemprop="name">How many joules are required to heat 15.2 g of water at 35.0°C to 70.0°C? </h1> | null | 2223.76 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine how much heat is required to increase the temperature of your sample of water from <mathjax>#35.0^@"C"#</mathjax> to <mathjax>#70.0^@"C"#</mathjax>, you need to know the value of water's <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>.</p>
<p>As you know, a substance's <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> tells you how much heat is required to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a specific heat of about <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, here's how you can think about what's going on here. in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you would need <mathjax>#4.18#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Likewise, in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#4.18^@"C"#</mathjax>, you'd need <mathjax>#(4.18 xx 4.18)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>In your case, you need to increase the temperature of <mathjax>#"15.2 g"#</mathjax> of water by <mathjax>#35.0^@"C"#</mathjax>, which tells you that you're going to need <mathjax>#(15.2 xx 35)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Mathematically, this is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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 change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q = 15.2 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (70.0 - 35.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "2223.76 J"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<blockquote>
<p><mathjax>#q = color(green)("2,220 J")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2,220 J"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine how much heat is required to increase the temperature of your sample of water from <mathjax>#35.0^@"C"#</mathjax> to <mathjax>#70.0^@"C"#</mathjax>, you need to know the value of water's <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>.</p>
<p>As you know, a substance's <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> tells you how much heat is required to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a specific heat of about <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, here's how you can think about what's going on here. in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you would need <mathjax>#4.18#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Likewise, in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#4.18^@"C"#</mathjax>, you'd need <mathjax>#(4.18 xx 4.18)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>In your case, you need to increase the temperature of <mathjax>#"15.2 g"#</mathjax> of water by <mathjax>#35.0^@"C"#</mathjax>, which tells you that you're going to need <mathjax>#(15.2 xx 35)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Mathematically, this is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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 change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q = 15.2 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (70.0 - 35.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "2223.76 J"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<blockquote>
<p><mathjax>#q = color(green)("2,220 J")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many joules are required to heat 15.2 g of water at 35.0°C to 70.0°C? </h1>
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Stefan V.
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Dec 14, 2015
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<div class="markdown"><p><mathjax>#"2,220 J"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine how much heat is required to increase the temperature of your sample of water from <mathjax>#35.0^@"C"#</mathjax> to <mathjax>#70.0^@"C"#</mathjax>, you need to know the value of water's <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>.</p>
<p>As you know, a substance's <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> tells you how much heat is required to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a specific heat of about <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, here's how you can think about what's going on here. in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you would need <mathjax>#4.18#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Likewise, in order to increase the temperature of <mathjax>#"4.18 g"#</mathjax> of water by <mathjax>#4.18^@"C"#</mathjax>, you'd need <mathjax>#(4.18 xx 4.18)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>In your case, you need to increase the temperature of <mathjax>#"15.2 g"#</mathjax> of water by <mathjax>#35.0^@"C"#</mathjax>, which tells you that you're going to need <mathjax>#(15.2 xx 35)#</mathjax> <strong>times more heat</strong> than water's specific heat value. </p>
<p>Mathematically, this is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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 change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q = 15.2 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (70.0 - 35.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = "2223.76 J"#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<blockquote>
<p><mathjax>#q = color(green)("2,220 J")#</mathjax></p>
</blockquote></div>
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</article> | How many joules are required to heat 15.2 g of water at 35.0°C to 70.0°C? | null |
2,859 | ac22443f-6ddd-11ea-8f5a-ccda262736ce | https://socratic.org/questions/a-2-0-liter-container-of-nitrogen-had-a-pressure-of-3-2-kpa-what-volume-would-be | 6.40 L | start physical_unit 5 5 volume l qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 10 11 pressure qc_end physical_unit 5 5 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen [IN] L"}] | [{"type":"physical unit","value":"6.40 L"}] | [{"type":"physical unit","value":"Volume1 [OF] nitrogen [=] \\pu{2.0 liter}"},{"type":"physical unit","value":"Pressure1 [OF] nitrogen [=] \\pu{3.2 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] nitrogen [=] \\pu{1.0 kPa}"}] | <h1 class="questionTitle" itemprop="name">A 2.0 liter container of nitrogen had a pressure of 3.2 kPa. What volume would be necessary to decrease the pressure to 1.0 kPa?</h1> | null | 6.40 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So we solve for <mathjax>#V_2=(P_1V_1)/P_2=(3.2*kPaxx2.0*L)/(1.0*kPa)#</mathjax></p>
<p><mathjax>#=6.4*L#</mathjax></p>
<p>Evidently the gas was in expandable piston. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>. This is old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>: <mathjax>#Pprop1/V#</mathjax> or <mathjax>#PV=k#</mathjax>, a constant. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So we solve for <mathjax>#V_2=(P_1V_1)/P_2=(3.2*kPaxx2.0*L)/(1.0*kPa)#</mathjax></p>
<p><mathjax>#=6.4*L#</mathjax></p>
<p>Evidently the gas was in expandable piston. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A 2.0 liter container of nitrogen had a pressure of 3.2 kPa. What volume would be necessary to decrease the pressure to 1.0 kPa?</h1>
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May 14, 2017
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<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>. This is old <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>: <mathjax>#Pprop1/V#</mathjax> or <mathjax>#PV=k#</mathjax>, a constant. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So we solve for <mathjax>#V_2=(P_1V_1)/P_2=(3.2*kPaxx2.0*L)/(1.0*kPa)#</mathjax></p>
<p><mathjax>#=6.4*L#</mathjax></p>
<p>Evidently the gas was in expandable piston. </p></div>
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</article> | A 2.0 liter container of nitrogen had a pressure of 3.2 kPa. What volume would be necessary to decrease the pressure to 1.0 kPa? | null |
2,860 | a8d70dd8-6ddd-11ea-b567-ccda262736ce | https://socratic.org/questions/59bbedd87c014966b0fe6e72 | 20 mL | start physical_unit 18 19 volume ml qc_end physical_unit 6 7 2 3 volume qc_end substance 10 11 qc_end end | [{"type":"physical unit","value":"Volume [OF] ammonia gas [IN] mL"}] | [{"type":"physical unit","value":"20 mL"}] | [{"type":"physical unit","value":"Volume [OF] nitrogen gas [=] \\pu{10 mL}"},{"type":"substance name","value":"Stoichiometric hydrogen"}] | <h1 class="questionTitle" itemprop="name">If a #10*mL# volume of nitrogen gas reacts with stoichiometric hydrogen, what is the likely volume of ammonia gas produced?</h1> | null | 20 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Would it not be <mathjax>#20*mL#</mathjax>?</p>
<p>Of course, we have assumed stoichiometric reaction (the which is very unlikely)......</p>
<p>While the stoichiometric coefficients relate to moles NOT volume, given constant temperature and pressure, VOLUME is proportional to number of moles. </p></div>
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<div class="markdown"><p><mathjax>#1/2N_2(g)+3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Would it not be <mathjax>#20*mL#</mathjax>?</p>
<p>Of course, we have assumed stoichiometric reaction (the which is very unlikely)......</p>
<p>While the stoichiometric coefficients relate to moles NOT volume, given constant temperature and pressure, VOLUME is proportional to number of moles. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If a #10*mL# volume of nitrogen gas reacts with stoichiometric hydrogen, what is the likely volume of ammonia gas produced?</h1>
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<div class="markdown"><p><mathjax>#1/2N_2(g)+3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Would it not be <mathjax>#20*mL#</mathjax>?</p>
<p>Of course, we have assumed stoichiometric reaction (the which is very unlikely)......</p>
<p>While the stoichiometric coefficients relate to moles NOT volume, given constant temperature and pressure, VOLUME is proportional to number of moles. </p></div>
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</article> | If a #10*mL# volume of nitrogen gas reacts with stoichiometric hydrogen, what is the likely volume of ammonia gas produced? | null |
2,861 | a992238a-6ddd-11ea-a0c2-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-4-50-mol-fe | 251.33 g | start physical_unit 7 7 mass g qc_end physical_unit 7 7 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] Fe [IN] g"}] | [{"type":"physical unit","value":"251.33 g"}] | [{"type":"physical unit","value":"Mole [OF] Fe [=] \\pu{4.50 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 4.50 mol #Fe#?</h1> | null | 251.33 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the mass of the given molar quantity?</p>
<p><mathjax>#4.50*molxx55.85*g*mol^-1#</mathjax> <mathjax>#~~#</mathjax> <mathjax>#250*g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Elemental iron has a molar mass of <mathjax>#55.85#</mathjax> <mathjax>#g*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the mass of the given molar quantity?</p>
<p><mathjax>#4.50*molxx55.85*g*mol^-1#</mathjax> <mathjax>#~~#</mathjax> <mathjax>#250*g#</mathjax></p></div>
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<div class="markdown"><p>Elemental iron has a molar mass of <mathjax>#55.85#</mathjax> <mathjax>#g*mol^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So the mass of the given molar quantity?</p>
<p><mathjax>#4.50*molxx55.85*g*mol^-1#</mathjax> <mathjax>#~~#</mathjax> <mathjax>#250*g#</mathjax></p></div>
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May 1, 2016
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<div class="markdown"><p>4.50 moles <mathjax>#xx#</mathjax> molar mass of Fe<br/>
4.50 <mathjax>#xx#</mathjax> 55.9 = ?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume that:<br/>
n - number of moles<br/>
m - mass of substance<br/>
M - molar mass (same as atomic weight on periodic table)<br/>
<mathjax>#n = m -: M#</mathjax></p>
<p>4.50 moles (n) has been given to you.</p>
<p>Now find the molar mass (M) of Fe. If you look into your periodic table, the molar mass of Fe is 55.9 g/mol.</p>
<p>To find the mass of substance (m), you need to flip over the formula, therefore, <mathjax>#m = M xx n#</mathjax><br/>
The mass (m) of Fe is:<br/>
[55.9 g/mol <mathjax>#xx#</mathjax> 4.50 moles] = 251.6 grams.<br/>
Note: It is rounded to the nearest one decimal place.</p>
<p>4.50 moles of Fe should give you 251.6 grams of Fe.</p></div>
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</article> | What is the mass of 4.50 mol #Fe#? | null |
2,862 | aade56c6-6ddd-11ea-aeeb-ccda262736ce | https://socratic.org/questions/what-is-the-freezing-point-of-a-nonionizing-antifreeze-solution-containing-388g- | -28.42 ℃ | start physical_unit 7 9 freezing_point_temperature °c qc_end physical_unit 15 15 11 12 mass qc_end physical_unit 20 20 17 18 mass qc_end end | [{"type":"physical unit","value":"Freezing point [OF] the nonionizing antifreeze solution [IN] ℃"}] | [{"type":"physical unit","value":"-28.42 ℃"}] | [{"type":"physical unit","value":"Mass [OF] C2H6O2 [=] \\pu{388 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{409 g}"}] | <h1 class="questionTitle" itemprop="name">What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol #C_2H_6O_2# and 409 g of water?</h1> | null | -28.42 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#∆T_b#</mathjax> =<mathjax># i*K_f*m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#∆T_b#</mathjax> = depression in freezing point</li>
<li><mathjax>#i #</mathjax>= vant hoff factor</li>
<li><mathjax>#K_f#</mathjax> = depression constant</li>
<li><mathjax>#m#</mathjax> = <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></li>
</ul>
<p>mass of ethylene = <mathjax>#388 g#</mathjax></p>
<p>molar mass =<mathjax># 62.07#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>no. of moles = <mathjax>#388#</mathjax>/<mathjax>#62.07#</mathjax> = <mathjax>#6.25#</mathjax> moles</p>
<p>mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> = <mathjax>#409 g#</mathjax> = <mathjax>#0.409 kg#</mathjax></p>
<p>molality = <mathjax>#6.25#</mathjax> moles/<mathjax>#0.409 kg#</mathjax> = <mathjax>#15.28 m#</mathjax></p>
<ul>
<li>
<p><mathjax>#i #</mathjax>= <mathjax>#1#</mathjax></p>
</li>
<li>
<p><mathjax>#T_i#</mathjax> =<mathjax># 0^oC#</mathjax> </p>
</li>
<li>
<p><mathjax>#K_f#</mathjax>= <mathjax>#1.86^ oC/m#</mathjax></p>
</li>
</ul>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#∆T_f = i*K_f*m#</mathjax><br/>
<mathjax>#∆T_f = 1.86^oC/m xx 15.28 m#</mathjax></p>
<p><mathjax>#∆T_f = 28.42^o C#</mathjax></p>
<p><mathjax>#T_f = 0^oC - 28.42^ oC = -28.42^ oC#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The freezing point = <mathjax># - 28.42^oC#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#∆T_b#</mathjax> =<mathjax># i*K_f*m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#∆T_b#</mathjax> = depression in freezing point</li>
<li><mathjax>#i #</mathjax>= vant hoff factor</li>
<li><mathjax>#K_f#</mathjax> = depression constant</li>
<li><mathjax>#m#</mathjax> = <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></li>
</ul>
<p>mass of ethylene = <mathjax>#388 g#</mathjax></p>
<p>molar mass =<mathjax># 62.07#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>no. of moles = <mathjax>#388#</mathjax>/<mathjax>#62.07#</mathjax> = <mathjax>#6.25#</mathjax> moles</p>
<p>mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> = <mathjax>#409 g#</mathjax> = <mathjax>#0.409 kg#</mathjax></p>
<p>molality = <mathjax>#6.25#</mathjax> moles/<mathjax>#0.409 kg#</mathjax> = <mathjax>#15.28 m#</mathjax></p>
<ul>
<li>
<p><mathjax>#i #</mathjax>= <mathjax>#1#</mathjax></p>
</li>
<li>
<p><mathjax>#T_i#</mathjax> =<mathjax># 0^oC#</mathjax> </p>
</li>
<li>
<p><mathjax>#K_f#</mathjax>= <mathjax>#1.86^ oC/m#</mathjax></p>
</li>
</ul>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#∆T_f = i*K_f*m#</mathjax><br/>
<mathjax>#∆T_f = 1.86^oC/m xx 15.28 m#</mathjax></p>
<p><mathjax>#∆T_f = 28.42^o C#</mathjax></p>
<p><mathjax>#T_f = 0^oC - 28.42^ oC = -28.42^ oC#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol #C_2H_6O_2# and 409 g of water?</h1>
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<div class="markdown"><p>The freezing point = <mathjax># - 28.42^oC#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#∆T_b#</mathjax> =<mathjax># i*K_f*m#</mathjax></p>
</blockquote>
<ul>
<li><mathjax>#∆T_b#</mathjax> = depression in freezing point</li>
<li><mathjax>#i #</mathjax>= vant hoff factor</li>
<li><mathjax>#K_f#</mathjax> = depression constant</li>
<li><mathjax>#m#</mathjax> = <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></li>
</ul>
<p>mass of ethylene = <mathjax>#388 g#</mathjax></p>
<p>molar mass =<mathjax># 62.07#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>no. of moles = <mathjax>#388#</mathjax>/<mathjax>#62.07#</mathjax> = <mathjax>#6.25#</mathjax> moles</p>
<p>mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> = <mathjax>#409 g#</mathjax> = <mathjax>#0.409 kg#</mathjax></p>
<p>molality = <mathjax>#6.25#</mathjax> moles/<mathjax>#0.409 kg#</mathjax> = <mathjax>#15.28 m#</mathjax></p>
<ul>
<li>
<p><mathjax>#i #</mathjax>= <mathjax>#1#</mathjax></p>
</li>
<li>
<p><mathjax>#T_i#</mathjax> =<mathjax># 0^oC#</mathjax> </p>
</li>
<li>
<p><mathjax>#K_f#</mathjax>= <mathjax>#1.86^ oC/m#</mathjax></p>
</li>
</ul>
<p>Therefore,</p>
<blockquote>
<p><mathjax>#∆T_f = i*K_f*m#</mathjax><br/>
<mathjax>#∆T_f = 1.86^oC/m xx 15.28 m#</mathjax></p>
<p><mathjax>#∆T_f = 28.42^o C#</mathjax></p>
<p><mathjax>#T_f = 0^oC - 28.42^ oC = -28.42^ oC#</mathjax></p>
</blockquote></div>
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</article> | What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol #C_2H_6O_2# and 409 g of water? | null |
2,863 | ac9ccf2e-6ddd-11ea-927f-ccda262736ce | https://socratic.org/questions/what-is-the-temperature-of-0-59-mol-of-gas-at-a-pressure-of-1-5-atm-and-a-volume | 399.68 K | start physical_unit 8 8 temperature k qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 8 8 13 14 pressure qc_end physical_unit 8 8 19 20 volume qc_end end | [{"type":"physical unit","value":"Temperature [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"399.68 K"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{0.59 mol}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{1.5 atm}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{12.9 L}"}] | <h1 class="questionTitle" itemprop="name">What is the temperature of 0.59 mol of gas at a pressure of 1.5 atm and a volume of 12.9 L ?</h1> | null | 399.68 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming it behaves as an ideal gas, you can say <mathjax>#Pv=nRT#</mathjax> where P is pressure, v is volume, n is number of moles, R is the gas constant and T is temperature in Kelvins</p>
<p>Note that pressure and volume can be expressed in a number of different units, so you need to ensure that you use the gas constant in a form that is consistent with the other units that you're working with. As we're working in atmospheres as a pressure unit and litres as the unit of volume you need the gas constant in units of <mathjax>#L*atm*K^-1*mol^-1#</mathjax>, and in these units the value of the gas constant is <mathjax>#0.082057#</mathjax></p>
<p>Now rearrange for T you have <mathjax>#T= (Pv)/(nR) = (1.5 * 12.9)/(0.59*0.082057) = 399.68 K#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>399.68 K or just over 126 celcius.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming it behaves as an ideal gas, you can say <mathjax>#Pv=nRT#</mathjax> where P is pressure, v is volume, n is number of moles, R is the gas constant and T is temperature in Kelvins</p>
<p>Note that pressure and volume can be expressed in a number of different units, so you need to ensure that you use the gas constant in a form that is consistent with the other units that you're working with. As we're working in atmospheres as a pressure unit and litres as the unit of volume you need the gas constant in units of <mathjax>#L*atm*K^-1*mol^-1#</mathjax>, and in these units the value of the gas constant is <mathjax>#0.082057#</mathjax></p>
<p>Now rearrange for T you have <mathjax>#T= (Pv)/(nR) = (1.5 * 12.9)/(0.59*0.082057) = 399.68 K#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the temperature of 0.59 mol of gas at a pressure of 1.5 atm and a volume of 12.9 L ?</h1>
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<div class="markdown"><p>399.68 K or just over 126 celcius.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming it behaves as an ideal gas, you can say <mathjax>#Pv=nRT#</mathjax> where P is pressure, v is volume, n is number of moles, R is the gas constant and T is temperature in Kelvins</p>
<p>Note that pressure and volume can be expressed in a number of different units, so you need to ensure that you use the gas constant in a form that is consistent with the other units that you're working with. As we're working in atmospheres as a pressure unit and litres as the unit of volume you need the gas constant in units of <mathjax>#L*atm*K^-1*mol^-1#</mathjax>, and in these units the value of the gas constant is <mathjax>#0.082057#</mathjax></p>
<p>Now rearrange for T you have <mathjax>#T= (Pv)/(nR) = (1.5 * 12.9)/(0.59*0.082057) = 399.68 K#</mathjax></p></div>
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</article> | What is the temperature of 0.59 mol of gas at a pressure of 1.5 atm and a volume of 12.9 L ? | null |
2,864 | ac28d118-6ddd-11ea-a188-ccda262736ce | https://socratic.org/questions/5817f2147c0149405b64cabd | 200.00 mL | start physical_unit 4 4 volume ml qc_end physical_unit 4 4 1 2 volume qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 14 15 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"200.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{300 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{0.500 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{0.750 atm}"}] | <h1 class="questionTitle" itemprop="name">A #300*mL# volume gas at #0.500*atm# in a piston, is compressed to #0.750*atm#. What is the new volume?</h1> | null | 200.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity of gas, at constant temperature, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant........</p>
<p>And it follows that <mathjax>#P_1V_1=P_2V_2#</mathjax>, and.........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume as required.....</p>
<p>Here <mathjax>#V_2=(300*mLxx0.500*atm)/(0.750*atm)#</mathjax></p>
<p><mathjax>#=200*mL#</mathjax>, volume is reduced as expected..........</p></div>
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<div class="markdown"><p><mathjax>#V_2=200*mL#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity of gas, at constant temperature, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant........</p>
<p>And it follows that <mathjax>#P_1V_1=P_2V_2#</mathjax>, and.........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume as required.....</p>
<p>Here <mathjax>#V_2=(300*mLxx0.500*atm)/(0.750*atm)#</mathjax></p>
<p><mathjax>#=200*mL#</mathjax>, volume is reduced as expected..........</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A #300*mL# volume gas at #0.500*atm# in a piston, is compressed to #0.750*atm#. What is the new volume?</h1>
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anor277
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<div class="markdown"><p><mathjax>#V_2=200*mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Old <mathjax>#"Boyle's Law"#</mathjax> holds that for a given quantity of gas, at constant temperature, <mathjax>#Pprop1/V#</mathjax>, and thus <mathjax>#PV=k#</mathjax>, where <mathjax>#k#</mathjax> is some constant........</p>
<p>And it follows that <mathjax>#P_1V_1=P_2V_2#</mathjax>, and.........</p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume as required.....</p>
<p>Here <mathjax>#V_2=(300*mLxx0.500*atm)/(0.750*atm)#</mathjax></p>
<p><mathjax>#=200*mL#</mathjax>, volume is reduced as expected..........</p></div>
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</article> | A #300*mL# volume gas at #0.500*atm# in a piston, is compressed to #0.750*atm#. What is the new volume? | null |
2,865 | ab2f0674-6ddd-11ea-ac03-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-containing-83-potassium-and-17-oxyge | K2O | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"K2O"}] | [{"type":"physical unit","value":"Percent [OF] potassium in the compound [=] \\pu{83%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{17%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound containing 83% potassium and 17% oxygen?</h1> | null | K2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The following method serves for every compound:</p>
<p>In each 100 g of the compound you have 83 g of potassium and 17 g of oxygen.</p>
<p>To determine the empiriacal formula you have to pass these masses into mols</p>
<p><mathjax>#83/39.1=2.12 mol K#</mathjax> and <mathjax>#17/16=1.06 mol O#</mathjax></p>
<p>Now divide all the numbers by the smallest you have obatined:</p>
<p><mathjax>#2.12/1.06=2.0 K#</mathjax> and <mathjax>#1.06/1.06=1 mol O#</mathjax></p>
<p>So it is <mathjax>#K_2O#</mathjax></p>
<p>At this stage you can have numbers that are not integer. In this case you would have to multiply the resulsts by an integer in a way the final result is a proportion of integers:</p>
<p>Imagine you had obtained 1.5 K for 1 O. You just had to multiply by 2 to get the integer proportion of 3:1.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#K_2O#</mathjax>,</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The following method serves for every compound:</p>
<p>In each 100 g of the compound you have 83 g of potassium and 17 g of oxygen.</p>
<p>To determine the empiriacal formula you have to pass these masses into mols</p>
<p><mathjax>#83/39.1=2.12 mol K#</mathjax> and <mathjax>#17/16=1.06 mol O#</mathjax></p>
<p>Now divide all the numbers by the smallest you have obatined:</p>
<p><mathjax>#2.12/1.06=2.0 K#</mathjax> and <mathjax>#1.06/1.06=1 mol O#</mathjax></p>
<p>So it is <mathjax>#K_2O#</mathjax></p>
<p>At this stage you can have numbers that are not integer. In this case you would have to multiply the resulsts by an integer in a way the final result is a proportion of integers:</p>
<p>Imagine you had obtained 1.5 K for 1 O. You just had to multiply by 2 to get the integer proportion of 3:1.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound containing 83% potassium and 17% oxygen?</h1>
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José F.
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<div class="markdown"><p><mathjax>#K_2O#</mathjax>,</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The following method serves for every compound:</p>
<p>In each 100 g of the compound you have 83 g of potassium and 17 g of oxygen.</p>
<p>To determine the empiriacal formula you have to pass these masses into mols</p>
<p><mathjax>#83/39.1=2.12 mol K#</mathjax> and <mathjax>#17/16=1.06 mol O#</mathjax></p>
<p>Now divide all the numbers by the smallest you have obatined:</p>
<p><mathjax>#2.12/1.06=2.0 K#</mathjax> and <mathjax>#1.06/1.06=1 mol O#</mathjax></p>
<p>So it is <mathjax>#K_2O#</mathjax></p>
<p>At this stage you can have numbers that are not integer. In this case you would have to multiply the resulsts by an integer in a way the final result is a proportion of integers:</p>
<p>Imagine you had obtained 1.5 K for 1 O. You just had to multiply by 2 to get the integer proportion of 3:1.</p></div>
</div>
</div>
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</article> | What is the empirical formula of a compound containing 83% potassium and 17% oxygen? | null |
2,866 | ad01d4a6-6ddd-11ea-829a-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-0-208g-of-mg | 8.55 × 10^(-3) moles | start physical_unit 8 8 mole mol qc_end physical_unit 8 8 5 6 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] Mg [IN] moles"}] | [{"type":"physical unit","value":"8.55 × 10^(-3) moles"}] | [{"type":"physical unit","value":"Mass [OF] Mg [=] \\pu{0.208 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 0.208g of Mg?</h1> | null | 8.55 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#"Mg"#</mathjax> is <mathjax>#"24.305 g/mol"#</mathjax>, so</p>
<p><mathjax>#"0.208 g Mg" /("24.305 g/mol") = 8.55*10^-3#</mathjax> <mathjax>#"moles"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#8.55*10^-3#</mathjax> <mathjax>#"moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#"Mg"#</mathjax> is <mathjax>#"24.305 g/mol"#</mathjax>, so</p>
<p><mathjax>#"0.208 g Mg" /("24.305 g/mol") = 8.55*10^-3#</mathjax> <mathjax>#"moles"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 0.208g of Mg?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#8.55*10^-3#</mathjax> <mathjax>#"moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#"Mg"#</mathjax> is <mathjax>#"24.305 g/mol"#</mathjax>, so</p>
<p><mathjax>#"0.208 g Mg" /("24.305 g/mol") = 8.55*10^-3#</mathjax> <mathjax>#"moles"#</mathjax></p></div>
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</article> | How many moles are in 0.208g of Mg? | null |
2,867 | abd111f4-6ddd-11ea-8e3f-ccda262736ce | https://socratic.org/questions/the-ph-of-a-fruit-juice-is-3-2-what-is-the-hydrogen-ion-concentration-of-the-jui | 6 × 10^(-4) M | start physical_unit 10 11 concentration mol/l qc_end physical_unit 3 4 6 6 ph qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydrogen ion [IN] M"}] | [{"type":"physical unit","value":"6 × 10^(-4) M"}] | [{"type":"physical unit","value":"pH [OF] fruit juice [=] \\pu{3.2}"}] | <h1 class="questionTitle" itemprop="name">The pH of a fruit juice is 3.2. What is the hydrogen ion concentration of the juice? </h1> | null | 6 × 10^(-4) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <mathjax>#["H"^+]#</mathjax> of a solution, given its <mathjax>#"pH"#</mathjax>.</p>
<p>We can use the equation</p>
<p><mathjax>#["H"^+] = 10^(-"pH")#</mathjax></p>
<p>to find the hydrogen ion concentration.</p>
<p>We thus have</p>
<p><mathjax>#["H"^+] = 10^-3.2 = color(red)(6xx10^-4M#</mathjax></p>
<p>Since the <mathjax>#"pH"#</mathjax> is rounded to one <strong>decimal place</strong>, the hydrogen ion concentration is rounded to one <strong>significant figure</strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#["H"^+] = 6xx10^-4M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <mathjax>#["H"^+]#</mathjax> of a solution, given its <mathjax>#"pH"#</mathjax>.</p>
<p>We can use the equation</p>
<p><mathjax>#["H"^+] = 10^(-"pH")#</mathjax></p>
<p>to find the hydrogen ion concentration.</p>
<p>We thus have</p>
<p><mathjax>#["H"^+] = 10^-3.2 = color(red)(6xx10^-4M#</mathjax></p>
<p>Since the <mathjax>#"pH"#</mathjax> is rounded to one <strong>decimal place</strong>, the hydrogen ion concentration is rounded to one <strong>significant figure</strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">The pH of a fruit juice is 3.2. What is the hydrogen ion concentration of the juice? </h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#["H"^+] = 6xx10^-4M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <mathjax>#["H"^+]#</mathjax> of a solution, given its <mathjax>#"pH"#</mathjax>.</p>
<p>We can use the equation</p>
<p><mathjax>#["H"^+] = 10^(-"pH")#</mathjax></p>
<p>to find the hydrogen ion concentration.</p>
<p>We thus have</p>
<p><mathjax>#["H"^+] = 10^-3.2 = color(red)(6xx10^-4M#</mathjax></p>
<p>Since the <mathjax>#"pH"#</mathjax> is rounded to one <strong>decimal place</strong>, the hydrogen ion concentration is rounded to one <strong>significant figure</strong>.</p></div>
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</article> | The pH of a fruit juice is 3.2. What is the hydrogen ion concentration of the juice? | null |
2,868 | abe652a2-6ddd-11ea-b59b-ccda262736ce | https://socratic.org/questions/how-many-moles-of-h-2so-4-are-required-to-completely-neutralize-0-10-moles-of-ca | 0.10 moles | start physical_unit 4 4 mole mol qc_end c_other OTHER qc_end physical_unit 13 13 10 11 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] H2SO4 [IN] moles"}] | [{"type":"physical unit","value":"0.10 moles"}] | [{"type":"other","value":"Completely neutralize."},{"type":"physical unit","value":"Mole [OF] Ca(OH)2 [=] \\pu{0.10 moles}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #H_2SO_4# are required to completely neutralize 0.10 moles of #Ca(OH)_2#? </h1> | null | 0.10 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When the reaction occurs CaSO4 and water (2 molecules) formed.</p>
<p>It means for one mole of sulphuric acid, one mole of Ca(OH)2 reacts.</p>
<p>If you have 0.1 mol of Ca(OH)2, you will need 0.1 mol of sulphuric acid. Or for 7.4 grams Ca(OH)2 you need 9.8 grams (or 0.1 mol) of H2SO4.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>0.10 moles or 9.8 grams.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When the reaction occurs CaSO4 and water (2 molecules) formed.</p>
<p>It means for one mole of sulphuric acid, one mole of Ca(OH)2 reacts.</p>
<p>If you have 0.1 mol of Ca(OH)2, you will need 0.1 mol of sulphuric acid. Or for 7.4 grams Ca(OH)2 you need 9.8 grams (or 0.1 mol) of H2SO4.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #H_2SO_4# are required to completely neutralize 0.10 moles of #Ca(OH)_2#? </h1>
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<div class="markdown"><p>0.10 moles or 9.8 grams.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When the reaction occurs CaSO4 and water (2 molecules) formed.</p>
<p>It means for one mole of sulphuric acid, one mole of Ca(OH)2 reacts.</p>
<p>If you have 0.1 mol of Ca(OH)2, you will need 0.1 mol of sulphuric acid. Or for 7.4 grams Ca(OH)2 you need 9.8 grams (or 0.1 mol) of H2SO4.</p></div>
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</article> | How many moles of #H_2SO_4# are required to completely neutralize 0.10 moles of #Ca(OH)_2#? | null |
2,869 | ab069876-6ddd-11ea-8774-ccda262736ce | https://socratic.org/questions/what-is-k-a-of-zncl-2 | 1.94 × 10^(-8) | start physical_unit 4 4 ka none qc_end chemical_equation 4 4 qc_end end | [{"type":"physical unit","value":"Ka [OF] ZnCl2"}] | [{"type":"physical unit","value":"1.94 × 10^(-8)"}] | [{"type":"chemical equation","value":"ZnCl2"}] | <h1 class="questionTitle" itemprop="name">What is #K_a# of #"ZnCl"_2# ?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>I know its small but what is the exact . Tell me the source too </p></div>
</h2>
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</div> | 1.94 × 10^(-8) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.5 * 10^(-10)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#2.5 * 10^(-10)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
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Professor Sam
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<div class="markdown"><p>The Ka is nearly <mathjax>#1.936E-8#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_8/manual.pdf" rel="nofollow" target="_blank">http://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_8/manual.pdf</a></p>
<p>In this website I checked that pH of 0.1M of ZnCl2 is 4.4<br/>
I tried to solve for Ka (<mathjax>#H^+#</mathjax> = <mathjax>#10^-4.4#</mathjax>)<br/>
<mathjax>#sqrt(0.1x) = H^+ or 10^-4.4#</mathjax><br/>
or<br/>
<mathjax>#0.1x = 0.000044^2#</mathjax></p>
<p><mathjax>#0.1x= 1.936E-9#</mathjax></p>
<p><mathjax>#x = Ka = 1.936E-8 = \frac{1.936E-9}{0.1}#</mathjax></p>
<p>You can solve this by ICE table. Answer would be the same</p>
<p>The Ka is <mathjax>#1.936E-8..#</mathjax> and If you check in this website what is the Ka of Zn(H2O)6 it would be the same . Just they have rounded it off to simpler numbers. These dont matter much</p></div>
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</article> | What is #K_a# of #"ZnCl"_2# ? |
I know its small but what is the exact . Tell me the source too
|
2,870 | ad0387c7-6ddd-11ea-877c-ccda262736ce | https://socratic.org/questions/what-is-the-formula-of-aluminum-carbonate | Al2(CO3)3 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] aluminum carbonate [IN] default"}] | [{"type":"chemical equation","value":"Al2(CO3)3"}] | [{"type":"substance name","value":"Aluminum carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the formula of aluminum carbonate? </h1> | null | Al2(CO3)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminium generally has a charge of +3 <br/>
The <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> of Aluminium is <mathjax># 1s^2 2s^2 2p^6 3s^2 3p^1#</mathjax><br/>
= <br/>
The 3p electron and the 2 3s electrons are outer valance electrons and are lost to form the +3 charge. </p>
<p>The Carbonate ion has a charge of -2 </p>
<p><mathjax>#1 C =1 xx +4 #</mathjax><br/>
<mathjax>#3 O = 3 xx -2 = -6#</mathjax></p>
<p><mathjax># +4 + -6 = -2 #</mathjax></p>
<p>Chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero </p>
<p><mathjax># 2 xx (+3) + 3 xx ( -2) = 0 #</mathjax><br/>
<mathjax># + 6 + -6 = 0#</mathjax></p>
<p>so the formula is </p>
<p><mathjax># Al_2 (CO_3)_3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax># Al_2CO_3)_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminium generally has a charge of +3 <br/>
The <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> of Aluminium is <mathjax># 1s^2 2s^2 2p^6 3s^2 3p^1#</mathjax><br/>
= <br/>
The 3p electron and the 2 3s electrons are outer valance electrons and are lost to form the +3 charge. </p>
<p>The Carbonate ion has a charge of -2 </p>
<p><mathjax>#1 C =1 xx +4 #</mathjax><br/>
<mathjax>#3 O = 3 xx -2 = -6#</mathjax></p>
<p><mathjax># +4 + -6 = -2 #</mathjax></p>
<p>Chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero </p>
<p><mathjax># 2 xx (+3) + 3 xx ( -2) = 0 #</mathjax><br/>
<mathjax># + 6 + -6 = 0#</mathjax></p>
<p>so the formula is </p>
<p><mathjax># Al_2 (CO_3)_3#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the formula of aluminum carbonate? </h1>
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David Drayer
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May 6, 2018
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<div class="markdown"><p><mathjax># Al_2CO_3)_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminium generally has a charge of +3 <br/>
The <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/electron-configuration">electron configuration</a> of Aluminium is <mathjax># 1s^2 2s^2 2p^6 3s^2 3p^1#</mathjax><br/>
= <br/>
The 3p electron and the 2 3s electrons are outer valance electrons and are lost to form the +3 charge. </p>
<p>The Carbonate ion has a charge of -2 </p>
<p><mathjax>#1 C =1 xx +4 #</mathjax><br/>
<mathjax>#3 O = 3 xx -2 = -6#</mathjax></p>
<p><mathjax># +4 + -6 = -2 #</mathjax></p>
<p>Chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of zero </p>
<p><mathjax># 2 xx (+3) + 3 xx ( -2) = 0 #</mathjax><br/>
<mathjax># + 6 + -6 = 0#</mathjax></p>
<p>so the formula is </p>
<p><mathjax># Al_2 (CO_3)_3#</mathjax></p></div>
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Camilleon
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May 6, 2018
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<div class="markdown"><p><mathjax>#Al_2#</mathjax><mathjax>#(CO_3)_3#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Aluminium - <mathjax>#Al^3#</mathjax><br/>
Carbonate - <mathjax>#CO_3^-2#</mathjax></p>
<p>Because the charge of aluminium is 3 and the charge of carbonate is -2, you need to multiply each to balance out at 6. So, "swap and drop":</p>
<p>2<mathjax>#Al^3#</mathjax> & 3<mathjax>#CO_3^-2#</mathjax></p>
<p><mathjax>#Al_2#</mathjax><mathjax>#(CO_3)_3#</mathjax></p></div>
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</article> | What is the formula of aluminum carbonate? | null |
2,871 | ab0bef88-6ddd-11ea-a084-ccda262736ce | https://socratic.org/questions/59274df6b72cff4a62ca48d2 | 18.00 mL | start physical_unit 20 20 volume ml qc_end physical_unit 5 6 8 9 concentration qc_end physical_unit 16 16 14 15 concentration qc_end end | [{"type":"physical unit","value":"Volume [OF] acid [IN] mL"}] | [{"type":"physical unit","value":"18.00 mL"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide solution [=] \\pu{60 mL}"},{"type":"physical unit","value":"Concentration [OF] sodium hydroxide solution [=] \\pu{0.15 mol/L}"},{"type":"physical unit","value":"Concentration [OF] HCl(aq) solution [=] \\pu{0.500 mol/L}"}] | <h1 class="questionTitle" itemprop="name">A #60*mL# volume of sodium hydroxide at #0.15*mol*L^-1# concentration is titrated with #0.500*mol*L^-1# #HCl(aq)#. What volume of acid is required?</h1> | null | 18.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, there is certainly 1:1 <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> between acid and base, i.e.:</p>
<p><mathjax>#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#</mathjax></p>
<p>And please note that the product <mathjax>#C_1V_1="moles"*L^-1xxL#</mathjax>, i.e. an ANSWER in moles as required...........</p>
<p>And so if <mathjax>#C_1V_1=C_2V_2#</mathjax>, then............</p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(60*mLxx0.15*mol*L^-1)/(0.500*mol*L^-1)=18*mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use the relationship <mathjax>#C_1V_1=C_2V_2#</mathjax>, where <mathjax>#C_i#</mathjax>, and <mathjax>#V_i#</mathjax> are the individual concentrations/volumes. But FIRST you need a stoichiometric equation. Eventually I gets under <mathjax>#20*mL#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, there is certainly 1:1 <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> between acid and base, i.e.:</p>
<p><mathjax>#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#</mathjax></p>
<p>And please note that the product <mathjax>#C_1V_1="moles"*L^-1xxL#</mathjax>, i.e. an ANSWER in moles as required...........</p>
<p>And so if <mathjax>#C_1V_1=C_2V_2#</mathjax>, then............</p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(60*mLxx0.15*mol*L^-1)/(0.500*mol*L^-1)=18*mL#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A #60*mL# volume of sodium hydroxide at #0.15*mol*L^-1# concentration is titrated with #0.500*mol*L^-1# #HCl(aq)#. What volume of acid is required?</h1>
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anor277
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May 25, 2017
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<div class="markdown"><p>We use the relationship <mathjax>#C_1V_1=C_2V_2#</mathjax>, where <mathjax>#C_i#</mathjax>, and <mathjax>#V_i#</mathjax> are the individual concentrations/volumes. But FIRST you need a stoichiometric equation. Eventually I gets under <mathjax>#20*mL#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, there is certainly 1:1 <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> between acid and base, i.e.:</p>
<p><mathjax>#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#</mathjax></p>
<p>And please note that the product <mathjax>#C_1V_1="moles"*L^-1xxL#</mathjax>, i.e. an ANSWER in moles as required...........</p>
<p>And so if <mathjax>#C_1V_1=C_2V_2#</mathjax>, then............</p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(60*mLxx0.15*mol*L^-1)/(0.500*mol*L^-1)=18*mL#</mathjax></p></div>
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</article> | A #60*mL# volume of sodium hydroxide at #0.15*mol*L^-1# concentration is titrated with #0.500*mol*L^-1# #HCl(aq)#. What volume of acid is required? | null |
2,872 | ab2805a2-6ddd-11ea-8495-ccda262736ce | https://socratic.org/questions/how-many-atoms-are-there-in-a-molecule-of-ammonium-sulfate | 15 | start physical_unit 2 2 number none qc_end end | [{"type":"physical unit","value":"Number [OF] atoms"}] | [{"type":"physical unit","value":"15"}] | [{"type":"physical unit","value":"Number [OF] ammonium sulfate molecule [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How many atoms are there in a molecule of ammonium sulfate?</h1> | null | 15 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Formula of Ammonium sulphate <mathjax>#"= (NH"_4")"_2"SO"_4#</mathjax></p>
<p>So, in a single molecule of Ammonium sulphate there are</p>
<ul>
<li><mathjax>#2#</mathjax> Nitrogen atoms</li>
<li><mathjax>#8#</mathjax> Hydrogen atoms</li>
<li><mathjax>#1#</mathjax> Sulphur atom</li>
<li><mathjax>#4#</mathjax> Oxygen atoms </li>
</ul>
<p>Total atoms<mathjax>#= 2 + 8 + 1 + 4 = 15#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#15#</mathjax> atoms</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Formula of Ammonium sulphate <mathjax>#"= (NH"_4")"_2"SO"_4#</mathjax></p>
<p>So, in a single molecule of Ammonium sulphate there are</p>
<ul>
<li><mathjax>#2#</mathjax> Nitrogen atoms</li>
<li><mathjax>#8#</mathjax> Hydrogen atoms</li>
<li><mathjax>#1#</mathjax> Sulphur atom</li>
<li><mathjax>#4#</mathjax> Oxygen atoms </li>
</ul>
<p>Total atoms<mathjax>#= 2 + 8 + 1 + 4 = 15#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many atoms are there in a molecule of ammonium sulfate?</h1>
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<div class="markdown"><p><mathjax>#15#</mathjax> atoms</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Formula of Ammonium sulphate <mathjax>#"= (NH"_4")"_2"SO"_4#</mathjax></p>
<p>So, in a single molecule of Ammonium sulphate there are</p>
<ul>
<li><mathjax>#2#</mathjax> Nitrogen atoms</li>
<li><mathjax>#8#</mathjax> Hydrogen atoms</li>
<li><mathjax>#1#</mathjax> Sulphur atom</li>
<li><mathjax>#4#</mathjax> Oxygen atoms </li>
</ul>
<p>Total atoms<mathjax>#= 2 + 8 + 1 + 4 = 15#</mathjax></p></div>
</div>
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</article> | How many atoms are there in a molecule of ammonium sulfate? | null |
2,873 | ac7dd136-6ddd-11ea-bfc3-ccda262736ce | https://socratic.org/questions/a-gas-at-362-k-occupies-a-volume-of-0-67-l-at-what-temperature-will-the-volume-i | 605.13 K | start physical_unit 1 1 temperature k qc_end physical_unit 1 1 3 4 temperature qc_end physical_unit 1 1 9 10 volume qc_end physical_unit 1 1 19 20 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"605.13 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{362 K}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.67 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{1.12 L}"}] | <h1 class="questionTitle" itemprop="name">A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?</h1> | null | 605.13 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.</p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>where <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#T_1#</mathjax> is the initial temperature, <mathjax>#V_2#</mathjax> is the final volume, and <mathjax>#"T_2"#</mathjax> is the final temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#T_1="362 K"#</mathjax><br/>
<mathjax>#V_1="0.67 L"#</mathjax><br/>
<mathjax>#V_2="1.12 L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, then substitute the known values into the equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2#</mathjax><mathjax>#color(white)(.) "K"#</mathjax> <br/>
(<mathjax>#"605 K"#</mathjax> rounded to two significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The temperature at which the volume increases to <mathjax>#"1.12 L"#</mathjax> is <mathjax>#6.1xx10^2#</mathjax><mathjax>#color(white)(.)"K"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.</p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>where <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#T_1#</mathjax> is the initial temperature, <mathjax>#V_2#</mathjax> is the final volume, and <mathjax>#"T_2"#</mathjax> is the final temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#T_1="362 K"#</mathjax><br/>
<mathjax>#V_1="0.67 L"#</mathjax><br/>
<mathjax>#V_2="1.12 L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, then substitute the known values into the equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2#</mathjax><mathjax>#color(white)(.) "K"#</mathjax> <br/>
(<mathjax>#"605 K"#</mathjax> rounded to two significant figures)</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?</h1>
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<div class="markdown"><p>The temperature at which the volume increases to <mathjax>#"1.12 L"#</mathjax> is <mathjax>#6.1xx10^2#</mathjax><mathjax>#color(white)(.)"K"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a></strong> , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.</p>
<p><strong>The equation to use is:</strong></p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>where <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#T_1#</mathjax> is the initial temperature, <mathjax>#V_2#</mathjax> is the final volume, and <mathjax>#"T_2"#</mathjax> is the final temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#T_1="362 K"#</mathjax><br/>
<mathjax>#V_1="0.67 L"#</mathjax><br/>
<mathjax>#V_2="1.12 L"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, then substitute the known values into the equation and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2#</mathjax><mathjax>#color(white)(.) "K"#</mathjax> <br/>
(<mathjax>#"605 K"#</mathjax> rounded to two significant figures)</p></div>
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</article> | A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L? | null |
2,874 | a9a3fbf6-6ddd-11ea-9cd4-ccda262736ce | https://socratic.org/questions/what-is-the-poh-of-a-8-4x10-11-m-oh-solution | 10.08 | start physical_unit 11 11 poh none qc_end physical_unit 10 10 6 9 molarity qc_end end | [{"type":"physical unit","value":"pOH [OF] the solution"}] | [{"type":"physical unit","value":"10.08"}] | [{"type":"physical unit","value":"Molarity [OF] OH- [=] \\pu{8.4 × 10^(-11) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pOH of a #8.4x10^-11 M# #OH^-# solution?</h1> | null | 10.08 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#pOH#</mathjax> of a solution is defined in terms of the concentration of hydroxide ions in solution as follows:</p>
<p><mathjax>#pOH=-log[OH^(-)]#</mathjax>.</p>
<p>In this question, <mathjax>#[OH^(-)]=8.4xx10^(-11)M#</mathjax>, thus the <mathjax>#pOH#</mathjax> can be calculated by:</p>
<p><mathjax>#pOH=-log(8.4xx10^(-11))=10.0757~=10.08#</mathjax></p>
<p>Note that if the concentration is given with <strong>2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></strong> , the <mathjax>#pOH#</mathjax> value is written with <strong>two decimal places</strong>.</p></div>
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<div class="markdown"><p><mathjax>#pOH=10.08#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#pOH#</mathjax> of a solution is defined in terms of the concentration of hydroxide ions in solution as follows:</p>
<p><mathjax>#pOH=-log[OH^(-)]#</mathjax>.</p>
<p>In this question, <mathjax>#[OH^(-)]=8.4xx10^(-11)M#</mathjax>, thus the <mathjax>#pOH#</mathjax> can be calculated by:</p>
<p><mathjax>#pOH=-log(8.4xx10^(-11))=10.0757~=10.08#</mathjax></p>
<p>Note that if the concentration is given with <strong>2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></strong> , the <mathjax>#pOH#</mathjax> value is written with <strong>two decimal places</strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pOH of a #8.4x10^-11 M# #OH^-# solution?</h1>
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<div class="markdown"><p><mathjax>#pOH=10.08#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#pOH#</mathjax> of a solution is defined in terms of the concentration of hydroxide ions in solution as follows:</p>
<p><mathjax>#pOH=-log[OH^(-)]#</mathjax>.</p>
<p>In this question, <mathjax>#[OH^(-)]=8.4xx10^(-11)M#</mathjax>, thus the <mathjax>#pOH#</mathjax> can be calculated by:</p>
<p><mathjax>#pOH=-log(8.4xx10^(-11))=10.0757~=10.08#</mathjax></p>
<p>Note that if the concentration is given with <strong>2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></strong> , the <mathjax>#pOH#</mathjax> value is written with <strong>two decimal places</strong>.</p></div>
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</article> | What is the pOH of a #8.4x10^-11 M# #OH^-# solution? | null |
2,875 | ac90a940-6ddd-11ea-8ca7-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-3-9-10-8-m-oh-solution | 7.08 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NaOH solution"}] | [{"type":"physical unit","value":"7.08"}] | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{3.9 × 10^(-8) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #3.9*10^-8# #M# #NaOH# solution? </h1> | null | 7.08 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
</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> is 7.08.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #3.9*10^-8# #M# #NaOH# solution? </h1>
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Ernest Z.
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is 7.08.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>The base is so dilute that we must also consider the <mathjax>#"OH"^"-"#</mathjax> that comes from the autoionization of water.</p>
<p><mathjax>#"H"_2"O" ⇌ "H"^+ + "OH"^"-"#</mathjax></p>
<p><mathjax>#K_w = ["H"^+] ["OH"^"-"] = 1.00 × 10^"-14"#</mathjax></p>
<blockquote></blockquote>
<p>We start by noting that charges must balance.</p>
<p>If the <mathjax>#"OH"^"-"#</mathjax> comes from <mathjax>#"NaOH"#</mathjax>, then</p>
<p><mathjax>#["OH"^"-"] = ["H"^+] + ["Na"^+]#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#"H"^+#</mathjax> comes from the autoionization of water, so</p>
<p><mathjax>#["H"^+] = (1.00 × 10^"-14")/["OH"^"-"]#</mathjax></p>
<p>Also, <mathjax>#["Na"^+] = 3.9 × 10^-8"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#["OH"^"-"] = (1.00 × 10^"-14")/(["OH"^"-"]) + 3.9 × 10^-8"#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 = 1.00 × 10^"-14" + 3.9 × 10^"-8"["OH"^"-"]#</mathjax></p>
<p><mathjax>#["OH"^"-"]^2 - 3.9 × 10^"-8"["OH"^"-"] - 1.00 × 10^"-14" = 0#</mathjax></p>
<p>Solving the quadratic, we get</p>
<p><mathjax>#["OH"^"-"] = 1.21 × 10^"-7"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"pOH" = "-log"(1.21 × 10^"-7") = 6.92#</mathjax></p>
<p><mathjax>#"pH" = "14.00 - pH" = "14.00 - 6.92" = 7.08#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#sf(pH=7.08)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We would predict that as this is such a low concentration the <mathjax>#sf(pH)#</mathjax> will be close to 7.</p>
<p>This is an incredibly low concentration, so we must take into account the <mathjax>#sf(OH^-)#</mathjax> ions which arise from the auto - ionisation of water:</p>
<p><mathjax>#sf(H_2O_((l))rightleftharpoonsH_((aq))^(+)+OH_((aq))^-)#</mathjax></p>
<p>For which:</p>
<p><mathjax>#sf(K_w=1xx10^(-14)" ""mol"^(2)."l"^(-2) #</mathjax> at <mathjax>#sf(25^@"C"#</mathjax>.</p>
<p>This means that the concentration of <mathjax>#sf(OH^-#</mathjax> ions is <mathjax>#sf(10^(-7)"M")#</mathjax> arising from the water.</p>
<p>A tiny amount of <mathjax>#sf(OH^-)#</mathjax> ions is added such that the concentration is <mathjax>#sf(3.9xx10^(-8)"M"#</mathjax>.</p>
<p>It might be reasoned, therefore, that the total <mathjax>#sf(OH^-)#</mathjax> ion concentration will be <mathjax>#sf((3.9xx10^(-8))+10^(-7)"M"#</mathjax>.</p>
<p>However this will not be the <mathjax>#sf(OH^-)#</mathjax> ion concentration <strong>at equilibrium.</strong></p>
<p>By adding those ions we have disturbed a system which is at equilibrium.</p>
<p>Le Chatelier's Principle tells us that the system will respond by opposing that change, thus restoring the equilibrium which represents the lowest energy state.</p>
<p>The reaction quotient <mathjax>#sf(Q)#</mathjax> is given by:</p>
<p><mathjax>#sf(Q=[H_((aq))^(+)][OH_((aq))^-]=10^(-7)xx[(3.9xx10^(-8)+10^(-7))])#</mathjax></p>
<p><mathjax>#=sf(1.39xx10^(-14)" ""mol"^2."l"^(-2))#</mathjax></p>
<p>This shows that <mathjax>#sf(Q>K)#</mathjax> which confirms that the equilibrium will shift to the left. More <mathjax>#sf(H^+)#</mathjax> will be consumed by the extra <mathjax>#sf(OH^-)#</mathjax> ions to produce more water.</p>
<p>To calculate the <mathjax>#sf(pH)#</mathjax> we need to get the concentration of <mathjax>#sf(H^+)#</mathjax> ions at equilibrium which we can do by setting up an <strong>ICE</strong> table:</p>
<p><mathjax>#" "sf(H_2O" "rightleftharpoons" "H^(+)" "+" "OH^(-))#</mathjax></p>
<p><mathjax>#sf(color(red)(I)" "--" "10^(-7)" "1.39xx10^(-7))#</mathjax></p>
<p><mathjax>#sf(color(red)(C)" "--" " -x" " -x)#</mathjax></p>
<p><mathjax>#sf(color(red)(E)" "--" "(10^(-7)-x)" "(1.39xx10^(-7)-x))#</mathjax></p>
<p>So:</p>
<p><mathjax>#sf(K_(w)=(10^(-7)-x)(1.39xx10^(-7))=10^(-14))#</mathjax></p>
<p>This simplifies to:</p>
<p><mathjax>#sf(x^2-(2.39xx10^(-7))x+0.39xx10^(-14)=0)#</mathjax></p>
<p>This is a quadratic equation which can be solved using the quadratic formula.</p>
<p>Discarding the absurd root we get:</p>
<p><mathjax>#sf(x=0.1765xx10^(-7)"mol/l")#</mathjax></p>
<p>So:</p>
<p><mathjax>#sf([H_(eqm)^(+)]=(1xx10^(-7)-0.1765xx10^(-7))=0.824xx10^(-7)"mol/l")#</mathjax></p>
<p><mathjax>#sf(pH=-log[H_(eqm)^+]=-log[0.824xx10^(-7)]#</mathjax></p>
<p><mathjax>#sf(color(red)(pH=7.08))#</mathjax></p>
<p>You may see this more generally described as "The Common Ion Effect" . <mathjax>#sf(OH^-)#</mathjax>being the common ion in question. </p></div>
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</article> | What is the pH of a #3.9*10^-8# #M# #NaOH# solution? | null |
2,876 | ac579c08-6ddd-11ea-adcc-ccda262736ce | https://socratic.org/questions/using-the-equation-fe-2o-3-2al-2fe-al-2o-3-how-many-grams-of-al-are-needed-to-co | 45.62 grams | start physical_unit 6 6 mass g qc_end physical_unit 3 3 23 24 mass qc_end chemical_equation 3 11 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] Al [IN] grams"}] | [{"type":"physical unit","value":"45.62 grams"}] | [{"type":"physical unit","value":"Mass [OF] Fe2O3 [=] \\pu{135 grams}"},{"type":"chemical equation","value":"Fe2O3 + 2 Al -> 2 Fe + Al2O3"},{"type":"other","value":"Completely react."}] | <h1 class="questionTitle" itemprop="name">Using the equation, #Fe_2O_3 + 2Al -> 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?</h1> | null | 45.62 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The coefficients in the equation provide a molar ratio between the chemical species in the equation, so you must first convert the mass of iron (III) oxide to moles of iron (III) oxide by multiplying it by the inverse of its molar mass. </p>
<p>Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of <mathjax>#Fe_2 O_3#</mathjax>. </p>
<p>Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g <mathjax>#Fe_2 O_3#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#135g Fe_2 O_3 * (1 mol Fe_2 O_3) / (159.69g Fe_2 O_3) * (2 mol Al) / (1 mol Fe_2 O_3) * (26.98g Al) / (1 mol Al) = 45.6g Al#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The coefficients in the equation provide a molar ratio between the chemical species in the equation, so you must first convert the mass of iron (III) oxide to moles of iron (III) oxide by multiplying it by the inverse of its molar mass. </p>
<p>Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of <mathjax>#Fe_2 O_3#</mathjax>. </p>
<p>Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g <mathjax>#Fe_2 O_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Using the equation, #Fe_2O_3 + 2Al -> 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?</h1>
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<div class="markdown"><p><mathjax>#135g Fe_2 O_3 * (1 mol Fe_2 O_3) / (159.69g Fe_2 O_3) * (2 mol Al) / (1 mol Fe_2 O_3) * (26.98g Al) / (1 mol Al) = 45.6g Al#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The coefficients in the equation provide a molar ratio between the chemical species in the equation, so you must first convert the mass of iron (III) oxide to moles of iron (III) oxide by multiplying it by the inverse of its molar mass. </p>
<p>Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of <mathjax>#Fe_2 O_3#</mathjax>. </p>
<p>Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g <mathjax>#Fe_2 O_3#</mathjax>.</p></div>
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</article> | Using the equation, #Fe_2O_3 + 2Al -> 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#? | null |
2,877 | ac2be38d-6ddd-11ea-ae4b-ccda262736ce | https://socratic.org/questions/what-is-the-formula-of-lithium-nitride | Li3N | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] lithium nitride [IN] default"}] | [{"type":"chemical equation","value":"Li3N"}] | [{"type":"substance name","value":"Lithium nitride"}] | <h1 class="questionTitle" itemprop="name">What is the formula of lithium nitride?</h1> | null | Li3N | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium nitride contains the <mathjax>#N^(3-)#</mathjax> ion. And lithium is one of the few materials that will directly fix dinitrogen:</p>
<p><mathjax>#3Li + 1/2N_2 rarr Li_3N#</mathjax></p>
<p>What do you think would happen if you mixed lithium nitride with water?</p></div>
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<div class="markdown"><p><mathjax>#Li_3N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium nitride contains the <mathjax>#N^(3-)#</mathjax> ion. And lithium is one of the few materials that will directly fix dinitrogen:</p>
<p><mathjax>#3Li + 1/2N_2 rarr Li_3N#</mathjax></p>
<p>What do you think would happen if you mixed lithium nitride with water?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula of lithium nitride?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Li_3N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Lithium nitride contains the <mathjax>#N^(3-)#</mathjax> ion. And lithium is one of the few materials that will directly fix dinitrogen:</p>
<p><mathjax>#3Li + 1/2N_2 rarr Li_3N#</mathjax></p>
<p>What do you think would happen if you mixed lithium nitride with water?</p></div>
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</article> | What is the formula of lithium nitride? | null |
2,878 | ac7b1258-6ddd-11ea-8b24-ccda262736ce | https://socratic.org/questions/how-many-moles-of-pb-no-3-2-are-required-if-12-moles-of-al-no-3-3-are-produced | 36.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 8 9 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] Pb(NO3)2 [IN] moles"}] | [{"type":"physical unit","value":"36.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] Al2(NO3)3 [=] \\pu{12 moles}"}] | <h1 class="questionTitle" itemprop="name">How many moles of # Pb(NO_3)_2# are required if 12 moles of # Al(NO_3)_3# are produced?</h1> | null | 36.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So if <mathjax>#12*mol#</mathjax> of aluminum nitrate are produced, clearly, from the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the given reaction, <mathjax>#6#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#"lead nitrate"#</mathjax> were required. But don't trust my arithmetic!</p>
<p>I chose aluminum sulfate because it is a fact that both lead nitrate and aluminum sulfate are reasonably soluble in aqueous solution, whereas lead sulfate is fairly insoluble. Lead sulfate should thus precipitate out. Lead can thus be separated from aluminum with this metathesis reaction. </p></div>
</div>
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<div class="markdown"><p><mathjax>#3Pb(NO_3)_2(aq) + 3Al_2(SO_4)_3(aq)rarr 3PbSO_4(s)darr + 6Al(NO_3)_3(aq)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So if <mathjax>#12*mol#</mathjax> of aluminum nitrate are produced, clearly, from the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the given reaction, <mathjax>#6#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#"lead nitrate"#</mathjax> were required. But don't trust my arithmetic!</p>
<p>I chose aluminum sulfate because it is a fact that both lead nitrate and aluminum sulfate are reasonably soluble in aqueous solution, whereas lead sulfate is fairly insoluble. Lead sulfate should thus precipitate out. Lead can thus be separated from aluminum with this metathesis reaction. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of # Pb(NO_3)_2# are required if 12 moles of # Al(NO_3)_3# are produced?</h1>
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<div class="markdown"><p><mathjax>#3Pb(NO_3)_2(aq) + 3Al_2(SO_4)_3(aq)rarr 3PbSO_4(s)darr + 6Al(NO_3)_3(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So if <mathjax>#12*mol#</mathjax> of aluminum nitrate are produced, clearly, from the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the given reaction, <mathjax>#6#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#"lead nitrate"#</mathjax> were required. But don't trust my arithmetic!</p>
<p>I chose aluminum sulfate because it is a fact that both lead nitrate and aluminum sulfate are reasonably soluble in aqueous solution, whereas lead sulfate is fairly insoluble. Lead sulfate should thus precipitate out. Lead can thus be separated from aluminum with this metathesis reaction. </p></div>
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</article> | How many moles of # Pb(NO_3)_2# are required if 12 moles of # Al(NO_3)_3# are produced? | null |
2,879 | ac5c4147-6ddd-11ea-a040-ccda262736ce | https://socratic.org/questions/if-you-evaporated-453-ml-of-a-6-5-m-solution-of-iron-ii-nitrite-what-mass-of-iro | 435.36 g | start physical_unit 11 13 mass g qc_end physical_unit 9 9 3 4 volume qc_end physical_unit 11 13 7 8 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] iron (II) nitrite [IN] g"}] | [{"type":"physical unit","value":"435.36 g"}] | [{"type":"physical unit","value":"Volume [OF] iron (II) nitrite solution [=] \\pu{453 mL}"},{"type":"physical unit","value":"Molarity [OF] iron (II) nitrite solution [=] \\pu{6.5 M}"}] | <h1 class="questionTitle" itemprop="name">If you evaporated 453. mL of a 6.5 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover?</h1> | null | 435.36 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that heating the iron(II) nitrite solution will <em>evaporate</em> the water and leave behind the solid. </p>
<p>In order to find the mass of the solid that can be recovered from this solution, you need to use the definition of <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and the <strong>molar mass</strong> of iron(II) nitrite, <mathjax>#"Fe"("NO"_2)_2#</mathjax>.</p>
<p>As you know, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> tells you how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Since you know the molarity and volume of the solution, you can solve for the <em>number of moles</em> of iron(II) nitrite present in solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume of the solution from <em>milliliters</em> to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n_(Fe(NO_2)_2) = "6.5 mol" color(red)(cancel(color(black)("L"^(-1)))) * 453. * 10^(-3)color(red)(cancel(color(black)("L"))) = "2.9445 moles Fe"("NO"_2)_2#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>molar mass</strong> of iron(II) nitrite to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.9445color(red)(cancel(color(black)("moles Fe"("NO"_2)_2))) * overbrace("147.856 g"/(1color(red)(cancel(color(black)("mole Fe"("NO"_2)_2)))))^(color(brown)("molar mass of iron(II) nitrite")) = "435.36 g"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_(Fe(NO_2)_2) = color(green)(|bar(ul(color(white)(a/a)"440 g Fe"("NO"_2)_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>The way I see it, this is more of a <strong>conceptual problem</strong> because I'm not even sure that you can have a</em> <mathjax>#"6.5-M"#</mathjax>, <mathjax>#"453. mL"#</mathjax> <em>aqueous solution of iron(II) nitrite</em>.</p>
<p><em>I know that iron(II) nitrite is soluble in aqueous solution, but I don't know its solubility at room temperature per</em> <mathjax>#"100 g"#</mathjax> <em>of water</em>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"440 g Fe("NO"_2)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that heating the iron(II) nitrite solution will <em>evaporate</em> the water and leave behind the solid. </p>
<p>In order to find the mass of the solid that can be recovered from this solution, you need to use the definition of <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and the <strong>molar mass</strong> of iron(II) nitrite, <mathjax>#"Fe"("NO"_2)_2#</mathjax>.</p>
<p>As you know, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> tells you how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Since you know the molarity and volume of the solution, you can solve for the <em>number of moles</em> of iron(II) nitrite present in solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume of the solution from <em>milliliters</em> to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n_(Fe(NO_2)_2) = "6.5 mol" color(red)(cancel(color(black)("L"^(-1)))) * 453. * 10^(-3)color(red)(cancel(color(black)("L"))) = "2.9445 moles Fe"("NO"_2)_2#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>molar mass</strong> of iron(II) nitrite to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.9445color(red)(cancel(color(black)("moles Fe"("NO"_2)_2))) * overbrace("147.856 g"/(1color(red)(cancel(color(black)("mole Fe"("NO"_2)_2)))))^(color(brown)("molar mass of iron(II) nitrite")) = "435.36 g"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_(Fe(NO_2)_2) = color(green)(|bar(ul(color(white)(a/a)"440 g Fe"("NO"_2)_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>The way I see it, this is more of a <strong>conceptual problem</strong> because I'm not even sure that you can have a</em> <mathjax>#"6.5-M"#</mathjax>, <mathjax>#"453. mL"#</mathjax> <em>aqueous solution of iron(II) nitrite</em>.</p>
<p><em>I know that iron(II) nitrite is soluble in aqueous solution, but I don't know its solubility at room temperature per</em> <mathjax>#"100 g"#</mathjax> <em>of water</em>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If you evaporated 453. mL of a 6.5 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-12T14:47:26" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"440 g Fe("NO"_2)_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The idea here is that heating the iron(II) nitrite solution will <em>evaporate</em> the water and leave behind the solid. </p>
<p>In order to find the mass of the solid that can be recovered from this solution, you need to use the definition of <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> and the <strong>molar mass</strong> of iron(II) nitrite, <mathjax>#"Fe"("NO"_2)_2#</mathjax>.</p>
<p>As you know, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> tells you how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Since you know the molarity and volume of the solution, you can solve for the <em>number of moles</em> of iron(II) nitrite present in solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>Do not</strong> forget to convert the volume of the solution from <em>milliliters</em> to <em>liters</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#n_(Fe(NO_2)_2) = "6.5 mol" color(red)(cancel(color(black)("L"^(-1)))) * 453. * 10^(-3)color(red)(cancel(color(black)("L"))) = "2.9445 moles Fe"("NO"_2)_2#</mathjax></p>
</blockquote>
<p>Finally, use the <strong>molar mass</strong> of iron(II) nitrite to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.9445color(red)(cancel(color(black)("moles Fe"("NO"_2)_2))) * overbrace("147.856 g"/(1color(red)(cancel(color(black)("mole Fe"("NO"_2)_2)))))^(color(brown)("molar mass of iron(II) nitrite")) = "435.36 g"#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_(Fe(NO_2)_2) = color(green)(|bar(ul(color(white)(a/a)"440 g Fe"("NO"_2)_2color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>The way I see it, this is more of a <strong>conceptual problem</strong> because I'm not even sure that you can have a</em> <mathjax>#"6.5-M"#</mathjax>, <mathjax>#"453. mL"#</mathjax> <em>aqueous solution of iron(II) nitrite</em>.</p>
<p><em>I know that iron(II) nitrite is soluble in aqueous solution, but I don't know its solubility at room temperature per</em> <mathjax>#"100 g"#</mathjax> <em>of water</em>.</p></div>
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</article> | If you evaporated 453. mL of a 6.5 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover? | null |
2,880 | a8b3329a-6ddd-11ea-8cd7-ccda262736ce | https://socratic.org/questions/the-volume-of-a-sample-of-a-gas-at-273-c-is-200-0-l-if-the-volume-is-decreased-t | 0 ℃ | start physical_unit 7 7 temperature °c qc_end physical_unit 7 7 9 10 temperature qc_end c_other OTHER qc_end physical_unit 7 7 12 13 volume qc_end physical_unit 7 7 20 21 volume qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas sample [IN] ℃"}] | [{"type":"physical unit","value":"0 ℃"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{273 ℃}"},{"type":"other","value":"Constant pressure"},{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{200.0 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas sample [=] \\pu{100.0 L}"}] | <h1 class="questionTitle" itemprop="name">The volume of a sample of a gas at 273°C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas? </h1> | null | 0 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that when pressure is held constant, the temperature and volume of a gas are directly proportional, so that if one goes up, so does the other. Note: The temperature needs to be in Kelvins.</p>
<p>The equation for Charles' Law is <mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_1="200.0 L"#</mathjax><br/>
<mathjax>#T_1="273"^"o""C+273=546 K"#</mathjax><br/>
<mathjax>#V_2="100.0 L"#</mathjax><br/>
<mathjax>#T_2="?"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(100.0cancel"L"xx546"K")/(200cancel"L")=#</mathjax></p>
<p><mathjax>#T_2="273 K"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#T_2="273 K"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that when pressure is held constant, the temperature and volume of a gas are directly proportional, so that if one goes up, so does the other. Note: The temperature needs to be in Kelvins.</p>
<p>The equation for Charles' Law is <mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_1="200.0 L"#</mathjax><br/>
<mathjax>#T_1="273"^"o""C+273=546 K"#</mathjax><br/>
<mathjax>#V_2="100.0 L"#</mathjax><br/>
<mathjax>#T_2="?"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(100.0cancel"L"xx546"K")/(200cancel"L")=#</mathjax></p>
<p><mathjax>#T_2="273 K"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The volume of a sample of a gas at 273°C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas? </h1>
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<div class="markdown"><p><mathjax>#T_2="273 K"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> states that when pressure is held constant, the temperature and volume of a gas are directly proportional, so that if one goes up, so does the other. Note: The temperature needs to be in Kelvins.</p>
<p>The equation for Charles' Law is <mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#V_1="200.0 L"#</mathjax><br/>
<mathjax>#T_1="273"^"o""C+273=546 K"#</mathjax><br/>
<mathjax>#V_2="100.0 L"#</mathjax><br/>
<mathjax>#T_2="?"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(100.0cancel"L"xx546"K")/(200cancel"L")=#</mathjax></p>
<p><mathjax>#T_2="273 K"#</mathjax></p></div>
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</article> | The volume of a sample of a gas at 273°C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas? | null |
2,881 | ad1165b4-6ddd-11ea-8f45-ccda262736ce | https://socratic.org/questions/what-is-specific-heat-of-steam | 2.01 kJ/(kg * ℃) | start physical_unit 5 5 specific_heat kj/(kg_·_°c) qc_end substance 5 5 qc_end end | [{"type":"physical unit","value":"Specific heat [OF] steam [IN] kJ/(kg * ℃)"}] | [{"type":"physical unit","value":"2.01 kJ/(kg * ℃)"}] | [{"type":"substance name","value":"Steam"}] | <h1 class="questionTitle" itemprop="name">What is specific heat of steam? </h1> | null | 2.01 kJ/(kg * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The specific heat of an object is the amount of heat required to raise the temperature of a specific amount of a substance by 1 °C.</p>
<p>Thus, it takes 2.010 kJ of heat to raise the temperature of 1 kg of steam by 1 °C.</p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of steam is <mathjax>#"2.010 kJ·°C"^"-1""kg"^"-1"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The specific heat of an object is the amount of heat required to raise the temperature of a specific amount of a substance by 1 °C.</p>
<p>Thus, it takes 2.010 kJ of heat to raise the temperature of 1 kg of steam by 1 °C.</p></div>
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<h1 class="questionTitle" itemprop="name">What is specific heat of steam? </h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of steam is <mathjax>#"2.010 kJ·°C"^"-1""kg"^"-1"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>The specific heat of an object is the amount of heat required to raise the temperature of a specific amount of a substance by 1 °C.</p>
<p>Thus, it takes 2.010 kJ of heat to raise the temperature of 1 kg of steam by 1 °C.</p></div>
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</article> | What is specific heat of steam? | null |
2,882 | a95c3bba-6ddd-11ea-9c9f-ccda262736ce | https://socratic.org/questions/59a2b83fb72cff6f4984a1a2 | 23.21 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 4 5 volume qc_end physical_unit 8 8 16 17 pressure qc_end physical_unit 8 8 19 20 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"23.21 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{22.5 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{748 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{725 mmHg}"}] | <h1 class="questionTitle" itemprop="name">The pressure on a #22.5*L# volume of gas enclosed in a piston is REDUCED from #748*mm*Hg# to #725*mm*Hg#. What is the new volume?</h1> | null | 23.21 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so....<mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume. Agreed?</p>
<p><mathjax>#V_2=(22.5*Lxx748*mm*Hg)/(725*mm*Hg)=??*L#</mathjax></p></div>
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<div class="markdown"><p>Well, <mathjax>#P_1V_1=P_2V_2#</mathjax>.......and the VOLUME should INCREASE BECAUSE YOU DECREASE the PRESSURE......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so....<mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume. Agreed?</p>
<p><mathjax>#V_2=(22.5*Lxx748*mm*Hg)/(725*mm*Hg)=??*L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The pressure on a #22.5*L# volume of gas enclosed in a piston is REDUCED from #748*mm*Hg# to #725*mm*Hg#. What is the new volume?</h1>
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<div class="markdown"><p>Well, <mathjax>#P_1V_1=P_2V_2#</mathjax>.......and the VOLUME should INCREASE BECAUSE YOU DECREASE the PRESSURE......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so....<mathjax>#V_2=(P_1V_1)/P_2#</mathjax>, which clearly has the units of volume. Agreed?</p>
<p><mathjax>#V_2=(22.5*Lxx748*mm*Hg)/(725*mm*Hg)=??*L#</mathjax></p></div>
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</article> | The pressure on a #22.5*L# volume of gas enclosed in a piston is REDUCED from #748*mm*Hg# to #725*mm*Hg#. What is the new volume? | null |
2,883 | a8f816d0-6ddd-11ea-919e-ccda262736ce | https://socratic.org/questions/57d9930911ef6b7d4dbed6b8 | 1.70 | start physical_unit 8 8 ph none qc_end physical_unit 10 10 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] HCl(aq) solution"}] | [{"type":"physical unit","value":"1.70"}] | [{"type":"physical unit","value":"Molarity [OF] HCl(aq) solution [=] \\pu{0.020 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# for a #0.020*mol*L^-1# solution of #HCl(aq)#?</h1> | null | 1.70 | <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(0.02)#</mathjax>, given that we assume hydrochloric acid dissociates completely to give stoichiometric <mathjax>#H_3O^+(aq)#</mathjax>:</p>
<p><mathjax>#HCl(aq) +H_2O(l) rightleftharpoons H_3O^+ + Cl^-#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=1.70#</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(0.02)#</mathjax>, given that we assume hydrochloric acid dissociates completely to give stoichiometric <mathjax>#H_3O^+(aq)#</mathjax>:</p>
<p><mathjax>#HCl(aq) +H_2O(l) rightleftharpoons H_3O^+ + Cl^-#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# for a #0.020*mol*L^-1# solution of #HCl(aq)#?</h1>
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<div class="markdown"><p><mathjax>#pH=1.70#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(0.02)#</mathjax>, given that we assume hydrochloric acid dissociates completely to give stoichiometric <mathjax>#H_3O^+(aq)#</mathjax>:</p>
<p><mathjax>#HCl(aq) +H_2O(l) rightleftharpoons H_3O^+ + Cl^-#</mathjax></p></div>
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</article> | What is the #pH# for a #0.020*mol*L^-1# solution of #HCl(aq)#? | null |
2,884 | a98ad092-6ddd-11ea-937d-ccda262736ce | https://socratic.org/questions/590a1ad211ef6b71ad15fdb1 | BaCO3(s) ->[\Delta] BaO(s) + CO2(g) | start chemical_equation qc_end substance 2 3 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the decompose"}] | [{"type":"chemical equation","value":"BaCO3(s) ->[\\Delta] BaO(s) + CO2(g)"}] | [{"type":"substance name","value":"barium carbonate"}] | <h1 class="questionTitle" itemprop="name">How does barium carbonate decompose under heating?</h1> | null | BaCO3(s) ->[\Delta] BaO(s) + CO2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the equation balanced with respect to mass and charge? If it is not, then we cannot accept it as a manifestation of physical reality. </p></div>
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<div class="markdown"><p><mathjax>#BaCO_3(s) + Delta rarr BaO(s) + CO_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the equation balanced with respect to mass and charge? If it is not, then we cannot accept it as a manifestation of physical reality. </p></div>
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<h1 class="questionTitle" itemprop="name">How does barium carbonate decompose under heating?</h1>
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<div class="markdown"><p><mathjax>#BaCO_3(s) + Delta rarr BaO(s) + CO_2(g)uarr#</mathjax></p></div>
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<div class="markdown"><p>Is the equation balanced with respect to mass and charge? If it is not, then we cannot accept it as a manifestation of physical reality. </p></div>
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</article> | How does barium carbonate decompose under heating? | null |
2,885 | ab04b52c-6ddd-11ea-a653-ccda262736ce | https://socratic.org/questions/a-solution-is-prepared-by-dissolving-5-6-g-of-koh-is-250-ml-of-it-calculate-the- | 0.40 M | start physical_unit 19 20 molarity mol/l qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 1 1 11 12 volume qc_end physical_unit 9 9 23 25 molar_mass qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"0.40 M"}] | [{"type":"physical unit","value":"Mass [OF] KOH [=] \\pu{5.6 g}"},{"type":"physical unit","value":"Volume [OF] KOH solution [=] \\pu{250 mL}"},{"type":"physical unit","value":"Molar mass [OF] KOH [=] \\pu{56 g/ mol}"}] | <h1 class="questionTitle" itemprop="name">A solution is prepared by dissolving 5.6 g of KOH is 250 ml of it. Calculate the molarity of the solution?(molar mass KOH: 56 g/ mol) [Ans. : 419M]
</h1> | null | 0.40 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for molar concentration <mathjax>#c#</mathjax> is</p>
<p><mathjax>#c = "moles"/"litres"#</mathjax></p>
<p><mathjax>#"Moles" = 5.6 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.0998 mol KOH"#</mathjax></p>
<p><mathjax>#"Litres" = 250 "mL" × "1 L"/(1000 "mL") = "0.2500 L"#</mathjax></p>
<p><mathjax>#c = "0.0998 mol"/"0.2500 L" = "0.40 mol/L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#c = "0.40 mol/L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for molar concentration <mathjax>#c#</mathjax> is</p>
<p><mathjax>#c = "moles"/"litres"#</mathjax></p>
<p><mathjax>#"Moles" = 5.6 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.0998 mol KOH"#</mathjax></p>
<p><mathjax>#"Litres" = 250 "mL" × "1 L"/(1000 "mL") = "0.2500 L"#</mathjax></p>
<p><mathjax>#c = "0.0998 mol"/"0.2500 L" = "0.40 mol/L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A solution is prepared by dissolving 5.6 g of KOH is 250 ml of it. Calculate the molarity of the solution?(molar mass KOH: 56 g/ mol) [Ans. : 419M]
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Ernest Z.
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<div class="markdown"><p><mathjax>#c = "0.40 mol/L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for molar concentration <mathjax>#c#</mathjax> is</p>
<p><mathjax>#c = "moles"/"litres"#</mathjax></p>
<p><mathjax>#"Moles" = 5.6 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.0998 mol KOH"#</mathjax></p>
<p><mathjax>#"Litres" = 250 "mL" × "1 L"/(1000 "mL") = "0.2500 L"#</mathjax></p>
<p><mathjax>#c = "0.0998 mol"/"0.2500 L" = "0.40 mol/L"#</mathjax></p></div>
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</article> | A solution is prepared by dissolving 5.6 g of KOH is 250 ml of it. Calculate the molarity of the solution?(molar mass KOH: 56 g/ mol) [Ans. : 419M]
| null |
2,886 | ab2f2d9d-6ddd-11ea-a6ef-ccda262736ce | https://socratic.org/questions/a-first-order-reaction-has-an-arrhenius-constant-for-the-reaction-is-9-9-x-1010- | 12 Celsius | start physical_unit 9 10 temperature °c qc_end c_other OTHER qc_end physical_unit 9 10 12 15 constant_r qc_end physical_unit 9 10 21 22 activation_barrier qc_end physical_unit 9 10 34 35 reaction_rate_constant qc_end end | [{"type":"physical unit","value":"Temperature [OF] the reaction [IN] Celsius"}] | [{"type":"physical unit","value":"12 Celsius"}] | [{"type":"other","value":"First order reaction."},{"type":"physical unit","value":"Arrhenius constant [OF] the reaction [=] \\pu{9.9 × 10^10 sec^(-1)}"},{"type":"physical unit","value":"Activation energy [OF] the reaction [=] \\pu{62.0 kJ}"},{"type":"physical unit","value":"Rate constant [OF] the reaction [=] \\pu{0.434 sec^(-1)}"}] | <h1 class="questionTitle" itemprop="name">A first order reaction has an Arrhenius constant for the reaction is #9.9 xx 10^10 sec^ -1#, and an activation energy of #"62.0 kJ"#. At what temperature (in Celsius) is the rate constant equal to #0.434 sec^-1#?</h1> | null | 12 Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> at which the reaction takes place</p>
<p>Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature <em>in degrees Celsius</em> at which the reaction has that particular rate constant. </p>
<p>Your strategy here will be to use the Arrhenius equation to solve for <mathjax>#T#</mathjax>, the absolute temperature at which the reaction takes place, then use the conversion factor </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>to go from Kelvin to degrees Celsius. </p>
<p>So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation</p>
<blockquote>
<p><mathjax>#k/A = "exp"(-E_a/(RT))#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation</p>
<blockquote>
<p><mathjax>#ln(k/A) = ln["exp"(-E_a/(RT))]#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#ln(k/A) = -E_a/(RT)#</mathjax></p>
</blockquote>
<p>Rearrange to isolate <mathjax>#T#</mathjax> on one side of the equation</p>
<blockquote>
<p><mathjax>#T = -E_a/R * 1/ln(k/A)#</mathjax></p>
</blockquote>
<p>Plug in your values to get - <strong>do not</strong> forget to convert the activation energy from <em>kilojoules</em> to <em>joules</em></p>
<blockquote>
<p><mathjax>#T = - (62.0 * 10^3color(red)(cancel(color(black)("J"))))/(8.314color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1)) * 1/(ln( (0.434 color(red)(cancel(color(black)("s"^(-1)))))/(9.9 * 10^(10)color(red)(cancel(color(black)("s"^(-1))))))#</mathjax></p>
<p><mathjax>#T = "285.14 K"#</mathjax></p>
</blockquote>
<p>Expressed in degrees Celsius, the temperature will be </p>
<blockquote>
<p><mathjax>#t[""^@"C"] = "285.14 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)12^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the pre-exponential factor. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#12^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> at which the reaction takes place</p>
<p>Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature <em>in degrees Celsius</em> at which the reaction has that particular rate constant. </p>
<p>Your strategy here will be to use the Arrhenius equation to solve for <mathjax>#T#</mathjax>, the absolute temperature at which the reaction takes place, then use the conversion factor </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>to go from Kelvin to degrees Celsius. </p>
<p>So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation</p>
<blockquote>
<p><mathjax>#k/A = "exp"(-E_a/(RT))#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation</p>
<blockquote>
<p><mathjax>#ln(k/A) = ln["exp"(-E_a/(RT))]#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#ln(k/A) = -E_a/(RT)#</mathjax></p>
</blockquote>
<p>Rearrange to isolate <mathjax>#T#</mathjax> on one side of the equation</p>
<blockquote>
<p><mathjax>#T = -E_a/R * 1/ln(k/A)#</mathjax></p>
</blockquote>
<p>Plug in your values to get - <strong>do not</strong> forget to convert the activation energy from <em>kilojoules</em> to <em>joules</em></p>
<blockquote>
<p><mathjax>#T = - (62.0 * 10^3color(red)(cancel(color(black)("J"))))/(8.314color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1)) * 1/(ln( (0.434 color(red)(cancel(color(black)("s"^(-1)))))/(9.9 * 10^(10)color(red)(cancel(color(black)("s"^(-1))))))#</mathjax></p>
<p><mathjax>#T = "285.14 K"#</mathjax></p>
</blockquote>
<p>Expressed in degrees Celsius, the temperature will be </p>
<blockquote>
<p><mathjax>#t[""^@"C"] = "285.14 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)12^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the pre-exponential factor. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A first order reaction has an Arrhenius constant for the reaction is #9.9 xx 10^10 sec^ -1#, and an activation energy of #"62.0 kJ"#. At what temperature (in Celsius) is the rate constant equal to #0.434 sec^-1#?</h1>
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<div class="markdown"><p><mathjax>#12^@"C"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong>Arrhenius equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#k#</mathjax> - the <em>rate constant</em> for a given reaction<br/>
<mathjax>#A#</mathjax> - the <em>pre-exponential factor</em>, specific to a given reaction<br/>
<mathjax>#E_a#</mathjax> - the <em>activation energy</em> of the reaction<br/>
<mathjax>#R#</mathjax> - the universal gas constant, useful here as <mathjax>#8.314"J mol"^(-1)"K"^(-1)#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> at which the reaction takes place</p>
<p>Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature <em>in degrees Celsius</em> at which the reaction has that particular rate constant. </p>
<p>Your strategy here will be to use the Arrhenius equation to solve for <mathjax>#T#</mathjax>, the absolute temperature at which the reaction takes place, then use the conversion factor </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>to go from Kelvin to degrees Celsius. </p>
<p>So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation</p>
<blockquote>
<p><mathjax>#k/A = "exp"(-E_a/(RT))#</mathjax></p>
</blockquote>
<p>Take the natural log of both sides of the equation</p>
<blockquote>
<p><mathjax>#ln(k/A) = ln["exp"(-E_a/(RT))]#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#ln(k/A) = -E_a/(RT)#</mathjax></p>
</blockquote>
<p>Rearrange to isolate <mathjax>#T#</mathjax> on one side of the equation</p>
<blockquote>
<p><mathjax>#T = -E_a/R * 1/ln(k/A)#</mathjax></p>
</blockquote>
<p>Plug in your values to get - <strong>do not</strong> forget to convert the activation energy from <em>kilojoules</em> to <em>joules</em></p>
<blockquote>
<p><mathjax>#T = - (62.0 * 10^3color(red)(cancel(color(black)("J"))))/(8.314color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1)) * 1/(ln( (0.434 color(red)(cancel(color(black)("s"^(-1)))))/(9.9 * 10^(10)color(red)(cancel(color(black)("s"^(-1))))))#</mathjax></p>
<p><mathjax>#T = "285.14 K"#</mathjax></p>
</blockquote>
<p>Expressed in degrees Celsius, the temperature will be </p>
<blockquote>
<p><mathjax>#t[""^@"C"] = "285.14 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)12^@"C"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the pre-exponential factor. </p></div>
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</article> | A first order reaction has an Arrhenius constant for the reaction is #9.9 xx 10^10 sec^ -1#, and an activation energy of #"62.0 kJ"#. At what temperature (in Celsius) is the rate constant equal to #0.434 sec^-1#? | null |
2,887 | aa14eea7-6ddd-11ea-ba6c-ccda262736ce | https://socratic.org/questions/58dd468311ef6b232962a977 | 0.24 molar | start physical_unit 7 7 mole mol/l qc_end physical_unit 7 7 9 10 pressure qc_end physical_unit 7 7 13 14 temperature qc_end physical_unit 7 7 21 22 volume qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] molar"}] | [{"type":"physical unit","value":"0.24 molar"}] | [{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{742 mmHg}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{298.0 K}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{6.0 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molar quantity of a gas at #742*mm*Hg# pressure, at #298.0*K# temperature, confined to a volume of #6.0*L#?</h1> | null | 0.24 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old ideal gas equation, i.e. <mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)=((742*cancel(mm*Hg))/(760*cancel(mm*Hg*atm^-1))xx6.0*cancelL)/(0.0821*cancel(L*atm*K^-1)*mol^-1xx298.0*cancelK)#</mathjax></p>
<p><mathjax>#=0.240*mol#</mathjax></p>
<p>What mass of <mathjax>#HC-=N#</mathjax> is present? Would you expect <mathjax>#HC-=N#</mathjax> to behave ideally? Why or why not? What hazards would be associated with this experiment? </p>
<p>Note the dimensional consistency of our calculation. We sought an answer in <mathjax>#"moles"#</mathjax>; the calculation gave us an answer in <mathjax>#"moles"#</mathjax> or at least <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax>, so perhaps we are doing something right (for once!). </p>
<p>The key knowledge is the idea that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. And thus we may use a unit of length, the height of a mercury column, to give a useful and reproducible measurement of pressure.</p>
<p>Note that these days, mercury columns are a bit of a rarity. Has your teacher shown you a mercury column?</p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#n=0.240*mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old ideal gas equation, i.e. <mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)=((742*cancel(mm*Hg))/(760*cancel(mm*Hg*atm^-1))xx6.0*cancelL)/(0.0821*cancel(L*atm*K^-1)*mol^-1xx298.0*cancelK)#</mathjax></p>
<p><mathjax>#=0.240*mol#</mathjax></p>
<p>What mass of <mathjax>#HC-=N#</mathjax> is present? Would you expect <mathjax>#HC-=N#</mathjax> to behave ideally? Why or why not? What hazards would be associated with this experiment? </p>
<p>Note the dimensional consistency of our calculation. We sought an answer in <mathjax>#"moles"#</mathjax>; the calculation gave us an answer in <mathjax>#"moles"#</mathjax> or at least <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax>, so perhaps we are doing something right (for once!). </p>
<p>The key knowledge is the idea that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. And thus we may use a unit of length, the height of a mercury column, to give a useful and reproducible measurement of pressure.</p>
<p>Note that these days, mercury columns are a bit of a rarity. Has your teacher shown you a mercury column?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molar quantity of a gas at #742*mm*Hg# pressure, at #298.0*K# temperature, confined to a volume of #6.0*L#?</h1>
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<div class="markdown"><p><mathjax>#n=0.240*mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the old ideal gas equation, i.e. <mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)=((742*cancel(mm*Hg))/(760*cancel(mm*Hg*atm^-1))xx6.0*cancelL)/(0.0821*cancel(L*atm*K^-1)*mol^-1xx298.0*cancelK)#</mathjax></p>
<p><mathjax>#=0.240*mol#</mathjax></p>
<p>What mass of <mathjax>#HC-=N#</mathjax> is present? Would you expect <mathjax>#HC-=N#</mathjax> to behave ideally? Why or why not? What hazards would be associated with this experiment? </p>
<p>Note the dimensional consistency of our calculation. We sought an answer in <mathjax>#"moles"#</mathjax>; the calculation gave us an answer in <mathjax>#"moles"#</mathjax> or at least <mathjax>#1/(mol^-1)=1/(1/(mol))=mol#</mathjax>, so perhaps we are doing something right (for once!). </p>
<p>The key knowledge is the idea that <mathjax>#1*atm#</mathjax> will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. And thus we may use a unit of length, the height of a mercury column, to give a useful and reproducible measurement of pressure.</p>
<p>Note that these days, mercury columns are a bit of a rarity. Has your teacher shown you a mercury column?</p></div>
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</article> | What is the molar quantity of a gas at #742*mm*Hg# pressure, at #298.0*K# temperature, confined to a volume of #6.0*L#? | null |
2,888 | aad41164-6ddd-11ea-acdd-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-if-its-empirical-formula-is-cfbro-an | C3F3Br3O3 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 7 7 19 20 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C3F3Br3O3"}] | [{"type":"other","value":"Its empirical formula is CFBrO"},{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{381.01 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound if its empirical formula is #CFBrO# and its molar mass is 381.01 g/mol? </h1> | null | C3F3Br3O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of compound (<mathjax>#381.01#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>) has been provided for you.</p>
<p>Now you need to calculate the molar mass of <mathjax>#CFBrO#</mathjax>. <br/>
The molar mass of:<br/>
<mathjax>#C = 12.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#F = 19.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#Br = 79.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#O = 16.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>The molar mass of empirical formula, therefore (<mathjax>#CFBrO#</mathjax>) is:<br/>
<mathjax>#12.0 + 19.0 + 79.9 + 16.0#</mathjax> = <mathjax>#126.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>.</p>
<p>In order to get the molecular formula of <mathjax>#CFBrO#</mathjax>,<br/>
divide molar mass of compound by molar mass of empirical formula, like this: <mathjax>#"molar mass of compound"/" molar mass of empirical formula"#</mathjax> = <mathjax>#"whole number"#</mathjax></p>
<p>i.e. <mathjax>#381.01/126.9#</mathjax> = <mathjax>#3.00#</mathjax><em> <br/>
</em>Rounded to nearest whole number</p>
<p>Now, multiply the whole number by the Empirical Formula.<br/>
Like this <mathjax>#(CFBrO)#</mathjax><mathjax>#xx#</mathjax> <mathjax>#3#</mathjax> = <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p>
<p>Therefore your molecular formula is <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#(CFBrO)#</mathjax><mathjax>#xx#</mathjax> <mathjax>#3#</mathjax> = ??</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of compound (<mathjax>#381.01#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>) has been provided for you.</p>
<p>Now you need to calculate the molar mass of <mathjax>#CFBrO#</mathjax>. <br/>
The molar mass of:<br/>
<mathjax>#C = 12.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#F = 19.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#Br = 79.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#O = 16.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>The molar mass of empirical formula, therefore (<mathjax>#CFBrO#</mathjax>) is:<br/>
<mathjax>#12.0 + 19.0 + 79.9 + 16.0#</mathjax> = <mathjax>#126.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>.</p>
<p>In order to get the molecular formula of <mathjax>#CFBrO#</mathjax>,<br/>
divide molar mass of compound by molar mass of empirical formula, like this: <mathjax>#"molar mass of compound"/" molar mass of empirical formula"#</mathjax> = <mathjax>#"whole number"#</mathjax></p>
<p>i.e. <mathjax>#381.01/126.9#</mathjax> = <mathjax>#3.00#</mathjax><em> <br/>
</em>Rounded to nearest whole number</p>
<p>Now, multiply the whole number by the Empirical Formula.<br/>
Like this <mathjax>#(CFBrO)#</mathjax><mathjax>#xx#</mathjax> <mathjax>#3#</mathjax> = <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p>
<p>Therefore your molecular formula is <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound if its empirical formula is #CFBrO# and its molar mass is 381.01 g/mol? </h1>
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<div class="markdown"><p><mathjax>#(CFBrO)#</mathjax><mathjax>#xx#</mathjax> <mathjax>#3#</mathjax> = ??</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Molar mass of compound (<mathjax>#381.01#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>) has been provided for you.</p>
<p>Now you need to calculate the molar mass of <mathjax>#CFBrO#</mathjax>. <br/>
The molar mass of:<br/>
<mathjax>#C = 12.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#F = 19.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#Br = 79.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax><br/>
<mathjax>#O = 16.0#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax></p>
<p>The molar mass of empirical formula, therefore (<mathjax>#CFBrO#</mathjax>) is:<br/>
<mathjax>#12.0 + 19.0 + 79.9 + 16.0#</mathjax> = <mathjax>#126.9#</mathjax> <mathjax>#g#</mathjax>/<mathjax>#mol#</mathjax>.</p>
<p>In order to get the molecular formula of <mathjax>#CFBrO#</mathjax>,<br/>
divide molar mass of compound by molar mass of empirical formula, like this: <mathjax>#"molar mass of compound"/" molar mass of empirical formula"#</mathjax> = <mathjax>#"whole number"#</mathjax></p>
<p>i.e. <mathjax>#381.01/126.9#</mathjax> = <mathjax>#3.00#</mathjax><em> <br/>
</em>Rounded to nearest whole number</p>
<p>Now, multiply the whole number by the Empirical Formula.<br/>
Like this <mathjax>#(CFBrO)#</mathjax><mathjax>#xx#</mathjax> <mathjax>#3#</mathjax> = <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p>
<p>Therefore your molecular formula is <mathjax>#C_3F_3Br_3O_3#</mathjax>.</p></div>
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</article> | What is the molecular formula of a compound if its empirical formula is #CFBrO# and its molar mass is 381.01 g/mol? | null |
2,889 | a86ea22e-6ddd-11ea-91fd-ccda262736ce | https://socratic.org/questions/58554f5db72cff6a47c8776d | CaCO3(s) + deltaH -> CaO(s) + CO2(g); DeltaH = +178kJ/mol | start chemical_equation qc_end physical_unit 4 5 20 21 enthalpy qc_end physical_unit 11 12 7 8 mole qc_end substance 14 15 qc_end substance 17 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the formation"}] | [{"type":"chemical equation","value":"CaCO3(s) + deltaH -> CaO(s) + CO2(g); DeltaH = +178kJ/mol"}] | [{"type":"physical unit","value":"Enthalpy [OF] the formation [=] \\pu{+178 kJ}"},{"type":"physical unit","value":"Mole [OF] calcium oxide [=] \\pu{1 mole}"},{"type":"substance name","value":"Carbon dioxide"},{"type":"substance name","value":"Calcium carbonate"}] | <h1 class="questionTitle" itemprop="name">The enthalpy associated with the formation of one mole quantities of calcium oxide, and carbon dioxide, from calcium carbonate is #+178*kJ#. How do we represent this reaction symbolically?</h1> | null | CaCO3(s) + deltaH -> CaO(s) + CO2(g); DeltaH = +178kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100*g)/(100.09*g*mol^-1)=1*mol.#</mathjax></p>
<p>And thus for:</p>
<p><mathjax>#CaCO_3(s) + DeltaH rarr CaO(s) + CO_2(g)uarr;#</mathjax> <mathjax>#DeltaH=178*kJ*mol^-1" of reaction as written"#</mathjax></p>
<p>Given that it is endothermic reaction, we quote <mathjax>#DeltaH_"rxn"=+178*kJ*mol^-1#</mathjax>.</p>
<p>I can think of one industry where the reaction is utilized on a monumental scale. Can you identify it?</p></div>
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<div class="markdown"><p><mathjax>#CaCO_3(s) + DeltaH rarr CaO(s) + CO_2(g)uarr#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100*g)/(100.09*g*mol^-1)=1*mol.#</mathjax></p>
<p>And thus for:</p>
<p><mathjax>#CaCO_3(s) + DeltaH rarr CaO(s) + CO_2(g)uarr;#</mathjax> <mathjax>#DeltaH=178*kJ*mol^-1" of reaction as written"#</mathjax></p>
<p>Given that it is endothermic reaction, we quote <mathjax>#DeltaH_"rxn"=+178*kJ*mol^-1#</mathjax>.</p>
<p>I can think of one industry where the reaction is utilized on a monumental scale. Can you identify it?</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The enthalpy associated with the formation of one mole quantities of calcium oxide, and carbon dioxide, from calcium carbonate is #+178*kJ#. How do we represent this reaction symbolically?</h1>
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<div class="markdown"><p><mathjax>#CaCO_3(s) + DeltaH rarr CaO(s) + CO_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of calcium carbonate"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100*g)/(100.09*g*mol^-1)=1*mol.#</mathjax></p>
<p>And thus for:</p>
<p><mathjax>#CaCO_3(s) + DeltaH rarr CaO(s) + CO_2(g)uarr;#</mathjax> <mathjax>#DeltaH=178*kJ*mol^-1" of reaction as written"#</mathjax></p>
<p>Given that it is endothermic reaction, we quote <mathjax>#DeltaH_"rxn"=+178*kJ*mol^-1#</mathjax>.</p>
<p>I can think of one industry where the reaction is utilized on a monumental scale. Can you identify it?</p></div>
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</article> | The enthalpy associated with the formation of one mole quantities of calcium oxide, and carbon dioxide, from calcium carbonate is #+178*kJ#. How do we represent this reaction symbolically? | null |
2,890 | ab1b46d8-6ddd-11ea-9a3a-ccda262736ce | https://socratic.org/questions/a-0-110-m-h-2so-4-solution-is-used-to-neutralize-10-0-ml-of-0-085-m-naoh-what-vo | 3.86 mL | start physical_unit 18 19 volume ml qc_end physical_unit 3 4 1 2 molarity qc_end physical_unit 14 14 12 13 molarity qc_end physical_unit 4 4 9 10 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] the acid [IN] mL"}] | [{"type":"physical unit","value":"3.86 mL"}] | [{"type":"physical unit","value":"Molarity [OF] H2SO4 solution [=] \\pu{0.110 M}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.085 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{10.0 mL}"}] | <h1 class="questionTitle" itemprop="name">A 0.110 M #H_2SO_4# solution is used to neutralize 10.0 mL of 0.085 M #NaOH#. What volume of the acid, in mL, is required to neutralize the base?</h1> | null | 3.86 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So what we have to do is use <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to find the volume of acid needed to neutralise the base. </p>
<p><strong>1:</strong> We need a balanced chemical equation: </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O"(l)#</mathjax></p>
</blockquote>
<p>So what we can find first is the moles of <mathjax>#"NaOH"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>2:</strong> Solve for moles. Note that I changed 10 mL to 0.01 L.</p>
<blockquote>
<p><mathjax>#n_1 = cv#</mathjax> </p>
<p><mathjax># = "0.085 mol/L (0.010 L)"#</mathjax></p>
<p><mathjax># = 8.5xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p><strong>3:</strong> Now we transfer the moles of <mathjax>#"NaOH"#</mathjax> to <mathjax>#"H"_2"SO"_4"#</mathjax>. </p>
<p>To do that, we divide by <mathjax>#"NaOH"#</mathjax>'s coefficient, and multiply by <mathjax>#"H"_2"SO"_4#</mathjax>'s. </p>
<blockquote>
<p><mathjax>#n_2 = (8.5xx10^-4 "mol")/2#</mathjax></p>
<p><mathjax># = 4.25xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p>There are <mathjax>#4.25xx10^-4#</mathjax> mol of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><strong>4:</strong> With <mathjax>#"H"_2"SO"_4#</mathjax>'s moles and concentration (given), we can solve for volume. </p>
<blockquote>
<p><mathjax>#v = n_"2"/c#</mathjax></p>
<p><mathjax># = (4.25xx10^-4 "mol")/("0.110 mol/L")#</mathjax></p>
<p><mathjax># "= 0.003,863,636"#</mathjax> L = <mathjax>#"3.863,636"#</mathjax> mL</p>
</blockquote>
<p>Significant digits tells us to round the volume to 3 decimal places:</p>
<p><mathjax>#3.86#</mathjax> mL.</p>
<p>Therefore, you need <mathjax>#3.86#</mathjax> mL of <mathjax>#"H"_2"SO"_4#</mathjax> to neutralise the base. </p>
<p>Hope this helps :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>You need 3.86 mL of <mathjax>#"H"_2"SO"_4#</mathjax> to neutralise the base. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So what we have to do is use <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to find the volume of acid needed to neutralise the base. </p>
<p><strong>1:</strong> We need a balanced chemical equation: </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O"(l)#</mathjax></p>
</blockquote>
<p>So what we can find first is the moles of <mathjax>#"NaOH"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>2:</strong> Solve for moles. Note that I changed 10 mL to 0.01 L.</p>
<blockquote>
<p><mathjax>#n_1 = cv#</mathjax> </p>
<p><mathjax># = "0.085 mol/L (0.010 L)"#</mathjax></p>
<p><mathjax># = 8.5xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p><strong>3:</strong> Now we transfer the moles of <mathjax>#"NaOH"#</mathjax> to <mathjax>#"H"_2"SO"_4"#</mathjax>. </p>
<p>To do that, we divide by <mathjax>#"NaOH"#</mathjax>'s coefficient, and multiply by <mathjax>#"H"_2"SO"_4#</mathjax>'s. </p>
<blockquote>
<p><mathjax>#n_2 = (8.5xx10^-4 "mol")/2#</mathjax></p>
<p><mathjax># = 4.25xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p>There are <mathjax>#4.25xx10^-4#</mathjax> mol of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><strong>4:</strong> With <mathjax>#"H"_2"SO"_4#</mathjax>'s moles and concentration (given), we can solve for volume. </p>
<blockquote>
<p><mathjax>#v = n_"2"/c#</mathjax></p>
<p><mathjax># = (4.25xx10^-4 "mol")/("0.110 mol/L")#</mathjax></p>
<p><mathjax># "= 0.003,863,636"#</mathjax> L = <mathjax>#"3.863,636"#</mathjax> mL</p>
</blockquote>
<p>Significant digits tells us to round the volume to 3 decimal places:</p>
<p><mathjax>#3.86#</mathjax> mL.</p>
<p>Therefore, you need <mathjax>#3.86#</mathjax> mL of <mathjax>#"H"_2"SO"_4#</mathjax> to neutralise the base. </p>
<p>Hope this helps :)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 0.110 M #H_2SO_4# solution is used to neutralize 10.0 mL of 0.085 M #NaOH#. What volume of the acid, in mL, is required to neutralize the base?</h1>
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<div class="markdown"><p>You need 3.86 mL of <mathjax>#"H"_2"SO"_4#</mathjax> to neutralise the base. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So what we have to do is use <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to find the volume of acid needed to neutralise the base. </p>
<p><strong>1:</strong> We need a balanced chemical equation: </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O"(l)#</mathjax></p>
</blockquote>
<p>So what we can find first is the moles of <mathjax>#"NaOH"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>2:</strong> Solve for moles. Note that I changed 10 mL to 0.01 L.</p>
<blockquote>
<p><mathjax>#n_1 = cv#</mathjax> </p>
<p><mathjax># = "0.085 mol/L (0.010 L)"#</mathjax></p>
<p><mathjax># = 8.5xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p><strong>3:</strong> Now we transfer the moles of <mathjax>#"NaOH"#</mathjax> to <mathjax>#"H"_2"SO"_4"#</mathjax>. </p>
<p>To do that, we divide by <mathjax>#"NaOH"#</mathjax>'s coefficient, and multiply by <mathjax>#"H"_2"SO"_4#</mathjax>'s. </p>
<blockquote>
<p><mathjax>#n_2 = (8.5xx10^-4 "mol")/2#</mathjax></p>
<p><mathjax># = 4.25xx10^-4 "mol"#</mathjax></p>
</blockquote>
<p>There are <mathjax>#4.25xx10^-4#</mathjax> mol of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><strong>4:</strong> With <mathjax>#"H"_2"SO"_4#</mathjax>'s moles and concentration (given), we can solve for volume. </p>
<blockquote>
<p><mathjax>#v = n_"2"/c#</mathjax></p>
<p><mathjax># = (4.25xx10^-4 "mol")/("0.110 mol/L")#</mathjax></p>
<p><mathjax># "= 0.003,863,636"#</mathjax> L = <mathjax>#"3.863,636"#</mathjax> mL</p>
</blockquote>
<p>Significant digits tells us to round the volume to 3 decimal places:</p>
<p><mathjax>#3.86#</mathjax> mL.</p>
<p>Therefore, you need <mathjax>#3.86#</mathjax> mL of <mathjax>#"H"_2"SO"_4#</mathjax> to neutralise the base. </p>
<p>Hope this helps :)</p></div>
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</article> | A 0.110 M #H_2SO_4# solution is used to neutralize 10.0 mL of 0.085 M #NaOH#. What volume of the acid, in mL, is required to neutralize the base? | null |
2,891 | a8e97703-6ddd-11ea-b0fb-ccda262736ce | https://socratic.org/questions/5933f8a67c0149397b1a83df | S2O3^2− + 2H+ -> S + H2O + SO2 | start chemical_equation qc_end chemical_equation 4 4 qc_end substance 8 8 qc_end substance 10 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"S2O3^2− + 2H+ -> S + H2O + SO2"}] | [{"type":"chemical equation","value":"S2O3^2−"},{"type":"substance name","value":"Sulfur"},{"type":"substance name","value":"Sulfur dioxide"}] | <h1 class="questionTitle" itemprop="name">How does #"thiosulfate ion"#, #S_2O_3^(2-)# give rise to sulfur and sulfur dioxide upon disproportionation?</h1> | null | S2O3^2− + 2H+ -> S + H2O + SO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thiosulfate, <mathjax>#S_2O_3^(2-)#</mathjax> has formal <mathjax>#stackrel(-II)S#</mathjax> and <mathjax>#stackrel(+VI)S#</mathjax> oxidation states, i.e. an average oxidation state of <mathjax>#(-II+VI)/2=+II#</mathjax>.</p>
<p>It is reduced to <mathjax>#stackrel(0)S#</mathjax>, i.e. zerovalent sulfur, and oxidized to <mathjax>#stackrel(+IV)"SO"_2#</mathjax>.</p>
<p>And thus reduction............:</p>
<p><mathjax>#S_2O_3^(2-) +6H^(+)+ 4e^(-) rarr 2S +3H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And oxidation............:</p>
<p><mathjax>#S_2O_3^(2-) +H_2O rarr 2SO_2 +2H^(+)+4e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>For both <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax>, charge and mass are balanced as is absolutely required. Are they balanced? Don't trust my arithmetic.</p>
<p>We add <mathjax>#(i) + (ii)#</mathjax> to remove the electrons...........</p>
<p><mathjax>#2S_2O_3^(2-) +4H^(+) rarr 2S +2H_2O +2SO_2#</mathjax> </p>
<p>And of course we could halve this, as per Nam D.'s suggestion..</p>
<p><mathjax>#underbrace(S_2O_3^(2-) +2H^(+))_("sulfurous acid"*H_2SO_3) rarr S +H_2O +SO_2#</mathjax> </p>
<p>Again, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, <mathjax>#2xxS#</mathjax>, <mathjax>#3xxO#</mathjax>, and <mathjax>#2xxH#</mathjax>.........so balanced with respect to mass and charge.</p>
<p>What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of <mathjax>#SO_2#</mathjax>. </p>
<p>Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give <mathjax>#H_2S_2O_3#</mathjax>. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. <mathjax>#S(-II)#</mathjax>. And so we got <mathjax>#S(VI+)#</mathjax> and <mathjax>#S(-II)#</mathjax>...the average oxidation state is still <mathjax>#S(+II)#</mathjax>. And we compare this with <mathjax>#SO_4^(2-)#</mathjax>, i.e. <mathjax>#S(VI+)+4xxO(-II)#</mathjax>...</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This is a disproportionation reaction........</p>
<p><mathjax>#S_2O_3^(2-) +2H^(+) rarr S +SO_2+H_2O#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thiosulfate, <mathjax>#S_2O_3^(2-)#</mathjax> has formal <mathjax>#stackrel(-II)S#</mathjax> and <mathjax>#stackrel(+VI)S#</mathjax> oxidation states, i.e. an average oxidation state of <mathjax>#(-II+VI)/2=+II#</mathjax>.</p>
<p>It is reduced to <mathjax>#stackrel(0)S#</mathjax>, i.e. zerovalent sulfur, and oxidized to <mathjax>#stackrel(+IV)"SO"_2#</mathjax>.</p>
<p>And thus reduction............:</p>
<p><mathjax>#S_2O_3^(2-) +6H^(+)+ 4e^(-) rarr 2S +3H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And oxidation............:</p>
<p><mathjax>#S_2O_3^(2-) +H_2O rarr 2SO_2 +2H^(+)+4e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>For both <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax>, charge and mass are balanced as is absolutely required. Are they balanced? Don't trust my arithmetic.</p>
<p>We add <mathjax>#(i) + (ii)#</mathjax> to remove the electrons...........</p>
<p><mathjax>#2S_2O_3^(2-) +4H^(+) rarr 2S +2H_2O +2SO_2#</mathjax> </p>
<p>And of course we could halve this, as per Nam D.'s suggestion..</p>
<p><mathjax>#underbrace(S_2O_3^(2-) +2H^(+))_("sulfurous acid"*H_2SO_3) rarr S +H_2O +SO_2#</mathjax> </p>
<p>Again, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, <mathjax>#2xxS#</mathjax>, <mathjax>#3xxO#</mathjax>, and <mathjax>#2xxH#</mathjax>.........so balanced with respect to mass and charge.</p>
<p>What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of <mathjax>#SO_2#</mathjax>. </p>
<p>Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give <mathjax>#H_2S_2O_3#</mathjax>. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. <mathjax>#S(-II)#</mathjax>. And so we got <mathjax>#S(VI+)#</mathjax> and <mathjax>#S(-II)#</mathjax>...the average oxidation state is still <mathjax>#S(+II)#</mathjax>. And we compare this with <mathjax>#SO_4^(2-)#</mathjax>, i.e. <mathjax>#S(VI+)+4xxO(-II)#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">How does #"thiosulfate ion"#, #S_2O_3^(2-)# give rise to sulfur and sulfur dioxide upon disproportionation?</h1>
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<div class="markdown"><p>This is a disproportionation reaction........</p>
<p><mathjax>#S_2O_3^(2-) +2H^(+) rarr S +SO_2+H_2O#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Thiosulfate, <mathjax>#S_2O_3^(2-)#</mathjax> has formal <mathjax>#stackrel(-II)S#</mathjax> and <mathjax>#stackrel(+VI)S#</mathjax> oxidation states, i.e. an average oxidation state of <mathjax>#(-II+VI)/2=+II#</mathjax>.</p>
<p>It is reduced to <mathjax>#stackrel(0)S#</mathjax>, i.e. zerovalent sulfur, and oxidized to <mathjax>#stackrel(+IV)"SO"_2#</mathjax>.</p>
<p>And thus reduction............:</p>
<p><mathjax>#S_2O_3^(2-) +6H^(+)+ 4e^(-) rarr 2S +3H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And oxidation............:</p>
<p><mathjax>#S_2O_3^(2-) +H_2O rarr 2SO_2 +2H^(+)+4e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>For both <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax>, charge and mass are balanced as is absolutely required. Are they balanced? Don't trust my arithmetic.</p>
<p>We add <mathjax>#(i) + (ii)#</mathjax> to remove the electrons...........</p>
<p><mathjax>#2S_2O_3^(2-) +4H^(+) rarr 2S +2H_2O +2SO_2#</mathjax> </p>
<p>And of course we could halve this, as per Nam D.'s suggestion..</p>
<p><mathjax>#underbrace(S_2O_3^(2-) +2H^(+))_("sulfurous acid"*H_2SO_3) rarr S +H_2O +SO_2#</mathjax> </p>
<p>Again, if mass and charge ARE NOT BALANCED, then we CANNOT accept it as a representation of chemical reality. Looking at it again, we gots both LHS and RHS neutral, LHS and RHS, <mathjax>#2xxS#</mathjax>, <mathjax>#3xxO#</mathjax>, and <mathjax>#2xxH#</mathjax>.........so balanced with respect to mass and charge.</p>
<p>What would you observe in this reaction? Well, probably (i) the precipitate of a fine white powder of elemental sulfur, and (ii) maybe the foul odour of <mathjax>#SO_2#</mathjax>. </p>
<p>Just as a comment, I like to think of sulfurous acid as the SULFUR analogue of sulfuric acid, where ONE sulfur has replaced ONE oxygen to give <mathjax>#H_2S_2O_3#</mathjax>. And this sulfur assumes THE SAME oxidation state as the oxygen it replaces, i.e. <mathjax>#S(-II)#</mathjax>. And so we got <mathjax>#S(VI+)#</mathjax> and <mathjax>#S(-II)#</mathjax>...the average oxidation state is still <mathjax>#S(+II)#</mathjax>. And we compare this with <mathjax>#SO_4^(2-)#</mathjax>, i.e. <mathjax>#S(VI+)+4xxO(-II)#</mathjax>...</p></div>
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</article> | How does #"thiosulfate ion"#, #S_2O_3^(2-)# give rise to sulfur and sulfur dioxide upon disproportionation? | null |
2,892 | a8b1fcd2-6ddd-11ea-9e3a-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-of-nitric-acid-if-0-216-of-barium-hydroxide-i | 0.1 mol/L | start physical_unit 8 9 molarity mol/l qc_end physical_unit 14 15 11 12 mass qc_end physical_unit 8 9 20 21 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] nitric acid solution [IN] mol/L"}] | [{"type":"physical unit","value":"0.1 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] barium hydroxide [=] \\pu{0.216 g}"},{"type":"physical unit","value":"Volume [OF] nitric acid [=] \\pu{20.00 mL}"}] | <h1 class="questionTitle" itemprop="name"> What is the molarity of a solution of nitric acid if 0.216 of barium hydroxide is required to neutralize 20.00mL of nitric acid?</h1> | null | 0.1 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#0.216#</mathjax> <mathjax>#g#</mathjax> <mathjax>#Ba(OH)_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.216*g)/(171.34*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.26xx10^(-3)#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Ba(OH)_2#</mathjax>.</p>
<p>Now, of course I need a chemical equation:</p>
<p><mathjax>#Ba(OH)_2(aq) + 2HNO_3 rarr Ba(NO_3)_2(aq) + 2H_2O(l)#</mathjax>.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation above (what do I mean here?), in <mathjax>#20.00#</mathjax> <mathjax>#mL#</mathjax> solution, there were <mathjax>#2.52xx10^(-3)#</mathjax> moles of nitric acid present.</p>
<p>So concentration of the <mathjax>#HNO_3(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.52xx10^(-3) mol)/(20.00xx10^(-3) L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.1*mol*L^-1#</mathjax>. You'll have to work out the actual figure; my calculator is never handy.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>I assume that it is <mathjax>#0.216#</mathjax> <mathjax>#g#</mathjax> of barium hydroxide, and <mathjax>#20.00#</mathjax> <mathjax>#mL#</mathjax> nitric acid.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#0.216#</mathjax> <mathjax>#g#</mathjax> <mathjax>#Ba(OH)_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.216*g)/(171.34*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.26xx10^(-3)#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Ba(OH)_2#</mathjax>.</p>
<p>Now, of course I need a chemical equation:</p>
<p><mathjax>#Ba(OH)_2(aq) + 2HNO_3 rarr Ba(NO_3)_2(aq) + 2H_2O(l)#</mathjax>.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation above (what do I mean here?), in <mathjax>#20.00#</mathjax> <mathjax>#mL#</mathjax> solution, there were <mathjax>#2.52xx10^(-3)#</mathjax> moles of nitric acid present.</p>
<p>So concentration of the <mathjax>#HNO_3(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.52xx10^(-3) mol)/(20.00xx10^(-3) L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.1*mol*L^-1#</mathjax>. You'll have to work out the actual figure; my calculator is never handy.</p></div>
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<h1 class="questionTitle" itemprop="name"> What is the molarity of a solution of nitric acid if 0.216 of barium hydroxide is required to neutralize 20.00mL of nitric acid?</h1>
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<div class="markdown"><p>I assume that it is <mathjax>#0.216#</mathjax> <mathjax>#g#</mathjax> of barium hydroxide, and <mathjax>#20.00#</mathjax> <mathjax>#mL#</mathjax> nitric acid.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#0.216#</mathjax> <mathjax>#g#</mathjax> <mathjax>#Ba(OH)_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.216*g)/(171.34*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.26xx10^(-3)#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Ba(OH)_2#</mathjax>.</p>
<p>Now, of course I need a chemical equation:</p>
<p><mathjax>#Ba(OH)_2(aq) + 2HNO_3 rarr Ba(NO_3)_2(aq) + 2H_2O(l)#</mathjax>.</p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation above (what do I mean here?), in <mathjax>#20.00#</mathjax> <mathjax>#mL#</mathjax> solution, there were <mathjax>#2.52xx10^(-3)#</mathjax> moles of nitric acid present.</p>
<p>So concentration of the <mathjax>#HNO_3(aq)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.52xx10^(-3) mol)/(20.00xx10^(-3) L)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#0.1*mol*L^-1#</mathjax>. You'll have to work out the actual figure; my calculator is never handy.</p></div>
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</article> | What is the molarity of a solution of nitric acid if 0.216 of barium hydroxide is required to neutralize 20.00mL of nitric acid? | null |
2,893 | ac0e8cf6-6ddd-11ea-a8d7-ccda262736ce | https://socratic.org/questions/what-volume-of-nh-3-at-stp-is-produced-if-25-0-g-of-n-2-is-reacted-with-an-exces | 40.0 L | start physical_unit 3 3 volume l qc_end c_other STP qc_end physical_unit 12 12 9 10 mass qc_end c_other OTHER qc_end chemical_equation 23 29 qc_end end | [{"type":"physical unit","value":"Volume [OF] NH3 [IN] L"}] | [{"type":"physical unit","value":"40.0 L"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Mass [OF] N2 [=] \\pu{25.0 g}"},{"type":"other","value":"Excess of H2."},{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"}] | <h1 class="questionTitle" itemprop="name">What volume of #"NH"_3# at STP is produced if #25.0 "g"# of #"N"_2# is reacted with an excess of #"H"_2# in the reaction #"N"_2 + 3"H"_2 -> 2"NH"_3#?</h1> | null | 40.0 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1 "mol"#</mathjax> of <mathjax>#"N"_2#</mathjax> has mass of <mathjax>#28.0 "g"#</mathjax></p>
<p>Therefore, <mathjax>#25.0"g"#</mathjax> of <mathjax>#"N"_2#</mathjax> in moles is</p>
<blockquote>
<p><mathjax>#frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>Each molecule of <mathjax>#"N"_2#</mathjax> will produce 2 molecules of <mathjax>#"NH"_3#</mathjax>. </p>
<p>Therefore, the 2 times the amount of <mathjax>#"NH"_3#</mathjax> will be produced.</p>
<blockquote>
<p><mathjax>#2 xx 0.893 "mol" = 1.79 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>One mole of an ideal gas has a volume of <mathjax>#22.4"L"#</mathjax> at STP.</p>
<p>Therefore, the volume occupied by <mathjax>#1.79 "mol"#</mathjax> of <mathjax>#"NH"_3#</mathjax> is</p>
<blockquote>
<p><mathjax>#1.79 "mol" xx 22.4 "L/mol" = 40.0 "L"#</mathjax></p>
</blockquote>
<h2></h2>
<p>That's it. But bear in mind that you need to somehow remove the excess <mathjax>#"H"_2#</mathjax> before the container will have a volume of <mathjax>#40.0 "L"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#40.0"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1 "mol"#</mathjax> of <mathjax>#"N"_2#</mathjax> has mass of <mathjax>#28.0 "g"#</mathjax></p>
<p>Therefore, <mathjax>#25.0"g"#</mathjax> of <mathjax>#"N"_2#</mathjax> in moles is</p>
<blockquote>
<p><mathjax>#frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>Each molecule of <mathjax>#"N"_2#</mathjax> will produce 2 molecules of <mathjax>#"NH"_3#</mathjax>. </p>
<p>Therefore, the 2 times the amount of <mathjax>#"NH"_3#</mathjax> will be produced.</p>
<blockquote>
<p><mathjax>#2 xx 0.893 "mol" = 1.79 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>One mole of an ideal gas has a volume of <mathjax>#22.4"L"#</mathjax> at STP.</p>
<p>Therefore, the volume occupied by <mathjax>#1.79 "mol"#</mathjax> of <mathjax>#"NH"_3#</mathjax> is</p>
<blockquote>
<p><mathjax>#1.79 "mol" xx 22.4 "L/mol" = 40.0 "L"#</mathjax></p>
</blockquote>
<h2></h2>
<p>That's it. But bear in mind that you need to somehow remove the excess <mathjax>#"H"_2#</mathjax> before the container will have a volume of <mathjax>#40.0 "L"#</mathjax>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of #"NH"_3# at STP is produced if #25.0 "g"# of #"N"_2# is reacted with an excess of #"H"_2# in the reaction #"N"_2 + 3"H"_2 -> 2"NH"_3#?</h1>
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<div class="markdown"><p><mathjax>#40.0"L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1 "mol"#</mathjax> of <mathjax>#"N"_2#</mathjax> has mass of <mathjax>#28.0 "g"#</mathjax></p>
<p>Therefore, <mathjax>#25.0"g"#</mathjax> of <mathjax>#"N"_2#</mathjax> in moles is</p>
<blockquote>
<p><mathjax>#frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>Each molecule of <mathjax>#"N"_2#</mathjax> will produce 2 molecules of <mathjax>#"NH"_3#</mathjax>. </p>
<p>Therefore, the 2 times the amount of <mathjax>#"NH"_3#</mathjax> will be produced.</p>
<blockquote>
<p><mathjax>#2 xx 0.893 "mol" = 1.79 "mol"#</mathjax></p>
</blockquote>
<h2></h2>
<p>One mole of an ideal gas has a volume of <mathjax>#22.4"L"#</mathjax> at STP.</p>
<p>Therefore, the volume occupied by <mathjax>#1.79 "mol"#</mathjax> of <mathjax>#"NH"_3#</mathjax> is</p>
<blockquote>
<p><mathjax>#1.79 "mol" xx 22.4 "L/mol" = 40.0 "L"#</mathjax></p>
</blockquote>
<h2></h2>
<p>That's it. But bear in mind that you need to somehow remove the excess <mathjax>#"H"_2#</mathjax> before the container will have a volume of <mathjax>#40.0 "L"#</mathjax>.</p></div>
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</article> | What volume of #"NH"_3# at STP is produced if #25.0 "g"# of #"N"_2# is reacted with an excess of #"H"_2# in the reaction #"N"_2 + 3"H"_2 -> 2"NH"_3#? | null |
2,894 | ac0f28ba-6ddd-11ea-b678-ccda262736ce | https://socratic.org/questions/if-3-38-g-of-cuno-3-is-dissolved-in-water-to-make-a-0-450-m-solution-what-is-the | 0.04 L | start physical_unit 14 14 volume l qc_end physical_unit 4 4 1 2 mass qc_end substance 8 8 qc_end physical_unit 4 4 12 13 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] CuNO3 solution [IN] L"}] | [{"type":"physical unit","value":"0.04 L"}] | [{"type":"physical unit","value":"Mass [OF] CuNO3 [=] \\pu{3.38 g}"},{"type":"substance name","value":"Water"},{"type":"physical unit","value":"Molarity [OF] CuNO3 solution [=] \\pu{0.450 M}"}] | <h1 class="questionTitle" itemprop="name">If 3.38 g of #CuNO_3# is dissolved in water to make a 0.450 M solution, what is the volume of the solution? </h1> | null | 0.04 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(3.38*g)/(187.56*g*mol^-1)xx1/(0.450*mol*L^-1)=0.040*L#</mathjax></p>
<p>Are you happy with this treatment? All I have done is manipulated the quotient, to give me an answer in terms of <mathjax>#"volume"#</mathjax>.</p>
<p>Note that if they really asked for <mathjax>#"cuprous nitrate"#</mathjax>, <mathjax>#CuNO_3#</mathjax>, they should have known better. I doubt that you could isolate <mathjax>#"cuprous nitrate"#</mathjax>.</p></div>
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<div class="markdown"><p>I will assume you mean <mathjax>#"cupric nitrate"#</mathjax>, <mathjax>#Cu(NO_3)_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(3.38*g)/(187.56*g*mol^-1)xx1/(0.450*mol*L^-1)=0.040*L#</mathjax></p>
<p>Are you happy with this treatment? All I have done is manipulated the quotient, to give me an answer in terms of <mathjax>#"volume"#</mathjax>.</p>
<p>Note that if they really asked for <mathjax>#"cuprous nitrate"#</mathjax>, <mathjax>#CuNO_3#</mathjax>, they should have known better. I doubt that you could isolate <mathjax>#"cuprous nitrate"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">If 3.38 g of #CuNO_3# is dissolved in water to make a 0.450 M solution, what is the volume of the solution? </h1>
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anor277
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<span class="dateCreated" datetime="2017-02-06T17:48:42" itemprop="dateCreated">
Feb 6, 2017
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<div class="markdown"><p>I will assume you mean <mathjax>#"cupric nitrate"#</mathjax>, <mathjax>#Cu(NO_3)_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"Volume of solution"="Moles of solute"/"Concentration"#</mathjax></p>
<p><mathjax>#=(3.38*g)/(187.56*g*mol^-1)xx1/(0.450*mol*L^-1)=0.040*L#</mathjax></p>
<p>Are you happy with this treatment? All I have done is manipulated the quotient, to give me an answer in terms of <mathjax>#"volume"#</mathjax>.</p>
<p>Note that if they really asked for <mathjax>#"cuprous nitrate"#</mathjax>, <mathjax>#CuNO_3#</mathjax>, they should have known better. I doubt that you could isolate <mathjax>#"cuprous nitrate"#</mathjax>.</p></div>
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</article> | If 3.38 g of #CuNO_3# is dissolved in water to make a 0.450 M solution, what is the volume of the solution? | null |
2,895 | ac12f939-6ddd-11ea-8952-ccda262736ce | https://socratic.org/questions/a-given-volume-of-a-buffer-solution-contains-6-85-x-10-3-mol-of-the-weak-acid-hy | 3.42 | start physical_unit 16 16 pka none qc_end physical_unit 16 16 8 11 mole qc_end physical_unit 25 25 18 21 mole qc_end physical_unit 29 31 33 33 ph qc_end end | [{"type":"physical unit","value":"pKa [OF] HY"}] | [{"type":"physical unit","value":"3.42"}] | [{"type":"physical unit","value":"Mole [OF] HY [=] \\pu{6.85 × 10^(-3) mol}"},{"type":"physical unit","value":"Mole [OF] NaY [=] \\pu{2.98 × 10^(-3) mol}"},{"type":"physical unit","value":"pH [OF] the buffer solution [=] \\pu{3.78}"}] | <h1 class="questionTitle" itemprop="name">A given volume of a buffer solution contains #6.85 x 10^-3# mol of the weak acid HY and #2.98 x 10^-3# mol of the salt NaY. The pH of the buffer solution is 3.78. How do you calculate the value of #pKa# for the acid HY at this temperature?</h1> | null | 3.42 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we were quoted molar quantities of the acid and its conjugate base....NOT concentrations...but given we have the quotient...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax>...and in the buffer expression we get....</p>
<p><mathjax>#log_10{((2.98.xx10^-3*mol)/(cancel"some volume"))/((6.85xx10^-3*mol)/cancel("some volume"))}#</mathjax>...that is the volumes cancel out....so WE DO NOT NEED TO KNOW THE VOLUME....</p>
<p>And so we solve for <mathjax>#pK_a#</mathjax> in the buffer equation....</p>
<p><mathjax>#pK_a=pH-log_10{(2.98xx10^-3*mol)/(6.85xx10^-3*mol)}#</mathjax></p>
<p><mathjax>#=3.78-underbrace(log_10{(2.98)/(6.85)})_"-0.361"=3.42#</mathjax></p>
<p><mathjax>#pH_"solution"#</mathjax> is more acidic than <mathjax>#pK_a#</mathjax> given that the parent acid is present in GREATER concentration.....to what would <mathjax>#pH#</mathjax> be equal if <mathjax>#[HY]-=[Y^-]#</mathjax>?</p></div>
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<div class="markdown"><p>Well, the <a href="https://socratic.org/questions/58d3dc28b72cff2e15eb8836">buffer equation</a> holds that.... <mathjax>#pH=pK_a+log_10{[[Y^-]]/[[HY]]}#</mathjax>...</p>
<p>...we finally get <mathjax>#pK_a=3.42...#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we were quoted molar quantities of the acid and its conjugate base....NOT concentrations...but given we have the quotient...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax>...and in the buffer expression we get....</p>
<p><mathjax>#log_10{((2.98.xx10^-3*mol)/(cancel"some volume"))/((6.85xx10^-3*mol)/cancel("some volume"))}#</mathjax>...that is the volumes cancel out....so WE DO NOT NEED TO KNOW THE VOLUME....</p>
<p>And so we solve for <mathjax>#pK_a#</mathjax> in the buffer equation....</p>
<p><mathjax>#pK_a=pH-log_10{(2.98xx10^-3*mol)/(6.85xx10^-3*mol)}#</mathjax></p>
<p><mathjax>#=3.78-underbrace(log_10{(2.98)/(6.85)})_"-0.361"=3.42#</mathjax></p>
<p><mathjax>#pH_"solution"#</mathjax> is more acidic than <mathjax>#pK_a#</mathjax> given that the parent acid is present in GREATER concentration.....to what would <mathjax>#pH#</mathjax> be equal if <mathjax>#[HY]-=[Y^-]#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">A given volume of a buffer solution contains #6.85 x 10^-3# mol of the weak acid HY and #2.98 x 10^-3# mol of the salt NaY. The pH of the buffer solution is 3.78. How do you calculate the value of #pKa# for the acid HY at this temperature?</h1>
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<div class="markdown"><p>Well, the <a href="https://socratic.org/questions/58d3dc28b72cff2e15eb8836">buffer equation</a> holds that.... <mathjax>#pH=pK_a+log_10{[[Y^-]]/[[HY]]}#</mathjax>...</p>
<p>...we finally get <mathjax>#pK_a=3.42...#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we were quoted molar quantities of the acid and its conjugate base....NOT concentrations...but given we have the quotient...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax>...and in the buffer expression we get....</p>
<p><mathjax>#log_10{((2.98.xx10^-3*mol)/(cancel"some volume"))/((6.85xx10^-3*mol)/cancel("some volume"))}#</mathjax>...that is the volumes cancel out....so WE DO NOT NEED TO KNOW THE VOLUME....</p>
<p>And so we solve for <mathjax>#pK_a#</mathjax> in the buffer equation....</p>
<p><mathjax>#pK_a=pH-log_10{(2.98xx10^-3*mol)/(6.85xx10^-3*mol)}#</mathjax></p>
<p><mathjax>#=3.78-underbrace(log_10{(2.98)/(6.85)})_"-0.361"=3.42#</mathjax></p>
<p><mathjax>#pH_"solution"#</mathjax> is more acidic than <mathjax>#pK_a#</mathjax> given that the parent acid is present in GREATER concentration.....to what would <mathjax>#pH#</mathjax> be equal if <mathjax>#[HY]-=[Y^-]#</mathjax>?</p></div>
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</article> | A given volume of a buffer solution contains #6.85 x 10^-3# mol of the weak acid HY and #2.98 x 10^-3# mol of the salt NaY. The pH of the buffer solution is 3.78. How do you calculate the value of #pKa# for the acid HY at this temperature? | null |
2,896 | a92e88ca-6ddd-11ea-8286-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-200-m-solution-of-formic-acid | 2.23 | start physical_unit 8 8 ph none qc_end physical_unit 10 11 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] formic acid solution"}] | [{"type":"physical unit","value":"2.23"}] | [{"type":"physical unit","value":"Molarity [OF] formic acid solution [=] \\pu{0.200 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a .200 M solution of formic acid?</h1> | null | 2.23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HC(=O)OH + H_2O rightleftharpoons H_3O^+ + HC(=O)O^-#</mathjax></p>
<p><mathjax>#K_a=([H_3O^+][HCO_2^-])/[[HC(=O)OH]]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#x_1^2/(0.200-x_1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4#</mathjax>. </p>
<p>Of course I had to look up <mathjax>#K_a" formic acid"#</mathjax> on the interwebz.</p>
<p>If, <mathjax>#x_1 <0.200#</mathjax>, then <mathjax>#(0.200-x_1)~=0.200#</mathjax></p>
<p>Thus <mathjax>#x_1^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4xx(0.200)#</mathjax></p>
<p><mathjax>#x_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.95xx10^-3#</mathjax></p>
<p><mathjax>#x_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.95xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p><mathjax>#x_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.86xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p>So successive approximations, leads us to <mathjax>#[H_3O^+]=5.86xx10^-3*mol*L^-1#</mathjax>.</p>
<p><mathjax>#pH=-log_10[H_3O^+]=-log_(10){5.86xx10^-3}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.23#</mathjax></p>
<p>As carboxylic acids go, this is a fairly low <mathjax>#pH#</mathjax>. Why should formic acid be more acidic than say propionic acid?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>And <mathjax>#pK_a#</mathjax> for formic acid is?</p>
<p><mathjax>#pH=2.23#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HC(=O)OH + H_2O rightleftharpoons H_3O^+ + HC(=O)O^-#</mathjax></p>
<p><mathjax>#K_a=([H_3O^+][HCO_2^-])/[[HC(=O)OH]]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#x_1^2/(0.200-x_1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4#</mathjax>. </p>
<p>Of course I had to look up <mathjax>#K_a" formic acid"#</mathjax> on the interwebz.</p>
<p>If, <mathjax>#x_1 <0.200#</mathjax>, then <mathjax>#(0.200-x_1)~=0.200#</mathjax></p>
<p>Thus <mathjax>#x_1^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4xx(0.200)#</mathjax></p>
<p><mathjax>#x_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.95xx10^-3#</mathjax></p>
<p><mathjax>#x_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.95xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p><mathjax>#x_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.86xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p>So successive approximations, leads us to <mathjax>#[H_3O^+]=5.86xx10^-3*mol*L^-1#</mathjax>.</p>
<p><mathjax>#pH=-log_10[H_3O^+]=-log_(10){5.86xx10^-3}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.23#</mathjax></p>
<p>As carboxylic acids go, this is a fairly low <mathjax>#pH#</mathjax>. Why should formic acid be more acidic than say propionic acid?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a .200 M solution of formic acid?</h1>
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anor277
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<div class="markdown"><p>And <mathjax>#pK_a#</mathjax> for formic acid is?</p>
<p><mathjax>#pH=2.23#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HC(=O)OH + H_2O rightleftharpoons H_3O^+ + HC(=O)O^-#</mathjax></p>
<p><mathjax>#K_a=([H_3O^+][HCO_2^-])/[[HC(=O)OH]]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#x_1^2/(0.200-x_1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4#</mathjax>. </p>
<p>Of course I had to look up <mathjax>#K_a" formic acid"#</mathjax> on the interwebz.</p>
<p>If, <mathjax>#x_1 <0.200#</mathjax>, then <mathjax>#(0.200-x_1)~=0.200#</mathjax></p>
<p>Thus <mathjax>#x_1^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.77xx10^-4xx(0.200)#</mathjax></p>
<p><mathjax>#x_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.95xx10^-3#</mathjax></p>
<p><mathjax>#x_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.95xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p><mathjax>#x_3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#sqrt(1.77xx10^-4xx(0.200-5.86xx10^-3))#</mathjax><mathjax>#=#</mathjax> <mathjax>#5.86xx10^-3#</mathjax></p>
<p>So successive approximations, leads us to <mathjax>#[H_3O^+]=5.86xx10^-3*mol*L^-1#</mathjax>.</p>
<p><mathjax>#pH=-log_10[H_3O^+]=-log_(10){5.86xx10^-3}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.23#</mathjax></p>
<p>As carboxylic acids go, this is a fairly low <mathjax>#pH#</mathjax>. Why should formic acid be more acidic than say propionic acid?</p></div>
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</article> | What is the pH of a .200 M solution of formic acid? | null |
2,897 | ab050aa6-6ddd-11ea-a775-ccda262736ce | https://socratic.org/questions/in-the-equation-mg-2hcl-mgcl-2-h-2-what-volume-of-hydrogen-at-stp-is-produced-fr | 23.07 L | start physical_unit 14 14 volume l qc_end physical_unit 3 3 23 24 mass qc_end physical_unit 6 6 31 32 mass qc_end chemical_equation 3 10 qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrogen [IN] L"}] | [{"type":"physical unit","value":"23.07 L"}] | [{"type":"physical unit","value":"Mass [OF] Mg [=] \\pu{50.0 g}"},{"type":"physical unit","value":"Mass [OF] HCl [=] \\pu{75 g}"},{"type":"chemical equation","value":"Mg + 2 HCl -> MgCl2 + H2"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">In the equation #Mg + 2HCl -> MgCl_2 + H_2#, what volume of hydrogen at STP is produced from the reaction of 50.0 g of #Mg# and the equivalent of 75 g of #HCl#?</h1> | null | 23.07 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of magnesium: <mathjax>#(50.0*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.06#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Moles of hydrogen chloride gas: <mathjax>#(75.0*g)/(36.2*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.07#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires 2 equiv of gas per equiv of metal. </p>
<p>So, the <mathjax>#1#</mathjax> mole of gas evolves, (conveniently, as at <mathjax>#STP#</mathjax> the molar volume is <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax>).</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#1.03#</mathjax> mol of dihydrogen gas will evolve, with a volume slightly over <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax> at <mathjax>#STP#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of magnesium: <mathjax>#(50.0*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.06#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Moles of hydrogen chloride gas: <mathjax>#(75.0*g)/(36.2*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.07#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires 2 equiv of gas per equiv of metal. </p>
<p>So, the <mathjax>#1#</mathjax> mole of gas evolves, (conveniently, as at <mathjax>#STP#</mathjax> the molar volume is <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax>).</p></div>
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<h1 class="questionTitle" itemprop="name">In the equation #Mg + 2HCl -> MgCl_2 + H_2#, what volume of hydrogen at STP is produced from the reaction of 50.0 g of #Mg# and the equivalent of 75 g of #HCl#?</h1>
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<div class="markdown"><p><mathjax>#1.03#</mathjax> mol of dihydrogen gas will evolve, with a volume slightly over <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax> at <mathjax>#STP#</mathjax>.</p></div>
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<div class="markdown"><p>Moles of magnesium: <mathjax>#(50.0*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.06#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Moles of hydrogen chloride gas: <mathjax>#(75.0*g)/(36.2*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.07#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires 2 equiv of gas per equiv of metal. </p>
<p>So, the <mathjax>#1#</mathjax> mole of gas evolves, (conveniently, as at <mathjax>#STP#</mathjax> the molar volume is <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax>).</p></div>
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</article> | In the equation #Mg + 2HCl -> MgCl_2 + H_2#, what volume of hydrogen at STP is produced from the reaction of 50.0 g of #Mg# and the equivalent of 75 g of #HCl#? | null |
2,898 | a9678ea2-6ddd-11ea-b634-ccda262736ce | https://socratic.org/questions/what-is-the-mole-fraction-of-kcl-in-a-mixture-of-0-564g-nacl-1-52g-kcl-and-0-857 | 0.40 | start physical_unit 6 9 mole_fraction none qc_end physical_unit 13 13 11 12 mass qc_end physical_unit 6 6 14 15 mass qc_end physical_unit 20 20 18 19 mass qc_end end | [{"type":"physical unit","value":"Mole fraction [OF] KCl in a mixture"}] | [{"type":"physical unit","value":"0.40"}] | [{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{0.564 g}"},{"type":"physical unit","value":"Mass [OF] KCl [=] \\pu{1.52 g}"},{"type":"physical unit","value":"Mass [OF] LiCl [=] \\pu{0.857 g}"}] | <h1 class="questionTitle" itemprop="name">What is the mole fraction of #KCl# in a mixture of 0.564g #NaCl#, 1.52g #KCl# and 0.857g #LiCl#?</h1> | null | 0.40 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Mole fraction of component"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of component"/"Moles of all components"#</mathjax></p>
<p><mathjax>#"Moles of KCl"=(1.52*g)/(74.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0204*mol#</mathjax></p>
<p><mathjax>#"Moles of NaCl"=(0.564*g)/(58.44*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.65xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Moles of LiCl"=(0.857*g)/(42.39*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0202*mol#</mathjax></p>
<p><mathjax>#"Mole fraction of KCl"=(0.0204*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>Just as a check, the sum of <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fractions of all the components should give unity.</p>
<p><mathjax>#"Mole fraction of NaCl"=(9.65xx10^-3*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.192#</mathjax></p>
<p><mathjax>#"Mole fraction of LiCl"=(0.0202*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>The sum of the mole fractions do give unity as required. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"Mole fraction of KCl"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Mole fraction of component"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of component"/"Moles of all components"#</mathjax></p>
<p><mathjax>#"Moles of KCl"=(1.52*g)/(74.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0204*mol#</mathjax></p>
<p><mathjax>#"Moles of NaCl"=(0.564*g)/(58.44*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.65xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Moles of LiCl"=(0.857*g)/(42.39*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0202*mol#</mathjax></p>
<p><mathjax>#"Mole fraction of KCl"=(0.0204*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>Just as a check, the sum of <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fractions of all the components should give unity.</p>
<p><mathjax>#"Mole fraction of NaCl"=(9.65xx10^-3*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.192#</mathjax></p>
<p><mathjax>#"Mole fraction of LiCl"=(0.0202*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>The sum of the mole fractions do give unity as required. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the mole fraction of #KCl# in a mixture of 0.564g #NaCl#, 1.52g #KCl# and 0.857g #LiCl#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Mole fraction of KCl"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Mole fraction of component"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of component"/"Moles of all components"#</mathjax></p>
<p><mathjax>#"Moles of KCl"=(1.52*g)/(74.55*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0204*mol#</mathjax></p>
<p><mathjax>#"Moles of NaCl"=(0.564*g)/(58.44*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#9.65xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Moles of LiCl"=(0.857*g)/(42.39*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0202*mol#</mathjax></p>
<p><mathjax>#"Mole fraction of KCl"=(0.0204*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>Just as a check, the sum of <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fractions of all the components should give unity.</p>
<p><mathjax>#"Mole fraction of NaCl"=(9.65xx10^-3*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.192#</mathjax></p>
<p><mathjax>#"Mole fraction of LiCl"=(0.0202*mol)/((0.0202+0.00965+0.0204)*mol)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.402#</mathjax></p>
<p>The sum of the mole fractions do give unity as required. </p></div>
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</article> | What is the mole fraction of #KCl# in a mixture of 0.564g #NaCl#, 1.52g #KCl# and 0.857g #LiCl#? | null |
2,899 | a8d00834-6ddd-11ea-accd-ccda262736ce | https://socratic.org/questions/5473b86c581e2a14bf83682d | 1.62 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 8 8 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"1.62 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] Fe3(PO4)2 [=] \\pu{120.23 g}"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms are in #"120.23 g Fe"_3("PO"_4)_2"#?
</h1> | null | 1.62 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are eight moles of oxygen atoms in each mole of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. To answer this question, you must convert the given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. Then multiply the moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> times <mathjax>#(8"mol O)/(1 mol Fe"_3("PO"_4)_2")#</mathjax> to determine the number of moles of oxygen. Then multiply the moles of oxygen times <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms O"#</mathjax>. </p>
<p><strong>Given/Known:</strong><br/>
mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#"120.23 g"#</mathjax><br/>
molar mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax># = #</mathjax><mathjax>#"357.478g/mol"#</mathjax><br/>
1 mole <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> contains 8 moles oxygen atoms<br/>
1 mole O atoms = <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
number of oxygen atoms in <mathjax>#"120.23g Fe"_3("PO"_4)_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p><strong>1. Convert given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles.</strong></p>
<p><mathjax>#120.23color(red)cancel(color(black)("g Fe"_3("PO"_4)_2))xx(1 "mol Fe"_3("PO"_4)_2)/(357.478color(red)cancel(color(black)("g Fe"_3("PO"_4)_2)))= "0.336328 mol Fe"_3("PO"_4)_2"#</mathjax></p>
<p><strong>2. Convert moles <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles O.</strong></p>
<p><mathjax>#0.336328color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2))xx(8"mol O")/(1color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2)))="2.690624 mol O"#</mathjax></p>
<p><strong>3. Convert moles O to atoms O.</strong></p>
<p><mathjax>#2.690624color(red)cancel(color(black)("mol O"))xx(6.022xx10^23"atoms O")/(1color(red)cancel(color(black)("mol O")))=1.6203xx10^24"atoms O"#</mathjax> (answer rounded to five <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>)</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"120.23g"#</mathjax> <mathjax>#"Fe"_3("PO"_4)_2#</mathjax> contains <mathjax>#"1.6203 x 10"^24"#</mathjax> <mathjax>#"atoms O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are eight moles of oxygen atoms in each mole of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. To answer this question, you must convert the given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. Then multiply the moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> times <mathjax>#(8"mol O)/(1 mol Fe"_3("PO"_4)_2")#</mathjax> to determine the number of moles of oxygen. Then multiply the moles of oxygen times <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms O"#</mathjax>. </p>
<p><strong>Given/Known:</strong><br/>
mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#"120.23 g"#</mathjax><br/>
molar mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax># = #</mathjax><mathjax>#"357.478g/mol"#</mathjax><br/>
1 mole <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> contains 8 moles oxygen atoms<br/>
1 mole O atoms = <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
number of oxygen atoms in <mathjax>#"120.23g Fe"_3("PO"_4)_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p><strong>1. Convert given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles.</strong></p>
<p><mathjax>#120.23color(red)cancel(color(black)("g Fe"_3("PO"_4)_2))xx(1 "mol Fe"_3("PO"_4)_2)/(357.478color(red)cancel(color(black)("g Fe"_3("PO"_4)_2)))= "0.336328 mol Fe"_3("PO"_4)_2"#</mathjax></p>
<p><strong>2. Convert moles <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles O.</strong></p>
<p><mathjax>#0.336328color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2))xx(8"mol O")/(1color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2)))="2.690624 mol O"#</mathjax></p>
<p><strong>3. Convert moles O to atoms O.</strong></p>
<p><mathjax>#2.690624color(red)cancel(color(black)("mol O"))xx(6.022xx10^23"atoms O")/(1color(red)cancel(color(black)("mol O")))=1.6203xx10^24"atoms O"#</mathjax> (answer rounded to five <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>)</p></div>
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<h1 class="questionTitle" itemprop="name">How many oxygen atoms are in #"120.23 g Fe"_3("PO"_4)_2"#?
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<div class="markdown"><p><mathjax>#"120.23g"#</mathjax> <mathjax>#"Fe"_3("PO"_4)_2#</mathjax> contains <mathjax>#"1.6203 x 10"^24"#</mathjax> <mathjax>#"atoms O"#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are eight moles of oxygen atoms in each mole of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. To answer this question, you must convert the given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax>. Then multiply the moles of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> times <mathjax>#(8"mol O)/(1 mol Fe"_3("PO"_4)_2")#</mathjax> to determine the number of moles of oxygen. Then multiply the moles of oxygen times <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms O"#</mathjax>. </p>
<p><strong>Given/Known:</strong><br/>
mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax>#=#</mathjax><mathjax>#"120.23 g"#</mathjax><br/>
molar mass <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax><mathjax># = #</mathjax><mathjax>#"357.478g/mol"#</mathjax><br/>
1 mole <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> contains 8 moles oxygen atoms<br/>
1 mole O atoms = <mathjax>#6.022xx"10"^23#</mathjax> <mathjax>#"atoms"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
number of oxygen atoms in <mathjax>#"120.23g Fe"_3("PO"_4)_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p><strong>1. Convert given mass of <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles.</strong></p>
<p><mathjax>#120.23color(red)cancel(color(black)("g Fe"_3("PO"_4)_2))xx(1 "mol Fe"_3("PO"_4)_2)/(357.478color(red)cancel(color(black)("g Fe"_3("PO"_4)_2)))= "0.336328 mol Fe"_3("PO"_4)_2"#</mathjax></p>
<p><strong>2. Convert moles <mathjax>#"Fe"_3("PO"_4)_2"#</mathjax> to moles O.</strong></p>
<p><mathjax>#0.336328color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2))xx(8"mol O")/(1color(red)cancel(color(black)("mol Fe"_3("PO"_4)_2)))="2.690624 mol O"#</mathjax></p>
<p><strong>3. Convert moles O to atoms O.</strong></p>
<p><mathjax>#2.690624color(red)cancel(color(black)("mol O"))xx(6.022xx10^23"atoms O")/(1color(red)cancel(color(black)("mol O")))=1.6203xx10^24"atoms O"#</mathjax> (answer rounded to five <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>)</p></div>
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</article> | How many oxygen atoms are in #"120.23 g Fe"_3("PO"_4)_2"#?
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