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800 | ac61231a-6ddd-11ea-ba5e-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-ch4-o2-co2-h2o | CH4 + 2 O2 -> CO2 + 2 H2O | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"CH4 + 2 O2 -> CO2 + 2 H2O"}] | [{"type":"chemical equation","value":"CH4 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation: CH4 + O2 --> CO2 + H2O?</h1> | null | CH4 + 2 O2 -> CO2 + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p>First look at the C atoms. At first glance they are balanced with 1 C on each side.</p>
<p>Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the <mathjax>#"H"_2"O""#</mathjax>. We now have 4 H atoms on both sides.</p>
<p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the <mathjax>#"O"_2"#</mathjax>. We now have 4 O atoms on both sides.</p>
<p><mathjax>#"CH"_4" + 2O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"CH"_4" + 2O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p>First look at the C atoms. At first glance they are balanced with 1 C on each side.</p>
<p>Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the <mathjax>#"H"_2"O""#</mathjax>. We now have 4 H atoms on both sides.</p>
<p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the <mathjax>#"O"_2"#</mathjax>. We now have 4 O atoms on both sides.</p>
<p><mathjax>#"CH"_4" + 2O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How would you balance the following equation: CH4 + O2 --> CO2 + H2O?</h1>
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<div class="markdown"><p><mathjax>#"CH"_4" + 2O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p>First look at the C atoms. At first glance they are balanced with 1 C on each side.</p>
<p>Now look at the H atoms. They are not balanced. There are 4 H atoms on the left side and 2 H atoms on the right. Place a coefficient of 2 in front of the <mathjax>#"H"_2"O""#</mathjax>. We now have 4 H atoms on both sides.</p>
<p><mathjax>#"CH"_4" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>Now look at the O atoms. They are not balanced. There are 2 O atoms on the left side and 4 on the right. Place a coefficient of 2 in front of the <mathjax>#"O"_2"#</mathjax>. We now have 4 O atoms on both sides.</p>
<p><mathjax>#"CH"_4" + 2O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 2H"_2"O"#</mathjax></p>
<p>The equation is now balanced. Each side has 1 C atom, 4 H atoms, and 4 O atoms.</p></div>
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</article> | How would you balance the following equation: CH4 + O2 --> CO2 + H2O? | null |
801 | ab51700a-6ddd-11ea-86d7-ccda262736ce | https://socratic.org/questions/a-volume-of-helium-occupies-11-0-l-at-a-pressure-of-98-0-kpa-what-is-the-new-vol | 12.51 L | start physical_unit 3 3 volume l qc_end physical_unit 3 3 5 6 volume qc_end physical_unit 3 3 11 12 pressure qc_end physical_unit 3 3 23 24 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] helium [IN] L"}] | [{"type":"physical unit","value":"12.51 L"}] | [{"type":"physical unit","value":"Volume1 [OF] helium [=] \\pu{11.0 L}"},{"type":"physical unit","value":"Pressure1 [OF] helium [=] \\pu{98.0 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] helium [=] \\pu{86.2 kPa}"}] | <h1 class="questionTitle" itemprop="name"> A volume of helium occupies 11.0 L at a pressure of 98.0 kPa. What is the new volume if the pressure drops to 86.2 kPa?</h1> | null | 12.51 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1= P_2V_2#</mathjax></p>
<p>Pressure, <mathjax>#P_1=98.0 kPa#</mathjax></p>
<p>Volume, <mathjax>#V_1=11.0L#</mathjax></p>
<p>Pressure, <mathjax>#P_2=86.2kPa#</mathjax></p>
<p>Volume, <mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=98.0*11.0/86.2=12.5L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The new volume is <mathjax>#=12.5L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1= P_2V_2#</mathjax></p>
<p>Pressure, <mathjax>#P_1=98.0 kPa#</mathjax></p>
<p>Volume, <mathjax>#V_1=11.0L#</mathjax></p>
<p>Pressure, <mathjax>#P_2=86.2kPa#</mathjax></p>
<p>Volume, <mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=98.0*11.0/86.2=12.5L#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name"> A volume of helium occupies 11.0 L at a pressure of 98.0 kPa. What is the new volume if the pressure drops to 86.2 kPa?</h1>
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<div class="markdown"><p>The new volume is <mathjax>#=12.5L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></p>
<p><mathjax>#P_1V_1= P_2V_2#</mathjax></p>
<p>Pressure, <mathjax>#P_1=98.0 kPa#</mathjax></p>
<p>Volume, <mathjax>#V_1=11.0L#</mathjax></p>
<p>Pressure, <mathjax>#P_2=86.2kPa#</mathjax></p>
<p>Volume, <mathjax>#V_2=(P_1V_1)/P_2#</mathjax></p>
<p><mathjax>#=98.0*11.0/86.2=12.5L#</mathjax></p></div>
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</article> | A volume of helium occupies 11.0 L at a pressure of 98.0 kPa. What is the new volume if the pressure drops to 86.2 kPa? | null |
802 | ac385692-6ddd-11ea-bcd7-ccda262736ce | https://socratic.org/questions/the-average-mass-of-a-dime-coin-is-2-28-g-and-the-mass-of-an-automobile-is-2-0-x | 6.9 × 10^17 | start physical_unit 26 26 number none qc_end physical_unit 5 6 8 9 mass qc_end physical_unit 15 15 17 20 mass qc_end physical_unit 37 37 34 35 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] automobiles"}] | [{"type":"physical unit","value":"6.9 × 10^17"}] | [{"type":"physical unit","value":"Average mass [OF] a dime coin [=] \\pu{2.28 g}"},{"type":"physical unit","value":"Mass [OF] an automobile [=] \\pu{2.0 × 10^3 kg}"},{"type":"physical unit","value":"Mole [OF] dimes [=] \\pu{1.0 mole}"},{"type":"other","value":"Automobiles' total mass is the same as dimes."}] | <h1 class="questionTitle" itemprop="name">The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes?</h1> | null | 6.9 × 10^17 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</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>#6.9 * 10^(17)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</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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<h1 class="questionTitle" itemprop="name">The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes?</h1>
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Stefan V.
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Aug 20, 2017
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<div class="markdown"><p><mathjax>#6.9 * 10^(17)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you should know that in order to have <mathjax>#1#</mathjax> <strong>mole</strong> of dime coins, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> <strong>dime coins</strong>.</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 mole of dime coins" = 6.022 * 10^(23)color(white)(.)"dime coins"))) ->#</mathjax> <em><strong>Avogadro's constant</strong></em></p>
</blockquote>
<p>Now, you know that the mass of single dime coin is equal to </p>
<blockquote>
<p><mathjax>#2.28 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 2.28 * 10^(-3)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>Use the mass of a single dime coin to calculate the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins</p>
<blockquote>
<p><mathjax>#6.022 * 10^(23)color(red)(cancel(color(black)("dime coins"))) * (2.28 * 10^(-3)color(white)(.)"kg")/(1color(red)(cancel(color(black)("dime coin")))) = 1.373 * 10^(19)color(white)(.)"kg"#</mathjax></p>
</blockquote>
<p>The mass of an automobile is equal to <mathjax>#2.0 * 10^3#</mathjax> <mathjax>#"kg"#</mathjax>, so you can say that the number of automobiles that will be <strong>equal</strong> to the mass of <mathjax>#1.0#</mathjax> <strong>mole</strong> of dime coins will be</p>
<blockquote>
<p><mathjax>#1.373 * 10^(19) color(red)(cancel(color(black)("kg"))) * "1 automobile"/(2.0 * 10^3 color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(6.9 * 10^(17)color(white)(.)"automobiles")))#</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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SCooke
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<div class="markdown"><p><mathjax># 6.86 xx 10^17#</mathjax> cars</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#2.28 (grams)/(unit) * 6.022 x 10^23#</mathjax> units/mole<mathjax># = 13.7 xx 10^23#</mathjax> grams/mole<br/>
<mathjax>#2.0 xx 10^3 kg * 1000g/"kg" = 2.0 xx 10^6 g/(car)#</mathjax><br/>
<mathjax>#(13.7 xx 10^23)/(2.0 xx 10^6) = 6.86 xx 10^17#</mathjax> cars</p></div>
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</article> | The average mass of a dime coin is #"2.28 g"# and the mass of an automobile is #2.0 * 10^3# #"kg"#. What is the number of automobiles whose total mass is the same as #1.0# mole of dimes? | null |
803 | a8a65e18-6ddd-11ea-9352-ccda262736ce | https://socratic.org/questions/finding-the-empirical-formula-of-iron-oxide | Fe3O4 | start chemical_formula qc_end physical_unit 28 31 26 27 mass qc_end c_other OTHER qc_end physical_unit 5 6 45 46 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] iron oxide [IN] empirical"}] | [{"type":"chemical equation","value":"Fe3O4"}] | [{"type":"physical unit","value":"Mass [OF] piece of iron metal [=] \\pu{85.65 g}"},{"type":"other","value":"Burn the piece of iron metal in air."},{"type":"physical unit","value":"Mass [OF] iron oxide [=] \\pu{118.37 g}"}] | <h1 class="questionTitle" itemprop="name">Finding the empirical formula of iron oxide?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Sam and Bob are conducting an investigation to determine the empirical formula of iron oxide. They start with an 85.65g piece of iron metal and burn it in air. The mass of the iron oxide produced is 118.37g. Sam thinks the empirical formula of iron oxide is Fe3O4 and Bob thinks it is FeO. Use what you know about how to determine the empirical formula of a compound. </p></div>
</h2>
</div>
</div> | Fe3O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, if you assume that <strong>all the mass</strong> of iron took part in the reaction, you can say that the difference between the mass of the <em>product</em> and the mass of <em>iron</em> will represent the mass of <strong>oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"product" = m_"iron" + m_"oxygen"#</mathjax></p>
<p><mathjax>#m_"oxygen" = "118.37 g" - "85.65 g" = "32.72 g"#</mathjax></p>
</blockquote>
<p>Now that you know how much iron and how much oxygen you have in the iron oxide, you can use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses to figure out how many <strong>moles</strong> of each you have. </p>
<p>For iron, you will have</p>
<blockquote>
<p><mathjax>#85.65color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.534 moles Fe"#</mathjax></p>
</blockquote>
<p>For oxygen, you will have</p>
<blockquote>
<p><mathjax>#32.72color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = "2.045 moles O"#</mathjax></p>
</blockquote>
<p>To determine the <em>empirical formula</em> of the iron oxide, you need to find the <strong>smallest whole number ratio</strong> that exists between the two elements in the compound. </p>
<p>To do that ,divide both values by <em>the smallest one</em>.</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.534color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.045color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1.333#</mathjax></p>
</blockquote>
<p>To get the smallest <em>whole number</em> ratio that exists between the two elements, multiply both values by <mathjax>#color(blue)(3)#</mathjax>. This will help you get rid of the decimal fraction for oxygen. </p>
<p>The <em>empirical formula</em> of the iron oxide will thus be </p>
<blockquote>
<p><mathjax>#("Fe"_1"O"_1.333)_color(blue)(3) implies color(green)("Fe"_3"O"_4)#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Fe"_3"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, if you assume that <strong>all the mass</strong> of iron took part in the reaction, you can say that the difference between the mass of the <em>product</em> and the mass of <em>iron</em> will represent the mass of <strong>oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"product" = m_"iron" + m_"oxygen"#</mathjax></p>
<p><mathjax>#m_"oxygen" = "118.37 g" - "85.65 g" = "32.72 g"#</mathjax></p>
</blockquote>
<p>Now that you know how much iron and how much oxygen you have in the iron oxide, you can use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses to figure out how many <strong>moles</strong> of each you have. </p>
<p>For iron, you will have</p>
<blockquote>
<p><mathjax>#85.65color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.534 moles Fe"#</mathjax></p>
</blockquote>
<p>For oxygen, you will have</p>
<blockquote>
<p><mathjax>#32.72color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = "2.045 moles O"#</mathjax></p>
</blockquote>
<p>To determine the <em>empirical formula</em> of the iron oxide, you need to find the <strong>smallest whole number ratio</strong> that exists between the two elements in the compound. </p>
<p>To do that ,divide both values by <em>the smallest one</em>.</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.534color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.045color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1.333#</mathjax></p>
</blockquote>
<p>To get the smallest <em>whole number</em> ratio that exists between the two elements, multiply both values by <mathjax>#color(blue)(3)#</mathjax>. This will help you get rid of the decimal fraction for oxygen. </p>
<p>The <em>empirical formula</em> of the iron oxide will thus be </p>
<blockquote>
<p><mathjax>#("Fe"_1"O"_1.333)_color(blue)(3) implies color(green)("Fe"_3"O"_4)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Finding the empirical formula of iron oxide?</h1>
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<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Sam and Bob are conducting an investigation to determine the empirical formula of iron oxide. They start with an 85.65g piece of iron metal and burn it in air. The mass of the iron oxide produced is 118.37g. Sam thinks the empirical formula of iron oxide is Fe3O4 and Bob thinks it is FeO. Use what you know about how to determine the empirical formula of a compound. </p></div>
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Stefan V.
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Nov 21, 2015
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<div class="markdown"><p><mathjax>#"Fe"_3"O"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, if you assume that <strong>all the mass</strong> of iron took part in the reaction, you can say that the difference between the mass of the <em>product</em> and the mass of <em>iron</em> will represent the mass of <strong>oxygen</strong>. </p>
<blockquote>
<p><mathjax>#m_"product" = m_"iron" + m_"oxygen"#</mathjax></p>
<p><mathjax>#m_"oxygen" = "118.37 g" - "85.65 g" = "32.72 g"#</mathjax></p>
</blockquote>
<p>Now that you know how much iron and how much oxygen you have in the iron oxide, you can use the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' respective molar masses to figure out how many <strong>moles</strong> of each you have. </p>
<p>For iron, you will have</p>
<blockquote>
<p><mathjax>#85.65color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "1.534 moles Fe"#</mathjax></p>
</blockquote>
<p>For oxygen, you will have</p>
<blockquote>
<p><mathjax>#32.72color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = "2.045 moles O"#</mathjax></p>
</blockquote>
<p>To determine the <em>empirical formula</em> of the iron oxide, you need to find the <strong>smallest whole number ratio</strong> that exists between the two elements in the compound. </p>
<p>To do that ,divide both values by <em>the smallest one</em>.</p>
<blockquote>
<p><mathjax>#"For Fe: " (1.534color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (2.045color(red)(cancel(color(black)("moles"))))/(1.534color(red)(cancel(color(black)("moles")))) = 1.333#</mathjax></p>
</blockquote>
<p>To get the smallest <em>whole number</em> ratio that exists between the two elements, multiply both values by <mathjax>#color(blue)(3)#</mathjax>. This will help you get rid of the decimal fraction for oxygen. </p>
<p>The <em>empirical formula</em> of the iron oxide will thus be </p>
<blockquote>
<p><mathjax>#("Fe"_1"O"_1.333)_color(blue)(3) implies color(green)("Fe"_3"O"_4)#</mathjax></p>
</blockquote></div>
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</article> | Finding the empirical formula of iron oxide? |
Sam and Bob are conducting an investigation to determine the empirical formula of iron oxide. They start with an 85.65g piece of iron metal and burn it in air. The mass of the iron oxide produced is 118.37g. Sam thinks the empirical formula of iron oxide is Fe3O4 and Bob thinks it is FeO. Use what you know about how to determine the empirical formula of a compound.
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804 | a91f54e6-6ddd-11ea-9006-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-n-in-ncl3 | +3 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] N"}] | [{"type":"physical unit","value":"+3"}] | [{"type":"chemical equation","value":"NCl3"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of N in NCl3? </h1> | null | +3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Step by step;</p>
<ol>
<li>1) Evaluate the oxidation number for other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>/element; In<br/>
this case <mathjax>#Cl#</mathjax></li>
</ol>
<p><mathjax>#Cl ^-#</mathjax> The charge on its ion is -1 <mathjax>#=>#</mathjax> that oxidation no. is -1</p>
<p>2) Set up an equation;</p>
<p>We find that the total charge on the compound is 0 ie. it is neutral<br/>
So use this formula;</p>
<p><strong><mathjax>#(n_a)(c _ a ) + (n_b )(c _b) =#</mathjax> Total <em>charge</em> on the compound</strong></p>
<p><em>Where c represents charge or oxidation number .n represents the number of atoms of that species present in the given compound</em></p>
<p><strong>Disclaimer ; Charge and oxidation number are not exactly one another.But for sake of simplicity using them as one another</strong></p>
<p>3) solve the equation</p>
<p><mathjax>#1(c) + 3(-1) = 0#</mathjax><br/>
<mathjax>#1(c) -3 = 0#</mathjax><br/>
<mathjax>#1(c) =3 #</mathjax><br/>
<mathjax>#=>(c) =3#</mathjax></p>
<p><mathjax>#therefore#</mathjax> the oxidation number of N is +3..</p></div>
</div>
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<div class="markdown"><p>It is + 3</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Step by step;</p>
<ol>
<li>1) Evaluate the oxidation number for other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>/element; In<br/>
this case <mathjax>#Cl#</mathjax></li>
</ol>
<p><mathjax>#Cl ^-#</mathjax> The charge on its ion is -1 <mathjax>#=>#</mathjax> that oxidation no. is -1</p>
<p>2) Set up an equation;</p>
<p>We find that the total charge on the compound is 0 ie. it is neutral<br/>
So use this formula;</p>
<p><strong><mathjax>#(n_a)(c _ a ) + (n_b )(c _b) =#</mathjax> Total <em>charge</em> on the compound</strong></p>
<p><em>Where c represents charge or oxidation number .n represents the number of atoms of that species present in the given compound</em></p>
<p><strong>Disclaimer ; Charge and oxidation number are not exactly one another.But for sake of simplicity using them as one another</strong></p>
<p>3) solve the equation</p>
<p><mathjax>#1(c) + 3(-1) = 0#</mathjax><br/>
<mathjax>#1(c) -3 = 0#</mathjax><br/>
<mathjax>#1(c) =3 #</mathjax><br/>
<mathjax>#=>(c) =3#</mathjax></p>
<p><mathjax>#therefore#</mathjax> the oxidation number of N is +3..</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of N in NCl3? </h1>
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<div class="markdown"><p>It is + 3</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Step by step;</p>
<ol>
<li>1) Evaluate the oxidation number for other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>/element; In<br/>
this case <mathjax>#Cl#</mathjax></li>
</ol>
<p><mathjax>#Cl ^-#</mathjax> The charge on its ion is -1 <mathjax>#=>#</mathjax> that oxidation no. is -1</p>
<p>2) Set up an equation;</p>
<p>We find that the total charge on the compound is 0 ie. it is neutral<br/>
So use this formula;</p>
<p><strong><mathjax>#(n_a)(c _ a ) + (n_b )(c _b) =#</mathjax> Total <em>charge</em> on the compound</strong></p>
<p><em>Where c represents charge or oxidation number .n represents the number of atoms of that species present in the given compound</em></p>
<p><strong>Disclaimer ; Charge and oxidation number are not exactly one another.But for sake of simplicity using them as one another</strong></p>
<p>3) solve the equation</p>
<p><mathjax>#1(c) + 3(-1) = 0#</mathjax><br/>
<mathjax>#1(c) -3 = 0#</mathjax><br/>
<mathjax>#1(c) =3 #</mathjax><br/>
<mathjax>#=>(c) =3#</mathjax></p>
<p><mathjax>#therefore#</mathjax> the oxidation number of N is +3..</p></div>
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<div class="markdown"><p>The oxidation number of N in the compound <mathjax>#"NCl"_3"#</mathjax> is +3.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"NCl"_3"#</mathjax></p>
<p>The sum of <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in a neutral compound must equal zero.</p>
<p>The oxidation number of group 17/VIIA <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> is usually <mathjax>#-1#</mathjax>. Since there are three chlorine atoms in the compound, the total oxidation number for chlorine is <mathjax>#-3#</mathjax>. Since the sum of the oxidation numbers must equal <mathjax>#0#</mathjax>, the oxidation number of nitrogen in the compound is <mathjax>#+3#</mathjax>.</p>
<p><mathjax>#1(+3) + 3(-1)=#</mathjax></p>
<p><mathjax>#(+3)+(-3)=0#</mathjax></p></div>
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</article> | What is the oxidation number of N in NCl3? | null |
805 | a8f61add-6ddd-11ea-9b3b-ccda262736ce | https://socratic.org/questions/when-magnesium-ions-form-compounds-with-bromide-ions-how-many-magnesium-ion-s-ar | 1 | start physical_unit 10 11 number none qc_end substance 6 7 qc_end end | [{"type":"physical unit","value":"Number [OF] magnesium ion(s)"}] | [{"type":"physical unit","value":"1"}] | [{"type":"substance name","value":"Bromide ions"}] | <h1 class="questionTitle" itemprop="name">When magnesium ions form compounds with bromide ions, how many magnesium ion(s) are needed to form a neutral compound?</h1> | null | 1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>and bromine forms a <mathjax>#Br^(-)#</mathjax> ion...........</p>
<p>And so, if these ions are to make music together, and form a NEUTRAL ionic solid, we need 2 equiv of anion, and 1 equiv of metal ion:</p>
<p><mathjax>#Mg^(2+) + 2Br^(-) rarr MgBr_2darr#</mathjax></p>
<p>All I have done here is to conserve mass and charge. Are they conserved?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well magnesium forms a <mathjax>#Mg^(2+)#</mathjax> ion...........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>and bromine forms a <mathjax>#Br^(-)#</mathjax> ion...........</p>
<p>And so, if these ions are to make music together, and form a NEUTRAL ionic solid, we need 2 equiv of anion, and 1 equiv of metal ion:</p>
<p><mathjax>#Mg^(2+) + 2Br^(-) rarr MgBr_2darr#</mathjax></p>
<p>All I have done here is to conserve mass and charge. Are they conserved?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">When magnesium ions form compounds with bromide ions, how many magnesium ion(s) are needed to form a neutral compound?</h1>
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<div class="markdown"><p>Well magnesium forms a <mathjax>#Mg^(2+)#</mathjax> ion...........</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>and bromine forms a <mathjax>#Br^(-)#</mathjax> ion...........</p>
<p>And so, if these ions are to make music together, and form a NEUTRAL ionic solid, we need 2 equiv of anion, and 1 equiv of metal ion:</p>
<p><mathjax>#Mg^(2+) + 2Br^(-) rarr MgBr_2darr#</mathjax></p>
<p>All I have done here is to conserve mass and charge. Are they conserved?</p></div>
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</article> | When magnesium ions form compounds with bromide ions, how many magnesium ion(s) are needed to form a neutral compound? | null |
806 | ad0b716e-6ddd-11ea-a881-ccda262736ce | https://socratic.org/questions/a-solution-of-barium-hydroxide-ba-oh-2-contains-4-285-grams-of-barium-hydroxide- | 2.50 M | start physical_unit 5 5 molarity mol/l qc_end physical_unit 5 5 7 8 mass qc_end physical_unit 1 1 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] Ba(OH)2 solution [IN] M"}] | [{"type":"physical unit","value":"2.50 M"}] | [{"type":"physical unit","value":"Mass [OF] Ba(OH)2 [=] \\pu{4.285 grams}"},{"type":"physical unit","value":"Volume [OF] Ba(OH)2 solution [=] \\pu{100 mL}"}] | <h1 class="questionTitle" itemprop="name">A solution of barium hydroxide, #Ba(OH)_2# contains 4.285 grams of barium hydroxide in 100 mL of solution. What is the molarity of the solution?</h1> | null | 2.50 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know the molar mass of barium hydroxide is 171.34 (you can google this or just add the mass of each atom in the molecule). Knowing that <mathjax>#molarity = (mols) / (liters)#</mathjax> we need to turn that 4.285 g Ba(OH)2 into mols. </p>
<p>This is done by dividing 4.285 g by the 171.34 g molar mass giving us 0.2500 mol of Ba(OH)2.</p>
<p>Plugging this all back into the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> we get <mathjax>#molarity = (0.2500 mol)/(0.100 L) #</mathjax> giving us a molarity of 2.5 M</p></div>
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<div class="markdown"><p>2.5 M </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know the molar mass of barium hydroxide is 171.34 (you can google this or just add the mass of each atom in the molecule). Knowing that <mathjax>#molarity = (mols) / (liters)#</mathjax> we need to turn that 4.285 g Ba(OH)2 into mols. </p>
<p>This is done by dividing 4.285 g by the 171.34 g molar mass giving us 0.2500 mol of Ba(OH)2.</p>
<p>Plugging this all back into the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> we get <mathjax>#molarity = (0.2500 mol)/(0.100 L) #</mathjax> giving us a molarity of 2.5 M</p></div>
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<h1 class="questionTitle" itemprop="name">A solution of barium hydroxide, #Ba(OH)_2# contains 4.285 grams of barium hydroxide in 100 mL of solution. What is the molarity of the solution?</h1>
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samsanit
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May 9, 2018
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<div class="markdown"><p>2.5 M </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know the molar mass of barium hydroxide is 171.34 (you can google this or just add the mass of each atom in the molecule). Knowing that <mathjax>#molarity = (mols) / (liters)#</mathjax> we need to turn that 4.285 g Ba(OH)2 into mols. </p>
<p>This is done by dividing 4.285 g by the 171.34 g molar mass giving us 0.2500 mol of Ba(OH)2.</p>
<p>Plugging this all back into the equation for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> we get <mathjax>#molarity = (0.2500 mol)/(0.100 L) #</mathjax> giving us a molarity of 2.5 M</p></div>
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</article> | A solution of barium hydroxide, #Ba(OH)_2# contains 4.285 grams of barium hydroxide in 100 mL of solution. What is the molarity of the solution? | null |
807 | ab0c64e6-6ddd-11ea-935c-ccda262736ce | https://socratic.org/questions/using-the-following-data-how-do-you-calculate-the-ksp-value-for-each-solid-a-the | 9.9 × 10^(-55) | start physical_unit 16 16 equilibrium_constant_k none qc_end physical_unit 16 16 18 21 solubility qc_end end | [{"type":"physical unit","value":"Ksp value [OF] Pb3(PO4)2"}] | [{"type":"physical unit","value":"9.9 × 10^(-55)"}] | [{"type":"physical unit","value":"Solubility [OF] Pb3(PO4)2 [=] \\pu{6.2 × 10^(-12) mol/L}"}] | <h1 class="questionTitle" itemprop="name">Using the following data, how do you calculate the Ksp value given that the solubility of #"Pb"_3("PO"_4)_2# is #6.2xx10^(-12) "mol/L"# ?
</h1> | null | 9.9 × 10^(-55) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a solution of <em>lead(II) phosphate</em>, an <strong>insoluble</strong> ionic compound that does not dissociate completely when dissolved in water. </p>
<p>The first thing to do here is write the equilibrium reaction that describes the <strong>partial dissociation</strong> of the salt. </p>
<blockquote>
<p><mathjax>#"Pb"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons color(blue)(3)"Pb"_ ((aq))^(2+) + color(red)(2)"PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of lead(II) phosphate that dissociates in solution you get <mathjax>#3#</mathjax> <strong>moles</strong> of lead(II) cations, <mathjax>#"Pb"^(2+)#</mathjax>, and <mathjax>#3#</mathjax> <strong>moles</strong> of phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax>.</p>
<p>Now, the <strong>molar solubility</strong> of lead(II) phosphate in water is said to be equal to <mathjax>#6.2 * 10^(-12)"mol L"^(-1)#</mathjax>.</p>
<p>This means that in <strong>one liter</strong> of water, presumably at room temperature, you can only hope to dissolve <mathjax>#6.2 * 10^(-12)#</mathjax> <strong>moles</strong> of lead(II) phosphate. </p>
<p>The concentration of the resulting ions will be </p>
<blockquote>
<p><mathjax>#["Pb"^(2+)] = color(blue)(3) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.86 * 10^(-11)"mol L"^(-1)#</mathjax></p>
<p><mathjax>#["PO"_4^(3-)] = color(red)(2) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.24 * 10^(-11)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>The <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Pb"^(2+)]^color(blue)(3) * ["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the concentrations of the two ions to get </p>
<blockquote>
<p><mathjax>#K_(sp) = (1.86 * 10^(-11))^color(blue)(3) * (1.24 * 10^(-11))^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(9.9 * 10^(-55))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to tow <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 molar solubility of lead(II) phosphate. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#K_(sp) = 9.9 * 10^(-55)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a solution of <em>lead(II) phosphate</em>, an <strong>insoluble</strong> ionic compound that does not dissociate completely when dissolved in water. </p>
<p>The first thing to do here is write the equilibrium reaction that describes the <strong>partial dissociation</strong> of the salt. </p>
<blockquote>
<p><mathjax>#"Pb"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons color(blue)(3)"Pb"_ ((aq))^(2+) + color(red)(2)"PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of lead(II) phosphate that dissociates in solution you get <mathjax>#3#</mathjax> <strong>moles</strong> of lead(II) cations, <mathjax>#"Pb"^(2+)#</mathjax>, and <mathjax>#3#</mathjax> <strong>moles</strong> of phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax>.</p>
<p>Now, the <strong>molar solubility</strong> of lead(II) phosphate in water is said to be equal to <mathjax>#6.2 * 10^(-12)"mol L"^(-1)#</mathjax>.</p>
<p>This means that in <strong>one liter</strong> of water, presumably at room temperature, you can only hope to dissolve <mathjax>#6.2 * 10^(-12)#</mathjax> <strong>moles</strong> of lead(II) phosphate. </p>
<p>The concentration of the resulting ions will be </p>
<blockquote>
<p><mathjax>#["Pb"^(2+)] = color(blue)(3) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.86 * 10^(-11)"mol L"^(-1)#</mathjax></p>
<p><mathjax>#["PO"_4^(3-)] = color(red)(2) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.24 * 10^(-11)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>The <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Pb"^(2+)]^color(blue)(3) * ["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the concentrations of the two ions to get </p>
<blockquote>
<p><mathjax>#K_(sp) = (1.86 * 10^(-11))^color(blue)(3) * (1.24 * 10^(-11))^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(9.9 * 10^(-55))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to tow <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 molar solubility of lead(II) phosphate. </p></div>
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<h1 class="questionTitle" itemprop="name">Using the following data, how do you calculate the Ksp value given that the solubility of #"Pb"_3("PO"_4)_2# is #6.2xx10^(-12) "mol/L"# ?
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Stefan V.
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Jun 26, 2016
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<div class="markdown"><p><mathjax>#K_(sp) = 9.9 * 10^(-55)#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that you're dealing with a solution of <em>lead(II) phosphate</em>, an <strong>insoluble</strong> ionic compound that does not dissociate completely when dissolved in water. </p>
<p>The first thing to do here is write the equilibrium reaction that describes the <strong>partial dissociation</strong> of the salt. </p>
<blockquote>
<p><mathjax>#"Pb"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons color(blue)(3)"Pb"_ ((aq))^(2+) + color(red)(2)"PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>for every mole</strong> of lead(II) phosphate that dissociates in solution you get <mathjax>#3#</mathjax> <strong>moles</strong> of lead(II) cations, <mathjax>#"Pb"^(2+)#</mathjax>, and <mathjax>#3#</mathjax> <strong>moles</strong> of phosphate anions, <mathjax>#"PO"_4^(3-)#</mathjax>.</p>
<p>Now, the <strong>molar solubility</strong> of lead(II) phosphate in water is said to be equal to <mathjax>#6.2 * 10^(-12)"mol L"^(-1)#</mathjax>.</p>
<p>This means that in <strong>one liter</strong> of water, presumably at room temperature, you can only hope to dissolve <mathjax>#6.2 * 10^(-12)#</mathjax> <strong>moles</strong> of lead(II) phosphate. </p>
<p>The concentration of the resulting ions will be </p>
<blockquote>
<p><mathjax>#["Pb"^(2+)] = color(blue)(3) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.86 * 10^(-11)"mol L"^(-1)#</mathjax></p>
<p><mathjax>#["PO"_4^(3-)] = color(red)(2) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.24 * 10^(-11)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>The <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Pb"^(2+)]^color(blue)(3) * ["PO"_4^(3-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the concentrations of the two ions to get </p>
<blockquote>
<p><mathjax>#K_(sp) = (1.86 * 10^(-11))^color(blue)(3) * (1.24 * 10^(-11))^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(9.9 * 10^(-55))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to tow <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 molar solubility of lead(II) phosphate. </p></div>
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</article> | Using the following data, how do you calculate the Ksp value given that the solubility of #"Pb"_3("PO"_4)_2# is #6.2xx10^(-12) "mol/L"# ?
| null |
808 | ac216868-6ddd-11ea-a210-ccda262736ce | https://socratic.org/questions/what-coefficient-would-the-o-2-have-after-balancing-c-2h-8-o-2-co-2-h-2o | 4 | start physical_unit 4 4 coefficient none qc_end chemical_equation 8 14 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] O2"}] | [{"type":"physical unit","value":"4"}] | [{"type":"chemical equation","value":"C2H8 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2H_8+O_2rarrCO_2+H_2O#</mathjax></p>
<p>First, I advise you to make a brief list of values for each side:</p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 2<br/>
O = 3</p>
<p>When <a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">balancing chemical equations</a>, we should always try to balance the term with the most <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved or one with an odd number of atoms. A single element is usually left until the end because it can be balanced without affecting other elements.</p>
<p>In this case, the oxygen on the RHS has only 3. If we multiply by 4, we are able to balance the hydrogen and get the oxygen even. </p>
<p><mathjax>#C_2H_8+O_2rarrCO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 8<br/>
O = 6</p>
<p>Let's multiply the <mathjax>#CO_2#</mathjax> 2 to balance to carbon:</p>
<p><mathjax>#C_2H_8+O_2rarr2CO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 2<br/>
H = 8<br/>
O = 8</p>
<p>Finally, if we balance the oxygen by multiplying the <mathjax>#O_2#</mathjax> by 4, we are done:</p>
<p><mathjax>#C_2H_8+4O_2rarr2CO_2+4H_2O#</mathjax></p>
<p>Answer = 4</p></div>
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<div class="markdown"><p>4</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2H_8+O_2rarrCO_2+H_2O#</mathjax></p>
<p>First, I advise you to make a brief list of values for each side:</p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 2<br/>
O = 3</p>
<p>When <a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">balancing chemical equations</a>, we should always try to balance the term with the most <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved or one with an odd number of atoms. A single element is usually left until the end because it can be balanced without affecting other elements.</p>
<p>In this case, the oxygen on the RHS has only 3. If we multiply by 4, we are able to balance the hydrogen and get the oxygen even. </p>
<p><mathjax>#C_2H_8+O_2rarrCO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 8<br/>
O = 6</p>
<p>Let's multiply the <mathjax>#CO_2#</mathjax> 2 to balance to carbon:</p>
<p><mathjax>#C_2H_8+O_2rarr2CO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 2<br/>
H = 8<br/>
O = 8</p>
<p>Finally, if we balance the oxygen by multiplying the <mathjax>#O_2#</mathjax> by 4, we are done:</p>
<p><mathjax>#C_2H_8+4O_2rarr2CO_2+4H_2O#</mathjax></p>
<p>Answer = 4</p></div>
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<h1 class="questionTitle" itemprop="name">What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#?</h1>
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<div class="markdown"><p>4</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#C_2H_8+O_2rarrCO_2+H_2O#</mathjax></p>
<p>First, I advise you to make a brief list of values for each side:</p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 2<br/>
O = 3</p>
<p>When <a href="https://socratic.org/chemistry/chemical-reactions/balancing-chemical-equations">balancing chemical equations</a>, we should always try to balance the term with the most <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved or one with an odd number of atoms. A single element is usually left until the end because it can be balanced without affecting other elements.</p>
<p>In this case, the oxygen on the RHS has only 3. If we multiply by 4, we are able to balance the hydrogen and get the oxygen even. </p>
<p><mathjax>#C_2H_8+O_2rarrCO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 1<br/>
H = 8<br/>
O = 6</p>
<p>Let's multiply the <mathjax>#CO_2#</mathjax> 2 to balance to carbon:</p>
<p><mathjax>#C_2H_8+O_2rarr2CO_2+4H_2O#</mathjax></p>
<p><strong>LHS</strong> </p>
<p>C = 2 <br/>
H = 8<br/>
O = 2</p>
<p><strong>RHS</strong> </p>
<p>C = 2<br/>
H = 8<br/>
O = 8</p>
<p>Finally, if we balance the oxygen by multiplying the <mathjax>#O_2#</mathjax> by 4, we are done:</p>
<p><mathjax>#C_2H_8+4O_2rarr2CO_2+4H_2O#</mathjax></p>
<p>Answer = 4</p></div>
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</article> | What coefficient would the #O_2# have after balancing #C_2H_8 + O_2 -> CO_2 + H_2O#? | null |
809 | a94c7188-6ddd-11ea-b3ff-ccda262736ce | https://socratic.org/questions/58f3df47b72cff7e37a30875 | 0.45 g | start physical_unit 11 12 mass g qc_end physical_unit 6 7 2 3 mass qc_end substance 21 22 qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium iodide [IN] g"}] | [{"type":"physical unit","value":"0.45 g"}] | [{"type":"physical unit","value":"Mass [OF] lead nitrate [=] \\pu{0.450 g}"},{"type":"substance name","value":"Iodide salt"}] | <h1 class="questionTitle" itemprop="name">Given a #0.450*g# mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt? </h1> | null | 0.45 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb(NO_3)_2(aq) + 2KI(aq) rarr PbI_2(s)darr + 2KNO_3(aq)#</mathjax></p>
<p><mathjax>#"Lead iodide"#</mathjax> precipitates from aqueous solution as a bright yellow powder. And it is pretty insoluble stuff. </p>
<p>With respect to <mathjax>#Pb(NO_3)_2#</mathjax>, we have a molar quantity of <mathjax>#(0.450*g)/(331.20*g*mol^-1)=1.36xx10^-3*mol#</mathjax> .</p>
<p>And thus, for metathesis, we need a TWICE molar quantity of <mathjax>#"potassium iodide"#</mathjax>, i.e. <mathjax>#2xx1.36xx10^-3*molxx166.0*g*mol^-1=0.451*g#</mathjax> <mathjax>#KI#</mathjax>. </p>
<p>Anyway, if I have misinterpreted your problem, send me a beatdown. </p></div>
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<div class="markdown"><p>I think you mean to ask how much <mathjax>#KI#</mathjax> you need to precipitate the <mathjax>#Pb^(2+)#</mathjax> out from aqueous solution..............</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb(NO_3)_2(aq) + 2KI(aq) rarr PbI_2(s)darr + 2KNO_3(aq)#</mathjax></p>
<p><mathjax>#"Lead iodide"#</mathjax> precipitates from aqueous solution as a bright yellow powder. And it is pretty insoluble stuff. </p>
<p>With respect to <mathjax>#Pb(NO_3)_2#</mathjax>, we have a molar quantity of <mathjax>#(0.450*g)/(331.20*g*mol^-1)=1.36xx10^-3*mol#</mathjax> .</p>
<p>And thus, for metathesis, we need a TWICE molar quantity of <mathjax>#"potassium iodide"#</mathjax>, i.e. <mathjax>#2xx1.36xx10^-3*molxx166.0*g*mol^-1=0.451*g#</mathjax> <mathjax>#KI#</mathjax>. </p>
<p>Anyway, if I have misinterpreted your problem, send me a beatdown. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Given a #0.450*g# mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt? </h1>
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<div class="markdown"><p>I think you mean to ask how much <mathjax>#KI#</mathjax> you need to precipitate the <mathjax>#Pb^(2+)#</mathjax> out from aqueous solution..............</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Pb(NO_3)_2(aq) + 2KI(aq) rarr PbI_2(s)darr + 2KNO_3(aq)#</mathjax></p>
<p><mathjax>#"Lead iodide"#</mathjax> precipitates from aqueous solution as a bright yellow powder. And it is pretty insoluble stuff. </p>
<p>With respect to <mathjax>#Pb(NO_3)_2#</mathjax>, we have a molar quantity of <mathjax>#(0.450*g)/(331.20*g*mol^-1)=1.36xx10^-3*mol#</mathjax> .</p>
<p>And thus, for metathesis, we need a TWICE molar quantity of <mathjax>#"potassium iodide"#</mathjax>, i.e. <mathjax>#2xx1.36xx10^-3*molxx166.0*g*mol^-1=0.451*g#</mathjax> <mathjax>#KI#</mathjax>. </p>
<p>Anyway, if I have misinterpreted your problem, send me a beatdown. </p></div>
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Douglas K.
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<div class="markdown"><p><mathjax>#0.451" g of "KI#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given: <mathjax>#0.450" g "Pb(NO_3)_2#</mathjax></p>
<p>We begin the factor-label method by writing the given over 1:</p>
<p><mathjax>#(0.450" g "Pb(NO_3)_2)/1#</mathjax></p>
<p>We look up the molar mass of <mathjax>#Pb(NO_3)_2#</mathjax> and add it as a factor of moles per gram to the conversion:</p>
<p><mathjax>#(0.450" g "Pb(NO_3)_2)/1(1" mole "Pb(NO_3)_2)/(331.2" g "Pb(NO_3)_2)#</mathjax></p>
<p>Please observe how the units cancel:</p>
<p><mathjax>#(0.450cancel(" g "Pb(NO_3)_2))/1(1" mole "Pb(NO_3)_2)/(331.2cancel(" g "Pb(NO_3)_2))#</mathjax></p>
<p>From the balanced equation, we observe that 2 moles of <mathjax>#KI#</mathjax> is needed for 1 mole of <mathjax>#Pb(NO_3)_2#</mathjax>, therefore, we add this as a factor to the conversion:</p>
<p><mathjax>#(0.450cancel(" g "Pb(NO_3)_2))/1(1" mole "Pb(NO_3)_2)/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(1" mole "Pb(NO_3)_2)#</mathjax></p>
<p>Again, please observe how we have cancelled the units as we progress toward the conversion:</p>
<p><mathjax>#(0.450cancel(" g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(cancel(1" mole "Pb(NO_3)_2))#</mathjax></p>
<p>Finally, we look up the molar mass of <mathjax>#KI#</mathjax> and add it as a factor of grams per mole to the conversion:</p>
<p><mathjax>#(0.450cancel(" g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(cancel(1" mole "Pb(NO_3)_2))(166.0" g "KI)/(1" mole "KI)#</mathjax></p>
<p>With a final cancellation of units, please observe that the only units that remain are grams of Potassium Iodide:</p>
<p><mathjax>#(0.450cancel(" g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2cancel(" moles "KI))/(cancel(1" mole "Pb(NO_3)_2))(166.0" g "KI)/(1cancel(" mole "KI))#</mathjax></p>
<p>We merely perform the multiplications and divisions to obtain the answer:</p>
<p><mathjax>#0.450/1(1/331.2)(2/1)(166.0" g "KI)/1= 0.451" g "KI#</mathjax></p></div>
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</article> | Given a #0.450*g# mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt? | null |
810 | ad081664-6ddd-11ea-975d-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-buffer-system-that-contains-0-170-m-hydrocyanic-acid-hcn-and | 9.13 | start physical_unit 6 7 ph none qc_end physical_unit 14 14 10 11 molarity qc_end physical_unit 20 20 16 17 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the buffer system"}] | [{"type":"physical unit","value":"9.13"}] | [{"type":"physical unit","value":"Molarity [OF] HCN solution [=] \\pu{0.170 M}"},{"type":"physical unit","value":"Molarity [OF] NaCN solution [=] \\pu{0.140 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a buffer system that contains 0.170 M hydrocyanic acid (HCN) and 0.140 M sodium cyanide (NaCN)?</h1> | null | 9.13 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer equation....</a> </p>
<p>And here....</p>
<p><mathjax>#pH=pK_a+log_10{[[A^-]]/[[HA]]}#</mathjax></p>
<p>And so in this context....</p>
<p><mathjax>#pH=9.21+log_10{[[N-=C^-]]/[[HC-=N]]}#</mathjax></p>
<p><mathjax>#pH=9.21+underbrace(log_10{[[0.140*mol*L^-1]]/[[0.170*mol*L^-1]]})_(-0.084)=9.13.#</mathjax></p>
<p>AS always in these buffers, the <mathjax>#pH#</mathjax> moves in the direction of the more concentrated species....here the acid is more concentrated, and the <mathjax>#pH#</mathjax> decreases with respect to <mathjax>#pK_a#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#pH=9.13...#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer equation....</a> </p>
<p>And here....</p>
<p><mathjax>#pH=pK_a+log_10{[[A^-]]/[[HA]]}#</mathjax></p>
<p>And so in this context....</p>
<p><mathjax>#pH=9.21+log_10{[[N-=C^-]]/[[HC-=N]]}#</mathjax></p>
<p><mathjax>#pH=9.21+underbrace(log_10{[[0.140*mol*L^-1]]/[[0.170*mol*L^-1]]})_(-0.084)=9.13.#</mathjax></p>
<p>AS always in these buffers, the <mathjax>#pH#</mathjax> moves in the direction of the more concentrated species....here the acid is more concentrated, and the <mathjax>#pH#</mathjax> decreases with respect to <mathjax>#pK_a#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a buffer system that contains 0.170 M hydrocyanic acid (HCN) and 0.140 M sodium cyanide (NaCN)?</h1>
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<div class="markdown"><p><mathjax>#pH=9.13...#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer equation....</a> </p>
<p>And here....</p>
<p><mathjax>#pH=pK_a+log_10{[[A^-]]/[[HA]]}#</mathjax></p>
<p>And so in this context....</p>
<p><mathjax>#pH=9.21+log_10{[[N-=C^-]]/[[HC-=N]]}#</mathjax></p>
<p><mathjax>#pH=9.21+underbrace(log_10{[[0.140*mol*L^-1]]/[[0.170*mol*L^-1]]})_(-0.084)=9.13.#</mathjax></p>
<p>AS always in these buffers, the <mathjax>#pH#</mathjax> moves in the direction of the more concentrated species....here the acid is more concentrated, and the <mathjax>#pH#</mathjax> decreases with respect to <mathjax>#pK_a#</mathjax>.</p></div>
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</article> | What is the pH of a buffer system that contains 0.170 M hydrocyanic acid (HCN) and 0.140 M sodium cyanide (NaCN)? | null |
811 | a9784b7a-6ddd-11ea-935b-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-containing-c-h-and-o-if-combustion-o | CH2O | start chemical_formula qc_end c_other OTHER qc_end physical_unit 18 19 15 16 mass qc_end physical_unit 24 24 21 22 mass qc_end physical_unit 29 29 26 27 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH2O"}] | [{"type":"other","value":"a compound containing C,H, and O"},{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{3.69 g}"},{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{5.40 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{2.22 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound containing #C, H#, and #O# if combustion of 3.69 g of the compound yields 5.40 g of #CO_2# and 2.22 g of #H_2O#?</h1> | null | CH2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ONLY the <mathjax>#C#</mathjax> and the <mathjax>#H#</mathjax> of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)</p>
<p><mathjax>#"Moles (i) and mass (ii) of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.40*g)/(44.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.123*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#1.47*g*C#</mathjax></p>
<p><mathjax>#"Moles (i) and mass (ii) of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2xx(2.22*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.247*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#0.249*g*H#</mathjax>. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.</p>
<p>And now, finally, we work out the percentage composition of <mathjax>#C,H,O#</mathjax> with respect to the original sample:</p>
<p><mathjax>#%C#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.47*g)/(3.69*g)xx100%=40.00%#</mathjax></p>
<p><mathjax>#%H#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.249*g)/(3.69*g)xx100%=6.75%#</mathjax></p>
<p><mathjax>#%O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100-39.84-6.75)%=53.25%#</mathjax></p>
<p>Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus <mathjax>#O%#</mathjax> is the percentage balance.</p>
<p>After all this wrok we start again. We assume that there were <mathjax>#100*g#</mathjax> of compound. And from this we work out the empirical formula. </p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol*C#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.75*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.70*mol*H#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(53.41*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.34*mol*H#</mathjax>.</p>
<p>If we divide thru by the smallest molar quantity, we get, an empirical formula of <mathjax>#CH_2O#</mathjax>. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>We get finally an empirical formula of <mathjax>#CH_2O#</mathjax>; I think the question is suspect. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ONLY the <mathjax>#C#</mathjax> and the <mathjax>#H#</mathjax> of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)</p>
<p><mathjax>#"Moles (i) and mass (ii) of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.40*g)/(44.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.123*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#1.47*g*C#</mathjax></p>
<p><mathjax>#"Moles (i) and mass (ii) of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2xx(2.22*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.247*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#0.249*g*H#</mathjax>. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.</p>
<p>And now, finally, we work out the percentage composition of <mathjax>#C,H,O#</mathjax> with respect to the original sample:</p>
<p><mathjax>#%C#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.47*g)/(3.69*g)xx100%=40.00%#</mathjax></p>
<p><mathjax>#%H#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.249*g)/(3.69*g)xx100%=6.75%#</mathjax></p>
<p><mathjax>#%O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100-39.84-6.75)%=53.25%#</mathjax></p>
<p>Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus <mathjax>#O%#</mathjax> is the percentage balance.</p>
<p>After all this wrok we start again. We assume that there were <mathjax>#100*g#</mathjax> of compound. And from this we work out the empirical formula. </p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol*C#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.75*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.70*mol*H#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(53.41*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.34*mol*H#</mathjax>.</p>
<p>If we divide thru by the smallest molar quantity, we get, an empirical formula of <mathjax>#CH_2O#</mathjax>. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound containing #C, H#, and #O# if combustion of 3.69 g of the compound yields 5.40 g of #CO_2# and 2.22 g of #H_2O#?</h1>
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<span class="dateCreated" datetime="2016-09-09T06:22:15" itemprop="dateCreated">
Sep 9, 2016
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<div class="markdown"><p>We get finally an empirical formula of <mathjax>#CH_2O#</mathjax>; I think the question is suspect. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>ONLY the <mathjax>#C#</mathjax> and the <mathjax>#H#</mathjax> of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)</p>
<p><mathjax>#"Moles (i) and mass (ii) of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(5.40*g)/(44.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.123*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#1.47*g*C#</mathjax></p>
<p><mathjax>#"Moles (i) and mass (ii) of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2xx(2.22*g)/(18.01*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.247*mol#</mathjax> <mathjax>#-=#</mathjax> <mathjax>#0.249*g*H#</mathjax>. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.</p>
<p>And now, finally, we work out the percentage composition of <mathjax>#C,H,O#</mathjax> with respect to the original sample:</p>
<p><mathjax>#%C#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.47*g)/(3.69*g)xx100%=40.00%#</mathjax></p>
<p><mathjax>#%H#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.249*g)/(3.69*g)xx100%=6.75%#</mathjax></p>
<p><mathjax>#%O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(100-39.84-6.75)%=53.25%#</mathjax></p>
<p>Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus <mathjax>#O%#</mathjax> is the percentage balance.</p>
<p>After all this wrok we start again. We assume that there were <mathjax>#100*g#</mathjax> of compound. And from this we work out the empirical formula. </p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33*mol*C#</mathjax>.</p>
<p><mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(6.75*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6.70*mol*H#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(53.41*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.34*mol*H#</mathjax>.</p>
<p>If we divide thru by the smallest molar quantity, we get, an empirical formula of <mathjax>#CH_2O#</mathjax>. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar. </p></div>
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</article> | What is the empirical formula of a compound containing #C, H#, and #O# if combustion of 3.69 g of the compound yields 5.40 g of #CO_2# and 2.22 g of #H_2O#? | null |
812 | ac0bf0ca-6ddd-11ea-8df2-ccda262736ce | https://socratic.org/questions/how-many-grams-of-carbon-monoxide-can-you-produce-from-3-grams-of-carbon-and-an- | 7 grams | start physical_unit 4 5 mass g qc_end physical_unit 4 4 10 11 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon monoxide [IN] grams"}] | [{"type":"physical unit","value":"7 grams"}] | [{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{3 grams}"},{"type":"other","value":"Excess oxygen."}] | <h1 class="questionTitle" itemprop="name">How many grams of carbon monoxide can you produce from 3 grams of carbon and an excess of oxygen?</h1> | null | 7 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation:</p>
<p><mathjax>#C(s) +1/2O_2(g) rarr CO(g)#</mathjax></p>
<p>And now (ii) we need the equivalent quantity of <mathjax>#C#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/4*"mole"#</mathjax>.</p>
<p>So mass of carbon monoxide <mathjax>#=#</mathjax> <mathjax>#1/4*molxx28*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7*g#</mathjax>.</p></div>
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<div class="markdown"><p>About <mathjax>#7#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#CO#</mathjax>,</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a stoichiometric equation:</p>
<p><mathjax>#C(s) +1/2O_2(g) rarr CO(g)#</mathjax></p>
<p>And now (ii) we need the equivalent quantity of <mathjax>#C#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/4*"mole"#</mathjax>.</p>
<p>So mass of carbon monoxide <mathjax>#=#</mathjax> <mathjax>#1/4*molxx28*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of carbon monoxide can you produce from 3 grams of carbon and an excess of oxygen?</h1>
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<div class="markdown"><p>About <mathjax>#7#</mathjax> <mathjax>#g#</mathjax> of <mathjax>#CO#</mathjax>,</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We need (i) a stoichiometric equation:</p>
<p><mathjax>#C(s) +1/2O_2(g) rarr CO(g)#</mathjax></p>
<p>And now (ii) we need the equivalent quantity of <mathjax>#C#</mathjax>.</p>
<p><mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1/4*"mole"#</mathjax>.</p>
<p>So mass of carbon monoxide <mathjax>#=#</mathjax> <mathjax>#1/4*molxx28*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7*g#</mathjax>.</p></div>
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</article> | How many grams of carbon monoxide can you produce from 3 grams of carbon and an excess of oxygen? | null |
813 | aaf5940b-6ddd-11ea-a3c2-ccda262736ce | https://socratic.org/questions/water-is-decomposed-by-electrolysis-to-form-the-gaseous-products-hydrogen-h-2-an | 2 H2O -> 2 H2 + O2 | start chemical_equation qc_end chemical_equation 11 11 qc_end chemical_equation 14 14 qc_end substance 0 0 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"2 H2O -> 2 H2 + O2"}] | [{"type":"chemical equation","value":"H2"},{"type":"chemical equation","value":"O2"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">Water is decomposed by electrolysis to form the gaseous products hydrogen, #H_2#, and oxygen, #O_2#. How do you write the balanced equation for this reaction?</h1> | null | 2 H2O -> 2 H2 + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One molecule of water <mathjax>#"H"_2"O"#</mathjax> can give 2 Hydrogen and 1 Oxygen atom, which is sufficient to make 1 <mathjax>#"H"_2#</mathjax> molecule and <mathjax>#1/2#</mathjax> of a molecule of <mathjax>#"O"_2#</mathjax>. You can write it as</p>
<blockquote>
<p><mathjax>#"H"_2"O" -> "H"_2 + 1/2"O"_2#</mathjax></p>
</blockquote>
<p>However, there is no such thing as half a molecule. To avoid fractions, multiply everywhere by 2. So it becomes </p>
<blockquote>
<p><mathjax>#2"H"_2"O" -> 2"H"_2 + "O"_2#</mathjax></p>
</blockquote>
<p>This means that you can make 2 molecules of <mathjax>#"H"_2#</mathjax> and 1 molecule of <mathjax>#"O"_2#</mathjax> with 2 molecules of <mathjax>#"H"_2"O"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#2"H"_2"O" -> 2"H"_2 + "O"_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One molecule of water <mathjax>#"H"_2"O"#</mathjax> can give 2 Hydrogen and 1 Oxygen atom, which is sufficient to make 1 <mathjax>#"H"_2#</mathjax> molecule and <mathjax>#1/2#</mathjax> of a molecule of <mathjax>#"O"_2#</mathjax>. You can write it as</p>
<blockquote>
<p><mathjax>#"H"_2"O" -> "H"_2 + 1/2"O"_2#</mathjax></p>
</blockquote>
<p>However, there is no such thing as half a molecule. To avoid fractions, multiply everywhere by 2. So it becomes </p>
<blockquote>
<p><mathjax>#2"H"_2"O" -> 2"H"_2 + "O"_2#</mathjax></p>
</blockquote>
<p>This means that you can make 2 molecules of <mathjax>#"H"_2#</mathjax> and 1 molecule of <mathjax>#"O"_2#</mathjax> with 2 molecules of <mathjax>#"H"_2"O"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Water is decomposed by electrolysis to form the gaseous products hydrogen, #H_2#, and oxygen, #O_2#. How do you write the balanced equation for this reaction?</h1>
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<div class="markdown"><p><mathjax>#2"H"_2"O" -> 2"H"_2 + "O"_2#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One molecule of water <mathjax>#"H"_2"O"#</mathjax> can give 2 Hydrogen and 1 Oxygen atom, which is sufficient to make 1 <mathjax>#"H"_2#</mathjax> molecule and <mathjax>#1/2#</mathjax> of a molecule of <mathjax>#"O"_2#</mathjax>. You can write it as</p>
<blockquote>
<p><mathjax>#"H"_2"O" -> "H"_2 + 1/2"O"_2#</mathjax></p>
</blockquote>
<p>However, there is no such thing as half a molecule. To avoid fractions, multiply everywhere by 2. So it becomes </p>
<blockquote>
<p><mathjax>#2"H"_2"O" -> 2"H"_2 + "O"_2#</mathjax></p>
</blockquote>
<p>This means that you can make 2 molecules of <mathjax>#"H"_2#</mathjax> and 1 molecule of <mathjax>#"O"_2#</mathjax> with 2 molecules of <mathjax>#"H"_2"O"#</mathjax>.</p></div>
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</article> | Water is decomposed by electrolysis to form the gaseous products hydrogen, #H_2#, and oxygen, #O_2#. How do you write the balanced equation for this reaction? | null |
814 | ab21888c-6ddd-11ea-b15d-ccda262736ce | https://socratic.org/questions/how-many-moles-of-gas-are-contained-in-1-25-l-at-303-k-and-1-10-atm | 0.06 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 volume qc_end physical_unit 4 4 11 12 temperature qc_end physical_unit 4 4 14 15 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] gas [IN] moles"}] | [{"type":"physical unit","value":"0.06 moles"}] | [{"type":"physical unit","value":"Volume [OF] gas [=] \\pu{1.25 L}"},{"type":"physical unit","value":"Temperature [OF] gas [=] \\pu{303 K}"},{"type":"physical unit","value":"Pressure [OF] gas [=] \\pu{1.10 atm}"}] | <h1 class="questionTitle" itemprop="name">How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm?</h1> | null | 0.06 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure <strong>must</strong> match those used in the expression of <mathjax>#R#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaacolor(Red)("Need")aaaaaaaaaaaaaaacolor(blue)("Have")aaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaa)color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())#</mathjax></p>
</blockquote>
<p>Rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#n#</mathjax> </p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.10 color(red)(cancel(color(black)("atm"))) * 1.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.0553 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure <strong>must</strong> match those used in the expression of <mathjax>#R#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaacolor(Red)("Need")aaaaaaaaaaaaaaacolor(blue)("Have")aaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaa)color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())#</mathjax></p>
</blockquote>
<p>Rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#n#</mathjax> </p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.10 color(red)(cancel(color(black)("atm"))) * 1.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-17T19:20:50" itemprop="dateCreated">
Jun 17, 2016
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<div class="markdown"><p><mathjax>#"0.0553 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>Here you have</p>
<blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
</blockquote>
<p>Now, take a look at the units used in the expression of the universal gas constant. The units you have for volume, temperature, and pressure <strong>must</strong> match those used in the expression of <mathjax>#R#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaacolor(Red)("Need")aaaaaaaaaaaaaaacolor(blue)("Have")aaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaa)color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaa)#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Liters, L")aaaaaaaaaaaaacolor(black)("Liters, L")aaaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Kelvin, K")aaaaaaaaaaaacolor(black)("Kelvin, K")aaaaaaaaaacolor(green)(sqrt())#</mathjax><br/>
<mathjax>#color(white)(aaaaaaacolor(black)("Atmospheres, atm")aaaaacolor(black)("Atmospheres, atm")aaaacolor(green)(sqrt())#</mathjax></p>
</blockquote>
<p>Rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to solve for <mathjax>#n#</mathjax> </p>
<blockquote>
<p><mathjax>#PV = nRT implies n = (PV)/(RT)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#n = (1.10 color(red)(cancel(color(black)("atm"))) * 1.25color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 303color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0553 moles")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many moles of gas are contained in 1.25 L at 303 K and 1.10 atm? | null |
815 | aa07cd3a-6ddd-11ea-998f-ccda262736ce | https://socratic.org/questions/58ac9f48b72cff4efe4558d5 | C7H5O3SN | start chemical_formula qc_end substance 15 15 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the organic compound [IN] empirical"}] | [{"type":"chemical equation","value":"C7H5O3SN"}] | [{"type":"physical unit","value":"Percentage [OF] C in the organic compound [=] \\pu{45.90%}"},{"type":"physical unit","value":"Percentage [OF] H in the organic compound [=] \\pu{2.75%}"},{"type":"physical unit","value":"Percentage [OF] N in the organic compound [=] \\pu{7.65%}"},{"type":"physical unit","value":"Percentage [OF] S in the organic compound [=] \\pu{17.50%}"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">An organic compound contains #45.90%# #C#; #2.75%# #H#; #7.65%# #N#; #17.50%# #S#; and the balance oxygen. What is the empirical formula of the compound?</h1> | null | C7H5O3SN | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>
<p>First, list all of your <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. In this case they are carbon, hydrogen, oxygen, sulphur and nitrogen.</p>
</li>
<li>
<p>Assume that you have 100g of this compound. Consequently, you can assume that you have in grams the percentage you are given for each element. For example, you can assume that you have 45.9g of carbon, 2.75g of hydrogen, and so on. </p>
</li>
<li>
<p>Now, you get out your periodic table and look for the <mathjax>#A_r#</mathjax> of each element. For carbon it's 12.01, hydrogen is 1.01 and so on.</p>
</li>
<li>
<p>Find the number of moles you have for of each element. You do this by dividing the mass in grams by the <mathjax>#A_r#</mathjax> for each element. So, we have:<br/>
<mathjax>#45.9/12.01=3.82#</mathjax> moles of carbon,</p>
</li>
</ol>
<p><mathjax>#2.75/1.01=2.72#</mathjax> moles of hydrogen, </p>
<p><mathjax>#26.2/16=1.64#</mathjax> moles of oxygen, </p>
<p><mathjax>#17.5/32.07=0.55#</mathjax> moles of sulphur, </p>
<p><mathjax>#7.65/14=0.55#</mathjax> moles of nitrogen.</p>
<ol>
<li>Lastly, you just have to divide all of the values we just calculated by the smallest number of moles that there was. So in this case the smallest is 0.55. I have rounded the numbers to integers because they have to be integers in chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. For instance, you can't have 2.5 atoms.</li>
</ol>
<p>Carbon: <mathjax>#3.82/0.55=7#</mathjax></p>
<p>Hydrogen: <mathjax>#2.72/0.55=5#</mathjax></p>
<p>Oxygen: <mathjax>#1.64/0.55=3#</mathjax></p>
<p>Sulphur: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>Nitrogen: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>So the empirical formula is <mathjax>#C_7H_5O_3SN#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#C_7H_5O_3SN#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><ol>
<li>
<p>First, list all of your <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. In this case they are carbon, hydrogen, oxygen, sulphur and nitrogen.</p>
</li>
<li>
<p>Assume that you have 100g of this compound. Consequently, you can assume that you have in grams the percentage you are given for each element. For example, you can assume that you have 45.9g of carbon, 2.75g of hydrogen, and so on. </p>
</li>
<li>
<p>Now, you get out your periodic table and look for the <mathjax>#A_r#</mathjax> of each element. For carbon it's 12.01, hydrogen is 1.01 and so on.</p>
</li>
<li>
<p>Find the number of moles you have for of each element. You do this by dividing the mass in grams by the <mathjax>#A_r#</mathjax> for each element. So, we have:<br/>
<mathjax>#45.9/12.01=3.82#</mathjax> moles of carbon,</p>
</li>
</ol>
<p><mathjax>#2.75/1.01=2.72#</mathjax> moles of hydrogen, </p>
<p><mathjax>#26.2/16=1.64#</mathjax> moles of oxygen, </p>
<p><mathjax>#17.5/32.07=0.55#</mathjax> moles of sulphur, </p>
<p><mathjax>#7.65/14=0.55#</mathjax> moles of nitrogen.</p>
<ol>
<li>Lastly, you just have to divide all of the values we just calculated by the smallest number of moles that there was. So in this case the smallest is 0.55. I have rounded the numbers to integers because they have to be integers in chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. For instance, you can't have 2.5 atoms.</li>
</ol>
<p>Carbon: <mathjax>#3.82/0.55=7#</mathjax></p>
<p>Hydrogen: <mathjax>#2.72/0.55=5#</mathjax></p>
<p>Oxygen: <mathjax>#1.64/0.55=3#</mathjax></p>
<p>Sulphur: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>Nitrogen: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>So the empirical formula is <mathjax>#C_7H_5O_3SN#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">An organic compound contains #45.90%# #C#; #2.75%# #H#; #7.65%# #N#; #17.50%# #S#; and the balance oxygen. What is the empirical formula of the compound?</h1>
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Hydro G.
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<div class="markdown"><p>The empirical formula is <mathjax>#C_7H_5O_3SN#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><ol>
<li>
<p>First, list all of your <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. In this case they are carbon, hydrogen, oxygen, sulphur and nitrogen.</p>
</li>
<li>
<p>Assume that you have 100g of this compound. Consequently, you can assume that you have in grams the percentage you are given for each element. For example, you can assume that you have 45.9g of carbon, 2.75g of hydrogen, and so on. </p>
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<li>
<p>Now, you get out your periodic table and look for the <mathjax>#A_r#</mathjax> of each element. For carbon it's 12.01, hydrogen is 1.01 and so on.</p>
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<li>
<p>Find the number of moles you have for of each element. You do this by dividing the mass in grams by the <mathjax>#A_r#</mathjax> for each element. So, we have:<br/>
<mathjax>#45.9/12.01=3.82#</mathjax> moles of carbon,</p>
</li>
</ol>
<p><mathjax>#2.75/1.01=2.72#</mathjax> moles of hydrogen, </p>
<p><mathjax>#26.2/16=1.64#</mathjax> moles of oxygen, </p>
<p><mathjax>#17.5/32.07=0.55#</mathjax> moles of sulphur, </p>
<p><mathjax>#7.65/14=0.55#</mathjax> moles of nitrogen.</p>
<ol>
<li>Lastly, you just have to divide all of the values we just calculated by the smallest number of moles that there was. So in this case the smallest is 0.55. I have rounded the numbers to integers because they have to be integers in chemical <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. For instance, you can't have 2.5 atoms.</li>
</ol>
<p>Carbon: <mathjax>#3.82/0.55=7#</mathjax></p>
<p>Hydrogen: <mathjax>#2.72/0.55=5#</mathjax></p>
<p>Oxygen: <mathjax>#1.64/0.55=3#</mathjax></p>
<p>Sulphur: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>Nitrogen: <mathjax>#0.55/0.55=1#</mathjax></p>
<p>So the empirical formula is <mathjax>#C_7H_5O_3SN#</mathjax>.</p></div>
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anor277
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<div class="markdown"><p>We get an empirical formula of <mathjax>#C_7H_5NO_3S#</mathjax>. And the empirical formula represents the simplest whole number ratio that relates constituent atoms in a species. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As with all these problems, we assume a mass of <mathjax>#100*g#</mathjax>, and work out the molar quantities of each element from the given percentages in this mass, and normalize the result:</p>
<p><mathjax>#"Moles of carbon"=(45.90*g)/(12.011*g*mol^-1)=3.82*mol.#</mathjax></p>
<p><mathjax>#"Moles of hydrogen"=(2.75*g)/(1.00794*g*mol^-1)=2.73*mol.#</mathjax></p>
<p><mathjax>#"Moles of nitrogen"=(7.65*g)/(14.01*g*mol^-1)=0.546*mol.#</mathjax></p>
<p><mathjax>#"Moles of sulfur"=(17.50*g)/(32.06*g*mol^-1)=0.546*mol.#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(26.20*g)/(15.999*g*mol^-1)=1.64*mol.#</mathjax></p>
<p>And now we divide thru by the smallest molar quantity, that of nitrogen, and sulfur. We do this to normalize the result:</p>
<p><mathjax>#"Carbon atoms"=(3.82*mol)/(0.546*mol)=7.00#</mathjax></p>
<p><mathjax>#"Hydrogen atoms"=(2.73*mol)/(0.546*mol)=5.00#</mathjax></p>
<p><mathjax>#"Nitrogen atoms"=(0.546*mol)/(0.546*mol^-1)=1.00#</mathjax></p>
<p><mathjax>#"Moles of sulfur"=(0.546*mol)/(0.546*mol^-1)=1.00#</mathjax></p>
<p><mathjax>#"Moles of oxygen"=(1.64*mol)/(0.546*mol^-1)=3.00#</mathjax></p>
<p>Note that normally we would not be given a percentage oxygen content (why not? because oxygen is experimentally hard to measure); we would calculate the <mathjax>#%O#</mathjax> from the difference of the other percentages from 100%.</p>
<p>And thus we get an empirical formula of <mathjax>#C_7H_5NO_3S#</mathjax>.</p>
<p>Note the <mathjax>#"molecular formula"#</mathjax> is a simple whole number multiple of the <mathjax>#"empirical formula"#</mathjax>. We need a <mathjax>#"molecular mass determination"#</mathjax> before we make this assessment. For completeness, we know that the molecular weight of saccharin is <mathjax>#183.15*g*mol^-1#</mathjax>.</p>
<p>The <mathjax>#"molecular formula"#</mathjax> is ALWAYS a whole number multiple of the <mathjax>#"empirical formula"#</mathjax>.</p>
<p>Thus<mathjax>#"......................................................."#</mathjax> </p>
<p><mathjax>#183.15*g*mol^-1=nxx{7xx12.011 + 5xx1.00794+14.01+32.06+3xx15.999}*g*mol^-1=183.15*g*mol^-1xxn#</mathjax>. </p>
<p>Clearly <mathjax>#n=1#</mathjax>, and the empirical formula is the same as the molecular formula, i.e. <mathjax>#C_7H_5NO_3S#</mathjax>.</p>
<p>Good question. I am stealing it for my A2 class. </p></div>
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</article> | An organic compound contains #45.90%# #C#; #2.75%# #H#; #7.65%# #N#; #17.50%# #S#; and the balance oxygen. What is the empirical formula of the compound? | null |
816 | ab27954b-6ddd-11ea-8e07-ccda262736ce | https://socratic.org/questions/if-the-specific-heat-of-water-is-4-18j-g-c-how-many-joules-of-heat-are-given-off | 418.00 joules | start physical_unit 5 5 heat_energy j qc_end physical_unit 5 5 7 10 specific_heat qc_end physical_unit 5 5 20 21 mass qc_end physical_unit 5 5 26 27 temperature qc_end physical_unit 5 5 29 30 temperature qc_end end | [{"type":"physical unit","value":"Given off heat [OF] water [IN] joules"}] | [{"type":"physical unit","value":"418.00 joules"}] | [{"type":"physical unit","value":"Specific heat [OF] water [=] \\pu{4.18 J/(g * ℃)}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{5.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{45 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">If the specific heat of water is 4.18J/g°C, how many joules of heat are given off when 5.0 g of water cool from 45°C to 25°C? </h1> | null | 418.00 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#W=mcDeltaT#</mathjax></p>
<p><mathjax>#=5gxx4,18J//g.^@Cxx20^@C#</mathjax></p>
<p><mathjax>#=418J#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#418J#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#W=mcDeltaT#</mathjax></p>
<p><mathjax>#=5gxx4,18J//g.^@Cxx20^@C#</mathjax></p>
<p><mathjax>#=418J#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If the specific heat of water is 4.18J/g°C, how many joules of heat are given off when 5.0 g of water cool from 45°C to 25°C? </h1>
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<div class="markdown"><p><mathjax>#418J#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#W=mcDeltaT#</mathjax></p>
<p><mathjax>#=5gxx4,18J//g.^@Cxx20^@C#</mathjax></p>
<p><mathjax>#=418J#</mathjax></p></div>
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</article> | If the specific heat of water is 4.18J/g°C, how many joules of heat are given off when 5.0 g of water cool from 45°C to 25°C? | null |
817 | ac345ea6-6ddd-11ea-8d83-ccda262736ce | https://socratic.org/questions/one-liter-of-sulfur-vapor-s8-g-at-600-c-and-1-00-atm-is-burned-in-excess-pure-o2-1 | 7.14 g | start physical_unit 32 33 mass g qc_end physical_unit 5 5 0 1 volume qc_end physical_unit 5 5 8 9 temperature qc_end physical_unit 5 5 11 12 pressure qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] SO2 gas [IN] g"}] | [{"type":"physical unit","value":"7.14 g"}] | [{"type":"physical unit","value":"Volume [OF] S8(g) [=] \\pu{1 liter}"},{"type":"physical unit","value":"Temperature [OF] S8(g) [=] \\pu{600 ℃}"},{"type":"physical unit","value":"Pressure [OF] S8(g) [=] \\pu{1.00 atm}"},{"type":"other","value":"Excess pure O2."},{"type":"other","value":"Measured at the same temperature and pressure."}] | <h1 class="questionTitle" itemprop="name">One liter of sulfur vapor, S8(g) , at 600 ˚ C and 1.00 atm is burned in excess pure O2 to give SO2, measured at the same temperature and pressure. What mass of SO2 gas is obtained?</h1> | null | 7.14 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Don't we all love <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>? The best way to start these types of problems is with a balanced equation:</p>
<p><mathjax>#"S"_8(s) + 8"O"_2(g) -> 8"SO"_2(g)#</mathjax></p>
<p>Okay, now we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is number of moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature.</p>
<p>Since we have excess (unlimited) oxygen gas, we can ignore that for this problem. Let's just focus on <mathjax>#"S"_8#</mathjax> and <mathjax>#"SO"_2#</mathjax>. First, we must find the number of moles of S8. Let's use the ideal gas law:</p>
<p><mathjax>#P = 1.00 atm#</mathjax></p>
<p><mathjax>#V = 1 L#</mathjax></p>
<p><mathjax>#R = 0.0821 (L*atm)/(mol*K)#</mathjax></p>
<p><mathjax>#T = 600 ˚ C = 873 K#</mathjax></p>
<p><mathjax>#(1.00 atm)(1 L) = n(0.0821 (L*atm)/(mol*K))(873 K)#</mathjax></p>
<p><mathjax>#n = 0.014#</mathjax> mols</p>
<p>Now, we can use stoichiometry to find the number of moles of <mathjax>#"SO"_2#</mathjax>. From our balanced equation, we find that, for every mole of <mathjax>#"S"_8#</mathjax>, we get <mathjax>#8#</mathjax> moles of <mathjax>#"SO"_2#</mathjax>:</p>
<p><mathjax>#("0.014 mol S"_8) xx ("8 mol SO"_2)/("1 mol S"_8) = "0.11 moles SO"_2#</mathjax></p>
<p>Using <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, we can determine that the molar mass of <mathjax>#"SO"_2#</mathjax> is:</p>
<p><mathjax>#32 g/(mol) +16 g/(mol) *2 = 64 g/(mol)#</mathjax></p>
<p>Finally, using the number of moles and the molar mass, we can determine the mass of <mathjax>#"SO"_2#</mathjax> produced:</p>
<p><mathjax>#0.11 mol * 64 g/(mol) = 7.14 g#</mathjax></p>
<p>Hope this helps!</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"7.14 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Don't we all love <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>? The best way to start these types of problems is with a balanced equation:</p>
<p><mathjax>#"S"_8(s) + 8"O"_2(g) -> 8"SO"_2(g)#</mathjax></p>
<p>Okay, now we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is number of moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature.</p>
<p>Since we have excess (unlimited) oxygen gas, we can ignore that for this problem. Let's just focus on <mathjax>#"S"_8#</mathjax> and <mathjax>#"SO"_2#</mathjax>. First, we must find the number of moles of S8. Let's use the ideal gas law:</p>
<p><mathjax>#P = 1.00 atm#</mathjax></p>
<p><mathjax>#V = 1 L#</mathjax></p>
<p><mathjax>#R = 0.0821 (L*atm)/(mol*K)#</mathjax></p>
<p><mathjax>#T = 600 ˚ C = 873 K#</mathjax></p>
<p><mathjax>#(1.00 atm)(1 L) = n(0.0821 (L*atm)/(mol*K))(873 K)#</mathjax></p>
<p><mathjax>#n = 0.014#</mathjax> mols</p>
<p>Now, we can use stoichiometry to find the number of moles of <mathjax>#"SO"_2#</mathjax>. From our balanced equation, we find that, for every mole of <mathjax>#"S"_8#</mathjax>, we get <mathjax>#8#</mathjax> moles of <mathjax>#"SO"_2#</mathjax>:</p>
<p><mathjax>#("0.014 mol S"_8) xx ("8 mol SO"_2)/("1 mol S"_8) = "0.11 moles SO"_2#</mathjax></p>
<p>Using <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, we can determine that the molar mass of <mathjax>#"SO"_2#</mathjax> is:</p>
<p><mathjax>#32 g/(mol) +16 g/(mol) *2 = 64 g/(mol)#</mathjax></p>
<p>Finally, using the number of moles and the molar mass, we can determine the mass of <mathjax>#"SO"_2#</mathjax> produced:</p>
<p><mathjax>#0.11 mol * 64 g/(mol) = 7.14 g#</mathjax></p>
<p>Hope this helps!</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">One liter of sulfur vapor, S8(g) , at 600 ˚ C and 1.00 atm is burned in excess pure O2 to give SO2, measured at the same temperature and pressure. What mass of SO2 gas is obtained?</h1>
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Joshua Z.
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Stefan V.
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Nov 20, 2017
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<div class="markdown"><p><mathjax>#"7.14 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Don't we all love <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>? The best way to start these types of problems is with a balanced equation:</p>
<p><mathjax>#"S"_8(s) + 8"O"_2(g) -> 8"SO"_2(g)#</mathjax></p>
<p>Okay, now we can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV = nRT#</mathjax></p>
<p>where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is number of moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the temperature.</p>
<p>Since we have excess (unlimited) oxygen gas, we can ignore that for this problem. Let's just focus on <mathjax>#"S"_8#</mathjax> and <mathjax>#"SO"_2#</mathjax>. First, we must find the number of moles of S8. Let's use the ideal gas law:</p>
<p><mathjax>#P = 1.00 atm#</mathjax></p>
<p><mathjax>#V = 1 L#</mathjax></p>
<p><mathjax>#R = 0.0821 (L*atm)/(mol*K)#</mathjax></p>
<p><mathjax>#T = 600 ˚ C = 873 K#</mathjax></p>
<p><mathjax>#(1.00 atm)(1 L) = n(0.0821 (L*atm)/(mol*K))(873 K)#</mathjax></p>
<p><mathjax>#n = 0.014#</mathjax> mols</p>
<p>Now, we can use stoichiometry to find the number of moles of <mathjax>#"SO"_2#</mathjax>. From our balanced equation, we find that, for every mole of <mathjax>#"S"_8#</mathjax>, we get <mathjax>#8#</mathjax> moles of <mathjax>#"SO"_2#</mathjax>:</p>
<p><mathjax>#("0.014 mol S"_8) xx ("8 mol SO"_2)/("1 mol S"_8) = "0.11 moles SO"_2#</mathjax></p>
<p>Using <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, we can determine that the molar mass of <mathjax>#"SO"_2#</mathjax> is:</p>
<p><mathjax>#32 g/(mol) +16 g/(mol) *2 = 64 g/(mol)#</mathjax></p>
<p>Finally, using the number of moles and the molar mass, we can determine the mass of <mathjax>#"SO"_2#</mathjax> produced:</p>
<p><mathjax>#0.11 mol * 64 g/(mol) = 7.14 g#</mathjax></p>
<p>Hope this helps!</p></div>
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</article> | One liter of sulfur vapor, S8(g) , at 600 ˚ C and 1.00 atm is burned in excess pure O2 to give SO2, measured at the same temperature and pressure. What mass of SO2 gas is obtained? | null |
818 | abf33a48-6ddd-11ea-8840-ccda262736ce | https://socratic.org/questions/what-formula-represents-strontium-phosphate | Sr3(PO4)2 | start chemical_formula qc_end substance 3 4 qc_end end | [{"type":"other","value":"Chemical Formula [OF] strontium phosphate [IN] default"}] | [{"type":"chemical equation","value":"Sr3(PO4)2"}] | [{"type":"substance name","value":"Strontium phosphate"}] | <h1 class="questionTitle" itemprop="name">What formula represents strontium phosphate?</h1> | null | Sr3(PO4)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The metal ion Strontium has an ionic charge of +2, </p>
<p>The polyatomic ion Phosphate has an ionic charge of -3, <mathjax>#PO_4^-3#</mathjax></p>
<p>In order for the ionic molecule strontium phosphate to form a bond the charges need to be equal and opposite.</p>
<p>Therefore, we will need 3 strontium ions</p>
<p><mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax></p>
<p>for a total ionic charge of +6<br/>
to bond with 2 phosphate ions</p>
<p><mathjax>#PO_4^-3#</mathjax><br/>
<mathjax>#PO_4^-3#</mathjax></p>
<p>for a total ionic charge of -6</p>
<p>to form the molecule</p>
<p><mathjax>#Sr_3(PO_4)_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/rSzXND59ptc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Sr_3(PO_4)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The metal ion Strontium has an ionic charge of +2, </p>
<p>The polyatomic ion Phosphate has an ionic charge of -3, <mathjax>#PO_4^-3#</mathjax></p>
<p>In order for the ionic molecule strontium phosphate to form a bond the charges need to be equal and opposite.</p>
<p>Therefore, we will need 3 strontium ions</p>
<p><mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax></p>
<p>for a total ionic charge of +6<br/>
to bond with 2 phosphate ions</p>
<p><mathjax>#PO_4^-3#</mathjax><br/>
<mathjax>#PO_4^-3#</mathjax></p>
<p>for a total ionic charge of -6</p>
<p>to form the molecule</p>
<p><mathjax>#Sr_3(PO_4)_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/rSzXND59ptc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What formula represents strontium phosphate?</h1>
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BRIAN M.
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Mar 17, 2016
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<div class="markdown"><p><mathjax>#Sr_3(PO_4)_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The metal ion Strontium has an ionic charge of +2, </p>
<p>The polyatomic ion Phosphate has an ionic charge of -3, <mathjax>#PO_4^-3#</mathjax></p>
<p>In order for the ionic molecule strontium phosphate to form a bond the charges need to be equal and opposite.</p>
<p>Therefore, we will need 3 strontium ions</p>
<p><mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax><br/>
<mathjax>#Sr^+2#</mathjax></p>
<p>for a total ionic charge of +6<br/>
to bond with 2 phosphate ions</p>
<p><mathjax>#PO_4^-3#</mathjax><br/>
<mathjax>#PO_4^-3#</mathjax></p>
<p>for a total ionic charge of -6</p>
<p>to form the molecule</p>
<p><mathjax>#Sr_3(PO_4)_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/rSzXND59ptc?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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</article> | What formula represents strontium phosphate? | null |
819 | a9396bca-6ddd-11ea-9758-ccda262736ce | https://socratic.org/questions/592be345b72cff02377dd363 | 4 MnO4- + 12 H+ + 5 H3CCH2OH -> 5 H3CCO2H + 4 Mn^2+ + 11 H2O(l) | start chemical_equation qc_end substance 7 7 qc_end substance 9 10 qc_end substance 12 13 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"4 MnO4- + 12 H+ + 5 H3CCH2OH -> 5 H3CCO2H + 4 Mn^2+ + 11 H2O(l)"}] | [{"type":"substance name","value":"Ethanol"},{"type":"substance name","value":"Acetic acid"},{"type":"substance name","value":"Potassium permanganate"},{"type":"other","value":"Use the oxidation number method."}] | <h1 class="questionTitle" itemprop="name">How do we represent the oxidation of ethanol to acetic acid by potassium permanganate using the oxidation number method?</h1> | null | 4 MnO4- + 12 H+ + 5 H3CCH2OH -> 5 H3CCO2H + 4 Mn^2+ + 11 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Just to expand on the subject of oxidation of alcohols, the method of oxidation number, and gain or loss of electrons, can be utilized in the scenario, provided that we know how to assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> for the individual carbons in a given compound.</p>
<p>For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:</p>
<p><mathjax>#H_3stackrel(-III)C-stackrel(-I)CH_2OH+H_2OrarrH_3stackrel(-III)C-stackrel(+III)C(=O)OH +4H^(+) + 4e^(-) " (i)"#</mathjax></p>
<p>And something, here <mathjax>#MnO_4^(-)#</mathjax> is reduced down to <mathjax>#Mn^(+2)#</mathjax>:</p>
<p><mathjax>#stackrel(VII+)"Mn"rarrstackrel(+II)"Mn"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+)+ 5e^(-) rarr"Mn"^(2+)+"4H"_2"O"(l)#</mathjax> <mathjax>#"(ii)"#</mathjax></p>
<p>Deep-red purple permanganate ion is reduced down to almost colourless <mathjax>#Mn^(2+)#</mathjax>. </p>
<p>And to give the final redox equation, we cross multiply <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> so that electrons DO NOT appear in the final equation:</p>
<p>And thus <mathjax>#5xx(i)+4xx(ii)#</mathjax> gives...............</p>
<p><mathjax>#4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)#</mathjax></p>
<p>Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality. Note that the reaction contains a built-in indicator. <mathjax>#MnO_4^-#</mathjax> is deep purple; <mathjax>#Mn^(2+)#</mathjax> is almost colourless. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Just to expand on the subject of oxidation of alcohols, the method of oxidation number, and gain or loss of electrons, can be utilized in the scenario, provided that we know how to assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> for the individual carbons in a given compound.</p>
<p>For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:</p>
<p><mathjax>#H_3stackrel(-III)C-stackrel(-I)CH_2OH+H_2OrarrH_3stackrel(-III)C-stackrel(+III)C(=O)OH +4H^(+) + 4e^(-) " (i)"#</mathjax></p>
<p>And something, here <mathjax>#MnO_4^(-)#</mathjax> is reduced down to <mathjax>#Mn^(+2)#</mathjax>:</p>
<p><mathjax>#stackrel(VII+)"Mn"rarrstackrel(+II)"Mn"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+)+ 5e^(-) rarr"Mn"^(2+)+"4H"_2"O"(l)#</mathjax> <mathjax>#"(ii)"#</mathjax></p>
<p>Deep-red purple permanganate ion is reduced down to almost colourless <mathjax>#Mn^(2+)#</mathjax>. </p>
<p>And to give the final redox equation, we cross multiply <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> so that electrons DO NOT appear in the final equation:</p>
<p>And thus <mathjax>#5xx(i)+4xx(ii)#</mathjax> gives...............</p>
<p><mathjax>#4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)#</mathjax></p>
<p>Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality. Note that the reaction contains a built-in indicator. <mathjax>#MnO_4^-#</mathjax> is deep purple; <mathjax>#Mn^(2+)#</mathjax> is almost colourless. </p></div>
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<h1 class="questionTitle" itemprop="name">How do we represent the oxidation of ethanol to acetic acid by potassium permanganate using the oxidation number method?</h1>
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<div class="markdown"><p><mathjax>#4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Just to expand on the subject of oxidation of alcohols, the method of oxidation number, and gain or loss of electrons, can be utilized in the scenario, provided that we know how to assign <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> for the individual carbons in a given compound.</p>
<p>For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:</p>
<p><mathjax>#H_3stackrel(-III)C-stackrel(-I)CH_2OH+H_2OrarrH_3stackrel(-III)C-stackrel(+III)C(=O)OH +4H^(+) + 4e^(-) " (i)"#</mathjax></p>
<p>And something, here <mathjax>#MnO_4^(-)#</mathjax> is reduced down to <mathjax>#Mn^(+2)#</mathjax>:</p>
<p><mathjax>#stackrel(VII+)"Mn"rarrstackrel(+II)"Mn"#</mathjax></p>
<p><mathjax>#MnO_4^(-) + 8H^(+)+ 5e^(-) rarr"Mn"^(2+)+"4H"_2"O"(l)#</mathjax> <mathjax>#"(ii)"#</mathjax></p>
<p>Deep-red purple permanganate ion is reduced down to almost colourless <mathjax>#Mn^(2+)#</mathjax>. </p>
<p>And to give the final redox equation, we cross multiply <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> so that electrons DO NOT appear in the final equation:</p>
<p>And thus <mathjax>#5xx(i)+4xx(ii)#</mathjax> gives...............</p>
<p><mathjax>#4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)#</mathjax></p>
<p>Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality. Note that the reaction contains a built-in indicator. <mathjax>#MnO_4^-#</mathjax> is deep purple; <mathjax>#Mn^(2+)#</mathjax> is almost colourless. </p></div>
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</article> | How do we represent the oxidation of ethanol to acetic acid by potassium permanganate using the oxidation number method? | null |
820 | ac48b1b4-6ddd-11ea-b1ad-ccda262736ce | https://socratic.org/questions/a250-ml-sample-of-gas-at-1-00-atm-and-20-degrees-celsius-has-the-temperature-inc | 0.53 atm | start physical_unit 3 5 pressure atm qc_end physical_unit 3 5 1 2 volume qc_end physical_unit 3 5 7 8 pressure qc_end physical_unit 3 5 10 12 temperature qc_end physical_unit 3 5 18 20 temperature qc_end physical_unit 3 5 26 27 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] gas sample [IN] atm"}] | [{"type":"physical unit","value":"0.53 atm "}] | [{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{250 mL}"},{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{1.00 atm }"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{20 degrees Celsius}"},{"type":"physical unit","value":"Temperature2 [OF] gas sample [=] \\pu{40 degrees Celsius }"},{"type":"physical unit","value":"Volume2 [OF] gas sample [=] \\pu{500 mL}"}] | <h1 class="questionTitle" itemprop="name">A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure? </h1> | null | 0.53 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, given a constant quantity of gas.</p>
<p>And thus, <mathjax>#(P_1xxV_1xxT_2)/(V_2xxT_1)=P_2#</mathjax>, even given this format, we can see the problem is dimensionally sound. We do need to use <mathjax>#"absolute temperature."#</mathjax></p>
<p><mathjax>#P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, given a constant quantity of gas.</p>
<p>And thus, <mathjax>#(P_1xxV_1xxT_2)/(V_2xxT_1)=P_2#</mathjax>, even given this format, we can see the problem is dimensionally sound. We do need to use <mathjax>#"absolute temperature."#</mathjax></p>
<p><mathjax>#P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure? </h1>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, given a constant quantity of gas.</p>
<p>And thus, <mathjax>#(P_1xxV_1xxT_2)/(V_2xxT_1)=P_2#</mathjax>, even given this format, we can see the problem is dimensionally sound. We do need to use <mathjax>#"absolute temperature."#</mathjax></p>
<p><mathjax>#P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.#</mathjax></p></div>
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</article> | A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure? | null |
821 | aa3cc793-6ddd-11ea-be17-ccda262736ce | https://socratic.org/questions/a-balloon-has-a-volume-of-2-9-l-at-320-kelvin-if-the-temperature-is-raised-to-34 | 3.11 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 17 18 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] L"}] | [{"type":"physical unit","value":"3.11 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{2.9 L}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{320 Kelvin}"},{"type":"physical unit","value":"Temperature2 [OF] the balloon [=] \\pu{343 Kelvin}"}] | <h1 class="questionTitle" itemprop="name">A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be?</h1> | null | 3.11 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#3.1084375L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be?</h1>
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<div class="markdown"><p><mathjax>#3.1084375L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use unitary method </p>
<p>We could use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> if we were given other information like moles of gas and pressure but since we dont have them we can use unitary method</p>
<p>So this question can be represented as</p>
<p><mathjax>#"2.9L" / "320K" * 343K = 3.1084375L#</mathjax></p>
<p>Now we know the volume so let us find Pressure<br/>
using the gas law <mathjax>#PV = nRT#</mathjax> </p>
<p><mathjax>#V = "(nRT)"/P#</mathjax></p>
<p><mathjax># 3.1084375L = "(n * 0.0821 * 343K)"/P#</mathjax></p>
<p><mathjax>#P = "28.1603n"/{3.1084375#</mathjax></p></div>
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</article> | A balloon has a volume of 2.9 L at 320 Kelvin. If the temperature is raised to 343 Kelvin, what will its volume be? | null |
822 | aa3af338-6ddd-11ea-a841-ccda262736ce | https://socratic.org/questions/what-is-the-ka-of-an-acid-of-6m-which-has-a-ph-of-2-5-at-274-5k | 2 × 10^(-6) | start physical_unit 6 6 ka none qc_end physical_unit 6 6 8 9 molarity qc_end physical_unit 6 6 15 15 ph qc_end physical_unit 6 6 17 18 temperature qc_end end | [{"type":"physical unit","value":"Ka [OF] the acid"}] | [{"type":"physical unit","value":"2 × 10^(-6)"}] | [{"type":"physical unit","value":"Molarity [OF] the acid [=] \\pu{6 M}"},{"type":"physical unit","value":"pH [OF] the acid [=] \\pu{2.5}"},{"type":"physical unit","value":"Temperature [OF] the acid [=] \\pu{277.5 K }"}] | <h1 class="questionTitle" itemprop="name">What is the Ka of an acid of 6M which has a pH of 2.5 at 277.5K ?</h1> | <div class="questionDetailsContainer">
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>I would argue that the temperature has nothing to do with the value of <mathjax>#K_text(a)#</mathjax>.</p>
<p>In this problem, it is the <mathjax>#"pH"#</mathjax> and the initial concentration of the acid that determine the value of <mathjax>#K_text(a)#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(mml)6color(white)(mmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(ml)"6 -"color(white)(l)xcolor(white)(mmmmml)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.5"#</mathjax></p>
<p>∴ <mathjax>#["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#K_text(a) = 2 × 10^"-6"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>I would argue that the temperature has nothing to do with the value of <mathjax>#K_text(a)#</mathjax>.</p>
<p>In this problem, it is the <mathjax>#"pH"#</mathjax> and the initial concentration of the acid that determine the value of <mathjax>#K_text(a)#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(mml)6color(white)(mmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(ml)"6 -"color(white)(l)xcolor(white)(mmmmml)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.5"#</mathjax></p>
<p>∴ <mathjax>#["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the Ka of an acid of 6M which has a pH of 2.5 at 277.5K ?</h1>
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<div class="markdown"><p><mathjax>#K_text(a) = 2 × 10^"-6"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>I would argue that the temperature has nothing to do with the value of <mathjax>#K_text(a)#</mathjax>.</p>
<p>In this problem, it is the <mathjax>#"pH"#</mathjax> and the initial concentration of the acid that determine the value of <mathjax>#K_text(a)#</mathjax>.</p>
<blockquote></blockquote>
<p>Let's set up an ICE table to solve this problem.</p>
<p><mathjax>#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(mml)6color(white)(mmmmmml)0color(white)(mmm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(ml)"6 -"color(white)(l)xcolor(white)(mmmmml)xcolor(white)(mmll)x#</mathjax></p>
<blockquote></blockquote>
<p>We must use the <mathjax>#"pH"#</mathjax> to calculate the value of <mathjax>#x#</mathjax>.</p>
<p><mathjax>#"pH = 2.5"#</mathjax></p>
<p>∴ <mathjax>#["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x#</mathjax></p>
<blockquote></blockquote>
<p>The <mathjax>#K_"a"#</mathjax> expression is:</p>
<p><mathjax>#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6"#</mathjax></p></div>
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</article> | What is the Ka of an acid of 6M which has a pH of 2.5 at 277.5K ? | |
823 | a8d1dd3a-6ddd-11ea-942d-ccda262736ce | https://socratic.org/questions/how-do-you-determine-the-empirical-formula-of-a-compound-that-has-the-following- | CH | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] empirical"}] | [{"type":"chemical equation","value":"CH"}] | [{"type":"physical unit","value":"Percent [OF] C in the compound [=] \\pu{92.3 percent}"},{"type":"physical unit","value":"Percent [OF] H in the compound [=] \\pu{7.7 percent}"}] | <h1 class="questionTitle" itemprop="name">How do you determine the empirical formula of a compound that has the following composition: 92.3 percent C and 7.7 percent H?</h1> | null | CH | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And then we work out the molar composition.</p>
<p>Thus <mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(92.3*g)/(12.011*g*mol^-1)=7.68*mol#</mathjax>.</p>
<p>And <mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.7*g)/(1.00794*g*mol^-1)=7.64*mol#</mathjax>.</p>
<p>Clearly, there is <mathjax>#1:1#</mathjax> molar equivalence with respect to <mathjax>#C:H#</mathjax>.</p>
<p>The molecular formula is a multiple of the empirical formula, but we have no handle on that here. </p></div>
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<div class="markdown"><p>As with all these problems, we assume that there are <mathjax>#100*g#</mathjax> of unknown compound........to give an empirical formula of <mathjax>#CH#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And then we work out the molar composition.</p>
<p>Thus <mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(92.3*g)/(12.011*g*mol^-1)=7.68*mol#</mathjax>.</p>
<p>And <mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.7*g)/(1.00794*g*mol^-1)=7.64*mol#</mathjax>.</p>
<p>Clearly, there is <mathjax>#1:1#</mathjax> molar equivalence with respect to <mathjax>#C:H#</mathjax>.</p>
<p>The molecular formula is a multiple of the empirical formula, but we have no handle on that here. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you determine the empirical formula of a compound that has the following composition: 92.3 percent C and 7.7 percent H?</h1>
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<div class="markdown"><p>As with all these problems, we assume that there are <mathjax>#100*g#</mathjax> of unknown compound........to give an empirical formula of <mathjax>#CH#</mathjax>.</p></div>
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<div class="markdown"><p>And then we work out the molar composition.</p>
<p>Thus <mathjax>#"Moles of carbon"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(92.3*g)/(12.011*g*mol^-1)=7.68*mol#</mathjax>.</p>
<p>And <mathjax>#"Moles of hydrogen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.7*g)/(1.00794*g*mol^-1)=7.64*mol#</mathjax>.</p>
<p>Clearly, there is <mathjax>#1:1#</mathjax> molar equivalence with respect to <mathjax>#C:H#</mathjax>.</p>
<p>The molecular formula is a multiple of the empirical formula, but we have no handle on that here. </p></div>
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</article> | How do you determine the empirical formula of a compound that has the following composition: 92.3 percent C and 7.7 percent H? | null |
824 | acc96048-6ddd-11ea-b397-ccda262736ce | https://socratic.org/questions/what-mass-of-naoh-is-contained-in-250-0-ml-of-a-0-400-m-sodium-hydroxide-solutio | 4.00 g | start physical_unit 3 3 mass g qc_end physical_unit 13 15 7 8 volume qc_end physical_unit 13 15 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] NaOH [IN] g"}] | [{"type":"physical unit","value":"4.00 g"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide solution [=] \\pu{250.0 mL}"},{"type":"physical unit","value":"Molarity [OF] sodium hydroxide solution [=] \\pu{0.400 M}"}] | <h1 class="questionTitle" itemprop="name">What mass of #NaOH# is contained in 250.0 mL of a 0.400 M sodium hydroxide solution?</h1> | null | 4.00 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume of solution"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume"xx"concentration "=#</mathjax></p>
<p><mathjax>#0.250*Lxx0.400*mol*L^-1#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#0.100*mol#</mathjax></p>
<p><mathjax>#"Mass of sodium hydroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.100*molxx40.0*g*mol^-1#</mathjax> </p>
<p><mathjax>#=??*g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Approx. 4 g of NaOH."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume of solution"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume"xx"concentration "=#</mathjax></p>
<p><mathjax>#0.250*Lxx0.400*mol*L^-1#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#0.100*mol#</mathjax></p>
<p><mathjax>#"Mass of sodium hydroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.100*molxx40.0*g*mol^-1#</mathjax> </p>
<p><mathjax>#=??*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of #NaOH# is contained in 250.0 mL of a 0.400 M sodium hydroxide solution?</h1>
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<div class="markdown"><p><mathjax>#"Approx. 4 g of NaOH."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles"/"Volume of solution"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Volume"xx"concentration "=#</mathjax></p>
<p><mathjax>#0.250*Lxx0.400*mol*L^-1#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#0.100*mol#</mathjax></p>
<p><mathjax>#"Mass of sodium hydroxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.100*molxx40.0*g*mol^-1#</mathjax> </p>
<p><mathjax>#=??*g#</mathjax></p></div>
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<div class="markdown"><p>3.998 grams NaOH </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> of a solution is determined by the comparison of the moles of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> compared to the volume of the solution in liters. </p>
<p>The equation of the formula is <mathjax>#M = (mol)/L#</mathjax></p>
<p>In order to find the mass we first need to solve for the moles. </p>
<p><mathjax>#mol = M/L#</mathjax></p>
<p>M = 0.400<br/>
moles = ?<br/>
L = 250.0 mL or 0.25 L</p>
<p><mathjax>#0.400 M (0.25 L) = (? mol)#</mathjax></p>
<p><mathjax># 0.100 mol of NaOH#</mathjax></p>
<p>Now we can calculate the grams be converting moles to grams. </p>
<p><mathjax>#Na = 22.98 g#</mathjax><br/>
<mathjax>#O = 15.99 g#</mathjax><br/>
<mathjax>#H = 1.01 g#</mathjax></p>
<p><mathjax>#NaOH = 39.98 g#</mathjax></p>
<p><mathjax>#0.100 cancel(mol NaOH) x (39.98 g NaOH)/( 1 cancel(molNaOH)#</mathjax></p>
<p>3.998 grams NaOH </p></div>
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</article> | What mass of #NaOH# is contained in 250.0 mL of a 0.400 M sodium hydroxide solution? | null |
825 | a9e3c7be-6ddd-11ea-ad1d-ccda262736ce | https://socratic.org/questions/the-incandescent-white-of-a-fireworks-display-is-caused-by-the-reaction-of-phosp | 4 P + 5 O2 -> P4O10 | start chemical_equation qc_end substance 13 13 qc_end chemical_equation 15 15 qc_end chemical_equation 18 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"4 P + 5 O2 -> P4O10"}] | [{"type":"substance name","value":"Phosphorous"},{"type":"chemical equation","value":"O2"},{"type":"chemical equation","value":"P4O10"}] | <h1 class="questionTitle" itemprop="name">The incandescent white of a fireworks display is caused by the reaction of phosphorous with #O_2# to give #P_4O_10#. What is the balanced chemical equation for the reaction?</h1> | null | 4 P + 5 O2 -> P4O10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note that phosphorous is <mathjax>#P#</mathjax>. <br/>
<mathjax>#therefore P+O_2rarrP_4O_10#</mathjax></p>
<p>We have 1 P on the LHS, and 4 on the RHS.<br/>
We have 2 O on the LHS, and 10 on the RHS. </p>
<p>We cannot balance a chemical equation by adding, subtracting or dividing atoms, the only way to balance is by multiplication. </p>
<p>Let's multiply the LHS by <mathjax>#5#</mathjax> to balance the oxygen.</p>
<p><mathjax>#P+5O_2rarrP_4O_10#</mathjax></p>
<p>Now let's multiply the phosphorus on the LHS by <mathjax>#4#</mathjax>. </p>
<p><mathjax>#4P+5O_2rarrP_4O_10#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#4P+5O_2rarrP_4O_10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Note that phosphorous is <mathjax>#P#</mathjax>. <br/>
<mathjax>#therefore P+O_2rarrP_4O_10#</mathjax></p>
<p>We have 1 P on the LHS, and 4 on the RHS.<br/>
We have 2 O on the LHS, and 10 on the RHS. </p>
<p>We cannot balance a chemical equation by adding, subtracting or dividing atoms, the only way to balance is by multiplication. </p>
<p>Let's multiply the LHS by <mathjax>#5#</mathjax> to balance the oxygen.</p>
<p><mathjax>#P+5O_2rarrP_4O_10#</mathjax></p>
<p>Now let's multiply the phosphorus on the LHS by <mathjax>#4#</mathjax>. </p>
<p><mathjax>#4P+5O_2rarrP_4O_10#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">The incandescent white of a fireworks display is caused by the reaction of phosphorous with #O_2# to give #P_4O_10#. What is the balanced chemical equation for the reaction?</h1>
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<div class="markdown"><p><mathjax>#4P+5O_2rarrP_4O_10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Note that phosphorous is <mathjax>#P#</mathjax>. <br/>
<mathjax>#therefore P+O_2rarrP_4O_10#</mathjax></p>
<p>We have 1 P on the LHS, and 4 on the RHS.<br/>
We have 2 O on the LHS, and 10 on the RHS. </p>
<p>We cannot balance a chemical equation by adding, subtracting or dividing atoms, the only way to balance is by multiplication. </p>
<p>Let's multiply the LHS by <mathjax>#5#</mathjax> to balance the oxygen.</p>
<p><mathjax>#P+5O_2rarrP_4O_10#</mathjax></p>
<p>Now let's multiply the phosphorus on the LHS by <mathjax>#4#</mathjax>. </p>
<p><mathjax>#4P+5O_2rarrP_4O_10#</mathjax></p></div>
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</article> | The incandescent white of a fireworks display is caused by the reaction of phosphorous with #O_2# to give #P_4O_10#. What is the balanced chemical equation for the reaction? | null |
826 | a945de91-6ddd-11ea-b591-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-hydrogen-oxide | H2O | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] hydrogen oxide [IN] default"}] | [{"type":"chemical equation","value":"H2O"}] | [{"type":"substance name","value":"Hydrogen oxide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for hydrogen oxide? </h1> | null | H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>From the 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> hydrogen has the value of 1, and oxygene - 2, so the formula has to be <mathjax>#H_2O#</mathjax></p></div>
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<div class="markdown"><p>The formula for <strong><em>di</em></strong><em>hydrogen</em> <strong><em>mon</em></strong><em>oxide</em> is <mathjax>#H_2O#</mathjax>. See explanation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>From the 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> hydrogen has the value of 1, and oxygene - 2, so the formula has to be <mathjax>#H_2O#</mathjax></p></div>
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<div class="markdown"><p>The formula for <strong><em>di</em></strong><em>hydrogen</em> <strong><em>mon</em></strong><em>oxide</em> is <mathjax>#H_2O#</mathjax>. See explanation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>From the 2 <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> hydrogen has the value of 1, and oxygene - 2, so the formula has to be <mathjax>#H_2O#</mathjax></p></div>
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</article> | What is the chemical formula for hydrogen oxide? | null |
827 | aa3c3537-6ddd-11ea-8a45-ccda262736ce | https://socratic.org/questions/58ca1274b72cff4665a6befe | 14 | start physical_unit 2 3 number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Number [OF] structural isomers"}] | [{"type":"physical unit","value":"14"}] | [{"type":"chemical equation","value":"C5H12O"}] | <h1 class="questionTitle" itemprop="name">How many structural isomers can a formula of #C_5H_12O# generate?</h1> | null | 14 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And these 14 isomers are alcohols, and ethers. And this reflects the fiendish connectivity of the carbon atom. You start with a relatively small, modest formula of <mathjax>#C_5H_12O#</mathjax>, and you end up with an array of different structure, all with distinct physical, and chemical properties. </p></div>
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<div class="markdown"><p>Well, I can count at least <mathjax>#"14"..........."structural isomers...."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And these 14 isomers are alcohols, and ethers. And this reflects the fiendish connectivity of the carbon atom. You start with a relatively small, modest formula of <mathjax>#C_5H_12O#</mathjax>, and you end up with an array of different structure, all with distinct physical, and chemical properties. </p></div>
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<div class="markdown"><p>Well, I can count at least <mathjax>#"14"..........."structural isomers...."#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And these 14 isomers are alcohols, and ethers. And this reflects the fiendish connectivity of the carbon atom. You start with a relatively small, modest formula of <mathjax>#C_5H_12O#</mathjax>, and you end up with an array of different structure, all with distinct physical, and chemical properties. </p></div>
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</article> | How many structural isomers can a formula of #C_5H_12O# generate? | null |
828 | aa4b5a36-6ddd-11ea-b569-ccda262736ce | https://socratic.org/questions/a-100-0-ml-sample-of-1-00-m-hcl-is-titrated-with-1-00-m-naoh-what-is-the-ph-of-t | 7.00 | start physical_unit 19 20 ph none qc_end physical_unit 20 20 1 2 volume qc_end physical_unit 20 20 1 2 volume qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 13 13 5 6 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"7.00"}] | [{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{1.00 M}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{1.00 M}"}] | <h1 class="questionTitle" itemprop="name">A 100.0-mL sample of 1.00 M #HCl# is titrated with 1.00 M #NaOH#. What is the pH of the solution after 100.0 mL of #NaOH# have been added to the acid? </h1> | null | 7.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This titration involves hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, and sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, so right from the start you should expect to have a <em>neutral solution</em> at <strong>equivalence point</strong>. </p>
<p>In other words, when the solution contains enough strong base to neutralize <strong>all the acid</strong>, its <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> should be equal to that of <em>pure water</em>. </p>
<p>Hydrochloric acid will ionize completely in aqueous solution to produce hydronium cations, <mathjax>#"H"_3"O"^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Similarly, sodium hydroxide will ionize completely in aqueous solution to produce hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydronium cations and the hydroxide anions will neutralize each other to produce water</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, it takes <strong>equal numbers of moles</strong> of hydronium cations and hydroxide anions to have a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>. </p>
<p>In your case, you have <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> hydrochloric acid solution. You're titrating this with <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> sodium hydroxide solution. </p>
<p>Since both solution have the <strong>same volume</strong> and the <strong>same concentration</strong>, it follows that they also have the <strong>same number of moles</strong> of hydronium cations and hydroxide anions. </p>
<p>Therefore, mixing these two solutions will result in a <em><strong>complete neutralization</strong></em>. </p>
<p>As a result, the pH of the solution will be equal to that of water, which at room temperature is <mathjax>#7#</mathjax>. </p>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"pH" = 7#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This titration involves hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, and sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, so right from the start you should expect to have a <em>neutral solution</em> at <strong>equivalence point</strong>. </p>
<p>In other words, when the solution contains enough strong base to neutralize <strong>all the acid</strong>, its <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> should be equal to that of <em>pure water</em>. </p>
<p>Hydrochloric acid will ionize completely in aqueous solution to produce hydronium cations, <mathjax>#"H"_3"O"^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Similarly, sodium hydroxide will ionize completely in aqueous solution to produce hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydronium cations and the hydroxide anions will neutralize each other to produce water</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, it takes <strong>equal numbers of moles</strong> of hydronium cations and hydroxide anions to have a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>. </p>
<p>In your case, you have <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> hydrochloric acid solution. You're titrating this with <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> sodium hydroxide solution. </p>
<p>Since both solution have the <strong>same volume</strong> and the <strong>same concentration</strong>, it follows that they also have the <strong>same number of moles</strong> of hydronium cations and hydroxide anions. </p>
<p>Therefore, mixing these two solutions will result in a <em><strong>complete neutralization</strong></em>. </p>
<p>As a result, the pH of the solution will be equal to that of water, which at room temperature is <mathjax>#7#</mathjax>. </p>
<p>
<iframe src="https://www.youtube.com/embed/VADhF7M8bJ8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">A 100.0-mL sample of 1.00 M #HCl# is titrated with 1.00 M #NaOH#. What is the pH of the solution after 100.0 mL of #NaOH# have been added to the acid? </h1>
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<div class="markdown"><p><mathjax>#"pH" = 7#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This titration involves hydrochloric acid, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, and sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, so right from the start you should expect to have a <em>neutral solution</em> at <strong>equivalence point</strong>. </p>
<p>In other words, when the solution contains enough strong base to neutralize <strong>all the acid</strong>, its <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> should be equal to that of <em>pure water</em>. </p>
<p>Hydrochloric acid will ionize completely in aqueous solution to produce hydronium cations, <mathjax>#"H"_3"O"^(+)#</mathjax></p>
<blockquote>
<p><mathjax>#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Similarly, sodium hydroxide will ionize completely in aqueous solution to produce hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>When these two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed, the hydronium cations and the hydroxide anions will neutralize each other to produce water</p>
<blockquote>
<p><mathjax>#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>As you can see, it takes <strong>equal numbers of moles</strong> of hydronium cations and hydroxide anions to have a <em>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></em>. </p>
<p>In your case, you have <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> hydrochloric acid solution. You're titrating this with <mathjax>#"100.0 mL"#</mathjax> of <mathjax>#"1.0 M"#</mathjax> sodium hydroxide solution. </p>
<p>Since both solution have the <strong>same volume</strong> and the <strong>same concentration</strong>, it follows that they also have the <strong>same number of moles</strong> of hydronium cations and hydroxide anions. </p>
<p>Therefore, mixing these two solutions will result in a <em><strong>complete neutralization</strong></em>. </p>
<p>As a result, the pH of the solution will be equal to that of water, which at room temperature is <mathjax>#7#</mathjax>. </p>
<p>
<iframe src="https://www.youtube.com/embed/VADhF7M8bJ8?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | A 100.0-mL sample of 1.00 M #HCl# is titrated with 1.00 M #NaOH#. What is the pH of the solution after 100.0 mL of #NaOH# have been added to the acid? | null |
829 | abe28210-6ddd-11ea-8c70-ccda262736ce | https://socratic.org/questions/a-sample-of-3-0-grams-of-an-ideal-gas-at-127-c-and-1-0-atmosphere-pressure-has-a | 66 g/mol | start physical_unit 28 29 molar_mass g/mol qc_end physical_unit 30 32 3 4 mass qc_end physical_unit 30 32 10 11 temperature qc_end physical_unit 30 32 13 14 pressure qc_end physical_unit 30 32 20 21 volume qc_end physical_unit 30 32 36 41 constant_r qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"66 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the ideal gas sample [=] \\pu{3.0 grams}"},{"type":"physical unit","value":"Temperature [OF] the ideal gas sample [=] \\pu{127 ℃}"},{"type":"physical unit","value":"Pressure [OF] the ideal gas sample [=] \\pu{1.0 atmosphere}"},{"type":"physical unit","value":"Volume [OF] the ideal gas sample [=] \\pu{1.5 liters}"},{"type":"physical unit","value":"Constant R [OF] the ideal gas [=] \\pu{0.08 (L * atm)/(mole * K)}"}] | <h1 class="questionTitle" itemprop="name">A sample of 3.0 grams of an ideal gas at 127°C and 1.0 atmosphere pressure has a volume of 1.5 liters. What is the molar mass of the gas?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The ideal gas constant R, is 0.08 (L<em>atm)/(mole</em>K)?</p></div>
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</div> | 66 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal Gas equation to give us the molar quantity.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax>, </p>
<p>i.e. <mathjax>#"Mass"/"Molar mass"="Amount of substance in moles"= (PV)/(RT)#</mathjax>.</p>
<p>And thus <mathjax>#"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of gas"/(PV)xxRT=#</mathjax></p>
<p><mathjax>#(3.0*gxx0.0821*cancelL*atm*cancel(K^-1)*mol^-1xx400cancelK)/(1*cancel(atm)xx1.5*cancelL)#</mathjax></p>
<p><mathjax>#=??g*mol^-1#</mathjax>.</p>
<p>This is dimensionally consistent, in that all the units EXCEPT <mathjax>#g*mol^-1#</mathjax> cancel out.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Molar mass"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#66*g*mol^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal Gas equation to give us the molar quantity.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax>, </p>
<p>i.e. <mathjax>#"Mass"/"Molar mass"="Amount of substance in moles"= (PV)/(RT)#</mathjax>.</p>
<p>And thus <mathjax>#"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of gas"/(PV)xxRT=#</mathjax></p>
<p><mathjax>#(3.0*gxx0.0821*cancelL*atm*cancel(K^-1)*mol^-1xx400cancelK)/(1*cancel(atm)xx1.5*cancelL)#</mathjax></p>
<p><mathjax>#=??g*mol^-1#</mathjax>.</p>
<p>This is dimensionally consistent, in that all the units EXCEPT <mathjax>#g*mol^-1#</mathjax> cancel out.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of 3.0 grams of an ideal gas at 127°C and 1.0 atmosphere pressure has a volume of 1.5 liters. What is the molar mass of the gas?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Molar mass"#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#66*g*mol^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal Gas equation to give us the molar quantity.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax>, </p>
<p>i.e. <mathjax>#"Mass"/"Molar mass"="Amount of substance in moles"= (PV)/(RT)#</mathjax>.</p>
<p>And thus <mathjax>#"Molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of gas"/(PV)xxRT=#</mathjax></p>
<p><mathjax>#(3.0*gxx0.0821*cancelL*atm*cancel(K^-1)*mol^-1xx400cancelK)/(1*cancel(atm)xx1.5*cancelL)#</mathjax></p>
<p><mathjax>#=??g*mol^-1#</mathjax>.</p>
<p>This is dimensionally consistent, in that all the units EXCEPT <mathjax>#g*mol^-1#</mathjax> cancel out.</p></div>
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</article> | A sample of 3.0 grams of an ideal gas at 127°C and 1.0 atmosphere pressure has a volume of 1.5 liters. What is the molar mass of the gas? |
The ideal gas constant R, is 0.08 (Latm)/(moleK)?
|
830 | aa76e62e-6ddd-11ea-aabc-ccda262736ce | https://socratic.org/questions/if-a-gas-pressure-gauge-reads-15-mm-hg-what-is-the-pressure-in-atmospheres | 0.02 atmospheres | start physical_unit 2 2 pressure atm qc_end physical_unit 2 2 6 7 pressure qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] atmospheres"}] | [{"type":"physical unit","value":"0.02 atmospheres"}] | [{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{15 mmHg}"}] | <h1 class="questionTitle" itemprop="name">If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? </h1> | null | 0.02 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>0,020 atmospheres </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? </h1>
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<div class="markdown"><p>0,020 atmospheres </p></div>
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<div class="markdown"><p>760 mm Hg = 1 atmosphere. </p>
<p><mathjax># 15/760 = atmosphere#</mathjax></p>
<p># 15/760 = 0.0197368 atmospheres rounding off to 2 significant digits gives </p>
<p>0.020 atmospheres </p></div>
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<div class="markdown"><p><mathjax>#0.020#</mathjax> <mathjax>#"atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To convert between millimeters Mercury (<mathjax>#"mm Hg"#</mathjax>) and atmospheres (<mathjax>#"atm"#</mathjax>), we use the conversion factor</p>
<p><mathjax>#(1"atm")/(760"mm Hg")#</mathjax></p>
<p>That is, <mathjax>#1#</mathjax> atmosphere equals <mathjax>#760#</mathjax> millimeters Mercury.</p>
<p>Using <em>dimensional analysis</em>, we can convert the given pressure:</p>
<p><mathjax>#15#</mathjax> <mathjax>#cancel("mm Hg")((1"atm")/(760cancel("mm Hg"))) = color(red)(0.020#</mathjax> <mathjax>#color(red)("atm"#</mathjax></p>
<p><mathjax>#15#</mathjax> <mathjax>#"mm Hg"#</mathjax> is equal to <mathjax>#0.020#</mathjax> <mathjax>#"atm"#</mathjax>, rounded to <mathjax>#color(blue)(2#</mathjax> <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, the amount given in the problem.</p></div>
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</article> | If a gas pressure gauge reads 15 mm Hg, what is the pressure in atmospheres? | null |
831 | aabe968c-6ddd-11ea-a2f2-ccda262736ce | https://socratic.org/questions/5732a3e011ef6b683c6b75cf | 5.10 × 10^(-3) molar | start physical_unit 4 6 mole mol/l qc_end physical_unit 15 16 11 12 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] the metal salt [IN] molar"}] | [{"type":"physical unit","value":"5.10 × 10^(-3) molar"}] | [{"type":"physical unit","value":"Mass [OF] sodium dichromate [=] \\pu{0.667 g}"}] | <h1 class="questionTitle" itemprop="name">What molar quantity of the metal salt are present in a #0.667*g# mass of sodium dichromate?</h1> | null | 5.10 × 10^(-3) molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of sodium dichromate" = (0.667*g)/(261.97*g*mol )=2.55xx10^-3 *mol#</mathjax>.</p>
<p>Given the composition of sodium dichromate, <mathjax>#Na_2Cr_2O_7#</mathjax>, there are twice this molar quantity of sodium ions. How many oxygen atoms?</p></div>
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<div class="markdown"><p><mathjax>#"5.10 mmol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of sodium dichromate" = (0.667*g)/(261.97*g*mol )=2.55xx10^-3 *mol#</mathjax>.</p>
<p>Given the composition of sodium dichromate, <mathjax>#Na_2Cr_2O_7#</mathjax>, there are twice this molar quantity of sodium ions. How many oxygen atoms?</p></div>
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<div class="markdown"><p><mathjax>#"5.10 mmol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of sodium dichromate" = (0.667*g)/(261.97*g*mol )=2.55xx10^-3 *mol#</mathjax>.</p>
<p>Given the composition of sodium dichromate, <mathjax>#Na_2Cr_2O_7#</mathjax>, there are twice this molar quantity of sodium ions. How many oxygen atoms?</p></div>
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</article> | What molar quantity of the metal salt are present in a #0.667*g# mass of sodium dichromate? | null |
832 | a86aa94a-6ddd-11ea-bf9d-ccda262736ce | https://socratic.org/questions/can-you-write-and-balance-the-equation-for-the-complete-combustion-of-ethane-c-2 | 2 C2H6 + 7O2 -> 4 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 13 13 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 C2H6 + 7O2 -> 4 CO2 + 6 H2O "}] | [{"type":"chemical equation","value":"C2H6"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">Can you write and balance the equation for the complete combustion of ethane, #C_2H_6#?</h1> | null | 2 C2H6 + 7O2 -> 4 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a combustion reaction with a hydrocarbon in the reactant side you will always have <mathjax>#O_2#</mathjax> as another reactant. As you will always have <mathjax>#CO_2#</mathjax> and <mathjax>#H_2O#</mathjax> as the products.</p>
<p>Knowing that much you can set up your reaction equation..</p>
<p><mathjax>#C_2H_6 + O_2#</mathjax><mathjax>#->#</mathjax> <mathjax>#CO_2 + H_2O#</mathjax></p>
<p>Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.</p>
<p>Start with the <mathjax>#C#</mathjax> atoms first and move to the <mathjax>#H#</mathjax> atoms next. It's easier to leave the <mathjax>#O_2#</mathjax> to the last, it has a way to alter the equation.</p>
<p>Initially, you would arrive at this, before the <mathjax>#O_2#</mathjax> has been balanced:</p>
<p><mathjax>#C_2H_6 + O_2 -> 2CO_2 + 3H_2O#</mathjax> </p>
<p>But, as you can see, you have an odd amount of <mathjax>#O_2#</mathjax> on the product side. In this case, you have to find the common factor of the amount of <mathjax>#O#</mathjax> on the product side and 2, Because of the <mathjax>#O_2#</mathjax> diatom. Therefore, 14 would be the lowest common factor of 2 and 7. </p></div>
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<div class="markdown"><p><mathjax>#2C_2H_6+7O_2#</mathjax> <mathjax>#->#</mathjax> <mathjax>#4O_2+6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a combustion reaction with a hydrocarbon in the reactant side you will always have <mathjax>#O_2#</mathjax> as another reactant. As you will always have <mathjax>#CO_2#</mathjax> and <mathjax>#H_2O#</mathjax> as the products.</p>
<p>Knowing that much you can set up your reaction equation..</p>
<p><mathjax>#C_2H_6 + O_2#</mathjax><mathjax>#->#</mathjax> <mathjax>#CO_2 + H_2O#</mathjax></p>
<p>Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.</p>
<p>Start with the <mathjax>#C#</mathjax> atoms first and move to the <mathjax>#H#</mathjax> atoms next. It's easier to leave the <mathjax>#O_2#</mathjax> to the last, it has a way to alter the equation.</p>
<p>Initially, you would arrive at this, before the <mathjax>#O_2#</mathjax> has been balanced:</p>
<p><mathjax>#C_2H_6 + O_2 -> 2CO_2 + 3H_2O#</mathjax> </p>
<p>But, as you can see, you have an odd amount of <mathjax>#O_2#</mathjax> on the product side. In this case, you have to find the common factor of the amount of <mathjax>#O#</mathjax> on the product side and 2, Because of the <mathjax>#O_2#</mathjax> diatom. Therefore, 14 would be the lowest common factor of 2 and 7. </p></div>
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<h1 class="questionTitle" itemprop="name">Can you write and balance the equation for the complete combustion of ethane, #C_2H_6#?</h1>
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<div class="markdown"><p><mathjax>#2C_2H_6+7O_2#</mathjax> <mathjax>#->#</mathjax> <mathjax>#4O_2+6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a combustion reaction with a hydrocarbon in the reactant side you will always have <mathjax>#O_2#</mathjax> as another reactant. As you will always have <mathjax>#CO_2#</mathjax> and <mathjax>#H_2O#</mathjax> as the products.</p>
<p>Knowing that much you can set up your reaction equation..</p>
<p><mathjax>#C_2H_6 + O_2#</mathjax><mathjax>#->#</mathjax> <mathjax>#CO_2 + H_2O#</mathjax></p>
<p>Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.</p>
<p>Start with the <mathjax>#C#</mathjax> atoms first and move to the <mathjax>#H#</mathjax> atoms next. It's easier to leave the <mathjax>#O_2#</mathjax> to the last, it has a way to alter the equation.</p>
<p>Initially, you would arrive at this, before the <mathjax>#O_2#</mathjax> has been balanced:</p>
<p><mathjax>#C_2H_6 + O_2 -> 2CO_2 + 3H_2O#</mathjax> </p>
<p>But, as you can see, you have an odd amount of <mathjax>#O_2#</mathjax> on the product side. In this case, you have to find the common factor of the amount of <mathjax>#O#</mathjax> on the product side and 2, Because of the <mathjax>#O_2#</mathjax> diatom. Therefore, 14 would be the lowest common factor of 2 and 7. </p></div>
</div>
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<div class="markdown"><p><mathjax>#C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(g)#</mathjax></p></div>
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<div class="markdown"><p>Any hydrocarbon combusts completely to give carbon dioxide and water. Is it balanced? The oxygen bears a half-coefficient. How could I remove it?</p>
<p>Can you represent the combustion of propane and butane with similar reactions?</p></div>
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</article> | Can you write and balance the equation for the complete combustion of ethane, #C_2H_6#? | null |
833 | a8fa1289-6ddd-11ea-9fb6-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-calcium-hydroxide | Ca(OH)2 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] calcium hydroxide [IN] default"}] | [{"type":"chemical equation","value":"Ca(OH)2"}] | [{"type":"substance name","value":"Calcium hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for calcium hydroxide?</h1> | null | Ca(OH)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To make a neutral species, we simply balance the charges. To balance the charge of the metal ion, <mathjax>#Ca^(2+)#</mathjax>, we need 2 hydroxide ions, <mathjax>#2xxOH^-#</mathjax>.</p>
<p>We get the formula <mathjax>#Ca(OH)_2#</mathjax>. Capisce?</p>
<p>Note that once you know the formula for calcium hydroxide, you know the formula for the hydroxides of ALL the Group 2 metals, i.e. <mathjax>#Ba(OH)_2#</mathjax>, <mathjax>#Sr(OH)_2#</mathjax>. Are you with me?</p></div>
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<div class="markdown"><p>Calcium is from Group 2 and commonly forms the <mathjax>#Ca^(2+)#</mathjax> ion. Hydroxide ion, <mathjax>#HO^-#</mathjax>, is singly, negatively charged. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To make a neutral species, we simply balance the charges. To balance the charge of the metal ion, <mathjax>#Ca^(2+)#</mathjax>, we need 2 hydroxide ions, <mathjax>#2xxOH^-#</mathjax>.</p>
<p>We get the formula <mathjax>#Ca(OH)_2#</mathjax>. Capisce?</p>
<p>Note that once you know the formula for calcium hydroxide, you know the formula for the hydroxides of ALL the Group 2 metals, i.e. <mathjax>#Ba(OH)_2#</mathjax>, <mathjax>#Sr(OH)_2#</mathjax>. Are you with me?</p></div>
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<div class="markdown"><p>Calcium is from Group 2 and commonly forms the <mathjax>#Ca^(2+)#</mathjax> ion. Hydroxide ion, <mathjax>#HO^-#</mathjax>, is singly, negatively charged. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>To make a neutral species, we simply balance the charges. To balance the charge of the metal ion, <mathjax>#Ca^(2+)#</mathjax>, we need 2 hydroxide ions, <mathjax>#2xxOH^-#</mathjax>.</p>
<p>We get the formula <mathjax>#Ca(OH)_2#</mathjax>. Capisce?</p>
<p>Note that once you know the formula for calcium hydroxide, you know the formula for the hydroxides of ALL the Group 2 metals, i.e. <mathjax>#Ba(OH)_2#</mathjax>, <mathjax>#Sr(OH)_2#</mathjax>. Are you with me?</p></div>
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</article> | What is the formula for calcium hydroxide? | null |
834 | aa0cfd58-6ddd-11ea-808a-ccda262736ce | https://socratic.org/questions/how-much-volume-would-63-9-g-of-carbon-monoxide-take-up | 32.60 L | start physical_unit 7 8 volume l qc_end physical_unit 7 8 4 5 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] carbon monoxide [IN] L"}] | [{"type":"physical unit","value":"32.60 L"}] | [{"type":"physical unit","value":"Mass [OF] carbon monoxide [=] \\pu{63.9 g}"}] | <h1 class="questionTitle" itemprop="name">How much volume would 63.9 g of carbon monoxide take up? </h1> | null | 32.60 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://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 volume of <mathjax>#"63.9 g"#</mathjax> carbon dioxide at STP is <mathjax>#"32.6 L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">How much volume would 63.9 g of carbon monoxide take up? </h1>
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<div class="markdown"><p>The volume of <mathjax>#"63.9 g"#</mathjax> carbon dioxide at STP is <mathjax>#"32.6 L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Temperature affects volume. Since no temperature was given, I'm going to use STP.</p>
<p>The <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide can be used to answer this question.</p>
<p>We have been given the mass of <mathjax>#"CO"_2"#</mathjax> gas. We can look up the density at STP, which is <mathjax>#"1.96 g/L"#</mathjax>. <a href="http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm" rel="nofollow" target="_blank">http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm</a></p>
<p>We now have density and mass. </p>
<p><strong>The Density Formula</strong></p>
<p><mathjax>#"density"="mass"/"volume"#</mathjax></p>
<p>To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.</p>
<p><mathjax>#"volume"="mass"/"density"#</mathjax></p>
<p><mathjax>#"volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L"#</mathjax> rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p>51.1 liters at STP</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of Carbon Monoxide <mathjax>#CO#</mathjax> is 28 grams/mole</p>
<p>1 x C = 1 x 12 = 12 <br/>
1 x O = 1 x 16 = 16<br/>
Total = 28 grams/ mole </p>
<p>The number of moles is found by dividing the mass by the grams/mole </p>
<p><mathjax># 63.9/28 = 2.28 #</mathjax> moles.</p>
<p>At STP ( Standard Temperature Pressure) The molar volume is 22.4 liters/mole </p>
<p>to find the volume multiple the number of moles time the molar volume. </p>
<p># 2.28 xx 22.4 = 51.1 liters. </p></div>
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</article> | How much volume would 63.9 g of carbon monoxide take up? | null |
835 | a94b8725-6ddd-11ea-8e39-ccda262736ce | https://socratic.org/questions/sucrose-is-a-nonvolatile-nonionizing-solute-on-water-determine-the-vapor-pressur | 0.63 torr | start physical_unit 19 20 vapor_pressure torr qc_end physical_unit 19 20 17 18 molarity qc_end physical_unit 19 20 13 14 temperature qc_end physical_unit 31 32 13 14 temperature qc_end physical_unit 31 32 37 38 vapor_pressure qc_end end | [{"type":"physical unit","value":"Vapor pressure [OF] sucrose solution [IN] torr"}] | [{"type":"physical unit","value":"0.63 torr"}] | [{"type":"physical unit","value":"Molarity [OF] sucrose solution [=] \\pu{1.5 M}"},{"type":"physical unit","value":"Temperature [OF] sucrose solution [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature [OF] pure water [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Vapor pressure [OF] pure water [=] \\pu{23.8 torr}"}] | <h1 class="questionTitle" itemprop="name">Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure at #25^@"C"# of the #"1.5-m"# sucrose solution. Assume that the solution behaves ideally. The vapor pressure of pure water at #25^"C"# is #"23.8 torr"# ?</h1> | null | 0.63 torr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that when you're dealing with a <strong>nonvolatile</strong> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, you can calculate the vapor pressure of the solution by using the <strong>mole fraction</strong> of the solute and the vapor pressure of the <strong>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> at the given temperature--think <strong>Raoult's Law</strong> here. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_"sol" = chi_"sucrose" * P_ ("H"_ 2"O")^@)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_"sol"#</mathjax> is the vapor pressure of the solution</li>
<li><mathjax>#chi_"sucrose"#</mathjax> is <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sucrose</li>
<li><mathjax>#P_ ("H"_ 2"O")^@#</mathjax> is the vapor pressure of water at <mathjax>#25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>Now, you know that your solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of <mathjax>#"1.5 m"#</mathjax>, which means that you get <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose, the solute, for every <mathjax>#"1 kg"#</mathjax> of water, the solvent. </p>
<p>To make the calculations easier, let's pick a sample of this solution that contains exactly <mathjax>#"1 kg"#</mathjax> of water. Consequently, this sample will also contain <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose.</p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water, which is equal to <mathjax>#10^3#</mathjax> <mathjax>#"g"#</mathjax>, to <em>moles</em></p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sucrose is calculated by dividing the number of moles of sucrose by the <strong>total number of moles</strong> present in the solution.</p>
<blockquote>
<p><mathjax>#chi_"sucrose" = (1.5 color(red)(cancel(color(black)("moles"))))/((1.5 + 55.51)color(red)(cancel(color(black)("moles")))) = 0.0263#</mathjax></p>
</blockquote>
<p>You can thus say that the vapor pressure of the solution will be equal to</p>
<blockquote>
<p><mathjax>#P_"sol" = 0.0263 * "23.8 torr" = color(darkgreen)(ul(color(black)("0.63 torr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molality of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.63 torr"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that when you're dealing with a <strong>nonvolatile</strong> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, you can calculate the vapor pressure of the solution by using the <strong>mole fraction</strong> of the solute and the vapor pressure of the <strong>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> at the given temperature--think <strong>Raoult's Law</strong> here. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_"sol" = chi_"sucrose" * P_ ("H"_ 2"O")^@)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_"sol"#</mathjax> is the vapor pressure of the solution</li>
<li><mathjax>#chi_"sucrose"#</mathjax> is <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sucrose</li>
<li><mathjax>#P_ ("H"_ 2"O")^@#</mathjax> is the vapor pressure of water at <mathjax>#25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>Now, you know that your solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of <mathjax>#"1.5 m"#</mathjax>, which means that you get <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose, the solute, for every <mathjax>#"1 kg"#</mathjax> of water, the solvent. </p>
<p>To make the calculations easier, let's pick a sample of this solution that contains exactly <mathjax>#"1 kg"#</mathjax> of water. Consequently, this sample will also contain <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose.</p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water, which is equal to <mathjax>#10^3#</mathjax> <mathjax>#"g"#</mathjax>, to <em>moles</em></p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sucrose is calculated by dividing the number of moles of sucrose by the <strong>total number of moles</strong> present in the solution.</p>
<blockquote>
<p><mathjax>#chi_"sucrose" = (1.5 color(red)(cancel(color(black)("moles"))))/((1.5 + 55.51)color(red)(cancel(color(black)("moles")))) = 0.0263#</mathjax></p>
</blockquote>
<p>You can thus say that the vapor pressure of the solution will be equal to</p>
<blockquote>
<p><mathjax>#P_"sol" = 0.0263 * "23.8 torr" = color(darkgreen)(ul(color(black)("0.63 torr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molality of the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure at #25^@"C"# of the #"1.5-m"# sucrose solution. Assume that the solution behaves ideally. The vapor pressure of pure water at #25^"C"# is #"23.8 torr"# ?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-02-24T23:12:47" itemprop="dateCreated">
Feb 24, 2018
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<div class="markdown"><p><mathjax>#"0.63 torr"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that when you're dealing with a <strong>nonvolatile</strong> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, you can calculate the vapor pressure of the solution by using the <strong>mole fraction</strong> of the solute and the vapor pressure of the <strong>pure <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> at the given temperature--think <strong>Raoult's Law</strong> here. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(P_"sol" = chi_"sucrose" * P_ ("H"_ 2"O")^@)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P_"sol"#</mathjax> is the vapor pressure of the solution</li>
<li><mathjax>#chi_"sucrose"#</mathjax> is <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> fraction of sucrose</li>
<li><mathjax>#P_ ("H"_ 2"O")^@#</mathjax> is the vapor pressure of water at <mathjax>#25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>Now, you know that your solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of <mathjax>#"1.5 m"#</mathjax>, which means that you get <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose, the solute, for every <mathjax>#"1 kg"#</mathjax> of water, the solvent. </p>
<p>To make the calculations easier, let's pick a sample of this solution that contains exactly <mathjax>#"1 kg"#</mathjax> of water. Consequently, this sample will also contain <mathjax>#1.5#</mathjax> <strong>moles</strong> of sucrose.</p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water, which is equal to <mathjax>#10^3#</mathjax> <mathjax>#"g"#</mathjax>, to <em>moles</em></p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The <strong>mole fraction</strong> of sucrose is calculated by dividing the number of moles of sucrose by the <strong>total number of moles</strong> present in the solution.</p>
<blockquote>
<p><mathjax>#chi_"sucrose" = (1.5 color(red)(cancel(color(black)("moles"))))/((1.5 + 55.51)color(red)(cancel(color(black)("moles")))) = 0.0263#</mathjax></p>
</blockquote>
<p>You can thus say that the vapor pressure of the solution will be equal to</p>
<blockquote>
<p><mathjax>#P_"sol" = 0.0263 * "23.8 torr" = color(darkgreen)(ul(color(black)("0.63 torr")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molality of the solution. </p></div>
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</article> | Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure at #25^@"C"# of the #"1.5-m"# sucrose solution. Assume that the solution behaves ideally. The vapor pressure of pure water at #25^"C"# is #"23.8 torr"# ? | null |
836 | ab4a3ac2-6ddd-11ea-951a-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-oxygen-in-hydrogen-peroxide | -1 | start physical_unit 6 6 oxidation_number none qc_end substance 8 9 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] oxygen"}] | [{"type":"physical unit","value":"-1"}] | [{"type":"substance name","value":"Hydrogen peroxide"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in hydrogen peroxide?</h1> | null | -1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxidation number is the charge left on the central atom, when all the bonds are broken with the charge assigned to most electronegative atom.</p>
<p>With <mathjax>#H-O-O-H#</mathjax>, we break a <mathjax>#H-O#</mathjax> bond, and get <mathjax>#H^+#</mathjax> and <mathjax>#""^(-)O#</mathjax>. However, if we break an <mathjax>#O-O#</mathjax> bond the 2 electrons in the <mathjax>#O-O#</mathjax> bond are SHARED between the oxygen atoms, to give <mathjax>#2xx*O^-#</mathjax>; i.e. the oxygen atoms necessarily have the same <a href="https://socratic.org/chemistry/bonding-basics/electronegativity-and-bonding">electronegativity</a>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>In hydrogen peroxide, the oxidation number of each oxygen is <mathjax>#-I#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxidation number is the charge left on the central atom, when all the bonds are broken with the charge assigned to most electronegative atom.</p>
<p>With <mathjax>#H-O-O-H#</mathjax>, we break a <mathjax>#H-O#</mathjax> bond, and get <mathjax>#H^+#</mathjax> and <mathjax>#""^(-)O#</mathjax>. However, if we break an <mathjax>#O-O#</mathjax> bond the 2 electrons in the <mathjax>#O-O#</mathjax> bond are SHARED between the oxygen atoms, to give <mathjax>#2xx*O^-#</mathjax>; i.e. the oxygen atoms necessarily have the same <a href="https://socratic.org/chemistry/bonding-basics/electronegativity-and-bonding">electronegativity</a>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in hydrogen peroxide?</h1>
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<div class="markdown"><p>In hydrogen peroxide, the oxidation number of each oxygen is <mathjax>#-I#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Oxidation number is the charge left on the central atom, when all the bonds are broken with the charge assigned to most electronegative atom.</p>
<p>With <mathjax>#H-O-O-H#</mathjax>, we break a <mathjax>#H-O#</mathjax> bond, and get <mathjax>#H^+#</mathjax> and <mathjax>#""^(-)O#</mathjax>. However, if we break an <mathjax>#O-O#</mathjax> bond the 2 electrons in the <mathjax>#O-O#</mathjax> bond are SHARED between the oxygen atoms, to give <mathjax>#2xx*O^-#</mathjax>; i.e. the oxygen atoms necessarily have the same <a href="https://socratic.org/chemistry/bonding-basics/electronegativity-and-bonding">electronegativity</a>.</p></div>
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</article> | What is the oxidation number of oxygen in hydrogen peroxide? | null |
837 | abab4608-6ddd-11ea-ac57-ccda262736ce | https://socratic.org/questions/how-many-grams-of-oxygen-are-in-6-15times10-23-formula-units-of-nh-4-2so-4 | 65.4 grams | start physical_unit 4 4 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] grams"}] | [{"type":"physical unit","value":"65.4 grams"}] | [{"type":"physical unit","value":"Number [OF] (NH4)2SO4 formula units [=] \\pu{6.15 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many grams of oxygen are in #6.15times10^23# formula units of #(NH_4)_2SO_4#?</h1> | null | 65.4 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of oxygen in <mathjax>#6.15xx10^23#</mathjax> <mathjax>#"formula units (NH"_4")"_2"SO"_4#</mathjax>.</p>
<p>To do this, we recognize that there are <mathjax>#4#</mathjax> <mathjax>#"atoms O"#</mathjax> per formula unit of ammonium sulfate:</p>
<p><mathjax>#6.15xx10^23cancel("formula units (NH"_4")"_2"SO"_4)((4color(white)(l)"atoms O")/(1cancel("formula unit (NH"_4")"_2"SO"_4)))#</mathjax></p>
<p><mathjax>#= 2.46xx10^24#</mathjax> <mathjax>#"atoms O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we use <strong>Avogadro's number</strong> to find the number of <em>moles</em> of <mathjax>#"O"#</mathjax>:</p>
<p><mathjax>#2.46xx10^24cancel("atoms O")((1color(white)(l)"mol O")/(6.022xx10^23cancel("atoms O"))) = ul(4.085color(white)(l)"mol O"#</mathjax></p>
<p>Finally, we use the <strong>molar mass</strong> of oxygen (<mathjax>#15.999#</mathjax> <mathjax>#"g/mol"#</mathjax>) to convert from moles to grams:</p>
<p><mathjax>#4.085cancel("mol O")((15.999color(white)(l)"g O")/(1cancel("mol O"))) = color(red)(ul(65.4color(white)(l)"g O"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#65.4#</mathjax> <mathjax>#"g O"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of oxygen in <mathjax>#6.15xx10^23#</mathjax> <mathjax>#"formula units (NH"_4")"_2"SO"_4#</mathjax>.</p>
<p>To do this, we recognize that there are <mathjax>#4#</mathjax> <mathjax>#"atoms O"#</mathjax> per formula unit of ammonium sulfate:</p>
<p><mathjax>#6.15xx10^23cancel("formula units (NH"_4")"_2"SO"_4)((4color(white)(l)"atoms O")/(1cancel("formula unit (NH"_4")"_2"SO"_4)))#</mathjax></p>
<p><mathjax>#= 2.46xx10^24#</mathjax> <mathjax>#"atoms O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we use <strong>Avogadro's number</strong> to find the number of <em>moles</em> of <mathjax>#"O"#</mathjax>:</p>
<p><mathjax>#2.46xx10^24cancel("atoms O")((1color(white)(l)"mol O")/(6.022xx10^23cancel("atoms O"))) = ul(4.085color(white)(l)"mol O"#</mathjax></p>
<p>Finally, we use the <strong>molar mass</strong> of oxygen (<mathjax>#15.999#</mathjax> <mathjax>#"g/mol"#</mathjax>) to convert from moles to grams:</p>
<p><mathjax>#4.085cancel("mol O")((15.999color(white)(l)"g O")/(1cancel("mol O"))) = color(red)(ul(65.4color(white)(l)"g O"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#65.4#</mathjax> <mathjax>#"g O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of oxygen in <mathjax>#6.15xx10^23#</mathjax> <mathjax>#"formula units (NH"_4")"_2"SO"_4#</mathjax>.</p>
<p>To do this, we recognize that there are <mathjax>#4#</mathjax> <mathjax>#"atoms O"#</mathjax> per formula unit of ammonium sulfate:</p>
<p><mathjax>#6.15xx10^23cancel("formula units (NH"_4")"_2"SO"_4)((4color(white)(l)"atoms O")/(1cancel("formula unit (NH"_4")"_2"SO"_4)))#</mathjax></p>
<p><mathjax>#= 2.46xx10^24#</mathjax> <mathjax>#"atoms O"#</mathjax></p>
<blockquote></blockquote>
<p>Now, we use <strong>Avogadro's number</strong> to find the number of <em>moles</em> of <mathjax>#"O"#</mathjax>:</p>
<p><mathjax>#2.46xx10^24cancel("atoms O")((1color(white)(l)"mol O")/(6.022xx10^23cancel("atoms O"))) = ul(4.085color(white)(l)"mol O"#</mathjax></p>
<p>Finally, we use the <strong>molar mass</strong> of oxygen (<mathjax>#15.999#</mathjax> <mathjax>#"g/mol"#</mathjax>) to convert from moles to grams:</p>
<p><mathjax>#4.085cancel("mol O")((15.999color(white)(l)"g O")/(1cancel("mol O"))) = color(red)(ul(65.4color(white)(l)"g O"#</mathjax></p></div>
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</article> | How many grams of oxygen are in #6.15times10^23# formula units of #(NH_4)_2SO_4#? | null |
838 | a91f7bd5-6ddd-11ea-9e96-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-carbon-in-the-ion-co3-2 | +4 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"oxidation number [OF] carbon"}] | [{"type":"physical unit","value":"+4"}] | [{"type":"chemical equation","value":"(CO3)^2-"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the ion (Co3)2-?</h1> | null | +4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The sum of the formal <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in an a complex or compound ion must sum to the charge of the ion. Here the charge is <mathjax>#-2#</mathjax>. Oxygen typically has an oxidation number of <mathjax>#-II#</mathjax>, and it does here. So <mathjax>#3 xx -2 + C_(ON) = -2#</mathjax>. And what therefore is <mathjax>#C_(ON)#</mathjax>?</p>
<p>What is the oxidation number of carbon in oxalate ion, <mathjax>#C_2O_4^(2-)#</mathjax>?</p></div>
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<div class="markdown"><p>The carbon is fully oxidized. Therefore, it is <mathjax>#C, IV^+#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sum of the formal <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in an a complex or compound ion must sum to the charge of the ion. Here the charge is <mathjax>#-2#</mathjax>. Oxygen typically has an oxidation number of <mathjax>#-II#</mathjax>, and it does here. So <mathjax>#3 xx -2 + C_(ON) = -2#</mathjax>. And what therefore is <mathjax>#C_(ON)#</mathjax>?</p>
<p>What is the oxidation number of carbon in oxalate ion, <mathjax>#C_2O_4^(2-)#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of carbon in the ion (Co3)2-?</h1>
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<div class="markdown"><p>The carbon is fully oxidized. Therefore, it is <mathjax>#C, IV^+#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sum of the formal <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in an a complex or compound ion must sum to the charge of the ion. Here the charge is <mathjax>#-2#</mathjax>. Oxygen typically has an oxidation number of <mathjax>#-II#</mathjax>, and it does here. So <mathjax>#3 xx -2 + C_(ON) = -2#</mathjax>. And what therefore is <mathjax>#C_(ON)#</mathjax>?</p>
<p>What is the oxidation number of carbon in oxalate ion, <mathjax>#C_2O_4^(2-)#</mathjax>?</p></div>
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<div class="markdown"><p>The oxidation number of the carbon is +4.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxygen normally has an oxidation state of -2. </p>
<p>Oxygen = three ions each with a charge of -2 = 3 x -2 = -6<br/>
The overall charge of the polyatomic ion is -2</p>
<p>This means that the oxidation number of the carbon must be +4.</p>
<p>Total charge = +4 + -6 = -2</p>
<p>Here is a video which discusses this topic by discussing how to determine <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the bicarbonate and dichromate ions.</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>Video from: <a href="https://mrpauller.weebly.com/" rel="nofollow">Noel Pauller</a> </p>
<p>Hope this helps!</p></div>
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</article> | What is the oxidation number of carbon in the ion (Co3)2-? | null |
839 | ace8f094-6ddd-11ea-8a42-ccda262736ce | https://socratic.org/questions/how-many-moles-of-kci-are-required-to-prepare-1-50-l-of-5-4-m-kcl | 8.10 moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] KCl [IN] moles"}] | [{"type":"physical unit","value":"8.10 moles"}] | [{"type":"physical unit","value":"Molarity [OF] KCl solution [=] \\pu{5.4 M}"},{"type":"physical unit","value":"Volume [OF] KCl solution [=] \\pu{1.50 L}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #KCI# are required to prepare 1.50 L of 5.4 M #KCl#?</h1> | null | 8.10 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>Since the concentration was specified to be <mathjax>#5.4#</mathjax> <mathjax>#mol•L^(-1)#</mathjax>, and the volume was specified to be <mathjax>#1.5#</mathjax> <mathjax>#L#</mathjax>, the given molar quantity is appropriate for this concentration. </p>
<p>As a word of caution, I don't know for a moment whether potassium chloride is soluble enough to give these concentrations. Check on the web to see the solubility of potassium chloride in aqueous solution. If this question quotes an unreasonable solubility, then it was entirely inappropriate to ask.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Why <mathjax>#8.1#</mathjax> <mathjax>#"moles"#</mathjax> of potassium chloride.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>Since the concentration was specified to be <mathjax>#5.4#</mathjax> <mathjax>#mol•L^(-1)#</mathjax>, and the volume was specified to be <mathjax>#1.5#</mathjax> <mathjax>#L#</mathjax>, the given molar quantity is appropriate for this concentration. </p>
<p>As a word of caution, I don't know for a moment whether potassium chloride is soluble enough to give these concentrations. Check on the web to see the solubility of potassium chloride in aqueous solution. If this question quotes an unreasonable solubility, then it was entirely inappropriate to ask.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #KCI# are required to prepare 1.50 L of 5.4 M #KCl#?</h1>
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<div class="markdown"><p>Why <mathjax>#8.1#</mathjax> <mathjax>#"moles"#</mathjax> of potassium chloride.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="moles of solute"/"volume of solution"#</mathjax></p>
<p>Since the concentration was specified to be <mathjax>#5.4#</mathjax> <mathjax>#mol•L^(-1)#</mathjax>, and the volume was specified to be <mathjax>#1.5#</mathjax> <mathjax>#L#</mathjax>, the given molar quantity is appropriate for this concentration. </p>
<p>As a word of caution, I don't know for a moment whether potassium chloride is soluble enough to give these concentrations. Check on the web to see the solubility of potassium chloride in aqueous solution. If this question quotes an unreasonable solubility, then it was entirely inappropriate to ask.</p></div>
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</article> | How many moles of #KCI# are required to prepare 1.50 L of 5.4 M #KCl#? | null |
840 | ab27954a-6ddd-11ea-ad88-ccda262736ce | https://socratic.org/questions/a-tire-has-a-pressure-of-780-mmhg-at-27-deg-c-if-the-tire-is-heated-to-47-deg-c- | 832.00 mmHg | start physical_unit 13 14 pressure mmhg qc_end physical_unit 13 14 6 7 pressure qc_end physical_unit 13 14 9 11 temperature qc_end physical_unit 13 14 18 20 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the tire [IN] mmHg"}] | [{"type":"physical unit","value":"832.00 mmHg"}] | [{"type":"physical unit","value":"Pressure1 [OF] the tire [=] \\pu{780 mmHg}"},{"type":"physical unit","value":"Temperature1 [OF] the tire [=] \\pu{27 deg C}"},{"type":"physical unit","value":"Temperature2 [OF] the tire [=] \\pu{47 deg C}"}] | <h1 class="questionTitle" itemprop="name">A tire has a pressure of 780 mmHg at 27 deg C. If the tire is heated to 47 deg C on hot pavement, what will the new pressure be?</h1> | null | 832.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#27^oC=27+273K=300K#</mathjax><br/>
<mathjax>#47^oC=47+273K=320K#</mathjax></p>
<p>The equation above can be re-written to: <br/>
<mathjax>#p_2=T_2/T_1xxp_1=320/300xx780=832mmHg#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use <mathjax>#(p_1)/T_1=(p_2)/T_2#</mathjax>, with <mathjax>#T#</mathjax> in <mathjax>#K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#27^oC=27+273K=300K#</mathjax><br/>
<mathjax>#47^oC=47+273K=320K#</mathjax></p>
<p>The equation above can be re-written to: <br/>
<mathjax>#p_2=T_2/T_1xxp_1=320/300xx780=832mmHg#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A tire has a pressure of 780 mmHg at 27 deg C. If the tire is heated to 47 deg C on hot pavement, what will the new pressure be?</h1>
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<div class="markdown"><p>We use <mathjax>#(p_1)/T_1=(p_2)/T_2#</mathjax>, with <mathjax>#T#</mathjax> in <mathjax>#K#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#27^oC=27+273K=300K#</mathjax><br/>
<mathjax>#47^oC=47+273K=320K#</mathjax></p>
<p>The equation above can be re-written to: <br/>
<mathjax>#p_2=T_2/T_1xxp_1=320/300xx780=832mmHg#</mathjax></p></div>
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</article> | A tire has a pressure of 780 mmHg at 27 deg C. If the tire is heated to 47 deg C on hot pavement, what will the new pressure be? | null |
841 | aa485bcc-6ddd-11ea-adde-ccda262736ce | https://socratic.org/questions/if-the-ph-of-a-solution-is-7-6-what-is-the-poh | 6.40 | start physical_unit 5 5 poh none qc_end physical_unit 5 5 7 7 ph qc_end end | [{"type":"physical unit","value":"pOH [OF] the solution"}] | [{"type":"physical unit","value":"6.40"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{7.6}"}] | <h1 class="questionTitle" itemprop="name">If the pH of a solution is 7.6, what is the pOH?</h1> | null | 6.40 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
<p>So if <mathjax>#"pH" = 7.6#</mathjax> then <mathjax>#"pOH" = 14 - "pH" = 6.4#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#6.4#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
<p>So if <mathjax>#"pH" = 7.6#</mathjax> then <mathjax>#"pOH" = 14 - "pH" = 6.4#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If the pH of a solution is 7.6, what is the pOH?</h1>
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<div class="markdown"><p><mathjax>#6.4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
<p>So if <mathjax>#"pH" = 7.6#</mathjax> then <mathjax>#"pOH" = 14 - "pH" = 6.4#</mathjax></p></div>
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</article> | If the pH of a solution is 7.6, what is the pOH? | null |
842 | a919751e-6ddd-11ea-8ce3-ccda262736ce | https://socratic.org/questions/100ml-of-5mol-l-1-nitric-acid-was-added-to-excess-copper-what-mass-of-copper-rea | 7.9 g | start physical_unit 11 11 mass g qc_end chemical_equation 17 25 qc_end physical_unit 5 6 0 1 volume qc_end physical_unit 5 6 3 4 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] copper [IN] g"}] | [{"type":"physical unit","value":"7.9 g"}] | [{"type":"chemical equation","value":"Cu + 4 HNO3 -> 2 NO2 + Cu(NO3)2"},{"type":"physical unit","value":"Volume [OF] nitric acid [=] \\pu{100 mL}"},{"type":"physical unit","value":"Molarity [OF] nitric acid [=] \\pu{5 mol/L}"},{"type":"other","value":"Excess copper."}] | <h1 class="questionTitle" itemprop="name">100mL of 5mol L-1 nitric acid was added to excess copper. What mass of copper reacted?
Cu +4HNO3 -> 2NO2 + Cu(NO3)2</h1> | null | 7.9 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Start with the balanced equation.</strong></p>
<p><mathjax>#"Cu" + "4HNO"_3 → "2NO"_2 + "Cu"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"HNO"_3#</mathjax></strong>.</p>
<p><mathjax>#"Moles of HNO"_3 = 0.100 color(red)(cancel(color(black)("L HNO"_3))) × ("5 mol HNO"_3)/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "0.50 mol HNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Cu"#</mathjax></strong></p>
<p><mathjax>#"Moles of Cu" = 0.50 color(red)(cancel(color(black)("mol HNO"_3))) × "1 mol Cu"/(4color(red)(cancel(color(black)("mol HNO"_3)))) = "0.12 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"Cu"#</mathjax></strong>.</p>
<p><mathjax>#"Mass of Cu" = 0.12 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "7.9 g Cu"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the concentration of the nitric acid.</p>
<p>However, I calculated the answer to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> for you.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><blockquote></blockquote>
<p>The mass of <mathjax>#"Cu"#</mathjax> was d) 7.9 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Start with the balanced equation.</strong></p>
<p><mathjax>#"Cu" + "4HNO"_3 → "2NO"_2 + "Cu"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"HNO"_3#</mathjax></strong>.</p>
<p><mathjax>#"Moles of HNO"_3 = 0.100 color(red)(cancel(color(black)("L HNO"_3))) × ("5 mol HNO"_3)/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "0.50 mol HNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Cu"#</mathjax></strong></p>
<p><mathjax>#"Moles of Cu" = 0.50 color(red)(cancel(color(black)("mol HNO"_3))) × "1 mol Cu"/(4color(red)(cancel(color(black)("mol HNO"_3)))) = "0.12 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"Cu"#</mathjax></strong>.</p>
<p><mathjax>#"Mass of Cu" = 0.12 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "7.9 g Cu"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the concentration of the nitric acid.</p>
<p>However, I calculated the answer to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> for you.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">100mL of 5mol L-1 nitric acid was added to excess copper. What mass of copper reacted?
Cu +4HNO3 -> 2NO2 + Cu(NO3)2</h1>
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Ernest Z.
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Aug 5, 2017
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<p>The mass of <mathjax>#"Cu"#</mathjax> was d) 7.9 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Start with the balanced equation.</strong></p>
<p><mathjax>#"Cu" + "4HNO"_3 → "2NO"_2 + "Cu"("NO"_3)_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the moles of <mathjax>#"HNO"_3#</mathjax></strong>.</p>
<p><mathjax>#"Moles of HNO"_3 = 0.100 color(red)(cancel(color(black)("L HNO"_3))) × ("5 mol HNO"_3)/(1 color(red)(cancel(color(black)("L HNO"_3)))) = "0.50 mol HNO"_3#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3. Calculate the moles of <mathjax>#"Cu"#</mathjax></strong></p>
<p><mathjax>#"Moles of Cu" = 0.50 color(red)(cancel(color(black)("mol HNO"_3))) × "1 mol Cu"/(4color(red)(cancel(color(black)("mol HNO"_3)))) = "0.12 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"Cu"#</mathjax></strong>.</p>
<p><mathjax>#"Mass of Cu" = 0.12 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "7.9 g Cu"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the concentration of the nitric acid.</p>
<p>However, I calculated the answer to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> for you.</p></div>
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</article> | 100mL of 5mol L-1 nitric acid was added to excess copper. What mass of copper reacted?
Cu +4HNO3 -> 2NO2 + Cu(NO3)2 | null |
843 | aa0dc15d-6ddd-11ea-9531-ccda262736ce | https://socratic.org/questions/58d9e4147c01490782a8b8ee | 10 | start physical_unit 9 10 ph none qc_end physical_unit 9 10 3 4 [h3o+] qc_end end | [{"type":"physical unit","value":"pH [OF] this solution"}] | [{"type":"physical unit","value":"10"}] | [{"type":"physical unit","value":"[H3O+] [OF] this solution [=] \\pu{10^(-10) mol/L}"}] | <h1 class="questionTitle" itemprop="name">If #[H_3O^+]-=10^(-10)*mol*L^-1#, what is #pH# of this solution?</h1> | null | 10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus, here, <mathjax>#pH=-log_(10)(10^(-10))=-(-10)=10.#</mathjax> </p>
<p>Given the definition of the logarithmic function, <mathjax>#log_ab=c#</mathjax>, <mathjax>#c#</mathjax> is that power to which we raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>, then <mathjax>#log_(10)10^x=x#</mathjax>. Logs were introduced in the days BEFORE the advent of cheap electronic calculators; scientists, and economists, and engineers, and A level students, used to use <mathjax>#"log tables"#</mathjax>, and <mathjax>#"slide rules"#</mathjax> which made complex calculations a bit less tiresome.</p>
<p>In water, under standard conditions, we can show that <mathjax>#pH+pOH=14#</mathjax>. What is <mathjax>#[HO^-]#</mathjax> in the given example?</p>
<p>The use of <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> allows the use of <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer solutions</a> to moderate acidity and alkalinity. </p></div>
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<div class="markdown"><p><mathjax>#pH=10#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus, here, <mathjax>#pH=-log_(10)(10^(-10))=-(-10)=10.#</mathjax> </p>
<p>Given the definition of the logarithmic function, <mathjax>#log_ab=c#</mathjax>, <mathjax>#c#</mathjax> is that power to which we raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>, then <mathjax>#log_(10)10^x=x#</mathjax>. Logs were introduced in the days BEFORE the advent of cheap electronic calculators; scientists, and economists, and engineers, and A level students, used to use <mathjax>#"log tables"#</mathjax>, and <mathjax>#"slide rules"#</mathjax> which made complex calculations a bit less tiresome.</p>
<p>In water, under standard conditions, we can show that <mathjax>#pH+pOH=14#</mathjax>. What is <mathjax>#[HO^-]#</mathjax> in the given example?</p>
<p>The use of <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> allows the use of <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer solutions</a> to moderate acidity and alkalinity. </p></div>
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<h1 class="questionTitle" itemprop="name">If #[H_3O^+]-=10^(-10)*mol*L^-1#, what is #pH# of this solution?</h1>
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<div class="markdown"><p><mathjax>#pH=10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>By definition, <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>And thus, here, <mathjax>#pH=-log_(10)(10^(-10))=-(-10)=10.#</mathjax> </p>
<p>Given the definition of the logarithmic function, <mathjax>#log_ab=c#</mathjax>, <mathjax>#c#</mathjax> is that power to which we raise the base <mathjax>#a#</mathjax> to get <mathjax>#b#</mathjax>, then <mathjax>#log_(10)10^x=x#</mathjax>. Logs were introduced in the days BEFORE the advent of cheap electronic calculators; scientists, and economists, and engineers, and A level students, used to use <mathjax>#"log tables"#</mathjax>, and <mathjax>#"slide rules"#</mathjax> which made complex calculations a bit less tiresome.</p>
<p>In water, under standard conditions, we can show that <mathjax>#pH+pOH=14#</mathjax>. What is <mathjax>#[HO^-]#</mathjax> in the given example?</p>
<p>The use of <mathjax>#pH#</mathjax> and <mathjax>#pOH#</mathjax> allows the use of <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">buffer solutions</a> to moderate acidity and alkalinity. </p></div>
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</article> | If #[H_3O^+]-=10^(-10)*mol*L^-1#, what is #pH# of this solution? | null |
844 | ac45f69c-6ddd-11ea-bd1d-ccda262736ce | https://socratic.org/questions/what-is-the-approximate-temperature-of-a-0-50-mol-sample-of-gas-at-750-mm-hg-and | 290 K | start physical_unit 9 11 temperature k qc_end physical_unit 9 11 7 8 mole qc_end physical_unit 9 11 13 14 pressure qc_end physical_unit 9 11 19 20 volume qc_end end | [{"type":"physical unit","value":"Temperature [OF] gas sample [IN] K"}] | [{"type":"physical unit","value":"290 K"}] | [{"type":"physical unit","value":"Mole [OF] gas sample [=] \\pu{0.50 mol}"},{"type":"physical unit","value":"Pressure [OF] gas sample [=] \\pu{750 mmHg}"},{"type":"physical unit","value":"Volume [OF] gas sample [=] \\pu{12 L}"}] | <h1 class="questionTitle" itemprop="name">What is the approximate temperature of a 0.50 mol sample of gas at 750 mm Hg and a volume of 12 L?</h1> | null | 290 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation: </p>
<p><mathjax>#T=(PV)/(nR)#</mathjax></p>
<p><mathjax>#T=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx12*L)/(0.50*molxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p>The key to understanding these problems is correct interpretation of the pressure reading. It is a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. So even though we report the <mathjax>#"length"#</mathjax> of the column, this actually relates to the <mathjax>#"pressure"#</mathjax>, and I used the conversion,</p>
<p><mathjax>#P=(750*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*atm#</mathjax>.</p>
<p>Pressure measurement by means of a mercury column was very common back in the day. These days, laboratories tend to want want to get rid of mercury columns because it is an absolute horror to clean up mercury spills. As a student, I found mercury absolutely fascinating, so I contributed to the problem.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#T~=290*K#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation: </p>
<p><mathjax>#T=(PV)/(nR)#</mathjax></p>
<p><mathjax>#T=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx12*L)/(0.50*molxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p>The key to understanding these problems is correct interpretation of the pressure reading. It is a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. So even though we report the <mathjax>#"length"#</mathjax> of the column, this actually relates to the <mathjax>#"pressure"#</mathjax>, and I used the conversion,</p>
<p><mathjax>#P=(750*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*atm#</mathjax>.</p>
<p>Pressure measurement by means of a mercury column was very common back in the day. These days, laboratories tend to want want to get rid of mercury columns because it is an absolute horror to clean up mercury spills. As a student, I found mercury absolutely fascinating, so I contributed to the problem.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the approximate temperature of a 0.50 mol sample of gas at 750 mm Hg and a volume of 12 L?</h1>
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<div class="markdown"><p><mathjax>#T~=290*K#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the Ideal Gas Equation: </p>
<p><mathjax>#T=(PV)/(nR)#</mathjax></p>
<p><mathjax>#T=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx12*L)/(0.50*molxx0.0821*L*atm*K^-1*mol^-1)#</mathjax></p>
<p>The key to understanding these problems is correct interpretation of the pressure reading. It is a fact that <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. So even though we report the <mathjax>#"length"#</mathjax> of the column, this actually relates to the <mathjax>#"pressure"#</mathjax>, and I used the conversion,</p>
<p><mathjax>#P=(750*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#1*atm#</mathjax>.</p>
<p>Pressure measurement by means of a mercury column was very common back in the day. These days, laboratories tend to want want to get rid of mercury columns because it is an absolute horror to clean up mercury spills. As a student, I found mercury absolutely fascinating, so I contributed to the problem.</p></div>
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</article> | What is the approximate temperature of a 0.50 mol sample of gas at 750 mm Hg and a volume of 12 L? | null |
845 | abb2c626-6ddd-11ea-9c50-ccda262736ce | https://socratic.org/questions/the-heat-of-vaporization-of-alcohol-is-879-j-g-what-is-the-energy-in-j-required- | 3955.5 J | start physical_unit 22 22 energy j qc_end physical_unit 22 22 18 19 mass qc_end physical_unit 5 5 7 8 enthalpy_of_vaporization qc_end end | [{"type":"physical unit","value":"Required energy [OF] C2H5OH [IN] J"}] | [{"type":"physical unit","value":"3955.5 J"}] | [{"type":"physical unit","value":"Mass [OF] C2H5OH [=] \\pu{4.50 g}"},{"type":"physical unit","value":"Heat of vaporization [OF] alcohol [=] \\pu{879 J/g}"}] | <h1 class="questionTitle" itemprop="name">The heat of vaporization of alcohol is 879 #Jg^-1#. What is the energy (in J) required to vaporize #4.50# #g# of alcohol, #C_2H_5OH#?</h1> | null | 3955.5 J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly straightforward question. Each gram of alcohol requires <mathjax>#879#</mathjax> <mathjax>#J#</mathjax> to vaporise it, and there are <mathjax>#4.5#</mathjax> <mathjax>#g#</mathjax>, so we simply multiply.</p>
<p>Notice the units: <mathjax>#Jg^-1*g = J#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#879#</mathjax> joules per gram times <mathjax>#4.50#</mathjax> grams equals <mathjax>#3955.5#</mathjax> joules</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly straightforward question. Each gram of alcohol requires <mathjax>#879#</mathjax> <mathjax>#J#</mathjax> to vaporise it, and there are <mathjax>#4.5#</mathjax> <mathjax>#g#</mathjax>, so we simply multiply.</p>
<p>Notice the units: <mathjax>#Jg^-1*g = J#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The heat of vaporization of alcohol is 879 #Jg^-1#. What is the energy (in J) required to vaporize #4.50# #g# of alcohol, #C_2H_5OH#?</h1>
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<div class="markdown"><p><mathjax>#879#</mathjax> joules per gram times <mathjax>#4.50#</mathjax> grams equals <mathjax>#3955.5#</mathjax> joules</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly straightforward question. Each gram of alcohol requires <mathjax>#879#</mathjax> <mathjax>#J#</mathjax> to vaporise it, and there are <mathjax>#4.5#</mathjax> <mathjax>#g#</mathjax>, so we simply multiply.</p>
<p>Notice the units: <mathjax>#Jg^-1*g = J#</mathjax></p></div>
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</article> | The heat of vaporization of alcohol is 879 #Jg^-1#. What is the energy (in J) required to vaporize #4.50# #g# of alcohol, #C_2H_5OH#? | null |
846 | aa5ffd47-6ddd-11ea-9799-ccda262736ce | https://socratic.org/questions/5958abdbb72cff185f00fbf8 | 3.61 × 10^24 | start physical_unit 7 8 number none qc_end physical_unit 4 4 1 2 mole qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"3.61 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] glucose [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">In one mole of glucose, how many oxygen atoms are present?</h1> | null | 3.61 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6xx"moles of glucose"#</mathjax>. Why? Because CLEARLY, each mole of glucose, <mathjax>#C_6H_12O_6#</mathjax> contains <mathjax>#6*mol#</mathjax> oxygen atoms. </p>
<p>Note that I am labouring the point that we deal with <strong>OXYGEN ATOMS</strong> , <strong>NOT OXYGEN MOLECULES</strong>.</p>
<p><mathjax>#96*g#</mathjax> oxygen is a molar quantity of <mathjax>#(96*g)/(16*g*mol^-1)=6*mol#</mathjax> with respect to oxygen atoms.</p>
<p>And of course if there are <mathjax>#6*mol#</mathjax> oxygen atoms, there must be <mathjax>#1*mol#</mathjax> glucose. Agreed?</p>
<p>We are not quite finished because we were asked for the mass, and so we multiply the molar quantity by the molar mass of glucose.....</p>
<p><mathjax>#1*molxx180.16*g*mol^-1=180.16*g#</mathjax>.</p>
<p>Capisce?</p></div>
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<div class="markdown"><p>In glucose, <mathjax>#C_6H_12O_6#</mathjax>.....I gets <mathjax>#180.16*g#</mathjax> of glucose.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6xx"moles of glucose"#</mathjax>. Why? Because CLEARLY, each mole of glucose, <mathjax>#C_6H_12O_6#</mathjax> contains <mathjax>#6*mol#</mathjax> oxygen atoms. </p>
<p>Note that I am labouring the point that we deal with <strong>OXYGEN ATOMS</strong> , <strong>NOT OXYGEN MOLECULES</strong>.</p>
<p><mathjax>#96*g#</mathjax> oxygen is a molar quantity of <mathjax>#(96*g)/(16*g*mol^-1)=6*mol#</mathjax> with respect to oxygen atoms.</p>
<p>And of course if there are <mathjax>#6*mol#</mathjax> oxygen atoms, there must be <mathjax>#1*mol#</mathjax> glucose. Agreed?</p>
<p>We are not quite finished because we were asked for the mass, and so we multiply the molar quantity by the molar mass of glucose.....</p>
<p><mathjax>#1*molxx180.16*g*mol^-1=180.16*g#</mathjax>.</p>
<p>Capisce?</p></div>
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<h1 class="questionTitle" itemprop="name">In one mole of glucose, how many oxygen atoms are present?</h1>
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<div class="markdown"><p>In glucose, <mathjax>#C_6H_12O_6#</mathjax>.....I gets <mathjax>#180.16*g#</mathjax> of glucose.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of oxygen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#6xx"moles of glucose"#</mathjax>. Why? Because CLEARLY, each mole of glucose, <mathjax>#C_6H_12O_6#</mathjax> contains <mathjax>#6*mol#</mathjax> oxygen atoms. </p>
<p>Note that I am labouring the point that we deal with <strong>OXYGEN ATOMS</strong> , <strong>NOT OXYGEN MOLECULES</strong>.</p>
<p><mathjax>#96*g#</mathjax> oxygen is a molar quantity of <mathjax>#(96*g)/(16*g*mol^-1)=6*mol#</mathjax> with respect to oxygen atoms.</p>
<p>And of course if there are <mathjax>#6*mol#</mathjax> oxygen atoms, there must be <mathjax>#1*mol#</mathjax> glucose. Agreed?</p>
<p>We are not quite finished because we were asked for the mass, and so we multiply the molar quantity by the molar mass of glucose.....</p>
<p><mathjax>#1*molxx180.16*g*mol^-1=180.16*g#</mathjax>.</p>
<p>Capisce?</p></div>
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</article> | In one mole of glucose, how many oxygen atoms are present? | null |
847 | ac343792-6ddd-11ea-b41d-ccda262736ce | https://socratic.org/questions/for-the-reaction-represented-by-the-equation-cl-2-2kbr-2kcl-br-2-how-many-grams- | 187.94 grams | start physical_unit 20 21 mass g qc_end chemical_equation 7 15 qc_end physical_unit 30 30 26 27 mass qc_end physical_unit 32 33 26 27 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium chloride [IN] grams"}] | [{"type":"physical unit","value":"187.94 grams"}] | [{"type":"chemical equation","value":"Cl2 + 2 KBr -> 2 KCl + Br2"},{"type":"physical unit","value":"Mass [OF] chlorine [=] \\pu{300 g}"},{"type":"physical unit","value":"Mass [OF] potassium bromide [=] \\pu{300 g}"}] | <h1 class="questionTitle" itemprop="name">For the reaction represented by the equation #Cl_2 + 2KBr -> 2KCl + Br_2#, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?</h1> | null | 187.94 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a limiting reactant problem so we need to see which will produce the least amount of moles of potassium chloride then use that to find how many grams.</p>
<p><mathjax>#300.0 g Cl_2(1 mol Cl_2) /(70.906 g) = 4.231 mol Cl_2(2 mol KCl)/(1 mol Cl_2)=8.462 mol KCl#</mathjax></p>
<p><mathjax>#300.0 g KBr(1 mol KBr ) /(119.002 g) = 2.521 mol KBr(2 mol KCl)/(2 mol KBr)=2.521 mol KCl#</mathjax></p>
<p>The limiting reactant is KBr from this we will determine the amount of potassium chloride.</p>
<p><mathjax>#2.521 mol KCl(74.5513 g/(mol)) = 187.941 g #</mathjax> </p>
<p>To the correct number of sig figs this is rounded to 188 g</p></div>
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<div class="markdown"><p>188 g of Potassium Chloride</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a limiting reactant problem so we need to see which will produce the least amount of moles of potassium chloride then use that to find how many grams.</p>
<p><mathjax>#300.0 g Cl_2(1 mol Cl_2) /(70.906 g) = 4.231 mol Cl_2(2 mol KCl)/(1 mol Cl_2)=8.462 mol KCl#</mathjax></p>
<p><mathjax>#300.0 g KBr(1 mol KBr ) /(119.002 g) = 2.521 mol KBr(2 mol KCl)/(2 mol KBr)=2.521 mol KCl#</mathjax></p>
<p>The limiting reactant is KBr from this we will determine the amount of potassium chloride.</p>
<p><mathjax>#2.521 mol KCl(74.5513 g/(mol)) = 187.941 g #</mathjax> </p>
<p>To the correct number of sig figs this is rounded to 188 g</p></div>
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<h1 class="questionTitle" itemprop="name">For the reaction represented by the equation #Cl_2 + 2KBr -> 2KCl + Br_2#, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?</h1>
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<div class="markdown"><p>188 g of Potassium Chloride</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a limiting reactant problem so we need to see which will produce the least amount of moles of potassium chloride then use that to find how many grams.</p>
<p><mathjax>#300.0 g Cl_2(1 mol Cl_2) /(70.906 g) = 4.231 mol Cl_2(2 mol KCl)/(1 mol Cl_2)=8.462 mol KCl#</mathjax></p>
<p><mathjax>#300.0 g KBr(1 mol KBr ) /(119.002 g) = 2.521 mol KBr(2 mol KCl)/(2 mol KBr)=2.521 mol KCl#</mathjax></p>
<p>The limiting reactant is KBr from this we will determine the amount of potassium chloride.</p>
<p><mathjax>#2.521 mol KCl(74.5513 g/(mol)) = 187.941 g #</mathjax> </p>
<p>To the correct number of sig figs this is rounded to 188 g</p></div>
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</article> | For the reaction represented by the equation #Cl_2 + 2KBr -> 2KCl + Br_2#, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide? | null |
848 | ab41d0cc-6ddd-11ea-b44b-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-32-39-percent-sodium-2 | Na2SO4 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Na2SO4"}] | [{"type":"physical unit","value":"Percent [OF] sodium in the compound [=] \\pu{32.39 percent}"},{"type":"physical unit","value":"Percent [OF] sulfur in the compound [=] \\pu{22.53 percent }"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{45.07 percent }"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?</h1> | null | Na2SO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, convert percentages into grams. </p>
<p><mathjax># 32.39 % = 32.39 g#</mathjax></p>
<p><mathjax>#22.53% = 22.53 g#</mathjax></p>
<p><mathjax>#45.07% = 45.07 g#</mathjax></p>
<p>Second, convert the grams into moles. </p>
<p><mathjax>#"32.39 g"/("22.99 g/mol Na")= "1.409 mol Na"#</mathjax></p>
<p><mathjax>#"22.53 g"/"32.06 g/mol S" = "0.7027 mol S"#</mathjax></p>
<p><mathjax>#"45.07 g"/"16.00 g/mol O" = "2.817 mol O"#</mathjax></p>
<p>Third, convert the moles into a ratio of atoms. </p>
<p><mathjax>#"2.817 mol O"/"0.7027 mol S" = "4.009 O"/"1 S"#</mathjax></p>
<p><mathjax>#"1.409 mol Na"/"0.7027 mol S" = "2.005 Na"/"1 S"#</mathjax></p>
<p>Rounding to integers, the ratio is 1 S: 2 Na : 4 O </p>
<p><mathjax>#Na_2 S O_4#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Na_2SO_4#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, convert percentages into grams. </p>
<p><mathjax># 32.39 % = 32.39 g#</mathjax></p>
<p><mathjax>#22.53% = 22.53 g#</mathjax></p>
<p><mathjax>#45.07% = 45.07 g#</mathjax></p>
<p>Second, convert the grams into moles. </p>
<p><mathjax>#"32.39 g"/("22.99 g/mol Na")= "1.409 mol Na"#</mathjax></p>
<p><mathjax>#"22.53 g"/"32.06 g/mol S" = "0.7027 mol S"#</mathjax></p>
<p><mathjax>#"45.07 g"/"16.00 g/mol O" = "2.817 mol O"#</mathjax></p>
<p>Third, convert the moles into a ratio of atoms. </p>
<p><mathjax>#"2.817 mol O"/"0.7027 mol S" = "4.009 O"/"1 S"#</mathjax></p>
<p><mathjax>#"1.409 mol Na"/"0.7027 mol S" = "2.005 Na"/"1 S"#</mathjax></p>
<p>Rounding to integers, the ratio is 1 S: 2 Na : 4 O </p>
<p><mathjax>#Na_2 S O_4#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?</h1>
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Ernest Z.
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<div class="markdown"><p><mathjax>#Na_2SO_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, convert percentages into grams. </p>
<p><mathjax># 32.39 % = 32.39 g#</mathjax></p>
<p><mathjax>#22.53% = 22.53 g#</mathjax></p>
<p><mathjax>#45.07% = 45.07 g#</mathjax></p>
<p>Second, convert the grams into moles. </p>
<p><mathjax>#"32.39 g"/("22.99 g/mol Na")= "1.409 mol Na"#</mathjax></p>
<p><mathjax>#"22.53 g"/"32.06 g/mol S" = "0.7027 mol S"#</mathjax></p>
<p><mathjax>#"45.07 g"/"16.00 g/mol O" = "2.817 mol O"#</mathjax></p>
<p>Third, convert the moles into a ratio of atoms. </p>
<p><mathjax>#"2.817 mol O"/"0.7027 mol S" = "4.009 O"/"1 S"#</mathjax></p>
<p><mathjax>#"1.409 mol Na"/"0.7027 mol S" = "2.005 Na"/"1 S"#</mathjax></p>
<p>Rounding to integers, the ratio is 1 S: 2 Na : 4 O </p>
<p><mathjax>#Na_2 S O_4#</mathjax></p></div>
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</article> | What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen? | null |
849 | aca96f40-6ddd-11ea-ae16-ccda262736ce | https://socratic.org/questions/a-sample-of-gas-has-a-volume-of-250-c-m-3-at-28-5-c-and-95-9-kpa-what-volume-wil | 0.22 L | start physical_unit 19 20 volume l qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 11 12 temperature qc_end physical_unit 1 3 14 15 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.22 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{250 cm^3}"},{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{28.5 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{95.9 kPa}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 250 c#m^3# at 28.5 °C and 95.9 kPa. What volume will the gas occupy at STP?</h1> | null | 0.22 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p>Use the equation from 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>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="95.9 kPa"#</mathjax><br/>
<mathjax>#V_1="250 cm"^3"=250 mL"xx(1 "L")/(1000 "mL")="0.250 L"#</mathjax><br/>
<mathjax>#T_1="28.5"^"o""C"+273.15="301.7 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#"V"_2=((95.9 "kPa") * (0.250"L") * (273.15"K"))/((301.7"K")*(100"kPa"))="0.217 L"#</mathjax></p></div>
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<div class="markdown"><p>The volume of the gas at STP will be <mathjax>#"0.217 L"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p>Use the equation from 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>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="95.9 kPa"#</mathjax><br/>
<mathjax>#V_1="250 cm"^3"=250 mL"xx(1 "L")/(1000 "mL")="0.250 L"#</mathjax><br/>
<mathjax>#T_1="28.5"^"o""C"+273.15="301.7 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#"V"_2=((95.9 "kPa") * (0.250"L") * (273.15"K"))/((301.7"K")*(100"kPa"))="0.217 L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A sample of gas has a volume of 250 c#m^3# at 28.5 °C and 95.9 kPa. What volume will the gas occupy at STP?</h1>
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<div class="markdown"><p>The volume of the gas at STP will be <mathjax>#"0.217 L"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"STP=273.15 K and 100 kPa"#</mathjax></p>
<p>Use the equation from 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>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P_1="95.9 kPa"#</mathjax><br/>
<mathjax>#V_1="250 cm"^3"=250 mL"xx(1 "L")/(1000 "mL")="0.250 L"#</mathjax><br/>
<mathjax>#T_1="28.5"^"o""C"+273.15="301.7 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax> and solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#"V"_2=((95.9 "kPa") * (0.250"L") * (273.15"K"))/((301.7"K")*(100"kPa"))="0.217 L"#</mathjax></p></div>
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</article> | A sample of gas has a volume of 250 c#m^3# at 28.5 °C and 95.9 kPa. What volume will the gas occupy at STP? | null |
850 | abf13e86-6ddd-11ea-8009-ccda262736ce | https://socratic.org/questions/when-56-6-g-of-calcium-is-reacted-with-nitrogen-gas-32-4-g-of-calcium-nitride-is | 46.60% | start physical_unit 13 14 percent_yield none qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 13 14 10 11 mass qc_end chemical_equation 28 33 qc_end end | [{"type":"physical unit","value":"Percent yield [OF] calcium nitride"}] | [{"type":"physical unit","value":"46.60%"}] | [{"type":"physical unit","value":"Mass [OF] calcium [=] \\pu{56.6 g}"},{"type":"physical unit","value":"Mass [OF] calcium nitride [=] \\pu{32.4 g}"},{"type":"chemical equation","value":"3 Ca(s) + N2(g) -> Ca3N2(s)"}] | <h1 class="questionTitle" itemprop="name">When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is produced. What is the percent yield of calcium nitride in the reaction #3Ca(s) + N_2(g) -> Ca_3N_2(s)#?</h1> | null | 46.60% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#3Ca(s) + N_2(g) rarr Ca_3N_2(s)#</mathjax></p>
<p><mathjax>#"Moles of calcium metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(56.6*g)/(40.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol#</mathjax>.</p>
<p><mathjax>#"Moles of calcium nitride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.4*g)/(148.25*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax>.</p>
<p>Given the equation, <mathjax>#0.470*mol#</mathjax> calcium nitride would be obtained if the yield were quantitative.</p>
<p><mathjax>#"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.219*mol)/(0.470*mol)xx100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"% yield"#</mathjax> <mathjax>#<#</mathjax> <mathjax>#50%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#3Ca(s) + N_2(g) rarr Ca_3N_2(s)#</mathjax></p>
<p><mathjax>#"Moles of calcium metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(56.6*g)/(40.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol#</mathjax>.</p>
<p><mathjax>#"Moles of calcium nitride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.4*g)/(148.25*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax>.</p>
<p>Given the equation, <mathjax>#0.470*mol#</mathjax> calcium nitride would be obtained if the yield were quantitative.</p>
<p><mathjax>#"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.219*mol)/(0.470*mol)xx100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is produced. What is the percent yield of calcium nitride in the reaction #3Ca(s) + N_2(g) -> Ca_3N_2(s)#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"% yield"#</mathjax> <mathjax>#<#</mathjax> <mathjax>#50%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#3Ca(s) + N_2(g) rarr Ca_3N_2(s)#</mathjax></p>
<p><mathjax>#"Moles of calcium metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(56.6*g)/(40.08*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol#</mathjax>.</p>
<p><mathjax>#"Moles of calcium nitride"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(32.4*g)/(148.25*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax>.</p>
<p>Given the equation, <mathjax>#0.470*mol#</mathjax> calcium nitride would be obtained if the yield were quantitative.</p>
<p><mathjax>#"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.219*mol)/(0.470*mol)xx100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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</article> | When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is produced. What is the percent yield of calcium nitride in the reaction #3Ca(s) + N_2(g) -> Ca_3N_2(s)#? | null |
851 | abc87914-6ddd-11ea-8d56-ccda262736ce | https://socratic.org/questions/what-volume-will-a-balloon-occupy-at-1-0-atm-if-the-balloon-has-a-volume-of-7-6- | 28.88 L | start physical_unit 10 11 volume l qc_end physical_unit 10 11 7 8 pressure qc_end physical_unit 10 11 16 17 volume qc_end physical_unit 10 11 19 20 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] L"}] | [{"type":"physical unit","value":"28.88 L"}] | [{"type":"physical unit","value":"Pressure2 [OF] the balloon [=] \\pu{1.0 atm}"},{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{7.6 L}"},{"type":"physical unit","value":"Pressure1 [OF] the balloon [=] \\pu{3.8 atm}"}] | <h1 class="questionTitle" itemprop="name">What volume will a balloon occupy at 1.0 atm, if the balloon has a volume of 7.6 L at 3.8 atm? </h1> | null | 28.88 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant -- this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>In other words, when temperature and number of moles of gas are kept constant, <strong>increasing</strong> the pressure will cause the volume to <strong>decrease</strong>; similarly, <strong>decreasing</strong> the pressure will cause the volume to <strong>increase</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)color(black)(P_1 * V_1 = P_2 * V_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> are the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> are the pressure and volume of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = P_2 * V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (3.8 color(red)(cancel(color(black)("atm"))))/(1.0color(red)(cancel(color(black)("atm")))) * "7.6 L" = color(green)(bar(ul(|color(white)(a/a)color(black)("29 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>As you can see, <em>decreasing</em> the pressure of the gas caused its volume to <strong>increase</strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"29 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant -- this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>In other words, when temperature and number of moles of gas are kept constant, <strong>increasing</strong> the pressure will cause the volume to <strong>decrease</strong>; similarly, <strong>decreasing</strong> the pressure will cause the volume to <strong>increase</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)color(black)(P_1 * V_1 = P_2 * V_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> are the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> are the pressure and volume of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = P_2 * V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (3.8 color(red)(cancel(color(black)("atm"))))/(1.0color(red)(cancel(color(black)("atm")))) * "7.6 L" = color(green)(bar(ul(|color(white)(a/a)color(black)("29 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>As you can see, <em>decreasing</em> the pressure of the gas caused its volume to <strong>increase</strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume will a balloon occupy at 1.0 atm, if the balloon has a volume of 7.6 L at 3.8 atm? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-10-29T00:02:30" itemprop="dateCreated">
Oct 29, 2016
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<div class="markdown"><p><mathjax>#"29 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember here is that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when temperature and number of moles of gas are kept constant -- this is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong>.</p>
<p>In other words, when temperature and number of moles of gas are kept constant, <strong>increasing</strong> the pressure will cause the volume to <strong>decrease</strong>; similarly, <strong>decreasing</strong> the pressure will cause the volume to <strong>increase</strong>. </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)color(black)(P_1 * V_1 = P_2 * V_2)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> are the pressure and volume of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> are the pressure and volume of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#P_1 * V_1 = P_2 * V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#V_2 = (3.8 color(red)(cancel(color(black)("atm"))))/(1.0color(red)(cancel(color(black)("atm")))) * "7.6 L" = color(green)(bar(ul(|color(white)(a/a)color(black)("29 L")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p>
<p>As you can see, <em>decreasing</em> the pressure of the gas caused its volume to <strong>increase</strong>.</p></div>
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</article> | What volume will a balloon occupy at 1.0 atm, if the balloon has a volume of 7.6 L at 3.8 atm? | null |
852 | aa36d446-6ddd-11ea-9137-ccda262736ce | https://socratic.org/questions/a-40-g-block-of-ice-is-cooled-to-76-c-and-is-then-added-to-640-g-of-water-in-an- | 19.54 ℃ | start physical_unit 38 39 temperature °c qc_end physical_unit 5 5 1 2 mass qc_end physical_unit 5 5 9 10 temperature qc_end physical_unit 19 19 16 17 mass qc_end physical_unit 24 25 22 23 mass qc_end physical_unit 38 39 30 31 temperature qc_end physical_unit 5 5 54 55 temperature qc_end c_other OTHER qc_end physical_unit 24 24 87 90 specific_heat qc_end physical_unit 24 24 83 85 temperature qc_end physical_unit 5 5 69 72 specific_heat qc_end end | [{"type":"physical unit","value":"Temperature [OF] the system [IN] ℃"}] | [{"type":"physical unit","value":"19.54 ℃"}] | [{"type":"physical unit","value":"Mass [OF] ice [=] \\pu{40 g}"},{"type":"physical unit","value":"Temperature1 [OF] ice [=] \\pu{-76 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{640 g}"},{"type":"physical unit","value":"Mass [OF] copper calorimeter [=] \\pu{80 g}"},{"type":"physical unit","value":"Temperature1 [OF] the system [=] \\pu{28 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] ice [=] \\pu{0 ℃}"},{"type":"other","value":"0.500 cal/(g * ℃) = 2090 J/(kg * ℃). "},{"type":"physical unit","value":"Specific heat [OF] copper [=] \\pu{0.9201731 calorie/(g * ℃)}"},{"type":"physical unit","value":"Temperature [OF] copper [=] \\pu{25 degree celcius}"},{"type":"physical unit","value":"Specific heat [OF] ice [=] \\pu{0.500 cal/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">A 40 g block of ice is cooled to -76°C. and is then added to 640 g of water in an 80 g copper calorimeter at a temperature of 28°C. What is the final temperature of the system consisting of the ice, water, and calorimeter?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. specific heat of copper at 25 degree celcius is 0.9201731 calorie/g C</p></div>
</h2>
</div>
</div> | 19.54 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that the heat <strong>absorbed</strong> by the ice will be <strong>equal</strong> to the heat <strong>given off</strong> by the liquid water and the calorimeter. </p>
<blockquote>
<p><mathjax>#color(blue)(q_"gained" = - q_"given off")#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(q_"ice" = -(q_"water" + q_"copper")" " " "color(orange)("(*)")#</mathjax></p>
</blockquote>
<p>The <em>minus sign</em> is used here because heat <strong>given off</strong> carries a minus sign. </p>
<p>Now, the trick is to realize that your sample of ice will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#-76^@"C"#</mathjax> <em>to <strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>The sample of liquid water will go from</p>
<blockquote>
<ul>
<li><em><strong>liquid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax> </li>
</ul>
</blockquote>
<p>Similarly, the calorimeter will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>solid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>Now, the equation that you can use looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Let's start with the heat <strong>absorbed</strong> by the ice. In order to heat the ice from solid at <mathjax>#-76^@"C"#</mathjax> to solid at <mathjax>#0^@"C"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#q_"ice 1" = m_"ice" * c_"ice" * DeltaT_"ice 1"#</mathjax></p>
</blockquote>
<p>Convert the <em><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of ice from <em>joules per kilogram Celsius</em> to <em>joules per gram Celsius</em> first</p>
<blockquote>
<p><mathjax>#2090 "J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = "2.090 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#q_"ice 1" = 40 color(red)(cancel(color(black)("g"))) * "2.090 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-76)]color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 1" = "6353.6 J"#</mathjax></p>
</blockquote>
<p>In order to calculate the heat absorbed when you go from <em>solid</em> ice at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> water at <mathjax>#0^@"C"#</mathjax>, you must use water's <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, <mathjax>#DeltaH_"fus"#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J g"^(-1)#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>You can thus say that the heat needed to convert the solid ice to liquid water at its melting point, <mathjax>#q_"ice 2"#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "13,342 J"#</mathjax></p>
</blockquote>
<p>Next, calculate the heat needed to warm the water from liquid at <mathjax>#0^@"C"#</mathjax> to liquid at the final temperature <mathjax>#T_f#</mathjax> by using water's <strong>specific heat</strong></p>
<blockquote>
<p><mathjax>#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#q_"ice 3" = 40 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 3" = (167.4 * T_f)" J"#</mathjax></p>
</blockquote>
<p>Now focus on the heat <strong>given off</strong> by the water that was initially present in the calorimeter. You will have</p>
<blockquote>
<p><mathjax>#q_"water" = m_"water" * c_"water" * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#q_"water" = 640 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = 2677.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Finally, calculate the heat <strong>given off</strong> by the calorimeter itself. Make sure that you convert the specific heat of copper from <em>calories per gram Celsius</em> to <em>joules per gram Celsius</em></p>
<blockquote>
<p><mathjax>#0.09201731 color(red)(cancel(color(black)("cal")))/("g" ""^@"C") * "4.184 J"/(1color(red)(cancel(color(black)("cal")))) = "0.385 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>The heat given off by the calorimeter will thus be </p>
<blockquote>
<p><mathjax>#q_"copper" = m_"copper" * c_"copper" * DeltaT_"copper"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q_"copper" = 80 color(red)(cancel(color(black)("g"))) * "0.385 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28) color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"copper" = 30.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Now it's time to put all this together using equation <mathjax>#color(orange)("(*)")#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#overbrace((q_"ice 1" + q_"ice 2" + q_"ice 3"))^(color(blue)("total heat absorbed by the ice")) = "6353.6 J" + "13,342 J" + (164.7 * T_f)" J"#</mathjax></p>
<p><mathjax>#q_"ice total" = (19696 + 164.7 * T_f)" J"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#(19696 + 164.7 * T_f) color(red)(cancel(color(black)("J"))) = -[2677.8 * (T_f - 28) + 30.8 * (T_f - 28)]color(red)(cancel(color(black)("J")))#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#T_f#</mathjax>, the final temperature of the ice + water + calorimeter system</p>
<blockquote>
<p><mathjax>#19696 + 164.7 * T_f = -2677.8 * T_ f + 74978.4 - 30.8 * T_f + 862.4#</mathjax></p>
<p><mathjax>#2873.3 * T_f = 56144.8 implies T_f = 55144.8/2873.3 = 19.5402#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the final temperature of the system will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(T_f = 19.5^@"C")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that your values only justify one sig fig for the final temperature.. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#19.5^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that the heat <strong>absorbed</strong> by the ice will be <strong>equal</strong> to the heat <strong>given off</strong> by the liquid water and the calorimeter. </p>
<blockquote>
<p><mathjax>#color(blue)(q_"gained" = - q_"given off")#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(q_"ice" = -(q_"water" + q_"copper")" " " "color(orange)("(*)")#</mathjax></p>
</blockquote>
<p>The <em>minus sign</em> is used here because heat <strong>given off</strong> carries a minus sign. </p>
<p>Now, the trick is to realize that your sample of ice will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#-76^@"C"#</mathjax> <em>to <strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>The sample of liquid water will go from</p>
<blockquote>
<ul>
<li><em><strong>liquid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax> </li>
</ul>
</blockquote>
<p>Similarly, the calorimeter will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>solid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>Now, the equation that you can use looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Let's start with the heat <strong>absorbed</strong> by the ice. In order to heat the ice from solid at <mathjax>#-76^@"C"#</mathjax> to solid at <mathjax>#0^@"C"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#q_"ice 1" = m_"ice" * c_"ice" * DeltaT_"ice 1"#</mathjax></p>
</blockquote>
<p>Convert the <em><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of ice from <em>joules per kilogram Celsius</em> to <em>joules per gram Celsius</em> first</p>
<blockquote>
<p><mathjax>#2090 "J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = "2.090 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#q_"ice 1" = 40 color(red)(cancel(color(black)("g"))) * "2.090 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-76)]color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 1" = "6353.6 J"#</mathjax></p>
</blockquote>
<p>In order to calculate the heat absorbed when you go from <em>solid</em> ice at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> water at <mathjax>#0^@"C"#</mathjax>, you must use water's <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, <mathjax>#DeltaH_"fus"#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J g"^(-1)#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>You can thus say that the heat needed to convert the solid ice to liquid water at its melting point, <mathjax>#q_"ice 2"#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "13,342 J"#</mathjax></p>
</blockquote>
<p>Next, calculate the heat needed to warm the water from liquid at <mathjax>#0^@"C"#</mathjax> to liquid at the final temperature <mathjax>#T_f#</mathjax> by using water's <strong>specific heat</strong></p>
<blockquote>
<p><mathjax>#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#q_"ice 3" = 40 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 3" = (167.4 * T_f)" J"#</mathjax></p>
</blockquote>
<p>Now focus on the heat <strong>given off</strong> by the water that was initially present in the calorimeter. You will have</p>
<blockquote>
<p><mathjax>#q_"water" = m_"water" * c_"water" * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#q_"water" = 640 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = 2677.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Finally, calculate the heat <strong>given off</strong> by the calorimeter itself. Make sure that you convert the specific heat of copper from <em>calories per gram Celsius</em> to <em>joules per gram Celsius</em></p>
<blockquote>
<p><mathjax>#0.09201731 color(red)(cancel(color(black)("cal")))/("g" ""^@"C") * "4.184 J"/(1color(red)(cancel(color(black)("cal")))) = "0.385 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>The heat given off by the calorimeter will thus be </p>
<blockquote>
<p><mathjax>#q_"copper" = m_"copper" * c_"copper" * DeltaT_"copper"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q_"copper" = 80 color(red)(cancel(color(black)("g"))) * "0.385 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28) color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"copper" = 30.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Now it's time to put all this together using equation <mathjax>#color(orange)("(*)")#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#overbrace((q_"ice 1" + q_"ice 2" + q_"ice 3"))^(color(blue)("total heat absorbed by the ice")) = "6353.6 J" + "13,342 J" + (164.7 * T_f)" J"#</mathjax></p>
<p><mathjax>#q_"ice total" = (19696 + 164.7 * T_f)" J"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#(19696 + 164.7 * T_f) color(red)(cancel(color(black)("J"))) = -[2677.8 * (T_f - 28) + 30.8 * (T_f - 28)]color(red)(cancel(color(black)("J")))#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#T_f#</mathjax>, the final temperature of the ice + water + calorimeter system</p>
<blockquote>
<p><mathjax>#19696 + 164.7 * T_f = -2677.8 * T_ f + 74978.4 - 30.8 * T_f + 862.4#</mathjax></p>
<p><mathjax>#2873.3 * T_f = 56144.8 implies T_f = 55144.8/2873.3 = 19.5402#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the final temperature of the system will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(T_f = 19.5^@"C")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that your values only justify one sig fig for the final temperature.. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 40 g block of ice is cooled to -76°C. and is then added to 640 g of water in an 80 g copper calorimeter at a temperature of 28°C. What is the final temperature of the system consisting of the ice, water, and calorimeter?</h1>
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<div class="markdown"><p>Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. specific heat of copper at 25 degree celcius is 0.9201731 calorie/g C</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-08-21T11:31:54" itemprop="dateCreated">
Aug 21, 2016
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<div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#19.5^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>!! LONG ANSWER !!</strong></p>
<p>The idea here is that the heat <strong>absorbed</strong> by the ice will be <strong>equal</strong> to the heat <strong>given off</strong> by the liquid water and the calorimeter. </p>
<blockquote>
<p><mathjax>#color(blue)(q_"gained" = - q_"given off")#</mathjax></p>
</blockquote>
<p>This is equivalent to</p>
<blockquote>
<p><mathjax>#color(blue)(q_"ice" = -(q_"water" + q_"copper")" " " "color(orange)("(*)")#</mathjax></p>
</blockquote>
<p>The <em>minus sign</em> is used here because heat <strong>given off</strong> carries a minus sign. </p>
<p>Now, the trick is to realize that your sample of ice will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#-76^@"C"#</mathjax> <em>to <strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>solid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax></li>
<li><em><strong>liquid</strong> at</em> <mathjax>#0^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>The sample of liquid water will go from</p>
<blockquote>
<ul>
<li><em><strong>liquid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>liquid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax> </li>
</ul>
</blockquote>
<p>Similarly, the calorimeter will go from</p>
<blockquote>
<ul>
<li><em><strong>solid</strong> at</em> <mathjax>#28^@"C"#</mathjax> <em>to <strong>solid</strong> at a final temperature</em> <mathjax>#T_f#</mathjax></li>
</ul>
</blockquote>
<p>Now, the equation that you can use looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Let's start with the heat <strong>absorbed</strong> by the ice. In order to heat the ice from solid at <mathjax>#-76^@"C"#</mathjax> to solid at <mathjax>#0^@"C"#</mathjax>, you must provide it with </p>
<blockquote>
<p><mathjax>#q_"ice 1" = m_"ice" * c_"ice" * DeltaT_"ice 1"#</mathjax></p>
</blockquote>
<p>Convert the <em><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></em> of ice from <em>joules per kilogram Celsius</em> to <em>joules per gram Celsius</em> first</p>
<blockquote>
<p><mathjax>#2090 "J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = "2.090 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#q_"ice 1" = 40 color(red)(cancel(color(black)("g"))) * "2.090 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-76)]color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 1" = "6353.6 J"#</mathjax></p>
</blockquote>
<p>In order to calculate the heat absorbed when you go from <em>solid</em> ice at <mathjax>#0^@"C"#</mathjax> to <em>liquid</em> water at <mathjax>#0^@"C"#</mathjax>, you must use water's <strong><a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of fusion</strong>, <mathjax>#DeltaH_"fus"#</mathjax>, which is equal to </p>
<blockquote>
<p><mathjax>#DeltaH_"fus" = "333.55 J g"^(-1)#</mathjax></p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Enthalpy_of_fusion" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Enthalpy_of_fusion</a></p>
<p>You can thus say that the heat needed to convert the solid ice to liquid water at its melting point, <mathjax>#q_"ice 2"#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#40 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "13,342 J"#</mathjax></p>
</blockquote>
<p>Next, calculate the heat needed to warm the water from liquid at <mathjax>#0^@"C"#</mathjax> to liquid at the final temperature <mathjax>#T_f#</mathjax> by using water's <strong>specific heat</strong></p>
<blockquote>
<p><mathjax>#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#q_"ice 3" = 40 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 0)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"ice 3" = (167.4 * T_f)" J"#</mathjax></p>
</blockquote>
<p>Now focus on the heat <strong>given off</strong> by the water that was initially present in the calorimeter. You will have</p>
<blockquote>
<p><mathjax>#q_"water" = m_"water" * c_"water" * DeltaT_"water"#</mathjax></p>
</blockquote>
<p>Plug in your values to get </p>
<blockquote>
<p><mathjax>#q_"water" = 640 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = 2677.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Finally, calculate the heat <strong>given off</strong> by the calorimeter itself. Make sure that you convert the specific heat of copper from <em>calories per gram Celsius</em> to <em>joules per gram Celsius</em></p>
<blockquote>
<p><mathjax>#0.09201731 color(red)(cancel(color(black)("cal")))/("g" ""^@"C") * "4.184 J"/(1color(red)(cancel(color(black)("cal")))) = "0.385 J g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>The heat given off by the calorimeter will thus be </p>
<blockquote>
<p><mathjax>#q_"copper" = m_"copper" * c_"copper" * DeltaT_"copper"#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#q_"copper" = 80 color(red)(cancel(color(black)("g"))) * "0.385 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28) color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"copper" = 30.8 * (T_f - 28)" J"#</mathjax></p>
</blockquote>
<p>Now it's time to put all this together using equation <mathjax>#color(orange)("(*)")#</mathjax>. You will have</p>
<blockquote>
<p><mathjax>#overbrace((q_"ice 1" + q_"ice 2" + q_"ice 3"))^(color(blue)("total heat absorbed by the ice")) = "6353.6 J" + "13,342 J" + (164.7 * T_f)" J"#</mathjax></p>
<p><mathjax>#q_"ice total" = (19696 + 164.7 * T_f)" J"#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#(19696 + 164.7 * T_f) color(red)(cancel(color(black)("J"))) = -[2677.8 * (T_f - 28) + 30.8 * (T_f - 28)]color(red)(cancel(color(black)("J")))#</mathjax></p>
</blockquote>
<p>All you have to do now is solve for <mathjax>#T_f#</mathjax>, the final temperature of the ice + water + calorimeter system</p>
<blockquote>
<p><mathjax>#19696 + 164.7 * T_f = -2677.8 * T_ f + 74978.4 - 30.8 * T_f + 862.4#</mathjax></p>
<p><mathjax>#2873.3 * T_f = 56144.8 implies T_f = 55144.8/2873.3 = 19.5402#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the final temperature of the system will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(T_f = 19.5^@"C")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that your values only justify one sig fig for the final temperature.. </p></div>
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</article> | A 40 g block of ice is cooled to -76°C. and is then added to 640 g of water in an 80 g copper calorimeter at a temperature of 28°C. What is the final temperature of the system consisting of the ice, water, and calorimeter? |
Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. specific heat of copper at 25 degree celcius is 0.9201731 calorie/g C
|
853 | a86d4193-6ddd-11ea-9c20-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-that-describes-the-dissociation-of-the-st | Mg(OH)2(s) <=> Mg^2+(aq) + 2OH-(aq) | start chemical_equation qc_end chemical_equation 16 16 qc_end substance 18 18 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the dissociation"}] | [{"type":"chemical equation","value":"Mg(OH)2(s) <=> Mg^2+(aq) + 2OH-(aq)"}] | [{"type":"chemical equation","value":"Mg(OH)2"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation that describes the dissociation of the strong base magnesium hydroxide, Mg(OH)2 in water?</h1> | null | Mg(OH)2(s) <=> Mg^2+(aq) + 2OH-(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax># "Mg(OH)"_2#</mathjax> is a <a href="http://socratic.org/questions/what-is-a-strong-base">strong base</a>.</p>
<p><mathjax>#"Mg"#</mathjax> is a Group 2 Metal, so it form <mathjax>#"Mg"^"2+"#</mathjax> ions. Each ion pairs up with two <mathjax>#"OH"^"-"#</mathjax> ions to form the neutral compound, <mathjax># "Mg(OH)"_2#</mathjax>.</p>
<p>Magnesium hydroxide is "insoluble", so only a small amount it goes into solution. But every bit that goes into solution dissociates into <mathjax>#"Mg"^"2+"#</mathjax> ions and <mathjax>#"OH"^"-"#</mathjax> ions.</p>
<blockquote></blockquote>
<p>There is a <a href="http://socratic.org/chemistry/phases-of-matter/dynamic-equilibrium-of-phase-changes">dynamic equilibrium</a> between the solid compound and the ions in solution.</p>
<p><mathjax>#"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#</mathjax></p>
<p>The <mathjax>#"Mg"^"2+"#</mathjax> ions and the <mathjax>#"OH"^"-"#</mathjax> ions are leaving the surface of the solid at the same rate as they return from the solution to the surface of the solid.</p>
<p>Here's a video showing the process.</p>
<p>
<iframe src="https://www.youtube.com/embed/Vgw1fYBYW3g?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The equation is <mathjax>#"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax># "Mg(OH)"_2#</mathjax> is a <a href="http://socratic.org/questions/what-is-a-strong-base">strong base</a>.</p>
<p><mathjax>#"Mg"#</mathjax> is a Group 2 Metal, so it form <mathjax>#"Mg"^"2+"#</mathjax> ions. Each ion pairs up with two <mathjax>#"OH"^"-"#</mathjax> ions to form the neutral compound, <mathjax># "Mg(OH)"_2#</mathjax>.</p>
<p>Magnesium hydroxide is "insoluble", so only a small amount it goes into solution. But every bit that goes into solution dissociates into <mathjax>#"Mg"^"2+"#</mathjax> ions and <mathjax>#"OH"^"-"#</mathjax> ions.</p>
<blockquote></blockquote>
<p>There is a <a href="http://socratic.org/chemistry/phases-of-matter/dynamic-equilibrium-of-phase-changes">dynamic equilibrium</a> between the solid compound and the ions in solution.</p>
<p><mathjax>#"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#</mathjax></p>
<p>The <mathjax>#"Mg"^"2+"#</mathjax> ions and the <mathjax>#"OH"^"-"#</mathjax> ions are leaving the surface of the solid at the same rate as they return from the solution to the surface of the solid.</p>
<p>Here's a video showing the process.</p>
<p>
<iframe src="https://www.youtube.com/embed/Vgw1fYBYW3g?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation that describes the dissociation of the strong base magnesium hydroxide, Mg(OH)2 in water?</h1>
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Ernest Z.
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Jul 14, 2014
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<div class="markdown"><p>The equation is <mathjax>#"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax># "Mg(OH)"_2#</mathjax> is a <a href="http://socratic.org/questions/what-is-a-strong-base">strong base</a>.</p>
<p><mathjax>#"Mg"#</mathjax> is a Group 2 Metal, so it form <mathjax>#"Mg"^"2+"#</mathjax> ions. Each ion pairs up with two <mathjax>#"OH"^"-"#</mathjax> ions to form the neutral compound, <mathjax># "Mg(OH)"_2#</mathjax>.</p>
<p>Magnesium hydroxide is "insoluble", so only a small amount it goes into solution. But every bit that goes into solution dissociates into <mathjax>#"Mg"^"2+"#</mathjax> ions and <mathjax>#"OH"^"-"#</mathjax> ions.</p>
<blockquote></blockquote>
<p>There is a <a href="http://socratic.org/chemistry/phases-of-matter/dynamic-equilibrium-of-phase-changes">dynamic equilibrium</a> between the solid compound and the ions in solution.</p>
<p><mathjax>#"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"#</mathjax></p>
<p>The <mathjax>#"Mg"^"2+"#</mathjax> ions and the <mathjax>#"OH"^"-"#</mathjax> ions are leaving the surface of the solid at the same rate as they return from the solution to the surface of the solid.</p>
<p>Here's a video showing the process.</p>
<p>
<iframe src="https://www.youtube.com/embed/Vgw1fYBYW3g?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | What is the balanced chemical equation that describes the dissociation of the strong base magnesium hydroxide, Mg(OH)2 in water? | null |
854 | ac3262e4-6ddd-11ea-81e8-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-molar-solubility-in-grams-100ml-of-calcium-iodate-in-water- | 0.25 grams/100 mL | start physical_unit 10 11 molar_solubility g/100_ml qc_end physical_unit 10 11 20 22 equilibrium_constant_k qc_end substance 13 13 qc_end end | [{"type":"physical unit","value":"Molar solubility [OF] calcium iodate [IN] grams/100 mL"}] | [{"type":"physical unit","value":"0.25 grams/100 mL"}] | [{"type":"physical unit","value":"Temperature [OF] calcium iodate solution [=] \\pu{25 degrees Celsius}"},{"type":"physical unit","value":"Ksp [OF] calcium iodate [=] \\pu{7.1 × 10^(-7)}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = #7.1 x 10^-7#</h1> | null | 0.25 grams/100 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We write out the dissolution reaction:</p>
<p><mathjax>#Ca(IO_3)_2(s) rightleftharpoonsCa^(2+) + 2IO_3^-#</mathjax></p>
<p>And then we write out the equilibrium expression, in which <mathjax>#Ca(IO_3)_2#</mathjax>, as a solid, does not appear:</p>
<p><mathjax>#K_(sp)=[Ca^(2+)][IO_3^-]^2=7.1xx10^-7#</mathjax></p>
<p>If we dub the solubility of calcium iodate under these conditions as <mathjax>#S#</mathjax>, then <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>. </p>
<p>This expression follows the equilibrium equation: each equiv of salt that dissolves gives 1 equiv of <mathjax>#Ca^(2+)#</mathjax>, but 2 equiv iodate ion.</p>
<p>Thus <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7.1xx10^-7#</mathjax></p>
<p><mathjax>#S=""^3sqrt{(7.1xx10^-7}/4)#</mathjax></p>
<p>This gives an answer in <mathjax>#mol*L^-1#</mathjax>. You will have to use the formula mass of calcium iodate, <mathjax>#389.88*g*mol^-1#</mathjax>, to give an answer in the units required. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"Solubility of calcium iodate "~=0.25*g*10^-1L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We write out the dissolution reaction:</p>
<p><mathjax>#Ca(IO_3)_2(s) rightleftharpoonsCa^(2+) + 2IO_3^-#</mathjax></p>
<p>And then we write out the equilibrium expression, in which <mathjax>#Ca(IO_3)_2#</mathjax>, as a solid, does not appear:</p>
<p><mathjax>#K_(sp)=[Ca^(2+)][IO_3^-]^2=7.1xx10^-7#</mathjax></p>
<p>If we dub the solubility of calcium iodate under these conditions as <mathjax>#S#</mathjax>, then <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>. </p>
<p>This expression follows the equilibrium equation: each equiv of salt that dissolves gives 1 equiv of <mathjax>#Ca^(2+)#</mathjax>, but 2 equiv iodate ion.</p>
<p>Thus <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7.1xx10^-7#</mathjax></p>
<p><mathjax>#S=""^3sqrt{(7.1xx10^-7}/4)#</mathjax></p>
<p>This gives an answer in <mathjax>#mol*L^-1#</mathjax>. You will have to use the formula mass of calcium iodate, <mathjax>#389.88*g*mol^-1#</mathjax>, to give an answer in the units required. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = #7.1 x 10^-7#</h1>
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<div class="markdown"><p><mathjax>#"Solubility of calcium iodate "~=0.25*g*10^-1L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We write out the dissolution reaction:</p>
<p><mathjax>#Ca(IO_3)_2(s) rightleftharpoonsCa^(2+) + 2IO_3^-#</mathjax></p>
<p>And then we write out the equilibrium expression, in which <mathjax>#Ca(IO_3)_2#</mathjax>, as a solid, does not appear:</p>
<p><mathjax>#K_(sp)=[Ca^(2+)][IO_3^-]^2=7.1xx10^-7#</mathjax></p>
<p>If we dub the solubility of calcium iodate under these conditions as <mathjax>#S#</mathjax>, then <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax>. </p>
<p>This expression follows the equilibrium equation: each equiv of salt that dissolves gives 1 equiv of <mathjax>#Ca^(2+)#</mathjax>, but 2 equiv iodate ion.</p>
<p>Thus <mathjax>#K_(sp)=(S)(2S)^2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4S^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#7.1xx10^-7#</mathjax></p>
<p><mathjax>#S=""^3sqrt{(7.1xx10^-7}/4)#</mathjax></p>
<p>This gives an answer in <mathjax>#mol*L^-1#</mathjax>. You will have to use the formula mass of calcium iodate, <mathjax>#389.88*g*mol^-1#</mathjax>, to give an answer in the units required. </p></div>
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</article> | How do you calculate Molar Solubility in grams/100mL of Calcium iodate in water at 25 degrees Celsius? Ksp = #7.1 x 10^-7# | null |
855 | a8652c10-6ddd-11ea-962d-ccda262736ce | https://socratic.org/questions/how-many-gold-atoms-are-contained-in-0-650-grams-of-gold | 1.99 × 10^21 | start physical_unit 2 3 number none qc_end physical_unit 2 2 7 8 mass qc_end end | [{"type":"physical unit","value":"Number [OF] gold atoms"}] | [{"type":"physical unit","value":"1.99 × 10^21"}] | [{"type":"physical unit","value":"Mass [OF] gold [=] \\pu{0.650 grams}"}] | <h1 class="questionTitle" itemprop="name"> How many gold atoms are contained in 0.650 grams of gold?</h1> | null | 1.99 × 10^21 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Calculate the moles of gold by dividing the given mass by its molar mass, <mathjax>#"196.966569 g/mol"#</mathjax> (atomic weight on periodic table in g/mol). </p>
<p><mathjax>#(0.650cancel"g Au")/(196.966569 cancel"g"/"mol" "Au")="0.00330 mol Au atoms"#</mathjax></p>
<p><mathjax>#1 "mol atoms"=6.022xx10^23 "atoms"#</mathjax></p>
<p>Multiply the calculated mol Au times <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#0.00330cancel"mol Au"xx(6.022xx10^23"atoms")/(1cancel"mol")=1.99xx10^21 "atoms Au"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"0.650 g Au"#</mathjax> contain <mathjax>#1.99xx10^21"atoms"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Calculate the moles of gold by dividing the given mass by its molar mass, <mathjax>#"196.966569 g/mol"#</mathjax> (atomic weight on periodic table in g/mol). </p>
<p><mathjax>#(0.650cancel"g Au")/(196.966569 cancel"g"/"mol" "Au")="0.00330 mol Au atoms"#</mathjax></p>
<p><mathjax>#1 "mol atoms"=6.022xx10^23 "atoms"#</mathjax></p>
<p>Multiply the calculated mol Au times <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#0.00330cancel"mol Au"xx(6.022xx10^23"atoms")/(1cancel"mol")=1.99xx10^21 "atoms Au"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name"> How many gold atoms are contained in 0.650 grams of gold?</h1>
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<div class="markdown"><p><mathjax>#"0.650 g Au"#</mathjax> contain <mathjax>#1.99xx10^21"atoms"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Calculate the moles of gold by dividing the given mass by its molar mass, <mathjax>#"196.966569 g/mol"#</mathjax> (atomic weight on periodic table in g/mol). </p>
<p><mathjax>#(0.650cancel"g Au")/(196.966569 cancel"g"/"mol" "Au")="0.00330 mol Au atoms"#</mathjax></p>
<p><mathjax>#1 "mol atoms"=6.022xx10^23 "atoms"#</mathjax></p>
<p>Multiply the calculated mol Au times <mathjax>#(6.022xx10^23"atoms")/(1"mol")#</mathjax>.</p>
<p><mathjax>#0.00330cancel"mol Au"xx(6.022xx10^23"atoms")/(1cancel"mol")=1.99xx10^21 "atoms Au"#</mathjax></p></div>
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</article> | How many gold atoms are contained in 0.650 grams of gold? | null |
856 | ac999934-6ddd-11ea-b4fc-ccda262736ce | https://socratic.org/questions/115-ml-of-a-2-35-m-potassium-fluoride-solution-is-diluted-with-1-28l-of-water-wh | 0.19 molarity | start physical_unit 6 8 concentration mol/l qc_end physical_unit 6 8 0 1 volume qc_end physical_unit 6 8 4 5 concentration qc_end physical_unit 15 15 12 13 volume qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] potassium fluoride solution [IN] molarity"}] | [{"type":"physical unit","value":"0.19 molarity"}] | [{"type":"physical unit","value":"Volume1 [OF] potassium fluoride solution [=] \\pu{115 mL}"},{"type":"physical unit","value":"Concentration1 [OF] potassium fluoride solution [=] \\pu{2.35 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{1.28 L}"}] | <h1 class="questionTitle" itemprop="name">115 mL of a 2.35 M potassium fluoride solution is diluted with 1.28L of water. What is the new concentration in molarity? </h1> | null | 0.19 molarity | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#(0.115*Lxx2.35*mol*L^-1)/(0.115*L+1.28*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.270*mol)/(1.395*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.194*mol*L^-1#</mathjax>.</p>
<p>Please note that the wording of the problem said that the initial <mathjax>#0.115*L#</mathjax> volume was diluted with <mathjax>#1.28*L#</mathjax> FRESH <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">SOLVENT</a>. Had it said <mathjax>#"diluted to 1.28"*L#</mathjax>, the volume would change accordingly. Do you see why I make the distinction?</p>
<p>Such a solution would be slightly basic. Why?</p></div>
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<div class="markdown"><p><mathjax>#[KF]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.194*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#(0.115*Lxx2.35*mol*L^-1)/(0.115*L+1.28*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.270*mol)/(1.395*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.194*mol*L^-1#</mathjax>.</p>
<p>Please note that the wording of the problem said that the initial <mathjax>#0.115*L#</mathjax> volume was diluted with <mathjax>#1.28*L#</mathjax> FRESH <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">SOLVENT</a>. Had it said <mathjax>#"diluted to 1.28"*L#</mathjax>, the volume would change accordingly. Do you see why I make the distinction?</p>
<p>Such a solution would be slightly basic. Why?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">115 mL of a 2.35 M potassium fluoride solution is diluted with 1.28L of water. What is the new concentration in molarity? </h1>
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<div class="markdown"><p><mathjax>#[KF]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.194*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#(0.115*Lxx2.35*mol*L^-1)/(0.115*L+1.28*L)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(0.270*mol)/(1.395*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.194*mol*L^-1#</mathjax>.</p>
<p>Please note that the wording of the problem said that the initial <mathjax>#0.115*L#</mathjax> volume was diluted with <mathjax>#1.28*L#</mathjax> FRESH <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">SOLVENT</a>. Had it said <mathjax>#"diluted to 1.28"*L#</mathjax>, the volume would change accordingly. Do you see why I make the distinction?</p>
<p>Such a solution would be slightly basic. Why?</p></div>
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</article> | 115 mL of a 2.35 M potassium fluoride solution is diluted with 1.28L of water. What is the new concentration in molarity? | null |
857 | acab6ad5-6ddd-11ea-a0c8-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-for-1-propanol-and-sodium-1 | C3H7OH(l) + Na(s) -> C3H7O- + 1/2 H2(g) | start chemical_equation qc_end substance 7 7 qc_end substance 9 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced chemical equation"}] | [{"type":"chemical equation","value":"C3H7OH(l) + Na(s) -> C3H7O- + 1/2 H2(g)"}] | [{"type":"substance name","value":"1-propanol"},{"type":"substance name","value":"Sodium"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for 1-propanol and sodium? </h1> | null | C3H7OH(l) + Na(s) -> C3H7O- + 1/2 H2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Interestingly, for both methanol and ethanol, this reaction would rapidly go to completion. The reaction with propanol would be much more sluggish. Why?</p>
<p>This is precisely the same type of reaction as when sodium reacts (much more rapidly) with water. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_3C-CH_2CH_2OH(l) + Na(s) rarr#</mathjax> <mathjax>#H_3C-CH_2CH_2O^(-)Na^(+) + 1/2H_2(g)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Interestingly, for both methanol and ethanol, this reaction would rapidly go to completion. The reaction with propanol would be much more sluggish. Why?</p>
<p>This is precisely the same type of reaction as when sodium reacts (much more rapidly) with water. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for 1-propanol and sodium? </h1>
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anor277
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<div class="markdown"><p><mathjax>#H_3C-CH_2CH_2OH(l) + Na(s) rarr#</mathjax> <mathjax>#H_3C-CH_2CH_2O^(-)Na^(+) + 1/2H_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Interestingly, for both methanol and ethanol, this reaction would rapidly go to completion. The reaction with propanol would be much more sluggish. Why?</p>
<p>This is precisely the same type of reaction as when sodium reacts (much more rapidly) with water. </p></div>
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</article> | What is the balanced chemical equation for 1-propanol and sodium? | null |
858 | aa895ddf-6ddd-11ea-a66b-ccda262736ce | https://socratic.org/questions/for-the-decomposition-reaction-2ag-2co-3-4ag-2co-2-o-2-how-many-moles-of-reactan | 3.00 moles | start physical_unit 18 18 mole mol qc_end physical_unit 8 8 25 26 mole qc_end chemical_equation 4 13 qc_end end | [{"type":"physical unit","value":"Mole [OF] reactant [IN] moles"}] | [{"type":"physical unit","value":"3.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] Ag [=] \\pu{6.0 moles}"},{"type":"chemical equation","value":"2 Ag2CO3 -> 4 Ag + 2 CO2 + O2"}] | <h1 class="questionTitle" itemprop="name">For the decomposition reaction #2Ag_2CO_3->4Ag+2CO_2+O_2#, how many moles of reactant undergo decomposition in order to produce 6.0 moles of #Ag#?</h1> | null | 3.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above correctly represents thermal decomposition of a metal carbonate. The carbonate gives metal oxide and carbon dioxide under STRONG heating. You are not going to get silver metal unless you get it all into solution and add some reductant (maybe copper). Of course, you could do this in a forge, but it would be very difficult to smelt; and you'd still have to add a reductant.</p>
<p>Instead, we could make silver nitrate:</p>
<p><mathjax>#Ag_2O + 2HNO_3(aq) rarr 2AgNO_3(aq) + H_2O(l)#</mathjax></p>
<p>Silver nitrate would be relatively easy to reduce. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Ag_2CO_3 + Delta rarr Ag_2O + CO_2(g)uarr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above correctly represents thermal decomposition of a metal carbonate. The carbonate gives metal oxide and carbon dioxide under STRONG heating. You are not going to get silver metal unless you get it all into solution and add some reductant (maybe copper). Of course, you could do this in a forge, but it would be very difficult to smelt; and you'd still have to add a reductant.</p>
<p>Instead, we could make silver nitrate:</p>
<p><mathjax>#Ag_2O + 2HNO_3(aq) rarr 2AgNO_3(aq) + H_2O(l)#</mathjax></p>
<p>Silver nitrate would be relatively easy to reduce. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">For the decomposition reaction #2Ag_2CO_3->4Ag+2CO_2+O_2#, how many moles of reactant undergo decomposition in order to produce 6.0 moles of #Ag#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Ag_2CO_3 + Delta rarr Ag_2O + CO_2(g)uarr#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction above correctly represents thermal decomposition of a metal carbonate. The carbonate gives metal oxide and carbon dioxide under STRONG heating. You are not going to get silver metal unless you get it all into solution and add some reductant (maybe copper). Of course, you could do this in a forge, but it would be very difficult to smelt; and you'd still have to add a reductant.</p>
<p>Instead, we could make silver nitrate:</p>
<p><mathjax>#Ag_2O + 2HNO_3(aq) rarr 2AgNO_3(aq) + H_2O(l)#</mathjax></p>
<p>Silver nitrate would be relatively easy to reduce. </p></div>
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<div class="markdown"><p>2/4 = x/6<br/>
Solve for 'x'.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From your balanced reaction equation, possible or not, you take the coefficients of the molecules as their molar ratios. Therefore, your reaction shows that 2 moles of silver carbonate decompose into 4 moles of silver, 2 moles of carbon dioxide, and one mole of oxygen.</p>
<p>To find the ratio you want - moles of silver carbonate to produce 6 moles of silver, you just use a ratio equivalence with these values.<br/>
2/4 = x/6<br/>
Solve for 'x'.</p></div>
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</article> | For the decomposition reaction #2Ag_2CO_3->4Ag+2CO_2+O_2#, how many moles of reactant undergo decomposition in order to produce 6.0 moles of #Ag#? | null |
859 | a94c7e62-6ddd-11ea-beb7-ccda262736ce | https://socratic.org/questions/if-1g-sample-of-limestone-was-dissolved-in-acid-to-give-0-38-g-if-co2-if-limesto | 86.4% | start physical_unit 28 31 percent none qc_end chemical_equation 32 41 qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 15 15 12 13 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Percentage [OF] CaCO3 in the limestone"}] | [{"type":"physical unit","value":"86.4%"}] | [{"type":"chemical equation","value":"CaCO3(s) + 2 HCl(aq) -> CO2(g) + CaCl2(aq) + H2O(l)"},{"type":"physical unit","value":"Mass [OF] limestone sample [=] \\pu{1 g}"},{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{0.38 g}"},{"type":"other","value":"Limestone contains no carbonate other than CaCO3."}] | <h1 class="questionTitle" itemprop="name">If 1g sample of limestone was dissolved in acid to give 0.38 g of #"CO"_2#. If limestone contains no carbonate other than #"CaCO"_3#, find the percentage of #"CaCO"_3# in the limestone?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"CaCO"_3(s) + 2"HCl"(aq) -> "CO"2(g)+ "CaCl"_2(aq) + "H"_2"O"(l)#</mathjax></p></div>
</h2>
</div>
</div> | 86.4% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that the problem provides you with <em>grams</em> of carbon dioxide, but that the balanced chemical equation uses <strong>moles</strong>, so right from the start, you know that you must convert grams to moles <em>or vice versa</em>.</p>
<blockquote>
<p><mathjax>#"CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#</mathjax></p>
</blockquote>
<p>You know that <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate reacts to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. Use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert this mole ratio to a <strong>gram ratio</strong>.</p>
<p>You know that</p>
<blockquote>
<p><mathjax>#M_ ("M CaCO"_ 3) = "100.1 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_ ("M CO"_ 2) = "44.01 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that when <mathjax>#"100.1 g"#</mathjax> of calcium carbonate take part in the reaction, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate, the reaction produces <mathjax>#"44.01 g"#</mathjax> of carbon dioxide, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>You thus have</p>
<blockquote>
<p><mathjax>#overbrace("1 mole CaCO"_3/"1 mole CO"_2)^(color(blue)("mole ratio")) = overbrace("100.1 g CaCO"_3/"44.01 g CO"_2)^(color(purple)("gram ratio"))#</mathjax></p>
</blockquote>
<p>Use the gram ratio to determine how many grams of calcium carbonate were needed to produce <mathjax>#"0.38 g"#</mathjax> of carbon dioxide</p>
<blockquote>
<p><mathjax>#0.38 color(red)(cancel(color(black)("g CO"_2))) * "100.1 g CaCO"_3/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "0.864 g CaCO"_3#</mathjax></p>
</blockquote>
<p>To find the percentage of calcium carbonate in the sample of limestone, simply divide the mass of calcium carbonate by the total mass of the limestone and multiply the result by <mathjax>#100%#</mathjax></p>
<blockquote>
<p><mathjax>#"% CaCO"_3 = (0.864 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(86%)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of limestone. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"86% CaCO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that the problem provides you with <em>grams</em> of carbon dioxide, but that the balanced chemical equation uses <strong>moles</strong>, so right from the start, you know that you must convert grams to moles <em>or vice versa</em>.</p>
<blockquote>
<p><mathjax>#"CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#</mathjax></p>
</blockquote>
<p>You know that <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate reacts to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. Use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert this mole ratio to a <strong>gram ratio</strong>.</p>
<p>You know that</p>
<blockquote>
<p><mathjax>#M_ ("M CaCO"_ 3) = "100.1 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_ ("M CO"_ 2) = "44.01 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that when <mathjax>#"100.1 g"#</mathjax> of calcium carbonate take part in the reaction, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate, the reaction produces <mathjax>#"44.01 g"#</mathjax> of carbon dioxide, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>You thus have</p>
<blockquote>
<p><mathjax>#overbrace("1 mole CaCO"_3/"1 mole CO"_2)^(color(blue)("mole ratio")) = overbrace("100.1 g CaCO"_3/"44.01 g CO"_2)^(color(purple)("gram ratio"))#</mathjax></p>
</blockquote>
<p>Use the gram ratio to determine how many grams of calcium carbonate were needed to produce <mathjax>#"0.38 g"#</mathjax> of carbon dioxide</p>
<blockquote>
<p><mathjax>#0.38 color(red)(cancel(color(black)("g CO"_2))) * "100.1 g CaCO"_3/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "0.864 g CaCO"_3#</mathjax></p>
</blockquote>
<p>To find the percentage of calcium carbonate in the sample of limestone, simply divide the mass of calcium carbonate by the total mass of the limestone and multiply the result by <mathjax>#100%#</mathjax></p>
<blockquote>
<p><mathjax>#"% CaCO"_3 = (0.864 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(86%)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of limestone. </p></div>
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<h1 class="questionTitle" itemprop="name">If 1g sample of limestone was dissolved in acid to give 0.38 g of #"CO"_2#. If limestone contains no carbonate other than #"CaCO"_3#, find the percentage of #"CaCO"_3# in the limestone?</h1>
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<div class="markdown"><p><mathjax>#"CaCO"_3(s) + 2"HCl"(aq) -> "CO"2(g)+ "CaCl"_2(aq) + "H"_2"O"(l)#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2017-04-25T15:08:52" itemprop="dateCreated">
Apr 25, 2017
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<div class="markdown"><p><mathjax>#"86% CaCO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that the problem provides you with <em>grams</em> of carbon dioxide, but that the balanced chemical equation uses <strong>moles</strong>, so right from the start, you know that you must convert grams to moles <em>or vice versa</em>.</p>
<blockquote>
<p><mathjax>#"CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr#</mathjax></p>
</blockquote>
<p>You know that <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate reacts to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. Use the <strong>molar masses</strong> of the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> to convert this mole ratio to a <strong>gram ratio</strong>.</p>
<p>You know that</p>
<blockquote>
<p><mathjax>#M_ ("M CaCO"_ 3) = "100.1 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_ ("M CO"_ 2) = "44.01 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>This means that when <mathjax>#"100.1 g"#</mathjax> of calcium carbonate take part in the reaction, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of calcium carbonate, the reaction produces <mathjax>#"44.01 g"#</mathjax> of carbon dioxide, the equivalent of <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>You thus have</p>
<blockquote>
<p><mathjax>#overbrace("1 mole CaCO"_3/"1 mole CO"_2)^(color(blue)("mole ratio")) = overbrace("100.1 g CaCO"_3/"44.01 g CO"_2)^(color(purple)("gram ratio"))#</mathjax></p>
</blockquote>
<p>Use the gram ratio to determine how many grams of calcium carbonate were needed to produce <mathjax>#"0.38 g"#</mathjax> of carbon dioxide</p>
<blockquote>
<p><mathjax>#0.38 color(red)(cancel(color(black)("g CO"_2))) * "100.1 g CaCO"_3/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "0.864 g CaCO"_3#</mathjax></p>
</blockquote>
<p>To find the percentage of calcium carbonate in the sample of limestone, simply divide the mass of calcium carbonate by the total mass of the limestone and multiply the result by <mathjax>#100%#</mathjax></p>
<blockquote>
<p><mathjax>#"% CaCO"_3 = (0.864 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("g")))) xx 100% = color(darkgreen)(ul(color(black)(86%)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of limestone. </p></div>
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</article> | If 1g sample of limestone was dissolved in acid to give 0.38 g of #"CO"_2#. If limestone contains no carbonate other than #"CaCO"_3#, find the percentage of #"CaCO"_3# in the limestone? |
#"CaCO"_3(s) + 2"HCl"(aq) -> "CO"2(g)+ "CaCl"_2(aq) + "H"_2"O"(l)#
|
860 | a912e564-6ddd-11ea-a62a-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-400-g-of-cuso-4-in-4-00-l-of-solution | 0.63 M | start physical_unit 8 8 molarity mol/l qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 13 13 10 11 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] CuSO4 solution [IN] M"}] | [{"type":"physical unit","value":"0.63 M"}] | [{"type":"physical unit","value":"Mass [OF] CuSO4 [=] \\pu{400 g}"},{"type":"physical unit","value":"Volume [OF] CuSO4 solution [=] \\pu{4.00 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of 400 g of #CuSO_4# in 4.00 L of solution?</h1> | null | 0.63 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular weight of <mathjax>#CuSO_"4"#</mathjax> is 159.609 g/mol, so 1 mole of <mathjax>#CuSO_"4"#</mathjax> weighs 159.609 g.</p>
<p>You have 400 g of <mathjax>#CuSO_"4"#</mathjax>, hence you have:</p>
<p><mathjax>#"400 g" / "159.609 g/mol" = 2.5061243 " moles"#</mathjax></p>
<p>This 2.5061243 moles is in 4 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is moles per liter, so:</p>
<p><mathjax>#"2.5061243 moles" / "4 liters" = 0.627 " Molar"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>0.627 M</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular weight of <mathjax>#CuSO_"4"#</mathjax> is 159.609 g/mol, so 1 mole of <mathjax>#CuSO_"4"#</mathjax> weighs 159.609 g.</p>
<p>You have 400 g of <mathjax>#CuSO_"4"#</mathjax>, hence you have:</p>
<p><mathjax>#"400 g" / "159.609 g/mol" = 2.5061243 " moles"#</mathjax></p>
<p>This 2.5061243 moles is in 4 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is moles per liter, so:</p>
<p><mathjax>#"2.5061243 moles" / "4 liters" = 0.627 " Molar"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of 400 g of #CuSO_4# in 4.00 L of solution?</h1>
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<div class="markdown"><p>0.627 M</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molecular weight of <mathjax>#CuSO_"4"#</mathjax> is 159.609 g/mol, so 1 mole of <mathjax>#CuSO_"4"#</mathjax> weighs 159.609 g.</p>
<p>You have 400 g of <mathjax>#CuSO_"4"#</mathjax>, hence you have:</p>
<p><mathjax>#"400 g" / "159.609 g/mol" = 2.5061243 " moles"#</mathjax></p>
<p>This 2.5061243 moles is in 4 liters.</p>
<p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is moles per liter, so:</p>
<p><mathjax>#"2.5061243 moles" / "4 liters" = 0.627 " Molar"#</mathjax></p></div>
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</article> | What is the molarity of 400 g of #CuSO_4# in 4.00 L of solution? | null |
861 | a8951fee-6ddd-11ea-9e3e-ccda262736ce | https://socratic.org/questions/581165ac7c014952676d91d4 | 4.91 L | start physical_unit 23 24 volume l qc_end physical_unit 23 24 6 7 volume qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"4.91 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{4.5 L}"},{"type":"other","value":"STP"},{"type":"other","value":"SATP"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water?</h1> | null | 4.91 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH~=13#</mathjax></p></div>
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<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water?</h1>
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<div class="markdown"><p><mathjax>#pH~=13#</mathjax></p></div>
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<div class="markdown"><p>We need to find <mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of hydroxide"/"Volume of water"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.20*mol*L^-1#</mathjax>.</p>
<p>Now <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(0.20)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-0.70)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax>.</p>
<p>But we know (or should know) that in water at <mathjax>#298*K#</mathjax>, <mathjax>#pH+pOH=14#</mathjax>.</p>
<p>Thus <mathjax>#pH=14-0.70=?????????#</mathjax></p></div>
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</article> | What is the #pH# of an aqueous solution prepared from a #600*mg# mass of #NaOH# dissolved in #75*mL# of water? | null |
862 | aa1563f1-6ddd-11ea-bfb7-ccda262736ce | https://socratic.org/questions/57e5e4be7c01497cd89c95e0 | H2SO4 + Na2CO3(aq) -> Na2SO4(aq) + CO2(g) ^ + H2O(l) | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"H2SO4 + Na2CO3(aq) -> Na2SO4(aq) + CO2(g) ^ + H2O(l)"}] | [{"type":"substance name","value":"Sulfuric acid"},{"type":"substance name","value":"Sodium carbonate"}] | <h1 class="questionTitle" itemprop="name">How does sulfuric acid react with sodium carbonate?</h1> | null | H2SO4 + Na2CO3(aq) -> Na2SO4(aq) + CO2(g) ^ + H2O(l) | <div class="answerDescription">
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<div class="markdown"><p>Is this balanced? Do not trust my arithmetic!</p></div>
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<div class="markdown"><p><mathjax>#H_2SO_4(aq) + Na_2CO_3(aq) rarr Na_2SO_4(aq) + CO_2(g) uarr+ H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is this balanced? Do not trust my arithmetic!</p></div>
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<div class="markdown"><p><mathjax>#H_2SO_4(aq) + Na_2CO_3(aq) rarr Na_2SO_4(aq) + CO_2(g) uarr+ H_2O(l)#</mathjax></p></div>
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</article> | How does sulfuric acid react with sodium carbonate? | null |
863 | acaa3967-6ddd-11ea-82d5-ccda262736ce | https://socratic.org/questions/578667ce11ef6b289eb5bcf8 | 46.00 amu | start physical_unit 6 6 molecular_weight amu qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] ^14CO2 [IN] amu"}] | [{"type":"physical unit","value":"46.00 amu"}] | [{"type":"chemical equation","value":"^14CO2"}] | <h1 class="questionTitle" itemprop="name">What is the molecular mass of #""^14CO_2#?</h1> | null | 46.00 amu | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Garden variety carbon dioxide has a mass of 44 amu. Label the carbon to give <mathjax>#""^(14)C#</mathjax> and the molecular mass increases by 2 amu. Capisce??</p></div>
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<div class="markdown"><p><mathjax>#14+16+16=??"amu"#</mathjax></p></div>
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<div class="markdown"><p>Garden variety carbon dioxide has a mass of 44 amu. Label the carbon to give <mathjax>#""^(14)C#</mathjax> and the molecular mass increases by 2 amu. Capisce??</p></div>
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<div class="markdown"><p><mathjax>#14+16+16=??"amu"#</mathjax></p></div>
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<div class="markdown"><p>Garden variety carbon dioxide has a mass of 44 amu. Label the carbon to give <mathjax>#""^(14)C#</mathjax> and the molecular mass increases by 2 amu. Capisce??</p></div>
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<div class="markdown"><p>Mass of one molecule of <mathjax>#color(white)(i)^14"CO"_2#</mathjax> <mathjax>#="Mass of one atom of" color(white)(i)^14"C"#</mathjax><mathjax>#+2xx"Average atomic mass of Oxygen atom"#</mathjax><br/>
<mathjax>#=14.003241+2xx15.9994=46.002041#</mathjax> amu</p></div>
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</article> | What is the molecular mass of #""^14CO_2#? | null |
864 | a8e8b4e3-6ddd-11ea-9f1e-ccda262736ce | https://socratic.org/questions/570189d711ef6b4c23382e8b | C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g) | start chemical_equation qc_end c_other OTHER qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g)"}] | [{"type":"other","value":"Complete combustion."},{"type":"substance name","value":"Octane"}] | <h1 class="questionTitle" itemprop="name">How do we represent the complete combustion of octane?</h1> | null | C8H18(g) + 25/2 O2(g) -> 8 CO2(g) + 9 H2O(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_8H_18(g) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p>
<p>Balance in the order: <mathjax>#C#</mathjax>; <mathjax>#H#</mathjax>; <mathjax>#O#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do we represent the complete combustion of octane?</h1>
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anor277
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<div class="markdown"><p><mathjax>#C_8H_18(g) + 25/2O_2(g) rarr 8CO_2(g) + 9H_2O(g)#</mathjax></p>
<p>Balance in the order: <mathjax>#C#</mathjax>; <mathjax>#H#</mathjax>; <mathjax>#O#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the carbons, then the hydrogens, then finally the oxygens. If you like you can double the entire reaction, to remove the non-stoichiometric coefficient on dioxygen. There is no need to do so.</p>
<p>Is this equation balanced? Does garbage in equal garbage out? If you made a cash transaction with a large bank note, would you immediately know whether you had been short-changed or not? I think you would know, and this is the basis of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. You have to try to develop this facility with chemical reactions, which are (MUCH much larger) atomic and molecular transactions rather than cash transactions. </p>
<p>Try it with nonane, <mathjax>#C_9H_20#</mathjax>.</p></div>
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</article> | How do we represent the complete combustion of octane? | null |
865 | a8f138fe-6ddd-11ea-b6f0-ccda262736ce | https://socratic.org/questions/if-11-grams-of-carbon-dioxide-is-produced-how-much-methane-ch4-must-have-reacted | 0.25 moles | start physical_unit 11 11 mole mol qc_end physical_unit 4 5 1 2 mass qc_end chemical_equation 18 26 qc_end end | [{"type":"physical unit","value":"Mole [OF] CH4 [IN] moles"}] | [{"type":"physical unit","value":"0.25 moles"}] | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [=] \\pu{11 grams}"},{"type":"chemical equation","value":"CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l)"}] | <h1 class="questionTitle" itemprop="name">If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#</mathjax></p></div>
</h2>
</div>
</div> | 0.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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<div class="markdown"><p>From the balanced equation determine the ratio of moles of carbon dioxide produced to methane reacted, then from the molar mass of carbon dioxide get the moles of carbon dioxide and multiply by the methane to <mathjax>#"CO"_2#</mathjax> ratio.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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<h1 class="questionTitle" itemprop="name">If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles?</h1>
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<div class="markdown"><p><mathjax>#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#</mathjax></p></div>
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Stefan V.
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<div class="markdown"><p>From the balanced equation determine the ratio of moles of carbon dioxide produced to methane reacted, then from the molar mass of carbon dioxide get the moles of carbon dioxide and multiply by the methane to <mathjax>#"CO"_2#</mathjax> ratio.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The balanced equation shows that <mathjax>#1.0#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with excess oxygen to form <mathjax>#1.0#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. </p>
<p>Calculating the molar mass of <mathjax>#"CO"_2#</mathjax>, we see that <mathjax>#"C"#</mathjax> is <mathjax>#"12 g/mol"#</mathjax> and <mathjax>#"O"#</mathjax> is <mathjax>#"16 g/mol"#</mathjax>, so <mathjax>#"CO"_2#</mathjax> has a <mathjax>#M_M#</mathjax> of <mathjax>#"44 g/mol"#</mathjax>. </p>
<p>Dividing <mathjax>#"11 g"#</mathjax> <mathjax>#"CO"_2#</mathjax> by <mathjax>#"44 g/mol"#</mathjax> <mathjax>#"CO"_2#</mathjax> gives <mathjax>#0.25#</mathjax> mol <mathjax>#"CO"_2#</mathjax>. The <mathjax>#1:1#</mathjax> ratio of methane to <mathjax>#"CO"_2#</mathjax> means that <mathjax>#0.25#</mathjax> mol <mathjax>#"CH"_4#</mathjax> originally reacted.</p></div>
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Stefan V.
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<div class="markdown"><p>The answer has to be <mathjax>#0.25#</mathjax> mol.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Methane undergoes a combustion reaction as follow:</p>
<p><mathjax>#"CH"_4 +2"O"_2 -> "CO"_2 + 2"H"_2"O"#</mathjax></p>
<p><mathjax>#n_("CO"_2)=m/M_r = "11 g"/((12.01+(2xx16)) \ "g/mol") = "0.25 mol"#</mathjax></p>
<p>According to the balanced equation, <mathjax>#1#</mathjax> mole of <mathjax>#"CH"_4#</mathjax> reacts with <mathjax>#2#</mathjax> moles of <mathjax>#"O"_2#</mathjax> to produce <mathjax>#1#</mathjax> mole of <mathjax>#"CO"_2#</mathjax>. That means</p>
<p><mathjax>#"1 mol" ("CO"_2) = "1 mol" ("CH"_4)#</mathjax></p>
<p>Therefore </p>
<p><mathjax>#n_("CH"_4) = "0.25 mol"#</mathjax></p></div>
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</article> | If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles? |
#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#
|
866 | a85dc2e2-6ddd-11ea-9b22-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-three-moles-of-ethane | 90.21 g | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] ethane [IN] g"}] | [{"type":"physical unit","value":"90.21 g"}] | [{"type":"physical unit","value":"Mole [OF] ethane [=] \\pu{3 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of three moles of ethane?</h1> | null | 90.21 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus the mass of <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3*cancel(mol)xx30.07*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Ethane, "H_3C-CH_3#</mathjax> has a molar mass of <mathjax>#30.07*g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus the mass of <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3*cancel(mol)xx30.07*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Ethane, "H_3C-CH_3#</mathjax> has a molar mass of <mathjax>#30.07*g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus the mass of <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3*cancel(mol)xx30.07*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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</article> | What is the mass of three moles of ethane? | null |
867 | ac3b5418-6ddd-11ea-9138-ccda262736ce | https://socratic.org/questions/how-many-grams-is-5-72-10-21-atoms-of-li | 0.66 grams | start physical_unit 9 9 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] Li [IN] grams"}] | [{"type":"physical unit","value":"0.66 grams"}] | [{"type":"physical unit","value":"Number [OF] Li atoms [=] \\pu{5.72 × 10^21}"}] | <h1 class="questionTitle" itemprop="name">How many grams is #5.72 * 10^21# atoms of #Li#?</h1> | null | 0.66 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>firstly, we have to calculate no. of moles <br/>
i.e. no. of moles = no. of atoms * <mathjax>#N_A(Avagadro's Number)#</mathjax><br/>
no. of moles = <mathjax>#5.72*10^21*6.023*10^23#</mathjax><br/>
no. of moles = <mathjax>#34.45*10^44=3.445*10^45#</mathjax><br/>
mass of Li = molar mass * no. of moles<br/>
=<mathjax>#6*3.445*10^45=2.067*10^46#</mathjax> gm</p></div>
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<div class="markdown"><p><mathjax>#2.067*10^46#</mathjax> gm</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>firstly, we have to calculate no. of moles <br/>
i.e. no. of moles = no. of atoms * <mathjax>#N_A(Avagadro's Number)#</mathjax><br/>
no. of moles = <mathjax>#5.72*10^21*6.023*10^23#</mathjax><br/>
no. of moles = <mathjax>#34.45*10^44=3.445*10^45#</mathjax><br/>
mass of Li = molar mass * no. of moles<br/>
=<mathjax>#6*3.445*10^45=2.067*10^46#</mathjax> gm</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams is #5.72 * 10^21# atoms of #Li#?</h1>
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<div class="markdown"><p><mathjax>#2.067*10^46#</mathjax> gm</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>firstly, we have to calculate no. of moles <br/>
i.e. no. of moles = no. of atoms * <mathjax>#N_A(Avagadro's Number)#</mathjax><br/>
no. of moles = <mathjax>#5.72*10^21*6.023*10^23#</mathjax><br/>
no. of moles = <mathjax>#34.45*10^44=3.445*10^45#</mathjax><br/>
mass of Li = molar mass * no. of moles<br/>
=<mathjax>#6*3.445*10^45=2.067*10^46#</mathjax> gm</p></div>
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</article> | How many grams is #5.72 * 10^21# atoms of #Li#? | null |
868 | a917eeae-6ddd-11ea-a9de-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-joules-of-heat-energy-released-when-20-grams-of-water-is-c | 836 joules | start physical_unit 14 14 heat_energy j qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 14 14 18 19 temperature qc_end physical_unit 14 14 21 22 temperature qc_end end | [{"type":"physical unit","value":"Released heat energy [OF] water [IN] joules"}] | [{"type":"physical unit","value":"836 joules"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{20 grams}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{293 K}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{283 K}"}] | <h1 class="questionTitle" itemprop="name">What is the number of joules of heat energy released when 20 grams of water is cooled from 293 K to 283 K?</h1> | null | 836 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula <mathjax>#q = mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p>The specific heat capacity of water is <mathjax>#4.18 J/g * K#</mathjax>. </p>
<p><mathjax>#q = 20(4.18)(293 - 283)#</mathjax><br/>
<mathjax>#q = 20(4.18)(10)#</mathjax><br/>
<mathjax>#q = 836#</mathjax></p>
<p>836 joules of heat energy are released.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>836 J</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula <mathjax>#q = mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p>The specific heat capacity of water is <mathjax>#4.18 J/g * K#</mathjax>. </p>
<p><mathjax>#q = 20(4.18)(293 - 283)#</mathjax><br/>
<mathjax>#q = 20(4.18)(10)#</mathjax><br/>
<mathjax>#q = 836#</mathjax></p>
<p>836 joules of heat energy are released.</p></div>
</div>
</div>
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<div class="markdown"><p>836 J</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula <mathjax>#q = mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p>The specific heat capacity of water is <mathjax>#4.18 J/g * K#</mathjax>. </p>
<p><mathjax>#q = 20(4.18)(293 - 283)#</mathjax><br/>
<mathjax>#q = 20(4.18)(10)#</mathjax><br/>
<mathjax>#q = 836#</mathjax></p>
<p>836 joules of heat energy are released.</p></div>
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</article> | What is the number of joules of heat energy released when 20 grams of water is cooled from 293 K to 283 K? | null |
869 | ac097f70-6ddd-11ea-8bdd-ccda262736ce | https://socratic.org/questions/how-many-grams-of-potassium-chlorate-decompose-to-potassium-chloride-and-797-ml- | 1.95 grams | start physical_unit 4 5 mass g qc_end physical_unit 14 14 11 12 volume qc_end chemical_equation 22 32 qc_end end | [{"type":"physical unit","value":"Mass [OF] potassium chlorate [IN] grams"}] | [{"type":"physical unit","value":"1.95 grams"}] | [{"type":"physical unit","value":"Volume [OF] O2 [=] \\pu{797 mL}"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{128 degrees Celsius}"},{"type":"physical unit","value":"Pressure [OF] the reaction [=] \\pu{747 torr}"},{"type":"chemical equation","value":"2 KClO3 (s) -> 2 KCl (s) + 3 O2 (g)"}] | <h1 class="questionTitle" itemprop="name">How many grams of potassium chlorate decompose to potassium chloride and 797 mL of O2 at 128 degrees Celsius and 747 torr? 2KClO3 (s) = 2KCl (s) + 3O2 (g)</h1> | null | 1.95 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of <mathjax>#"KClO"_3#</mathjax> that decomposes to give a certain amount of <mathjax>#"O"_2#</mathjax>. </p>
<p>Let's use the ideal gas equation to find the moles of <mathjax>#"O"_2#</mathjax> that form. </p>
<p><mathjax>#P = 747cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 0.983#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#V= 797cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.797#</mathjax> <mathjax>#"L"#</mathjax></p>
<p><mathjax>#T = 128^"o""C" + 273 = 401#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((0.983cancel("atm"))(0.797cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(401cancel("K"))) = 0.0238#</mathjax> <mathjax>#"mol O"_2#</mathjax></p>
<p>Now, we'll use the coefficients of the chemical equation to find the relative number of moles of <mathjax>#"KClO"_3#</mathjax> that reacted:</p>
<p><mathjax>#0.0238cancel("mol O"_2)((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = 0.0159#</mathjax> <mathjax>#"mol KClO"_3#</mathjax></p>
<p>Finally, we'll use the <em>molar mass</em> of potassium chlorate (<mathjax>#122.55#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#0.0159cancel("mol KClO"_3)((122.55color(white)(l)"g KClO"_3)/(1cancel("mol O"_2))) = color(blue)(1.95#</mathjax> <mathjax>#color(blue)("g KClO"_3#</mathjax></p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#1.95#</mathjax> <mathjax>#"g KClO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of <mathjax>#"KClO"_3#</mathjax> that decomposes to give a certain amount of <mathjax>#"O"_2#</mathjax>. </p>
<p>Let's use the ideal gas equation to find the moles of <mathjax>#"O"_2#</mathjax> that form. </p>
<p><mathjax>#P = 747cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 0.983#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#V= 797cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.797#</mathjax> <mathjax>#"L"#</mathjax></p>
<p><mathjax>#T = 128^"o""C" + 273 = 401#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((0.983cancel("atm"))(0.797cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(401cancel("K"))) = 0.0238#</mathjax> <mathjax>#"mol O"_2#</mathjax></p>
<p>Now, we'll use the coefficients of the chemical equation to find the relative number of moles of <mathjax>#"KClO"_3#</mathjax> that reacted:</p>
<p><mathjax>#0.0238cancel("mol O"_2)((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = 0.0159#</mathjax> <mathjax>#"mol KClO"_3#</mathjax></p>
<p>Finally, we'll use the <em>molar mass</em> of potassium chlorate (<mathjax>#122.55#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#0.0159cancel("mol KClO"_3)((122.55color(white)(l)"g KClO"_3)/(1cancel("mol O"_2))) = color(blue)(1.95#</mathjax> <mathjax>#color(blue)("g KClO"_3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of potassium chlorate decompose to potassium chloride and 797 mL of O2 at 128 degrees Celsius and 747 torr? 2KClO3 (s) = 2KCl (s) + 3O2 (g)</h1>
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Nathan L.
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<span class="dateCreated" datetime="2017-07-08T03:28:45" itemprop="dateCreated">
Jul 8, 2017
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<div class="markdown"><p><mathjax>#1.95#</mathjax> <mathjax>#"g KClO"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the mass, in <mathjax>#"g"#</mathjax>, of <mathjax>#"KClO"_3#</mathjax> that decomposes to give a certain amount of <mathjax>#"O"_2#</mathjax>. </p>
<p>Let's use the ideal gas equation to find the moles of <mathjax>#"O"_2#</mathjax> that form. </p>
<p><mathjax>#P = 747cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 0.983#</mathjax> <mathjax>#"atm"#</mathjax></p>
<p><mathjax>#V= 797cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.797#</mathjax> <mathjax>#"L"#</mathjax></p>
<p><mathjax>#T = 128^"o""C" + 273 = 401#</mathjax> <mathjax>#"K"#</mathjax></p>
<p>Plugging in known values, we have</p>
<p><mathjax>#n = (PV)/(RT) = ((0.983cancel("atm"))(0.797cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(401cancel("K"))) = 0.0238#</mathjax> <mathjax>#"mol O"_2#</mathjax></p>
<p>Now, we'll use the coefficients of the chemical equation to find the relative number of moles of <mathjax>#"KClO"_3#</mathjax> that reacted:</p>
<p><mathjax>#0.0238cancel("mol O"_2)((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = 0.0159#</mathjax> <mathjax>#"mol KClO"_3#</mathjax></p>
<p>Finally, we'll use the <em>molar mass</em> of potassium chlorate (<mathjax>#122.55#</mathjax> <mathjax>#"g/mol"#</mathjax>) to find the number of grams that reacted:</p>
<p><mathjax>#0.0159cancel("mol KClO"_3)((122.55color(white)(l)"g KClO"_3)/(1cancel("mol O"_2))) = color(blue)(1.95#</mathjax> <mathjax>#color(blue)("g KClO"_3#</mathjax></p></div>
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</article> | How many grams of potassium chlorate decompose to potassium chloride and 797 mL of O2 at 128 degrees Celsius and 747 torr? 2KClO3 (s) = 2KCl (s) + 3O2 (g) | null |
870 | aa7ded7b-6ddd-11ea-9f79-ccda262736ce | https://socratic.org/questions/how-many-joules-are-needed-to-boil-59-8-g-of-ethanol-if-the-heat-of-vaporization | 51069.20 joules | start physical_unit 10 10 energy j qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 17 18 enthalpy_of_vaporization qc_end end | [{"type":"physical unit","value":"Needed energy [OF] ethanol [IN] joules"}] | [{"type":"physical unit","value":"51069.20 joules"}] | [{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{59.8 g}"},{"type":"physical unit","value":"Heat of vaporization [OF] ethanol [=] \\pu{854 J/g}"}] | <h1 class="questionTitle" itemprop="name">How many joules are needed to boil 59.8 g of ethanol if the heat of vaporization is 854 J/g?</h1> | null | 51069.20 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <strong>heat of vaporization</strong> formula: <mathjax>#q = m * H_v#</mathjax></p>
<p>The heat of vaporization of ethanol is 854 J/g (given in the question), which means that it takes 854 joules of energy to vaporize one gram of ethanol.</p>
<p>Since we have 59.8 grams of ethanol, plug 59.8 into the formula and multiply by 854.</p>
<p><mathjax>#q = (59.8)(854)#</mathjax><br/>
<mathjax>#q = 51069.2 #</mathjax> J</p>
<p>That's your answer!</p></div>
</div>
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<div class="markdown"><p>51,069.2 J</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <strong>heat of vaporization</strong> formula: <mathjax>#q = m * H_v#</mathjax></p>
<p>The heat of vaporization of ethanol is 854 J/g (given in the question), which means that it takes 854 joules of energy to vaporize one gram of ethanol.</p>
<p>Since we have 59.8 grams of ethanol, plug 59.8 into the formula and multiply by 854.</p>
<p><mathjax>#q = (59.8)(854)#</mathjax><br/>
<mathjax>#q = 51069.2 #</mathjax> J</p>
<p>That's your answer!</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many joules are needed to boil 59.8 g of ethanol if the heat of vaporization is 854 J/g?</h1>
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Michelle
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<div class="markdown"><p>51,069.2 J</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <strong>heat of vaporization</strong> formula: <mathjax>#q = m * H_v#</mathjax></p>
<p>The heat of vaporization of ethanol is 854 J/g (given in the question), which means that it takes 854 joules of energy to vaporize one gram of ethanol.</p>
<p>Since we have 59.8 grams of ethanol, plug 59.8 into the formula and multiply by 854.</p>
<p><mathjax>#q = (59.8)(854)#</mathjax><br/>
<mathjax>#q = 51069.2 #</mathjax> J</p>
<p>That's your answer!</p></div>
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</article> | How many joules are needed to boil 59.8 g of ethanol if the heat of vaporization is 854 J/g? | null |
871 | ab3768c8-6ddd-11ea-bbf6-ccda262736ce | https://socratic.org/questions/589818ca7c01495303d2c558 | 3.38 g | start physical_unit 5 6 mass g qc_end physical_unit 5 6 16 17 concentration qc_end end | [{"type":"physical unit","value":"Mass [OF] lactic acid [IN] g"}] | [{"type":"physical unit","value":"3.38 g"}] | [{"type":"physical unit","value":"Volume [OF] lactic acid solution [=] \\pu{1.5 L}"},{"type":"physical unit","value":"Concentration [OF] lactic acid solution [=] \\pu{0.0250 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of lactic acid in a #1.5*L# volume of lactic acid at #0.0250*mol*L^-1# concentration?</h1> | null | 3.38 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hence<mathjax># "mass"/"molar mass"="moles"="volume"xx"concentration"#</mathjax></p>
<p>And thus.........</p>
<p><mathjax>#"mass"="volume"xx"concentration"xx"molar mass"#</mathjax></p>
<p><mathjax>#1.500*cancelLxx0.0250*cancel(mol)*cancel(L^-1)xx74.09*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Is this method clear enuff?</p></div>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles"/"Volume"#</mathjax>. I get a mass of under <mathjax>#3*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hence<mathjax># "mass"/"molar mass"="moles"="volume"xx"concentration"#</mathjax></p>
<p>And thus.........</p>
<p><mathjax>#"mass"="volume"xx"concentration"xx"molar mass"#</mathjax></p>
<p><mathjax>#1.500*cancelLxx0.0250*cancel(mol)*cancel(L^-1)xx74.09*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Is this method clear enuff?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of lactic acid in a #1.5*L# volume of lactic acid at #0.0250*mol*L^-1# concentration?</h1>
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Feb 6, 2017
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<div class="markdown"><p><mathjax>#"Concentration"="Moles"/"Volume"#</mathjax>. I get a mass of under <mathjax>#3*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Hence<mathjax># "mass"/"molar mass"="moles"="volume"xx"concentration"#</mathjax></p>
<p>And thus.........</p>
<p><mathjax>#"mass"="volume"xx"concentration"xx"molar mass"#</mathjax></p>
<p><mathjax>#1.500*cancelLxx0.0250*cancel(mol)*cancel(L^-1)xx74.09*g*cancel(mol^-1)=??g#</mathjax></p>
<p>Is this method clear enuff?</p></div>
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</article> | What is the mass of lactic acid in a #1.5*L# volume of lactic acid at #0.0250*mol*L^-1# concentration? | null |
872 | a9dc6b5a-6ddd-11ea-8e16-ccda262736ce | https://socratic.org/questions/58d851fc11ef6b3bb4d52051 | 1.24 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 9 10 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] water [IN] moles"}] | [{"type":"physical unit","value":"1.24 moles"}] | [{"type":"physical unit","value":"Mass [OF] B2H6 [=] \\pu{17.2 g}"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion?</h1> | null | 1.24 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p><mathjax>#1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion?</h1>
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<div class="markdown"><p><mathjax>#1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#B_2H_6 + O_2 -> HBO_2 + H_2O#</mathjax></p>
<p>Balanced: <mathjax>#B_2H_6 + 3O_2 -> 2HBO_2 + 2H_2O#</mathjax></p>
<p>Molar mass of <mathjax>#B_2H_6 = 2(10.81) + 6(1.008) = 27.668 " g/mol"#</mathjax></p>
<p>Molar mass of <mathjax>#H_2O = 2(1.008) + (15.999) = 18.015 " g/mol"#</mathjax></p>
<p>Moles of <mathjax>#B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"#</mathjax> </p>
<p>From the balanced equation 0.6217 mol of <mathjax>#B_2H_6#</mathjax> produces <mathjax>#2(0.6217) = 1.24 " mol" #</mathjax> of <mathjax>#H_2O#</mathjax></p>
<p><mathjax>#0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O#</mathjax> (rounded to two significant figures)</p></div>
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May 19, 2018
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<div class="markdown"><p>The complete combustion of <mathjax>#"17.2 g B"_2"H"_6"#</mathjax> produces <mathjax>#"2.01 mol H"_2"O"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"B"_2"H"_6("g") + "3O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"B"_2"O"_3("s") + "3H"_2"O(g)"#</mathjax></p>
<p>We first need to determine mol <mathjax>#"B"_2"H"_6"#</mathjax> by dividing its given mass by its molar mass <mathjax>#("25.652 g/mol")#</mathjax>. Divide by multiplying by the reciprocal of its molar mass (mol/g).</p>
<p><mathjax>#17.2color(red)cancel(color(black)("g B"_2"H"_6))xx(1"mol B"_2"H"_6)/(25.652color(red)cancel(color(black)("g B"_2"H"_6)))="0.671 mol B"_2"H"_6"#</mathjax></p>
<p>To determine mol <mathjax>#"H"_2"O"#</mathjax> produced, multiply mol <mathjax>#"B"_2"H"_6"#</mathjax> by the mol ratio between <mathjax>#"H"_2"O"#</mathjax> and <mathjax>#"B"_2"H"_6"#</mathjax> from the balanced equation, with <mathjax>#"H"_2"O"#</mathjax> in the numerator.</p>
<p><mathjax>#0.671color(red)cancel(color(black)("mol B"_2"H"_6))xx(3"mol H"_2"O")/(1color(red)cancel(color(black)("mol B"_2"H"_6)))="2.01 mol H"_2"O"#</mathjax></p>
<p>The complete combustion of <mathjax>#"17.2 g B"_2"H"_6"#</mathjax> produces <mathjax>#"2.01 mol H"_2"O"#</mathjax>.</p></div>
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</article> | How many moles of water can be produced when #"17.2 g B"_2"H"_6"# undergoes complete combustion? | null |
873 | abaa6c54-6ddd-11ea-b431-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-dichloroacetic-acid-if-the-emperical-formula-is | CHCl2COOH | start chemical_formula qc_end c_other OTHER qc_end physical_unit 6 7 20 21 molecular_weight qc_end end | [{"type":"other","value":"Chemical Formula [OF] dichloroacetic acid [IN] molecular"}] | [{"type":"chemical equation","value":"CHCl2COOH"}] | [{"type":"other","value":"The emperical formula dichloroacetic acid is CHOCl."},{"type":"physical unit","value":"Gram molecular mass [OF] dichloroacetic acid [=] \\pu{129.0 g}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of dichloroacetic acid if the emperical formula is CHOCl and the gram molecular mass is 129.0 g?</h1> | null | CHCl2COOH | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, <mathjax>#(CHOCl)xxn#</mathjax> <mathjax>#=#</mathjax> <mathjax>#MF#</mathjax>;</p>
<p>i.e. <mathjax>#(12.011 + 1.00794 + 15.999 + 35.45)xxn*g*mol^-1 = 129.0*g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>. The molecular formula, a multiple of the empirical formula, is <mathjax>#C_2H_2Cl_2O_2#</mathjax>, or rather <mathjax>#Cl_2CH-C(=O)OH#</mathjax>.</p></div>
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<div class="markdown"><p>The molecular formula is always a multiple of the empirical formula. <mathjax>#MF#</mathjax> <mathjax>#=#</mathjax> <mathjax>#nxxEF#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, <mathjax>#(CHOCl)xxn#</mathjax> <mathjax>#=#</mathjax> <mathjax>#MF#</mathjax>;</p>
<p>i.e. <mathjax>#(12.011 + 1.00794 + 15.999 + 35.45)xxn*g*mol^-1 = 129.0*g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>. The molecular formula, a multiple of the empirical formula, is <mathjax>#C_2H_2Cl_2O_2#</mathjax>, or rather <mathjax>#Cl_2CH-C(=O)OH#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of dichloroacetic acid if the emperical formula is CHOCl and the gram molecular mass is 129.0 g?</h1>
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<div class="markdown"><p>The molecular formula is always a multiple of the empirical formula. <mathjax>#MF#</mathjax> <mathjax>#=#</mathjax> <mathjax>#nxxEF#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So, <mathjax>#(CHOCl)xxn#</mathjax> <mathjax>#=#</mathjax> <mathjax>#MF#</mathjax>;</p>
<p>i.e. <mathjax>#(12.011 + 1.00794 + 15.999 + 35.45)xxn*g*mol^-1 = 129.0*g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>. The molecular formula, a multiple of the empirical formula, is <mathjax>#C_2H_2Cl_2O_2#</mathjax>, or rather <mathjax>#Cl_2CH-C(=O)OH#</mathjax>.</p></div>
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<div class="markdown"><p>The molecular formula for dichloroacetic acid is <mathjax>#"C"_2"H"_2"O"_2"Cl"_2"#</mathjax> or <mathjax>#"CHCl"_2"COOH"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The gram molecular mass is a multiple of the empirical formula mass.</p>
<p>Determine the empirical formula mass for <mathjax>#"CHOCl"#</mathjax> by multiplying the subscript of each element times its molar mass (atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol).</p>
<p><mathjax>#(1xx12.0107 "g/mol")+(1xx1.00794"g/mol")+(1xx15.999"g/mol")+(1xx35.453"g/mol")="64.564 g"#</mathjax></p>
<p>To determine the multiplication factor, <mathjax>#n#</mathjax>, divide the gram molecular mass by the empirical formula mass.</p>
<p><mathjax>#n=(129cancel"g/mol")/(64.546cancel"g/mol")=2.00#</mathjax></p>
<p>To determine the molecular formula for dichloroacetic acid, multiply the empirical formula times <mathjax>#2#</mathjax>.</p>
<p>The molecular formula for dichloroacetic acid is <mathjax>#"C"_2"H"_2"O"_2"Cl"_2"#</mathjax> or <mathjax>#"CHCl"_2"COOH"#</mathjax>.</p></div>
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</article> | What is the molecular formula of dichloroacetic acid if the emperical formula is CHOCl and the gram molecular mass is 129.0 g? | null |
874 | acb26ff9-6ddd-11ea-aca9-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-6-0-liters-at-a-pressure-of-380-mm-hg-the-pressure-is-incr | 3.00 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 12 13 pressure qc_end physical_unit 1 1 20 21 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"3.00 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{6.0 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{380 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{760 mmHg}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 6.0 liters at a pressure of 380 mm Hg. The pressure is increased to gas 760 mm Hg at the same temperature. What is its new volume? </h1> | null | 3.00 L | <div class="answerDescription">
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<div class="markdown"><p><mathjax>#V_2-=(P_1V_1)/P_2=((380*mm*Hg)/(760*mm*Hg*atm^-1)xx6.0*L)/((760*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2-=(P_1V_1)/P_2=((380*mm*Hg)/(760*mm*Hg*atm^-1)xx6.0*L)/((760*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 6.0 liters at a pressure of 380 mm Hg. The pressure is increased to gas 760 mm Hg at the same temperature. What is its new volume? </h1>
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_2-=(P_1V_1)/P_2=((380*mm*Hg)/(760*mm*Hg*atm^-1)xx6.0*L)/((760*mm*Hg)/(760*mm*Hg*atm^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</article> | A gas has a volume of 6.0 liters at a pressure of 380 mm Hg. The pressure is increased to gas 760 mm Hg at the same temperature. What is its new volume? | null |
875 | aad203ec-6ddd-11ea-8a61-ccda262736ce | https://socratic.org/questions/the-concentration-of-io-3-ions-in-a-pure-saturated-solution-of-ba-io-3-2-is-1-06 | 5.96 × 10^(-10) | start physical_unit 11 11 equilibrium_constant_k none qc_end physical_unit 3 4 13 16 concentration qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Ksp [OF] Ba(IO3)2"}] | [{"type":"physical unit","value":"5.96 × 10^(-10)"}] | [{"type":"physical unit","value":"Concentration [OF] IO3- ions [=] \\pu{1.06 × 10^(-3) M}"},{"type":"other","value":"Pure, saturated solution of Ba(IO3)2."}] | <h1 class="questionTitle" itemprop="name">The concentration of #"IO"_3^(-)# ions in a pure, saturated solution of #"Ba"("IO"_3)_2# is #1.06xx10^(-3)"M"#. What is the #K_(sp)# for #"Ba"("IO"_3)_2#?
</h1> | null | 5.96 × 10^(-10) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that a <strong>saturated</strong> solution of barium iodate, <mathjax>#"Ba"("IO"_3)_2#</mathjax>, has a concentration of iodate anions, <mathjax>#"IO"_3^(-)#</mathjax>, equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>The dissociation equilibrium for barium iodate in aqueous solution looks like this</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of barium iodate that dissociates produces <mathjax>#1#</mathjax> <strong>mole</strong> of barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of iodate anions. </p>
<p>This means that in a saturated barium iodate solution, the concentration of iodate anions will be <strong>twice as high</strong> as the concentration of barium cations. </p>
<p>This means that the latter will be equal to </p>
<blockquote>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]#</mathjax></p>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this dissociation equilibrium is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = 5.96 * 10^(-10)"M"^3#</mathjax></p>
</blockquote>
<p>Solubility product constants are usually given <em>without added units</em>, which means that you answer is </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div class="markdown"><p><mathjax>#K_(sp) = 5.96 * 10^(-10)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that a <strong>saturated</strong> solution of barium iodate, <mathjax>#"Ba"("IO"_3)_2#</mathjax>, has a concentration of iodate anions, <mathjax>#"IO"_3^(-)#</mathjax>, equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>The dissociation equilibrium for barium iodate in aqueous solution looks like this</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of barium iodate that dissociates produces <mathjax>#1#</mathjax> <strong>mole</strong> of barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of iodate anions. </p>
<p>This means that in a saturated barium iodate solution, the concentration of iodate anions will be <strong>twice as high</strong> as the concentration of barium cations. </p>
<p>This means that the latter will be equal to </p>
<blockquote>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]#</mathjax></p>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this dissociation equilibrium is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = 5.96 * 10^(-10)"M"^3#</mathjax></p>
</blockquote>
<p>Solubility product constants are usually given <em>without added units</em>, which means that you answer is </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The concentration of #"IO"_3^(-)# ions in a pure, saturated solution of #"Ba"("IO"_3)_2# is #1.06xx10^(-3)"M"#. What is the #K_(sp)# for #"Ba"("IO"_3)_2#?
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Stefan V.
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<div class="markdown"><p><mathjax>#K_(sp) = 5.96 * 10^(-10)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem tells you that a <strong>saturated</strong> solution of barium iodate, <mathjax>#"Ba"("IO"_3)_2#</mathjax>, has a concentration of iodate anions, <mathjax>#"IO"_3^(-)#</mathjax>, equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to <mathjax>#1.06 * 10^(-3)"M"#</mathjax>. </p>
<p>The dissociation equilibrium for barium iodate in aqueous solution looks like this</p>
<blockquote>
<p><mathjax>#"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of barium iodate that dissociates produces <mathjax>#1#</mathjax> <strong>mole</strong> of barium cations, <mathjax>#"Ba"^(2+)#</mathjax>, and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of iodate anions. </p>
<p>This means that in a saturated barium iodate solution, the concentration of iodate anions will be <strong>twice as high</strong> as the concentration of barium cations. </p>
<p>This means that the latter will be equal to </p>
<blockquote>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]#</mathjax></p>
<p><mathjax>#["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"#</mathjax></p>
</blockquote>
<p>Now, the <em>solubility product constant</em>, <mathjax>#K_(sp)#</mathjax>, for this dissociation equilibrium is defined as </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)#</mathjax></p>
<p><mathjax>#K_(sp) = 5.96 * 10^(-10)"M"^3#</mathjax></p>
</blockquote>
<p>Solubility product constants are usually given <em>without added units</em>, which means that you answer is </p>
<blockquote>
<p><mathjax>#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | The concentration of #"IO"_3^(-)# ions in a pure, saturated solution of #"Ba"("IO"_3)_2# is #1.06xx10^(-3)"M"#. What is the #K_(sp)# for #"Ba"("IO"_3)_2#?
| null |
876 | a9784b7c-6ddd-11ea-9367-ccda262736ce | https://socratic.org/questions/what-is-the-specific-heat-of-ice-at-0-oc | 37.9 J/(mol * K) | start physical_unit 6 6 specific_heat j/(mol_·_k) qc_end physical_unit 6 6 8 9 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] ice [IN] J/(mol * K)"}] | [{"type":"physical unit","value":"37.9 J/(mol * K)"}] | [{"type":"physical unit","value":"Temperature [OF] ice [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the specific heat of ice at #0^oC#?</h1> | null | 37.9 J/(mol * K) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You might also try using the "Search" on this site for previous answers. This one was obtained from:<br/>
<a href="/questions/what-is-the-specific-heat-capacity-of-ice-water-and-steam">https://socratic.org/questions/what-is-the-specific-heat-capacity-of-ice-water-and-steam</a><br/>
- Probably obtained from here:<br/>
<a href="https://www.engineeringtoolbox.com/water-thermal-properties-d_162.html" rel="nofollow" target="_blank">https://www.engineeringtoolbox.com/water-thermal-properties-d_162.html</a></p>
<p>Also, often an initial Internet Search will quickly turn up physical constants such as this one.</p></div>
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<div class="markdown"><p>2.106 J/g K or 37.9 J/mol K</p></div>
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<h1 class="questionTitle" itemprop="name">What is the specific heat of ice at #0^oC#?</h1>
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<div class="markdown"><p>2.106 J/g K or 37.9 J/mol K</p></div>
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<div class="markdown"><p><mathjax>#2.108 "kJ"/(kg*K)#</mathjax></p></div>
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<div class="markdown"><p>You might also try using the "Search" on this site for previous answers. This one was obtained from:<br/>
<a href="/questions/what-is-the-specific-heat-capacity-of-ice-water-and-steam">https://socratic.org/questions/what-is-the-specific-heat-capacity-of-ice-water-and-steam</a><br/>
- Probably obtained from here:<br/>
<a href="https://www.engineeringtoolbox.com/water-thermal-properties-d_162.html" rel="nofollow" target="_blank">https://www.engineeringtoolbox.com/water-thermal-properties-d_162.html</a></p>
<p>Also, often an initial Internet Search will quickly turn up physical constants such as this one.</p></div>
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</article> | What is the specific heat of ice at #0^oC#? | null |
877 | ac7397ee-6ddd-11ea-825a-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-if-given-the-following-percent-composition-62-1-c- | C12H12NO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"C12H12NO3"}] | [{"type":"physical unit","value":"Percent [OF] C in the compound [=] \\pu{62.1%}"},{"type":"physical unit","value":"Percent [OF] H in the compound [=] \\pu{5.21%}"},{"type":"physical unit","value":"Percent [OF] N in the compound [=] \\pu{12.1%}"},{"type":"physical unit","value":"Percent [OF] O in the compound [=] \\pu{20.7%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula, if given the following percent composition: 62.1% C, 5.21%H, 12.1% N, and 20.7% O?</h1> | null | C12H12NO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Your percentages don't add up to exactly 100 %, but they are within the limits of experimental error.</p>
<blockquote></blockquote>
<p>Assume that you have 100 g of the compound.</p>
<p>Then you have 62.1 g of <mathjax>#"C"#</mathjax>, 5.21 g of <mathjax>#"H"#</mathjax>, 12.1 g of <mathjax>#"N"#</mathjax>, and 20.7 g of <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of C" = 62.1color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "5.171 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 5.21 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "5.169 mol H"#</mathjax></p>
<p><mathjax>#"Moles of N" = 12.1 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = "0.8637 mol N"#</mathjax></p>
<p><mathjax>#"Moles of O" = 20.7 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "1.294 mol O"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"0.8637"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio" color(white)(m)×2color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXX)62.1 color(white)(Xmll)"5.171"
color(white)(Xll)5.987color(white)(m) 11.97color(white)(Xmm)12#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXll)5.21 color(white)(mm)"5.169" color(white)(Xll)5.985color(white)(m)11.97 color(white)(XXX)12#</mathjax><br/>
<mathjax>#color(white)(ll)"N" color(white)(XXXml)12.1 color(white)(mml)"0.8637" color(white)(Xl)1color(white)(mmml)2 color(white)(XXXmlll)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(XXXXl)20.7 color(white)(mml)"1.294" color(white)(Xll)1.498color(white)(mll)2.998 color(white)(XXX)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_12"H"_12"NO"_3#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#"C"_12"H"_12"NO"_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Your percentages don't add up to exactly 100 %, but they are within the limits of experimental error.</p>
<blockquote></blockquote>
<p>Assume that you have 100 g of the compound.</p>
<p>Then you have 62.1 g of <mathjax>#"C"#</mathjax>, 5.21 g of <mathjax>#"H"#</mathjax>, 12.1 g of <mathjax>#"N"#</mathjax>, and 20.7 g of <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of C" = 62.1color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "5.171 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 5.21 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "5.169 mol H"#</mathjax></p>
<p><mathjax>#"Moles of N" = 12.1 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = "0.8637 mol N"#</mathjax></p>
<p><mathjax>#"Moles of O" = 20.7 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "1.294 mol O"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"0.8637"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio" color(white)(m)×2color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXX)62.1 color(white)(Xmll)"5.171"
color(white)(Xll)5.987color(white)(m) 11.97color(white)(Xmm)12#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXll)5.21 color(white)(mm)"5.169" color(white)(Xll)5.985color(white)(m)11.97 color(white)(XXX)12#</mathjax><br/>
<mathjax>#color(white)(ll)"N" color(white)(XXXml)12.1 color(white)(mml)"0.8637" color(white)(Xl)1color(white)(mmml)2 color(white)(XXXmlll)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(XXXXl)20.7 color(white)(mml)"1.294" color(white)(Xll)1.498color(white)(mll)2.998 color(white)(XXX)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_12"H"_12"NO"_3#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula, if given the following percent composition: 62.1% C, 5.21%H, 12.1% N, and 20.7% O?</h1>
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Ernest Z.
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<div class="markdown"><p>The empirical formula is <mathjax>#"C"_12"H"_12"NO"_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Your percentages don't add up to exactly 100 %, but they are within the limits of experimental error.</p>
<blockquote></blockquote>
<p>Assume that you have 100 g of the compound.</p>
<p>Then you have 62.1 g of <mathjax>#"C"#</mathjax>, 5.21 g of <mathjax>#"H"#</mathjax>, 12.1 g of <mathjax>#"N"#</mathjax>, and 20.7 g of <mathjax>#"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>Our job is to calculate the ratio of the moles of each element.</p>
<p><mathjax>#"Moles of C" = 62.1color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "5.171 mol C"#</mathjax></p>
<p><mathjax>#"Moles of H" = 5.21 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "5.169 mol H"#</mathjax></p>
<p><mathjax>#"Moles of N" = 12.1 color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = "0.8637 mol N"#</mathjax></p>
<p><mathjax>#"Moles of O" = 20.7 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "1.294 mol O"#</mathjax></p>
<blockquote></blockquote>
<p>To get the molar ratio, we divide each number of moles by the smallest number<br/>
(<mathjax>#"0.8637"#</mathjax>).</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio" color(white)(m)×2color(white)(mll)"Integers"#</mathjax><br/>
<mathjax>#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXX)62.1 color(white)(Xmll)"5.171"
color(white)(Xll)5.987color(white)(m) 11.97color(white)(Xmm)12#</mathjax><br/>
<mathjax>#color(white)(ll)"H" color(white)(XXXXll)5.21 color(white)(mm)"5.169" color(white)(Xll)5.985color(white)(m)11.97 color(white)(XXX)12#</mathjax><br/>
<mathjax>#color(white)(ll)"N" color(white)(XXXml)12.1 color(white)(mml)"0.8637" color(white)(Xl)1color(white)(mmml)2 color(white)(XXXmlll)2#</mathjax><br/>
<mathjax>#color(white)(ll)"O" color(white)(XXXXl)20.7 color(white)(mml)"1.294" color(white)(Xll)1.498color(white)(mll)2.998 color(white)(XXX)3#</mathjax></p>
<p>The empirical formula is <mathjax>#"C"_12"H"_12"NO"_3#</mathjax>.</p></div>
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</article> | What is the empirical formula, if given the following percent composition: 62.1% C, 5.21%H, 12.1% N, and 20.7% O? | null |
878 | a8377ffb-6ddd-11ea-b87c-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-4-5-gr-of-co-2 | 2.3 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 5 6 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 [IN] L"}] | [{"type":"physical unit","value":"2.3 L"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{4.5 g}"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 4.5 g of #"CO_2"#?</h1> | null | 2.3 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 4.5 g of #"CO_2"#?</h1>
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<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>, the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of carbon dioxide is <mathjax>#"0.001977 g/cm"^3"#</mathjax>.<br/>
<a href="http://www.engineeringtoolbox.com/gas-density-d_158.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/gas-density-d_158.html</a></p>
<p>We can use the given mass and known density to determine the volume of <mathjax>#"CO"_2"#</mathjax> at STP. Divide the mass by the density.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"#</mathjax> (rounded to two significant figures)</p></div>
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<div class="markdown"><p>The volume of <mathjax>#"4.5 g CO"_2"#</mathjax> is <mathjax>#"2300 cm"^3"#</mathjax> at STP.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can also use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to solve this problem.<br/>
The equation is <mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is the number of moles, and <mathjax>#R#</mathjax> is the gas constant.</p>
<p>In this problem, <mathjax>#T#</mathjax> and <mathjax>#P#</mathjax> are at STP, <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"1 atm"#</mathjax>.</p>
<p>Determine moles of <mathjax>#"CO"_2"#</mathjax> by dividing the given mass by the molar mass.</p>
<p><mathjax>#4.5cancel"g CO"_2xx(1"mol CO"_2)/(44.01cancel"g CO"_2)="0.10225 mol CO"_2"#</mathjax></p>
<p>I am leaving some guard digits to reduce rounding errors.</p>
<p><strong>Ideal Gas Law</strong></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="1 atm"#</mathjax><br/>
<mathjax>#n="0.10225 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Volume, <mathjax>#V#</mathjax></p>
<p><strong>Equation</strong><br/>
<mathjax>#P_1V_1=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V#</mathjax> and solve.</p>
<p><mathjax>#V=(nRT)/(P)#</mathjax></p>
<p><mathjax>#V=((0.10225cancel"mol"xx0.082057338 "L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K"))/(1cancel"atm")="2.3 L"#</mathjax> (rounded to two significant figures)</p>
<p>Convert to <mathjax>#"cm"^3"#</mathjax>.</p>
<p><mathjax>#2.3cancel"L"xx(1000cancel"mL")/(1cancel"L")xx(1"cm"^3)/(1cancel"mL")="2300 cm"^3"#</mathjax></p></div>
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</article> | What is the volume of 4.5 g of #"CO_2"#? | null |
879 | acae034a-6ddd-11ea-b66c-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-moles-in-197-g-as-2o-3 | 9.96 × 10^(-4) moles | start physical_unit 9 9 mole mol qc_end physical_unit 9 9 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] As2O3 [IN] moles"}] | [{"type":"physical unit","value":"9.96 × 10^(-4) moles"}] | [{"type":"physical unit","value":"Mass [OF] As2O3 [=] \\pu{0.197 g}"}] | <h1 class="questionTitle" itemprop="name">What is the number of moles in .197 g #As_2O_3#?</h1> | null | 9.96 × 10^(-4) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Do you mean 0.197 g?</p>
<p>Assume that:<br/>
n = number of moles<br/>
m = mass of substance<br/>
M = molar mass (equivalent to atomic weight on periodic table)</p>
<p>n = m <mathjax>#-:#</mathjax>M</p>
<p>0.197 g (mass) of <mathjax>#As_2O_3#</mathjax> is provided for you.</p>
<p>The next step is to find the molar mass (M). First thing first, you should know what <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> are involved. As you can see, arsenic and oxygen are involved.</p>
<p>If you refer to your periodic table, the molar mass of arsenic is 74.9 g/mol and oxygen is 16.0 g/mol. </p>
<p>Since there are 2 arsenic and 3 oxygen involved, the molar mass is <mathjax>#[2 xx 74.9 + 3 xx 16.0]#</mathjax> = 197.8 g/mol.</p>
<p>You've now found the molar mass of <mathjax>#As_2O_3#</mathjax>.</p>
<p>The final step is to find out the number of moles (n). Looking back at the <mathjax># n = m -: M#</mathjax> formula, the n is now the unknown.</p>
<p>The number of moles present in <mathjax>#As_2O_3#</mathjax> is:<br/>
n = 0.197 moles <mathjax>#-:#</mathjax>197.8 g/mol = 0.000996 moles. Note that this answer is rounded to 3 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Do you mean 0.197 g?<br/>
Then, <br/>
number of moles = 0.197 <mathjax>#-:#</mathjax>molar mass</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Do you mean 0.197 g?</p>
<p>Assume that:<br/>
n = number of moles<br/>
m = mass of substance<br/>
M = molar mass (equivalent to atomic weight on periodic table)</p>
<p>n = m <mathjax>#-:#</mathjax>M</p>
<p>0.197 g (mass) of <mathjax>#As_2O_3#</mathjax> is provided for you.</p>
<p>The next step is to find the molar mass (M). First thing first, you should know what <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> are involved. As you can see, arsenic and oxygen are involved.</p>
<p>If you refer to your periodic table, the molar mass of arsenic is 74.9 g/mol and oxygen is 16.0 g/mol. </p>
<p>Since there are 2 arsenic and 3 oxygen involved, the molar mass is <mathjax>#[2 xx 74.9 + 3 xx 16.0]#</mathjax> = 197.8 g/mol.</p>
<p>You've now found the molar mass of <mathjax>#As_2O_3#</mathjax>.</p>
<p>The final step is to find out the number of moles (n). Looking back at the <mathjax># n = m -: M#</mathjax> formula, the n is now the unknown.</p>
<p>The number of moles present in <mathjax>#As_2O_3#</mathjax> is:<br/>
n = 0.197 moles <mathjax>#-:#</mathjax>197.8 g/mol = 0.000996 moles. Note that this answer is rounded to 3 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of moles in .197 g #As_2O_3#?</h1>
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<div class="markdown"><p>Do you mean 0.197 g?<br/>
Then, <br/>
number of moles = 0.197 <mathjax>#-:#</mathjax>molar mass</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Do you mean 0.197 g?</p>
<p>Assume that:<br/>
n = number of moles<br/>
m = mass of substance<br/>
M = molar mass (equivalent to atomic weight on periodic table)</p>
<p>n = m <mathjax>#-:#</mathjax>M</p>
<p>0.197 g (mass) of <mathjax>#As_2O_3#</mathjax> is provided for you.</p>
<p>The next step is to find the molar mass (M). First thing first, you should know what <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> are involved. As you can see, arsenic and oxygen are involved.</p>
<p>If you refer to your periodic table, the molar mass of arsenic is 74.9 g/mol and oxygen is 16.0 g/mol. </p>
<p>Since there are 2 arsenic and 3 oxygen involved, the molar mass is <mathjax>#[2 xx 74.9 + 3 xx 16.0]#</mathjax> = 197.8 g/mol.</p>
<p>You've now found the molar mass of <mathjax>#As_2O_3#</mathjax>.</p>
<p>The final step is to find out the number of moles (n). Looking back at the <mathjax># n = m -: M#</mathjax> formula, the n is now the unknown.</p>
<p>The number of moles present in <mathjax>#As_2O_3#</mathjax> is:<br/>
n = 0.197 moles <mathjax>#-:#</mathjax>197.8 g/mol = 0.000996 moles. Note that this answer is rounded to 3 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</article> | What is the number of moles in .197 g #As_2O_3#? | null |
880 | aa3ceecc-6ddd-11ea-94ac-ccda262736ce | https://socratic.org/questions/what-mass-of-carbon-dioxide-is-produced-from-the-complete-combustion-of-8-50-10- | 3.15 × 10^(-3) g | start physical_unit 3 4 mass g qc_end physical_unit 17 17 12 15 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] g"}] | [{"type":"physical unit","value":"3.15 × 10^(-3) g"}] | [{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{8.50 × 10^(-3) g}"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">What mass of carbon dioxide is produced from the complete combustion of #8.50*10^-3# #g# of methane?</h1> | null | 3.15 × 10^(-3) g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation for the combustion of Methane is as follows.</p>
<p>C<mathjax>#H_4#</mathjax> + <mathjax>#O_2#</mathjax> ------> C<mathjax>#O_2#</mathjax> + 2 <mathjax>#H_2#</mathjax>O </p>
<p>One mole of Methane on burning produces one mole of Carbon Dioxide or using molar masses , I can write,</p>
<p>16 g of Methane produces 44 g of Carbon dioxide<br/>
using ratio and proportions , we can get the answer;</p>
<p>mass of C<mathjax>#O_2#</mathjax> = [ 16g C<mathjax>#H_4#</mathjax> / 44 g C<mathjax># O_2#</mathjax> ] x 8.50 x <mathjax>#10^-3#</mathjax> of C<mathjax>#H_4#</mathjax></p>
<pre><code> = 0.37 x 8.50 x #10^-3# g
= 3.145 X # 10^-3# g
</code></pre></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>3.145 X <mathjax># 10^-3#</mathjax> g </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation for the combustion of Methane is as follows.</p>
<p>C<mathjax>#H_4#</mathjax> + <mathjax>#O_2#</mathjax> ------> C<mathjax>#O_2#</mathjax> + 2 <mathjax>#H_2#</mathjax>O </p>
<p>One mole of Methane on burning produces one mole of Carbon Dioxide or using molar masses , I can write,</p>
<p>16 g of Methane produces 44 g of Carbon dioxide<br/>
using ratio and proportions , we can get the answer;</p>
<p>mass of C<mathjax>#O_2#</mathjax> = [ 16g C<mathjax>#H_4#</mathjax> / 44 g C<mathjax># O_2#</mathjax> ] x 8.50 x <mathjax>#10^-3#</mathjax> of C<mathjax>#H_4#</mathjax></p>
<pre><code> = 0.37 x 8.50 x #10^-3# g
= 3.145 X # 10^-3# g
</code></pre></div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of carbon dioxide is produced from the complete combustion of #8.50*10^-3# #g# of methane?</h1>
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Manish Bhardwaj
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<div class="markdown"><p>3.145 X <mathjax># 10^-3#</mathjax> g </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation for the combustion of Methane is as follows.</p>
<p>C<mathjax>#H_4#</mathjax> + <mathjax>#O_2#</mathjax> ------> C<mathjax>#O_2#</mathjax> + 2 <mathjax>#H_2#</mathjax>O </p>
<p>One mole of Methane on burning produces one mole of Carbon Dioxide or using molar masses , I can write,</p>
<p>16 g of Methane produces 44 g of Carbon dioxide<br/>
using ratio and proportions , we can get the answer;</p>
<p>mass of C<mathjax>#O_2#</mathjax> = [ 16g C<mathjax>#H_4#</mathjax> / 44 g C<mathjax># O_2#</mathjax> ] x 8.50 x <mathjax>#10^-3#</mathjax> of C<mathjax>#H_4#</mathjax></p>
<pre><code> = 0.37 x 8.50 x #10^-3# g
= 3.145 X # 10^-3# g
</code></pre></div>
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12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2
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</article> | What mass of carbon dioxide is produced from the complete combustion of #8.50*10^-3# #g# of methane? | null |
881 | a8d7f854-6ddd-11ea-b63f-ccda262736ce | https://socratic.org/questions/how-many-moles-of-water-are-made-from-the-complete-reaction-of-2-2-moles-of-oxyg | 4.4 moles | start physical_unit 4 4 mole mol qc_end physical_unit 15 16 12 13 mole qc_end chemical_equation 23 29 qc_end end | [{"type":"physical unit","value":"Mole [OF] water [IN] moles"}] | [{"type":"physical unit","value":"4.4 moles"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{2.2 moles}"},{"type":"chemical equation","value":"2 H2 + O2 -> 2 H2O"}] | <h1 class="questionTitle" itemprop="name">How many moles of water are made from the complete reaction of 2.2 moles of oxygen gas with hydrogen gas in the reaction #2H_2 + O_2 -> 2H_2O#? </h1> | null | 4.4 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation says that for every <mathjax>#1O_2#</mathjax>, there are <mathjax>#2H_2O#</mathjax> produced.</p>
<p>Therefore, if you have <mathjax>#2.2O_2#</mathjax>, you must have <mathjax>#2.2 * 2H_2O#</mathjax>, or <mathjax>#4.4#</mathjax> moles of <mathjax>#H_2O#</mathjax>. </p></div>
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<div class="markdown"><p><mathjax>#4.4molH_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation says that for every <mathjax>#1O_2#</mathjax>, there are <mathjax>#2H_2O#</mathjax> produced.</p>
<p>Therefore, if you have <mathjax>#2.2O_2#</mathjax>, you must have <mathjax>#2.2 * 2H_2O#</mathjax>, or <mathjax>#4.4#</mathjax> moles of <mathjax>#H_2O#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of water are made from the complete reaction of 2.2 moles of oxygen gas with hydrogen gas in the reaction #2H_2 + O_2 -> 2H_2O#? </h1>
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<div class="markdown"><p><mathjax>#4.4molH_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation says that for every <mathjax>#1O_2#</mathjax>, there are <mathjax>#2H_2O#</mathjax> produced.</p>
<p>Therefore, if you have <mathjax>#2.2O_2#</mathjax>, you must have <mathjax>#2.2 * 2H_2O#</mathjax>, or <mathjax>#4.4#</mathjax> moles of <mathjax>#H_2O#</mathjax>. </p></div>
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</article> | How many moles of water are made from the complete reaction of 2.2 moles of oxygen gas with hydrogen gas in the reaction #2H_2 + O_2 -> 2H_2O#? | null |
882 | abce0a71-6ddd-11ea-8664-ccda262736ce | https://socratic.org/questions/59c5bc8011ef6b28381543d7 | 7.9 × 10^20 moles | start physical_unit 4 4 mole mol qc_end physical_unit 17 17 9 16 size qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon [IN] moles"}] | [{"type":"physical unit","value":"7.9 × 10^20 moles"}] | [{"type":"physical unit","value":"Size [OF] diamond [=] \\pu{3.0 mm by 3.00 mm by 0.5 mm}"}] | <h1 class="questionTitle" itemprop="name">How many moles of carbon are present in a 3.0 mm by 3.00 mm by 0.5 mm diamond that's cut in the shape of a rectangular prism?</h1> | null | 7.9 × 10^20 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <em>volume</em> of the sample by using the fact that you're dealing with a <em>rectangular prism</em>, which has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("volume" = l xx w xx h)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#h#</mathjax> is the height of the prism</li>
<li><mathjax>#l#</mathjax> is its length</li>
<li><mathjax>#w#</mathjax> is its width</li>
</ul>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"volume" = "3.0 mm" * "3.0 mm" * "0.50 mm"#</mathjax></p>
<p><mathjax>#"volume" = "4.5 mm"^3#</mathjax></p>
</blockquote>
<p>Next, convert the volume of the prism to <em>cubic centimeters</em> by using the fact that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 cm" = 10color(white)(.)"mm")))#</mathjax></p>
</blockquote>
<p>Since you have</p>
<blockquote>
<p><mathjax>#"4.5 mm"^3 = 4.5 * "1 mm" * "1 mm" * "1 mm"#</mathjax></p>
</blockquote>
<p>you will end up with</p>
<blockquote>
<p><mathjax>#4.5 * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) = 4.5 * 10^(-3)color(white)(.)"cm"^3#</mathjax></p>
</blockquote>
<p>Next, use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of diamond to calculate the <em>mass</em> of the sample</p>
<blockquote>
<p><mathjax>#4.5 * 10^(-3) color(red)(cancel(color(black)("cm"^3))) * "3.52 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.584 * 10^(-2)color(white)(.)"g"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar mass</strong> of carbon to find the number of <em>moles</em> of carbon present in the sample</p>
<blockquote>
<p><mathjax>#1.584 * 10^(-2) color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.319 * 10^(-3)color(white)(.)"moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find the number of <em>atoms</em> of carbon present in the sample, use <strong>Avogadro's constant</strong></p>
<blockquote>
<p><mathjax>#1.319 * 10^(-3) color(red)(cancel(color(black)("moles C"))) * (6.022 * 10^(23)color(white)(.)"atoms C")/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)(7.9 * 10^(20)color(white)(.)"atoms C")))#</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>#7.9 * 10^(20)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <em>volume</em> of the sample by using the fact that you're dealing with a <em>rectangular prism</em>, which has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("volume" = l xx w xx h)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#h#</mathjax> is the height of the prism</li>
<li><mathjax>#l#</mathjax> is its length</li>
<li><mathjax>#w#</mathjax> is its width</li>
</ul>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"volume" = "3.0 mm" * "3.0 mm" * "0.50 mm"#</mathjax></p>
<p><mathjax>#"volume" = "4.5 mm"^3#</mathjax></p>
</blockquote>
<p>Next, convert the volume of the prism to <em>cubic centimeters</em> by using the fact that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 cm" = 10color(white)(.)"mm")))#</mathjax></p>
</blockquote>
<p>Since you have</p>
<blockquote>
<p><mathjax>#"4.5 mm"^3 = 4.5 * "1 mm" * "1 mm" * "1 mm"#</mathjax></p>
</blockquote>
<p>you will end up with</p>
<blockquote>
<p><mathjax>#4.5 * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) = 4.5 * 10^(-3)color(white)(.)"cm"^3#</mathjax></p>
</blockquote>
<p>Next, use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of diamond to calculate the <em>mass</em> of the sample</p>
<blockquote>
<p><mathjax>#4.5 * 10^(-3) color(red)(cancel(color(black)("cm"^3))) * "3.52 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.584 * 10^(-2)color(white)(.)"g"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar mass</strong> of carbon to find the number of <em>moles</em> of carbon present in the sample</p>
<blockquote>
<p><mathjax>#1.584 * 10^(-2) color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.319 * 10^(-3)color(white)(.)"moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find the number of <em>atoms</em> of carbon present in the sample, use <strong>Avogadro's constant</strong></p>
<blockquote>
<p><mathjax>#1.319 * 10^(-3) color(red)(cancel(color(black)("moles C"))) * (6.022 * 10^(23)color(white)(.)"atoms C")/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)(7.9 * 10^(20)color(white)(.)"atoms C")))#</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">How many moles of carbon are present in a 3.0 mm by 3.00 mm by 0.5 mm diamond that's cut in the shape of a rectangular prism?</h1>
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Stefan V.
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Sep 25, 2017
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<div class="markdown"><p><mathjax>#7.9 * 10^(20)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing that you need to do here is to figure out the <em>volume</em> of the sample by using the fact that you're dealing with a <em>rectangular prism</em>, which has</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("volume" = l xx w xx h)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#h#</mathjax> is the height of the prism</li>
<li><mathjax>#l#</mathjax> is its length</li>
<li><mathjax>#w#</mathjax> is its width</li>
</ul>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#"volume" = "3.0 mm" * "3.0 mm" * "0.50 mm"#</mathjax></p>
<p><mathjax>#"volume" = "4.5 mm"^3#</mathjax></p>
</blockquote>
<p>Next, convert the volume of the prism to <em>cubic centimeters</em> by using the fact that</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 cm" = 10color(white)(.)"mm")))#</mathjax></p>
</blockquote>
<p>Since you have</p>
<blockquote>
<p><mathjax>#"4.5 mm"^3 = 4.5 * "1 mm" * "1 mm" * "1 mm"#</mathjax></p>
</blockquote>
<p>you will end up with</p>
<blockquote>
<p><mathjax>#4.5 * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) = 4.5 * 10^(-3)color(white)(.)"cm"^3#</mathjax></p>
</blockquote>
<p>Next, use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of diamond to calculate the <em>mass</em> of the sample</p>
<blockquote>
<p><mathjax>#4.5 * 10^(-3) color(red)(cancel(color(black)("cm"^3))) * "3.52 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.584 * 10^(-2)color(white)(.)"g"#</mathjax></p>
</blockquote>
<p>Next, use the <strong>molar mass</strong> of carbon to find the number of <em>moles</em> of carbon present in the sample</p>
<blockquote>
<p><mathjax>#1.584 * 10^(-2) color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.319 * 10^(-3)color(white)(.)"moles C"#</mathjax></p>
</blockquote>
<p>Finally, to find the number of <em>atoms</em> of carbon present in the sample, use <strong>Avogadro's constant</strong></p>
<blockquote>
<p><mathjax>#1.319 * 10^(-3) color(red)(cancel(color(black)("moles C"))) * (6.022 * 10^(23)color(white)(.)"atoms C")/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)(7.9 * 10^(20)color(white)(.)"atoms C")))#</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> | How many moles of carbon are present in a 3.0 mm by 3.00 mm by 0.5 mm diamond that's cut in the shape of a rectangular prism? | null |
883 | a8ad3afa-6ddd-11ea-b5c1-ccda262736ce | https://socratic.org/questions/how-many-moles-of-nitrogen-n-2-are-there-in-4-3-10-23-n-2-molecules | 0.71 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 8 10 number qc_end end | [{"type":"physical unit","value":"Mole [OF] N2 [IN] moles"}] | [{"type":"physical unit","value":"0.71 moles"}] | [{"type":"physical unit","value":"Number [OF] molecules [=] \\pu{4.3 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">How many moles of nitrogen, #N_2#, are there in #4.3*10^23# #N_2# molecules?</h1> | null | 0.71 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many <em>moles</em> of nitrogen gas would contain that many molecules, you need to know how many molecules of a substance are needed in order to make <strong>one mole</strong>. </p>
<p>If you know how many molecules of a substance you get in <strong>one mole</strong>, you can use this as a <strong>conversion factor</strong> to help you find the answer. </p>
<p>The number of molecules that are needed in order to have <strong>one mole</strong> of a substance is known as <strong>Avogadro's number</strong> and is equal to <mathjax>#6.022 * 10^(23)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->#</mathjax> <em><strong>Avogadro's number</strong></em></p>
</blockquote>
<p>This is true <em>regardless</em> of the substance you're dealing with. Simply put, in order to have a mole of <em>anything</em>, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> pieces of that <em>anything</em>.</p>
<p>Use Avogadro's number as a conversion factor to get</p>
<blockquote>
<p><mathjax>#4.3 * 10^(23)color(red)(cancel(color(black)("molec. N"_2))) * overbrace("1 mole N"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molec. N"_2)))))^(color(purple)("Avogadro's number")) = "0.71405 moles N"_2#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of nitrogen, the answer will be </p>
<blockquote>
<p><mathjax>#"no. of moles N"_2 = color(green)(|bar(ul(color(white)(a/a)0.71color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/Pft2CASl0M0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.71 moles N"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many <em>moles</em> of nitrogen gas would contain that many molecules, you need to know how many molecules of a substance are needed in order to make <strong>one mole</strong>. </p>
<p>If you know how many molecules of a substance you get in <strong>one mole</strong>, you can use this as a <strong>conversion factor</strong> to help you find the answer. </p>
<p>The number of molecules that are needed in order to have <strong>one mole</strong> of a substance is known as <strong>Avogadro's number</strong> and is equal to <mathjax>#6.022 * 10^(23)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->#</mathjax> <em><strong>Avogadro's number</strong></em></p>
</blockquote>
<p>This is true <em>regardless</em> of the substance you're dealing with. Simply put, in order to have a mole of <em>anything</em>, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> pieces of that <em>anything</em>.</p>
<p>Use Avogadro's number as a conversion factor to get</p>
<blockquote>
<p><mathjax>#4.3 * 10^(23)color(red)(cancel(color(black)("molec. N"_2))) * overbrace("1 mole N"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molec. N"_2)))))^(color(purple)("Avogadro's number")) = "0.71405 moles N"_2#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of nitrogen, the answer will be </p>
<blockquote>
<p><mathjax>#"no. of moles N"_2 = color(green)(|bar(ul(color(white)(a/a)0.71color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/Pft2CASl0M0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of nitrogen, #N_2#, are there in #4.3*10^23# #N_2# molecules?</h1>
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<div class="markdown"><p><mathjax>#"0.71 moles N"_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many <em>moles</em> of nitrogen gas would contain that many molecules, you need to know how many molecules of a substance are needed in order to make <strong>one mole</strong>. </p>
<p>If you know how many molecules of a substance you get in <strong>one mole</strong>, you can use this as a <strong>conversion factor</strong> to help you find the answer. </p>
<p>The number of molecules that are needed in order to have <strong>one mole</strong> of a substance is known as <strong>Avogadro's number</strong> and is equal to <mathjax>#6.022 * 10^(23)#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->#</mathjax> <em><strong>Avogadro's number</strong></em></p>
</blockquote>
<p>This is true <em>regardless</em> of the substance you're dealing with. Simply put, in order to have a mole of <em>anything</em>, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> pieces of that <em>anything</em>.</p>
<p>Use Avogadro's number as a conversion factor to get</p>
<blockquote>
<p><mathjax>#4.3 * 10^(23)color(red)(cancel(color(black)("molec. N"_2))) * overbrace("1 mole N"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molec. N"_2)))))^(color(purple)("Avogadro's number")) = "0.71405 moles N"_2#</mathjax></p>
</blockquote>
<p>Rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of molecules of nitrogen, the answer will be </p>
<blockquote>
<p><mathjax>#"no. of moles N"_2 = color(green)(|bar(ul(color(white)(a/a)0.71color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/Pft2CASl0M0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How many moles of nitrogen, #N_2#, are there in #4.3*10^23# #N_2# molecules? | null |
884 | aa871386-6ddd-11ea-871b-ccda262736ce | https://socratic.org/questions/what-is-the-poh-of-a-4-5x10-3-m-oh-solution | 2.35 | start physical_unit 10 11 poh none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pOH [OF] OH- solution"}] | [{"type":"physical unit","value":"2.35"}] | [{"type":"physical unit","value":"Molarity [OF] OH- solution [=] \\pu{4.5 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pOH of a #4.5x10^-3 M# #OH^-# solution?</h1> | null | 2.35 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(4.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pOH=2.35#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(4.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution?</p></div>
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anor277
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<div class="markdown"><p><mathjax>#pOH=2.35#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#-log_10(4.5xx10^-3)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>What is the <mathjax>#pH#</mathjax> of this solution?</p></div>
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</article> | What is the pOH of a #4.5x10^-3 M# #OH^-# solution? | null |
885 | a8d363b4-6ddd-11ea-b772-ccda262736ce | https://socratic.org/questions/calculate-the-weight-of-carbon-dioxide-that-contain-same-number-of-oxygen-atom-a | 2.2 g | start physical_unit 4 5 mass g qc_end physical_unit 21 21 18 19 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] g"}] | [{"type":"physical unit","value":"2.2 g"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{1.8 g}"},{"type":"other","value":"Carbon dioxide that contains the same number of atoms of oxygen as present in water."}] | <h1 class="questionTitle" itemprop="name">Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in #"1.8 g"# of water? </h1> | null | 2.2 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of <em>moles</em> of oxygen present in <mathjax>#"1.8 g"#</mathjax> of water.</p>
<p>As you know, one <strong>molecule</strong> of water contains two atoms of hydrogen and one atom of oxygen. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen. </p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#1.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0color(red)(cancel(color(black)("g")))) = "0.10 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can thus say that this sample contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.10 moles O"#</mathjax></p>
</blockquote>
<p>Now, for every <mathjax>#1#</mathjax> <strong>molecule</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, you get <mathjax>#2#</mathjax> atoms of oxygen, so <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide will contain <mathjax>#2#</mathjax> <strong>moles</strong> of atoms of oxygen. </p>
<p>This means that the number of moles of oxygen present in your sample of water is enough to get you </p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moless O")))) = "0.050 moles CO"_2" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of moles of carbon dioxide to <em>moles</em>, use the <strong>molar mass</strong> of carbon dioxide.</p>
<blockquote>
<p><mathjax>#0.050 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("2.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of water. </p>
<h2></h2>
<p>Notice that you didn't even have to convert the mass of water to <em>atoms of oxygen</em> because a <strong>mole</strong> of oxygen is simply a very large collection of atoms of oxygen, <mathjax>#6.022 * 10^(23)#</mathjax>, or <mathjax>#N_"A"#</mathjax>, to be precise <mathjax>#->#</mathjax> this is known as <strong>Avogadro's constant</strong>. </p>
<p>So if you want to double-check your answer, you can say that your sample of water contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * (N_"A" quad "atoms of O")/(1 color(red)(cancel(color(black)("mole O")))) = (0.10 * N_"A") quad "atoms of O"#</mathjax></p>
</blockquote>
<p>The same logic applies--you need <mathjax>#2#</mathjax> atoms of oxygen to get <mathjax>#1#</mathjax> molecule of carbon dioxide, so the number of atoms of oxygen present in your sample will get you </p>
<blockquote>
<p><mathjax>#(0.10 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O")))) = (0.050 * N_"A") quad "molecules CO"_2#</mathjax></p>
</blockquote>
<p>To convert the number of <em>molecules</em> of carbon dioxide to <em>moles</em>, use <strong>Avogadro's constant</strong> again.</p>
<blockquote>
<p><mathjax>#(0.050 * color(blue)(cancel(color(black)(N_ "A")))) color(red)(cancel(color(black)("molecules CO"_ 2))) * "1 mole CO"_ 2/(color(blue)(cancel(color(black)(N_ "A"))) color(red)(cancel(color(black)("molecules CO"_ 2)))) = "0.050 moles CO"_2#</mathjax></p>
</blockquote>
<p>This brings you right back to <mathjax>#color(darkorange)("(*)")#</mathjax>, so once again, the answer will be <mathjax>#"2.2 g"#</mathjax> of carbon dioxide.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.2 g CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of <em>moles</em> of oxygen present in <mathjax>#"1.8 g"#</mathjax> of water.</p>
<p>As you know, one <strong>molecule</strong> of water contains two atoms of hydrogen and one atom of oxygen. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen. </p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#1.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0color(red)(cancel(color(black)("g")))) = "0.10 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can thus say that this sample contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.10 moles O"#</mathjax></p>
</blockquote>
<p>Now, for every <mathjax>#1#</mathjax> <strong>molecule</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, you get <mathjax>#2#</mathjax> atoms of oxygen, so <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide will contain <mathjax>#2#</mathjax> <strong>moles</strong> of atoms of oxygen. </p>
<p>This means that the number of moles of oxygen present in your sample of water is enough to get you </p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moless O")))) = "0.050 moles CO"_2" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of moles of carbon dioxide to <em>moles</em>, use the <strong>molar mass</strong> of carbon dioxide.</p>
<blockquote>
<p><mathjax>#0.050 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("2.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of water. </p>
<h2></h2>
<p>Notice that you didn't even have to convert the mass of water to <em>atoms of oxygen</em> because a <strong>mole</strong> of oxygen is simply a very large collection of atoms of oxygen, <mathjax>#6.022 * 10^(23)#</mathjax>, or <mathjax>#N_"A"#</mathjax>, to be precise <mathjax>#->#</mathjax> this is known as <strong>Avogadro's constant</strong>. </p>
<p>So if you want to double-check your answer, you can say that your sample of water contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * (N_"A" quad "atoms of O")/(1 color(red)(cancel(color(black)("mole O")))) = (0.10 * N_"A") quad "atoms of O"#</mathjax></p>
</blockquote>
<p>The same logic applies--you need <mathjax>#2#</mathjax> atoms of oxygen to get <mathjax>#1#</mathjax> molecule of carbon dioxide, so the number of atoms of oxygen present in your sample will get you </p>
<blockquote>
<p><mathjax>#(0.10 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O")))) = (0.050 * N_"A") quad "molecules CO"_2#</mathjax></p>
</blockquote>
<p>To convert the number of <em>molecules</em> of carbon dioxide to <em>moles</em>, use <strong>Avogadro's constant</strong> again.</p>
<blockquote>
<p><mathjax>#(0.050 * color(blue)(cancel(color(black)(N_ "A")))) color(red)(cancel(color(black)("molecules CO"_ 2))) * "1 mole CO"_ 2/(color(blue)(cancel(color(black)(N_ "A"))) color(red)(cancel(color(black)("molecules CO"_ 2)))) = "0.050 moles CO"_2#</mathjax></p>
</blockquote>
<p>This brings you right back to <mathjax>#color(darkorange)("(*)")#</mathjax>, so once again, the answer will be <mathjax>#"2.2 g"#</mathjax> of carbon dioxide.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in #"1.8 g"# of water? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"2.2 g CO"_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by calculating the number of <em>moles</em> of oxygen present in <mathjax>#"1.8 g"#</mathjax> of water.</p>
<p>As you know, one <strong>molecule</strong> of water contains two atoms of hydrogen and one atom of oxygen. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen. </p>
<p>Use the <strong>molar mass</strong> of water to convert the mass of water to <em>moles</em>.</p>
<blockquote>
<p><mathjax>#1.8 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0color(red)(cancel(color(black)("g")))) = "0.10 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>You can thus say that this sample contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole O"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.10 moles O"#</mathjax></p>
</blockquote>
<p>Now, for every <mathjax>#1#</mathjax> <strong>molecule</strong> of carbon dioxide, <mathjax>#"CO"_2#</mathjax>, you get <mathjax>#2#</mathjax> atoms of oxygen, so <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide will contain <mathjax>#2#</mathjax> <strong>moles</strong> of atoms of oxygen. </p>
<p>This means that the number of moles of oxygen present in your sample of water is enough to get you </p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moless O")))) = "0.050 moles CO"_2" " " "color(darkorange)("(*)")#</mathjax></p>
</blockquote>
<p>Finally, to convert the number of moles of carbon dioxide to <em>moles</em>, use the <strong>molar mass</strong> of carbon dioxide.</p>
<blockquote>
<p><mathjax>#0.050 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("2.2 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of water. </p>
<h2></h2>
<p>Notice that you didn't even have to convert the mass of water to <em>atoms of oxygen</em> because a <strong>mole</strong> of oxygen is simply a very large collection of atoms of oxygen, <mathjax>#6.022 * 10^(23)#</mathjax>, or <mathjax>#N_"A"#</mathjax>, to be precise <mathjax>#->#</mathjax> this is known as <strong>Avogadro's constant</strong>. </p>
<p>So if you want to double-check your answer, you can say that your sample of water contains</p>
<blockquote>
<p><mathjax>#0.10 color(red)(cancel(color(black)("moles O"))) * (N_"A" quad "atoms of O")/(1 color(red)(cancel(color(black)("mole O")))) = (0.10 * N_"A") quad "atoms of O"#</mathjax></p>
</blockquote>
<p>The same logic applies--you need <mathjax>#2#</mathjax> atoms of oxygen to get <mathjax>#1#</mathjax> molecule of carbon dioxide, so the number of atoms of oxygen present in your sample will get you </p>
<blockquote>
<p><mathjax>#(0.10 * N_"A") color(red)(cancel(color(black)("atoms O"))) * "1 molecule CO"_2/(2color(red)(cancel(color(black)("atoms O")))) = (0.050 * N_"A") quad "molecules CO"_2#</mathjax></p>
</blockquote>
<p>To convert the number of <em>molecules</em> of carbon dioxide to <em>moles</em>, use <strong>Avogadro's constant</strong> again.</p>
<blockquote>
<p><mathjax>#(0.050 * color(blue)(cancel(color(black)(N_ "A")))) color(red)(cancel(color(black)("molecules CO"_ 2))) * "1 mole CO"_ 2/(color(blue)(cancel(color(black)(N_ "A"))) color(red)(cancel(color(black)("molecules CO"_ 2)))) = "0.050 moles CO"_2#</mathjax></p>
</blockquote>
<p>This brings you right back to <mathjax>#color(darkorange)("(*)")#</mathjax>, so once again, the answer will be <mathjax>#"2.2 g"#</mathjax> of carbon dioxide.</p></div>
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</article> | Calculate the mass of carbon dioxide that contains the same number of atoms of oxygen as present in #"1.8 g"# of water? | null |
886 | aa31cb12-6ddd-11ea-a571-ccda262736ce | https://socratic.org/questions/58edfdce11ef6b7e9f9fde4f | 5 | start physical_unit 2 3 number none qc_end chemical_equation 5 5 qc_end end | [{"type":"physical unit","value":"Number [OF] structural isomers"}] | [{"type":"physical unit","value":"5"}] | [{"type":"chemical equation","value":"C6H14"}] | <h1 class="questionTitle" itemprop="name">How many structural isomers of #C_6H_14# are there?</h1> | null | 5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And these are.......</p>
<p><mathjax>#(i)#</mathjax> <mathjax>#"n-hexane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH"_2"CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(ii)#</mathjax> <mathjax>#"2-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iii)#</mathjax> <mathjax>#"3-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH(CH"_3")CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iv)#</mathjax> <mathjax>#"2,2-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"C(CH"_3")"_2"CH"_3#</mathjax></p>
<p><mathjax>#(v)#</mathjax> <mathjax>#"2,3-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH(CH"_3")CH"_3#</mathjax></p>
<p>The way to remember these is to sit down with a molecular model set, and actually make models of each isomer. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>I count 5 structural isomers of hexane, <mathjax>#C_6H_14#</mathjax>..........</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And these are.......</p>
<p><mathjax>#(i)#</mathjax> <mathjax>#"n-hexane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH"_2"CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(ii)#</mathjax> <mathjax>#"2-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iii)#</mathjax> <mathjax>#"3-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH(CH"_3")CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iv)#</mathjax> <mathjax>#"2,2-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"C(CH"_3")"_2"CH"_3#</mathjax></p>
<p><mathjax>#(v)#</mathjax> <mathjax>#"2,3-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH(CH"_3")CH"_3#</mathjax></p>
<p>The way to remember these is to sit down with a molecular model set, and actually make models of each isomer. </p></div>
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<h1 class="questionTitle" itemprop="name">How many structural isomers of #C_6H_14# are there?</h1>
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<div class="markdown"><p>I count 5 structural isomers of hexane, <mathjax>#C_6H_14#</mathjax>..........</p></div>
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<div class="markdown"><p>And these are.......</p>
<p><mathjax>#(i)#</mathjax> <mathjax>#"n-hexane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH"_2"CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(ii)#</mathjax> <mathjax>#"2-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH"_2"CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iii)#</mathjax> <mathjax>#"3-methylpentane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"CH(CH"_3")CH"_2"CH"_3#</mathjax></p>
<p><mathjax>#(iv)#</mathjax> <mathjax>#"2,2-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH"_2"C(CH"_3")"_2"CH"_3#</mathjax></p>
<p><mathjax>#(v)#</mathjax> <mathjax>#"2,3-dimethylbutane"#</mathjax>, <mathjax>#"H"_3"CCH(CH"_3")CH(CH"_3")CH"_3#</mathjax></p>
<p>The way to remember these is to sit down with a molecular model set, and actually make models of each isomer. </p></div>
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</article> | How many structural isomers of #C_6H_14# are there? | null |
887 | aad63fd8-6ddd-11ea-9662-ccda262736ce | https://socratic.org/questions/how-many-chloride-ions-would-combine-with-an-al-3-ion-to-form-aluminum-chloride | 3 | start physical_unit 2 3 number none qc_end chemical_equation 8 8 qc_end substance 12 13 qc_end end | [{"type":"physical unit","value":"Number [OF] chloride ions"}] | [{"type":"physical unit","value":"3"}] | [{"type":"chemical equation","value":"Al^3+"},{"type":"substance name","value":"Aluminum chloride"}] | <h1 class="questionTitle" itemprop="name">How many chloride ions would combine with an #Al^(3+)# ion to form aluminum chloride?</h1> | null | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Al has a charge of +3. Cl has a charge of -1. Hence 3 chloride ions are required to neutralize charge of +3 on aluminium and the compound formed is <mathjax># AlCl_3#</mathjax>.</p></div>
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<div class="markdown"><p>3</p></div>
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<div class="markdown"><p>Al has a charge of +3. Cl has a charge of -1. Hence 3 chloride ions are required to neutralize charge of +3 on aluminium and the compound formed is <mathjax># AlCl_3#</mathjax>.</p></div>
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<div class="markdown"><p>3</p></div>
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<div class="markdown"><p>Al has a charge of +3. Cl has a charge of -1. Hence 3 chloride ions are required to neutralize charge of +3 on aluminium and the compound formed is <mathjax># AlCl_3#</mathjax>.</p></div>
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<div class="markdown"><p>Three chloride ions are required to bond with one aluminum ion.</p></div>
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<div class="markdown"><p>Chloride ions have a <mathjax>#1^-#</mathjax> charge with the formula <mathjax>#"Cl"^(-)"#</mathjax>. Aluminum ions have a <mathjax>#3^(+)"#</mathjax> charge with the formula <mathjax>#"Al"^(3+)"#</mathjax>. </p>
<p>An ionic compound must be neutral, which means the positive and negative charges must be equal. So we need three chloride ions to bond with one aluminum ion, and the formula for aluminum chloride is <mathjax>#"AlCl"_3"#</mathjax>.</p></div>
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</article> | How many chloride ions would combine with an #Al^(3+)# ion to form aluminum chloride? | null |
888 | a9476507-6ddd-11ea-8bb4-ccda262736ce | https://socratic.org/questions/if-3-5-l-of-a-4-8-m-srcl-2-solution-is-diluted-to-45-l-what-is-the-molarity-of-t | 0.37 M | start physical_unit 19 21 molarity mol/l qc_end physical_unit 7 8 1 2 volume qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 7 8 12 13 volume qc_end end | [{"type":"physical unit","value":"Molarity2 [OF] the diluted solution [IN] M"}] | [{"type":"physical unit","value":"0.37 M"}] | [{"type":"physical unit","value":"Volume1 [OF] SrCl2 solution [=] \\pu{3.5 L}"},{"type":"physical unit","value":"Molarity1 [OF] SrCl2 solution [=] \\pu{4.8 M}"},{"type":"physical unit","value":"Volume2 [OF] SrCl2 solution [=] \\pu{45 L}"}] | <h1 class="questionTitle" itemprop="name">If 3.5 L of a 4.8 M #SrCl_2# solution is diluted to 45 L, what is the molarity of the diluted solution?</h1> | null | 0.37 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#c_1V_1=c_2V_2#</mathjax><br/>
<mathjax>#c_1=4.8M#</mathjax><br/>
<mathjax>#V_1=3.5L#</mathjax><br/>
<mathjax>#c_2=?#</mathjax><br/>
<mathjax>#V_2=45L#</mathjax></p>
<p><mathjax>#4.5xx3.5=x xx45#</mathjax><br/>
<mathjax>#x=0.3733M#</mathjax></p>
<p>Hope it helps :)</p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a><mathjax>#=0.37333M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#c_1V_1=c_2V_2#</mathjax><br/>
<mathjax>#c_1=4.8M#</mathjax><br/>
<mathjax>#V_1=3.5L#</mathjax><br/>
<mathjax>#c_2=?#</mathjax><br/>
<mathjax>#V_2=45L#</mathjax></p>
<p><mathjax>#4.5xx3.5=x xx45#</mathjax><br/>
<mathjax>#x=0.3733M#</mathjax></p>
<p>Hope it helps :)</p></div>
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<h1 class="questionTitle" itemprop="name">If 3.5 L of a 4.8 M #SrCl_2# solution is diluted to 45 L, what is the molarity of the diluted solution?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a><mathjax>#=0.37333M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#c_1V_1=c_2V_2#</mathjax><br/>
<mathjax>#c_1=4.8M#</mathjax><br/>
<mathjax>#V_1=3.5L#</mathjax><br/>
<mathjax>#c_2=?#</mathjax><br/>
<mathjax>#V_2=45L#</mathjax></p>
<p><mathjax>#4.5xx3.5=x xx45#</mathjax><br/>
<mathjax>#x=0.3733M#</mathjax></p>
<p>Hope it helps :)</p></div>
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</article> | If 3.5 L of a 4.8 M #SrCl_2# solution is diluted to 45 L, what is the molarity of the diluted solution? | null |
889 | a8d427b6-6ddd-11ea-ac9d-ccda262736ce | https://socratic.org/questions/how-many-hydrogen-molecules-are-in-2-75-l-of-h-2-gas-at-stp | 7.39 × 10^22 | start physical_unit 2 3 number none qc_end physical_unit 9 10 6 7 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Number [OF] hydrogen molecules"}] | [{"type":"physical unit","value":"7.39 × 10^22"}] | [{"type":"physical unit","value":"Volume [OF] H2 gas [=] \\pu{2.75 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many hydrogen molecules are in 2.75 L of #H_2# gas at STP?</h1> | null | 7.39 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> is <mathjax>#22.4*L#</mathjax>.</p>
<p>We assume (reasonably) that dihydrogen behaves ideally, and mulitply the quotient of volumes by <mathjax>#"Avogadro's number, " N_A, 6.022xx10^23*mol^-1#</mathjax> to give an answer in numbers of hydrogen molecules. </p>
<p>How many hydrogen atoms in such a quantity?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#(2.75*L)/(22.4*L*mol^-1)xx6.022xx10^23" hydrogen molecules mol"^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??"hydrogen molecules"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> is <mathjax>#22.4*L#</mathjax>.</p>
<p>We assume (reasonably) that dihydrogen behaves ideally, and mulitply the quotient of volumes by <mathjax>#"Avogadro's number, " N_A, 6.022xx10^23*mol^-1#</mathjax> to give an answer in numbers of hydrogen molecules. </p>
<p>How many hydrogen atoms in such a quantity?</p></div>
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<h1 class="questionTitle" itemprop="name">How many hydrogen molecules are in 2.75 L of #H_2# gas at STP?</h1>
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anor277
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<div class="markdown"><p><mathjax>#(2.75*L)/(22.4*L*mol^-1)xx6.022xx10^23" hydrogen molecules mol"^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??"hydrogen molecules"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> is <mathjax>#22.4*L#</mathjax>.</p>
<p>We assume (reasonably) that dihydrogen behaves ideally, and mulitply the quotient of volumes by <mathjax>#"Avogadro's number, " N_A, 6.022xx10^23*mol^-1#</mathjax> to give an answer in numbers of hydrogen molecules. </p>
<p>How many hydrogen atoms in such a quantity?</p></div>
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</article> | How many hydrogen molecules are in 2.75 L of #H_2# gas at STP? | null |
890 | a99d1e40-6ddd-11ea-a1ab-ccda262736ce | https://socratic.org/questions/you-transfer-a-sample-of-gas-at-17-c-from-a-volume-of-4-81-l-and-1-10-atm-to-a-c | 6.64 L | start physical_unit 3 5 volume l qc_end physical_unit 3 5 13 14 volume qc_end physical_unit 3 5 16 17 pressure qc_end physical_unit 3 5 7 8 temperature qc_end physical_unit 3 5 16 17 pressure qc_end physical_unit 3 5 22 23 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas sample [IN] L"}] | [{"type":"physical unit","value":"6.64 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{4.81 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{1.10 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{17 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas sample [=] \\pu{1.10 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the gas sample [=] \\pu{37 ℃}"}] | <h1 class="questionTitle" itemprop="name">You transfer a sample of gas at 17°C from a volume of 4.81 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 atm. What is the new volume of the gas?</h1> | null | 6.64 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An important thing to notice here is that, unless the question is mistyped, the <strong>pressure</strong> of the gas is being kept <strong>constant</strong>.</p>
<p>If you take into account the fact that the <em>number of moles</em> of gas is constant as well, you can say that the <em>change in volume</em> will only depend on the <em>change in temperature</em>. </p>
<p>As you know, when number of moles of gas and pressure are kept constant, volume and temperature have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch11.html" src="https://useruploads.socratic.org/I7WzHAJgS1TNJYK7maRw_FG11_17.JPG"/> </p>
<p>So, when temperature <strong>Increases</strong>, the volume <strong>increases</strong> as well. Likewise, when temperature <strong>decreases</strong>, the volume <strong>decreases</strong> as well. </p>
<p>In your case, the temperature increased from <mathjax>#17^@"C"#</mathjax> to <mathjax>#37^@"C"#</mathjax>, which means that you can expect the volume of the gas to <em>increase</em>. </p>
<p>Mathematically, <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(V_1/T_1 = V_2/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax> - <strong>no not</strong> forget that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>! </p>
<blockquote>
<p><mathjax>#V_2 = T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>In your case, this will be equal to </p>
<blockquote>
<p><mathjax>#V_2 = ( (273.15 + 17) color(red)(cancel(color(black)("K"))))/( (273.15 + 37) color(red)(cancel(color(black)("K")))) * "4.81 L"#</mathjax></p>
<p><mathjax>#V_2 = "6.641 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the temperatures of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.6 L")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>If the pressure of the gas is <strong>not constant</strong>, you can solve the problem by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</em></p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - <em>the pressure, volume, and temperature of the gas at an initial state</em><br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - <em>the pressure, volume, and temperature of the gas at a final state</em></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"6.6 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An important thing to notice here is that, unless the question is mistyped, the <strong>pressure</strong> of the gas is being kept <strong>constant</strong>.</p>
<p>If you take into account the fact that the <em>number of moles</em> of gas is constant as well, you can say that the <em>change in volume</em> will only depend on the <em>change in temperature</em>. </p>
<p>As you know, when number of moles of gas and pressure are kept constant, volume and temperature have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch11.html" src="https://useruploads.socratic.org/I7WzHAJgS1TNJYK7maRw_FG11_17.JPG"/> </p>
<p>So, when temperature <strong>Increases</strong>, the volume <strong>increases</strong> as well. Likewise, when temperature <strong>decreases</strong>, the volume <strong>decreases</strong> as well. </p>
<p>In your case, the temperature increased from <mathjax>#17^@"C"#</mathjax> to <mathjax>#37^@"C"#</mathjax>, which means that you can expect the volume of the gas to <em>increase</em>. </p>
<p>Mathematically, <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(V_1/T_1 = V_2/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax> - <strong>no not</strong> forget that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>! </p>
<blockquote>
<p><mathjax>#V_2 = T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>In your case, this will be equal to </p>
<blockquote>
<p><mathjax>#V_2 = ( (273.15 + 17) color(red)(cancel(color(black)("K"))))/( (273.15 + 37) color(red)(cancel(color(black)("K")))) * "4.81 L"#</mathjax></p>
<p><mathjax>#V_2 = "6.641 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the temperatures of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.6 L")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>If the pressure of the gas is <strong>not constant</strong>, you can solve the problem by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</em></p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - <em>the pressure, volume, and temperature of the gas at an initial state</em><br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - <em>the pressure, volume, and temperature of the gas at a final state</em></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">You transfer a sample of gas at 17°C from a volume of 4.81 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 atm. What is the new volume of the gas?</h1>
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Stefan V.
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Jan 14, 2016
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<div class="markdown"><p><mathjax>#"6.6 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>An important thing to notice here is that, unless the question is mistyped, the <strong>pressure</strong> of the gas is being kept <strong>constant</strong>.</p>
<p>If you take into account the fact that the <em>number of moles</em> of gas is constant as well, you can say that the <em>change in volume</em> will only depend on the <em>change in temperature</em>. </p>
<p>As you know, when number of moles of gas and pressure are kept constant, volume and temperature have a <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p><img alt="http://wps.prenhall.com/wps/media/objects/476/488316/ch11.html" src="https://useruploads.socratic.org/I7WzHAJgS1TNJYK7maRw_FG11_17.JPG"/> </p>
<p>So, when temperature <strong>Increases</strong>, the volume <strong>increases</strong> as well. Likewise, when temperature <strong>decreases</strong>, the volume <strong>decreases</strong> as well. </p>
<p>In your case, the temperature increased from <mathjax>#17^@"C"#</mathjax> to <mathjax>#37^@"C"#</mathjax>, which means that you can expect the volume of the gas to <em>increase</em>. </p>
<p>Mathematically, <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a> is expressed as </p>
<blockquote>
<p><mathjax>#color(blue)(V_1/T_1 = V_2/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final state</p>
<p>Rearrange to solve for <mathjax>#V_2#</mathjax> - <strong>no not</strong> forget that the temperature of the gas <strong>must</strong> be expressed in <em>Kelvin</em>! </p>
<blockquote>
<p><mathjax>#V_2 = T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>In your case, this will be equal to </p>
<blockquote>
<p><mathjax>#V_2 = ( (273.15 + 17) color(red)(cancel(color(black)("K"))))/( (273.15 + 37) color(red)(cancel(color(black)("K")))) * "4.81 L"#</mathjax></p>
<p><mathjax>#V_2 = "6.641 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the temperatures of the gas, the answer will be </p>
<blockquote>
<p><mathjax>#V_2 = color(green)("6.6 L")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>If the pressure of the gas is <strong>not constant</strong>, you can solve the problem by using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation</em></p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - <em>the pressure, volume, and temperature of the gas at an initial state</em><br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - <em>the pressure, volume, and temperature of the gas at a final state</em></p></div>
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</article> | You transfer a sample of gas at 17°C from a volume of 4.81 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 atm. What is the new volume of the gas? | null |
891 | a9eb6753-6ddd-11ea-a617-ccda262736ce | https://socratic.org/questions/how-would-you-balance-this-equation-pb-no-3-2-pb-no-2-no-2-o-2 | Pb(NO3)2 -> Pb + 2 NO2 + O2 | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"Pb(NO3)2 -> Pb + 2 NO2 + O2"}] | [{"type":"chemical equation","value":"Pb(NO3)2 -> Pb + NO2 + O2"}] | <h1 class="questionTitle" itemprop="name">How would you balance this equation: #?Pb(NO_3)_2 → ?Pb + ?NO_2 + ?O_2#?</h1> | null | Pb(NO3)2 -> Pb + 2 NO2 + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this reaction first we have to see the number of moles of each element.</p>
<p><mathjax>#Pb(NO_3)_2 rarr Pb + O_2 + NO_2#</mathjax></p>
<p>(Pb = 1) (Pb = 1)</p>
<p><mathjax>#(N = 2)#</mathjax> <mathjax>#(O = cancel(4)^6)#</mathjax></p>
<p>(O = 6 ) <mathjax>#(N = cancel(1)^2)#</mathjax></p>
<p>First try adding 3 as the coefficient in front of <mathjax>#O_2#</mathjax> and 2 in front of <mathjax>#NO_2#</mathjax> but that will be wrong as 2NO2 has 4 oxygen atoms which we dont have, so to make a balanced equation we must remove 3 from O2 and add 2 in front of <mathjax>#NO_2#</mathjax></p>
<p>= <mathjax>#Pb(NO_3)_2 rarr Pb + O2 + 2NO_2#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Pb(NO_3)_2 rarr Pb + O2 + 2NO_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this reaction first we have to see the number of moles of each element.</p>
<p><mathjax>#Pb(NO_3)_2 rarr Pb + O_2 + NO_2#</mathjax></p>
<p>(Pb = 1) (Pb = 1)</p>
<p><mathjax>#(N = 2)#</mathjax> <mathjax>#(O = cancel(4)^6)#</mathjax></p>
<p>(O = 6 ) <mathjax>#(N = cancel(1)^2)#</mathjax></p>
<p>First try adding 3 as the coefficient in front of <mathjax>#O_2#</mathjax> and 2 in front of <mathjax>#NO_2#</mathjax> but that will be wrong as 2NO2 has 4 oxygen atoms which we dont have, so to make a balanced equation we must remove 3 from O2 and add 2 in front of <mathjax>#NO_2#</mathjax></p>
<p>= <mathjax>#Pb(NO_3)_2 rarr Pb + O2 + 2NO_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance this equation: #?Pb(NO_3)_2 → ?Pb + ?NO_2 + ?O_2#?</h1>
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<div class="markdown"><p><mathjax>#Pb(NO_3)_2 rarr Pb + O2 + 2NO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To balance this reaction first we have to see the number of moles of each element.</p>
<p><mathjax>#Pb(NO_3)_2 rarr Pb + O_2 + NO_2#</mathjax></p>
<p>(Pb = 1) (Pb = 1)</p>
<p><mathjax>#(N = 2)#</mathjax> <mathjax>#(O = cancel(4)^6)#</mathjax></p>
<p>(O = 6 ) <mathjax>#(N = cancel(1)^2)#</mathjax></p>
<p>First try adding 3 as the coefficient in front of <mathjax>#O_2#</mathjax> and 2 in front of <mathjax>#NO_2#</mathjax> but that will be wrong as 2NO2 has 4 oxygen atoms which we dont have, so to make a balanced equation we must remove 3 from O2 and add 2 in front of <mathjax>#NO_2#</mathjax></p>
<p>= <mathjax>#Pb(NO_3)_2 rarr Pb + O2 + 2NO_2#</mathjax></p></div>
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</article> | How would you balance this equation: #?Pb(NO_3)_2 → ?Pb + ?NO_2 + ?O_2#? | null |
892 | a91d5908-6ddd-11ea-8368-ccda262736ce | https://socratic.org/questions/a-compound-contains-1-10-mol-of-k-055-mol-of-te-and-1-65-mol-o-what-is-the-simpl | K2TeO3 | start chemical_formula qc_end physical_unit 6 6 3 4 mole qc_end physical_unit 14 14 12 13 mole qc_end end | [{"type":"other","value":"Chemical Formula [OF] this compound [IN] default"}] | [{"type":"chemical equation","value":"K2TeO3"}] | [{"type":"physical unit","value":"Mole [OF] K [=] \\pu{1.10 mol}"},{"type":"physical unit","value":"Mole [OF] Te [=] \\pu{0.55 mol}"},{"type":"physical unit","value":"Mole [OF] O [=] \\pu{1.65 mol}"}] | <h1 class="questionTitle" itemprop="name">A compound contains 1.10 mol of K, 055 mol of Te, and 1.65 mol O. What is the simplest formula of this compound?</h1> | null | K2TeO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide thru by the SMALLEST molar quantity, that of tellurium, to get an empirical formula.</p>
<p><mathjax>#"Moles of tellurium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.55*mol)/(0.55*mol)=1#</mathjax></p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.10*mol)/(0.55*mol)=2#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.65*mol)/(0.55*mol)=3#</mathjax></p>
<p>And thus the empirical formula is <mathjax>#K_2TeO_3#</mathjax>.</p>
<p>All I have done is to divide thru by the SMALLEST MOLAR QUANTITY.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>If we assume you meant <mathjax>#0.55*mol#</mathjax> <mathjax>#"tellurium"#</mathjax>, we get an <mathjax>#"empirical formula"#</mathjax> of <mathjax>#K_2TeO_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide thru by the SMALLEST molar quantity, that of tellurium, to get an empirical formula.</p>
<p><mathjax>#"Moles of tellurium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.55*mol)/(0.55*mol)=1#</mathjax></p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.10*mol)/(0.55*mol)=2#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.65*mol)/(0.55*mol)=3#</mathjax></p>
<p>And thus the empirical formula is <mathjax>#K_2TeO_3#</mathjax>.</p>
<p>All I have done is to divide thru by the SMALLEST MOLAR QUANTITY.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A compound contains 1.10 mol of K, 055 mol of Te, and 1.65 mol O. What is the simplest formula of this compound?</h1>
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<div class="markdown"><p>If we assume you meant <mathjax>#0.55*mol#</mathjax> <mathjax>#"tellurium"#</mathjax>, we get an <mathjax>#"empirical formula"#</mathjax> of <mathjax>#K_2TeO_3#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We divide thru by the SMALLEST molar quantity, that of tellurium, to get an empirical formula.</p>
<p><mathjax>#"Moles of tellurium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.55*mol)/(0.55*mol)=1#</mathjax></p>
<p><mathjax>#"Moles of potassium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.10*mol)/(0.55*mol)=2#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.65*mol)/(0.55*mol)=3#</mathjax></p>
<p>And thus the empirical formula is <mathjax>#K_2TeO_3#</mathjax>.</p>
<p>All I have done is to divide thru by the SMALLEST MOLAR QUANTITY.</p></div>
</div>
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</article> | A compound contains 1.10 mol of K, 055 mol of Te, and 1.65 mol O. What is the simplest formula of this compound? | null |
893 | a837f71b-6ddd-11ea-a435-ccda262736ce | https://socratic.org/questions/if-it-takes-54-ml-of-0-100-m-naoh-to-neutralize-125-ml-of-an-hcl-solution-what-i | 0.04 M | start physical_unit 15 16 concentration mol/l qc_end physical_unit 8 8 6 7 concentration qc_end physical_unit 16 16 3 4 volume qc_end physical_unit 15 16 11 12 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] HCl solution [IN] M"}] | [{"type":"physical unit","value":"0.04 M"}] | [{"type":"physical unit","value":"Concentration [OF] NaOH solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{54 mL}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{125 mL}"}] | <h1 class="questionTitle" itemprop="name">If it takes 54 mL of 0.100 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl?</h1> | null | 0.04 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:</p>
<p><strong>Acid+Base ->Salt+ Water</strong></p>
<p>Hence:<br/>
<mathjax>#HCl(aq)+NaOH(aq) -> NaCl(aq) + H_2O(l)#</mathjax></p>
<p>So our acid and base are in a 1:1 molar ratio in this case- so an equal amount of <mathjax>#NaOH#</mathjax> must have reacted with <mathjax>#HCl#</mathjax> for the solution to neutralize.</p>
<p>Using the concentration formula:</p>
<p><mathjax>#c=(n)/(v)#</mathjax></p>
<p><mathjax>#c#</mathjax>=concentration in <mathjax>#mol dm^-3#</mathjax><br/>
<mathjax>#n#</mathjax>=number of moles of substance dissolved in solution volume <mathjax>#(v)#</mathjax><br/>
<mathjax>#v#</mathjax>=volume of the solution in liters - <mathjax>#dm^3#</mathjax></p>
<p>We were given concentration and volume of <mathjax>#NaOH#</mathjax>, so we can find its number of moles:</p>
<p><mathjax>#0.1=(n)/0.054#</mathjax></p>
<p><mathjax>#n=0.0054#</mathjax> mol</p>
<p>Hence, this must be the number of moles of <mathjax>#HCl#</mathjax> found in the 125-milliliter solution, as we established that they reacted in a 1:1 ratio.</p>
<p>So:</p>
<p><mathjax>#0.0054/(0.125)=c#</mathjax></p>
<p><mathjax>#c=0.0432 mol dm^-3#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#c=0.0432 mol dm^-3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:</p>
<p><strong>Acid+Base ->Salt+ Water</strong></p>
<p>Hence:<br/>
<mathjax>#HCl(aq)+NaOH(aq) -> NaCl(aq) + H_2O(l)#</mathjax></p>
<p>So our acid and base are in a 1:1 molar ratio in this case- so an equal amount of <mathjax>#NaOH#</mathjax> must have reacted with <mathjax>#HCl#</mathjax> for the solution to neutralize.</p>
<p>Using the concentration formula:</p>
<p><mathjax>#c=(n)/(v)#</mathjax></p>
<p><mathjax>#c#</mathjax>=concentration in <mathjax>#mol dm^-3#</mathjax><br/>
<mathjax>#n#</mathjax>=number of moles of substance dissolved in solution volume <mathjax>#(v)#</mathjax><br/>
<mathjax>#v#</mathjax>=volume of the solution in liters - <mathjax>#dm^3#</mathjax></p>
<p>We were given concentration and volume of <mathjax>#NaOH#</mathjax>, so we can find its number of moles:</p>
<p><mathjax>#0.1=(n)/0.054#</mathjax></p>
<p><mathjax>#n=0.0054#</mathjax> mol</p>
<p>Hence, this must be the number of moles of <mathjax>#HCl#</mathjax> found in the 125-milliliter solution, as we established that they reacted in a 1:1 ratio.</p>
<p>So:</p>
<p><mathjax>#0.0054/(0.125)=c#</mathjax></p>
<p><mathjax>#c=0.0432 mol dm^-3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If it takes 54 mL of 0.100 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl?</h1>
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Oliver S.
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<span class="dateCreated" datetime="2018-04-08T13:11:48" itemprop="dateCreated">
Apr 8, 2018
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<div class="markdown"><p><mathjax>#c=0.0432 mol dm^-3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:</p>
<p><strong>Acid+Base ->Salt+ Water</strong></p>
<p>Hence:<br/>
<mathjax>#HCl(aq)+NaOH(aq) -> NaCl(aq) + H_2O(l)#</mathjax></p>
<p>So our acid and base are in a 1:1 molar ratio in this case- so an equal amount of <mathjax>#NaOH#</mathjax> must have reacted with <mathjax>#HCl#</mathjax> for the solution to neutralize.</p>
<p>Using the concentration formula:</p>
<p><mathjax>#c=(n)/(v)#</mathjax></p>
<p><mathjax>#c#</mathjax>=concentration in <mathjax>#mol dm^-3#</mathjax><br/>
<mathjax>#n#</mathjax>=number of moles of substance dissolved in solution volume <mathjax>#(v)#</mathjax><br/>
<mathjax>#v#</mathjax>=volume of the solution in liters - <mathjax>#dm^3#</mathjax></p>
<p>We were given concentration and volume of <mathjax>#NaOH#</mathjax>, so we can find its number of moles:</p>
<p><mathjax>#0.1=(n)/0.054#</mathjax></p>
<p><mathjax>#n=0.0054#</mathjax> mol</p>
<p>Hence, this must be the number of moles of <mathjax>#HCl#</mathjax> found in the 125-milliliter solution, as we established that they reacted in a 1:1 ratio.</p>
<p>So:</p>
<p><mathjax>#0.0054/(0.125)=c#</mathjax></p>
<p><mathjax>#c=0.0432 mol dm^-3#</mathjax></p></div>
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</article> | If it takes 54 mL of 0.100 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl? | null |
894 | aad25210-6ddd-11ea-9066-ccda262736ce | https://socratic.org/questions/592a900eb72cff0235a60eef | 5 NO3- + 1/2 I2 + 4 H+ -> 5 NO2 + IO3- + 2 H2O | start chemical_equation qc_end substance 2 2 qc_end substance 5 6 qc_end substance 8 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"5 NO3- + 1/2 I2 + 4 H+ -> 5 NO2 + IO3- + 2 H2O"}] | [{"type":"substance name","value":"Iodine"},{"type":"substance name","value":"Iodate ion"},{"type":"substance name","value":"Nitric acid"}] | <h1 class="questionTitle" itemprop="name">How is iodine oxidized to iodate ion by nitric acid?</h1> | null | 5 NO3- + 1/2 I2 + 4 H+ -> 5 NO2 + IO3- + 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental iodine is oxidized up to iodate, <mathjax>#IO_3^(-)#</mathjax>......</p>
<p><mathjax>#1/2I_2(s) + 3H_2O(l) rarr IO_3^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And nitric acid is reduced to <mathjax>#NO_2#</mathjax>:</p>
<p><mathjax>#stackrel(V)NO_3^(-) rarr stackrel(IV)NO_2#</mathjax>, i.e.</p>
<p><mathjax>#NO_3^(-) +2H^(+) +e^(-) rarrNO_2 +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And in the usual way we cross-multiply the individual redox equations, so that electrons do not appear in the final reaction.</p>
<p><mathjax>#(i) + 5xx(ii)=#</mathjax></p>
<p><mathjax>#5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)#</mathjax></p>
<p>Which looks balanced with respect to mass and charge to me. Is it? All care taken but no responsibility admitted! </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>With conc. nitric acid...........</p>
<p><mathjax>#5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l) #</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental iodine is oxidized up to iodate, <mathjax>#IO_3^(-)#</mathjax>......</p>
<p><mathjax>#1/2I_2(s) + 3H_2O(l) rarr IO_3^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And nitric acid is reduced to <mathjax>#NO_2#</mathjax>:</p>
<p><mathjax>#stackrel(V)NO_3^(-) rarr stackrel(IV)NO_2#</mathjax>, i.e.</p>
<p><mathjax>#NO_3^(-) +2H^(+) +e^(-) rarrNO_2 +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And in the usual way we cross-multiply the individual redox equations, so that electrons do not appear in the final reaction.</p>
<p><mathjax>#(i) + 5xx(ii)=#</mathjax></p>
<p><mathjax>#5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)#</mathjax></p>
<p>Which looks balanced with respect to mass and charge to me. Is it? All care taken but no responsibility admitted! </p></div>
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<h1 class="questionTitle" itemprop="name">How is iodine oxidized to iodate ion by nitric acid?</h1>
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<div class="markdown"><p>With conc. nitric acid...........</p>
<p><mathjax>#5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l) #</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental iodine is oxidized up to iodate, <mathjax>#IO_3^(-)#</mathjax>......</p>
<p><mathjax>#1/2I_2(s) + 3H_2O(l) rarr IO_3^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And nitric acid is reduced to <mathjax>#NO_2#</mathjax>:</p>
<p><mathjax>#stackrel(V)NO_3^(-) rarr stackrel(IV)NO_2#</mathjax>, i.e.</p>
<p><mathjax>#NO_3^(-) +2H^(+) +e^(-) rarrNO_2 +H_2O(l)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And in the usual way we cross-multiply the individual redox equations, so that electrons do not appear in the final reaction.</p>
<p><mathjax>#(i) + 5xx(ii)=#</mathjax></p>
<p><mathjax>#5NO_3^(-) +1/2I_2(s) +4H^(+)rarr5NO_2 +IO_3^(-)+2H_2O(l)#</mathjax></p>
<p>Which looks balanced with respect to mass and charge to me. Is it? All care taken but no responsibility admitted! </p></div>
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</article> | How is iodine oxidized to iodate ion by nitric acid? | null |
895 | aa78e200-6ddd-11ea-9554-ccda262736ce | https://socratic.org/questions/a-sample-of-metal-with-a-specific-heat-of-502-j-g-c-is-heated-to-100-0-c-and-the | 459 g | start physical_unit 44 45 mass g qc_end physical_unit 1 3 9 12 specific_heat qc_end physical_unit 1 3 16 17 temperature qc_end physical_unit 24 26 22 23 mass qc_end physical_unit 24 26 28 29 temperature qc_end physical_unit 34 35 37 38 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] the metal [IN] g"}] | [{"type":"physical unit","value":"459 g"}] | [{"type":"physical unit","value":"Specific heat [OF] the metal sample [=] \\pu{0.502 J/(g * ℃)}"},{"type":"physical unit","value":"Temperature1 [OF] the metal sample [=] \\pu{100.0 ℃}"},{"type":"physical unit","value":"Mass [OF] water sample [=] \\pu{50.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] water sample [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the system [=] \\pu{76.2 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of metal with a specific heat of .502 J/g°c is heated to 100.0°C and then placed a 50.0 g sample of water at 20.0°C. The final temperature of the system is 76.2°C. What is the mass of the metal used in process?</h1> | null | 459 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must identify the heat transfers that are happening here.</p>
<p>One is the heat transferred from the metal as it cools (<mathjax>#q_1#</mathjax>).</p>
<p>The second is the heat transferred to the water as it warms (<mathjax>#q_2#</mathjax>).</p>
<p>Per the <strong>Law of Conservation of Energy</strong>, the sum of the two heat transfers must be zero.</p>
<blockquote>
<blockquote>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
</blockquote>
</blockquote>
<p>The formula for the heat absorbed by or released from a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat<br/>
<mathjax>#m#</mathjax> is the mass of the substance<br/>
<mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the material<br/>
<mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>This gives us</p>
<p><mathjax>#color(white)(ml)q_1 color(white)(mm)+color(white)(mll) q_2 color(white)(mm)= 0#</mathjax></p>
<p><mathjax>#m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem, we have</p>
<p><mathjax>#m_1color(white)(l) = ?#</mathjax><br/>
<mathjax>#c_1color(white)(m) = "0.502 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_1 = T_"f" - T_"i" = "(76.2 - 100.0) °C" = "-23.8 °C"#</mathjax></p>
<p><mathjax>#m_2color(white)(l) = "50.0 g"#</mathjax><br/>
<mathjax>#c_2color(white)(m)= "4.184 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_2 = T_"f" - T_"i" = "(76.2 - 20.0) °C" = "26.2 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = m_1 × "0.502 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × ("-23.8" color(red)(cancel(color(black)("°C")))) = "-11.95"color(white)(l)m_1 color(white)(l)"J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = "50.0 g" × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 26.2 color(red)(cancel(color(black)("°C"))) = "5481 J"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = "-11.95"m_1 color(red)(cancel(color(black)("J")))·"g"^"-1" + 5481 color(red)(cancel(color(black)("J"))) = 0#</mathjax></p>
<p><mathjax>#m_1 = 5481/("11.95 g"^"-1") = "459 g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The mass of the metal is 459 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must identify the heat transfers that are happening here.</p>
<p>One is the heat transferred from the metal as it cools (<mathjax>#q_1#</mathjax>).</p>
<p>The second is the heat transferred to the water as it warms (<mathjax>#q_2#</mathjax>).</p>
<p>Per the <strong>Law of Conservation of Energy</strong>, the sum of the two heat transfers must be zero.</p>
<blockquote>
<blockquote>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
</blockquote>
</blockquote>
<p>The formula for the heat absorbed by or released from a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat<br/>
<mathjax>#m#</mathjax> is the mass of the substance<br/>
<mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the material<br/>
<mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>This gives us</p>
<p><mathjax>#color(white)(ml)q_1 color(white)(mm)+color(white)(mll) q_2 color(white)(mm)= 0#</mathjax></p>
<p><mathjax>#m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem, we have</p>
<p><mathjax>#m_1color(white)(l) = ?#</mathjax><br/>
<mathjax>#c_1color(white)(m) = "0.502 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_1 = T_"f" - T_"i" = "(76.2 - 100.0) °C" = "-23.8 °C"#</mathjax></p>
<p><mathjax>#m_2color(white)(l) = "50.0 g"#</mathjax><br/>
<mathjax>#c_2color(white)(m)= "4.184 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_2 = T_"f" - T_"i" = "(76.2 - 20.0) °C" = "26.2 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = m_1 × "0.502 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × ("-23.8" color(red)(cancel(color(black)("°C")))) = "-11.95"color(white)(l)m_1 color(white)(l)"J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = "50.0 g" × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 26.2 color(red)(cancel(color(black)("°C"))) = "5481 J"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = "-11.95"m_1 color(red)(cancel(color(black)("J")))·"g"^"-1" + 5481 color(red)(cancel(color(black)("J"))) = 0#</mathjax></p>
<p><mathjax>#m_1 = 5481/("11.95 g"^"-1") = "459 g"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of metal with a specific heat of .502 J/g°c is heated to 100.0°C and then placed a 50.0 g sample of water at 20.0°C. The final temperature of the system is 76.2°C. What is the mass of the metal used in process?</h1>
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<div class="markdown"><p>The mass of the metal is 459 g.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We must identify the heat transfers that are happening here.</p>
<p>One is the heat transferred from the metal as it cools (<mathjax>#q_1#</mathjax>).</p>
<p>The second is the heat transferred to the water as it warms (<mathjax>#q_2#</mathjax>).</p>
<p>Per the <strong>Law of Conservation of Energy</strong>, the sum of the two heat transfers must be zero.</p>
<blockquote>
<blockquote>
<p><mathjax>#q_1 + q_2 = 0#</mathjax></p>
</blockquote>
</blockquote>
<p>The formula for the heat absorbed by or released from a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat<br/>
<mathjax>#m#</mathjax> is the mass of the substance<br/>
<mathjax>#c#</mathjax> is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the material<br/>
<mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>This gives us</p>
<p><mathjax>#color(white)(ml)q_1 color(white)(mm)+color(white)(mll) q_2 color(white)(mm)= 0#</mathjax></p>
<p><mathjax>#m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0#</mathjax></p>
<blockquote></blockquote>
<p>In this problem, we have</p>
<p><mathjax>#m_1color(white)(l) = ?#</mathjax><br/>
<mathjax>#c_1color(white)(m) = "0.502 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_1 = T_"f" - T_"i" = "(76.2 - 100.0) °C" = "-23.8 °C"#</mathjax></p>
<p><mathjax>#m_2color(white)(l) = "50.0 g"#</mathjax><br/>
<mathjax>#c_2color(white)(m)= "4.184 J·°C"^"-1""g"^"-1"#</mathjax><br/>
<mathjax>#ΔT_2 = T_"f" - T_"i" = "(76.2 - 20.0) °C" = "26.2 °C"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q_1 = m_1 × "0.502 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × ("-23.8" color(red)(cancel(color(black)("°C")))) = "-11.95"color(white)(l)m_1 color(white)(l)"J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = "50.0 g" × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × 26.2 color(red)(cancel(color(black)("°C"))) = "5481 J"#</mathjax></p>
<p><mathjax>#q_1 + q_2 = "-11.95"m_1 color(red)(cancel(color(black)("J")))·"g"^"-1" + 5481 color(red)(cancel(color(black)("J"))) = 0#</mathjax></p>
<p><mathjax>#m_1 = 5481/("11.95 g"^"-1") = "459 g"#</mathjax></p></div>
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</article> | A sample of metal with a specific heat of .502 J/g°c is heated to 100.0°C and then placed a 50.0 g sample of water at 20.0°C. The final temperature of the system is 76.2°C. What is the mass of the metal used in process? | null |
896 | ace11440-6ddd-11ea-93e2-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-a-gas-in-890ml-at-21-c-and-750-mmhg | 0.04 moles | start physical_unit 6 6 mole mol qc_end physical_unit 6 6 14 15 pressure qc_end physical_unit 6 6 11 12 temperature qc_end physical_unit 6 6 8 9 volume qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] moles"}] | [{"type":"physical unit","value":"0.04 moles"}] | [{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{750 mmHg}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{21 ℃}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{890 mL}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in a gas in 890mL at 21 °C and 750 mmHg?</h1> | null | 0.04 moles | <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/ideal-gas-law">ideal gas law</a>. The formula is <mathjax>#PV=nRT#</mathjax>, where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the Kelvin temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="750 mmHg"="750 Torr"#</mathjax><br/>
<mathjax>#V=890cancel"mL"xx(1"L")/(1000cancel"mL")="0.89 L"#</mathjax><br/>
<mathjax>#R="62.363577 L Torr K"^(-1) "mol"^(-1)#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="21"^@"C"+273.15"=294 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#n#</mathjax>. Substitute the given/known values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(750cancel"Torr" * 0.89cancel"L")/((62.363577 cancel"L" cancel"Torr" cancel"K"^(-1)" mol"^(-1))xx(294cancel"K"))="0.036 mol"#</mathjax> rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#n="0.036 mol"#</mathjax></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/ideal-gas-law">ideal gas law</a>. The formula is <mathjax>#PV=nRT#</mathjax>, where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the Kelvin temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="750 mmHg"="750 Torr"#</mathjax><br/>
<mathjax>#V=890cancel"mL"xx(1"L")/(1000cancel"mL")="0.89 L"#</mathjax><br/>
<mathjax>#R="62.363577 L Torr K"^(-1) "mol"^(-1)#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="21"^@"C"+273.15"=294 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#n#</mathjax>. Substitute the given/known values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(750cancel"Torr" * 0.89cancel"L")/((62.363577 cancel"L" cancel"Torr" cancel"K"^(-1)" mol"^(-1))xx(294cancel"K"))="0.036 mol"#</mathjax> rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in a gas in 890mL at 21 °C and 750 mmHg?</h1>
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<span class="dateCreated" datetime="2016-03-03T06:33:49" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#n="0.036 mol"#</mathjax></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/ideal-gas-law">ideal gas law</a>. The formula is <mathjax>#PV=nRT#</mathjax>, where <mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is the Kelvin temperature.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="750 mmHg"="750 Torr"#</mathjax><br/>
<mathjax>#V=890cancel"mL"xx(1"L")/(1000cancel"mL")="0.89 L"#</mathjax><br/>
<mathjax>#R="62.363577 L Torr K"^(-1) "mol"^(-1)#</mathjax><br/>
<a href="https://en.m.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.m.wikipedia.org/wiki/Gas_constant</a><br/>
<mathjax>#T="21"^@"C"+273.15"=294 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#n#</mathjax>. Substitute the given/known values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(750cancel"Torr" * 0.89cancel"L")/((62.363577 cancel"L" cancel"Torr" cancel"K"^(-1)" mol"^(-1))xx(294cancel"K"))="0.036 mol"#</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 many moles are in a gas in 890mL at 21 °C and 750 mmHg? | null |
897 | a9bc9bc6-6ddd-11ea-8898-ccda262736ce | https://socratic.org/questions/how-would-you-balance-the-following-equation-nh3-o2-no2-h2o | 4 NH3 + 7 O2 -> 4 NO2 + 6 H2O | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"4 NH3 + 7 O2 -> 4 NO2 + 6 H2O"}] | [{"type":"chemical equation","value":"NH3 + O2 -> NO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How would you balance the following equation: NH3 + O2-->NO2 +H2O?</h1> | null | 4 NH3 + 7 O2 -> 4 NO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms of each element must be the same on left and right hand side of equation.<br/>
We may only add balancing numbers in front (ie molecules) in order to achieve this.</p></div>
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<div class="markdown"><p><mathjax>#4NH_3+7O_2->4NO_2+6H_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms of each element must be the same on left and right hand side of equation.<br/>
We may only add balancing numbers in front (ie molecules) in order to achieve this.</p></div>
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Trevor Ryan.
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<div class="markdown"><p><mathjax>#4NH_3+7O_2->4NO_2+6H_2O#</mathjax></p></div>
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<div class="markdown"><p>The number of atoms of each element must be the same on left and right hand side of equation.<br/>
We may only add balancing numbers in front (ie molecules) in order to achieve this.</p></div>
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</article> | How would you balance the following equation: NH3 + O2-->NO2 +H2O? | null |
898 | ac785358-6ddd-11ea-aaa5-ccda262736ce | https://socratic.org/questions/a-sample-of-h-2o-with-a-mass-of-46-0-grams-has-a-temperature-of-100-c-how-many-j | 103960.00 joules | start physical_unit 1 3 heat_energy j qc_end physical_unit 1 3 8 9 mass qc_end physical_unit 1 3 14 15 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Necessary energy [OF] H2O sample [IN] joules"}] | [{"type":"physical unit","value":"103960.00 joules"}] | [{"type":"physical unit","value":"Mass [OF] H2O sample [=] \\pu{46.0 grams}"},{"type":"physical unit","value":"Temperature [OF] H2O sample [=] \\pu{100 ℃.}"},{"type":"other","value":"Boil the water."}] | <h1 class="questionTitle" itemprop="name">A sample of #H_2O# with a mass of 46.0 grams has a temperature of 100°C. How many joules of energy are necessary to boil the water?</h1> | null | 103960.00 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat of vaporization of water, <mathjax>#DeltaH_("vap")~=2.26*kJ*g^-1#</mathjax>. Your heat of vaporization is out by three order of magnitude, This is all too easy to do of course.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>If you said <mathjax>#Delta=2.09*J*g^-1xx46.0*g#</mathjax> you would be wrong.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat of vaporization of water, <mathjax>#DeltaH_("vap")~=2.26*kJ*g^-1#</mathjax>. Your heat of vaporization is out by three order of magnitude, This is all too easy to do of course.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of #H_2O# with a mass of 46.0 grams has a temperature of 100°C. How many joules of energy are necessary to boil the water?</h1>
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anor277
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Jun 23, 2016
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<div class="markdown"><p>If you said <mathjax>#Delta=2.09*J*g^-1xx46.0*g#</mathjax> you would be wrong.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The heat of vaporization of water, <mathjax>#DeltaH_("vap")~=2.26*kJ*g^-1#</mathjax>. Your heat of vaporization is out by three order of magnitude, This is all too easy to do of course.</p></div>
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</article> | A sample of #H_2O# with a mass of 46.0 grams has a temperature of 100°C. How many joules of energy are necessary to boil the water? | null |
899 | a9c7422a-6ddd-11ea-bf1c-ccda262736ce | https://socratic.org/questions/a-sample-of-an-unknown-metal-chlorate-weighing-5-837-g-is-heated-until-all-of-th | 75.00% | start physical_unit 34 38 percent none qc_end physical_unit 1 6 8 9 mass qc_end c_other OTHER qc_end physical_unit 20 21 27 28 mass qc_end end | [{"type":"physical unit","value":"Percentage [OF] oxygen in this metal chlorate"}] | [{"type":"physical unit","value":"75.00%"}] | [{"type":"physical unit","value":"Weight [OF] an unknown metal chlorate sample [=] \\pu{5.837 g}"},{"type":"other","value":"A sample of an unknown metal chlorate is heated until all of the oxygen is driven off."},{"type":"physical unit","value":"Weight [OF] the residue [=] \\pu{1.459 g}"}] | <h1 class="questionTitle" itemprop="name">A sample of an unknown metal chlorate, weighing 5.837 g, is heated until all of the oxygen is driven off. The residue remaining in the container weighs 1.459 g. What is the percentage of oxygen in this metal chlorate?</h1> | null | 75.00% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>While I have given the stoichiometric equation for the decomposition of the metal chlorate, this reaction is simpler than it seems. We know that a mass of <mathjax>#5.837-1.459=4.378#</mathjax> <mathjax>#g#</mathjax> was lost during the reaction. This must represent a mass of oxygen.</p>
<p>Given that we started with a mass of <mathjax>#5.837#</mathjax> <mathjax>#g#</mathjax>, this represents a percentage of <mathjax>#(4.378*g)/(5,837*g) xx 100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> in the initial sample.</p></div>
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<div class="markdown"><p><mathjax>#M_n(ClO_3)_m(s) + Delta rarr M_nCl_m(s) + (3m)/2O_2(g)#</mathjax></p></div>
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<div class="markdown"><p>While I have given the stoichiometric equation for the decomposition of the metal chlorate, this reaction is simpler than it seems. We know that a mass of <mathjax>#5.837-1.459=4.378#</mathjax> <mathjax>#g#</mathjax> was lost during the reaction. This must represent a mass of oxygen.</p>
<p>Given that we started with a mass of <mathjax>#5.837#</mathjax> <mathjax>#g#</mathjax>, this represents a percentage of <mathjax>#(4.378*g)/(5,837*g) xx 100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> in the initial sample.</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of an unknown metal chlorate, weighing 5.837 g, is heated until all of the oxygen is driven off. The residue remaining in the container weighs 1.459 g. What is the percentage of oxygen in this metal chlorate?</h1>
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anor277
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<div class="markdown"><p><mathjax>#M_n(ClO_3)_m(s) + Delta rarr M_nCl_m(s) + (3m)/2O_2(g)#</mathjax></p></div>
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<div class="markdown"><p>While I have given the stoichiometric equation for the decomposition of the metal chlorate, this reaction is simpler than it seems. We know that a mass of <mathjax>#5.837-1.459=4.378#</mathjax> <mathjax>#g#</mathjax> was lost during the reaction. This must represent a mass of oxygen.</p>
<p>Given that we started with a mass of <mathjax>#5.837#</mathjax> <mathjax>#g#</mathjax>, this represents a percentage of <mathjax>#(4.378*g)/(5,837*g) xx 100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> in the initial sample.</p></div>
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</article> | A sample of an unknown metal chlorate, weighing 5.837 g, is heated until all of the oxygen is driven off. The residue remaining in the container weighs 1.459 g. What is the percentage of oxygen in this metal chlorate? | null |
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