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1,100 | ac2b477a-6ddd-11ea-8ef3-ccda262736ce | https://socratic.org/questions/a-given-mass-of-oxygen-occupies-200-ml-when-the-pressure-is-400-mm-of-hg-what-vo | 200.00 mL | start physical_unit 17 18 volume ml qc_end physical_unit 4 4 6 7 volume qc_end physical_unit 4 4 12 13 pressure qc_end physical_unit 4 4 26 27 pressure qc_end c_other constant_temperature 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] oxygen [=] \\pu{200 mL}"},{"type":"physical unit","value":"Pressure1 [OF] oxygen [=] \\pu{400 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] oxygen [=] \\pu{200 mmHg}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A given mass of oxygen occupies 200 ml when the pressure is 400 mm of Hg. What volume will the gas occupy when the pressure is raised to 200 mmHg, provided the temperature remains constant?</h1> | null | 200.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we know that <mathjax>#1*atm#</mathjax> pressure will support of column of mercury <mathjax>#760*mm#</mathjax> high. A pressure of LESS than one atmosphere will support a column of mercury LESS than this length. </p>
<p>So here <mathjax>#P_1=(400*mm)/(760*mm*Hg*atm^-1)=0.526*atm#</mathjax></p>
<p>And <mathjax>#P_2=(200*mm)/(760*mm*Hg*atm^-1)=0.263*atm#</mathjax></p>
<p>So <mathjax>#P_2#</mathjax> was half of <mathjax>#P_1#</mathjax>.</p>
<p>And we plug in the numbers, and expect reasonably that <mathjax>#V_2#</mathjax> WILL INCREASE, given that we reduce the pressure, the force per unit area on the gas:</p>
<p><mathjax>#V_2=(0.526*atmxx200*mL)/(0.263*atm)=400*mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>........<mathjax>#V_2=400*mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we know that <mathjax>#1*atm#</mathjax> pressure will support of column of mercury <mathjax>#760*mm#</mathjax> high. A pressure of LESS than one atmosphere will support a column of mercury LESS than this length. </p>
<p>So here <mathjax>#P_1=(400*mm)/(760*mm*Hg*atm^-1)=0.526*atm#</mathjax></p>
<p>And <mathjax>#P_2=(200*mm)/(760*mm*Hg*atm^-1)=0.263*atm#</mathjax></p>
<p>So <mathjax>#P_2#</mathjax> was half of <mathjax>#P_1#</mathjax>.</p>
<p>And we plug in the numbers, and expect reasonably that <mathjax>#V_2#</mathjax> WILL INCREASE, given that we reduce the pressure, the force per unit area on the gas:</p>
<p><mathjax>#V_2=(0.526*atmxx200*mL)/(0.263*atm)=400*mL#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A given mass of oxygen occupies 200 ml when the pressure is 400 mm of Hg. What volume will the gas occupy when the pressure is raised to 200 mmHg, provided the temperature remains constant?</h1>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>........<mathjax>#V_2=400*mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now we know that <mathjax>#1*atm#</mathjax> pressure will support of column of mercury <mathjax>#760*mm#</mathjax> high. A pressure of LESS than one atmosphere will support a column of mercury LESS than this length. </p>
<p>So here <mathjax>#P_1=(400*mm)/(760*mm*Hg*atm^-1)=0.526*atm#</mathjax></p>
<p>And <mathjax>#P_2=(200*mm)/(760*mm*Hg*atm^-1)=0.263*atm#</mathjax></p>
<p>So <mathjax>#P_2#</mathjax> was half of <mathjax>#P_1#</mathjax>.</p>
<p>And we plug in the numbers, and expect reasonably that <mathjax>#V_2#</mathjax> WILL INCREASE, given that we reduce the pressure, the force per unit area on the gas:</p>
<p><mathjax>#V_2=(0.526*atmxx200*mL)/(0.263*atm)=400*mL#</mathjax></p></div>
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</article> | A given mass of oxygen occupies 200 ml when the pressure is 400 mm of Hg. What volume will the gas occupy when the pressure is raised to 200 mmHg, provided the temperature remains constant? | null |
1,101 | aad18ed2-6ddd-11ea-ab16-ccda262736ce | https://socratic.org/questions/when-2-5-liters-of-oxygen-gas-reacts-with-excess-methane-how-many-liters-of-carb | 1.25 liters | start physical_unit 14 15 volume l qc_end physical_unit 4 5 1 2 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon dioxide [IN] liters"}] | [{"type":"physical unit","value":"1.25 liters"}] | [{"type":"physical unit","value":"Volume [OF] oxygen gas [=] \\pu{2.5 liters}"},{"type":"other","value":"Excess methane."}] | <h1 class="questionTitle" itemprop="name">When 2.5 liters of oxygen gas reacts with excess methane how many liters of carbon dioxide is produced?</h1> | null | 1.25 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#</mathjax></p>
<p>Given equal pressures and common temperatures, equal volumes of gases contain an equal number molecules. The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the combustion reaction demands that there is TWICE the molar quantity of dioxygen to methane, and thus twice the volume.</p>
<p>Of course, under these conditions of limited oxidant, incomplete combustion of methane to carbon monoxide or carbon as soot may occur. How would you represent this reaction?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.25 litres"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#</mathjax></p>
<p>Given equal pressures and common temperatures, equal volumes of gases contain an equal number molecules. The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the combustion reaction demands that there is TWICE the molar quantity of dioxygen to methane, and thus twice the volume.</p>
<p>Of course, under these conditions of limited oxidant, incomplete combustion of methane to carbon monoxide or carbon as soot may occur. How would you represent this reaction?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">When 2.5 liters of oxygen gas reacts with excess methane how many liters of carbon dioxide is produced?</h1>
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<div class="markdown"><p><mathjax>#"1.25 litres"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l)#</mathjax></p>
<p>Given equal pressures and common temperatures, equal volumes of gases contain an equal number molecules. The <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the combustion reaction demands that there is TWICE the molar quantity of dioxygen to methane, and thus twice the volume.</p>
<p>Of course, under these conditions of limited oxidant, incomplete combustion of methane to carbon monoxide or carbon as soot may occur. How would you represent this reaction?</p></div>
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</article> | When 2.5 liters of oxygen gas reacts with excess methane how many liters of carbon dioxide is produced? | null |
1,102 | abefb877-6ddd-11ea-94cd-ccda262736ce | https://socratic.org/questions/a-solution-has-a-oh-of-1-10-2-what-is-the-poh-of-this-solution | 2.00 | start physical_unit 15 16 poh none qc_end physical_unit 15 16 6 9 [oh-] qc_end end | [{"type":"physical unit","value":"pOH [OF] this solution"}] | [{"type":"physical unit","value":"2.00"}] | [{"type":"physical unit","value":"[OH-] [OF] this solution [=] \\pu{1 × 10^(-2) mol/L}"}] | <h1 class="questionTitle" itemprop="name">A solution has a #[OH^-]# of #1*10^-2#. What is the #pOH# of this solution?</h1> | null | 2.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[HO^-]#</mathjax> (i.e. just as <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>.</p>
<p>Since know that when <mathjax>#log_ab#</mathjax> <mathjax>#=c#</mathjax>, then, by definition, <mathjax>#a^c#</mathjax> <mathjax>#=#</mathjax> <mathjax>#b#</mathjax>, thus <mathjax>#-log_10(1xx10^-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2#</mathjax>. </p>
<p>Given that in water at <mathjax>#298*K#</mathjax>, <mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>, what is <mathjax>#pH#</mathjax> of this solution. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[HO^-]#</mathjax> (i.e. just as <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>.</p>
<p>Since know that when <mathjax>#log_ab#</mathjax> <mathjax>#=c#</mathjax>, then, by definition, <mathjax>#a^c#</mathjax> <mathjax>#=#</mathjax> <mathjax>#b#</mathjax>, thus <mathjax>#-log_10(1xx10^-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2#</mathjax>. </p>
<p>Given that in water at <mathjax>#298*K#</mathjax>, <mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>, what is <mathjax>#pH#</mathjax> of this solution. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A solution has a #[OH^-]# of #1*10^-2#. What is the #pOH# of this solution?</h1>
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<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[HO^-]#</mathjax> (i.e. just as <mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H_3O^+]#</mathjax>.</p>
<p>Since know that when <mathjax>#log_ab#</mathjax> <mathjax>#=c#</mathjax>, then, by definition, <mathjax>#a^c#</mathjax> <mathjax>#=#</mathjax> <mathjax>#b#</mathjax>, thus <mathjax>#-log_10(1xx10^-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2#</mathjax>. </p>
<p>Given that in water at <mathjax>#298*K#</mathjax>, <mathjax>#K_w#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^-14#</mathjax>, what is <mathjax>#pH#</mathjax> of this solution. </p></div>
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</article> | A solution has a #[OH^-]# of #1*10^-2#. What is the #pOH# of this solution? | null |
1,103 | abfb2aef-6ddd-11ea-b1a4-ccda262736ce | https://socratic.org/questions/a-balloon-has-a-volume-of-2-9-l-at-320-kelvin-if-the-temperature-is-raised-to-34 | 3.11 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 17 18 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] L"}] | [{"type":"physical unit","value":"3.11 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{2.9 L}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{320 Kelvin}"},{"type":"physical unit","value":"Temperature2 [OF] the balloon [=] \\pu{343 Kelvin}"}] | <h1 class="questionTitle" itemprop="name">A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be?</h1> | null | 3.11 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3.1084375L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be?</h1>
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<div class="markdown"><p><mathjax>#3.1084375L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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</article> | A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be? | null |
1,104 | ac4a2f5d-6ddd-11ea-b083-ccda262736ce | https://socratic.org/questions/57356cb711ef6b67888f34d5 | (NH4)2Cr2O7 | start chemical_formula qc_end substance 2 3 qc_end end | [{"type":"other","value":"Chemical Formula [OF] ammonium dichromate [IN] default"}] | [{"type":"chemical equation","value":"(NH4)2Cr2O7"}] | [{"type":"substance name","value":"Ammonium dichromate"}] | <h1 class="questionTitle" itemprop="name">What is #"ammonium dichromate?"#</h1> | null | (NH4)2Cr2O7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonium dichromate is the ammonium salt of the dichromate ion, <mathjax>#Cr_2O_7^(2-)#</mathjax>, which is a powerful oxidant, and used extensively for this purpose in organic chemistry. </p></div>
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<div class="markdown"><p>Well, it is crystalline <mathjax>#(NH_4)_2Cr_2O_7#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ammonium dichromate is the ammonium salt of the dichromate ion, <mathjax>#Cr_2O_7^(2-)#</mathjax>, which is a powerful oxidant, and used extensively for this purpose in organic chemistry. </p></div>
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<h1 class="questionTitle" itemprop="name">What is #"ammonium dichromate?"#</h1>
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<div class="markdown"><p>Well, it is crystalline <mathjax>#(NH_4)_2Cr_2O_7#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Ammonium dichromate is the ammonium salt of the dichromate ion, <mathjax>#Cr_2O_7^(2-)#</mathjax>, which is a powerful oxidant, and used extensively for this purpose in organic chemistry. </p></div>
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</article> | What is #"ammonium dichromate?"# | null |
1,105 | ac330199-6ddd-11ea-8401-ccda262736ce | https://socratic.org/questions/how-would-you-balance-al-fe2o3-al2o3-fe | 2 Al + Fe2O3 -> Al2O3 + 2 Fe | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Al + Fe2O3 -> Al2O3 + 2 Fe"}] | [{"type":"chemical equation","value":"Al + Fe2O3 -> Al2O3 + Fe"}] | <h1 class="questionTitle" itemprop="name">How would you balance:
Al + Fe2O3 → Al2O3 + Fe?</h1> | null | 2 Al + Fe2O3 -> Al2O3 + 2 Fe | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One must first balance the metals before balancing non-metals such as Oxygen and Hydrogen (Hydrogen isn't involved in this case).</p>
<p>So therefore on the right hand side you see 2 Aluminium atoms and thus one must add a 2 before the Aluminium on the left hand side.</p>
<p>Secondly the Iron must be balanced, on the left it can be seen that there are 2 Iron atoms in <mathjax>#Fe_2O_3#</mathjax> and thus a 2 must be added on the right hand side in order for them to balance.</p>
<p>Lastly the Oxygens must be balanced but in this case they already are at 3 on each side and thus the whole equation is now balanced.</p>
<p>Hope I helped :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2Al#</mathjax> + <mathjax>#Fe_2O_3#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Al_2O_3#</mathjax> + <mathjax>#2Fe#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One must first balance the metals before balancing non-metals such as Oxygen and Hydrogen (Hydrogen isn't involved in this case).</p>
<p>So therefore on the right hand side you see 2 Aluminium atoms and thus one must add a 2 before the Aluminium on the left hand side.</p>
<p>Secondly the Iron must be balanced, on the left it can be seen that there are 2 Iron atoms in <mathjax>#Fe_2O_3#</mathjax> and thus a 2 must be added on the right hand side in order for them to balance.</p>
<p>Lastly the Oxygens must be balanced but in this case they already are at 3 on each side and thus the whole equation is now balanced.</p>
<p>Hope I helped :)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you balance:
Al + Fe2O3 → Al2O3 + Fe?</h1>
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Rogan V.
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<span class="dateCreated" datetime="2015-11-18T06:27:14" itemprop="dateCreated">
Nov 18, 2015
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<div class="markdown"><p><mathjax>#2Al#</mathjax> + <mathjax>#Fe_2O_3#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Al_2O_3#</mathjax> + <mathjax>#2Fe#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One must first balance the metals before balancing non-metals such as Oxygen and Hydrogen (Hydrogen isn't involved in this case).</p>
<p>So therefore on the right hand side you see 2 Aluminium atoms and thus one must add a 2 before the Aluminium on the left hand side.</p>
<p>Secondly the Iron must be balanced, on the left it can be seen that there are 2 Iron atoms in <mathjax>#Fe_2O_3#</mathjax> and thus a 2 must be added on the right hand side in order for them to balance.</p>
<p>Lastly the Oxygens must be balanced but in this case they already are at 3 on each side and thus the whole equation is now balanced.</p>
<p>Hope I helped :)</p></div>
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</article> | How would you balance:
Al + Fe2O3 → Al2O3 + Fe? | null |
1,106 | ac2771f6-6ddd-11ea-83bb-ccda262736ce | https://socratic.org/questions/what-is-the-poh-of-a-7-9x10-4-m-oh-solution | 10.90 | start physical_unit 10 11 poh none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pOH [OF] OH- solution"}] | [{"type":"physical unit","value":"10.90"}] | [{"type":"physical unit","value":"Molarity [OF] OH- solution [=] \\pu{7.9 × 10^(-4) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pOH of a #7.9x10^-4 M# #OH^-# solution?</h1> | null | 10.90 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.9xx10^-4)#</mathjax><br/>
<mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of the solution?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.9xx10^-4)#</mathjax><br/>
<mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of the solution?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pOH of a #7.9x10^-4 M# #OH^-# solution?</h1>
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anor277
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Aug 31, 2016
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<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(7.9xx10^-4)#</mathjax><br/>
<mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of the solution?</p></div>
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</article> | What is the pOH of a #7.9x10^-4 M# #OH^-# solution? | null |
1,107 | ab5ce562-6ddd-11ea-ac1a-ccda262736ce | https://socratic.org/questions/2-c8h18-l-25-o2-g-16-co2-g-18-h2o-g-if-you-burned-one-gallon-of-gas-c8h18-approx | 6762.72 liters | start physical_unit 26 27 volume l qc_end chemical_equation 0 10 qc_end physical_unit 1 1 14 15 volume qc_end physical_unit 1 1 20 21 mass qc_end physical_unit 26 27 35 36 temperature qc_end physical_unit 26 27 41 42 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon dioxide [IN] liters"}] | [{"type":"physical unit","value":"6762.72 liters"}] | [{"type":"chemical equation","value":"2 C8H18(l) + 25 O2(g) -> 16 CO2(g) + 18 H2O(g)"},{"type":"physical unit","value":"Volume [OF] C8H18 [=] \\pu{1 gallon}"},{"type":"physical unit","value":"mass [OF] C8H18 [=] \\pu{4000 grams}"},{"type":"physical unit","value":"Temperature [OF] carbon dioxide [=] \\pu{21.0 ℃}"},{"type":"physical unit","value":"Pressure [OF] carbon dioxide [=] \\pu{1.00 atm}"}] | <h1 class="questionTitle" itemprop="name">2 C8H18(l) + 25 O2(g) #-># 16 CO2(g) + 18 H2O(g)
If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 21.0°C and a pressure of 1.00 atm?</h1> | null | 6762.72 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One gallon of octane produces approximately 7000 L of carbon dioxide.</p>
<p><strong>Note:</strong> At <mathjax>#"20"^("o")"C"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of octane, <mathjax>#"C"_8"H"_18#</mathjax>, is 0.70300 g/mL, and the mass in grams in one gallon would be 2661 g. <a href="http://blueskymodel.org/gallon-gas" rel="nofollow" target="_blank">http://blueskymodel.org/gallon-gas</a> At <mathjax>#"15"^("o")"C"#</mathjax>, the density of octane is 0.91786 g/mL (<a href="http://www.simetric.co.uk/si_liquids.htm" rel="nofollow" target="_blank">http://www.simetric.co.uk/si_liquids.htm</a>), and the mass of one gallon of octane would be 3474 g, which is approximately 4000 g. </p>
<p>Since the temperature in the problem is <mathjax>#"21"^("o")"C"#</mathjax>, I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.</p>
<p><strong>Step1.</strong> You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. <strong>Step 2.</strong> Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between octane and carbon dioxide in the balanced equation. <strong>Step 3.</strong> Then use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the volume in liters of carbon dioxide that can be formed.</p>
<p><strong>Known/Given:</strong><br/>
mass of octane = approximately 4000 g<br/>
temperature: <mathjax>#"21.0"^("o")"C"#</mathjax> + <mathjax>#"273.15"#</mathjax> = <mathjax>#"294.2 K"#</mathjax><br/>
pressure = 1.00 atm<br/>
gas constant <mathjax>#"R"#</mathjax> = <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
molar mass of octane, <mathjax>#"C"_8"H"_18#</mathjax> = <mathjax>#"114.232 g/mol"#</mathjax><br/>
mole/mole ratio for <mathjax>#"C"_8"H"_18#</mathjax> and <mathjax>#"CO"_2"#</mathjax> = <mathjax>#"2 mol" "C"_8"H"_18#</mathjax>/<mathjax>#"16 mol" "CO"_2"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
volume of <mathjax>#"CO"_2"#</mathjax></p>
<p><strong>Balanced Chemical Equation:</strong><br/>
<mathjax>#"2 C"_8"H"_18("l")"#</mathjax> + <mathjax>#"25 O"_2("g")"#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#"16 O"_2("g")#</mathjax> + <mathjax>#"18 H"_2"O"("g")#</mathjax></p>
<p><strong>Ideal Gas Law:</strong><br/>
<mathjax>#"PV"#</mathjax> = <mathjax>#"nRT"#</mathjax></p>
<p><strong>Step 1. Determine Moles of Octane in One Gallon</strong><br/>
<mathjax>#"4000 g C"_8"H"_18"#</mathjax> x <mathjax>#"1 mol"/"114.232 g"#</mathjax> = <mathjax>#"35.0164 mol C"_8"H"_18"#</mathjax> </p>
<p><strong>Step 2. Moles of Carbon Dioxide Produced by One Gallon of Octane</strong><br/>
Multiply moles of octane times the mole/mole ratio between octane and carbon dioxide, so that carbon dioxide is in the numerator.</p>
<p><mathjax>#"35.0164 mol C"_8"H"_18#</mathjax> x (<mathjax>#"16 mol" "CO"_2"#</mathjax>/<mathjax>#"2 mol" "C"_8"H"_18#</mathjax>) = <mathjax>#"280.1312 mol" "CO"_2"#</mathjax> </p>
<p><strong>Step 3. Calculate Volume of <mathjax>#"CO"_2#</mathjax> Produced from One Gallon of Octane using the Ideal Gas Law, PV = nRT.</strong><br/>
P = 1.00 atm<br/>
n = 280.1312 mol <mathjax>#"CO"_2"#</mathjax><br/>
R= <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
T = <mathjax>#"294.2 K"#</mathjax></p>
<p><mathjax>#"V"#</mathjax> = <mathjax>#"nRT"/"P"#</mathjax> = <mathjax>#"(280.1312 x 0.08205736 x 294.2)"/"1.00 atm"#</mathjax> = <mathjax>#"6763 L"#</mathjax> = <mathjax>#"7000 L"#</mathjax> due to only one significant figure in 4000 g.</p></div>
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<div class="markdown"><p>7000 L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One gallon of octane produces approximately 7000 L of carbon dioxide.</p>
<p><strong>Note:</strong> At <mathjax>#"20"^("o")"C"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of octane, <mathjax>#"C"_8"H"_18#</mathjax>, is 0.70300 g/mL, and the mass in grams in one gallon would be 2661 g. <a href="http://blueskymodel.org/gallon-gas" rel="nofollow" target="_blank">http://blueskymodel.org/gallon-gas</a> At <mathjax>#"15"^("o")"C"#</mathjax>, the density of octane is 0.91786 g/mL (<a href="http://www.simetric.co.uk/si_liquids.htm" rel="nofollow" target="_blank">http://www.simetric.co.uk/si_liquids.htm</a>), and the mass of one gallon of octane would be 3474 g, which is approximately 4000 g. </p>
<p>Since the temperature in the problem is <mathjax>#"21"^("o")"C"#</mathjax>, I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.</p>
<p><strong>Step1.</strong> You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. <strong>Step 2.</strong> Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between octane and carbon dioxide in the balanced equation. <strong>Step 3.</strong> Then use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the volume in liters of carbon dioxide that can be formed.</p>
<p><strong>Known/Given:</strong><br/>
mass of octane = approximately 4000 g<br/>
temperature: <mathjax>#"21.0"^("o")"C"#</mathjax> + <mathjax>#"273.15"#</mathjax> = <mathjax>#"294.2 K"#</mathjax><br/>
pressure = 1.00 atm<br/>
gas constant <mathjax>#"R"#</mathjax> = <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
molar mass of octane, <mathjax>#"C"_8"H"_18#</mathjax> = <mathjax>#"114.232 g/mol"#</mathjax><br/>
mole/mole ratio for <mathjax>#"C"_8"H"_18#</mathjax> and <mathjax>#"CO"_2"#</mathjax> = <mathjax>#"2 mol" "C"_8"H"_18#</mathjax>/<mathjax>#"16 mol" "CO"_2"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
volume of <mathjax>#"CO"_2"#</mathjax></p>
<p><strong>Balanced Chemical Equation:</strong><br/>
<mathjax>#"2 C"_8"H"_18("l")"#</mathjax> + <mathjax>#"25 O"_2("g")"#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#"16 O"_2("g")#</mathjax> + <mathjax>#"18 H"_2"O"("g")#</mathjax></p>
<p><strong>Ideal Gas Law:</strong><br/>
<mathjax>#"PV"#</mathjax> = <mathjax>#"nRT"#</mathjax></p>
<p><strong>Step 1. Determine Moles of Octane in One Gallon</strong><br/>
<mathjax>#"4000 g C"_8"H"_18"#</mathjax> x <mathjax>#"1 mol"/"114.232 g"#</mathjax> = <mathjax>#"35.0164 mol C"_8"H"_18"#</mathjax> </p>
<p><strong>Step 2. Moles of Carbon Dioxide Produced by One Gallon of Octane</strong><br/>
Multiply moles of octane times the mole/mole ratio between octane and carbon dioxide, so that carbon dioxide is in the numerator.</p>
<p><mathjax>#"35.0164 mol C"_8"H"_18#</mathjax> x (<mathjax>#"16 mol" "CO"_2"#</mathjax>/<mathjax>#"2 mol" "C"_8"H"_18#</mathjax>) = <mathjax>#"280.1312 mol" "CO"_2"#</mathjax> </p>
<p><strong>Step 3. Calculate Volume of <mathjax>#"CO"_2#</mathjax> Produced from One Gallon of Octane using the Ideal Gas Law, PV = nRT.</strong><br/>
P = 1.00 atm<br/>
n = 280.1312 mol <mathjax>#"CO"_2"#</mathjax><br/>
R= <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
T = <mathjax>#"294.2 K"#</mathjax></p>
<p><mathjax>#"V"#</mathjax> = <mathjax>#"nRT"/"P"#</mathjax> = <mathjax>#"(280.1312 x 0.08205736 x 294.2)"/"1.00 atm"#</mathjax> = <mathjax>#"6763 L"#</mathjax> = <mathjax>#"7000 L"#</mathjax> due to only one significant figure in 4000 g.</p></div>
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<h1 class="questionTitle" itemprop="name">2 C8H18(l) + 25 O2(g) #-># 16 CO2(g) + 18 H2O(g)
If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 21.0°C and a pressure of 1.00 atm?</h1>
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<div class="markdown"><p>7000 L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One gallon of octane produces approximately 7000 L of carbon dioxide.</p>
<p><strong>Note:</strong> At <mathjax>#"20"^("o")"C"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of octane, <mathjax>#"C"_8"H"_18#</mathjax>, is 0.70300 g/mL, and the mass in grams in one gallon would be 2661 g. <a href="http://blueskymodel.org/gallon-gas" rel="nofollow" target="_blank">http://blueskymodel.org/gallon-gas</a> At <mathjax>#"15"^("o")"C"#</mathjax>, the density of octane is 0.91786 g/mL (<a href="http://www.simetric.co.uk/si_liquids.htm" rel="nofollow" target="_blank">http://www.simetric.co.uk/si_liquids.htm</a>), and the mass of one gallon of octane would be 3474 g, which is approximately 4000 g. </p>
<p>Since the temperature in the problem is <mathjax>#"21"^("o")"C"#</mathjax>, I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.</p>
<p><strong>Step1.</strong> You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. <strong>Step 2.</strong> Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between octane and carbon dioxide in the balanced equation. <strong>Step 3.</strong> Then use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the volume in liters of carbon dioxide that can be formed.</p>
<p><strong>Known/Given:</strong><br/>
mass of octane = approximately 4000 g<br/>
temperature: <mathjax>#"21.0"^("o")"C"#</mathjax> + <mathjax>#"273.15"#</mathjax> = <mathjax>#"294.2 K"#</mathjax><br/>
pressure = 1.00 atm<br/>
gas constant <mathjax>#"R"#</mathjax> = <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
molar mass of octane, <mathjax>#"C"_8"H"_18#</mathjax> = <mathjax>#"114.232 g/mol"#</mathjax><br/>
mole/mole ratio for <mathjax>#"C"_8"H"_18#</mathjax> and <mathjax>#"CO"_2"#</mathjax> = <mathjax>#"2 mol" "C"_8"H"_18#</mathjax>/<mathjax>#"16 mol" "CO"_2"#</mathjax></p>
<p><strong>Unknown:</strong><br/>
volume of <mathjax>#"CO"_2"#</mathjax></p>
<p><strong>Balanced Chemical Equation:</strong><br/>
<mathjax>#"2 C"_8"H"_18("l")"#</mathjax> + <mathjax>#"25 O"_2("g")"#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#"16 O"_2("g")#</mathjax> + <mathjax>#"18 H"_2"O"("g")#</mathjax></p>
<p><strong>Ideal Gas Law:</strong><br/>
<mathjax>#"PV"#</mathjax> = <mathjax>#"nRT"#</mathjax></p>
<p><strong>Step 1. Determine Moles of Octane in One Gallon</strong><br/>
<mathjax>#"4000 g C"_8"H"_18"#</mathjax> x <mathjax>#"1 mol"/"114.232 g"#</mathjax> = <mathjax>#"35.0164 mol C"_8"H"_18"#</mathjax> </p>
<p><strong>Step 2. Moles of Carbon Dioxide Produced by One Gallon of Octane</strong><br/>
Multiply moles of octane times the mole/mole ratio between octane and carbon dioxide, so that carbon dioxide is in the numerator.</p>
<p><mathjax>#"35.0164 mol C"_8"H"_18#</mathjax> x (<mathjax>#"16 mol" "CO"_2"#</mathjax>/<mathjax>#"2 mol" "C"_8"H"_18#</mathjax>) = <mathjax>#"280.1312 mol" "CO"_2"#</mathjax> </p>
<p><strong>Step 3. Calculate Volume of <mathjax>#"CO"_2#</mathjax> Produced from One Gallon of Octane using the Ideal Gas Law, PV = nRT.</strong><br/>
P = 1.00 atm<br/>
n = 280.1312 mol <mathjax>#"CO"_2"#</mathjax><br/>
R= <mathjax>#"0.08205736 L atm K"^(-1)"mol"^(-1)#</mathjax><br/>
T = <mathjax>#"294.2 K"#</mathjax></p>
<p><mathjax>#"V"#</mathjax> = <mathjax>#"nRT"/"P"#</mathjax> = <mathjax>#"(280.1312 x 0.08205736 x 294.2)"/"1.00 atm"#</mathjax> = <mathjax>#"6763 L"#</mathjax> = <mathjax>#"7000 L"#</mathjax> due to only one significant figure in 4000 g.</p></div>
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</article> | 2 C8H18(l) + 25 O2(g) #-># 16 CO2(g) + 18 H2O(g)
If you burned one gallon of gas (C8H18) (approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 21.0°C and a pressure of 1.00 atm? | null |
1,108 | ab64a23a-6ddd-11ea-89ae-ccda262736ce | https://socratic.org/questions/the-pressure-in-a-container-is-8-atm-at-a-temperature-of-75-c-if-the-temperature | 6.85 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 4 4 6 7 pressure qc_end physical_unit 4 4 12 13 temperature qc_end physical_unit 4 4 20 21 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the container [IN] atm"}] | [{"type":"physical unit","value":"6.85 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the container [=] \\pu{8 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the container [=] \\pu{75 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the container [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">The pressure in a container is 8 atm at a temperature of 75°C. If the temperature is changed to 25°C what would be the new pressure?</h1> | null | 6.85 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a></p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><mathjax>#P_1=8 atm#</mathjax></p>
<p><mathjax>#T_1=75+273=348K#</mathjax></p>
<p><mathjax>#T_2=25+273=298K#</mathjax></p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=298/348*8#</mathjax></p>
<p><mathjax>#=6.85 atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new pressure is <mathjax>#=6.85 atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a></p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><mathjax>#P_1=8 atm#</mathjax></p>
<p><mathjax>#T_1=75+273=348K#</mathjax></p>
<p><mathjax>#T_2=25+273=298K#</mathjax></p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=298/348*8#</mathjax></p>
<p><mathjax>#=6.85 atm#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The pressure in a container is 8 atm at a temperature of 75°C. If the temperature is changed to 25°C what would be the new pressure?</h1>
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<div class="markdown"><p>The new pressure is <mathjax>#=6.85 atm#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We apply <a href="https://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a></p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><mathjax>#P_1=8 atm#</mathjax></p>
<p><mathjax>#T_1=75+273=348K#</mathjax></p>
<p><mathjax>#T_2=25+273=298K#</mathjax></p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=298/348*8#</mathjax></p>
<p><mathjax>#=6.85 atm#</mathjax></p></div>
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</article> | The pressure in a container is 8 atm at a temperature of 75°C. If the temperature is changed to 25°C what would be the new pressure? | null |
1,109 | a908d4b9-6ddd-11ea-b383-ccda262736ce | https://socratic.org/questions/how-many-grams-of-nacl-are-present-in-14-65-moles | 856.15 grams | start physical_unit 4 4 mass g qc_end physical_unit 4 4 8 9 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] NaCl [IN] grams"}] | [{"type":"physical unit","value":"856.15 grams"}] | [{"type":"physical unit","value":"Mole [OF] NaCl [=] \\pu{14.65 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of NaCl are present in 14.65 moles?</h1> | null | 856.15 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have a formula mass in <mathjax>#g*mol^-1#</mathjax>, and a <mathjax>#14.65#</mathjax> <mathjax>#mol#</mathjax> quantity of <mathjax>#NaCl#</mathjax>.</p>
<p>The mass is simply the product: <mathjax>#14.65*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#???g#</mathjax>.</p>
<p>When we use units in the calculation like this, we introduce an extra safeguard in that if the answer simplifies to moles we know we have not divided when we should have multiplied.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>In <mathjax>#1#</mathjax> mol of <mathjax>#NaCl#</mathjax> there are 58.44 g. So <mathjax>#58.44#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#????#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have a formula mass in <mathjax>#g*mol^-1#</mathjax>, and a <mathjax>#14.65#</mathjax> <mathjax>#mol#</mathjax> quantity of <mathjax>#NaCl#</mathjax>.</p>
<p>The mass is simply the product: <mathjax>#14.65*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#???g#</mathjax>.</p>
<p>When we use units in the calculation like this, we introduce an extra safeguard in that if the answer simplifies to moles we know we have not divided when we should have multiplied.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of NaCl are present in 14.65 moles?</h1>
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<div class="markdown"><p>In <mathjax>#1#</mathjax> mol of <mathjax>#NaCl#</mathjax> there are 58.44 g. So <mathjax>#58.44#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#????#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We have a formula mass in <mathjax>#g*mol^-1#</mathjax>, and a <mathjax>#14.65#</mathjax> <mathjax>#mol#</mathjax> quantity of <mathjax>#NaCl#</mathjax>.</p>
<p>The mass is simply the product: <mathjax>#14.65*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#???g#</mathjax>.</p>
<p>When we use units in the calculation like this, we introduce an extra safeguard in that if the answer simplifies to moles we know we have not divided when we should have multiplied.</p></div>
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</article> | How many grams of NaCl are present in 14.65 moles? | null |
1,110 | aa3a56c8-6ddd-11ea-b987-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-if-0-350-mols-of-vitamin-c-is-dissolved-in-water-to-gi | 3.50 M | start physical_unit 8 9 concentration mol/l qc_end physical_unit 8 9 5 6 mole qc_end physical_unit 18 18 21 22 volume qc_end substance 13 13 qc_end end | [{"type":"physical unit","value":"Concentration [OF] vitamin C solution [IN] M"}] | [{"type":"physical unit","value":"3.50 M"}] | [{"type":"physical unit","value":"Mole [OF] vitamin C [=] \\pu{0.350 mols}"},{"type":"physical unit","value":"Volume [OF] vitamin C solution [=] \\pu{100.00 mLs}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the concentration if 0.350 mols of vitamin C is dissolved in water to give a final solution volume of 100.00 mLs?</h1> | null | 3.50 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) "Molarity" = "moles"/"litres" = n/Vcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#n = "0.350 mol"#</mathjax><br/>
<mathjax>#V = "100.00 mL" = "0.100 00 L"#</mathjax></p>
<p>∴ <mathjax>#"Molarity" = "0.350 mol"/"0.100 00 L" = "3.50 mol/L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The concentration is 3.50 mol/L.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) "Molarity" = "moles"/"litres" = n/Vcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#n = "0.350 mol"#</mathjax><br/>
<mathjax>#V = "100.00 mL" = "0.100 00 L"#</mathjax></p>
<p>∴ <mathjax>#"Molarity" = "0.350 mol"/"0.100 00 L" = "3.50 mol/L"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the concentration if 0.350 mols of vitamin C is dissolved in water to give a final solution volume of 100.00 mLs?</h1>
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Ernest Z.
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<div class="markdown"><p>The concentration is 3.50 mol/L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) "Molarity" = "moles"/"litres" = n/Vcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem,</p>
<p><mathjax>#n = "0.350 mol"#</mathjax><br/>
<mathjax>#V = "100.00 mL" = "0.100 00 L"#</mathjax></p>
<p>∴ <mathjax>#"Molarity" = "0.350 mol"/"0.100 00 L" = "3.50 mol/L"#</mathjax></p></div>
</div>
</div>
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</article> | What is the concentration if 0.350 mols of vitamin C is dissolved in water to give a final solution volume of 100.00 mLs? | null |
1,111 | a8722d38-6ddd-11ea-8d00-ccda262736ce | https://socratic.org/questions/a-patient-with-respiratory-alkalosis-has-a-blood-plasma-ph-of-7-58-what-is-the-h | 2.6 × 10^(-8) mol/L | start physical_unit 17 19 [h3o+] mol/l qc_end physical_unit 7 8 11 11 ph qc_end end | [{"type":"physical unit","value":"[H3O+] [OF] the blood plasma [IN] mol/L"}] | [{"type":"physical unit","value":"2.6 × 10^(-8) mol/L"}] | [{"type":"physical unit","value":"pH [OF] blood plasma [=] \\pu{7.58}"}] | <h1 class="questionTitle" itemprop="name">A patient with respiratory alkalosis has a blood plasma pH of 7.58. What is the #[H_3O^+]# of the blood plasma?</h1> | null | 2.6 × 10^(-8) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-7.58)*mol*L^-1=2.6xx10^-8*mol*L^-1#</mathjax>.</p>
<p><mathjax>#[HO^-]#</mathjax>, so I presume, is elevated with this condition. What is <mathjax>#[HO^-]#</mathjax> here? What do you think it should be?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax>. Here <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-7.58)*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-7.58)*mol*L^-1=2.6xx10^-8*mol*L^-1#</mathjax>.</p>
<p><mathjax>#[HO^-]#</mathjax>, so I presume, is elevated with this condition. What is <mathjax>#[HO^-]#</mathjax> here? What do you think it should be?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A patient with respiratory alkalosis has a blood plasma pH of 7.58. What is the #[H_3O^+]# of the blood plasma?</h1>
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<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax>. Here <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-7.58)*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here <mathjax>#[H_3O^+]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^(-7.58)*mol*L^-1=2.6xx10^-8*mol*L^-1#</mathjax>.</p>
<p><mathjax>#[HO^-]#</mathjax>, so I presume, is elevated with this condition. What is <mathjax>#[HO^-]#</mathjax> here? What do you think it should be?</p></div>
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</article> | A patient with respiratory alkalosis has a blood plasma pH of 7.58. What is the #[H_3O^+]# of the blood plasma? | null |
1,112 | a901a97e-6ddd-11ea-a41b-ccda262736ce | https://socratic.org/questions/how-do-you-convert-5-3-10-25-molecules-of-co-2-to-moles | 88 moles | start physical_unit 9 9 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] CO2 [IN] moles"}] | [{"type":"physical unit","value":"88 moles"}] | [{"type":"physical unit","value":"Number [OF] CO2 molecules [=] \\pu{5.3 × 10^25}"}] | <h1 class="questionTitle" itemprop="name">How do you convert #5.3 * 10^25# molecules of #CO_2# to moles?</h1> | null | 88 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to go from molecules of <mathjax>#CO_2#</mathjax> to moles of <mathjax>#CO_2#</mathjax>, we have to use the following relationship:<img alt="slideplayer.com" src="https://useruploads.socratic.org/uj5mYngTzyfBdbl7r61h_slide_11.jpg"/> </p>
<p>I usually set up <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions/solving-mole-problems---dimensional-analysis">dimensional analysis</a> questions like this:</p>
<p>Quantity sought = Quantity given x conversion factor</p>
<ul>
<li>Quantity sought <mathjax>#rarr#</mathjax> mol <mathjax>#CO_2#</mathjax></li>
<li>Quantity given <mathjax>#rarr#</mathjax> <mathjax>#5.3xx10^(25)#</mathjax> molecules</li>
<li>Conversion factor <mathjax>#rarr#</mathjax> <mathjax>#(1mol)/(6.02xx10^(23))#</mathjax></li>
</ul>
<p>Now we just plug the values into the format I have above (make sure your units cancel out as you are going through this process, that's how you know if you are doing the calculations correctly or not):</p>
<p>moles <mathjax>#CO_2#</mathjax> = <mathjax>#5.3xx10^(25)cancel "molecules" xx #</mathjax><mathjax>#(1mol)/(6.02xx10^(23)cancel"molecules"#</mathjax></p>
<p>moles <mathjax>#CO_2#</mathjax> = <strong>88 mol</strong></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>There are <strong>88 moles</strong> of <mathjax>#CO_2#</mathjax> in <mathjax>#5.3xx10^(25)#</mathjax> molecules of <mathjax>#CO_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to go from molecules of <mathjax>#CO_2#</mathjax> to moles of <mathjax>#CO_2#</mathjax>, we have to use the following relationship:<img alt="slideplayer.com" src="https://useruploads.socratic.org/uj5mYngTzyfBdbl7r61h_slide_11.jpg"/> </p>
<p>I usually set up <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions/solving-mole-problems---dimensional-analysis">dimensional analysis</a> questions like this:</p>
<p>Quantity sought = Quantity given x conversion factor</p>
<ul>
<li>Quantity sought <mathjax>#rarr#</mathjax> mol <mathjax>#CO_2#</mathjax></li>
<li>Quantity given <mathjax>#rarr#</mathjax> <mathjax>#5.3xx10^(25)#</mathjax> molecules</li>
<li>Conversion factor <mathjax>#rarr#</mathjax> <mathjax>#(1mol)/(6.02xx10^(23))#</mathjax></li>
</ul>
<p>Now we just plug the values into the format I have above (make sure your units cancel out as you are going through this process, that's how you know if you are doing the calculations correctly or not):</p>
<p>moles <mathjax>#CO_2#</mathjax> = <mathjax>#5.3xx10^(25)cancel "molecules" xx #</mathjax><mathjax>#(1mol)/(6.02xx10^(23)cancel"molecules"#</mathjax></p>
<p>moles <mathjax>#CO_2#</mathjax> = <strong>88 mol</strong></p></div>
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<h1 class="questionTitle" itemprop="name">How do you convert #5.3 * 10^25# molecules of #CO_2# to moles?</h1>
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<div class="markdown"><p>There are <strong>88 moles</strong> of <mathjax>#CO_2#</mathjax> in <mathjax>#5.3xx10^(25)#</mathjax> molecules of <mathjax>#CO_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to go from molecules of <mathjax>#CO_2#</mathjax> to moles of <mathjax>#CO_2#</mathjax>, we have to use the following relationship:<img alt="slideplayer.com" src="https://useruploads.socratic.org/uj5mYngTzyfBdbl7r61h_slide_11.jpg"/> </p>
<p>I usually set up <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions/solving-mole-problems---dimensional-analysis">dimensional analysis</a> questions like this:</p>
<p>Quantity sought = Quantity given x conversion factor</p>
<ul>
<li>Quantity sought <mathjax>#rarr#</mathjax> mol <mathjax>#CO_2#</mathjax></li>
<li>Quantity given <mathjax>#rarr#</mathjax> <mathjax>#5.3xx10^(25)#</mathjax> molecules</li>
<li>Conversion factor <mathjax>#rarr#</mathjax> <mathjax>#(1mol)/(6.02xx10^(23))#</mathjax></li>
</ul>
<p>Now we just plug the values into the format I have above (make sure your units cancel out as you are going through this process, that's how you know if you are doing the calculations correctly or not):</p>
<p>moles <mathjax>#CO_2#</mathjax> = <mathjax>#5.3xx10^(25)cancel "molecules" xx #</mathjax><mathjax>#(1mol)/(6.02xx10^(23)cancel"molecules"#</mathjax></p>
<p>moles <mathjax>#CO_2#</mathjax> = <strong>88 mol</strong></p></div>
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</article> | How do you convert #5.3 * 10^25# molecules of #CO_2# to moles? | null |
1,113 | aa7a1a95-6ddd-11ea-989e-ccda262736ce | https://socratic.org/questions/a-chemical-reaction-between-acetic-acid-and-calcium-carbonate-produces-0-76-mol- | 17.02 liters | start physical_unit 20 20 volume l qc_end physical_unit 13 13 10 11 mole qc_end c_other STP qc_end substance 4 5 qc_end substance 7 8 qc_end end | [{"type":"physical unit","value":"Volume [OF] gas [IN] liters"}] | [{"type":"physical unit","value":"17.02 liters"}] | [{"type":"physical unit","value":"Mole [OF] CO2 [=] \\pu{0.76 mol}"},{"type":"other","value":"STP"},{"type":"substance name","value":"Acetic acid"},{"type":"substance name","value":"Calcium carbonate"}] | <h1 class="questionTitle" itemprop="name">A chemical reaction between acetic acid and calcium carbonate produces 0.76 mol of #CO_2# at STP. How many liters of gas were produced?</h1> | null | 17.02 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I mole of gas occupies 22.4 L at STP<br/>
So 0.76mole will occupy <mathjax>#22.4xx0,76L~~17L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#~~17L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I mole of gas occupies 22.4 L at STP<br/>
So 0.76mole will occupy <mathjax>#22.4xx0,76L~~17L#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A chemical reaction between acetic acid and calcium carbonate produces 0.76 mol of #CO_2# at STP. How many liters of gas were produced?</h1>
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<div class="markdown"><p><mathjax>#~~17L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I mole of gas occupies 22.4 L at STP<br/>
So 0.76mole will occupy <mathjax>#22.4xx0,76L~~17L#</mathjax></p></div>
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</article> | A chemical reaction between acetic acid and calcium carbonate produces 0.76 mol of #CO_2# at STP. How many liters of gas were produced? | null |
1,114 | abfdbf7c-6ddd-11ea-8094-ccda262736ce | https://socratic.org/questions/specific-heat-of-ice-in-j-kg-k | 2.06 × 10^3 J/(kg * K) | start physical_unit 3 3 specific_heat j/(kg_·_k) qc_end substance 3 3 qc_end end | [{"type":"physical unit","value":"Specific heat [OF] ice [IN] J/(kg * K)"}] | [{"type":"physical unit","value":"2.06 × 10^3 J/(kg * K)"}] | [{"type":"substance name","value":"Ice"}] | <h1 class="questionTitle" itemprop="name">Specific heat of ice in #"J/kg K"#?</h1> | null | 2.06 × 10^3 J/(kg * K) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of ice expressed in <em>joules per gram Kelvin</em>, <mathjax>#"J g"^(-1)"K"^(-1)#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#c_"ice" = "2.06 J g"^(-1)"K"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of <strong>ice</strong> by <mathjax>#"1 K"#</mathjax> you must provide it with <mathjax>#"2.06 J"#</mathjax> of heat. </p>
<p>Your goal here is to determine the specific heat of ice in <em>joules per <strong>kilogram</strong> Kelvin</em>, <mathjax>#"J kg"^(-1)"K"^(-1)#</mathjax>, which essentially tells you how much heat is required in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>. </p>
<p>The conversion factor that takes you from <em>grams</em> to <em>kilograms</em> is</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You can use this conversion factor to get</p>
<blockquote>
<p><mathjax>#2.06"J"/(color(red)(cancel(color(black)("g"))) * "K") * (10^3color(red)(cancel(color(black)("g"))))/("1 kg") = color(green)(|bar(ul(color(white)(a/a)color(black)(2.06 * 10^3color(white)(a) "J kg"^(-1)"K"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#2.06 * 10^3"J" = "2060 J"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2060 J kg"^(-1)"K"^(-1)#</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 <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of ice expressed in <em>joules per gram Kelvin</em>, <mathjax>#"J g"^(-1)"K"^(-1)#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#c_"ice" = "2.06 J g"^(-1)"K"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of <strong>ice</strong> by <mathjax>#"1 K"#</mathjax> you must provide it with <mathjax>#"2.06 J"#</mathjax> of heat. </p>
<p>Your goal here is to determine the specific heat of ice in <em>joules per <strong>kilogram</strong> Kelvin</em>, <mathjax>#"J kg"^(-1)"K"^(-1)#</mathjax>, which essentially tells you how much heat is required in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>. </p>
<p>The conversion factor that takes you from <em>grams</em> to <em>kilograms</em> is</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You can use this conversion factor to get</p>
<blockquote>
<p><mathjax>#2.06"J"/(color(red)(cancel(color(black)("g"))) * "K") * (10^3color(red)(cancel(color(black)("g"))))/("1 kg") = color(green)(|bar(ul(color(white)(a/a)color(black)(2.06 * 10^3color(white)(a) "J kg"^(-1)"K"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#2.06 * 10^3"J" = "2060 J"#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Specific heat of ice in #"J/kg K"#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-17T23:27:57" itemprop="dateCreated">
Jun 17, 2016
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<div class="markdown"><p><mathjax>#"2060 J kg"^(-1)"K"^(-1)#</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 <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of ice expressed in <em>joules per gram Kelvin</em>, <mathjax>#"J g"^(-1)"K"^(-1)#</mathjax>, which is listed as being equal to </p>
<blockquote>
<p><mathjax>#c_"ice" = "2.06 J g"^(-1)"K"^(-1)#</mathjax></p>
</blockquote>
<p>This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of <strong>ice</strong> by <mathjax>#"1 K"#</mathjax> you must provide it with <mathjax>#"2.06 J"#</mathjax> of heat. </p>
<p>Your goal here is to determine the specific heat of ice in <em>joules per <strong>kilogram</strong> Kelvin</em>, <mathjax>#"J kg"^(-1)"K"^(-1)#</mathjax>, which essentially tells you how much heat is required in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>. </p>
<p>The conversion factor that takes you from <em>grams</em> to <em>kilograms</em> is</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You can use this conversion factor to get</p>
<blockquote>
<p><mathjax>#2.06"J"/(color(red)(cancel(color(black)("g"))) * "K") * (10^3color(red)(cancel(color(black)("g"))))/("1 kg") = color(green)(|bar(ul(color(white)(a/a)color(black)(2.06 * 10^3color(white)(a) "J kg"^(-1)"K"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, in order to increase the temperature of <mathjax>#"1 kg"#</mathjax> of ice by <mathjax>#"1 K"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#2.06 * 10^3"J" = "2060 J"#</mathjax></p>
</blockquote></div>
</div>
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</article> | Specific heat of ice in #"J/kg K"#? | null |
1,115 | abe6a0a3-6ddd-11ea-8c61-ccda262736ce | https://socratic.org/questions/how-many-ml-of-a-2-00-m-nabr-solution-are-needed-to-make-200-0-ml-of-0-500m-nabr | 50.00 mL | start physical_unit 7 8 volume ml qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 7 8 13 14 volume qc_end physical_unit 7 8 16 17 molarity qc_end end | [{"type":"physical unit","value":"Volume1 [OF] NaBr solution [IN] mL"}] | [{"type":"physical unit","value":"50.00 mL"}] | [{"type":"physical unit","value":"Molarity1 [OF] NaBr solution [=] \\pu{2.00 M}"},{"type":"physical unit","value":"Volume2 [OF] NaBr solution [=] \\pu{200.0 mL}"},{"type":"physical unit","value":"Molarity2 [OF] NaBr solution [=] \\pu{0.500 M}"}] | <h1 class="questionTitle" itemprop="name">How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr? </h1> | null | 50.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Number of moles"/"Volume of solution"#</mathjax>...and so we use this quotient appropriately to find the THIRD value if the other two values are specified. </p>
<p>We want a <mathjax>#200*mL#</mathjax> volume of <mathjax>#0.500*mol*L^-1#</mathjax> <mathjax>#NaBr#</mathjax>...</p>
<p><mathjax>#n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.#</mathjax></p>
<p>And <mathjax>#2.00*mol*L^-1#</mathjax> <mathjax>#NaBr(aq)#</mathjax> is available...so....</p>
<p><mathjax>#(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Approx. <mathjax>#50.0*mL#</mathjax>....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Number of moles"/"Volume of solution"#</mathjax>...and so we use this quotient appropriately to find the THIRD value if the other two values are specified. </p>
<p>We want a <mathjax>#200*mL#</mathjax> volume of <mathjax>#0.500*mol*L^-1#</mathjax> <mathjax>#NaBr#</mathjax>...</p>
<p><mathjax>#n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.#</mathjax></p>
<p>And <mathjax>#2.00*mol*L^-1#</mathjax> <mathjax>#NaBr(aq)#</mathjax> is available...so....</p>
<p><mathjax>#(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr? </h1>
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anor277
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<span class="dateCreated" datetime="2018-05-13T06:25:43" itemprop="dateCreated">
May 13, 2018
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<div class="markdown"><p>Approx. <mathjax>#50.0*mL#</mathjax>....</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Number of moles"/"Volume of solution"#</mathjax>...and so we use this quotient appropriately to find the THIRD value if the other two values are specified. </p>
<p>We want a <mathjax>#200*mL#</mathjax> volume of <mathjax>#0.500*mol*L^-1#</mathjax> <mathjax>#NaBr#</mathjax>...</p>
<p><mathjax>#n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.#</mathjax></p>
<p>And <mathjax>#2.00*mol*L^-1#</mathjax> <mathjax>#NaBr(aq)#</mathjax> is available...so....</p>
<p><mathjax>#(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL#</mathjax>.</p></div>
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May 13, 2018
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<div class="markdown"><p><mathjax>#"50.0 mL"#</mathjax> of the <mathjax>#"2.00 M NaBr"#</mathjax> solution is needed to make <mathjax>#"200.0 mL"#</mathjax> of a <mathjax>#"0.500 M NaBr"#</mathjax> solution.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Use the dilution equation:</strong></p>
<p><mathjax>#C_1V_1=C_2V_2#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#C_1#</mathjax> is the initial concentration, <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#C_2#</mathjax> is the final concentration, and <mathjax>#V_2#</mathjax> is the final volume.</p>
<p><strong>Known</strong></p>
<p><mathjax>#C_1="2.00 M"#</mathjax></p>
<p><mathjax>#C_2="0.500 M"#</mathjax></p>
<p><mathjax>#V_2=200.0"mL"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#V_1#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#V_1#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#V_1=(C_2V_2)/C_1#</mathjax></p>
<p><mathjax>#V_1=(0.500color(red)cancel(color(black)("M"))xx"200.0 mL"
)/(0.200color(red)cancel(color(black)("M")))= "50.0 mL"#</mathjax> (rounded to three significant figures)</p>
<p><mathjax>#"50.0 mL"#</mathjax> of the <mathjax>#"2.00 M NaBr"#</mathjax> solution is needed to make <mathjax>#"200.0 mL"#</mathjax> of a <mathjax>#"0.500 M NaBr"#</mathjax> solution.</p></div>
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</article> | How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr? | null |
1,116 | abe1cf54-6ddd-11ea-819c-ccda262736ce | https://socratic.org/questions/570189d711ef6b4c23382e8b | C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g) | start chemical_equation qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g)"}] | [{"type":"other","value":"Complete combustion."},{"type":"substance name","value":"Octane"}] | <h1 class="questionTitle" itemprop="name">How do we represent the complete combustion of octane?</h1> | null | C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_8H_18(g) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p>
<p>Balance in the order: <mathjax>#C#</mathjax>; <mathjax>#H#</mathjax>; <mathjax>#O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do we represent the complete combustion of octane?</h1>
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anor277
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Apr 10, 2016
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<div class="markdown"><p><mathjax>#C_8H_18(g) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p>
<p>Balance in the order: <mathjax>#C#</mathjax>; <mathjax>#H#</mathjax>; <mathjax>#O#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
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</article> | How do we represent the complete combustion of octane? | null |
1,117 | ab1f8cee-6ddd-11ea-8f13-ccda262736ce | https://socratic.org/questions/how-many-grams-of-water-are-produced-when-propane-c3h8-burns-with-12-0-l-of-oxyg | 7.72 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 12 13 volume qc_end chemical_equation 9 9 qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] grams"}] | [{"type":"physical unit","value":"7.72 grams"}] | [{"type":"physical unit","value":"Volume [OF] oxygen [=] \\pu{12.0 L}"},{"type":"chemical equation","value":"C3H8"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many grams of water are produced when propane (C3H8) burns with 12.0 L of oxygen at STP?</h1> | null | 7.72 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The NIE for the complete combustion of propane is:</p>
<p><mathjax>#"C"_3 "H"_8 + 5 "O"_2 -> 3 "CO"_2 + 4 "H"_2"O"#</mathjax></p>
<p>The problem doesn't state how much propane is burned, so we're going to assume that there is enough to use up all the oxygen. To find out how much that is, we need to convert liters of <mathjax>#"O"_2#</mathjax> to moles of <mathjax>#"O"_2#</mathjax> using the fact that at STP, a mole of gas is equivalent to 22.4 liters.</p>
<p><mathjax>#12.0 cancel("L") * (1 "mol O"_2)/(22.4 cancel("L")) = 0.536 "mol O"_2#</mathjax></p>
<p>The molar ratio of <mathjax>#"O"_2#</mathjax> to <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#5/4#</mathjax> and the molecular weight of <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#18.0"g"#</mathjax>, so:</p>
<p><mathjax>#0.536 cancel("mol O"_2) * (4 cancel("mol H"_2"O"))/(5 cancel("mol O"_2)) * (18.0"g H"_2"O")/(1 cancel("mol H"_2"O")) = 7.72 "g H"_2"O"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#7.72 "g H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The NIE for the complete combustion of propane is:</p>
<p><mathjax>#"C"_3 "H"_8 + 5 "O"_2 -> 3 "CO"_2 + 4 "H"_2"O"#</mathjax></p>
<p>The problem doesn't state how much propane is burned, so we're going to assume that there is enough to use up all the oxygen. To find out how much that is, we need to convert liters of <mathjax>#"O"_2#</mathjax> to moles of <mathjax>#"O"_2#</mathjax> using the fact that at STP, a mole of gas is equivalent to 22.4 liters.</p>
<p><mathjax>#12.0 cancel("L") * (1 "mol O"_2)/(22.4 cancel("L")) = 0.536 "mol O"_2#</mathjax></p>
<p>The molar ratio of <mathjax>#"O"_2#</mathjax> to <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#5/4#</mathjax> and the molecular weight of <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#18.0"g"#</mathjax>, so:</p>
<p><mathjax>#0.536 cancel("mol O"_2) * (4 cancel("mol H"_2"O"))/(5 cancel("mol O"_2)) * (18.0"g H"_2"O")/(1 cancel("mol H"_2"O")) = 7.72 "g H"_2"O"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of water are produced when propane (C3H8) burns with 12.0 L of oxygen at STP?</h1>
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<div class="markdown"><p><mathjax>#7.72 "g H"_2"O"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The NIE for the complete combustion of propane is:</p>
<p><mathjax>#"C"_3 "H"_8 + 5 "O"_2 -> 3 "CO"_2 + 4 "H"_2"O"#</mathjax></p>
<p>The problem doesn't state how much propane is burned, so we're going to assume that there is enough to use up all the oxygen. To find out how much that is, we need to convert liters of <mathjax>#"O"_2#</mathjax> to moles of <mathjax>#"O"_2#</mathjax> using the fact that at STP, a mole of gas is equivalent to 22.4 liters.</p>
<p><mathjax>#12.0 cancel("L") * (1 "mol O"_2)/(22.4 cancel("L")) = 0.536 "mol O"_2#</mathjax></p>
<p>The molar ratio of <mathjax>#"O"_2#</mathjax> to <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#5/4#</mathjax> and the molecular weight of <mathjax>#"H"_2"O"#</mathjax> is <mathjax>#18.0"g"#</mathjax>, so:</p>
<p><mathjax>#0.536 cancel("mol O"_2) * (4 cancel("mol H"_2"O"))/(5 cancel("mol O"_2)) * (18.0"g H"_2"O")/(1 cancel("mol H"_2"O")) = 7.72 "g H"_2"O"#</mathjax></p></div>
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</article> | How many grams of water are produced when propane (C3H8) burns with 12.0 L of oxygen at STP? | null |
1,118 | ab2b0ed2-6ddd-11ea-8c3e-ccda262736ce | https://socratic.org/questions/how-many-moles-of-solute-particles-are-present-in-1-ml-exact-of-aqueous-0-0060-m | 1.80 × 10^(-5) moles | start physical_unit 4 5 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] solute particles [IN] moles"}] | [{"type":"physical unit","value":"1.80 × 10^(-5) moles"}] | [{"type":"physical unit","value":"Volume [OF] Ba(OH)2 aqueous solution [=] \\pu{1 mL}"},{"type":"physical unit","value":"Molarity [OF] Ba(OH)2 aqueous solution [=] \\pu{0.0060 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of solute particles are present in 1 mL (exact) of aqueous 0.0060 M #Ba(OH)_2#?</h1> | null | 1.80 × 10^(-5) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium hydroxide is an electrolyte that gives <mathjax>#Ba^(2+)#</mathjax> and <mathjax>#2 " equiv "HO^-#</mathjax> ions upon dissolution.</p>
<p><mathjax>#"Volume "(L)xx"concentration "(mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"moles, amount of subtance."#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3xx1xx10^-3cancelLxx 6.0xx10^-3*mol*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#18xx10^-6* mol#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium hydroxide is an electrolyte that gives <mathjax>#Ba^(2+)#</mathjax> and <mathjax>#2 " equiv "HO^-#</mathjax> ions upon dissolution.</p>
<p><mathjax>#"Volume "(L)xx"concentration "(mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"moles, amount of subtance."#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of solute particles are present in 1 mL (exact) of aqueous 0.0060 M #Ba(OH)_2#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#3xx1xx10^-3cancelLxx 6.0xx10^-3*mol*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#18xx10^-6* mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Barium hydroxide is an electrolyte that gives <mathjax>#Ba^(2+)#</mathjax> and <mathjax>#2 " equiv "HO^-#</mathjax> ions upon dissolution.</p>
<p><mathjax>#"Volume "(L)xx"concentration "(mol*L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"moles, amount of subtance."#</mathjax></p></div>
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</article> | How many moles of solute particles are present in 1 mL (exact) of aqueous 0.0060 M #Ba(OH)_2#? | null |
1,119 | a875f798-6ddd-11ea-b98e-ccda262736ce | https://socratic.org/questions/for-the-following-reaction-what-is-the-maximum-mass-of-hcl-that-can-be-obtained- | 35.0 g | start physical_unit 10 10 mass g qc_end physical_unit 20 20 18 19 mass qc_end physical_unit 25 25 22 23 mass qc_end chemical_equation 26 34 qc_end end | [{"type":"physical unit","value":"maximum mass [OF] HCl [IN] g"}] | [{"type":"physical unit","value":"35.0 g"}] | [{"type":"physical unit","value":"mass [OF] BCl3 [=] \\pu{37.5 g}"},{"type":"physical unit","value":"mass [OF] H2O [=] \\pu{60.0 g}"},{"type":"chemical equation","value":"BCl3(g) + 3 H2O(l) -> H3BO3(s) + 3 HCl(g)"}] | <h1 class="questionTitle" itemprop="name">For the following reaction, what is the maximum mass of #"HCl"# that can be obtained by the reaction #"37.5 g"# #"BCl"_3# and #"60.0 g"# of #"H"_2"O"#? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"BCl"_3 (g) + 3"H"_2"O"(l) -> "H"_3"BO"_3(s) + 3"HCl" (g)#</mathjax></p></div>
</h2>
</div>
</div> | 35.0 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by determining which reactant, if any, acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>You know from the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"BCl"_ (3(g)) + color(blue)(3)"H"_ 2"O"_ ((l)) -> "H"_ 3"BO"_ (3(s)) + 3"HCl"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of boron trichloride consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of water, so use the <strong>molar masses</strong> of the two reactants to convert the masses to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#37.5 color(red)(cancel(color(black)("g"))) * "1 mole BCl"_3/(117.17color(red)(cancel(color(black)("g")))) = "0.32005 moles BCl"_3#</mathjax></p>
<p><mathjax>#60.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "3.3306 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, pick one of the two reactants and check to see if you have enough moles of the other reactant to ensure that <strong>all the moles</strong> of the first reactant take part in the reaction. </p>
<p>For boron trichloride, you have</p>
<blockquote>
<p><mathjax>#0.32005color(red)(cancel(color(black)("moles BCl"_3))) * (color(blue)(3)color(white)(.)"moles H"_2"O")/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96015 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, in order for all the moles of boron trichloride to take part in the reaction, you need <mathjax>#0.96015#</mathjax> <strong>moles</strong> of water. Since you have more than enough moles of water to ensure that this happens</p>
<blockquote>
<p><mathjax>#overbrace("3.3306 moles H"_2"O")^(color(blue)("what you have")) " " > " " overbrace("0.96015 moles H"_2"O")^(color(blue)("what you need"))#</mathjax></p>
</blockquote>
<p>you can say that water is <em>in excess</em>, which implies that boron trichloride acts as a <strong>limiting reagent</strong>, i.e. it gets completely consumed <em>before</em> all the moles of water get the chance to react. </p>
<p>You now know that <mathjax>#0.32005#</mathjax> <strong>moles</strong> of boron trichloride take part in the reaction, which implies that the reaction produces</p>
<blockquote>
<p><mathjax>#0.32005 color(red)(cancel(color(black)("moles BCl"_3))) * "3 moles HCl"/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96014 moles HCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrogen chloride</p>
<blockquote>
<p><mathjax>#0.96015 color(red)(cancel(color(black)("moles HCl"))) * "36.461 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(darkgreen)(ul(color(black)("35.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"35.0 g HCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by determining which reactant, if any, acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>You know from the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"BCl"_ (3(g)) + color(blue)(3)"H"_ 2"O"_ ((l)) -> "H"_ 3"BO"_ (3(s)) + 3"HCl"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of boron trichloride consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of water, so use the <strong>molar masses</strong> of the two reactants to convert the masses to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#37.5 color(red)(cancel(color(black)("g"))) * "1 mole BCl"_3/(117.17color(red)(cancel(color(black)("g")))) = "0.32005 moles BCl"_3#</mathjax></p>
<p><mathjax>#60.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "3.3306 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, pick one of the two reactants and check to see if you have enough moles of the other reactant to ensure that <strong>all the moles</strong> of the first reactant take part in the reaction. </p>
<p>For boron trichloride, you have</p>
<blockquote>
<p><mathjax>#0.32005color(red)(cancel(color(black)("moles BCl"_3))) * (color(blue)(3)color(white)(.)"moles H"_2"O")/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96015 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, in order for all the moles of boron trichloride to take part in the reaction, you need <mathjax>#0.96015#</mathjax> <strong>moles</strong> of water. Since you have more than enough moles of water to ensure that this happens</p>
<blockquote>
<p><mathjax>#overbrace("3.3306 moles H"_2"O")^(color(blue)("what you have")) " " > " " overbrace("0.96015 moles H"_2"O")^(color(blue)("what you need"))#</mathjax></p>
</blockquote>
<p>you can say that water is <em>in excess</em>, which implies that boron trichloride acts as a <strong>limiting reagent</strong>, i.e. it gets completely consumed <em>before</em> all the moles of water get the chance to react. </p>
<p>You now know that <mathjax>#0.32005#</mathjax> <strong>moles</strong> of boron trichloride take part in the reaction, which implies that the reaction produces</p>
<blockquote>
<p><mathjax>#0.32005 color(red)(cancel(color(black)("moles BCl"_3))) * "3 moles HCl"/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96014 moles HCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrogen chloride</p>
<blockquote>
<p><mathjax>#0.96015 color(red)(cancel(color(black)("moles HCl"))) * "36.461 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(darkgreen)(ul(color(black)("35.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">For the following reaction, what is the maximum mass of #"HCl"# that can be obtained by the reaction #"37.5 g"# #"BCl"_3# and #"60.0 g"# of #"H"_2"O"#? </h1>
<div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"BCl"_3 (g) + 3"H"_2"O"(l) -> "H"_3"BO"_3(s) + 3"HCl" (g)#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2017-06-13T23:48:24" itemprop="dateCreated">
Jun 13, 2017
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<div class="markdown"><p><mathjax>#"35.0 g HCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by determining which reactant, if any, acts as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>You know from the balanced chemical equation</p>
<blockquote>
<p><mathjax>#"BCl"_ (3(g)) + color(blue)(3)"H"_ 2"O"_ ((l)) -> "H"_ 3"BO"_ (3(s)) + 3"HCl"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of boron trichloride consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of water, so use the <strong>molar masses</strong> of the two reactants to convert the masses to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#37.5 color(red)(cancel(color(black)("g"))) * "1 mole BCl"_3/(117.17color(red)(cancel(color(black)("g")))) = "0.32005 moles BCl"_3#</mathjax></p>
<p><mathjax>#60.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "3.3306 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, pick one of the two reactants and check to see if you have enough moles of the other reactant to ensure that <strong>all the moles</strong> of the first reactant take part in the reaction. </p>
<p>For boron trichloride, you have</p>
<blockquote>
<p><mathjax>#0.32005color(red)(cancel(color(black)("moles BCl"_3))) * (color(blue)(3)color(white)(.)"moles H"_2"O")/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96015 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, in order for all the moles of boron trichloride to take part in the reaction, you need <mathjax>#0.96015#</mathjax> <strong>moles</strong> of water. Since you have more than enough moles of water to ensure that this happens</p>
<blockquote>
<p><mathjax>#overbrace("3.3306 moles H"_2"O")^(color(blue)("what you have")) " " > " " overbrace("0.96015 moles H"_2"O")^(color(blue)("what you need"))#</mathjax></p>
</blockquote>
<p>you can say that water is <em>in excess</em>, which implies that boron trichloride acts as a <strong>limiting reagent</strong>, i.e. it gets completely consumed <em>before</em> all the moles of water get the chance to react. </p>
<p>You now know that <mathjax>#0.32005#</mathjax> <strong>moles</strong> of boron trichloride take part in the reaction, which implies that the reaction produces</p>
<blockquote>
<p><mathjax>#0.32005 color(red)(cancel(color(black)("moles BCl"_3))) * "3 moles HCl"/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96014 moles HCl"#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrogen chloride</p>
<blockquote>
<p><mathjax>#0.96015 color(red)(cancel(color(black)("moles HCl"))) * "36.461 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(darkgreen)(ul(color(black)("35.0 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | For the following reaction, what is the maximum mass of #"HCl"# that can be obtained by the reaction #"37.5 g"# #"BCl"_3# and #"60.0 g"# of #"H"_2"O"#? |
#"BCl"_3 (g) + 3"H"_2"O"(l) -> "H"_3"BO"_3(s) + 3"HCl" (g)#
|
1,120 | aa38a95c-6ddd-11ea-8e1b-ccda262736ce | https://socratic.org/questions/a-10-g-sample-of-a-compound-contains-4-00g-c-0-667g-h-and-5-33g-o-what-is-the-mo | C6H12O6 | start chemical_formula qc_end physical_unit 10 10 8 9 mass qc_end physical_unit 13 13 11 12 mass qc_end physical_unit 17 17 15 16 mass qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 6 6 27 28 molecular_weight qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound sample [IN] molecular"}] | [{"type":"chemical equation","value":"C6H12O6"}] | [{"type":"physical unit","value":"Mass [OF] C [=] \\pu{4.00 g}"},{"type":"physical unit","value":"Mass [OF] H [=] \\pu{0.667 g}"},{"type":"physical unit","value":"Mass [OF] O [=] \\pu{5.33 g}"},{"type":"physical unit","value":"Mass [OF] the compound sample [=] \\pu{10 g}"},{"type":"physical unit","value":"Molecular mass [OF] the compound sample [=] \\pu{180.156 g/mol}"}] | <h1 class="questionTitle" itemprop="name">A 10 g sample of a compound contains 4.00g #C#, 0.667g #H#, and 5.33g #O#. What is the molecular formula?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The molecular mass is 180.156 g/mol.</p></div>
</h2>
</div>
</div> | C6H12O6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must first determine the <strong>empirical formula</strong> of the compound.</p>
<p>We must figure out the moles of <mathjax>#"C"#</mathjax>, <mathjax>#"H"#</mathjax>, and <mathjax>#"O"#</mathjax> and then calculate their ratio.</p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of each element.</strong></p>
<p><mathjax>#"Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C"#</mathjax></p>
<p><mathjax>#"Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula.</strong></p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"#</mathjax><br/>
<mathjax>#stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331
color(white)(Xm)1color(white)(mmmmml)1)#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXm)0.667 color(white)(mml)0.6617 color(white)(Xm)1.987 color(white)(mmml)2#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXXm)5.33 color(white)(mmm)0.3331 color(white)(Xm)1.000 color(white)(mmml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the empirical formula mass</strong></p>
<p>The empirical formula mass of <mathjax>#"CH"_2"O"#</mathjax> is 30.03 u.</p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the Molecular Formula</strong></p>
<p>The molecular formula is 6 times the empirical formula.</p>
<p>The molecular formula is <mathjax>#("CH"_2"O")_6 = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molecular formula is <mathjax>#"C"_6"H"_12"O"_6#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must first determine the <strong>empirical formula</strong> of the compound.</p>
<p>We must figure out the moles of <mathjax>#"C"#</mathjax>, <mathjax>#"H"#</mathjax>, and <mathjax>#"O"#</mathjax> and then calculate their ratio.</p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of each element.</strong></p>
<p><mathjax>#"Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C"#</mathjax></p>
<p><mathjax>#"Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula.</strong></p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"#</mathjax><br/>
<mathjax>#stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331
color(white)(Xm)1color(white)(mmmmml)1)#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXm)0.667 color(white)(mml)0.6617 color(white)(Xm)1.987 color(white)(mmml)2#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXXm)5.33 color(white)(mmm)0.3331 color(white)(Xm)1.000 color(white)(mmml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the empirical formula mass</strong></p>
<p>The empirical formula mass of <mathjax>#"CH"_2"O"#</mathjax> is 30.03 u.</p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the Molecular Formula</strong></p>
<p>The molecular formula is 6 times the empirical formula.</p>
<p>The molecular formula is <mathjax>#("CH"_2"O")_6 = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 10 g sample of a compound contains 4.00g #C#, 0.667g #H#, and 5.33g #O#. What is the molecular formula?</h1>
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Ernest Z.
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Nov 29, 2016
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<div class="markdown"><p>The molecular formula is <mathjax>#"C"_6"H"_12"O"_6#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must first determine the <strong>empirical formula</strong> of the compound.</p>
<p>We must figure out the moles of <mathjax>#"C"#</mathjax>, <mathjax>#"H"#</mathjax>, and <mathjax>#"O"#</mathjax> and then calculate their ratio.</p>
<blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of each element.</strong></p>
<p><mathjax>#"Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C"#</mathjax></p>
<p><mathjax>#"Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the empirical formula.</strong></p>
<p>From this point on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"#</mathjax><br/>
<mathjax>#stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331
color(white)(Xm)1color(white)(mmmmml)1)#</mathjax><br/>
<mathjax>#color(white)(m)"H" color(white)(XXXXm)0.667 color(white)(mml)0.6617 color(white)(Xm)1.987 color(white)(mmml)2#</mathjax><br/>
<mathjax>#color(white)(m)"O" color(white)(XXXXm)5.33 color(white)(mmm)0.3331 color(white)(Xm)1.000 color(white)(mmml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the empirical formula mass</strong></p>
<p>The empirical formula mass of <mathjax>#"CH"_2"O"#</mathjax> is 30.03 u.</p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the molecular mass.</strong></p>
<p>The molecular mass must be an integral multiple multiple of the empirical formula mass.</p>
<p><mathjax>#"MM" = n × "EFM"#</mathjax></p>
<p><mathjax>#n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the Molecular Formula</strong></p>
<p>The molecular formula is 6 times the empirical formula.</p>
<p>The molecular formula is <mathjax>#("CH"_2"O")_6 = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
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</article> | A 10 g sample of a compound contains 4.00g #C#, 0.667g #H#, and 5.33g #O#. What is the molecular formula? |
The molecular mass is 180.156 g/mol.
|
1,121 | aa71b8bb-6ddd-11ea-81e1-ccda262736ce | https://socratic.org/questions/571f82b211ef6b1298d65f9a | 1.20 × 10^24 | start physical_unit 8 9 number none qc_end physical_unit 18 19 14 15 mass qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"1.20 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] sulfuric acid [=] \\pu{49.0 g}"}] | <h1 class="questionTitle" itemprop="name">In terms of the oxygen atom, how many oxygen atoms are associated with a #49.0*g# mass of #"sulfuric acid"#?</h1> | null | 1.20 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We work out (i) the molar quantity of sulfuric acid; and then (ii) mulitply this number by <mathjax>#4N_A#</mathjax>.</p>
<p>(i) <mathjax>#"Moles of sulfuric acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(49.0*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.50*mol#</mathjax>.</p>
<p>(ii) Because there are 4 atoms of oxygen per formula unit of sulfuric acid, there are <mathjax>#1/2xx4xxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2N_A#</mathjax> <mathjax>#"oxygen atoms"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's number, "6.022xx10^23*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We work out (i) the molar quantity of sulfuric acid; and then (ii) mulitply this number by <mathjax>#4N_A#</mathjax>.</p>
<p>(i) <mathjax>#"Moles of sulfuric acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(49.0*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.50*mol#</mathjax>.</p>
<p>(ii) Because there are 4 atoms of oxygen per formula unit of sulfuric acid, there are <mathjax>#1/2xx4xxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2N_A#</mathjax> <mathjax>#"oxygen atoms"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In terms of the oxygen atom, how many oxygen atoms are associated with a #49.0*g# mass of #"sulfuric acid"#?</h1>
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<div class="markdown"><p><mathjax>#2xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's number, "6.022xx10^23*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We work out (i) the molar quantity of sulfuric acid; and then (ii) mulitply this number by <mathjax>#4N_A#</mathjax>.</p>
<p>(i) <mathjax>#"Moles of sulfuric acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(49.0*g)/(98.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.50*mol#</mathjax>.</p>
<p>(ii) Because there are 4 atoms of oxygen per formula unit of sulfuric acid, there are <mathjax>#1/2xx4xxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2N_A#</mathjax> <mathjax>#"oxygen atoms"#</mathjax></p></div>
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</article> | In terms of the oxygen atom, how many oxygen atoms are associated with a #49.0*g# mass of #"sulfuric acid"#? | null |
1,122 | a859dfa8-6ddd-11ea-8dc0-ccda262736ce | https://socratic.org/questions/elemental-analysis-of-a-compound-showed-that-it-consisted-of-81-82-carbon-and-18 | 8 | start physical_unit 19 20 number none qc_end end | [{"type":"physical unit","value":"Number [OF] hydrogen atoms"}] | [{"type":"physical unit","value":"8"}] | [{"type":"physical unit","value":"Percent by mass [OF] carbon in the compound [=] \\pu{81.82%}"},{"type":"physical unit","value":"Percent by mass [OF] hydrogen in the compound [=] \\pu{18.18%}"}] | <h1 class="questionTitle" itemprop="name">Elemental analysis of a compound showed that it consisted of 81.82% carbon and 18.18% hydrogen by mass, how many hydrogen atoms appear in the empirical formula of the compound?</h1> | null | 8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this kind of problem, we have to assume that you have 100g unknown sample (since the percentages add up to 100%).</p>
<p>Thus, masses are C = 81.82 g; H = 18.18 g</p>
<p>Since chemical formulas deal with number of moles more than their respective weights, we need to multiply the masses with their respective atomic weights to get the number of moles.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#81.82 cancel g#</mathjax> x <mathjax>#"1 mol"/(12.01 cancel g)#</mathjax> = 6.81 mol <mathjax>#C#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#18.18 cancel g#</mathjax> x <mathjax>#"1 mol"/(1.01 cancel g)#</mathjax> = 18 mol <mathjax>#H#</mathjax></p>
<p>Next, to get the ratio of atoms to each other, we would need to divide the number of moles by the <em>smallest number of mole</em>.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(6.81 cancel "mol")/(6.81 cancel "mol")#</mathjax> = 1</p>
<p><mathjax>#H#</mathjax> = <mathjax>#(18.18 cancel "mol")/(6.81 cancel "mol")#</mathjax> = <mathjax>#color (red) (2.64)#</mathjax></p>
<p>Now, since the number of <mathjax>#H#</mathjax> atoms is too far to round off, we need to find a factor that we can multiply to BOTH atoms to get the accurate ratio. In this case, the factor is 3.</p>
<p><mathjax>#C#</mathjax> = 1 x 3 = <mathjax>#3#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.64 x 3 = <mathjax>#7.92~~ 8#</mathjax></p>
<p>Therefore, the empirical formula is <mathjax>#C_3H_8#</mathjax>.</p></div>
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<div class="markdown"><p>8 <mathjax>#H#</mathjax> atoms</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this kind of problem, we have to assume that you have 100g unknown sample (since the percentages add up to 100%).</p>
<p>Thus, masses are C = 81.82 g; H = 18.18 g</p>
<p>Since chemical formulas deal with number of moles more than their respective weights, we need to multiply the masses with their respective atomic weights to get the number of moles.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#81.82 cancel g#</mathjax> x <mathjax>#"1 mol"/(12.01 cancel g)#</mathjax> = 6.81 mol <mathjax>#C#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#18.18 cancel g#</mathjax> x <mathjax>#"1 mol"/(1.01 cancel g)#</mathjax> = 18 mol <mathjax>#H#</mathjax></p>
<p>Next, to get the ratio of atoms to each other, we would need to divide the number of moles by the <em>smallest number of mole</em>.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(6.81 cancel "mol")/(6.81 cancel "mol")#</mathjax> = 1</p>
<p><mathjax>#H#</mathjax> = <mathjax>#(18.18 cancel "mol")/(6.81 cancel "mol")#</mathjax> = <mathjax>#color (red) (2.64)#</mathjax></p>
<p>Now, since the number of <mathjax>#H#</mathjax> atoms is too far to round off, we need to find a factor that we can multiply to BOTH atoms to get the accurate ratio. In this case, the factor is 3.</p>
<p><mathjax>#C#</mathjax> = 1 x 3 = <mathjax>#3#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.64 x 3 = <mathjax>#7.92~~ 8#</mathjax></p>
<p>Therefore, the empirical formula is <mathjax>#C_3H_8#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Elemental analysis of a compound showed that it consisted of 81.82% carbon and 18.18% hydrogen by mass, how many hydrogen atoms appear in the empirical formula of the compound?</h1>
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<div class="markdown"><p>8 <mathjax>#H#</mathjax> atoms</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>For this kind of problem, we have to assume that you have 100g unknown sample (since the percentages add up to 100%).</p>
<p>Thus, masses are C = 81.82 g; H = 18.18 g</p>
<p>Since chemical formulas deal with number of moles more than their respective weights, we need to multiply the masses with their respective atomic weights to get the number of moles.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#81.82 cancel g#</mathjax> x <mathjax>#"1 mol"/(12.01 cancel g)#</mathjax> = 6.81 mol <mathjax>#C#</mathjax></p>
<p><mathjax>#H#</mathjax> = <mathjax>#18.18 cancel g#</mathjax> x <mathjax>#"1 mol"/(1.01 cancel g)#</mathjax> = 18 mol <mathjax>#H#</mathjax></p>
<p>Next, to get the ratio of atoms to each other, we would need to divide the number of moles by the <em>smallest number of mole</em>.</p>
<p><mathjax>#C#</mathjax> = <mathjax>#(6.81 cancel "mol")/(6.81 cancel "mol")#</mathjax> = 1</p>
<p><mathjax>#H#</mathjax> = <mathjax>#(18.18 cancel "mol")/(6.81 cancel "mol")#</mathjax> = <mathjax>#color (red) (2.64)#</mathjax></p>
<p>Now, since the number of <mathjax>#H#</mathjax> atoms is too far to round off, we need to find a factor that we can multiply to BOTH atoms to get the accurate ratio. In this case, the factor is 3.</p>
<p><mathjax>#C#</mathjax> = 1 x 3 = <mathjax>#3#</mathjax></p>
<p><mathjax>#H#</mathjax> = 2.64 x 3 = <mathjax>#7.92~~ 8#</mathjax></p>
<p>Therefore, the empirical formula is <mathjax>#C_3H_8#</mathjax>.</p></div>
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</article> | Elemental analysis of a compound showed that it consisted of 81.82% carbon and 18.18% hydrogen by mass, how many hydrogen atoms appear in the empirical formula of the compound? | null |
1,123 | a965b9e8-6ddd-11ea-9610-ccda262736ce | https://socratic.org/questions/what-formula-represents-lead-ll-chromate | PbCrO4 | start chemical_formula qc_end substance 3 4 qc_end end | [{"type":"other","value":"Chemical Formula [OF] lead(ll) chromate [IN] default"}] | [{"type":"chemical equation","value":"PbCrO4"}] | [{"type":"substance name","value":"Lead(ll) chromate"}] | <h1 class="questionTitle" itemprop="name">What formula represents lead(ll) chromate?</h1> | null | PbCrO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Canary yellow lead <a href="https://en.wikipedia.org/wiki/Lead%28II%29_chromate" rel="nofollow">chromate</a> is formed from <mathjax>#Pb^(2+)#</mathjax> and <mathjax>#CrO_4^(2-)#</mathjax>, whose 1:1 combination of course forms a neutral salt. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of the metal ions in <mathjax>#PbCrO_4#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#PbCrO_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Canary yellow lead <a href="https://en.wikipedia.org/wiki/Lead%28II%29_chromate" rel="nofollow">chromate</a> is formed from <mathjax>#Pb^(2+)#</mathjax> and <mathjax>#CrO_4^(2-)#</mathjax>, whose 1:1 combination of course forms a neutral salt. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of the metal ions in <mathjax>#PbCrO_4#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">What formula represents lead(ll) chromate?</h1>
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anor277
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<div class="markdown"><p><mathjax>#PbCrO_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Canary yellow lead <a href="https://en.wikipedia.org/wiki/Lead%28II%29_chromate" rel="nofollow">chromate</a> is formed from <mathjax>#Pb^(2+)#</mathjax> and <mathjax>#CrO_4^(2-)#</mathjax>, whose 1:1 combination of course forms a neutral salt. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of the metal ions in <mathjax>#PbCrO_4#</mathjax>?</p></div>
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1,124 | a86ec8a4-6ddd-11ea-85be-ccda262736ce | https://socratic.org/questions/if-you-need-to-prepare-1-5-l-of-a-1-5-m-solution-of-nitric-acid-how-many-millili | 250 milliliters | start physical_unit 13 14 volume ml qc_end physical_unit 11 11 5 6 volume qc_end physical_unit 13 14 9 10 molarity qc_end physical_unit 13 14 19 20 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitric acid [IN] milliliters"}] | [{"type":"physical unit","value":"250 milliliters"}] | [{"type":"physical unit","value":"Volume1 [OF] nitric acid solution [=] \\pu{1.5 L}"},{"type":"physical unit","value":"Molarity1 [OF] nitric acid solution [=] \\pu{1.5 M}"},{"type":"physical unit","value":"Molarity2 [OF] nitric acid [=] \\pu{9.0 M}"}] | <h1 class="questionTitle" itemprop="name">If you need to prepare 1.5 L of a 1.5 M solution of nitric acid, how many milliliters of 9.0 M nitric acid will you have to dilute to 1.5 L? </h1> | null | 250 milliliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want a solution whose volume is <mathjax>#1.50*L#</mathjax>, and whose concentration is <mathjax>#1.50*mol*L^-1#</mathjax>....</p>
<p>And this represents a molar quantity of <mathjax>#1.50*Lxx1.50*mol*L^-1=2.25*mol#</mathjax> with respect to nitric acid...</p>
<p>We have <mathjax>#9.0*mol*L^-1#</mathjax> nitric available.....and so we take the quotient....</p>
<p><mathjax>#(2.25*mol)/(9.0*mol*L^-1)xx1000*mL*L^-1-=250*mL#</mathjax>...</p>
<p>And so you add <mathjax>#250*mL#</mathjax> nitric acid to <mathjax>#1250*mL#</mathjax> of distilled water. </p>
<p>We remember that <a href="https://socratic.org/questions/in-what-order-should-strong-acids-and-water-be-mixed?source=search">if you spit in acid, it spits back.</a> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, we must remember to add acid to water....we need <mathjax>#250*mL#</mathjax> of the starting acid. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want a solution whose volume is <mathjax>#1.50*L#</mathjax>, and whose concentration is <mathjax>#1.50*mol*L^-1#</mathjax>....</p>
<p>And this represents a molar quantity of <mathjax>#1.50*Lxx1.50*mol*L^-1=2.25*mol#</mathjax> with respect to nitric acid...</p>
<p>We have <mathjax>#9.0*mol*L^-1#</mathjax> nitric available.....and so we take the quotient....</p>
<p><mathjax>#(2.25*mol)/(9.0*mol*L^-1)xx1000*mL*L^-1-=250*mL#</mathjax>...</p>
<p>And so you add <mathjax>#250*mL#</mathjax> nitric acid to <mathjax>#1250*mL#</mathjax> of distilled water. </p>
<p>We remember that <a href="https://socratic.org/questions/in-what-order-should-strong-acids-and-water-be-mixed?source=search">if you spit in acid, it spits back.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">If you need to prepare 1.5 L of a 1.5 M solution of nitric acid, how many milliliters of 9.0 M nitric acid will you have to dilute to 1.5 L? </h1>
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<div class="markdown"><p>Well, we must remember to add acid to water....we need <mathjax>#250*mL#</mathjax> of the starting acid. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want a solution whose volume is <mathjax>#1.50*L#</mathjax>, and whose concentration is <mathjax>#1.50*mol*L^-1#</mathjax>....</p>
<p>And this represents a molar quantity of <mathjax>#1.50*Lxx1.50*mol*L^-1=2.25*mol#</mathjax> with respect to nitric acid...</p>
<p>We have <mathjax>#9.0*mol*L^-1#</mathjax> nitric available.....and so we take the quotient....</p>
<p><mathjax>#(2.25*mol)/(9.0*mol*L^-1)xx1000*mL*L^-1-=250*mL#</mathjax>...</p>
<p>And so you add <mathjax>#250*mL#</mathjax> nitric acid to <mathjax>#1250*mL#</mathjax> of distilled water. </p>
<p>We remember that <a href="https://socratic.org/questions/in-what-order-should-strong-acids-and-water-be-mixed?source=search">if you spit in acid, it spits back.</a> </p></div>
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</article> | If you need to prepare 1.5 L of a 1.5 M solution of nitric acid, how many milliliters of 9.0 M nitric acid will you have to dilute to 1.5 L? | null |
1,125 | a8e9c2fb-6ddd-11ea-b11d-ccda262736ce | https://socratic.org/questions/what-is-the-steric-number-for-the-o-atom-in-water | 4 | start physical_unit 7 8 steric_number none qc_end substance 10 10 qc_end end | [{"type":"physical unit","value":"Steric number [OF] O atom"}] | [{"type":"physical unit","value":"4"}] | [{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the steric number for the #"O"# atom in water?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An atom's <strong>steric number</strong> tells you how many <em>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></em> surround the atom in a given molecule. </p>
<p>A region of electron density is simply </p>
<blockquote>
<ul>
<li><em>a <strong>single</strong>, <strong>double</strong>, or <strong>triple bond</strong></em> <mathjax>#->#</mathjax> <em>all three count as <strong>one region</strong> of electron dnesity</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>This means that in order to find an atom's steric number, all you have to do is see how many lone pairs of electrons it has and to count the <em>number of bonds</em> it forms with other atoms. </p>
<p>A good starting point here will be to draw the <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong> of water, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>Each water molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>atom</strong> of oxygen. The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong></p>
<blockquote>
<ul>
<li><em><strong>one</strong> from each of the two hydrogen atoms</em></li>
<li><em><strong>six</strong> from oxygen</em></li>
</ul>
</blockquote>
<p>The oxygen atom will be the <em>central atom</em>. It will form two <strong>single bonds</strong>, one with each hydrogen atom, that account for <mathjax>#4#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and has <mathjax>#2#</mathjax> <strong>lone pairs of electrons</strong> that account for the other <mathjax>#4#</mathjax> valence electrons. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="https://simple.wikipedia.org/wiki/Lewis_structure" src="https://useruploads.socratic.org/OzuSVFotR4u29y7KxHWv_220px-Water-2D-flat.png"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, the oxygen atom is surrounded by <mathjax>#4#</mathjax> <em>regions of electron density</em>, <mathjax>#2#</mathjax> single bonds and <mathjax>#2#</mathjax> lone pairs of electrons. </p>
<p>As a result, the atom's <strong>steric number</strong> will be equal to <mathjax>#4#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An atom's <strong>steric number</strong> tells you how many <em>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></em> surround the atom in a given molecule. </p>
<p>A region of electron density is simply </p>
<blockquote>
<ul>
<li><em>a <strong>single</strong>, <strong>double</strong>, or <strong>triple bond</strong></em> <mathjax>#->#</mathjax> <em>all three count as <strong>one region</strong> of electron dnesity</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>This means that in order to find an atom's steric number, all you have to do is see how many lone pairs of electrons it has and to count the <em>number of bonds</em> it forms with other atoms. </p>
<p>A good starting point here will be to draw the <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong> of water, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>Each water molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>atom</strong> of oxygen. The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong></p>
<blockquote>
<ul>
<li><em><strong>one</strong> from each of the two hydrogen atoms</em></li>
<li><em><strong>six</strong> from oxygen</em></li>
</ul>
</blockquote>
<p>The oxygen atom will be the <em>central atom</em>. It will form two <strong>single bonds</strong>, one with each hydrogen atom, that account for <mathjax>#4#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and has <mathjax>#2#</mathjax> <strong>lone pairs of electrons</strong> that account for the other <mathjax>#4#</mathjax> valence electrons. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="https://simple.wikipedia.org/wiki/Lewis_structure" src="https://useruploads.socratic.org/OzuSVFotR4u29y7KxHWv_220px-Water-2D-flat.png"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, the oxygen atom is surrounded by <mathjax>#4#</mathjax> <em>regions of electron density</em>, <mathjax>#2#</mathjax> single bonds and <mathjax>#2#</mathjax> lone pairs of electrons. </p>
<p>As a result, the atom's <strong>steric number</strong> will be equal to <mathjax>#4#</mathjax>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the steric number for the #"O"# atom in water?</h1>
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Stefan V.
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Jun 29, 2016
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<div class="markdown"><p><mathjax>#4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An atom's <strong>steric number</strong> tells you how many <em>regions of electron <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></em> surround the atom in a given molecule. </p>
<p>A region of electron density is simply </p>
<blockquote>
<ul>
<li><em>a <strong>single</strong>, <strong>double</strong>, or <strong>triple bond</strong></em> <mathjax>#->#</mathjax> <em>all three count as <strong>one region</strong> of electron dnesity</em></li>
<li><em>a <strong>lone pair of electrons</strong></em></li>
</ul>
</blockquote>
<p>This means that in order to find an atom's steric number, all you have to do is see how many lone pairs of electrons it has and to count the <em>number of bonds</em> it forms with other atoms. </p>
<p>A good starting point here will be to draw the <strong><a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/how-to-draw-lewis-structures">Lewis structure</a></strong> of water, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>Each water molecule contains <mathjax>#2#</mathjax> <strong>atoms</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>atom</strong> of oxygen. The molecule will have a total of <mathjax>#8#</mathjax> <strong><a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a></strong></p>
<blockquote>
<ul>
<li><em><strong>one</strong> from each of the two hydrogen atoms</em></li>
<li><em><strong>six</strong> from oxygen</em></li>
</ul>
</blockquote>
<p>The oxygen atom will be the <em>central atom</em>. It will form two <strong>single bonds</strong>, one with each hydrogen atom, that account for <mathjax>#4#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and has <mathjax>#2#</mathjax> <strong>lone pairs of electrons</strong> that account for the other <mathjax>#4#</mathjax> valence electrons. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="https://simple.wikipedia.org/wiki/Lewis_structure" src="https://useruploads.socratic.org/OzuSVFotR4u29y7KxHWv_220px-Water-2D-flat.png"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>As you can see, the oxygen atom is surrounded by <mathjax>#4#</mathjax> <em>regions of electron density</em>, <mathjax>#2#</mathjax> single bonds and <mathjax>#2#</mathjax> lone pairs of electrons. </p>
<p>As a result, the atom's <strong>steric number</strong> will be equal to <mathjax>#4#</mathjax>. </p></div>
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</article> | What is the steric number for the #"O"# atom in water? | null |
1,126 | ac9550fc-6ddd-11ea-ac05-ccda262736ce | https://socratic.org/questions/how-many-eggs-are-in-one-mole-of-eggs | 6.02 × 10^23 | start physical_unit 2 2 number none qc_end physical_unit 2 2 5 6 mole qc_end end | [{"type":"physical unit","value":"Number [OF] eggs"}] | [{"type":"physical unit","value":"6.02 × 10^23"}] | [{"type":"physical unit","value":"Mole [OF] eggs [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name"> How many eggs are in one mole of eggs?</h1> | null | 6.02 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1" mole"#</mathjax> of any substance consists of <mathjax>#6.022xx10^23#</mathjax> individual items of that substance. It has the property that <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00#</mathjax> <mathjax>#g#</mathjax> precisely. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is thus the link between the micro world of atoms and molecules to the macro world of grams and litres. </p>
<p>You specified a mole of eggs. There are thus <mathjax>#6.022xx10^23#</mathjax> eggses. How many dozen eggs is this quantity?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#N_A/12#</mathjax> <mathjax>#"dozen"#</mathjax>, where <mathjax>#N_A=6.022xx10^23#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1" mole"#</mathjax> of any substance consists of <mathjax>#6.022xx10^23#</mathjax> individual items of that substance. It has the property that <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00#</mathjax> <mathjax>#g#</mathjax> precisely. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is thus the link between the micro world of atoms and molecules to the macro world of grams and litres. </p>
<p>You specified a mole of eggs. There are thus <mathjax>#6.022xx10^23#</mathjax> eggses. How many dozen eggs is this quantity?</p></div>
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<h1 class="questionTitle" itemprop="name"> How many eggs are in one mole of eggs?</h1>
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<div class="markdown"><p><mathjax>#N_A/12#</mathjax> <mathjax>#"dozen"#</mathjax>, where <mathjax>#N_A=6.022xx10^23#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1" mole"#</mathjax> of any substance consists of <mathjax>#6.022xx10^23#</mathjax> individual items of that substance. It has the property that <mathjax>#6.022xx10^23#</mathjax> individual <mathjax>#""^12C#</mathjax> atoms have a mass of <mathjax>#12.00#</mathjax> <mathjax>#g#</mathjax> precisely. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is thus the link between the micro world of atoms and molecules to the macro world of grams and litres. </p>
<p>You specified a mole of eggs. There are thus <mathjax>#6.022xx10^23#</mathjax> eggses. How many dozen eggs is this quantity?</p></div>
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</article> | How many eggs are in one mole of eggs? | null |
1,127 | a8bccf6a-6ddd-11ea-814f-ccda262736ce | https://socratic.org/questions/5919bc4611ef6b726b967f97 | 11 g/L | start physical_unit 2 3 concentration g/l qc_end physical_unit 2 3 6 7 concentration qc_end physical_unit 2 3 9 10 osmotic_pressure qc_end physical_unit 2 3 25 26 osmotic_pressure qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] aqueous solution [IN] g/L"}] | [{"type":"physical unit","value":"11 g/L"}] | [{"type":"physical unit","value":"Concentration1 [OF] aqueous solution [=] \\pu{36 g/L}"},{"type":"physical unit","value":"Osmotic pressure1 [OF] aqueous solution [=] \\pu{4.98 bar}"},{"type":"physical unit","value":"Osmotic pressure2 [OF] aqueous solution [=] \\pu{1.52 bar}"}] | <h1 class="questionTitle" itemprop="name">If an aqueous solution starts at #"36 g/L"# and #"4.98 bar"#, what is the new concentration in #"g/L"# needed to accomplish an osmotic pressure of #"1.52 bar"#?</h1> | null | 11 g/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The osmotic pressure is directly proportional to the concentration, when the temperature is kept constant (as it is in this case).</p>
<p>If a concentration of <mathjax>#36#</mathjax> <mathjax>#gL^-1#</mathjax> yields an osmotic pressure of <mathjax>#4.98#</mathjax> bar, then a concentration of <mathjax>#x#</mathjax> <mathjax>#gL^-1#</mathjax> will yield an osmotic pressure of <mathjax>#1.52#</mathjax> bar.</p>
<p>After that it's just a matter of solving for <mathjax>#x#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This is essentially a proportional reasoning question: the concentration will be <mathjax>#(1.52)/(4.98)xx 36#</mathjax> <mathjax>#gL^-1#</mathjax> = <mathjax>#11#</mathjax> <mathjax>#gL^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The osmotic pressure is directly proportional to the concentration, when the temperature is kept constant (as it is in this case).</p>
<p>If a concentration of <mathjax>#36#</mathjax> <mathjax>#gL^-1#</mathjax> yields an osmotic pressure of <mathjax>#4.98#</mathjax> bar, then a concentration of <mathjax>#x#</mathjax> <mathjax>#gL^-1#</mathjax> will yield an osmotic pressure of <mathjax>#1.52#</mathjax> bar.</p>
<p>After that it's just a matter of solving for <mathjax>#x#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If an aqueous solution starts at #"36 g/L"# and #"4.98 bar"#, what is the new concentration in #"g/L"# needed to accomplish an osmotic pressure of #"1.52 bar"#?</h1>
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<div class="markdown"><p>This is essentially a proportional reasoning question: the concentration will be <mathjax>#(1.52)/(4.98)xx 36#</mathjax> <mathjax>#gL^-1#</mathjax> = <mathjax>#11#</mathjax> <mathjax>#gL^-1#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The osmotic pressure is directly proportional to the concentration, when the temperature is kept constant (as it is in this case).</p>
<p>If a concentration of <mathjax>#36#</mathjax> <mathjax>#gL^-1#</mathjax> yields an osmotic pressure of <mathjax>#4.98#</mathjax> bar, then a concentration of <mathjax>#x#</mathjax> <mathjax>#gL^-1#</mathjax> will yield an osmotic pressure of <mathjax>#1.52#</mathjax> bar.</p>
<p>After that it's just a matter of solving for <mathjax>#x#</mathjax>.</p></div>
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Truong-Son N.
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<div class="markdown"><blockquote>
<p><mathjax>#D_2 = "11 g/L"#</mathjax></p>
</blockquote>
<hr/>
<p><strong>Osmotic pressure</strong> <mathjax>#Pi#</mathjax> is given by:</p>
<blockquote>
<p><mathjax>#bb(Pi = icRT)#</mathjax></p>
<p>where:</p>
<ul>
<li><mathjax>#i#</mathjax> is the van't Hoff factor. For non-electrolytes, <mathjax>#i = 1#</mathjax>, as they hardly dissociate.</li>
<li><mathjax>#c#</mathjax> is the concentration in the appropriate units.</li>
<li><mathjax>#R#</mathjax> is the universal gas constant.</li>
<li><mathjax>#T#</mathjax> is the temperature in <mathjax>#"K"#</mathjax>.</li>
</ul>
</blockquote>
<p>Given an osmotic pressure in <mathjax>#"bar"#</mathjax>, the units of <mathjax>#c#</mathjax> must be:</p>
<blockquote>
<p><mathjax>#cancel"bar"/(("L"cdotcancel"bar""/mol"cdotcancel"K") xx cancel"K")#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#"mol/L"#</mathjax></p>
</blockquote>
<p>Converting to a concentration in <mathjax>#"g/L"#</mathjax> (i.e. mass concentration) would mean that:</p>
<blockquote>
<p><mathjax>#PiM = icMRT -= iDRT#</mathjax>,</p>
<p>where <mathjax>#M#</mathjax> is the molar mass in <mathjax>#"g/mol"#</mathjax>, and <mathjax>#D#</mathjax> is the mass concentration in <mathjax>#"g/L"#</mathjax>.</p>
<p>(This can be seen as analogous to the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<blockquote>
<p><mathjax>#P = n/VRT#</mathjax></p>
<p><mathjax>#=> PM = (nM)/VRT = DRT#</mathjax>.)</p>
</blockquote>
</blockquote>
<p>Given two states with the same temperature and van't Hoff factor (due to the same solute):</p>
<blockquote>
<p><mathjax>#Pi_1M = iD_1RT#</mathjax><br/>
<mathjax>#Pi_2M = iD_2RT#</mathjax></p>
</blockquote>
<p>Thus, the new concentration is gotten as follows:</p>
<blockquote>
<p><mathjax>#Pi_1/D_1 = Pi_2/D_2 = (iRT)/M#</mathjax></p>
<p><mathjax>#=> color(blue)(D_2) = D_1 (Pi_2/Pi_1)#</mathjax></p>
<p><mathjax>#= "36 g"/"L" xx ("1.52 bar"/"4.98 bar")#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#color(blue)("11 g/L")#</mathjax></p>
</blockquote></div>
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</article> | If an aqueous solution starts at #"36 g/L"# and #"4.98 bar"#, what is the new concentration in #"g/L"# needed to accomplish an osmotic pressure of #"1.52 bar"#? | null |
1,128 | aaafbbbc-6ddd-11ea-bf51-ccda262736ce | https://socratic.org/questions/how-do-you-balance-the-following-equation-ca-o-2-cao | 2 Ca + O2 -> 2 CaO | start chemical_equation qc_end chemical_equation 7 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Ca + O2 -> 2 CaO"}] | [{"type":"chemical equation","value":"Ca + O2 -> CaO"}] | <h1 class="questionTitle" itemprop="name">How do you balance the following equation: #Ca + O_2 -> CaO#?</h1> | null | 2 Ca + O2 -> 2 CaO | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Make sure there's the same number of atoms on each side of the equation,</p>
<p>On the left side there are <mathjax>#2#</mathjax> <mathjax>#O#</mathjax> atoms and only <mathjax>#1#</mathjax> <mathjax>#O#</mathjax> atom on the right.</p>
<p>Therefore, </p>
<p><mathjax>#Ca+1/2O_2rarrCaO#</mathjax></p>
<p>This is the balanced reaction (it's ok to have fractions as the coefficient).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Ca+1/2O_2rarrCaO#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Make sure there's the same number of atoms on each side of the equation,</p>
<p>On the left side there are <mathjax>#2#</mathjax> <mathjax>#O#</mathjax> atoms and only <mathjax>#1#</mathjax> <mathjax>#O#</mathjax> atom on the right.</p>
<p>Therefore, </p>
<p><mathjax>#Ca+1/2O_2rarrCaO#</mathjax></p>
<p>This is the balanced reaction (it's ok to have fractions as the coefficient).</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance the following equation: #Ca + O_2 -> CaO#?</h1>
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David L.
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Apr 10, 2017
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<div class="markdown"><p><mathjax>#Ca+1/2O_2rarrCaO#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Make sure there's the same number of atoms on each side of the equation,</p>
<p>On the left side there are <mathjax>#2#</mathjax> <mathjax>#O#</mathjax> atoms and only <mathjax>#1#</mathjax> <mathjax>#O#</mathjax> atom on the right.</p>
<p>Therefore, </p>
<p><mathjax>#Ca+1/2O_2rarrCaO#</mathjax></p>
<p>This is the balanced reaction (it's ok to have fractions as the coefficient).</p></div>
</div>
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</article> | How do you balance the following equation: #Ca + O_2 -> CaO#? | null |
1,129 | aafa7600-6ddd-11ea-946c-ccda262736ce | https://socratic.org/questions/consider-1-60-g-of-naphthalene-c10h8-is-dissolved-in-20-0-g-of-benzene-the-freez | 4.3 ℃/(kg * mol) | start physical_unit 12 12 freezing_point_temperature °c/(kg_·_mol) qc_end physical_unit 5 5 1 2 mass qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 19 20 freezing_point_temperature qc_end physical_unit 26 27 29 30 freezing_point_temperature qc_end end | [{"type":"physical unit","value":"Molal freezing point constant [OF] benzene [IN] ℃/(kg * mol)"}] | [{"type":"physical unit","value":"4.3 ℃/(kg * mol)"}] | [{"type":"physical unit","value":"Mass [OF] C10H8 [=] \\pu{1.60 g}"},{"type":"physical unit","value":"Mass [OF] benzene [=] \\pu{20.0 g}"},{"type":"physical unit","value":"Freezing point [OF] benzene [=] \\pu{5.5 ℃}"},{"type":"physical unit","value":"Freezing point [OF] the mixture [=] \\pu{2.8 ℃}"}] | <h1 class="questionTitle" itemprop="name">Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the
mixture is 2.8oC. What is the molal freezing point constant for benzene?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of<br/>
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the<br/>
mixture is 2.8oC. What is the molal freezing point constant for benzene?</p></div>
</h2>
</div>
</div> | 4.3 ℃/(kg * mol) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for calculating freezing point depression is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> is the decrease in freezing point</p>
<p><mathjax>#K_"f"#</mathjax> is the molal freezing point depression constant</p>
<p><mathjax>#"m"#</mathjax> is the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<blockquote></blockquote>
<p>We can rearrange the above formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#K_"f" = (ΔT_"f")/m#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#ΔT_"f" = T_"f"^° -T_"f" = "5.5 °C – 2.8 °C" = "2.7 °C"#</mathjax></p>
<p><mathjax>#"Moles of C"_10"H"_8 = 1.60 color(red)(cancel(color(black)("g C"_10"H"_8))) × ("1 mol C"_10"H"_8)/(128.17 color(red)(cancel(color(black)("g C"_10"H"_8)))) = "0.012 48 mol C"_10"H"_8#</mathjax></p>
<p><mathjax>#"Molality" = "moles of solute"/"kilograms of solvent" = "0.012 48 mol"/"0.0200 kg" = "0.6242 mol/kg"#</mathjax></p>
<p><mathjax>#K_"f" = (ΔT_"f")/m = "2.7 °C"/"0.6242 mol/kg" = "4.3 °C·kg·mol"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>The actual value of <mathjax>#K_"f"#</mathjax> for benzene is <mathjax>#"5.1 °C·kg·mol"^"-1"#</mathjax>.</p>
<p>Are you sure that the observed freezing point wasn't 2.3 °C?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Your experimental molal freezing point depression constant for benzene is <mathjax>#"4.3 °C·kg·mol"^"-1"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for calculating freezing point depression is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> is the decrease in freezing point</p>
<p><mathjax>#K_"f"#</mathjax> is the molal freezing point depression constant</p>
<p><mathjax>#"m"#</mathjax> is the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<blockquote></blockquote>
<p>We can rearrange the above formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#K_"f" = (ΔT_"f")/m#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#ΔT_"f" = T_"f"^° -T_"f" = "5.5 °C – 2.8 °C" = "2.7 °C"#</mathjax></p>
<p><mathjax>#"Moles of C"_10"H"_8 = 1.60 color(red)(cancel(color(black)("g C"_10"H"_8))) × ("1 mol C"_10"H"_8)/(128.17 color(red)(cancel(color(black)("g C"_10"H"_8)))) = "0.012 48 mol C"_10"H"_8#</mathjax></p>
<p><mathjax>#"Molality" = "moles of solute"/"kilograms of solvent" = "0.012 48 mol"/"0.0200 kg" = "0.6242 mol/kg"#</mathjax></p>
<p><mathjax>#K_"f" = (ΔT_"f")/m = "2.7 °C"/"0.6242 mol/kg" = "4.3 °C·kg·mol"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>The actual value of <mathjax>#K_"f"#</mathjax> for benzene is <mathjax>#"5.1 °C·kg·mol"^"-1"#</mathjax>.</p>
<p>Are you sure that the observed freezing point wasn't 2.3 °C?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the
mixture is 2.8oC. What is the molal freezing point constant for benzene?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of<br/>
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the<br/>
mixture is 2.8oC. What is the molal freezing point constant for benzene?</p></div>
</h2>
</div>
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Ernest Z.
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May 11, 2016
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<div class="markdown"><p>Your experimental molal freezing point depression constant for benzene is <mathjax>#"4.3 °C·kg·mol"^"-1"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for calculating freezing point depression is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> is the decrease in freezing point</p>
<p><mathjax>#K_"f"#</mathjax> is the molal freezing point depression constant</p>
<p><mathjax>#"m"#</mathjax> is the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution.</p>
<blockquote></blockquote>
<p>We can rearrange the above formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#K_"f" = (ΔT_"f")/m#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#ΔT_"f" = T_"f"^° -T_"f" = "5.5 °C – 2.8 °C" = "2.7 °C"#</mathjax></p>
<p><mathjax>#"Moles of C"_10"H"_8 = 1.60 color(red)(cancel(color(black)("g C"_10"H"_8))) × ("1 mol C"_10"H"_8)/(128.17 color(red)(cancel(color(black)("g C"_10"H"_8)))) = "0.012 48 mol C"_10"H"_8#</mathjax></p>
<p><mathjax>#"Molality" = "moles of solute"/"kilograms of solvent" = "0.012 48 mol"/"0.0200 kg" = "0.6242 mol/kg"#</mathjax></p>
<p><mathjax>#K_"f" = (ΔT_"f")/m = "2.7 °C"/"0.6242 mol/kg" = "4.3 °C·kg·mol"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>The actual value of <mathjax>#K_"f"#</mathjax> for benzene is <mathjax>#"5.1 °C·kg·mol"^"-1"#</mathjax>.</p>
<p>Are you sure that the observed freezing point wasn't 2.3 °C?</p></div>
</div>
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</article> | Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the
mixture is 2.8oC. What is the molal freezing point constant for benzene? |
Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the
mixture is 2.8oC. What is the molal freezing point constant for benzene?
|
1,130 | aa9fcbc7-6ddd-11ea-b90b-ccda262736ce | https://socratic.org/questions/how-many-moles-of-fe-2o-3-are-in-171-g-of-the-compound | 1.07 moles | start physical_unit 4 4 mole mol qc_end physical_unit 10 11 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] Fe2O3 [IN] moles"}] | [{"type":"physical unit","value":"1.07 moles"}] | [{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{171 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #Fe_2O_3# are in 171 g of the compound? </h1> | null | 1.07 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the molar quantity we divide the given mass by the molar mass, i.e. <mathjax>#(171*g)/(159.69* g·mol^-1)=??mol#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>A bit over <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#"ferric oxide."#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the molar quantity we divide the given mass by the molar mass, i.e. <mathjax>#(171*g)/(159.69* g·mol^-1)=??mol#</mathjax>.</p></div>
</div>
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anor277
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<div class="markdown"><p>A bit over <mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of <mathjax>#"ferric oxide."#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To get the molar quantity we divide the given mass by the molar mass, i.e. <mathjax>#(171*g)/(159.69* g·mol^-1)=??mol#</mathjax>.</p></div>
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</article> | How many moles of #Fe_2O_3# are in 171 g of the compound? | null |
1,131 | a9eca5ca-6ddd-11ea-89ab-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-ammonium-tetraborate | (NH4)2B4O7 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] ammonium tetraborate [IN] default"}] | [{"type":"chemical equation","value":"(NH4)2B4O7"}] | [{"type":"substance name","value":"Ammonium tetraborate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for ammonium tetraborate? </h1> | null | (NH4)2B4O7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(NH_4)_2B_4O_7#</mathjax><br/>
Ion tetraboric is<mathjax>#B_4O_7^(2-)#</mathjax> since B have oxidation number = +3 and O =-2<br/>
ammonium is <mathjax>#NH_4^+#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#(NH_4)_2B_4O_7#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(NH_4)_2B_4O_7#</mathjax><br/>
Ion tetraboric is<mathjax>#B_4O_7^(2-)#</mathjax> since B have oxidation number = +3 and O =-2<br/>
ammonium is <mathjax>#NH_4^+#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for ammonium tetraborate? </h1>
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<div class="markdown"><p><mathjax>#(NH_4)_2B_4O_7#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(NH_4)_2B_4O_7#</mathjax><br/>
Ion tetraboric is<mathjax>#B_4O_7^(2-)#</mathjax> since B have oxidation number = +3 and O =-2<br/>
ammonium is <mathjax>#NH_4^+#</mathjax></p></div>
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</article> | What is the chemical formula for ammonium tetraborate? | null |
1,132 | aad3d8ba-6ddd-11ea-903b-ccda262736ce | https://socratic.org/questions/5880e4ff11ef6b445903ee00 | 4.00 cm^3 | start physical_unit 14 14 volume cm^3 qc_end end | [{"type":"physical unit","value":"Volume [OF] solution [IN] cm^3"}] | [{"type":"physical unit","value":"4.00 cm^3"}] | [{"type":"physical unit","value":"Concentration [OF] the drag [=] \\pu{0.50 g/cm^3}"},{"type":"physical unit","value":"Mass [OF] the drag [=] \\pu{2 g}"}] | <h1 class="questionTitle" itemprop="name">You have a drug that has a concentration of #0.50*g*cm^-3#. What volume of solution is delivered for a #2*g# dose?</h1> | null | 4.00 cm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given a concentration in <mathjax>#g*cm^-3#</mathjax>. The product, <mathjax>#"Volume "xx" concentration"#</mathjax> gives the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in the drug.</p>
<p>This is clear dimensionally, when the units of volume and concentration cancel out to give an answer with the units of mass, i.e. <mathjax>#g#</mathjax>:</p>
<p><mathjax>#0.50*g*cancel(cm^-3)xx4*cancel(cm^3)=2*g#</mathjax>.</p>
<p>Is this clear? As a chemist, I usually do not care too much about errors. You get a calculation or dilution wrong, and you simply say <mathjax>#"oops"#</mathjax>, and repeat the experiment (and hopefully conceal your mistake from your supervisor!). As a nurse or a medico, you get a dose wrong, and you might injure someone.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Then you would administer a <mathjax>#4*mL#</mathjax> or <mathjax>#4*cm^3#</mathjax> dose. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given a concentration in <mathjax>#g*cm^-3#</mathjax>. The product, <mathjax>#"Volume "xx" concentration"#</mathjax> gives the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in the drug.</p>
<p>This is clear dimensionally, when the units of volume and concentration cancel out to give an answer with the units of mass, i.e. <mathjax>#g#</mathjax>:</p>
<p><mathjax>#0.50*g*cancel(cm^-3)xx4*cancel(cm^3)=2*g#</mathjax>.</p>
<p>Is this clear? As a chemist, I usually do not care too much about errors. You get a calculation or dilution wrong, and you simply say <mathjax>#"oops"#</mathjax>, and repeat the experiment (and hopefully conceal your mistake from your supervisor!). As a nurse or a medico, you get a dose wrong, and you might injure someone.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">You have a drug that has a concentration of #0.50*g*cm^-3#. What volume of solution is delivered for a #2*g# dose?</h1>
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<div class="markdown"><p>Then you would administer a <mathjax>#4*mL#</mathjax> or <mathjax>#4*cm^3#</mathjax> dose. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given a concentration in <mathjax>#g*cm^-3#</mathjax>. The product, <mathjax>#"Volume "xx" concentration"#</mathjax> gives the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> in the drug.</p>
<p>This is clear dimensionally, when the units of volume and concentration cancel out to give an answer with the units of mass, i.e. <mathjax>#g#</mathjax>:</p>
<p><mathjax>#0.50*g*cancel(cm^-3)xx4*cancel(cm^3)=2*g#</mathjax>.</p>
<p>Is this clear? As a chemist, I usually do not care too much about errors. You get a calculation or dilution wrong, and you simply say <mathjax>#"oops"#</mathjax>, and repeat the experiment (and hopefully conceal your mistake from your supervisor!). As a nurse or a medico, you get a dose wrong, and you might injure someone.</p></div>
</div>
</div>
</div>
</div>
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<a href="https://socratic.org/answers/366894" itemprop="url">Answer link</a>
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</article> | You have a drug that has a concentration of #0.50*g*cm^-3#. What volume of solution is delivered for a #2*g# dose? | null |
1,133 | ad1663af-6ddd-11ea-8381-ccda262736ce | https://socratic.org/questions/56fbe7d17c014942feaf03b6 | 6.57 × 10^22 | start physical_unit 2 3 number none qc_end physical_unit 10 11 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] chromium atoms"}] | [{"type":"physical unit","value":"6.57 × 10^22"}] | [{"type":"physical unit","value":"Mass [OF] chromium metal [=] \\pu{5.67 g}"}] | <h1 class="questionTitle" itemprop="name">How many chromium atoms in a #5.67*g# mass of chromium metal?</h1> | null | 6.57 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of elemental chromium is <mathjax>#52.00*g*mol^-1#</mathjax>. What does this mean? It means that if there are <mathjax>#52.00*g#</mathjax> of chromium metal, there are <mathjax>#"Avogadro's number"#</mathjax> of chromium atoms, i.e. <mathjax>#N_A#</mathjax> chromium atoms, or <mathjax>#6.022xx10^23#</mathjax> individual atoms. </p>
<p>Here, I have used <mathjax>#N_A#</mathjax> as I would a dozen, or a score, or a gross, or any other collective number. </p>
<p>So the number of atoms?</p>
<p><mathjax>#(5.67*g)/(52.00*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many chromium atoms?"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>In <mathjax>#5.67*g#</mathjax> chromium metal?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of elemental chromium is <mathjax>#52.00*g*mol^-1#</mathjax>. What does this mean? It means that if there are <mathjax>#52.00*g#</mathjax> of chromium metal, there are <mathjax>#"Avogadro's number"#</mathjax> of chromium atoms, i.e. <mathjax>#N_A#</mathjax> chromium atoms, or <mathjax>#6.022xx10^23#</mathjax> individual atoms. </p>
<p>Here, I have used <mathjax>#N_A#</mathjax> as I would a dozen, or a score, or a gross, or any other collective number. </p>
<p>So the number of atoms?</p>
<p><mathjax>#(5.67*g)/(52.00*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many chromium atoms?"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many chromium atoms in a #5.67*g# mass of chromium metal?</h1>
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<div class="markdown"><p>In <mathjax>#5.67*g#</mathjax> chromium metal?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of elemental chromium is <mathjax>#52.00*g*mol^-1#</mathjax>. What does this mean? It means that if there are <mathjax>#52.00*g#</mathjax> of chromium metal, there are <mathjax>#"Avogadro's number"#</mathjax> of chromium atoms, i.e. <mathjax>#N_A#</mathjax> chromium atoms, or <mathjax>#6.022xx10^23#</mathjax> individual atoms. </p>
<p>Here, I have used <mathjax>#N_A#</mathjax> as I would a dozen, or a score, or a gross, or any other collective number. </p>
<p>So the number of atoms?</p>
<p><mathjax>#(5.67*g)/(52.00*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many chromium atoms?"#</mathjax></p></div>
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</article> | How many chromium atoms in a #5.67*g# mass of chromium metal? | null |
1,134 | a8de7926-6ddd-11ea-a91b-ccda262736ce | https://socratic.org/questions/a-sample-of-an-unknown-metal-has-a-mass-of-120-7-g-as-the-sample-cools-from-90-5 | 0.90 J/(g * ℃) | start physical_unit 13 14 specific_heat j/(°c_·_g) qc_end physical_unit 1 5 10 11 mass qc_end physical_unit 13 14 17 18 temperature qc_end physical_unit 13 14 20 21 temperature qc_end physical_unit 13 14 24 25 heat_energy qc_end end | [{"type":"physical unit","value":"Specific heat [OF] the sample [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.90 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] an unknown metal sample [=] \\pu{120.7 g}"},{"type":"physical unit","value":"Temperature1 [OF] the sample [=] \\pu{90.5 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the sample [=] \\pu{25.7 ℃}"},{"type":"physical unit","value":"Released energy [OF] the sample [=] \\pu{7020 J}"}] | <h1 class="questionTitle" itemprop="name">A sample of an unknown metal has a mass of 120.7 g. As the sample cools from 90.5 °C to 25.7 °C, it releases 7020 J of energy. What is the specific heat of the sample? </h1> | null | 0.90 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat must be absorbed or lost in order for <mathjax>#"1 g"#</mathjax> of that substance to experience a <mathjax>#1^@"C"#</mathjax> <em>temperature change</em>. </p>
<p>The equation that establishes a relationship between heat absorbed / lost and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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>Now, it is important to realize that the value of <mathjax>#q#</mathjax> <strong>must come out negative</strong> for samples that experience a <strong>decrease in temperature</strong>. </p>
<p>From a thermodynamic point of view, <strong>heat lost</strong> always carries a <em>negative sign</em>, so you need to keep that in mind when plugging in your values. </p>
<p>Simply put, when <mathjax>#"7020 J"#</mathjax> of energy are <strong>released</strong>, the heat lost is written as</p>
<blockquote>
<p><mathjax>#q = -"7020 J"#</mathjax></p>
</blockquote>
<p>With this being said, plug in your values into the above equation and solve for <mathjax>#c#</mathjax>, the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the metal</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = (-"7020 J")/("120.7 g" * (25.7 - 90.5)^@"C")#</mathjax></p>
<p><mathjax>#c = (color(red)(cancel(color(black)(-)))"7020 J")/(color(red)(cancel(color(black)(-)))"7821.36 g" ""^@"C") = 0.8975 "J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures of the metal, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)(0.898"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#c = 0.898"J"/("g" ""^@"C")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat must be absorbed or lost in order for <mathjax>#"1 g"#</mathjax> of that substance to experience a <mathjax>#1^@"C"#</mathjax> <em>temperature change</em>. </p>
<p>The equation that establishes a relationship between heat absorbed / lost and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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>Now, it is important to realize that the value of <mathjax>#q#</mathjax> <strong>must come out negative</strong> for samples that experience a <strong>decrease in temperature</strong>. </p>
<p>From a thermodynamic point of view, <strong>heat lost</strong> always carries a <em>negative sign</em>, so you need to keep that in mind when plugging in your values. </p>
<p>Simply put, when <mathjax>#"7020 J"#</mathjax> of energy are <strong>released</strong>, the heat lost is written as</p>
<blockquote>
<p><mathjax>#q = -"7020 J"#</mathjax></p>
</blockquote>
<p>With this being said, plug in your values into the above equation and solve for <mathjax>#c#</mathjax>, the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the metal</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = (-"7020 J")/("120.7 g" * (25.7 - 90.5)^@"C")#</mathjax></p>
<p><mathjax>#c = (color(red)(cancel(color(black)(-)))"7020 J")/(color(red)(cancel(color(black)(-)))"7821.36 g" ""^@"C") = 0.8975 "J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures of the metal, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)(0.898"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of an unknown metal has a mass of 120.7 g. As the sample cools from 90.5 °C to 25.7 °C, it releases 7020 J of energy. What is the specific heat of the sample? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-18T12:13:50" itemprop="dateCreated">
Dec 18, 2015
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<div class="markdown"><p><mathjax>#c = 0.898"J"/("g" ""^@"C")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat must be absorbed or lost in order for <mathjax>#"1 g"#</mathjax> of that substance to experience a <mathjax>#1^@"C"#</mathjax> <em>temperature change</em>. </p>
<p>The equation that establishes a relationship between heat absorbed / lost and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/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>Now, it is important to realize that the value of <mathjax>#q#</mathjax> <strong>must come out negative</strong> for samples that experience a <strong>decrease in temperature</strong>. </p>
<p>From a thermodynamic point of view, <strong>heat lost</strong> always carries a <em>negative sign</em>, so you need to keep that in mind when plugging in your values. </p>
<p>Simply put, when <mathjax>#"7020 J"#</mathjax> of energy are <strong>released</strong>, the heat lost is written as</p>
<blockquote>
<p><mathjax>#q = -"7020 J"#</mathjax></p>
</blockquote>
<p>With this being said, plug in your values into the above equation and solve for <mathjax>#c#</mathjax>, the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the metal</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies c = q/(m * DeltaT)#</mathjax></p>
<p><mathjax>#c = (-"7020 J")/("120.7 g" * (25.7 - 90.5)^@"C")#</mathjax></p>
<p><mathjax>#c = (color(red)(cancel(color(black)(-)))"7020 J")/(color(red)(cancel(color(black)(-)))"7821.36 g" ""^@"C") = 0.8975 "J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the two temperatures of the metal, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)(0.898"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote></div>
</div>
</div>
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</article> | A sample of an unknown metal has a mass of 120.7 g. As the sample cools from 90.5 °C to 25.7 °C, it releases 7020 J of energy. What is the specific heat of the sample? | null |
1,135 | ab1b46d9-6ddd-11ea-b036-ccda262736ce | https://socratic.org/questions/what-would-be-the-compound-formula-of-strontium-bromide | SrBr2 | start chemical_formula qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] Strontium Bromide [IN] default"}] | [{"type":"chemical equation","value":"SrBr2"}] | [{"type":"substance name","value":"Strontium Bromide"}] | <h1 class="questionTitle" itemprop="name">What would be the compound formula of Strontium Bromide?</h1> | null | SrBr2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It has two <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> and commonly forms <mathjax>#Sr^(2+)#</mathjax> ion. Bromine is a non-metal with seven <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and commonly forms the <mathjax>#Br^-#</mathjax> ion. </p>
<p>When strontium and bromine make music together, we conceive of a redox process...</p>
<p><mathjax>#Sr(s) +Br_2(l) rarr SrBr_2(s)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, strontium is an <mathjax>#"alkaline earth..."#</mathjax> from <mathjax>#"Group 2"#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>..</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It has two <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> and commonly forms <mathjax>#Sr^(2+)#</mathjax> ion. Bromine is a non-metal with seven <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and commonly forms the <mathjax>#Br^-#</mathjax> ion. </p>
<p>When strontium and bromine make music together, we conceive of a redox process...</p>
<p><mathjax>#Sr(s) +Br_2(l) rarr SrBr_2(s)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What would be the compound formula of Strontium Bromide?</h1>
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<div class="markdown"><p>Well, strontium is an <mathjax>#"alkaline earth..."#</mathjax> from <mathjax>#"Group 2"#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>..</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It has two <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> and commonly forms <mathjax>#Sr^(2+)#</mathjax> ion. Bromine is a non-metal with seven <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, and commonly forms the <mathjax>#Br^-#</mathjax> ion. </p>
<p>When strontium and bromine make music together, we conceive of a redox process...</p>
<p><mathjax>#Sr(s) +Br_2(l) rarr SrBr_2(s)#</mathjax></p></div>
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</article> | What would be the compound formula of Strontium Bromide? | null |
1,136 | ab0153f1-6ddd-11ea-9667-ccda262736ce | https://socratic.org/questions/water-can-be-formed-from-the-stoichiometric-reaction-of-hydrogen-with-oxygen-2h- | 5.62 grams | start physical_unit 18 18 mass g qc_end chemical_equation 12 18 qc_end physical_unit 15 15 23 24 mass qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] H2O [IN] grams"}] | [{"type":"physical unit","value":"5.62 grams"}] | [{"type":"chemical equation","value":"2 H2(g) + O2(g) -> 2 H2O(g)"},{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{5.0 g}"},{"type":"other","value":"Complete reaction."},{"type":"other","value":"Excess hydrogen."}] | <h1 class="questionTitle" itemprop="name">Water can be formed from the stoichiometric reaction of hydrogen with oxygen: #2H_2(g) + O_2(g) + 2H_2O(g)#. A complete reaction of 5.0 g of #O_2# with excess hydrogen produces how many grams of #H_2O#?</h1> | null | 5.62 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(g)#</mathjax></p>
<p>We started with <mathjax>#(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"#</mathjax>.</p>
<p>Given excess dihydrogen, this molar quantity thus gives <mathjax>#2xx0.156*mol#</mathjax> water.</p>
<p>And thus, <mathjax>#2xx0.156*molxx18.01*g*mol^-1~=6*g#</mathjax> water are evolved. Is this reaction exothermic or endothermic?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Your equation specifies that <mathjax>#32*g#</mathjax> of dioxygen and <mathjax>#4*g#</mathjax> of dihydrogen give <mathjax>#36*g#</mathjax> of water upon reaction, so.......... </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(g)#</mathjax></p>
<p>We started with <mathjax>#(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"#</mathjax>.</p>
<p>Given excess dihydrogen, this molar quantity thus gives <mathjax>#2xx0.156*mol#</mathjax> water.</p>
<p>And thus, <mathjax>#2xx0.156*molxx18.01*g*mol^-1~=6*g#</mathjax> water are evolved. Is this reaction exothermic or endothermic?</p></div>
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<h1 class="questionTitle" itemprop="name">Water can be formed from the stoichiometric reaction of hydrogen with oxygen: #2H_2(g) + O_2(g) + 2H_2O(g)#. A complete reaction of 5.0 g of #O_2# with excess hydrogen produces how many grams of #H_2O#?</h1>
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<div class="markdown"><p>Your equation specifies that <mathjax>#32*g#</mathjax> of dioxygen and <mathjax>#4*g#</mathjax> of dihydrogen give <mathjax>#36*g#</mathjax> of water upon reaction, so.......... </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(g)#</mathjax></p>
<p>We started with <mathjax>#(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"#</mathjax>.</p>
<p>Given excess dihydrogen, this molar quantity thus gives <mathjax>#2xx0.156*mol#</mathjax> water.</p>
<p>And thus, <mathjax>#2xx0.156*molxx18.01*g*mol^-1~=6*g#</mathjax> water are evolved. Is this reaction exothermic or endothermic?</p></div>
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</article> | Water can be formed from the stoichiometric reaction of hydrogen with oxygen: #2H_2(g) + O_2(g) + 2H_2O(g)#. A complete reaction of 5.0 g of #O_2# with excess hydrogen produces how many grams of #H_2O#? | null |
1,137 | ace8b1fe-6ddd-11ea-ba1f-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-50-g-of-o | 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] O [IN] moles"}] | [{"type":"physical unit","value":"3 moles"}] | [{"type":"physical unit","value":"Mass [OF] O [=] \\pu{50 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 50 g of O?</h1> | null | 3 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To determine the number of moles in a given mass of an element, divide its given mass by its molar mass.</p>
<p>The mass of one mole (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>The molar mass of oxygen is <mathjax>#"15.999 g/mol"#</mathjax>.</p>
<p><mathjax>#50cancel"g O"xx(1"mol O")/(15.999cancel"g O")="3 mol O"#</mathjax> (rounded to one significant figure)</p></div>
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<div class="markdown"><p>There are 3 moles of O in 50 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To determine the number of moles in a given mass of an element, divide its given mass by its molar mass.</p>
<p>The mass of one mole (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>The molar mass of oxygen is <mathjax>#"15.999 g/mol"#</mathjax>.</p>
<p><mathjax>#50cancel"g O"xx(1"mol O")/(15.999cancel"g O")="3 mol O"#</mathjax> (rounded to one significant figure)</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 50 g of O?</h1>
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<div class="markdown"><p>There are 3 moles of O in 50 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To determine the number of moles in a given mass of an element, divide its given mass by its molar mass.</p>
<p>The mass of one mole (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>The molar mass of oxygen is <mathjax>#"15.999 g/mol"#</mathjax>.</p>
<p><mathjax>#50cancel"g O"xx(1"mol O")/(15.999cancel"g O")="3 mol O"#</mathjax> (rounded to one significant figure)</p></div>
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</article> | How many moles are in 50 g of O? | null |
1,138 | aad3ffcc-6ddd-11ea-b3f1-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-with-a-mass-of-180-grams-and-an-empi | C6H12O6 | start chemical_formula qc_end physical_unit 7 7 12 13 mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C6H12O6"}] | [{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{180 grams}"},{"type":"other","value":"The compound has an empirical formula CH2O."}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound with a mass of 180 grams and an empirical formula of #CH_2O#?</h1> | null | C6H12O6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, calculate the <strong>formula mass</strong> of the empirical formula:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"CH"_2"O"#</mathjax>)<mathjax># = 12 + 2xx1 + 16 = 30#</mathjax></p>
<p>Next we need to find what I personally refer to as the <strong>multiplier</strong>: it is not given an official name, but it is the number that we will <em>multiply</em> the <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> (the subscript numbers) in the empirical formula by to give the <strong>molecular formula</strong>. Observe this simple equation:</p>
<p><mathjax>#"multiplier"#</mathjax> <mathjax>#= ("M"_r("molecular formula"))/("M"_r("CH"_2"O"))#</mathjax></p>
<p><mathjax>#=> "multiplier"#</mathjax> <mathjax>#= 180/30 = 6#</mathjax></p>
<p>Now, put that multiplier into effect:</p>
<p><mathjax>#"molecular formula"#</mathjax> <mathjax>#= "C"_ (1xx6) "H"_ (2xx6) "O"_ (1xx6) = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"C"_6"H"_12"O"_6#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, calculate the <strong>formula mass</strong> of the empirical formula:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"CH"_2"O"#</mathjax>)<mathjax># = 12 + 2xx1 + 16 = 30#</mathjax></p>
<p>Next we need to find what I personally refer to as the <strong>multiplier</strong>: it is not given an official name, but it is the number that we will <em>multiply</em> the <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> (the subscript numbers) in the empirical formula by to give the <strong>molecular formula</strong>. Observe this simple equation:</p>
<p><mathjax>#"multiplier"#</mathjax> <mathjax>#= ("M"_r("molecular formula"))/("M"_r("CH"_2"O"))#</mathjax></p>
<p><mathjax>#=> "multiplier"#</mathjax> <mathjax>#= 180/30 = 6#</mathjax></p>
<p>Now, put that multiplier into effect:</p>
<p><mathjax>#"molecular formula"#</mathjax> <mathjax>#= "C"_ (1xx6) "H"_ (2xx6) "O"_ (1xx6) = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound with a mass of 180 grams and an empirical formula of #CH_2O#?</h1>
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<div class="markdown"><p><mathjax>#"C"_6"H"_12"O"_6#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, calculate the <strong>formula mass</strong> of the empirical formula:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"CH"_2"O"#</mathjax>)<mathjax># = 12 + 2xx1 + 16 = 30#</mathjax></p>
<p>Next we need to find what I personally refer to as the <strong>multiplier</strong>: it is not given an official name, but it is the number that we will <em>multiply</em> the <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> (the subscript numbers) in the empirical formula by to give the <strong>molecular formula</strong>. Observe this simple equation:</p>
<p><mathjax>#"multiplier"#</mathjax> <mathjax>#= ("M"_r("molecular formula"))/("M"_r("CH"_2"O"))#</mathjax></p>
<p><mathjax>#=> "multiplier"#</mathjax> <mathjax>#= 180/30 = 6#</mathjax></p>
<p>Now, put that multiplier into effect:</p>
<p><mathjax>#"molecular formula"#</mathjax> <mathjax>#= "C"_ (1xx6) "H"_ (2xx6) "O"_ (1xx6) = "C"_6"H"_12"O"_6#</mathjax>.</p></div>
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</article> | What is the molecular formula of a compound with a mass of 180 grams and an empirical formula of #CH_2O#? | null |
1,139 | a92c195c-6ddd-11ea-b06f-ccda262736ce | https://socratic.org/questions/if-the-initial-volume-of-the-bubbles-in-a-diver-s-blood-is-15-ml-and-the-initial | 191.25 mL | start physical_unit 5 6 volume ml qc_end physical_unit 5 6 12 13 volume qc_end physical_unit 5 6 19 20 pressure qc_end physical_unit 5 6 34 35 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the bubbles [IN] mL"}] | [{"type":"physical unit","value":"191.25 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the bubbles [=] \\pu{15 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the bubbles [=] \\pu{12.75 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the bubbles [=] \\pu{1.00 atm}"}] | <h1 class="questionTitle" itemprop="name">If the initial volume of the bubbles in a diver's blood is 15 mL and the initial pressure is 12.75 atm, what is the volume of the bubbles when the diver has surfaced to 1.00 atm pressure?</h1> | null | 191.25 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>. Thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax>.</p>
<p>And thus, <mathjax>#(12.75*atm)/(1*atm)xx15*mL#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Of course, the temperature might have changed between depth and surface, but we have no way of interrogating this change. That diver is very deep <mathjax>#(>100*m)#</mathjax>. Sports divers would never go near that depth. </p></div>
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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>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>. Thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax>.</p>
<p>And thus, <mathjax>#(12.75*atm)/(1*atm)xx15*mL#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Of course, the temperature might have changed between depth and surface, but we have no way of interrogating this change. That diver is very deep <mathjax>#(>100*m)#</mathjax>. Sports divers would never go near that depth. </p></div>
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<h1 class="questionTitle" itemprop="name">If the initial volume of the bubbles in a diver's blood is 15 mL and the initial pressure is 12.75 atm, what is the volume of the bubbles when the diver has surfaced to 1.00 atm pressure?</h1>
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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>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>. Thus <mathjax>#V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_1V_1)/P_2#</mathjax>.</p>
<p>And thus, <mathjax>#(12.75*atm)/(1*atm)xx15*mL#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Of course, the temperature might have changed between depth and surface, but we have no way of interrogating this change. That diver is very deep <mathjax>#(>100*m)#</mathjax>. Sports divers would never go near that depth. </p></div>
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</article> | If the initial volume of the bubbles in a diver's blood is 15 mL and the initial pressure is 12.75 atm, what is the volume of the bubbles when the diver has surfaced to 1.00 atm pressure? | null |
1,140 | abea60d0-6ddd-11ea-bf6a-ccda262736ce | https://socratic.org/questions/when-burning-180-g-glucose-in-the-presence-of-the-192-g-of-oxygen-water-and-carb | 263.62 g | start physical_unit 16 17 mass g qc_end physical_unit 4 4 2 3 mass qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 14 14 21 22 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] g"}] | [{"type":"physical unit","value":"263.62 g"}] | [{"type":"physical unit","value":"Mass [OF] glucose [=] \\pu{180 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{192 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{108 g}"}] | <h1 class="questionTitle" itemprop="name">When burning 180 g glucose in the presence of the 192 g of oxygen, water and carbon dioxide are produced. If 108 g of water is produced, how much carbon dioxide is produced? </h1> | null | 263.62 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass of <mathjax>#"CO"_2#</mathjax> produced in a given reaction, given that <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> is formed.</p>
<p>What we can do first is write the balanced chemical equation for this reaction:</p>
<p><mathjax>#"C"_6"H"_12"O"_6(s) + 6"O"_2(g) rarr 6"CO"_2(g) + 6"H"_2"O"(g)#</mathjax></p>
<p>We really only needed the amount of water produced, because that would already factor in the limiting reactant and <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a>, so we'll use <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> in our calculations.</p>
<blockquote></blockquote>
<p>Convert mass of water to moles using its <strong>molar mass</strong>:</p>
<p><mathjax>#108cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(5.99#</mathjax> <mathjax>#color(red)("mol H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, using the coefficients of the chemical equation, we can figure out the relative number of moles of <mathjax>#"CO"_2#</mathjax> that form:</p>
<p><mathjax>#color(red)(5.99)cancel(color(red)("mol H"_2"O"))((6color(white)(l)"mol CO"_2)/(6cancel("mol H"_2"O"))) = 5.99color(white)(l)"mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Finally, we use the molar mass of carbon dioxide to find the number of grams:</p>
<p><mathjax>#5.99cancel("mol CO"_2)((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(ul(264color(white)(l)"g CO"_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#264#</mathjax> <mathjax>#"g CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass of <mathjax>#"CO"_2#</mathjax> produced in a given reaction, given that <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> is formed.</p>
<p>What we can do first is write the balanced chemical equation for this reaction:</p>
<p><mathjax>#"C"_6"H"_12"O"_6(s) + 6"O"_2(g) rarr 6"CO"_2(g) + 6"H"_2"O"(g)#</mathjax></p>
<p>We really only needed the amount of water produced, because that would already factor in the limiting reactant and <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a>, so we'll use <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> in our calculations.</p>
<blockquote></blockquote>
<p>Convert mass of water to moles using its <strong>molar mass</strong>:</p>
<p><mathjax>#108cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(5.99#</mathjax> <mathjax>#color(red)("mol H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, using the coefficients of the chemical equation, we can figure out the relative number of moles of <mathjax>#"CO"_2#</mathjax> that form:</p>
<p><mathjax>#color(red)(5.99)cancel(color(red)("mol H"_2"O"))((6color(white)(l)"mol CO"_2)/(6cancel("mol H"_2"O"))) = 5.99color(white)(l)"mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Finally, we use the molar mass of carbon dioxide to find the number of grams:</p>
<p><mathjax>#5.99cancel("mol CO"_2)((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(ul(264color(white)(l)"g CO"_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When burning 180 g glucose in the presence of the 192 g of oxygen, water and carbon dioxide are produced. If 108 g of water is produced, how much carbon dioxide is produced? </h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-07-29T19:42:40" itemprop="dateCreated">
Jul 29, 2017
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<div class="markdown"><p><mathjax>#264#</mathjax> <mathjax>#"g CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass of <mathjax>#"CO"_2#</mathjax> produced in a given reaction, given that <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> is formed.</p>
<p>What we can do first is write the balanced chemical equation for this reaction:</p>
<p><mathjax>#"C"_6"H"_12"O"_6(s) + 6"O"_2(g) rarr 6"CO"_2(g) + 6"H"_2"O"(g)#</mathjax></p>
<p>We really only needed the amount of water produced, because that would already factor in the limiting reactant and <a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a>, so we'll use <mathjax>#108#</mathjax> <mathjax>#"g H"_2"O"#</mathjax> in our calculations.</p>
<blockquote></blockquote>
<p>Convert mass of water to moles using its <strong>molar mass</strong>:</p>
<p><mathjax>#108cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(5.99#</mathjax> <mathjax>#color(red)("mol H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, using the coefficients of the chemical equation, we can figure out the relative number of moles of <mathjax>#"CO"_2#</mathjax> that form:</p>
<p><mathjax>#color(red)(5.99)cancel(color(red)("mol H"_2"O"))((6color(white)(l)"mol CO"_2)/(6cancel("mol H"_2"O"))) = 5.99color(white)(l)"mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p>Finally, we use the molar mass of carbon dioxide to find the number of grams:</p>
<p><mathjax>#5.99cancel("mol CO"_2)((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(ul(264color(white)(l)"g CO"_2#</mathjax></p></div>
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</article> | When burning 180 g glucose in the presence of the 192 g of oxygen, water and carbon dioxide are produced. If 108 g of water is produced, how much carbon dioxide is produced? | null |
1,141 | ac9cf4ba-6ddd-11ea-a898-ccda262736ce | https://socratic.org/questions/how-do-you-balance-hbr-o-2-br-2-h-2o | 4 HBr + O2 -> 2 Br2 + 2 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"4 HBr + O2 -> 2 Br2 + 2 H2O"}] | [{"type":"chemical equation","value":"HBr + O2 -> Br2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #HBr + O_2 -> Br_2 + H_2O#?</h1> | null | 4 HBr + O2 -> 2 Br2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #HBr + O_2 -> Br_2 + H_2O#?</h1>
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Trevor Ryan.
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<div class="markdown"><p><mathjax>#4HBr+O_2->2Br_2+2H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Add molecules (balancing numbers in front) to ensure that the number of atoms of each element is the same on the left and right hand sides of the equation.</p></div>
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A08
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<div class="markdown"><p><mathjax>#4"HBr"+"O"_2→2"Br"_2+2"H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This equation is one of those where you need to balance O at the very beginning. </p>
<p>Otherwise as a thumb-rule one leaves H and O to taken up at the last.</p>
<p>Observe that in the unbalanced equation, on the reactants side there are 2 atoms of oxygen, whereas there is single oxygen atom on the products side.</p>
<p>Therefore to balance O on both sides, we need to put 2 in front of water molecule. This fixes 4 hydrogen atoms on the right side of the equation.</p>
<p>To balance 4H atoms in the products, we need 4 molecules HBr on the reactants side.</p>
<p>Now we are left with balancing of bromine atoms. Clearly to balance 4 Br atoms on the reactants side, we need to put 2 in front of <mathjax>#" Br"_2 ,#</mathjax> to balance the complete equation.</p></div>
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</article> | How do you balance #HBr + O_2 -> Br_2 + H_2O#? | null |
1,142 | acb3bf62-6ddd-11ea-82bb-ccda262736ce | https://socratic.org/questions/a-balloon-contains-2-58-10-24-molecules-of-methane-gas-how-many-moles-of-methane | 4.28 moles | start physical_unit 8 8 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] methane [IN] moles"}] | [{"type":"physical unit","value":"4.28 moles"}] | [{"type":"physical unit","value":"Number [OF] methane molecules [=] \\pu{2.58 × 10^24}"}] | <h1 class="questionTitle" itemprop="name">A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon?</h1> | null | 4.28 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
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<div class="markdown"><p>A farmer collects 150 eggs; how many dozen eggs does he collect? </p></div>
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<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon?</h1>
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<span class="dateCreated" datetime="2016-03-24T22:27:53" itemprop="dateCreated">
Mar 24, 2016
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<div class="markdown"><p>A farmer collects 150 eggs; how many dozen eggs does he collect? </p></div>
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<div class="markdown"><p><mathjax>#("150 eggs")/("12 eggs per dozen")#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12#</mathjax> <mathjax>#1/2#</mathjax> <mathjax>#dozen#</mathjax></p>
<p><mathjax>#(2.58xx10^24" molecules")/(6.022xx10^23*"molecules"*mol^-1)#</mathjax>. This is approx. <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>?</p></div>
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<span class="dateCreated" datetime="2016-03-25T00:59:06" itemprop="dateCreated">
Mar 25, 2016
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<div class="markdown"><p><mathjax>#2.58xx10^24"molecules CH"_4"#</mathjax> contains <mathjax>#4.28"mol CH"_4"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#1 "mole CH"_4=6.022xx10^23 "molecules CH"_4"#</mathjax></p>
<p>In order to convert molecules to moles of a substance, divide the number of given molecules by <mathjax>#6.022xx10^23 "molecules/mol"#</mathjax>.</p>
<p><mathjax>#2.58xx10^24cancel"molecules CH"_4xx(1"mol CH"_4)/(6.022xx10^23cancel"molecules CH"_4)="4.28 moles CH"_4"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | A balloon contains #2.58 * 10^24# molecules of methane gas. How many moles of methane are in this balloon? | null |
1,143 | aca0c4e6-6ddd-11ea-a1c1-ccda262736ce | https://socratic.org/questions/how-many-moles-of-na-2co-3-are-there-in-10-0-l-of-2-0-m-solution | 20.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 13 13 8 9 volume qc_end physical_unit 4 4 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] Na2CO3 [IN] moles"}] | [{"type":"physical unit","value":"20.00 moles"}] | [{"type":"physical unit","value":"Volume [OF] Na2CO3 solution [=] \\pu{10.0 L}"},{"type":"physical unit","value":"Molarity [OF] Na2CO3 solution [=] \\pu{2.0 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #Na_2CO_3# are there in 10.0 L of 2.0 M solution?</h1> | null | 20.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by this equation:</p>
<p><img alt="www.clutchprep.com" src="https://useruploads.socratic.org/risoIJ8wTrmcOMzHhMeR_vpKmwpMTS22trbAYf6LK_Screen%2520Shot%25202014-08-06%2520at%25201.54.03%2520AM.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of solution, both of which have good units.</p>
<p>Let's rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a><mathjax># = L solutionxxMolarity#</mathjax></p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (10.0 L) (2.0M) = 20. moles</p></div>
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<div class="markdown"><p>There are <strong>20. mol</strong> of <mathjax>#Na_2CO_3#</mathjax> in 10.0L of 2.0M solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by this equation:</p>
<p><img alt="www.clutchprep.com" src="https://useruploads.socratic.org/risoIJ8wTrmcOMzHhMeR_vpKmwpMTS22trbAYf6LK_Screen%2520Shot%25202014-08-06%2520at%25201.54.03%2520AM.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of solution, both of which have good units.</p>
<p>Let's rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a><mathjax># = L solutionxxMolarity#</mathjax></p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (10.0 L) (2.0M) = 20. moles</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #Na_2CO_3# are there in 10.0 L of 2.0 M solution?</h1>
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<div class="markdown"><p>There are <strong>20. mol</strong> of <mathjax>#Na_2CO_3#</mathjax> in 10.0L of 2.0M solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by this equation:</p>
<p><img alt="www.clutchprep.com" src="https://useruploads.socratic.org/risoIJ8wTrmcOMzHhMeR_vpKmwpMTS22trbAYf6LK_Screen%2520Shot%25202014-08-06%2520at%25201.54.03%2520AM.png"/> </p>
<p>In our case, we already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of solution, both of which have good units.</p>
<p>Let's rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a><mathjax># = L solutionxxMolarity#</mathjax></p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (10.0 L) (2.0M) = 20. moles</p></div>
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</article> | How many moles of #Na_2CO_3# are there in 10.0 L of 2.0 M solution? | null |
1,144 | a8b703fe-6ddd-11ea-9434-ccda262736ce | https://socratic.org/questions/557e128c581e2a2c4ddb0a43 | 0.03 M | start physical_unit 5 7 concentration mol/l qc_end physical_unit 5 7 9 10 volume qc_end physical_unit 19 21 14 15 volume qc_end physical_unit 19 21 17 18 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydrochloric acid solution [IN] M"}] | [{"type":"physical unit","value":"0.03 M"}] | [{"type":"physical unit","value":"Volume [OF] hydrochloric acid solution [=] \\pu{6.68 ml}"},{"type":"physical unit","value":"Volume [OF] potassium hydroxide solution [=] \\pu{13.90 ml}"},{"type":"physical unit","value":"Concentration [OF] potassium hydroxide solution [=] \\pu{0.0161 M}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of hydrochloric acid solution if 6.68 ml is neutralised by 13.90 ml of 0.0161 M potassium hydroxide solution ?</h1> | null | 0.03 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> in which hydrochloric acid, a strong acid, reacts with potassium hydroxide, a strong base, to produce potassium chloride, a salt, and water. </p>
<p>The balanced chemical equation looks like this </p>
<p><mathjax>#HCl_((aq)) + KOH_((aq)) -> KCl_((aq)) + H_2O_((l))#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between hydrochloric acid and potassium hydroxide. This means that, in order to get a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>, you need <strong>equal numbers of moles</strong> of acid and of base. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per <strong>liters</strong> of solution, you can use the volume and molarity of the potassium hydroxide solution to determine how many moles of base were needed</p>
<p><mathjax>#C = n/V => n = C * V#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.0161 M" * 13.90 * 10^(-3)"L" = "0.0002238 moles"#</mathjax> <mathjax>#KOH#</mathjax></p>
<p>This means that your hydrochloric acid solution contained </p>
<p><mathjax>#n_(HCl) = n_(KOH) = "0.0002238 moles"#</mathjax> <mathjax>#HCl#</mathjax></p>
<p>Now use the volume of the hydrochloric solution and the number of moles of <mathjax>#HCl#</mathjax> it contains to determine its molarity</p>
<p><mathjax>#C_(HCl) = n_(HCl)/V_(HCl) = "0.0002238 moles"/(6.68 * 10^(-3)"L")= color(green)("0.0335 M")#</mathjax></p>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the molarity of the potassium hydroxide solution. </p></div>
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid solution was <strong>0.0335 M</strong>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> in which hydrochloric acid, a strong acid, reacts with potassium hydroxide, a strong base, to produce potassium chloride, a salt, and water. </p>
<p>The balanced chemical equation looks like this </p>
<p><mathjax>#HCl_((aq)) + KOH_((aq)) -> KCl_((aq)) + H_2O_((l))#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between hydrochloric acid and potassium hydroxide. This means that, in order to get a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>, you need <strong>equal numbers of moles</strong> of acid and of base. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per <strong>liters</strong> of solution, you can use the volume and molarity of the potassium hydroxide solution to determine how many moles of base were needed</p>
<p><mathjax>#C = n/V => n = C * V#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.0161 M" * 13.90 * 10^(-3)"L" = "0.0002238 moles"#</mathjax> <mathjax>#KOH#</mathjax></p>
<p>This means that your hydrochloric acid solution contained </p>
<p><mathjax>#n_(HCl) = n_(KOH) = "0.0002238 moles"#</mathjax> <mathjax>#HCl#</mathjax></p>
<p>Now use the volume of the hydrochloric solution and the number of moles of <mathjax>#HCl#</mathjax> it contains to determine its molarity</p>
<p><mathjax>#C_(HCl) = n_(HCl)/V_(HCl) = "0.0002238 moles"/(6.68 * 10^(-3)"L")= color(green)("0.0335 M")#</mathjax></p>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the molarity of the potassium hydroxide solution. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of hydrochloric acid solution if 6.68 ml is neutralised by 13.90 ml of 0.0161 M potassium hydroxide solution ?</h1>
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the hydrochloric acid solution was <strong>0.0335 M</strong>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a> in which hydrochloric acid, a strong acid, reacts with potassium hydroxide, a strong base, to produce potassium chloride, a salt, and water. </p>
<p>The balanced chemical equation looks like this </p>
<p><mathjax>#HCl_((aq)) + KOH_((aq)) -> KCl_((aq)) + H_2O_((l))#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between hydrochloric acid and potassium hydroxide. This means that, in order to get a <em>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>, you need <strong>equal numbers of moles</strong> of acid and of base. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per <strong>liters</strong> of solution, you can use the volume and molarity of the potassium hydroxide solution to determine how many moles of base were needed</p>
<p><mathjax>#C = n/V => n = C * V#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.0161 M" * 13.90 * 10^(-3)"L" = "0.0002238 moles"#</mathjax> <mathjax>#KOH#</mathjax></p>
<p>This means that your hydrochloric acid solution contained </p>
<p><mathjax>#n_(HCl) = n_(KOH) = "0.0002238 moles"#</mathjax> <mathjax>#HCl#</mathjax></p>
<p>Now use the volume of the hydrochloric solution and the number of moles of <mathjax>#HCl#</mathjax> it contains to determine its molarity</p>
<p><mathjax>#C_(HCl) = n_(HCl)/V_(HCl) = "0.0002238 moles"/(6.68 * 10^(-3)"L")= color(green)("0.0335 M")#</mathjax></p>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the molarity of the potassium hydroxide solution. </p></div>
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<div class="markdown"><p>The concentration of HCl = <mathjax>#0.033"mol/l"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#HCl_((aq))+KOH_((aq))rarrKCl_((aq))+H_2O_((l))#</mathjax></p>
<p><mathjax>#c=n/v#</mathjax></p>
<p><mathjax>#n=cv#</mathjax></p>
<p>So no. moles KOH = <mathjax>#(13.90xx0.0161)/(1000)=2.23xx10^(-4)#</mathjax></p>
<p>From the equation we can see that they react in a molar ratio of 1:1.</p>
<p>This means that the no. moles HCl must be the same:</p>
<p><mathjax>#nHCl=2.23xx10^(-4)#</mathjax></p>
<p>Since <mathjax>#c=n/v#</mathjax>:</p>
<p><mathjax>#[HCl]= (2.23xx10^(-4))/(6.68xx10^(-3))=0.033"mol/l"#</mathjax></p></div>
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</article> | What is the concentration of hydrochloric acid solution if 6.68 ml is neutralised by 13.90 ml of 0.0161 M potassium hydroxide solution ? | null |
1,145 | a8431c52-6ddd-11ea-9c41-ccda262736ce | https://socratic.org/questions/a-gas-occupies-a-volume-of-0-2-l-at-25-kpa-what-volume-will-the-gas-occupy-at-2- | 2 L | start physical_unit 14 15 volume l qc_end physical_unit 14 15 6 7 volume qc_end physical_unit 14 15 9 10 pressure qc_end physical_unit 14 15 18 19 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"2 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.2 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{25 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{2.5 kPa}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies a volume of 0.2 L at 25 kPa. What volume will the gas occupy at 2.5 kPa?</h1> | null | 2 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, assuming, of course, that the temperature and the number of moles of gas <strong>remain unchanged</strong>, you can say that the volume of the gas will <strong>increase</strong> as pressure changes from <mathjax>#"25 kPa"#</mathjax> to <mathjax>#"2.5 kPa"#</mathjax>.</p>
<p>That is the case because when the temperature and the number of moles of gas remain constant, the pressure of a gas varies <strong>indirectly</strong> with its volume, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="http://javiergaslaws.weebly.com/boyles-law.html" src="https://useruploads.socratic.org/LCbn8Q9hSwGeD1oz9yem_9790896_orig.jpeg"/> </p>
<p>In other words, when pressure <strong>decreases</strong> by factor <mathjax>#k#</mathjax>, volume <strong>increases</strong> by the same factor <mathjax>#k#</mathjax>.</p>
<blockquote>
<p><mathjax>#P * V = k#</mathjax></p>
</blockquote>
<p>This implies that you can write </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (25 color(red)(cancel(color(black)("kPa"))))/(2.5color(red)(cancel(color(black)("kPa")))) * "0.2 L" = color(darkgreen)(ul(color(black)("2 L")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the initial volume of the gas. </p>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. Moreover, it increased by <em>the same factor</em>!</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, assuming, of course, that the temperature and the number of moles of gas <strong>remain unchanged</strong>, you can say that the volume of the gas will <strong>increase</strong> as pressure changes from <mathjax>#"25 kPa"#</mathjax> to <mathjax>#"2.5 kPa"#</mathjax>.</p>
<p>That is the case because when the temperature and the number of moles of gas remain constant, the pressure of a gas varies <strong>indirectly</strong> with its volume, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="http://javiergaslaws.weebly.com/boyles-law.html" src="https://useruploads.socratic.org/LCbn8Q9hSwGeD1oz9yem_9790896_orig.jpeg"/> </p>
<p>In other words, when pressure <strong>decreases</strong> by factor <mathjax>#k#</mathjax>, volume <strong>increases</strong> by the same factor <mathjax>#k#</mathjax>.</p>
<blockquote>
<p><mathjax>#P * V = k#</mathjax></p>
</blockquote>
<p>This implies that you can write </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (25 color(red)(cancel(color(black)("kPa"))))/(2.5color(red)(cancel(color(black)("kPa")))) * "0.2 L" = color(darkgreen)(ul(color(black)("2 L")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the initial volume of the gas. </p>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. Moreover, it increased by <em>the same factor</em>!</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A gas occupies a volume of 0.2 L at 25 kPa. What volume will the gas occupy at 2.5 kPa?</h1>
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Stefan V.
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Mar 15, 2017
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<div class="markdown"><p><mathjax>#"2 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, assuming, of course, that the temperature and the number of moles of gas <strong>remain unchanged</strong>, you can say that the volume of the gas will <strong>increase</strong> as pressure changes from <mathjax>#"25 kPa"#</mathjax> to <mathjax>#"2.5 kPa"#</mathjax>.</p>
<p>That is the case because when the temperature and the number of moles of gas remain constant, the pressure of a gas varies <strong>indirectly</strong> with its volume, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p><img alt="http://javiergaslaws.weebly.com/boyles-law.html" src="https://useruploads.socratic.org/LCbn8Q9hSwGeD1oz9yem_9790896_orig.jpeg"/> </p>
<p>In other words, when pressure <strong>decreases</strong> by factor <mathjax>#k#</mathjax>, volume <strong>increases</strong> by the same factor <mathjax>#k#</mathjax>.</p>
<blockquote>
<p><mathjax>#P * V = k#</mathjax></p>
</blockquote>
<p>This implies that you can write </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_1V_1 = P_2V_2)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> represent the pressure and volume of the gas at an initial state</li>
<li><mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> represent the pressure and volume of the gas at a final state</li>
</ul>
</blockquote>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (25 color(red)(cancel(color(black)("kPa"))))/(2.5color(red)(cancel(color(black)("kPa")))) * "0.2 L" = color(darkgreen)(ul(color(black)("2 L")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> you have for the initial volume of the gas. </p>
<p>As predicted, the volume of the gas <strong>increased</strong> as a result of the <em>decrease</em> in pressure. Moreover, it increased by <em>the same factor</em>!</p></div>
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</article> | A gas occupies a volume of 0.2 L at 25 kPa. What volume will the gas occupy at 2.5 kPa? | null |
1,146 | abe2d02f-6ddd-11ea-9663-ccda262736ce | https://socratic.org/questions/a-sample-of-sulfur-having-a-mass-of-1-28-g-combines-with-oxygen-to-form-a-compou | SO3 | start chemical_formula qc_end physical_unit 1 3 8 9 mass qc_end substance 12 12 qc_end physical_unit 29 30 21 22 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"SO3"}] | [{"type":"physical unit","value":"Mass [OF] sulfur sample [=] \\pu{1.28 g}"},{"type":"substance name","value":"Oxygen"},{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{3.20 g}"}] | <h1 class="questionTitle" itemprop="name">A sample of sulfur having a mass of 1.28 g combines with oxygen to form a compound with a mass of 3.20g. What is the empirical formula of the compound?</h1> | null | SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your compound will only contain <em>sulfur</em> and <em>oxygen</em>, you can conclude that the difference between the <em>mass of sulfur</em> and the mass of the <em>final compound</em> will represent the <strong>mass of oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"compound" = m_"oxygen" + m_"sulfur"#</mathjax></p>
</blockquote>
<p>In your case, the mass of oxygen will be equal to </p>
<blockquote>
<p><mathjax>#m_"oxygen" = "3.20 g" - "1.28 g" = "1.92 g"#</mathjax></p>
</blockquote>
<p>So, your compound contains </p>
<blockquote>
<ul>
<li><mathjax>#"1.28 g"#</mathjax> <em>of sulfur</em></li>
<li><mathjax>#"1.92 g"#</mathjax> <em>of oxygen</em></li>
</ul>
</blockquote>
<p>Your next step will be to use the <strong>molar masses</strong> of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you have in this <mathjax>#"3.-20 g"#</mathjax> sample of compound. </p>
<blockquote>
<p><mathjax>#"For O: " 1.92 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.1200 moles O"#</mathjax></p>
<p><mathjax>#"For S: " 1.28 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.03992 moles S"#</mathjax></p>
</blockquote>
<p>In order to determine the <a href="https://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of the compound, you need to find the <strong>smallest whole number ratio</strong> that exists between these two elements in the compound. </p>
<p>In order to do that, divide both values by <em>the smallest one</em> to get </p>
<blockquote>
<p><mathjax>#"For O: " (0.1200 color(red)(cancel(color(black)("moles"))))/(0.03992color(red)(cancel(color(black)("moles")))) = 3.01 ~~ 3#</mathjax></p>
<p><mathjax>#"For S: " (0.03992color(red)(cancel(color(black)("moles"))))/(0.03992 color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:3#</mathjax> mole ratio is the smallest possible while using whole numbers, the empirical formula of the compound will be </p>
<blockquote>
<p><mathjax>#"S"_1"O"_3 implies color(green)("SO"_3)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"SO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your compound will only contain <em>sulfur</em> and <em>oxygen</em>, you can conclude that the difference between the <em>mass of sulfur</em> and the mass of the <em>final compound</em> will represent the <strong>mass of oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"compound" = m_"oxygen" + m_"sulfur"#</mathjax></p>
</blockquote>
<p>In your case, the mass of oxygen will be equal to </p>
<blockquote>
<p><mathjax>#m_"oxygen" = "3.20 g" - "1.28 g" = "1.92 g"#</mathjax></p>
</blockquote>
<p>So, your compound contains </p>
<blockquote>
<ul>
<li><mathjax>#"1.28 g"#</mathjax> <em>of sulfur</em></li>
<li><mathjax>#"1.92 g"#</mathjax> <em>of oxygen</em></li>
</ul>
</blockquote>
<p>Your next step will be to use the <strong>molar masses</strong> of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you have in this <mathjax>#"3.-20 g"#</mathjax> sample of compound. </p>
<blockquote>
<p><mathjax>#"For O: " 1.92 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.1200 moles O"#</mathjax></p>
<p><mathjax>#"For S: " 1.28 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.03992 moles S"#</mathjax></p>
</blockquote>
<p>In order to determine the <a href="https://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of the compound, you need to find the <strong>smallest whole number ratio</strong> that exists between these two elements in the compound. </p>
<p>In order to do that, divide both values by <em>the smallest one</em> to get </p>
<blockquote>
<p><mathjax>#"For O: " (0.1200 color(red)(cancel(color(black)("moles"))))/(0.03992color(red)(cancel(color(black)("moles")))) = 3.01 ~~ 3#</mathjax></p>
<p><mathjax>#"For S: " (0.03992color(red)(cancel(color(black)("moles"))))/(0.03992 color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:3#</mathjax> mole ratio is the smallest possible while using whole numbers, the empirical formula of the compound will be </p>
<blockquote>
<p><mathjax>#"S"_1"O"_3 implies color(green)("SO"_3)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of sulfur having a mass of 1.28 g combines with oxygen to form a compound with a mass of 3.20g. What is the empirical formula of the compound?</h1>
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<div class="answer" id="205433" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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Stefan V.
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<span class="dateCreated" datetime="2015-12-29T23:41:39" itemprop="dateCreated">
Dec 29, 2015
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<div class="markdown"><p><mathjax>#"SO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since your compound will only contain <em>sulfur</em> and <em>oxygen</em>, you can conclude that the difference between the <em>mass of sulfur</em> and the mass of the <em>final compound</em> will represent the <strong>mass of oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"compound" = m_"oxygen" + m_"sulfur"#</mathjax></p>
</blockquote>
<p>In your case, the mass of oxygen will be equal to </p>
<blockquote>
<p><mathjax>#m_"oxygen" = "3.20 g" - "1.28 g" = "1.92 g"#</mathjax></p>
</blockquote>
<p>So, your compound contains </p>
<blockquote>
<ul>
<li><mathjax>#"1.28 g"#</mathjax> <em>of sulfur</em></li>
<li><mathjax>#"1.92 g"#</mathjax> <em>of oxygen</em></li>
</ul>
</blockquote>
<p>Your next step will be to use the <strong>molar masses</strong> of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you have in this <mathjax>#"3.-20 g"#</mathjax> sample of compound. </p>
<blockquote>
<p><mathjax>#"For O: " 1.92 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.1200 moles O"#</mathjax></p>
<p><mathjax>#"For S: " 1.28 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.03992 moles S"#</mathjax></p>
</blockquote>
<p>In order to determine the <a href="https://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of the compound, you need to find the <strong>smallest whole number ratio</strong> that exists between these two elements in the compound. </p>
<p>In order to do that, divide both values by <em>the smallest one</em> to get </p>
<blockquote>
<p><mathjax>#"For O: " (0.1200 color(red)(cancel(color(black)("moles"))))/(0.03992color(red)(cancel(color(black)("moles")))) = 3.01 ~~ 3#</mathjax></p>
<p><mathjax>#"For S: " (0.03992color(red)(cancel(color(black)("moles"))))/(0.03992 color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:3#</mathjax> mole ratio is the smallest possible while using whole numbers, the empirical formula of the compound will be </p>
<blockquote>
<p><mathjax>#"S"_1"O"_3 implies color(green)("SO"_3)#</mathjax></p>
</blockquote></div>
</div>
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</article> | A sample of sulfur having a mass of 1.28 g combines with oxygen to form a compound with a mass of 3.20g. What is the empirical formula of the compound? | null |
1,147 | ac1dd50b-6ddd-11ea-b31d-ccda262736ce | https://socratic.org/questions/a-100g-ice-cube-at-0-c-is-placed-in-650g-of-water-at-25-c-what-is-the-final-temp | 11.03 ℃ | start physical_unit 25 26 temperature °c qc_end physical_unit 3 4 6 7 temperature qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 3 4 1 2 mass qc_end physical_unit 14 14 11 12 mass qc_end end | [{"type":"physical unit","value":"Temperature [OF] the mixture [IN] ℃"}] | [{"type":"physical unit","value":"11.03 ℃"}] | [{"type":"physical unit","value":"Temperature [OF] ice cube [=] \\pu{0.0 ℃}"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Mass [OF] ice cube [=] \\pu{100.0 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{650 g}"}] | <h1 class="questionTitle" itemprop="name">A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture?</h1> | null | 11.03 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As far as solving this problem goes, it is <strong>very important</strong> that you <strong>do not</strong> forget to account for the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a> underwent by the <em>solid</em> water at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The heat needed to <em>melt</em> the solid at its melting point will come from the <em>warmer</em> water sample. This means that you have </p>
<blockquote>
<p><mathjax>#q_1 + q_2 = - q_3" " " "color(purple)((1))#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q_1#</mathjax> - the heat absorbed by the solid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_2#</mathjax> - the heat absorbed by the liquid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_3#</mathjax> - the heat lost by the warmer water sample</p>
<p>The two equations that you will use are</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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>and</p>
<blockquote>
<p><mathjax>#color(blue)(q = n * DeltaH_"fus")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#n#</mathjax> - the <em>number of moles</em> of water<br/>
<mathjax>#DeltaH_"fus"#</mathjax> - the <em>molar heat of fusion</em> of water, equal to <mathjax>#"6.01 kJ/mol"#</mathjax></p>
<p>Use water's molar mass to find how many <em>moles</em> of water you have in the <mathjax>#"100.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, how much heat is <strong>needed</strong> to allow the sample to go from <em>solid</em> at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>?</p>
<blockquote>
<p><mathjax>#q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"#</mathjax></p>
</blockquote>
<p>This means that equation <mathjax>#color(purple)((1))#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + q_2 = -q_3#</mathjax></p>
</blockquote>
<p>The minus sign for <mathjax>#q_3#</mathjax> is used because <strong>heat lost</strong> carries a negative sign. </p>
<p>So, if <mathjax>#T_"f"#</mathjax> is the final temperature of the water, you can say that</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>More specifically, you have</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>Convert the <em>joules</em> to <em>kilojoules</em> to get</p>
<blockquote>
<p><mathjax>#33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#0.418 * T_"f" + 2.717 * T_"f" = 67.925 - 33.36#</mathjax></p>
<p><mathjax>#T_"f" = 34.565/(0.418 + 2.717) = 11.026^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of warmer water, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)(11^@"C")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#11^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As far as solving this problem goes, it is <strong>very important</strong> that you <strong>do not</strong> forget to account for the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a> underwent by the <em>solid</em> water at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The heat needed to <em>melt</em> the solid at its melting point will come from the <em>warmer</em> water sample. This means that you have </p>
<blockquote>
<p><mathjax>#q_1 + q_2 = - q_3" " " "color(purple)((1))#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q_1#</mathjax> - the heat absorbed by the solid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_2#</mathjax> - the heat absorbed by the liquid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_3#</mathjax> - the heat lost by the warmer water sample</p>
<p>The two equations that you will use are</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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>and</p>
<blockquote>
<p><mathjax>#color(blue)(q = n * DeltaH_"fus")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#n#</mathjax> - the <em>number of moles</em> of water<br/>
<mathjax>#DeltaH_"fus"#</mathjax> - the <em>molar heat of fusion</em> of water, equal to <mathjax>#"6.01 kJ/mol"#</mathjax></p>
<p>Use water's molar mass to find how many <em>moles</em> of water you have in the <mathjax>#"100.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, how much heat is <strong>needed</strong> to allow the sample to go from <em>solid</em> at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>?</p>
<blockquote>
<p><mathjax>#q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"#</mathjax></p>
</blockquote>
<p>This means that equation <mathjax>#color(purple)((1))#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + q_2 = -q_3#</mathjax></p>
</blockquote>
<p>The minus sign for <mathjax>#q_3#</mathjax> is used because <strong>heat lost</strong> carries a negative sign. </p>
<p>So, if <mathjax>#T_"f"#</mathjax> is the final temperature of the water, you can say that</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>More specifically, you have</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>Convert the <em>joules</em> to <em>kilojoules</em> to get</p>
<blockquote>
<p><mathjax>#33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#0.418 * T_"f" + 2.717 * T_"f" = 67.925 - 33.36#</mathjax></p>
<p><mathjax>#T_"f" = 34.565/(0.418 + 2.717) = 11.026^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of warmer water, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)(11^@"C")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-26T12:49:00" itemprop="dateCreated">
Nov 26, 2015
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<div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#11^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As far as solving this problem goes, it is <strong>very important</strong> that you <strong>do not</strong> forget to account for the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/introduction-to-phase-changes">phase change</a> underwent by the <em>solid</em> water at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>. </p>
<p>The heat needed to <em>melt</em> the solid at its melting point will come from the <em>warmer</em> water sample. This means that you have </p>
<blockquote>
<p><mathjax>#q_1 + q_2 = - q_3" " " "color(purple)((1))#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q_1#</mathjax> - the heat absorbed by the solid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_2#</mathjax> - the heat absorbed by the liquid at <mathjax>#0^@"C"#</mathjax><br/>
<mathjax>#q_3#</mathjax> - the heat lost by the warmer water sample</p>
<p>The two equations that you will use are</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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>and</p>
<blockquote>
<p><mathjax>#color(blue)(q = n * DeltaH_"fus")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed<br/>
<mathjax>#n#</mathjax> - the <em>number of moles</em> of water<br/>
<mathjax>#DeltaH_"fus"#</mathjax> - the <em>molar heat of fusion</em> of water, equal to <mathjax>#"6.01 kJ/mol"#</mathjax></p>
<p>Use water's molar mass to find how many <em>moles</em> of water you have in the <mathjax>#"100.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>So, how much heat is <strong>needed</strong> to allow the sample to go from <em>solid</em> at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#0^@"C"#</mathjax>?</p>
<blockquote>
<p><mathjax>#q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"#</mathjax></p>
</blockquote>
<p>This means that equation <mathjax>#color(purple)((1))#</mathjax> becomes</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + q_2 = -q_3#</mathjax></p>
</blockquote>
<p>The minus sign for <mathjax>#q_3#</mathjax> is used because <strong>heat lost</strong> carries a negative sign. </p>
<p>So, if <mathjax>#T_"f"#</mathjax> is the final temperature of the water, you can say that</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>More specifically, you have</p>
<blockquote>
<p><mathjax>#"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>Convert the <em>joules</em> to <em>kilojoules</em> to get</p>
<blockquote>
<p><mathjax>#33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#0.418 * T_"f" + 2.717 * T_"f" = 67.925 - 33.36#</mathjax></p>
<p><mathjax>#T_"f" = 34.565/(0.418 + 2.717) = 11.026^@"C"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of warmer water, the answer will be </p>
<blockquote>
<p><mathjax>#T_"f" = color(green)(11^@"C")#</mathjax></p>
</blockquote></div>
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</article> | A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture? | null |
1,148 | a9bcc2c6-6ddd-11ea-833a-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-form-for-cobalt-ii-hydrogen-carbonate | Co(HCO3)2 | start chemical_formula qc_end substance 6 9 qc_end end | [{"type":"other","value":"Chemical Formula [OF] cobalt (II) hydrogen carbonate [IN] default"}] | [{"type":"chemical equation","value":"Co(HCO3)2"}] | [{"type":"substance name","value":"Cobalt (II) hydrogen carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for cobalt (II) hydrogen carbonate?</h1> | null | Co(HCO3)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The name tells you that this is an ionic compound, made of cobalt(II) ions and hydrogen carbonate ions.</p>
<p>The symbol for the cobalt(II) ion is <mathjax>#"Co"^"2+"#</mathjax>.</p>
<blockquote></blockquote>
<p>The symbol for the hydrogen carbonate ion is <mathjax>#"HCO"_3^"-"#</mathjax> (you have memorized the formulas of the <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>, haven't you?).</p>
<p>So, the formula contains <mathjax>#"Co"^"2+", "HCO"_3^"-"#</mathjax>.</p>
<blockquote></blockquote>
<p>We need two "-" charges to balance the "2+" charges on the <mathjax>#"Co"#</mathjax>.</p>
<p>Thus, we need two hydrogen carbonate ions, so we enclose the formula in parentheses followed by a subscript 2.</p>
<p>The formula is <mathjax>#"Co"("HCO"_3)_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The chemical formula for cobalt(II) hydrogen carbonate is <mathjax>#"Co"("HCO"_3)_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The name tells you that this is an ionic compound, made of cobalt(II) ions and hydrogen carbonate ions.</p>
<p>The symbol for the cobalt(II) ion is <mathjax>#"Co"^"2+"#</mathjax>.</p>
<blockquote></blockquote>
<p>The symbol for the hydrogen carbonate ion is <mathjax>#"HCO"_3^"-"#</mathjax> (you have memorized the formulas of the <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>, haven't you?).</p>
<p>So, the formula contains <mathjax>#"Co"^"2+", "HCO"_3^"-"#</mathjax>.</p>
<blockquote></blockquote>
<p>We need two "-" charges to balance the "2+" charges on the <mathjax>#"Co"#</mathjax>.</p>
<p>Thus, we need two hydrogen carbonate ions, so we enclose the formula in parentheses followed by a subscript 2.</p>
<p>The formula is <mathjax>#"Co"("HCO"_3)_2#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the chemical formula for cobalt (II) hydrogen carbonate?</h1>
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Ernest Z.
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Feb 10, 2017
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<div class="markdown"><p>The chemical formula for cobalt(II) hydrogen carbonate is <mathjax>#"Co"("HCO"_3)_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The name tells you that this is an ionic compound, made of cobalt(II) ions and hydrogen carbonate ions.</p>
<p>The symbol for the cobalt(II) ion is <mathjax>#"Co"^"2+"#</mathjax>.</p>
<blockquote></blockquote>
<p>The symbol for the hydrogen carbonate ion is <mathjax>#"HCO"_3^"-"#</mathjax> (you have memorized the formulas of the <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>, haven't you?).</p>
<p>So, the formula contains <mathjax>#"Co"^"2+", "HCO"_3^"-"#</mathjax>.</p>
<blockquote></blockquote>
<p>We need two "-" charges to balance the "2+" charges on the <mathjax>#"Co"#</mathjax>.</p>
<p>Thus, we need two hydrogen carbonate ions, so we enclose the formula in parentheses followed by a subscript 2.</p>
<p>The formula is <mathjax>#"Co"("HCO"_3)_2#</mathjax></p></div>
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</article> | What is the chemical formula for cobalt (II) hydrogen carbonate? | null |
1,149 | a847160c-6ddd-11ea-b15a-ccda262736ce | https://socratic.org/questions/a-gaseous-compound-composed-of-sulfur-and-oxygen-which-is-linked-to-the-formatio | 81.31 g/mol | start physical_unit 31 32 molar_mass g/mol qc_end physical_unit 1 2 21 22 density qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molar mass [OF] this gas [IN] g/mol"}] | [{"type":"physical unit","value":"81.31 g/mol"}] | [{"type":"physical unit","value":"Density [OF] the gaseous compound [=] \\pu{3.58 g/L}"},{"type":"other","value":"STP"},{"type":"other","value":"A gaseous compound composed of sulfur and oxygen."}] | <h1 class="questionTitle" itemprop="name">A gaseous compound composed of sulfur and oxygen, which is linked to the formation of acid rain, has a density of 3.58 g/L at STP. What Is the molar mass of this gas? </h1> | null | 81.31 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be pick a sample of this gas and use the definition of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to help you find the number of moles it contains. </p>
<p>To make the calculations easier, let's say that we're going to pick a <mathjax>#"1.00-L"#</mathjax> sample of this gas. </p>
<p>As you know, <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.71 L"#</mathjax> under STP conditions, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, if <em>one mole</em> of this gas will occupy <mathjax>#"22.71 L"#</mathjax> at STP, it follows that our <mathjax>#"1.00-L"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L"))) * "1 mole"/(22.71 color(red)(cancel(color(black)("L")))) = "0.04403 moles"#</mathjax></p>
</blockquote>
<p>According to the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, this <mathjax>#"1.00-L"#</mathjax> sample will contain <mathjax>#3.58#</mathjax> grams of this unknown gas. As you know, molar mass is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass" = "mass in grams"/"number of moles")#</mathjax></p>
</blockquote>
<p>This means that the gas' molar mass will be </p>
<blockquote>
<p><mathjax>#M_"M" = "3.58 g"/"0.04403 moles" = color(green)("81.3 g/mol")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>It is very likely that this problem meant for you to use the old definition of STP, which is a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</em></p>
<p><em>In this case, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP is equal to 22.4 L. This in turn will make the molar mass of the gas equal to <mathjax>#"80.2 g/mol"#</mathjax>.</em></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"81.3 g/mol "#</mathjax> (or possibly <mathjax>#"80.2 g/mol"#</mathjax>)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be pick a sample of this gas and use the definition of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to help you find the number of moles it contains. </p>
<p>To make the calculations easier, let's say that we're going to pick a <mathjax>#"1.00-L"#</mathjax> sample of this gas. </p>
<p>As you know, <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.71 L"#</mathjax> under STP conditions, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, if <em>one mole</em> of this gas will occupy <mathjax>#"22.71 L"#</mathjax> at STP, it follows that our <mathjax>#"1.00-L"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L"))) * "1 mole"/(22.71 color(red)(cancel(color(black)("L")))) = "0.04403 moles"#</mathjax></p>
</blockquote>
<p>According to the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, this <mathjax>#"1.00-L"#</mathjax> sample will contain <mathjax>#3.58#</mathjax> grams of this unknown gas. As you know, molar mass is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass" = "mass in grams"/"number of moles")#</mathjax></p>
</blockquote>
<p>This means that the gas' molar mass will be </p>
<blockquote>
<p><mathjax>#M_"M" = "3.58 g"/"0.04403 moles" = color(green)("81.3 g/mol")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>It is very likely that this problem meant for you to use the old definition of STP, which is a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</em></p>
<p><em>In this case, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP is equal to 22.4 L. This in turn will make the molar mass of the gas equal to <mathjax>#"80.2 g/mol"#</mathjax>.</em></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A gaseous compound composed of sulfur and oxygen, which is linked to the formation of acid rain, has a density of 3.58 g/L at STP. What Is the molar mass of this gas? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"81.3 g/mol "#</mathjax> (or possibly <mathjax>#"80.2 g/mol"#</mathjax>)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be pick a sample of this gas and use the definition of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a> to help you find the number of moles it contains. </p>
<p>To make the calculations easier, let's say that we're going to pick a <mathjax>#"1.00-L"#</mathjax> sample of this gas. </p>
<p>As you know, <strong>one mole</strong> of any ideal gas occupies exactly <mathjax>#"22.71 L"#</mathjax> under STP conditions, which are defined as a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, if <em>one mole</em> of this gas will occupy <mathjax>#"22.71 L"#</mathjax> at STP, it follows that our <mathjax>#"1.00-L"#</mathjax> sample will contain </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L"))) * "1 mole"/(22.71 color(red)(cancel(color(black)("L")))) = "0.04403 moles"#</mathjax></p>
</blockquote>
<p>According to the given <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a>, this <mathjax>#"1.00-L"#</mathjax> sample will contain <mathjax>#3.58#</mathjax> grams of this unknown gas. As you know, molar mass is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass" = "mass in grams"/"number of moles")#</mathjax></p>
</blockquote>
<p>This means that the gas' molar mass will be </p>
<blockquote>
<p><mathjax>#M_"M" = "3.58 g"/"0.04403 moles" = color(green)("81.3 g/mol")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>It is very likely that this problem meant for you to use the old definition of STP, which is a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</em></p>
<p><em>In this case, the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP is equal to 22.4 L. This in turn will make the molar mass of the gas equal to <mathjax>#"80.2 g/mol"#</mathjax>.</em></p></div>
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</article> | A gaseous compound composed of sulfur and oxygen, which is linked to the formation of acid rain, has a density of 3.58 g/L at STP. What Is the molar mass of this gas? | null |
1,150 | ac101375-6ddd-11ea-9287-ccda262736ce | https://socratic.org/questions/aluminum-oxide-is-decomposed-to-extract-pure-aluminum-from-the-mineral-if-6-7-to | 3.62 tons | start physical_unit 6 7 mass t qc_end physical_unit 0 1 12 13 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] pure aluminum [IN] tons"}] | [{"type":"physical unit","value":"3.62 tons"}] | [{"type":"physical unit","value":"Mass [OF] aluminum oxide [=] \\pu{6.7 tons}"},{"type":"other","value":"Aluminum oxide is decomposed to extract pure aluminum from the mineral."}] | <h1 class="questionTitle" itemprop="name">Aluminum oxide is decomposed to extract pure aluminum from the mineral. If 6.7 tons of aluminum oxide is mined, calculate the mass of pure aluminum produced?</h1> | null | 3.62 tons | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms of aluminum (in pure quality). The ratio is <mathjax>#=54/102=0.529#</mathjax></p>
<p>For 6.7 tons it means <mathjax>#0.529*6.7=3.544#</mathjax> tons or <mathjax>#3544#</mathjax> kg of aluminum you can get. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Al2O3 when only Al is wanted, you get 3.618 tons.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms of aluminum (in pure quality). The ratio is <mathjax>#=54/102=0.529#</mathjax></p>
<p>For 6.7 tons it means <mathjax>#0.529*6.7=3.544#</mathjax> tons or <mathjax>#3544#</mathjax> kg of aluminum you can get. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Aluminum oxide is decomposed to extract pure aluminum from the mineral. If 6.7 tons of aluminum oxide is mined, calculate the mass of pure aluminum produced?</h1>
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<div class="markdown"><p>Al2O3 when only Al is wanted, you get 3.618 tons.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms of aluminum (in pure quality). The ratio is <mathjax>#=54/102=0.529#</mathjax></p>
<p>For 6.7 tons it means <mathjax>#0.529*6.7=3.544#</mathjax> tons or <mathjax>#3544#</mathjax> kg of aluminum you can get. </p></div>
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</article> | Aluminum oxide is decomposed to extract pure aluminum from the mineral. If 6.7 tons of aluminum oxide is mined, calculate the mass of pure aluminum produced? | null |
1,151 | ac064d82-6ddd-11ea-a83c-ccda262736ce | https://socratic.org/questions/at-standard-temperature-and-pressure-a-given-sample-of-water-vapor-occupies-a-vo | 0.12 moles | start physical_unit 9 10 mole mol qc_end c_other OTHER qc_end physical_unit 9 10 15 16 volume qc_end end | [{"type":"physical unit","value":"Mole [OF] water vapor [IN] moles"}] | [{"type":"physical unit","value":"0.12 moles"}] | [{"type":"other","value":"At standard temperature and pressure."},{"type":"physical unit","value":"Volume [OF] water vapor [=] \\pu{2.80 L}"}] | <h1 class="questionTitle" itemprop="name">At standard temperature and pressure, a given sample of water vapor occupies a volume of 2.8 0 L. How many moles of water vapor are present?</h1> | null | 0.12 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that an ideal gas kept under <strong>Standard Temperature and Pressure</strong> (<strong>STP</strong>) conditions has a <em>molar volume</em> equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In other words, when pressure is set to <mathjax>#"100 kPa"#</mathjax> and temperature to <mathjax>#0^@"C"#</mathjax>, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L"#</mathjax> - this is known as the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>So, if your sample of water vapor occupies <mathjax>#"2.80 L"#</mathjax> at STP, it follows that it must contain </p>
<blockquote>
<p><mathjax>#2.80 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.12335 moles"#</mathjax></p>
</blockquote>
<p>of water vapor. You need to round this off to three <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 volume of water vapor</p>
<blockquote>
<p><mathjax>#"no. of moles of water vapor" = color(green)(|bar(ul(color(white)(a/a)"0.123 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>It's worth noting that many textbooks and online sources still list STP conditions as a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>Under these conditions, one mole of any ideal gas occupies <mathjax>#"22.4 L"#</mathjax>. If this is the value given to you for the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP, simply redo the calculation using <mathjax>#"22.4 L"#</mathjax> instead of <mathjax>#"22.7 L"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.123 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that an ideal gas kept under <strong>Standard Temperature and Pressure</strong> (<strong>STP</strong>) conditions has a <em>molar volume</em> equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In other words, when pressure is set to <mathjax>#"100 kPa"#</mathjax> and temperature to <mathjax>#0^@"C"#</mathjax>, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L"#</mathjax> - this is known as the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>So, if your sample of water vapor occupies <mathjax>#"2.80 L"#</mathjax> at STP, it follows that it must contain </p>
<blockquote>
<p><mathjax>#2.80 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.12335 moles"#</mathjax></p>
</blockquote>
<p>of water vapor. You need to round this off to three <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 volume of water vapor</p>
<blockquote>
<p><mathjax>#"no. of moles of water vapor" = color(green)(|bar(ul(color(white)(a/a)"0.123 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>It's worth noting that many textbooks and online sources still list STP conditions as a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>Under these conditions, one mole of any ideal gas occupies <mathjax>#"22.4 L"#</mathjax>. If this is the value given to you for the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP, simply redo the calculation using <mathjax>#"22.4 L"#</mathjax> instead of <mathjax>#"22.7 L"#</mathjax>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At standard temperature and pressure, a given sample of water vapor occupies a volume of 2.8 0 L. How many moles of water vapor are present?</h1>
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Stefan V.
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Mar 26, 2016
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<div class="markdown"><p><mathjax>#"0.123 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that an ideal gas kept under <strong>Standard Temperature and Pressure</strong> (<strong>STP</strong>) conditions has a <em>molar volume</em> equal to <mathjax>#"22.7 L"#</mathjax>. </p>
<p>In other words, when pressure is set to <mathjax>#"100 kPa"#</mathjax> and temperature to <mathjax>#0^@"C"#</mathjax>, <strong>one mole</strong> of any ideal gas occupies <mathjax>#"22.7 L"#</mathjax> - this is known as the <strong><a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a></strong>. </p>
<p>So, if your sample of water vapor occupies <mathjax>#"2.80 L"#</mathjax> at STP, it follows that it must contain </p>
<blockquote>
<p><mathjax>#2.80 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.12335 moles"#</mathjax></p>
</blockquote>
<p>of water vapor. You need to round this off to three <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 volume of water vapor</p>
<blockquote>
<p><mathjax>#"no. of moles of water vapor" = color(green)(|bar(ul(color(white)(a/a)"0.123 moles"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>It's worth noting that many textbooks and online sources still list STP conditions as a pressure of <mathjax>#"1 atm"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>. </p>
<p>Under these conditions, one mole of any ideal gas occupies <mathjax>#"22.4 L"#</mathjax>. If this is the value given to you for the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> at STP, simply redo the calculation using <mathjax>#"22.4 L"#</mathjax> instead of <mathjax>#"22.7 L"#</mathjax>. </p></div>
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</article> | At standard temperature and pressure, a given sample of water vapor occupies a volume of 2.8 0 L. How many moles of water vapor are present? | null |
1,152 | abc8c790-6ddd-11ea-ba43-ccda262736ce | https://socratic.org/questions/a-buffer-solution-is-1-30-m-in-nh-3-and-1-20-m-in-nh-4cl-if-0-120-moles-of-naoh- | 9.31 | start physical_unit 1 2 ph none qc_end physical_unit 7 7 4 5 molarity qc_end physical_unit 12 12 9 10 molarity qc_end physical_unit 17 17 14 15 mole qc_end physical_unit 24 25 21 22 volume qc_end c_other OTHER qc_end physical_unit 7 7 39 41 kb qc_end end | [{"type":"physical unit","value":"pH [OF] buffer solution"}] | [{"type":"physical unit","value":"9.31"}] | [{"type":"physical unit","value":"Molarity [OF] NH3 solution [=] \\pu{1.30 M}"},{"type":"physical unit","value":"Molarity [OF] NH4Cl solution [=] \\pu{1.20 M}"},{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{0.120 moles}"},{"type":"physical unit","value":"Volume [OF] the buffer [=] \\pu{1.00 L}"},{"type":"other","value":"Assume the volume remains constant."},{"type":"physical unit","value":"Kb [OF] NH3 [=] \\pu{1.8 × 10^(-5)}"}] | <h1 class="questionTitle" itemprop="name">A buffer solution is 1.30 M in #NH_3# and 1.20 M in #NH_4Cl#. If 0.120 moles of #NaOH# are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of #NH_3 = 1.8 xx 10^(-5)#.</h1> | null | 9.31 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, and ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, the salt of its <strong>conjugate acid</strong>, the ammonium cation, <mathjax>#"NH"_4^(+)#</mathjax>. </p>
<p><em>Sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, si a <strong>strong base</strong> that dissociates completely in aqueous solution to form sodium cations, which are of no interest here, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>.</p>
<p>The hydroxide anions will react with the ammonium cations to form ammonia and water <mathjax>#->#</mathjax> think <strong><a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> reaction. </p>
<blockquote>
<p><mathjax>#"NH"_text(4(aq])^(+) + "OH"_text((aq])^(-) -> "NH"_text(3(aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Since you're dealing with a <mathjax>#"1.0-L"#</mathjax> solution, you can treat <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>number of moles</em> <strong>interchangeably</strong>. </p>
<p>This means that your initial solution contains <mathjax>#1.30#</mathjax> <strong>moles</strong> of ammonia and <mathjax>#1.20#</mathjax> <strong>moles</strong> of ammonium cation (ammonium chloride dissociates in a <mathjax>#1:1#</mathjax> mole ratio to form ammonium cations and chloride anions). </p>
<p>Now, notice that the ammonium cations and the hydroxide anions react in a <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>. This tells you that the reaction consumes <strong>equal numbers of moles</strong> of each reactant. </p>
<p>Moreover, for every mole of ammonium cations that reacts with one mole of hydroxide anions, <strong>one mole</strong> of ammonia is formed. </p>
<p>Since you're adding <mathjax>#0.120#</mathjax> <strong>moles</strong> of hydroxide buffer solution, you can say that the <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction will <strong>completely consume</strong> the added hydroxide anions and leave you with </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = 0 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(NH_4^(+)) = 1.30 - 0.120 = "1.18 moles NH"_4^(+)#</mathjax></p>
<p><mathjax>#n_(NH_3) = 1.20 + 0.120 = "1.32 moles NH"_3#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is assumed to be constant, you can say that the resulting solution will have</p>
<blockquote>
<p><mathjax>#["NH"_4^(+)] = "1.18 M" " "#</mathjax> and <mathjax>#" " ["NH"_3] = "1.32 M"#</mathjax></p>
</blockquote>
<p>Now, to calculate the pOH of the solution, use the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))|))#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find the pOH of the solution</p>
<blockquote>
<p><mathjax>#"pOH" = - log(1.8 * 10^(-5)) + log( (1.18 color(red)(cancel(color(black)("M"))))/(1.32color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.74 + log(1.18/1.32) = 4.69#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution, use the equation</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|))) ->#</mathjax> <em>true for aqueous <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature</em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 4.69 = color(green)(|bar(ul(color(white)(a/a)9.31color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 9.31#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, and ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, the salt of its <strong>conjugate acid</strong>, the ammonium cation, <mathjax>#"NH"_4^(+)#</mathjax>. </p>
<p><em>Sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, si a <strong>strong base</strong> that dissociates completely in aqueous solution to form sodium cations, which are of no interest here, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>.</p>
<p>The hydroxide anions will react with the ammonium cations to form ammonia and water <mathjax>#->#</mathjax> think <strong><a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> reaction. </p>
<blockquote>
<p><mathjax>#"NH"_text(4(aq])^(+) + "OH"_text((aq])^(-) -> "NH"_text(3(aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Since you're dealing with a <mathjax>#"1.0-L"#</mathjax> solution, you can treat <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>number of moles</em> <strong>interchangeably</strong>. </p>
<p>This means that your initial solution contains <mathjax>#1.30#</mathjax> <strong>moles</strong> of ammonia and <mathjax>#1.20#</mathjax> <strong>moles</strong> of ammonium cation (ammonium chloride dissociates in a <mathjax>#1:1#</mathjax> mole ratio to form ammonium cations and chloride anions). </p>
<p>Now, notice that the ammonium cations and the hydroxide anions react in a <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>. This tells you that the reaction consumes <strong>equal numbers of moles</strong> of each reactant. </p>
<p>Moreover, for every mole of ammonium cations that reacts with one mole of hydroxide anions, <strong>one mole</strong> of ammonia is formed. </p>
<p>Since you're adding <mathjax>#0.120#</mathjax> <strong>moles</strong> of hydroxide buffer solution, you can say that the <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction will <strong>completely consume</strong> the added hydroxide anions and leave you with </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = 0 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(NH_4^(+)) = 1.30 - 0.120 = "1.18 moles NH"_4^(+)#</mathjax></p>
<p><mathjax>#n_(NH_3) = 1.20 + 0.120 = "1.32 moles NH"_3#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is assumed to be constant, you can say that the resulting solution will have</p>
<blockquote>
<p><mathjax>#["NH"_4^(+)] = "1.18 M" " "#</mathjax> and <mathjax>#" " ["NH"_3] = "1.32 M"#</mathjax></p>
</blockquote>
<p>Now, to calculate the pOH of the solution, use the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))|))#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find the pOH of the solution</p>
<blockquote>
<p><mathjax>#"pOH" = - log(1.8 * 10^(-5)) + log( (1.18 color(red)(cancel(color(black)("M"))))/(1.32color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.74 + log(1.18/1.32) = 4.69#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution, use the equation</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|))) ->#</mathjax> <em>true for aqueous <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature</em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 4.69 = color(green)(|bar(ul(color(white)(a/a)9.31color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A buffer solution is 1.30 M in #NH_3# and 1.20 M in #NH_4Cl#. If 0.120 moles of #NaOH# are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of #NH_3 = 1.8 xx 10^(-5)#.</h1>
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<div class="markdown"><p><mathjax>#"pH" = 9.31#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains ammonia, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, and ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, the salt of its <strong>conjugate acid</strong>, the ammonium cation, <mathjax>#"NH"_4^(+)#</mathjax>. </p>
<p><em>Sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, si a <strong>strong base</strong> that dissociates completely in aqueous solution to form sodium cations, which are of no interest here, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax>.</p>
<p>The hydroxide anions will react with the ammonium cations to form ammonia and water <mathjax>#->#</mathjax> think <strong><a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> reaction. </p>
<blockquote>
<p><mathjax>#"NH"_text(4(aq])^(+) + "OH"_text((aq])^(-) -> "NH"_text(3(aq]) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Since you're dealing with a <mathjax>#"1.0-L"#</mathjax> solution, you can treat <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>number of moles</em> <strong>interchangeably</strong>. </p>
<p>This means that your initial solution contains <mathjax>#1.30#</mathjax> <strong>moles</strong> of ammonia and <mathjax>#1.20#</mathjax> <strong>moles</strong> of ammonium cation (ammonium chloride dissociates in a <mathjax>#1:1#</mathjax> mole ratio to form ammonium cations and chloride anions). </p>
<p>Now, notice that the ammonium cations and the hydroxide anions react in a <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a></strong>. This tells you that the reaction consumes <strong>equal numbers of moles</strong> of each reactant. </p>
<p>Moreover, for every mole of ammonium cations that reacts with one mole of hydroxide anions, <strong>one mole</strong> of ammonia is formed. </p>
<p>Since you're adding <mathjax>#0.120#</mathjax> <strong>moles</strong> of hydroxide buffer solution, you can say that the <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction will <strong>completely consume</strong> the added hydroxide anions and leave you with </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = 0 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_(NH_4^(+)) = 1.30 - 0.120 = "1.18 moles NH"_4^(+)#</mathjax></p>
<p><mathjax>#n_(NH_3) = 1.20 + 0.120 = "1.32 moles NH"_3#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is assumed to be constant, you can say that the resulting solution will have</p>
<blockquote>
<p><mathjax>#["NH"_4^(+)] = "1.18 M" " "#</mathjax> and <mathjax>#" " ["NH"_3] = "1.32 M"#</mathjax></p>
</blockquote>
<p>Now, to calculate the pOH of the solution, use the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))|))#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find the pOH of the solution</p>
<blockquote>
<p><mathjax>#"pOH" = - log(1.8 * 10^(-5)) + log( (1.18 color(red)(cancel(color(black)("M"))))/(1.32color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.74 + log(1.18/1.32) = 4.69#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution, use the equation</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|))) ->#</mathjax> <em>true for aqueous <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> at room temperature</em></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#"pH" = 14 - 4.69 = color(green)(|bar(ul(color(white)(a/a)9.31color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | A buffer solution is 1.30 M in #NH_3# and 1.20 M in #NH_4Cl#. If 0.120 moles of #NaOH# are added to 1.00 L of the buffer, what is its pH? Assume the volume remains constant. Kb of #NH_3 = 1.8 xx 10^(-5)#. | null |
1,153 | ac36109f-6ddd-11ea-8b1b-ccda262736ce | https://socratic.org/questions/how-many-li-atoms-are-in-5-8-moles-of-li | 3.49 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 2 2 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] Li atoms"}] | [{"type":"physical unit","value":"3.49 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] Li [=] \\pu{5.8 moles}"}] | <h1 class="questionTitle" itemprop="name">How many Li atoms are in 5.8 moles of Li?</h1> | null | 3.49 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#5.8xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number, " 6.022xx10^23*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many Li atoms are in 5.8 moles of Li?</h1>
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<div class="markdown"><p><mathjax>#5.8xxN_A#</mathjax>, where <mathjax>#N_A="Avogadro's Number, " 6.022xx10^23*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And <mathjax>#5.8*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>What is the mass of this quantity of lithium atoms? Hint, 1 mole has a mass of <mathjax>#6.94*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#3.49*10^24#</mathjax> atoms Li</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Consider:<img alt="enter image source here" src="https://useruploads.socratic.org/bZS7TxCUTuWGviVQXz3W_Atoms%20Li.png"/> </p></div>
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</article> | How many Li atoms are in 5.8 moles of Li? | null |
1,154 | a901338c-6ddd-11ea-bbf5-ccda262736ce | https://socratic.org/questions/what-weight-of-solute-is-needed-to-prepare-350ml-of-0-6m-na2co2 | 22.26 g | start physical_unit 3 3 mass g qc_end physical_unit 13 13 11 12 molarity qc_end end | [{"type":"physical unit","value":"Weight [OF] solute [IN] g"}] | [{"type":"physical unit","value":"22.26 g"}] | [{"type":"physical unit","value":"Volume [OF] Na2CO3 solution [=] \\pu{350 mL}"},{"type":"physical unit","value":"Molarity [OF] Na2CO3 solution [=] \\pu{0.6 M}"}] | <h1 class="questionTitle" itemprop="name">What weight of solute is needed to prepare 350mL of 0.6M Na2CO2?</h1> | null | 22.26 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> needed to prepare <mathjax>#350#</mathjax> <mathjax>#"mL"#</mathjax> of a <mathjax>#0.6#</mathjax> <mathjax>#M#</mathjax> solution. </p>
<blockquote></blockquote>
<p>First, let's use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" ""molarity" = "mol solute"/"L solution"" ")|)#</mathjax></p>
</blockquote>
<p>We're given:</p>
<ul>
<li>
<p><mathjax>#"molarity" = 0.6#</mathjax> <mathjax>#M#</mathjax></p>
</li>
<li>
<p>volume<mathjax>#= 350#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= 0.350#</mathjax> <mathjax>#"L"#</mathjax></p>
<blockquote></blockquote>
</li>
</ul>
<p>Let's rearrange the equation to solve for the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molarity")("L solution")#</mathjax></p>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#"mol Na"_2"CO"_2 = (0.6"mol"/(cancel("L")))(0.350cancel("L")) = color(red)(ul(0.21color(white)(l)"mol Na"_2"CO"_2#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now, we can use the <strong>molar mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> (<mathjax>#89.989#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of <em>grams</em>:</p>
<blockquote>
<p><mathjax>#color(red)(0.21)cancel(color(red)("mol Na"_2"CO"_2))((89.989color(white)(l)"g Na"_2"CO"_2)/(1cancel("mol Na"_2"CO"_2))) = color(blue)(ulbar(|stackrel(" ")(" "18.9color(white)(l)"g Na"_2"CO"_2" ")|)#</mathjax></p>
</blockquote>
<p>I'll leave it up to you (or your instructor) as to how many <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> there should be. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#18.9#</mathjax> <mathjax>#"g Na"_2"CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> needed to prepare <mathjax>#350#</mathjax> <mathjax>#"mL"#</mathjax> of a <mathjax>#0.6#</mathjax> <mathjax>#M#</mathjax> solution. </p>
<blockquote></blockquote>
<p>First, let's use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" ""molarity" = "mol solute"/"L solution"" ")|)#</mathjax></p>
</blockquote>
<p>We're given:</p>
<ul>
<li>
<p><mathjax>#"molarity" = 0.6#</mathjax> <mathjax>#M#</mathjax></p>
</li>
<li>
<p>volume<mathjax>#= 350#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= 0.350#</mathjax> <mathjax>#"L"#</mathjax></p>
<blockquote></blockquote>
</li>
</ul>
<p>Let's rearrange the equation to solve for the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molarity")("L solution")#</mathjax></p>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#"mol Na"_2"CO"_2 = (0.6"mol"/(cancel("L")))(0.350cancel("L")) = color(red)(ul(0.21color(white)(l)"mol Na"_2"CO"_2#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now, we can use the <strong>molar mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> (<mathjax>#89.989#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of <em>grams</em>:</p>
<blockquote>
<p><mathjax>#color(red)(0.21)cancel(color(red)("mol Na"_2"CO"_2))((89.989color(white)(l)"g Na"_2"CO"_2)/(1cancel("mol Na"_2"CO"_2))) = color(blue)(ulbar(|stackrel(" ")(" "18.9color(white)(l)"g Na"_2"CO"_2" ")|)#</mathjax></p>
</blockquote>
<p>I'll leave it up to you (or your instructor) as to how many <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> there should be. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What weight of solute is needed to prepare 350mL of 0.6M Na2CO2?</h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-08-06T12:58:54" itemprop="dateCreated">
Aug 6, 2017
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<div class="markdown"><p><mathjax>#18.9#</mathjax> <mathjax>#"g Na"_2"CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> needed to prepare <mathjax>#350#</mathjax> <mathjax>#"mL"#</mathjax> of a <mathjax>#0.6#</mathjax> <mathjax>#M#</mathjax> solution. </p>
<blockquote></blockquote>
<p>First, let's use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" ""molarity" = "mol solute"/"L solution"" ")|)#</mathjax></p>
</blockquote>
<p>We're given:</p>
<ul>
<li>
<p><mathjax>#"molarity" = 0.6#</mathjax> <mathjax>#M#</mathjax></p>
</li>
<li>
<p>volume<mathjax>#= 350#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= 0.350#</mathjax> <mathjax>#"L"#</mathjax></p>
<blockquote></blockquote>
</li>
</ul>
<p>Let's rearrange the equation to solve for the moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molarity")("L solution")#</mathjax></p>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#"mol Na"_2"CO"_2 = (0.6"mol"/(cancel("L")))(0.350cancel("L")) = color(red)(ul(0.21color(white)(l)"mol Na"_2"CO"_2#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now, we can use the <strong>molar mass</strong> of <mathjax>#"Na"_2"CO"_2#</mathjax> (<mathjax>#89.989#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of <em>grams</em>:</p>
<blockquote>
<p><mathjax>#color(red)(0.21)cancel(color(red)("mol Na"_2"CO"_2))((89.989color(white)(l)"g Na"_2"CO"_2)/(1cancel("mol Na"_2"CO"_2))) = color(blue)(ulbar(|stackrel(" ")(" "18.9color(white)(l)"g Na"_2"CO"_2" ")|)#</mathjax></p>
</blockquote>
<p>I'll leave it up to you (or your instructor) as to how many <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> there should be. </p></div>
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anor277
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<span class="dateCreated" datetime="2017-08-06T13:03:50" itemprop="dateCreated">
Aug 6, 2017
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<div class="markdown"><p>Do you mean <mathjax>#Na_2CO_2#</mathjax> or <mathjax>#Na_2CO_3#</mathjax>......??</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"#</mathjax></p>
<p>Thus <mathjax>#0.6*mol*L^-1="moles of solute"/"350 mL"#</mathjax>....</p>
<p>and.................... <mathjax>#"mols of solute"=0.6*mol*L^-1xx350*mLxx10^-3*L*mL^-1=0.21*mol#</mathjax></p>
<p>And this represents a mass of <mathjax>#0.21*molxx105.99*g*mol^-1#</mathjax></p>
<p><mathjax>#=22.26*g#</mathjax> with respect to <mathjax>#"sodium carbonate,"#</mathjax> <mathjax>#Na_2CO_3#</mathjax>. There ain't no such beast as <mathjax>#Na_2CO_2#</mathjax>.</p></div>
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</article> | What weight of solute is needed to prepare 350mL of 0.6M Na2CO2? | null |
1,155 | a8de024b-6ddd-11ea-870f-ccda262736ce | https://socratic.org/questions/581ab59b7c01492e4784bd6e | 0.83 atm | start physical_unit 1 1 pressure atm qc_end physical_unit 1 1 10 11 volume qc_end physical_unit 1 1 16 17 pressure qc_end physical_unit 1 1 23 24 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"0.83 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{19.7 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{0.95 atm}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{22.5 L}"}] | <h1 class="questionTitle" itemprop="name">A gas is placed in a piston whose volume was #19.7*L# under a pressure of #0.95*atm#. If the volume were reduced to #22.5*L#, what resultant pressure would develop?</h1> | null | 0.83 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the final pressure of the system we can use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:<br/>
<img alt="http://slideplayer.com/slide/5281881/" src="https://useruploads.socratic.org/orXRcSMaT7CLysJjBVJM_slide_4.jpg"/> </p>
<p>Based on the information you've given me, we know the following variables:</p>
<p><mathjax>#P_1#</mathjax> = 0.95 atm<br/>
<mathjax>#V_1#</mathjax> = 19.7 L<br/>
<mathjax>#V_2#</mathjax> = 22.5 L<br/>
<mathjax>#P_2#</mathjax> = <mathjax>#x#</mathjax></p>
<p>Since we don't know what <mathjax>#P_2#</mathjax> is, the equation has to be rearranged by dividing both sides by <mathjax>#V_2#</mathjax> to get <mathjax>#P_2#</mathjax> by itself:</p>
<p><mathjax>#P_2 = (P_1xxV_1)/V_2#</mathjax></p>
<p>Now, just plug the given values into the equation:</p>
<p><mathjax>#P_2 = (0.95 atm xx19.7cancelL)/(22.5cancelL)#</mathjax></p>
<p><mathjax>#P_2#</mathjax> = 0.83 atm</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The pressure of the system is 0.83 atm</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the final pressure of the system we can use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:<br/>
<img alt="http://slideplayer.com/slide/5281881/" src="https://useruploads.socratic.org/orXRcSMaT7CLysJjBVJM_slide_4.jpg"/> </p>
<p>Based on the information you've given me, we know the following variables:</p>
<p><mathjax>#P_1#</mathjax> = 0.95 atm<br/>
<mathjax>#V_1#</mathjax> = 19.7 L<br/>
<mathjax>#V_2#</mathjax> = 22.5 L<br/>
<mathjax>#P_2#</mathjax> = <mathjax>#x#</mathjax></p>
<p>Since we don't know what <mathjax>#P_2#</mathjax> is, the equation has to be rearranged by dividing both sides by <mathjax>#V_2#</mathjax> to get <mathjax>#P_2#</mathjax> by itself:</p>
<p><mathjax>#P_2 = (P_1xxV_1)/V_2#</mathjax></p>
<p>Now, just plug the given values into the equation:</p>
<p><mathjax>#P_2 = (0.95 atm xx19.7cancelL)/(22.5cancelL)#</mathjax></p>
<p><mathjax>#P_2#</mathjax> = 0.83 atm</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas is placed in a piston whose volume was #19.7*L# under a pressure of #0.95*atm#. If the volume were reduced to #22.5*L#, what resultant pressure would develop?</h1>
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<div class="markdown"><p>The pressure of the system is 0.83 atm</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To find the final pressure of the system we can use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>:<br/>
<img alt="http://slideplayer.com/slide/5281881/" src="https://useruploads.socratic.org/orXRcSMaT7CLysJjBVJM_slide_4.jpg"/> </p>
<p>Based on the information you've given me, we know the following variables:</p>
<p><mathjax>#P_1#</mathjax> = 0.95 atm<br/>
<mathjax>#V_1#</mathjax> = 19.7 L<br/>
<mathjax>#V_2#</mathjax> = 22.5 L<br/>
<mathjax>#P_2#</mathjax> = <mathjax>#x#</mathjax></p>
<p>Since we don't know what <mathjax>#P_2#</mathjax> is, the equation has to be rearranged by dividing both sides by <mathjax>#V_2#</mathjax> to get <mathjax>#P_2#</mathjax> by itself:</p>
<p><mathjax>#P_2 = (P_1xxV_1)/V_2#</mathjax></p>
<p>Now, just plug the given values into the equation:</p>
<p><mathjax>#P_2 = (0.95 atm xx19.7cancelL)/(22.5cancelL)#</mathjax></p>
<p><mathjax>#P_2#</mathjax> = 0.83 atm</p></div>
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</article> | A gas is placed in a piston whose volume was #19.7*L# under a pressure of #0.95*atm#. If the volume were reduced to #22.5*L#, what resultant pressure would develop? | null |
1,156 | ad2437a8-6ddd-11ea-a7ef-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-for-heptane-c7h16-burning-in-oxygen-to-make-carbon | C7H16 + 11 O2 -> 7 CO2 + 8 H2O | start chemical_equation qc_end chemical_equation 7 7 qc_end substance 10 10 qc_end substance 13 14 qc_end substance 16 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the burning equation"}] | [{"type":"chemical equation","value":"C7H16 + 11 O2 -> 7 CO2 + 8 H2O"}] | [{"type":"chemical equation","value":"C7H16"},{"type":"substance name","value":"oxygen"},{"type":"substance name","value":"carbon dioxide"},{"type":"substance name","value":"water"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation for heptane (C7H16) burning in oxygen to make carbon dioxide and water?</h1> | null | C7H16 + 11 O2 -> 7 CO2 + 8 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</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>#"C"_7"H"_16#</mathjax>. We put a <mathjax>#1#</mathjax> in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have fixed 7 <mathjax>#"C"#</mathjax> atoms on the left-hand side, so we need 7 <mathjax>#"C"#</mathjax> atoms on the right-hand side. We put a <mathjax>#7#</mathjax> in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> yet because we have <em>two</em> formulas that contain <mathjax>#"O"#</mathjax> and lack coefficients. So we balance <mathjax>#"H"#</mathjax> instead.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed 16 <mathjax>#"H"#</mathjax> atoms on the left-hand side, so we need 16 <mathjax>#"H"#</mathjax> atoms on the right-hand side. We put an <mathjax>#8#</mathjax> in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Now we can balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We have fixed 22 <mathjax>#"O"#</mathjax> atoms on the right-hand side: 14 from the <mathjax>#"CO"_2#</mathjax> and 8 from the <mathjax>#"H"_2"O"#</mathjax>. We put an <mathjax>#11#</mathjax> in front of the <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(mmll)"C"color(white)(mmmmml)7color(white)(mmmmmmmmll)7#</mathjax><br/>
<mathjax>#color(white)(mmll)"H"color(white)(mmmmll)16color(white)(mmmmmmmm)16#</mathjax><br/>
<mathjax>#color(white)(mmll)"O"color(white)(mmmmll)22color(white)(mmmmmmmm)22#</mathjax></p>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The balanced equation is <mathjax>#"C"_7"H"_16 + "11O"_2 → "7CO"_2 + "8H"_2"O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</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>#"C"_7"H"_16#</mathjax>. We put a <mathjax>#1#</mathjax> in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have fixed 7 <mathjax>#"C"#</mathjax> atoms on the left-hand side, so we need 7 <mathjax>#"C"#</mathjax> atoms on the right-hand side. We put a <mathjax>#7#</mathjax> in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> yet because we have <em>two</em> formulas that contain <mathjax>#"O"#</mathjax> and lack coefficients. So we balance <mathjax>#"H"#</mathjax> instead.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed 16 <mathjax>#"H"#</mathjax> atoms on the left-hand side, so we need 16 <mathjax>#"H"#</mathjax> atoms on the right-hand side. We put an <mathjax>#8#</mathjax> in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Now we can balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We have fixed 22 <mathjax>#"O"#</mathjax> atoms on the right-hand side: 14 from the <mathjax>#"CO"_2#</mathjax> and 8 from the <mathjax>#"H"_2"O"#</mathjax>. We put an <mathjax>#11#</mathjax> in front of the <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(mmll)"C"color(white)(mmmmml)7color(white)(mmmmmmmmll)7#</mathjax><br/>
<mathjax>#color(white)(mmll)"H"color(white)(mmmmll)16color(white)(mmmmmmmm)16#</mathjax><br/>
<mathjax>#color(white)(mmll)"O"color(white)(mmmmll)22color(white)(mmmmmmmm)22#</mathjax></p>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation for heptane (C7H16) burning in oxygen to make carbon dioxide and water?</h1>
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Ernest Z.
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May 28, 2014
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<div class="markdown"><p>The balanced equation is <mathjax>#"C"_7"H"_16 + "11O"_2 → "7CO"_2 + "8H"_2"O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You follow a systematic procedure to balance the equation.</p>
<p>Start with the unbalanced equation:</p>
<p><mathjax>#"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</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>#"C"_7"H"_16#</mathjax>. We put a <mathjax>#1#</mathjax> in front of it to remind ourselves that the coefficient is now fixed.</p>
<p>We start with</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → "CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"C"#</mathjax>:</strong></p>
<p>We have fixed 7 <mathjax>#"C"#</mathjax> atoms on the left-hand side, so we need 7 <mathjax>#"C"#</mathjax> atoms on the right-hand side. We put a <mathjax>#7#</mathjax> in front of the <mathjax>#"CO"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We can't balance <mathjax>#"O"#</mathjax> yet because we have <em>two</em> formulas that contain <mathjax>#"O"#</mathjax> and lack coefficients. So we balance <mathjax>#"H"#</mathjax> instead.</p>
<blockquote></blockquote>
<p><strong>Balance <mathjax>#"H"#</mathjax>:</strong></p>
<p>We have fixed 16 <mathjax>#"H"#</mathjax> atoms on the left-hand side, so we need 16 <mathjax>#"H"#</mathjax> atoms on the right-hand side. We put an <mathjax>#8#</mathjax> in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + "O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Now we can balance <mathjax>#"O"#</mathjax>:</strong></p>
<p>We have fixed 22 <mathjax>#"O"#</mathjax> atoms on the right-hand side: 14 from the <mathjax>#"CO"_2#</mathjax> and 8 from the <mathjax>#"H"_2"O"#</mathjax>. We put an <mathjax>#11#</mathjax> in front of the <mathjax>#"O"_2#</mathjax>.</p>
<p><mathjax>#color(red)(1)"C"_7"H"_16 + color(teal)(11)"O"_2 → color(blue)(7)"CO"_2 + color(green)(8)"H"_2"O"#</mathjax></p>
<p>Every formula now has a fixed coefficient. We should have a balanced equation.</p>
<blockquote></blockquote>
<p>Let’s check:</p>
<p><mathjax>#color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side"#</mathjax><br/>
<mathjax>#color(white)(mmll)"C"color(white)(mmmmml)7color(white)(mmmmmmmmll)7#</mathjax><br/>
<mathjax>#color(white)(mmll)"H"color(white)(mmmmll)16color(white)(mmmmmmmm)16#</mathjax><br/>
<mathjax>#color(white)(mmll)"O"color(white)(mmmmll)22color(white)(mmmmmmmm)22#</mathjax></p>
<p>All atoms balance. The balanced equation is</p>
<p><mathjax>#"C"_7"H"_16 + 11"O"_2 → 7"CO"_2 + 8"H"_2"O"#</mathjax></p></div>
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</article> | What is the balanced equation for heptane (C7H16) burning in oxygen to make carbon dioxide and water? | null |
1,157 | a8e3f69b-6ddd-11ea-aeeb-ccda262736ce | https://socratic.org/questions/if-9-01-g-of-be-reacts-with-70-90-g-of-cl-2-how-much-becl-2-will-be-produced | 80 g | start physical_unit 13 13 mass g qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] BeCl2 [IN] g"}] | [{"type":"physical unit","value":"80 g"}] | [{"type":"physical unit","value":"Mass [OF] Be [=] \\pu{9.01 g}"},{"type":"physical unit","value":"Mass [OF] Cl2 [=] \\pu{70.90 g}"}] | <h1 class="questionTitle" itemprop="name">If 9.01 g of #Be# reacts with 70.90 g of #Cl_2#, how much #BeCl_2# will be produced?</h1> | null | 80 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given above a stoichiometric equation. <mathjax>#1#</mathjax> mole of beryllium metal is oxidized by <mathjax>#1#</mathjax> mole of chlorine gas to give 1 mole of beryllium chloride.</p>
<p>Given that we start with <mathjax>#1#</mathjax> mol of beryllium and <mathjax>#1#</mathjax> mole of chlorine gas, how much <mathjax>#BeCl_2#</mathjax> would we isolate upon complete reaction. My guess is about 80 g. Can you refine this estimate?</p></div>
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<div class="markdown"><p><mathjax>#Be(s) + Cl_2(g) rarr BeCl_2(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given above a stoichiometric equation. <mathjax>#1#</mathjax> mole of beryllium metal is oxidized by <mathjax>#1#</mathjax> mole of chlorine gas to give 1 mole of beryllium chloride.</p>
<p>Given that we start with <mathjax>#1#</mathjax> mol of beryllium and <mathjax>#1#</mathjax> mole of chlorine gas, how much <mathjax>#BeCl_2#</mathjax> would we isolate upon complete reaction. My guess is about 80 g. Can you refine this estimate?</p></div>
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<h1 class="questionTitle" itemprop="name">If 9.01 g of #Be# reacts with 70.90 g of #Cl_2#, how much #BeCl_2# will be produced?</h1>
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<div class="markdown"><p><mathjax>#Be(s) + Cl_2(g) rarr BeCl_2(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have been given above a stoichiometric equation. <mathjax>#1#</mathjax> mole of beryllium metal is oxidized by <mathjax>#1#</mathjax> mole of chlorine gas to give 1 mole of beryllium chloride.</p>
<p>Given that we start with <mathjax>#1#</mathjax> mol of beryllium and <mathjax>#1#</mathjax> mole of chlorine gas, how much <mathjax>#BeCl_2#</mathjax> would we isolate upon complete reaction. My guess is about 80 g. Can you refine this estimate?</p></div>
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</article> | If 9.01 g of #Be# reacts with 70.90 g of #Cl_2#, how much #BeCl_2# will be produced? | null |
1,158 | a84bd0d0-6ddd-11ea-9b04-ccda262736ce | https://socratic.org/questions/what-is-the-h-concentration-if-the-poh-is-7-08 | 10^(-6.92) mol/L | start physical_unit 3 3 concentration mol/l qc_end end | [{"type":"physical unit","value":"concentration [OF] [H+] [IN] mol/L"}] | [{"type":"physical unit","value":"10^(-6.92) mol/L"}] | [{"type":"physical unit","value":"pOH [OF] solution [=] \\pu{7.08}"}] | <h1 class="questionTitle" itemprop="name">What is the [H+] concentration if the pOH is 7.08? </h1> | null | 10^(-6.92) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution under standard conditions, <mathjax>#pH + pOH =14#</mathjax>.</p>
<p>See this earlier <a href="https://socratic.org/questions/what-is-the-ph-of-a-0-0001-m-solution-of-hno3-poh">answer</a> .</p></div>
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<div class="markdown"><p>Well, <mathjax>#pH=6.92#</mathjax>, so <mathjax>#[H^+]=10^(-6.92)mol*L^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution under standard conditions, <mathjax>#pH + pOH =14#</mathjax>.</p>
<p>See this earlier <a href="https://socratic.org/questions/what-is-the-ph-of-a-0-0001-m-solution-of-hno3-poh">answer</a> .</p></div>
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<div class="markdown"><p>Well, <mathjax>#pH=6.92#</mathjax>, so <mathjax>#[H^+]=10^(-6.92)mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In aqueous solution under standard conditions, <mathjax>#pH + pOH =14#</mathjax>.</p>
<p>See this earlier <a href="https://socratic.org/questions/what-is-the-ph-of-a-0-0001-m-solution-of-hno3-poh">answer</a> .</p></div>
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</article> | What is the [H+] concentration if the pOH is 7.08? | null |
1,159 | acde559d-6ddd-11ea-9299-ccda262736ce | https://socratic.org/questions/5523407b581e2a3f066ffa7e | 1.15 M | start physical_unit 8 10 molarity mol/l qc_end physical_unit 8 10 5 6 volume qc_end physical_unit 8 10 12 13 temperature qc_end end | [{"type":"physical unit","value":"Molarity [OF] glacial acetic acid [IN] M"}] | [{"type":"physical unit","value":"1.15 M"}] | [{"type":"physical unit","value":"Volume [OF] glacial acetic acid [=] \\pu{19.00 cm^3}"},{"type":"physical unit","value":"Temperature [OF] glacial acetic acid [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of #"19.00 cm"^3"# of glacial acetic acid at #"25"^@"C"#?</h1> | null | 1.15 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of your solution will be <strong>1.146 M</strong>.</p>
<p>You need the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid, or anhydrous acetic acid, to be able to solve this problem. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case glacial acetic acid, divided by <em>liters of solution</em>, you need to know what the mass of dissolved glacial acetic acid is. </p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid is listed as <mathjax>#"1.05 g/mL"#</mathjax> (see here: <a href="http://www.csudh.edu/oliver/chemdata/acid-str.htm" rel="nofollow" target="_blank">http://www.csudh.edu/oliver/chemdata/acid-str.htm</a>), which means that you dissolve a mass of </p>
<p><mathjax>#19.00cancel("mL") * "1.05 g"/(1cancel("mL")) = "19.95 g acetic acid"#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>One cubic centimeter is equal to 1 mL, so I'll just use mL for volume without doing an actual conversion.</em></p>
<p>Now use acetic acid's molar mass to see how many moles you have </p>
<p><mathjax>#19.95cancel("g") * "1 mole"/(60.05cancel("g")) = "0.33222 moles"#</mathjax></p>
<p>Finally, use the volume of the solution to determine its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></p>
<p><mathjax>#C = n/V = "0.33222 moles"/(290.0 * 10^(-3)"L") = "1.14559 M"#</mathjax></p>
<p>Rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<p><mathjax>#C = color(green)("1.146 M")#</mathjax></p></div>
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<div class="markdown"><p>C=1.146 M</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/solutions-and-their-behavior/molarity">molarity</a> of your solution will be <strong>1.146 M</strong>.</p>
<p>You need the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid, or anhydrous acetic acid, to be able to solve this problem. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case glacial acetic acid, divided by <em>liters of solution</em>, you need to know what the mass of dissolved glacial acetic acid is. </p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid is listed as <mathjax>#"1.05 g/mL"#</mathjax> (see here: <a href="http://www.csudh.edu/oliver/chemdata/acid-str.htm" rel="nofollow" target="_blank">http://www.csudh.edu/oliver/chemdata/acid-str.htm</a>), which means that you dissolve a mass of </p>
<p><mathjax>#19.00cancel("mL") * "1.05 g"/(1cancel("mL")) = "19.95 g acetic acid"#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>One cubic centimeter is equal to 1 mL, so I'll just use mL for volume without doing an actual conversion.</em></p>
<p>Now use acetic acid's molar mass to see how many moles you have </p>
<p><mathjax>#19.95cancel("g") * "1 mole"/(60.05cancel("g")) = "0.33222 moles"#</mathjax></p>
<p>Finally, use the volume of the solution to determine its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></p>
<p><mathjax>#C = n/V = "0.33222 moles"/(290.0 * 10^(-3)"L") = "1.14559 M"#</mathjax></p>
<p>Rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<p><mathjax>#C = color(green)("1.146 M")#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of #"19.00 cm"^3"# of glacial acetic acid at #"25"^@"C"#?</h1>
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Apr 7, 2015
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<div class="markdown"><p>C=1.146 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of your solution will be <strong>1.146 M</strong>.</p>
<p>You need the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid, or anhydrous acetic acid, to be able to solve this problem. </p>
<p>Since <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case glacial acetic acid, divided by <em>liters of solution</em>, you need to know what the mass of dissolved glacial acetic acid is. </p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of glacial acetic acid is listed as <mathjax>#"1.05 g/mL"#</mathjax> (see here: <a href="http://www.csudh.edu/oliver/chemdata/acid-str.htm" rel="nofollow" target="_blank">http://www.csudh.edu/oliver/chemdata/acid-str.htm</a>), which means that you dissolve a mass of </p>
<p><mathjax>#19.00cancel("mL") * "1.05 g"/(1cancel("mL")) = "19.95 g acetic acid"#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>One cubic centimeter is equal to 1 mL, so I'll just use mL for volume without doing an actual conversion.</em></p>
<p>Now use acetic acid's molar mass to see how many moles you have </p>
<p><mathjax>#19.95cancel("g") * "1 mole"/(60.05cancel("g")) = "0.33222 moles"#</mathjax></p>
<p>Finally, use the volume of the solution to determine its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></p>
<p><mathjax>#C = n/V = "0.33222 moles"/(290.0 * 10^(-3)"L") = "1.14559 M"#</mathjax></p>
<p>Rounded to four <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be</p>
<p><mathjax>#C = color(green)("1.146 M")#</mathjax></p></div>
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<div class="markdown"><p>The molarity will be 1.1445 mol/L or 1.1445 M.</p>
<p>Because you were not given the molarity of the glacial acetic acid, you will have to calculate it using its volume, density, and molar mass. This is why your teacher gave you the temperature (density changes with temperature). The density will have to come from a resource. The molecular formula for glacial acetic acid is <mathjax>#"CH"_3"COOH"#</mathjax>, so you can calculate its molar mass, but I looked it up. </p>
<p>The molarity of a substance is the moles of solute per liter of solution. The symbol for molarity is M, and the units are mol/L. <br/>
Note: I will abbreviate glacial acetic acid as GA.</p>
<p><strong>GIVEN/KNOWN:</strong><br/>
<mathjax>#V_1=19.00"cm^3#</mathjax><br/>
<mathjax>#"density of GA at 25"^("o")"C"=1.049"g/cm"^3#</mathjax><br/>
<a href="http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US" rel="nofollow" target="_blank">http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US</a><br/>
<mathjax>#"molar mass of GA"="60.05g/mol"#</mathjax><br/>
<a href="http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US" rel="nofollow" target="_blank">http://www.sigmaaldrich.com/catalog/product/sial/537020?lang=en&region=US</a><br/>
<mathjax>#V_2=290.0 "cm"^3"=290.0mL=0.2900L#</mathjax></p>
<p><strong>UNKNOWN:</strong><br/>
<mathjax>#M_"GA"#</mathjax></p>
<p><strong>SOLUTION:</strong></p>
<p><strong>Mass of GA.</strong><br/>
Determine the mass of GA using its given volume and its density.<br/>
mass = density x volume = <mathjax>#(1.049 g)/(1 cancel(cm^3))xx19.00 cancel(cm^3)#</mathjax> = <mathjax>#"19.931 g"#</mathjax><br/>
<strong>mass of GA = <mathjax>#"19.931 g"#</mathjax></strong></p>
<p><strong>Moles of GA</strong> <br/>
Determine the moles of GA by using its mass and molar mass of 60.05 g/mol.<br/>
<mathjax>#19.931 cancel("g")xx(1 "mol")/(60.05 cancel("g")#</mathjax> = <mathjax>#"0.33191 mol GA"#</mathjax></p>
<p><strong>Moles of GA = <mathjax>#"0.33191 mol"#</mathjax></strong></p>
<p><strong>Molarity of GA.</strong><br/>
Determine the molarity of GA by dividing the mol GA by the liters of the solution. </p>
<p><mathjax>#M = "mol/L"#</mathjax>= <mathjax>#"0.33191 mol"/"0.2900 L"= 1.1445 "mol/L"#</mathjax></p></div>
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</article> | What is the molarity of #"19.00 cm"^3"# of glacial acetic acid at #"25"^@"C"#? | null |
1,160 | a9e39dd8-6ddd-11ea-b41b-ccda262736ce | https://socratic.org/questions/what-is-the-change-in-volume-when-4-8-l-sample-of-n-2-g-is-heated-from-388-0-c-t | 2.26 L | start physical_unit 9 11 volume l qc_end physical_unit 9 11 7 8 volume qc_end physical_unit 9 11 15 16 temperature qc_end physical_unit 9 11 18 19 temperature qc_end end | [{"type":"physical unit","value":"Change in volume [OF] N2(g) sample [IN] L"}] | [{"type":"physical unit","value":"2.26 L"}] | [{"type":"physical unit","value":"Volume1 [OF] N2(g) sample [=] \\pu{4.8 L}"},{"type":"physical unit","value":"Temperature1 [OF] N2(g) sample [=] \\pu{388.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] N2(g) sample [=] \\pu{700 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the change in volume when 4.8 L sample of #N_2(g)# is heated from 388.0°C to 700°C?</h1> | null | 2.26 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#VpropT#</mathjax>, then <mathjax>#V=kT#</mathjax>, and given the conditions (constant pressure, and constant amount of gas), we can solve for <mathjax>#k#</mathjax>:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, where <mathjax>#T="temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=(4.8*L)/(661*cancelK)xx973*cancelK
=??L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>We use <mathjax>#"Charles's Law"#</mathjax>, <mathjax>#VpropT#</mathjax>, we get <mathjax>#V_2~=7L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#VpropT#</mathjax>, then <mathjax>#V=kT#</mathjax>, and given the conditions (constant pressure, and constant amount of gas), we can solve for <mathjax>#k#</mathjax>:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, where <mathjax>#T="temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=(4.8*L)/(661*cancelK)xx973*cancelK
=??L#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the change in volume when 4.8 L sample of #N_2(g)# is heated from 388.0°C to 700°C?</h1>
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<div class="markdown"><p>We use <mathjax>#"Charles's Law"#</mathjax>, <mathjax>#VpropT#</mathjax>, we get <mathjax>#V_2~=7L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If <mathjax>#VpropT#</mathjax>, then <mathjax>#V=kT#</mathjax>, and given the conditions (constant pressure, and constant amount of gas), we can solve for <mathjax>#k#</mathjax>:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax>, where <mathjax>#T="temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#V_2=V_1/T_1xxT_2#</mathjax></p>
<p><mathjax>#=(4.8*L)/(661*cancelK)xx973*cancelK
=??L#</mathjax></p></div>
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</article> | What is the change in volume when 4.8 L sample of #N_2(g)# is heated from 388.0°C to 700°C? | null |
1,161 | ac9c8314-6ddd-11ea-9baa-ccda262736ce | https://socratic.org/questions/what-is-the-final-temperature-of-water-if-380-grams-of-water-at-36-c-absorb-4788 | 66.14 ℃ | start physical_unit 6 6 temperature °c qc_end physical_unit 6 6 13 14 temperature qc_end physical_unit 6 6 8 9 mass qc_end physical_unit 6 6 16 17 heat_energy qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"66.14 ℃"}] | [{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{36 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{380 grams}"},{"type":"physical unit","value":"Absorbed energy [OF] water [=] \\pu{47880 joules}"}] | <h1 class="questionTitle" itemprop="name">What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?</h1> | null | 66.14 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you need to know the value of water's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, which is usually given to be <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. </p>
<p>A substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is required to increase the temperature of <mathjax>#"1.0 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat <br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - its specific heat<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined s <em>final temperature</em> minus <em>initial temperature</em></p>
<p>So, in order to find the final temperature of the water sample, you need to first find the <em>change</em> in temperature, <mathjax>#DeltaT#</mathjax>. Since heat was <strong>absorbed</strong> by the water, you <strong>must</strong> have a positive value for <mathjax>#DeltaT#</mathjax>. </p>
<p>Plug in your values into the above equation to get</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies DeltaT = q/(m * c)#</mathjax></p>
<p><mathjax>#DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"#</mathjax></p>
</blockquote>
<p>Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that</p>
<blockquote>
<p><mathjax>#DeltaT = T_"final" - T_"initial" implies T_"final" = T_"initial" + DeltaT#</mathjax></p>
<p><mathjax>#T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial temperature and the mass of the sample. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#66^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you need to know the value of water's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, which is usually given to be <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. </p>
<p>A substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is required to increase the temperature of <mathjax>#"1.0 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat <br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - its specific heat<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined s <em>final temperature</em> minus <em>initial temperature</em></p>
<p>So, in order to find the final temperature of the water sample, you need to first find the <em>change</em> in temperature, <mathjax>#DeltaT#</mathjax>. Since heat was <strong>absorbed</strong> by the water, you <strong>must</strong> have a positive value for <mathjax>#DeltaT#</mathjax>. </p>
<p>Plug in your values into the above equation to get</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies DeltaT = q/(m * c)#</mathjax></p>
<p><mathjax>#DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"#</mathjax></p>
</blockquote>
<p>Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that</p>
<blockquote>
<p><mathjax>#DeltaT = T_"final" - T_"initial" implies T_"final" = T_"initial" + DeltaT#</mathjax></p>
<p><mathjax>#T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial temperature and the mass of the sample. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-17T12:41:07" itemprop="dateCreated">
Nov 17, 2015
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<div class="markdown"><p><mathjax>#66^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you need to know the value of water's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, which is usually given to be <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. </p>
<p>A substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is required to increase the temperature of <mathjax>#"1.0 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat <br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - its specific heat<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined s <em>final temperature</em> minus <em>initial temperature</em></p>
<p>So, in order to find the final temperature of the water sample, you need to first find the <em>change</em> in temperature, <mathjax>#DeltaT#</mathjax>. Since heat was <strong>absorbed</strong> by the water, you <strong>must</strong> have a positive value for <mathjax>#DeltaT#</mathjax>. </p>
<p>Plug in your values into the above equation to get</p>
<blockquote>
<p><mathjax>#q = m * c * DeltaT implies DeltaT = q/(m * c)#</mathjax></p>
<p><mathjax>#DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"#</mathjax></p>
</blockquote>
<p>Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that</p>
<blockquote>
<p><mathjax>#DeltaT = T_"final" - T_"initial" implies T_"final" = T_"initial" + DeltaT#</mathjax></p>
<p><mathjax>#T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial temperature and the mass of the sample. </p></div>
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</article> | What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy? | null |
1,162 | aa5f1299-6ddd-11ea-99fd-ccda262736ce | https://socratic.org/questions/a-sample-of-neon-is-collected-at-2-7-atm-and-12-0-degrees-c-it-has-a-volume-of-2 | 5.82 L | start physical_unit 26 27 volume l qc_end physical_unit 1 3 7 8 pressure qc_end physical_unit 1 3 10 12 temperature qc_end physical_unit 1 3 18 19 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] this gas [IN] L"}] | [{"type":"physical unit","value":"5.82 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] neon sample [=] \\pu{2.7 atm}"},{"type":"physical unit","value":"Temperature1 [OF] neon sample [=] \\pu{12.0 degrees C}"},{"type":"physical unit","value":"Volume1 [OF] neon sample [=] \\pu{2.25 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of neon is collected at 2.7 atm and 12.0 degrees C. It has a volume of 2.25 L. What would be the volume of this gas at STP?</h1> | null | 5.82 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.7*atmxx2.25*Lxx273.15*K)/(285.15*Kxx1*atm)#</mathjax></p>
<p><mathjax>#~=#</mathjax> <mathjax>#6*L#</mathjax></p>
<p>This volume increase is reasonable as we have only marginally changed temperature, but have reduced pressure by a third. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Standard temperature and pressure"#</mathjax> specifies <mathjax>#1*atm#</mathjax> pressure, and a temperature of <mathjax>#273.15*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.7*atmxx2.25*Lxx273.15*K)/(285.15*Kxx1*atm)#</mathjax></p>
<p><mathjax>#~=#</mathjax> <mathjax>#6*L#</mathjax></p>
<p>This volume increase is reasonable as we have only marginally changed temperature, but have reduced pressure by a third. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of neon is collected at 2.7 atm and 12.0 degrees C. It has a volume of 2.25 L. What would be the volume of this gas at STP?</h1>
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anor277
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<span class="dateCreated" datetime="2016-09-23T15:46:49" itemprop="dateCreated">
Sep 23, 2016
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<div class="markdown"><p><mathjax>#"Standard temperature and pressure"#</mathjax> specifies <mathjax>#1*atm#</mathjax> pressure, and a temperature of <mathjax>#273.15*K#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p>And thus <mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(2.7*atmxx2.25*Lxx273.15*K)/(285.15*Kxx1*atm)#</mathjax></p>
<p><mathjax>#~=#</mathjax> <mathjax>#6*L#</mathjax></p>
<p>This volume increase is reasonable as we have only marginally changed temperature, but have reduced pressure by a third. </p></div>
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</article> | A sample of neon is collected at 2.7 atm and 12.0 degrees C. It has a volume of 2.25 L. What would be the volume of this gas at STP? | null |
1,163 | ac5933e4-6ddd-11ea-9c87-ccda262736ce | https://socratic.org/questions/what-would-be-the-volume-of-6-00-g-of-helium-gas-at-stp | 33.6 L | start physical_unit 9 10 volume l qc_end physical_unit 9 10 6 7 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] helium gas [IN] L"}] | [{"type":"physical unit","value":"33.6 L"}] | [{"type":"physical unit","value":"Mass [OF] helium gas [=] \\pu{6.00 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What would be the volume of 6.00 g of helium gas at STP?</h1> | null | 33.6 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this question, we'll need to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<p><mathjax>#pV = nRT#</mathjax>,<br/>
where <mathjax>#p#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is the number of moles <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvin.</p>
<p>The question already gives us the values for <mathjax>#p#</mathjax> and <mathjax>#T#</mathjax>, because helium is at STP. This means that temperature is <mathjax>#"273.15 K"#</mathjax> and pressure is <mathjax>#"1 atm"#</mathjax>. </p>
<p>We also already know the gas constant. In our case, we'll use the value of <mathjax>#"0.08206 L atm/K mol"#</mathjax> since these units fit the units of our given values the best. </p>
<p>We can find the value for <mathjax>#n#</mathjax> by dividing the mass of helium gas by its molar mass: </p>
<p><mathjax>#n = "number of moles" = "mass of sample"/"molar mass"#</mathjax><br/>
<mathjax>#= "6.00 g"/"4.00 g/mol" = "1.50 mol"#</mathjax></p>
<p>Now, we can just plug all of these values in and solve for <mathjax>#V#</mathjax>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p><mathjax>#V = (nRT)/p = ("1.50 mol" xx "0.08206 L atm/K mol" xx "273.15 K")/"1 atm"#</mathjax></p>
<p><mathjax>#"= 33.6 L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"33.6 L"#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this question, we'll need to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<p><mathjax>#pV = nRT#</mathjax>,<br/>
where <mathjax>#p#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is the number of moles <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvin.</p>
<p>The question already gives us the values for <mathjax>#p#</mathjax> and <mathjax>#T#</mathjax>, because helium is at STP. This means that temperature is <mathjax>#"273.15 K"#</mathjax> and pressure is <mathjax>#"1 atm"#</mathjax>. </p>
<p>We also already know the gas constant. In our case, we'll use the value of <mathjax>#"0.08206 L atm/K mol"#</mathjax> since these units fit the units of our given values the best. </p>
<p>We can find the value for <mathjax>#n#</mathjax> by dividing the mass of helium gas by its molar mass: </p>
<p><mathjax>#n = "number of moles" = "mass of sample"/"molar mass"#</mathjax><br/>
<mathjax>#= "6.00 g"/"4.00 g/mol" = "1.50 mol"#</mathjax></p>
<p>Now, we can just plug all of these values in and solve for <mathjax>#V#</mathjax>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p><mathjax>#V = (nRT)/p = ("1.50 mol" xx "0.08206 L atm/K mol" xx "273.15 K")/"1 atm"#</mathjax></p>
<p><mathjax>#"= 33.6 L"#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What would be the volume of 6.00 g of helium gas at STP?</h1>
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zhirou
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<div class="markdown"><p><mathjax>#"33.6 L"#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To answer this question, we'll need to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<p><mathjax>#pV = nRT#</mathjax>,<br/>
where <mathjax>#p#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is the number of moles <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvin.</p>
<p>The question already gives us the values for <mathjax>#p#</mathjax> and <mathjax>#T#</mathjax>, because helium is at STP. This means that temperature is <mathjax>#"273.15 K"#</mathjax> and pressure is <mathjax>#"1 atm"#</mathjax>. </p>
<p>We also already know the gas constant. In our case, we'll use the value of <mathjax>#"0.08206 L atm/K mol"#</mathjax> since these units fit the units of our given values the best. </p>
<p>We can find the value for <mathjax>#n#</mathjax> by dividing the mass of helium gas by its molar mass: </p>
<p><mathjax>#n = "number of moles" = "mass of sample"/"molar mass"#</mathjax><br/>
<mathjax>#= "6.00 g"/"4.00 g/mol" = "1.50 mol"#</mathjax></p>
<p>Now, we can just plug all of these values in and solve for <mathjax>#V#</mathjax>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p><mathjax>#V = (nRT)/p = ("1.50 mol" xx "0.08206 L atm/K mol" xx "273.15 K")/"1 atm"#</mathjax></p>
<p><mathjax>#"= 33.6 L"#</mathjax></p></div>
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<div class="markdown"><p>33.6 litres</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Assuming that helium behaves as an ideal gas, we know that a mole of ideal gas occupies 22.4 litres at STP.</p>
<p>The atomic weight of helium is 4 g/mol, so 6 g is 1.5 moles. </p>
<p>The volume occupied at STP, therefore, is 22.4 x 1.5 = 33.6 litres.</p></div>
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</article> | What would be the volume of 6.00 g of helium gas at STP? | null |
1,164 | ac63bb79-6ddd-11ea-a1b5-ccda262736ce | https://socratic.org/questions/595721fa7c014916caf95b71 | 9.56 | start physical_unit 5 5 ph none qc_end physical_unit 13 14 9 10 mass qc_end physical_unit 22 22 18 19 volume qc_end end | [{"type":"physical unit","value":"pH [OF] barium hydroxide solution"}] | [{"type":"physical unit","value":"9.56"}] | [{"type":"physical unit","value":"Mass [OF] barium hydroxide [=] \\pu{313 mg}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{1 L}"}] | <h1 class="questionTitle" itemprop="name">What is #pH# of a solution prepared from a #313*mg# mass of barium hydroxide dissolved in a #1*L# volume of water...?</h1> | null | 9.56 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH+pOH=14#</mathjax>, and thus.......</p>
<p><mathjax>#pOH=14-pH#</mathjax>, and <mathjax>#pH=14-pOH#</mathjax>......</p>
<p>But <mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=-log_10{((2xx3.13xx10^-3*g)/(171.34*g*mol^-1))/(1*L)}#</mathjax></p>
<p><mathjax>#=-log_10{3.65xx10^-5}=-(-4.44)=4.44#</mathjax></p>
<p>And so <mathjax>#pH=14-pOH=14-4.44=9.56#</mathjax></p>
<p>Why did I double the <mathjax>#[Ba(OH)_2]#</mathjax> to get <mathjax>#[HO^-]#</mathjax>?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pH=9.56#</mathjax>.......</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH+pOH=14#</mathjax>, and thus.......</p>
<p><mathjax>#pOH=14-pH#</mathjax>, and <mathjax>#pH=14-pOH#</mathjax>......</p>
<p>But <mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=-log_10{((2xx3.13xx10^-3*g)/(171.34*g*mol^-1))/(1*L)}#</mathjax></p>
<p><mathjax>#=-log_10{3.65xx10^-5}=-(-4.44)=4.44#</mathjax></p>
<p>And so <mathjax>#pH=14-pOH=14-4.44=9.56#</mathjax></p>
<p>Why did I double the <mathjax>#[Ba(OH)_2]#</mathjax> to get <mathjax>#[HO^-]#</mathjax>?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is #pH# of a solution prepared from a #313*mg# mass of barium hydroxide dissolved in a #1*L# volume of water...?</h1>
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<div class="markdown"><p><mathjax>#pH=9.56#</mathjax>.......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, <mathjax>#pH+pOH=14#</mathjax>, and thus.......</p>
<p><mathjax>#pOH=14-pH#</mathjax>, and <mathjax>#pH=14-pOH#</mathjax>......</p>
<p>But <mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=-log_10{((2xx3.13xx10^-3*g)/(171.34*g*mol^-1))/(1*L)}#</mathjax></p>
<p><mathjax>#=-log_10{3.65xx10^-5}=-(-4.44)=4.44#</mathjax></p>
<p>And so <mathjax>#pH=14-pOH=14-4.44=9.56#</mathjax></p>
<p>Why did I double the <mathjax>#[Ba(OH)_2]#</mathjax> to get <mathjax>#[HO^-]#</mathjax>?</p></div>
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</article> | What is #pH# of a solution prepared from a #313*mg# mass of barium hydroxide dissolved in a #1*L# volume of water...? | null |
1,165 | a8c903fa-6ddd-11ea-a6e5-ccda262736ce | https://socratic.org/questions/if-1-mmole-of-naoh-is-added-into-a-solution-in-which-there-is-8-mmoles-of-aspart | 10.1 | start physical_unit 40 41 ph none qc_end physical_unit 4 4 1 2 mole qc_end physical_unit 17 18 14 15 mole qc_end physical_unit 17 18 22 22 charge qc_end physical_unit 17 18 25 26 mole qc_end physical_unit 17 18 33 33 charge qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"10.1"}] | [{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{1 mmole}"},{"type":"physical unit","value":"Mole1 [OF] aspartic acid [=] \\pu{8 mmoles}"},{"type":"physical unit","value":"Charge1 [OF] aspartic acid [=] \\pu{-1}"},{"type":"physical unit","value":"Mole2 [OF] aspartic acid [=] \\pu{10 moles}"},{"type":"physical unit","value":"Charge2 [OF] aspartic acid [=] \\pu{-2}"}] | <h1 class="questionTitle" itemprop="name">If 1 mmole of NaOH is added into a solution in which there is 8 mmoles of aspartic acid with a net -1 charge and 10 moles of aspartic acid with a ned -2 charge, what would the pH of the solution be?</h1> | null | 10.1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>! LONG ANSWER !!</strong></p>
<p>The idea here is that <em>aspartic acid</em>, <mathjax>#"C"_4"H"_7"NO"_4#</mathjax>, is a <strong>triprotic acid</strong>, which means that it contains three acidic protons that it can release in solution. </p>
<p>For simplicity, I'll use <mathjax>#"H"_3"A"^(+)#</mathjax> as the notation for aspartic acid. Now, these acidic protons will come off at <em>different</em> <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> values. </p>
<p>These pH values will depend on the acid's <mathjax>#pK_a#</mathjax> values, which are listed as </p>
<blockquote>
<p><mathjax>#{(pK_(a1) = 1.99), (pK_(a2) = 3.90), (pK_(a3) = 9.90) :}#</mathjax></p>
</blockquote>
<p><a href="http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm" rel="nofollow" target="_blank">http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm</a></p>
<p>Here's how a molecule of aspartic acid looks like when <strong>fully protonated</strong></p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/yuuBz16ZSviI9AkpqbI2_aspartic.acid.jpg"/> </p>
<p>This is the form you get at pH values that are below <mathjax>#pK_(a1)#</mathjax>. Notice that the molecule carries an overall <mathjax>#(color(red)(1+))#</mathjax> charge from the protonated <mathjax>#-"NH"_3^(+)#</mathjax> group. </p>
<p>Here's how a <a href="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" rel="nofollow">titration</a> would look like for the titration of aspartic acid with a <strong>strong base</strong><br/>
<img alt="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" src="https://useruploads.socratic.org/bdz78JaQlykMGuYpTXP3_Chap6-titration-aspartate.gif"/> </p>
<p>As the pH of the solution increases from <mathjax>#pK_(a1)#</mathjax> to just below <mathjax>#pK_(a2)#</mathjax>, a proton from one of the two carboxyl groups starts to comes off, leaving behind an <mathjax>#alpha#</mathjax>-<strong>carboxilic acid group</strong>, <mathjax>#"COO"^(-)#</mathjax>. </p>
<p>At this point, the molecule is <strong>neutral</strong>, since the <mathjax>#(color(red)(1+))#</mathjax> charge on the protonated amino group is balanced by the <mathjax>#(color(red)(1-))#</mathjax> charge on the <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>The same thing happens as the pH increases from <mathjax>#pK_(a2)#</mathjax> to just below <mathjax>#pK_(a3)#</mathjax> - the proton from the second carboxyl group comes off, leaving behind a second <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(1-))#</mathjax> charge, since you now get <strong>two</strong> <mathjax>#(color(red)(1-))#</mathjax> charges and only <strong>one</strong> <mathjax>#(color(red)(1+))#</mathjax> charge. </p>
<p>Finally, at pH values that exceed <mathjax>#pK_(a3)#</mathjax>, the last acidic proton comes off, leaving behind the <em>amine group</em>, <mathjax>#-"NH"_2#</mathjax>.</p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(2-))#</mathjax> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>. </p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/tou3GykYSuOTXdZniya1_aspartic.acid_A.JPG"/> </p>
<p>So, you know that your solution contains the <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"A"^(2-)#</mathjax> forms of the molecule. More specifically, it contains <mathjax>#"8 mmoles"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"10 mmoles"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p>Since <mathjax>#"A"^(2-)#</mathjax> is the <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a> of <mathjax>#"HA"^(-)#</mathjax>, you can say that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that contains a <strong>weak acid</strong>, <mathjax>#"HA"^(-)#</mathjax>, and its conjugate base. </p>
<p>This means that you can use the <strong>Henderson - Hasselbalch equation</strong> to help you figure out the pH of the solution</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>Since <mathjax>#"HA"^(-)#</mathjax> is your acidic molecule, you will need to use <mathjax>#pK_(a3)#</mathjax> in the above equation. </p>
<p>Right from the start, you can tell that the pH of the solution <strong>before</strong> adding the base is higher than <mathjax>#pK_(a3)#</mathjax>, since you have more moles of conjguate base than you have of weak acid. </p>
<p>Now, sodium hydroxide is a <strong>strong base</strong> that dissociates in a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> to produce <em>hydroxide anions</em>, <mathjax>#"OH"^(-)#</mathjax>, in solution.</p>
<p>The hydroxide anions will then <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> the <mathjax>#"HA"^(-)#</mathjax> molecules to produce water and <mathjax>#"A"^(2-)#</mathjax>, the conjugate base of the acid form.</p>
<blockquote>
<p><mathjax>#"HA"_text((aq])^(-) + "OH"_text((aq])^(-) -> "A"_text((aq])^(2-) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>You have <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between all the species that are taking part in the reaction. </p>
<p>This means that <mathjax>#"1 mmole"#</mathjax> of hydroxide anions will consume <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and produce <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p><strong>After</strong> the reaction is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = "0 moles" ->#</mathjax> <em>completely consumed by the reaction</em></p>
<p><mathjax>#n_("HA"^(-)) = "8 mmoles" - "1 mmole" = "7 mmoles"#</mathjax></p>
<p><mathjax>#n_(A^(2-)) = "10 mmoles" + "1 mmole" = "11 mmoles"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"pH" = pK_(a3) + log( (["A"^(2-)])/(["HA"^(-)]))#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is the same for all species, you can use the number of moles in the H - H equation.</p>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"pH" = 9.90 + log( (11 color(red)(cancel(color(black)("mmoles"))))/(7color(red)(cancel(color(black)("mmoles"))))) = color(green)(10.1)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 10.1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>! LONG ANSWER !!</strong></p>
<p>The idea here is that <em>aspartic acid</em>, <mathjax>#"C"_4"H"_7"NO"_4#</mathjax>, is a <strong>triprotic acid</strong>, which means that it contains three acidic protons that it can release in solution. </p>
<p>For simplicity, I'll use <mathjax>#"H"_3"A"^(+)#</mathjax> as the notation for aspartic acid. Now, these acidic protons will come off at <em>different</em> <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> values. </p>
<p>These pH values will depend on the acid's <mathjax>#pK_a#</mathjax> values, which are listed as </p>
<blockquote>
<p><mathjax>#{(pK_(a1) = 1.99), (pK_(a2) = 3.90), (pK_(a3) = 9.90) :}#</mathjax></p>
</blockquote>
<p><a href="http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm" rel="nofollow" target="_blank">http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm</a></p>
<p>Here's how a molecule of aspartic acid looks like when <strong>fully protonated</strong></p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/yuuBz16ZSviI9AkpqbI2_aspartic.acid.jpg"/> </p>
<p>This is the form you get at pH values that are below <mathjax>#pK_(a1)#</mathjax>. Notice that the molecule carries an overall <mathjax>#(color(red)(1+))#</mathjax> charge from the protonated <mathjax>#-"NH"_3^(+)#</mathjax> group. </p>
<p>Here's how a <a href="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" rel="nofollow">titration</a> would look like for the titration of aspartic acid with a <strong>strong base</strong><br/>
<img alt="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" src="https://useruploads.socratic.org/bdz78JaQlykMGuYpTXP3_Chap6-titration-aspartate.gif"/> </p>
<p>As the pH of the solution increases from <mathjax>#pK_(a1)#</mathjax> to just below <mathjax>#pK_(a2)#</mathjax>, a proton from one of the two carboxyl groups starts to comes off, leaving behind an <mathjax>#alpha#</mathjax>-<strong>carboxilic acid group</strong>, <mathjax>#"COO"^(-)#</mathjax>. </p>
<p>At this point, the molecule is <strong>neutral</strong>, since the <mathjax>#(color(red)(1+))#</mathjax> charge on the protonated amino group is balanced by the <mathjax>#(color(red)(1-))#</mathjax> charge on the <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>The same thing happens as the pH increases from <mathjax>#pK_(a2)#</mathjax> to just below <mathjax>#pK_(a3)#</mathjax> - the proton from the second carboxyl group comes off, leaving behind a second <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(1-))#</mathjax> charge, since you now get <strong>two</strong> <mathjax>#(color(red)(1-))#</mathjax> charges and only <strong>one</strong> <mathjax>#(color(red)(1+))#</mathjax> charge. </p>
<p>Finally, at pH values that exceed <mathjax>#pK_(a3)#</mathjax>, the last acidic proton comes off, leaving behind the <em>amine group</em>, <mathjax>#-"NH"_2#</mathjax>.</p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(2-))#</mathjax> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>. </p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/tou3GykYSuOTXdZniya1_aspartic.acid_A.JPG"/> </p>
<p>So, you know that your solution contains the <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"A"^(2-)#</mathjax> forms of the molecule. More specifically, it contains <mathjax>#"8 mmoles"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"10 mmoles"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p>Since <mathjax>#"A"^(2-)#</mathjax> is the <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a> of <mathjax>#"HA"^(-)#</mathjax>, you can say that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that contains a <strong>weak acid</strong>, <mathjax>#"HA"^(-)#</mathjax>, and its conjugate base. </p>
<p>This means that you can use the <strong>Henderson - Hasselbalch equation</strong> to help you figure out the pH of the solution</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>Since <mathjax>#"HA"^(-)#</mathjax> is your acidic molecule, you will need to use <mathjax>#pK_(a3)#</mathjax> in the above equation. </p>
<p>Right from the start, you can tell that the pH of the solution <strong>before</strong> adding the base is higher than <mathjax>#pK_(a3)#</mathjax>, since you have more moles of conjguate base than you have of weak acid. </p>
<p>Now, sodium hydroxide is a <strong>strong base</strong> that dissociates in a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> to produce <em>hydroxide anions</em>, <mathjax>#"OH"^(-)#</mathjax>, in solution.</p>
<p>The hydroxide anions will then <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> the <mathjax>#"HA"^(-)#</mathjax> molecules to produce water and <mathjax>#"A"^(2-)#</mathjax>, the conjugate base of the acid form.</p>
<blockquote>
<p><mathjax>#"HA"_text((aq])^(-) + "OH"_text((aq])^(-) -> "A"_text((aq])^(2-) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>You have <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between all the species that are taking part in the reaction. </p>
<p>This means that <mathjax>#"1 mmole"#</mathjax> of hydroxide anions will consume <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and produce <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p><strong>After</strong> the reaction is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = "0 moles" ->#</mathjax> <em>completely consumed by the reaction</em></p>
<p><mathjax>#n_("HA"^(-)) = "8 mmoles" - "1 mmole" = "7 mmoles"#</mathjax></p>
<p><mathjax>#n_(A^(2-)) = "10 mmoles" + "1 mmole" = "11 mmoles"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"pH" = pK_(a3) + log( (["A"^(2-)])/(["HA"^(-)]))#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is the same for all species, you can use the number of moles in the H - H equation.</p>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"pH" = 9.90 + log( (11 color(red)(cancel(color(black)("mmoles"))))/(7color(red)(cancel(color(black)("mmoles"))))) = color(green)(10.1)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If 1 mmole of NaOH is added into a solution in which there is 8 mmoles of aspartic acid with a net -1 charge and 10 moles of aspartic acid with a ned -2 charge, what would the pH of the solution be?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-11T13:09:32" itemprop="dateCreated">
Jan 11, 2016
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<div class="markdown"><p><mathjax>#"pH" = 10.1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>! LONG ANSWER !!</strong></p>
<p>The idea here is that <em>aspartic acid</em>, <mathjax>#"C"_4"H"_7"NO"_4#</mathjax>, is a <strong>triprotic acid</strong>, which means that it contains three acidic protons that it can release in solution. </p>
<p>For simplicity, I'll use <mathjax>#"H"_3"A"^(+)#</mathjax> as the notation for aspartic acid. Now, these acidic protons will come off at <em>different</em> <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> values. </p>
<p>These pH values will depend on the acid's <mathjax>#pK_a#</mathjax> values, which are listed as </p>
<blockquote>
<p><mathjax>#{(pK_(a1) = 1.99), (pK_(a2) = 3.90), (pK_(a3) = 9.90) :}#</mathjax></p>
</blockquote>
<p><a href="http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm" rel="nofollow" target="_blank">http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm</a></p>
<p>Here's how a molecule of aspartic acid looks like when <strong>fully protonated</strong></p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/yuuBz16ZSviI9AkpqbI2_aspartic.acid.jpg"/> </p>
<p>This is the form you get at pH values that are below <mathjax>#pK_(a1)#</mathjax>. Notice that the molecule carries an overall <mathjax>#(color(red)(1+))#</mathjax> charge from the protonated <mathjax>#-"NH"_3^(+)#</mathjax> group. </p>
<p>Here's how a <a href="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" rel="nofollow">titration</a> would look like for the titration of aspartic acid with a <strong>strong base</strong><br/>
<img alt="http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-aspartate.html" src="https://useruploads.socratic.org/bdz78JaQlykMGuYpTXP3_Chap6-titration-aspartate.gif"/> </p>
<p>As the pH of the solution increases from <mathjax>#pK_(a1)#</mathjax> to just below <mathjax>#pK_(a2)#</mathjax>, a proton from one of the two carboxyl groups starts to comes off, leaving behind an <mathjax>#alpha#</mathjax>-<strong>carboxilic acid group</strong>, <mathjax>#"COO"^(-)#</mathjax>. </p>
<p>At this point, the molecule is <strong>neutral</strong>, since the <mathjax>#(color(red)(1+))#</mathjax> charge on the protonated amino group is balanced by the <mathjax>#(color(red)(1-))#</mathjax> charge on the <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>The same thing happens as the pH increases from <mathjax>#pK_(a2)#</mathjax> to just below <mathjax>#pK_(a3)#</mathjax> - the proton from the second carboxyl group comes off, leaving behind a second <mathjax>#alpha#</mathjax>-carboxilic acid group. </p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(1-))#</mathjax> charge, since you now get <strong>two</strong> <mathjax>#(color(red)(1-))#</mathjax> charges and only <strong>one</strong> <mathjax>#(color(red)(1+))#</mathjax> charge. </p>
<p>Finally, at pH values that exceed <mathjax>#pK_(a3)#</mathjax>, the last acidic proton comes off, leaving behind the <em>amine group</em>, <mathjax>#-"NH"_2#</mathjax>.</p>
<p>At this point, the molecule carries a <mathjax>#(color(red)(2-))#</mathjax> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>. </p>
<p><img alt="http://chemicalsareyourfriends.com/posts/synthetic-sweeteners-chemicals-that-are-finger-lickin-sweet/" src="https://useruploads.socratic.org/tou3GykYSuOTXdZniya1_aspartic.acid_A.JPG"/> </p>
<p>So, you know that your solution contains the <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"A"^(2-)#</mathjax> forms of the molecule. More specifically, it contains <mathjax>#"8 mmoles"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and <mathjax>#"10 mmoles"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p>Since <mathjax>#"A"^(2-)#</mathjax> is the <a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a> of <mathjax>#"HA"^(-)#</mathjax>, you can say that you're dealing with a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that contains a <strong>weak acid</strong>, <mathjax>#"HA"^(-)#</mathjax>, and its conjugate base. </p>
<p>This means that you can use the <strong>Henderson - Hasselbalch equation</strong> to help you figure out the pH of the solution</p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>Since <mathjax>#"HA"^(-)#</mathjax> is your acidic molecule, you will need to use <mathjax>#pK_(a3)#</mathjax> in the above equation. </p>
<p>Right from the start, you can tell that the pH of the solution <strong>before</strong> adding the base is higher than <mathjax>#pK_(a3)#</mathjax>, since you have more moles of conjguate base than you have of weak acid. </p>
<p>Now, sodium hydroxide is a <strong>strong base</strong> that dissociates in a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> to produce <em>hydroxide anions</em>, <mathjax>#"OH"^(-)#</mathjax>, in solution.</p>
<p>The hydroxide anions will then <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralize</a> the <mathjax>#"HA"^(-)#</mathjax> molecules to produce water and <mathjax>#"A"^(2-)#</mathjax>, the conjugate base of the acid form.</p>
<blockquote>
<p><mathjax>#"HA"_text((aq])^(-) + "OH"_text((aq])^(-) -> "A"_text((aq])^(2-) + "H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>You have <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between all the species that are taking part in the reaction. </p>
<p>This means that <mathjax>#"1 mmole"#</mathjax> of hydroxide anions will consume <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"HA"^(-)#</mathjax> and produce <mathjax>#"1 mmol"#</mathjax> of <mathjax>#"A"^(2-)#</mathjax>. </p>
<p><strong>After</strong> the reaction is complete, the solution will contain </p>
<blockquote>
<p><mathjax>#n_(OH^(-)) = "0 moles" ->#</mathjax> <em>completely consumed by the reaction</em></p>
<p><mathjax>#n_("HA"^(-)) = "8 mmoles" - "1 mmole" = "7 mmoles"#</mathjax></p>
<p><mathjax>#n_(A^(2-)) = "10 mmoles" + "1 mmole" = "11 mmoles"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#"pH" = pK_(a3) + log( (["A"^(2-)])/(["HA"^(-)]))#</mathjax></p>
</blockquote>
<p>Since the volume of the buffer is the same for all species, you can use the number of moles in the H - H equation.</p>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"pH" = 9.90 + log( (11 color(red)(cancel(color(black)("mmoles"))))/(7color(red)(cancel(color(black)("mmoles"))))) = color(green)(10.1)#</mathjax></p>
</blockquote></div>
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</article> | If 1 mmole of NaOH is added into a solution in which there is 8 mmoles of aspartic acid with a net -1 charge and 10 moles of aspartic acid with a ned -2 charge, what would the pH of the solution be? | null |
1,166 | a8bccf69-6ddd-11ea-ad89-ccda262736ce | https://socratic.org/questions/what-volume-is-occupied-by-0-35-mol-of-helium-gas-at-satp | 7.84 L | start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mole qc_end c_other SATP qc_end end | [{"type":"physical unit","value":"Volume [OF] helium gas [IN] L"}] | [{"type":"physical unit","value":"7.84 L"}] | [{"type":"physical unit","value":"Mole [OF] helium gas [=] \\pu{0.35 mol}"},{"type":"other","value":"SATP"}] | <h1 class="questionTitle" itemprop="name">What volume is occupied by 0.35 mol of helium gas at SATP?</h1> | null | 7.84 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will assume that you mean <mathjax>#"STP (Standard Temperature and Pressure)"#</mathjax>.</p>
<p>Since we know that <mathjax>#"1 mole"#</mathjax> of an ideal gas occupies <mathjax>#"22.4 L"#</mathjax> at <mathjax>#"STP"#</mathjax>, then can establish the following relationships:</p>
<p><mathjax>#color(white)(------)color(orange)[("22.4 L")/("1 mole") or ("1 mole")/("22.4 L")#</mathjax></p>
<p>Knowing that, we can solve for the volume occupied by the helium gas,</p>
<p><mathjax>#(0.35 cancel"mole")/(1) * ("22.4 L")/(1 cancel"mole") = color(blue)"7.84 L"#</mathjax></p>
<p>as <mathjax>#color(blue)"7.84 L"#</mathjax></p>
<p><mathjax>#--------------------#</mathjax></p>
<p>The answer makes sense because if <mathjax>#"1 mole"#</mathjax> occupies <mathjax>#"22.4 L"#</mathjax>, then <mathjax>#"0.35 moles"#</mathjax> (roughly a third of the moles) will occupy a third of <mathjax>#"22.4 L" ("if we round down 22.4 L to 21 L")#</mathjax></p>
<p><mathjax>#color(white)(--------)21/3 ~~"7 L"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"7.84 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will assume that you mean <mathjax>#"STP (Standard Temperature and Pressure)"#</mathjax>.</p>
<p>Since we know that <mathjax>#"1 mole"#</mathjax> of an ideal gas occupies <mathjax>#"22.4 L"#</mathjax> at <mathjax>#"STP"#</mathjax>, then can establish the following relationships:</p>
<p><mathjax>#color(white)(------)color(orange)[("22.4 L")/("1 mole") or ("1 mole")/("22.4 L")#</mathjax></p>
<p>Knowing that, we can solve for the volume occupied by the helium gas,</p>
<p><mathjax>#(0.35 cancel"mole")/(1) * ("22.4 L")/(1 cancel"mole") = color(blue)"7.84 L"#</mathjax></p>
<p>as <mathjax>#color(blue)"7.84 L"#</mathjax></p>
<p><mathjax>#--------------------#</mathjax></p>
<p>The answer makes sense because if <mathjax>#"1 mole"#</mathjax> occupies <mathjax>#"22.4 L"#</mathjax>, then <mathjax>#"0.35 moles"#</mathjax> (roughly a third of the moles) will occupy a third of <mathjax>#"22.4 L" ("if we round down 22.4 L to 21 L")#</mathjax></p>
<p><mathjax>#color(white)(--------)21/3 ~~"7 L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume is occupied by 0.35 mol of helium gas at SATP?</h1>
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<div class="markdown"><p><mathjax>#"7.84 L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I will assume that you mean <mathjax>#"STP (Standard Temperature and Pressure)"#</mathjax>.</p>
<p>Since we know that <mathjax>#"1 mole"#</mathjax> of an ideal gas occupies <mathjax>#"22.4 L"#</mathjax> at <mathjax>#"STP"#</mathjax>, then can establish the following relationships:</p>
<p><mathjax>#color(white)(------)color(orange)[("22.4 L")/("1 mole") or ("1 mole")/("22.4 L")#</mathjax></p>
<p>Knowing that, we can solve for the volume occupied by the helium gas,</p>
<p><mathjax>#(0.35 cancel"mole")/(1) * ("22.4 L")/(1 cancel"mole") = color(blue)"7.84 L"#</mathjax></p>
<p>as <mathjax>#color(blue)"7.84 L"#</mathjax></p>
<p><mathjax>#--------------------#</mathjax></p>
<p>The answer makes sense because if <mathjax>#"1 mole"#</mathjax> occupies <mathjax>#"22.4 L"#</mathjax>, then <mathjax>#"0.35 moles"#</mathjax> (roughly a third of the moles) will occupy a third of <mathjax>#"22.4 L" ("if we round down 22.4 L to 21 L")#</mathjax></p>
<p><mathjax>#color(white)(--------)21/3 ~~"7 L"#</mathjax></p></div>
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</article> | What volume is occupied by 0.35 mol of helium gas at SATP? | null |
1,167 | aa9fcbc6-6ddd-11ea-aaa5-ccda262736ce | https://socratic.org/questions/what-is-the-total-mass-of-0-75-mol-of-so-2 | 64.07 g | start physical_unit 9 9 mass g qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] SO2 [IN] g"}] | [{"type":"physical unit","value":"64.07 g"}] | [{"type":"physical unit","value":"Mole [OF] SO2 [=] \\pu{0.75 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the total mass of 0.75 mol of #SO_2#?</h1> | null | 64.07 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, if there are <mathjax>#0.75*mol#</mathjax>, the mass is <mathjax>#0.75*cancel(mol)xx64.07*g*cancel(mol^-1)~=48*g#</mathjax>.</p>
<p>What is the mass of <mathjax>#1/2*mol#</mathjax> of stuff?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, one mole of sulfur dioxide has a mass of <mathjax>#64.07*g#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, if there are <mathjax>#0.75*mol#</mathjax>, the mass is <mathjax>#0.75*cancel(mol)xx64.07*g*cancel(mol^-1)~=48*g#</mathjax>.</p>
<p>What is the mass of <mathjax>#1/2*mol#</mathjax> of stuff?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the total mass of 0.75 mol of #SO_2#?</h1>
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<div class="markdown"><p>Well, one mole of sulfur dioxide has a mass of <mathjax>#64.07*g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, if there are <mathjax>#0.75*mol#</mathjax>, the mass is <mathjax>#0.75*cancel(mol)xx64.07*g*cancel(mol^-1)~=48*g#</mathjax>.</p>
<p>What is the mass of <mathjax>#1/2*mol#</mathjax> of stuff?</p></div>
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</article> | What is the total mass of 0.75 mol of #SO_2#? | null |
1,168 | ad2792a8-6ddd-11ea-b677-ccda262736ce | https://socratic.org/questions/how-would-you-find-the-molecular-formula-for-a-compound-with-molar-mass-180-amu- | C6H12O6 | start chemical_formula qc_end physical_unit 9 9 13 14 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C6H12O6"}] | [{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{180 g/mol}"},{"type":"physical unit","value":"Percent [OF] carbon in the compound [=] \\pu{40%}"},{"type":"physical unit","value":"Percent [OF] hydrogen in the compound [=] \\pu{6.67%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{53.3%}"}] | <h1 class="questionTitle" itemprop="name">How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen? </h1> | null | C6H12O6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume 100 g of compound.</p>
<p>Elemental composition is divided thru by the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each element:</p>
<p><mathjax>#C: (40*cancelg)/(12.011*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p><mathjax>#H: (6.67*cancelg)/(1.00794*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6,62*mol#</mathjax>.</p>
<p><mathjax>#O: (53.3*cancelg)/(15.99*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p>So the empirical formula is <mathjax>#CH_2O#</mathjax>, after we divide thru by the lowest quotient.</p>
<p>Now we know that the molecular formula is always a whole number mulitple of the empirical formula:</p>
<p>i.e. <mathjax>#(12.011+2xx1.0074+15.99)_n*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#180*g*mol^-1.#</mathjax></p>
<p>Thus <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6#</mathjax>, and the molecular formula is <mathjax>#C_6H_12O_6#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_6H_12O_6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume 100 g of compound.</p>
<p>Elemental composition is divided thru by the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each element:</p>
<p><mathjax>#C: (40*cancelg)/(12.011*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p><mathjax>#H: (6.67*cancelg)/(1.00794*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6,62*mol#</mathjax>.</p>
<p><mathjax>#O: (53.3*cancelg)/(15.99*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p>So the empirical formula is <mathjax>#CH_2O#</mathjax>, after we divide thru by the lowest quotient.</p>
<p>Now we know that the molecular formula is always a whole number mulitple of the empirical formula:</p>
<p>i.e. <mathjax>#(12.011+2xx1.0074+15.99)_n*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#180*g*mol^-1.#</mathjax></p>
<p>Thus <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6#</mathjax>, and the molecular formula is <mathjax>#C_6H_12O_6#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen? </h1>
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anor277
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<div class="markdown"><p><mathjax>#C_6H_12O_6#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume 100 g of compound.</p>
<p>Elemental composition is divided thru by the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of each element:</p>
<p><mathjax>#C: (40*cancelg)/(12.011*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p><mathjax>#H: (6.67*cancelg)/(1.00794*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6,62*mol#</mathjax>.</p>
<p><mathjax>#O: (53.3*cancelg)/(15.99*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol#</mathjax>.</p>
<p>So the empirical formula is <mathjax>#CH_2O#</mathjax>, after we divide thru by the lowest quotient.</p>
<p>Now we know that the molecular formula is always a whole number mulitple of the empirical formula:</p>
<p>i.e. <mathjax>#(12.011+2xx1.0074+15.99)_n*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#180*g*mol^-1.#</mathjax></p>
<p>Thus <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6#</mathjax>, and the molecular formula is <mathjax>#C_6H_12O_6#</mathjax>.</p></div>
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</article> | How would you find the molecular formula for a compound with molar mass 180 amu, that is composed of 40% carbon 6,67% hydrogen 53.3 oxygen? | null |
1,169 | accd0a22-6ddd-11ea-8a99-ccda262736ce | https://socratic.org/questions/what-volume-will-454-grams-1-lb-of-hydrogen-occupy-at-1-05-atm-and-25-o-c | 5.24 × 10^3 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 3 4 mass qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 8 8 11 12 pressure qc_end physical_unit 8 8 14 15 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] hydrogen [IN] L"}] | [{"type":"physical unit","value":"5.24 × 10^3 L"}] | [{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{454 grams}"},{"type":"physical unit","value":"Mass [OF] hydrogen [=] \\pu{1 lb}"},{"type":"physical unit","value":"Pressure [OF] hydrogen [=] \\pu{1.05 atm}"},{"type":"physical unit","value":"Temperature [OF] hydrogen [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">What volume will 454 grams (1 lb) of hydrogen occupy at 1.05 atm and 25 #"^o#C?</h1> | null | 5.24 × 10^3 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the volume occupied by <mathjax>#454#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#"H"_2#</mathjax> at <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal-gas equation</strong>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#p#</mathjax> is the <em>pressure</em>, in units of <mathjax>#"atm"#</mathjax> (given as <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the <em>volume</em> the gas occupies, in units of <mathjax>#"L"#</mathjax> (we'll be finding this)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of <em>moles</em> of the gas present (we'll need to convert the given mass in grams to moles)</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the <em>absolute temperature</em> of the gas, in units of <mathjax>#"K"#</mathjax> (given <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>)</p>
</li>
</ul>
<p>We need to do some conversions:</p>
<p>Let's find the number of <strong>moles</strong> of <mathjax>#"H"_2#</mathjax> present via the <em>molar mass</em> of <mathjax>#"H"_2#</mathjax> (<mathjax>#2.02#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225#</mathjax> <mathjax>#"mol H"_2#</mathjax></p>
<p>The temperature in <mathjax>#"K"#</mathjax> is</p>
<p><mathjax>#25^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, and solving for the volume, <mathjax>#V#</mathjax>, we have</p>
<p><mathjax>#V = (nRT)/p = ((225cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm")))#</mathjax></p>
<p><mathjax>#= color(red)(5.24xx10^3#</mathjax> <mathjax>#color(red)("L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V = 5.24 xx 10^3#</mathjax> <mathjax>#"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the volume occupied by <mathjax>#454#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#"H"_2#</mathjax> at <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal-gas equation</strong>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#p#</mathjax> is the <em>pressure</em>, in units of <mathjax>#"atm"#</mathjax> (given as <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the <em>volume</em> the gas occupies, in units of <mathjax>#"L"#</mathjax> (we'll be finding this)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of <em>moles</em> of the gas present (we'll need to convert the given mass in grams to moles)</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the <em>absolute temperature</em> of the gas, in units of <mathjax>#"K"#</mathjax> (given <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>)</p>
</li>
</ul>
<p>We need to do some conversions:</p>
<p>Let's find the number of <strong>moles</strong> of <mathjax>#"H"_2#</mathjax> present via the <em>molar mass</em> of <mathjax>#"H"_2#</mathjax> (<mathjax>#2.02#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225#</mathjax> <mathjax>#"mol H"_2#</mathjax></p>
<p>The temperature in <mathjax>#"K"#</mathjax> is</p>
<p><mathjax>#25^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, and solving for the volume, <mathjax>#V#</mathjax>, we have</p>
<p><mathjax>#V = (nRT)/p = ((225cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm")))#</mathjax></p>
<p><mathjax>#= color(red)(5.24xx10^3#</mathjax> <mathjax>#color(red)("L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume will 454 grams (1 lb) of hydrogen occupy at 1.05 atm and 25 #"^o#C?</h1>
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<div class="markdown"><p><mathjax>#V = 5.24 xx 10^3#</mathjax> <mathjax>#"L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the volume occupied by <mathjax>#454#</mathjax> <mathjax>#"g"#</mathjax> of <mathjax>#"H"_2#</mathjax> at <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>.</p>
<p>To do this, we can use the <strong>ideal-gas equation</strong>:</p>
<p><mathjax>#pV = nRT#</mathjax></p>
<p>where</p>
<ul>
<li>
<p><mathjax>#p#</mathjax> is the <em>pressure</em>, in units of <mathjax>#"atm"#</mathjax> (given as <mathjax>#1.05#</mathjax> <mathjax>#"atm"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the <em>volume</em> the gas occupies, in units of <mathjax>#"L"#</mathjax> (we'll be finding this)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of <em>moles</em> of the gas present (we'll need to convert the given mass in grams to moles)</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.082057("L"·"atm")/("mol"·"K")#</mathjax></p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the <em>absolute temperature</em> of the gas, in units of <mathjax>#"K"#</mathjax> (given <mathjax>#25#</mathjax> <mathjax>#""^"o""C"#</mathjax>)</p>
</li>
</ul>
<p>We need to do some conversions:</p>
<p>Let's find the number of <strong>moles</strong> of <mathjax>#"H"_2#</mathjax> present via the <em>molar mass</em> of <mathjax>#"H"_2#</mathjax> (<mathjax>#2.02#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225#</mathjax> <mathjax>#"mol H"_2#</mathjax></p>
<p>The temperature in <mathjax>#"K"#</mathjax> is</p>
<p><mathjax>#25^"o""C" + 273 = 298#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, and solving for the volume, <mathjax>#V#</mathjax>, we have</p>
<p><mathjax>#V = (nRT)/p = ((225cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm")))#</mathjax></p>
<p><mathjax>#= color(red)(5.24xx10^3#</mathjax> <mathjax>#color(red)("L"#</mathjax></p></div>
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</article> | What volume will 454 grams (1 lb) of hydrogen occupy at 1.05 atm and 25 #"^o#C? | null |
1,170 | aafa2810-6ddd-11ea-ae8a-ccda262736ce | https://socratic.org/questions/what-mass-of-iron-ll-sulfate-heptahydrate-would-completely-react-with-approximat | 0.14 g | start physical_unit 3 5 mass g qc_end c_other OTHER qc_end physical_unit 16 16 14 15 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] iron(ll) sulfate heptahydrate [IN] g"}] | [{"type":"physical unit","value":"0.14 g"}] | [{"type":"other","value":"Completely react."},{"type":"physical unit","value":"Volume [OF] KMnO4 solution [=] \\pu{10 ml}"},{"type":"physical unit","value":"Molarity [OF] KMnO4 solution [=] \\pu{0.010 M}"}] | <h1 class="questionTitle" itemprop="name">What mass of iron(ll) sulfate heptahydrate would completely react with approximately 10 ml of 0.010 M #KMnO_4#?</h1> | null | 0.14 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation is as given, and shows that 5 equiv of <mathjax>#Fe^(2+)#</mathjax> will react with 1 equiv of permanganate in acidic media.</p>
<p>Because <mathjax>#"Mn"^(2+)#</mathjax> is almost completely colourless, the endpoint is signalled by the persistence of the deep crimson colour of permanganate ion. This is an easy endpoint to vizualize. </p>
<p><mathjax>#"Moles of permanganate"=10xx10^-3Lxx0.010*mol*L^-1=1.0xx10^-4*mol.#</mathjax> </p>
<p>Because of the molar equivalence given in the reaction, 5 equiv of ferrous ion will be oxidized. And so we mulitply this molar quantity by the formula mass of <mathjax>#"iron(II) sulfate heptahydrate"#</mathjax></p>
<p><mathjax>#5xx1.0xx10^-4*molxx278.02*"g"*"mol"^-1=??g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"5Fe"^(2+) + "MnO"_4^(-) + 8"H"^+ → "5Fe"^(3+) + "Mn"^(2+) + 4"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation is as given, and shows that 5 equiv of <mathjax>#Fe^(2+)#</mathjax> will react with 1 equiv of permanganate in acidic media.</p>
<p>Because <mathjax>#"Mn"^(2+)#</mathjax> is almost completely colourless, the endpoint is signalled by the persistence of the deep crimson colour of permanganate ion. This is an easy endpoint to vizualize. </p>
<p><mathjax>#"Moles of permanganate"=10xx10^-3Lxx0.010*mol*L^-1=1.0xx10^-4*mol.#</mathjax> </p>
<p>Because of the molar equivalence given in the reaction, 5 equiv of ferrous ion will be oxidized. And so we mulitply this molar quantity by the formula mass of <mathjax>#"iron(II) sulfate heptahydrate"#</mathjax></p>
<p><mathjax>#5xx1.0xx10^-4*molxx278.02*"g"*"mol"^-1=??g#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of iron(ll) sulfate heptahydrate would completely react with approximately 10 ml of 0.010 M #KMnO_4#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"5Fe"^(2+) + "MnO"_4^(-) + 8"H"^+ → "5Fe"^(3+) + "Mn"^(2+) + 4"H"_2"O"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The stoichiometric equation is as given, and shows that 5 equiv of <mathjax>#Fe^(2+)#</mathjax> will react with 1 equiv of permanganate in acidic media.</p>
<p>Because <mathjax>#"Mn"^(2+)#</mathjax> is almost completely colourless, the endpoint is signalled by the persistence of the deep crimson colour of permanganate ion. This is an easy endpoint to vizualize. </p>
<p><mathjax>#"Moles of permanganate"=10xx10^-3Lxx0.010*mol*L^-1=1.0xx10^-4*mol.#</mathjax> </p>
<p>Because of the molar equivalence given in the reaction, 5 equiv of ferrous ion will be oxidized. And so we mulitply this molar quantity by the formula mass of <mathjax>#"iron(II) sulfate heptahydrate"#</mathjax></p>
<p><mathjax>#5xx1.0xx10^-4*molxx278.02*"g"*"mol"^-1=??g#</mathjax></p></div>
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</article> | What mass of iron(ll) sulfate heptahydrate would completely react with approximately 10 ml of 0.010 M #KMnO_4#? | null |
1,171 | abe9a9ae-6ddd-11ea-a0fc-ccda262736ce | https://socratic.org/questions/571dec3c7c01496e3214ef9e | 66686.57 g/mol | start physical_unit 4 4 molar_mass g/mol qc_end physical_unit 7 8 6 6 number qc_end end | [{"type":"physical unit","value":"Molar mass [OF] hemoglobin [IN] g/mol"}] | [{"type":"physical unit","value":"66686.57 g/mol"}] | [{"type":"physical unit","value":"Number [OF] iron atoms [=] \\pu{4}"},{"type":"physical unit","value":"Percent by mass [OF] iron atoms in hemoglobin [=] \\pu{0.335%}"},{"type":"physical unit","value":"Number [OF] hemoglobin molecule [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">If one molecule of hemoglobin contains 4 iron atoms, and they constitute #0.335%# by mass of hemoglobin, what is the molar mass of hemoglobin?</h1> | null | 66686.57 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let molecular mass of heamoglobin molecule be <mathjax>#=M#</mathjax></p>
<p>One molecule of heamoglobin contains <mathjax>#4"Fe"#</mathjax> atoms of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> </p>
<blockquote>
<p><mathjax>#=55.85" each"#</mathjax>,</p>
</blockquote>
<p>Total mass of iron in one molecule of heamoglobin</p>
<blockquote>
<p><mathjax>#=4xx55.85=223.4#</mathjax></p>
</blockquote>
<p><strong>To calculate the percentage of iron</strong>:</p>
<blockquote>
<p><mathjax>#M#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4#</mathjax></p>
<p><mathjax>#100#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4/Mxx100#</mathjax> </p>
</blockquote>
<p>Equating to the given value, </p>
<blockquote>
<p><mathjax>#223.4/Mxx100=0.335#</mathjax></p>
</blockquote>
<p>Solving for <mathjax>#M#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=223.4/0.335xx100#</mathjax></p>
<p><mathjax>#=66686.57gmol^-1#</mathjax>, rounded to two decimal places.<br/>
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.</p>
</blockquote>
<p>At some places the iron percentage in heomoglobin is given as <mathjax>#0.3335%#</mathjax>. Using the published average atomic mass of iron as <mathjax>#55.847#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=66982.9085gmol^-1#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#66686.57gmol^-1#</mathjax>, rounded to two decimal places.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let molecular mass of heamoglobin molecule be <mathjax>#=M#</mathjax></p>
<p>One molecule of heamoglobin contains <mathjax>#4"Fe"#</mathjax> atoms of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> </p>
<blockquote>
<p><mathjax>#=55.85" each"#</mathjax>,</p>
</blockquote>
<p>Total mass of iron in one molecule of heamoglobin</p>
<blockquote>
<p><mathjax>#=4xx55.85=223.4#</mathjax></p>
</blockquote>
<p><strong>To calculate the percentage of iron</strong>:</p>
<blockquote>
<p><mathjax>#M#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4#</mathjax></p>
<p><mathjax>#100#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4/Mxx100#</mathjax> </p>
</blockquote>
<p>Equating to the given value, </p>
<blockquote>
<p><mathjax>#223.4/Mxx100=0.335#</mathjax></p>
</blockquote>
<p>Solving for <mathjax>#M#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=223.4/0.335xx100#</mathjax></p>
<p><mathjax>#=66686.57gmol^-1#</mathjax>, rounded to two decimal places.<br/>
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.</p>
</blockquote>
<p>At some places the iron percentage in heomoglobin is given as <mathjax>#0.3335%#</mathjax>. Using the published average atomic mass of iron as <mathjax>#55.847#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=66982.9085gmol^-1#</mathjax></p>
</blockquote></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">If one molecule of hemoglobin contains 4 iron atoms, and they constitute #0.335%# by mass of hemoglobin, what is the molar mass of hemoglobin?</h1>
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<div class="markdown"><p><mathjax>#66686.57gmol^-1#</mathjax>, rounded to two decimal places.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let molecular mass of heamoglobin molecule be <mathjax>#=M#</mathjax></p>
<p>One molecule of heamoglobin contains <mathjax>#4"Fe"#</mathjax> atoms of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> </p>
<blockquote>
<p><mathjax>#=55.85" each"#</mathjax>,</p>
</blockquote>
<p>Total mass of iron in one molecule of heamoglobin</p>
<blockquote>
<p><mathjax>#=4xx55.85=223.4#</mathjax></p>
</blockquote>
<p><strong>To calculate the percentage of iron</strong>:</p>
<blockquote>
<p><mathjax>#M#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4#</mathjax></p>
<p><mathjax>#100#</mathjax> amu of heamoglobin has iron<mathjax>#=223.4/Mxx100#</mathjax> </p>
</blockquote>
<p>Equating to the given value, </p>
<blockquote>
<p><mathjax>#223.4/Mxx100=0.335#</mathjax></p>
</blockquote>
<p>Solving for <mathjax>#M#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=223.4/0.335xx100#</mathjax></p>
<p><mathjax>#=66686.57gmol^-1#</mathjax>, rounded to two decimal places.<br/>
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.</p>
</blockquote>
<p>At some places the iron percentage in heomoglobin is given as <mathjax>#0.3335%#</mathjax>. Using the published average atomic mass of iron as <mathjax>#55.847#</mathjax>,</p>
<blockquote>
<p><mathjax>#M=66982.9085gmol^-1#</mathjax></p>
</blockquote></div>
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</article> | If one molecule of hemoglobin contains 4 iron atoms, and they constitute #0.335%# by mass of hemoglobin, what is the molar mass of hemoglobin? | null |
1,172 | a8ab8d46-6ddd-11ea-9c8b-ccda262736ce | https://socratic.org/questions/how-many-moles-are-present-in-25cm3-of-0-2m-sodium-hydroxide-solution | 5.00 × 10^(−3) moles | start physical_unit 11 12 mole mol qc_end physical_unit 11 13 6 7 volume qc_end physical_unit 11 13 9 10 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] sodium hydroxide [IN] moles"}] | [{"type":"physical unit","value":"5.00 × 10^(−3) moles"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide solution [=] \\pu{25 cm^3}"},{"type":"physical unit","value":"Molarity [OF] sodium hydroxide solution [=] \\pu{0.2 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution?</h1> | null | 5.00 × 10^(−3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5.0xx10^-3*mol#</mathjax> with respect to <mathjax>#NaOH#</mathjax>....</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution?</h1>
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anor277
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Apr 23, 2018
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<div class="markdown"><p><mathjax>#5.0xx10^-3*mol#</mathjax> with respect to <mathjax>#NaOH#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition...<mathjax>#"concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>And thus <mathjax>#"moles of solute"="concentration"xx"volume of solution"#</mathjax></p>
<p>And....</p>
<p><mathjax>#0.20*mol*L^-1xx25*cm^3xx10^-3*L*cm^-3=5.0xx10^-3*mol#</mathjax></p>
<p>What is <mathjax>#pH#</mathjax> of this solution?</p></div>
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Dominic B.
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.005 moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Base equation </p>
<p><mathjax>#"concentration" = "moles"/("volume" ("in dm"^3)#</mathjax></p>
<p>Here</p>
<ul>
<li>concentration <mathjax>#= \ "0.2 M"#</mathjax></li>
<li>volume <mathjax>#= "25 cm"^3 = (25/1000) = "0.025 dm"^3#</mathjax></li>
<li>moles <mathjax>#=x#</mathjax></li>
</ul>
<p>Therefore </p>
<p><mathjax>#0.2= x/0.025#</mathjax></p>
<p>Multiply by <mathjax>#0.025#</mathjax> on both sides to leave <mathjax>#x#</mathjax> by itself</p>
<p><mathjax>#- 0.2xx0.025=x#</mathjax></p>
<p><mathjax>#x = "0.005 moles" = 5xx 10^(-3) \ "moles"#</mathjax></p></div>
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</article> | How many moles are present in #"25 cm"^3# of #"0.2 M"# sodium hydroxide solution? | null |
1,173 | ab136312-6ddd-11ea-9f11-ccda262736ce | https://socratic.org/questions/a-sample-of-ammonium-nitrate-having-a-mass-of-3-88-grams-of-water-is-dissolved-i | +23.80 kJ/mol | start physical_unit 3 4 molar_enthalpy_of_dissolution kj/mol qc_end physical_unit 1 4 9 10 mass qc_end physical_unit 12 12 16 17 mass qc_end physical_unit 12 12 27 28 temperature qc_end physical_unit 12 12 30 31 temperature qc_end end | [{"type":"physical unit","value":"Molar enthalpy of dissolution [OF] ammonium nitrate [IN] kJ/mol"}] | [{"type":"physical unit","value":"+23.80 kJ/mol"}] | [{"type":"physical unit","value":"Mass [OF] ammonium nitrate sample [=] \\pu{3.88 grams}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{60.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{23.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{18.4 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from 23.0°C to 18.4°C. What is the molar enthalpy of dissolution for ammonium nitrate?</h1> | null | +23.80 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even before doing any calculations, you can look at the change in temperature measured for the water to say that you can expect the <em>standard molar <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of dissolution</em> of ammonium nitrate to be <strong>positive</strong>. </p>
<p>Since adding the ammonium nitrate salt to the water results in a <strong>decrease in temperature</strong>, you can conclude that the dissolution of ammonium nitrate <strong>absorbs heat</strong> from the surroundings <mathjax>#->#</mathjax> you're dealing with an <a href="https://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic reaction</a>. </p>
<p>So, your strategy here will be to use the mass of the water, its <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and its change in temperature to determine how much heat was absorbed by the dissolution reaction. </p>
<p>Your go-to equation will be </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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><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 = 60.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (18.4 - 23.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = -"1153.7 J"#</mathjax></p>
</blockquote>
<p>Now, don't be confused by the <em>negative sign</em>. If you look at things from <strong>water's perspective</strong>, you can say that water <strong>gives off</strong> heat, hence the negative sign. </p>
<p>Now, the standard molar enthalpy of dissolution is usually expressed in <em>kilojoules per mole</em>. Use ammonium nitrate's molar mass to determine how many moles you get in that sample </p>
<blockquote>
<p><mathjax>#3.88 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.04color(red)(cancel(color(black)("g")))) = "0.04848 moles NH"_4"NO"_3#</mathjax></p>
</blockquote>
<p>Now, if the dissolution of <mathjax>#0.04848#</mathjax> moles of ammonium nitrate required <mathjax>#"1153.7 J"#</mathjax> of heat, it follows that the dissolution of <strong>one mole</strong> of the compound will require </p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)("mole"))) * "1153.7 J"/(0.04848color(red)(cancel(color(black)("moles")))) = "23797.4 J"#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em> and 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>#DeltaH_"sol"^@ = color(green)(+"23.8 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#+"23.8 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even before doing any calculations, you can look at the change in temperature measured for the water to say that you can expect the <em>standard molar <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of dissolution</em> of ammonium nitrate to be <strong>positive</strong>. </p>
<p>Since adding the ammonium nitrate salt to the water results in a <strong>decrease in temperature</strong>, you can conclude that the dissolution of ammonium nitrate <strong>absorbs heat</strong> from the surroundings <mathjax>#->#</mathjax> you're dealing with an <a href="https://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic reaction</a>. </p>
<p>So, your strategy here will be to use the mass of the water, its <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and its change in temperature to determine how much heat was absorbed by the dissolution reaction. </p>
<p>Your go-to equation will be </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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><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 = 60.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (18.4 - 23.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = -"1153.7 J"#</mathjax></p>
</blockquote>
<p>Now, don't be confused by the <em>negative sign</em>. If you look at things from <strong>water's perspective</strong>, you can say that water <strong>gives off</strong> heat, hence the negative sign. </p>
<p>Now, the standard molar enthalpy of dissolution is usually expressed in <em>kilojoules per mole</em>. Use ammonium nitrate's molar mass to determine how many moles you get in that sample </p>
<blockquote>
<p><mathjax>#3.88 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.04color(red)(cancel(color(black)("g")))) = "0.04848 moles NH"_4"NO"_3#</mathjax></p>
</blockquote>
<p>Now, if the dissolution of <mathjax>#0.04848#</mathjax> moles of ammonium nitrate required <mathjax>#"1153.7 J"#</mathjax> of heat, it follows that the dissolution of <strong>one mole</strong> of the compound will require </p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)("mole"))) * "1153.7 J"/(0.04848color(red)(cancel(color(black)("moles")))) = "23797.4 J"#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em> and 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>#DeltaH_"sol"^@ = color(green)(+"23.8 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from 23.0°C to 18.4°C. What is the molar enthalpy of dissolution for ammonium nitrate?</h1>
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Stefan V.
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Dec 2, 2015
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<div class="markdown"><p><mathjax>#+"23.8 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Even before doing any calculations, you can look at the change in temperature measured for the water to say that you can expect the <em>standard molar <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of dissolution</em> of ammonium nitrate to be <strong>positive</strong>. </p>
<p>Since adding the ammonium nitrate salt to the water results in a <strong>decrease in temperature</strong>, you can conclude that the dissolution of ammonium nitrate <strong>absorbs heat</strong> from the surroundings <mathjax>#->#</mathjax> you're dealing with an <a href="https://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic reaction</a>. </p>
<p>So, your strategy here will be to use the mass of the water, its <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a>, and its change in temperature to determine how much heat was absorbed by the dissolution reaction. </p>
<p>Your go-to equation will be </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 water, equal to <mathjax>#4.18 "J"/("g" ""^@"C")#</mathjax><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 = 60.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (18.4 - 23.0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q = -"1153.7 J"#</mathjax></p>
</blockquote>
<p>Now, don't be confused by the <em>negative sign</em>. If you look at things from <strong>water's perspective</strong>, you can say that water <strong>gives off</strong> heat, hence the negative sign. </p>
<p>Now, the standard molar enthalpy of dissolution is usually expressed in <em>kilojoules per mole</em>. Use ammonium nitrate's molar mass to determine how many moles you get in that sample </p>
<blockquote>
<p><mathjax>#3.88 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.04color(red)(cancel(color(black)("g")))) = "0.04848 moles NH"_4"NO"_3#</mathjax></p>
</blockquote>
<p>Now, if the dissolution of <mathjax>#0.04848#</mathjax> moles of ammonium nitrate required <mathjax>#"1153.7 J"#</mathjax> of heat, it follows that the dissolution of <strong>one mole</strong> of the compound will require </p>
<blockquote>
<p><mathjax>#1color(red)(cancel(color(black)("mole"))) * "1153.7 J"/(0.04848color(red)(cancel(color(black)("moles")))) = "23797.4 J"#</mathjax></p>
</blockquote>
<p>Expressed in <em>kilojoules per mole</em> and 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>#DeltaH_"sol"^@ = color(green)(+"23.8 kJ/mol")#</mathjax></p>
</blockquote></div>
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</article> | A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from 23.0°C to 18.4°C. What is the molar enthalpy of dissolution for ammonium nitrate? | null |
1,174 | a87645a6-6ddd-11ea-8a03-ccda262736ce | https://socratic.org/questions/5901e2d17c01495fc5ba3208 | 50 g | start physical_unit 21 21 mass g qc_end physical_unit 9 10 6 7 mass qc_end physical_unit 14 14 12 13 mass qc_end physical_unit 4 4 16 17 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] ice [IN] g"}] | [{"type":"physical unit","value":"50 g"}] | [{"type":"physical unit","value":"Mass [OF] ethylene glycol [=] \\pu{62 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{250 g}"},{"type":"physical unit","value":"Temperature [OF] the solution [=] \\pu{-9.3 ℃}"}] | <h1 class="questionTitle" itemprop="name">If I cool a solution containing 62 g of ethylene glycol in 250 g water to
-9.3 °C, what mass of ice will separate out?</h1> | null | 50 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression <mathjax>#ΔT_text(f)#</mathjax> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ΔT_text(f) = iK_text(f)bcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#i =#</mathjax> the van't Hoff <mathjax>#i#</mathjax> factor<br/>
<mathjax>#K_text(f) =#</mathjax> the freezing point depression constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#b =#</mathjax> the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></p>
<p>We can rearrange this expression to get</p>
<blockquote>
<blockquote>
<p><mathjax>#b = (ΔT_text(f))/(iK_text(f)#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem, </p>
<p><mathjax>#ΔT_text(f) = "9.3 °C"#</mathjax><br/>
<mathjax>#i=1#</mathjax>, because ethylene glycol is a nonelectrolyte<br/>
<mathjax>#K_text(f) = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#b = (9.3 color(red)(cancel(color(black)("°C"))))/(1 × 1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "5.00 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of ethylene glycol" = 62 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "0.999 mol EG"#</mathjax></p>
<blockquote></blockquote>
<p>So, the allowed amount of water is</p>
<p><mathjax>#"Mass of water" = 0.999 color(red)(cancel(color(black)("mol EG"))) × "1 kg water"/(5 color(red)(cancel(color(black)("mol EG")))) = "0.200 kg water" = "200 g water"#</mathjax></p>
<p>You started with 250 g of water, so the other 50 g must be present as ice.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The amount of ice that will separate out is 50 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression <mathjax>#ΔT_text(f)#</mathjax> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ΔT_text(f) = iK_text(f)bcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#i =#</mathjax> the van't Hoff <mathjax>#i#</mathjax> factor<br/>
<mathjax>#K_text(f) =#</mathjax> the freezing point depression constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#b =#</mathjax> the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></p>
<p>We can rearrange this expression to get</p>
<blockquote>
<blockquote>
<p><mathjax>#b = (ΔT_text(f))/(iK_text(f)#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem, </p>
<p><mathjax>#ΔT_text(f) = "9.3 °C"#</mathjax><br/>
<mathjax>#i=1#</mathjax>, because ethylene glycol is a nonelectrolyte<br/>
<mathjax>#K_text(f) = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#b = (9.3 color(red)(cancel(color(black)("°C"))))/(1 × 1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "5.00 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of ethylene glycol" = 62 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "0.999 mol EG"#</mathjax></p>
<blockquote></blockquote>
<p>So, the allowed amount of water is</p>
<p><mathjax>#"Mass of water" = 0.999 color(red)(cancel(color(black)("mol EG"))) × "1 kg water"/(5 color(red)(cancel(color(black)("mol EG")))) = "0.200 kg water" = "200 g water"#</mathjax></p>
<p>You started with 250 g of water, so the other 50 g must be present as ice.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If I cool a solution containing 62 g of ethylene glycol in 250 g water to
-9.3 °C, what mass of ice will separate out?</h1>
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<div class="markdown"><p>The amount of ice that will separate out is 50 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression <mathjax>#ΔT_text(f)#</mathjax> is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ΔT_text(f) = iK_text(f)bcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#i =#</mathjax> the van't Hoff <mathjax>#i#</mathjax> factor<br/>
<mathjax>#K_text(f) =#</mathjax> the freezing point depression constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#b =#</mathjax> the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></p>
<p>We can rearrange this expression to get</p>
<blockquote>
<blockquote>
<p><mathjax>#b = (ΔT_text(f))/(iK_text(f)#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem, </p>
<p><mathjax>#ΔT_text(f) = "9.3 °C"#</mathjax><br/>
<mathjax>#i=1#</mathjax>, because ethylene glycol is a nonelectrolyte<br/>
<mathjax>#K_text(f) = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#b = (9.3 color(red)(cancel(color(black)("°C"))))/(1 × 1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "5.00 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#"Moles of ethylene glycol" = 62 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "0.999 mol EG"#</mathjax></p>
<blockquote></blockquote>
<p>So, the allowed amount of water is</p>
<p><mathjax>#"Mass of water" = 0.999 color(red)(cancel(color(black)("mol EG"))) × "1 kg water"/(5 color(red)(cancel(color(black)("mol EG")))) = "0.200 kg water" = "200 g water"#</mathjax></p>
<p>You started with 250 g of water, so the other 50 g must be present as ice.</p></div>
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</article> | If I cool a solution containing 62 g of ethylene glycol in 250 g water to
-9.3 °C, what mass of ice will separate out? | null |
1,175 | a91381d2-6ddd-11ea-912b-ccda262736ce | https://socratic.org/questions/if-a-25-00g-piece-of-copper-c-s-0-385j-g-c-at-115-0-c-is-added-to-50-00g-of-wate | 28.96 ℃ | start physical_unit 38 41 temperature °c qc_end physical_unit 6 6 2 3 mass qc_end physical_unit 6 6 14 15 temperature qc_end physical_unit 6 6 9 12 specific_heat_capacity qc_end physical_unit 22 22 25 28 specific_heat_capacity qc_end physical_unit 22 22 19 20 mass qc_end c_other OTHER qc_end physical_unit 22 22 30 31 temperature qc_end end | [{"type":"physical unit","value":"Temperature [OF] the object and water [IN] ℃"}] | [{"type":"physical unit","value":"28.96 ℃"}] | [{"type":"physical unit","value":"Mass [OF] copper [=] \\pu{25.00 g}"},{"type":"physical unit","value":"Temperature [OF] copper [=] \\pu{115.0 ℃}"},{"type":"physical unit","value":"Cs [OF] copper [=] \\pu{0.385 J/(g * ℃)}"},{"type":"physical unit","value":"Cs [OF] water [=] \\pu{4.184 J/(g * ℃)}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{50.00 g}"},{"type":"other","value":"The object and water thermally equilibrate."},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{25.00 ℃}"}] | <h1 class="questionTitle" itemprop="name">If a #25.00g# piece of copper (#C_s = (0.385J)/(g*°C)#) at #115.0°C# is added to #50.00g# of water (#C_s = (4.184J)/(g*°C)#) at #25.00°C#, what is the final temperature once the object and water thermally equilibrate?</h1> | null | 28.96 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Have a look:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/4yug4iURgOEXTdHW46sA_heat.jpeg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>I got almost <mathjax>#29^@C#</mathjax> but check my maths anyway!</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Have a look:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/4yug4iURgOEXTdHW46sA_heat.jpeg"/> </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If a #25.00g# piece of copper (#C_s = (0.385J)/(g*°C)#) at #115.0°C# is added to #50.00g# of water (#C_s = (4.184J)/(g*°C)#) at #25.00°C#, what is the final temperature once the object and water thermally equilibrate?</h1>
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<div class="markdown"><p>I got almost <mathjax>#29^@C#</mathjax> but check my maths anyway!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Have a look:<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/4yug4iURgOEXTdHW46sA_heat.jpeg"/> </p></div>
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</article> | If a #25.00g# piece of copper (#C_s = (0.385J)/(g*°C)#) at #115.0°C# is added to #50.00g# of water (#C_s = (4.184J)/(g*°C)#) at #25.00°C#, what is the final temperature once the object and water thermally equilibrate? | null |
1,176 | ab4c848c-6ddd-11ea-a701-ccda262736ce | https://socratic.org/questions/if-given-2c-4h-10-g-13o-2-g-8co-2-g-10h-20-g-and-deltah-5750-kj-how-do-you-deter | -2875 kJ/mol | start physical_unit 28 28 molar_heat kj/mol qc_end chemical_equation 2 12 qc_end end | [{"type":"physical unit","value":"Molar enthalpy of combustion [OF] butane [IN] kJ/mol"}] | [{"type":"physical unit","value":"-2875 kJ/mol"}] | [{"type":"chemical equation","value":"2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)"},{"type":"physical unit","value":"DeltaH [OF] the reaction [=] \\pu{-5750 kJ}"}] | <h1 class="questionTitle" itemprop="name">If given #2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)# and #DeltaH# = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?</h1> | null | -2875 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"molar enthalpy of combustion"#</mathjax> is the energy associated with the complete combustion of 1 mole of hydrocarbon to give carbon dioxide and water. The given equation quotes <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> output,</p>
<p><mathjax>#"PER MOLE OF REACTION AS WRITTEN"#</mathjax></p>
<p>And thus it represents TWICE the enthalpy of combustion of butane because TWO moles of butane were combusted.</p>
<p><mathjax>#DeltaH""^@""_"combustion"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(-5750*kJ*mol^-1)/2= -2875*kJ*mol^-1#</mathjax></p>
<p>Of course, certain reference standards are specified (the most annoying of which is <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax>, but that's another story!).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaH""^@""_"combustion"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-2875*kJ*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"molar enthalpy of combustion"#</mathjax> is the energy associated with the complete combustion of 1 mole of hydrocarbon to give carbon dioxide and water. The given equation quotes <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> output,</p>
<p><mathjax>#"PER MOLE OF REACTION AS WRITTEN"#</mathjax></p>
<p>And thus it represents TWICE the enthalpy of combustion of butane because TWO moles of butane were combusted.</p>
<p><mathjax>#DeltaH""^@""_"combustion"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(-5750*kJ*mol^-1)/2= -2875*kJ*mol^-1#</mathjax></p>
<p>Of course, certain reference standards are specified (the most annoying of which is <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax>, but that's another story!).</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If given #2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)# and #DeltaH# = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?</h1>
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anor277
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<div class="markdown"><p><mathjax>#DeltaH""^@""_"combustion"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-2875*kJ*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#"molar enthalpy of combustion"#</mathjax> is the energy associated with the complete combustion of 1 mole of hydrocarbon to give carbon dioxide and water. The given equation quotes <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> output,</p>
<p><mathjax>#"PER MOLE OF REACTION AS WRITTEN"#</mathjax></p>
<p>And thus it represents TWICE the enthalpy of combustion of butane because TWO moles of butane were combusted.</p>
<p><mathjax>#DeltaH""^@""_"combustion"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(-5750*kJ*mol^-1)/2= -2875*kJ*mol^-1#</mathjax></p>
<p>Of course, certain reference standards are specified (the most annoying of which is <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax>, but that's another story!).</p></div>
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</article> | If given #2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)# and #DeltaH# = -5750 kJ. How do you determine the molar enthalpy of combustion of butane? | null |
1,177 | a9653c70-6ddd-11ea-b66f-ccda262736ce | https://socratic.org/questions/what-is-the-formula-of-ammonium-carbonate | (NH4)2CO3 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] ammonium carbonate [IN] default"}] | [{"type":"chemical equation","value":"(NH4)2CO3"}] | [{"type":"substance name","value":"Ammonium carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the formula of ammonium carbonate? </h1> | null | (NH4)2CO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to recognize here is the fact that you're dealing with two <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a></strong>, one which acts as <strong>cation</strong> and one which acts as <strong>anion</strong>. </p>
<p><img alt="https://chemistrybytes.com/welcome/concepts/chemistry-fundamentals/polyatomic-ions/" src="https://useruploads.socratic.org/Hm5yjG9UQGKOQoan3m1V_polyatomicions.jpg"/> </p>
<p>The name of the <em>cation</em> is always added first to the name of the ionic compound. Likewise, the cation is added first to the compound's chemical formula. In this case, you know that you have <em>ammonium</em>, <mathjax>#"NH"_4^(+)#</mathjax>, as the <strong>cation</strong>. </p>
<p>The name of the <em>anion</em> follows the name of the cation. In this case, you know that you have the <em>carbonate ion</em>, <mathjax>#"CO"_3^(2-)#</mathjax>, as the <strong>anion</strong>. </p>
<p>Now, notice that the anion carries a <mathjax>#2-#</mathjax> charge. As you know, <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> <strong>must be electrically neutral</strong>, meaning that the overall positive charge coming from the cation must be <strong>balanced</strong> by the overall negative charge coming from the anion. </p>
<p>In this case, you need <strong>two</strong> ammonium cations to balance the <mathjax>#2-#</mathjax> charge of the carbonate anion. You will thus have</p>
<blockquote>
<p><mathjax>#2 xx ["NH"_4^(+)]" "#</mathjax> and <mathjax>#" "1 xx ["CO"_3^(2-)]#</mathjax></p>
</blockquote>
<p>which means that the chemical formula for this compound will be </p>
<blockquote>
<p><mathjax>#("NH"_4)_2"CO"_3 ->#</mathjax> <em><strong>ammonium carbonate</strong></em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#("NH"_4)_2"CO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to recognize here is the fact that you're dealing with two <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a></strong>, one which acts as <strong>cation</strong> and one which acts as <strong>anion</strong>. </p>
<p><img alt="https://chemistrybytes.com/welcome/concepts/chemistry-fundamentals/polyatomic-ions/" src="https://useruploads.socratic.org/Hm5yjG9UQGKOQoan3m1V_polyatomicions.jpg"/> </p>
<p>The name of the <em>cation</em> is always added first to the name of the ionic compound. Likewise, the cation is added first to the compound's chemical formula. In this case, you know that you have <em>ammonium</em>, <mathjax>#"NH"_4^(+)#</mathjax>, as the <strong>cation</strong>. </p>
<p>The name of the <em>anion</em> follows the name of the cation. In this case, you know that you have the <em>carbonate ion</em>, <mathjax>#"CO"_3^(2-)#</mathjax>, as the <strong>anion</strong>. </p>
<p>Now, notice that the anion carries a <mathjax>#2-#</mathjax> charge. As you know, <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> <strong>must be electrically neutral</strong>, meaning that the overall positive charge coming from the cation must be <strong>balanced</strong> by the overall negative charge coming from the anion. </p>
<p>In this case, you need <strong>two</strong> ammonium cations to balance the <mathjax>#2-#</mathjax> charge of the carbonate anion. You will thus have</p>
<blockquote>
<p><mathjax>#2 xx ["NH"_4^(+)]" "#</mathjax> and <mathjax>#" "1 xx ["CO"_3^(2-)]#</mathjax></p>
</blockquote>
<p>which means that the chemical formula for this compound will be </p>
<blockquote>
<p><mathjax>#("NH"_4)_2"CO"_3 ->#</mathjax> <em><strong>ammonium carbonate</strong></em></p>
</blockquote></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the formula of ammonium carbonate? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#("NH"_4)_2"CO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to recognize here is the fact that you're dealing with two <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a></strong>, one which acts as <strong>cation</strong> and one which acts as <strong>anion</strong>. </p>
<p><img alt="https://chemistrybytes.com/welcome/concepts/chemistry-fundamentals/polyatomic-ions/" src="https://useruploads.socratic.org/Hm5yjG9UQGKOQoan3m1V_polyatomicions.jpg"/> </p>
<p>The name of the <em>cation</em> is always added first to the name of the ionic compound. Likewise, the cation is added first to the compound's chemical formula. In this case, you know that you have <em>ammonium</em>, <mathjax>#"NH"_4^(+)#</mathjax>, as the <strong>cation</strong>. </p>
<p>The name of the <em>anion</em> follows the name of the cation. In this case, you know that you have the <em>carbonate ion</em>, <mathjax>#"CO"_3^(2-)#</mathjax>, as the <strong>anion</strong>. </p>
<p>Now, notice that the anion carries a <mathjax>#2-#</mathjax> charge. As you know, <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> <strong>must be electrically neutral</strong>, meaning that the overall positive charge coming from the cation must be <strong>balanced</strong> by the overall negative charge coming from the anion. </p>
<p>In this case, you need <strong>two</strong> ammonium cations to balance the <mathjax>#2-#</mathjax> charge of the carbonate anion. You will thus have</p>
<blockquote>
<p><mathjax>#2 xx ["NH"_4^(+)]" "#</mathjax> and <mathjax>#" "1 xx ["CO"_3^(2-)]#</mathjax></p>
</blockquote>
<p>which means that the chemical formula for this compound will be </p>
<blockquote>
<p><mathjax>#("NH"_4)_2"CO"_3 ->#</mathjax> <em><strong>ammonium carbonate</strong></em></p>
</blockquote></div>
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</article> | What is the formula of ammonium carbonate? | null |
1,178 | a9fc07cc-6ddd-11ea-95da-ccda262736ce | https://socratic.org/questions/the-freezing-point-of-an-aqueous-solution-that-contains-a-nonelectrolyte-is-40-o | 2.2 mol/kg | start physical_unit 20 21 molality mol/kg qc_end physical_unit 5 6 12 13 freezing_point_temperature qc_end physical_unit 5 6 24 25 kf qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molal concentration [OF] the solution [IN] mol/kg"}] | [{"type":"physical unit","value":"2.2 mol/kg"}] | [{"type":"physical unit","value":"Freezing point [OF] the aqueous solution [=] \\pu{-4.0 ℃}"},{"type":"physical unit","value":"Kf [OF] the aqueous solution [=] \\pu{1.86 ℃/m}"},{"type":"other","value":"The aqueous solution contains a nonelectrolyte."}] | <h1 class="questionTitle" itemprop="name">The freezing point of an aqueous solution that contains a nonelectrolyte is #-4.0^@"C"#. What is the molal concentration of the solution? #Kf = 1.86^@"C/m"#</h1> | null | 2.2 mol/kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the freezing point of the <strong>solution</strong> to calculate the <em>freezing-point depression</em>, <mathjax>#DeltaT_f#</mathjax>. </p>
<p>As you know, the <strong>freezing-point depression</strong> is a measure of how <strong>low</strong> the freezing point of a solution is compared with the freezing point of the <em>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>You're dealing with an <em><strong>aqueous solution</strong></em>, so right from the start you know that the solvent is water. Pure water freezes at <mathjax>#0^@"C"#</mathjax> under normal pressure.</p>
<p>Your solution freezes at <mathjax>#-4.0^@"C"#</mathjax>. This means that the freezing-point depression account for a <mathjax>#4.0^@"C"#</mathjax> <strong>decrease</strong> in the freezing temperature of the solution compared with that of pure water</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T_"f sol" = T_"f pure solvent" - DeltaT_f)color(white)(a/a)|)))#</mathjax></p>
<p><mathjax>#-4.0^@"C" = - DeltaT_f implies DeltaT_f = 4.0^@"C"#</mathjax></p>
</blockquote>
<p>Now, the freezing-point depression is calculated sing the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<p>Your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is a <strong>non-electrolyte</strong>, which means that its molecules <em><strong>do not</strong></em> dissociate in aqueous solution. This implies that the van't Hoff factor will be equal to <mathjax>#i=1#</mathjax>. </p>
<p>Rearrange the above equation to solve for <mathjax>#b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies b = (DeltaT_f)/(i * K_f)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#b = (4.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.2 mol kg"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the freezing point of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.2 mol kg"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the freezing point of the <strong>solution</strong> to calculate the <em>freezing-point depression</em>, <mathjax>#DeltaT_f#</mathjax>. </p>
<p>As you know, the <strong>freezing-point depression</strong> is a measure of how <strong>low</strong> the freezing point of a solution is compared with the freezing point of the <em>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>You're dealing with an <em><strong>aqueous solution</strong></em>, so right from the start you know that the solvent is water. Pure water freezes at <mathjax>#0^@"C"#</mathjax> under normal pressure.</p>
<p>Your solution freezes at <mathjax>#-4.0^@"C"#</mathjax>. This means that the freezing-point depression account for a <mathjax>#4.0^@"C"#</mathjax> <strong>decrease</strong> in the freezing temperature of the solution compared with that of pure water</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T_"f sol" = T_"f pure solvent" - DeltaT_f)color(white)(a/a)|)))#</mathjax></p>
<p><mathjax>#-4.0^@"C" = - DeltaT_f implies DeltaT_f = 4.0^@"C"#</mathjax></p>
</blockquote>
<p>Now, the freezing-point depression is calculated sing the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<p>Your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is a <strong>non-electrolyte</strong>, which means that its molecules <em><strong>do not</strong></em> dissociate in aqueous solution. This implies that the van't Hoff factor will be equal to <mathjax>#i=1#</mathjax>. </p>
<p>Rearrange the above equation to solve for <mathjax>#b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies b = (DeltaT_f)/(i * K_f)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#b = (4.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.2 mol kg"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the freezing point of the solution. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The freezing point of an aqueous solution that contains a nonelectrolyte is #-4.0^@"C"#. What is the molal concentration of the solution? #Kf = 1.86^@"C/m"#</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"2.2 mol kg"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the freezing point of the <strong>solution</strong> to calculate the <em>freezing-point depression</em>, <mathjax>#DeltaT_f#</mathjax>. </p>
<p>As you know, the <strong>freezing-point depression</strong> is a measure of how <strong>low</strong> the freezing point of a solution is compared with the freezing point of the <em>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>. </p>
<p>You're dealing with an <em><strong>aqueous solution</strong></em>, so right from the start you know that the solvent is water. Pure water freezes at <mathjax>#0^@"C"#</mathjax> under normal pressure.</p>
<p>Your solution freezes at <mathjax>#-4.0^@"C"#</mathjax>. This means that the freezing-point depression account for a <mathjax>#4.0^@"C"#</mathjax> <strong>decrease</strong> in the freezing temperature of the solution compared with that of pure water</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(T_"f sol" = T_"f pure solvent" - DeltaT_f)color(white)(a/a)|)))#</mathjax></p>
<p><mathjax>#-4.0^@"C" = - DeltaT_f implies DeltaT_f = 4.0^@"C"#</mathjax></p>
</blockquote>
<p>Now, the freezing-point depression is calculated sing the equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#DeltaT_f#</mathjax> - the freezing-point depression;<br/>
<mathjax>#i#</mathjax> - the <em>van't Hoff factor</em><br/>
<mathjax>#K_f#</mathjax> - the <em>cryoscopic constant</em> of the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>;<br/>
<mathjax>#b#</mathjax> - the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<p>Your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> is a <strong>non-electrolyte</strong>, which means that its molecules <em><strong>do not</strong></em> dissociate in aqueous solution. This implies that the van't Hoff factor will be equal to <mathjax>#i=1#</mathjax>. </p>
<p>Rearrange the above equation to solve for <mathjax>#b#</mathjax></p>
<blockquote>
<p><mathjax>#DeltaT_f = i * K_f * b implies b = (DeltaT_f)/(i * K_f)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#b = (4.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.2 mol kg"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the freezing point of the solution. </p></div>
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</article> | The freezing point of an aqueous solution that contains a nonelectrolyte is #-4.0^@"C"#. What is the molal concentration of the solution? #Kf = 1.86^@"C/m"# | null |
1,179 | acab6ad3-6ddd-11ea-8daf-ccda262736ce | https://socratic.org/questions/5945229cb72cff162155983a | 1.44 × 10^24 | start physical_unit 2 2 number none qc_end physical_unit 11 12 7 8 mass qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"1.44 × 10^24"}] | [{"type":"physical unit","value":"Mass [OF] nitride anion [=] \\pu{4.2 g}"}] | <h1 class="questionTitle" itemprop="name">How many electrons are associated with a #4.2*g# mass of nitride anion?</h1> | null | 1.44 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well in one formula unit of <mathjax>#"nitride anion........."#</mathjax>, there are <mathjax>#3xx7+3=24*"electrons"#</mathjax>.</p>
<p>And in <mathjax>#"one mole"#</mathjax> of anything there are <mathjax>#6.022xx10^23#</mathjax> individual items, or there are <mathjax>#6.022xx10^23*"particles"*mol^-1#</mathjax>.</p>
<p><mathjax>#"Moles of nitride anion"=(4.2*g)/(42.03*g*mol^-1)=1.0xx10^-1*mol#</mathjax>.</p>
<p>And thus <mathjax>#"number of electrons.................."#</mathjax></p>
<p><mathjax>#=1xx10^-1*molxx24*"electrons"xx6.022xx10^23*mol^-1#</mathjax></p>
<p><mathjax>#=1.45xx10^24*"electrons............."#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well in one mole of nitride anion.........there are <mathjax>#1.44xx10^24#</mathjax> <mathjax>#"electrons."#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well in one formula unit of <mathjax>#"nitride anion........."#</mathjax>, there are <mathjax>#3xx7+3=24*"electrons"#</mathjax>.</p>
<p>And in <mathjax>#"one mole"#</mathjax> of anything there are <mathjax>#6.022xx10^23#</mathjax> individual items, or there are <mathjax>#6.022xx10^23*"particles"*mol^-1#</mathjax>.</p>
<p><mathjax>#"Moles of nitride anion"=(4.2*g)/(42.03*g*mol^-1)=1.0xx10^-1*mol#</mathjax>.</p>
<p>And thus <mathjax>#"number of electrons.................."#</mathjax></p>
<p><mathjax>#=1xx10^-1*molxx24*"electrons"xx6.022xx10^23*mol^-1#</mathjax></p>
<p><mathjax>#=1.45xx10^24*"electrons............."#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many electrons are associated with a #4.2*g# mass of nitride anion?</h1>
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anor277
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<div class="markdown"><p>Well in one mole of nitride anion.........there are <mathjax>#1.44xx10^24#</mathjax> <mathjax>#"electrons."#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well in one formula unit of <mathjax>#"nitride anion........."#</mathjax>, there are <mathjax>#3xx7+3=24*"electrons"#</mathjax>.</p>
<p>And in <mathjax>#"one mole"#</mathjax> of anything there are <mathjax>#6.022xx10^23#</mathjax> individual items, or there are <mathjax>#6.022xx10^23*"particles"*mol^-1#</mathjax>.</p>
<p><mathjax>#"Moles of nitride anion"=(4.2*g)/(42.03*g*mol^-1)=1.0xx10^-1*mol#</mathjax>.</p>
<p>And thus <mathjax>#"number of electrons.................."#</mathjax></p>
<p><mathjax>#=1xx10^-1*molxx24*"electrons"xx6.022xx10^23*mol^-1#</mathjax></p>
<p><mathjax>#=1.45xx10^24*"electrons............."#</mathjax></p></div>
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</article> | How many electrons are associated with a #4.2*g# mass of nitride anion? | null |
1,180 | aa898477-6ddd-11ea-8572-ccda262736ce | https://socratic.org/questions/734-ml-of-a-1-13-m-calcium-sulfate-caso-4-solution-is-diluted-with-2-56-l-of-wat | 0.25 molarity | start physical_unit 8 8 concentration mol/l qc_end physical_unit 9 9 0 1 volume qc_end physical_unit 8 8 4 5 concentration qc_end physical_unit 16 16 13 14 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] CaSO4 solution [IN] molarity"}] | [{"type":"physical unit","value":"0.25 molarity"}] | [{"type":"physical unit","value":"Volume1 [OF] CaSO4 solution [=] \\pu{734 mL}"},{"type":"physical unit","value":"Concentration1 [OF] CaSO4 solution [=] \\pu{1.13 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{2.56 L}"}] | <h1 class="questionTitle" itemprop="name">734 mL of a 1.13 M calcium sulfate (#CaSO_4#) solution is diluted with 2.56 L of water. What is the new concentration in molarity?</h1> | null | 0.25 molarity | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol*L^-1#</mathjax></p>
<p>How was the calculation performed?</p>
<p>You have a known volume of calcium sulfate with known concentration. We work out the initial molar quantity:</p>
<p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>.</p>
<p>Given the molar quantity, we divide this by the new volume:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax></p>
<p>Note that the question said it is diluted with (and not to) <mathjax>#2.56*L#</mathjax>; this is why the volumes are additive. I don't think this question is entirely realistic in that calcium sulfate is not particularly soluble in water. They should have made it calcium nitrate or calcium acetate. </p></div>
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<div class="markdown"><p><mathjax>#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol*L^-1#</mathjax></p>
<p>How was the calculation performed?</p>
<p>You have a known volume of calcium sulfate with known concentration. We work out the initial molar quantity:</p>
<p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>.</p>
<p>Given the molar quantity, we divide this by the new volume:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax></p>
<p>Note that the question said it is diluted with (and not to) <mathjax>#2.56*L#</mathjax>; this is why the volumes are additive. I don't think this question is entirely realistic in that calcium sulfate is not particularly soluble in water. They should have made it calcium nitrate or calcium acetate. </p></div>
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<h1 class="questionTitle" itemprop="name">734 mL of a 1.13 M calcium sulfate (#CaSO_4#) solution is diluted with 2.56 L of water. What is the new concentration in molarity?</h1>
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anor277
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<span class="dateCreated" datetime="2016-04-20T03:54:16" itemprop="dateCreated">
Apr 20, 2016
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<div class="markdown"><p><mathjax>#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2=(0.734*Lxx1.13*mol*L^-1)/(0.734*L+2.56*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol*L^-1#</mathjax></p>
<p>How was the calculation performed?</p>
<p>You have a known volume of calcium sulfate with known concentration. We work out the initial molar quantity:</p>
<p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>.</p>
<p>Given the molar quantity, we divide this by the new volume:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume"#</mathjax></p>
<p>Note that the question said it is diluted with (and not to) <mathjax>#2.56*L#</mathjax>; this is why the volumes are additive. I don't think this question is entirely realistic in that calcium sulfate is not particularly soluble in water. They should have made it calcium nitrate or calcium acetate. </p></div>
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</article> | 734 mL of a 1.13 M calcium sulfate (#CaSO_4#) solution is diluted with 2.56 L of water. What is the new concentration in molarity? | null |
1,181 | a9f689b6-6ddd-11ea-a197-ccda262736ce | https://socratic.org/questions/how-much-energy-in-joules-must-be-absorbed-if-all-the-o-3-molecules-in-the-sampl | 8.85 × 10^(-3) joules | start physical_unit 11 12 heat_energy j qc_end c_other OTHER qc_end physical_unit 37 38 40 41 wavelength qc_end chemical_equation 59 65 qc_end physical_unit 11 11 67 68 volume qc_end physical_unit 11 11 73 74 temperature qc_end physical_unit 11 11 76 77 pressure qc_end physical_unit 11 11 79 80 molarity qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] O3 molecules [IN] joules"}] | [{"type":"physical unit","value":"8.85 × 10^(-3) joules"}] | [{"type":"other","value":"Assume that each photon absorbed causes one O3 molecule to dissociate."},{"type":"physical unit","value":"Wavelength [OF] the radiation [=] \\pu{254 nm}"},{"type":"chemical equation","value":"O3 + hv -> O2 + O"},{"type":"physical unit","value":"Volume [OF] O3 sample [=] \\pu{1.85 L}"},{"type":"physical unit","value":"Temperature [OF] O3 sample [=] \\pu{22 ℃}"},{"type":"physical unit","value":"Pressure [OF] O3 sample [=] \\pu{748 mmHg}"},{"type":"physical unit","value":"Molarity [OF] O3 [=] \\pu{0.25 ppm}"}] | <h1 class="questionTitle" itemprop="name">How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm.</h1> | <div class="questionDetailsContainer">
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<p>Ozone, <mathjax>#O_3#</mathjax>, absorbs ultraviolet radiation and dissociates into <mathjax>#O_2#</mathjax> molecules and <mathjax>#O#</mathjax> atoms:<br/>
<mathjax>#O_3+h\nu→O_2+O.#</mathjax><br/>
A 1.85-<mathjax>#L#</mathjax> sample of air at 22<mathjax>#°C#</mathjax> and 748 <mathjax>#mmHg#</mathjax> contains 0.25 ppm of <mathjax>#O_3#</mathjax>.</p>
<p><em>My thoughts</em></p>
<p>I assume "parts per million" (ppm) is "<mathjax>#(mL)/L#</mathjax>"?<br/>
Also, I don't really know what section I should be posting this in, sorry.</p></div>
</h2>
</div>
</div> | 8.85 × 10^(-3) joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>8.9 mJ</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm.</h1>
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<p>Ozone, <mathjax>#O_3#</mathjax>, absorbs ultraviolet radiation and dissociates into <mathjax>#O_2#</mathjax> molecules and <mathjax>#O#</mathjax> atoms:<br/>
<mathjax>#O_3+h\nu→O_2+O.#</mathjax><br/>
A 1.85-<mathjax>#L#</mathjax> sample of air at 22<mathjax>#°C#</mathjax> and 748 <mathjax>#mmHg#</mathjax> contains 0.25 ppm of <mathjax>#O_3#</mathjax>.</p>
<p><em>My thoughts</em></p>
<p>I assume "parts per million" (ppm) is "<mathjax>#(mL)/L#</mathjax>"?<br/>
Also, I don't really know what section I should be posting this in, sorry.</p></div>
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Andy Wolff
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<span class="dateCreated" datetime="2016-11-20T03:57:58" itemprop="dateCreated">
Nov 20, 2016
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<div class="markdown"><p>8.9 mJ</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a very difficult problem. What makes it difficult is the number of steps involved in getting to the answer.</p>
<p>The outline of how we are going to solve the problem is this:<br/>
1) figure out how much energy, in Joules, a single photon of 254 nm light has.<br/>
2) figure out how many ozone molecules you have in the container. This is even more complicated because "ppm" can refer to mg/kg or <mathjax>#mu#</mathjax>l/l. Fortunately, the convention for gas is <mathjax>#mu#</mathjax>l/l, so we will go with that.<br/>
3) multiply the two together to get the answer.</p>
<p>So, step 1: we will use E = hc/ <mathjax>#lambda#</mathjax> where E is in Joules, h is Plank's constant (<mathjax>#6.62607 xx 10^(-34) J*s#</mathjax>), c is the speed of light in meters/sec (299,792,458 m/sec), and <mathjax>#lambda#</mathjax> is 254 nm (<mathjax>#254 xx 10^(-9)#</mathjax> meters). Each photon thus has an energy of <mathjax>#7.82065 xx 10^(-19)#</mathjax> J.</p>
<p>Reality check: is this number reasonable?<br/>
(Yes! A Joule is a significant amount of heat--enough that you can feel it on your skin. But one photon has a very small amount of energy. so a very small number is what we are expecting for this step of the problem.)</p>
<p>Step 2: We will rearrange the ideal gas equation to determine how many moles of gas are present. </p>
<p><mathjax>#m_(gas) = (P_(atm) * V_l) / (R_((l atm)/(mol K)) * T_K)#</mathjax> </p>
<p>We will then take 0.25 ppm(vol) of that for the number of moles of ozone present.</p>
<p><mathjax>#m_(gas) = (((748 mmHg)/(760 mmHg))*1.85 liters)/(0.082057_((l atm)/(mol K)) * (22+273.15)K) = 7.5180 xx 10^(-2) m#</mathjax> </p>
<p>Reality check: we know at STP that 1 mole of gas occupies 22.414 liters. We have calculated a little less than 1/10 of a mole here, and that fits since the volume is a little less than 2 liters and the pressure is a little less than 1 atm. </p>
<p>So far, so good!</p>
<p>So how many molecules of ozone do we have?</p>
<p><mathjax>#"molecules " O_3 = 6.022 xx 10^(23) " molecules " O_3/("mole " O_3) xx 7.5180 xx 10^(-2)" mole " O_3 xx 0.25 xx 10^(-6)"(ppm of " O_3 " in the sample) " = 1.1318 xx 10^(16) " Ozone molecules"#</mathjax></p>
<p>Now to multiply the part 1 answer by the part 2 answer:</p>
<p><mathjax>#7.82065 xx 10^(-19)" J/photon" xx " 1 photon"/" Ozone molecule" xx 1.1318 xx 10^(16) " Ozone molecules" = 8.85166 xx 10^(-3) " joules"#</mathjax></p>
<p>For this question, the limiting number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> is 2 from the 0.25 ppm ozone concentration. Thus the final answer is</p>
<p>8.9 mJ.</p></div>
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Andy Wolff
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Nov 29, 2016
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<div class="markdown"><p><strong>WARNING! Long answer!</strong> The energy absorbed is 8.9 mJ.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1: Calculate the number of molecules in the container.</strong></p>
<p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> is </p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) PV=nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange the formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P = 748 color(red)(cancel(color(black)("mmHg"))) × ("1 atm")/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9842 atm"#</mathjax></p>
<p><mathjax>#V = "1.85 L"#</mathjax></p>
<p><mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax></p>
<p><mathjax>#T = "22 °C" = "295.15 K"#</mathjax></p>
<p><mathjax>#n = (0.9842 color(red)(cancel(color(black)("atm"))) × 1.85 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 295.15 color(red)(cancel(color(black)("K")))) = "0.075 18 mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the number of gas molecules</strong></p>
<p><mathjax>#"Number of gas molecules" = "0.075 18" color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23 color(white)(l)"molecules")/(1 color(red)(cancel(color(black)("mol"))))#</mathjax></p>
<p><mathjax>#= 4.527 × 10^22 color(white)(l)"molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the number of <mathjax>#"O"_3#</mathjax> molecules.</strong></p>
<p><mathjax>#"Number of"color(white)(l) "O"_3 color(white)(l)"molecules" = 4.527 × 10^22 color(red)(cancel(color(black)("gas molecules"))) × ("0.25 O"_3 color(white)(l)"molecule")/(10^6 color(red)(cancel(color(black)("gas molecules")))) = 1.13 × 10^16 color(white)(l)"O"_3 color(white)(l)"molecules"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the energy absorbed by 1 <mathjax>#"O"_3#</mathjax> molecule.</strong></p>
<p>The formula for the energy <mathjax>#E#</mathjax> of a quantum is</p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) E = (hc)/λ color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#h#</mathjax> = <a href="https://socratic.org/chemistry/the-bohr-model-of-the-atom/calculations-with-plancks-constant-and-frequency">Planck's constant</a><br/>
<mathjax>#c#</mathjax> = the speed of light<br/>
<mathjax>#λ#</mathjax> = the wavelength of the light</p>
<p><mathjax>#E = (6.626 × 10^"-34" "J"·color(red)(cancel(color(black)("s"))) × 2.998 × 10^8 color(red)(cancel(color(black)("m·s"^"-1"))))/(254 × 10^"-9" color(red)(cancel(color(black)("m")))) = 7.821 × 10^"-19"color(white)(l) "J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 5. Calculate the energy absorbed by all the <mathjax>#"O"_3#</mathjax> molecules</strong></p>
<p><mathjax>#"Total energy" = 1.13 × 10^16 color(red)(cancel(color(black)("O"_3color(white)(l) "molecules"))) × (7.821 × 10^"-19"color(white)(l) "J")/(1 color(red)(cancel(color(black)("O"_3color(white)(l) "molecule")))) = 8.9 × 10^"-3"color(white)(l) "J" = "8.9 mJ"#</mathjax></p></div>
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</article> | How much energy, in joules, must be absorbed if all the #O_3# molecules in the sample of air are to dissociate?
Assume that each photon absorbed causes one #O_3# molecule to dissociate, and that the wavelength of the radiation is 254 nm. |
Information provided with question
Ozone, #O_3#, absorbs ultraviolet radiation and dissociates into #O_2# molecules and #O# atoms:
#O_3+h\nu→O_2+O.#
A 1.85-#L# sample of air at 22#°C# and 748 #mmHg# contains 0.25 ppm of #O_3#.
My thoughts
I assume "parts per million" (ppm) is "#(mL)/L#"?
Also, I don't really know what section I should be posting this in, sorry.
|
1,182 | a86c08d8-6ddd-11ea-bcbe-ccda262736ce | https://socratic.org/questions/if-980-kj-of-energy-are-added-to-6-2-l-of-water-at-291-k-what-wil-the-final-temp | 328.81 K | start physical_unit 21 22 temperature k qc_end physical_unit 21 22 1 2 energy qc_end physical_unit 21 22 8 9 volume qc_end physical_unit 21 22 13 14 temperature qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the water [IN] K"}] | [{"type":"physical unit","value":"328.81 K"}] | [{"type":"physical unit","value":"Added energy [OF] the water [=] \\pu{980 kJ}"},{"type":"physical unit","value":"Volume [OF] the water [=] \\pu{6.2 L}"},{"type":"physical unit","value":"Temperature1 [OF] the water [=] \\pu{291 K}"}] | <h1 class="questionTitle" itemprop="name">If 980 kJ of energy are added to 6.2 L of water at 291 K, what wil the final temperature of the water be?</h1> | null | 328.81 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation <mathjax>#q=mcDeltaT#</mathjax>, where <mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature. The mass of water is 1000g per litre, so <mathjax>#6.2L = 6200g#</mathjax>. The specific heat capacity of water is <mathjax>#~~4.18 J*g^-1*K^-1#</mathjax>, so:<br/>
<mathjax>#980,000 = 6,200 * 4.18 * DeltaT#</mathjax><br/>
<mathjax>#DeltaT = \frac{980,000}{6,200 * 4.18} = 37.81K#</mathjax><br/>
<mathjax>#:. temperature = 291 + 37.81 = 328.81K#</mathjax> .<br/>
But realistically it would never be this high as energy will escape to the surroundings. This is only a theoretical value.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>328.81k</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation <mathjax>#q=mcDeltaT#</mathjax>, where <mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature. The mass of water is 1000g per litre, so <mathjax>#6.2L = 6200g#</mathjax>. The specific heat capacity of water is <mathjax>#~~4.18 J*g^-1*K^-1#</mathjax>, so:<br/>
<mathjax>#980,000 = 6,200 * 4.18 * DeltaT#</mathjax><br/>
<mathjax>#DeltaT = \frac{980,000}{6,200 * 4.18} = 37.81K#</mathjax><br/>
<mathjax>#:. temperature = 291 + 37.81 = 328.81K#</mathjax> .<br/>
But realistically it would never be this high as energy will escape to the surroundings. This is only a theoretical value.</p></div>
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<h1 class="questionTitle" itemprop="name">If 980 kJ of energy are added to 6.2 L of water at 291 K, what wil the final temperature of the water be?</h1>
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<div class="markdown"><p>328.81k</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation <mathjax>#q=mcDeltaT#</mathjax>, where <mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature. The mass of water is 1000g per litre, so <mathjax>#6.2L = 6200g#</mathjax>. The specific heat capacity of water is <mathjax>#~~4.18 J*g^-1*K^-1#</mathjax>, so:<br/>
<mathjax>#980,000 = 6,200 * 4.18 * DeltaT#</mathjax><br/>
<mathjax>#DeltaT = \frac{980,000}{6,200 * 4.18} = 37.81K#</mathjax><br/>
<mathjax>#:. temperature = 291 + 37.81 = 328.81K#</mathjax> .<br/>
But realistically it would never be this high as energy will escape to the surroundings. This is only a theoretical value.</p></div>
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</article> | If 980 kJ of energy are added to 6.2 L of water at 291 K, what wil the final temperature of the water be? | null |
1,183 | ac30dc5b-6ddd-11ea-b019-ccda262736ce | https://socratic.org/questions/how-many-grams-of-cobalt-ii-chloride-are-needed-to-react-completely-with-68-9-ml | 1.64 grams | start physical_unit 4 5 mass g qc_end c_other OTHER qc_end physical_unit 17 18 12 13 volume qc_end physical_unit 17 18 15 16 molarity qc_end chemical_equation 26 31 qc_end end | [{"type":"physical unit","value":"Mass [OF] cobalt(II) chloride [IN] grams"}] | [{"type":"physical unit","value":"1.64 grams"}] | [{"type":"other","value":"React completely."},{"type":"physical unit","value":"Volume [OF] KOH solution [=] \\pu{68.9 mL}"},{"type":"physical unit","value":"Molarity [OF] KOH solution [=] \\pu{0.366 M}"},{"type":"chemical equation","value":"Co^2+(aq) + 2 OH-(aq) -> Co(OH)2(s)"}] | <h1 class="questionTitle" itemprop="name">How many grams of cobalt(II) chloride are needed to react completely with 68.9 mL of .366 M KOH solution, if the equation for the reaction is #Co^(2+) (aq) + 2OH^(-)(aq) -> Co(OH)_2(s)#?</h1> | null | 1.64 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy for this problem will be to use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <em>potassium hydroxide</em> solution, <mathjax>#"KOH"#</mathjax>, to find the number of <em>moles</em> of potassium hydroxide that take part in the reaction. </p>
<p>Once you know how many moles of potassium hydroxide you have, you can use the <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine the number of moles of cobalt(II) chloride needed for the reaction. </p>
<p>So, molarity is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by <em>liters</em> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>This means that the number of moles of potassium hydroxide will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.366 M" * 68.9 * 10^(-3)"L" = "0.02522 moles KOH"#</mathjax></p>
</blockquote>
<p>Use the <mathjax>#1:2#</mathjax> mole ratio that exists between the two compounds to get the number of moles of cobalt(II) chloride needed for the reaction</p>
<blockquote>
<p><mathjax>#0.02522 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole CoCl"_2)/(2color(red)(cancel(color(black)("moles KOH")))) = "0.01261 moles CoCl"_2#</mathjax></p>
</blockquote>
<p>Finally, use cobalt(II) chloride's <strong>molar mass</strong> to determine how many grams would contain this many moles </p>
<blockquote>
<p><mathjax>#0.01261 color(red)(cancel(color(black)("moles"))) * "129.839 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.64 g CoCl"_2)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.64 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy for this problem will be to use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <em>potassium hydroxide</em> solution, <mathjax>#"KOH"#</mathjax>, to find the number of <em>moles</em> of potassium hydroxide that take part in the reaction. </p>
<p>Once you know how many moles of potassium hydroxide you have, you can use the <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine the number of moles of cobalt(II) chloride needed for the reaction. </p>
<p>So, molarity is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by <em>liters</em> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>This means that the number of moles of potassium hydroxide will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.366 M" * 68.9 * 10^(-3)"L" = "0.02522 moles KOH"#</mathjax></p>
</blockquote>
<p>Use the <mathjax>#1:2#</mathjax> mole ratio that exists between the two compounds to get the number of moles of cobalt(II) chloride needed for the reaction</p>
<blockquote>
<p><mathjax>#0.02522 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole CoCl"_2)/(2color(red)(cancel(color(black)("moles KOH")))) = "0.01261 moles CoCl"_2#</mathjax></p>
</blockquote>
<p>Finally, use cobalt(II) chloride's <strong>molar mass</strong> to determine how many grams would contain this many moles </p>
<blockquote>
<p><mathjax>#0.01261 color(red)(cancel(color(black)("moles"))) * "129.839 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.64 g CoCl"_2)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of cobalt(II) chloride are needed to react completely with 68.9 mL of .366 M KOH solution, if the equation for the reaction is #Co^(2+) (aq) + 2OH^(-)(aq) -> Co(OH)_2(s)#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-27T19:42:29" itemprop="dateCreated">
Nov 27, 2015
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<div class="markdown"><p><mathjax>#"1.64 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy for this problem will be to use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <em>potassium hydroxide</em> solution, <mathjax>#"KOH"#</mathjax>, to find the number of <em>moles</em> of potassium hydroxide that take part in the reaction. </p>
<p>Once you know how many moles of potassium hydroxide you have, you can use the <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to determine the number of moles of cobalt(II) chloride needed for the reaction. </p>
<p>So, molarity is defined as moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> divided by <em>liters</em> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
</blockquote>
<p>This means that the number of moles of potassium hydroxide will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n_(KOH) = "0.366 M" * 68.9 * 10^(-3)"L" = "0.02522 moles KOH"#</mathjax></p>
</blockquote>
<p>Use the <mathjax>#1:2#</mathjax> mole ratio that exists between the two compounds to get the number of moles of cobalt(II) chloride needed for the reaction</p>
<blockquote>
<p><mathjax>#0.02522 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole CoCl"_2)/(2color(red)(cancel(color(black)("moles KOH")))) = "0.01261 moles CoCl"_2#</mathjax></p>
</blockquote>
<p>Finally, use cobalt(II) chloride's <strong>molar mass</strong> to determine how many grams would contain this many moles </p>
<blockquote>
<p><mathjax>#0.01261 color(red)(cancel(color(black)("moles"))) * "129.839 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.64 g CoCl"_2)#</mathjax></p>
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</article> | How many grams of cobalt(II) chloride are needed to react completely with 68.9 mL of .366 M KOH solution, if the equation for the reaction is #Co^(2+) (aq) + 2OH^(-)(aq) -> Co(OH)_2(s)#? | null |
1,184 | ab5ee08c-6ddd-11ea-9591-ccda262736ce | https://socratic.org/questions/if-a-student-needs-to-make-exactly-2-5-liters-of-a-1-25-m-solution-of-acetic-aci | 0.26 L | start physical_unit 21 22 volume l qc_end physical_unit 21 22 19 20 molarity qc_end physical_unit 13 13 7 8 volume qc_end physical_unit 15 16 11 12 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] stock solution [IN] L"}] | [{"type":"physical unit","value":"0.26 L"}] | [{"type":"physical unit","value":"Molarity2 [OF] stock solution [=] \\pu{12.0 M}"},{"type":"physical unit","value":"Volume1 [OF] acetic acid solution [=] \\pu{2.5 liters}"},{"type":"physical unit","value":"Molarity1 [OF] acetic acid solution [=] \\pu{1.25 M}"}] | <h1 class="questionTitle" itemprop="name">If a student needs to make exactly 2.5 liters of a 1.25 M solution of acetic acid from the 12.0 M stock solution in the chemistry closet, what volume of the stock solution should be used?</h1> | null | 0.26 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> to figure out the volume of <em>stock solution</em> needed to make <mathjax>#"2.5 L"#</mathjax> of <em>diluted solution</em>.</p>
<p>As you know, you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>, i.e. by <em>diluting</em> it. </p>
<p>The important thing to remember here is that if you increase the volume of a solution by a specific factor, the concentration <strong>must decrease</strong> by the same factor. </p>
<p>This factor is called the <em>dilution factor</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"stock" = c_"stock"/c_"diluted"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution must go from <mathjax>#"12.0 M"#</mathjax>, which is what you have for the stock solution, to <mathjax>#"1.25 M"#</mathjax>. </p>
<p>The dilution factor will thus be </p>
<blockquote>
<p><mathjax>#"D.F." = (12.0 color(red)(cancel(color(black)("M"))))/(1.25color(red)(cancel(color(black)("M")))) = 9.6#</mathjax></p>
</blockquote>
<p>This means that the volume of the stock solution used for the dilution was </p>
<blockquote>
<p><mathjax>#"D.F." = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "2.5 L"/9.6 = "0.26 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em> and rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("260 mL")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"260 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> to figure out the volume of <em>stock solution</em> needed to make <mathjax>#"2.5 L"#</mathjax> of <em>diluted solution</em>.</p>
<p>As you know, you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>, i.e. by <em>diluting</em> it. </p>
<p>The important thing to remember here is that if you increase the volume of a solution by a specific factor, the concentration <strong>must decrease</strong> by the same factor. </p>
<p>This factor is called the <em>dilution factor</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"stock" = c_"stock"/c_"diluted"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution must go from <mathjax>#"12.0 M"#</mathjax>, which is what you have for the stock solution, to <mathjax>#"1.25 M"#</mathjax>. </p>
<p>The dilution factor will thus be </p>
<blockquote>
<p><mathjax>#"D.F." = (12.0 color(red)(cancel(color(black)("M"))))/(1.25color(red)(cancel(color(black)("M")))) = 9.6#</mathjax></p>
</blockquote>
<p>This means that the volume of the stock solution used for the dilution was </p>
<blockquote>
<p><mathjax>#"D.F." = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "2.5 L"/9.6 = "0.26 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em> and rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("260 mL")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If a student needs to make exactly 2.5 liters of a 1.25 M solution of acetic acid from the 12.0 M stock solution in the chemistry closet, what volume of the stock solution should be used?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-01T00:10:15" itemprop="dateCreated">
Jul 1, 2016
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<div class="markdown"><p><mathjax>#"260 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you can use the <strong><a href="https://socratic.org/questions/how-do-you-calculate-dilution-factor">dilution factor</a></strong> to figure out the volume of <em>stock solution</em> needed to make <mathjax>#"2.5 L"#</mathjax> of <em>diluted solution</em>.</p>
<p>As you know, you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its volume while keeping the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>constant</strong>, i.e. by <em>diluting</em> it. </p>
<p>The important thing to remember here is that if you increase the volume of a solution by a specific factor, the concentration <strong>must decrease</strong> by the same factor. </p>
<p>This factor is called the <em>dilution factor</em></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"stock" = c_"stock"/c_"diluted"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of the solution must go from <mathjax>#"12.0 M"#</mathjax>, which is what you have for the stock solution, to <mathjax>#"1.25 M"#</mathjax>. </p>
<p>The dilution factor will thus be </p>
<blockquote>
<p><mathjax>#"D.F." = (12.0 color(red)(cancel(color(black)("M"))))/(1.25color(red)(cancel(color(black)("M")))) = 9.6#</mathjax></p>
</blockquote>
<p>This means that the volume of the stock solution used for the dilution was </p>
<blockquote>
<p><mathjax>#"D.F." = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"D.F."#</mathjax></p>
<p><mathjax>#V_"stock" = "2.5 L"/9.6 = "0.26 L"#</mathjax></p>
</blockquote>
<p>Expressed in <em>milliliters</em> and rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("260 mL")color(white)(a/a)|)))#</mathjax></p>
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</article> | If a student needs to make exactly 2.5 liters of a 1.25 M solution of acetic acid from the 12.0 M stock solution in the chemistry closet, what volume of the stock solution should be used? | null |
1,185 | ab209edc-6ddd-11ea-b9d7-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-the-k-atom-in-kclo4 | +1 | start physical_unit 7 8 oxidation_state none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] K atom"}] | [{"type":"physical unit","value":"+1"}] | [{"type":"chemical equation","value":"KClO4"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of the K atom in KClO4?
</h1> | null | +1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Rules to remember when trying to find out the oxidation state of an element:</p>
<p>(1) The total charge of a stable compound is always equal to zero (meaning no charge). </p>
<p>For example, the <mathjax>#H_2O#</mathjax> molecule exists as a neutrally charged substance. </p>
<p>(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of <mathjax>#NO_3^"-1"#</mathjax> ion is -1).</p>
<p>(3) All <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from Group 1A has an oxidation state of +1 (e.g. <mathjax>#Na^"+1"#</mathjax>, <mathjax>#Li^"+1"#</mathjax>). All Group 2A and 3A <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have an oxidation state of +2 and +3, respectively. (e.g. <mathjax>#Ca^"2+"#</mathjax>, <mathjax>#Mg^"2+"#</mathjax>, <mathjax>#Al^"3+"#</mathjax>)</p>
<p>(4) Oxygen always have a charge -2 except for peroxide ion (<mathjax>#O_2^"2-"#</mathjax>) which has a charge of -1.</p>
<p>(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of <mathjax>#HCl#</mathjax>) and always have a -1 charge if it is bonded with a metal (as in <mathjax>#AlH_3#</mathjax>).</p>
<p>For your question, <mathjax>#KClO_4#</mathjax>, based on rule 3, the answer is +1.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>+1</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Rules to remember when trying to find out the oxidation state of an element:</p>
<p>(1) The total charge of a stable compound is always equal to zero (meaning no charge). </p>
<p>For example, the <mathjax>#H_2O#</mathjax> molecule exists as a neutrally charged substance. </p>
<p>(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of <mathjax>#NO_3^"-1"#</mathjax> ion is -1).</p>
<p>(3) All <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from Group 1A has an oxidation state of +1 (e.g. <mathjax>#Na^"+1"#</mathjax>, <mathjax>#Li^"+1"#</mathjax>). All Group 2A and 3A <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have an oxidation state of +2 and +3, respectively. (e.g. <mathjax>#Ca^"2+"#</mathjax>, <mathjax>#Mg^"2+"#</mathjax>, <mathjax>#Al^"3+"#</mathjax>)</p>
<p>(4) Oxygen always have a charge -2 except for peroxide ion (<mathjax>#O_2^"2-"#</mathjax>) which has a charge of -1.</p>
<p>(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of <mathjax>#HCl#</mathjax>) and always have a -1 charge if it is bonded with a metal (as in <mathjax>#AlH_3#</mathjax>).</p>
<p>For your question, <mathjax>#KClO_4#</mathjax>, based on rule 3, the answer is +1.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of the K atom in KClO4?
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<div class="markdown"><p>+1</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Rules to remember when trying to find out the oxidation state of an element:</p>
<p>(1) The total charge of a stable compound is always equal to zero (meaning no charge). </p>
<p>For example, the <mathjax>#H_2O#</mathjax> molecule exists as a neutrally charged substance. </p>
<p>(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of <mathjax>#NO_3^"-1"#</mathjax> ion is -1).</p>
<p>(3) All <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from Group 1A has an oxidation state of +1 (e.g. <mathjax>#Na^"+1"#</mathjax>, <mathjax>#Li^"+1"#</mathjax>). All Group 2A and 3A <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> have an oxidation state of +2 and +3, respectively. (e.g. <mathjax>#Ca^"2+"#</mathjax>, <mathjax>#Mg^"2+"#</mathjax>, <mathjax>#Al^"3+"#</mathjax>)</p>
<p>(4) Oxygen always have a charge -2 except for peroxide ion (<mathjax>#O_2^"2-"#</mathjax>) which has a charge of -1.</p>
<p>(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of <mathjax>#HCl#</mathjax>) and always have a -1 charge if it is bonded with a metal (as in <mathjax>#AlH_3#</mathjax>).</p>
<p>For your question, <mathjax>#KClO_4#</mathjax>, based on rule 3, the answer is +1.</p></div>
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</article> | What is the oxidation state of the K atom in KClO4?
| null |
1,186 | ab437e9e-6ddd-11ea-b07d-ccda262736ce | https://socratic.org/questions/how-many-liters-of-pure-acid-should-be-added-to-25-liters-of-a-60-solution-of-ac | 15 liters | start physical_unit 4 5 volume l qc_end physical_unit 22 23 10 11 volume qc_end physical_unit 5 5 14 14 percent qc_end physical_unit 5 5 21 21 percent qc_end end | [{"type":"physical unit","value":"Added volume [OF] pure acid [IN] liters"}] | [{"type":"physical unit","value":"15 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] acid solution [=] \\pu{25 liters}"},{"type":"physical unit","value":"Percent1 [OF] acid in solution [=] \\pu{60%}"},{"type":"physical unit","value":"Percent2 [OF] acid in solution [=] \\pu{75%}"}] | <h1 class="questionTitle" itemprop="name">How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution?</h1> | null | 15 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea behind this problem is that you're adding <em>pure acid</em> to a solution that is <strong>60% v/v</strong> acid, so the added acid will change the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the starting solution by increasing the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> <strong>by the same amount</strong>. </p>
<p>So, start by calculating how much acid you have in your starting solution</p>
<p><mathjax>#25color(red)(cancel(color(black)("L solution"))) * "60 L acid"/(100color(red)(cancel(color(black)("L solution")))) = "15 L acid"#</mathjax></p>
<p>Let's say that <mathjax>#x#</mathjax> denotes the volume of pure acid that you must add to your starting solution. Since you're dealing with <em>pure acid</em>, the volume of acid will <em>Increase</em> by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"acid" = 15 + x#</mathjax></p>
<p>At the same time, the volume of the solution will also incrase by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"sol" = 25 + x#</mathjax></p>
<p>This means that the target solution's percent concetration by volume will be equal to</p>
<p><mathjax>#V_"acid"/V_"soL" * 100 = 75%#</mathjax></p>
<p><mathjax>#((15 + x)color(red)(cancel(color(black)("L"))))/((25 + x)color(red)(cancel(color(black)("L")))) * 100 = 75%#</mathjax></p>
<p>Solve this equation for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#(15+x) * 100 = (25 + x) * 75#</mathjax></p>
<p><mathjax>#1500 + 100x = 1875 + 75x#</mathjax></p>
<p><mathjax>#25x = 375 implies x = 375/25 = color(green)("15 L")#</mathjax></p>
<p>So, if you add <strong>15 L</strong> of pure acid to 25 L of 60% v/v acid solution, you'll get <strong>40 L</strong> of 75% v/v acid solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You need to add <strong>15 L</strong> of pure acid.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea behind this problem is that you're adding <em>pure acid</em> to a solution that is <strong>60% v/v</strong> acid, so the added acid will change the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the starting solution by increasing the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> <strong>by the same amount</strong>. </p>
<p>So, start by calculating how much acid you have in your starting solution</p>
<p><mathjax>#25color(red)(cancel(color(black)("L solution"))) * "60 L acid"/(100color(red)(cancel(color(black)("L solution")))) = "15 L acid"#</mathjax></p>
<p>Let's say that <mathjax>#x#</mathjax> denotes the volume of pure acid that you must add to your starting solution. Since you're dealing with <em>pure acid</em>, the volume of acid will <em>Increase</em> by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"acid" = 15 + x#</mathjax></p>
<p>At the same time, the volume of the solution will also incrase by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"sol" = 25 + x#</mathjax></p>
<p>This means that the target solution's percent concetration by volume will be equal to</p>
<p><mathjax>#V_"acid"/V_"soL" * 100 = 75%#</mathjax></p>
<p><mathjax>#((15 + x)color(red)(cancel(color(black)("L"))))/((25 + x)color(red)(cancel(color(black)("L")))) * 100 = 75%#</mathjax></p>
<p>Solve this equation for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#(15+x) * 100 = (25 + x) * 75#</mathjax></p>
<p><mathjax>#1500 + 100x = 1875 + 75x#</mathjax></p>
<p><mathjax>#25x = 375 implies x = 375/25 = color(green)("15 L")#</mathjax></p>
<p>So, if you add <strong>15 L</strong> of pure acid to 25 L of 60% v/v acid solution, you'll get <strong>40 L</strong> of 75% v/v acid solution. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution?</h1>
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<div class="markdown"><p>You need to add <strong>15 L</strong> of pure acid.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea behind this problem is that you're adding <em>pure acid</em> to a solution that is <strong>60% v/v</strong> acid, so the added acid will change the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the starting solution by increasing the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and the volume of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> <strong>by the same amount</strong>. </p>
<p>So, start by calculating how much acid you have in your starting solution</p>
<p><mathjax>#25color(red)(cancel(color(black)("L solution"))) * "60 L acid"/(100color(red)(cancel(color(black)("L solution")))) = "15 L acid"#</mathjax></p>
<p>Let's say that <mathjax>#x#</mathjax> denotes the volume of pure acid that you must add to your starting solution. Since you're dealing with <em>pure acid</em>, the volume of acid will <em>Increase</em> by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"acid" = 15 + x#</mathjax></p>
<p>At the same time, the volume of the solution will also incrase by <mathjax>#x#</mathjax></p>
<p><mathjax>#V_"sol" = 25 + x#</mathjax></p>
<p>This means that the target solution's percent concetration by volume will be equal to</p>
<p><mathjax>#V_"acid"/V_"soL" * 100 = 75%#</mathjax></p>
<p><mathjax>#((15 + x)color(red)(cancel(color(black)("L"))))/((25 + x)color(red)(cancel(color(black)("L")))) * 100 = 75%#</mathjax></p>
<p>Solve this equation for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#(15+x) * 100 = (25 + x) * 75#</mathjax></p>
<p><mathjax>#1500 + 100x = 1875 + 75x#</mathjax></p>
<p><mathjax>#25x = 375 implies x = 375/25 = color(green)("15 L")#</mathjax></p>
<p>So, if you add <strong>15 L</strong> of pure acid to 25 L of 60% v/v acid solution, you'll get <strong>40 L</strong> of 75% v/v acid solution. </p></div>
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</article> | How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution? | null |
1,187 | abe0d480-6ddd-11ea-8eb6-ccda262736ce | https://socratic.org/questions/a-compound-has-a-molar-mass-of-142-00g-mol-and-the-percent-composition-is-50-7-c | C6H6O4 | start chemical_formula qc_end physical_unit 1 1 7 8 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C6H6O4"}] | [{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{142.00 g/mol}"},{"type":"physical unit","value":"Percent composition [OF] Carbon in the compound [=] \\pu{50.7%}"},{"type":"physical unit","value":"Percent composition [OF] Hydrogen in the compound [=] \\pu{4.2%}"},{"type":"physical unit","value":"Percent composition [OF] Oxygen in the compound [=] \\pu{45.1%}"}] | <h1 class="questionTitle" itemprop="name">A compound has a molar mass of 142.00g/mol and the percent composition is 50.7% Carbon, 4.2% Hydrogen and 45.1% Oxygen. What is the molecular formula?</h1> | null | C6H6O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time a problem provides you with a compound's <strong>molar mass</strong>, you can use a little trick to help you determine the compound's <em>molecular mass</em> faster. </p>
<p>More specifically, instead of determining the <em>empirical formula</em> first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether. </p>
<p>As you know, a compound's molar mass tells you what the mass of <strong>one mole </strong> of that substance is. In this case, one mole of your compound has a mass of <mathjax>#"142.00 g"#</mathjax>. </p>
<p>This means that if you pick a sample of <mathjax>#"142.00 g"#</mathjax> of this compound, you can use its <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to get the <strong>exact</strong> number of moles of each element you get <em>per mole</em> of compound. </p>
<p>So, you know that this compound's <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is as follows</p>
<blockquote>
<ul>
<li><em>carbon</em> <mathjax>#-> 50.7%#</mathjax></li>
<li><em>hydrogen</em> <mathjax>#-> 4.2%#</mathjax></li>
<li><em>oxygen</em> <mathjax>#-> 45.1%#</mathjax></li>
</ul>
</blockquote>
<p>This means that it contains </p>
<blockquote>
<p><mathjax>#142.00 color(red)(cancel(color(black)("g compound"))) * "50.7 g C"/(100color(red)(cancel(color(black)("g compound")))) = "71.994 g C"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "4.2 g H"/(100color(red)(cancel(color(black)("g compound")))) = "5.964 g H"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "45.1 g O"/(100color(red)(cancel(color(black)("g compound")))) = "64.042 g O"#</mathjax></p>
</blockquote>
<p>Now all you have to do is use the molar masses of these three <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you get in <strong>one mole</strong> of your compound</p>
<blockquote>
<p><mathjax>#"For C: " 71.994 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 5.994 ~~ 6#</mathjax></p>
<p><mathjax>#"For H: " 5.964 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 5.92 ~~6#</mathjax></p>
<p><mathjax>#"For O: " 64.042 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 4.003 ~~4#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for this compound will be </p>
<blockquote>
<p><mathjax>#"C"_6"H"_6"O"_4#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>I recommend finding the empirical formula first, then using the molar mass to find the molecular formula. The result will be exactly the same.</em></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_6"H"_6"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time a problem provides you with a compound's <strong>molar mass</strong>, you can use a little trick to help you determine the compound's <em>molecular mass</em> faster. </p>
<p>More specifically, instead of determining the <em>empirical formula</em> first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether. </p>
<p>As you know, a compound's molar mass tells you what the mass of <strong>one mole </strong> of that substance is. In this case, one mole of your compound has a mass of <mathjax>#"142.00 g"#</mathjax>. </p>
<p>This means that if you pick a sample of <mathjax>#"142.00 g"#</mathjax> of this compound, you can use its <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to get the <strong>exact</strong> number of moles of each element you get <em>per mole</em> of compound. </p>
<p>So, you know that this compound's <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is as follows</p>
<blockquote>
<ul>
<li><em>carbon</em> <mathjax>#-> 50.7%#</mathjax></li>
<li><em>hydrogen</em> <mathjax>#-> 4.2%#</mathjax></li>
<li><em>oxygen</em> <mathjax>#-> 45.1%#</mathjax></li>
</ul>
</blockquote>
<p>This means that it contains </p>
<blockquote>
<p><mathjax>#142.00 color(red)(cancel(color(black)("g compound"))) * "50.7 g C"/(100color(red)(cancel(color(black)("g compound")))) = "71.994 g C"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "4.2 g H"/(100color(red)(cancel(color(black)("g compound")))) = "5.964 g H"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "45.1 g O"/(100color(red)(cancel(color(black)("g compound")))) = "64.042 g O"#</mathjax></p>
</blockquote>
<p>Now all you have to do is use the molar masses of these three <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you get in <strong>one mole</strong> of your compound</p>
<blockquote>
<p><mathjax>#"For C: " 71.994 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 5.994 ~~ 6#</mathjax></p>
<p><mathjax>#"For H: " 5.964 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 5.92 ~~6#</mathjax></p>
<p><mathjax>#"For O: " 64.042 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 4.003 ~~4#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for this compound will be </p>
<blockquote>
<p><mathjax>#"C"_6"H"_6"O"_4#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>I recommend finding the empirical formula first, then using the molar mass to find the molecular formula. The result will be exactly the same.</em></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A compound has a molar mass of 142.00g/mol and the percent composition is 50.7% Carbon, 4.2% Hydrogen and 45.1% Oxygen. What is the molecular formula?</h1>
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Stefan V.
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Dec 12, 2015
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<div class="markdown"><p><mathjax>#"C"_6"H"_6"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every time a problem provides you with a compound's <strong>molar mass</strong>, you can use a little trick to help you determine the compound's <em>molecular mass</em> faster. </p>
<p>More specifically, instead of determining the <em>empirical formula</em> first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether. </p>
<p>As you know, a compound's molar mass tells you what the mass of <strong>one mole </strong> of that substance is. In this case, one mole of your compound has a mass of <mathjax>#"142.00 g"#</mathjax>. </p>
<p>This means that if you pick a sample of <mathjax>#"142.00 g"#</mathjax> of this compound, you can use its <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to get the <strong>exact</strong> number of moles of each element you get <em>per mole</em> of compound. </p>
<p>So, you know that this compound's <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> is as follows</p>
<blockquote>
<ul>
<li><em>carbon</em> <mathjax>#-> 50.7%#</mathjax></li>
<li><em>hydrogen</em> <mathjax>#-> 4.2%#</mathjax></li>
<li><em>oxygen</em> <mathjax>#-> 45.1%#</mathjax></li>
</ul>
</blockquote>
<p>This means that it contains </p>
<blockquote>
<p><mathjax>#142.00 color(red)(cancel(color(black)("g compound"))) * "50.7 g C"/(100color(red)(cancel(color(black)("g compound")))) = "71.994 g C"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "4.2 g H"/(100color(red)(cancel(color(black)("g compound")))) = "5.964 g H"#</mathjax></p>
<p><mathjax>#142.00color(red)(cancel(color(black)("g compound"))) * "45.1 g O"/(100color(red)(cancel(color(black)("g compound")))) = "64.042 g O"#</mathjax></p>
</blockquote>
<p>Now all you have to do is use the molar masses of these three <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to determine how many <em>moles</em> of each you get in <strong>one mole</strong> of your compound</p>
<blockquote>
<p><mathjax>#"For C: " 71.994 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 5.994 ~~ 6#</mathjax></p>
<p><mathjax>#"For H: " 5.964 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 5.92 ~~6#</mathjax></p>
<p><mathjax>#"For O: " 64.042 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 4.003 ~~4#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> for this compound will be </p>
<blockquote>
<p><mathjax>#"C"_6"H"_6"O"_4#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>I recommend finding the empirical formula first, then using the molar mass to find the molecular formula. The result will be exactly the same.</em></p></div>
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</article> | A compound has a molar mass of 142.00g/mol and the percent composition is 50.7% Carbon, 4.2% Hydrogen and 45.1% Oxygen. What is the molecular formula? | null |
1,188 | a9ce3549-6ddd-11ea-8cff-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-30-grams-of-h-3po-4 | 0.31 moles | start physical_unit 9 9 mole mol qc_end physical_unit 9 9 6 7 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] H3PO4 [IN] moles"}] | [{"type":"physical unit","value":"0.31 moles"}] | [{"type":"physical unit","value":"Mass [OF] H3PO4 [=] \\pu{30 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in 30 grams of #H_3PO_4#?</h1> | null | 0.31 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles = Mass / Molar mass.</p>
<p><mathjax>#n=m/(Mr)#</mathjax></p>
<p><mathjax>#=(30g)/((1xx3)+31+(16xx4)#</mathjax></p>
<p><mathjax>#=0,3061 mol#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#0,3061mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles = Mass / Molar mass.</p>
<p><mathjax>#n=m/(Mr)#</mathjax></p>
<p><mathjax>#=(30g)/((1xx3)+31+(16xx4)#</mathjax></p>
<p><mathjax>#=0,3061 mol#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles are there in 30 grams of #H_3PO_4#?</h1>
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<div class="markdown"><p><mathjax>#0,3061mol#</mathjax></p></div>
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<div class="markdown"><p>Moles = Mass / Molar mass.</p>
<p><mathjax>#n=m/(Mr)#</mathjax></p>
<p><mathjax>#=(30g)/((1xx3)+31+(16xx4)#</mathjax></p>
<p><mathjax>#=0,3061 mol#</mathjax></p></div>
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</article> | How many moles are there in 30 grams of #H_3PO_4#? | null |
1,189 | ac8fbd02-6ddd-11ea-9bf4-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-36-8-g-of-carbon-monoxide-at-stp | 29.4 L | start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon monoxide [IN] L"}] | [{"type":"physical unit","value":"29.4 L"}] | [{"type":"physical unit","value":"Mass [OF] carbon monoxide [=] \\pu{36.8 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 36.8 g of carbon monoxide at STP?</h1> | null | 29.4 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Let us first find the number of mole of carbon monoxide in <mathjax>#36.8g#</mathjax>:</p>
<p><mathjax>#n=m/(MM)=(36.8cancel(g))/(28.0cancel(g)/(mol))=1.31mol#</mathjax></p>
<p>At STP, the molar volume is equal to <mathjax>#22.4L/(mol)#</mathjax>, therefore, the volume that will be occupied by <mathjax>#36.8g#</mathjax> carbon monoxide is:</p>
<p><mathjax>#V=1.31cancel(mol)xx(22.4L)/(1cancel(mol))=29.4L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V=29.4L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let us first find the number of mole of carbon monoxide in <mathjax>#36.8g#</mathjax>:</p>
<p><mathjax>#n=m/(MM)=(36.8cancel(g))/(28.0cancel(g)/(mol))=1.31mol#</mathjax></p>
<p>At STP, the molar volume is equal to <mathjax>#22.4L/(mol)#</mathjax>, therefore, the volume that will be occupied by <mathjax>#36.8g#</mathjax> carbon monoxide is:</p>
<p><mathjax>#V=1.31cancel(mol)xx(22.4L)/(1cancel(mol))=29.4L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 36.8 g of carbon monoxide at STP?</h1>
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<div class="markdown"><p><mathjax>#V=29.4L#</mathjax></p></div>
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<div class="markdown"><p>Let us first find the number of mole of carbon monoxide in <mathjax>#36.8g#</mathjax>:</p>
<p><mathjax>#n=m/(MM)=(36.8cancel(g))/(28.0cancel(g)/(mol))=1.31mol#</mathjax></p>
<p>At STP, the molar volume is equal to <mathjax>#22.4L/(mol)#</mathjax>, therefore, the volume that will be occupied by <mathjax>#36.8g#</mathjax> carbon monoxide is:</p>
<p><mathjax>#V=1.31cancel(mol)xx(22.4L)/(1cancel(mol))=29.4L#</mathjax></p></div>
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</article> | What is the volume of 36.8 g of carbon monoxide at STP? | null |
1,190 | ab21888d-6ddd-11ea-a4e6-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-two-moles-of-oxygen-gas-o2 | 64.00 grams | start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 gas [IN] grams"}] | [{"type":"physical unit","value":"64.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] O2 gas [=] \\pu{2 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of two moles of oxygen gas(O2)?</h1> | null | 64.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of oxygen is <mathjax>#"15.9994 g"#</mathjax>, but because oxygen is a diatomic molecule, it is bonded with itself. So, </p>
<p><mathjax>#"15.9994 g" * 2 = "31.9988 g"#</mathjax> </p>
<p>per mole of oxygen. From here, the answer is simple. Just multiply <mathjax>#2#</mathjax> by <mathjax>#"31.9988 g"#</mathjax> because there are two moles of oxygen. So, the answer is <mathjax>#"63.9976 g"#</mathjax>..</p></div>
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<div class="markdown"><p><mathjax>#"63.9976 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of oxygen is <mathjax>#"15.9994 g"#</mathjax>, but because oxygen is a diatomic molecule, it is bonded with itself. So, </p>
<p><mathjax>#"15.9994 g" * 2 = "31.9988 g"#</mathjax> </p>
<p>per mole of oxygen. From here, the answer is simple. Just multiply <mathjax>#2#</mathjax> by <mathjax>#"31.9988 g"#</mathjax> because there are two moles of oxygen. So, the answer is <mathjax>#"63.9976 g"#</mathjax>..</p></div>
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Stefan V.
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<div class="markdown"><p><mathjax>#"63.9976 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>One mole of oxygen is <mathjax>#"15.9994 g"#</mathjax>, but because oxygen is a diatomic molecule, it is bonded with itself. So, </p>
<p><mathjax>#"15.9994 g" * 2 = "31.9988 g"#</mathjax> </p>
<p>per mole of oxygen. From here, the answer is simple. Just multiply <mathjax>#2#</mathjax> by <mathjax>#"31.9988 g"#</mathjax> because there are two moles of oxygen. So, the answer is <mathjax>#"63.9976 g"#</mathjax>..</p></div>
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</article> | What is the mass of two moles of oxygen gas(O2)? | null |
1,191 | a8c30fe3-6ddd-11ea-9431-ccda262736ce | https://socratic.org/questions/the-ph-of-a-0-100-m-solution-of-an-aqueous-weak-acid-ha-is-4-what-is-the-ka-for- | 1.0 × 10^(-7) | start physical_unit 20 22 ka none qc_end physical_unit 6 6 14 14 ph qc_end physical_unit 12 12 4 5 molarity qc_end end | [{"type":"physical unit","value":"Ka [OF] the weak acid"}] | [{"type":"physical unit","value":"1.0 × 10^(-7)"}] | [{"type":"physical unit","value":"pH [OF] HA solution [=] \\pu{4}"},{"type":"physical unit","value":"Molarity [OF] HA solution [=] \\pu{0.100 M}"}] | <h1 class="questionTitle" itemprop="name">The pH of a 0.100 M solution of an aqueous weak acid (#HA#) is 4. What is the #Ka# for the weak acid?</h1> | null | 1.0 × 10^(-7) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation for the <em>partial ionization</em> of the acid</p>
<blockquote>
<p><mathjax>#"HA"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p>Notice that you have <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> across the board. <strong>For every mole</strong> of acid <strong>that ionizes</strong> in aqueous solution, you get <strong>one mole</strong> of its <strong><a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a></strong> and <strong>one mole</strong> of hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>.</p>
<p>In other words, the equation produces <strong>equal concentrations</strong> of conjugate base and hydronium ions. </p>
<p>Now, you can use the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to calculate the <strong>equilibrium concentration</strong> of the hydronium ions. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log(["H"_3"O"^(+)]) implies ["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the pH of the solution is equal to <mathjax>#4#</mathjax>, which means that you'll have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>By definition, the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_a = (["A"^(-)] * ["H"_3"O"^(+)])/(["HA"])#</mathjax></p>
</blockquote>
<p>The expression for the acid dissociation constant is written using <strong>equilibrium concentrations</strong>. So, if the reaction produced a concentration of hydronium ions equal to <mathjax>#10^(-4)"M"#</mathjax>, it follows that it also produced a concentration of conjugate base equal to <mathjax>#10^(-4)"M"#</mathjax>.</p>
<p>Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you can <em>approximate</em> it to be <em>constant</em>. </p>
<p>This means that the acid dissociation constant for this acid will be </p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote>
<p>This is the underlying concept behind an <strong>ICE table</strong></p>
<blockquote>
<p><mathjax>#" ""HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "A"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.100" " " " " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.100-x" " " " " " " " " " " " " "x" " " " " " " " " " "x#</mathjax></p>
<p>Here <mathjax>#x#</mathjax> represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know that <mathjax>#x = 10^(-4)#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/(0.100 - 10^(-4))#</mathjax></p>
</blockquote>
<p>Once again, you can use</p>
<blockquote>
<p><mathjax>#0.100 - 10^(-4) = 0.0999 ~~ 0.100#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#K_a = 10^(-8)/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.0 * 10^(-7)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation for the <em>partial ionization</em> of the acid</p>
<blockquote>
<p><mathjax>#"HA"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p>Notice that you have <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> across the board. <strong>For every mole</strong> of acid <strong>that ionizes</strong> in aqueous solution, you get <strong>one mole</strong> of its <strong><a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a></strong> and <strong>one mole</strong> of hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>.</p>
<p>In other words, the equation produces <strong>equal concentrations</strong> of conjugate base and hydronium ions. </p>
<p>Now, you can use the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to calculate the <strong>equilibrium concentration</strong> of the hydronium ions. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log(["H"_3"O"^(+)]) implies ["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the pH of the solution is equal to <mathjax>#4#</mathjax>, which means that you'll have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>By definition, the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_a = (["A"^(-)] * ["H"_3"O"^(+)])/(["HA"])#</mathjax></p>
</blockquote>
<p>The expression for the acid dissociation constant is written using <strong>equilibrium concentrations</strong>. So, if the reaction produced a concentration of hydronium ions equal to <mathjax>#10^(-4)"M"#</mathjax>, it follows that it also produced a concentration of conjugate base equal to <mathjax>#10^(-4)"M"#</mathjax>.</p>
<p>Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you can <em>approximate</em> it to be <em>constant</em>. </p>
<p>This means that the acid dissociation constant for this acid will be </p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote>
<p>This is the underlying concept behind an <strong>ICE table</strong></p>
<blockquote>
<p><mathjax>#" ""HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "A"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.100" " " " " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.100-x" " " " " " " " " " " " " "x" " " " " " " " " " "x#</mathjax></p>
<p>Here <mathjax>#x#</mathjax> represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know that <mathjax>#x = 10^(-4)#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/(0.100 - 10^(-4))#</mathjax></p>
</blockquote>
<p>Once again, you can use</p>
<blockquote>
<p><mathjax>#0.100 - 10^(-4) = 0.0999 ~~ 0.100#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#K_a = 10^(-8)/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The pH of a 0.100 M solution of an aqueous weak acid (#HA#) is 4. What is the #Ka# for the weak acid?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-02-13T14:01:27" itemprop="dateCreated">
Feb 13, 2016
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<div>
<div class="markdown"><p><mathjax>#1.0 * 10^(-7)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing a balanced chemical equation for the <em>partial ionization</em> of the acid</p>
<blockquote>
<p><mathjax>#"HA"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p>Notice that you have <mathjax>#1:1#</mathjax> <strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> across the board. <strong>For every mole</strong> of acid <strong>that ionizes</strong> in aqueous solution, you get <strong>one mole</strong> of its <strong><a href="http://socratic.org/chemistry/acids-and-bases/conjugate-acids-and-conjugate-bases">conjugate base</a></strong> and <strong>one mole</strong> of hydronium ions, <mathjax>#"H"_3"O"^(+)#</mathjax>.</p>
<p>In other words, the equation produces <strong>equal concentrations</strong> of conjugate base and hydronium ions. </p>
<p>Now, you can use the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution to calculate the <strong>equilibrium concentration</strong> of the hydronium ions. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = - log(["H"_3"O"^(+)]) implies ["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the pH of the solution is equal to <mathjax>#4#</mathjax>, which means that you'll have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>By definition, the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_a = (["A"^(-)] * ["H"_3"O"^(+)])/(["HA"])#</mathjax></p>
</blockquote>
<p>The expression for the acid dissociation constant is written using <strong>equilibrium concentrations</strong>. So, if the reaction produced a concentration of hydronium ions equal to <mathjax>#10^(-4)"M"#</mathjax>, it follows that it also produced a concentration of conjugate base equal to <mathjax>#10^(-4)"M"#</mathjax>.</p>
<p>Because the initial concentration of the acid is considerably higher than the concentrations of the conjugate base and hydronium ions, you can <em>approximate</em> it to be <em>constant</em>. </p>
<p>This means that the acid dissociation constant for this acid will be </p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote>
<p>This is the underlying concept behind an <strong>ICE table</strong></p>
<blockquote>
<p><mathjax>#" ""HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "A"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "0.100" " " " " " " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.100-x" " " " " " " " " " " " " "x" " " " " " " " " " "x#</mathjax></p>
<p>Here <mathjax>#x#</mathjax> represents the equilibrium concentration for the conjugate acid and hydronium ions. Since you know that <mathjax>#x = 10^(-4)#</mathjax>, you will have</p>
<blockquote>
<p><mathjax>#K_a = (10^(-4) * 10^(-4))/(0.100 - 10^(-4))#</mathjax></p>
</blockquote>
<p>Once again, you can use</p>
<blockquote>
<p><mathjax>#0.100 - 10^(-4) = 0.0999 ~~ 0.100#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#K_a = 10^(-8)/0.100 = color(green)(1.0 * 10^(-7))#</mathjax></p>
</blockquote></div>
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</article> | The pH of a 0.100 M solution of an aqueous weak acid (#HA#) is 4. What is the #Ka# for the weak acid? | null |
1,192 | aaf08b0b-6ddd-11ea-bad5-ccda262736ce | https://socratic.org/questions/a-175-ml-sample-of-neon-had-its-pressure-changed-from-75-kpa-to-150-kpa-what-is- | 87.50 mL | start physical_unit 3 5 volume ml qc_end physical_unit 3 5 11 12 pressure qc_end physical_unit 3 5 14 15 pressure qc_end physical_unit 3 5 1 2 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] neon sample [IN] mL"}] | [{"type":"physical unit","value":"87.50 mL"}] | [{"type":"physical unit","value":"Pressure1 [OF] neon sample [=] \\pu{75 KPa}"},{"type":"physical unit","value":"Pressure2 [OF] neon sample [=] \\pu{150 KPa}"},{"type":"physical unit","value":"Volume1 [OF] neon sample [=] \\pu{175 mL}"}] | <h1 class="questionTitle" itemprop="name">A 175 mL sample of neon had its pressure changed from 75 KPa to 150 KPa. What is its new volume?</h1> | null | 87.50 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a> states that pressure is inversely proportional to volume for a fixed quantity of an ideal gas at a constant temperature.</p>
<p>The mathematical formula is : <mathjax>#V_2=(P_1V_1)/(P2)#</mathjax></p>
<p>Here we have: <mathjax>#P_1=75KPa; P_2=150KPa; V_1=175mL#</mathjax></p>
<p>Then: <mathjax>#V_2= (75KPa*175mL)/(150KPa)#</mathjax></p>
<p><mathjax>#V_2= (cancel(75KPa)1*175mL)/(cancel(150KPa)2)#</mathjax></p>
<p><mathjax>#V_2=87.5mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new volume of the neon is: <mathjax>#V_2=87.5mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a> states that pressure is inversely proportional to volume for a fixed quantity of an ideal gas at a constant temperature.</p>
<p>The mathematical formula is : <mathjax>#V_2=(P_1V_1)/(P2)#</mathjax></p>
<p>Here we have: <mathjax>#P_1=75KPa; P_2=150KPa; V_1=175mL#</mathjax></p>
<p>Then: <mathjax>#V_2= (75KPa*175mL)/(150KPa)#</mathjax></p>
<p><mathjax>#V_2= (cancel(75KPa)1*175mL)/(cancel(150KPa)2)#</mathjax></p>
<p><mathjax>#V_2=87.5mL#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A 175 mL sample of neon had its pressure changed from 75 KPa to 150 KPa. What is its new volume?</h1>
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EET-AP
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<div class="markdown"><p>The new volume of the neon is: <mathjax>#V_2=87.5mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a> states that pressure is inversely proportional to volume for a fixed quantity of an ideal gas at a constant temperature.</p>
<p>The mathematical formula is : <mathjax>#V_2=(P_1V_1)/(P2)#</mathjax></p>
<p>Here we have: <mathjax>#P_1=75KPa; P_2=150KPa; V_1=175mL#</mathjax></p>
<p>Then: <mathjax>#V_2= (75KPa*175mL)/(150KPa)#</mathjax></p>
<p><mathjax>#V_2= (cancel(75KPa)1*175mL)/(cancel(150KPa)2)#</mathjax></p>
<p><mathjax>#V_2=87.5mL#</mathjax></p></div>
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</article> | A 175 mL sample of neon had its pressure changed from 75 KPa to 150 KPa. What is its new volume? | null |
1,193 | aa7140b6-6ddd-11ea-9d23-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-the-compound-consisting-of-63-manganese-and-37 | MnO2 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"MnO2"}] | [{"type":"physical unit","value":"Percent by mass [OF] manganese in the compound [=] \\pu{63%}"},{"type":"physical unit","value":"Percent by mass [OF] oxygen in the compound [=] \\pu{37%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound consisting of 63% manganese and 37% oxygen by mass? </h1> | null | MnO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all empirical formula calculations, a starting mass of <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> is assumed. </p>
<p>There are thus <mathjax>#(63*g)/(54.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.15*mol#</mathjax> <mathjax>#Mn#</mathjax>, and <mathjax>#(37*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.132*mol#</mathjax> with respect to <mathjax>#O#</mathjax>.</p>
<p>Now, we simply divide thru by the LOWEST molar quantity, which here is the metal to give an empirical formula of <mathjax>#MnO_2", manganese(IV) oxide"#</mathjax>.</p>
<p>AS always, the empirical formula is the simplest, whole number ratio defining constituent atoms in a species. </p></div>
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<div class="markdown"><p><mathjax>#MnO_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all empirical formula calculations, a starting mass of <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> is assumed. </p>
<p>There are thus <mathjax>#(63*g)/(54.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.15*mol#</mathjax> <mathjax>#Mn#</mathjax>, and <mathjax>#(37*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.132*mol#</mathjax> with respect to <mathjax>#O#</mathjax>.</p>
<p>Now, we simply divide thru by the LOWEST molar quantity, which here is the metal to give an empirical formula of <mathjax>#MnO_2", manganese(IV) oxide"#</mathjax>.</p>
<p>AS always, the empirical formula is the simplest, whole number ratio defining constituent atoms in a species. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula for the compound consisting of 63% manganese and 37% oxygen by mass? </h1>
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<div class="markdown"><p><mathjax>#MnO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all empirical formula calculations, a starting mass of <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> is assumed. </p>
<p>There are thus <mathjax>#(63*g)/(54.94*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.15*mol#</mathjax> <mathjax>#Mn#</mathjax>, and <mathjax>#(37*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.132*mol#</mathjax> with respect to <mathjax>#O#</mathjax>.</p>
<p>Now, we simply divide thru by the LOWEST molar quantity, which here is the metal to give an empirical formula of <mathjax>#MnO_2", manganese(IV) oxide"#</mathjax>.</p>
<p>AS always, the empirical formula is the simplest, whole number ratio defining constituent atoms in a species. </p></div>
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</article> | What is the empirical formula for the compound consisting of 63% manganese and 37% oxygen by mass? | null |
1,194 | abe2d02d-6ddd-11ea-a59f-ccda262736ce | https://socratic.org/questions/533668a402bf34573b4e66fb | 919.81 grams | start physical_unit 3 4 mass g qc_end physical_unit 9 10 6 7 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] grams"}] | [{"type":"physical unit","value":"919.81 grams"}] | [{"type":"physical unit","value":"Mass [OF] lithium oxide [=] \\pu{1.00 kg}"}] | <h1 class="questionTitle" itemprop="name">What mass of carbon dioxide can 1.00 kg of lithium oxide absorb?</h1> | null | 919.81 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#"2LiOH" + "CO"_2 → "Li"_2"CO"_3 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>You must make the following conversions:</p>
<p><mathjax>#"mass of LiOH" → "moles of LiOH" → "moles of CO"_2 → "mass of CO"_2#</mathjax></p>
<p><mathjax>#1.00 × 10^3 color(red)(cancel(color(black)("g LiOH"))) × "1 mol LiOH"/(23.95 color(red)(cancel(color(black)("g LiOH")))) = "41.8 mol LiOH"#</mathjax></p>
<p><mathjax>#41.8 color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol CO"_2)/(2 color(red)(cancel(color(black)("mol LiOH")))) = "20.9 mol CO"_2#</mathjax></p>
<p><mathjax>#20.9 color(red)(cancel(color(black)("mol CO"_2))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "920 g CO"_2#</mathjax></p></div>
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<div class="markdown"><p>The LiOH can absorb 920 g of carbon dioxide.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#"2LiOH" + "CO"_2 → "Li"_2"CO"_3 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>You must make the following conversions:</p>
<p><mathjax>#"mass of LiOH" → "moles of LiOH" → "moles of CO"_2 → "mass of CO"_2#</mathjax></p>
<p><mathjax>#1.00 × 10^3 color(red)(cancel(color(black)("g LiOH"))) × "1 mol LiOH"/(23.95 color(red)(cancel(color(black)("g LiOH")))) = "41.8 mol LiOH"#</mathjax></p>
<p><mathjax>#41.8 color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol CO"_2)/(2 color(red)(cancel(color(black)("mol LiOH")))) = "20.9 mol CO"_2#</mathjax></p>
<p><mathjax>#20.9 color(red)(cancel(color(black)("mol CO"_2))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "920 g CO"_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of carbon dioxide can 1.00 kg of lithium oxide absorb?</h1>
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<div class="markdown"><p>The LiOH can absorb 920 g of carbon dioxide.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The balanced equation is</p>
<p><mathjax>#"2LiOH" + "CO"_2 → "Li"_2"CO"_3 + "H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p>You must make the following conversions:</p>
<p><mathjax>#"mass of LiOH" → "moles of LiOH" → "moles of CO"_2 → "mass of CO"_2#</mathjax></p>
<p><mathjax>#1.00 × 10^3 color(red)(cancel(color(black)("g LiOH"))) × "1 mol LiOH"/(23.95 color(red)(cancel(color(black)("g LiOH")))) = "41.8 mol LiOH"#</mathjax></p>
<p><mathjax>#41.8 color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol CO"_2)/(2 color(red)(cancel(color(black)("mol LiOH")))) = "20.9 mol CO"_2#</mathjax></p>
<p><mathjax>#20.9 color(red)(cancel(color(black)("mol CO"_2))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "920 g CO"_2#</mathjax></p></div>
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</article> | What mass of carbon dioxide can 1.00 kg of lithium oxide absorb? | null |
1,195 | ac0b2d06-6ddd-11ea-9099-ccda262736ce | https://socratic.org/questions/58fe28b411ef6b7a0b7fe57e | 0.06 molar | start physical_unit 4 5 mole mol/l qc_end physical_unit 4 4 11 12 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] potassium bromide [IN] molar"}] | [{"type":"physical unit","value":"0.06 molar"}] | [{"type":"physical unit","value":"Mass [OF] potassium [=] \\pu{2.45 g}"}] | <h1 class="questionTitle" itemprop="name">What molar quantity of #"potassium bromide"# results from oxidation of a #2.45*g# mass of potassium? </h1> | null | 0.06 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus the molar quantity of potassium bromide is PRECISELY EQUIVALENT to the molar quantity of potassium metal.........</p>
<p><mathjax>#"Moles of potassium"=(2.45*g)/(39.10*g*mol^-1)=0.0626*mol#</mathjax>. </p>
<p>And thus the molar quantity of <mathjax>#0.0626#</mathjax> WITH RESPECT to <mathjax>#"potassium bromide"#</mathjax> is the same......... </p>
<p>Given quantitative yield, what mass of the salt would result?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, we interrogate the stoichiometric reaction:</p>
<p><mathjax>#K(s) + 1/2Br_2(l) rarr KBr(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus the molar quantity of potassium bromide is PRECISELY EQUIVALENT to the molar quantity of potassium metal.........</p>
<p><mathjax>#"Moles of potassium"=(2.45*g)/(39.10*g*mol^-1)=0.0626*mol#</mathjax>. </p>
<p>And thus the molar quantity of <mathjax>#0.0626#</mathjax> WITH RESPECT to <mathjax>#"potassium bromide"#</mathjax> is the same......... </p>
<p>Given quantitative yield, what mass of the salt would result?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What molar quantity of #"potassium bromide"# results from oxidation of a #2.45*g# mass of potassium? </h1>
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anor277
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Apr 24, 2017
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<div class="markdown"><p>Well, we interrogate the stoichiometric reaction:</p>
<p><mathjax>#K(s) + 1/2Br_2(l) rarr KBr(s)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus the molar quantity of potassium bromide is PRECISELY EQUIVALENT to the molar quantity of potassium metal.........</p>
<p><mathjax>#"Moles of potassium"=(2.45*g)/(39.10*g*mol^-1)=0.0626*mol#</mathjax>. </p>
<p>And thus the molar quantity of <mathjax>#0.0626#</mathjax> WITH RESPECT to <mathjax>#"potassium bromide"#</mathjax> is the same......... </p>
<p>Given quantitative yield, what mass of the salt would result?</p></div>
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</article> | What molar quantity of #"potassium bromide"# results from oxidation of a #2.45*g# mass of potassium? | null |
1,196 | acb7b96e-6ddd-11ea-b82f-ccda262736ce | https://socratic.org/questions/5637fc4411ef6b73ec4d50d7 | 6.07 cm^3 | start physical_unit 1 1 volume cm^3 qc_end physical_unit 1 1 6 9 pressure qc_end physical_unit 1 1 18 21 pressure qc_end physical_unit 1 1 11 12 temperature qc_end physical_unit 1 1 23 24 temperature qc_end physical_unit 1 1 3 4 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] cm^3"}] | [{"type":"physical unit","value":"6.07 cm^3"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{67 cm^3}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{9.38 × 10^4 Pa}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{22 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{10.6 × 10^5 Pa}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{29 ℃}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies #"67. cm"^3# at #9.38 × 10^4color(white)(l)"Pa"# and 22 °C. What is its volume at #10.6 × 10^5color(white)(l)"Pa"# and 29 °C?</h1> | null | 6.07 cm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, it's always a good idea to start by making a note of what information is being provided by the problem. </p>
<p>In your case, you know that the initial sample of gas </p>
<blockquote>
<ul>
<li><em>occupies a volume equal to</em> <mathjax>#"67 cm"^3#</mathjax></li>
<li><em>has a temperature of</em> <mathjax>#22^@"C"#</mathjax></li>
<li><em>has a pressure of</em> <mathjax>#9.38 * 10^4"Pa"#</mathjax></li>
</ul>
</blockquote>
<p>You then go on to change the temperature to <mathjax>#29^@"C"#</mathjax> and the pressure to <mathjax>#10.6 * 10^5"Pa"#</mathjax>. </p>
<p>Notice that no mention of number of moles was made. This means that you can assume it to be <strong>constant</strong>. So, if you start from the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, you can say that</p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1 ->#</mathjax> the initial state of the gas</p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2 ->#</mathjax> the final state of the gas</p>
</blockquote>
<p>Since <mathjax>#n#</mathjax> is <em>constant</em>, and <mathjax>#R#</mathjax> is the <em>universal gas constant</em>, you can rearrange these equations to isolate these two constant terms on one side</p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = n * R" "#</mathjax> and <mathjax>#" "(P_2 * V_2)/T_2 = n * R#</mathjax></p>
</blockquote>
<p>Notice that you have two expressions that are equal to the same value, <mathjax>#n * R#</mathjax>. This means that they are equal to each other as well. </p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2 ->#</mathjax> the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
</blockquote>
<p>Now all you have to do is rearrange this to solve for <mathjax>#V_2#</mathjax>, the volume of the gas at the final state. </p>
<p>Look what happens if you divide both sides of the equation by <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1/P_2 * V_1/T_1 = (color(red)(cancel(color(black)(P_2))) * V_2)/(T_2 * color(red)(cancel(color(black)(P_2))))#</mathjax></p>
<p><mathjax>#P_1/P_2 * V_1/T_1 = V_2/T_2#</mathjax></p>
</blockquote>
<p>Now multiply both sides by <mathjax>#T_2#</mathjax> to get <mathjax>#V_2#</mathjax> alone one one side of the equation</p>
<blockquote>
<p><mathjax>#P_1/P_2 * T_2/T_1 * V_1 = V_2/color(red)(cancel(color(black)(T_2))) * color(red)(cancel(color(black)(T_2)))#</mathjax></p>
</blockquote>
<p>Finally, you got </p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V_2#</mathjax> - <strong>do not</strong> foget to convert the temperature from degrees Celsius to Kelvin!</p>
<blockquote>
<p><mathjax>#V_2 = (9.38 * 10^4color(red)(cancel(color(black)("Pa"))))/(10.6 * 10^5color(red)(cancel(color(black)("Pa")))) * ((273.15 + 29)color(red)(cancel(color(black)("K"))))/((273.15 + 22)color(red)(cancel(color(black)("K")))) * "67 cm"^3#</mathjax></p>
<p><mathjax>#V_2 = "6.0695 cm"^3#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.1 cm"^3)#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/bftkRnTcFj8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"6.1 cm"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, it's always a good idea to start by making a note of what information is being provided by the problem. </p>
<p>In your case, you know that the initial sample of gas </p>
<blockquote>
<ul>
<li><em>occupies a volume equal to</em> <mathjax>#"67 cm"^3#</mathjax></li>
<li><em>has a temperature of</em> <mathjax>#22^@"C"#</mathjax></li>
<li><em>has a pressure of</em> <mathjax>#9.38 * 10^4"Pa"#</mathjax></li>
</ul>
</blockquote>
<p>You then go on to change the temperature to <mathjax>#29^@"C"#</mathjax> and the pressure to <mathjax>#10.6 * 10^5"Pa"#</mathjax>. </p>
<p>Notice that no mention of number of moles was made. This means that you can assume it to be <strong>constant</strong>. So, if you start from the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, you can say that</p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1 ->#</mathjax> the initial state of the gas</p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2 ->#</mathjax> the final state of the gas</p>
</blockquote>
<p>Since <mathjax>#n#</mathjax> is <em>constant</em>, and <mathjax>#R#</mathjax> is the <em>universal gas constant</em>, you can rearrange these equations to isolate these two constant terms on one side</p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = n * R" "#</mathjax> and <mathjax>#" "(P_2 * V_2)/T_2 = n * R#</mathjax></p>
</blockquote>
<p>Notice that you have two expressions that are equal to the same value, <mathjax>#n * R#</mathjax>. This means that they are equal to each other as well. </p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2 ->#</mathjax> the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
</blockquote>
<p>Now all you have to do is rearrange this to solve for <mathjax>#V_2#</mathjax>, the volume of the gas at the final state. </p>
<p>Look what happens if you divide both sides of the equation by <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1/P_2 * V_1/T_1 = (color(red)(cancel(color(black)(P_2))) * V_2)/(T_2 * color(red)(cancel(color(black)(P_2))))#</mathjax></p>
<p><mathjax>#P_1/P_2 * V_1/T_1 = V_2/T_2#</mathjax></p>
</blockquote>
<p>Now multiply both sides by <mathjax>#T_2#</mathjax> to get <mathjax>#V_2#</mathjax> alone one one side of the equation</p>
<blockquote>
<p><mathjax>#P_1/P_2 * T_2/T_1 * V_1 = V_2/color(red)(cancel(color(black)(T_2))) * color(red)(cancel(color(black)(T_2)))#</mathjax></p>
</blockquote>
<p>Finally, you got </p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V_2#</mathjax> - <strong>do not</strong> foget to convert the temperature from degrees Celsius to Kelvin!</p>
<blockquote>
<p><mathjax>#V_2 = (9.38 * 10^4color(red)(cancel(color(black)("Pa"))))/(10.6 * 10^5color(red)(cancel(color(black)("Pa")))) * ((273.15 + 29)color(red)(cancel(color(black)("K"))))/((273.15 + 22)color(red)(cancel(color(black)("K")))) * "67 cm"^3#</mathjax></p>
<p><mathjax>#V_2 = "6.0695 cm"^3#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.1 cm"^3)#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/bftkRnTcFj8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas occupies #"67. cm"^3# at #9.38 × 10^4color(white)(l)"Pa"# and 22 °C. What is its volume at #10.6 × 10^5color(white)(l)"Pa"# and 29 °C?</h1>
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Stefan V.
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Nov 3, 2015
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<div class="markdown"><p><mathjax>#"6.1 cm"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, it's always a good idea to start by making a note of what information is being provided by the problem. </p>
<p>In your case, you know that the initial sample of gas </p>
<blockquote>
<ul>
<li><em>occupies a volume equal to</em> <mathjax>#"67 cm"^3#</mathjax></li>
<li><em>has a temperature of</em> <mathjax>#22^@"C"#</mathjax></li>
<li><em>has a pressure of</em> <mathjax>#9.38 * 10^4"Pa"#</mathjax></li>
</ul>
</blockquote>
<p>You then go on to change the temperature to <mathjax>#29^@"C"#</mathjax> and the pressure to <mathjax>#10.6 * 10^5"Pa"#</mathjax>. </p>
<p>Notice that no mention of number of moles was made. This means that you can assume it to be <strong>constant</strong>. So, if you start from the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, you can say that</p>
<blockquote>
<p><mathjax>#P_1 * V_1 = n * R * T_1 ->#</mathjax> the initial state of the gas</p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#P_2 * V_2 = n * R * T_2 ->#</mathjax> the final state of the gas</p>
</blockquote>
<p>Since <mathjax>#n#</mathjax> is <em>constant</em>, and <mathjax>#R#</mathjax> is the <em>universal gas constant</em>, you can rearrange these equations to isolate these two constant terms on one side</p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = n * R" "#</mathjax> and <mathjax>#" "(P_2 * V_2)/T_2 = n * R#</mathjax></p>
</blockquote>
<p>Notice that you have two expressions that are equal to the same value, <mathjax>#n * R#</mathjax>. This means that they are equal to each other as well. </p>
<blockquote>
<p><mathjax>#(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2 ->#</mathjax> the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</p>
</blockquote>
<p>Now all you have to do is rearrange this to solve for <mathjax>#V_2#</mathjax>, the volume of the gas at the final state. </p>
<p>Look what happens if you divide both sides of the equation by <mathjax>#P_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1/P_2 * V_1/T_1 = (color(red)(cancel(color(black)(P_2))) * V_2)/(T_2 * color(red)(cancel(color(black)(P_2))))#</mathjax></p>
<p><mathjax>#P_1/P_2 * V_1/T_1 = V_2/T_2#</mathjax></p>
</blockquote>
<p>Now multiply both sides by <mathjax>#T_2#</mathjax> to get <mathjax>#V_2#</mathjax> alone one one side of the equation</p>
<blockquote>
<p><mathjax>#P_1/P_2 * T_2/T_1 * V_1 = V_2/color(red)(cancel(color(black)(T_2))) * color(red)(cancel(color(black)(T_2)))#</mathjax></p>
</blockquote>
<p>Finally, you got </p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>Now plug in your values and solve for <mathjax>#V_2#</mathjax> - <strong>do not</strong> foget to convert the temperature from degrees Celsius to Kelvin!</p>
<blockquote>
<p><mathjax>#V_2 = (9.38 * 10^4color(red)(cancel(color(black)("Pa"))))/(10.6 * 10^5color(red)(cancel(color(black)("Pa")))) * ((273.15 + 29)color(red)(cancel(color(black)("K"))))/((273.15 + 22)color(red)(cancel(color(black)("K")))) * "67 cm"^3#</mathjax></p>
<p><mathjax>#V_2 = "6.0695 cm"^3#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the initial volume of the gas</p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.1 cm"^3)#</mathjax></p>
</blockquote>
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Ernest Z.
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<span class="dateCreated" datetime="2015-11-03T01:53:06" itemprop="dateCreated">
Nov 3, 2015
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<div class="markdown"><p>The new volume will be <mathjax>#color(blue)("6.1 cm"^3)#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a> equation,</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<blockquote></blockquote>
<p>Let's start by listing our given information.</p>
<p><mathjax>#P_1 = 9.38 × 10^4 "Pa"#</mathjax>; <mathjax>#V_1 = "67. cm"^3#</mathjax>; <mathjax>#T_1 = "22 °C =(22 + 273.15) K = 295 K"#</mathjax><br/>
<mathjax>#P_2 = 10.6 × 10^5 "Pa"#</mathjax>; <mathjax>#V_2 = "?"#</mathjax>; <mathjax>#T_2 = "29 °C = (29 + 273.15) K = 302 K"#</mathjax></p>
<blockquote></blockquote>
<p>Now we must rearrange the Combined Gas Law equation to get <mathjax>#V_2#</mathjax> by itself.</p>
<p>We'll take it step by step.</p>
<blockquote></blockquote>
<p><strong>Step 1.</strong> Multiply both sides by <mathjax>#T_2#</mathjax>.</p>
<p><mathjax>#(P_1V_1)/T_1 × T_2 = (P_2V_2)/(color(red)(cancel(color(black)(T_1)))) × color(red)(cancel(color(black)(T_2)#</mathjax></p>
<p><mathjax>#(P_1V_1T_2)/T_1 = P_2V_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Divide both sides by <mathjax>#P_2#</mathjax>.</p>
<p><mathjax>#(P_1V_1T_2)/(P_2T_1) = (color(red)(cancel(color(black)(P_2)))V_2)/color(red)(cancel(color(black)(P_2)#</mathjax></p>
<p><mathjax>#(P_1V_1T_2)/(P_2T_1) = V_2#</mathjax> or <mathjax>#V_2 = (P_1V_1T_2)/(P_2T_1)#</mathjax></p>
<blockquote></blockquote>
<p>Now we insert the values into the equation.</p>
<p><mathjax>#V_2 = (P_1V_1T_2)/(P_2T_1) = (9.38 × 10^4 color(red)(cancel(color(black)("Pa"))) × "67. cm"^ 3 × 302 color(red)(cancel(color(black)("K"))))/( 10.6 × 10^5 color(red)(cancel(color(black)("Pa"))) × 295 color(red)(cancel(color(black)("K")))) = "6.1 cm"^3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Check</strong>: The temperature doesn't change much, but the pressure increases by about ten-fold.</p>
<p>The new volume should be about one-tenth of the original volume, or about <mathjax>#"7 cm"^3#</mathjax>.</p></div>
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</article> | A gas occupies #"67. cm"^3# at #9.38 × 10^4color(white)(l)"Pa"# and 22 °C. What is its volume at #10.6 × 10^5color(white)(l)"Pa"# and 29 °C? | null |
1,197 | aa1b65fe-6ddd-11ea-b39d-ccda262736ce | https://socratic.org/questions/in-the-equation-al-2-so-4-3-6naoh-2al-oh-3-3na-2so-4-what-is-the-mole-ratio-of-n | 3:1 | start physical_unit 19 21 mole_fraction none qc_end chemical_equation 3 12 qc_end end | [{"type":"physical unit","value":"Mole ratio [OF] NaOH to Al(OH)3"}] | [{"type":"physical unit","value":"3:1"}] | [{"type":"chemical equation","value":"Al2(SO4)3 + 6 NaOH -> 2 Al(OH)3 + 3 NaSO4"}] | <h1 class="questionTitle" itemprop="name">In the equation #Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4#, what is the mole ratio of #NaOH# to #Al(OH)_3#?</h1> | null | 3:1 | <div class="answerDescription">
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<div class="markdown"><p>The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before <mathjax>#NaOH#</mathjax> is 6 and that before <mathjax>#Al(OH)_3#</mathjax> is 2. <br/>
So the required mole ratio is 6:2=3:1</p></div>
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<div class="markdown"><p>3:1</p></div>
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<div class="markdown"><p>The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before <mathjax>#NaOH#</mathjax> is 6 and that before <mathjax>#Al(OH)_3#</mathjax> is 2. <br/>
So the required mole ratio is 6:2=3:1</p></div>
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<h1 class="questionTitle" itemprop="name">In the equation #Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4#, what is the mole ratio of #NaOH# to #Al(OH)_3#?</h1>
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<div class="markdown"><p>3:1</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before <mathjax>#NaOH#</mathjax> is 6 and that before <mathjax>#Al(OH)_3#</mathjax> is 2. <br/>
So the required mole ratio is 6:2=3:1</p></div>
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</article> | In the equation #Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4#, what is the mole ratio of #NaOH# to #Al(OH)_3#? | null |
1,198 | a8d5404a-6ddd-11ea-82ef-ccda262736ce | https://socratic.org/questions/lactic-acid-has-one-acidic-hydrogen-a-0-1-m-solution-of-lactic-acid-has-a-ph-of- | 1.4 × 10^(-4) | start physical_unit 0 1 ka none qc_end physical_unit 0 1 7 8 molarity qc_end physical_unit 9 9 17 17 ph qc_end end | [{"type":"physical unit","value":"Ka [OF] lactic acid"}] | [{"type":"physical unit","value":"1.4 × 10^(-4)"}] | [{"type":"physical unit","value":"Molarity [OF] lactic acid solution [=] \\pu{0.1 M}"},{"type":"physical unit","value":"pH [OF] lactic acid solution [=] \\pu{2.44}"},{"type":"physical unit","value":"Number [OF] acidic hydrogen in lactic acid [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">Lactic acid has one acidic hydrogen. A 0.1 M solution of lactic acid has a pH of 2.44. Calculate Ka for lactic acid?</h1> | null | 1.4 × 10^(-4) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You don't actually need to know the molecular formula for <em>lactic acid</em> in order to solve this problem. </p>
<p>You are told that lactic acid has <strong>one acidic proton</strong>, which means that you can represent it as <mathjax>#"HA"#</mathjax>.</p>
<p>So, you know that the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of a solution that is <mathjax>#"0.10 M"#</mathjax> lactic acid is equal to <mathjax>#2.44#</mathjax>. As you know, a solution's pH is simply a measure of the <strong>concentration</strong> of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = -log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>This means that you can use the given pH to determine the concentration of hydronium ions in this solution</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-2.44) = 3.63 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>As you know, a <em>weak acid</em> <strong>does not</strong> dissociate completely in aqueous solution to form hydronium ions and the conjugate base of the acid. </p>
<p>Instead, an <a href="https://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium</a> is established between the <em>unprotonated</em> molecules and the two <em>ions</em>.</p>
<p>Use an <strong>ICE table</strong> to help you determine the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, for this weak acid. Keep in mind that the <strong>equilibrium concentration</strong> of hydronium ions must be equal to <mathjax>#3.63 * 10^(-3)"M"#</mathjax>, since that corresponds to the pH of the solution. </p>
<blockquote>
<p><mathjax>#" " "HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.10" " " " " " " " " " " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.10-x" " " " " " " " " " " " " " " "x" " " " " " "3.63 * 10^(-3)#</mathjax></p>
<p>Since you have </p>
<blockquote>
<p><mathjax>#x = 3.63 * 10^(-3)#</mathjax></p>
</blockquote>
<p>you can say that the equilibrium concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["A"^(-)] = 3.63 * 10^(-3)"M"#</mathjax></p>
<p><mathjax>#["HA"] = 0.10 - 3.63 * 10^(-3) = "0.09637 M"#</mathjax></p>
</blockquote>
<p>By definition, the acid dissociation constant for this reaction will be </p>
<blockquote>
<p><mathjax>#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#</mathjax></p>
</blockquote>
<p>Plug in your values to get - I'll leave the acid dissociation constant <em>unitless</em> </p>
<blockquote>
<p><mathjax>#K_a = (3.63 * 10^(-3) * 3.63 * 10^(-3))/0.09637 = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the acid. </p>
<p>The listed value for lactic acid's acid dissociation constant is </p>
<blockquote>
<p><mathjax>#K_a = 1.38 * 10^(-4)#</mathjax></p>
</blockquote>
<p>so this is an excellent result. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.4 * 10^(-4)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You don't actually need to know the molecular formula for <em>lactic acid</em> in order to solve this problem. </p>
<p>You are told that lactic acid has <strong>one acidic proton</strong>, which means that you can represent it as <mathjax>#"HA"#</mathjax>.</p>
<p>So, you know that the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of a solution that is <mathjax>#"0.10 M"#</mathjax> lactic acid is equal to <mathjax>#2.44#</mathjax>. As you know, a solution's pH is simply a measure of the <strong>concentration</strong> of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = -log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>This means that you can use the given pH to determine the concentration of hydronium ions in this solution</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-2.44) = 3.63 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>As you know, a <em>weak acid</em> <strong>does not</strong> dissociate completely in aqueous solution to form hydronium ions and the conjugate base of the acid. </p>
<p>Instead, an <a href="https://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium</a> is established between the <em>unprotonated</em> molecules and the two <em>ions</em>.</p>
<p>Use an <strong>ICE table</strong> to help you determine the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, for this weak acid. Keep in mind that the <strong>equilibrium concentration</strong> of hydronium ions must be equal to <mathjax>#3.63 * 10^(-3)"M"#</mathjax>, since that corresponds to the pH of the solution. </p>
<blockquote>
<p><mathjax>#" " "HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.10" " " " " " " " " " " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.10-x" " " " " " " " " " " " " " " "x" " " " " " "3.63 * 10^(-3)#</mathjax></p>
<p>Since you have </p>
<blockquote>
<p><mathjax>#x = 3.63 * 10^(-3)#</mathjax></p>
</blockquote>
<p>you can say that the equilibrium concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["A"^(-)] = 3.63 * 10^(-3)"M"#</mathjax></p>
<p><mathjax>#["HA"] = 0.10 - 3.63 * 10^(-3) = "0.09637 M"#</mathjax></p>
</blockquote>
<p>By definition, the acid dissociation constant for this reaction will be </p>
<blockquote>
<p><mathjax>#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#</mathjax></p>
</blockquote>
<p>Plug in your values to get - I'll leave the acid dissociation constant <em>unitless</em> </p>
<blockquote>
<p><mathjax>#K_a = (3.63 * 10^(-3) * 3.63 * 10^(-3))/0.09637 = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the acid. </p>
<p>The listed value for lactic acid's acid dissociation constant is </p>
<blockquote>
<p><mathjax>#K_a = 1.38 * 10^(-4)#</mathjax></p>
</blockquote>
<p>so this is an excellent result. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Lactic acid has one acidic hydrogen. A 0.1 M solution of lactic acid has a pH of 2.44. Calculate Ka for lactic acid?</h1>
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Stefan V.
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Dec 27, 2015
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<div class="markdown"><p><mathjax>#1.4 * 10^(-4)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You don't actually need to know the molecular formula for <em>lactic acid</em> in order to solve this problem. </p>
<p>You are told that lactic acid has <strong>one acidic proton</strong>, which means that you can represent it as <mathjax>#"HA"#</mathjax>.</p>
<p>So, you know that the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of a solution that is <mathjax>#"0.10 M"#</mathjax> lactic acid is equal to <mathjax>#2.44#</mathjax>. As you know, a solution's pH is simply a measure of the <strong>concentration</strong> of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = -log(["H"_3"O"^(+)])#</mathjax></p>
</blockquote>
<p>This means that you can use the given pH to determine the concentration of hydronium ions in this solution</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH")#</mathjax></p>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-2.44) = 3.63 * 10^(-3)"M"#</mathjax></p>
</blockquote>
<p>As you know, a <em>weak acid</em> <strong>does not</strong> dissociate completely in aqueous solution to form hydronium ions and the conjugate base of the acid. </p>
<p>Instead, an <a href="https://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium</a> is established between the <em>unprotonated</em> molecules and the two <em>ions</em>.</p>
<p>Use an <strong>ICE table</strong> to help you determine the <em>acid dissociation constant</em>, <mathjax>#K_a#</mathjax>, for this weak acid. Keep in mind that the <strong>equilibrium concentration</strong> of hydronium ions must be equal to <mathjax>#3.63 * 10^(-3)"M"#</mathjax>, since that corresponds to the pH of the solution. </p>
<blockquote>
<p><mathjax>#" " "HA"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " "0.10" " " " " " " " " " " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)#</mathjax><br/>
<mathjax>#color(purple)("E")" "0.10-x" " " " " " " " " " " " " " " "x" " " " " " "3.63 * 10^(-3)#</mathjax></p>
<p>Since you have </p>
<blockquote>
<p><mathjax>#x = 3.63 * 10^(-3)#</mathjax></p>
</blockquote>
<p>you can say that the equilibrium concentrations of the weak acid and of the conjugate base will be </p>
<blockquote>
<p><mathjax>#["A"^(-)] = 3.63 * 10^(-3)"M"#</mathjax></p>
<p><mathjax>#["HA"] = 0.10 - 3.63 * 10^(-3) = "0.09637 M"#</mathjax></p>
</blockquote>
<p>By definition, the acid dissociation constant for this reaction will be </p>
<blockquote>
<p><mathjax>#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#</mathjax></p>
</blockquote>
<p>Plug in your values to get - I'll leave the acid dissociation constant <em>unitless</em> </p>
<blockquote>
<p><mathjax>#K_a = (3.63 * 10^(-3) * 3.63 * 10^(-3))/0.09637 = color(green)(1.4 * 10^(-4))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the acid. </p>
<p>The listed value for lactic acid's acid dissociation constant is </p>
<blockquote>
<p><mathjax>#K_a = 1.38 * 10^(-4)#</mathjax></p>
</blockquote>
<p>so this is an excellent result. </p></div>
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</article> | Lactic acid has one acidic hydrogen. A 0.1 M solution of lactic acid has a pH of 2.44. Calculate Ka for lactic acid? | null |
1,199 | aa2c4710-6ddd-11ea-a023-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-43-38-sodium-11-33-car | Na2CO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Na2CO3"}] | [{"type":"physical unit","value":"Percent [OF] sodium in the compound [=] \\pu{43.38%}"},{"type":"physical unit","value":"Percent [OF] carbon in the compound [=] \\pu{11.33%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{45.29%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen? </h1> | null | Na2CO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all divide the percentages with their <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a><br/>
<mathjax>#43.38/23=1.89 , 11.33/12=0.94 , 45.29/16=2.83#</mathjax><br/>
Now divide every value with the smallest value <br/>
<mathjax>#1.88/0.94=2, 0.94/0.94=1, 2.83/0.94=3#</mathjax><br/>
put these ratios by their respected <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a><br/>
so the empirical formula is <mathjax>#Na_2CO_3#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Na_2CO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all divide the percentages with their <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a><br/>
<mathjax>#43.38/23=1.89 , 11.33/12=0.94 , 45.29/16=2.83#</mathjax><br/>
Now divide every value with the smallest value <br/>
<mathjax>#1.88/0.94=2, 0.94/0.94=1, 2.83/0.94=3#</mathjax><br/>
put these ratios by their respected <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a><br/>
so the empirical formula is <mathjax>#Na_2CO_3#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen? </h1>
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Abdul Sammad
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<div class="markdown"><p><mathjax>#Na_2CO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First of all divide the percentages with their <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a><br/>
<mathjax>#43.38/23=1.89 , 11.33/12=0.94 , 45.29/16=2.83#</mathjax><br/>
Now divide every value with the smallest value <br/>
<mathjax>#1.88/0.94=2, 0.94/0.94=1, 2.83/0.94=3#</mathjax><br/>
put these ratios by their respected <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a><br/>
so the empirical formula is <mathjax>#Na_2CO_3#</mathjax></p></div>
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</article> | What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen? | null |
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