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1,300 | a9b98f18-6ddd-11ea-b052-ccda262736ce | https://socratic.org/questions/how-much-boiling-water-would-you-need-to-raise-the-bath-to-body-temperature-abou | 4.8 kg | start physical_unit 2 3 mass kg qc_end physical_unit 3 3 50 51 temperature qc_end physical_unit 3 3 15 16 temperature qc_end physical_unit 3 3 44 45 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] boiling water [IN] kg"}] | [{"type":"physical unit","value":"4.8 kg"}] | [{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{37 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{25 kg}"},{"type":"other","value":"Assume that no heat is transferred to the surrounding environment."}] | <h1 class="questionTitle" itemprop="name">How much boiling water would you need to raise the bath to body temperature (about 37 ∘C)? Assume that no heat is transferred to the surrounding environment.
Express your answer to two significant figures and include the appropriate units.</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>You fill your bathtub with 25 kg of room-temperature water (about 25 ∘C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.</p></div>
</h2>
</div>
</div> | 4.8 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the heat <strong>given off</strong> by the boiling water will be <strong>equal</strong> to the heat <strong>absorbed</strong> by the room-temperature sample.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q_"absorbed" = -q_"given off")))" " " "color(darkorange)("(*)")#</mathjax></p>
<blockquote>
<p>The minus sign is used here because, by convention, <em>heat given off</em> carries a minus sign. </p>
</blockquote>
</blockquote>
<p>Another assumption that you have to make is that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of <em>liquid water</em> is constant regardless of the temperature of the liquid water. </p>
<p>In other words, you need to have</p>
<blockquote>
<p><mathjax>#c_ ("liquid water at 25"^@"C") = c_ ("liquid water at 100"^@"C")#</mathjax></p>
</blockquote>
<p>Now, your tool of choice here will be the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c_"liquid water" * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q#</mathjax> is the heat absorbed or given off</li>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c_"liquid water"#</mathjax> is the <strong>specific heat</strong> of liquid water</li>
<li><mathjax>#DeltaT#</mathjax> is the <strong>change in temperature</strong>, calculated as the difference between the <em>final temperature</em> and the <em>initial temperature</em> of the sample</li>
</ul>
</blockquote>
<p>So, you know that you have</p>
<blockquote>
<p><mathjax>#q_"absorbed" = m_1 * c_"liquid water" * DeltaT_"warming"#</mathjax></p>
</blockquote>
<p>for the room-temperature water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"warming" = 37^@"C" - 25^@"C" = 12^@"C"#</mathjax></p>
</blockquote>
<p>Similarly, you have</p>
<blockquote>
<p><mathjax>#q_"given off" = m_2 * c_"liquid water" * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>for the boiling water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"cooling" = 37^@"C" - 100^@"C" = -63^@"C"#</mathjax></p>
</blockquote>
<p>Use equation <mathjax>#color(darkorange)("(*)")#</mathjax> to get</p>
<blockquote>
<p><mathjax>#m_1 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"warming" = - m_2 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"cooling"#</mathjax></p>
<p><mathjax>#m_1 * DeltaT_"warming" = - m_2 * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_2 = (DeltaT_"warming")/(-DeltaT_"cooling") * m_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#m_2 = (12 color(red)(cancel(color(black)(""^@"C"))))/(-(-63color(red)(cancel(color(black)(""^@"C"))))) * "25 kg" = color(darkgreen)(ul(color(black)("4.8 kg")))#</mathjax></p>
<blockquote>
<p>Notice that you need the minus sign to cancel out the minus sign coming from the change in temperature.</p>
</blockquote>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, if you add <mathjax>#"4.8 kg"#</mathjax> of liquid water at <mathjax>#100^@"C"#</mathjax> to <mathjax>#"25 kg"#</mathjax> of water at <mathjax>#25^@"C"#</mathjax>, you will end up with a mixture that has a final temperature of <mathjax>#37^@"C"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"4.8 kg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the heat <strong>given off</strong> by the boiling water will be <strong>equal</strong> to the heat <strong>absorbed</strong> by the room-temperature sample.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q_"absorbed" = -q_"given off")))" " " "color(darkorange)("(*)")#</mathjax></p>
<blockquote>
<p>The minus sign is used here because, by convention, <em>heat given off</em> carries a minus sign. </p>
</blockquote>
</blockquote>
<p>Another assumption that you have to make is that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of <em>liquid water</em> is constant regardless of the temperature of the liquid water. </p>
<p>In other words, you need to have</p>
<blockquote>
<p><mathjax>#c_ ("liquid water at 25"^@"C") = c_ ("liquid water at 100"^@"C")#</mathjax></p>
</blockquote>
<p>Now, your tool of choice here will be the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c_"liquid water" * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q#</mathjax> is the heat absorbed or given off</li>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c_"liquid water"#</mathjax> is the <strong>specific heat</strong> of liquid water</li>
<li><mathjax>#DeltaT#</mathjax> is the <strong>change in temperature</strong>, calculated as the difference between the <em>final temperature</em> and the <em>initial temperature</em> of the sample</li>
</ul>
</blockquote>
<p>So, you know that you have</p>
<blockquote>
<p><mathjax>#q_"absorbed" = m_1 * c_"liquid water" * DeltaT_"warming"#</mathjax></p>
</blockquote>
<p>for the room-temperature water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"warming" = 37^@"C" - 25^@"C" = 12^@"C"#</mathjax></p>
</blockquote>
<p>Similarly, you have</p>
<blockquote>
<p><mathjax>#q_"given off" = m_2 * c_"liquid water" * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>for the boiling water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"cooling" = 37^@"C" - 100^@"C" = -63^@"C"#</mathjax></p>
</blockquote>
<p>Use equation <mathjax>#color(darkorange)("(*)")#</mathjax> to get</p>
<blockquote>
<p><mathjax>#m_1 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"warming" = - m_2 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"cooling"#</mathjax></p>
<p><mathjax>#m_1 * DeltaT_"warming" = - m_2 * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_2 = (DeltaT_"warming")/(-DeltaT_"cooling") * m_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#m_2 = (12 color(red)(cancel(color(black)(""^@"C"))))/(-(-63color(red)(cancel(color(black)(""^@"C"))))) * "25 kg" = color(darkgreen)(ul(color(black)("4.8 kg")))#</mathjax></p>
<blockquote>
<p>Notice that you need the minus sign to cancel out the minus sign coming from the change in temperature.</p>
</blockquote>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, if you add <mathjax>#"4.8 kg"#</mathjax> of liquid water at <mathjax>#100^@"C"#</mathjax> to <mathjax>#"25 kg"#</mathjax> of water at <mathjax>#25^@"C"#</mathjax>, you will end up with a mixture that has a final temperature of <mathjax>#37^@"C"#</mathjax>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much boiling water would you need to raise the bath to body temperature (about 37 ∘C)? Assume that no heat is transferred to the surrounding environment.
Express your answer to two significant figures and include the appropriate units.</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>You fill your bathtub with 25 kg of room-temperature water (about 25 ∘C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-01-05T10:26:13" itemprop="dateCreated">
Jan 5, 2018
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<div>
<div class="markdown"><p><mathjax>#"4.8 kg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the heat <strong>given off</strong> by the boiling water will be <strong>equal</strong> to the heat <strong>absorbed</strong> by the room-temperature sample.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q_"absorbed" = -q_"given off")))" " " "color(darkorange)("(*)")#</mathjax></p>
<blockquote>
<p>The minus sign is used here because, by convention, <em>heat given off</em> carries a minus sign. </p>
</blockquote>
</blockquote>
<p>Another assumption that you have to make is that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of <em>liquid water</em> is constant regardless of the temperature of the liquid water. </p>
<p>In other words, you need to have</p>
<blockquote>
<p><mathjax>#c_ ("liquid water at 25"^@"C") = c_ ("liquid water at 100"^@"C")#</mathjax></p>
</blockquote>
<p>Now, your tool of choice here will be the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c_"liquid water" * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q#</mathjax> is the heat absorbed or given off</li>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c_"liquid water"#</mathjax> is the <strong>specific heat</strong> of liquid water</li>
<li><mathjax>#DeltaT#</mathjax> is the <strong>change in temperature</strong>, calculated as the difference between the <em>final temperature</em> and the <em>initial temperature</em> of the sample</li>
</ul>
</blockquote>
<p>So, you know that you have</p>
<blockquote>
<p><mathjax>#q_"absorbed" = m_1 * c_"liquid water" * DeltaT_"warming"#</mathjax></p>
</blockquote>
<p>for the room-temperature water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"warming" = 37^@"C" - 25^@"C" = 12^@"C"#</mathjax></p>
</blockquote>
<p>Similarly, you have</p>
<blockquote>
<p><mathjax>#q_"given off" = m_2 * c_"liquid water" * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>for the boiling water, which has</p>
<blockquote>
<p><mathjax>#DeltaT_"cooling" = 37^@"C" - 100^@"C" = -63^@"C"#</mathjax></p>
</blockquote>
<p>Use equation <mathjax>#color(darkorange)("(*)")#</mathjax> to get</p>
<blockquote>
<p><mathjax>#m_1 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"warming" = - m_2 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"cooling"#</mathjax></p>
<p><mathjax>#m_1 * DeltaT_"warming" = - m_2 * DeltaT_"cooling"#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#m_2 = (DeltaT_"warming")/(-DeltaT_"cooling") * m_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#m_2 = (12 color(red)(cancel(color(black)(""^@"C"))))/(-(-63color(red)(cancel(color(black)(""^@"C"))))) * "25 kg" = color(darkgreen)(ul(color(black)("4.8 kg")))#</mathjax></p>
<blockquote>
<p>Notice that you need the minus sign to cancel out the minus sign coming from the change in temperature.</p>
</blockquote>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, if you add <mathjax>#"4.8 kg"#</mathjax> of liquid water at <mathjax>#100^@"C"#</mathjax> to <mathjax>#"25 kg"#</mathjax> of water at <mathjax>#25^@"C"#</mathjax>, you will end up with a mixture that has a final temperature of <mathjax>#37^@"C"#</mathjax>. </p></div>
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</article> | How much boiling water would you need to raise the bath to body temperature (about 37 ∘C)? Assume that no heat is transferred to the surrounding environment.
Express your answer to two significant figures and include the appropriate units. |
You fill your bathtub with 25 kg of room-temperature water (about 25 ∘C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.
|
1,301 | aaf6a603-6ddd-11ea-9a32-ccda262736ce | https://socratic.org/questions/the-gas-in-a-container-is-at-a-pressure-of-3-00-atm-at-25-c-directions-on-the-co | 3.27 atm | start physical_unit 0 1 pressure atm qc_end physical_unit 0 1 10 11 pressure qc_end physical_unit 0 1 13 14 temperature qc_end physical_unit 0 1 33 34 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"3.27 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{3.00 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{52 ℃}"}] | <h1 class="questionTitle" itemprop="name">The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container wam the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?</h1> | null | 3.27 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>The initial pressure, <mathjax>#P_1=3.00atm#</mathjax></p>
<p>The initial temperature <mathjax>#T_1=25+273=298K#</mathjax></p>
<p>The final temperature <mathjax>#T_2=52+273=325K#</mathjax></p>
<p>The final pressure is</p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=325/298*3=3.27atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The pressure is <mathjax>#=3.27atm#</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>The initial pressure, <mathjax>#P_1=3.00atm#</mathjax></p>
<p>The initial temperature <mathjax>#T_1=25+273=298K#</mathjax></p>
<p>The final temperature <mathjax>#T_2=52+273=325K#</mathjax></p>
<p>The final pressure is</p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=325/298*3=3.27atm#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container wam the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?</h1>
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Narad T.
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<div class="markdown"><p>The pressure is <mathjax>#=3.27atm#</mathjax></p></div>
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<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>The initial pressure, <mathjax>#P_1=3.00atm#</mathjax></p>
<p>The initial temperature <mathjax>#T_1=25+273=298K#</mathjax></p>
<p>The final temperature <mathjax>#T_2=52+273=325K#</mathjax></p>
<p>The final pressure is</p>
<p><mathjax>#P_2=T_2/T_1*P_1#</mathjax></p>
<p><mathjax>#=325/298*3=3.27atm#</mathjax></p></div>
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</article> | The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container wam the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C? | null |
1,302 | a914ba10-6ddd-11ea-90a0-ccda262736ce | https://socratic.org/questions/if-1-mole-of-h2-gas-was-collected-under-294-2k-and-746-7-mmhg-what-volume-will-i | 24.58 L | start physical_unit 4 5 volume l qc_end physical_unit 4 5 1 2 mole qc_end physical_unit 4 5 9 10 temperature qc_end physical_unit 4 5 12 13 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] H2 gas [IN] L"}] | [{"type":"physical unit","value":"24.58 L"}] | [{"type":"physical unit","value":"Mole [OF] H2 gas [=] \\pu{1 mole}"},{"type":"physical unit","value":"Temperature [OF] H2 gas [=] \\pu{294.2 K}"},{"type":"physical unit","value":"Pressure [OF] H2 gas [=] \\pu{746.7 mmHg}"}] | <h1 class="questionTitle" itemprop="name">If 1 mole of H2 gas was collected under 294.2K and 746.7 mmHg, what volume will it occupy?</h1> | null | 24.58 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving these problems is to recall that <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A column of mercury may thus be used to measure pressures up to <mathjax>#1*atm#</mathjax>.</p>
<p>And so <mathjax>#V=(1*molxx0.0821*(L*atm)/(K*mol)xx294.2*K)/((746.7*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=??L#</mathjax>. It should be about <mathjax>#25*L#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, <mathjax>#V=(nRT)/P#</mathjax>, and of course, we must choose the appropriate gas constant; <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax> is used here. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The key to solving these problems is to recall that <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A column of mercury may thus be used to measure pressures up to <mathjax>#1*atm#</mathjax>.</p>
<p>And so <mathjax>#V=(1*molxx0.0821*(L*atm)/(K*mol)xx294.2*K)/((746.7*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=??L#</mathjax>. It should be about <mathjax>#25*L#</mathjax>. </p></div>
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anor277
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<div class="markdown"><p>Well, <mathjax>#V=(nRT)/P#</mathjax>, and of course, we must choose the appropriate gas constant; <mathjax>#R=0.0821*L*atm*K^-1*mol^-1#</mathjax> is used here. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The key to solving these problems is to recall that <mathjax>#1*atm#</mathjax> of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A column of mercury may thus be used to measure pressures up to <mathjax>#1*atm#</mathjax>.</p>
<p>And so <mathjax>#V=(1*molxx0.0821*(L*atm)/(K*mol)xx294.2*K)/((746.7*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax></p>
<p><mathjax>#=??L#</mathjax>. It should be about <mathjax>#25*L#</mathjax>. </p></div>
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</article> | If 1 mole of H2 gas was collected under 294.2K and 746.7 mmHg, what volume will it occupy? | null |
1,303 | a8b4b902-6ddd-11ea-817c-ccda262736ce | https://socratic.org/questions/another-compound-found-on-mars-contains-iron-and-oxygen-the-compound-contains-70 | Fe2O3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] this compound [IN] empirical"}] | [{"type":"chemical equation","value":"Fe2O3"}] | [{"type":"physical unit","value":"Percent by mass [OF] iron in this compound [=] \\pu{70%}"},{"type":"physical unit","value":"Percent by mass [OF] oxygen in this compound [=] \\pu{30%}"}] | <h1 class="questionTitle" itemprop="name">Another compound found on Mars contains iron and oxygen. The compound contains 70% by mass of iron and 30% by mass of oxygen. How do you calculate the empirical formula of this compound?</h1> | null | Fe2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your ultimate goal here is to figure out the <strong>smallest whole number ratio</strong> that exists between iron and oxygen in this unknown compound, i.e. the <strong>empirical formula</strong> of the compound. </p>
<p>Start by converting the <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to <em>grams</em> by using a <mathjax>#"100-g"#</mathjax> sample of compound. </p>
<p>You will have</p>
<blockquote>
<ul>
<li><mathjax>#"70% Fe " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 70 g Fe"#</mathjax></li>
<li><mathjax>#"30% O " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 30 g O"#</mathjax></li>
</ul>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to convert the masses to <em>moles</em></p>
<blockquote>
<p><mathjax>#"For Fe: " 70 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.2535 moles Fe"#</mathjax></p>
<p><mathjax>#"For O: " 30 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.8750 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between iron and oxygen, divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.2535 color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (1.8750color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1.496 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Now, you are looking for the <strong>smallest whole number ratio</strong> that exists between the two elements, so multiply both values by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"Fe"_((2 * 1))"O"_ ((2 * 1.5))#</mathjax></p>
</blockquote>
<p>You can thus say that the empirical formula of the compound is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Fe"_2"O"_3))) ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Fe"_2"O"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your ultimate goal here is to figure out the <strong>smallest whole number ratio</strong> that exists between iron and oxygen in this unknown compound, i.e. the <strong>empirical formula</strong> of the compound. </p>
<p>Start by converting the <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to <em>grams</em> by using a <mathjax>#"100-g"#</mathjax> sample of compound. </p>
<p>You will have</p>
<blockquote>
<ul>
<li><mathjax>#"70% Fe " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 70 g Fe"#</mathjax></li>
<li><mathjax>#"30% O " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 30 g O"#</mathjax></li>
</ul>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to convert the masses to <em>moles</em></p>
<blockquote>
<p><mathjax>#"For Fe: " 70 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.2535 moles Fe"#</mathjax></p>
<p><mathjax>#"For O: " 30 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.8750 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between iron and oxygen, divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.2535 color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (1.8750color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1.496 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Now, you are looking for the <strong>smallest whole number ratio</strong> that exists between the two elements, so multiply both values by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"Fe"_((2 * 1))"O"_ ((2 * 1.5))#</mathjax></p>
</blockquote>
<p>You can thus say that the empirical formula of the compound is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Fe"_2"O"_3))) ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Another compound found on Mars contains iron and oxygen. The compound contains 70% by mass of iron and 30% by mass of oxygen. How do you calculate the empirical formula of this compound?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"Fe"_2"O"_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your ultimate goal here is to figure out the <strong>smallest whole number ratio</strong> that exists between iron and oxygen in this unknown compound, i.e. the <strong>empirical formula</strong> of the compound. </p>
<p>Start by converting the <a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> to <em>grams</em> by using a <mathjax>#"100-g"#</mathjax> sample of compound. </p>
<p>You will have</p>
<blockquote>
<ul>
<li><mathjax>#"70% Fe " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 70 g Fe"#</mathjax></li>
<li><mathjax>#"30% O " stackrel(color(white)(acolor(blue)("100 g sample")aaa))(->) " 30 g O"#</mathjax></li>
</ul>
</blockquote>
<p>Next, use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> to convert the masses to <em>moles</em></p>
<blockquote>
<p><mathjax>#"For Fe: " 70 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.2535 moles Fe"#</mathjax></p>
<p><mathjax>#"For O: " 30 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.8750 moles O"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between iron and oxygen, divide both values by the <em>smallest one</em>. </p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.2535 color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (1.8750color(red)(cancel(color(black)("moles"))))/(1.2535color(red)(cancel(color(black)("moles")))) = 1.496 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Now, you are looking for the <strong>smallest whole number ratio</strong> that exists between the two elements, so multiply both values by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#"Fe"_((2 * 1))"O"_ ((2 * 1.5))#</mathjax></p>
</blockquote>
<p>You can thus say that the empirical formula of the compound is</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("Fe"_2"O"_3))) ->#</mathjax> <em><strong>empirical formula</strong></em></p>
</blockquote></div>
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</article> | Another compound found on Mars contains iron and oxygen. The compound contains 70% by mass of iron and 30% by mass of oxygen. How do you calculate the empirical formula of this compound? | null |
1,304 | ab2d0abe-6ddd-11ea-ad90-ccda262736ce | https://socratic.org/questions/a-170-00-g-sample-of-an-unidentified-compound-contains-29-84-g-sodium-67-49-chro | Na2Cr2O7 | start chemical_formula qc_end physical_unit 6 7 1 2 mass qc_end physical_unit 11 11 9 10 mass qc_end physical_unit 14 14 12 13 mass qc_end physical_unit 18 18 16 17 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Na2Cr2O7"}] | [{"type":"physical unit","value":"Mass [OF] unidentified compound sample [=] \\pu{170.00 g}"},{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{29.84 g}"},{"type":"physical unit","value":"Mass [OF] chromium [=] \\pu{67.49 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{72.67 g}"}] | <h1 class="questionTitle" itemprop="name">A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 chromium, and 72.67 g oxygen. What is the compound's empirical formula?</h1> | null | Na2Cr2O7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So you have 29.84g of Sodium, 67.49g of chromium and 72.67g of oxygen</p>
<p>You would first find the number of mole of each substance in each given mass using:</p>
<p><mathjax>#mols = (Mass)/(Molar Mass)#</mathjax></p>
<p><mathjax>#n(Na) =( 29.84g)/(22.9897g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(Cr) = (67.49g)/(51.996g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(O) = (72.67g)/(16g) = 4.54 mols#</mathjax></p>
<p>Then you find the molar ratios between each element by dividing the moles by the lowest moles calculated</p>
<p>Molar ratio (Na) = <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (Cr) <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (O) <mathjax>#4.54/1.298 = 3.5#</mathjax></p>
<p>However 3.5 is not a whole number, to make a whole number we multiply it by 2 however if we do that we have to multiply each mol by 2 as well.</p>
<p>Molar ratio (Na) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (Cr) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (O) = <mathjax>#3.5*2 = 7#</mathjax></p>
<p>Then we can write the empirical formula</p>
<p><mathjax>#Na_2Cr_2O_7#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#Na_2Cr_2O_7#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So you have 29.84g of Sodium, 67.49g of chromium and 72.67g of oxygen</p>
<p>You would first find the number of mole of each substance in each given mass using:</p>
<p><mathjax>#mols = (Mass)/(Molar Mass)#</mathjax></p>
<p><mathjax>#n(Na) =( 29.84g)/(22.9897g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(Cr) = (67.49g)/(51.996g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(O) = (72.67g)/(16g) = 4.54 mols#</mathjax></p>
<p>Then you find the molar ratios between each element by dividing the moles by the lowest moles calculated</p>
<p>Molar ratio (Na) = <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (Cr) <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (O) <mathjax>#4.54/1.298 = 3.5#</mathjax></p>
<p>However 3.5 is not a whole number, to make a whole number we multiply it by 2 however if we do that we have to multiply each mol by 2 as well.</p>
<p>Molar ratio (Na) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (Cr) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (O) = <mathjax>#3.5*2 = 7#</mathjax></p>
<p>Then we can write the empirical formula</p>
<p><mathjax>#Na_2Cr_2O_7#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 chromium, and 72.67 g oxygen. What is the compound's empirical formula?</h1>
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<div class="markdown"><p><mathjax>#Na_2Cr_2O_7#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So you have 29.84g of Sodium, 67.49g of chromium and 72.67g of oxygen</p>
<p>You would first find the number of mole of each substance in each given mass using:</p>
<p><mathjax>#mols = (Mass)/(Molar Mass)#</mathjax></p>
<p><mathjax>#n(Na) =( 29.84g)/(22.9897g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(Cr) = (67.49g)/(51.996g) = 1.298 mols#</mathjax></p>
<p><mathjax>#n(O) = (72.67g)/(16g) = 4.54 mols#</mathjax></p>
<p>Then you find the molar ratios between each element by dividing the moles by the lowest moles calculated</p>
<p>Molar ratio (Na) = <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (Cr) <mathjax>#1.298/1.298 = 1#</mathjax></p>
<p>Molar ratio (O) <mathjax>#4.54/1.298 = 3.5#</mathjax></p>
<p>However 3.5 is not a whole number, to make a whole number we multiply it by 2 however if we do that we have to multiply each mol by 2 as well.</p>
<p>Molar ratio (Na) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (Cr) = <mathjax>#1*2 = 2#</mathjax></p>
<p>Molar ratio (O) = <mathjax>#3.5*2 = 7#</mathjax></p>
<p>Then we can write the empirical formula</p>
<p><mathjax>#Na_2Cr_2O_7#</mathjax></p></div>
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</article> | A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 chromium, and 72.67 g oxygen. What is the compound's empirical formula? | null |
1,305 | a9a761ae-6ddd-11ea-b229-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-7-5-10-78-atoms-of-zinc | 1.25 × 10^55 moles | start physical_unit 10 10 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] zinc [IN] moles"}] | [{"type":"physical unit","value":"1.25 × 10^55 moles"}] | [{"type":"physical unit","value":"Number [OF] zinc atoms [=] \\pu{7.5 × 10^78}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in #7.5*10^78# atoms of zinc?</h1> | null | 1.25 × 10^55 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So this approximately <mathjax>#10^55#</mathjax> <mathjax>#"moles"#</mathjax>. Given that world production of <a href="https://en.wikipedia.org/wiki/Zinc" rel="nofollow">zinc</a> was approx <mathjax>#1xx10^9#</mathjax> <mathjax>#g#</mathjax>, in 2015, approx., <mathjax>#10^7#</mathjax> moles, this amount probably represents more of the metal than exists on Earth, and possibly more than the entire amount of zinc present in the solar system. So this is not a serious question.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Hmm <mathjax>#(7.5xx10^(78) " zinc atoms")/(6.022xx10^23" zinc atoms "mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??? mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So this approximately <mathjax>#10^55#</mathjax> <mathjax>#"moles"#</mathjax>. Given that world production of <a href="https://en.wikipedia.org/wiki/Zinc" rel="nofollow">zinc</a> was approx <mathjax>#1xx10^9#</mathjax> <mathjax>#g#</mathjax>, in 2015, approx., <mathjax>#10^7#</mathjax> moles, this amount probably represents more of the metal than exists on Earth, and possibly more than the entire amount of zinc present in the solar system. So this is not a serious question.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in #7.5*10^78# atoms of zinc?</h1>
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<div class="markdown"><p>Hmm <mathjax>#(7.5xx10^(78) " zinc atoms")/(6.022xx10^23" zinc atoms "mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??? mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So this approximately <mathjax>#10^55#</mathjax> <mathjax>#"moles"#</mathjax>. Given that world production of <a href="https://en.wikipedia.org/wiki/Zinc" rel="nofollow">zinc</a> was approx <mathjax>#1xx10^9#</mathjax> <mathjax>#g#</mathjax>, in 2015, approx., <mathjax>#10^7#</mathjax> moles, this amount probably represents more of the metal than exists on Earth, and possibly more than the entire amount of zinc present in the solar system. So this is not a serious question.</p></div>
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</article> | How many moles are in #7.5*10^78# atoms of zinc? | null |
1,306 | acfdd846-6ddd-11ea-84db-ccda262736ce | https://socratic.org/questions/when-oxygen-combines-with-any-alkali-metal-m-what-is-the-formula-of-the-compound | M2O | start chemical_formula qc_end substance 1 1 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"M2O"}] | [{"type":"substance name","value":"Oxygen"},{"type":"other","value":"Combine with any alkali metal M."}] | <h1 class="questionTitle" itemprop="name">When oxygen combines with any alkali metal, M, what is the formula of the compound produced usually?</h1> | null | M2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this, we can take a look at the <strong>ionic charges</strong> of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved.</p>
<p>Since oxygen is in group 16, it will have <mathjax>#color(red)(6#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/zgilTiMGSVKqsra8FzAe_2000px-Electron_shell_008_Oxygen_-_no_label.svg.png"/> </p>
<p>Recall that an atom will want to gain or lose electrons to fulfill the <em><a href="https://socratic.org/chemistry/bonding-basics/electrons-in-bonding-and-the-octet-rule">octet rule</a></em>; it wants to do whatever is easiest to obtain <mathjax>#8#</mathjax> electrons in its outermost shell.</p>
<p>For an oxygen atom, the easiest way to get <mathjax>#8#</mathjax> electrons in the outer shell is to obtain <mathjax>#2#</mathjax> more electrons. This will in turn cause the <mathjax>#sfcolor(blue)("oxygen"#</mathjax> to have an ionic charge of <mathjax>#color(blue)(2-#</mathjax>, because there are <mathjax>#color(blue)(2#</mathjax> more electrons than protons now.</p>
<p><mathjax>#-----#</mathjax></p>
<p>For the alkali metals, which are located in group 1 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, they each have <mathjax>#1#</mathjax> valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/i1D37ATShieRF0jBTGSH_2000px-Electron_shell_019_Potassium_-_no_label.svg.png"/> </p>
<p>In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol <mathjax>#color(red)("M"#</mathjax> for the alkali metal) to have an ionic charge of <mathjax>#color(red)(1+#</mathjax>, because there is <mathjax>#color(red)(1#</mathjax> less electron than the number of protons.</p>
<blockquote></blockquote>
<p><strong>The formula</strong>:</p>
<p>The compound the alkali metal <mathjax>#color(red)("M"#</mathjax> and oxygen <mathjax>#color(blue)("O"#</mathjax> form must be <em>electrically neutral</em>. That is to say, the charges must balance out to <mathjax>#0#</mathjax>. </p>
<p>The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to <strong>put the charge of the anion (the negative ion, in this case is <mathjax>#"O"^(2-)#</mathjax>), <mathjax>#2-#</mathjax> as the <em>subscript</em> of the cation, and the charge of the cation (the positive ion, in this case <mathjax>#"M"^+#</mathjax>) as the subscript of the *anion</strong>*. Here's what I mean, using sodium (<mathjax>#"Na"#</mathjax>) as an example:</p>
<p><img alt="http://web.fscj.edu" src="https://useruploads.socratic.org/30JmgfTZQQ24Jzzuxa49_slide816.gif"/> </p>
<p><mathjax>#sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"#</mathjax><br/>
<mathjax>#sfcolor(green)(" is M"_2"O"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The basic formula is <mathjax>#color(red)("M"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this, we can take a look at the <strong>ionic charges</strong> of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved.</p>
<p>Since oxygen is in group 16, it will have <mathjax>#color(red)(6#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/zgilTiMGSVKqsra8FzAe_2000px-Electron_shell_008_Oxygen_-_no_label.svg.png"/> </p>
<p>Recall that an atom will want to gain or lose electrons to fulfill the <em><a href="https://socratic.org/chemistry/bonding-basics/electrons-in-bonding-and-the-octet-rule">octet rule</a></em>; it wants to do whatever is easiest to obtain <mathjax>#8#</mathjax> electrons in its outermost shell.</p>
<p>For an oxygen atom, the easiest way to get <mathjax>#8#</mathjax> electrons in the outer shell is to obtain <mathjax>#2#</mathjax> more electrons. This will in turn cause the <mathjax>#sfcolor(blue)("oxygen"#</mathjax> to have an ionic charge of <mathjax>#color(blue)(2-#</mathjax>, because there are <mathjax>#color(blue)(2#</mathjax> more electrons than protons now.</p>
<p><mathjax>#-----#</mathjax></p>
<p>For the alkali metals, which are located in group 1 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, they each have <mathjax>#1#</mathjax> valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/i1D37ATShieRF0jBTGSH_2000px-Electron_shell_019_Potassium_-_no_label.svg.png"/> </p>
<p>In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol <mathjax>#color(red)("M"#</mathjax> for the alkali metal) to have an ionic charge of <mathjax>#color(red)(1+#</mathjax>, because there is <mathjax>#color(red)(1#</mathjax> less electron than the number of protons.</p>
<blockquote></blockquote>
<p><strong>The formula</strong>:</p>
<p>The compound the alkali metal <mathjax>#color(red)("M"#</mathjax> and oxygen <mathjax>#color(blue)("O"#</mathjax> form must be <em>electrically neutral</em>. That is to say, the charges must balance out to <mathjax>#0#</mathjax>. </p>
<p>The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to <strong>put the charge of the anion (the negative ion, in this case is <mathjax>#"O"^(2-)#</mathjax>), <mathjax>#2-#</mathjax> as the <em>subscript</em> of the cation, and the charge of the cation (the positive ion, in this case <mathjax>#"M"^+#</mathjax>) as the subscript of the *anion</strong>*. Here's what I mean, using sodium (<mathjax>#"Na"#</mathjax>) as an example:</p>
<p><img alt="http://web.fscj.edu" src="https://useruploads.socratic.org/30JmgfTZQQ24Jzzuxa49_slide816.gif"/> </p>
<p><mathjax>#sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"#</mathjax><br/>
<mathjax>#sfcolor(green)(" is M"_2"O"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When oxygen combines with any alkali metal, M, what is the formula of the compound produced usually?</h1>
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Nathan L.
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Jun 15, 2017
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<div class="markdown"><p>The basic formula is <mathjax>#color(red)("M"_2"O"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this, we can take a look at the <strong>ionic charges</strong> of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved.</p>
<p>Since oxygen is in group 16, it will have <mathjax>#color(red)(6#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/zgilTiMGSVKqsra8FzAe_2000px-Electron_shell_008_Oxygen_-_no_label.svg.png"/> </p>
<p>Recall that an atom will want to gain or lose electrons to fulfill the <em><a href="https://socratic.org/chemistry/bonding-basics/electrons-in-bonding-and-the-octet-rule">octet rule</a></em>; it wants to do whatever is easiest to obtain <mathjax>#8#</mathjax> electrons in its outermost shell.</p>
<p>For an oxygen atom, the easiest way to get <mathjax>#8#</mathjax> electrons in the outer shell is to obtain <mathjax>#2#</mathjax> more electrons. This will in turn cause the <mathjax>#sfcolor(blue)("oxygen"#</mathjax> to have an ionic charge of <mathjax>#color(blue)(2-#</mathjax>, because there are <mathjax>#color(blue)(2#</mathjax> more electrons than protons now.</p>
<p><mathjax>#-----#</mathjax></p>
<p>For the alkali metals, which are located in group 1 of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, they each have <mathjax>#1#</mathjax> valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:</p>
<p><img alt="upload.wikimedia.org" src="https://useruploads.socratic.org/i1D37ATShieRF0jBTGSH_2000px-Electron_shell_019_Potassium_-_no_label.svg.png"/> </p>
<p>In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol <mathjax>#color(red)("M"#</mathjax> for the alkali metal) to have an ionic charge of <mathjax>#color(red)(1+#</mathjax>, because there is <mathjax>#color(red)(1#</mathjax> less electron than the number of protons.</p>
<blockquote></blockquote>
<p><strong>The formula</strong>:</p>
<p>The compound the alkali metal <mathjax>#color(red)("M"#</mathjax> and oxygen <mathjax>#color(blue)("O"#</mathjax> form must be <em>electrically neutral</em>. That is to say, the charges must balance out to <mathjax>#0#</mathjax>. </p>
<p>The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to <strong>put the charge of the anion (the negative ion, in this case is <mathjax>#"O"^(2-)#</mathjax>), <mathjax>#2-#</mathjax> as the <em>subscript</em> of the cation, and the charge of the cation (the positive ion, in this case <mathjax>#"M"^+#</mathjax>) as the subscript of the *anion</strong>*. Here's what I mean, using sodium (<mathjax>#"Na"#</mathjax>) as an example:</p>
<p><img alt="http://web.fscj.edu" src="https://useruploads.socratic.org/30JmgfTZQQ24Jzzuxa49_slide816.gif"/> </p>
<p><mathjax>#sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"#</mathjax><br/>
<mathjax>#sfcolor(green)(" is M"_2"O"#</mathjax>.</p></div>
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</article> | When oxygen combines with any alkali metal, M, what is the formula of the compound produced usually? | null |
1,307 | abd0a0b0-6ddd-11ea-a228-ccda262736ce | https://socratic.org/questions/a-sample-of-oxygen-gas-occupies-a-volume-of-250-ml-at-100-kpa-pressure-what-volu | 125.00 mL | start physical_unit 1 4 volume ml qc_end physical_unit 1 4 9 10 volume qc_end physical_unit 1 4 12 13 pressure qc_end physical_unit 1 4 21 22 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] oxygen gas sample [IN] mL"}] | [{"type":"physical unit","value":"125.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] oxygen gas sample [=] \\pu{250 mL}"},{"type":"physical unit","value":"Pressure1 [OF] oxygen gas sample [=] \\pu{100 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] oxygen gas sample [=] \\pu{200 kPa}"}] | <h1 class="questionTitle" itemprop="name">A sample of oxygen gas occupies a volume of 250. mL at 100 kPa pressure. What volume will it occupy at 200 kPa pressure?</h1> | null | 125.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas varies inversely with pressure, as long as temperature and amount are kept constant. The equation to use for this law is <mathjax>#P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P_1#</mathjax> is the initial pressure, <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#P_2#</mathjax> is the final pressure, and <mathjax>#V_2#</mathjax> is the final volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="100 kPa O"_2"#</mathjax><br/>
<mathjax>#V_1="250 mL O"_2"#</mathjax><br/>
<mathjax>#P_2="200 kPa O"_2"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=(100 cancel"kPa"xx250 "mL")/(200 cancel"kPa")="125 mL O"_2"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new volume of <mathjax>#"O"_2"#</mathjax> is <mathjax>#"125 mL"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas varies inversely with pressure, as long as temperature and amount are kept constant. The equation to use for this law is <mathjax>#P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P_1#</mathjax> is the initial pressure, <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#P_2#</mathjax> is the final pressure, and <mathjax>#V_2#</mathjax> is the final volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="100 kPa O"_2"#</mathjax><br/>
<mathjax>#V_1="250 mL O"_2"#</mathjax><br/>
<mathjax>#P_2="200 kPa O"_2"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=(100 cancel"kPa"xx250 "mL")/(200 cancel"kPa")="125 mL O"_2"#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of oxygen gas occupies a volume of 250. mL at 100 kPa pressure. What volume will it occupy at 200 kPa pressure?</h1>
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May 2, 2016
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<div class="markdown"><p>The new volume of <mathjax>#"O"_2"#</mathjax> is <mathjax>#"125 mL"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a gas varies inversely with pressure, as long as temperature and amount are kept constant. The equation to use for this law is <mathjax>#P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P_1#</mathjax> is the initial pressure, <mathjax>#V_1#</mathjax> is the initial volume, <mathjax>#P_2#</mathjax> is the final pressure, and <mathjax>#V_2#</mathjax> is the final volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="100 kPa O"_2"#</mathjax><br/>
<mathjax>#V_1="250 mL O"_2"#</mathjax><br/>
<mathjax>#P_2="200 kPa O"_2"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#V_2=(100 cancel"kPa"xx250 "mL")/(200 cancel"kPa")="125 mL O"_2"#</mathjax></p></div>
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</article> | A sample of oxygen gas occupies a volume of 250. mL at 100 kPa pressure. What volume will it occupy at 200 kPa pressure? | null |
1,308 | ab44691c-6ddd-11ea-b9ca-ccda262736ce | https://socratic.org/questions/566788a011ef6b308239202d | 0.21 nm | start physical_unit 35 37 distance nm qc_end physical_unit 13 15 17 17 angle_of_the_reflected_beam qc_end physical_unit 20 23 28 29 wavelength qc_end end | [{"type":"physical unit","value":"Separation [OF] the crystal planes [IN] nm"}] | [{"type":"physical unit","value":"0.21 nm"}] | [{"type":"physical unit","value":"Angle of the reflected beam [OF] 1st order diffraction [=] \\pu{21.9°}"},{"type":"physical unit","value":"Wavelength [OF] incident X - ray [=] \\pu{1.57 Å}"}] | <h1 class="questionTitle" itemprop="name">In X - Ray diffraction analysis the angle of the reflected beam for 1st order diffraction is #sf(21.9^@)#. If the incident X - ray has a wavelength of #1.57color(white)(x)stackrel(@)"A"#what is the separation of the crystal planes ?</h1> | null | 0.21 nm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question is based on X - ray diffraction.</p>
<p><img alt="chemistryviews.org" src="https://useruploads.socratic.org/InOsR1I2RI2YVbAvqmWV_Braggs_law%281%29.gif"/> </p>
<p>The incident X - ray beams are reflected off the crystal layers as shown in the graphic. </p>
<p>After reflection the beams interfere with each other to produce a pattern which can be detected with a photographic plate.</p>
<p>If the X - rays are in phase constructive interference occurs:</p>
<p><img alt="thescienceclassroom.wikispaces.com" src="https://thescienceclassroom.wikispaces.com/file/view/constructive_interference2.jpg/227715678/293x185/constructive_interference2.jpg"/></p>
<p>This produces a pattern on the plate enabling <mathjax>#theta#</mathjax> to be measured.</p>
<p>The condition for constructive interference to occur is that the path difference of the two beams must be a whole number of wavelengths.</p>
<p>From the graphic you can see that this path difference is <mathjax>#dsintheta#</mathjax> going in + <mathjax>#dsintheta#</mathjax> coming out.</p>
<p>Since this must be equal to a whole number of wavelengths <mathjax>#lambda#</mathjax> we can write:</p>
<p><mathjax>#nlambda=2dsintheta#</mathjax></p>
<p>Where <mathjax>#n#</mathjax> is an integer.</p>
<p>In this case <mathjax>#n=1#</mathjax> so we can write:</p>
<p><mathjax>#d=(lambda)/(2sintheta)#</mathjax></p>
<p><mathjax>#:.d=(1.57)/(2xx0.373)=2.10color(white)(x)stackrel(@)"A"#</mathjax></p>
<p>The <mathjax>#stackrel(@)("A")"ngstrom"#</mathjax> unit is not strictly SI. This is equal to <mathjax>#0.210color(white)(x)"nm"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.10color(white)(x)stackrel(@)"A"#</mathjax></p>
<p>or <mathjax>#0.210color(white)(x)"nm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question is based on X - ray diffraction.</p>
<p><img alt="chemistryviews.org" src="https://useruploads.socratic.org/InOsR1I2RI2YVbAvqmWV_Braggs_law%281%29.gif"/> </p>
<p>The incident X - ray beams are reflected off the crystal layers as shown in the graphic. </p>
<p>After reflection the beams interfere with each other to produce a pattern which can be detected with a photographic plate.</p>
<p>If the X - rays are in phase constructive interference occurs:</p>
<p><img alt="thescienceclassroom.wikispaces.com" src="https://thescienceclassroom.wikispaces.com/file/view/constructive_interference2.jpg/227715678/293x185/constructive_interference2.jpg"/></p>
<p>This produces a pattern on the plate enabling <mathjax>#theta#</mathjax> to be measured.</p>
<p>The condition for constructive interference to occur is that the path difference of the two beams must be a whole number of wavelengths.</p>
<p>From the graphic you can see that this path difference is <mathjax>#dsintheta#</mathjax> going in + <mathjax>#dsintheta#</mathjax> coming out.</p>
<p>Since this must be equal to a whole number of wavelengths <mathjax>#lambda#</mathjax> we can write:</p>
<p><mathjax>#nlambda=2dsintheta#</mathjax></p>
<p>Where <mathjax>#n#</mathjax> is an integer.</p>
<p>In this case <mathjax>#n=1#</mathjax> so we can write:</p>
<p><mathjax>#d=(lambda)/(2sintheta)#</mathjax></p>
<p><mathjax>#:.d=(1.57)/(2xx0.373)=2.10color(white)(x)stackrel(@)"A"#</mathjax></p>
<p>The <mathjax>#stackrel(@)("A")"ngstrom"#</mathjax> unit is not strictly SI. This is equal to <mathjax>#0.210color(white)(x)"nm"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">In X - Ray diffraction analysis the angle of the reflected beam for 1st order diffraction is #sf(21.9^@)#. If the incident X - ray has a wavelength of #1.57color(white)(x)stackrel(@)"A"#what is the separation of the crystal planes ?</h1>
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<div class="markdown"><p><mathjax>#2.10color(white)(x)stackrel(@)"A"#</mathjax></p>
<p>or <mathjax>#0.210color(white)(x)"nm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This question is based on X - ray diffraction.</p>
<p><img alt="chemistryviews.org" src="https://useruploads.socratic.org/InOsR1I2RI2YVbAvqmWV_Braggs_law%281%29.gif"/> </p>
<p>The incident X - ray beams are reflected off the crystal layers as shown in the graphic. </p>
<p>After reflection the beams interfere with each other to produce a pattern which can be detected with a photographic plate.</p>
<p>If the X - rays are in phase constructive interference occurs:</p>
<p><img alt="thescienceclassroom.wikispaces.com" src="https://thescienceclassroom.wikispaces.com/file/view/constructive_interference2.jpg/227715678/293x185/constructive_interference2.jpg"/></p>
<p>This produces a pattern on the plate enabling <mathjax>#theta#</mathjax> to be measured.</p>
<p>The condition for constructive interference to occur is that the path difference of the two beams must be a whole number of wavelengths.</p>
<p>From the graphic you can see that this path difference is <mathjax>#dsintheta#</mathjax> going in + <mathjax>#dsintheta#</mathjax> coming out.</p>
<p>Since this must be equal to a whole number of wavelengths <mathjax>#lambda#</mathjax> we can write:</p>
<p><mathjax>#nlambda=2dsintheta#</mathjax></p>
<p>Where <mathjax>#n#</mathjax> is an integer.</p>
<p>In this case <mathjax>#n=1#</mathjax> so we can write:</p>
<p><mathjax>#d=(lambda)/(2sintheta)#</mathjax></p>
<p><mathjax>#:.d=(1.57)/(2xx0.373)=2.10color(white)(x)stackrel(@)"A"#</mathjax></p>
<p>The <mathjax>#stackrel(@)("A")"ngstrom"#</mathjax> unit is not strictly SI. This is equal to <mathjax>#0.210color(white)(x)"nm"#</mathjax></p></div>
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</article> | In X - Ray diffraction analysis the angle of the reflected beam for 1st order diffraction is #sf(21.9^@)#. If the incident X - ray has a wavelength of #1.57color(white)(x)stackrel(@)"A"#what is the separation of the crystal planes ? | null |
1,309 | ac8eeb5e-6ddd-11ea-b55c-ccda262736ce | https://socratic.org/questions/if-a-2-60g-sample-of-titanium-metal-chemically-combines-with-chlorine-gas-to-for | TiCl4 | start chemical_formula qc_end physical_unit 4 7 2 3 mass qc_end physical_unit 19 20 15 16 mass qc_end substance 11 12 qc_end end | [{"type":"other","value":"Chemical Formula [OF] titanium chloride [IN] empirical"}] | [{"type":"chemical equation","value":"TiCl4"}] | [{"type":"physical unit","value":"Mass [OF] titanium metal sample [=] \\pu{2.60 g}"},{"type":"physical unit","value":"Mass [OF] titanium chloride [=] \\pu{10.31 g}"},{"type":"substance name","value":"Chlorine gas"}] | <h1 class="questionTitle" itemprop="name">If a 2.60g sample of titanium metal chemically combines with chlorine gas to form 10.31g of a titanium chloride, what is the empirical formula product?
</h1> | null | TiCl4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the difference between the mass of the <em>titanium chloride</em> and the mass of titanium metal will represent the mass of <em>chlorine</em>. </p>
<p>This means that you can find how much chlorine was needed to form that much titanium chloride by</p>
<blockquote>
<p><mathjax>#m_"titanium chloride" = m_"chlorine" + m_"titanium"#</mathjax></p>
<p><mathjax>#n_"chlorine" = "10.31 g" - "2.60 g" = "7.71 g Cl"#</mathjax></p>
</blockquote>
<p>So, your titanium chloride contains <mathjax>#"2.60 g"#</mathjax> of titanium and <mathjax>#"7.71 g"#</mathjax> of chlorine. To get the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you need to first establish how many <em>moles</em> of each element you get in that sample. </p>
<p>To do that, use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses</p>
<blockquote>
<p><mathjax>#"For Ti: " (2.60color(red)(cancel(color(black)("g"))))/(47.867color(red)(cancel(color(black)("g")))/"mol") = "0.054317 moles Ti"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"For Cl: " (7.71color(red)(cancel(color(black)("g"))))/(35.453color(red)(cancel(color(black)("g")))/"mol") = "0.21747 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the sample, divide both numbers of moles by the <em>smallest</em> of the two</p>
<blockquote>
<p><mathjax>#"For Ti: " (0.054317color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.21747color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 4.0037 ~~ 4#</mathjax></p>
</blockquote>
<p>This means that the <em>empirical formula</em> will be </p>
<blockquote>
<p><mathjax>#"Ti"_1"Cl"_4#</mathjax>, or <mathjax>#color(green)("TiCl"_4)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"TiCl"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the difference between the mass of the <em>titanium chloride</em> and the mass of titanium metal will represent the mass of <em>chlorine</em>. </p>
<p>This means that you can find how much chlorine was needed to form that much titanium chloride by</p>
<blockquote>
<p><mathjax>#m_"titanium chloride" = m_"chlorine" + m_"titanium"#</mathjax></p>
<p><mathjax>#n_"chlorine" = "10.31 g" - "2.60 g" = "7.71 g Cl"#</mathjax></p>
</blockquote>
<p>So, your titanium chloride contains <mathjax>#"2.60 g"#</mathjax> of titanium and <mathjax>#"7.71 g"#</mathjax> of chlorine. To get the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you need to first establish how many <em>moles</em> of each element you get in that sample. </p>
<p>To do that, use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses</p>
<blockquote>
<p><mathjax>#"For Ti: " (2.60color(red)(cancel(color(black)("g"))))/(47.867color(red)(cancel(color(black)("g")))/"mol") = "0.054317 moles Ti"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"For Cl: " (7.71color(red)(cancel(color(black)("g"))))/(35.453color(red)(cancel(color(black)("g")))/"mol") = "0.21747 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the sample, divide both numbers of moles by the <em>smallest</em> of the two</p>
<blockquote>
<p><mathjax>#"For Ti: " (0.054317color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.21747color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 4.0037 ~~ 4#</mathjax></p>
</blockquote>
<p>This means that the <em>empirical formula</em> will be </p>
<blockquote>
<p><mathjax>#"Ti"_1"Cl"_4#</mathjax>, or <mathjax>#color(green)("TiCl"_4)#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">If a 2.60g sample of titanium metal chemically combines with chlorine gas to form 10.31g of a titanium chloride, what is the empirical formula product?
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Stefan V.
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Oct 20, 2015
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<div class="markdown"><p><mathjax>#"TiCl"_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the difference between the mass of the <em>titanium chloride</em> and the mass of titanium metal will represent the mass of <em>chlorine</em>. </p>
<p>This means that you can find how much chlorine was needed to form that much titanium chloride by</p>
<blockquote>
<p><mathjax>#m_"titanium chloride" = m_"chlorine" + m_"titanium"#</mathjax></p>
<p><mathjax>#n_"chlorine" = "10.31 g" - "2.60 g" = "7.71 g Cl"#</mathjax></p>
</blockquote>
<p>So, your titanium chloride contains <mathjax>#"2.60 g"#</mathjax> of titanium and <mathjax>#"7.71 g"#</mathjax> of chlorine. To get the compound's <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> you need to first establish how many <em>moles</em> of each element you get in that sample. </p>
<p>To do that, use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses</p>
<blockquote>
<p><mathjax>#"For Ti: " (2.60color(red)(cancel(color(black)("g"))))/(47.867color(red)(cancel(color(black)("g")))/"mol") = "0.054317 moles Ti"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#"For Cl: " (7.71color(red)(cancel(color(black)("g"))))/(35.453color(red)(cancel(color(black)("g")))/"mol") = "0.21747 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between the two elements in the sample, divide both numbers of moles by the <em>smallest</em> of the two</p>
<blockquote>
<p><mathjax>#"For Ti: " (0.054317color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.21747color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 4.0037 ~~ 4#</mathjax></p>
</blockquote>
<p>This means that the <em>empirical formula</em> will be </p>
<blockquote>
<p><mathjax>#"Ti"_1"Cl"_4#</mathjax>, or <mathjax>#color(green)("TiCl"_4)#</mathjax></p>
</blockquote></div>
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</article> | If a 2.60g sample of titanium metal chemically combines with chlorine gas to form 10.31g of a titanium chloride, what is the empirical formula product?
| null |
1,310 | ac994b1d-6ddd-11ea-af5e-ccda262736ce | https://socratic.org/questions/57f7a78c11ef6b05b6c8a365 | 8.68 | start physical_unit 8 8 ph none qc_end physical_unit 10 11 6 7 molarity qc_end physical_unit 16 17 19 19 pka qc_end end | [{"type":"physical unit","value":"pH [OF] sodium ethanoate solution"}] | [{"type":"physical unit","value":"8.68"}] | [{"type":"physical unit","value":"Molarity [OF] sodium ethanoate solution [=] \\pu{0.05 M}"},{"type":"physical unit","value":"pKa value [OF] ethanoic acid [=] \\pu{4.66}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.05 M solution of sodium ethanoate ?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>The <mathjax>#sf(pK_a)#</mathjax> value for ethanoic acid is 4.66.</p></div>
</h2>
</div>
</div> | 8.68 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ethanoate is the salt of a weak acid and strong base so undergoes hydrolysis:</p>
<p><mathjax>#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#</mathjax></p>
<p><mathjax>#sf(K_b=([CH_3COOH][OH^(-)])/([CH_3COO^(-)])#</mathjax></p>
<p>Note that these refer to equilibrium concentrations.</p>
<p>We can find the <mathjax>#sf(pOH)#</mathjax> hence the <mathjax>#sf(pH)#</mathjax> if we can get the value of <mathjax>#sf(pK_b)#</mathjax>.</p>
<p><mathjax>#sf(pK_a+pK_b=14)#</mathjax> at <mathjax>#sf(25^@C)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pK_b=14-4.66=9.34)#</mathjax></p>
<p>Because the equilibrium is well to the left we can assume that the equilibrium concentration of <mathjax>#sf(CH_3COO^(-)#</mathjax> is very close to 0.05 M.</p>
<p>By rearranging the expression for <mathjax>#sf(K_b)#</mathjax> and taking negative logs of both sides we get:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>Where <mathjax>#sf(b)#</mathjax> is the concentration of the co - base.</p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(pOH=1/2[9.34-log0.05]=1/2[9.34-(-1.3)]=5.32)#</mathjax></p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=14-5.32=8.68)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#sf(pH=8.68)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ethanoate is the salt of a weak acid and strong base so undergoes hydrolysis:</p>
<p><mathjax>#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#</mathjax></p>
<p><mathjax>#sf(K_b=([CH_3COOH][OH^(-)])/([CH_3COO^(-)])#</mathjax></p>
<p>Note that these refer to equilibrium concentrations.</p>
<p>We can find the <mathjax>#sf(pOH)#</mathjax> hence the <mathjax>#sf(pH)#</mathjax> if we can get the value of <mathjax>#sf(pK_b)#</mathjax>.</p>
<p><mathjax>#sf(pK_a+pK_b=14)#</mathjax> at <mathjax>#sf(25^@C)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pK_b=14-4.66=9.34)#</mathjax></p>
<p>Because the equilibrium is well to the left we can assume that the equilibrium concentration of <mathjax>#sf(CH_3COO^(-)#</mathjax> is very close to 0.05 M.</p>
<p>By rearranging the expression for <mathjax>#sf(K_b)#</mathjax> and taking negative logs of both sides we get:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>Where <mathjax>#sf(b)#</mathjax> is the concentration of the co - base.</p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(pOH=1/2[9.34-log0.05]=1/2[9.34-(-1.3)]=5.32)#</mathjax></p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=14-5.32=8.68)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.05 M solution of sodium ethanoate ?</h1>
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<div class="markdown"><p>The <mathjax>#sf(pK_a)#</mathjax> value for ethanoic acid is 4.66.</p></div>
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<div class="markdown"><p><mathjax>#sf(pH=8.68)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ethanoate is the salt of a weak acid and strong base so undergoes hydrolysis:</p>
<p><mathjax>#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#</mathjax></p>
<p><mathjax>#sf(K_b=([CH_3COOH][OH^(-)])/([CH_3COO^(-)])#</mathjax></p>
<p>Note that these refer to equilibrium concentrations.</p>
<p>We can find the <mathjax>#sf(pOH)#</mathjax> hence the <mathjax>#sf(pH)#</mathjax> if we can get the value of <mathjax>#sf(pK_b)#</mathjax>.</p>
<p><mathjax>#sf(pK_a+pK_b=14)#</mathjax> at <mathjax>#sf(25^@C)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pK_b=14-4.66=9.34)#</mathjax></p>
<p>Because the equilibrium is well to the left we can assume that the equilibrium concentration of <mathjax>#sf(CH_3COO^(-)#</mathjax> is very close to 0.05 M.</p>
<p>By rearranging the expression for <mathjax>#sf(K_b)#</mathjax> and taking negative logs of both sides we get:</p>
<p><mathjax>#sf(pOH=1/2[pK_b-logb])#</mathjax></p>
<p>Where <mathjax>#sf(b)#</mathjax> is the concentration of the co - base.</p>
<p>Putting in the numbers:</p>
<p><mathjax>#sf(pOH=1/2[9.34-log0.05]=1/2[9.34-(-1.3)]=5.32)#</mathjax></p>
<p><mathjax>#sf(pOH+pH=14)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=14-5.32=8.68)#</mathjax></p></div>
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</article> | What is the pH of a 0.05 M solution of sodium ethanoate ? |
The #sf(pK_a)# value for ethanoic acid is 4.66.
|
1,311 | acd442f4-6ddd-11ea-a728-ccda262736ce | https://socratic.org/questions/calcium-chloride-dissolves-in-water-to-form-calcium-and-chloride-ions-producing- | 21.6 ℃ | start physical_unit 4 4 temperature °c qc_end physical_unit 0 1 22 23 mole qc_end physical_unit 4 4 30 31 volume qc_end physical_unit 4 4 39 40 temperature qc_end substance 0 0 qc_end substance 9 10 qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"21.6 ℃"}] | [{"type":"physical unit","value":"Produced heat [OF] the reaction [=] \\pu{83 kJ per mole}"},{"type":"physical unit","value":"Mole [OF] calcium chloride [=] \\pu{0.02 moles}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{150 ml}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{19 ℃}"},{"type":"substance name","value":"Calcium"},{"type":"substance name","value":"Chloride ions"}] | <h1 class="questionTitle" itemprop="name">Calcium chloride dissolves in water to form calcium and chloride ions, producing 83kJ of heat per mole. Calculate final temp. when 0.02 moles of calcium chloride is dissolved into 150ml of water at a temperature initially of 19 degrees?</h1> | null | 21.6 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<p>The process of dissolving <mathjax>#CaCl_2#</mathjax> in releases <mathjax>#(8.3xx10^4"J")/("mole of "CaCl_2)#</mathjax></p>
<p>Compute the heat energy released for <mathjax>#0.02"mole of "CaCl_2#</mathjax>:</p>
<p><mathjax>#(8.3xx10^4"J")/(cancel("mole of "CaCl_2))(0.02cancel("mole of "CaCl_2))/1 = 1660"J"#</mathjax></p>
<p>This heat energy must be absorbed by the water. The reference <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" rel="nofollow">Specific Heat</a> gives us a useful equation:</p>
<p><mathjax>#Q = cmDeltaT#</mathjax></p>
<p>Substitute <mathjax>#1660"J"#</mathjax> for Q:</p>
<p><mathjax>#1660"J" = cmDeltaT#</mathjax></p>
<p>The reference, also, gives us a value for c:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")mDeltaT#</mathjax></p>
<p>Substitute 150g for m:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")DeltaT#</mathjax></p>
<p>Substitute <mathjax>#T_"final" - 19^@"C"#</mathjax> for <mathjax>#DeltaT#</mathjax>:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")(T_"final" - 19^@"C")#</mathjax></p>
<p>Solve for <mathjax>#T_"final"#</mathjax></p>
<p><mathjax>#T_"final" = (1660"J")/((4.186"J")/("g "^@"C")(150"g")) + 19^@"C"#</mathjax></p>
<p><mathjax>#T_"final" = 21.6^@"C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#T_"final" = 21.6^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<p>The process of dissolving <mathjax>#CaCl_2#</mathjax> in releases <mathjax>#(8.3xx10^4"J")/("mole of "CaCl_2)#</mathjax></p>
<p>Compute the heat energy released for <mathjax>#0.02"mole of "CaCl_2#</mathjax>:</p>
<p><mathjax>#(8.3xx10^4"J")/(cancel("mole of "CaCl_2))(0.02cancel("mole of "CaCl_2))/1 = 1660"J"#</mathjax></p>
<p>This heat energy must be absorbed by the water. The reference <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" rel="nofollow">Specific Heat</a> gives us a useful equation:</p>
<p><mathjax>#Q = cmDeltaT#</mathjax></p>
<p>Substitute <mathjax>#1660"J"#</mathjax> for Q:</p>
<p><mathjax>#1660"J" = cmDeltaT#</mathjax></p>
<p>The reference, also, gives us a value for c:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")mDeltaT#</mathjax></p>
<p>Substitute 150g for m:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")DeltaT#</mathjax></p>
<p>Substitute <mathjax>#T_"final" - 19^@"C"#</mathjax> for <mathjax>#DeltaT#</mathjax>:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")(T_"final" - 19^@"C")#</mathjax></p>
<p>Solve for <mathjax>#T_"final"#</mathjax></p>
<p><mathjax>#T_"final" = (1660"J")/((4.186"J")/("g "^@"C")(150"g")) + 19^@"C"#</mathjax></p>
<p><mathjax>#T_"final" = 21.6^@"C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Calcium chloride dissolves in water to form calcium and chloride ions, producing 83kJ of heat per mole. Calculate final temp. when 0.02 moles of calcium chloride is dissolved into 150ml of water at a temperature initially of 19 degrees?</h1>
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<div class="markdown"><p><mathjax>#T_"final" = 21.6^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<p>The process of dissolving <mathjax>#CaCl_2#</mathjax> in releases <mathjax>#(8.3xx10^4"J")/("mole of "CaCl_2)#</mathjax></p>
<p>Compute the heat energy released for <mathjax>#0.02"mole of "CaCl_2#</mathjax>:</p>
<p><mathjax>#(8.3xx10^4"J")/(cancel("mole of "CaCl_2))(0.02cancel("mole of "CaCl_2))/1 = 1660"J"#</mathjax></p>
<p>This heat energy must be absorbed by the water. The reference <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" rel="nofollow">Specific Heat</a> gives us a useful equation:</p>
<p><mathjax>#Q = cmDeltaT#</mathjax></p>
<p>Substitute <mathjax>#1660"J"#</mathjax> for Q:</p>
<p><mathjax>#1660"J" = cmDeltaT#</mathjax></p>
<p>The reference, also, gives us a value for c:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")mDeltaT#</mathjax></p>
<p>Substitute 150g for m:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")DeltaT#</mathjax></p>
<p>Substitute <mathjax>#T_"final" - 19^@"C"#</mathjax> for <mathjax>#DeltaT#</mathjax>:</p>
<p><mathjax>#1660"J" = (4.186"J")/("g "^@"C")(150"g")(T_"final" - 19^@"C")#</mathjax></p>
<p>Solve for <mathjax>#T_"final"#</mathjax></p>
<p><mathjax>#T_"final" = (1660"J")/((4.186"J")/("g "^@"C")(150"g")) + 19^@"C"#</mathjax></p>
<p><mathjax>#T_"final" = 21.6^@"C"#</mathjax></p></div>
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</article> | Calcium chloride dissolves in water to form calcium and chloride ions, producing 83kJ of heat per mole. Calculate final temp. when 0.02 moles of calcium chloride is dissolved into 150ml of water at a temperature initially of 19 degrees? | null |
1,312 | ab519aaf-6ddd-11ea-893b-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-5-42-cm-3-of-table-sugar | 3.79 grams | start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] table sugar [IN] grams"}] | [{"type":"physical unit","value":"3.79 grams"}] | [{"type":"physical unit","value":"Volume [OF] table sugar [=] \\pu{5.42 cm^3}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #5.42# #cm^3# of table sugar?</h1> | null | 3.79 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In handier units this is: <mathjax>#700*cancel(kg)*cancel(m^-3)xx10^3*g*cancel(kg^-1)xx10^-6*cancel(m^(3))*cm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.7*g*cm^-3#</mathjax>, or <mathjax>#0.7*g*mL#</mathjax>.</p>
<p>So (finally) the answer to your question is <mathjax>#5.42*cancel(cm^-3)xx0.7*g*cancel(cm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4*g#</mathjax> approximately!</p>
<p>It is a fact that sugar DOES NOT dissolve in hexanes. Given that <mathjax>#rho_"hexanes"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.67*g*mL#</mathjax>, should sugar float in hexanes? Can you account for the fact that sucrose, <mathjax>#C_12H_22O_11#</mathjax> is insoluble in hexanes?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>From <a href="http://www.sugartech.co.za/density/index.php" rel="nofollow">this site</a> I learnt that the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of table sugar is <mathjax>#700#</mathjax> <mathjax>#kg*m^-3#</mathjax>. So the answer is a bit under <mathjax>#4#</mathjax> <mathjax>#g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In handier units this is: <mathjax>#700*cancel(kg)*cancel(m^-3)xx10^3*g*cancel(kg^-1)xx10^-6*cancel(m^(3))*cm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.7*g*cm^-3#</mathjax>, or <mathjax>#0.7*g*mL#</mathjax>.</p>
<p>So (finally) the answer to your question is <mathjax>#5.42*cancel(cm^-3)xx0.7*g*cancel(cm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4*g#</mathjax> approximately!</p>
<p>It is a fact that sugar DOES NOT dissolve in hexanes. Given that <mathjax>#rho_"hexanes"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.67*g*mL#</mathjax>, should sugar float in hexanes? Can you account for the fact that sucrose, <mathjax>#C_12H_22O_11#</mathjax> is insoluble in hexanes?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of #5.42# #cm^3# of table sugar?</h1>
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anor277
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Feb 27, 2016
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<div class="markdown"><p>From <a href="http://www.sugartech.co.za/density/index.php" rel="nofollow">this site</a> I learnt that the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of table sugar is <mathjax>#700#</mathjax> <mathjax>#kg*m^-3#</mathjax>. So the answer is a bit under <mathjax>#4#</mathjax> <mathjax>#g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In handier units this is: <mathjax>#700*cancel(kg)*cancel(m^-3)xx10^3*g*cancel(kg^-1)xx10^-6*cancel(m^(3))*cm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.7*g*cm^-3#</mathjax>, or <mathjax>#0.7*g*mL#</mathjax>.</p>
<p>So (finally) the answer to your question is <mathjax>#5.42*cancel(cm^-3)xx0.7*g*cancel(cm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4*g#</mathjax> approximately!</p>
<p>It is a fact that sugar DOES NOT dissolve in hexanes. Given that <mathjax>#rho_"hexanes"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.67*g*mL#</mathjax>, should sugar float in hexanes? Can you account for the fact that sucrose, <mathjax>#C_12H_22O_11#</mathjax> is insoluble in hexanes?</p></div>
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</article> | What is the mass of #5.42# #cm^3# of table sugar? | null |
1,313 | ac6dcb90-6ddd-11ea-98ab-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-iodine-in-ni-3-1 | 32.15% | start physical_unit 8 10 percent_composition none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Percentage composition [OF] Iodine in NI3"}] | [{"type":"physical unit","value":"32.15%"}] | [{"type":"chemical equation","value":"NI3"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the percentage composition of Iodine in #NI_3#?</h1> | null | 32.15% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the percent composition of an element in a compound, you need to divide the molar mass (MM) of the element by the MM of the compound then multiply by 100.</p>
<p><mathjax>#%"composition of element"=("MM of element")/("MM of compound")xx100#</mathjax></p>
<p><mathjax>#"MM of Iodine=126.90447 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Iodine" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Iodine</a></p>
<p><mathjax>#"MM of NI"_3="394.719 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Nitrogen_triiodide" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Nitrogen_triiodide</a></p>
<p><strong>Percent Composition of Iodine in Nitrogen Triiodide</strong></p>
<p><mathjax>#%"composition I"=(126.90447 cancel"g/mol")/(394.719 cancel"g/mol")xx100="32.1506%"~~32%"#</mathjax></p></div>
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of iodine in nitrogen triioxide is <mathjax>#32.1506%~~32%"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the percent composition of an element in a compound, you need to divide the molar mass (MM) of the element by the MM of the compound then multiply by 100.</p>
<p><mathjax>#%"composition of element"=("MM of element")/("MM of compound")xx100#</mathjax></p>
<p><mathjax>#"MM of Iodine=126.90447 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Iodine" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Iodine</a></p>
<p><mathjax>#"MM of NI"_3="394.719 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Nitrogen_triiodide" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Nitrogen_triiodide</a></p>
<p><strong>Percent Composition of Iodine in Nitrogen Triiodide</strong></p>
<p><mathjax>#%"composition I"=(126.90447 cancel"g/mol")/(394.719 cancel"g/mol")xx100="32.1506%"~~32%"#</mathjax></p></div>
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<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of iodine in nitrogen triioxide is <mathjax>#32.1506%~~32%"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the percent composition of an element in a compound, you need to divide the molar mass (MM) of the element by the MM of the compound then multiply by 100.</p>
<p><mathjax>#%"composition of element"=("MM of element")/("MM of compound")xx100#</mathjax></p>
<p><mathjax>#"MM of Iodine=126.90447 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Iodine" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Iodine</a></p>
<p><mathjax>#"MM of NI"_3="394.719 g/mol"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Nitrogen_triiodide" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Nitrogen_triiodide</a></p>
<p><strong>Percent Composition of Iodine in Nitrogen Triiodide</strong></p>
<p><mathjax>#%"composition I"=(126.90447 cancel"g/mol")/(394.719 cancel"g/mol")xx100="32.1506%"~~32%"#</mathjax></p></div>
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</article> | How do you calculate the percentage composition of Iodine in #NI_3#? | null |
1,314 | acad1920-6ddd-11ea-8dda-ccda262736ce | https://socratic.org/questions/if-98-5-mol-of-an-ideal-gas-is-at-1-73-atm-and-477-k-what-is-the-volume-of-the-g | 2229.7 L | start physical_unit 19 20 volume l qc_end physical_unit 19 20 1 2 mole qc_end physical_unit 19 20 12 13 temperature qc_end physical_unit 19 20 9 10 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"2229.7 L"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{98.5 mol}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{477 K}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{1.73 atm}"}] | <h1 class="questionTitle" itemprop="name">If 98.5 mol of an ideal gas is at 1.73 atm and 477 K, what is the volume of the gas? </h1> | null | 2229.7 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax>.</p>
<p>The volume then could be obtained after rearranging the aforementioned expression as: </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p>Therefore, <mathjax>#=>V=(98.5cancel(mol)xx0.0821(L*cancel(atm))/(cancel(K)*cancel(mol))xx477cancel(K))/(1.73cancel(atm))=2229.7L~=2230L#</mathjax> (rounded to 3 significant figures)</p></div>
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<div>
<div class="markdown"><p><mathjax>#V=2230L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax>.</p>
<p>The volume then could be obtained after rearranging the aforementioned expression as: </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p>Therefore, <mathjax>#=>V=(98.5cancel(mol)xx0.0821(L*cancel(atm))/(cancel(K)*cancel(mol))xx477cancel(K))/(1.73cancel(atm))=2229.7L~=2230L#</mathjax> (rounded to 3 significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">If 98.5 mol of an ideal gas is at 1.73 atm and 477 K, what is the volume of the gas? </h1>
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<div class="markdown"><p><mathjax>#V=2230L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax>.</p>
<p>The volume then could be obtained after rearranging the aforementioned expression as: </p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p>Therefore, <mathjax>#=>V=(98.5cancel(mol)xx0.0821(L*cancel(atm))/(cancel(K)*cancel(mol))xx477cancel(K))/(1.73cancel(atm))=2229.7L~=2230L#</mathjax> (rounded to 3 significant figures)</p></div>
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</article> | If 98.5 mol of an ideal gas is at 1.73 atm and 477 K, what is the volume of the gas? | null |
1,315 | ac2734dd-6ddd-11ea-ab14-ccda262736ce | https://socratic.org/questions/in-a-2-5-m-cacl-2-solution-what-is-the-molarity-of-the-calcium-ions | 2.50 M | start physical_unit 12 13 molarity mol/l qc_end physical_unit 4 5 2 3 molarity qc_end end | [{"type":"physical unit","value":"Molarity [OF] the calcium ions [IN] M"}] | [{"type":"physical unit","value":"2.50 M"}] | [{"type":"physical unit","value":"Molarity [OF] CaCl2 solution [=] \\pu{2.5 M}"}] | <h1 class="questionTitle" itemprop="name">In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions?</h1> | null | 2.50 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<div class="markdown"><p>AS the name says on the tin, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of calcium ion is <mathjax>#2.5*mol*L^-1#</mathjax>....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<h1 class="questionTitle" itemprop="name">In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions?</h1>
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<div class="markdown"><p>AS the name says on the tin, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of calcium ion is <mathjax>#2.5*mol*L^-1#</mathjax>....</p></div>
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<div class="markdown"><p>And in one litre of solution there are <mathjax>#1*Lxx2.5*mol*L^-1xx6.022xx10^23*mol^-1#</mathjax> individual calcium ions.....What is the concentration of chloride ions?</p></div>
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<div class="markdown"><p><mathjax>#2.5M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the number of moles in a liter. If the solution is <mathjax>#2.5M#</mathjax>, the molarity of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> is the same basic amount, modified by the number of atoms of each element in the original compound.</p>
<p>In this case, we have one <mathjax>#Ca#</mathjax> atom and two <mathjax>#Cl#</mathjax> atoms in the compound. So, we will have <mathjax>#2.5M Ca^(+2)#</mathjax> and <mathjax>#2.5 xx 2 = 5M Cl^(-1)#</mathjax> ions in solution.</p></div>
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</article> | In a 2.5 M #CaCl_2# solution, what is the molarity of the calcium ions? | null |
1,316 | ac1f5b40-6ddd-11ea-98aa-ccda262736ce | https://socratic.org/questions/59400b767c0149794a24c51d | 82.4% | start physical_unit 9 9 percent_yield none qc_end physical_unit 4 4 1 2 mass qc_end chemical_equation 11 11 qc_end chemical_equation 13 13 qc_end end | [{"type":"physical unit","value":"Percentage yield [OF] the reaction"}] | [{"type":"physical unit","value":"82.4%"}] | [{"type":"physical unit","value":"Mass [OF] ammonia [=] \\pu{14 g}"},{"type":"chemical equation","value":"N2"},{"type":"chemical equation","value":"H2"}] | <h1 class="questionTitle" itemprop="name">If #14*g# of ammonia was isolated from a reaction between #N_2#, and #H_2#, what is the percentage yield?</h1> | null | 82.4% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"% yield"="moles of products"/"moles of reactants"xx100%#</mathjax></p>
<p>And thus quotient would have to be appropriately modified for the given reaction:</p>
<p><mathjax>#1/2N_2(g) + 3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax>.....</p>
<p>i.e. <mathjax>#"yield"="Moles of ammonia"/(1/2xx"moles of dinitrogen")#</mathjax> OR</p>
<p><mathjax>#"yield"="Moles of ammonia"/(3/2xx"moles of dihydrogen")#</mathjax> </p>
<p>Now we were given that <mathjax>#14*g#</mathjax> of ammonia were isolated. We were NOT given the mass of dihydrogen or dinitrogen reactant, so we cannot address yield. </p></div>
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<div class="markdown"><p>Unknown..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"% yield"="moles of products"/"moles of reactants"xx100%#</mathjax></p>
<p>And thus quotient would have to be appropriately modified for the given reaction:</p>
<p><mathjax>#1/2N_2(g) + 3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax>.....</p>
<p>i.e. <mathjax>#"yield"="Moles of ammonia"/(1/2xx"moles of dinitrogen")#</mathjax> OR</p>
<p><mathjax>#"yield"="Moles of ammonia"/(3/2xx"moles of dihydrogen")#</mathjax> </p>
<p>Now we were given that <mathjax>#14*g#</mathjax> of ammonia were isolated. We were NOT given the mass of dihydrogen or dinitrogen reactant, so we cannot address yield. </p></div>
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<h1 class="questionTitle" itemprop="name">If #14*g# of ammonia was isolated from a reaction between #N_2#, and #H_2#, what is the percentage yield?</h1>
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<div class="markdown"><p>Unknown..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"% yield"="moles of products"/"moles of reactants"xx100%#</mathjax></p>
<p>And thus quotient would have to be appropriately modified for the given reaction:</p>
<p><mathjax>#1/2N_2(g) + 3/2H_2(g) rightleftharpoons NH_3(g)#</mathjax>.....</p>
<p>i.e. <mathjax>#"yield"="Moles of ammonia"/(1/2xx"moles of dinitrogen")#</mathjax> OR</p>
<p><mathjax>#"yield"="Moles of ammonia"/(3/2xx"moles of dihydrogen")#</mathjax> </p>
<p>Now we were given that <mathjax>#14*g#</mathjax> of ammonia were isolated. We were NOT given the mass of dihydrogen or dinitrogen reactant, so we cannot address yield. </p></div>
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<div class="markdown"><p>Assuming an attempt/intent to synthesize 1 mole <mathjax>#NH_3#</mathjax>, then %yield = <mathjax>#82.4% Yield#</mathjax> w/w</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#%Yield = ("Lab Yield"/"Theoritical Yield")xx100%#</mathjax></p>
<p>For the sake of showing how %Yield is calculated, assume the intent is to synthesize 1 mole ammonia. Then the theoretical Yield from equation stoichiometry ...</p>
<p><mathjax>#3/2H_2(g)#</mathjax> + <mathjax>#1/2N_2(g)#</mathjax> => <mathjax>#NH_3(g)#</mathjax> <br/>
=> Theoretical yield from reaction = 1 mole <mathjax>#NH_3(g)#</mathjax> = 17 grams <mathjax>#NH_3(g)#</mathjax></p>
<p>Given Lab Yield = 14 grams <mathjax>#NH_3(g)#</mathjax></p>
<p>=> <mathjax>#"%"Yield"#</mathjax> = <mathjax>#14/17xx 100% = 82.4%#</mathjax> w/w </p></div>
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</article> | If #14*g# of ammonia was isolated from a reaction between #N_2#, and #H_2#, what is the percentage yield? | null |
1,317 | a88bd2d8-6ddd-11ea-894f-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-of-an-unknown-substance-if-2000-j-energy-are-required- | 100.00 J/(g * ℃) | start physical_unit 6 8 specific_heat j/(°c_·_g) qc_end end | [{"type":"physical unit","value":"Specific heat [OF] an unknown substance [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"100.00 J/(g * ℃)"}] | [{"type":"physical unit","value":"Required energy [OF] n unknown substance [=] \\pu{2000 J}"},{"type":"physical unit","value":"Raised temperature [OF] n unknown substance [=] \\pu{5 ℃}"},{"type":"physical unit","value":"mass [OF] n unknown substance [=] \\pu{4 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat of an unknown substance if 2000 J energy are required to raise the temperature of 4 grams of the substance 5°C?</h1> | null | 100.00 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're looking for the equation that relates all of these variables:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is the energy in Joules (J), <mathjax>#m#</mathjax> is the mass, <mathjax>#C_s#</mathjax> is the specific heat, and<mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p><mathjax>#2.000*10^3J = 4g * 5°C*C_s#</mathjax></p>
<p><mathjax>#C_s = (100J)/(g*°C)#</mathjax></p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#C_s = (100J)/(g*°C)#</mathjax></p>
<p>In English: it requires 100J of energy to raise one gram of this substance one degree celsius/kelvin.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're looking for the equation that relates all of these variables:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is the energy in Joules (J), <mathjax>#m#</mathjax> is the mass, <mathjax>#C_s#</mathjax> is the specific heat, and<mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p><mathjax>#2.000*10^3J = 4g * 5°C*C_s#</mathjax></p>
<p><mathjax>#C_s = (100J)/(g*°C)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the specific heat of an unknown substance if 2000 J energy are required to raise the temperature of 4 grams of the substance 5°C?</h1>
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<div class="markdown"><p><mathjax>#C_s = (100J)/(g*°C)#</mathjax></p>
<p>In English: it requires 100J of energy to raise one gram of this substance one degree celsius/kelvin.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're looking for the equation that relates all of these variables:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax>,</p>
<p>where <mathjax>#q#</mathjax> is the energy in Joules (J), <mathjax>#m#</mathjax> is the mass, <mathjax>#C_s#</mathjax> is the specific heat, and<mathjax>#Delta T#</mathjax> is the change in temperature.</p>
<p><mathjax>#2.000*10^3J = 4g * 5°C*C_s#</mathjax></p>
<p><mathjax>#C_s = (100J)/(g*°C)#</mathjax></p></div>
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</article> | What is the specific heat of an unknown substance if 2000 J energy are required to raise the temperature of 4 grams of the substance 5°C? | null |
1,318 | a840f5ac-6ddd-11ea-80ec-ccda262736ce | https://socratic.org/questions/570ca15d11ef6b6517580685 | 0.99 atm | start physical_unit 13 13 pressure atm qc_end end | [{"type":"physical unit","value":"Pressure [OF] atmospheres [IN] atm"}] | [{"type":"physical unit","value":"0.99 atm"}] | [{"type":"physical unit","value":"Reported pressure [OF] gas [=] \\pu{750 mmHg}"}] | <h1 class="questionTitle" itemprop="name">If the pressure is reported as #750*mm*Hg#, what is the pressure in atmospheres?</h1> | null | 0.99 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It makes sense to preconvert the pressure measurement ot atmospheres. So <mathjax>#P_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(750*mm*Hg)/(760mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#atm#</mathjax></p>
<p>Since amount of gas is constant, use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_2V_2)/T_2#</mathjax>, and solve for <mathjax>#V_2#</mathjax>. </p>
<p>You are only justified in quoting an answer to 1 decimal place, as this was the accuracy of the initial conditions. </p>
<p><a href="https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA" rel="nofollow" target="_blank">https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA</a></p></div>
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<div>
<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> <mathjax>#Hg#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It makes sense to preconvert the pressure measurement ot atmospheres. So <mathjax>#P_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(750*mm*Hg)/(760mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#atm#</mathjax></p>
<p>Since amount of gas is constant, use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_2V_2)/T_2#</mathjax>, and solve for <mathjax>#V_2#</mathjax>. </p>
<p>You are only justified in quoting an answer to 1 decimal place, as this was the accuracy of the initial conditions. </p>
<p><a href="https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA" rel="nofollow" target="_blank">https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA</a></p></div>
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<h1 class="questionTitle" itemprop="name">If the pressure is reported as #750*mm*Hg#, what is the pressure in atmospheres?</h1>
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Denise Granger
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<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#760#</mathjax> <mathjax>#mm#</mathjax> <mathjax>#Hg#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>It makes sense to preconvert the pressure measurement ot atmospheres. So <mathjax>#P_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(750*mm*Hg)/(760mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#atm#</mathjax></p>
<p>Since amount of gas is constant, use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_2V_2)/T_2#</mathjax>, and solve for <mathjax>#V_2#</mathjax>. </p>
<p>You are only justified in quoting an answer to 1 decimal place, as this was the accuracy of the initial conditions. </p>
<p><a href="https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA" rel="nofollow" target="_blank">https://www.youtube.com/playlist?list=PL3hPm0ZdYhyxbtpjC_Pue2ewlVlIPEDAA</a></p></div>
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</article> | If the pressure is reported as #750*mm*Hg#, what is the pressure in atmospheres? | null |
1,319 | a842b05c-6ddd-11ea-9dbb-ccda262736ce | https://socratic.org/questions/calculate-the-heat-required-to-raise-the-temperature-of-2-0-kg-of-ice-at-25-c-to | 4.0 × 10^5 J | start physical_unit 12 12 heat_energy j qc_end physical_unit 12 12 14 15 temperature qc_end physical_unit 12 12 17 18 temperature qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Required heat [OF] ice [IN] J"}] | [{"type":"physical unit","value":"4.0 × 10^5 J"}] | [{"type":"physical unit","value":"Temperature1 [OF] ice [=] \\pu{-25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] ice [=] \\pu{70 ℃}"},{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{2.0 kg}"}] | <h1 class="questionTitle" itemprop="name">Calculate the heat required to raise the temperature of 2.0 kg of ice at -25˚C to 70.˚C? </h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>(I keep on getting a value of <mathjax>#4.0 * 10^5#</mathjax> but the answer key says, <mathjax>#1.4 * 10^6 J#</mathjax>.</p></div>
</h2>
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</div> | 4.0 × 10^5 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#q = mC_sDeltaT#</mathjax></p>
<p><mathjax>#q = (2.0*10^3g) * (2.06J/(g*°C)) * 95°C#</mathjax><br/>
<mathjax>#q = 4.0*10^5J#</mathjax></p>
<p><mathjax>#1.4*10^6J = (2.0*10^3g) * C_s * 95°C#</mathjax></p>
<p>If the answer key is "right", then the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of ice is <mathjax>#7.368J*g^-1*°C^-1#</mathjax>. That doesn't sound accurate, so perhaps the key is wrong—maybe I'm missing something that someone else can catch!</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#q = 4.0*10^5J#</mathjax></p>
<p>I have a feeling you're right.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#q = mC_sDeltaT#</mathjax></p>
<p><mathjax>#q = (2.0*10^3g) * (2.06J/(g*°C)) * 95°C#</mathjax><br/>
<mathjax>#q = 4.0*10^5J#</mathjax></p>
<p><mathjax>#1.4*10^6J = (2.0*10^3g) * C_s * 95°C#</mathjax></p>
<p>If the answer key is "right", then the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of ice is <mathjax>#7.368J*g^-1*°C^-1#</mathjax>. That doesn't sound accurate, so perhaps the key is wrong—maybe I'm missing something that someone else can catch!</p></div>
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<h1 class="questionTitle" itemprop="name">Calculate the heat required to raise the temperature of 2.0 kg of ice at -25˚C to 70.˚C? </h1>
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<div class="markdown"><p>(I keep on getting a value of <mathjax>#4.0 * 10^5#</mathjax> but the answer key says, <mathjax>#1.4 * 10^6 J#</mathjax>.</p></div>
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Al E.
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<div class="markdown"><p><mathjax>#q = 4.0*10^5J#</mathjax></p>
<p>I have a feeling you're right.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#q = mC_sDeltaT#</mathjax></p>
<p><mathjax>#q = (2.0*10^3g) * (2.06J/(g*°C)) * 95°C#</mathjax><br/>
<mathjax>#q = 4.0*10^5J#</mathjax></p>
<p><mathjax>#1.4*10^6J = (2.0*10^3g) * C_s * 95°C#</mathjax></p>
<p>If the answer key is "right", then the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of ice is <mathjax>#7.368J*g^-1*°C^-1#</mathjax>. That doesn't sound accurate, so perhaps the key is wrong—maybe I'm missing something that someone else can catch!</p></div>
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Ernest Z.
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Jul 3, 2017
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<div class="markdown"><p>The heat required is <mathjax>#1.40 × 10^6color(white)(l) "J"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>A typical heating curve of water is shown below.</p>
<p><img alt="www.kentchemistry.com" src="http://www.kentchemistry.com/images/links/matter/aim10.5.jpg"/></p>
<p>There are three separate heat transfers involved in this problem:</p>
<ul>
<li><mathjax>#q_1#</mathjax> = heat required to warm the ice from -25 °C to 0 °C (Point <strong>A</strong> to <strong>B</strong> in the diagram)</li>
<li><mathjax>#q_2#</mathjax> = heat required to melt the ice to water at 0 °C (Point <strong>B</strong> to <strong>C</strong>)</li>
<li><mathjax>#q_3#</mathjax> = heat required to warm the water from 0 °C to 70. °C (From Point <strong>C</strong> part way to Point <strong>D</strong>)</li>
</ul>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 = mc_1ΔT_1 + mΔ_text(fus)H + mc_3ΔT_3#</mathjax></p>
<p>where</p>
<p><mathjax>#q_1, q_2,#</mathjax> and <mathjax>#q_3#</mathjax> are the heats involved in each step</p>
<p><mathjax>#m#</mathjax> is the mass of the sample</p>
<p><mathjax>#ΔTcolor(white)(m) = T_"f" -T_"i"#</mathjax></p>
<p><mathjax>#c_1color(white)(mm) = "the specific heat capacity of ice" = "2.03 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_3color(white)(mm) = "the specific heat capacity of water" = "4.179 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(fus)H = "the enthalpy of fusion of ice" = 3.33 × 10^5 color(white)(l)"J·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_1#</mathjax></p>
<p><mathjax>#ΔT_1 = "0 °C - (-25 °C)" = "25 °C"#</mathjax></p>
<p><mathjax>#q_1 = mc_1ΔT_1 = 2.0 × 10^3 color(red)(cancel(color(black)("g"))) × 2.03 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 25 color(red)(cancel(color(black)("°C"))) = 1.02 × 10^5color(white)(l) "J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_2#</mathjax></p>
<p><mathjax>#q_2 = 2.0 color(red)(cancel(color(black)("kg"))) × 3.33 × 10^5color(white)(l) "J"·color(red)(cancel(color(black)("kg"^"-1"))) = 6.66 × 10^5color(white)(l) "J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_3#</mathjax></p>
<p><mathjax>#ΔT = "70. °C - 0 °C" = "70. °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 2.0 × 10^3 color(red)(cancel(color(black)("g"))) × 4.179 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 70. color(red)(cancel(color(black)("°C"))) = 5.85 × 10^5color(white)(l) "J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 = (1.02 + 6.66 + 5.85) × 10^5color(white)(l) "J" = 14.0 × 10^5color(white)(l) "J"#</mathjax></p>
<p><mathjax>#= 1.40 × 10^6color(white)(l) "J"#</mathjax></p>
<p>The process absorbs <mathjax>#1.40 × 10^6color(white)(l) "J"#</mathjax> of heat.</p></div>
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</article> | Calculate the heat required to raise the temperature of 2.0 kg of ice at -25˚C to 70.˚C? |
(I keep on getting a value of #4.0 * 10^5# but the answer key says, #1.4 * 10^6 J#.
|
1,320 | a8487590-6ddd-11ea-9592-ccda262736ce | https://socratic.org/questions/56e76e2b7c014905ee84bda4 | 57.16 g | start physical_unit 3 4 mass g qc_end physical_unit 11 11 9 10 mass qc_end chemical_equation 18 22 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] g"}] | [{"type":"physical unit","value":"57.16 g"}] | [{"type":"physical unit","value":"Mass [OF] C [=] \\pu{15.6 g}"},{"type":"chemical equation","value":"C + O2 -> CO2"},{"type":"other","value":"Oxygen gas is in excess."}] | <h1 class="questionTitle" itemprop="name">What mass of carbon dioxide will be produced by #"15.6 g C"# if oxygen gas is in excess? #"C + O"_2"##rarr##"CO"_2"#</h1> | null | 57.16 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"+"O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2"#</mathjax></p>
<p>Since there is oxygen gas left over, it is the reactant in excess. This means that carbon is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, so we will use the given mass of carbon and <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to determine how much carbon dioxide can be produced.</p>
<p>We need the molar masses of <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, as well as <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>.</p>
<p><strong>Molar Masses</strong><br/>
<mathjax>#"C":#</mathjax><mathjax>#"12.011 g/mol"#</mathjax> (periodic table)<br/>
<mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax> <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong><br/>
<mathjax>#"1 mol C":#</mathjax><mathjax>#"1 mol CO"_2"#</mathjax></p>
<p><strong>Solution</strong><br/>
We will need to make the following conversions using stoichiometry:</p>
<p><mathjax>#color(red)"mass C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)"moles C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)"moles CO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)"mass CO"_2"#</mathjax></p>
<p><mathjax>#color(red)(15.6"C")xx(color(blue)(1"mol C"))/(color(blue)(12.011"g C"))xx(color(green)(1"mol CO"_2))/(color(green)(1"mol C"))xx(color(purple)(44.0095"g CO"_2))/(color(purple)("mol CO"_2))="5.72 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><strong>Note:</strong> Some people will leave out the mole ratio since it is <mathjax>#1:1#</mathjax>, but I think it is important to show how we get from moles of carbon to moles of carbon dioxide.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"15.6 g C"#</mathjax> will produce <mathjax>#"5.72 g CO"_2"#</mathjax> when oxygen gas is present in excess.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"+"O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2"#</mathjax></p>
<p>Since there is oxygen gas left over, it is the reactant in excess. This means that carbon is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, so we will use the given mass of carbon and <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to determine how much carbon dioxide can be produced.</p>
<p>We need the molar masses of <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, as well as <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>.</p>
<p><strong>Molar Masses</strong><br/>
<mathjax>#"C":#</mathjax><mathjax>#"12.011 g/mol"#</mathjax> (periodic table)<br/>
<mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax> <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong><br/>
<mathjax>#"1 mol C":#</mathjax><mathjax>#"1 mol CO"_2"#</mathjax></p>
<p><strong>Solution</strong><br/>
We will need to make the following conversions using stoichiometry:</p>
<p><mathjax>#color(red)"mass C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)"moles C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)"moles CO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)"mass CO"_2"#</mathjax></p>
<p><mathjax>#color(red)(15.6"C")xx(color(blue)(1"mol C"))/(color(blue)(12.011"g C"))xx(color(green)(1"mol CO"_2))/(color(green)(1"mol C"))xx(color(purple)(44.0095"g CO"_2))/(color(purple)("mol CO"_2))="5.72 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><strong>Note:</strong> Some people will leave out the mole ratio since it is <mathjax>#1:1#</mathjax>, but I think it is important to show how we get from moles of carbon to moles of carbon dioxide.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What mass of carbon dioxide will be produced by #"15.6 g C"# if oxygen gas is in excess? #"C + O"_2"##rarr##"CO"_2"#</h1>
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<span class="dateCreated" datetime="2016-03-15T08:11:36" itemprop="dateCreated">
Mar 15, 2016
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<div class="markdown"><p><mathjax>#"15.6 g C"#</mathjax> will produce <mathjax>#"5.72 g CO"_2"#</mathjax> when oxygen gas is present in excess.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"C"+"O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2"#</mathjax></p>
<p>Since there is oxygen gas left over, it is the reactant in excess. This means that carbon is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, so we will use the given mass of carbon and <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> to determine how much carbon dioxide can be produced.</p>
<p>We need the molar masses of <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, as well as <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"C"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>.</p>
<p><strong>Molar Masses</strong><br/>
<mathjax>#"C":#</mathjax><mathjax>#"12.011 g/mol"#</mathjax> (periodic table)<br/>
<mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax> <br/>
<a href="https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong><br/>
<mathjax>#"1 mol C":#</mathjax><mathjax>#"1 mol CO"_2"#</mathjax></p>
<p><strong>Solution</strong><br/>
We will need to make the following conversions using stoichiometry:</p>
<p><mathjax>#color(red)"mass C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(blue)"moles C"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(green)"moles CO"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#color(purple)"mass CO"_2"#</mathjax></p>
<p><mathjax>#color(red)(15.6"C")xx(color(blue)(1"mol C"))/(color(blue)(12.011"g C"))xx(color(green)(1"mol CO"_2))/(color(green)(1"mol C"))xx(color(purple)(44.0095"g CO"_2))/(color(purple)("mol CO"_2))="5.72 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p><strong>Note:</strong> Some people will leave out the mole ratio since it is <mathjax>#1:1#</mathjax>, but I think it is important to show how we get from moles of carbon to moles of carbon dioxide.</p></div>
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</article> | What mass of carbon dioxide will be produced by #"15.6 g C"# if oxygen gas is in excess? #"C + O"_2"##rarr##"CO"_2"# | null |
1,321 | ab27da81-6ddd-11ea-a3c9-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-in-torr-that-a-0-44-g-sample-of-carbon-dioxide-gas-will-exe | 31.92 torr | start physical_unit 10 14 pressure torr qc_end physical_unit 10 14 8 9 mass qc_end physical_unit 10 14 21 22 temperature qc_end physical_unit 10 14 29 30 volume qc_end end | [{"type":"physical unit","value":"Pressure [OF] carbon dioxide gas sample [IN] torr"}] | [{"type":"physical unit","value":"31.92 torr"}] | [{"type":"physical unit","value":"Mass [OF] carbon dioxide gas sample [=] \\pu{0.44 g}"},{"type":"physical unit","value":"Temperature [OF] carbon dioxide gas sample [=] \\pu{46.2 ℃}"},{"type":"physical unit","value":"Volume [OF] carbon dioxide gas sample [=] \\pu{5.00 L}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure in torr that a 0.44-g sample of carbon dioxide gas will exert at a temperature of 46.2°C when it occupies a volume of 5.00 L? </h1> | null | 31.92 torr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem we will use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation:</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>To find pressure we divided both sides by <mathjax>#V#</mathjax></p>
<p><mathjax>#(PcancelV)/cancelV=(nRT)/V#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>First we need to convert the temperature to K so we add <mathjax>#212#</mathjax></p>
<p><mathjax>#46.2°C +212K=258.2K#</mathjax></p>
<p>Now we also need to find the number of moles first we find the molar mass of <mathjax>#CO_2#</mathjax>, which is 44. We take our <mathjax>#.44#</mathjax> and divided by <mathjax>#44#</mathjax>.</p>
<p><mathjax>#("gram sample of carbon dioxide")/("molar mass of " CO_2)#</mathjax></p>
<p><mathjax>#(.44 " grams")/(44 " Molar mass")=.01 " moles"#</mathjax></p>
<p>Now we have:</p>
<p><mathjax>#n=.01#</mathjax></p>
<p><mathjax>#R=0.0806#</mathjax> This is the universal gas constant</p>
<p><mathjax>#T=258.2K#</mathjax></p>
<p><mathjax>#V=5.00L#</mathjax></p>
<p>Let's plug it in:</p>
<p><mathjax>#P=(.01"moles"(0.0806(Lxxatm)/(molxxK)(258.2K)))/(5.00L)#</mathjax></p>
<p><mathjax>#P=.042 " atm"#</mathjax></p>
<p>To find the torr you multiply:</p>
<p><mathjax>#.042xx760=31.92#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#31.92#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem we will use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation:</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>To find pressure we divided both sides by <mathjax>#V#</mathjax></p>
<p><mathjax>#(PcancelV)/cancelV=(nRT)/V#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>First we need to convert the temperature to K so we add <mathjax>#212#</mathjax></p>
<p><mathjax>#46.2°C +212K=258.2K#</mathjax></p>
<p>Now we also need to find the number of moles first we find the molar mass of <mathjax>#CO_2#</mathjax>, which is 44. We take our <mathjax>#.44#</mathjax> and divided by <mathjax>#44#</mathjax>.</p>
<p><mathjax>#("gram sample of carbon dioxide")/("molar mass of " CO_2)#</mathjax></p>
<p><mathjax>#(.44 " grams")/(44 " Molar mass")=.01 " moles"#</mathjax></p>
<p>Now we have:</p>
<p><mathjax>#n=.01#</mathjax></p>
<p><mathjax>#R=0.0806#</mathjax> This is the universal gas constant</p>
<p><mathjax>#T=258.2K#</mathjax></p>
<p><mathjax>#V=5.00L#</mathjax></p>
<p>Let's plug it in:</p>
<p><mathjax>#P=(.01"moles"(0.0806(Lxxatm)/(molxxK)(258.2K)))/(5.00L)#</mathjax></p>
<p><mathjax>#P=.042 " atm"#</mathjax></p>
<p>To find the torr you multiply:</p>
<p><mathjax>#.042xx760=31.92#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the pressure in torr that a 0.44-g sample of carbon dioxide gas will exert at a temperature of 46.2°C when it occupies a volume of 5.00 L? </h1>
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Marvin V.
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Jul 23, 2017
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<div class="markdown"><p><mathjax>#31.92#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem we will use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation:</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p>To find pressure we divided both sides by <mathjax>#V#</mathjax></p>
<p><mathjax>#(PcancelV)/cancelV=(nRT)/V#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>First we need to convert the temperature to K so we add <mathjax>#212#</mathjax></p>
<p><mathjax>#46.2°C +212K=258.2K#</mathjax></p>
<p>Now we also need to find the number of moles first we find the molar mass of <mathjax>#CO_2#</mathjax>, which is 44. We take our <mathjax>#.44#</mathjax> and divided by <mathjax>#44#</mathjax>.</p>
<p><mathjax>#("gram sample of carbon dioxide")/("molar mass of " CO_2)#</mathjax></p>
<p><mathjax>#(.44 " grams")/(44 " Molar mass")=.01 " moles"#</mathjax></p>
<p>Now we have:</p>
<p><mathjax>#n=.01#</mathjax></p>
<p><mathjax>#R=0.0806#</mathjax> This is the universal gas constant</p>
<p><mathjax>#T=258.2K#</mathjax></p>
<p><mathjax>#V=5.00L#</mathjax></p>
<p>Let's plug it in:</p>
<p><mathjax>#P=(.01"moles"(0.0806(Lxxatm)/(molxxK)(258.2K)))/(5.00L)#</mathjax></p>
<p><mathjax>#P=.042 " atm"#</mathjax></p>
<p>To find the torr you multiply:</p>
<p><mathjax>#.042xx760=31.92#</mathjax></p></div>
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</article> | What is the pressure in torr that a 0.44-g sample of carbon dioxide gas will exert at a temperature of 46.2°C when it occupies a volume of 5.00 L? | null |
1,322 | aa417374-6ddd-11ea-9121-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-8-4x10-6-m-h-solution | 5.08 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] H+ solution"}] | [{"type":"physical unit","value":"5.08"}] | [{"type":"physical unit","value":"Molarity [OF] H+ solution [=] \\pu{8.4 × 10^(-6) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #8.4x10^-6 M# #H^+# solution?</h1> | null | 5.08 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydronium ions that are in the solution:</p>
<p><mathjax>#pH = -log (8.4xx10^(-6)M) = 5.08#</mathjax> </p>
<p><mathjax>#color(blue)("Note")#</mathjax>: <mathjax>#H_3O^(+)#</mathjax> is the same as <mathjax>#H^(+)#</mathjax>. This is because we're assuming that water molecules interact with <mathjax>#H^(+)#</mathjax> in solution. </p></div>
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<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> = 5.08</p></div>
</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/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydronium ions that are in the solution:</p>
<p><mathjax>#pH = -log (8.4xx10^(-6)M) = 5.08#</mathjax> </p>
<p><mathjax>#color(blue)("Note")#</mathjax>: <mathjax>#H_3O^(+)#</mathjax> is the same as <mathjax>#H^(+)#</mathjax>. This is because we're assuming that water molecules interact with <mathjax>#H^(+)#</mathjax> in solution. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #8.4x10^-6 M# #H^+# solution?</h1>
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<div class="markdown"><p><mathjax>#pH#</mathjax> = 5.08</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydronium ions that are in the solution:</p>
<p><mathjax>#pH = -log (8.4xx10^(-6)M) = 5.08#</mathjax> </p>
<p><mathjax>#color(blue)("Note")#</mathjax>: <mathjax>#H_3O^(+)#</mathjax> is the same as <mathjax>#H^(+)#</mathjax>. This is because we're assuming that water molecules interact with <mathjax>#H^(+)#</mathjax> in solution. </p></div>
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</article> | What is the pH of a #8.4x10^-6 M# #H^+# solution? | null |
1,323 | acab6ad2-6ddd-11ea-b2cc-ccda262736ce | https://socratic.org/questions/5a40599a7c014940a71858d2 | 11.70 | start physical_unit 6 7 ph none qc_end physical_unit 10 10 13 16 concentration qc_end physical_unit 6 7 18 19 temperature qc_end end | [{"type":"physical unit","value":"pH [OF] the aqueous solution"}] | [{"type":"physical unit","value":"11.70"}] | [{"type":"physical unit","value":"Concentration [OF] hydroxid [=] \\pu{5 × 10^(-3) M}"},{"type":"physical unit","value":"Temperature [OF] the aqueous solution [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the #"pH"# for an aqueous solution with a hydroxide concentration of #5 xx 10^(-3) "M"# at #25^@ "C"#?</h1> | null | 11.70 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#"pOH" = - log["OH"^-]#</mathjax></p>
</blockquote>
<p><mathjax>#"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30#</mathjax></p>
<p>Now,</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p><mathjax>#"pH" = 14 - "pOH" = 14 - 2.30 = 11.7#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#11.7#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#"pOH" = - log["OH"^-]#</mathjax></p>
</blockquote>
<p><mathjax>#"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30#</mathjax></p>
<p>Now,</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p><mathjax>#"pH" = 14 - "pOH" = 14 - 2.30 = 11.7#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the #"pH"# for an aqueous solution with a hydroxide concentration of #5 xx 10^(-3) "M"# at #25^@ "C"#?</h1>
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<div class="markdown"><p><mathjax>#11.7#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote>
<p><mathjax>#"pOH" = - log["OH"^-]#</mathjax></p>
</blockquote>
<p><mathjax>#"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30#</mathjax></p>
<p>Now,</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p><mathjax>#"pH" = 14 - "pOH" = 14 - 2.30 = 11.7#</mathjax></p></div>
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Chirag Mehta
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Truong-Son N.
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Dec 26, 2017
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<div class="markdown"><p><mathjax>#"pH" ~~ color(blue)(11.7)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#color (indigo)(pH=-log [H_3O^+])#</mathjax><br/>
<mathjax>#color (indigo)(pOH=-log [OH^-])#</mathjax><br/>
<mathjax>#color(cyan)([OH^-]=0.005)#</mathjax><br/>
Here, <br/>
<mathjax>#pOH=-log [0.005]#</mathjax><br/>
<mathjax>#pOH=-log [5×10^-3]#</mathjax><br/>
<mathjax>#pOH=2.3010299957#</mathjax></p>
<p><mathjax>#color (green)(pH=14-pOH)#</mathjax></p>
<p><mathjax>#ph=14-2.301 =11.699#</mathjax> <mathjax>#~~#</mathjax> <mathjax>#bb11.7#</mathjax></p></div>
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</article> | What is the #"pH"# for an aqueous solution with a hydroxide concentration of #5 xx 10^(-3) "M"# at #25^@ "C"#? | null |
1,324 | ab069875-6ddd-11ea-9c6c-ccda262736ce | https://socratic.org/questions/58c03660b72cff20114173f4 | 1.27 × 10^(-11) M | start physical_unit 5 5 [oh-] mol/l qc_end physical_unit 5 5 10 10 ph qc_end end | [{"type":"physical unit","value":"[HO-] [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"1.27 × 10^(-11) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{3.10}"}] | <h1 class="questionTitle" itemprop="name">What is #[HO^-]# of a solution for which #pH=3.10#? </h1> | null | 1.27 × 10^(-11) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship <mathjax>#14=pH+pOH#</mathjax></p>
<p>Now <mathjax>#pH=-log_10(7.90xx10^-4)=-(-3.10)=3.10#</mathjax></p>
<p>And <mathjax>#pOH=14-3.10=10.80#</mathjax></p>
<p>And we can now take antilogarithms, </p>
<p><mathjax>#[HO^-]=10^(-10.8)=1.27xx10^-11*mol*L^-1#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#[HO^-]=1.27xx10^-11*mol*L^-1#</mathjax>.............</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship <mathjax>#14=pH+pOH#</mathjax></p>
<p>Now <mathjax>#pH=-log_10(7.90xx10^-4)=-(-3.10)=3.10#</mathjax></p>
<p>And <mathjax>#pOH=14-3.10=10.80#</mathjax></p>
<p>And we can now take antilogarithms, </p>
<p><mathjax>#[HO^-]=10^(-10.8)=1.27xx10^-11*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#[HO^-]=1.27xx10^-11*mol*L^-1#</mathjax>.............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship <mathjax>#14=pH+pOH#</mathjax></p>
<p>Now <mathjax>#pH=-log_10(7.90xx10^-4)=-(-3.10)=3.10#</mathjax></p>
<p>And <mathjax>#pOH=14-3.10=10.80#</mathjax></p>
<p>And we can now take antilogarithms, </p>
<p><mathjax>#[HO^-]=10^(-10.8)=1.27xx10^-11*mol*L^-1#</mathjax></p></div>
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</article> | What is #[HO^-]# of a solution for which #pH=3.10#? | null |
1,325 | acae0349-6ddd-11ea-81b0-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-number-of-moles-of-potassium-hydroxide-that-must-be-dis | 0.50 moles | start physical_unit 9 10 mole mol qc_end physical_unit 19 19 20 21 volume qc_end physical_unit 9 10 23 24 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] potassium hydroxide [IN] moles"}] | [{"type":"physical unit","value":"0.50 moles"}] | [{"type":"physical unit","value":"Volume [OF] potassium hydroxide solution [=] \\pu{500 cm^3}"},{"type":"physical unit","value":"Molarity [OF] potassium hydroxide solution [=] \\pu{1 mol/dm^3}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 #cm^3# of 1 #mol##/##dm^3#?</h1> | null | 0.50 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given that <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*dm^-3#</mathjax>, and that the volume of this solution is <mathjax>#500*cm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*dm^3#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"concentration "xx" volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*cancel(dm^-3)xx0.500*cancel(dm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*mol#</mathjax>.</p>
<p>What is the mass of the <mathjax>#KOH#</mathjax> used here?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use the relationship, <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>. <mathjax>#0.500*mol#</mathjax> <mathjax>#KOH#</mathjax> are required. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given that <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*dm^-3#</mathjax>, and that the volume of this solution is <mathjax>#500*cm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*dm^3#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"concentration "xx" volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*cancel(dm^-3)xx0.500*cancel(dm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*mol#</mathjax>.</p>
<p>What is the mass of the <mathjax>#KOH#</mathjax> used here?</p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 #cm^3# of 1 #mol##/##dm^3#?</h1>
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<div class="markdown"><p>We use the relationship, <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>. <mathjax>#0.500*mol#</mathjax> <mathjax>#KOH#</mathjax> are required. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given that <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*dm^-3#</mathjax>, and that the volume of this solution is <mathjax>#500*cm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*dm^3#</mathjax>.</p>
<p>Thus <mathjax>#"moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"concentration "xx" volume"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1*mol*cancel(dm^-3)xx0.500*cancel(dm^3)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.500*mol#</mathjax>.</p>
<p>What is the mass of the <mathjax>#KOH#</mathjax> used here?</p></div>
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</article> | How do you calculate the number of moles of potassium hydroxide that must be dissolved to make the following solution: 500 #cm^3# of 1 #mol##/##dm^3#? | null |
1,326 | aa260579-6ddd-11ea-9f3f-ccda262736ce | https://socratic.org/questions/what-are-the-steps-to-solve-this | 25.00 cm^3 | start physical_unit 17 19 volume cm^3 qc_end physical_unit 17 19 5 6 molarity qc_end physical_unit 17 19 15 16 molarity qc_end physical_unit 17 19 12 13 volume qc_end end | [{"type":"physical unit","value":"Volume1 [OF] sodium chloride solution [IN] cm^3"}] | [{"type":"physical unit","value":"25.00 cm^3"}] | [{"type":"physical unit","value":"Molarity1 [OF] sodium chloride solution [=] \\pu{2.0 mol/dm^3}"},{"type":"physical unit","value":"Molarity2 [OF] sodium chloride solution [=] \\pu{0.2 mol/dm^3}"},{"type":"physical unit","value":"Volume2 [OF] sodium chloride solution [=] \\pu{250 cm^3}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of #2.0 # #mol# #dm^-3 # sodium chloride needed to prepare #250cm^3# of #0.2 ## mol# #dm^-3# sodium chloride solution ?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What are the steps to solve this ?</p></div>
</h2>
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</div> | 25.00 cm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the ratio that exists between the concentration of the <strong>stock solution</strong> and the concentration of the <strong>diluted solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = (2.0 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.2color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(10)#</mathjax></p>
</blockquote>
<p>This ratio gives you the <strong>dilution factor</strong>, <mathjax>#"DF"#</mathjax>.</p>
<p>Now, the thing to remember about the dilution factor is that can also be calculated by taking the ratio that exists between the volume of the <strong>diluted solution</strong> and the volume of the <strong>stock solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the diluted solution has a volume of <mathjax>#"250 cm"^(3)#</mathjax>, which means that the stock solution must have a volume of</p>
<blockquote>
<p><mathjax>#V_"stock" = "250 cm"^3/color(blue)(10) = color(darkgreen)(ul(color(black)("25 cm"^3)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have one significant figure for the concentration of the diluted solution. </p>
<p>So, in order to prepare this solution, you need to take <mathjax>#"25 cm"^3#</mathjax> of the <mathjax>#"2.0-mol dm"^(-3)#</mathjax> sodium chloride solution and add <em>enough water</em> to get the total volume of the solution to <mathjax>#"250 cm"^3#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"25 cm"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the ratio that exists between the concentration of the <strong>stock solution</strong> and the concentration of the <strong>diluted solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = (2.0 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.2color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(10)#</mathjax></p>
</blockquote>
<p>This ratio gives you the <strong>dilution factor</strong>, <mathjax>#"DF"#</mathjax>.</p>
<p>Now, the thing to remember about the dilution factor is that can also be calculated by taking the ratio that exists between the volume of the <strong>diluted solution</strong> and the volume of the <strong>stock solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the diluted solution has a volume of <mathjax>#"250 cm"^(3)#</mathjax>, which means that the stock solution must have a volume of</p>
<blockquote>
<p><mathjax>#V_"stock" = "250 cm"^3/color(blue)(10) = color(darkgreen)(ul(color(black)("25 cm"^3)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have one significant figure for the concentration of the diluted solution. </p>
<p>So, in order to prepare this solution, you need to take <mathjax>#"25 cm"^3#</mathjax> of the <mathjax>#"2.0-mol dm"^(-3)#</mathjax> sodium chloride solution and add <em>enough water</em> to get the total volume of the solution to <mathjax>#"250 cm"^3#</mathjax>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the volume of #2.0 # #mol# #dm^-3 # sodium chloride needed to prepare #250cm^3# of #0.2 ## mol# #dm^-3# sodium chloride solution ?</h1>
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Stefan V.
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Sep 12, 2017
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<div class="markdown"><p><mathjax>#"25 cm"^3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the ratio that exists between the concentration of the <strong>stock solution</strong> and the concentration of the <strong>diluted solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = (2.0 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.2color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(10)#</mathjax></p>
</blockquote>
<p>This ratio gives you the <strong>dilution factor</strong>, <mathjax>#"DF"#</mathjax>.</p>
<p>Now, the thing to remember about the dilution factor is that can also be calculated by taking the ratio that exists between the volume of the <strong>diluted solution</strong> and the volume of the <strong>stock solution</strong>.</p>
<blockquote>
<p><mathjax>#"DF" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the diluted solution has a volume of <mathjax>#"250 cm"^(3)#</mathjax>, which means that the stock solution must have a volume of</p>
<blockquote>
<p><mathjax>#V_"stock" = "250 cm"^3/color(blue)(10) = color(darkgreen)(ul(color(black)("25 cm"^3)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you have one significant figure for the concentration of the diluted solution. </p>
<p>So, in order to prepare this solution, you need to take <mathjax>#"25 cm"^3#</mathjax> of the <mathjax>#"2.0-mol dm"^(-3)#</mathjax> sodium chloride solution and add <em>enough water</em> to get the total volume of the solution to <mathjax>#"250 cm"^3#</mathjax>. </p></div>
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</article> | What is the volume of #2.0 # #mol# #dm^-3 # sodium chloride needed to prepare #250cm^3# of #0.2 ## mol# #dm^-3# sodium chloride solution ? |
What are the steps to solve this ?
|
1,327 | abbd661a-6ddd-11ea-a0a1-ccda262736ce | https://socratic.org/questions/how-do-you-balance-naoh-cu-no-3-2-cu-oh-2-nano-3 | 2 NaOH + Cu(NO3)2 -> Cu(OH)2 + 2 NaNO3 | 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 NaOH + Cu(NO3)2 -> Cu(OH)2 + 2 NaNO3"}] | [{"type":"chemical equation","value":"NaOH + Cu(NO3)2 -> Cu(OH)2 + NaNO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #NaOH + Cu(NO_3)_2 -> Cu(OH)_2 + NaNO_3#?</h1> | null | 2 NaOH + Cu(NO3)2 -> Cu(OH)2 + 2 NaNO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation is stoiciometrically balanced. It is balanced with respect to MASS and to charge. We could write the net ionic equation as :</p>
<p><mathjax>#Cu^(2+) + 2OH^(-) rarr Cu(OH)_2(s)darr#</mathjax></p>
<p>This represents the macroscopic, observable chemical change. All hydroxides are soluble (save for those of the alkali metals). Thus, the sodium counterion is simply along for the ride in the reaction. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You have almost done it:</p>
<p><mathjax>#2NaOH(aq) + Cu(NO_3)_2(aq) rarr Cu(OH)_2darr + 2NaNO_3(aq)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation is stoiciometrically balanced. It is balanced with respect to MASS and to charge. We could write the net ionic equation as :</p>
<p><mathjax>#Cu^(2+) + 2OH^(-) rarr Cu(OH)_2(s)darr#</mathjax></p>
<p>This represents the macroscopic, observable chemical change. All hydroxides are soluble (save for those of the alkali metals). Thus, the sodium counterion is simply along for the ride in the reaction. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #NaOH + Cu(NO_3)_2 -> Cu(OH)_2 + NaNO_3#?</h1>
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anor277
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Mar 12, 2016
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<div class="markdown"><p>You have almost done it:</p>
<p><mathjax>#2NaOH(aq) + Cu(NO_3)_2(aq) rarr Cu(OH)_2darr + 2NaNO_3(aq)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation is stoiciometrically balanced. It is balanced with respect to MASS and to charge. We could write the net ionic equation as :</p>
<p><mathjax>#Cu^(2+) + 2OH^(-) rarr Cu(OH)_2(s)darr#</mathjax></p>
<p>This represents the macroscopic, observable chemical change. All hydroxides are soluble (save for those of the alkali metals). Thus, the sodium counterion is simply along for the ride in the reaction. </p></div>
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</article> | How do you balance #NaOH + Cu(NO_3)_2 -> Cu(OH)_2 + NaNO_3#? | null |
1,328 | ad08d954-6ddd-11ea-a912-ccda262736ce | https://socratic.org/questions/how-many-moles-of-oxygen-are-produced-by-the-decomposition-of-six-moles-of-potas | 9.00 moles | start physical_unit 4 4 mole mol qc_end chemical_equation 19 26 qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"9.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] potassium chlorate [=] \\pu{6 moles}"},{"type":"chemical equation","value":"2 KClO3 -> 2 KCl + 3 O2"}] | <h1 class="questionTitle" itemprop="name">How many moles of oxygen are produced by the decomposition of six moles of potassium chlorate in the reaction #2KClO_3 -> 2KCl + 3O_2#?</h1> | null | 9.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the equation, <mathjax>#3/2#</mathjax> equiv dixoygen gas evolve from <mathjax>#1#</mathjax> equiv chlorate.</p>
<p>You started with <mathjax>#6#</mathjax> equiv chlorate, and thus <mathjax>#9#</mathjax> equiv dioxygen gas are evolved. This is a common undergrad experiment. I think you need to add some sort of manganese oxide catalyst. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#KClO_3(g) + Deltararr KCl(s) + 3/2O_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the equation, <mathjax>#3/2#</mathjax> equiv dixoygen gas evolve from <mathjax>#1#</mathjax> equiv chlorate.</p>
<p>You started with <mathjax>#6#</mathjax> equiv chlorate, and thus <mathjax>#9#</mathjax> equiv dioxygen gas are evolved. This is a common undergrad experiment. I think you need to add some sort of manganese oxide catalyst. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of oxygen are produced by the decomposition of six moles of potassium chlorate in the reaction #2KClO_3 -> 2KCl + 3O_2#?</h1>
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<div class="markdown"><p><mathjax>#KClO_3(g) + Deltararr KCl(s) + 3/2O_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>From the equation, <mathjax>#3/2#</mathjax> equiv dixoygen gas evolve from <mathjax>#1#</mathjax> equiv chlorate.</p>
<p>You started with <mathjax>#6#</mathjax> equiv chlorate, and thus <mathjax>#9#</mathjax> equiv dioxygen gas are evolved. This is a common undergrad experiment. I think you need to add some sort of manganese oxide catalyst. </p></div>
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</article> | How many moles of oxygen are produced by the decomposition of six moles of potassium chlorate in the reaction #2KClO_3 -> 2KCl + 3O_2#? | null |
1,329 | a8ab6670-6ddd-11ea-a967-ccda262736ce | https://socratic.org/questions/how-many-ml-of-a-2-00m-h-2so-4-are-needed-to-provide-250-mole-of-h-2so-4 | 125.00 mL | start physical_unit 7 7 volume ml qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 7 7 12 13 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] H2SO4 [IN] mL"}] | [{"type":"physical unit","value":"125.00 mL"}] | [{"type":"physical unit","value":"Molarity [OF] H2SO4 [=] \\pu{2.00 M}"},{"type":"physical unit","value":"Mole [OF] H2SO4 [=] \\pu{0.250 mole}"}] | <h1 class="questionTitle" itemprop="name"> How many mL of a 2.00M #H_2SO_4# are needed to provide .250 mole of #H_2SO_4#?</h1> | null | 125.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>. This can be appreciated dimensionally, i.e. <mathjax>#mol*L^-1xxL=mol!#</mathjax>.</p>
<p>We require <mathjax>#0.250*mol#</mathjax>, and so we divide the required number of moles by the concentration to get the volume:</p>
<p><mathjax>#(0.250*cancel(mol))/(2.00*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p></div>
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<div class="markdown"><p>Over <mathjax>#100*mL#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>. This can be appreciated dimensionally, i.e. <mathjax>#mol*L^-1xxL=mol!#</mathjax>.</p>
<p>We require <mathjax>#0.250*mol#</mathjax>, and so we divide the required number of moles by the concentration to get the volume:</p>
<p><mathjax>#(0.250*cancel(mol))/(2.00*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name"> How many mL of a 2.00M #H_2SO_4# are needed to provide .250 mole of #H_2SO_4#?</h1>
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<div class="markdown"><p>Over <mathjax>#100*mL#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume"#</mathjax>. This can be appreciated dimensionally, i.e. <mathjax>#mol*L^-1xxL=mol!#</mathjax>.</p>
<p>We require <mathjax>#0.250*mol#</mathjax>, and so we divide the required number of moles by the concentration to get the volume:</p>
<p><mathjax>#(0.250*cancel(mol))/(2.00*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mL#</mathjax></p></div>
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</article> | How many mL of a 2.00M #H_2SO_4# are needed to provide .250 mole of #H_2SO_4#? | null |
1,330 | acff6e5c-6ddd-11ea-92c7-ccda262736ce | https://socratic.org/questions/what-is-the-resulting-temperature-if-a-sample-of-gas-began-with-a-temperature-of | 308.42 K | start physical_unit 7 9 temperature k qc_end physical_unit 7 9 15 16 temperature qc_end physical_unit 7 9 20 21 pressure qc_end physical_unit 7 9 17 18 volume qc_end physical_unit 7 9 25 26 volume qc_end physical_unit 7 9 32 33 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas sample [IN] K"}] | [{"type":"physical unit","value":"308.42 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{760 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{1 liter}"},{"type":"physical unit","value":"Volume2 [OF] the gas sample [=] \\pu{800 mL}"},{"type":"physical unit","value":"Pressure2 [OF] the gas sample [=] \\pu{1000 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What is the resulting temperature if a sample of gas began with a temperature of 20 C, 1 liter, and 760 mmHg and now occupies 800 mL and has a pressure of 1000 mmHg?</h1> | null | 308.42 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new temperature of a gas after it is subjected to changes in pressure and volume.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "(P_1V_1)/(T_1) = (P_2V_2)/(T_2)" ")|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#P_1#</mathjax> is the <strong>original pressure</strong> (given as <mathjax>#760#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_1#</mathjax> is the <strong>original volume</strong> (given as <mathjax>#1#</mathjax> <mathjax>#"L"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#T_1#</mathjax> is the <strong>original <em>absolute</em> temperature</strong>, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#20#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#+ 273 = ul(298color(white)(l)"K"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#P_2#</mathjax> is the <strong>final pressure</strong> (given as <mathjax>#1000#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_2#</mathjax> is the <strong>final volume</strong> (given as <mathjax>#800#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= ul(0.800color(white)(l)"L"#</mathjax>) (units must be consistent, so convert this to liters)</p>
</li>
<li>
<p><mathjax>#T_2#</mathjax> is the <strong>final <em>absolute</em> temperature</strong> (<strong><em>what we're trying to find</em></strong>)</p>
</li>
</ul>
<p>Let's rearrange this equation to solve for the final temperature, <mathjax>#T_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#T_2 = (P_2V_2T_1)/(P_1V_1)#</mathjax></p>
</blockquote>
<p>Plugging in the above values:</p>
<blockquote>
<p><mathjax>#T_2 = ((1000cancel("mm Hg"))(0.800cancel("L"))(298color(white)(l)"K"))/((760cancel("mm Hg"))(1cancel("L"))) = color(red)(ulbar(|stackrel(" ")(" "317color(white)(l)"K"" ")|)#</mathjax></p>
</blockquote>
<p>The final temperature of the gas is thus <mathjax>#color(red)(317#</mathjax> <mathjax>#sfcolor(red)("kelvin"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#T_2 = 317#</mathjax> <mathjax>#"K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new temperature of a gas after it is subjected to changes in pressure and volume.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "(P_1V_1)/(T_1) = (P_2V_2)/(T_2)" ")|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#P_1#</mathjax> is the <strong>original pressure</strong> (given as <mathjax>#760#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_1#</mathjax> is the <strong>original volume</strong> (given as <mathjax>#1#</mathjax> <mathjax>#"L"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#T_1#</mathjax> is the <strong>original <em>absolute</em> temperature</strong>, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#20#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#+ 273 = ul(298color(white)(l)"K"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#P_2#</mathjax> is the <strong>final pressure</strong> (given as <mathjax>#1000#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_2#</mathjax> is the <strong>final volume</strong> (given as <mathjax>#800#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= ul(0.800color(white)(l)"L"#</mathjax>) (units must be consistent, so convert this to liters)</p>
</li>
<li>
<p><mathjax>#T_2#</mathjax> is the <strong>final <em>absolute</em> temperature</strong> (<strong><em>what we're trying to find</em></strong>)</p>
</li>
</ul>
<p>Let's rearrange this equation to solve for the final temperature, <mathjax>#T_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#T_2 = (P_2V_2T_1)/(P_1V_1)#</mathjax></p>
</blockquote>
<p>Plugging in the above values:</p>
<blockquote>
<p><mathjax>#T_2 = ((1000cancel("mm Hg"))(0.800cancel("L"))(298color(white)(l)"K"))/((760cancel("mm Hg"))(1cancel("L"))) = color(red)(ulbar(|stackrel(" ")(" "317color(white)(l)"K"" ")|)#</mathjax></p>
</blockquote>
<p>The final temperature of the gas is thus <mathjax>#color(red)(317#</mathjax> <mathjax>#sfcolor(red)("kelvin"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the resulting temperature if a sample of gas began with a temperature of 20 C, 1 liter, and 760 mmHg and now occupies 800 mL and has a pressure of 1000 mmHg?</h1>
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<div class="markdown"><p><mathjax>#T_2 = 317#</mathjax> <mathjax>#"K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new temperature of a gas after it is subjected to changes in pressure and volume.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></strong>:</p>
<blockquote>
<p><mathjax>#ulbar(|stackrel(" ")(" "(P_1V_1)/(T_1) = (P_2V_2)/(T_2)" ")|)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#P_1#</mathjax> is the <strong>original pressure</strong> (given as <mathjax>#760#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_1#</mathjax> is the <strong>original volume</strong> (given as <mathjax>#1#</mathjax> <mathjax>#"L"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#T_1#</mathjax> is the <strong>original <em>absolute</em> temperature</strong>, which is</p>
</li>
</ul>
<blockquote>
<p><mathjax>#20#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#+ 273 = ul(298color(white)(l)"K"#</mathjax></p>
</blockquote>
<ul>
<li>
<p><mathjax>#P_2#</mathjax> is the <strong>final pressure</strong> (given as <mathjax>#1000#</mathjax> <mathjax>#"mm Hg"#</mathjax>)</p>
</li>
<li>
<p><mathjax>#V_2#</mathjax> is the <strong>final volume</strong> (given as <mathjax>#800#</mathjax> <mathjax>#"mL"#</mathjax> <mathjax>#= ul(0.800color(white)(l)"L"#</mathjax>) (units must be consistent, so convert this to liters)</p>
</li>
<li>
<p><mathjax>#T_2#</mathjax> is the <strong>final <em>absolute</em> temperature</strong> (<strong><em>what we're trying to find</em></strong>)</p>
</li>
</ul>
<p>Let's rearrange this equation to solve for the final temperature, <mathjax>#T_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#T_2 = (P_2V_2T_1)/(P_1V_1)#</mathjax></p>
</blockquote>
<p>Plugging in the above values:</p>
<blockquote>
<p><mathjax>#T_2 = ((1000cancel("mm Hg"))(0.800cancel("L"))(298color(white)(l)"K"))/((760cancel("mm Hg"))(1cancel("L"))) = color(red)(ulbar(|stackrel(" ")(" "317color(white)(l)"K"" ")|)#</mathjax></p>
</blockquote>
<p>The final temperature of the gas is thus <mathjax>#color(red)(317#</mathjax> <mathjax>#sfcolor(red)("kelvin"#</mathjax>.</p></div>
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</article> | What is the resulting temperature if a sample of gas began with a temperature of 20 C, 1 liter, and 760 mmHg and now occupies 800 mL and has a pressure of 1000 mmHg? | null |
1,331 | a9041f1c-6ddd-11ea-a3b7-ccda262736ce | https://socratic.org/questions/58a76e0db72cff26617ada6f | 13.00 L | start physical_unit 3 3 volume l qc_end physical_unit 13 14 10 11 volume qc_end physical_unit 13 14 17 18 temperature qc_end physical_unit 13 14 20 21 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] ammonia [IN] L"}] | [{"type":"physical unit","value":"13.00 L"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen gas [=] \\pu{19.5 L}"},{"type":"physical unit","value":"Temperature [OF] hydrogen gas [=] \\pu{93 ℃}"},{"type":"physical unit","value":"Pressure [OF] hydrogen gas [=] \\pu{48.7 kPa}"}] | <h1 class="questionTitle" itemprop="name">What volume of ammonia is produced by the reaction when #"19.5 L"# of hydrogen gas react at #93^@"C"# and #"48.7 kPa"# ?</h1> | null | 13.00 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that <strong>all the chemical species</strong> that take part in this reaction are kept <strong>under the same conditions</strong> for pressure and temperature. </p>
<p>This means that the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exists between the chemical species that take part in the reaction will be <strong>equivalent</strong> to <strong>volume ratios</strong>. </p>
<p>The balanced chemical equation that describes this reaction looks like this </p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + color(blue)(3)"H"_ (2(g)) -> color(darkorange)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydrogen gas that take part in the reaction, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>When the reaction takes place at constant temperature and pressure, you can say that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>liters</strong> of hydrogen gas that react, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>liters</strong> of ammonia. </p>
<p>Assuming that nitrogen gas is <em>in excess</em>, you can say that the given volume of hydrogen gas will produce</p>
<blockquote>
<p><mathjax>#19.5 color(red)(cancel(color(black)("L H"_2))) * (color(darkorange)(2)color(white)(.)"moles NH"_3)/(color(blue)(3)color(red)(cancel(color(black)("L H"_2)))) = color(darkgreen)(ul(color(black)("13.0 L NH"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the volume of hydrogen gas.</p>
<p>Keep in mind that this volume of ammonia is produced at a temperature of <mathjax>#93.0^@"C"#</mathjax> and a pressure of <mathjax>#"48.7 kPa"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"13.0 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that <strong>all the chemical species</strong> that take part in this reaction are kept <strong>under the same conditions</strong> for pressure and temperature. </p>
<p>This means that the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exists between the chemical species that take part in the reaction will be <strong>equivalent</strong> to <strong>volume ratios</strong>. </p>
<p>The balanced chemical equation that describes this reaction looks like this </p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + color(blue)(3)"H"_ (2(g)) -> color(darkorange)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydrogen gas that take part in the reaction, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>When the reaction takes place at constant temperature and pressure, you can say that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>liters</strong> of hydrogen gas that react, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>liters</strong> of ammonia. </p>
<p>Assuming that nitrogen gas is <em>in excess</em>, you can say that the given volume of hydrogen gas will produce</p>
<blockquote>
<p><mathjax>#19.5 color(red)(cancel(color(black)("L H"_2))) * (color(darkorange)(2)color(white)(.)"moles NH"_3)/(color(blue)(3)color(red)(cancel(color(black)("L H"_2)))) = color(darkgreen)(ul(color(black)("13.0 L NH"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the volume of hydrogen gas.</p>
<p>Keep in mind that this volume of ammonia is produced at a temperature of <mathjax>#93.0^@"C"#</mathjax> and a pressure of <mathjax>#"48.7 kPa"#</mathjax>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of ammonia is produced by the reaction when #"19.5 L"# of hydrogen gas react at #93^@"C"# and #"48.7 kPa"# ?</h1>
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Stefan V.
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Feb 18, 2017
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<div class="markdown"><p><mathjax>#"13.0 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that <strong>all the chemical species</strong> that take part in this reaction are kept <strong>under the same conditions</strong> for pressure and temperature. </p>
<p>This means that the <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong> that exists between the chemical species that take part in the reaction will be <strong>equivalent</strong> to <strong>volume ratios</strong>. </p>
<p>The balanced chemical equation that describes this reaction looks like this </p>
<blockquote>
<p><mathjax>#"N"_ (2(g)) + color(blue)(3)"H"_ (2(g)) -> color(darkorange)(2)"NH"_ (3(g))#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of hydrogen gas that take part in the reaction, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>When the reaction takes place at constant temperature and pressure, you can say that <strong>for every</strong> <mathjax>#color(blue)(3)#</mathjax> <strong>liters</strong> of hydrogen gas that react, the reaction produces <mathjax>#color(darkorange)(2)#</mathjax> <strong>liters</strong> of ammonia. </p>
<p>Assuming that nitrogen gas is <em>in excess</em>, you can say that the given volume of hydrogen gas will produce</p>
<blockquote>
<p><mathjax>#19.5 color(red)(cancel(color(black)("L H"_2))) * (color(darkorange)(2)color(white)(.)"moles NH"_3)/(color(blue)(3)color(red)(cancel(color(black)("L H"_2)))) = color(darkgreen)(ul(color(black)("13.0 L NH"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the volume of hydrogen gas.</p>
<p>Keep in mind that this volume of ammonia is produced at a temperature of <mathjax>#93.0^@"C"#</mathjax> and a pressure of <mathjax>#"48.7 kPa"#</mathjax>.</p></div>
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</article> | What volume of ammonia is produced by the reaction when #"19.5 L"# of hydrogen gas react at #93^@"C"# and #"48.7 kPa"# ? | null |
1,332 | ab41f7de-6ddd-11ea-ab13-ccda262736ce | https://socratic.org/questions/if-a-mixture-of-gases-with-a-total-pressure-of-1-0-atm-contains-2-mol-he-4-mol-n | 0.25 atm | start physical_unit 15 15 partial_pressure atm qc_end physical_unit 15 15 13 14 mole qc_end physical_unit 18 18 16 17 mole qc_end physical_unit 22 22 13 14 mole qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] He [IN] atm"}] | [{"type":"physical unit","value":"0.25 atm"}] | [{"type":"physical unit","value":"Total pressure [OF] gases mixture [=] \\pu{1.0 atm}"},{"type":"physical unit","value":"Mole [OF] He [=] \\pu{2 mol}"},{"type":"physical unit","value":"Mole [OF] Ne [=] \\pu{4 mol}"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{2 mol}"}] | <h1 class="questionTitle" itemprop="name">If a mixture of gases with a total pressure of 1.0 atm contains 2 mol #He#, 4 mol #Ne#, and 2 mol #O_2#, what is the partial pressure of the #He# in the mixture?</h1> | null | 0.25 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture will depend on two things</p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> said gas has in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Dalton's Law of Partial Pressures</a> allows you to calculate the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas that's part of a gaseous mixture by using the number of moles of that gas and the <strong>total number of moles</strong> present in the mixture - this is known as the <em>mole fraction</em></p>
<blockquote>
<p><mathjax>#color(blue)(P_i - chi_i xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_i#</mathjax> - the partial pressure of gas <mathjax>#i#</mathjax><br/>
<mathjax>#chi_i#</mathjax> - its <strong>mole fraction</strong><br/>
<mathjax>#P_"total"#</mathjax> - the <em>total pressure</em> of the mixture</p>
<p>In your case, you know that the mixture contains </p>
<blockquote>
<ul>
<li><mathjax>#"2 moles"#</mathjax> <em>of helium</em></li>
<li><mathjax>#"4 moles"#</mathjax> <em>of neon</em></li>
<li><mathjax>#"2 moles"#</mathjax> <em>of oxygen gas</em></li>
</ul>
</blockquote>
<p>The *<em>total number of moles8</em> present in the mixture will thus be </p>
<blockquote>
<p><mathjax>#n_"total" = 2 + 4 + 2 = "8 moles"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> fraction of helium will be </p>
<blockquote>
<p><mathjax>#chi_(He) = (2 color(red)(cancel(color(black)("moles "))))/(8color(red)(cancel(color(black)("moles")))) = 1/4 = 0.25#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = chi_(He) xx P_"total"#</mathjax></p>
<p><mathjax>#P_(He) = 0.25 xx "1.0 atm" = color(green)("0.25 atm")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.25 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture will depend on two things</p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> said gas has in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Dalton's Law of Partial Pressures</a> allows you to calculate the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas that's part of a gaseous mixture by using the number of moles of that gas and the <strong>total number of moles</strong> present in the mixture - this is known as the <em>mole fraction</em></p>
<blockquote>
<p><mathjax>#color(blue)(P_i - chi_i xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_i#</mathjax> - the partial pressure of gas <mathjax>#i#</mathjax><br/>
<mathjax>#chi_i#</mathjax> - its <strong>mole fraction</strong><br/>
<mathjax>#P_"total"#</mathjax> - the <em>total pressure</em> of the mixture</p>
<p>In your case, you know that the mixture contains </p>
<blockquote>
<ul>
<li><mathjax>#"2 moles"#</mathjax> <em>of helium</em></li>
<li><mathjax>#"4 moles"#</mathjax> <em>of neon</em></li>
<li><mathjax>#"2 moles"#</mathjax> <em>of oxygen gas</em></li>
</ul>
</blockquote>
<p>The *<em>total number of moles8</em> present in the mixture will thus be </p>
<blockquote>
<p><mathjax>#n_"total" = 2 + 4 + 2 = "8 moles"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> fraction of helium will be </p>
<blockquote>
<p><mathjax>#chi_(He) = (2 color(red)(cancel(color(black)("moles "))))/(8color(red)(cancel(color(black)("moles")))) = 1/4 = 0.25#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = chi_(He) xx P_"total"#</mathjax></p>
<p><mathjax>#P_(He) = 0.25 xx "1.0 atm" = color(green)("0.25 atm")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If a mixture of gases with a total pressure of 1.0 atm contains 2 mol #He#, 4 mol #Ne#, and 2 mol #O_2#, what is the partial pressure of the #He# in the mixture?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-19T14:36:18" itemprop="dateCreated">
Jan 19, 2016
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<div class="markdown"><p><mathjax>#"0.25 atm"#</mathjax></p></div>
</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/the-behavior-of-gases/partial-pressure">partial pressure</a> of a gas that's part of a gaseous mixture will depend on two things</p>
<blockquote>
<ul>
<li><em>the <strong>mole fraction</strong> said gas has in the mixture</em></li>
<li><em>the <strong>total pressure</strong> of the mixture</em></li>
</ul>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">Dalton's Law of Partial Pressures</a> allows you to calculate the <a href="http://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of gas that's part of a gaseous mixture by using the number of moles of that gas and the <strong>total number of moles</strong> present in the mixture - this is known as the <em>mole fraction</em></p>
<blockquote>
<p><mathjax>#color(blue)(P_i - chi_i xx P_"total")" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_i#</mathjax> - the partial pressure of gas <mathjax>#i#</mathjax><br/>
<mathjax>#chi_i#</mathjax> - its <strong>mole fraction</strong><br/>
<mathjax>#P_"total"#</mathjax> - the <em>total pressure</em> of the mixture</p>
<p>In your case, you know that the mixture contains </p>
<blockquote>
<ul>
<li><mathjax>#"2 moles"#</mathjax> <em>of helium</em></li>
<li><mathjax>#"4 moles"#</mathjax> <em>of neon</em></li>
<li><mathjax>#"2 moles"#</mathjax> <em>of oxygen gas</em></li>
</ul>
</blockquote>
<p>The *<em>total number of moles8</em> present in the mixture will thus be </p>
<blockquote>
<p><mathjax>#n_"total" = 2 + 4 + 2 = "8 moles"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> fraction of helium will be </p>
<blockquote>
<p><mathjax>#chi_(He) = (2 color(red)(cancel(color(black)("moles "))))/(8color(red)(cancel(color(black)("moles")))) = 1/4 = 0.25#</mathjax></p>
</blockquote>
<p>Therefore, the partial pressure of helium will be </p>
<blockquote>
<p><mathjax>#P_(He) = chi_(He) xx P_"total"#</mathjax></p>
<p><mathjax>#P_(He) = 0.25 xx "1.0 atm" = color(green)("0.25 atm")#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>.</p></div>
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</article> | If a mixture of gases with a total pressure of 1.0 atm contains 2 mol #He#, 4 mol #Ne#, and 2 mol #O_2#, what is the partial pressure of the #He# in the mixture? | null |
1,333 | acb86846-6ddd-11ea-88e6-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-occupies-a-volume-of-50-0-milliliters-in-a-cylinder-with-a-movab | 41.25 mL | start physical_unit 1 3 volume ml qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 23 24 pressure qc_end physical_unit 1 3 29 30 temperature qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas sample [IN] mL"}] | [{"type":"physical unit","value":"41.25 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{50.0 milliliters}"},{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{0.90 atmosphere}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{298 K}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 atmosphere and the temperature is 298 K. What is the volume of the sample at STP?</h1> | null | 41.25 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.<br/>
<mathjax>#"STP=273.15 K and 1 atm"#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="0.90 atm"#</mathjax><br/>
<mathjax>#V_1="50.0 mL"#</mathjax><br/>
<mathjax>#T_1="298 K"#</mathjax><br/>
<mathjax>#P_2="1 atm"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(0.90cancel"atm"xx50.0"mL"xx273.15cancel"K")/(298cancel"K"xx1cancel"atm")="41 mL"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new volume is 41 mL.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.<br/>
<mathjax>#"STP=273.15 K and 1 atm"#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="0.90 atm"#</mathjax><br/>
<mathjax>#V_1="50.0 mL"#</mathjax><br/>
<mathjax>#T_1="298 K"#</mathjax><br/>
<mathjax>#P_2="1 atm"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(0.90cancel"atm"xx50.0"mL"xx273.15cancel"K")/(298cancel"K"xx1cancel"atm")="41 mL"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 atmosphere and the temperature is 298 K. What is the volume of the sample at STP?</h1>
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<div class="markdown"><p>The new volume is 41 mL.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.<br/>
<mathjax>#"STP=273.15 K and 1 atm"#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="0.90 atm"#</mathjax><br/>
<mathjax>#V_1="50.0 mL"#</mathjax><br/>
<mathjax>#T_1="298 K"#</mathjax><br/>
<mathjax>#P_2="1 atm"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=(0.90cancel"atm"xx50.0"mL"xx273.15cancel"K")/(298cancel"K"xx1cancel"atm")="41 mL"#</mathjax></p></div>
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</article> | A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 atmosphere and the temperature is 298 K. What is the volume of the sample at STP? | null |
1,334 | aa5eed74-6ddd-11ea-8b5c-ccda262736ce | https://socratic.org/questions/how-many-grams-of-naoh-are-in-4-35-moles | 174.00 grams | start physical_unit 4 4 mass g qc_end physical_unit 4 4 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] NaOH [IN] grams"}] | [{"type":"physical unit","value":"174.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{4.35 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of NaOH are in 4.35 moles? </h1> | null | 174.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you have <mathjax>#4.35#</mathjax> <mathjax>#mol#</mathjax>, then <mathjax>#4.35cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.00#</mathjax> <mathjax>#g*cancel(mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1#</mathjax> mole of sodium hydroxide has a mass of <mathjax>#40.00#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you have <mathjax>#4.35#</mathjax> <mathjax>#mol#</mathjax>, then <mathjax>#4.35cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.00#</mathjax> <mathjax>#g*cancel(mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1#</mathjax> mole of sodium hydroxide has a mass of <mathjax>#40.00#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If you have <mathjax>#4.35#</mathjax> <mathjax>#mol#</mathjax>, then <mathjax>#4.35cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.00#</mathjax> <mathjax>#g*cancel(mol^(-1))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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</article> | How many grams of NaOH are in 4.35 moles? | null |
1,335 | ac0699cb-6ddd-11ea-ae39-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-a-solution-expressed-in-g-100-ml-if-25-g-of-solute- | 62.5 g/100 mL | start physical_unit 6 6 concentration g/100_ml qc_end physical_unit 15 15 12 13 mass qc_end physical_unit 21 21 18 19 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] the solution [IN] g/100 mL"}] | [{"type":"physical unit","value":"62.5 g/100 mL"}] | [{"type":"physical unit","value":"Mass [OF] solute [=] \\pu{25 g}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{40 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of a solution, expressed in g/100 mL, if 25 g of solute dissolved in 40 mL of water?</h1> | null | 62.5 g/100 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want <mathjax>#"Concentration"="Mass of solute"/"100 mL volume of solvent"#</mathjax>.</p>
<p>We make the reasonable assumption that the volume of solution is equal to <mathjax>#40*mL#</mathjax>. </p>
<p>And so we set up the quotient.......</p>
<p><mathjax>#(25*g)/((40/100)*mL)=62.5*g*(100*mL)^-1#</mathjax>..........</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Solubility"-=62.5*g*100*mL^-1#</mathjax>.............</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want <mathjax>#"Concentration"="Mass of solute"/"100 mL volume of solvent"#</mathjax>.</p>
<p>We make the reasonable assumption that the volume of solution is equal to <mathjax>#40*mL#</mathjax>. </p>
<p>And so we set up the quotient.......</p>
<p><mathjax>#(25*g)/((40/100)*mL)=62.5*g*(100*mL)^-1#</mathjax>..........</p></div>
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<div class="markdown"><p><mathjax>#"Solubility"-=62.5*g*100*mL^-1#</mathjax>.............</p></div>
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<div class="markdown"><p>We want <mathjax>#"Concentration"="Mass of solute"/"100 mL volume of solvent"#</mathjax>.</p>
<p>We make the reasonable assumption that the volume of solution is equal to <mathjax>#40*mL#</mathjax>. </p>
<p>And so we set up the quotient.......</p>
<p><mathjax>#(25*g)/((40/100)*mL)=62.5*g*(100*mL)^-1#</mathjax>..........</p></div>
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</article> | What is the concentration of a solution, expressed in g/100 mL, if 25 g of solute dissolved in 40 mL of water? | null |
1,336 | ac53299e-6ddd-11ea-b390-ccda262736ce | https://socratic.org/questions/a-3-060-g-sample-of-a-mixture-was-analyzed-for-barium-ion-by-adding-a-small-exce | 5.27% | start physical_unit 56 59 mass_percent none qc_end physical_unit 33 34 43 44 mass qc_end c_other OTHER qc_end substance 10 11 qc_end end | [{"type":"physical unit","value":"Mass percentage [OF] barium in the sample"}] | [{"type":"physical unit","value":"5.27%"}] | [{"type":"physical unit","value":"Mass [OF] the mixture sample [=] \\pu{3.060 g}"},{"type":"physical unit","value":"Mass [OF] barium sulfate [=] \\pu{0.2745 g}"},{"type":"other","value":"Small excess of sulfuric acid."},{"type":"substance name","value":"Barium ion"}] | <h1 class="questionTitle" itemprop="name">A 3.060 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.The resultant reaction produced a precipitate of barium sulfate, which collected by filtration, washed, dried, weight?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>If 0.2745 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?</p></div>
</h2>
</div>
</div> | 5.27% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>For <mathjax>#"BaSO"_4#</mathjax>,</p>
<p><mathjax>#M_r = "137.33 + 32.06 + 64.00 = 233.39"#</mathjax></p>
<p>∴ <mathjax>#0.2745 color(red)(cancel(color(black)("g BaSO"_4))) × "137.33 g Ba"/(233.99 color(red)(cancel(color(black)("g BaSO"_4)))) = "0.1611 g Ba"#</mathjax></p>
<p><mathjax>#"% Ba" = ("0.1611" color(red)(cancel(color(black)("g"))))/(3.060 color(red)(cancel(color(black)("g")))) × 100 % = 5.265 %#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass percentage of barium was 5.265 %.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>For <mathjax>#"BaSO"_4#</mathjax>,</p>
<p><mathjax>#M_r = "137.33 + 32.06 + 64.00 = 233.39"#</mathjax></p>
<p>∴ <mathjax>#0.2745 color(red)(cancel(color(black)("g BaSO"_4))) × "137.33 g Ba"/(233.99 color(red)(cancel(color(black)("g BaSO"_4)))) = "0.1611 g Ba"#</mathjax></p>
<p><mathjax>#"% Ba" = ("0.1611" color(red)(cancel(color(black)("g"))))/(3.060 color(red)(cancel(color(black)("g")))) × 100 % = 5.265 %#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 3.060 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.The resultant reaction produced a precipitate of barium sulfate, which collected by filtration, washed, dried, weight?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>If 0.2745 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?</p></div>
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Ernest Z.
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<div class="markdown"><p>The mass percentage of barium was 5.265 %.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>For <mathjax>#"BaSO"_4#</mathjax>,</p>
<p><mathjax>#M_r = "137.33 + 32.06 + 64.00 = 233.39"#</mathjax></p>
<p>∴ <mathjax>#0.2745 color(red)(cancel(color(black)("g BaSO"_4))) × "137.33 g Ba"/(233.99 color(red)(cancel(color(black)("g BaSO"_4)))) = "0.1611 g Ba"#</mathjax></p>
<p><mathjax>#"% Ba" = ("0.1611" color(red)(cancel(color(black)("g"))))/(3.060 color(red)(cancel(color(black)("g")))) × 100 % = 5.265 %#</mathjax></p></div>
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</article> | A 3.060 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.The resultant reaction produced a precipitate of barium sulfate, which collected by filtration, washed, dried, weight? |
If 0.2745 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?
|
1,337 | acac7cb0-6ddd-11ea-a187-ccda262736ce | https://socratic.org/questions/how-many-liters-of-a-88-m-solution-can-be-made-with-25-5-grams-of-lithium-fluori | 1.10 liters | start physical_unit 6 6 volume l qc_end physical_unit 16 16 4 5 molarity qc_end physical_unit 16 16 11 12 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] LiF solution [IN] liters"}] | [{"type":"physical unit","value":"1.10 liters"}] | [{"type":"physical unit","value":"Molarity [OF] LiF solution [=] \\pu{0.88 M}"},{"type":"physical unit","value":"Mass [OF] LiF [=] \\pu{25.5 grams}"}] | <h1 class="questionTitle" itemprop="name">How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (#LiF#)?</h1> | null | 1.10 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> <mathjax>#M#</mathjax> is calculated by the equation comparing moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution </p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p>For this question we are given the Molarity 0.88M</p>
<p>We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride</p>
<p>We can convert the mass of LiF to moles by dividing by the molar mass of LiF</p>
<p>Li = 6.94<br/>
F = 19.0</p>
<p>LiF = 25.94 g/mole</p>
<p><mathjax>#25.2 cancel(grams) x (1 mol)/(25.94cancel(grams))#</mathjax> = <mathjax>#0.97#</mathjax> moles</p>
<p>Now we can take the the molarity and the moles and calculate the Liters of solution</p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p><mathjax>#ML= mol#</mathjax></p>
<p><mathjax>#L = (mol)/M#</mathjax></p>
<p><mathjax>#L = (0.97mol)/(0.88M)#</mathjax></p>
<p><mathjax>#L = 1.10 L#</mathjax> of solution</p>
<p>
<iframe src="https://www.youtube.com/embed/loL46MQt1vQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#L = 1.10 L#</mathjax> of solution</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> <mathjax>#M#</mathjax> is calculated by the equation comparing moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution </p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p>For this question we are given the Molarity 0.88M</p>
<p>We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride</p>
<p>We can convert the mass of LiF to moles by dividing by the molar mass of LiF</p>
<p>Li = 6.94<br/>
F = 19.0</p>
<p>LiF = 25.94 g/mole</p>
<p><mathjax>#25.2 cancel(grams) x (1 mol)/(25.94cancel(grams))#</mathjax> = <mathjax>#0.97#</mathjax> moles</p>
<p>Now we can take the the molarity and the moles and calculate the Liters of solution</p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p><mathjax>#ML= mol#</mathjax></p>
<p><mathjax>#L = (mol)/M#</mathjax></p>
<p><mathjax>#L = (0.97mol)/(0.88M)#</mathjax></p>
<p><mathjax>#L = 1.10 L#</mathjax> of solution</p>
<p>
<iframe src="https://www.youtube.com/embed/loL46MQt1vQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (#LiF#)?</h1>
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BRIAN M.
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<div class="markdown"><p><mathjax>#L = 1.10 L#</mathjax> of solution</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> <mathjax>#M#</mathjax> is calculated by the equation comparing moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> to liters of solution </p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p>For this question we are given the Molarity 0.88M</p>
<p>We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride</p>
<p>We can convert the mass of LiF to moles by dividing by the molar mass of LiF</p>
<p>Li = 6.94<br/>
F = 19.0</p>
<p>LiF = 25.94 g/mole</p>
<p><mathjax>#25.2 cancel(grams) x (1 mol)/(25.94cancel(grams))#</mathjax> = <mathjax>#0.97#</mathjax> moles</p>
<p>Now we can take the the molarity and the moles and calculate the Liters of solution</p>
<p><mathjax>#M=(mol)/L#</mathjax> </p>
<p><mathjax>#ML= mol#</mathjax></p>
<p><mathjax>#L = (mol)/M#</mathjax></p>
<p><mathjax>#L = (0.97mol)/(0.88M)#</mathjax></p>
<p><mathjax>#L = 1.10 L#</mathjax> of solution</p>
<p>
<iframe src="https://www.youtube.com/embed/loL46MQt1vQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (#LiF#)? | null |
1,338 | a9e1c889-6ddd-11ea-8ec5-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-hydrogen-in-49-2-g-ammonium-hydrogen-phosphite | 3.35 grams | start physical_unit 5 5 mass g qc_end physical_unit 9 11 7 8 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen [IN] grams"}] | [{"type":"physical unit","value":"3.35 grams"}] | [{"type":"physical unit","value":"Mass [OF] ammonium hydrogen phosphite [=] \\pu{49.2 g}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of hydrogen in 49.2 g ammonium hydrogen phosphite? </h1> | null | 3.35 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of ammonium hydrogen phosphate"#</mathjax> <mathjax>#(49.2*g)/(132.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.373*mol#</mathjax>.<br/>
Clearly, there are <mathjax>#9#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax> per <mathjax>#mol#</mathjax> <mathjax>#(NH_4)_2HPO_4#</mathjax>, thus <mathjax>#3.35*mol#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If you mean ammonium hydrogen phosphate, <mathjax>#(NH_4)_2HPO_4" (ammonium biphosphate)"#</mathjax>, then there are <mathjax>#3.35#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#"H atoms"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of ammonium hydrogen phosphate"#</mathjax> <mathjax>#(49.2*g)/(132.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.373*mol#</mathjax>.<br/>
Clearly, there are <mathjax>#9#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax> per <mathjax>#mol#</mathjax> <mathjax>#(NH_4)_2HPO_4#</mathjax>, thus <mathjax>#3.35*mol#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the mass of hydrogen in 49.2 g ammonium hydrogen phosphite? </h1>
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anor277
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<div class="markdown"><p>If you mean ammonium hydrogen phosphate, <mathjax>#(NH_4)_2HPO_4" (ammonium biphosphate)"#</mathjax>, then there are <mathjax>#3.35#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#"H atoms"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of ammonium hydrogen phosphate"#</mathjax> <mathjax>#(49.2*g)/(132.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.373*mol#</mathjax>.<br/>
Clearly, there are <mathjax>#9#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#H#</mathjax> per <mathjax>#mol#</mathjax> <mathjax>#(NH_4)_2HPO_4#</mathjax>, thus <mathjax>#3.35*mol#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax>.</p></div>
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</article> | What is the mass of hydrogen in 49.2 g ammonium hydrogen phosphite? | null |
1,339 | abb00764-6ddd-11ea-98d9-ccda262736ce | https://socratic.org/questions/582f383db72cff3e840630e7 | CaCl2(aq) + 2 AgNO3(aq) -> 2 AgCl(s) + Ca(NO3)2(aq) | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"CaCl2(aq) + 2 AgNO3(aq) -> 2 AgCl(s) + Ca(NO3)2(aq)"}] | [{"type":"substance name","value":"Silver nitrate"},{"type":"substance name","value":"Calcium chloride"}] | <h1 class="questionTitle" itemprop="name">How does silver nitrate react with calcium chloride?</h1> | null | CaCl2(aq) + 2 AgNO3(aq) -> 2 AgCl(s) + Ca(NO3)2(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The calcium and nitrate ions are along for the ride, and the net ionic equation is:</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CaCl_2(aq) + 2AgNO_3(aq) rarr 2AgCl(s)darr + Ca(NO_3)_2(aq)#</mathjax></p>
<p>For every reactant particle, is there a corresponding product particle?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The calcium and nitrate ions are along for the ride, and the net ionic equation is:</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How does silver nitrate react with calcium chloride?</h1>
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anor277
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Nov 18, 2016
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<div class="markdown"><p><mathjax>#CaCl_2(aq) + 2AgNO_3(aq) rarr 2AgCl(s)darr + Ca(NO_3)_2(aq)#</mathjax></p>
<p>For every reactant particle, is there a corresponding product particle?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The calcium and nitrate ions are along for the ride, and the net ionic equation is:</p>
<p><mathjax>#Ag^(+) + Cl^(-) rarr AgCl(s)darr#</mathjax></p></div>
</div>
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</article> | How does silver nitrate react with calcium chloride? | null |
1,340 | ac7aeb18-6ddd-11ea-bcb1-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-of-nano-3-84-99-g-mol-which-contains-11-8-gra | 1.49 M | start physical_unit 8 8 molarity mol/l qc_end physical_unit 8 8 9 10 molar_mass qc_end physical_unit 8 8 13 14 mass qc_end physical_unit 6 6 22 23 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaNO3 solution [IN] M"}] | [{"type":"physical unit","value":"1.49 M"}] | [{"type":"physical unit","value":"Molar mass [OF] NaNO3 [=] \\pu{84.99 g/mol}"},{"type":"physical unit","value":"Mass [OF] solute [=] \\pu{11.8 grams}"},{"type":"physical unit","value":"Volume [OF] NaNO3 solution [=] \\pu{93.9 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?</h1> | null | 1.49 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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<div class="markdown"><p>1.49 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL?</h1>
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MM
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<div class="markdown"><p>1.49 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Number of moles of sodium nitrate: <br/>
<mathjax>#n = m/(MW) = 11.8/85 = 0.13882 mol#</mathjax></p>
<p><mathjax>#(0.13882 mol)/ (93.9 cancel (mL)) times (1000 cancel(mL)) /(1 L) = 1.49 (mol)/L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Molarity"=1.48*mol*L^-1...................#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>For this problem.......</p>
<p><mathjax>#"Molarity"=((11.8*cancelg)/(84.99*cancelg*mol^-1))/(93.9*cancel(mL)xx10^-3*L*cancel(mL^-1))#</mathjax></p>
<p><mathjax>#=??*mol*L^-1#</mathjax>...........</p></div>
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</article> | What is the molarity of a solution of #NaNO_3# (84.99 g/mol) which contains 11.8 grams of solute in a total volume of 93.9 mL? | null |
1,341 | ac89c228-6ddd-11ea-95f5-ccda262736ce | https://socratic.org/questions/how-many-moles-of-nh-3-are-in-a-3-0-l-vessel-at-3-10-2-k-with-a-pressure-of-1-50 | 0.18 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 volume qc_end physical_unit 4 4 12 15 temperature qc_end physical_unit 4 4 20 21 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] NH3 [IN] moles"}] | [{"type":"physical unit","value":"0.18 moles"}] | [{"type":"physical unit","value":"Volume [OF] NH3 [=] \\pu{3.0 L}"},{"type":"physical unit","value":"Temperature [OF] NH3 [=] \\pu{3 × 10^2 K}"},{"type":"physical unit","value":"Pressure [OF] NH3 [=] \\pu{1.50 atm}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #NH_3# are in a 3.0 L vessel at #3*10^2# #K# with a pressure of 1.50 atm?</h1> | null | 0.18 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like the time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<blockquote>
<blockquote>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = "1.50 atm"#</mathjax><br/>
<mathjax>#V = "3.0 L"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3×10^2 color(white)(l) "K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#n = (PV)/(RT) = (1.50 color(red)(cancel(color(black)("atm"))) × 3.0 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1"))) "mol"^"-1" × 3 × 10^2 color(red)(cancel(color(black)("K")))) = "0.18 mol"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>There are 0.18 mol of <mathjax>#"NH"_3#</mathjax> in the vessel.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like the time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<blockquote>
<blockquote>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = "1.50 atm"#</mathjax><br/>
<mathjax>#V = "3.0 L"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3×10^2 color(white)(l) "K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#n = (PV)/(RT) = (1.50 color(red)(cancel(color(black)("atm"))) × 3.0 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1"))) "mol"^"-1" × 3 × 10^2 color(red)(cancel(color(black)("K")))) = "0.18 mol"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of #NH_3# are in a 3.0 L vessel at #3*10^2# #K# with a pressure of 1.50 atm?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-05-30T00:06:59" itemprop="dateCreated">
May 30, 2016
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<div class="markdown"><p>There are 0.18 mol of <mathjax>#"NH"_3#</mathjax> in the vessel.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This looks like the time to apply the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(PV = nRT)|)#</mathjax>,</p>
</blockquote>
</blockquote>
<p>where </p>
<ul>
<li><mathjax>#P#</mathjax> is the pressure</li>
<li><mathjax>#V#</mathjax> is the volume</li>
<li><mathjax>#n#</mathjax> is the number of moles</li>
<li><mathjax>#R#</mathjax> is the gas constant</li>
<li><mathjax>#T#</mathjax> is the temperature</li>
</ul>
<blockquote></blockquote>
<p>We can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to get</p>
<blockquote>
<blockquote>
<p><mathjax>#n = (PV)/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#P = "1.50 atm"#</mathjax><br/>
<mathjax>#V = "3.0 L"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = 3×10^2 color(white)(l) "K"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#n = (PV)/(RT) = (1.50 color(red)(cancel(color(black)("atm"))) × 3.0 color(red)(cancel(color(black)("L"))))/( "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1"))) "mol"^"-1" × 3 × 10^2 color(red)(cancel(color(black)("K")))) = "0.18 mol"#</mathjax></p></div>
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</article> | How many moles of #NH_3# are in a 3.0 L vessel at #3*10^2# #K# with a pressure of 1.50 atm? | null |
1,342 | a85d7952-6ddd-11ea-95f9-ccda262736ce | https://socratic.org/questions/how-many-gm-of-solid-naoh-must-be-added-to-100ml-of-a-buffer-solution-which-is-0 | 0.2 gm | start physical_unit 4 5 mass g qc_end physical_unit 13 15 10 11 volume qc_end physical_unit 22 23 18 19 molarity qc_end physical_unit 25 26 18 19 molarity qc_end physical_unit 15 15 33 33 ph qc_end physical_unit 23 23 33 33 pka qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] solid NaOH [IN] gm"}] | [{"type":"physical unit","value":"0.2 gm"}] | [{"type":"physical unit","value":"Volume [OF] a buffer solution [=] \\pu{100 ml}"},{"type":"physical unit","value":"Molarity [OF] acid HA [=] \\pu{0.1 M}"},{"type":"physical unit","value":"Molarity [OF] salt NaA [=] \\pu{0.1 M}"},{"type":"physical unit","value":"pH [OF] solution [=] \\pu{5.5}"},{"type":"physical unit","value":"pKa [OF] HA [=] \\pu{5}"},{"type":"other","value":"Antilog (0.5) = 3.16."}] | <h1 class="questionTitle" itemprop="name">How many gm of solid NaOH must be added to 100ml of a buffer solution which is 0.1M each w.r.t acid HA and salt NaA to make the pH of solution 5.5. Given #pK_a#(HA) = 5 (Use antilog (0.5) = 3.16?</h1> | null | 0.2 gm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the hydroxide anions delivered to the solution by the sodium hydroxide will react with the weak acid to produce the conjugate base in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>number of moles</strong> of weak acid and of conjugate base present in the buffer <em>before</em> the addition of the strong base is given by</p>
<blockquote>
<p><mathjax>#"moles HA" = "moles A"^(-) = "0.1 moles"/(10^3quad color(red)(cancel(color(black)("mL")))) * 100 color(red)(cancel(color(black)("mL")))#</mathjax></p>
<p><mathjax>#"moles HA = moles A"^(-) = "0.01 moles"#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> to be the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can say that after the reaction is complete, the resulting solution will contain</p>
<blockquote>
<p><mathjax>#(0.01 - x) quad "moles HA"#</mathjax></p>
<blockquote>
<p>The reaction will <strong>consume</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the weak acid.</p>
</blockquote>
<p><mathjax>#(0.01 + x) quad "moles A"^(-)#</mathjax></p>
<blockquote>
<p>The reaction will <strong>produce</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the conjugate base. </p>
</blockquote>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong>.</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["A"^(-)])/(["HA"]))#</mathjax></p>
</blockquote>
<p>Assuming that the volume of the buffer <em>does not change</em> upon the addition of the strong base, you can say that the volume of the solution <strong>after</strong> the strong base is added is equal to <mathjax>#"100 mL"#</mathjax>. </p>
<p>That said, the fact that the volume <strong>is the same</strong> for both the weak acid and the conjugate base allows you to treat the concentrations of the two chemical species and the number of moles <em>interchangeably</em>. </p>
<p>You can thus say that after the strong base is added, the <mathjax>#"pH"#</mathjax> of the solution will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log [ ((0.01 + x) color(red)(cancel(color(black)("moles"))))/((0.01 - x) color(red)(cancel(color(black)("moles"))))]#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#5.5 = 5 + log((0.01 +x)/(0.01 - x))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#log((0.01 + x)/(0.01 - x)) = 0.5#</mathjax></p>
</blockquote>
<p>To find the value of <mathjax>#x#</mathjax>, rewrite this as</p>
<blockquote>
<p><mathjax>#10^(log((0.01 + x)/(0.01-x))) = 10^(0.5)#</mathjax></p>
</blockquote>
<p>You will ned up with</p>
<blockquote>
<p><mathjax>#(0.01 + x)/(0.01 - x) = 3.16#</mathjax></p>
<p><mathjax>#0.01 + x = 0.0316 - 3.16x#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#x = (0.0316 - 0.01)/(1 + 3.16) = 0.00519#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> represents the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can use the <strong>molar mass</strong> of the compound to convert this to <em>grams</em>.</p>
<blockquote>
<p><mathjax>#0.0519 color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.2 g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the hydroxide anions delivered to the solution by the sodium hydroxide will react with the weak acid to produce the conjugate base in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>number of moles</strong> of weak acid and of conjugate base present in the buffer <em>before</em> the addition of the strong base is given by</p>
<blockquote>
<p><mathjax>#"moles HA" = "moles A"^(-) = "0.1 moles"/(10^3quad color(red)(cancel(color(black)("mL")))) * 100 color(red)(cancel(color(black)("mL")))#</mathjax></p>
<p><mathjax>#"moles HA = moles A"^(-) = "0.01 moles"#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> to be the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can say that after the reaction is complete, the resulting solution will contain</p>
<blockquote>
<p><mathjax>#(0.01 - x) quad "moles HA"#</mathjax></p>
<blockquote>
<p>The reaction will <strong>consume</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the weak acid.</p>
</blockquote>
<p><mathjax>#(0.01 + x) quad "moles A"^(-)#</mathjax></p>
<blockquote>
<p>The reaction will <strong>produce</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the conjugate base. </p>
</blockquote>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong>.</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["A"^(-)])/(["HA"]))#</mathjax></p>
</blockquote>
<p>Assuming that the volume of the buffer <em>does not change</em> upon the addition of the strong base, you can say that the volume of the solution <strong>after</strong> the strong base is added is equal to <mathjax>#"100 mL"#</mathjax>. </p>
<p>That said, the fact that the volume <strong>is the same</strong> for both the weak acid and the conjugate base allows you to treat the concentrations of the two chemical species and the number of moles <em>interchangeably</em>. </p>
<p>You can thus say that after the strong base is added, the <mathjax>#"pH"#</mathjax> of the solution will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log [ ((0.01 + x) color(red)(cancel(color(black)("moles"))))/((0.01 - x) color(red)(cancel(color(black)("moles"))))]#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#5.5 = 5 + log((0.01 +x)/(0.01 - x))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#log((0.01 + x)/(0.01 - x)) = 0.5#</mathjax></p>
</blockquote>
<p>To find the value of <mathjax>#x#</mathjax>, rewrite this as</p>
<blockquote>
<p><mathjax>#10^(log((0.01 + x)/(0.01-x))) = 10^(0.5)#</mathjax></p>
</blockquote>
<p>You will ned up with</p>
<blockquote>
<p><mathjax>#(0.01 + x)/(0.01 - x) = 3.16#</mathjax></p>
<p><mathjax>#0.01 + x = 0.0316 - 3.16x#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#x = (0.0316 - 0.01)/(1 + 3.16) = 0.00519#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> represents the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can use the <strong>molar mass</strong> of the compound to convert this to <em>grams</em>.</p>
<blockquote>
<p><mathjax>#0.0519 color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many gm of solid NaOH must be added to 100ml of a buffer solution which is 0.1M each w.r.t acid HA and salt NaA to make the pH of solution 5.5. Given #pK_a#(HA) = 5 (Use antilog (0.5) = 3.16?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-01-27T01:01:51" itemprop="dateCreated">
Jan 27, 2018
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<div class="markdown"><p><mathjax>#"0.2 g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the hydroxide anions delivered to the solution by the sodium hydroxide will react with the weak acid to produce the conjugate base in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>.</p>
<blockquote>
<p><mathjax>#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>The <strong>number of moles</strong> of weak acid and of conjugate base present in the buffer <em>before</em> the addition of the strong base is given by</p>
<blockquote>
<p><mathjax>#"moles HA" = "moles A"^(-) = "0.1 moles"/(10^3quad color(red)(cancel(color(black)("mL")))) * 100 color(red)(cancel(color(black)("mL")))#</mathjax></p>
<p><mathjax>#"moles HA = moles A"^(-) = "0.01 moles"#</mathjax></p>
</blockquote>
<p>Now, if you take <mathjax>#x#</mathjax> to be the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can say that after the reaction is complete, the resulting solution will contain</p>
<blockquote>
<p><mathjax>#(0.01 - x) quad "moles HA"#</mathjax></p>
<blockquote>
<p>The reaction will <strong>consume</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the weak acid.</p>
</blockquote>
<p><mathjax>#(0.01 + x) quad "moles A"^(-)#</mathjax></p>
<blockquote>
<p>The reaction will <strong>produce</strong> <mathjax>#x#</mathjax> <strong>moles</strong> of the conjugate base. </p>
</blockquote>
</blockquote>
<p>As you know, the <mathjax>#"pH"#</mathjax> of a weak acid-conjugate base buffer can be calculated using the <strong>Henderson - Hasselbalch equation</strong>.</p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log( (["A"^(-)])/(["HA"]))#</mathjax></p>
</blockquote>
<p>Assuming that the volume of the buffer <em>does not change</em> upon the addition of the strong base, you can say that the volume of the solution <strong>after</strong> the strong base is added is equal to <mathjax>#"100 mL"#</mathjax>. </p>
<p>That said, the fact that the volume <strong>is the same</strong> for both the weak acid and the conjugate base allows you to treat the concentrations of the two chemical species and the number of moles <em>interchangeably</em>. </p>
<p>You can thus say that after the strong base is added, the <mathjax>#"pH"#</mathjax> of the solution will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = "p"K_a + log [ ((0.01 + x) color(red)(cancel(color(black)("moles"))))/((0.01 - x) color(red)(cancel(color(black)("moles"))))]#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#5.5 = 5 + log((0.01 +x)/(0.01 - x))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#log((0.01 + x)/(0.01 - x)) = 0.5#</mathjax></p>
</blockquote>
<p>To find the value of <mathjax>#x#</mathjax>, rewrite this as</p>
<blockquote>
<p><mathjax>#10^(log((0.01 + x)/(0.01-x))) = 10^(0.5)#</mathjax></p>
</blockquote>
<p>You will ned up with</p>
<blockquote>
<p><mathjax>#(0.01 + x)/(0.01 - x) = 3.16#</mathjax></p>
<p><mathjax>#0.01 + x = 0.0316 - 3.16x#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#x = (0.0316 - 0.01)/(1 + 3.16) = 0.00519#</mathjax></p>
</blockquote>
<p>Since <mathjax>#x#</mathjax> represents the <strong>number of moles</strong> of sodium hydroxide added to the buffer, you can use the <strong>molar mass</strong> of the compound to convert this to <em>grams</em>.</p>
<blockquote>
<p><mathjax>#0.0519 color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values.</p></div>
</div>
</div>
</div>
</div>
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</article> | How many gm of solid NaOH must be added to 100ml of a buffer solution which is 0.1M each w.r.t acid HA and salt NaA to make the pH of solution 5.5. Given #pK_a#(HA) = 5 (Use antilog (0.5) = 3.16? | null |
1,343 | a95e37dc-6ddd-11ea-acfb-ccda262736ce | https://socratic.org/questions/580c61cb7c014919419aa81a | KMnO4 + 4 H2SO4 + 5 KCl -> MnSO4 + 5/2 Cl2 + 3 K2SO4 + 4 H2O | start chemical_equation qc_end substance 9 10 qc_end substance 12 13 qc_end substance 15 16 qc_end substance 21 22 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the redox equation"}] | [{"type":"chemical equation","value":"KMnO4 + 4 H2SO4 + 5 KCl -> MnSO4 + 5/2 Cl2 + 3 K2SO4 + 4 H2O"}] | [{"type":"substance name","value":"Potassium chloride"},{"type":"substance name","value":"Chlorine gas"},{"type":"substance name","value":"Potassium permanganate"},{"type":"substance name","value":"Sulfuric acid"}] | <h1 class="questionTitle" itemprop="name">What is the redox equation for the oxidation of potassium chloride to chlorine gas by potassium permanganate in the presence of sulfuric acid? </h1> | null | KMnO4 + 4 H2SO4 + 5 KCl -> MnSO4 + 5/2 Cl2 + 3 K2SO4 + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a redox reaction. Permanganate ion is reduced to <mathjax>#Mn^(2+)#</mathjax>, and chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#"Oxidation reaction:"#</mathjax></p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2 + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction reaction:"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>In each reaction, both charge and mass are balanced (are they?). And the equations must be balanced if they reflect chemical, i.e. physical reality. The overall reaction eliminates the electrons, and thus I take <mathjax>#5xx(i) + (ii)#</mathjax>.</p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5Cl^(-) rarr Mn^(2+) + 5/2Cl_2 +4H_2O#</mathjax> </p>
<p>This is the same reaction as given above, however, there, we used neutral entities not ions. </p>
<p>What we would observe in this reaction? The intense purple/red colour of permanganate ion would dissipate to give almost colourless <mathjax>#Mn^(2+)#</mathjax> ion. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#KMnO_4 + 4H_2SO_4 + 5KCl rarr MnSO_4 + 5/2Cl_2 +3K_2SO_4 +4H_2O#</mathjax> </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a redox reaction. Permanganate ion is reduced to <mathjax>#Mn^(2+)#</mathjax>, and chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#"Oxidation reaction:"#</mathjax></p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2 + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction reaction:"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>In each reaction, both charge and mass are balanced (are they?). And the equations must be balanced if they reflect chemical, i.e. physical reality. The overall reaction eliminates the electrons, and thus I take <mathjax>#5xx(i) + (ii)#</mathjax>.</p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5Cl^(-) rarr Mn^(2+) + 5/2Cl_2 +4H_2O#</mathjax> </p>
<p>This is the same reaction as given above, however, there, we used neutral entities not ions. </p>
<p>What we would observe in this reaction? The intense purple/red colour of permanganate ion would dissipate to give almost colourless <mathjax>#Mn^(2+)#</mathjax> ion. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the redox equation for the oxidation of potassium chloride to chlorine gas by potassium permanganate in the presence of sulfuric acid? </h1>
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<div class="markdown"><p><mathjax>#KMnO_4 + 4H_2SO_4 + 5KCl rarr MnSO_4 + 5/2Cl_2 +3K_2SO_4 +4H_2O#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a redox reaction. Permanganate ion is reduced to <mathjax>#Mn^(2+)#</mathjax>, and chloride ion is oxidized to chlorine gas:</p>
<p><mathjax>#"Oxidation reaction:"#</mathjax></p>
<p><mathjax>#Cl^(-) rarr 1/2Cl_2 + e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction reaction:"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>In each reaction, both charge and mass are balanced (are they?). And the equations must be balanced if they reflect chemical, i.e. physical reality. The overall reaction eliminates the electrons, and thus I take <mathjax>#5xx(i) + (ii)#</mathjax>.</p>
<p><mathjax>#MnO_4^(-) + 8H^(+) + 5Cl^(-) rarr Mn^(2+) + 5/2Cl_2 +4H_2O#</mathjax> </p>
<p>This is the same reaction as given above, however, there, we used neutral entities not ions. </p>
<p>What we would observe in this reaction? The intense purple/red colour of permanganate ion would dissipate to give almost colourless <mathjax>#Mn^(2+)#</mathjax> ion. </p></div>
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</article> | What is the redox equation for the oxidation of potassium chloride to chlorine gas by potassium permanganate in the presence of sulfuric acid? | null |
1,344 | aa1563f0-6ddd-11ea-89d5-ccda262736ce | https://socratic.org/questions/the-pressure-of-air-in-a-2-25-l-container-is-1-20-atm-what-is-the-new-pressure-i | 0.42 atm | start physical_unit 18 19 pressure atm qc_end physical_unit 3 3 10 11 pressure qc_end physical_unit 8 8 6 7 volume qc_end physical_unit 8 8 24 25 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the sample [IN] atm"}] | [{"type":"physical unit","value":"0.42 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] air [=] \\pu{1.20 atm}"},{"type":"physical unit","value":"Volume1 [OF] container [=] \\pu{2.25 L}"},{"type":"physical unit","value":"Volume2 [OF] container [=] \\pu{6.50 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant?</h1> | null | 0.42 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle's law, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*atmxx2.25*cancelL)/(6.50*cancelL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Using <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, <mathjax>#P_1V_1=P_2V_2#</mathjax>, <mathjax>#P_2~=0.40*atm#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle's law, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*atmxx2.25*cancelL)/(6.50*cancelL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant?</h1>
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<div class="markdown"><p>Using <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, <mathjax>#P_1V_1=P_2V_2#</mathjax>, <mathjax>#P_2~=0.40*atm#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle's law, <mathjax>#P_1V_1=P_2V_2#</mathjax>.</p>
<p><mathjax>#P_2 = (P_1V_1)/V_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.20*atmxx2.25*cancelL)/(6.50*cancelL)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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</article> | The pressure of air in a 2.25-L container is 1.20 atm. What is the new pressure if the sample is transferred to a 6.50-L container, assuming temperature is constant? | null |
1,345 | ac8ac97a-6ddd-11ea-b430-ccda262736ce | https://socratic.org/questions/56af2b637c0149170bab7319 | 1:1 | start physical_unit 30 32 ratio none qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Ratio [OF] oxygen atoms Al2(SO4)3:H2SO4"}] | [{"type":"physical unit","value":"1:1"}] | [{"type":"other","value":"There are equal numbers of sulfur atoms with respect to each salt."}] | <h1 class="questionTitle" itemprop="name">You are given quantities of aluminum sulfate, #Al_2(SO_4)_3#, and sulfuric acid, #H_2SO_4#. If there are EQUAL numbers of SULFUR atoms with respect to each salt, what is the ratio of oxygen atoms...#Al_2(SO_4)_3:H_2SO_4#?</h1> | null | 1:1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is given that both salts contain the same number of sulfur atoms. Thus we have <mathjax>#Al_2S_3O_12#</mathjax>, and necessarily <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#H_2SO_4#</mathjax>. There are <mathjax>#3#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid per equiv aluminum sufate, and necessarily <mathjax>#"12"#</mathjax> <mathjax>#"equiv"#</mathjax> of <mathjax>#O#</mathjax> atoms in the acid. And (if I have been doing my sums correctly), the ratio of oxygens is <mathjax>#1:1#</mathjax>.</p>
<p>This is a nasty question, which I suspect has been asked below A level. </p>
<p>PS You have just asked a bunch of questions. Do you think that you could spend some time formatting the questions properly? It makes interpreting the question a much easier proposition.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If the number of sulfur atoms are EQUAL, then the ratio of oxygen atoms, (<mathjax>#Al_2(SO_4)_3:H_2SO_4#</mathjax>) <mathjax>#=#</mathjax> <mathjax>#1:1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is given that both salts contain the same number of sulfur atoms. Thus we have <mathjax>#Al_2S_3O_12#</mathjax>, and necessarily <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#H_2SO_4#</mathjax>. There are <mathjax>#3#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid per equiv aluminum sufate, and necessarily <mathjax>#"12"#</mathjax> <mathjax>#"equiv"#</mathjax> of <mathjax>#O#</mathjax> atoms in the acid. And (if I have been doing my sums correctly), the ratio of oxygens is <mathjax>#1:1#</mathjax>.</p>
<p>This is a nasty question, which I suspect has been asked below A level. </p>
<p>PS You have just asked a bunch of questions. Do you think that you could spend some time formatting the questions properly? It makes interpreting the question a much easier proposition.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">You are given quantities of aluminum sulfate, #Al_2(SO_4)_3#, and sulfuric acid, #H_2SO_4#. If there are EQUAL numbers of SULFUR atoms with respect to each salt, what is the ratio of oxygen atoms...#Al_2(SO_4)_3:H_2SO_4#?</h1>
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<div class="markdown"><p>If the number of sulfur atoms are EQUAL, then the ratio of oxygen atoms, (<mathjax>#Al_2(SO_4)_3:H_2SO_4#</mathjax>) <mathjax>#=#</mathjax> <mathjax>#1:1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is given that both salts contain the same number of sulfur atoms. Thus we have <mathjax>#Al_2S_3O_12#</mathjax>, and necessarily <mathjax>#3#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#H_2SO_4#</mathjax>. There are <mathjax>#3#</mathjax> <mathjax>#"equiv"#</mathjax> sulfuric acid per equiv aluminum sufate, and necessarily <mathjax>#"12"#</mathjax> <mathjax>#"equiv"#</mathjax> of <mathjax>#O#</mathjax> atoms in the acid. And (if I have been doing my sums correctly), the ratio of oxygens is <mathjax>#1:1#</mathjax>.</p>
<p>This is a nasty question, which I suspect has been asked below A level. </p>
<p>PS You have just asked a bunch of questions. Do you think that you could spend some time formatting the questions properly? It makes interpreting the question a much easier proposition.</p></div>
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</article> | You are given quantities of aluminum sulfate, #Al_2(SO_4)_3#, and sulfuric acid, #H_2SO_4#. If there are EQUAL numbers of SULFUR atoms with respect to each salt, what is the ratio of oxygen atoms...#Al_2(SO_4)_3:H_2SO_4#? | null |
1,346 | ac7015ef-6ddd-11ea-a490-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-nh-3-produced-in-the-following-reaction-when-3-0-l-of-n-2- | 2.67 L | start physical_unit 5 5 volume l qc_end physical_unit 15 15 12 13 volume qc_end physical_unit 21 21 18 19 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] NH3 [IN] L"}] | [{"type":"physical unit","value":"2.67 L"}] | [{"type":"physical unit","value":"Volume [OF] N2 [=] \\pu{3.0 L}"},{"type":"physical unit","value":"Volume [OF] H2 [=] \\pu{4.0 L}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of #NH_3# produced in the following reaction when 3.0 L of #N_2# reacts with 4.0 L of #H_2#?</h1> | null | 2.67 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced reaction is the following:</p>
<p><mathjax>#N_2(g)+3H_2(g)->2NH_3(g)#</mathjax></p>
<p>We will assume that this reaction is happening at constant pressure and that the temperature is not changing through the entire process, therefore, we can say that: </p>
<p><mathjax>#PV=nRT=>V=underbrace((RT)/P)_(color(blue)("constant"))n=>#</mathjax><mathjax>#Valpha" n"#</mathjax></p>
<p>Therefore, now we can use volume the same way that we use number of mole in stoichiometric calculations:</p>
<p><mathjax>#?LNH_3=4.0LH_2xx(2LNH_3)/(3LH_2)=2.7LNH_3#</mathjax></p>
<p><mathjax>#?LNH_3=3.0LN_2xx(2LNH_3)/(1LN_2)=6.0LNH_3#</mathjax></p>
<p>Since <mathjax>#H_2#</mathjax> produces the smallest amount of <mathjax>#NH_3#</mathjax> therefore, it is the <strong>limiting reactant</strong> and the volume of <mathjax>#NH_3#</mathjax> produced is <mathjax>#2.7L#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#V_(NH_3)=2.7L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced reaction is the following:</p>
<p><mathjax>#N_2(g)+3H_2(g)->2NH_3(g)#</mathjax></p>
<p>We will assume that this reaction is happening at constant pressure and that the temperature is not changing through the entire process, therefore, we can say that: </p>
<p><mathjax>#PV=nRT=>V=underbrace((RT)/P)_(color(blue)("constant"))n=>#</mathjax><mathjax>#Valpha" n"#</mathjax></p>
<p>Therefore, now we can use volume the same way that we use number of mole in stoichiometric calculations:</p>
<p><mathjax>#?LNH_3=4.0LH_2xx(2LNH_3)/(3LH_2)=2.7LNH_3#</mathjax></p>
<p><mathjax>#?LNH_3=3.0LN_2xx(2LNH_3)/(1LN_2)=6.0LNH_3#</mathjax></p>
<p>Since <mathjax>#H_2#</mathjax> produces the smallest amount of <mathjax>#NH_3#</mathjax> therefore, it is the <strong>limiting reactant</strong> and the volume of <mathjax>#NH_3#</mathjax> produced is <mathjax>#2.7L#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of #NH_3# produced in the following reaction when 3.0 L of #N_2# reacts with 4.0 L of #H_2#?</h1>
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<div class="markdown"><p><mathjax>#V_(NH_3)=2.7L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced reaction is the following:</p>
<p><mathjax>#N_2(g)+3H_2(g)->2NH_3(g)#</mathjax></p>
<p>We will assume that this reaction is happening at constant pressure and that the temperature is not changing through the entire process, therefore, we can say that: </p>
<p><mathjax>#PV=nRT=>V=underbrace((RT)/P)_(color(blue)("constant"))n=>#</mathjax><mathjax>#Valpha" n"#</mathjax></p>
<p>Therefore, now we can use volume the same way that we use number of mole in stoichiometric calculations:</p>
<p><mathjax>#?LNH_3=4.0LH_2xx(2LNH_3)/(3LH_2)=2.7LNH_3#</mathjax></p>
<p><mathjax>#?LNH_3=3.0LN_2xx(2LNH_3)/(1LN_2)=6.0LNH_3#</mathjax></p>
<p>Since <mathjax>#H_2#</mathjax> produces the smallest amount of <mathjax>#NH_3#</mathjax> therefore, it is the <strong>limiting reactant</strong> and the volume of <mathjax>#NH_3#</mathjax> produced is <mathjax>#2.7L#</mathjax>.</p></div>
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</article> | What is the volume of #NH_3# produced in the following reaction when 3.0 L of #N_2# reacts with 4.0 L of #H_2#? | null |
1,347 | a8fd6247-6ddd-11ea-8993-ccda262736ce | https://socratic.org/questions/what-is-the-temperature-in-c-of-0-938-moles-of-cl-2-gas-with-a-pressure-of-3-24- | 118.7 ℃ | start physical_unit 10 11 temperature °c qc_end physical_unit 10 11 7 8 mole qc_end physical_unit 10 11 21 22 volume qc_end physical_unit 10 11 16 17 pressure qc_end end | [{"type":"physical unit","value":"Temperature [OF] Cl2 gas [IN] ℃"}] | [{"type":"physical unit","value":"118.7 ℃"}] | [{"type":"physical unit","value":"Mole [OF] Cl2 gas [=] \\pu{0.938 moles}"},{"type":"physical unit","value":"Volume [OF] Cl2 gas [=] \\pu{9.31 L}"},{"type":"physical unit","value":"Pressure [OF] Cl2 gas [=] \\pu{3.24 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the temperature, in °C, of 0.938 moles of #Cl_2# gas with a pressure of 3.24 atm and volume of 9.31 L?</h1> | null | 118.7 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV = nRT#</mathjax><br/>
P = pressure<br/>
V = volume in liters<br/>
n = moles<br/>
R = the universal gas constant, which is different depending on the unit of pressure<br/>
T = temperature in Kelvin</p>
<p>So for this question<br/>
P = 3.24 atm<br/>
V = 9.31 L<br/>
n = .938<br/>
R = .0821 <mathjax>#(L*atm)/(mol *K)#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax><br/>
(3.24)(9.31) = (.938)(.0821)(T)<br/>
30.16 = 0.077T<br/>
T = 391.7 K</p>
<p>To convert Kelvin to Celsius, subtract 273.<br/>
391.7 - 273 = 118.7 C</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Using <mathjax>#PV = nRT#</mathjax>, about 392 degrees Kelvin or 119 degrees Celsius. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV = nRT#</mathjax><br/>
P = pressure<br/>
V = volume in liters<br/>
n = moles<br/>
R = the universal gas constant, which is different depending on the unit of pressure<br/>
T = temperature in Kelvin</p>
<p>So for this question<br/>
P = 3.24 atm<br/>
V = 9.31 L<br/>
n = .938<br/>
R = .0821 <mathjax>#(L*atm)/(mol *K)#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax><br/>
(3.24)(9.31) = (.938)(.0821)(T)<br/>
30.16 = 0.077T<br/>
T = 391.7 K</p>
<p>To convert Kelvin to Celsius, subtract 273.<br/>
391.7 - 273 = 118.7 C</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the temperature, in °C, of 0.938 moles of #Cl_2# gas with a pressure of 3.24 atm and volume of 9.31 L?</h1>
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I.F.M
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<div class="markdown"><p>Using <mathjax>#PV = nRT#</mathjax>, about 392 degrees Kelvin or 119 degrees Celsius. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#PV = nRT#</mathjax><br/>
P = pressure<br/>
V = volume in liters<br/>
n = moles<br/>
R = the universal gas constant, which is different depending on the unit of pressure<br/>
T = temperature in Kelvin</p>
<p>So for this question<br/>
P = 3.24 atm<br/>
V = 9.31 L<br/>
n = .938<br/>
R = .0821 <mathjax>#(L*atm)/(mol *K)#</mathjax></p>
<p><mathjax>#PV = nRT#</mathjax><br/>
(3.24)(9.31) = (.938)(.0821)(T)<br/>
30.16 = 0.077T<br/>
T = 391.7 K</p>
<p>To convert Kelvin to Celsius, subtract 273.<br/>
391.7 - 273 = 118.7 C</p></div>
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</article> | What is the temperature, in °C, of 0.938 moles of #Cl_2# gas with a pressure of 3.24 atm and volume of 9.31 L? | null |
1,348 | a8e1895d-6ddd-11ea-a2d3-ccda262736ce | https://socratic.org/questions/how-many-moles-of-ethanol-are-represented-by-50-0-kg-of-ethanol | 1.09 × 10^3 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] ethanol [IN] moles"}] | [{"type":"physical unit","value":"1.09 × 10^3 moles"}] | [{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{50.0 kg}"}] | <h1 class="questionTitle" itemprop="name">How many moles of ethanol are represented by 50.0 kg of ethanol? </h1> | null | 1.09 × 10^3 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2H_5OH#</mathjax> is the formula for ethanol.</p>
<p><mathjax>#50.0kg((1000g)/(kg))((C_2H_5OH)/(46.08g)) = 1.09*10^3 mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1.09*10^3 mol#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2H_5OH#</mathjax> is the formula for ethanol.</p>
<p><mathjax>#50.0kg((1000g)/(kg))((C_2H_5OH)/(46.08g)) = 1.09*10^3 mol#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of ethanol are represented by 50.0 kg of ethanol? </h1>
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Al E.
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<div class="markdown"><p><mathjax>#1.09*10^3 mol#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#C_2H_5OH#</mathjax> is the formula for ethanol.</p>
<p><mathjax>#50.0kg((1000g)/(kg))((C_2H_5OH)/(46.08g)) = 1.09*10^3 mol#</mathjax></p></div>
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</article> | How many moles of ethanol are represented by 50.0 kg of ethanol? | null |
1,349 | ac09f479-6ddd-11ea-b853-ccda262736ce | https://socratic.org/questions/how-do-you-balance-c-2h-4-o-2-co-2-h-2o | C2H4 + 3 O2 -> 2 CO2 + 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":"C2H4 + 3 O2 -> 2 CO2 + 2 H2O"}] | [{"type":"chemical equation","value":"C2H4 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #C_2H_4 + O_2 -> CO_2 + H_2O#?</h1> | null | C2H4 + 3 O2 -> 2 CO2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Set the proportions so that:<br/>
<mathjax>#color(white)("XXX")color(red)(p)C_2H_4+color(red)(q)O_2 rarr color(red)(r)CO_2+color(red)(s)H_2O#</mathjax></p>
<p>C: [1]<mathjax>#color(white)("XXX")2p - r = 0#</mathjax><br/>
H: [2]<mathjax>#color(white)("XXX")4p-2s=0#</mathjax><br/>
O: [3]<mathjax>#color(white)("XXX")2q-2r-s=0#</mathjax></p>
<p>We have 3 equations in 4 unknowns which is insufficient to give a unique solution (but then we shouldn't have expected one).</p>
<p>We know that we want <mathjax>#p, q, r, s in ZZ#</mathjax><br/>
but if we temporarily relax this requirement we can determine relative proportions by setting one of them to a a constant and solving for the other values.</p>
<p>I (arbitrarily) chose <mathjax>#p=1#</mathjax></p>
<p>from [1] with <mathjax>#p=1#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")r=2#</mathjax><br/>
from [2] with <mathjax>#p=1#</mathjax><br/>
[5]<mathjax>#color(white)("XXX")s=2#</mathjax><br/>
from[3] with <mathjax>#r=2#</mathjax> and <mathjax>#s=2#</mathjax><br/>
[6]<mathjax>#color(white)("XXX")q=3#</mathjax></p>
<p>So<br/>
<mathjax>#color(white)("XXX")p:q:r:s = 1:3:2:2#</mathjax></p>
<p>(Normally at this point we would have some fractional results and need to multiply through be the LCM of the denominators,<br/>
but in this case we got lucky and have the integer values we need directly).</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#color(red)(1)C_2H_4 + color(red)(3)O_2 rarr color(red)(2)CO_2+color(red)(2)H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Set the proportions so that:<br/>
<mathjax>#color(white)("XXX")color(red)(p)C_2H_4+color(red)(q)O_2 rarr color(red)(r)CO_2+color(red)(s)H_2O#</mathjax></p>
<p>C: [1]<mathjax>#color(white)("XXX")2p - r = 0#</mathjax><br/>
H: [2]<mathjax>#color(white)("XXX")4p-2s=0#</mathjax><br/>
O: [3]<mathjax>#color(white)("XXX")2q-2r-s=0#</mathjax></p>
<p>We have 3 equations in 4 unknowns which is insufficient to give a unique solution (but then we shouldn't have expected one).</p>
<p>We know that we want <mathjax>#p, q, r, s in ZZ#</mathjax><br/>
but if we temporarily relax this requirement we can determine relative proportions by setting one of them to a a constant and solving for the other values.</p>
<p>I (arbitrarily) chose <mathjax>#p=1#</mathjax></p>
<p>from [1] with <mathjax>#p=1#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")r=2#</mathjax><br/>
from [2] with <mathjax>#p=1#</mathjax><br/>
[5]<mathjax>#color(white)("XXX")s=2#</mathjax><br/>
from[3] with <mathjax>#r=2#</mathjax> and <mathjax>#s=2#</mathjax><br/>
[6]<mathjax>#color(white)("XXX")q=3#</mathjax></p>
<p>So<br/>
<mathjax>#color(white)("XXX")p:q:r:s = 1:3:2:2#</mathjax></p>
<p>(Normally at this point we would have some fractional results and need to multiply through be the LCM of the denominators,<br/>
but in this case we got lucky and have the integer values we need directly).</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #C_2H_4 + O_2 -> CO_2 + H_2O#?</h1>
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<a class="topContributorPic" href="/users/alan-p"><img alt="" class="" src="https://graph.facebook.com/1409517466/picture?height=50&width=50" title=""/></a>
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Alan P.
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Dec 25, 2015
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<div class="markdown"><p><mathjax>#color(red)(1)C_2H_4 + color(red)(3)O_2 rarr color(red)(2)CO_2+color(red)(2)H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Set the proportions so that:<br/>
<mathjax>#color(white)("XXX")color(red)(p)C_2H_4+color(red)(q)O_2 rarr color(red)(r)CO_2+color(red)(s)H_2O#</mathjax></p>
<p>C: [1]<mathjax>#color(white)("XXX")2p - r = 0#</mathjax><br/>
H: [2]<mathjax>#color(white)("XXX")4p-2s=0#</mathjax><br/>
O: [3]<mathjax>#color(white)("XXX")2q-2r-s=0#</mathjax></p>
<p>We have 3 equations in 4 unknowns which is insufficient to give a unique solution (but then we shouldn't have expected one).</p>
<p>We know that we want <mathjax>#p, q, r, s in ZZ#</mathjax><br/>
but if we temporarily relax this requirement we can determine relative proportions by setting one of them to a a constant and solving for the other values.</p>
<p>I (arbitrarily) chose <mathjax>#p=1#</mathjax></p>
<p>from [1] with <mathjax>#p=1#</mathjax><br/>
[4]<mathjax>#color(white)("XXX")r=2#</mathjax><br/>
from [2] with <mathjax>#p=1#</mathjax><br/>
[5]<mathjax>#color(white)("XXX")s=2#</mathjax><br/>
from[3] with <mathjax>#r=2#</mathjax> and <mathjax>#s=2#</mathjax><br/>
[6]<mathjax>#color(white)("XXX")q=3#</mathjax></p>
<p>So<br/>
<mathjax>#color(white)("XXX")p:q:r:s = 1:3:2:2#</mathjax></p>
<p>(Normally at this point we would have some fractional results and need to multiply through be the LCM of the denominators,<br/>
but in this case we got lucky and have the integer values we need directly).</p></div>
</div>
</div>
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<a href="https://socratic.org/answers/203472" itemprop="url">Answer link</a>
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</article> | How do you balance #C_2H_4 + O_2 -> CO_2 + H_2O#? | null |
1,350 | a9a7afc7-6ddd-11ea-9d0c-ccda262736ce | https://socratic.org/questions/when-5-0-g-of-tin-reacts-with-hydrochloric-acid-the-mass-of-the-products-tin-chl | 3.1 grams | start physical_unit 7 8 mass g qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 12 13 19 20 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrochloric acid [IN] grams"}] | [{"type":"physical unit","value":"3.1 grams"}] | [{"type":"physical unit","value":"Mass [OF] tin [=] \\pu{5.0 g}"},{"type":"physical unit","value":"Mass [OF] the products [=] \\pu{8.1 g}"},{"type":"other","value":"The products are tin chloride and hydrogen."}] | <h1 class="questionTitle" itemprop="name">When 5.0 g of tin reacts with hydrochloric acid, the mass of the products, tin chloride and hydrogen, totals 8.1 g. How many grams of hydrochloric acid were used? </h1> | null | 3.1 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The law of conservation of matter/mass states that matter is neither created nor destroyed during a chemical equation in a closed system. So the total mass of the reactants equals the mass of the products.</p>
<p>So, if the mass of the products totals <mathjax>#"8.1 g"#</mathjax>, then you know that the total mass of the reactants must be equal to the total mass of the products. If the mass of the reactant tin is <mathjax>#"5.0 g"#</mathjax>, then the mass of the reactant hydrochloric acid is the difference between the total mass of the products and the mass of the reactant tin, as long as the reaction took place in a closed system.</p>
<p><mathjax>#"8.1 g"-"5.0 g"="3.1 g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass of hydrochloric acid that was used is <mathjax>#"3.1 g"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The law of conservation of matter/mass states that matter is neither created nor destroyed during a chemical equation in a closed system. So the total mass of the reactants equals the mass of the products.</p>
<p>So, if the mass of the products totals <mathjax>#"8.1 g"#</mathjax>, then you know that the total mass of the reactants must be equal to the total mass of the products. If the mass of the reactant tin is <mathjax>#"5.0 g"#</mathjax>, then the mass of the reactant hydrochloric acid is the difference between the total mass of the products and the mass of the reactant tin, as long as the reaction took place in a closed system.</p>
<p><mathjax>#"8.1 g"-"5.0 g"="3.1 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When 5.0 g of tin reacts with hydrochloric acid, the mass of the products, tin chloride and hydrogen, totals 8.1 g. How many grams of hydrochloric acid were used? </h1>
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<div class="markdown"><p>The mass of hydrochloric acid that was used is <mathjax>#"3.1 g"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The law of conservation of matter/mass states that matter is neither created nor destroyed during a chemical equation in a closed system. So the total mass of the reactants equals the mass of the products.</p>
<p>So, if the mass of the products totals <mathjax>#"8.1 g"#</mathjax>, then you know that the total mass of the reactants must be equal to the total mass of the products. If the mass of the reactant tin is <mathjax>#"5.0 g"#</mathjax>, then the mass of the reactant hydrochloric acid is the difference between the total mass of the products and the mass of the reactant tin, as long as the reaction took place in a closed system.</p>
<p><mathjax>#"8.1 g"-"5.0 g"="3.1 g"#</mathjax></p></div>
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</article> | When 5.0 g of tin reacts with hydrochloric acid, the mass of the products, tin chloride and hydrogen, totals 8.1 g. How many grams of hydrochloric acid were used? | null |
1,351 | ac9550fa-6ddd-11ea-9858-ccda262736ce | https://socratic.org/questions/if-you-start-with-5-5-grams-of-sodium-fluoride-how-many-grams-of-magnesium-fluor | 4.08 grams | start physical_unit 13 14 mass g qc_end physical_unit 7 8 4 5 mass qc_end chemical_equation 21 29 qc_end end | [{"type":"physical unit","value":"Mass [OF] magnesium fluoride [IN] grams"}] | [{"type":"physical unit","value":"4.08 grams"}] | [{"type":"physical unit","value":"Mass [OF] sodium fluoride [=] \\pu{5.5 grams}"},{"type":"chemical equation","value":"Mg + 2 NaF -> MgF2 + 2 Na"}] | <h1 class="questionTitle" itemprop="name">If you start with 5.5 grams of sodium fluoride, how many grams of magnesium fluoride will be produced in the reaction #Mg + 2NaF -> MgF_2 + 2Na#?</h1> | null | 4.08 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Mass of sodium fluoride = 5.5 g</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"moles" = "5.5 g"/"(22.99 + 18.998) g/mole"#</mathjax></p>
<p><mathjax>#"moles" = 5.5/41.998 "moles"#</mathjax></p>
<p><mathjax>#"moles" = "0.13096 moles of NaF"#</mathjax></p>
<blockquote></blockquote>
<p>For every 2 moles of <mathjax>#"NaF"#</mathjax>, there is 1 mole of <mathjax>#"MgF"_2#</mathjax>.</p>
<p>Using this knowledge, we can determine the amount of moles of <mathjax>#"MgF"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"0.13096 moles of NaF"/2 = "moles of MgF"_2#</mathjax></p>
<p><mathjax>#"moles of MgF"_2 = "0.06548 moles"#</mathjax></p>
<blockquote></blockquote>
<p>We are now going to convert this into grams.</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"0.06548 moles" = "mass"/"(24.305+2(18.998)) g/mole"#</mathjax></p>
<p><mathjax>#0.06548 = "mass"/"62.301 g"#</mathjax></p>
<p><mathjax>#"mass of MgF"_2 = "4.1 grams"#</mathjax></p>
<p>Therefore, there will be 4.1 grams of <mathjax>#"MgF"_2#</mathjax> produced.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>There will be 4.1 g of <mathjax>#"MgF"_2#</mathjax> produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Mass of sodium fluoride = 5.5 g</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"moles" = "5.5 g"/"(22.99 + 18.998) g/mole"#</mathjax></p>
<p><mathjax>#"moles" = 5.5/41.998 "moles"#</mathjax></p>
<p><mathjax>#"moles" = "0.13096 moles of NaF"#</mathjax></p>
<blockquote></blockquote>
<p>For every 2 moles of <mathjax>#"NaF"#</mathjax>, there is 1 mole of <mathjax>#"MgF"_2#</mathjax>.</p>
<p>Using this knowledge, we can determine the amount of moles of <mathjax>#"MgF"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"0.13096 moles of NaF"/2 = "moles of MgF"_2#</mathjax></p>
<p><mathjax>#"moles of MgF"_2 = "0.06548 moles"#</mathjax></p>
<blockquote></blockquote>
<p>We are now going to convert this into grams.</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"0.06548 moles" = "mass"/"(24.305+2(18.998)) g/mole"#</mathjax></p>
<p><mathjax>#0.06548 = "mass"/"62.301 g"#</mathjax></p>
<p><mathjax>#"mass of MgF"_2 = "4.1 grams"#</mathjax></p>
<p>Therefore, there will be 4.1 grams of <mathjax>#"MgF"_2#</mathjax> produced.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">If you start with 5.5 grams of sodium fluoride, how many grams of magnesium fluoride will be produced in the reaction #Mg + 2NaF -> MgF_2 + 2Na#?</h1>
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<div class="markdown"><p>There will be 4.1 g of <mathjax>#"MgF"_2#</mathjax> produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Mass of sodium fluoride = 5.5 g</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"moles" = "5.5 g"/"(22.99 + 18.998) g/mole"#</mathjax></p>
<p><mathjax>#"moles" = 5.5/41.998 "moles"#</mathjax></p>
<p><mathjax>#"moles" = "0.13096 moles of NaF"#</mathjax></p>
<blockquote></blockquote>
<p>For every 2 moles of <mathjax>#"NaF"#</mathjax>, there is 1 mole of <mathjax>#"MgF"_2#</mathjax>.</p>
<p>Using this knowledge, we can determine the amount of moles of <mathjax>#"MgF"_2#</mathjax>.</p>
<blockquote></blockquote>
<p><mathjax>#"0.13096 moles of NaF"/2 = "moles of MgF"_2#</mathjax></p>
<p><mathjax>#"moles of MgF"_2 = "0.06548 moles"#</mathjax></p>
<blockquote></blockquote>
<p>We are now going to convert this into grams.</p>
<p><mathjax>#"moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#"0.06548 moles" = "mass"/"(24.305+2(18.998)) g/mole"#</mathjax></p>
<p><mathjax>#0.06548 = "mass"/"62.301 g"#</mathjax></p>
<p><mathjax>#"mass of MgF"_2 = "4.1 grams"#</mathjax></p>
<p>Therefore, there will be 4.1 grams of <mathjax>#"MgF"_2#</mathjax> produced.</p></div>
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</article> | If you start with 5.5 grams of sodium fluoride, how many grams of magnesium fluoride will be produced in the reaction #Mg + 2NaF -> MgF_2 + 2Na#? | null |
1,352 | a9be021e-6ddd-11ea-82a5-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-the-ni-atom-in-ni-3-po-4-2 | +2 | start physical_unit 7 8 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] Ni atom"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"Ni3(PO4)2"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of the #Ni# atom in #Ni_3(PO_4)_2#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P#</mathjax> can be either <mathjax>#+3or+5#</mathjax><br/>
In this case it must be <mathjax>#+5#</mathjax> or it wouldn't work out with the rest.</p>
<p>So <mathjax>#PO_4=+5+4*(-2)=-3#</mathjax>, so it's <mathjax>#PO_4^(3-)#</mathjax></p>
<p>There are 2 <mathjax>#PO_4^(3-)#</mathjax>'s for a total charge of <mathjax>#-6#</mathjax></p>
<p>To balance that, the three <mathjax>#Ni#</mathjax>'s must have a charge of <mathjax>#+2#</mathjax> each.</p>
<p>So the oxidation number is <mathjax>#Ni^(2+)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>We first work out the charge of the <mathjax>#PO_4#</mathjax>-group.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P#</mathjax> can be either <mathjax>#+3or+5#</mathjax><br/>
In this case it must be <mathjax>#+5#</mathjax> or it wouldn't work out with the rest.</p>
<p>So <mathjax>#PO_4=+5+4*(-2)=-3#</mathjax>, so it's <mathjax>#PO_4^(3-)#</mathjax></p>
<p>There are 2 <mathjax>#PO_4^(3-)#</mathjax>'s for a total charge of <mathjax>#-6#</mathjax></p>
<p>To balance that, the three <mathjax>#Ni#</mathjax>'s must have a charge of <mathjax>#+2#</mathjax> each.</p>
<p>So the oxidation number is <mathjax>#Ni^(2+)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of the #Ni# atom in #Ni_3(PO_4)_2#?</h1>
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<span class="dateCreated" datetime="2016-05-23T15:36:03" itemprop="dateCreated">
May 23, 2016
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<div class="markdown"><p>We first work out the charge of the <mathjax>#PO_4#</mathjax>-group.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P#</mathjax> can be either <mathjax>#+3or+5#</mathjax><br/>
In this case it must be <mathjax>#+5#</mathjax> or it wouldn't work out with the rest.</p>
<p>So <mathjax>#PO_4=+5+4*(-2)=-3#</mathjax>, so it's <mathjax>#PO_4^(3-)#</mathjax></p>
<p>There are 2 <mathjax>#PO_4^(3-)#</mathjax>'s for a total charge of <mathjax>#-6#</mathjax></p>
<p>To balance that, the three <mathjax>#Ni#</mathjax>'s must have a charge of <mathjax>#+2#</mathjax> each.</p>
<p>So the oxidation number is <mathjax>#Ni^(2+)#</mathjax></p></div>
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</article> | What is the oxidation number of the #Ni# atom in #Ni_3(PO_4)_2#? | null |
1,353 | aba70e86-6ddd-11ea-a49a-ccda262736ce | https://socratic.org/questions/a-1-50-liter-flask-at-a-temperature-of-25-c-contains-a-mixture-of-158-moles-of-m | 4.76 atm | start physical_unit 33 34 total_pressure atm qc_end physical_unit 36 37 1 2 volume qc_end physical_unit 33 34 8 9 temperature qc_end physical_unit 17 17 14 15 mole qc_end physical_unit 21 21 18 19 mole qc_end physical_unit 26 26 23 24 mole qc_end end | [{"type":"physical unit","value":"Total pressure [OF] the mixture [IN] atm"}] | [{"type":"physical unit","value":"4.76 atm"}] | [{"type":"physical unit","value":"Volume [OF] the flask [=] \\pu{1.50 liter}"},{"type":"physical unit","value":"Temperature [OF] the mixture [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Mole [OF] methane [=] \\pu{0.158 moles}"},{"type":"physical unit","value":"Mole [OF] ethane [=] \\pu{0.09 moles}"},{"type":"physical unit","value":"Mole [OF] butane [=] \\pu{0.044 moles}"}] | <h1 class="questionTitle" itemprop="name">A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask?</h1> | null | 4.76 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's law of partial pressures holds that in a gaseous mixture, (i) the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of any component gas is the same as the pressure it would exert if it ALONE occupied the container; and (ii) that the total pressure is the sum of the individual partial pressures.</p>
<p>Thus <mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#P_"methane"+P_"ethane"+P_"butane"#</mathjax>, and if we (reasonably) assume ideality, then:</p>
<p><mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(n_"total"xx0.0821*L*atm*K^-1*mol^-1xx298*K)/(1.50*L)#</mathjax> </p>
<p>where <mathjax>#(n_"total"=n_"methane"+n_"ethane"+n_"butane")#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#(0.158+0.09+0.044)*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.292*mol#</mathjax></p>
<p>And so, <mathjax>#P_"total"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#5*atm#</mathjax>. You will be able to make a better estimate. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V~=5*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's law of partial pressures holds that in a gaseous mixture, (i) the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of any component gas is the same as the pressure it would exert if it ALONE occupied the container; and (ii) that the total pressure is the sum of the individual partial pressures.</p>
<p>Thus <mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#P_"methane"+P_"ethane"+P_"butane"#</mathjax>, and if we (reasonably) assume ideality, then:</p>
<p><mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(n_"total"xx0.0821*L*atm*K^-1*mol^-1xx298*K)/(1.50*L)#</mathjax> </p>
<p>where <mathjax>#(n_"total"=n_"methane"+n_"ethane"+n_"butane")#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#(0.158+0.09+0.044)*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.292*mol#</mathjax></p>
<p>And so, <mathjax>#P_"total"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#5*atm#</mathjax>. You will be able to make a better estimate. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask?</h1>
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Oct 8, 2016
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<div class="markdown"><p><mathjax>#P=(nRT)/V~=5*atm#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's law of partial pressures holds that in a gaseous mixture, (i) the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of any component gas is the same as the pressure it would exert if it ALONE occupied the container; and (ii) that the total pressure is the sum of the individual partial pressures.</p>
<p>Thus <mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#P_"methane"+P_"ethane"+P_"butane"#</mathjax>, and if we (reasonably) assume ideality, then:</p>
<p><mathjax>#P_"total"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(n_"total"xx0.0821*L*atm*K^-1*mol^-1xx298*K)/(1.50*L)#</mathjax> </p>
<p>where <mathjax>#(n_"total"=n_"methane"+n_"ethane"+n_"butane")#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#(0.158+0.09+0.044)*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.292*mol#</mathjax></p>
<p>And so, <mathjax>#P_"total"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#5*atm#</mathjax>. You will be able to make a better estimate. </p></div>
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</article> | A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask? | null |
1,354 | a8de51e8-6ddd-11ea-b702-ccda262736ce | https://socratic.org/questions/a-2-2-g-sample-of-quinone-c-6h-4o-2-was-burned-in-a-bomb-calorimeter-for-which-t | -2700 kJ/mol | start physical_unit 40 42 molar_heat kj/mol qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 6 6 20 21 heat_capacity qc_end physical_unit 25 26 29 30 temperature qc_end physical_unit 25 26 32 33 temperature qc_end end | [{"type":"physical unit","value":"Molar heat [OF] combustion of quinone [IN] kJ/mol"}] | [{"type":"physical unit","value":"-2700 kJ/mol"}] | [{"type":"physical unit","value":"Mass [OF] C6H4O2 sample [=] \\pu{2.2 g}"},{"type":"physical unit","value":"Total heat capacity [OF] C6H4O2 sample [=] \\pu{7854 J/℃}"},{"type":"physical unit","value":"Temperature1 [OF] the calorimeter [=] \\pu{23.44 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the calorimeter [=] \\pu{30.57 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 2.2 g sample of quinone #C_6H_4O_2# was burned in a bomb calorimeter for which the total heat capacity is 7854 J/°C. The temperature of the calorimeter increased from 23.44 C to 30.57 °C. What is the molar heat of combustion of quinone (in kJ/mol)?</h1> | null | -2700 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are two heat transfers involved.</p>
<p><mathjax>#"heat of combustion of quinone + heat gained by calorimeter = 0"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
<p><mathjax>#nΔ_ cH + C_"cal"ΔT = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem,</p>
<p><mathjax>#n = 2.2 color(red)(cancel(color(black)("g quinone"))) × "1 mol quinone"/(108.09 color(red)(cancel(color(black)("g quinone")))) = "0.0204 mol quinone"#</mathjax></p>
<p><mathjax>#C_"cal" = "7854 J°C"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_f - T_i = "30.57 °C - 23.44 °C" = "7.13 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = nΔ_cH = 0.0204 Δ_cH color(white)(l)"mol"#</mathjax></p>
<p><mathjax>#q_2 = C_"cal"ΔT = "7854 J" color(red)(cancel(color(black)("°C"^"-1"))) × 7.13 color(red)(cancel(color(black)("°C"))) = "56 000 J" = "56.0 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 + q_2 =0.0204 Δ_cHcolor(white)(l) "mol" + "56.0 kJ" = 0#</mathjax></p>
<p><mathjax>#Δ_cH = ("-56.0 kJ")/("0.0204 mol") = "-2700 kJ/mol"#</mathjax> (2 significant figures)</p></div>
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<div class="markdown"><p>The molar heat of combustion of quinone is -2700 kJ/mol.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are two heat transfers involved.</p>
<p><mathjax>#"heat of combustion of quinone + heat gained by calorimeter = 0"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
<p><mathjax>#nΔ_ cH + C_"cal"ΔT = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem,</p>
<p><mathjax>#n = 2.2 color(red)(cancel(color(black)("g quinone"))) × "1 mol quinone"/(108.09 color(red)(cancel(color(black)("g quinone")))) = "0.0204 mol quinone"#</mathjax></p>
<p><mathjax>#C_"cal" = "7854 J°C"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_f - T_i = "30.57 °C - 23.44 °C" = "7.13 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = nΔ_cH = 0.0204 Δ_cH color(white)(l)"mol"#</mathjax></p>
<p><mathjax>#q_2 = C_"cal"ΔT = "7854 J" color(red)(cancel(color(black)("°C"^"-1"))) × 7.13 color(red)(cancel(color(black)("°C"))) = "56 000 J" = "56.0 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 + q_2 =0.0204 Δ_cHcolor(white)(l) "mol" + "56.0 kJ" = 0#</mathjax></p>
<p><mathjax>#Δ_cH = ("-56.0 kJ")/("0.0204 mol") = "-2700 kJ/mol"#</mathjax> (2 significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">A 2.2 g sample of quinone #C_6H_4O_2# was burned in a bomb calorimeter for which the total heat capacity is 7854 J/°C. The temperature of the calorimeter increased from 23.44 C to 30.57 °C. What is the molar heat of combustion of quinone (in kJ/mol)?</h1>
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<div class="markdown"><p>The molar heat of combustion of quinone is -2700 kJ/mol.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are two heat transfers involved.</p>
<p><mathjax>#"heat of combustion of quinone + heat gained by calorimeter = 0"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
<p><mathjax>#nΔ_ cH + C_"cal"ΔT = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem,</p>
<p><mathjax>#n = 2.2 color(red)(cancel(color(black)("g quinone"))) × "1 mol quinone"/(108.09 color(red)(cancel(color(black)("g quinone")))) = "0.0204 mol quinone"#</mathjax></p>
<p><mathjax>#C_"cal" = "7854 J°C"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_f - T_i = "30.57 °C - 23.44 °C" = "7.13 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = nΔ_cH = 0.0204 Δ_cH color(white)(l)"mol"#</mathjax></p>
<p><mathjax>#q_2 = C_"cal"ΔT = "7854 J" color(red)(cancel(color(black)("°C"^"-1"))) × 7.13 color(red)(cancel(color(black)("°C"))) = "56 000 J" = "56.0 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 + q_2 =0.0204 Δ_cHcolor(white)(l) "mol" + "56.0 kJ" = 0#</mathjax></p>
<p><mathjax>#Δ_cH = ("-56.0 kJ")/("0.0204 mol") = "-2700 kJ/mol"#</mathjax> (2 significant figures)</p></div>
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</article> | A 2.2 g sample of quinone #C_6H_4O_2# was burned in a bomb calorimeter for which the total heat capacity is 7854 J/°C. The temperature of the calorimeter increased from 23.44 C to 30.57 °C. What is the molar heat of combustion of quinone (in kJ/mol)? | null |
1,355 | a8e2e52c-6ddd-11ea-80ac-ccda262736ce | https://socratic.org/questions/under-certain-conditions-of-temperature-and-pressure-hydrogen-gas-and-nitrogen-g | 39.7 g | start physical_unit 15 15 mass g qc_end physical_unit 7 8 24 25 mole qc_end physical_unit 10 11 31 32 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] ammonia [IN] g"}] | [{"type":"physical unit","value":"39.7 g"}] | [{"type":"physical unit","value":"Mole [OF] hydrogen gas [=] \\pu{3.50 moles}"},{"type":"physical unit","value":"Mole [OF] nitrogen gas [=] \\pu{5.00 moles}"}] | <h1 class="questionTitle" itemprop="name">Under certain conditions of temperature and pressure, hydrogen gas and nitrogen gas react to form ammonia gas. Determine the mass of ammonia produced if 3.50 moles of hydrogen has reacts with 5.00 moles of nitrogen gas?</h1> | null | 39.7 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by balancing the equation for the reaction between hydrogen <mathjax>#"H"_2#</mathjax> and nitrogen <mathjax>#"N"_2#</mathjax>, which produces ammonia <mathjax>#"NH"_3#</mathjax>:</p>
<p><mathjax>#color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3 (g)#</mathjax></p>
<p>From the coefficients <mathjax>#(n("N"_2))/(n("H"_2))=(color(darkgreen)(1))/(color(darkgreen)(3))#</mathjax> <br/>
meaning that all of the nitrogen and hydrogen molecules available would have been converted to ammonia if supplied at a <mathjax>#1:3#</mathjax> ratio; otherwise one of the reagents will be in excess.</p>
<p>The question states that, however, </p>
<p><mathjax>#(n'("N"_2))/( n' ("H"_2))=3.5/5.0=7/10#</mathjax>;</p>
<p><mathjax>#7/10 color(purple)(>) 1/3#</mathjax></p>
<p>meaning that the species represented in the numerator- nitrogen <mathjax>#"N"_2#</mathjax>- is in excess; The number of moles hence the mass of ammonia produced shall be calculated from the quantity of the <em>limiting</em> reactant- <mathjax>#"H"_2#</mathjax>- available. Also, from coefficients in the balanced equation:</p>
<p><mathjax>#(n("NH"_3))/(n("H"_2))=(color(darkgreen)(2))/(color(darkgreen)(3))#</mathjax> </p>
<p><mathjax>#n("NH"_3)=(color(darkgreen)(2))/(color(darkgreen)(3))*n("H"_2)#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=(color(darkgreen)(2))/(color(darkgreen)(3))*3.50 color(white)(l) "mol"#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=2.33 color(white)(l) "mol"#</mathjax> </p>
<p>Take <mathjax>#17.031 color(white)(l) g*"mol"^(-1)#</mathjax> as the molar mass <mathjax>#M#</mathjax> of <a href="https://en.wikipedia.org/wiki/Ammonia" rel="nofollow">ammonia</a> <mathjax>#"NH"_3#</mathjax>,</p>
<p><mathjax>#m("NH"_3)=n("NH"_3)*M("NH"_3)#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=2.33 color(white)(l) color(red)(cancel(color(black)("mol"))) * 17.031 color(white)(l) g*color(red)(cancel(color(black)("mol"^(-1))))#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=39.7 color(white)(l) "g"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#m("NH"_3)=39.7 color(white)(l) "g"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by balancing the equation for the reaction between hydrogen <mathjax>#"H"_2#</mathjax> and nitrogen <mathjax>#"N"_2#</mathjax>, which produces ammonia <mathjax>#"NH"_3#</mathjax>:</p>
<p><mathjax>#color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3 (g)#</mathjax></p>
<p>From the coefficients <mathjax>#(n("N"_2))/(n("H"_2))=(color(darkgreen)(1))/(color(darkgreen)(3))#</mathjax> <br/>
meaning that all of the nitrogen and hydrogen molecules available would have been converted to ammonia if supplied at a <mathjax>#1:3#</mathjax> ratio; otherwise one of the reagents will be in excess.</p>
<p>The question states that, however, </p>
<p><mathjax>#(n'("N"_2))/( n' ("H"_2))=3.5/5.0=7/10#</mathjax>;</p>
<p><mathjax>#7/10 color(purple)(>) 1/3#</mathjax></p>
<p>meaning that the species represented in the numerator- nitrogen <mathjax>#"N"_2#</mathjax>- is in excess; The number of moles hence the mass of ammonia produced shall be calculated from the quantity of the <em>limiting</em> reactant- <mathjax>#"H"_2#</mathjax>- available. Also, from coefficients in the balanced equation:</p>
<p><mathjax>#(n("NH"_3))/(n("H"_2))=(color(darkgreen)(2))/(color(darkgreen)(3))#</mathjax> </p>
<p><mathjax>#n("NH"_3)=(color(darkgreen)(2))/(color(darkgreen)(3))*n("H"_2)#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=(color(darkgreen)(2))/(color(darkgreen)(3))*3.50 color(white)(l) "mol"#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=2.33 color(white)(l) "mol"#</mathjax> </p>
<p>Take <mathjax>#17.031 color(white)(l) g*"mol"^(-1)#</mathjax> as the molar mass <mathjax>#M#</mathjax> of <a href="https://en.wikipedia.org/wiki/Ammonia" rel="nofollow">ammonia</a> <mathjax>#"NH"_3#</mathjax>,</p>
<p><mathjax>#m("NH"_3)=n("NH"_3)*M("NH"_3)#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=2.33 color(white)(l) color(red)(cancel(color(black)("mol"))) * 17.031 color(white)(l) g*color(red)(cancel(color(black)("mol"^(-1))))#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=39.7 color(white)(l) "g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Under certain conditions of temperature and pressure, hydrogen gas and nitrogen gas react to form ammonia gas. Determine the mass of ammonia produced if 3.50 moles of hydrogen has reacts with 5.00 moles of nitrogen gas?</h1>
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<div class="markdown"><p><mathjax>#m("NH"_3)=39.7 color(white)(l) "g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by balancing the equation for the reaction between hydrogen <mathjax>#"H"_2#</mathjax> and nitrogen <mathjax>#"N"_2#</mathjax>, which produces ammonia <mathjax>#"NH"_3#</mathjax>:</p>
<p><mathjax>#color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3 (g)#</mathjax></p>
<p>From the coefficients <mathjax>#(n("N"_2))/(n("H"_2))=(color(darkgreen)(1))/(color(darkgreen)(3))#</mathjax> <br/>
meaning that all of the nitrogen and hydrogen molecules available would have been converted to ammonia if supplied at a <mathjax>#1:3#</mathjax> ratio; otherwise one of the reagents will be in excess.</p>
<p>The question states that, however, </p>
<p><mathjax>#(n'("N"_2))/( n' ("H"_2))=3.5/5.0=7/10#</mathjax>;</p>
<p><mathjax>#7/10 color(purple)(>) 1/3#</mathjax></p>
<p>meaning that the species represented in the numerator- nitrogen <mathjax>#"N"_2#</mathjax>- is in excess; The number of moles hence the mass of ammonia produced shall be calculated from the quantity of the <em>limiting</em> reactant- <mathjax>#"H"_2#</mathjax>- available. Also, from coefficients in the balanced equation:</p>
<p><mathjax>#(n("NH"_3))/(n("H"_2))=(color(darkgreen)(2))/(color(darkgreen)(3))#</mathjax> </p>
<p><mathjax>#n("NH"_3)=(color(darkgreen)(2))/(color(darkgreen)(3))*n("H"_2)#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=(color(darkgreen)(2))/(color(darkgreen)(3))*3.50 color(white)(l) "mol"#</mathjax> <br/>
<mathjax>#color(white)(n("NH"_3))=2.33 color(white)(l) "mol"#</mathjax> </p>
<p>Take <mathjax>#17.031 color(white)(l) g*"mol"^(-1)#</mathjax> as the molar mass <mathjax>#M#</mathjax> of <a href="https://en.wikipedia.org/wiki/Ammonia" rel="nofollow">ammonia</a> <mathjax>#"NH"_3#</mathjax>,</p>
<p><mathjax>#m("NH"_3)=n("NH"_3)*M("NH"_3)#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=2.33 color(white)(l) color(red)(cancel(color(black)("mol"))) * 17.031 color(white)(l) g*color(red)(cancel(color(black)("mol"^(-1))))#</mathjax><br/>
<mathjax>#color(white)(m("NH"_3))=39.7 color(white)(l) "g"#</mathjax></p></div>
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</article> | Under certain conditions of temperature and pressure, hydrogen gas and nitrogen gas react to form ammonia gas. Determine the mass of ammonia produced if 3.50 moles of hydrogen has reacts with 5.00 moles of nitrogen gas? | null |
1,356 | a8aaa2d2-6ddd-11ea-af6c-ccda262736ce | https://socratic.org/questions/the-half-life-of-titanium-44-is-63-years-what-is-the-constant-k-in-the-decay-for | 0.01 years^(-1) | start physical_unit 3 3 equilibrium_constant_k years^(-1) qc_end physical_unit 3 3 5 6 half-life qc_end end | [{"type":"physical unit","value":"Constant k [OF] titanium-44 [IN] years^(-1)"}] | [{"type":"physical unit","value":"0.01 years^(-1)"}] | [{"type":"physical unit","value":"Half-life [OF] titanium-44 [=] \\pu{63 years}"}] | <h1 class="questionTitle" itemprop="name">The half-life of titanium-44 is 63 years. What is the constant k in the decay formula for the substance?</h1> | null | 0.01 years^(-1) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really need to use here is the fact that for a <strong>first-order reaction</strong> like radioactive decay, the <strong>integrated <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a></strong> takes the form</p>
<blockquote>
<p><mathjax>#ln(["A"]) - ln(["A"]_0) = - k * t#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#["A"]#</mathjax> is the <strong>concentration</strong> of the reactant after a given time <mathjax>#t#</mathjax> passes</li>
<li><mathjax>#["A"]_0#</mathjax> is the <strong>initial concentration</strong> of the reactant</li>
<li><mathjax>#k#</mathjax> is the <strong>rate constant</strong></li>
</ul>
</blockquote>
<p>Now, the <strong>half-life</strong> of a radioactive nuclide, <mathjax>#t_"1/2"#</mathjax>, tells you the time needed for exactly <strong>half</strong> of an initial sample of the nuclide to undergo radioactive decay.</p>
<p>This means that after the passing of <strong>one half-life</strong>, you have</p>
<blockquote>
<p><mathjax>#t = t_"1/2"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["A"] = 1/2 * [A"]_0#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the integrated rate law to get</p>
<blockquote>
<p><mathjax>#ln(1/2 * ["A"]_0) - ln(["A"]_0) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#ln( (1/2 * color(red)(cancel(color(black)(["A"]_0))))/color(red)(cancel(color(black)(["A"]_0)))) = - k * t_"1/2"#</mathjax></p>
<p><mathjax>#ln(1/2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>You can rewrite this as</p>
<blockquote>
<p><mathjax>#ln(1) - ln(2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>which will get you</p>
<blockquote>
<p><mathjax>#k = ln(2)/t_"1/2"#</mathjax></p>
</blockquote>
<p>Finally, plug in the value you have for the half-life of titanium-44 to find</p>
<blockquote>
<p><mathjax>#k = ln(2)/"63 years" = color(darkgreen)(ul(color(black)("0.011 years"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#k = "0.011 years"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really need to use here is the fact that for a <strong>first-order reaction</strong> like radioactive decay, the <strong>integrated <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a></strong> takes the form</p>
<blockquote>
<p><mathjax>#ln(["A"]) - ln(["A"]_0) = - k * t#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#["A"]#</mathjax> is the <strong>concentration</strong> of the reactant after a given time <mathjax>#t#</mathjax> passes</li>
<li><mathjax>#["A"]_0#</mathjax> is the <strong>initial concentration</strong> of the reactant</li>
<li><mathjax>#k#</mathjax> is the <strong>rate constant</strong></li>
</ul>
</blockquote>
<p>Now, the <strong>half-life</strong> of a radioactive nuclide, <mathjax>#t_"1/2"#</mathjax>, tells you the time needed for exactly <strong>half</strong> of an initial sample of the nuclide to undergo radioactive decay.</p>
<p>This means that after the passing of <strong>one half-life</strong>, you have</p>
<blockquote>
<p><mathjax>#t = t_"1/2"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["A"] = 1/2 * [A"]_0#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the integrated rate law to get</p>
<blockquote>
<p><mathjax>#ln(1/2 * ["A"]_0) - ln(["A"]_0) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#ln( (1/2 * color(red)(cancel(color(black)(["A"]_0))))/color(red)(cancel(color(black)(["A"]_0)))) = - k * t_"1/2"#</mathjax></p>
<p><mathjax>#ln(1/2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>You can rewrite this as</p>
<blockquote>
<p><mathjax>#ln(1) - ln(2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>which will get you</p>
<blockquote>
<p><mathjax>#k = ln(2)/t_"1/2"#</mathjax></p>
</blockquote>
<p>Finally, plug in the value you have for the half-life of titanium-44 to find</p>
<blockquote>
<p><mathjax>#k = ln(2)/"63 years" = color(darkgreen)(ul(color(black)("0.011 years"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The half-life of titanium-44 is 63 years. What is the constant k in the decay formula for the substance?</h1>
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Stefan V.
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Jan 14, 2018
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<div class="markdown"><p><mathjax>#k = "0.011 years"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you really need to use here is the fact that for a <strong>first-order reaction</strong> like radioactive decay, the <strong>integrated <a href="https://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a></strong> takes the form</p>
<blockquote>
<p><mathjax>#ln(["A"]) - ln(["A"]_0) = - k * t#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#["A"]#</mathjax> is the <strong>concentration</strong> of the reactant after a given time <mathjax>#t#</mathjax> passes</li>
<li><mathjax>#["A"]_0#</mathjax> is the <strong>initial concentration</strong> of the reactant</li>
<li><mathjax>#k#</mathjax> is the <strong>rate constant</strong></li>
</ul>
</blockquote>
<p>Now, the <strong>half-life</strong> of a radioactive nuclide, <mathjax>#t_"1/2"#</mathjax>, tells you the time needed for exactly <strong>half</strong> of an initial sample of the nuclide to undergo radioactive decay.</p>
<p>This means that after the passing of <strong>one half-life</strong>, you have</p>
<blockquote>
<p><mathjax>#t = t_"1/2"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["A"] = 1/2 * [A"]_0#</mathjax></p>
</blockquote>
<p>Plug this into the expression of the integrated rate law to get</p>
<blockquote>
<p><mathjax>#ln(1/2 * ["A"]_0) - ln(["A"]_0) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#k#</mathjax></p>
<blockquote>
<p><mathjax>#ln( (1/2 * color(red)(cancel(color(black)(["A"]_0))))/color(red)(cancel(color(black)(["A"]_0)))) = - k * t_"1/2"#</mathjax></p>
<p><mathjax>#ln(1/2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>You can rewrite this as</p>
<blockquote>
<p><mathjax>#ln(1) - ln(2) = - k * t_"1/2"#</mathjax></p>
</blockquote>
<p>which will get you</p>
<blockquote>
<p><mathjax>#k = ln(2)/t_"1/2"#</mathjax></p>
</blockquote>
<p>Finally, plug in the value you have for the half-life of titanium-44 to find</p>
<blockquote>
<p><mathjax>#k = ln(2)/"63 years" = color(darkgreen)(ul(color(black)("0.011 years"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | The half-life of titanium-44 is 63 years. What is the constant k in the decay formula for the substance? | null |
1,357 | aa469ef0-6ddd-11ea-9a39-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-molar-mass-for-bromine-trioxide | 127.90 g/mol | start physical_unit 8 9 molar_mass g/mol qc_end substance 8 9 qc_end end | [{"type":"physical unit","value":"Molar mass [OF] Bromine Trioxide [IN] g/mol"}] | [{"type":"physical unit","value":"127.90 g/mol"}] | [{"type":"substance name","value":"Bromine Trioxide"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the molar mass for Bromine Trioxide?</h1> | null | 127.90 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, <mathjax>#(79.9+3xx16.00)#</mathjax> <mathjax>#g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Can you tell me the formal oxidation number of <mathjax>#Br#</mathjax> in this compound if the oxidation state of <mathjax>#O#</mathjax> is (as usual) <mathjax>#-II#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Molar mass of <mathjax>#BrO_3#</mathjax>? The molar mass is the sum of the atomic masses, and expressed in grams per mole.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, <mathjax>#(79.9+3xx16.00)#</mathjax> <mathjax>#g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Can you tell me the formal oxidation number of <mathjax>#Br#</mathjax> in this compound if the oxidation state of <mathjax>#O#</mathjax> is (as usual) <mathjax>#-II#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">How would you calculate the molar mass for Bromine Trioxide?</h1>
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anor277
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<div class="markdown"><p>Molar mass of <mathjax>#BrO_3#</mathjax>? The molar mass is the sum of the atomic masses, and expressed in grams per mole.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, <mathjax>#(79.9+3xx16.00)#</mathjax> <mathjax>#g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Can you tell me the formal oxidation number of <mathjax>#Br#</mathjax> in this compound if the oxidation state of <mathjax>#O#</mathjax> is (as usual) <mathjax>#-II#</mathjax>?</p></div>
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</article> | How would you calculate the molar mass for Bromine Trioxide? | null |
1,358 | ac0560dc-6ddd-11ea-910a-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-289-moles-of-fecl-3-dissolved-in-120-ml-of-solution | 2.41 M | start physical_unit 8 8 molarity mol/l qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 14 14 11 12 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] FeCl3 solution [IN] M"}] | [{"type":"physical unit","value":"2.41 M"}] | [{"type":"physical unit","value":"Mole [OF] FeCl3 [=] \\pu{0.289 moles}"},{"type":"physical unit","value":"Volume [OF] FeCl3 solution [=] \\pu{120 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of .289 moles of #FeCl_3# dissolved in 120 mL of solution?</h1> | null | 2.41 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molarity=<mathjax>#"Moles of solute"/"Liters of solution"#</mathjax></p>
<p>Moles=<mathjax>#0.289#</mathjax></p>
<p>Solution=<mathjax>#120mL=0.12L#</mathjax></p>
<p>Therefore,</p>
<p>Molarity=<mathjax>#(0.289 "mol")/(0.12 L)~~(2.41 "mol")/L#</mathjax></p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a>=<mathjax>#(2.41"mol")/L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molarity=<mathjax>#"Moles of solute"/"Liters of solution"#</mathjax></p>
<p>Moles=<mathjax>#0.289#</mathjax></p>
<p>Solution=<mathjax>#120mL=0.12L#</mathjax></p>
<p>Therefore,</p>
<p>Molarity=<mathjax>#(0.289 "mol")/(0.12 L)~~(2.41 "mol")/L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of .289 moles of #FeCl_3# dissolved in 120 mL of solution?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a>=<mathjax>#(2.41"mol")/L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molarity=<mathjax>#"Moles of solute"/"Liters of solution"#</mathjax></p>
<p>Moles=<mathjax>#0.289#</mathjax></p>
<p>Solution=<mathjax>#120mL=0.12L#</mathjax></p>
<p>Therefore,</p>
<p>Molarity=<mathjax>#(0.289 "mol")/(0.12 L)~~(2.41 "mol")/L#</mathjax></p></div>
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</article> | What is the molarity of .289 moles of #FeCl_3# dissolved in 120 mL of solution? | null |
1,359 | ab690e31-6ddd-11ea-983f-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-397-grams-of-na-2so-4 | 2.79 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] Na2SO4 [IN] moles"}] | [{"type":"physical unit","value":"2.79 moles"}] | [{"type":"physical unit","value":"Mass [OF] Na2SO4 [=] \\pu{397 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in 397 grams of #Na_2SO_4#? </h1> | null | 2.79 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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<div class="markdown"><p>2.80</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are there in 397 grams of #Na_2SO_4#? </h1>
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<div class="markdown"><p>2.80</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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David L.
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<div class="markdown"><p><mathjax>#n_(Na_2SO_4)=2.79 mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of <mathjax>#Na_2SO_4 =142.04g/(mol)#</mathjax></p>
<p>Use the equation, </p>
<p><mathjax>#n_(Na_2SO_4)=m/M#</mathjax></p>
<p>where, </p>
<p><mathjax>#n=#</mathjax> number of moles<br/>
<mathjax>#m=#</mathjax>mass (in grams)<br/>
<mathjax>#M=#</mathjax>molar mass (in <mathjax>#(grams)/(mol)#</mathjax>)</p>
<p><mathjax>#n_(Na_2SO_4)=(397g)/(142.04(g/(mol)))#</mathjax></p>
<p><mathjax>#n_(Na_2SO_4)=2.79 mol#</mathjax></p></div>
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</article> | How many moles are there in 397 grams of #Na_2SO_4#? | null |
1,360 | ac874724-6ddd-11ea-97c1-ccda262736ce | https://socratic.org/questions/a-sample-of-ideal-gas-has-a-volume-of-325-l-at-13-60-c-and-1-60-atm-what-is-the- | 541 L | start physical_unit 22 23 volume l qc_end physical_unit 22 23 9 10 volume qc_end physical_unit 22 23 12 13 temperature qc_end physical_unit 22 23 15 16 pressure qc_end physical_unit 22 23 25 26 temperature qc_end physical_unit 22 23 28 29 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"541 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{325 L}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{13.60 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.60 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{23.60 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{0.994 atm}"}] | <h1 class="questionTitle" itemprop="name">A sample of ideal gas has a volume of 325 L at 13.60*C and 1.60 atm. What is the volume of the gas at 23.60°C and .994 atm? </h1> | null | 541 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And so we solve for <mathjax>#V_2=(P_1T_2V_1)/(P_2T_1)#</mathjax></p>
<p><mathjax>#=(1.60*atmxx296.75*Kxx325*L)/(0.994*atmxx286.8*K)#</mathjax></p>
<p><mathjax>#=541*L#</mathjax>.</p>
<p>Is my right arithmetic right? All care taken, but no responsibility accepted. Is it reasonable that the volume has increased so markedly? Why or why not?</p></div>
</div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#V_2=541*L#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And so we solve for <mathjax>#V_2=(P_1T_2V_1)/(P_2T_1)#</mathjax></p>
<p><mathjax>#=(1.60*atmxx296.75*Kxx325*L)/(0.994*atmxx286.8*K)#</mathjax></p>
<p><mathjax>#=541*L#</mathjax>.</p>
<p>Is my right arithmetic right? All care taken, but no responsibility accepted. Is it reasonable that the volume has increased so markedly? Why or why not?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of ideal gas has a volume of 325 L at 13.60*C and 1.60 atm. What is the volume of the gas at 23.60°C and .994 atm? </h1>
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anor277
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<span class="dateCreated" datetime="2017-05-08T20:41:10" itemprop="dateCreated">
May 8, 2017
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<div class="markdown"><p><mathjax>#V_2=541*L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>And so we solve for <mathjax>#V_2=(P_1T_2V_1)/(P_2T_1)#</mathjax></p>
<p><mathjax>#=(1.60*atmxx296.75*Kxx325*L)/(0.994*atmxx286.8*K)#</mathjax></p>
<p><mathjax>#=541*L#</mathjax>.</p>
<p>Is my right arithmetic right? All care taken, but no responsibility accepted. Is it reasonable that the volume has increased so markedly? Why or why not?</p></div>
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</article> | A sample of ideal gas has a volume of 325 L at 13.60*C and 1.60 atm. What is the volume of the gas at 23.60°C and .994 atm? | null |
1,361 | a9af9eff-6ddd-11ea-a553-ccda262736ce | https://socratic.org/questions/a-solution-contains-225-g-of-glucose-c-6h-12o-6-dissolved-in-enough-water-to-mak | 1.51 M | start physical_unit 23 24 molarity mol/l qc_end physical_unit 7 7 3 4 mass qc_end c_other OTHER qc_end physical_unit 1 1 14 15 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"1.51 M"}] | [{"type":"physical unit","value":"Mass [OF] C6H12O6 [=] \\pu{225 g}"},{"type":"other","value":"Enough water."},{"type":"physical unit","value":"Volume [OF] C6H12O6 solution [=] \\pu{0.825 L}"}] | <h1 class="questionTitle" itemprop="name">A solution contains 225 g of glucose, #C_6H_12O_6#, dissolved in enough water to make 0.825 L of solution. What is the molarity of the solution?</h1> | null | 1.51 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, we use the following equation:</p>
<p><img alt="www.chemteam.info" src="http://www.chemteam.info/Solutions/Molarity-Definition.GIF"/></p>
<p>The volume of the given solution has the proper units, but the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> does not. We are given the mass of glucose, not the number of moles. In order to find the number of moles of glucose, you would divide the given mass by the molecular weight of glucose, which is <mathjax>#"180.16 g/mol"#</mathjax>. </p>
<blockquote>
<p><mathjax>#"moles of glucose" = (225 cancel("g"))/(180.16 cancel("g")/"mol") = "1.25 mol"#</mathjax></p>
</blockquote>
<p>Now all we have to do is divide that value by the volume to obtain the molarity like so:</p>
<blockquote>
<p><mathjax>#"molarity" = "1.25 mol" / "0.825 L" = "1.51 molar"#</mathjax></p>
</blockquote>
<p><a href="http://www.chemteam.info/Solutions/Molarity.html" rel="nofollow">Here's a few more examples if you need more help!</a> </p></div>
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<div class="markdown"><p><mathjax>#"1.51 M"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, we use the following equation:</p>
<p><img alt="www.chemteam.info" src="http://www.chemteam.info/Solutions/Molarity-Definition.GIF"/></p>
<p>The volume of the given solution has the proper units, but the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> does not. We are given the mass of glucose, not the number of moles. In order to find the number of moles of glucose, you would divide the given mass by the molecular weight of glucose, which is <mathjax>#"180.16 g/mol"#</mathjax>. </p>
<blockquote>
<p><mathjax>#"moles of glucose" = (225 cancel("g"))/(180.16 cancel("g")/"mol") = "1.25 mol"#</mathjax></p>
</blockquote>
<p>Now all we have to do is divide that value by the volume to obtain the molarity like so:</p>
<blockquote>
<p><mathjax>#"molarity" = "1.25 mol" / "0.825 L" = "1.51 molar"#</mathjax></p>
</blockquote>
<p><a href="http://www.chemteam.info/Solutions/Molarity.html" rel="nofollow">Here's a few more examples if you need more help!</a> </p></div>
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<h1 class="questionTitle" itemprop="name">A solution contains 225 g of glucose, #C_6H_12O_6#, dissolved in enough water to make 0.825 L of solution. What is the molarity of the solution?</h1>
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<div class="markdown"><p><mathjax>#"1.51 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, we use the following equation:</p>
<p><img alt="www.chemteam.info" src="http://www.chemteam.info/Solutions/Molarity-Definition.GIF"/></p>
<p>The volume of the given solution has the proper units, but the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> does not. We are given the mass of glucose, not the number of moles. In order to find the number of moles of glucose, you would divide the given mass by the molecular weight of glucose, which is <mathjax>#"180.16 g/mol"#</mathjax>. </p>
<blockquote>
<p><mathjax>#"moles of glucose" = (225 cancel("g"))/(180.16 cancel("g")/"mol") = "1.25 mol"#</mathjax></p>
</blockquote>
<p>Now all we have to do is divide that value by the volume to obtain the molarity like so:</p>
<blockquote>
<p><mathjax>#"molarity" = "1.25 mol" / "0.825 L" = "1.51 molar"#</mathjax></p>
</blockquote>
<p><a href="http://www.chemteam.info/Solutions/Molarity.html" rel="nofollow">Here's a few more examples if you need more help!</a> </p></div>
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</article> | A solution contains 225 g of glucose, #C_6H_12O_6#, dissolved in enough water to make 0.825 L of solution. What is the molarity of the solution? | null |
1,362 | a8cde588-6ddd-11ea-ba39-ccda262736ce | https://socratic.org/questions/a-15-00-ml-solution-of-potassium-nitrate-kno3-was-diluted-to-125-0-ml-and-25-00- | 1.28 M | start physical_unit 41 43 molarity mol/l qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 3 3 11 12 volume qc_end physical_unit 3 3 14 15 volume qc_end physical_unit 3 3 23 26 volume qc_end physical_unit 30 32 34 35 concentration qc_end end | [{"type":"physical unit","value":"Molarity1 [OF] the original solution [IN] M"}] | [{"type":"physical unit","value":"1.28 M"}] | [{"type":"physical unit","value":"Volume1 [OF] KNO3 solution [=] \\pu{15.00 mL}"},{"type":"physical unit","value":"Volume2 [OF] KNO3 solution [=] \\pu{125.0 mL}"},{"type":"physical unit","value":"Volume3 [OF] KNO3 solution [=] \\pu{25.00 mL}"},{"type":"physical unit","value":"Volume4 [OF] KNO3 solution [=] \\pu{1.000 x 10^3 mL}"},{"type":"physical unit","value":"Concentration4 [OF] the final solution [=] \\pu{0.00383 M}"}] | <h1 class="questionTitle" itemprop="name">A 15.00 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution were then diluted to 1.000 x 10^3 mL. The concentration of the final solution is 0.00383 M. What is the molarity of the original solution?</h1> | null | 1.28 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solution 1: 15.00mL<br/>
Solution 2: 125.0mL<br/>
Solution 3: 25.00mL<br/>
Solution 4: 1000.mL</p>
<p><mathjax>#"mols Solution 4" = 1000.mL * 0.00383M = color(teal)(0.00383 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 3" = color(teal)(0.00383 mols)/(25.00mL) = color(purple)(0.1532M)#</mathjax></p>
<p><mathjax>#"mols Solution 2" = color(purple)(0.1532M) * 125.0mL = color(orchid)(0.01915 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 1" = color(orchid)(0.01915 mols)/(15.00mL) = color(salmon)(1.276666667M)#</mathjax></p>
<p><mathjax>#"Molarity of original solution" = color(salmon)(1.277M)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Molarity of original solution" = 1.277M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solution 1: 15.00mL<br/>
Solution 2: 125.0mL<br/>
Solution 3: 25.00mL<br/>
Solution 4: 1000.mL</p>
<p><mathjax>#"mols Solution 4" = 1000.mL * 0.00383M = color(teal)(0.00383 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 3" = color(teal)(0.00383 mols)/(25.00mL) = color(purple)(0.1532M)#</mathjax></p>
<p><mathjax>#"mols Solution 2" = color(purple)(0.1532M) * 125.0mL = color(orchid)(0.01915 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 1" = color(orchid)(0.01915 mols)/(15.00mL) = color(salmon)(1.276666667M)#</mathjax></p>
<p><mathjax>#"Molarity of original solution" = color(salmon)(1.277M)#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 15.00 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution were then diluted to 1.000 x 10^3 mL. The concentration of the final solution is 0.00383 M. What is the molarity of the original solution?</h1>
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<div class="markdown"><p><mathjax>#"Molarity of original solution" = 1.277M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Solution 1: 15.00mL<br/>
Solution 2: 125.0mL<br/>
Solution 3: 25.00mL<br/>
Solution 4: 1000.mL</p>
<p><mathjax>#"mols Solution 4" = 1000.mL * 0.00383M = color(teal)(0.00383 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 3" = color(teal)(0.00383 mols)/(25.00mL) = color(purple)(0.1532M)#</mathjax></p>
<p><mathjax>#"mols Solution 2" = color(purple)(0.1532M) * 125.0mL = color(orchid)(0.01915 mols)#</mathjax></p>
<p><mathjax>#"Molarity Solution 1" = color(orchid)(0.01915 mols)/(15.00mL) = color(salmon)(1.276666667M)#</mathjax></p>
<p><mathjax>#"Molarity of original solution" = color(salmon)(1.277M)#</mathjax></p></div>
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</article> | A 15.00 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution were then diluted to 1.000 x 10^3 mL. The concentration of the final solution is 0.00383 M. What is the molarity of the original solution? | null |
1,363 | ac0d8aa4-6ddd-11ea-b686-ccda262736ce | https://socratic.org/questions/what-is-the-percent-concentration-of-sodium-chloride-in-normal-saline | 0.90% | start physical_unit 6 10 molarity_percent none qc_end substance 6 7 qc_end end | [{"type":"physical unit","value":"Percent concentration [OF] sodium chloride in normal saline"}] | [{"type":"physical unit","value":"0.90%"}] | [{"type":"substance name","value":"Sodium chloride"}] | <h1 class="questionTitle" itemprop="name">What is the percent concentration of sodium chloride in normal saline?</h1> | null | 0.90% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is prepared by dissolving 9.0 g (154 mmol) of sodium chloride in water to a total volume of 1000 mL.</p>
<p>This means that normal saline solution contains <mathjax>#"154 mmol"//"L of Na"^+#</mathjax> ions and <mathjax>#"154 mmol"//"L of Cl"^"-"#</mathjax> ions. </p>
<blockquote></blockquote>
<p>Normal saline has many uses:</p>
<p><strong>Normal saline for injection</strong></p>
<p><img alt="Injection" src="https://useruploads.socratic.org/XOdcq08xSVuG3o3FrWv2_Injection.jpg"/><br/>
(from <a href="http://medimart.com" rel="nofollow" target="_blank">medimart.com</a>)</p>
<p>Normal saline for injection is used in medicine because it is isotonic with body fluids. This means that it will not cause fluid overloading or dehydration.</p>
<p>It maintains the concentration of sodium and chloride ions that the body requires.</p>
<p>It is also a stable medium for delivering most intravenous medications without incompatibility issues.</p>
<blockquote></blockquote>
<p><strong>Normal saline for irrigation</strong></p>
<p><img alt="Irrigation" src="https://useruploads.socratic.org/pFsqLxuiSOW4spTPQfK3_Irrigation.jpg"/> <br/>
(from <a href="http://www.guardianemsproducts.com" rel="nofollow" target="_blank">www.guardianemsproducts.com</a>)</p>
<p>Normal saline for irrigation is used for flushing wounds and skin abrasions, because it does not burn or sting when applied.</p>
<blockquote></blockquote>
<p><strong>Nasal Drops</strong></p>
<p><img alt="Nasal Drops" src="https://useruploads.socratic.org/ejz3AHeoREmsELW325LT_Nasal%20Drops.jpg"/> <br/>
(from <a href="http://betadinesolution.net" rel="nofollow" target="_blank">betadinesolution.net</a>)</p>
<p>Saline nasal washes soften and loosen the mucus, making it easier to wash out and clear the nasal passages.</p>
<blockquote></blockquote>
<p><strong>Eye drops</strong></p>
<p><img alt="Eye drops" src="https://useruploads.socratic.org/lFhcBjmOQrqfiT57HAqL_Eye%20Drops.jpg"/> <br/>
(from <a href="http://www.rakuten.com" rel="nofollow" target="_blank">www.rakuten.com</a>.my)</p>
<p>Eye drops are saline solutions that are used to administer many different drugs to the eye.</p>
<p>Eye drops with no medications simply lubricate the eye and replace tears.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Normal saline that is used in medicine has a concentration of 0.90% w/v of <mathjax>#"Na"#</mathjax>Cl in water. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is prepared by dissolving 9.0 g (154 mmol) of sodium chloride in water to a total volume of 1000 mL.</p>
<p>This means that normal saline solution contains <mathjax>#"154 mmol"//"L of Na"^+#</mathjax> ions and <mathjax>#"154 mmol"//"L of Cl"^"-"#</mathjax> ions. </p>
<blockquote></blockquote>
<p>Normal saline has many uses:</p>
<p><strong>Normal saline for injection</strong></p>
<p><img alt="Injection" src="https://useruploads.socratic.org/XOdcq08xSVuG3o3FrWv2_Injection.jpg"/><br/>
(from <a href="http://medimart.com" rel="nofollow" target="_blank">medimart.com</a>)</p>
<p>Normal saline for injection is used in medicine because it is isotonic with body fluids. This means that it will not cause fluid overloading or dehydration.</p>
<p>It maintains the concentration of sodium and chloride ions that the body requires.</p>
<p>It is also a stable medium for delivering most intravenous medications without incompatibility issues.</p>
<blockquote></blockquote>
<p><strong>Normal saline for irrigation</strong></p>
<p><img alt="Irrigation" src="https://useruploads.socratic.org/pFsqLxuiSOW4spTPQfK3_Irrigation.jpg"/> <br/>
(from <a href="http://www.guardianemsproducts.com" rel="nofollow" target="_blank">www.guardianemsproducts.com</a>)</p>
<p>Normal saline for irrigation is used for flushing wounds and skin abrasions, because it does not burn or sting when applied.</p>
<blockquote></blockquote>
<p><strong>Nasal Drops</strong></p>
<p><img alt="Nasal Drops" src="https://useruploads.socratic.org/ejz3AHeoREmsELW325LT_Nasal%20Drops.jpg"/> <br/>
(from <a href="http://betadinesolution.net" rel="nofollow" target="_blank">betadinesolution.net</a>)</p>
<p>Saline nasal washes soften and loosen the mucus, making it easier to wash out and clear the nasal passages.</p>
<blockquote></blockquote>
<p><strong>Eye drops</strong></p>
<p><img alt="Eye drops" src="https://useruploads.socratic.org/lFhcBjmOQrqfiT57HAqL_Eye%20Drops.jpg"/> <br/>
(from <a href="http://www.rakuten.com" rel="nofollow" target="_blank">www.rakuten.com</a>.my)</p>
<p>Eye drops are saline solutions that are used to administer many different drugs to the eye.</p>
<p>Eye drops with no medications simply lubricate the eye and replace tears.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the percent concentration of sodium chloride in normal saline?</h1>
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<div class="markdown"><p>Normal saline that is used in medicine has a concentration of 0.90% w/v of <mathjax>#"Na"#</mathjax>Cl in water. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is prepared by dissolving 9.0 g (154 mmol) of sodium chloride in water to a total volume of 1000 mL.</p>
<p>This means that normal saline solution contains <mathjax>#"154 mmol"//"L of Na"^+#</mathjax> ions and <mathjax>#"154 mmol"//"L of Cl"^"-"#</mathjax> ions. </p>
<blockquote></blockquote>
<p>Normal saline has many uses:</p>
<p><strong>Normal saline for injection</strong></p>
<p><img alt="Injection" src="https://useruploads.socratic.org/XOdcq08xSVuG3o3FrWv2_Injection.jpg"/><br/>
(from <a href="http://medimart.com" rel="nofollow" target="_blank">medimart.com</a>)</p>
<p>Normal saline for injection is used in medicine because it is isotonic with body fluids. This means that it will not cause fluid overloading or dehydration.</p>
<p>It maintains the concentration of sodium and chloride ions that the body requires.</p>
<p>It is also a stable medium for delivering most intravenous medications without incompatibility issues.</p>
<blockquote></blockquote>
<p><strong>Normal saline for irrigation</strong></p>
<p><img alt="Irrigation" src="https://useruploads.socratic.org/pFsqLxuiSOW4spTPQfK3_Irrigation.jpg"/> <br/>
(from <a href="http://www.guardianemsproducts.com" rel="nofollow" target="_blank">www.guardianemsproducts.com</a>)</p>
<p>Normal saline for irrigation is used for flushing wounds and skin abrasions, because it does not burn or sting when applied.</p>
<blockquote></blockquote>
<p><strong>Nasal Drops</strong></p>
<p><img alt="Nasal Drops" src="https://useruploads.socratic.org/ejz3AHeoREmsELW325LT_Nasal%20Drops.jpg"/> <br/>
(from <a href="http://betadinesolution.net" rel="nofollow" target="_blank">betadinesolution.net</a>)</p>
<p>Saline nasal washes soften and loosen the mucus, making it easier to wash out and clear the nasal passages.</p>
<blockquote></blockquote>
<p><strong>Eye drops</strong></p>
<p><img alt="Eye drops" src="https://useruploads.socratic.org/lFhcBjmOQrqfiT57HAqL_Eye%20Drops.jpg"/> <br/>
(from <a href="http://www.rakuten.com" rel="nofollow" target="_blank">www.rakuten.com</a>.my)</p>
<p>Eye drops are saline solutions that are used to administer many different drugs to the eye.</p>
<p>Eye drops with no medications simply lubricate the eye and replace tears.</p></div>
</div>
</div>
</div>
</div>
</div>
<div>
<a href="https://socratic.org/answers/103139" itemprop="url">Answer link</a>
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</article> | What is the percent concentration of sodium chloride in normal saline? | null |
1,364 | a919287a-6ddd-11ea-831e-ccda262736ce | https://socratic.org/questions/a-sample-of-a-pure-substance-with-a-mass-of-0-963g-contains-4-84x1-3-moles-what- | 200.21 amu | start physical_unit 23 24 formula_weight amu qc_end physical_unit 4 5 10 11 mass qc_end physical_unit 4 5 13 16 mole qc_end end | [{"type":"physical unit","value":"Formula weight [OF] the substance [IN] amu"}] | [{"type":"physical unit","value":"200.21 amu"}] | [{"type":"physical unit","value":"Mass [OF] pure substance sample [=] \\pu{0.963 g}"},{"type":"physical unit","value":"Mole [OF] pure substance sample [=] \\pu{4.84 x 10^(-3) moles}"}] | <h1 class="questionTitle" itemprop="name">A sample of a pure substance with a mass of 0.963g contains 4.84x1-^-3 moles. What is the formula weight of the substance comprising the sample?</h1> | null | 200.21 amu | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the <em>formula weight</em> of a pure substance, knowing that <mathjax>#4.81xx10^-3#</mathjax> <mathjax>#"mol"#</mathjax> has a mass of <mathjax>#0.963#</mathjax> <mathjax>#"g"#</mathjax>.</p>
<p>To calculate the<em> formula weight</em> (<em>relative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a></em>, <em>atomic weight</em>, etc.; they mean the same thing), we'll first calculate the *<em>molar mass</em> of the substance:</p>
<p><mathjax>#"molar mass" = "mass"/"mol"#</mathjax> <mathjax>#= (0.963color(white)(l)"g")/(4.81xx10^-3color(white)(l)"mol")#</mathjax></p>
<p><mathjax>#= color(red)(200.#</mathjax> <mathjax>#color(red)("g/mol"#</mathjax></p>
<p>Remember the important conecpt of the <strong>mole</strong>:</p>
<blockquote>
<p><strong><em>The mass in grams of one mole of a substance is equal to the mass in atomic mass units of one individual unit of that substance.</em></strong></p>
</blockquote>
<p>Therefore, we have</p>
<p><mathjax>#overbrace(200. color(white)(l)"g/mol")^"mass of one mole" = overbrace(color(blue)(200. color(white)(l)"amu"))^"mass of one unit"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#200.#</mathjax> <mathjax>#"amu"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the <em>formula weight</em> of a pure substance, knowing that <mathjax>#4.81xx10^-3#</mathjax> <mathjax>#"mol"#</mathjax> has a mass of <mathjax>#0.963#</mathjax> <mathjax>#"g"#</mathjax>.</p>
<p>To calculate the<em> formula weight</em> (<em>relative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a></em>, <em>atomic weight</em>, etc.; they mean the same thing), we'll first calculate the *<em>molar mass</em> of the substance:</p>
<p><mathjax>#"molar mass" = "mass"/"mol"#</mathjax> <mathjax>#= (0.963color(white)(l)"g")/(4.81xx10^-3color(white)(l)"mol")#</mathjax></p>
<p><mathjax>#= color(red)(200.#</mathjax> <mathjax>#color(red)("g/mol"#</mathjax></p>
<p>Remember the important conecpt of the <strong>mole</strong>:</p>
<blockquote>
<p><strong><em>The mass in grams of one mole of a substance is equal to the mass in atomic mass units of one individual unit of that substance.</em></strong></p>
</blockquote>
<p>Therefore, we have</p>
<p><mathjax>#overbrace(200. color(white)(l)"g/mol")^"mass of one mole" = overbrace(color(blue)(200. color(white)(l)"amu"))^"mass of one unit"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of a pure substance with a mass of 0.963g contains 4.84x1-^-3 moles. What is the formula weight of the substance comprising the sample?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#200.#</mathjax> <mathjax>#"amu"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We're asked to calculate the <em>formula weight</em> of a pure substance, knowing that <mathjax>#4.81xx10^-3#</mathjax> <mathjax>#"mol"#</mathjax> has a mass of <mathjax>#0.963#</mathjax> <mathjax>#"g"#</mathjax>.</p>
<p>To calculate the<em> formula weight</em> (<em>relative <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a></em>, <em>atomic weight</em>, etc.; they mean the same thing), we'll first calculate the *<em>molar mass</em> of the substance:</p>
<p><mathjax>#"molar mass" = "mass"/"mol"#</mathjax> <mathjax>#= (0.963color(white)(l)"g")/(4.81xx10^-3color(white)(l)"mol")#</mathjax></p>
<p><mathjax>#= color(red)(200.#</mathjax> <mathjax>#color(red)("g/mol"#</mathjax></p>
<p>Remember the important conecpt of the <strong>mole</strong>:</p>
<blockquote>
<p><strong><em>The mass in grams of one mole of a substance is equal to the mass in atomic mass units of one individual unit of that substance.</em></strong></p>
</blockquote>
<p>Therefore, we have</p>
<p><mathjax>#overbrace(200. color(white)(l)"g/mol")^"mass of one mole" = overbrace(color(blue)(200. color(white)(l)"amu"))^"mass of one unit"#</mathjax></p></div>
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</article> | A sample of a pure substance with a mass of 0.963g contains 4.84x1-^-3 moles. What is the formula weight of the substance comprising the sample? | null |
1,365 | a8722d39-6ddd-11ea-a28b-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-equation-for-this-redox-reaction-using-the-smallest- | 5 ClO^- + I2 + H2O -> 5 Cl- + 2 IO3^- + 2 H^+ | start chemical_equation qc_end chemical_equation 24 30 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] this redox reaction"}] | [{"type":"chemical equation","value":"5 ClO^- + I2 + H2O -> 5 Cl- + 2 IO3^- + 2 H^+"}] | [{"type":"chemical equation","value":"ClO^- + I2 -> Cl^- + IO3^-"},{"type":"other","value":"The reaction occurs in aqueous acidic solution."}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced equation for this redox reaction using the smallest whole number coefficients?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>The reaction occurs in aqueous acidic solution.</p>
<p><mathjax>#"ClO"^(-)+"I"_2->"Cl"^(-)+"IO"_3^-#</mathjax></p></div>
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</div> | 5 ClO^- + I2 + H2O -> 5 Cl- + 2 IO3^- + 2 H^+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hypochlorite is REDUCED.....<mathjax>#stackrel(+I)Clrarrstackrel(-I)Cl#</mathjax></p>
<p><mathjax>#ClO^(-) + 2H^+ + 2e^(-) rarr Cl^(-) +H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so tick.....</p>
<p>And elemental iodine is oxidized.....<mathjax>#stackrel(0)I_2rarrstackrel(+V)I#</mathjax></p>
<p><mathjax>#1/2I_2 + 3H_2O rarr IO_3^(-) +6H^(+) + 5e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced, so double tick.....</p>
<p>And so we take <mathjax>#5xx(i)+2xx(ii)#</mathjax>.........</p>
<p><mathjax>#I_2 + 5ClO^(-) +H_2O rarr 2IO_3^(-) +5Cl^(-) +2H^(+) #</mathjax></p>
<p>Is this balanced with respect to mass and charge? If it is not, then we may reject it out of hand!</p></div>
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<div class="markdown"><p>Well try.....</p>
<p><mathjax>#I_2 + 5ClO^(-) +H_2O rarr 2IO_3^(-) +5Cl^(-) +2H^(+) #</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hypochlorite is REDUCED.....<mathjax>#stackrel(+I)Clrarrstackrel(-I)Cl#</mathjax></p>
<p><mathjax>#ClO^(-) + 2H^+ + 2e^(-) rarr Cl^(-) +H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so tick.....</p>
<p>And elemental iodine is oxidized.....<mathjax>#stackrel(0)I_2rarrstackrel(+V)I#</mathjax></p>
<p><mathjax>#1/2I_2 + 3H_2O rarr IO_3^(-) +6H^(+) + 5e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced, so double tick.....</p>
<p>And so we take <mathjax>#5xx(i)+2xx(ii)#</mathjax>.........</p>
<p><mathjax>#I_2 + 5ClO^(-) +H_2O rarr 2IO_3^(-) +5Cl^(-) +2H^(+) #</mathjax></p>
<p>Is this balanced with respect to mass and charge? If it is not, then we may reject it out of hand!</p></div>
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<h1 class="questionTitle" itemprop="name">How do you write a balanced equation for this redox reaction using the smallest whole number coefficients?</h1>
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<div class="markdown"><p>The reaction occurs in aqueous acidic solution.</p>
<p><mathjax>#"ClO"^(-)+"I"_2->"Cl"^(-)+"IO"_3^-#</mathjax></p></div>
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<div class="markdown"><p>Well try.....</p>
<p><mathjax>#I_2 + 5ClO^(-) +H_2O rarr 2IO_3^(-) +5Cl^(-) +2H^(+) #</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Hypochlorite is REDUCED.....<mathjax>#stackrel(+I)Clrarrstackrel(-I)Cl#</mathjax></p>
<p><mathjax>#ClO^(-) + 2H^+ + 2e^(-) rarr Cl^(-) +H_2O#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so tick.....</p>
<p>And elemental iodine is oxidized.....<mathjax>#stackrel(0)I_2rarrstackrel(+V)I#</mathjax></p>
<p><mathjax>#1/2I_2 + 3H_2O rarr IO_3^(-) +6H^(+) + 5e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Charge and mass are balanced, so double tick.....</p>
<p>And so we take <mathjax>#5xx(i)+2xx(ii)#</mathjax>.........</p>
<p><mathjax>#I_2 + 5ClO^(-) +H_2O rarr 2IO_3^(-) +5Cl^(-) +2H^(+) #</mathjax></p>
<p>Is this balanced with respect to mass and charge? If it is not, then we may reject it out of hand!</p></div>
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</article> | How do you write a balanced equation for this redox reaction using the smallest whole number coefficients? |
The reaction occurs in aqueous acidic solution.
#"ClO"^(-)+"I"_2->"Cl"^(-)+"IO"_3^-#
|
1,366 | a8bc35e4-6ddd-11ea-8078-ccda262736ce | https://socratic.org/questions/596a30b911ef6b0e551fe52f | 1929 amperes | start physical_unit 2 4 electric_current amperes qc_end physical_unit 2 4 12 13 time qc_end physical_unit 2 4 21 22 temperature qc_end end | [{"type":"physical unit","value":"Electric current [OF] hydrogen/oxygen fuel cell [IN] amperes"}] | [{"type":"physical unit","value":"1929 amperes"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{67.2 liters}"},{"type":"physical unit","value":"Time consumed [OF] hydrogen/oxygen fuel cell [=] \\pu{5 minutes}"},{"type":"physical unit","value":"Temperature [OF] hydrogen/oxygen fuel cell [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Pressure [OF] hydrogen/oxygen fuel cell [=] \\pu{1 atmoshere}"}] | <h1 class="questionTitle" itemprop="name">In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ?</h1> | null | 1929 amperes | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")#</mathjax></p>
<p>Use the factor that converts Liters to Moles:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")(1"mol H"_2)/(22.4"L H"_2"@S.T.P.")#</mathjax></p>
<p>Please observe how the units cancel:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1"mol H"_2)/(22.4cancel("L H"_2"@S.T.P."))#</mathjax></p>
<p>We know that 1 mole of H<mathjax>#"_2#</mathjax> produces 2 moles of electrons:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2"mol e"^-)/(1cancel("mol H"_2))#</mathjax></p>
<p>Use <a href="https://en.wikipedia.org/wiki/Avogadro_constant" rel="nofollow">Avagadro's Constant</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23"e"^-)/(1cancel("mol e"^-))#</mathjax></p>
<p>Next we use the definition of a <a href="https://en.wikipedia.org/wiki/Coulomb" rel="nofollow">Coulomb</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))#</mathjax><br/>
<mathjax>#(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))#</mathjax></p>
<p>Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5cancel("min"))(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))#</mathjax><mathjax>#(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))(1cancel("min"))/(60"s")#</mathjax></p>
<p>Do the multiplication and division and the answer will be in Amperes:</p>
<p><mathjax>#1929"Amperes"#</mathjax></p>
<p>This seems reasonable, because the consumption rate of Hydrogen is quite high and the potential of a single cell is between 0.6 and 0.7 volts, thereby, making the output power about 1.3 kW. To practically do this, one would stack the cells in a series-parallel configuration.</p></div>
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<div class="markdown"><p><mathjax>#1929"Amperes"#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")#</mathjax></p>
<p>Use the factor that converts Liters to Moles:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")(1"mol H"_2)/(22.4"L H"_2"@S.T.P.")#</mathjax></p>
<p>Please observe how the units cancel:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1"mol H"_2)/(22.4cancel("L H"_2"@S.T.P."))#</mathjax></p>
<p>We know that 1 mole of H<mathjax>#"_2#</mathjax> produces 2 moles of electrons:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2"mol e"^-)/(1cancel("mol H"_2))#</mathjax></p>
<p>Use <a href="https://en.wikipedia.org/wiki/Avogadro_constant" rel="nofollow">Avagadro's Constant</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23"e"^-)/(1cancel("mol e"^-))#</mathjax></p>
<p>Next we use the definition of a <a href="https://en.wikipedia.org/wiki/Coulomb" rel="nofollow">Coulomb</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))#</mathjax><br/>
<mathjax>#(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))#</mathjax></p>
<p>Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5cancel("min"))(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))#</mathjax><mathjax>#(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))(1cancel("min"))/(60"s")#</mathjax></p>
<p>Do the multiplication and division and the answer will be in Amperes:</p>
<p><mathjax>#1929"Amperes"#</mathjax></p>
<p>This seems reasonable, because the consumption rate of Hydrogen is quite high and the potential of a single cell is between 0.6 and 0.7 volts, thereby, making the output power about 1.3 kW. To practically do this, one would stack the cells in a series-parallel configuration.</p></div>
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<h1 class="questionTitle" itemprop="name">In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ?</h1>
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Douglas K.
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<div class="markdown"><p><mathjax>#1929"Amperes"#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We are given:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")#</mathjax></p>
<p>Use the factor that converts Liters to Moles:</p>
<p><mathjax>#(67.2"L H"_2"@S.T.P.")/(5"min")(1"mol H"_2)/(22.4"L H"_2"@S.T.P.")#</mathjax></p>
<p>Please observe how the units cancel:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1"mol H"_2)/(22.4cancel("L H"_2"@S.T.P."))#</mathjax></p>
<p>We know that 1 mole of H<mathjax>#"_2#</mathjax> produces 2 moles of electrons:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2"mol e"^-)/(1cancel("mol H"_2))#</mathjax></p>
<p>Use <a href="https://en.wikipedia.org/wiki/Avogadro_constant" rel="nofollow">Avagadro's Constant</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23"e"^-)/(1cancel("mol e"^-))#</mathjax></p>
<p>Next we use the definition of a <a href="https://en.wikipedia.org/wiki/Coulomb" rel="nofollow">Coulomb</a> :</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))#</mathjax><br/>
<mathjax>#(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))#</mathjax></p>
<p>Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:</p>
<p><mathjax>#(67.2cancel("L H"_2"@S.T.P."))/(5cancel("min"))(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))#</mathjax><mathjax>#(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))(1cancel("min"))/(60"s")#</mathjax></p>
<p>Do the multiplication and division and the answer will be in Amperes:</p>
<p><mathjax>#1929"Amperes"#</mathjax></p>
<p>This seems reasonable, because the consumption rate of Hydrogen is quite high and the potential of a single cell is between 0.6 and 0.7 volts, thereby, making the output power about 1.3 kW. To practically do this, one would stack the cells in a series-parallel configuration.</p></div>
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Michael
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<div class="markdown"><p><mathjax>#sf(1930color(white)(x)A)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrogen is oxidised at the anode:</p>
<p><mathjax>#sf(H_(2(g))rarr2H_((aq))^(+)+2e)#</mathjax></p>
<p>This tells us that 1 mole of <mathjax>#sf(H_2)#</mathjax> produces 2 moles of electrons.</p>
<p>So <mathjax>#sf(22.4)#</mathjax> litres produce 2 moles of electrons.</p>
<p>So <mathjax>#sf(67.2)#</mathjax> litres produce <mathjax>#sf(2xx67.2/22.4=6)#</mathjax> moles of electrons.</p>
<p>The charge on 1 mole of electrons is given by the Faraday Constant and is equal to <mathjax>#sf(9.65xx10^4color(white)(x)"C/mol")#</mathjax>.</p>
<p>So the total charge <strong>Q</strong> produced is given by:</p>
<p><mathjax>#sf(Q=6xx9.65xx10^(4)=57.9xx10^(4)color(white)(x)C)#</mathjax></p>
<p>This is produced in 5 minutes which = 5 x 60 = 300 s.</p>
<p>Electric current <strong>I</strong> is the rate of flow of charge as given by:</p>
<p><mathjax>#sf(I=Q/t)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(I=(57.9xx10^(4))/(300)=1930color(white)(x)A)#</mathjax></p></div>
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</article> | In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ? | null |
1,367 | ab68e724-6ddd-11ea-afc1-ccda262736ce | https://socratic.org/questions/the-temperature-of-a-sample-of-water-changes-from-10-c-to-20-c-when-the-sample-a | 10 g | start physical_unit 15 16 mass g qc_end physical_unit 4 6 9 10 temperature qc_end physical_unit 4 6 12 13 temperature qc_end physical_unit 4 6 18 19 heat_energy qc_end end | [{"type":"physical unit","value":"Mass [OF] the sample [IN] g"}] | [{"type":"physical unit","value":"10 g"}] | [{"type":"physical unit","value":"Temperature1 [OF] water sample [=] \\pu{10 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water sample [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Absorbed heat [OF] water sample [=] \\pu{418 joules}"}] | <h1 class="questionTitle" itemprop="name">The temperature of a sample of water changes from 10° C to 20°C when the sample absorbs 418 joules of heat. What is the mass of the sample?</h1> | null | 10 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, just by inspecting the values given, you can say that the answer will be <mathjax>#"10 g"#</mathjax>.</p>
<p>Now, here's what that is the case. </p>
<p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of approximately <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, how much heat would be required to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>?</p>
<p>Well, you'd need <mathjax>#"4.18 J"#</mathjax> to increase it by <mathjax>#1^@"C"#</mathjax>, <strong>another</strong> <mathjax>#"4.18 J"#</mathjax> to increase it by <strong>another</strong> <mathjax>#1^@"C"#</mathjax>, and so on. This means that you'd need </p>
<blockquote>
<p><mathjax>#"4.18 J" xx 10 = "41.8 J"#</mathjax></p>
</blockquote>
<p>to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>. </p>
<p>Now look at the value given to you. If you need <mathjax>#"41.8 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>, what mass of water would require <mathjax>#10#</mathjax> <strong>times as much heat</strong> to increase its temperature by <mathjax>#10^@"C"#</mathjax>? </p>
<blockquote>
<p><mathjax>#"1 g" xx 10 = "10 g"#</mathjax></p>
</blockquote>
<p>And that's your answer. </p>
<p>Mathematically, you can calculate this by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#418 color(red)(cancel(color(black)("J"))) = m * 4.18color(red)(cancel(color(black)("J")))/("g" color(red)(cancel(color(black)(""^@"C")))) * (20 - 10)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#m = 418/(4.18 * 10) = "10 g"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"10 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, just by inspecting the values given, you can say that the answer will be <mathjax>#"10 g"#</mathjax>.</p>
<p>Now, here's what that is the case. </p>
<p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of approximately <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, how much heat would be required to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>?</p>
<p>Well, you'd need <mathjax>#"4.18 J"#</mathjax> to increase it by <mathjax>#1^@"C"#</mathjax>, <strong>another</strong> <mathjax>#"4.18 J"#</mathjax> to increase it by <strong>another</strong> <mathjax>#1^@"C"#</mathjax>, and so on. This means that you'd need </p>
<blockquote>
<p><mathjax>#"4.18 J" xx 10 = "41.8 J"#</mathjax></p>
</blockquote>
<p>to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>. </p>
<p>Now look at the value given to you. If you need <mathjax>#"41.8 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>, what mass of water would require <mathjax>#10#</mathjax> <strong>times as much heat</strong> to increase its temperature by <mathjax>#10^@"C"#</mathjax>? </p>
<blockquote>
<p><mathjax>#"1 g" xx 10 = "10 g"#</mathjax></p>
</blockquote>
<p>And that's your answer. </p>
<p>Mathematically, you can calculate this by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#418 color(red)(cancel(color(black)("J"))) = m * 4.18color(red)(cancel(color(black)("J")))/("g" color(red)(cancel(color(black)(""^@"C")))) * (20 - 10)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#m = 418/(4.18 * 10) = "10 g"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The temperature of a sample of water changes from 10° C to 20°C when the sample absorbs 418 joules of heat. What is the mass of the sample?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"10 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Right from the start, just by inspecting the values given, you can say that the answer will be <mathjax>#"10 g"#</mathjax>.</p>
<p>Now, here's what that is the case. </p>
<p>As you know, a substance's <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> tells you how much heat is needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>Water has a <em><a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of approximately <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>. This tells you that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide <mathjax>#"4.18 J"#</mathjax> of heat. </p>
<p>Now, how much heat would be required to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>?</p>
<p>Well, you'd need <mathjax>#"4.18 J"#</mathjax> to increase it by <mathjax>#1^@"C"#</mathjax>, <strong>another</strong> <mathjax>#"4.18 J"#</mathjax> to increase it by <strong>another</strong> <mathjax>#1^@"C"#</mathjax>, and so on. This means that you'd need </p>
<blockquote>
<p><mathjax>#"4.18 J" xx 10 = "41.8 J"#</mathjax></p>
</blockquote>
<p>to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>. </p>
<p>Now look at the value given to you. If you need <mathjax>#"41.8 J"#</mathjax> to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#10^@"C"#</mathjax>, what mass of water would require <mathjax>#10#</mathjax> <strong>times as much heat</strong> to increase its temperature by <mathjax>#10^@"C"#</mathjax>? </p>
<blockquote>
<p><mathjax>#"1 g" xx 10 = "10 g"#</mathjax></p>
</blockquote>
<p>And that's your answer. </p>
<p>Mathematically, you can calculate this by using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#418 color(red)(cancel(color(black)("J"))) = m * 4.18color(red)(cancel(color(black)("J")))/("g" color(red)(cancel(color(black)(""^@"C")))) * (20 - 10)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#m = 418/(4.18 * 10) = "10 g"#</mathjax></p>
</blockquote></div>
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</article> | The temperature of a sample of water changes from 10° C to 20°C when the sample absorbs 418 joules of heat. What is the mass of the sample? | null |
1,368 | aae911d1-6ddd-11ea-9ac6-ccda262736ce | https://socratic.org/questions/how-many-cl-ions-are-present-in-2-00-mol-of-kcl | 1.20 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 10 10 7 8 mole qc_end end | [{"type":"physical unit","value":"Number [OF] Cl- ions"}] | [{"type":"physical unit","value":"1.20 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{2.00 mol}"}] | <h1 class="questionTitle" itemprop="name">How many #Cl^-# ions are present in 2.00 mol of #KCl#?</h1> | null | 1.20 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The question specified 2 moles of potassium chloride. Clearly, there are 2 moles of chloride ions. How many moles of chloride ions in 95.2 g <mathjax>#MgCl_2#</mathjax>?</p></div>
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<div class="markdown"><p>In one 1 mole of potassium chloride. <mathjax>#KCl#</mathjax>, there are clearly 1 mole of potassium ions, and 1 mole of chloride ions. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The question specified 2 moles of potassium chloride. Clearly, there are 2 moles of chloride ions. How many moles of chloride ions in 95.2 g <mathjax>#MgCl_2#</mathjax>?</p></div>
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<div class="markdown"><p>In one 1 mole of potassium chloride. <mathjax>#KCl#</mathjax>, there are clearly 1 mole of potassium ions, and 1 mole of chloride ions. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The question specified 2 moles of potassium chloride. Clearly, there are 2 moles of chloride ions. How many moles of chloride ions in 95.2 g <mathjax>#MgCl_2#</mathjax>?</p></div>
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</article> | How many #Cl^-# ions are present in 2.00 mol of #KCl#? | null |
1,369 | ac3bc952-6ddd-11ea-8406-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-atm-of-3-5-moles-of-helium-at-50-c-in-a-rigid-container-who | 2.56 atm | start physical_unit 9 9 pressure atm qc_end physical_unit 9 9 6 7 mole qc_end physical_unit 9 9 11 12 temperature qc_end physical_unit 9 9 20 21 volume qc_end end | [{"type":"physical unit","value":"Pressure [OF] helium [IN] atm"}] | [{"type":"physical unit","value":"2.56 atm"}] | [{"type":"physical unit","value":"Mole [OF] helium [=] \\pu{3.5 moles}"},{"type":"physical unit","value":"Temperature [OF] helium [=] \\pu{-50 ℃}"},{"type":"physical unit","value":"Volume [OF] helium [=] \\pu{25.0 L}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure (atm) of 3.5 moles of helium at -50 °C in a rigid container whose volume is 25.0 L?</h1> | null | 2.56 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are given the number of moles, temperature, and volume of helium, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a> to determine the pressure.</p>
<p><img alt="slideplayer.com" src="https://useruploads.socratic.org/gHs2ECdiSxaNBiNU1nC8_slide_3.jpg"/> </p>
<ul>
<li>P can have units of atm, depending on the units of the universal gas constant</li>
<li>V must have units of liters</li>
<li>n should have units of moles</li>
<li>R has a value of 0.0821 with units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></li>
<li>T has units of Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the pressure of helium. Our known variables are n,V,R, and T. </p>
<p>The only issue is that we have to convert the temperature from centigrade to Kelvins. We can do that using the following conversion:</p>
<p><img alt="ww2010.atmos.uiuc.edu" src="https://useruploads.socratic.org/c8jNkioTcaJyiVYjHD0A_kel-cel.gif"/> </p>
<p>Therefore, <mathjax>#-50^(o)#</mathjax>C + 273 = 223K</p>
<p>Now we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to solve for P:</p>
<p><mathjax>#P=(nxxRxxT)/V#</mathjax></p>
<p><mathjax>#P = (3.5cancel"mol"xx(0.0821cancel"L"xx(atm)/cancel"mol"xxcancelK)xx223cancelK)/(25.0cancelL)#</mathjax></p>
<p><mathjax>#P = 2.56 atm#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Helium has a pressure of <strong>2.56 atm.</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are given the number of moles, temperature, and volume of helium, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a> to determine the pressure.</p>
<p><img alt="slideplayer.com" src="https://useruploads.socratic.org/gHs2ECdiSxaNBiNU1nC8_slide_3.jpg"/> </p>
<ul>
<li>P can have units of atm, depending on the units of the universal gas constant</li>
<li>V must have units of liters</li>
<li>n should have units of moles</li>
<li>R has a value of 0.0821 with units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></li>
<li>T has units of Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the pressure of helium. Our known variables are n,V,R, and T. </p>
<p>The only issue is that we have to convert the temperature from centigrade to Kelvins. We can do that using the following conversion:</p>
<p><img alt="ww2010.atmos.uiuc.edu" src="https://useruploads.socratic.org/c8jNkioTcaJyiVYjHD0A_kel-cel.gif"/> </p>
<p>Therefore, <mathjax>#-50^(o)#</mathjax>C + 273 = 223K</p>
<p>Now we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to solve for P:</p>
<p><mathjax>#P=(nxxRxxT)/V#</mathjax></p>
<p><mathjax>#P = (3.5cancel"mol"xx(0.0821cancel"L"xx(atm)/cancel"mol"xxcancelK)xx223cancelK)/(25.0cancelL)#</mathjax></p>
<p><mathjax>#P = 2.56 atm#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pressure (atm) of 3.5 moles of helium at -50 °C in a rigid container whose volume is 25.0 L?</h1>
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<div class="markdown"><p>Helium has a pressure of <strong>2.56 atm.</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are given the number of moles, temperature, and volume of helium, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a> to determine the pressure.</p>
<p><img alt="slideplayer.com" src="https://useruploads.socratic.org/gHs2ECdiSxaNBiNU1nC8_slide_3.jpg"/> </p>
<ul>
<li>P can have units of atm, depending on the units of the universal gas constant</li>
<li>V must have units of liters</li>
<li>n should have units of moles</li>
<li>R has a value of 0.0821 with units of <mathjax>#(Lxxatm)/ (molxxK)#</mathjax></li>
<li>T has units of Kelvins.</li>
</ul>
<p>Next, list your known and unknown variables. Our only unknown is the pressure of helium. Our known variables are n,V,R, and T. </p>
<p>The only issue is that we have to convert the temperature from centigrade to Kelvins. We can do that using the following conversion:</p>
<p><img alt="ww2010.atmos.uiuc.edu" src="https://useruploads.socratic.org/c8jNkioTcaJyiVYjHD0A_kel-cel.gif"/> </p>
<p>Therefore, <mathjax>#-50^(o)#</mathjax>C + 273 = 223K</p>
<p>Now we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to solve for P:</p>
<p><mathjax>#P=(nxxRxxT)/V#</mathjax></p>
<p><mathjax>#P = (3.5cancel"mol"xx(0.0821cancel"L"xx(atm)/cancel"mol"xxcancelK)xx223cancelK)/(25.0cancelL)#</mathjax></p>
<p><mathjax>#P = 2.56 atm#</mathjax></p></div>
</div>
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</article> | What is the pressure (atm) of 3.5 moles of helium at -50 °C in a rigid container whose volume is 25.0 L? | null |
1,370 | aa90b001-6ddd-11ea-8178-ccda262736ce | https://socratic.org/questions/how-do-you-convert-5-0g-of-nitrogen-dioxide-no-2-to-grams-of-ozone-gas-o-3 | 3.49 grams | start physical_unit 15 15 mass g qc_end physical_unit 9 9 4 5 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] O3 [IN] grams"}] | [{"type":"physical unit","value":"3.49 grams"}] | [{"type":"physical unit","value":"Mass [OF] NO2 [=] \\pu{5.0 g}"}] | <h1 class="questionTitle" itemprop="name">How do you convert 5.0g of nitrogen dioxide (#NO_2#) to grams of ozone gas (#O_3#)?</h1> | null | 3.49 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, first we need a stoichiometric equation:</p>
<p><mathjax>#3NO_2(g) rarr 2O_3(g) + N_2(g)#</mathjax></p>
<p>Please let me say that I do not know whether this reaction could be reliably accomplished. Certainly, we could assess the energy transfer of the reaction.</p>
<p>Given the equation, we see that 2 equiv of ozone derives from 3 equiv of nitric oxide:</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#NO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.0*g)/(46.01*g*mol^-1)=0.109*mol.#</mathjax></p>
<p>And thus with quantitative reaction we would get:</p>
<p><mathjax>#2/3xx0.109*molxx48.00*g*mol^-1=3.49*g#</mathjax> <mathjax>#"ozone gas"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>With difficulty........we would get <mathjax>#3.5*g#</mathjax> of ozone........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, first we need a stoichiometric equation:</p>
<p><mathjax>#3NO_2(g) rarr 2O_3(g) + N_2(g)#</mathjax></p>
<p>Please let me say that I do not know whether this reaction could be reliably accomplished. Certainly, we could assess the energy transfer of the reaction.</p>
<p>Given the equation, we see that 2 equiv of ozone derives from 3 equiv of nitric oxide:</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#NO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.0*g)/(46.01*g*mol^-1)=0.109*mol.#</mathjax></p>
<p>And thus with quantitative reaction we would get:</p>
<p><mathjax>#2/3xx0.109*molxx48.00*g*mol^-1=3.49*g#</mathjax> <mathjax>#"ozone gas"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you convert 5.0g of nitrogen dioxide (#NO_2#) to grams of ozone gas (#O_3#)?</h1>
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anor277
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<div class="markdown"><p>With difficulty........we would get <mathjax>#3.5*g#</mathjax> of ozone........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, first we need a stoichiometric equation:</p>
<p><mathjax>#3NO_2(g) rarr 2O_3(g) + N_2(g)#</mathjax></p>
<p>Please let me say that I do not know whether this reaction could be reliably accomplished. Certainly, we could assess the energy transfer of the reaction.</p>
<p>Given the equation, we see that 2 equiv of ozone derives from 3 equiv of nitric oxide:</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#NO_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.0*g)/(46.01*g*mol^-1)=0.109*mol.#</mathjax></p>
<p>And thus with quantitative reaction we would get:</p>
<p><mathjax>#2/3xx0.109*molxx48.00*g*mol^-1=3.49*g#</mathjax> <mathjax>#"ozone gas"#</mathjax>.</p></div>
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</article> | How do you convert 5.0g of nitrogen dioxide (#NO_2#) to grams of ozone gas (#O_3#)? | null |
1,371 | ac044f92-6ddd-11ea-9a5e-ccda262736ce | https://socratic.org/questions/my-car-has-an-internal-volume-of-2600-liters-assume-the-pressure-was-initially-7 | 850.74 mmHg | start physical_unit 1 1 pressure mmhg qc_end physical_unit 1 1 7 8 volume qc_end physical_unit 1 1 14 15 pressure qc_end physical_unit 1 1 26 27 temperature qc_end physical_unit 1 1 32 33 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the car [IN] mmHg"}] | [{"type":"physical unit","value":"850.74 mmHg"}] | [{"type":"physical unit","value":"Volume [OF] the car [=] \\pu{2600 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the car [=] \\pu{760 mmHg}"},{"type":"physical unit","value":"Temperature1 [OF] the car [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the car [=] \\pu{55 ℃}"}] | <h1 class="questionTitle" itemprop="name">My car has an internal volume of 2600 liters. Assume the pressure was initially 760 mm Hg. If the sun heats my car from a temperature of 20° C to a temperature of 55“ C, what will the pressure inside my car be? </h1> | null | 850.74 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume of the car and the number of moles of gas you have in the car <strong>remain unchanged</strong>.</p>
<p>When volume and number of moles are kept constant, pressure and temperatue have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>. </p>
<p>In simple terms, when the temperature of the gas <em>increases</em>, pressure <strong>increases</strong> as well. Likewise, when the temperature of the gas <em>decreases</em>, pressure <strong>decreases</strong> as well. </p>
<p><img alt="https://sites.google.com/site/tiyahgaslaws/gay-lussac-s-law" src="https://useruploads.socratic.org/E0jSnk9mQR2dtXMiZVav_685557087.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state</p>
<p>When plugging in your values into this equation, don't forget to convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>.</p>
<p>So, rearrange and solve for <mathjax>#P_2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#P_2 = T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = ((273.15 + 55)color(red)(cancel(color(black)("K"))))/((273.15 + 20)color(red)(cancel(color(black)("K")))) * "760 mmHg" = "850.74 mmHg"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, despite the fact that you only gave one sig fig for the initial temperature of the gas</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("850 mmHg")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/0Oq7bCSDPxE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"850 mmHg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume of the car and the number of moles of gas you have in the car <strong>remain unchanged</strong>.</p>
<p>When volume and number of moles are kept constant, pressure and temperatue have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>. </p>
<p>In simple terms, when the temperature of the gas <em>increases</em>, pressure <strong>increases</strong> as well. Likewise, when the temperature of the gas <em>decreases</em>, pressure <strong>decreases</strong> as well. </p>
<p><img alt="https://sites.google.com/site/tiyahgaslaws/gay-lussac-s-law" src="https://useruploads.socratic.org/E0jSnk9mQR2dtXMiZVav_685557087.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state</p>
<p>When plugging in your values into this equation, don't forget to convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>.</p>
<p>So, rearrange and solve for <mathjax>#P_2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#P_2 = T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = ((273.15 + 55)color(red)(cancel(color(black)("K"))))/((273.15 + 20)color(red)(cancel(color(black)("K")))) * "760 mmHg" = "850.74 mmHg"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, despite the fact that you only gave one sig fig for the initial temperature of the gas</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("850 mmHg")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/0Oq7bCSDPxE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">My car has an internal volume of 2600 liters. Assume the pressure was initially 760 mm Hg. If the sun heats my car from a temperature of 20° C to a temperature of 55“ C, what will the pressure inside my car be? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-02T01:30:25" itemprop="dateCreated">
Nov 2, 2015
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<div class="markdown"><p><mathjax>#"850 mmHg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the volume of the car and the number of moles of gas you have in the car <strong>remain unchanged</strong>.</p>
<p>When volume and number of moles are kept constant, pressure and temperatue have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>. </p>
<p>In simple terms, when the temperature of the gas <em>increases</em>, pressure <strong>increases</strong> as well. Likewise, when the temperature of the gas <em>decreases</em>, pressure <strong>decreases</strong> as well. </p>
<p><img alt="https://sites.google.com/site/tiyahgaslaws/gay-lussac-s-law" src="https://useruploads.socratic.org/E0jSnk9mQR2dtXMiZVav_685557087.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state</p>
<p>When plugging in your values into this equation, don't forget to convert the temperature from <em>degrees Celsius</em> to <em>Kelvin</em>.</p>
<p>So, rearrange and solve for <mathjax>#P_2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#P_2 = T_2/T_1 * P_1#</mathjax></p>
<p><mathjax>#P_2 = ((273.15 + 55)color(red)(cancel(color(black)("K"))))/((273.15 + 20)color(red)(cancel(color(black)("K")))) * "760 mmHg" = "850.74 mmHg"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to one <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a>, despite the fact that you only gave one sig fig for the initial temperature of the gas</p>
<blockquote>
<p><mathjax>#P_2 = color(green)("850 mmHg")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/0Oq7bCSDPxE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | My car has an internal volume of 2600 liters. Assume the pressure was initially 760 mm Hg. If the sun heats my car from a temperature of 20° C to a temperature of 55“ C, what will the pressure inside my car be? | null |
1,372 | ac95029a-6ddd-11ea-b4a7-ccda262736ce | https://socratic.org/questions/it-takes-80-cal-g-to-change-solid-water-at-0-c-to-liquid-water-how-many-calories | 240.00 calories | start physical_unit 6 7 heat_energy cal qc_end physical_unit 6 7 2 3 heat_energy qc_end c_other OTHER qc_end physical_unit 6 7 9 10 temperature qc_end physical_unit 6 7 22 23 mass qc_end end | [{"type":"physical unit","value":"Taken energy [OF] solid water [IN] calories"}] | [{"type":"physical unit","value":"240.00 calories"}] | [{"type":"physical unit","value":"Calorific value [OF] solid water [=] \\pu{80 cal/g}"},{"type":"other","value":"Change solid water to liquid water."},{"type":"physical unit","value":"Temperature [OF] solid water [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Mass [OF] solid water [=] \\pu{3 g}"}] | <h1 class="questionTitle" itemprop="name">It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water?</h1> | null | 240.00 calories | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the problem is providing you with a measure of how much heat is needed <strong>per gram</strong> of solid water, i.e. of <em>ice</em>, in order for a solid <mathjax>#->#</mathjax> liquid phase change to take place. </p>
<p>Simply put, you are given a measure of how much heat is required to melt <mathjax>#"1 g"#</mathjax> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, you can say that <mathjax>#"80 cal g"^(-1)#</mathjax> is equivalent to saying that it takes <mathjax>#"80 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of ice</strong> to convert it from solid at <mathjax>#0^@"C"#</mathjax> to lqiuid water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>This means that <mathjax>#"3 g"#</mathjax> of ice would require </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you only have one significant figure for the mass of ice. </p>
<p>So remember this notation </p>
<blockquote>
<p><mathjax>#color(blue)("80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g"^(-1))#</mathjax></p>
</blockquote>
<p>means that <strong>every</strong> <mathjax>#color(red)("1 g")#</mathjax> <strong>of ice</strong> at <mathjax>#0^@"C"#</mathjax> can be converted to liquid water at <mathjax>#0^@"C"#</mathjax> by providing it with <mathjax>#color(blue)("80 cal")#</mathjax> of heat. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"240 cal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the problem is providing you with a measure of how much heat is needed <strong>per gram</strong> of solid water, i.e. of <em>ice</em>, in order for a solid <mathjax>#->#</mathjax> liquid phase change to take place. </p>
<p>Simply put, you are given a measure of how much heat is required to melt <mathjax>#"1 g"#</mathjax> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, you can say that <mathjax>#"80 cal g"^(-1)#</mathjax> is equivalent to saying that it takes <mathjax>#"80 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of ice</strong> to convert it from solid at <mathjax>#0^@"C"#</mathjax> to lqiuid water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>This means that <mathjax>#"3 g"#</mathjax> of ice would require </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you only have one significant figure for the mass of ice. </p>
<p>So remember this notation </p>
<blockquote>
<p><mathjax>#color(blue)("80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g"^(-1))#</mathjax></p>
</blockquote>
<p>means that <strong>every</strong> <mathjax>#color(red)("1 g")#</mathjax> <strong>of ice</strong> at <mathjax>#0^@"C"#</mathjax> can be converted to liquid water at <mathjax>#0^@"C"#</mathjax> by providing it with <mathjax>#color(blue)("80 cal")#</mathjax> of heat. </p></div>
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<h1 class="questionTitle" itemprop="name">It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water?</h1>
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Stefan V.
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Mar 3, 2017
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<div class="markdown"><p><mathjax>#"240 cal"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the problem is providing you with a measure of how much heat is needed <strong>per gram</strong> of solid water, i.e. of <em>ice</em>, in order for a solid <mathjax>#->#</mathjax> liquid phase change to take place. </p>
<p>Simply put, you are given a measure of how much heat is required to melt <mathjax>#"1 g"#</mathjax> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>. </p>
<p>So, you can say that <mathjax>#"80 cal g"^(-1)#</mathjax> is equivalent to saying that it takes <mathjax>#"80 cal"#</mathjax> of heat <strong>for every</strong> <mathjax>#"1 g"#</mathjax> <strong>of ice</strong> to convert it from solid at <mathjax>#0^@"C"#</mathjax> to lqiuid water at <mathjax>#0^@"C"#</mathjax>. </p>
<p>This means that <mathjax>#"3 g"#</mathjax> of ice would require </p>
<blockquote>
<p><mathjax>#3 color(red)(cancel(color(black)("g"))) * "80 cal"/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("240 cal")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but don't forget that you only have one significant figure for the mass of ice. </p>
<p>So remember this notation </p>
<blockquote>
<p><mathjax>#color(blue)("80 cal")color(red)("/g") " "=" " color(blue)("80 cal")color(white)(.)color(red)("g"^(-1))#</mathjax></p>
</blockquote>
<p>means that <strong>every</strong> <mathjax>#color(red)("1 g")#</mathjax> <strong>of ice</strong> at <mathjax>#0^@"C"#</mathjax> can be converted to liquid water at <mathjax>#0^@"C"#</mathjax> by providing it with <mathjax>#color(blue)("80 cal")#</mathjax> of heat. </p></div>
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</article> | It takes 80 cal/g to change solid water at 0°C to liquid water. How many calories will it take to change 3 g of water at 0 °C to liquid water? | null |
1,373 | ab0a41da-6ddd-11ea-a981-ccda262736ce | https://socratic.org/questions/calculate-the-number-of-collisions-per-second-of-one-hydrogen-molecule-at-24-c-a | 3.46 × 10^10 | start physical_unit 4 4 number none qc_end physical_unit 9 10 12 13 temperature qc_end physical_unit 9 10 15 16 pressure qc_end physical_unit 20 22 24 25 diameter qc_end end | [{"type":"physical unit","value":"Number [OF] collisions"}] | [{"type":"physical unit","value":"3.46 × 10^10"}] | [{"type":"physical unit","value":"Time [OF] hydrogen molecule [=] \\pu{1 second}"},{"type":"physical unit","value":"Number [OF] hydrogen molecule [=] \\pu{1}"},{"type":"physical unit","value":"Temperature [OF] hydrogen molecule [=] \\pu{24 ℃}"},{"type":"physical unit","value":"Pressure [OF] hydrogen molecule [=] \\pu{2.00 bar}"},{"type":"physical unit","value":"Diameter [OF] a hydrogen molecule [=] \\pu{270 pm}"}] | <h1 class="questionTitle" itemprop="name">Calculate the number of collisions per second of one hydrogen molecule at 24 °C and 2.00 bar. The diameter of a hydrogen molecule is 270 pm?</h1> | null | 3.46 × 10^10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>According to <strong>Kinetic Molecular Theory</strong>, the collision frequency is equal to the root-mean-square velocity of the molecules divided by their mean free path.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ν = v_"rms"/λcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><strong>Root-mean-square velocity</strong></p>
<p>The formula relating the rms velocity to the temperature and molar mass is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) v_"rms" = sqrt((3RT)/M)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax># R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature<br/>
<mathjax>#M#</mathjax> = the molar mass</p>
<p>For <mathjax>#"H"_2#</mathjax> at 24 °C, </p>
<p><mathjax>#T = "(24 + 273.15) K" = "297.15 K"#</mathjax><br/>
<mathjax>#M = "2.016 g·mol"^"-1" = 2.016 × 10^"-3"color(white)(l) "kg·mol"^"-1"#</mathjax></p>
<p><mathjax>#v_"rms" = sqrt((3RT)/M) = sqrt((3 × 8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/( 2.016 × 10^"-3" color(red)(cancel(color(black)("kg·mol"^"-1")))) × ( 1 color(red)(cancel(color(black)("kg")))·"m"^2"s"^"-2")/(1 color(red)(cancel(color(black)("J"))))) = = "1917 m·s"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The mean free path</strong></p>
<p>If the molecules have diameter d, then we can use a circle of diameter <mathjax>#σ = 2d#</mathjax> to represent a molecule's <strong>effective collision area</strong>. </p>
<p><img alt="www.schoolphysics.co.uk" src="http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Kinetic%20theory%20of%20matter/text/Mean_free_path/images/1.png"/></p>
<p>For a hydrogen molecule, <mathjax>#σ = "289 pm"#</mathjax>.</p>
<p>The formula for the mean free path is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) λ = (RT)/(sqrt2πσ^2N_"A"P)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#R = 0".083 14 bar·L·K"^"-1""mol"^"-1" = 8.314 × 10^"-5"color(white)(l)"bar·m"^3·"K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "297.15 K"#</mathjax><br/>
<mathjax>#σ = "289 pm" = 289 × 10^"-12"color(white)(l) "m"#</mathjax><br/>
<mathjax>#N_"A" = 6.022 × 10^23color(white)(l) "mol"^"-1"#</mathjax><br/>
<mathjax>#P = "2.00 bar"#</mathjax></p>
<p><mathjax>#λ = (RT)/(sqrt2πσ^2N_"A"P) = (8.314 × 10^"-5"color(red)(cancel(color(black)("bar")))·stackrelcolor(blue)("m")(color(red)(cancel(color(black)("m"^3))))·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/(sqrt2π × (289 × 10^"-12" color(red)(cancel(color(black)("m"))))^2 × 6.022 × 10^23 color(red)(cancel(color(black)("mol"^"-1"))) × 2.00 color(red)(cancel(color(black)("bar"))))#</mathjax></p>
<p><mathjax>#= 5.52 × 10^"-8"color(white)(l) "m" = "55.2 nm"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Collision frequency</strong></p>
<p><mathjax>#ν = v_"rms"/λ = (1917 color(red)(cancel(color(black)("m")))·"s"^"-1")/(5.52 × 10^"-8" color(red)(cancel(color(black)("m")))) = 3.46 × 10^10color(white)(l) "s"^"-1"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>WARNING! Long answer!</strong> There are <mathjax>#3.46 × 10^10#</mathjax> collisions per second.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>According to <strong>Kinetic Molecular Theory</strong>, the collision frequency is equal to the root-mean-square velocity of the molecules divided by their mean free path.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ν = v_"rms"/λcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><strong>Root-mean-square velocity</strong></p>
<p>The formula relating the rms velocity to the temperature and molar mass is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) v_"rms" = sqrt((3RT)/M)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax># R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature<br/>
<mathjax>#M#</mathjax> = the molar mass</p>
<p>For <mathjax>#"H"_2#</mathjax> at 24 °C, </p>
<p><mathjax>#T = "(24 + 273.15) K" = "297.15 K"#</mathjax><br/>
<mathjax>#M = "2.016 g·mol"^"-1" = 2.016 × 10^"-3"color(white)(l) "kg·mol"^"-1"#</mathjax></p>
<p><mathjax>#v_"rms" = sqrt((3RT)/M) = sqrt((3 × 8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/( 2.016 × 10^"-3" color(red)(cancel(color(black)("kg·mol"^"-1")))) × ( 1 color(red)(cancel(color(black)("kg")))·"m"^2"s"^"-2")/(1 color(red)(cancel(color(black)("J"))))) = = "1917 m·s"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The mean free path</strong></p>
<p>If the molecules have diameter d, then we can use a circle of diameter <mathjax>#σ = 2d#</mathjax> to represent a molecule's <strong>effective collision area</strong>. </p>
<p><img alt="www.schoolphysics.co.uk" src="http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Kinetic%20theory%20of%20matter/text/Mean_free_path/images/1.png"/></p>
<p>For a hydrogen molecule, <mathjax>#σ = "289 pm"#</mathjax>.</p>
<p>The formula for the mean free path is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) λ = (RT)/(sqrt2πσ^2N_"A"P)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#R = 0".083 14 bar·L·K"^"-1""mol"^"-1" = 8.314 × 10^"-5"color(white)(l)"bar·m"^3·"K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "297.15 K"#</mathjax><br/>
<mathjax>#σ = "289 pm" = 289 × 10^"-12"color(white)(l) "m"#</mathjax><br/>
<mathjax>#N_"A" = 6.022 × 10^23color(white)(l) "mol"^"-1"#</mathjax><br/>
<mathjax>#P = "2.00 bar"#</mathjax></p>
<p><mathjax>#λ = (RT)/(sqrt2πσ^2N_"A"P) = (8.314 × 10^"-5"color(red)(cancel(color(black)("bar")))·stackrelcolor(blue)("m")(color(red)(cancel(color(black)("m"^3))))·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/(sqrt2π × (289 × 10^"-12" color(red)(cancel(color(black)("m"))))^2 × 6.022 × 10^23 color(red)(cancel(color(black)("mol"^"-1"))) × 2.00 color(red)(cancel(color(black)("bar"))))#</mathjax></p>
<p><mathjax>#= 5.52 × 10^"-8"color(white)(l) "m" = "55.2 nm"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Collision frequency</strong></p>
<p><mathjax>#ν = v_"rms"/λ = (1917 color(red)(cancel(color(black)("m")))·"s"^"-1")/(5.52 × 10^"-8" color(red)(cancel(color(black)("m")))) = 3.46 × 10^10color(white)(l) "s"^"-1"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Calculate the number of collisions per second of one hydrogen molecule at 24 °C and 2.00 bar. The diameter of a hydrogen molecule is 270 pm?</h1>
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<div class="markdown"><p><strong>WARNING! Long answer!</strong> There are <mathjax>#3.46 × 10^10#</mathjax> collisions per second.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>According to <strong>Kinetic Molecular Theory</strong>, the collision frequency is equal to the root-mean-square velocity of the molecules divided by their mean free path.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ν = v_"rms"/λcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><strong>Root-mean-square velocity</strong></p>
<p>The formula relating the rms velocity to the temperature and molar mass is:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) v_"rms" = sqrt((3RT)/M)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax># R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature<br/>
<mathjax>#M#</mathjax> = the molar mass</p>
<p>For <mathjax>#"H"_2#</mathjax> at 24 °C, </p>
<p><mathjax>#T = "(24 + 273.15) K" = "297.15 K"#</mathjax><br/>
<mathjax>#M = "2.016 g·mol"^"-1" = 2.016 × 10^"-3"color(white)(l) "kg·mol"^"-1"#</mathjax></p>
<p><mathjax>#v_"rms" = sqrt((3RT)/M) = sqrt((3 × 8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/( 2.016 × 10^"-3" color(red)(cancel(color(black)("kg·mol"^"-1")))) × ( 1 color(red)(cancel(color(black)("kg")))·"m"^2"s"^"-2")/(1 color(red)(cancel(color(black)("J"))))) = = "1917 m·s"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The mean free path</strong></p>
<p>If the molecules have diameter d, then we can use a circle of diameter <mathjax>#σ = 2d#</mathjax> to represent a molecule's <strong>effective collision area</strong>. </p>
<p><img alt="www.schoolphysics.co.uk" src="http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Kinetic%20theory%20of%20matter/text/Mean_free_path/images/1.png"/></p>
<p>For a hydrogen molecule, <mathjax>#σ = "289 pm"#</mathjax>.</p>
<p>The formula for the mean free path is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a) λ = (RT)/(sqrt2πσ^2N_"A"P)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#R = 0".083 14 bar·L·K"^"-1""mol"^"-1" = 8.314 × 10^"-5"color(white)(l)"bar·m"^3·"K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "297.15 K"#</mathjax><br/>
<mathjax>#σ = "289 pm" = 289 × 10^"-12"color(white)(l) "m"#</mathjax><br/>
<mathjax>#N_"A" = 6.022 × 10^23color(white)(l) "mol"^"-1"#</mathjax><br/>
<mathjax>#P = "2.00 bar"#</mathjax></p>
<p><mathjax>#λ = (RT)/(sqrt2πσ^2N_"A"P) = (8.314 × 10^"-5"color(red)(cancel(color(black)("bar")))·stackrelcolor(blue)("m")(color(red)(cancel(color(black)("m"^3))))·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 297.15 color(red)(cancel(color(black)("K"))))/(sqrt2π × (289 × 10^"-12" color(red)(cancel(color(black)("m"))))^2 × 6.022 × 10^23 color(red)(cancel(color(black)("mol"^"-1"))) × 2.00 color(red)(cancel(color(black)("bar"))))#</mathjax></p>
<p><mathjax>#= 5.52 × 10^"-8"color(white)(l) "m" = "55.2 nm"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Collision frequency</strong></p>
<p><mathjax>#ν = v_"rms"/λ = (1917 color(red)(cancel(color(black)("m")))·"s"^"-1")/(5.52 × 10^"-8" color(red)(cancel(color(black)("m")))) = 3.46 × 10^10color(white)(l) "s"^"-1"#</mathjax></p></div>
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</article> | Calculate the number of collisions per second of one hydrogen molecule at 24 °C and 2.00 bar. The diameter of a hydrogen molecule is 270 pm? | null |
1,374 | a924abec-6ddd-11ea-9fb3-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-397-grams-of-na-2so-4 | 2.79 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] Na2SO4 [IN] moles"}] | [{"type":"physical unit","value":"2.79 moles"}] | [{"type":"physical unit","value":"Mass [OF] Na2SO4 [=] \\pu{397 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in 397 grams of #Na_2SO_4#? </h1> | null | 2.79 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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<div class="markdown"><p>2.80</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are there in 397 grams of #Na_2SO_4#? </h1>
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<div class="markdown"><p>2.80</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms. </p>
<p>For <mathjax>#Na_2SO_4#</mathjax> this is <mathjax>#2 x 23 = 46#</mathjax> for the sodium, plus <mathjax>#4 x 16 = 64#</mathjax> for the oxygen, plus <mathjax>#32#</mathjax> for the sulfur. That adds up to a molecular weight of <mathjax>#142 (g/"mole")#</mathjax>. Now we divide the given number of grams by this value to derive the number of moles.<br/>
<mathjax>#(397/142)(g/(g/"mol")) = 2.80 "moles"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#n_(Na_2SO_4)=2.79 mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Molar mass of <mathjax>#Na_2SO_4 =142.04g/(mol)#</mathjax></p>
<p>Use the equation, </p>
<p><mathjax>#n_(Na_2SO_4)=m/M#</mathjax></p>
<p>where, </p>
<p><mathjax>#n=#</mathjax> number of moles<br/>
<mathjax>#m=#</mathjax>mass (in grams)<br/>
<mathjax>#M=#</mathjax>molar mass (in <mathjax>#(grams)/(mol)#</mathjax>)</p>
<p><mathjax>#n_(Na_2SO_4)=(397g)/(142.04(g/(mol)))#</mathjax></p>
<p><mathjax>#n_(Na_2SO_4)=2.79 mol#</mathjax></p></div>
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</article> | How many moles are there in 397 grams of #Na_2SO_4#? | null |
1,375 | abde17a8-6ddd-11ea-96bf-ccda262736ce | https://socratic.org/questions/if-60-2-grams-of-hg-combines-completely-with-24-0-grams-of-br-to-form-a-compound | 71.5% | start physical_unit 22 25 percent_composition none qc_end physical_unit 4 4 1 2 mass qc_end c_other OTHER qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Percent composition [OF] Hg in the compound"}] | [{"type":"physical unit","value":"71.5%"}] | [{"type":"physical unit","value":"Mass [OF] Hg [=] \\pu{60.2 grams}"},{"type":"other","value":"Combine completely."},{"type":"physical unit","value":"Mass [OF] Br [=] \\pu{24.0 grams}"}] | <h1 class="questionTitle" itemprop="name">If 60.2 grams of Hg combines completely with 24.0 grams of Br to form a compound, what is the percent composition of Hg in the compound?</h1> | null | 71.5% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%Hg="mass of mercury"/"mass of mercury and bromine"xx100%#</mathjax></p>
<p><mathjax>#=(60.2*g)/(60.2*g+24.0*g)xx100%=??%#</mathjax></p>
<p>In such a binary compound, if you have the percentage by mass of one component, you also have the percentage by mass of the other component. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#%Hg=71.5%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%Hg="mass of mercury"/"mass of mercury and bromine"xx100%#</mathjax></p>
<p><mathjax>#=(60.2*g)/(60.2*g+24.0*g)xx100%=??%#</mathjax></p>
<p>In such a binary compound, if you have the percentage by mass of one component, you also have the percentage by mass of the other component. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If 60.2 grams of Hg combines completely with 24.0 grams of Br to form a compound, what is the percent composition of Hg in the compound?</h1>
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anor277
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May 23, 2017
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<div class="markdown"><p><mathjax>#%Hg=71.5%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#%Hg="mass of mercury"/"mass of mercury and bromine"xx100%#</mathjax></p>
<p><mathjax>#=(60.2*g)/(60.2*g+24.0*g)xx100%=??%#</mathjax></p>
<p>In such a binary compound, if you have the percentage by mass of one component, you also have the percentage by mass of the other component. </p></div>
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</article> | If 60.2 grams of Hg combines completely with 24.0 grams of Br to form a compound, what is the percent composition of Hg in the compound? | null |
1,376 | a9f996d5-6ddd-11ea-9347-ccda262736ce | https://socratic.org/questions/what-is-the-standard-state-free-energy-for-this-reaction | -2.19 × 10^5 J/mol | start physical_unit 9 9 standard_free_energy j/mol qc_end chemical_equation 21 26 qc_end physical_unit 9 9 29 31 kf qc_end end | [{"type":"physical unit","value":"Standard state free energy [OF] the reaction [IN] J/mol"}] | [{"type":"physical unit","value":"-2.19 × 10^5 J/mol"}] | [{"type":"chemical equation","value":"Au+(aq) + 2 CN- <=> Au(CN)2"},{"type":"physical unit","value":"kf [OF] the reaction [=] \\pu{2 × 10^38}"}] | <h1 class="questionTitle" itemprop="name">What is the standard state free energy for this reaction?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>Gold ions form complexes with cyanide ion according to the equation:<br/>
Au+(aq) + 2 CN <-> Au(CN)2 kf=2x10^38</p>
<p>What is the standard state free energy for this reaction?<img alt="enter image source here" src="https://useruploads.socratic.org/qF1QgxRTSdWohjYqSBET_tHlko7.jpg"/> </p></div>
</h2>
</div>
</div> | -2.19 × 10^5 J/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation </p>
<h2><mathjax>#K_"eq" = e^(-(Delta G^ "o")/("R" cdot "T"))#</mathjax></h2>
<p>Relates the equilibrium constant of a reaction, <mathjax>#K_"eq"#</mathjax> to its change in <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs Free Energy</a> per mole reaction <mathjax>#Delta G^"o"#</mathjax>. <mathjax>#"R"#</mathjax> is the ideal gas constant and <mathjax>#"T"#</mathjax> the temperature in degree Kelvins. The question is asking for the Gibbs Energy Change under standard conditions. Thus <mathjax>#"T"=298 color(white)(l) "K"#</mathjax>.</p>
<p>Solve for <mathjax>#Delta G^"o"#</mathjax>:</p>
<p><mathjax>#Delta G^"o" = - ("R" * "T") * ln K_"eq" #</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = - 8.314 color(white)(l) "J" * "mol"^(-1) * color(red)(cancel(color(black)("K"^(-1)))) * 298 color(white)(l) color(red)(cancel(color(black)("K"))) * ln (2 xx 10^38)#</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = -2.19 xx 10^5 color(white)(l) "J" * "mol"^(-1)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Delta G^"o" = -2.19 xx 10^5 color(white)(l) "J" * "mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation </p>
<h2><mathjax>#K_"eq" = e^(-(Delta G^ "o")/("R" cdot "T"))#</mathjax></h2>
<p>Relates the equilibrium constant of a reaction, <mathjax>#K_"eq"#</mathjax> to its change in <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs Free Energy</a> per mole reaction <mathjax>#Delta G^"o"#</mathjax>. <mathjax>#"R"#</mathjax> is the ideal gas constant and <mathjax>#"T"#</mathjax> the temperature in degree Kelvins. The question is asking for the Gibbs Energy Change under standard conditions. Thus <mathjax>#"T"=298 color(white)(l) "K"#</mathjax>.</p>
<p>Solve for <mathjax>#Delta G^"o"#</mathjax>:</p>
<p><mathjax>#Delta G^"o" = - ("R" * "T") * ln K_"eq" #</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = - 8.314 color(white)(l) "J" * "mol"^(-1) * color(red)(cancel(color(black)("K"^(-1)))) * 298 color(white)(l) color(red)(cancel(color(black)("K"))) * ln (2 xx 10^38)#</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = -2.19 xx 10^5 color(white)(l) "J" * "mol"^(-1)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the standard state free energy for this reaction?</h1>
<div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Gold ions form complexes with cyanide ion according to the equation:<br/>
Au+(aq) + 2 CN <-> Au(CN)2 kf=2x10^38</p>
<p>What is the standard state free energy for this reaction?<img alt="enter image source here" src="https://useruploads.socratic.org/qF1QgxRTSdWohjYqSBET_tHlko7.jpg"/> </p></div>
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<div class="markdown"><p><mathjax>#Delta G^"o" = -2.19 xx 10^5 color(white)(l) "J" * "mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation </p>
<h2><mathjax>#K_"eq" = e^(-(Delta G^ "o")/("R" cdot "T"))#</mathjax></h2>
<p>Relates the equilibrium constant of a reaction, <mathjax>#K_"eq"#</mathjax> to its change in <a href="https://socratic.org/chemistry/thermochemistry/gibbs-free-energy">Gibbs Free Energy</a> per mole reaction <mathjax>#Delta G^"o"#</mathjax>. <mathjax>#"R"#</mathjax> is the ideal gas constant and <mathjax>#"T"#</mathjax> the temperature in degree Kelvins. The question is asking for the Gibbs Energy Change under standard conditions. Thus <mathjax>#"T"=298 color(white)(l) "K"#</mathjax>.</p>
<p>Solve for <mathjax>#Delta G^"o"#</mathjax>:</p>
<p><mathjax>#Delta G^"o" = - ("R" * "T") * ln K_"eq" #</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = - 8.314 color(white)(l) "J" * "mol"^(-1) * color(red)(cancel(color(black)("K"^(-1)))) * 298 color(white)(l) color(red)(cancel(color(black)("K"))) * ln (2 xx 10^38)#</mathjax><br/>
<mathjax>#color(white)(Delta G^"o") = -2.19 xx 10^5 color(white)(l) "J" * "mol"^(-1)#</mathjax></p></div>
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</article> | What is the standard state free energy for this reaction? |
Gold ions form complexes with cyanide ion according to the equation:
Au+(aq) + 2 CN <-> Au(CN)2 kf=2x10^38
What is the standard state free energy for this reaction?
|
1,377 | ac09f47a-6ddd-11ea-b875-ccda262736ce | https://socratic.org/questions/water-can-be-formed-according-to-the-equation-2h-2-o-2-2h-20-if-8-0-l-of-hydroge | 4.00 liters | start physical_unit 29 29 volume l qc_end chemical_equation 8 14 qc_end physical_unit 19 19 16 17 volume qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] liters"}] | [{"type":"physical unit","value":"4.00 liters"}] | [{"type":"chemical equation","value":"2 H2 + O2 -> 2 H2O"},{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{8.0 L}"},{"type":"other","value":"STP"},{"type":"other","value":"Complete reaction."}] | <h1 class="questionTitle" itemprop="name">Water can be formed according to the equation: #2H_2 + O_2 -> 2H_20#. If 8.0 L of hydrogen is reacted at STP, exactly how many liters of oxygen at STP would be need to allow complete reaction?</h1> | null | 4.00 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Avogadro's <a href="https://en.wikipedia.org/wiki/Avogadro%27s_law" rel="nofollow">gas law</a> holds that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (i.e. <mathjax>#Vpropn)#</mathjax>. It follows that different volumes of gases under equivalent conditions contain numbers of molecules proportional to their respective volumes. </p>
<p>Given <mathjax>#8.0*L#</mathjax> dihydrogen gas, <mathjax>#4.0*L#</mathjax> dioxygen gas are required; i.e. the respective gas volumes follow the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4.0*L#</mathjax> of dioxygen gas are required. </p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Avogadro's <a href="https://en.wikipedia.org/wiki/Avogadro%27s_law" rel="nofollow">gas law</a> holds that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (i.e. <mathjax>#Vpropn)#</mathjax>. It follows that different volumes of gases under equivalent conditions contain numbers of molecules proportional to their respective volumes. </p>
<p>Given <mathjax>#8.0*L#</mathjax> dihydrogen gas, <mathjax>#4.0*L#</mathjax> dioxygen gas are required; i.e. the respective gas volumes follow the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Water can be formed according to the equation: #2H_2 + O_2 -> 2H_20#. If 8.0 L of hydrogen is reacted at STP, exactly how many liters of oxygen at STP would be need to allow complete reaction?</h1>
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<div class="markdown"><p><mathjax>#4.0*L#</mathjax> of dioxygen gas are required. </p>
<p><mathjax>#2H_2(g) + O_2(g) rarr 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Avogadro's <a href="https://en.wikipedia.org/wiki/Avogadro%27s_law" rel="nofollow">gas law</a> holds that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (i.e. <mathjax>#Vpropn)#</mathjax>. It follows that different volumes of gases under equivalent conditions contain numbers of molecules proportional to their respective volumes. </p>
<p>Given <mathjax>#8.0*L#</mathjax> dihydrogen gas, <mathjax>#4.0*L#</mathjax> dioxygen gas are required; i.e. the respective gas volumes follow the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation. </p></div>
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</article> | Water can be formed according to the equation: #2H_2 + O_2 -> 2H_20#. If 8.0 L of hydrogen is reacted at STP, exactly how many liters of oxygen at STP would be need to allow complete reaction? | null |
1,378 | ab12d690-6ddd-11ea-b689-ccda262736ce | https://socratic.org/questions/how-do-you-balance-koh-bacl-2-kci-ba-oh-2 | 2 KOH + BaCl2 -> 2 KCl + Ba(OH)2 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 KOH + BaCl2 -> 2 KCl + Ba(OH)2"}] | [{"type":"chemical equation","value":"KOH + BaCl2 -> KCl + Ba(OH)2"}] | <h1 class="questionTitle" itemprop="name">How do you balance #KOH + BaCl_2 -> KCI + Ba(OH)_2#?</h1> | null | 2 KOH + BaCl2 -> 2 KCl + Ba(OH)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is the given equation stoichiometrically balanced? Does garbage in equal garbage out? In every chemical reaction mass is conserved, and the balanced equation, which after all is our attempt to model an actual chemical process, reflects this.</p></div>
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<div class="markdown"><p><mathjax>#2KOH + BaCl_2 rarr Ba(OH)_2 + 2KCl#</mathjax></p></div>
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<div class="markdown"><p>Is the given equation stoichiometrically balanced? Does garbage in equal garbage out? In every chemical reaction mass is conserved, and the balanced equation, which after all is our attempt to model an actual chemical process, reflects this.</p></div>
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<div class="markdown"><p><mathjax>#2KOH + BaCl_2 rarr Ba(OH)_2 + 2KCl#</mathjax></p></div>
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<div class="markdown"><p>Is the given equation stoichiometrically balanced? Does garbage in equal garbage out? In every chemical reaction mass is conserved, and the balanced equation, which after all is our attempt to model an actual chemical process, reflects this.</p></div>
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</article> | How do you balance #KOH + BaCl_2 -> KCI + Ba(OH)_2#? | null |
1,379 | ab4d69c0-6ddd-11ea-abed-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-phosphorus-in-hpo3 | +3 | start physical_unit 6 6 oxidation_state none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] phosphorus"}] | [{"type":"physical unit","value":"+3"}] | [{"type":"chemical equation","value":"HPO3"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of phosphorus in HPO3?</h1> | null | +3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Phophite (<mathjax>#PO_3^(3-)#</mathjax>) has a charge of -3, so I'm going to guess you meant to ask what the oxidation state of P was in <mathjax>#H_3PO_3#</mathjax></p>
<p>In <mathjax>#H_3PO_3#</mathjax> the oxygens will always have a -2 charge and hydrogen is +1.</p>
<p>So the total positive charge from Hydrogen is +3 (+1 x 3)<br/>
The total negative charge from Oxygen is -6 (-2 x 3)</p>
<p>The compound is electrically neutral, so the phosphorus must have an oxidation state of +3.</p>
<p>3H = +1 x 3 = +3<br/>
1P = +3 x 1 = +3<br/>
3O = -2 x 3 = -6</p>
<p>Total charge = 0</p>
<p>
<iframe src="https://www.youtube.com/embed/uy9ay4xrVTE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<iframe src="https://www.youtube.com/embed/H1c1bk6FXQg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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<div class="markdown"><p>+3</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Phophite (<mathjax>#PO_3^(3-)#</mathjax>) has a charge of -3, so I'm going to guess you meant to ask what the oxidation state of P was in <mathjax>#H_3PO_3#</mathjax></p>
<p>In <mathjax>#H_3PO_3#</mathjax> the oxygens will always have a -2 charge and hydrogen is +1.</p>
<p>So the total positive charge from Hydrogen is +3 (+1 x 3)<br/>
The total negative charge from Oxygen is -6 (-2 x 3)</p>
<p>The compound is electrically neutral, so the phosphorus must have an oxidation state of +3.</p>
<p>3H = +1 x 3 = +3<br/>
1P = +3 x 1 = +3<br/>
3O = -2 x 3 = -6</p>
<p>Total charge = 0</p>
<p>
<iframe src="https://www.youtube.com/embed/uy9ay4xrVTE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>
<iframe src="https://www.youtube.com/embed/H1c1bk6FXQg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of phosphorus in HPO3?</h1>
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<div class="markdown"><p>+3</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Phophite (<mathjax>#PO_3^(3-)#</mathjax>) has a charge of -3, so I'm going to guess you meant to ask what the oxidation state of P was in <mathjax>#H_3PO_3#</mathjax></p>
<p>In <mathjax>#H_3PO_3#</mathjax> the oxygens will always have a -2 charge and hydrogen is +1.</p>
<p>So the total positive charge from Hydrogen is +3 (+1 x 3)<br/>
The total negative charge from Oxygen is -6 (-2 x 3)</p>
<p>The compound is electrically neutral, so the phosphorus must have an oxidation state of +3.</p>
<p>3H = +1 x 3 = +3<br/>
1P = +3 x 1 = +3<br/>
3O = -2 x 3 = -6</p>
<p>Total charge = 0</p>
<p>
<iframe src="https://www.youtube.com/embed/uy9ay4xrVTE?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>
<iframe src="https://www.youtube.com/embed/H1c1bk6FXQg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<p>Hope this helps!</p></div>
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</article> | What is the oxidation state of phosphorus in HPO3? | null |
1,380 | acf757f6-6ddd-11ea-8320-ccda262736ce | https://socratic.org/questions/what-pressure-will-14-0-g-of-co-exert-in-a-3-5-l-container-at-75-c | 1.65 × 10^5 kPa | start physical_unit 6 6 pressure kpa qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 6 6 14 15 temperature qc_end physical_unit 12 12 10 11 volume qc_end end | [{"type":"physical unit","value":"Pressure [OF] CO [IN] kPa"}] | [{"type":"physical unit","value":"1.65 × 10^5 kPa"}] | [{"type":"physical unit","value":"Mass [OF] CO [=] \\pu{14.0 g }"},{"type":"physical unit","value":"Temperature [OF] CO [=] \\pu{75 ℃}"},{"type":"physical unit","value":"Volume [OF] container [=] \\pu{3.5 L}"}] | <h1 class="questionTitle" itemprop="name">What pressure will 14.0 g of CO exert in a 3.5 L container at 75°C? </h1> | null | 1.65 × 10^5 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal gas law</a>:<br/>
<mathjax>#PV=nRT#</mathjax><br/>
Rearrangement:<br/>
<mathjax>#P=(nRT)/V#</mathjax><br/>
Calculate amount of <mathjax>#"CO"_2#</mathjax>:<br/>
<mathjax>#n("CO"_2)=m/M=(14.0g)/(28gmol^-1)=2mol#</mathjax><br/>
Convert temperature to Kelvin:<br/>
<mathjax>#T=75+273=348K#</mathjax><br/>
Volume of container:<br/>
<mathjax>#V=3.5L#</mathjax><br/>
Ideal gas constant:<br/>
<mathjax>#R=8.314L"bar "K^-1 mol^-1#</mathjax><br/>
Substitute values into ideal gas law:<br/>
<mathjax>#P=(2molxx8.314L "bar " K^-1 mol^-1xx348K)/(3.5L)=1650 "bar (3 s.f.)"=165000kPa#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1650 "bar"#</mathjax> or <mathjax>#165000kPa#</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/ideal-gas-law">Ideal gas law</a>:<br/>
<mathjax>#PV=nRT#</mathjax><br/>
Rearrangement:<br/>
<mathjax>#P=(nRT)/V#</mathjax><br/>
Calculate amount of <mathjax>#"CO"_2#</mathjax>:<br/>
<mathjax>#n("CO"_2)=m/M=(14.0g)/(28gmol^-1)=2mol#</mathjax><br/>
Convert temperature to Kelvin:<br/>
<mathjax>#T=75+273=348K#</mathjax><br/>
Volume of container:<br/>
<mathjax>#V=3.5L#</mathjax><br/>
Ideal gas constant:<br/>
<mathjax>#R=8.314L"bar "K^-1 mol^-1#</mathjax><br/>
Substitute values into ideal gas law:<br/>
<mathjax>#P=(2molxx8.314L "bar " K^-1 mol^-1xx348K)/(3.5L)=1650 "bar (3 s.f.)"=165000kPa#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What pressure will 14.0 g of CO exert in a 3.5 L container at 75°C? </h1>
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Cameron G.
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<span class="dateCreated" datetime="2016-07-05T02:15:32" itemprop="dateCreated">
Jul 5, 2016
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<div class="markdown"><p><mathjax>#1650 "bar"#</mathjax> or <mathjax>#165000kPa#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal gas law</a>:<br/>
<mathjax>#PV=nRT#</mathjax><br/>
Rearrangement:<br/>
<mathjax>#P=(nRT)/V#</mathjax><br/>
Calculate amount of <mathjax>#"CO"_2#</mathjax>:<br/>
<mathjax>#n("CO"_2)=m/M=(14.0g)/(28gmol^-1)=2mol#</mathjax><br/>
Convert temperature to Kelvin:<br/>
<mathjax>#T=75+273=348K#</mathjax><br/>
Volume of container:<br/>
<mathjax>#V=3.5L#</mathjax><br/>
Ideal gas constant:<br/>
<mathjax>#R=8.314L"bar "K^-1 mol^-1#</mathjax><br/>
Substitute values into ideal gas law:<br/>
<mathjax>#P=(2molxx8.314L "bar " K^-1 mol^-1xx348K)/(3.5L)=1650 "bar (3 s.f.)"=165000kPa#</mathjax></p></div>
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</article> | What pressure will 14.0 g of CO exert in a 3.5 L container at 75°C? | null |
1,381 | a95e85d8-6ddd-11ea-b5c5-ccda262736ce | https://socratic.org/questions/the-latent-heat-of-fusion-for-ice-is-6-0-kj-mole-in-order-to-melt-36-g-of-ice-so | 12 kJ | start physical_unit 6 6 energy kj qc_end physical_unit 6 6 8 9 enthalpy_of_fusion qc_end physical_unit 6 6 14 15 mass qc_end physical_unit 6 6 21 22 temperature qc_end end | [{"type":"physical unit","value":"Required energy [OF] ice [IN] kJ"}] | [{"type":"physical unit","value":"12 kJ"}] | [{"type":"physical unit","value":"Latent heat of fusion [OF] ice [=] \\pu{6.0 kJ/mole}"},{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{36 g}"},{"type":"physical unit","value":"Temperature [OF] ice [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">The latent heat of fusion for ice is 6.0 kJ/mole. In order to melt 36 g of ice (solid #H_2O#) at 0°C, how much energy is required? </h1> | null | 12 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>molar latent heat of fusion</em>, which is an alternative name given to the <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, tells you how much heat is required in order to convert a specific amount of a given substance, either a <em>gram</em> or a <em>mole</em>, from <strong>solid</strong> at its melting point to <strong>liquid</strong> at its melting point. </p>
<p>Ice is said to have a molar enthalpy of fusion equal to</p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "6.0 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to melt <mathjax>#1#</mathjax> <strong>mole</strong> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>, you must supply it with <mathjax>#"6.0 kJ"#</mathjax> of heat. </p>
<p>Now, your sample of ice has a mass of <mathjax>#"36 g"#</mathjax>, so the first thing to do here is to convert it to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#36 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.998 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar enthalpy of fusion as a <em>conversion factor</em> to help you figure out how much heat must be supplied to your sample</p>
<blockquote>
<p><mathjax>#1.998 color(red)(cancel(color(black)("moles ice"))) * "6.0 kJ"/(1color(red)(cancel(color(black)("mole ice")))) = color(darkgreen)(ul(color(black)("12 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"12 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>molar latent heat of fusion</em>, which is an alternative name given to the <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, tells you how much heat is required in order to convert a specific amount of a given substance, either a <em>gram</em> or a <em>mole</em>, from <strong>solid</strong> at its melting point to <strong>liquid</strong> at its melting point. </p>
<p>Ice is said to have a molar enthalpy of fusion equal to</p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "6.0 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to melt <mathjax>#1#</mathjax> <strong>mole</strong> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>, you must supply it with <mathjax>#"6.0 kJ"#</mathjax> of heat. </p>
<p>Now, your sample of ice has a mass of <mathjax>#"36 g"#</mathjax>, so the first thing to do here is to convert it to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#36 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.998 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar enthalpy of fusion as a <em>conversion factor</em> to help you figure out how much heat must be supplied to your sample</p>
<blockquote>
<p><mathjax>#1.998 color(red)(cancel(color(black)("moles ice"))) * "6.0 kJ"/(1color(red)(cancel(color(black)("mole ice")))) = color(darkgreen)(ul(color(black)("12 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The latent heat of fusion for ice is 6.0 kJ/mole. In order to melt 36 g of ice (solid #H_2O#) at 0°C, how much energy is required? </h1>
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Stefan V.
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Jan 29, 2017
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<div class="markdown"><p><mathjax>#"12 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <em>molar latent heat of fusion</em>, which is an alternative name given to the <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, tells you how much heat is required in order to convert a specific amount of a given substance, either a <em>gram</em> or a <em>mole</em>, from <strong>solid</strong> at its melting point to <strong>liquid</strong> at its melting point. </p>
<p>Ice is said to have a molar enthalpy of fusion equal to</p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "6.0 kJ mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to melt <mathjax>#1#</mathjax> <strong>mole</strong> of ice at its normal melting point of <mathjax>#0^@"C"#</mathjax>, you must supply it with <mathjax>#"6.0 kJ"#</mathjax> of heat. </p>
<p>Now, your sample of ice has a mass of <mathjax>#"36 g"#</mathjax>, so the first thing to do here is to convert it to <em>moles</em> by using the <strong>molar mass</strong> of water</p>
<blockquote>
<p><mathjax>#36 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.998 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can now use the molar enthalpy of fusion as a <em>conversion factor</em> to help you figure out how much heat must be supplied to your sample</p>
<blockquote>
<p><mathjax>#1.998 color(red)(cancel(color(black)("moles ice"))) * "6.0 kJ"/(1color(red)(cancel(color(black)("mole ice")))) = color(darkgreen)(ul(color(black)("12 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | The latent heat of fusion for ice is 6.0 kJ/mole. In order to melt 36 g of ice (solid #H_2O#) at 0°C, how much energy is required? | null |
1,382 | ab99e740-6ddd-11ea-b43b-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-mass-of-a-1h-atom | 1.66 × 10^(-24) g | start physical_unit 8 9 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] ^1H atom [IN] g"}] | [{"type":"physical unit","value":"1.66 × 10^(-24) g"}] | [{"type":"physical unit","value":"Number [OF] ^1H atom [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the mass of a #""^1H# atom?</h1> | null | 1.66 × 10^(-24) g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that the mass of 1 mol of hydrogen is 1 gram. 1 mol of hydrogen is <mathjax>#6.022*10^23#</mathjax> atoms. To find the mass of 1 atom, we just have to divide.</p>
<p><mathjax>#(1g H)/(6.022*10^23 at oms H)#</mathjax>=<mathjax>#1.66*10^-24g/(at om#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1.66*10^-24g/(at om#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that the mass of 1 mol of hydrogen is 1 gram. 1 mol of hydrogen is <mathjax>#6.022*10^23#</mathjax> atoms. To find the mass of 1 atom, we just have to divide.</p>
<p><mathjax>#(1g H)/(6.022*10^23 at oms H)#</mathjax>=<mathjax>#1.66*10^-24g/(at om#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the mass of a #""^1H# atom?</h1>
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<div class="markdown"><p><mathjax>#1.66*10^-24g/(at om#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We know that the mass of 1 mol of hydrogen is 1 gram. 1 mol of hydrogen is <mathjax>#6.022*10^23#</mathjax> atoms. To find the mass of 1 atom, we just have to divide.</p>
<p><mathjax>#(1g H)/(6.022*10^23 at oms H)#</mathjax>=<mathjax>#1.66*10^-24g/(at om#</mathjax></p></div>
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</article> | How do you calculate the mass of a #""^1H# atom? | null |
1,383 | a8df4092-6ddd-11ea-9e4c-ccda262736ce | https://socratic.org/questions/a-given-mass-of-oxygen-at-room-temperature-occupies-a-volume-of-500-0-ml-at-1-50 | 5.00 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 4 4 15 16 pressure qc_end physical_unit 4 4 12 13 volume qc_end physical_unit 4 4 32 33 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] oxygen gas [IN] atm"}] | [{"type":"physical unit","value":"5.00 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] oxygen gas [=] \\pu{1.50 atm}"},{"type":"physical unit","value":"Volume1 [OF] oxygen gas [=] \\pu{500.0 mL}"},{"type":"physical unit","value":"Volume2 [OF] oxygen gas [=] \\pu{150.0 mL}"},{"type":"other","value":"Room temperature."}] | <h1 class="questionTitle" itemprop="name">A given mass of oxygen at room temperature occupies a volume of 500.0 mL at 1.50 atm pressure. What pressure must be applied to compress the gas to a volume of only 150.0 mL?</h1> | null | 5.00 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles are kept constant, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>This implies that <strong>decreasing</strong> the pressure by a given factor, let's say <mathjax>#k#</mathjax>, will cause the volume to <strong>increase</strong> by the <em><strong>same factor</strong></em> <mathjax>#k#</mathjax>. </p>
<p>Similarly, <strong>increasing</strong> the pressure by a factor <mathjax>#k#</mathjax> will cause the volume to <strong>decrease</strong> by the same factor <mathjax>#k#</mathjax>. </p>
<p><img alt="https://commons.wikimedia.org" src="https://useruploads.socratic.org/e8R7ffE7RumlwtxbIkXD_2314_Boyles_Law.jpg"/> </p>
<p>In your case, the volume decreases by a factor of </p>
<blockquote>
<p><mathjax>#(500.0color(red)(cancel(color(black)("mL"))))/(150.0color(red)(cancel(color(black)("mL")))) = color(blue)(10/3)#</mathjax></p>
</blockquote>
<p>This means that the pressure of the gas <strong>increased</strong> by a factor of <mathjax>#color(blue)(10/3)#</mathjax>, which would make its final value equal to </p>
<blockquote>
<p><mathjax>#P_"final" = color(blue)(10/3) * "1.50 atm" = color(green)(|bar(ul(color(white)(a/a)color(black)("5.00 atm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the initial pressure of the gas. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"5.00 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles are kept constant, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>This implies that <strong>decreasing</strong> the pressure by a given factor, let's say <mathjax>#k#</mathjax>, will cause the volume to <strong>increase</strong> by the <em><strong>same factor</strong></em> <mathjax>#k#</mathjax>. </p>
<p>Similarly, <strong>increasing</strong> the pressure by a factor <mathjax>#k#</mathjax> will cause the volume to <strong>decrease</strong> by the same factor <mathjax>#k#</mathjax>. </p>
<p><img alt="https://commons.wikimedia.org" src="https://useruploads.socratic.org/e8R7ffE7RumlwtxbIkXD_2314_Boyles_Law.jpg"/> </p>
<p>In your case, the volume decreases by a factor of </p>
<blockquote>
<p><mathjax>#(500.0color(red)(cancel(color(black)("mL"))))/(150.0color(red)(cancel(color(black)("mL")))) = color(blue)(10/3)#</mathjax></p>
</blockquote>
<p>This means that the pressure of the gas <strong>increased</strong> by a factor of <mathjax>#color(blue)(10/3)#</mathjax>, which would make its final value equal to </p>
<blockquote>
<p><mathjax>#P_"final" = color(blue)(10/3) * "1.50 atm" = color(green)(|bar(ul(color(white)(a/a)color(black)("5.00 atm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the initial pressure of the gas. </p></div>
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<h1 class="questionTitle" itemprop="name">A given mass of oxygen at room temperature occupies a volume of 500.0 mL at 1.50 atm pressure. What pressure must be applied to compress the gas to a volume of only 150.0 mL?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"5.00 atm"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles are kept constant, as described by <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>This implies that <strong>decreasing</strong> the pressure by a given factor, let's say <mathjax>#k#</mathjax>, will cause the volume to <strong>increase</strong> by the <em><strong>same factor</strong></em> <mathjax>#k#</mathjax>. </p>
<p>Similarly, <strong>increasing</strong> the pressure by a factor <mathjax>#k#</mathjax> will cause the volume to <strong>decrease</strong> by the same factor <mathjax>#k#</mathjax>. </p>
<p><img alt="https://commons.wikimedia.org" src="https://useruploads.socratic.org/e8R7ffE7RumlwtxbIkXD_2314_Boyles_Law.jpg"/> </p>
<p>In your case, the volume decreases by a factor of </p>
<blockquote>
<p><mathjax>#(500.0color(red)(cancel(color(black)("mL"))))/(150.0color(red)(cancel(color(black)("mL")))) = color(blue)(10/3)#</mathjax></p>
</blockquote>
<p>This means that the pressure of the gas <strong>increased</strong> by a factor of <mathjax>#color(blue)(10/3)#</mathjax>, which would make its final value equal to </p>
<blockquote>
<p><mathjax>#P_"final" = color(blue)(10/3) * "1.50 atm" = color(green)(|bar(ul(color(white)(a/a)color(black)("5.00 atm")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the initial pressure of the gas. </p></div>
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</article> | A given mass of oxygen at room temperature occupies a volume of 500.0 mL at 1.50 atm pressure. What pressure must be applied to compress the gas to a volume of only 150.0 mL? | null |
1,384 | a8a7e3a8-6ddd-11ea-89a8-ccda262736ce | https://socratic.org/questions/if-a-0-075-liter-solution-contains-0877-moles-of-cuco-4-what-is-the-molarity | 1.17 mol/L | start physical_unit 4 4 molarity mol/l qc_end physical_unit 4 4 2 3 volume qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Molarity [OF] solution [IN] mol/L"}] | [{"type":"physical unit","value":"1.17 mol/L"}] | [{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{0.075 liter}"},{"type":"physical unit","value":"Mole [OF] CuCO4 [=] \\pu{0.0877 moles}"}] | <h1 class="questionTitle" itemprop="name">If a 0.075 liter solution contains .0877 moles of #CuCO_4#, what is the molarity?</h1> | null | 1.17 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(0.0877*mol)/(0.075*L)=1.17*mol*L^-1#</mathjax>..............</p></div>
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<div class="markdown"><p>Well, let's make the solute <mathjax>#CuCl_2#</mathjax>...........Why? Because there ain't no such beast as <mathjax>#CuCO_4#</mathjax>. (There are copper carbonyls, but I don't think that's what you mean.)</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(0.0877*mol)/(0.075*L)=1.17*mol*L^-1#</mathjax>..............</p></div>
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<h1 class="questionTitle" itemprop="name">If a 0.075 liter solution contains .0877 moles of #CuCO_4#, what is the molarity?</h1>
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anor277
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<div class="markdown"><p>Well, let's make the solute <mathjax>#CuCl_2#</mathjax>...........Why? Because there ain't no such beast as <mathjax>#CuCO_4#</mathjax>. (There are copper carbonyls, but I don't think that's what you mean.)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Molarity"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(0.0877*mol)/(0.075*L)=1.17*mol*L^-1#</mathjax>..............</p></div>
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</article> | If a 0.075 liter solution contains .0877 moles of #CuCO_4#, what is the molarity? | null |
1,385 | a8549c26-6ddd-11ea-9a38-ccda262736ce | https://socratic.org/questions/what-is-k-eq-for-the-reaction-n-2-3h-2-2nh-3-if-the-equilibrium-concentrations-a | 4.5 | start physical_unit 4 5 equilibrium_constant_k none qc_end chemical_equation 6 12 qc_end physical_unit 18 18 20 21 equilibrium_concentration qc_end physical_unit 22 22 24 25 equilibrium_concentration qc_end physical_unit 27 27 29 30 equilibrium_concentration qc_end end | [{"type":"physical unit","value":"Keq [OF] the reaction"}] | [{"type":"physical unit","value":"4.5"}] | [{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"},{"type":"physical unit","value":"equilibrium concentration [OF] [NH3] [=] \\pu{3.0 M}"},{"type":"physical unit","value":"equilibrium concentration [OF] [N2] [=] \\pu{2.0 M}"},{"type":"physical unit","value":"equilibrium concentration [OF] [H2] [=] \\pu{1.0 M}"}] | <h1 class="questionTitle" itemprop="name">What is #K_(eq)# for the reaction #N_2+3H_2 -> 2NH_3# if the equilibrium concentrations are #[NH_3] = 3.0 M, [N_2] = 2.0M#, and #[H_2] = 1.0M#?</h1> | null | 4.5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <strong>equilibrium constant</strong> for a given chemical equilibrium depends on </p>
<ul>
<li><em>the <strong>equilibrium concentrations</strong> of the chemical species that take part in the reaction</em></li>
<li><em>the <strong>stoichiometric coefficients</strong> of these chemical species</em></li>
</ul>
<p>More specifically, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the <strong>products</strong> and the product of the equilibrium concentrations of the <strong>reactants</strong>, each raised to the power of their <em>stoichiometric coefficients</em>. </p>
<p>Your equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"N"_text(2(g]) + color(red)(3)"H"_text(2(g]) rightleftharpoons color(blue)(2)"NH"_text(3(g])#</mathjax></p>
</blockquote>
<p>By definition, the equilibrium constant for this reaction, <mathjax>#K_c#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_c = (["NH"_3]^color(blue)(2))/(["N"_2] * ["H"_2]^color(red)(3))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#["NH"_3]#</mathjax>, <mathjax>#["N"_2]#</mathjax>, <mathjax>#["H"_2]#</mathjax> represent the <strong>equilibrium concentrations</strong> of the three species. </p>
<p>In your case, you know that these concentrations are equal to </p>
<blockquote>
<p><mathjax>#["NH"_3] = "3.0 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "1.0 M"#</mathjax></p>
<p><mathjax>#["N"_2] = "2.0 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant will be equal to - keep in mind that equilibrium constant are usually used <em>without corresponding units</em></p>
<blockquote>
<p><mathjax>#K_c = 3.0^color(blue)(2)/(2.0 * 1.0^color(red)(3)) = 9/2 = color(green)(4.5)#</mathjax></p>
</blockquote>
<p>If you want, you can include the units in the calculation to get </p>
<blockquote>
<p><mathjax>#K_c = (3.0^color(blue)(2) "M"^color(blue)(2))/("2.0 M" * 1.0^color(red)(3)"M"^color(red)(3)) = "4.5 M"^(-2)#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#K_c = 4.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <strong>equilibrium constant</strong> for a given chemical equilibrium depends on </p>
<ul>
<li><em>the <strong>equilibrium concentrations</strong> of the chemical species that take part in the reaction</em></li>
<li><em>the <strong>stoichiometric coefficients</strong> of these chemical species</em></li>
</ul>
<p>More specifically, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the <strong>products</strong> and the product of the equilibrium concentrations of the <strong>reactants</strong>, each raised to the power of their <em>stoichiometric coefficients</em>. </p>
<p>Your equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"N"_text(2(g]) + color(red)(3)"H"_text(2(g]) rightleftharpoons color(blue)(2)"NH"_text(3(g])#</mathjax></p>
</blockquote>
<p>By definition, the equilibrium constant for this reaction, <mathjax>#K_c#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_c = (["NH"_3]^color(blue)(2))/(["N"_2] * ["H"_2]^color(red)(3))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#["NH"_3]#</mathjax>, <mathjax>#["N"_2]#</mathjax>, <mathjax>#["H"_2]#</mathjax> represent the <strong>equilibrium concentrations</strong> of the three species. </p>
<p>In your case, you know that these concentrations are equal to </p>
<blockquote>
<p><mathjax>#["NH"_3] = "3.0 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "1.0 M"#</mathjax></p>
<p><mathjax>#["N"_2] = "2.0 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant will be equal to - keep in mind that equilibrium constant are usually used <em>without corresponding units</em></p>
<blockquote>
<p><mathjax>#K_c = 3.0^color(blue)(2)/(2.0 * 1.0^color(red)(3)) = 9/2 = color(green)(4.5)#</mathjax></p>
</blockquote>
<p>If you want, you can include the units in the calculation to get </p>
<blockquote>
<p><mathjax>#K_c = (3.0^color(blue)(2) "M"^color(blue)(2))/("2.0 M" * 1.0^color(red)(3)"M"^color(red)(3)) = "4.5 M"^(-2)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is #K_(eq)# for the reaction #N_2+3H_2 -> 2NH_3# if the equilibrium concentrations are #[NH_3] = 3.0 M, [N_2] = 2.0M#, and #[H_2] = 1.0M#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#K_c = 4.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <strong>equilibrium constant</strong> for a given chemical equilibrium depends on </p>
<ul>
<li><em>the <strong>equilibrium concentrations</strong> of the chemical species that take part in the reaction</em></li>
<li><em>the <strong>stoichiometric coefficients</strong> of these chemical species</em></li>
</ul>
<p>More specifically, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the <strong>products</strong> and the product of the equilibrium concentrations of the <strong>reactants</strong>, each raised to the power of their <em>stoichiometric coefficients</em>. </p>
<p>Your equilibrium looks like this </p>
<blockquote>
<p><mathjax>#"N"_text(2(g]) + color(red)(3)"H"_text(2(g]) rightleftharpoons color(blue)(2)"NH"_text(3(g])#</mathjax></p>
</blockquote>
<p>By definition, the equilibrium constant for this reaction, <mathjax>#K_c#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#K_c = (["NH"_3]^color(blue)(2))/(["N"_2] * ["H"_2]^color(red)(3))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#["NH"_3]#</mathjax>, <mathjax>#["N"_2]#</mathjax>, <mathjax>#["H"_2]#</mathjax> represent the <strong>equilibrium concentrations</strong> of the three species. </p>
<p>In your case, you know that these concentrations are equal to </p>
<blockquote>
<p><mathjax>#["NH"_3] = "3.0 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "1.0 M"#</mathjax></p>
<p><mathjax>#["N"_2] = "2.0 M"#</mathjax></p>
</blockquote>
<p>This means that the equilibrium constant will be equal to - keep in mind that equilibrium constant are usually used <em>without corresponding units</em></p>
<blockquote>
<p><mathjax>#K_c = 3.0^color(blue)(2)/(2.0 * 1.0^color(red)(3)) = 9/2 = color(green)(4.5)#</mathjax></p>
</blockquote>
<p>If you want, you can include the units in the calculation to get </p>
<blockquote>
<p><mathjax>#K_c = (3.0^color(blue)(2) "M"^color(blue)(2))/("2.0 M" * 1.0^color(red)(3)"M"^color(red)(3)) = "4.5 M"^(-2)#</mathjax></p>
</blockquote></div>
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</article> | What is #K_(eq)# for the reaction #N_2+3H_2 -> 2NH_3# if the equilibrium concentrations are #[NH_3] = 3.0 M, [N_2] = 2.0M#, and #[H_2] = 1.0M#? | null |
1,386 | aa25de1c-6ddd-11ea-9240-ccda262736ce | https://socratic.org/questions/an-after-shave-lotion-contains-50-ethyl-alcohol-how-much-water-must-be-added-to- | 6.00 fluid ounces | start physical_unit 10 10 volume fluid_ounces qc_end physical_unit 3 3 15 17 volume qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] fluid ounces"}] | [{"type":"physical unit","value":"6.00 fluid ounces"}] | [{"type":"physical unit","value":"Volume [OF] lotion [=] \\pu{6 fluid ounces}"},{"type":"physical unit","value":"Percent [OF] ethyl alcohol in lotion [=] \\pu{50%}"},{"type":"physical unit","value":"Percent [OF] nonalcohol in lotion [=] \\pu{75%}"}] | <h1 class="questionTitle" itemprop="name">An after shave lotion contains 50% ethyl alcohol. How much water must be added to 6 fluid ounces of this lotion to reduce it to one which is 75% nonalcoholic?</h1> | null | 6.00 fluid ounces | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, your solution is <mathjax>#50% "v/v"#</mathjax>, which means that you get <strong>50 lf. oz</strong> of ethanol <em>for every</em> <strong>100 lf. oz</strong> of solution. </p>
<p>This means that your <strong>6-fl. oz</strong> sample will contain </p>
<p><mathjax>#6color(red)(cancel(color(black)("fl. oz. solution"))) * "50 fl. oz ethanol"/(100color(red)(cancel(color(black)("fl. oz. solution")))) = "3 fl. oz ethanol"#</mathjax></p>
<p>Now, your target solution must be <strong>75%</strong> nonalcoholic, which means that it must have a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by volume</a> of <strong>25%</strong>. </p>
<p>The important thing to ralize here is tha the volume of ethanol <strong>will be the same</strong> for both your starting solution, and for your target solution. </p>
<p>The only thing that will change will be the volume of solution. </p>
<p>This means that you can write </p>
<p><mathjax>#"3 fl. oz ethanol"/((6 + x)"fl. oz. solution") * 100 = 25%#</mathjax>, where</p>
<p><mathjax>#x#</mathjax> - the volume of water you need to add to the 50% v/v solution. </p>
<p>Rearrange this equation and solve for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#3 * 100 = 25 * (6+x)#</mathjax></p>
<p><mathjax>#300 = 150 + 25x => x= 150/25 = color(green)("6 fl. oz.")#</mathjax></p>
<p>SImply put, if you add enough water to double the volume of the initial solution, the percent concetration of ethanol will go from 50 to 25%. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You need to add <strong>6 fl. oz.</strong> of water. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, your solution is <mathjax>#50% "v/v"#</mathjax>, which means that you get <strong>50 lf. oz</strong> of ethanol <em>for every</em> <strong>100 lf. oz</strong> of solution. </p>
<p>This means that your <strong>6-fl. oz</strong> sample will contain </p>
<p><mathjax>#6color(red)(cancel(color(black)("fl. oz. solution"))) * "50 fl. oz ethanol"/(100color(red)(cancel(color(black)("fl. oz. solution")))) = "3 fl. oz ethanol"#</mathjax></p>
<p>Now, your target solution must be <strong>75%</strong> nonalcoholic, which means that it must have a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by volume</a> of <strong>25%</strong>. </p>
<p>The important thing to ralize here is tha the volume of ethanol <strong>will be the same</strong> for both your starting solution, and for your target solution. </p>
<p>The only thing that will change will be the volume of solution. </p>
<p>This means that you can write </p>
<p><mathjax>#"3 fl. oz ethanol"/((6 + x)"fl. oz. solution") * 100 = 25%#</mathjax>, where</p>
<p><mathjax>#x#</mathjax> - the volume of water you need to add to the 50% v/v solution. </p>
<p>Rearrange this equation and solve for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#3 * 100 = 25 * (6+x)#</mathjax></p>
<p><mathjax>#300 = 150 + 25x => x= 150/25 = color(green)("6 fl. oz.")#</mathjax></p>
<p>SImply put, if you add enough water to double the volume of the initial solution, the percent concetration of ethanol will go from 50 to 25%. </p></div>
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<h1 class="questionTitle" itemprop="name">An after shave lotion contains 50% ethyl alcohol. How much water must be added to 6 fluid ounces of this lotion to reduce it to one which is 75% nonalcoholic?</h1>
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<div class="markdown"><p>You need to add <strong>6 fl. oz.</strong> of water. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, your solution is <mathjax>#50% "v/v"#</mathjax>, which means that you get <strong>50 lf. oz</strong> of ethanol <em>for every</em> <strong>100 lf. oz</strong> of solution. </p>
<p>This means that your <strong>6-fl. oz</strong> sample will contain </p>
<p><mathjax>#6color(red)(cancel(color(black)("fl. oz. solution"))) * "50 fl. oz ethanol"/(100color(red)(cancel(color(black)("fl. oz. solution")))) = "3 fl. oz ethanol"#</mathjax></p>
<p>Now, your target solution must be <strong>75%</strong> nonalcoholic, which means that it must have a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by volume</a> of <strong>25%</strong>. </p>
<p>The important thing to ralize here is tha the volume of ethanol <strong>will be the same</strong> for both your starting solution, and for your target solution. </p>
<p>The only thing that will change will be the volume of solution. </p>
<p>This means that you can write </p>
<p><mathjax>#"3 fl. oz ethanol"/((6 + x)"fl. oz. solution") * 100 = 25%#</mathjax>, where</p>
<p><mathjax>#x#</mathjax> - the volume of water you need to add to the 50% v/v solution. </p>
<p>Rearrange this equation and solve for <mathjax>#x#</mathjax> to get </p>
<p><mathjax>#3 * 100 = 25 * (6+x)#</mathjax></p>
<p><mathjax>#300 = 150 + 25x => x= 150/25 = color(green)("6 fl. oz.")#</mathjax></p>
<p>SImply put, if you add enough water to double the volume of the initial solution, the percent concetration of ethanol will go from 50 to 25%. </p></div>
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</article> | An after shave lotion contains 50% ethyl alcohol. How much water must be added to 6 fluid ounces of this lotion to reduce it to one which is 75% nonalcoholic? | null |
1,387 | a91d8022-6ddd-11ea-bbc0-ccda262736ce | https://socratic.org/questions/a-175-0-g-sample-of-a-compound-contains-56-15-g-c-9-43-g-h3-74-81-g-o-13-11-g-n- | C5H10NO5Na2 | start chemical_formula qc_end physical_unit 3 6 1 2 mass qc_end physical_unit 10 10 8 9 mass qc_end physical_unit 13 13 11 12 mass qc_end physical_unit 15 15 13 14 mass qc_end physical_unit 17 17 15 16 mass qc_end physical_unit 20 20 18 19 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound sample [IN] empirical"}] | [{"type":"chemical equation","value":"C5H10NO5Na2"}] | [{"type":"physical unit","value":"Mass [OF] a compound sample [=] \\pu{175.0 g}"},{"type":"physical unit","value":"Mass [OF] C [=] \\pu{56.15 g}"},{"type":"physical unit","value":"Mass [OF] H [=] \\pu{9.43 g}"},{"type":"physical unit","value":"Mass [OF] O [=] \\pu{74.81 g}"},{"type":"physical unit","value":"Mass [OF] N [=] \\pu{13.11 g}"},{"type":"physical unit","value":"Mass [OF] Na [=] \\pu{21.49 g}"}] | <h1 class="questionTitle" itemprop="name">A 175.0 g sample of a compound contains 56.15 g #C#, 9.43 g #H3, 74.81 g #O#, 13.11 g #N#, and 21.49 g #Na.# What is its empirical formula?</h1> | null | C5H10NO5Na2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need the percentage composition of each element:</p>
<p><mathjax>#%C:#</mathjax> <mathjax>#(56.15*g)/(175.0*g)xx100%=32.1%#</mathjax></p>
<p><mathjax>#%H:#</mathjax> <mathjax>#(9.43*g)/(175*g)xx100%=5.39%#</mathjax></p>
<p><mathjax>#%O:#</mathjax> <mathjax>#(74.81*g)/(175*g)xx100%=42.8%#</mathjax></p>
<p><mathjax>#%N:#</mathjax> <mathjax>#(13.11*g)/(175*g)xx100%=7.50%#</mathjax></p>
<p><mathjax>#%Na:#</mathjax> <mathjax>#(21.49*g)/(175*g)xx100%=12.28%#</mathjax></p>
<p>Given this percentage makeup, we can work out an empirical formula by (i) assuming that there are <mathjax>#100*g#</mathjax> of unknown compound, and dividing the individual, elemental masses by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each component element:</p>
<p><mathjax>#C:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.1*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#H:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.39*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5.35*mol#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(42.8*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#N:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.50*g)/(14.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.535*mol#</mathjax>.</p>
<p><mathjax>#Na:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.49*g)/(22.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.935*mol#</mathjax>.</p>
<p>Now we divide each molar quantity thru by the SMALLEST MOLAR QUANTITY, that of nitrogen, to give an empirical formula of <mathjax>#C_5H_10NO_5Na_2#</mathjax>. Note that the sodium percentage was a bit out, but clearly we want WHOLE numbers, so a bit of rounding up is acceptable.</p>
<p>Now of course, I have done this the long way round, and made a real meal out of the the problem. I could have missed the step where the percentages are calculated, and been a bit more direct. I did it the long way, because often you are quoted percentage composition by mass in empirical formula problems. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>We calculate (i) the percentage composition, and then (ii) the emprical formula, to give an empirical formula of <mathjax>#C_5H_10NO_5Na_2#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need the percentage composition of each element:</p>
<p><mathjax>#%C:#</mathjax> <mathjax>#(56.15*g)/(175.0*g)xx100%=32.1%#</mathjax></p>
<p><mathjax>#%H:#</mathjax> <mathjax>#(9.43*g)/(175*g)xx100%=5.39%#</mathjax></p>
<p><mathjax>#%O:#</mathjax> <mathjax>#(74.81*g)/(175*g)xx100%=42.8%#</mathjax></p>
<p><mathjax>#%N:#</mathjax> <mathjax>#(13.11*g)/(175*g)xx100%=7.50%#</mathjax></p>
<p><mathjax>#%Na:#</mathjax> <mathjax>#(21.49*g)/(175*g)xx100%=12.28%#</mathjax></p>
<p>Given this percentage makeup, we can work out an empirical formula by (i) assuming that there are <mathjax>#100*g#</mathjax> of unknown compound, and dividing the individual, elemental masses by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each component element:</p>
<p><mathjax>#C:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.1*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#H:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.39*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5.35*mol#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(42.8*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#N:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.50*g)/(14.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.535*mol#</mathjax>.</p>
<p><mathjax>#Na:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.49*g)/(22.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.935*mol#</mathjax>.</p>
<p>Now we divide each molar quantity thru by the SMALLEST MOLAR QUANTITY, that of nitrogen, to give an empirical formula of <mathjax>#C_5H_10NO_5Na_2#</mathjax>. Note that the sodium percentage was a bit out, but clearly we want WHOLE numbers, so a bit of rounding up is acceptable.</p>
<p>Now of course, I have done this the long way round, and made a real meal out of the the problem. I could have missed the step where the percentages are calculated, and been a bit more direct. I did it the long way, because often you are quoted percentage composition by mass in empirical formula problems. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 175.0 g sample of a compound contains 56.15 g #C#, 9.43 g #H3, 74.81 g #O#, 13.11 g #N#, and 21.49 g #Na.# What is its empirical formula?</h1>
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<div class="markdown"><p>We calculate (i) the percentage composition, and then (ii) the emprical formula, to give an empirical formula of <mathjax>#C_5H_10NO_5Na_2#</mathjax>. </p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need the percentage composition of each element:</p>
<p><mathjax>#%C:#</mathjax> <mathjax>#(56.15*g)/(175.0*g)xx100%=32.1%#</mathjax></p>
<p><mathjax>#%H:#</mathjax> <mathjax>#(9.43*g)/(175*g)xx100%=5.39%#</mathjax></p>
<p><mathjax>#%O:#</mathjax> <mathjax>#(74.81*g)/(175*g)xx100%=42.8%#</mathjax></p>
<p><mathjax>#%N:#</mathjax> <mathjax>#(13.11*g)/(175*g)xx100%=7.50%#</mathjax></p>
<p><mathjax>#%Na:#</mathjax> <mathjax>#(21.49*g)/(175*g)xx100%=12.28%#</mathjax></p>
<p>Given this percentage makeup, we can work out an empirical formula by (i) assuming that there are <mathjax>#100*g#</mathjax> of unknown compound, and dividing the individual, elemental masses by the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC mass</a> of each component element:</p>
<p><mathjax>#C:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.1*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#H:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.39*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5.35*mol#</mathjax>.</p>
<p><mathjax>#O:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(42.8*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.67*mol#</mathjax>.</p>
<p><mathjax>#N:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.50*g)/(14.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.535*mol#</mathjax>.</p>
<p><mathjax>#Na:#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.49*g)/(22.99*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.935*mol#</mathjax>.</p>
<p>Now we divide each molar quantity thru by the SMALLEST MOLAR QUANTITY, that of nitrogen, to give an empirical formula of <mathjax>#C_5H_10NO_5Na_2#</mathjax>. Note that the sodium percentage was a bit out, but clearly we want WHOLE numbers, so a bit of rounding up is acceptable.</p>
<p>Now of course, I have done this the long way round, and made a real meal out of the the problem. I could have missed the step where the percentages are calculated, and been a bit more direct. I did it the long way, because often you are quoted percentage composition by mass in empirical formula problems. </p></div>
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</article> | A 175.0 g sample of a compound contains 56.15 g #C#, 9.43 g #H3, 74.81 g #O#, 13.11 g #N#, and 21.49 g #Na.# What is its empirical formula? | null |
1,388 | ac0bf0c8-6ddd-11ea-94fd-ccda262736ce | https://socratic.org/questions/how-many-liters-of-oxygen-are-required-to-react-completely-with-2-4-liters-of-hy | 1.20 liters | start physical_unit 4 4 volume l qc_end physical_unit 14 14 11 12 volume qc_end substance 17 17 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] liters"}] | [{"type":"physical unit","value":"1.20 liters"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{2.4 liters}"},{"type":"substance name","value":"Water"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? </h1> | null | 1.20 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the stoichiometric equation...</p>
<p><mathjax>#H_2(g) + 1/2O_2(g) rarrH_2O(l)#</mathjax></p>
<p>Given the Avogadro condition, equal volumes of gases contain an equal number of particles, stoichiometric equivalence demands addition of HALF the volume of dioxygen.</p></div>
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<div class="markdown"><p>Why...<mathjax>#1.2*L#</mathjax>...</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the stoichiometric equation...</p>
<p><mathjax>#H_2(g) + 1/2O_2(g) rarrH_2O(l)#</mathjax></p>
<p>Given the Avogadro condition, equal volumes of gases contain an equal number of particles, stoichiometric equivalence demands addition of HALF the volume of dioxygen.</p></div>
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<div class="markdown"><p>Why...<mathjax>#1.2*L#</mathjax>...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Look at the stoichiometric equation...</p>
<p><mathjax>#H_2(g) + 1/2O_2(g) rarrH_2O(l)#</mathjax></p>
<p>Given the Avogadro condition, equal volumes of gases contain an equal number of particles, stoichiometric equivalence demands addition of HALF the volume of dioxygen.</p></div>
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</article> | How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? | null |
1,389 | acf757f5-6ddd-11ea-8e76-ccda262736ce | https://socratic.org/questions/how-can-i-write-the-formula-for-aluminum-oxide | Al2O3 | start chemical_formula qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] aluminum oxide [IN] default"}] | [{"type":"chemical equation","value":"Al2O3"}] | [{"type":"substance name","value":"Aluminum oxide"}] | <h1 class="questionTitle" itemprop="name">How can I write the formula for aluminum oxide?</h1> | null | Al2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The correct answer is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>Let us see how we got the answer; Look at the electronic arrangement of Al and O atoms.</p>
<p>Al ( Z= 13) has 13 electrons with following electronic configuration. 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax>3<mathjax>#s^2#</mathjax>3<mathjax>#p^1#</mathjax></p>
<p>It loses three electron in its 3s and 3p subshell to achieve stability and forms ion <mathjax>#Al^(3+)#</mathjax>.</p>
<p><mathjax>#Al^(3+)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>O ( Z=8) on the other hand has eight electrons and wants to gain two electrons to achieve stable noble gas configuration.Oxygen atom on gaining two electrons forms negative oxide ion, <mathjax>#O^(2-)#</mathjax> ion.</p>
<p>O ( Z=7) = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^4#</mathjax></p>
<p><mathjax>#O^(2-)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes <mathjax>#Al^(3+)#</mathjax> ion and Oxygen atom after gaining two electrons becomes <mathjax>#O^(2-)#</mathjax> ion.</p>
<p><img alt="http://gallery4share.com/a/aluminum-ion-lewis-dot-structure.html" src="https://useruploads.socratic.org/fdem7yigQvykudaJaOs1_Ionic_compound_of_O_and_Al.jpg"/></p>
<p>2<mathjax>#Al^(3+)#</mathjax> and 3 <mathjax>#O^(2-)#</mathjax></p>
<p>or the formula is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/3AJVg8dAYJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The formula for aluminum oxide is <mathjax>#Al_2O_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The correct answer is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>Let us see how we got the answer; Look at the electronic arrangement of Al and O atoms.</p>
<p>Al ( Z= 13) has 13 electrons with following electronic configuration. 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax>3<mathjax>#s^2#</mathjax>3<mathjax>#p^1#</mathjax></p>
<p>It loses three electron in its 3s and 3p subshell to achieve stability and forms ion <mathjax>#Al^(3+)#</mathjax>.</p>
<p><mathjax>#Al^(3+)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>O ( Z=8) on the other hand has eight electrons and wants to gain two electrons to achieve stable noble gas configuration.Oxygen atom on gaining two electrons forms negative oxide ion, <mathjax>#O^(2-)#</mathjax> ion.</p>
<p>O ( Z=7) = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^4#</mathjax></p>
<p><mathjax>#O^(2-)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes <mathjax>#Al^(3+)#</mathjax> ion and Oxygen atom after gaining two electrons becomes <mathjax>#O^(2-)#</mathjax> ion.</p>
<p><img alt="http://gallery4share.com/a/aluminum-ion-lewis-dot-structure.html" src="https://useruploads.socratic.org/fdem7yigQvykudaJaOs1_Ionic_compound_of_O_and_Al.jpg"/></p>
<p>2<mathjax>#Al^(3+)#</mathjax> and 3 <mathjax>#O^(2-)#</mathjax></p>
<p>or the formula is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/3AJVg8dAYJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p>The formula for aluminum oxide is <mathjax>#Al_2O_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The correct answer is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>Let us see how we got the answer; Look at the electronic arrangement of Al and O atoms.</p>
<p>Al ( Z= 13) has 13 electrons with following electronic configuration. 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax>3<mathjax>#s^2#</mathjax>3<mathjax>#p^1#</mathjax></p>
<p>It loses three electron in its 3s and 3p subshell to achieve stability and forms ion <mathjax>#Al^(3+)#</mathjax>.</p>
<p><mathjax>#Al^(3+)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>O ( Z=8) on the other hand has eight electrons and wants to gain two electrons to achieve stable noble gas configuration.Oxygen atom on gaining two electrons forms negative oxide ion, <mathjax>#O^(2-)#</mathjax> ion.</p>
<p>O ( Z=7) = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^4#</mathjax></p>
<p><mathjax>#O^(2-)#</mathjax> = 1<mathjax>#s^2#</mathjax>2<mathjax>#s^2#</mathjax>2<mathjax>#p^6#</mathjax></p>
<p>Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes <mathjax>#Al^(3+)#</mathjax> ion and Oxygen atom after gaining two electrons becomes <mathjax>#O^(2-)#</mathjax> ion.</p>
<p><img alt="http://gallery4share.com/a/aluminum-ion-lewis-dot-structure.html" src="https://useruploads.socratic.org/fdem7yigQvykudaJaOs1_Ionic_compound_of_O_and_Al.jpg"/></p>
<p>2<mathjax>#Al^(3+)#</mathjax> and 3 <mathjax>#O^(2-)#</mathjax></p>
<p>or the formula is <mathjax>#Al_2O_3#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/3AJVg8dAYJk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How can I write the formula for aluminum oxide? | null |
1,390 | ac6e200a-6ddd-11ea-822e-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-7-744-grams-of-cobalt | 0.13 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] cobalt [IN] moles"}] | [{"type":"physical unit","value":"0.13 moles"}] | [{"type":"physical unit","value":"Mass [OF] cobalt [=] \\pu{7.744 grams}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 7.744 grams of cobalt? </h1> | null | 0.13 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar Mass"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(7.744*cancelg)/(58.93*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax></p></div>
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<div class="markdown"><p>A bit of a <mathjax>#1/10#</mathjax> of a mole. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar Mass"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(7.744*cancelg)/(58.93*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 7.744 grams of cobalt? </h1>
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<div class="markdown"><p>A bit of a <mathjax>#1/10#</mathjax> of a mole. </p></div>
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<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar Mass"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(7.744*cancelg)/(58.93*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax></p></div>
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</article> | How many moles are in 7.744 grams of cobalt? | null |
1,391 | abe679ab-6ddd-11ea-98b6-ccda262736ce | https://socratic.org/questions/what-is-the-value-of-ksp-for-tin-ii-sulfide-given-that-its-solubility-is-5-39-10 | 1.28 × 10^(-25) | start physical_unit 7 8 equilibrium_constant_k none qc_end physical_unit 7 8 14 19 solubility qc_end end | [{"type":"physical unit","value":"Ksp [OF] tin(II) sulfide"}] | [{"type":"physical unit","value":"1.28 × 10^(-25)"}] | [{"type":"physical unit","value":"Solubility [OF] tin(II) sulfide [=] \\pu{5.39 × 10^(-12) g/100 g H2O}"}] | <h1 class="questionTitle" itemprop="name">What is the value of #K_(sp)# for tin(II) sulfide given that its solubility is
#5.39 * 10^(−12)# #"g/100 g H"_2"O"# ?
</h1> | null | 1.28 × 10^(-25) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Tin(II) sulfide is considered <strong>insoluble</strong> in water, so you can write its dissociation equilibrium by taking into account the fact that a <strong>saturated solution</strong> of tin(II) sulfide contains the undissolved solid in equilibrium with the dissolved ions.</p>
<blockquote>
<p><mathjax>#"SnS"_ ((s)) rightleftharpoons "Sn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for tin(II) sulfide looks like this</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Sn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, you know that this salt has a solubility of <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> per <mathjax>#"100 g"#</mathjax> of water at an unspecified temperature. </p>
<p>This tells you that at that temperature, only <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 g"#</mathjax> of water <strong>will dissociate</strong> to produce tin(II) cations and sulfide anions in a <strong>saturated solution</strong> of tin(II) sulfide. </p>
<p>Since the problem made no mention of the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water, you can take it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax> and say that in a saturated solution of tin(II) sulfide, <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 mL"#</mathjax> of water will dissociate to produce tin(II) cations and sulfide anions. </p>
<p>Now, because the mass of tin(II) sulfide that dissociates is so small, you can say that the volume of solution will be equal to the volume of water. </p>
<p>Therefore, in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of this solution, only</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * (5.39 * 10^(-12) quad "g SnS")/(100color(red)(cancel(color(black)("mL solution")))) = 5.39 * 10^(-11) quad "g SnS"#</mathjax></p>
</blockquote>
<p>will dissociate to produce ions. Use the <strong>molar mass</strong> of tin(II) sulfide to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#5.39 * 10^(-11) color(red)(cancel(color(black)("g"))) * "1 mole SnS"/(150.76color(red)(cancel(color(black)("g")))) = 3.575 * 10^(-13) quad "moles SnS"#</mathjax></p>
</blockquote>
<p>Since each mole of tin(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of tin(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of sulfide anions, the saturated solution will contain </p>
<blockquote>
<p><mathjax>#["Sn"^(2+)] = ["S"^(2-)] = 3.575 * 10^(13) quad "M"#</mathjax></p>
</blockquote>
<p>You can thus say that the solubility product constant for tin(II) sulfide is equal to--I'll leave the value <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_(sp) = (3.575 * 10^(-13)) * (3.575 * 10^(-13)) = color(darkgreen)(ul(color(black)(1.28 * 10^(-25))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the solubility of the salt. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.28 * 10^(-25)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Tin(II) sulfide is considered <strong>insoluble</strong> in water, so you can write its dissociation equilibrium by taking into account the fact that a <strong>saturated solution</strong> of tin(II) sulfide contains the undissolved solid in equilibrium with the dissolved ions.</p>
<blockquote>
<p><mathjax>#"SnS"_ ((s)) rightleftharpoons "Sn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for tin(II) sulfide looks like this</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Sn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, you know that this salt has a solubility of <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> per <mathjax>#"100 g"#</mathjax> of water at an unspecified temperature. </p>
<p>This tells you that at that temperature, only <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 g"#</mathjax> of water <strong>will dissociate</strong> to produce tin(II) cations and sulfide anions in a <strong>saturated solution</strong> of tin(II) sulfide. </p>
<p>Since the problem made no mention of the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water, you can take it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax> and say that in a saturated solution of tin(II) sulfide, <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 mL"#</mathjax> of water will dissociate to produce tin(II) cations and sulfide anions. </p>
<p>Now, because the mass of tin(II) sulfide that dissociates is so small, you can say that the volume of solution will be equal to the volume of water. </p>
<p>Therefore, in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of this solution, only</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * (5.39 * 10^(-12) quad "g SnS")/(100color(red)(cancel(color(black)("mL solution")))) = 5.39 * 10^(-11) quad "g SnS"#</mathjax></p>
</blockquote>
<p>will dissociate to produce ions. Use the <strong>molar mass</strong> of tin(II) sulfide to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#5.39 * 10^(-11) color(red)(cancel(color(black)("g"))) * "1 mole SnS"/(150.76color(red)(cancel(color(black)("g")))) = 3.575 * 10^(-13) quad "moles SnS"#</mathjax></p>
</blockquote>
<p>Since each mole of tin(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of tin(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of sulfide anions, the saturated solution will contain </p>
<blockquote>
<p><mathjax>#["Sn"^(2+)] = ["S"^(2-)] = 3.575 * 10^(13) quad "M"#</mathjax></p>
</blockquote>
<p>You can thus say that the solubility product constant for tin(II) sulfide is equal to--I'll leave the value <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_(sp) = (3.575 * 10^(-13)) * (3.575 * 10^(-13)) = color(darkgreen)(ul(color(black)(1.28 * 10^(-25))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the solubility of the salt. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the value of #K_(sp)# for tin(II) sulfide given that its solubility is
#5.39 * 10^(−12)# #"g/100 g H"_2"O"# ?
</h1>
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Stefan V.
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Feb 25, 2018
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<div class="markdown"><p><mathjax>#1.28 * 10^(-25)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Tin(II) sulfide is considered <strong>insoluble</strong> in water, so you can write its dissociation equilibrium by taking into account the fact that a <strong>saturated solution</strong> of tin(II) sulfide contains the undissolved solid in equilibrium with the dissolved ions.</p>
<blockquote>
<p><mathjax>#"SnS"_ ((s)) rightleftharpoons "Sn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#</mathjax></p>
</blockquote>
<p>The <strong>solubility product constant</strong> for tin(II) sulfide looks like this</p>
<blockquote>
<p><mathjax>#K_(sp) = ["Sn"^(2+)] * ["S"^(2-)]#</mathjax></p>
</blockquote>
<p>Now, you know that this salt has a solubility of <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> per <mathjax>#"100 g"#</mathjax> of water at an unspecified temperature. </p>
<p>This tells you that at that temperature, only <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 g"#</mathjax> of water <strong>will dissociate</strong> to produce tin(II) cations and sulfide anions in a <strong>saturated solution</strong> of tin(II) sulfide. </p>
<p>Since the problem made no mention of the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water, you can take it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax> and say that in a saturated solution of tin(II) sulfide, <mathjax>#5.39 * 10^(-12)#</mathjax> <mathjax>#"g"#</mathjax> of tin(II) sulfide per <mathjax>#"100 mL"#</mathjax> of water will dissociate to produce tin(II) cations and sulfide anions. </p>
<p>Now, because the mass of tin(II) sulfide that dissociates is so small, you can say that the volume of solution will be equal to the volume of water. </p>
<p>Therefore, in <mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> of this solution, only</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * (5.39 * 10^(-12) quad "g SnS")/(100color(red)(cancel(color(black)("mL solution")))) = 5.39 * 10^(-11) quad "g SnS"#</mathjax></p>
</blockquote>
<p>will dissociate to produce ions. Use the <strong>molar mass</strong> of tin(II) sulfide to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#5.39 * 10^(-11) color(red)(cancel(color(black)("g"))) * "1 mole SnS"/(150.76color(red)(cancel(color(black)("g")))) = 3.575 * 10^(-13) quad "moles SnS"#</mathjax></p>
</blockquote>
<p>Since each mole of tin(II) sulfide <strong>that dissociates</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of tin(II) cations and <mathjax>#1#</mathjax> <strong>mole</strong> of sulfide anions, the saturated solution will contain </p>
<blockquote>
<p><mathjax>#["Sn"^(2+)] = ["S"^(2-)] = 3.575 * 10^(13) quad "M"#</mathjax></p>
</blockquote>
<p>You can thus say that the solubility product constant for tin(II) sulfide is equal to--I'll leave the value <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_(sp) = (3.575 * 10^(-13)) * (3.575 * 10^(-13)) = color(darkgreen)(ul(color(black)(1.28 * 10^(-25))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the solubility of the salt. </p></div>
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</article> | What is the value of #K_(sp)# for tin(II) sulfide given that its solubility is
#5.39 * 10^(−12)# #"g/100 g H"_2"O"# ?
| null |
1,392 | ac09cda2-6ddd-11ea-b9b2-ccda262736ce | https://socratic.org/questions/the-pressure-acting-on-500-cubic-meters-of-gas-is-reduced-from-4-atm-to-2-atm-if | 1000.00 cubic meters | start physical_unit 8 8 volume m^3 qc_end physical_unit 8 8 4 6 volume qc_end physical_unit 8 8 12 13 pressure qc_end physical_unit 8 8 15 16 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] cubic meters"}] | [{"type":"physical unit","value":"1000.00 cubic meters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{500 cubic meters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{4 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{2 atm}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume? </h1> | null | 1000.00 cubic meters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We will use <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that pressure and volume are inversely proportional at constant temperature. We can express this mathematically as:</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1="P"_2"V"_2#</mathjax></p>
</blockquote>
<p>Here, we see that</p>
<blockquote>
<p><mathjax>#"P"_1="4 atm"#</mathjax><br/>
<mathjax>#"P"_2="2 atm"#</mathjax></p>
<p><mathjax>#"V"_1="500 m"^3#</mathjax><br/>
<mathjax>#"V"_2="unknown"#</mathjax></p>
</blockquote>
<p>We want to manipulate Boyle's law to solve for <mathjax>#"V"_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#"V"_2=("P"_1"V"_1)/("P"_2)#</mathjax></p>
</blockquote>
<p>Plugging in what we know:</p>
<blockquote>
<p><mathjax>#"V"_2=("4 atm" * "500 m"^3)/("2 atm")#</mathjax></p>
<p><mathjax>#"V"_2=2 * "500 m"^3#</mathjax></p>
<p><mathjax>#"V"_2="1000 m"^3#</mathjax></p>
</blockquote>
<p>We can see that the volume increased from <mathjax>#"500 m"^3#</mathjax> to <mathjax>#"1000 m"^3#</mathjax>, so it doubled. This is paired with the pressure being halved, as should be the case in an inversely proportional relationship.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"V"_"new"="1000 m"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We will use <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that pressure and volume are inversely proportional at constant temperature. We can express this mathematically as:</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1="P"_2"V"_2#</mathjax></p>
</blockquote>
<p>Here, we see that</p>
<blockquote>
<p><mathjax>#"P"_1="4 atm"#</mathjax><br/>
<mathjax>#"P"_2="2 atm"#</mathjax></p>
<p><mathjax>#"V"_1="500 m"^3#</mathjax><br/>
<mathjax>#"V"_2="unknown"#</mathjax></p>
</blockquote>
<p>We want to manipulate Boyle's law to solve for <mathjax>#"V"_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#"V"_2=("P"_1"V"_1)/("P"_2)#</mathjax></p>
</blockquote>
<p>Plugging in what we know:</p>
<blockquote>
<p><mathjax>#"V"_2=("4 atm" * "500 m"^3)/("2 atm")#</mathjax></p>
<p><mathjax>#"V"_2=2 * "500 m"^3#</mathjax></p>
<p><mathjax>#"V"_2="1000 m"^3#</mathjax></p>
</blockquote>
<p>We can see that the volume increased from <mathjax>#"500 m"^3#</mathjax> to <mathjax>#"1000 m"^3#</mathjax>, so it doubled. This is paired with the pressure being halved, as should be the case in an inversely proportional relationship.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume? </h1>
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mason m
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<span class="dateCreated" datetime="2016-03-02T04:04:44" itemprop="dateCreated">
Mar 2, 2016
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<div class="markdown"><p><mathjax>#"V"_"new"="1000 m"^3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We will use <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that pressure and volume are inversely proportional at constant temperature. We can express this mathematically as:</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1="P"_2"V"_2#</mathjax></p>
</blockquote>
<p>Here, we see that</p>
<blockquote>
<p><mathjax>#"P"_1="4 atm"#</mathjax><br/>
<mathjax>#"P"_2="2 atm"#</mathjax></p>
<p><mathjax>#"V"_1="500 m"^3#</mathjax><br/>
<mathjax>#"V"_2="unknown"#</mathjax></p>
</blockquote>
<p>We want to manipulate Boyle's law to solve for <mathjax>#"V"_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#"V"_2=("P"_1"V"_1)/("P"_2)#</mathjax></p>
</blockquote>
<p>Plugging in what we know:</p>
<blockquote>
<p><mathjax>#"V"_2=("4 atm" * "500 m"^3)/("2 atm")#</mathjax></p>
<p><mathjax>#"V"_2=2 * "500 m"^3#</mathjax></p>
<p><mathjax>#"V"_2="1000 m"^3#</mathjax></p>
</blockquote>
<p>We can see that the volume increased from <mathjax>#"500 m"^3#</mathjax> to <mathjax>#"1000 m"^3#</mathjax>, so it doubled. This is paired with the pressure being halved, as should be the case in an inversely proportional relationship.</p></div>
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</article> | The pressure acting on 500 cubic meters of gas is reduced from 4 atm to 2 atm. If the temperature remains constant, what is the new volume? | null |
1,393 | a925bd94-6ddd-11ea-93b1-ccda262736ce | https://socratic.org/questions/how-do-you-write-an-equation-showing-how-hcl-which-is-a-molecular-substance-and- | HCl(g) + H2O(l) -> H3O+ + Cl- | start chemical_equation qc_end substance 23 23 qc_end chemical_equation 8 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"HCl(g) + H2O(l) -> H3O+ + Cl-"}] | [{"type":"substance name","value":"Water"},{"type":"chemical equation","value":"HCl"}] | <h1 class="questionTitle" itemprop="name">How do you write an equation showing how HCl, which is a molecular substance and strong acid, could give ions when added to water?</h1> | null | HCl(g) + H2O(l) -> H3O+ + Cl- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you say, hydrogen chloride is a room temperature gas; however, it is mighty soluble in water. And in water we conceive that it protonates the water molecular to give <mathjax>#H_3O^+#</mathjax>. <mathjax>#HCl#</mathjax> PROTONATES the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to form water-solvated hydronium ions and chloride ions......</p>
<p><mathjax>#H_3O^+#</mathjax>, the so-called hydronium ion, is our conception of a protonated water molecule. So far as anybody knows, it is a cluster of 2-4 water molecules with AN EXTRA <mathjax>#H^+#</mathjax> incorporated in the cluster, i.e. <mathjax>#H_5O_2^+#</mathjax> or <mathjax>#H_7O_3^+#</mathjax>. We call it <mathjax>#H_3O^+#</mathjax> or even <mathjax>#H^+#</mathjax> for convenience. The extra proton can rapidly move, or even tunnel if you like, from cluster to cluster.....</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#HCl(g) + H_2O(l) rarr H_3O^+ + Cl^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you say, hydrogen chloride is a room temperature gas; however, it is mighty soluble in water. And in water we conceive that it protonates the water molecular to give <mathjax>#H_3O^+#</mathjax>. <mathjax>#HCl#</mathjax> PROTONATES the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to form water-solvated hydronium ions and chloride ions......</p>
<p><mathjax>#H_3O^+#</mathjax>, the so-called hydronium ion, is our conception of a protonated water molecule. So far as anybody knows, it is a cluster of 2-4 water molecules with AN EXTRA <mathjax>#H^+#</mathjax> incorporated in the cluster, i.e. <mathjax>#H_5O_2^+#</mathjax> or <mathjax>#H_7O_3^+#</mathjax>. We call it <mathjax>#H_3O^+#</mathjax> or even <mathjax>#H^+#</mathjax> for convenience. The extra proton can rapidly move, or even tunnel if you like, from cluster to cluster.....</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you write an equation showing how HCl, which is a molecular substance and strong acid, could give ions when added to water?</h1>
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<div class="markdown"><p><mathjax>#HCl(g) + H_2O(l) rarr H_3O^+ + Cl^(-)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you say, hydrogen chloride is a room temperature gas; however, it is mighty soluble in water. And in water we conceive that it protonates the water molecular to give <mathjax>#H_3O^+#</mathjax>. <mathjax>#HCl#</mathjax> PROTONATES the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> to form water-solvated hydronium ions and chloride ions......</p>
<p><mathjax>#H_3O^+#</mathjax>, the so-called hydronium ion, is our conception of a protonated water molecule. So far as anybody knows, it is a cluster of 2-4 water molecules with AN EXTRA <mathjax>#H^+#</mathjax> incorporated in the cluster, i.e. <mathjax>#H_5O_2^+#</mathjax> or <mathjax>#H_7O_3^+#</mathjax>. We call it <mathjax>#H_3O^+#</mathjax> or even <mathjax>#H^+#</mathjax> for convenience. The extra proton can rapidly move, or even tunnel if you like, from cluster to cluster.....</p></div>
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</article> | How do you write an equation showing how HCl, which is a molecular substance and strong acid, could give ions when added to water? | null |
1,394 | abe4bbe5-6ddd-11ea-832a-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-exerted-by-32-00-g-of-oxygen-gas-in-a-20-l-container-at-30- | 1.24 atm | start physical_unit 9 10 pressure atm qc_end physical_unit 9 10 22 27 constant_r qc_end physical_unit 9 10 6 7 mass qc_end physical_unit 9 10 13 14 volume qc_end physical_unit 9 10 17 18 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] Oxygen gas [IN] atm"}] | [{"type":"physical unit","value":"1.24 atm"}] | [{"type":"physical unit","value":"Mass [OF] Oxygen gas [=] \\pu{32.00 g}"},{"type":"physical unit","value":"Volume [OF] Oxygen gas [=] \\pu{20 L}"},{"type":"physical unit","value":"Temperature [OF] Oxygen gas [=] \\pu{30.00 ℃}"},{"type":"physical unit","value":"R [OF] Oxygen gas [=] \\pu{0.0821 (L * atm)/(mol * K)}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure exerted by 32.00 g of Oxygen gas in a 20. L container at 30.00 C? Use R = 0.0821 L atm/mol k?</h1> | null | 1.24 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's determine the number of mole of oxygen gas.</p>
<p>Using <mathjax>#n = frac(m)(M)#</mathjax>:</p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(2 times 15.99))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(31.98))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#therefore n("O") = 1.00#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p>Then, let's convert the units of the given temperature to <mathjax>#"K"#</mathjax>:</p>
<p><mathjax>#Rightarrow T_((""^(@)"" "C")) = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow 30.00 = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow T_(("K")) = 303.15#</mathjax></p>
<p><mathjax>#therefore 30.00#</mathjax> <mathjax>#""^(@)"" "C"#</mathjax> <mathjax>#= 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Now, let's substitute all relevant values into the equation <mathjax>#P V = n R T#</mathjax>:</p>
<p><mathjax>#Rightarrow P times 20.0#</mathjax> <mathjax>#"L"#</mathjax> <mathjax>#= 1.00#</mathjax> <mathjax>#"mol"#</mathjax> <mathjax>#times 0.0821#</mathjax> <mathjax>#"L atm mol"^(- 1) "K"^(- 1)#</mathjax> <mathjax>#times 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#Rightarrow P times 20.0 = 1.00 times 0.0821#</mathjax> <mathjax>#"atm"#</mathjax> <mathjax>#times 303.15#</mathjax></p>
<p><mathjax>#Rightarrow P = (frac(0.0821 times 303.15)(20.0))#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#therefore P = 1.24#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p>Therefore, the pressure exerted by the oxygen gas is <mathjax>#1.24#</mathjax> <mathjax>#"atm"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#1.24#</mathjax> <mathjax>#"atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's determine the number of mole of oxygen gas.</p>
<p>Using <mathjax>#n = frac(m)(M)#</mathjax>:</p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(2 times 15.99))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(31.98))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#therefore n("O") = 1.00#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p>Then, let's convert the units of the given temperature to <mathjax>#"K"#</mathjax>:</p>
<p><mathjax>#Rightarrow T_((""^(@)"" "C")) = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow 30.00 = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow T_(("K")) = 303.15#</mathjax></p>
<p><mathjax>#therefore 30.00#</mathjax> <mathjax>#""^(@)"" "C"#</mathjax> <mathjax>#= 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Now, let's substitute all relevant values into the equation <mathjax>#P V = n R T#</mathjax>:</p>
<p><mathjax>#Rightarrow P times 20.0#</mathjax> <mathjax>#"L"#</mathjax> <mathjax>#= 1.00#</mathjax> <mathjax>#"mol"#</mathjax> <mathjax>#times 0.0821#</mathjax> <mathjax>#"L atm mol"^(- 1) "K"^(- 1)#</mathjax> <mathjax>#times 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#Rightarrow P times 20.0 = 1.00 times 0.0821#</mathjax> <mathjax>#"atm"#</mathjax> <mathjax>#times 303.15#</mathjax></p>
<p><mathjax>#Rightarrow P = (frac(0.0821 times 303.15)(20.0))#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#therefore P = 1.24#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p>Therefore, the pressure exerted by the oxygen gas is <mathjax>#1.24#</mathjax> <mathjax>#"atm"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pressure exerted by 32.00 g of Oxygen gas in a 20. L container at 30.00 C? Use R = 0.0821 L atm/mol k?</h1>
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Tazwar Sikder
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<div class="markdown"><p><mathjax>#1.24#</mathjax> <mathjax>#"atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, let's determine the number of mole of oxygen gas.</p>
<p>Using <mathjax>#n = frac(m)(M)#</mathjax>:</p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(2 times 15.99))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#Rightarrow n("O") = (frac(32.00)(31.98))#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p><mathjax>#therefore n("O") = 1.00#</mathjax> <mathjax>#"mol"#</mathjax></p>
<p>Then, let's convert the units of the given temperature to <mathjax>#"K"#</mathjax>:</p>
<p><mathjax>#Rightarrow T_((""^(@)"" "C")) = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow 30.00 = T_(("K")) - 273.15#</mathjax></p>
<p><mathjax>#Rightarrow T_(("K")) = 303.15#</mathjax></p>
<p><mathjax>#therefore 30.00#</mathjax> <mathjax>#""^(@)"" "C"#</mathjax> <mathjax>#= 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Now, let's substitute all relevant values into the equation <mathjax>#P V = n R T#</mathjax>:</p>
<p><mathjax>#Rightarrow P times 20.0#</mathjax> <mathjax>#"L"#</mathjax> <mathjax>#= 1.00#</mathjax> <mathjax>#"mol"#</mathjax> <mathjax>#times 0.0821#</mathjax> <mathjax>#"L atm mol"^(- 1) "K"^(- 1)#</mathjax> <mathjax>#times 303.15#</mathjax> <mathjax>#"K"#</mathjax></p>
<p><mathjax>#Rightarrow P times 20.0 = 1.00 times 0.0821#</mathjax> <mathjax>#"atm"#</mathjax> <mathjax>#times 303.15#</mathjax></p>
<p><mathjax>#Rightarrow P = (frac(0.0821 times 303.15)(20.0))#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#therefore P = 1.24#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p>Therefore, the pressure exerted by the oxygen gas is <mathjax>#1.24#</mathjax> <mathjax>#"atm"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#1.24#</mathjax> atm, rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note: mols and moles are the same things; formatting issues prevent me from using the same word throughout the answer<br/>
We will use the ideal gas equation, which is as follows:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p><mathjax>#P#</mathjax> is pressure (atm)<br/>
<mathjax>#V#</mathjax> is volume( liters)<br/>
<mathjax>#n#</mathjax> is moles of substance (<mathjax>#(Liters*atm)/(mols*K)#</mathjax>)<br/>
<mathjax>#R#</mathjax> is the constant <mathjax>#0.0821#</mathjax><br/>
<mathjax>#T#</mathjax> is temperature (kelvin).</p>
<p>We are solving for <mathjax>#P#</mathjax> in this equation, so the formula is reworked as so: <br/>
<mathjax>#P = (nRT)/V#</mathjax> <br/>
Now, we find the rest of the numbers to plug in</p>
<p><mathjax>#n#</mathjax>: We convert Grams of <mathjax>#O_"2"#</mathjax> to moles using dimensional analysis: <mathjax>#32.00g O_2 * (1 mol O_2)/(32.00g O_2) = 1 mol O_2#</mathjax></p>
<p><mathjax>#R#</mathjax>: We use the constant <mathjax>#0.0821#</mathjax> </p>
<p><mathjax>#T#</mathjax>: We convert the temperature from Celcius to Fahrenheit, using the formula <mathjax>#K = ^oC + 273#</mathjax><br/>
<mathjax>#K = 30 + 273 = 303K#</mathjax></p>
<p><mathjax>#V#</mathjax>: We use the given liters, which is <mathjax>#20L#</mathjax></p>
<p>Now, we solve. </p>
<p>First by plugging in numbers:</p>
<p><mathjax>#P = (nRT)/V#</mathjax> <br/>
=> <mathjax>#P = ((1 mol)(0.0821(Liters*atm)/(mols*K))(303K))/(20L)#</mathjax> </p>
<p>Which is the same as:</p>
<p><mathjax>#P = ((1 mol)(0.0821(Liters*atm))(303K))/(20L*mols*K)#</mathjax></p>
<p>Liters, moles, and Kelvin cancel out, giving us:</p>
<p><mathjax>#P = ((1)(0.0821 atm)(303))/(20)#</mathjax></p>
<p>Now we simplify to get our answer:</p>
<p><mathjax>#P = ((1)(0.0821 atm)(303))/(20)#</mathjax></p>
<p>=> <mathjax>#P = (24.88atm)/(20)#</mathjax></p>
<p>=> <mathjax>#P = 1.24 atm#</mathjax></p></div>
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<div class="markdown"><p>The pressure is 1.2 atm.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is an <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> problem. The equation is:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is a gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p>You have been given the mass of <mathjax>#"O"_2#</mathjax>, but the equation requires moles. Determine the mol <mathjax>#"O"_2#</mathjax> by multiplying its given mass by the inverse of its molar mass (31.998 g/mol).</p>
<p><mathjax>#32.00color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="1.000 mol O"_2"#</mathjax></p>
<p>You have been given temperature in degrees Celsius, but gas problems require the temperature to be in Kelvins. Convert <mathjax>#30.00^@"C"#</mathjax> to Kelvins by adding <mathjax>#273.15#</mathjax>.</p>
<p><mathjax>#30.00^@"C" + 273.15="303.15 K"#</mathjax></p>
<p><mathjax>#color(blue)("Now organize your data."#</mathjax></p>
<p><strong>Given/Known</strong></p>
<p><mathjax>#V="20. L"=2.0xx10^2color(white)(.)"L"#</mathjax></p>
<p><mathjax>#n="1.000 mol"#</mathjax></p>
<p><mathjax>#R="0.0821 L atm K"^(-1) "mol"^(-1)"#</mathjax></p>
<p><mathjax>#T="303.15 K"#</mathjax></p>
<p><mathjax>#color(blue)("Solution."#</mathjax><br/>
Rearrange the equation to isolate <mathjax>#P#</mathjax>. Insert your data and solve.</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=(1.000color(red)cancel(color(black)("mol"))xx0.0821color(red)cancel(color(black)("L")) "atm" color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx303.15color(red)cancel(color(black)("K")))/(2.0xx10^2color(red)cancel(color(black)("L")))="1.2 atm"#</mathjax> rounded to two significant figures)</p></div>
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</article> | What is the pressure exerted by 32.00 g of Oxygen gas in a 20. L container at 30.00 C? Use R = 0.0821 L atm/mol k? | null |
1,395 | ac2af881-6ddd-11ea-b222-ccda262736ce | https://socratic.org/questions/calculate-amount-of-formic-acid-ka-2-10-4-which-shoud-be-dissolved-in-10l-of-wat | 0.18 g | start physical_unit 3 4 mass g qc_end physical_unit 3 4 7 9 ka qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 22 22 26 26 ph qc_end end | [{"type":"physical unit","value":"Mass [OF] formic acid [IN] g"}] | [{"type":"physical unit","value":"0.18 g"}] | [{"type":"physical unit","value":"ka [OF] formic acid [=] \\pu{2 × 10^(-4)}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{10 L}"},{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{3.7}"}] | <h1 class="questionTitle" itemprop="name">Calculate amount of formic acid (ka=2*10-4) which shoud be dissolved in 10L of water to obtain a solution of pH=3.7?</h1> | null | 0.18 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the desired <mathjax>#["H"_3"O"^"+"]#</mathjax></strong></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-pH"color(white)(l)"mol/L" = 10^"-3.7"color(white)(l)"mol/L" = 2.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the concentration of formic acid</strong></p>
<p>We can use an ICE table to help with the calculation.</p>
<p><mathjax>#color(white)(mmmmmml)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)c color(white)(mmmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": color(white)(m)c"-"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x#</mathjax></p>
<p><mathjax>#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(c-x) = 2.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#x^2 = (c-x)(2.0 × 10^"-4") = 2.0 × 10^"-4"c - 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#(2.0 × 10^"-4")c = x^2 + 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#c = (x^2 + 2.0 × 10^"-4"x)/(2.0 ×10^"-4")#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#x = 2.0 × 10^"-4"#</mathjax></p>
<p>∴ <mathjax>#c = ((2.0 × 10^"-4")^color(red)(cancel(color(black)(2))) + (2.0 × 10^"-4")^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(2.0 × 10^"-4")))) = 4.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#c = 4.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of formic acid</strong></p>
<p><mathjax>#"Moles" = 10 color(red)(cancel(color(black)("L"))) × (4.0 × 10^"-4"color(white)(l)"mol")/(1 color(red)(cancel(color(black)("L")))) = 4.0 × 10^"-3"color(white)(l)"mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of formic acid</strong></p>
<p><mathjax>#"Mass" = 4.0 × 10^"-3" color(red)(cancel(color(black)("mol"))) × "46.03 g"/(1 color(red)(cancel(color(black)("mol")))) = "0.18 g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You should use 0.18 g of formic acid.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the desired <mathjax>#["H"_3"O"^"+"]#</mathjax></strong></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-pH"color(white)(l)"mol/L" = 10^"-3.7"color(white)(l)"mol/L" = 2.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the concentration of formic acid</strong></p>
<p>We can use an ICE table to help with the calculation.</p>
<p><mathjax>#color(white)(mmmmmml)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)c color(white)(mmmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": color(white)(m)c"-"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x#</mathjax></p>
<p><mathjax>#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(c-x) = 2.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#x^2 = (c-x)(2.0 × 10^"-4") = 2.0 × 10^"-4"c - 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#(2.0 × 10^"-4")c = x^2 + 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#c = (x^2 + 2.0 × 10^"-4"x)/(2.0 ×10^"-4")#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#x = 2.0 × 10^"-4"#</mathjax></p>
<p>∴ <mathjax>#c = ((2.0 × 10^"-4")^color(red)(cancel(color(black)(2))) + (2.0 × 10^"-4")^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(2.0 × 10^"-4")))) = 4.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#c = 4.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of formic acid</strong></p>
<p><mathjax>#"Moles" = 10 color(red)(cancel(color(black)("L"))) × (4.0 × 10^"-4"color(white)(l)"mol")/(1 color(red)(cancel(color(black)("L")))) = 4.0 × 10^"-3"color(white)(l)"mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of formic acid</strong></p>
<p><mathjax>#"Mass" = 4.0 × 10^"-3" color(red)(cancel(color(black)("mol"))) × "46.03 g"/(1 color(red)(cancel(color(black)("mol")))) = "0.18 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Calculate amount of formic acid (ka=2*10-4) which shoud be dissolved in 10L of water to obtain a solution of pH=3.7?</h1>
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<div class="markdown"><p>You should use 0.18 g of formic acid.</p></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 desired <mathjax>#["H"_3"O"^"+"]#</mathjax></strong></p>
<p><mathjax>#["H"_3"O"^"+"] = 10^"-pH"color(white)(l)"mol/L" = 10^"-3.7"color(white)(l)"mol/L" = 2.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the concentration of formic acid</strong></p>
<p>We can use an ICE table to help with the calculation.</p>
<p><mathjax>#color(white)(mmmmmml)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)c color(white)(mmmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": color(white)(m)c"-"xcolor(white)(mmmmmmm)xcolor(white)(mmm)x#</mathjax></p>
<p><mathjax>#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(c-x) = 2.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#x^2 = (c-x)(2.0 × 10^"-4") = 2.0 × 10^"-4"c - 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#(2.0 × 10^"-4")c = x^2 + 2.0 × 10^"-4"x#</mathjax></p>
<p><mathjax>#c = (x^2 + 2.0 × 10^"-4"x)/(2.0 ×10^"-4")#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#x = 2.0 × 10^"-4"#</mathjax></p>
<p>∴ <mathjax>#c = ((2.0 × 10^"-4")^color(red)(cancel(color(black)(2))) + (2.0 × 10^"-4")^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(2.0 × 10^"-4")))) = 4.0 × 10^"-4"#</mathjax></p>
<p><mathjax>#c = 4.0 × 10^"-4"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of formic acid</strong></p>
<p><mathjax>#"Moles" = 10 color(red)(cancel(color(black)("L"))) × (4.0 × 10^"-4"color(white)(l)"mol")/(1 color(red)(cancel(color(black)("L")))) = 4.0 × 10^"-3"color(white)(l)"mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4. Calculate the mass of formic acid</strong></p>
<p><mathjax>#"Mass" = 4.0 × 10^"-3" color(red)(cancel(color(black)("mol"))) × "46.03 g"/(1 color(red)(cancel(color(black)("mol")))) = "0.18 g"#</mathjax></p></div>
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</article> | Calculate amount of formic acid (ka=2*10-4) which shoud be dissolved in 10L of water to obtain a solution of pH=3.7? | null |
1,396 | aafc4af6-6ddd-11ea-ac42-ccda262736ce | https://socratic.org/questions/if-9-69-moles-of-an-ideal-gas-has-a-pressure-of-3-10-atm-and-a-volume-of-64-51-l | 251.37 K | start physical_unit 24 25 temperature k qc_end physical_unit 5 6 1 2 mole qc_end physical_unit 5 6 11 12 pressure qc_end physical_unit 5 6 17 18 volume qc_end end | [{"type":"physical unit","value":"Temperature [OF] the sample [IN] K"}] | [{"type":"physical unit","value":"251.37 K"}] | [{"type":"physical unit","value":"Mole [OF] ideal gas [=] \\pu{9.69 moles}"},{"type":"physical unit","value":"Pressure [OF] ideal gas [=] \\pu{3.10 atm}"},{"type":"physical unit","value":"Volume [OF] ideal gas [=] \\pu{64.51 L}"}] | <h1 class="questionTitle" itemprop="name">If 9.69 moles of an ideal gas has a pressure of 3.10 atm, and a volume of 64.51 L, what is the temperature of the sample? </h1> | null | 251.37 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>Before plugging in your values, make sure that the units given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, the volume is given in <em>liters</em> and the pressure in <em>atmospheres</em>, so you're good to go. </p>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T = (3.10 color(red)(cancel(color(black)("atm"))) * 64.51 color(red)(cancel(color(black)("L"))))/(9.69 color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(T = "251 K")))#</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>#"251 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>Before plugging in your values, make sure that the units given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, the volume is given in <em>liters</em> and the pressure in <em>atmospheres</em>, so you're good to go. </p>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T = (3.10 color(red)(cancel(color(black)("atm"))) * 64.51 color(red)(cancel(color(black)("L"))))/(9.69 color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(T = "251 K")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If 9.69 moles of an ideal gas has a pressure of 3.10 atm, and a volume of 64.51 L, what is the temperature of the sample? </h1>
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Stefan V.
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Dec 21, 2016
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<div class="markdown"><p><mathjax>#"251 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Rearrange the equation to solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>Before plugging in your values, make sure that the units given to you <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, the volume is given in <em>liters</em> and the pressure in <em>atmospheres</em>, so you're good to go. </p>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T = (3.10 color(red)(cancel(color(black)("atm"))) * 64.51 color(red)(cancel(color(black)("L"))))/(9.69 color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(T = "251 K")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If 9.69 moles of an ideal gas has a pressure of 3.10 atm, and a volume of 64.51 L, what is the temperature of the sample? | null |
1,397 | aad44eb7-6ddd-11ea-848e-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-ph-of-0-15-m-aqueous-solution-of-hydrazine | 10.59 | start physical_unit 10 10 ph none qc_end end | [{"type":"physical unit","value":"pH [OF] hydrazine aqueous solution"}] | [{"type":"physical unit","value":"10.59"}] | [{"type":"physical unit","value":"Molarity [OF] hydrazine aqueous solution [=] \\pu{0.15 M}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the pH of 0.15 M aqueous solution of hydrazine?
</h1> | null | 10.59 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...............</p>
<p><mathjax>#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#</mathjax></p>
<p><mathjax>#K_"eq"=([N_2H_5^+][HO^-])/([N_2H_4])=1.0xx10^-6#</mathjax>, and if <mathjax>#x*mol#</mathjax> of hydrazine associate, we can put in the numbers........</p>
<p><mathjax>#K_"eq"=1.0xx10^-6=((x)xx(x))/(0.15-x)=x^2/(0.15-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax> that we could solve exactly. However, because <mathjax>#K_"eq"#</mathjax> is so small, we are justified in making the approx. <mathjax>#0.15">>"x#</mathjax>, and we write.........</p>
<p><mathjax>#1.0xx10^-6~=x^2/0.15#</mathjax></p>
<p><mathjax>#x_1=sqrt(1.0xx10^-6xx0.15)=3.87xx10^-4#</mathjax>, this is indeed small compared to <mathjax>#0.15*mol*L^-1#</mathjax>, but now we have an approximation for <mathjax>#x#</mathjax> we can plug this value back into the equation to get <mathjax>#x^2#</mathjax>...</p>
<p><mathjax>#x_2=3.87xx10^-4*mol*L^-1#</mathjax>, and given the convergence of <mathjax>#x_1#</mathjax> and <mathjax>#x_2#</mathjax> we are satisfied that these values are as good as if we used the quadratic equation............</p>
<p>But we are not finished there.....we know that ..............</p>
<p><mathjax>#[N_2H_5^+]=[HO^-]=3.87xx10^-4*mol*L^-1#</mathjax>...</p>
<p>We DO NOT KNOW <mathjax>#pH#</mathjax>.</p>
<p>But in aqueous solution, <mathjax>#14=pH+pOH#</mathjax></p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(3.87xx10^-4)=3.41#</mathjax></p>
<p><mathjax>#pH=14-3.41=10.6#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We interrogate the reaction.........<br/>
<mathjax>#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#</mathjax> and finally get <mathjax>#pH=10.6........#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...............</p>
<p><mathjax>#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#</mathjax></p>
<p><mathjax>#K_"eq"=([N_2H_5^+][HO^-])/([N_2H_4])=1.0xx10^-6#</mathjax>, and if <mathjax>#x*mol#</mathjax> of hydrazine associate, we can put in the numbers........</p>
<p><mathjax>#K_"eq"=1.0xx10^-6=((x)xx(x))/(0.15-x)=x^2/(0.15-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax> that we could solve exactly. However, because <mathjax>#K_"eq"#</mathjax> is so small, we are justified in making the approx. <mathjax>#0.15">>"x#</mathjax>, and we write.........</p>
<p><mathjax>#1.0xx10^-6~=x^2/0.15#</mathjax></p>
<p><mathjax>#x_1=sqrt(1.0xx10^-6xx0.15)=3.87xx10^-4#</mathjax>, this is indeed small compared to <mathjax>#0.15*mol*L^-1#</mathjax>, but now we have an approximation for <mathjax>#x#</mathjax> we can plug this value back into the equation to get <mathjax>#x^2#</mathjax>...</p>
<p><mathjax>#x_2=3.87xx10^-4*mol*L^-1#</mathjax>, and given the convergence of <mathjax>#x_1#</mathjax> and <mathjax>#x_2#</mathjax> we are satisfied that these values are as good as if we used the quadratic equation............</p>
<p>But we are not finished there.....we know that ..............</p>
<p><mathjax>#[N_2H_5^+]=[HO^-]=3.87xx10^-4*mol*L^-1#</mathjax>...</p>
<p>We DO NOT KNOW <mathjax>#pH#</mathjax>.</p>
<p>But in aqueous solution, <mathjax>#14=pH+pOH#</mathjax></p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(3.87xx10^-4)=3.41#</mathjax></p>
<p><mathjax>#pH=14-3.41=10.6#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the pH of 0.15 M aqueous solution of hydrazine?
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<div class="markdown"><p>We interrogate the reaction.........<br/>
<mathjax>#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#</mathjax> and finally get <mathjax>#pH=10.6........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...............</p>
<p><mathjax>#H_2N-NH_2(aq) + H_2O(l)rightleftharpoonsH_3N^(+)-NH_2 +HO^-#</mathjax></p>
<p><mathjax>#K_"eq"=([N_2H_5^+][HO^-])/([N_2H_4])=1.0xx10^-6#</mathjax>, and if <mathjax>#x*mol#</mathjax> of hydrazine associate, we can put in the numbers........</p>
<p><mathjax>#K_"eq"=1.0xx10^-6=((x)xx(x))/(0.15-x)=x^2/(0.15-x)#</mathjax></p>
<p>This is a quadratic in <mathjax>#x#</mathjax> that we could solve exactly. However, because <mathjax>#K_"eq"#</mathjax> is so small, we are justified in making the approx. <mathjax>#0.15">>"x#</mathjax>, and we write.........</p>
<p><mathjax>#1.0xx10^-6~=x^2/0.15#</mathjax></p>
<p><mathjax>#x_1=sqrt(1.0xx10^-6xx0.15)=3.87xx10^-4#</mathjax>, this is indeed small compared to <mathjax>#0.15*mol*L^-1#</mathjax>, but now we have an approximation for <mathjax>#x#</mathjax> we can plug this value back into the equation to get <mathjax>#x^2#</mathjax>...</p>
<p><mathjax>#x_2=3.87xx10^-4*mol*L^-1#</mathjax>, and given the convergence of <mathjax>#x_1#</mathjax> and <mathjax>#x_2#</mathjax> we are satisfied that these values are as good as if we used the quadratic equation............</p>
<p>But we are not finished there.....we know that ..............</p>
<p><mathjax>#[N_2H_5^+]=[HO^-]=3.87xx10^-4*mol*L^-1#</mathjax>...</p>
<p>We DO NOT KNOW <mathjax>#pH#</mathjax>.</p>
<p>But in aqueous solution, <mathjax>#14=pH+pOH#</mathjax></p>
<p><mathjax>#pOH=-log_10[HO^-]=-log_10(3.87xx10^-4)=3.41#</mathjax></p>
<p><mathjax>#pH=14-3.41=10.6#</mathjax></p></div>
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</article> | How do you calculate the pH of 0.15 M aqueous solution of hydrazine?
| null |
1,398 | a8f1144c-6ddd-11ea-bb11-ccda262736ce | https://socratic.org/questions/how-hot-would-a-2-3-l-balloon-have-to-get-to-expand-to-a-volume-of-3-0-l-assumin | 390 K | start physical_unit 6 6 temperature k qc_end physical_unit 6 6 4 5 volume qc_end physical_unit 6 6 16 17 volume qc_end physical_unit 6 6 26 27 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] balloon [IN] K"}] | [{"type":"physical unit","value":"390 K"}] | [{"type":"physical unit","value":"Volume1 [OF] balloon [=] \\pu{2.3 L}"},{"type":"physical unit","value":"Volume2 [OF] balloon [=] \\pu{3.0 L}"},{"type":"physical unit","value":"Temperature1 [OF] balloon [=] \\pu{24 ℃}"},{"type":"other","value":"The pressure inside the balloon remains constant."}] | <h1 class="questionTitle" itemprop="name">How hot would a 2.3 L balloon have to get to expand to a volume of 3.0 L, assuming the initial temperature of the balloon is 24°C, and that the pressure inside the balloon remains constant?</h1> | null | 390 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>, which states that the volume of a gas held at constant pressure varies directly with the Kelvin temperature. The equation for this law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="2.3 L"#</mathjax><br/>
<mathjax>#T_1=24^@"C"+273.15="297 K"#</mathjax><br/>
<mathjax>#V_2="3.0 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, substitute the values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/(V_1)#</mathjax></p>
<p><mathjax>#T_2=((297"K")xx(3.0"L"))/(2.3"L")#</mathjax></p>
<p><mathjax>#T_2="390 K"#</mathjax> rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The temperature needs to be 390 K. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>, which states that the volume of a gas held at constant pressure varies directly with the Kelvin temperature. The equation for this law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="2.3 L"#</mathjax><br/>
<mathjax>#T_1=24^@"C"+273.15="297 K"#</mathjax><br/>
<mathjax>#V_2="3.0 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, substitute the values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/(V_1)#</mathjax></p>
<p><mathjax>#T_2=((297"K")xx(3.0"L"))/(2.3"L")#</mathjax></p>
<p><mathjax>#T_2="390 K"#</mathjax> rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How hot would a 2.3 L balloon have to get to expand to a volume of 3.0 L, assuming the initial temperature of the balloon is 24°C, and that the pressure inside the balloon remains constant?</h1>
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<div class="markdown"><p>The temperature needs to be 390 K. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a>, which states that the volume of a gas held at constant pressure varies directly with the Kelvin temperature. The equation for this law is:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="2.3 L"#</mathjax><br/>
<mathjax>#T_1=24^@"C"+273.15="297 K"#</mathjax><br/>
<mathjax>#V_2="3.0 L"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax>, substitute the values into the equation and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(T_1V_2)/(V_1)#</mathjax></p>
<p><mathjax>#T_2=((297"K")xx(3.0"L"))/(2.3"L")#</mathjax></p>
<p><mathjax>#T_2="390 K"#</mathjax> rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</article> | How hot would a 2.3 L balloon have to get to expand to a volume of 3.0 L, assuming the initial temperature of the balloon is 24°C, and that the pressure inside the balloon remains constant? | null |
1,399 | ac50e554-6ddd-11ea-97fd-ccda262736ce | https://socratic.org/questions/what-would-be-the-net-ionic-equation-of-a-solution-of-sodium-acetate-and-a-solut | 2 Hg+ + 2 Cl- -> Hg2Cl2 | start chemical_equation qc_end end | [{"type":"other","value":"Chemical Equation [OF] net ionic equation"}] | [{"type":"chemical equation","value":"2 Hg+ + 2 Cl- -> Hg2Cl2"}] | [{"type":"substance name","value":"Sodium acetate solution"},{"type":"substance name","value":"Calcium chloride solution"},{"type":"substance name","value":"Mercury(I) nitrate solution"}] | <h1 class="questionTitle" itemprop="name">What would be the net ionic equation of a solution of sodium acetate and a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I'm also having trouble figuring out what the product would be in the overall equation. </p></div>
</h2>
</div>
</div> | 2 Hg+ + 2 Cl- -> Hg2Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ion is always inert. calcium ions can react with several ions to. Calcium ions can react with several ions to form carbonate but not with nitrate or chloride. nitrate ions are only weak oxidants, but cannot oxidize Hg(I) to Hg(II). Cloride ions are weak reducing, but here don't react in this way. <mathjax>#Cl^-#</mathjax> can form precipitate with <mathjax>#Pb^(2+)#</mathjax> in cold water, with <mathjax>#Ag^+#</mathjax> and some less common cations (like <mathjax>#Tl^+#</mathjax>) . Acetate ions are weak base but in this solution cannot do nothing. Hg^+ dimerizes forming a precipitate with cloride ions.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2Hg^+ + 2Cl^- = Hg_2Cl_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ion is always inert. calcium ions can react with several ions to. Calcium ions can react with several ions to form carbonate but not with nitrate or chloride. nitrate ions are only weak oxidants, but cannot oxidize Hg(I) to Hg(II). Cloride ions are weak reducing, but here don't react in this way. <mathjax>#Cl^-#</mathjax> can form precipitate with <mathjax>#Pb^(2+)#</mathjax> in cold water, with <mathjax>#Ag^+#</mathjax> and some less common cations (like <mathjax>#Tl^+#</mathjax>) . Acetate ions are weak base but in this solution cannot do nothing. Hg^+ dimerizes forming a precipitate with cloride ions.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What would be the net ionic equation of a solution of sodium acetate and a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together?</h1>
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<div class="markdown"><p>I'm also having trouble figuring out what the product would be in the overall equation. </p></div>
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Andrea B.
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<div class="markdown"><p><mathjax>#2Hg^+ + 2Cl^- = Hg_2Cl_2#</mathjax></p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium ion is always inert. calcium ions can react with several ions to. Calcium ions can react with several ions to form carbonate but not with nitrate or chloride. nitrate ions are only weak oxidants, but cannot oxidize Hg(I) to Hg(II). Cloride ions are weak reducing, but here don't react in this way. <mathjax>#Cl^-#</mathjax> can form precipitate with <mathjax>#Pb^(2+)#</mathjax> in cold water, with <mathjax>#Ag^+#</mathjax> and some less common cations (like <mathjax>#Tl^+#</mathjax>) . Acetate ions are weak base but in this solution cannot do nothing. Hg^+ dimerizes forming a precipitate with cloride ions.</p></div>
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</article> | What would be the net ionic equation of a solution of sodium acetate and a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? |
I'm also having trouble figuring out what the product would be in the overall equation.
|
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