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600 | a91dcea6-6ddd-11ea-bb45-ccda262736ce | https://socratic.org/questions/how-many-grams-of-sodium-is-present-in-2-moles | 45.98 grams | start physical_unit 4 4 mass g qc_end physical_unit 4 4 8 9 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium [IN] grams"}] | [{"type":"physical unit","value":"45.98 grams"}] | [{"type":"physical unit","value":"Mole [OF] sodium [=] \\pu{2 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of sodium is present in 2 moles?</h1> | null | 45.98 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium metal has a molar mass of <mathjax>#22.99#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>Thus if there is a <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> quantity the mass is:</p>
<p><mathjax>#2*cancel(mol)xx22.99*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>How did I know that sodium metal has such a molar mass? Because I have access to a Periodic Table, which gives me the molar mass of not just sodium but of every other known element. </p>
<p>Can you take the Table into an exam? Of course you can, or at least 1 will be provided. In every exam in Physics and Chemistry you ever sit, you must be provided a copy of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>. It is up to you to learn how to use it effectively. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Approx. <mathjax>#46#</mathjax> <mathjax>#g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium metal has a molar mass of <mathjax>#22.99#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>Thus if there is a <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> quantity the mass is:</p>
<p><mathjax>#2*cancel(mol)xx22.99*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>How did I know that sodium metal has such a molar mass? Because I have access to a Periodic Table, which gives me the molar mass of not just sodium but of every other known element. </p>
<p>Can you take the Table into an exam? Of course you can, or at least 1 will be provided. In every exam in Physics and Chemistry you ever sit, you must be provided a copy of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>. It is up to you to learn how to use it effectively. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of sodium is present in 2 moles?</h1>
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anor277
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Apr 25, 2016
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<div class="markdown"><p>Approx. <mathjax>#46#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sodium metal has a molar mass of <mathjax>#22.99#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p>
<p>Thus if there is a <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax> quantity the mass is:</p>
<p><mathjax>#2*cancel(mol)xx22.99*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p>
<p>How did I know that sodium metal has such a molar mass? Because I have access to a Periodic Table, which gives me the molar mass of not just sodium but of every other known element. </p>
<p>Can you take the Table into an exam? Of course you can, or at least 1 will be provided. In every exam in Physics and Chemistry you ever sit, you must be provided a copy of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>. It is up to you to learn how to use it effectively. </p></div>
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Edmund Liew
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Apr 26, 2016
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<div class="markdown"><p>2 moles <mathjax>#xx#</mathjax> molar mass of sodium = ??</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assume that:<br/>
n = number of moles<br/>
m = mass of substance<br/>
M = molar mass (equivalent to atomic weight on the periodic table)</p>
<p><mathjax>#n = m -: M#</mathjax></p>
<p>2 moles (n) has been given to you.</p>
<p>The next step is to find the molar mass (M) of Na. If you refer to your periodic table, the molar mass of Na (sodium) is 23.0 g/mol. </p>
<p>Now you have to find the mass (m).<br/>
The mass of sodium is: <br/>
[m = 23.0 g/mol <mathjax>#xx#</mathjax> 2 moles] = 46.0 grams of sodium is present in 2 moles</p></div>
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</article> | How many grams of sodium is present in 2 moles? | null |
601 | ab023ea9-6ddd-11ea-b818-ccda262736ce | https://socratic.org/questions/how-many-grams-of-h-are-there-in-23-5-g-of-h-2o | 2.63 grams | start physical_unit 4 4 mass g qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] H [IN] grams"}] | [{"type":"physical unit","value":"2.63 grams"}] | [{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{23.5 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of H are there in 23.5 g of #H_2O#? </h1> | null | 2.63 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One way to approach this problem is to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water. </p>
<p>You know that one water molecule contains</p>
<blockquote>
<ul>
<li><em>one oxygen atom</em></li>
<li><em>two hydrogen atoms</em></li>
</ul>
</blockquote>
<p>This means that <em>every mole</em> of water molecules will contain <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen atoms. </p>
<p>To find the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water, use the molar mass of water and that of hydrogen</p>
<blockquote>
<p><mathjax>#(2 xx 1.00794color(red)(cancel(color(black)("g/mol"))))/(18.0153color(red)(cancel(color(black)("g/mol")))) xx 100 = 11.19%#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of water will contain a total of <mathjax>#"11.19 g"#</mathjax> of hydrogen. Since your sample has a mass of <mathjax>#"23.5 g"#</mathjax>, it follows that it will contain</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g water"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g water")))) = color(green)("2.63 g H"#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can first find the number of moles of water in get in that <mathjax>#"23.5 g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0153color(red)(cancel(color(black)("g")))) = "1.3044 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>SInce one molecule of water contais 2 atoms of hydrogen, you will get</p>
<blockquote>
<p><mathjax>#1.3044color(red)(cancel(color(black)("moles water"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole water")))) = "2.6088 moles H"#</mathjax></p>
</blockquote>
<p>Finally, use hydrogen's molar mass to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.6088color(red)(cancel(color(black)("moles"))) * "1.00794 g H"/(1color(red)(cancel(color(black)("mole")))) = color(green)("2.63 g H")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.63 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One way to approach this problem is to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water. </p>
<p>You know that one water molecule contains</p>
<blockquote>
<ul>
<li><em>one oxygen atom</em></li>
<li><em>two hydrogen atoms</em></li>
</ul>
</blockquote>
<p>This means that <em>every mole</em> of water molecules will contain <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen atoms. </p>
<p>To find the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water, use the molar mass of water and that of hydrogen</p>
<blockquote>
<p><mathjax>#(2 xx 1.00794color(red)(cancel(color(black)("g/mol"))))/(18.0153color(red)(cancel(color(black)("g/mol")))) xx 100 = 11.19%#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of water will contain a total of <mathjax>#"11.19 g"#</mathjax> of hydrogen. Since your sample has a mass of <mathjax>#"23.5 g"#</mathjax>, it follows that it will contain</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g water"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g water")))) = color(green)("2.63 g H"#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can first find the number of moles of water in get in that <mathjax>#"23.5 g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0153color(red)(cancel(color(black)("g")))) = "1.3044 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>SInce one molecule of water contais 2 atoms of hydrogen, you will get</p>
<blockquote>
<p><mathjax>#1.3044color(red)(cancel(color(black)("moles water"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole water")))) = "2.6088 moles H"#</mathjax></p>
</blockquote>
<p>Finally, use hydrogen's molar mass to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.6088color(red)(cancel(color(black)("moles"))) * "1.00794 g H"/(1color(red)(cancel(color(black)("mole")))) = color(green)("2.63 g H")#</mathjax></p>
</blockquote></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of H are there in 23.5 g of #H_2O#? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"2.63 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One way to approach this problem is to determine the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water. </p>
<p>You know that one water molecule contains</p>
<blockquote>
<ul>
<li><em>one oxygen atom</em></li>
<li><em>two hydrogen atoms</em></li>
</ul>
</blockquote>
<p>This means that <em>every mole</em> of water molecules will contain <strong>2 moles</strong> of hydrogen and <strong>1 mole</strong> of oxygen atoms. </p>
<p>To find the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of hydrogen in water, use the molar mass of water and that of hydrogen</p>
<blockquote>
<p><mathjax>#(2 xx 1.00794color(red)(cancel(color(black)("g/mol"))))/(18.0153color(red)(cancel(color(black)("g/mol")))) xx 100 = 11.19%#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of water will contain a total of <mathjax>#"11.19 g"#</mathjax> of hydrogen. Since your sample has a mass of <mathjax>#"23.5 g"#</mathjax>, it follows that it will contain</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g water"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g water")))) = color(green)("2.63 g H"#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, you can first find the number of moles of water in get in that <mathjax>#"23.5 g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#23.5color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.0153color(red)(cancel(color(black)("g")))) = "1.3044 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>SInce one molecule of water contais 2 atoms of hydrogen, you will get</p>
<blockquote>
<p><mathjax>#1.3044color(red)(cancel(color(black)("moles water"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole water")))) = "2.6088 moles H"#</mathjax></p>
</blockquote>
<p>Finally, use hydrogen's molar mass to find how many grams would contain this many moles</p>
<blockquote>
<p><mathjax>#2.6088color(red)(cancel(color(black)("moles"))) * "1.00794 g H"/(1color(red)(cancel(color(black)("mole")))) = color(green)("2.63 g H")#</mathjax></p>
</blockquote></div>
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</article> | How many grams of H are there in 23.5 g of #H_2O#? | null |
602 | aaabf523-6ddd-11ea-b9b7-ccda262736ce | https://socratic.org/questions/how-would-you-balance-albr3-k2so4-kbr-al2-so4-3 | 2 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3"}] | [{"type":"chemical equation","value":"AlBr3 + K2SO4 -> KBr + Al2(SO4)3"}] | <h1 class="questionTitle" itemprop="name">How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?
</h1> | null | 2 AlBr3 + 3 K2SO4 -> 6 KBr + Al2(SO4)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.</p>
<p>However, this leaves us with 6 potassium atoms on the reactant side and only 1 on the product side, so we multiply the potassium bromide by 6.</p>
<p>Now, we have the imbalance of 6 bromine atoms on the right and only 3 on the right, so we multiply the aluminum bromide on the right by 2. Finally, this leaves 2 aluminum atoms on the reactant side and on the product side, so we are done balancing.</p></div>
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<div class="markdown"><p><mathjax>#2AlBr_3+3K_2SO_4rarr6KBr+Al_2(SO_4)_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.</p>
<p>However, this leaves us with 6 potassium atoms on the reactant side and only 1 on the product side, so we multiply the potassium bromide by 6.</p>
<p>Now, we have the imbalance of 6 bromine atoms on the right and only 3 on the right, so we multiply the aluminum bromide on the right by 2. Finally, this leaves 2 aluminum atoms on the reactant side and on the product side, so we are done balancing.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?
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mason m
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<div class="markdown"><p><mathjax>#2AlBr_3+3K_2SO_4rarr6KBr+Al_2(SO_4)_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.</p>
<p>However, this leaves us with 6 potassium atoms on the reactant side and only 1 on the product side, so we multiply the potassium bromide by 6.</p>
<p>Now, we have the imbalance of 6 bromine atoms on the right and only 3 on the right, so we multiply the aluminum bromide on the right by 2. Finally, this leaves 2 aluminum atoms on the reactant side and on the product side, so we are done balancing.</p></div>
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</article> | How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?
| null |
603 | ad2521d0-6ddd-11ea-b8ce-ccda262736ce | https://socratic.org/questions/using-the-following-reaction-if-5-93-grams-of-sodium-cyanide-are-reacted-with-8- | 8.59 grams | start physical_unit 22 23 mass g qc_end chemical_equation 27 35 qc_end physical_unit 8 9 5 6 mass qc_end physical_unit 16 17 13 14 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium sulfate [IN] grams"}] | [{"type":"physical unit","value":"8.59 grams"}] | [{"type":"chemical equation","value":"2 NaCN + H2SO4 -> Na2SO4 + 2 HCN"},{"type":"physical unit","value":"Mass [OF] sodium cyanide [=] \\pu{5.93 grams}"},{"type":"physical unit","value":"Mass [OF] sulfuric acid [=] \\pu{8.45 grams}"}] | <h1 class="questionTitle" itemprop="name">Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? #2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN#</h1> | null | 8.59 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong>limiting reactant</strong> problem.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong></p>
<p><mathjax>#"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"NaCN"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 5.93 color(red)(cancel(color(black)("g NaCN"))) × (1 color(red)(cancel(color(black)("mol NaCN"))))/(49.01 color(red)(cancel(color(black)("g NaCN")))) × (1 "mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaCN")))) = "0.060 50 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"H"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 8.45 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) × (1 "mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.086 15 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Identify the limiting reactant</strong></p>
<p>The limiting reactant is <mathjax>#"NaCN"#</mathjax>, because it produces the fewest moles of product.</p>
<blockquote></blockquote>
<p><strong>5. Calculate the mass of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Mass of Na"_2"SO"_4 = "0.060 50" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "8.59 g Na"_2"SO"_4#</mathjax></p></div>
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<div>
<div class="markdown"><p>The reaction will produce 8.59 g of <mathjax>#"Na"_2"SO"_4#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong>limiting reactant</strong> problem.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong></p>
<p><mathjax>#"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"NaCN"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 5.93 color(red)(cancel(color(black)("g NaCN"))) × (1 color(red)(cancel(color(black)("mol NaCN"))))/(49.01 color(red)(cancel(color(black)("g NaCN")))) × (1 "mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaCN")))) = "0.060 50 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"H"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 8.45 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) × (1 "mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.086 15 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Identify the limiting reactant</strong></p>
<p>The limiting reactant is <mathjax>#"NaCN"#</mathjax>, because it produces the fewest moles of product.</p>
<blockquote></blockquote>
<p><strong>5. Calculate the mass of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Mass of Na"_2"SO"_4 = "0.060 50" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "8.59 g Na"_2"SO"_4#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? #2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN#</h1>
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<div class="markdown"><p>The reaction will produce 8.59 g of <mathjax>#"Na"_2"SO"_4#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This appears to be a <strong>limiting reactant</strong> problem.</p>
<blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong></p>
<p><mathjax>#"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"NaCN"#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 5.93 color(red)(cancel(color(black)("g NaCN"))) × (1 color(red)(cancel(color(black)("mol NaCN"))))/(49.01 color(red)(cancel(color(black)("g NaCN")))) × (1 "mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaCN")))) = "0.060 50 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"Na"_2"SO"_4#</mathjax> formed from the <mathjax>#"H"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Moles of Na"_2"SO"_4 = 8.45 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) × (1 "mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.086 15 mol Na"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Identify the limiting reactant</strong></p>
<p>The limiting reactant is <mathjax>#"NaCN"#</mathjax>, because it produces the fewest moles of product.</p>
<blockquote></blockquote>
<p><strong>5. Calculate the mass of <mathjax>#"Na"_2"SO"_4#</mathjax></strong></p>
<p><mathjax>#"Mass of Na"_2"SO"_4 = "0.060 50" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "8.59 g Na"_2"SO"_4#</mathjax></p></div>
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</article> | Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? #2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN# | null |
604 | a93d0ae8-6ddd-11ea-b183-ccda262736ce | https://socratic.org/questions/587cc2247c01490e5a25fb57 | 12.67 molar | start physical_unit 13 13 mole mol/l qc_end physical_unit 4 4 1 2 mole qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Mole [OF] ammonia [IN] molar"}] | [{"type":"physical unit","value":"12.67 molar"}] | [{"type":"physical unit","value":"Mole [OF] H2(g) [=] \\pu{19 mol}"},{"type":"chemical equation","value":"N2(g)"}] | <h1 class="questionTitle" itemprop="name">If #19*mol# of #H_2(g)# react with stoichiometric #N_2(g)#, what molar quantity of ammonia results?</h1> | null | 12.67 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric equation........</p>
<p><mathjax>#N_2(g) + 3H_2(g) stackrel("catalysis")rarr2NH_3(g)#</mathjax></p>
<p>And clearly, if <mathjax>#19*mol#</mathjax> dihydrogen react completely (which is NOT an altogether probable outcome) then by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> we get <mathjax>#19/3xx2*mol=12.7*mol#</mathjax> with respect to ammonia, <mathjax>#NH_3#</mathjax>.......</p>
<p>What is the significance of this ammonia synthesis? Do you think that this reaction is widely performed? </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>A bit under <mathjax>#13*mol#</mathjax> of ammonia..........</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric equation........</p>
<p><mathjax>#N_2(g) + 3H_2(g) stackrel("catalysis")rarr2NH_3(g)#</mathjax></p>
<p>And clearly, if <mathjax>#19*mol#</mathjax> dihydrogen react completely (which is NOT an altogether probable outcome) then by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> we get <mathjax>#19/3xx2*mol=12.7*mol#</mathjax> with respect to ammonia, <mathjax>#NH_3#</mathjax>.......</p>
<p>What is the significance of this ammonia synthesis? Do you think that this reaction is widely performed? </p></div>
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<h1 class="questionTitle" itemprop="name">If #19*mol# of #H_2(g)# react with stoichiometric #N_2(g)#, what molar quantity of ammonia results?</h1>
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<div class="markdown"><p>A bit under <mathjax>#13*mol#</mathjax> of ammonia..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric equation........</p>
<p><mathjax>#N_2(g) + 3H_2(g) stackrel("catalysis")rarr2NH_3(g)#</mathjax></p>
<p>And clearly, if <mathjax>#19*mol#</mathjax> dihydrogen react completely (which is NOT an altogether probable outcome) then by the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> we get <mathjax>#19/3xx2*mol=12.7*mol#</mathjax> with respect to ammonia, <mathjax>#NH_3#</mathjax>.......</p>
<p>What is the significance of this ammonia synthesis? Do you think that this reaction is widely performed? </p></div>
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</article> | If #19*mol# of #H_2(g)# react with stoichiometric #N_2(g)#, what molar quantity of ammonia results? | null |
605 | aacadaca-6ddd-11ea-a4c4-ccda262736ce | https://socratic.org/questions/what-is-the-of-c-in-c-6h-12o-6 | 40.00% | start physical_unit 5 7 percent none qc_end chemical_equation 7 7 qc_end end | [{"type":"physical unit","value":"% [OF] C in C6H12O6"}] | [{"type":"physical unit","value":"40.00%"}] | [{"type":"chemical equation","value":"C6H12O6"}] | <h1 class="questionTitle" itemprop="name">What is the % of #C# in #C_6H_12O_6#?</h1> | null | 40.00% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"% carbon"="Mass of carbon"/"Mass of glucose"xx100%#</mathjax></p>
<p><mathjax>#(6*xx12.011*g*mol^-1)/(180.16*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#=40%#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#40%#</mathjax>.............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"% carbon"="Mass of carbon"/"Mass of glucose"xx100%#</mathjax></p>
<p><mathjax>#(6*xx12.011*g*mol^-1)/(180.16*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#=40%#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#40%#</mathjax>.............</p></div>
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<div class="markdown"><p><mathjax>#"% carbon"="Mass of carbon"/"Mass of glucose"xx100%#</mathjax></p>
<p><mathjax>#(6*xx12.011*g*mol^-1)/(180.16*g*mol^-1)xx100%#</mathjax></p>
<p><mathjax>#=40%#</mathjax></p></div>
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</article> | What is the % of #C# in #C_6H_12O_6#? | null |
606 | a8bc35e6-6ddd-11ea-9e14-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-moles-in-432-g-ba-no-3-2 | 1.65 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] Ba(NO3)2 [IN] moles"}] | [{"type":"physical unit","value":"1.65 moles"}] | [{"type":"physical unit","value":"Mass [OF] Ba(NO3)2 [=] \\pu{432 g}"}] | <h1 class="questionTitle" itemprop="name">What is the number of moles in 432 g #Ba(NO_3)_2#?</h1> | null | 1.65 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I believe that one of the principal uses of barium nitrate is as an explosive. In the early atomic bombs (WWII era), barium nitrate and TNT were the so-called explosive lense used to assemble the fission reaction of uranium. </p></div>
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<div class="markdown"><p><mathjax>#(432*cancelg)/(261.38*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax></p></div>
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<div class="markdown"><p>I believe that one of the principal uses of barium nitrate is as an explosive. In the early atomic bombs (WWII era), barium nitrate and TNT were the so-called explosive lense used to assemble the fission reaction of uranium. </p></div>
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<div class="markdown"><p><mathjax>#(432*cancelg)/(261.38*cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*mol#</mathjax></p></div>
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<div class="markdown"><p>I believe that one of the principal uses of barium nitrate is as an explosive. In the early atomic bombs (WWII era), barium nitrate and TNT were the so-called explosive lense used to assemble the fission reaction of uranium. </p></div>
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</article> | What is the number of moles in 432 g #Ba(NO_3)_2#? | null |
607 | aa04721f-6ddd-11ea-b05a-ccda262736ce | https://socratic.org/questions/what-mass-of-aluminum-has-a-volume-of-6-3-cm-3 | 17.01 g | start physical_unit 3 3 mass g qc_end physical_unit 3 3 8 9 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum [IN] g"}] | [{"type":"physical unit","value":"17.01 g"}] | [{"type":"physical unit","value":"Volume [OF] aluminum [=] \\pu{6.3 cm^3}"}] | <h1 class="questionTitle" itemprop="name">What mass of aluminum has a volume of #6.3# #cm^3#?</h1> | null | 17.01 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can figure out this out by using aluminum's density. Density is the ratio of the amount of mass that can be fit into a certain volume.</p>
<p>The equation for density is <mathjax>#D = M / V#</mathjax>. </p>
<p><mathjax>#M#</mathjax> stands for mass, <mathjax>#V#</mathjax> stands for volume, and <mathjax>#D#</mathjax> stands for density.</p>
<p>According to <a href="http://chemistry.elmhurst.edu/vchembook/102aluminum.html" rel="nofollow">Elmhurst College</a>, the density of aluminum is <mathjax>#(2.7 \ grams)/(mL)#</mathjax>.</p>
<p>In order to use this density with a volume in <mathjax>#cm^3#</mathjax>, we need to convert it into <mathjax>#(grams)/(cm^3)#</mathjax>. Fortunately, <mathjax>#cm^3#</mathjax> and <mathjax>#mL#</mathjax> are equal to each other, so the density of aluminum is <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax></p>
<p>The equation for density can be rearranged to solve for mass: <mathjax>#V * D = M#</mathjax></p>
<p>To get the value of <mathjax>#M#</mathjax>, we can multiply <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> and <mathjax>#6.3 \ cm^3#</mathjax> together, and you are left with the mass of the aluminum.</p>
<p><mathjax>#6.3 \ cm^3 * (2.7 \ grams)/(cm^3) = 17.01 \ grams#</mathjax></p>
<p>The <mathjax>#cm^3#</mathjax> both cancel, and you are left with <mathjax>#grams#</mathjax>.</p>
<p>From there, we use significant figures to show the precise digits of our answer. Both <mathjax>#6.3 \ cm^3#</mathjax> and <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> have only 2 significant figures, so we round our answer to 2 significant figures as well.</p>
<p><mathjax>#17 \ grams#</mathjax> of aluminum is the final answer.</p></div>
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<div>
<div class="markdown"><p>The mass of aluminum with a volume of <mathjax>#6.3 \ cm^3#</mathjax> is <mathjax>#17 \ grams#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can figure out this out by using aluminum's density. Density is the ratio of the amount of mass that can be fit into a certain volume.</p>
<p>The equation for density is <mathjax>#D = M / V#</mathjax>. </p>
<p><mathjax>#M#</mathjax> stands for mass, <mathjax>#V#</mathjax> stands for volume, and <mathjax>#D#</mathjax> stands for density.</p>
<p>According to <a href="http://chemistry.elmhurst.edu/vchembook/102aluminum.html" rel="nofollow">Elmhurst College</a>, the density of aluminum is <mathjax>#(2.7 \ grams)/(mL)#</mathjax>.</p>
<p>In order to use this density with a volume in <mathjax>#cm^3#</mathjax>, we need to convert it into <mathjax>#(grams)/(cm^3)#</mathjax>. Fortunately, <mathjax>#cm^3#</mathjax> and <mathjax>#mL#</mathjax> are equal to each other, so the density of aluminum is <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax></p>
<p>The equation for density can be rearranged to solve for mass: <mathjax>#V * D = M#</mathjax></p>
<p>To get the value of <mathjax>#M#</mathjax>, we can multiply <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> and <mathjax>#6.3 \ cm^3#</mathjax> together, and you are left with the mass of the aluminum.</p>
<p><mathjax>#6.3 \ cm^3 * (2.7 \ grams)/(cm^3) = 17.01 \ grams#</mathjax></p>
<p>The <mathjax>#cm^3#</mathjax> both cancel, and you are left with <mathjax>#grams#</mathjax>.</p>
<p>From there, we use significant figures to show the precise digits of our answer. Both <mathjax>#6.3 \ cm^3#</mathjax> and <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> have only 2 significant figures, so we round our answer to 2 significant figures as well.</p>
<p><mathjax>#17 \ grams#</mathjax> of aluminum is the final answer.</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of aluminum has a volume of #6.3# #cm^3#?</h1>
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<div class="markdown"><p>The mass of aluminum with a volume of <mathjax>#6.3 \ cm^3#</mathjax> is <mathjax>#17 \ grams#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can figure out this out by using aluminum's density. Density is the ratio of the amount of mass that can be fit into a certain volume.</p>
<p>The equation for density is <mathjax>#D = M / V#</mathjax>. </p>
<p><mathjax>#M#</mathjax> stands for mass, <mathjax>#V#</mathjax> stands for volume, and <mathjax>#D#</mathjax> stands for density.</p>
<p>According to <a href="http://chemistry.elmhurst.edu/vchembook/102aluminum.html" rel="nofollow">Elmhurst College</a>, the density of aluminum is <mathjax>#(2.7 \ grams)/(mL)#</mathjax>.</p>
<p>In order to use this density with a volume in <mathjax>#cm^3#</mathjax>, we need to convert it into <mathjax>#(grams)/(cm^3)#</mathjax>. Fortunately, <mathjax>#cm^3#</mathjax> and <mathjax>#mL#</mathjax> are equal to each other, so the density of aluminum is <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax></p>
<p>The equation for density can be rearranged to solve for mass: <mathjax>#V * D = M#</mathjax></p>
<p>To get the value of <mathjax>#M#</mathjax>, we can multiply <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> and <mathjax>#6.3 \ cm^3#</mathjax> together, and you are left with the mass of the aluminum.</p>
<p><mathjax>#6.3 \ cm^3 * (2.7 \ grams)/(cm^3) = 17.01 \ grams#</mathjax></p>
<p>The <mathjax>#cm^3#</mathjax> both cancel, and you are left with <mathjax>#grams#</mathjax>.</p>
<p>From there, we use significant figures to show the precise digits of our answer. Both <mathjax>#6.3 \ cm^3#</mathjax> and <mathjax>#(2.7 \ grams)/(cm^3)#</mathjax> have only 2 significant figures, so we round our answer to 2 significant figures as well.</p>
<p><mathjax>#17 \ grams#</mathjax> of aluminum is the final answer.</p></div>
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</article> | What mass of aluminum has a volume of #6.3# #cm^3#? | null |
608 | aa3a322e-6ddd-11ea-9730-ccda262736ce | https://socratic.org/questions/how-many-moles-of-lithium-nitrate-are-in-256-ml-of-a-0-855-m-solution | 0.22 moles | start physical_unit 4 5 mole mol qc_end physical_unit 14 14 8 9 volume qc_end physical_unit 4 5 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] lithium nitrate [IN] moles"}] | [{"type":"physical unit","value":"0.22 moles"}] | [{"type":"physical unit","value":"Volume [OF] lithium nitrate solution [=] \\pu{256 mL}"},{"type":"physical unit","value":"Molarity [OF] lithium nitrate solution [=] \\pu{0.855 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of lithium nitrate are in 256 mL of a 0.855 M solution?</h1> | null | 0.22 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the formula below:<br/>
<img alt="dl.clackamas.edu" src="https://useruploads.socratic.org/gHLDsfPWS8yxWQUPID4F_soln13.jpg"/> </p>
<p>We already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. </p>
<p>We can convert 256 mL into liters by using the conversion factor 1000mL = 1L. When 256mL is divided by 1000 mL, we obtain a volume of 0.256 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (0.256 L) (0.855M) = 0.219 moles</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>There are <strong>0.219 mol of <mathjax>#LiNO_3#</mathjax></strong> in 256 mL of 0.855 M solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the formula below:<br/>
<img alt="dl.clackamas.edu" src="https://useruploads.socratic.org/gHLDsfPWS8yxWQUPID4F_soln13.jpg"/> </p>
<p>We already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. </p>
<p>We can convert 256 mL into liters by using the conversion factor 1000mL = 1L. When 256mL is divided by 1000 mL, we obtain a volume of 0.256 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (0.256 L) (0.855M) = 0.219 moles</p></div>
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<div class="markdown"><p>There are <strong>0.219 mol of <mathjax>#LiNO_3#</mathjax></strong> in 256 mL of 0.855 M solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the formula below:<br/>
<img alt="dl.clackamas.edu" src="https://useruploads.socratic.org/gHLDsfPWS8yxWQUPID4F_soln13.jpg"/> </p>
<p>We already have the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. </p>
<p>We can convert 256 mL into liters by using the conversion factor 1000mL = 1L. When 256mL is divided by 1000 mL, we obtain a volume of 0.256 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (0.256 L) (0.855M) = 0.219 moles</p></div>
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</article> | How many moles of lithium nitrate are in 256 mL of a 0.855 M solution? | null |
609 | aa1711a8-6ddd-11ea-a905-ccda262736ce | https://socratic.org/questions/h-2o-2h-2-o-2-what-is-the-percent-yield-of-o-2-if-10-2-g-of-o-2-is-produced-from | 67.5% | start physical_unit 6 6 percent_yield none qc_end chemical_equation 0 6 qc_end physical_unit 6 6 15 16 mass qc_end physical_unit 1 1 25 26 mass qc_end end | [{"type":"physical unit","value":"Percent yield [OF] O2"}] | [{"type":"physical unit","value":"67.5%"}] | [{"type":"chemical equation","value":"2 H2O -> 2 H2 + O2"},{"type":"physical unit","value":"Mass [OF] O2 [=] \\pu{10.2 g}"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{17.0 g}"}] | <h1 class="questionTitle" itemprop="name">#2H_2O -> 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#? </h1> | null | 67.5% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>that it takes <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. Use the <strong>molar masses</strong> of the two chemical species to convert this mole ratio to a gram ratio</p>
<blockquote>
<p><mathjax>#("2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)#</mathjax></p>
</blockquote>
<p>So, you know that every <mathjax>#"36.03 g"#</mathjax> of water that take part in the reaction produce <mathjax>#"32.0 g"#</mathjax> of oxygen gas.</p>
<p>This means that when <mathjax>#"17.0 g"#</mathjax> of water undergo decomposition, you should expect to get</p>
<blockquote>
<p><mathjax>#17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2#</mathjax></p>
</blockquote>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what is produced by a reaction that has a <mathjax>#100%#</mathjax> yield. </p>
<p>Now, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#"10.2 g"#</mathjax> of oxygen gas. To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong>, divide the actual yield by the theoretical yield and multiply the result by <mathjax>#100%#</mathjax>.</p>
<p>You should get</p>
<blockquote>
<p><mathjax>#"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))#</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>#67.5%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>that it takes <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. Use the <strong>molar masses</strong> of the two chemical species to convert this mole ratio to a gram ratio</p>
<blockquote>
<p><mathjax>#("2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)#</mathjax></p>
</blockquote>
<p>So, you know that every <mathjax>#"36.03 g"#</mathjax> of water that take part in the reaction produce <mathjax>#"32.0 g"#</mathjax> of oxygen gas.</p>
<p>This means that when <mathjax>#"17.0 g"#</mathjax> of water undergo decomposition, you should expect to get</p>
<blockquote>
<p><mathjax>#17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2#</mathjax></p>
</blockquote>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what is produced by a reaction that has a <mathjax>#100%#</mathjax> yield. </p>
<p>Now, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#"10.2 g"#</mathjax> of oxygen gas. To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong>, divide the actual yield by the theoretical yield and multiply the result by <mathjax>#100%#</mathjax>.</p>
<p>You should get</p>
<blockquote>
<p><mathjax>#"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))#</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">#2H_2O -> 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#67.5%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))#</mathjax></p>
</blockquote>
<p>that it takes <mathjax>#2#</mathjax> <strong>moles</strong> of water to produce <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas. Use the <strong>molar masses</strong> of the two chemical species to convert this mole ratio to a gram ratio</p>
<blockquote>
<p><mathjax>#("2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)#</mathjax></p>
</blockquote>
<p>So, you know that every <mathjax>#"36.03 g"#</mathjax> of water that take part in the reaction produce <mathjax>#"32.0 g"#</mathjax> of oxygen gas.</p>
<p>This means that when <mathjax>#"17.0 g"#</mathjax> of water undergo decomposition, you should expect to get</p>
<blockquote>
<p><mathjax>#17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2#</mathjax></p>
</blockquote>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what is produced by a reaction that has a <mathjax>#100%#</mathjax> yield. </p>
<p>Now, you know that the <strong>actual yield</strong> of the reaction is <mathjax>#"10.2 g"#</mathjax> of oxygen gas. To find the <strong><a href="https://socratic.org/chemistry/stoichiometry/percent-yield">percent yield</a></strong>, divide the actual yield by the theoretical yield and multiply the result by <mathjax>#100%#</mathjax>.</p>
<p>You should get</p>
<blockquote>
<p><mathjax>#"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))#</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> | #2H_2O -> 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#? | null |
610 | abd00126-6ddd-11ea-9800-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-equation-fe-no-3-3-kscn-fe-no-3-3-kscn | Fe(NO3)3 + 3 KSCN -> [Fe(SCN)3] + 3 KNO3 | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"Fe(NO3)3 + 3 KSCN -> [Fe(SCN)3] + 3 KNO3"}] | [{"type":"chemical equation","value":"Fe(NO3)3 + KSCN -> [Fe(SCN)3] + KNO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance this equation?
#Fe(NO_3)_3 + KSCN -> Fe(NO_3)_3 + KSCN#</h1> | null | Fe(NO3)3 + 3 KSCN -> [Fe(SCN)3] + 3 KNO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since,the reactants side is the same as product side,hence, no chemical reaction has taken place.</p>
<p>Hence,since it isn't even a chemical reaction it cant be ballanced.</p></div>
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<div class="markdown"><p>NO</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since,the reactants side is the same as product side,hence, no chemical reaction has taken place.</p>
<p>Hence,since it isn't even a chemical reaction it cant be ballanced.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance this equation?
#Fe(NO_3)_3 + KSCN -> Fe(NO_3)_3 + KSCN#</h1>
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Wricheek
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<div class="markdown"><p>NO</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Since,the reactants side is the same as product side,hence, no chemical reaction has taken place.</p>
<p>Hence,since it isn't even a chemical reaction it cant be ballanced.</p></div>
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anor277
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<div class="markdown"><p><mathjax>#Fe(NO_3)_3 + 3KSCN rarr [Fe(SCN)_3] + 3KNO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The tristhiocyanato iron species is a so-called complex ion. Both mass and charge are balanced in the equation. </p></div>
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Michael
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<div class="markdown"><p>You haven't actually written any products but I will describe the reaction below:</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Iron(III) ions form a blood-red complex with thiocyanate ions in a 1:1 molar ratio.</p>
<p><mathjax>#Fe(NO_3)_3#</mathjax> + <mathjax>#KSCNrarr[FeSCN]^(2+)+K^++3NO_3^-#</mathjax></p>
<p>The ionic equation which omits the spectator ions is:</p>
<p><mathjax>#Fe_((aq))^(3+)+SCN_((aq))^(-)rarr[Fe(SCN)]_((aq))^(2+)#</mathjax></p>
<p>The complex is very intensely coloured and this reaction is often used as a qualitative test for iron(III) ions.</p>
<p>Strictly speaking we have a ligand exchange reaction:</p>
<p><mathjax>#[Fe(H_2O)_6]^(3+)+SCN^(-)rarr[Fe(H_2O)_5SCN]^(2+)+H_2O#</mathjax> </p>
<p>The reaction is shown here:</p>
<p>
<iframe src="https://www.youtube.com/embed/7UQdPAXhLbA?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | How do you balance this equation?
#Fe(NO_3)_3 + KSCN -> Fe(NO_3)_3 + KSCN# | null |
611 | aae7d89e-6ddd-11ea-9d94-ccda262736ce | https://socratic.org/questions/a-steel-tank-with-a-volume-of-9-583-l-contains-n-2-gas-under-a-pressure-of-4-972 | 1.90 moles | start physical_unit 10 11 mole mol qc_end physical_unit 10 11 7 8 volume qc_end physical_unit 10 11 16 17 pressure qc_end physical_unit 10 11 19 20 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] N2 gas [IN] moles"}] | [{"type":"physical unit","value":"1.90 moles"}] | [{"type":"physical unit","value":"Volume [OF] N2 gas [=] \\pu{9.583 L}"},{"type":"physical unit","value":"Pressure [OF] N2 gas [=] \\pu{4.972 atm}"},{"type":"physical unit","value":"Temperature [OF] N2 gas [=] \\pu{31.8 ℃}"}] | <h1 class="questionTitle" itemprop="name">A steel tank with a volume of 9.583 L contains #N_2# gas under a pressure of 4.972 atm at 31.8°C. How many moles of #N_2# are in the tank?</h1> | null | 1.90 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For the old Ideal Gas equation, <mathjax>#n=(PV)/(RT)#</mathjax>...</p>
<p>And so <mathjax>#n=(4.972*atmxx9.583*L)/(0.0821*(L*atm)/(K*mol)*305*K)#</mathjax></p>
<p><mathjax>#=1.90*mol#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#n~=2*mol#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For the old Ideal Gas equation, <mathjax>#n=(PV)/(RT)#</mathjax>...</p>
<p>And so <mathjax>#n=(4.972*atmxx9.583*L)/(0.0821*(L*atm)/(K*mol)*305*K)#</mathjax></p>
<p><mathjax>#=1.90*mol#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A steel tank with a volume of 9.583 L contains #N_2# gas under a pressure of 4.972 atm at 31.8°C. How many moles of #N_2# are in the tank?</h1>
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<div class="markdown"><p><mathjax>#n~=2*mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>For the old Ideal Gas equation, <mathjax>#n=(PV)/(RT)#</mathjax>...</p>
<p>And so <mathjax>#n=(4.972*atmxx9.583*L)/(0.0821*(L*atm)/(K*mol)*305*K)#</mathjax></p>
<p><mathjax>#=1.90*mol#</mathjax></p></div>
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</article> | A steel tank with a volume of 9.583 L contains #N_2# gas under a pressure of 4.972 atm at 31.8°C. How many moles of #N_2# are in the tank? | null |
612 | a9c74228-6ddd-11ea-b1e8-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-oxygen | 32.00 g/mol | start physical_unit 6 6 molar_mass g/mol qc_end substance 6 6 qc_end end | [{"type":"physical unit","value":"Molar mass [OF] oxygen [IN] g/mol"}] | [{"type":"physical unit","value":"32.00 g/mol"}] | [{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of oxygen?</h1> | null | 32.00 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that the oxygen gas we breathe is the diatomic molecule, <mathjax>#O_2#</mathjax>; in fact most common elemental gases are diatomic (what are the exceptions to this generalization?).</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Of oxygen <mathjax>#"atoms"#</mathjax>, <mathjax>#16.00#</mathjax> <mathjax>#g*mol#</mathjax>. Of oxygen <mathjax>#"molecules"#</mathjax>, <mathjax>#32.00#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We know that the oxygen gas we breathe is the diatomic molecule, <mathjax>#O_2#</mathjax>; in fact most common elemental gases are diatomic (what are the exceptions to this generalization?).</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar mass of oxygen?</h1>
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<div class="markdown"><p>Of oxygen <mathjax>#"atoms"#</mathjax>, <mathjax>#16.00#</mathjax> <mathjax>#g*mol#</mathjax>. Of oxygen <mathjax>#"molecules"#</mathjax>, <mathjax>#32.00#</mathjax> <mathjax>#g*mol^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that the oxygen gas we breathe is the diatomic molecule, <mathjax>#O_2#</mathjax>; in fact most common elemental gases are diatomic (what are the exceptions to this generalization?).</p></div>
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</article> | What is the molar mass of oxygen? | null |
613 | a9be2940-6ddd-11ea-882b-ccda262736ce | https://socratic.org/questions/how-do-you-balance-mg-no-3-2-k-3po-4-mg-3-po-4-2-kno-3 | 3 Mg(NO3)2 + 2 K3PO4 -> Mg3(PO4)2 + 6 KNO3 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 Mg(NO3)2 + 2 K3PO4 -> Mg3(PO4)2 + 6 KNO3"}] | [{"type":"chemical equation","value":"Mg(NO3)2 + K3PO4 -> Mg3(PO4)2 + KNO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3#?</h1> | null | 3 Mg(NO3)2 + 2 K3PO4 -> Mg3(PO4)2 + 6 KNO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the given reaction stoichiometrically balanced? For every reactant particle, is there a corresponding product particle. If there is not, then the reaction cannot be accepted as a representation of chemical (i.e. physical) reality.</p></div>
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<div class="markdown"><p><mathjax>#3Mg(NO_3)_2 + 2K_3PO_4 rarr Mg_3(PO_4)_2 + 6KNO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is the given reaction stoichiometrically balanced? For every reactant particle, is there a corresponding product particle. If there is not, then the reaction cannot be accepted as a representation of chemical (i.e. physical) reality.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3#?</h1>
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<div class="markdown"><p><mathjax>#3Mg(NO_3)_2 + 2K_3PO_4 rarr Mg_3(PO_4)_2 + 6KNO_3#</mathjax></p></div>
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<div class="markdown"><p>Is the given reaction stoichiometrically balanced? For every reactant particle, is there a corresponding product particle. If there is not, then the reaction cannot be accepted as a representation of chemical (i.e. physical) reality.</p></div>
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<div class="markdown"><p><mathjax>#3Mg(NO_3)_2#</mathjax> + <mathjax>#2K_3PO_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Mg_3(PO_4)_2#</mathjax> + <mathjax>#)6KNO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's assume that the chemical reaction will progress as written.</p>
<p>Balancing ionic equations can be quite easy AS LONG AS YOU MEMORIZE YOUR CATIONS AND ANIONS. There are a number of ways to do this, so you have to find a way that actually works for you. For now, let's do a tally sheet of all ions involved.</p>
<p><mathjax>#Mg(NO_3)_2#</mathjax> + <mathjax>#K_3PO_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Mg_3(PO_4)_2#</mathjax> + <mathjax>#KNO_3#</mathjax> </p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2<br/>
<mathjax>#K^(1+)#</mathjax> = 3<br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1</p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1<br/>
<mathjax>#K^(1+)#</mathjax> = 1<br/>
<mathjax>#PO_4^(-3)#</mathjax> = 2</p>
<p>Let's balance the most complicated ion first: the <mathjax>#PO_4^(3-)#</mathjax>.</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2<br/>
<mathjax>#K^(1+)#</mathjax> = 3<br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x <mathjax>#color(red)2#</mathjax> = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1<br/>
<mathjax>#K^(1+)#</mathjax> = 1<br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p>Since the <mathjax>#PO_4^(3-)#</mathjax> ion is bonded to the <mathjax>#K^(1+)#</mathjax> ion, you need to multiply it by 2 as well.</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2<br/>
<mathjax>#K^(1+)#</mathjax> = 3 x <mathjax>#color(red)2#</mathjax> = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x <mathjax>#color(red)2#</mathjax> = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1<br/>
<mathjax>#K^(1+)#</mathjax> = 1<br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p><mathjax>#Mg(NO_3)_2#</mathjax> + <mathjax>#color(red)2K_3PO_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Mg_3(PO_4)_2#</mathjax> + <mathjax>#KNO_3#</mathjax> </p>
<p>Now notice that this move have created an imbalance in your <mathjax>#K^(1+)#</mathjax> ion with 6 ions on the left as opposed to only one ion on the right. To compensate, we have:</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2<br/>
<mathjax>#K^(1+)#</mathjax> = 3 x 2 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x 2 = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1<br/>
<mathjax>#K^(1+)#</mathjax> = 1 x <mathjax>#color(blue)6#</mathjax> = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p>Again, since the <mathjax>#K^(1+)#</mathjax> ion is bonded to the <mathjax>#NO_3^(-1)#</mathjax>, we need to also multiply it by 6.</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2 <br/>
<mathjax>#K^(1+)#</mathjax> = 3 x 2 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x 2 = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1 x <mathjax>#color(blue)6#</mathjax> = <strong>6</strong><br/>
<mathjax>#K^(1+)#</mathjax> = 1 x <mathjax>#color(blue)6#</mathjax> = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p><mathjax>#Mg(NO_3)_2#</mathjax> + <mathjax>#color(red)2K_3PO_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Mg_3(PO_4)_2#</mathjax> + <mathjax>#color(blue)6KNO_3#</mathjax> </p>
<p>Now your <mathjax>#NO_3^(-1)#</mathjax> ion is unbalanced with only 2 ions on the left as opposed to the 6 on the right. Thus,</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1 <br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2 x <mathjax>#color(green)3#</mathjax> = <strong>6</strong><br/>
<mathjax>#K^(1+)#</mathjax> = 3 x 2 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x 2 = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = 3<br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1 x 6 = <strong>6</strong><br/>
<mathjax>#K^(1+)#</mathjax> = 1 x 6 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p>Same as above the <mathjax>#NO_3^(-1)#</mathjax> ion is bonded with your <mathjax>#Mg^(2+)#</mathjax>. Therefore,</p>
<p>Left side:<br/>
<mathjax>#Mg^(2+)#</mathjax> = 1 x <mathjax>#color(green)3#</mathjax> ==<strong>3</strong><br/>
<mathjax>#NO_3^(-1)#</mathjax> = 2 x <mathjax>#color(green)3#</mathjax> = <strong>6</strong><br/>
<mathjax>#K^(1+)#</mathjax> = 3 x 2 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = 1 x 2 = <strong>2</strong></p>
<p>Right side:</p>
<p><mathjax>#Mg^(2+)#</mathjax> = <strong>3</strong><br/>
<mathjax>#NO_3^(-1)#</mathjax> = 1 x 6 = <strong>6</strong><br/>
<mathjax>#K^(1+)#</mathjax> = 1 x 6 = <strong>6</strong><br/>
<mathjax>#PO_4^(-3)#</mathjax> = <strong>2</strong></p>
<p>Final Answer:</p>
<p><mathjax>#color(green)3Mg(NO_3)_2#</mathjax> + <mathjax>#color(red)2K_3PO_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#Mg_3(PO_4)_2#</mathjax> + <mathjax>#color(blue)6KNO_3#</mathjax> (balanced)</p></div>
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</article> | How do you balance #Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3#? | null |
614 | aabe91ae-6ddd-11ea-baae-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-half-reaction-for-fe-s-2hcl-aq-fecl-2-aq-h-2-g | Fe - 2 e- -> Fe^2+ | start chemical_equation qc_end chemical_equation 7 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] oxidation half reaction"}] | [{"type":"physical unit","value":"Fe - 2 e- -> Fe^2+"}] | [{"type":"chemical equation","value":"Fe(s) + 2 HCl(aq) -> FeCl2(aq) + H2(g)"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation half reaction for #Fe(s) + 2HCl(aq) -> FeCl_2(aq) + H_2(g)#?</h1> | null | Fe - 2 e- -> Fe^2+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In this reaction Fe is being oxidized and H is being reduced. To check the balance of a redox reaction, make sure that the number of electrons exchanged are equal in both half-reactions.</p></div>
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<div class="markdown"><p><mathjax>#Fe - 2e^- → Fe^(2+)#</mathjax></p></div>
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<div class="markdown"><p>In this reaction Fe is being oxidized and H is being reduced. To check the balance of a redox reaction, make sure that the number of electrons exchanged are equal in both half-reactions.</p></div>
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<div class="markdown"><p><mathjax>#Fe - 2e^- → Fe^(2+)#</mathjax></p></div>
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<div class="markdown"><p>In this reaction Fe is being oxidized and H is being reduced. To check the balance of a redox reaction, make sure that the number of electrons exchanged are equal in both half-reactions.</p></div>
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</article> | What is the oxidation half reaction for #Fe(s) + 2HCl(aq) -> FeCl_2(aq) + H_2(g)#? | null |
615 | acab1cba-6ddd-11ea-88f8-ccda262736ce | https://socratic.org/questions/how-would-you-calculate-the-mass-in-ng-of-2-33-x-10-20-atoms-of-oxygen | 6.19 × 10^9 ng | start physical_unit 14 14 mass ng qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] ng"}] | [{"type":"physical unit","value":"6.19 × 10^9 ng"}] | [{"type":"physical unit","value":"Number [OF] oxygen atoms [=] \\pu{2.33 × 10^20}"}] | <h1 class="questionTitle" itemprop="name">How would you calculate the mass (in ng) of #2.33 xx 10^20 #atoms of oxygen?</h1> | null | 6.19 × 10^9 ng | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to determine the mass <em>in grams</em> of <mathjax>#2.33 * 10^(20)#</mathjax> atoms of oxygen.</p>
<p>To do that, use <strong>Avogadro's number</strong> to convert the number of atoms to <em>moles</em>, the <strong>molar mass</strong> of oxygen to convert the moles to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which acts as the definition of <strong>a mole</strong>, tells you that in order to have one mole of oxygen, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> atoms of oxygen. </p>
<p>In your case, the sample will be equivalent to </p>
<blockquote>
<p><mathjax>#2.33 * 10^(20) color(red)(cancel(color(black)("atoms of O"))) * "1 mole O"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 3.869 * 10^(-4)"moles O"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, oxygen has a <strong>molar mass</strong> of <mathjax>#"15.9994 g mol"^(-1)#</mathjax> as listed in the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
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<blockquote>
<p><img alt="https://infograph.venngage.com/p/51725/science-infograph" src="https://useruploads.socratic.org/v7GLGFV2QiWDqc7SsEgQ_265122-d8302cf29b1482443550f2bd1ba3d9a9.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#3.869 * 10^(-4) color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 6.190 * 10^(-3)"g"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, you can convert the mass from <em>grams</em> to <em>nanograms</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 g" = 10^(9)"ng")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will end up with </p>
<blockquote>
<p><mathjax>#6.190 * 10^(-3) color(red)(cancel(color(black)("g"))) * (10^(9)"ng")/(1color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.19 * 10^6"ng")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>#6.19 * 10^(9)"ng"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to determine the mass <em>in grams</em> of <mathjax>#2.33 * 10^(20)#</mathjax> atoms of oxygen.</p>
<p>To do that, use <strong>Avogadro's number</strong> to convert the number of atoms to <em>moles</em>, the <strong>molar mass</strong> of oxygen to convert the moles to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which acts as the definition of <strong>a mole</strong>, tells you that in order to have one mole of oxygen, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> atoms of oxygen. </p>
<p>In your case, the sample will be equivalent to </p>
<blockquote>
<p><mathjax>#2.33 * 10^(20) color(red)(cancel(color(black)("atoms of O"))) * "1 mole O"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 3.869 * 10^(-4)"moles O"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, oxygen has a <strong>molar mass</strong> of <mathjax>#"15.9994 g mol"^(-1)#</mathjax> as listed in the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="https://infograph.venngage.com/p/51725/science-infograph" src="https://useruploads.socratic.org/v7GLGFV2QiWDqc7SsEgQ_265122-d8302cf29b1482443550f2bd1ba3d9a9.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#3.869 * 10^(-4) color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 6.190 * 10^(-3)"g"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, you can convert the mass from <em>grams</em> to <em>nanograms</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 g" = 10^(9)"ng")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will end up with </p>
<blockquote>
<p><mathjax>#6.190 * 10^(-3) color(red)(cancel(color(black)("g"))) * (10^(9)"ng")/(1color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.19 * 10^6"ng")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>
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<h1 class="questionTitle" itemprop="name">How would you calculate the mass (in ng) of #2.33 xx 10^20 #atoms of oxygen?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-08-28T00:10:42" itemprop="dateCreated">
Aug 28, 2016
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<div class="markdown"><p><mathjax>#6.19 * 10^(9)"ng"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to determine the mass <em>in grams</em> of <mathjax>#2.33 * 10^(20)#</mathjax> atoms of oxygen.</p>
<p>To do that, use <strong>Avogadro's number</strong> to convert the number of atoms to <em>moles</em>, the <strong>molar mass</strong> of oxygen to convert the moles to <em>grams</em>. </p>
<p>So, <em>Avogadro's number</em>, which acts as the definition of <strong>a mole</strong>, tells you that in order to have one mole of oxygen, you need to have <mathjax>#6.022 * 10^(23)#</mathjax> atoms of oxygen. </p>
<p>In your case, the sample will be equivalent to </p>
<blockquote>
<p><mathjax>#2.33 * 10^(20) color(red)(cancel(color(black)("atoms of O"))) * "1 mole O"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 3.869 * 10^(-4)"moles O"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Now, oxygen has a <strong>molar mass</strong> of <mathjax>#"15.9994 g mol"^(-1)#</mathjax> as listed in the <em>Periodic Table of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">Elements</a></em></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="https://infograph.venngage.com/p/51725/science-infograph" src="https://useruploads.socratic.org/v7GLGFV2QiWDqc7SsEgQ_265122-d8302cf29b1482443550f2bd1ba3d9a9.jpg"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>This means that your sample will have a mass of </p>
<blockquote>
<p><mathjax>#3.869 * 10^(-4) color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= 6.190 * 10^(-3)"g"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Finally, you can convert the mass from <em>grams</em> to <em>nanograms</em> by using the conversion factor</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 g" = 10^(9)"ng")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will end up with </p>
<blockquote>
<p><mathjax>#6.190 * 10^(-3) color(red)(cancel(color(black)("g"))) * (10^(9)"ng")/(1color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.19 * 10^6"ng")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 would you calculate the mass (in ng) of #2.33 xx 10^20 #atoms of oxygen? | null |
616 | ad05d9b4-6ddd-11ea-939a-ccda262736ce | https://socratic.org/questions/what-s-the-covalent-compound-formula-for-oxalic-acid | C2H2O4 | start chemical_formula qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] oxalic acid [IN] default"}] | [{"type":"chemical equation","value":"C2H2O4"}] | [{"type":"substance name","value":"Oxalic acid"}] | <h1 class="questionTitle" itemprop="name">What s the covalent compound formula for oxalic acid?</h1> | null | C2H2O4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxalic acid"#</mathjax> is the simplest dicarboxylic acid, <mathjax>#HO(O=)C-C(=O)OH#</mathjax>; its salts are called <mathjax>#"oxalates"#</mathjax>, i.e. <mathjax>#C_2O_4^(2-)#</mathjax>, which is a chelating ligand. In the laboratory it is often used as the <mathjax>#"dihydrate"#</mathjax>, <mathjax>#C_2H_2O_4*2H_2O#</mathjax>. </p></div>
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<div class="markdown"><p>For <mathjax>#"oxalic acid, "C_2H_2O_4#</mathjax>?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxalic acid"#</mathjax> is the simplest dicarboxylic acid, <mathjax>#HO(O=)C-C(=O)OH#</mathjax>; its salts are called <mathjax>#"oxalates"#</mathjax>, i.e. <mathjax>#C_2O_4^(2-)#</mathjax>, which is a chelating ligand. In the laboratory it is often used as the <mathjax>#"dihydrate"#</mathjax>, <mathjax>#C_2H_2O_4*2H_2O#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What s the covalent compound formula for oxalic acid?</h1>
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anor277
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<div class="markdown"><p>For <mathjax>#"oxalic acid, "C_2H_2O_4#</mathjax>?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxalic acid"#</mathjax> is the simplest dicarboxylic acid, <mathjax>#HO(O=)C-C(=O)OH#</mathjax>; its salts are called <mathjax>#"oxalates"#</mathjax>, i.e. <mathjax>#C_2O_4^(2-)#</mathjax>, which is a chelating ligand. In the laboratory it is often used as the <mathjax>#"dihydrate"#</mathjax>, <mathjax>#C_2H_2O_4*2H_2O#</mathjax>. </p></div>
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</article> | What s the covalent compound formula for oxalic acid? | null |
617 | a8b22164-6ddd-11ea-a1d9-ccda262736ce | https://socratic.org/questions/what-is-the-minimum-energy-required-to-ionize-a-hydrogen-atom-in-the-n-3-state | 2.42 × 10^(-16) J | start physical_unit 9 10 energy j qc_end physical_unit 9 10 14 15 state_n qc_end end | [{"type":"physical unit","value":"Minimum energy [OF] hydrogen atom [IN] J"}] | [{"type":"physical unit","value":"2.42 × 10^(-16) J"}] | [{"type":"physical unit","value":"n [OF] hydrogen atom [=] \\pu{3 state}"}] | <h1 class="questionTitle" itemprop="name">What is the minimum energy required to ionize a hydrogen atom in the n= 3 state? </h1> | null | 2.42 × 10^(-16) J | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the Bohr Energy Equation for <mathjax>#n_i = 3 -> n_f = oo#</mathjax><br/>
<mathjax>#DeltaE_i#</mathjax> = -<mathjax>#A(1/n_f^2 - 1/n_i^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(1/oo^2 - 1/3^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(0 -0.111)J = 2.42x10^-19J#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.42x10^(-16)J#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the Bohr Energy Equation for <mathjax>#n_i = 3 -> n_f = oo#</mathjax><br/>
<mathjax>#DeltaE_i#</mathjax> = -<mathjax>#A(1/n_f^2 - 1/n_i^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(1/oo^2 - 1/3^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(0 -0.111)J = 2.42x10^-19J#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the minimum energy required to ionize a hydrogen atom in the n= 3 state? </h1>
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Doc048
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May 26, 2017
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<div class="markdown"><p><mathjax>#2.42x10^(-16)J#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the Bohr Energy Equation for <mathjax>#n_i = 3 -> n_f = oo#</mathjax><br/>
<mathjax>#DeltaE_i#</mathjax> = -<mathjax>#A(1/n_f^2 - 1/n_i^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(1/oo^2 - 1/3^2)#</mathjax> = <mathjax>#-2.18x10^(-18)J(0 -0.111)J = 2.42x10^-19J#</mathjax></p></div>
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</article> | What is the minimum energy required to ionize a hydrogen atom in the n= 3 state? | null |
618 | a8f3bc1e-6ddd-11ea-904c-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-with-the-following-percentages-75-69 | C7H9O | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] molecular"}] | [{"type":"chemical equation","value":"C7H9O"}] | [{"type":"physical unit","value":"Percentage [OF] C in the compound [=] \\pu{75.69%}"},{"type":"physical unit","value":"Percentage [OF] H in the compound [=] \\pu{8.80%}"},{"type":"physical unit","value":"Percentage [OF] O in the compound [=] \\pu{15.51%}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound with the following percentages: 75.69% C, 8.80% H, and 15.51% O?</h1> | null | C7H9O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I might be mistaken, but here's what I did.<br/>
Assume you have 100 g of the compund. Then you take these percentages as normal weight, meaning, you can use formula for mole number. You count it for every atom and then compare the results.</p>
<p>So:<br/>
<mathjax>#n = (m)/"M"#</mathjax> <br/>
For C it's 6,3, for H 8,8 and for O it's 0,96</p>
<p>6,3:8,8:0,96 /*100<br/>
630:880:96 /:96<br/>
7:9:1</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>C <mathjax>#C_7 H_9O#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I might be mistaken, but here's what I did.<br/>
Assume you have 100 g of the compund. Then you take these percentages as normal weight, meaning, you can use formula for mole number. You count it for every atom and then compare the results.</p>
<p>So:<br/>
<mathjax>#n = (m)/"M"#</mathjax> <br/>
For C it's 6,3, for H 8,8 and for O it's 0,96</p>
<p>6,3:8,8:0,96 /*100<br/>
630:880:96 /:96<br/>
7:9:1</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of a compound with the following percentages: 75.69% C, 8.80% H, and 15.51% O?</h1>
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Mc'Coob
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Jan 28, 2016
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<div class="markdown"><p>C <mathjax>#C_7 H_9O#</mathjax> </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I might be mistaken, but here's what I did.<br/>
Assume you have 100 g of the compund. Then you take these percentages as normal weight, meaning, you can use formula for mole number. You count it for every atom and then compare the results.</p>
<p>So:<br/>
<mathjax>#n = (m)/"M"#</mathjax> <br/>
For C it's 6,3, for H 8,8 and for O it's 0,96</p>
<p>6,3:8,8:0,96 /*100<br/>
630:880:96 /:96<br/>
7:9:1</p></div>
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</article> | What is the molecular formula of a compound with the following percentages: 75.69% C, 8.80% H, and 15.51% O? | null |
619 | ac876e8c-6ddd-11ea-ae17-ccda262736ce | https://socratic.org/questions/a-sample-of-hydrogen-gas-occupies-14-1-l-at-stp-how-many-moles-of-the-gas-are-pr | 0.63 moles | start physical_unit 14 15 mole mol qc_end physical_unit 1 4 6 7 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mole [OF] the gas [IN] moles"}] | [{"type":"physical unit","value":"0.63 moles"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen gas sample [=] \\pu{14.1 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?</h1> | null | 0.63 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For gases at STP, the molar volume equation/relationship can be used where <strong>1 mol</strong> of gas occupies <strong>22.4 L</strong> of volume.</p>
<p>Therefore, <br/>
<mathjax>#14.1cancelL \cdot (1mol)/(22.4cancelL) = 0.6294642857143 mol #</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#0.63 mol#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For gases at STP, the molar volume equation/relationship can be used where <strong>1 mol</strong> of gas occupies <strong>22.4 L</strong> of volume.</p>
<p>Therefore, <br/>
<mathjax>#14.1cancelL \cdot (1mol)/(22.4cancelL) = 0.6294642857143 mol #</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?</h1>
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Mark Keanu James E. Exconde
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<div class="markdown"><p><mathjax>#0.63 mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For gases at STP, the molar volume equation/relationship can be used where <strong>1 mol</strong> of gas occupies <strong>22.4 L</strong> of volume.</p>
<p>Therefore, <br/>
<mathjax>#14.1cancelL \cdot (1mol)/(22.4cancelL) = 0.6294642857143 mol #</mathjax></p></div>
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</article> | A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present? | null |
620 | a95cb61e-6ddd-11ea-a6f9-ccda262736ce | https://socratic.org/questions/how-many-calories-are-absorbed-when-500-grams-of-water-1-cal-g-c-is-heated-from- | 25000 calories | start physical_unit 9 9 heat_energy cal qc_end physical_unit 9 9 10 13 specific_heat qc_end physical_unit 9 9 6 7 mass qc_end physical_unit 9 9 19 21 temperature qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] water [IN] calories"}] | [{"type":"physical unit","value":"25000 calories"}] | [{"type":"physical unit","value":"specific heat [OF] water [=] \\pu{1 cal/(g * ℃)}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{500 grams}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{50 degrees Celsius}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{100 degrees Celsius}"}] | <h1 class="questionTitle" itemprop="name">How many calories are absorbed when 500 grams of water (1 cal/g c) is heated from 50 to 100 degrees Celsius? </h1> | null | 25000 calories | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of a substance tells you the amount of heat needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, water is said to have a specific heat equal to</p>
<blockquote>
<p><mathjax>#c_"water" = "1 cal g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax> ?</p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for 1"^@"C")) = "500 cal"#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by </p>
<blockquote>
<p><mathjax>#100^@"C" - 50^@"C" = 50^@"C"#</mathjax></p>
</blockquote>
<p>you need to provide it with</p>
<blockquote>
<p><mathjax>#50 color(red)(cancel(color(black)(""^@"C"))) * overbrace("500 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 500 g")) = "25,000 cal" = color(darkgreen)(ul(color(black)("30,000 cal")))#</mathjax></p>
</blockquote>
<p>The answer <strong>must</strong> be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p>
<p>Keep in mind that this much heat is needed in order to get <mathjax>#"500 g"#</mathjax> of water from <em>liquid</em> at <mathjax>#50^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#100^@"C"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"30,000 cal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of a substance tells you the amount of heat needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, water is said to have a specific heat equal to</p>
<blockquote>
<p><mathjax>#c_"water" = "1 cal g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax> ?</p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for 1"^@"C")) = "500 cal"#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by </p>
<blockquote>
<p><mathjax>#100^@"C" - 50^@"C" = 50^@"C"#</mathjax></p>
</blockquote>
<p>you need to provide it with</p>
<blockquote>
<p><mathjax>#50 color(red)(cancel(color(black)(""^@"C"))) * overbrace("500 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 500 g")) = "25,000 cal" = color(darkgreen)(ul(color(black)("30,000 cal")))#</mathjax></p>
</blockquote>
<p>The answer <strong>must</strong> be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p>
<p>Keep in mind that this much heat is needed in order to get <mathjax>#"500 g"#</mathjax> of water from <em>liquid</em> at <mathjax>#50^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#100^@"C"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many calories are absorbed when 500 grams of water (1 cal/g c) is heated from 50 to 100 degrees Celsius? </h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"30,000 cal"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of a substance tells you the amount of heat needed to increase the temperature of <mathjax>#"1 g"#</mathjax> of that substance by <mathjax>#1^@"C"#</mathjax>. </p>
<p>In your case, water is said to have a specific heat equal to</p>
<blockquote>
<p><mathjax>#c_"water" = "1 cal g"^(-1)""^@"C"^(-1)#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"1 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax>, you need to provide it with <mathjax>#"1 cal"#</mathjax> of heat. </p>
<p>So, how much heat would be needed to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by <mathjax>#1^@"C"#</mathjax> ?</p>
<blockquote>
<p><mathjax>#500 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for 1"^@"C")) = "500 cal"#</mathjax></p>
</blockquote>
<p>This means that in order to increase the temperature of <mathjax>#"500 g"#</mathjax> of water by </p>
<blockquote>
<p><mathjax>#100^@"C" - 50^@"C" = 50^@"C"#</mathjax></p>
</blockquote>
<p>you need to provide it with</p>
<blockquote>
<p><mathjax>#50 color(red)(cancel(color(black)(""^@"C"))) * overbrace("500 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 500 g")) = "25,000 cal" = color(darkgreen)(ul(color(black)("30,000 cal")))#</mathjax></p>
</blockquote>
<p>The answer <strong>must</strong> be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>, the number of sig figs you have for your values. </p>
<p>Keep in mind that this much heat is needed in order to get <mathjax>#"500 g"#</mathjax> of water from <em>liquid</em> at <mathjax>#50^@"C"#</mathjax> to <em>liquid</em> at <mathjax>#100^@"C"#</mathjax>.</p></div>
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</article> | How many calories are absorbed when 500 grams of water (1 cal/g c) is heated from 50 to 100 degrees Celsius? | null |
621 | aaa8e986-6ddd-11ea-9e56-ccda262736ce | https://socratic.org/questions/a-compound-is-found-to-contain-23-3-magnesium-30-7-sulfur-and-46-0-oxygen-what-i | MgSO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] this compound [IN] empirical"}] | [{"type":"chemical equation","value":"MgSO3"}] | [{"type":"physical unit","value":"Percent [OF] magnesium in this compound [=] \\pu{23.3%}"},{"type":"physical unit","value":"Percent [OF] sulfur in this compound [=] \\pu{30.7%}"},{"type":"physical unit","value":"Percent [OF] oxygen in this compound [=] \\pu{46.0%}"}] | <h1 class="questionTitle" itemprop="name">A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound? </h1> | null | MgSO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As is standard for these sorts of questions, we assume 100 g of substance, and use the percentage elemental compositions to calculate an empirical formula.</p>
<p><mathjax>#"Magnesium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(23.3*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Sulfur"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(30.7*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(46.0*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.86*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity <mathjax>#(0.958*mol)#</mathjax> to get and empirical formula of <mathjax>#MgSO_3#</mathjax>, magnesium sulfite. </p>
<p>Ordinarily you would not be given the percentage of oxygen by mass. You would be given the percentages of the metal and sulfur, and would have to assume that the balance was oxygen. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#MgSO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As is standard for these sorts of questions, we assume 100 g of substance, and use the percentage elemental compositions to calculate an empirical formula.</p>
<p><mathjax>#"Magnesium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(23.3*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Sulfur"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(30.7*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(46.0*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.86*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity <mathjax>#(0.958*mol)#</mathjax> to get and empirical formula of <mathjax>#MgSO_3#</mathjax>, magnesium sulfite. </p>
<p>Ordinarily you would not be given the percentage of oxygen by mass. You would be given the percentages of the metal and sulfur, and would have to assume that the balance was oxygen. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound? </h1>
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anor277
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<div class="markdown"><p><mathjax>#MgSO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As is standard for these sorts of questions, we assume 100 g of substance, and use the percentage elemental compositions to calculate an empirical formula.</p>
<p><mathjax>#"Magnesium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(23.3*g)/(24.31*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Sulfur"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(30.7*g)/(32.06*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.958*mol#</mathjax></p>
<p><mathjax>#"Oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(46.0*g)/(15.999*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.86*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity <mathjax>#(0.958*mol)#</mathjax> to get and empirical formula of <mathjax>#MgSO_3#</mathjax>, magnesium sulfite. </p>
<p>Ordinarily you would not be given the percentage of oxygen by mass. You would be given the percentages of the metal and sulfur, and would have to assume that the balance was oxygen. </p></div>
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</article> | A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound? | null |
622 | ac201f15-6ddd-11ea-85d8-ccda262736ce | https://socratic.org/questions/an-open-flask-sitting-in-a-lab-fridge-looks-empty-but-it-is-filled-with-a-mixtur | 1.32 × 10^23 | start physical_unit 39 40 number none qc_end physical_unit 22 23 26 27 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] gaseous molecules"}] | [{"type":"physical unit","value":"1.32 × 10^23"}] | [{"type":"physical unit","value":"Volume [OF] the flask [=] \\pu{2.50 L}"},{"type":"other","value":"Standard temperature and pressure."}] | <h1 class="questionTitle" itemprop="name">An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?</h1> | null | 1.32 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>: <mathjax>#n=(PV)/(RT)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5*L)/(0.0821*L*atm*K^-1*mol^-1xx278*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax></p>
<p>And thus we mulitply <mathjax>#0.219*mol#</mathjax> by <mathjax>#"Avogadro's number"#</mathjax>.</p>
<p><mathjax>#0.219*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>If the flask were at room temperature, would it contain the same number of molecules?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>If we assume that the temperature inside the fridge is <mathjax>#5#</mathjax> <mathjax>#""^@C#</mathjax>, there are <mathjax>#0.219*molxx6.022xx10^23*mol^-1#</mathjax> of dinitrogen and dioxygen molecules. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>: <mathjax>#n=(PV)/(RT)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5*L)/(0.0821*L*atm*K^-1*mol^-1xx278*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax></p>
<p>And thus we mulitply <mathjax>#0.219*mol#</mathjax> by <mathjax>#"Avogadro's number"#</mathjax>.</p>
<p><mathjax>#0.219*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>If the flask were at room temperature, would it contain the same number of molecules?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?</h1>
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<div class="markdown"><p>If we assume that the temperature inside the fridge is <mathjax>#5#</mathjax> <mathjax>#""^@C#</mathjax>, there are <mathjax>#0.219*molxx6.022xx10^23*mol^-1#</mathjax> of dinitrogen and dioxygen molecules. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a>: <mathjax>#n=(PV)/(RT)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1*atmxx5*L)/(0.0821*L*atm*K^-1*mol^-1xx278*K)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.219*mol#</mathjax></p>
<p>And thus we mulitply <mathjax>#0.219*mol#</mathjax> by <mathjax>#"Avogadro's number"#</mathjax>.</p>
<p><mathjax>#0.219*molxx6.022xx10^23*mol^-1=??#</mathjax></p>
<p>If the flask were at room temperature, would it contain the same number of molecules?</p></div>
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</article> | An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain? | null |
623 | ac6457b4-6ddd-11ea-8ec8-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-of-sulphur-trioxide | SO3 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] sulphur trioxide [IN] default"}] | [{"type":"chemical equation","value":"SO3"}] | [{"type":"substance name","value":"Sulphur trioxide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula of sulphur trioxide?</h1> | null | SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur trioxide: <mathjax>#SO_"3"#</mathjax>.</p>
<p><strong>How I came to this conclusion:</strong> </p>
<p>I noticed that the compound consists of non-metals, therefore it must be a covalent compound - meaning I use prefixes for <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. </p>
<p>Sulfur is <mathjax>#S#</mathjax>. Simple enough. Trioxide has the prefix of <em>tri</em>, meaning "3". Therefore there is 3 atoms of oxygen. Combine them and I get <mathjax>#SO_"3"#</mathjax>.</p>
<p>Don't be confused with sulfite: <mathjax>#(SO_"3")^(2-)#</mathjax> and sulfate: <mathjax>#(SO_"4")^(2-)#</mathjax></p>
<p>Hope this helps :)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Sulfur trioxide - <mathjax>#SO_"3"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur trioxide: <mathjax>#SO_"3"#</mathjax>.</p>
<p><strong>How I came to this conclusion:</strong> </p>
<p>I noticed that the compound consists of non-metals, therefore it must be a covalent compound - meaning I use prefixes for <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. </p>
<p>Sulfur is <mathjax>#S#</mathjax>. Simple enough. Trioxide has the prefix of <em>tri</em>, meaning "3". Therefore there is 3 atoms of oxygen. Combine them and I get <mathjax>#SO_"3"#</mathjax>.</p>
<p>Don't be confused with sulfite: <mathjax>#(SO_"3")^(2-)#</mathjax> and sulfate: <mathjax>#(SO_"4")^(2-)#</mathjax></p>
<p>Hope this helps :)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula of sulphur trioxide?</h1>
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<div class="markdown"><p>Sulfur trioxide - <mathjax>#SO_"3"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Sulfur trioxide: <mathjax>#SO_"3"#</mathjax>.</p>
<p><strong>How I came to this conclusion:</strong> </p>
<p>I noticed that the compound consists of non-metals, therefore it must be a covalent compound - meaning I use prefixes for <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>. </p>
<p>Sulfur is <mathjax>#S#</mathjax>. Simple enough. Trioxide has the prefix of <em>tri</em>, meaning "3". Therefore there is 3 atoms of oxygen. Combine them and I get <mathjax>#SO_"3"#</mathjax>.</p>
<p>Don't be confused with sulfite: <mathjax>#(SO_"3")^(2-)#</mathjax> and sulfate: <mathjax>#(SO_"4")^(2-)#</mathjax></p>
<p>Hope this helps :)</p></div>
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</article> | What is the chemical formula of sulphur trioxide? | null |
624 | a8b17964-6ddd-11ea-8185-ccda262736ce | https://socratic.org/questions/a-gas-occupies-246-ml-at-567-3-torr-and-24-0-c-when-the-pressure-is-changed-to-7 | 32.62 °C | start physical_unit 1 1 temperature °c qc_end physical_unit 1 1 3 4 volume qc_end physical_unit 1 1 6 7 pressure qc_end physical_unit 1 1 9 10 temperature qc_end physical_unit 1 1 17 18 pressure qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] °C"}] | [{"type":"physical unit","value":"32.62 °C"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{246 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{567.3 Torr}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{24.0 °C}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{771.0 Torr}"},{"type":"other","value":"maintain the same volume."}] | <h1 class="questionTitle" itemprop="name"> A gas occupies 246 mL at 567.3 Torr and 24.0 °C. When the pressure is changed to 771.0 Torr, what temperature is needed to maintain the same volume?
</h1> | null | 32.62 °C | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It's always a good idea to start with what you know. </p>
<p>You know that the volume of the container, which is given to you as <mathjax>#"246 mL"#</mathjax>, must <strong>remain constant</strong>. Since no mention of the number of moles of gas was made, you can safely assume that it remains <strong>constant</strong> as well. </p>
<p>This means that only temperature and pressure will change. </p>
<p>When <em>volume</em> and <em>number of moles</em> are kept constant, pressure and temperature have <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<p><img alt="http://www.ck12.org/book/CK-12-Chemistry-Intermediate/section/14.1/" src="https://useruploads.socratic.org/MHYkD6XCRlWohvZjvGYI_gay_lussac.jpg"/> </p>
<p>In other words, if pressure <strong>increases</strong>, temperature increases as well. If pressure <strong>decreases</strong>, temperature decreases as well. </p>
<p>Mathematically, this is written as</p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state;<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state.</p>
<p>In your case, you know that the pressue is <em>increasing</em> from <mathjax>#"567.3 torr"#</mathjax>, to <mathjax>#"771.0 torr"#</mathjax>. This means that you can expect the needed temperature to be <strong>higher</strong> than the initial <mathjax>#24.0^@"C"#</mathjax>. </p>
<p>Plug in your values and solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * T_1#</mathjax></p>
<p><mathjax>#T_2 = (771.0color(red)(cancel(color(black)("torr"))))/(567.3color(red)(cancel(color(black)("torr")))) * 24.0^@"C" = color(green)(32.6^@"C")#</mathjax></p>
</blockquote>
<p>As predict, the temperature must be higher than what it initially was. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#32.6^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It's always a good idea to start with what you know. </p>
<p>You know that the volume of the container, which is given to you as <mathjax>#"246 mL"#</mathjax>, must <strong>remain constant</strong>. Since no mention of the number of moles of gas was made, you can safely assume that it remains <strong>constant</strong> as well. </p>
<p>This means that only temperature and pressure will change. </p>
<p>When <em>volume</em> and <em>number of moles</em> are kept constant, pressure and temperature have <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<p><img alt="http://www.ck12.org/book/CK-12-Chemistry-Intermediate/section/14.1/" src="https://useruploads.socratic.org/MHYkD6XCRlWohvZjvGYI_gay_lussac.jpg"/> </p>
<p>In other words, if pressure <strong>increases</strong>, temperature increases as well. If pressure <strong>decreases</strong>, temperature decreases as well. </p>
<p>Mathematically, this is written as</p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state;<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state.</p>
<p>In your case, you know that the pressue is <em>increasing</em> from <mathjax>#"567.3 torr"#</mathjax>, to <mathjax>#"771.0 torr"#</mathjax>. This means that you can expect the needed temperature to be <strong>higher</strong> than the initial <mathjax>#24.0^@"C"#</mathjax>. </p>
<p>Plug in your values and solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * T_1#</mathjax></p>
<p><mathjax>#T_2 = (771.0color(red)(cancel(color(black)("torr"))))/(567.3color(red)(cancel(color(black)("torr")))) * 24.0^@"C" = color(green)(32.6^@"C")#</mathjax></p>
</blockquote>
<p>As predict, the temperature must be higher than what it initially was. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name"> A gas occupies 246 mL at 567.3 Torr and 24.0 °C. When the pressure is changed to 771.0 Torr, what temperature is needed to maintain the same volume?
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Stefan V.
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<span class="dateCreated" datetime="2015-10-23T18:23:06" itemprop="dateCreated">
Oct 23, 2015
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<div class="markdown"><p><mathjax>#32.6^@"C"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It's always a good idea to start with what you know. </p>
<p>You know that the volume of the container, which is given to you as <mathjax>#"246 mL"#</mathjax>, must <strong>remain constant</strong>. Since no mention of the number of moles of gas was made, you can safely assume that it remains <strong>constant</strong> as well. </p>
<p>This means that only temperature and pressure will change. </p>
<p>When <em>volume</em> and <em>number of moles</em> are kept constant, pressure and temperature have <strong>direct relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/gay-lussac-s-law">Gay Lussac's Law</a>.</p>
<p><img alt="http://www.ck12.org/book/CK-12-Chemistry-Intermediate/section/14.1/" src="https://useruploads.socratic.org/MHYkD6XCRlWohvZjvGYI_gay_lussac.jpg"/> </p>
<p>In other words, if pressure <strong>increases</strong>, temperature increases as well. If pressure <strong>decreases</strong>, temperature decreases as well. </p>
<p>Mathematically, this is written as</p>
<blockquote>
<p><mathjax>#P_1/T_1 = P_2/T_2" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure and temperature of the gas at an initial state;<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure and temperature of the gas at a final state.</p>
<p>In your case, you know that the pressue is <em>increasing</em> from <mathjax>#"567.3 torr"#</mathjax>, to <mathjax>#"771.0 torr"#</mathjax>. This means that you can expect the needed temperature to be <strong>higher</strong> than the initial <mathjax>#24.0^@"C"#</mathjax>. </p>
<p>Plug in your values and solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#T_2 = P_2/P_1 * T_1#</mathjax></p>
<p><mathjax>#T_2 = (771.0color(red)(cancel(color(black)("torr"))))/(567.3color(red)(cancel(color(black)("torr")))) * 24.0^@"C" = color(green)(32.6^@"C")#</mathjax></p>
</blockquote>
<p>As predict, the temperature must be higher than what it initially was. </p></div>
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</article> | A gas occupies 246 mL at 567.3 Torr and 24.0 °C. When the pressure is changed to 771.0 Torr, what temperature is needed to maintain the same volume?
| null |
625 | abb7f65c-6ddd-11ea-ab51-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-potassium-sulfide | K2S | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] potassium sulfide [IN] default"}] | [{"type":"chemical equation","value":"K2S"}] | [{"type":"substance name","value":"Potassium sulfide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for potassium sulfide? </h1> | null | K2S | <div class="answerDescription">
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">Solutions</a> of <mathjax>#K_2S#</mathjax> hydrolyze in water to give some <mathjax>#HO^-#</mathjax> and <mathjax>#HS^-#</mathjax> ions. </p></div>
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<div class="markdown"><p><mathjax>#K_2S#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">Solutions</a> of <mathjax>#K_2S#</mathjax> hydrolyze in water to give some <mathjax>#HO^-#</mathjax> and <mathjax>#HS^-#</mathjax> ions. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for potassium sulfide? </h1>
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anor277
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<div class="markdown"><p><mathjax>#K_2S#</mathjax></p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">Solutions</a> of <mathjax>#K_2S#</mathjax> hydrolyze in water to give some <mathjax>#HO^-#</mathjax> and <mathjax>#HS^-#</mathjax> ions. </p></div>
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</article> | What is the formula for potassium sulfide? | null |
626 | aa7962a8-6ddd-11ea-888a-ccda262736ce | https://socratic.org/questions/55862b80581e2a7bae15ca2e | 0.20 g | start physical_unit 2 2 mass g qc_end physical_unit 8 8 6 7 volume qc_end chemical_equation 15 21 qc_end end | [{"type":"physical unit","value":"Mass [OF] Ca [IN] g"}] | [{"type":"physical unit","value":"0.20 g"}] | [{"type":"physical unit","value":"Volume [OF] O2 [=] \\pu{56.8 mL}"},{"type":"chemical equation","value":"2 Ca + O2 -> 2 CaO"}] | <h1 class="questionTitle" itemprop="name">How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"2Ca+O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p></div>
</h2>
</div>
</div> | 0.20 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This reaction will consume <strong>0.201 g</strong> of calcium.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
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<h1 class="questionTitle" itemprop="name">How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#?</h1>
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<div class="markdown"><p><mathjax>#"2Ca+O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p></div>
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<div class="markdown"><p>This reaction will consume <strong>0.201 g</strong> of calcium.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
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<div class="markdown"><p><mathjax>#0.203"g Ca"#</mathjax> are consumed when <mathjax>#56.8"mL O"_2#</mathjax> reacts with calcium to produce calcium oxide, <mathjax>#"CaO"#</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>#"2Ca+O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio of <mathjax>#"Ca"#</mathjax> to <mathjax>#"O"_2#</mathjax> is <mathjax>#(2 "mol Ca")/(1 "mol O"_2")#</mathjax>.</p>
<p>In order to solve this problem, we need to convert the volume of oxygen to liters. Then convert the volume to moles using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a>. Next we will multiply the moles of oxygen times the mole ratio from the equation to get moles of calcium. Finally, we will multiply the moles of calcium times its molar mass to get the mass of calcium consumed.</p>
<p><strong>Convert volume to liters.</strong></p>
<p><mathjax>#56.8color(red)cancel(color(black)("mL O"_2))xx("1L O"_2)/(1000color(red)cancel(color(black)("mL O"_2")))#</mathjax><mathjax>#=0.0568"mol O"_2"#</mathjax></p>
<p><strong>Convert volume in liters to moles.</strong></p>
<p>The volume of one mole of a gas is its molar volume. At STP the molar volume of an ideal gas is <mathjax>#22.414"L/mol"#</mathjax>.</p>
<p><mathjax>#0.0568color(red)cancel(color(black)("L O"_2))xx(1"mol O"_2)/(22.414color(red)cancel(color(black)("L O"_2)))=0.0025341"mol O"_2#</mathjax></p>
<p><strong>Determine the moles of Ca consumed.</strong></p>
<p>Multiply the moles of <mathjax>#"O"_2#</mathjax> times the previously determined mole ratio between calcium and oxygen.</p>
<p><mathjax>#0.0025341color(red)cancel(color(black)("mol O"_2))xx("2mol Ca")/(1color(red)cancel(color(black)("mol O"_2)))=0.0050682"mol Ca"#</mathjax></p>
<p><strong>Convert moles of Ca consumed to mass of Ca consumed.</strong></p>
<p>We need the molar mass of <mathjax>#"Ca"#</mathjax>, which is <mathjax>#40.078"g/mol"#</mathjax>. (This is its atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole.)</p>
<p><mathjax>#0.0050682color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))=0.203"g Ca"#</mathjax> (rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>)</p></div>
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</article> | How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#? |
#"2Ca+O"_2"##rarr##"2CaO"#
|
627 | a8b248b8-6ddd-11ea-a4d8-ccda262736ce | https://socratic.org/questions/the-poh-of-a-solution-is-2-0-what-is-its-ph | 12 | start physical_unit 3 4 ph none qc_end physical_unit 3 4 6 6 poh qc_end end | [{"type":"physical unit","value":"pH [OF] a solution"}] | [{"type":"physical unit","value":"12"}] | [{"type":"physical unit","value":"pOH [OF] a solution [=] \\pu{2.0}"}] | <h1 class="questionTitle" itemprop="name">The pOH of a solution is 2.0. What is its pH?</h1> | null | 12 | <div class="answerDescription">
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<div class="markdown"><p>You know (or should know!) that <mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions.......</p>
<p>And thus <mathjax>#pH=14-2=??#</mathjax></p>
<p>See <a href="https://socratic.org/questions/what-is-the-ph-of-a-1-00-x-10-4-m-solution-of-lithium-hydroxide-solution#306871">here</a> for its derivation. </p></div>
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<div class="markdown"><p><mathjax>#pH=12#</mathjax>...........</p></div>
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<div class="markdown"><p>You know (or should know!) that <mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions.......</p>
<p>And thus <mathjax>#pH=14-2=??#</mathjax></p>
<p>See <a href="https://socratic.org/questions/what-is-the-ph-of-a-1-00-x-10-4-m-solution-of-lithium-hydroxide-solution#306871">here</a> for its derivation. </p></div>
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<h1 class="questionTitle" itemprop="name">The pOH of a solution is 2.0. What is its pH?</h1>
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<div class="markdown"><p><mathjax>#pH=12#</mathjax>...........</p></div>
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<div class="markdown"><p>You know (or should know!) that <mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions.......</p>
<p>And thus <mathjax>#pH=14-2=??#</mathjax></p>
<p>See <a href="https://socratic.org/questions/what-is-the-ph-of-a-1-00-x-10-4-m-solution-of-lithium-hydroxide-solution#306871">here</a> for its derivation. </p></div>
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</article> | The pOH of a solution is 2.0. What is its pH? | null |
628 | ab59fea6-6ddd-11ea-9259-ccda262736ce | https://socratic.org/questions/if-18-6-g-of-methanol-is-used-to-dissolve-2-68-g-of-hg-cn-2-what-is-the-molality | 0.60 mol/kg | start physical_unit 18 19 molality mol/kg qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 4 4 1 2 mass qc_end end | [{"type":"physical unit","value":"Molality [OF] the solution [IN] mol/kg"}] | [{"type":"physical unit","value":"0.60 mol/kg"}] | [{"type":"physical unit","value":"Mass [OF] Hg(CN)2 [=] \\pu{2.68 g}"},{"type":"physical unit","value":"Mass [OF] methanol [=] \\pu{18.6 g}"}] | <h1 class="questionTitle" itemprop="name">If 18.6 g of methanol is used to dissolve 2.68 g of #Hg(CN)_2#, what is the molality of the solution?</h1> | null | 0.60 mol/kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And here,</p>
<p><mathjax>#"Molality"=((2.68*g)/(252.65*g*mol^-1))/(18.6xx10^-3*kg)=??mol*kg^-1??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Molality"="Moles of solute"/"Kilograms of solvent"~=0.60*mol*kg^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And here,</p>
<p><mathjax>#"Molality"=((2.68*g)/(252.65*g*mol^-1))/(18.6xx10^-3*kg)=??mol*kg^-1??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If 18.6 g of methanol is used to dissolve 2.68 g of #Hg(CN)_2#, what is the molality of the solution?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Molality"="Moles of solute"/"Kilograms of solvent"~=0.60*mol*kg^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And here,</p>
<p><mathjax>#"Molality"=((2.68*g)/(252.65*g*mol^-1))/(18.6xx10^-3*kg)=??mol*kg^-1??#</mathjax></p></div>
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</article> | If 18.6 g of methanol is used to dissolve 2.68 g of #Hg(CN)_2#, what is the molality of the solution? | null |
629 | a9ec29ae-6ddd-11ea-be76-ccda262736ce | https://socratic.org/questions/how-many-kilojoules-are-released-when-111-g-of-ch-reacts-with-silicon-in-the-rea | 514 kilojoules | start physical_unit 14 15 heat_energy kj qc_end physical_unit 9 9 6 7 mass qc_end chemical_equation 16 21 qc_end physical_unit 14 15 24 25 deltah qc_end end | [{"type":"physical unit","value":"Released energy [OF] the reaction [IN] kilojoules"}] | [{"type":"physical unit","value":"514 kilojoules"}] | [{"type":"physical unit","value":"Mass [OF] Cl2 [=] \\pu{111 g}"},{"type":"chemical equation","value":"Si(s) + 2 Cl2(g) -> SiCl4(g)"},{"type":"physical unit","value":"DeltaH [OF] the reaction [=] \\pu{-657 kJ}"}] | <h1 class="questionTitle" itemprop="name">How many kilojoules are released when 111 g of #Cl_2# reacts with silicon in the reaction #Si(s)+2Cl_2(g) ->SiCl_4(g)# #ΔH = - 657kJ?#</h1> | null | 514 kilojoules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Notice that the problem provides you with the <strong>thermochemical equation</strong> for this <a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a>.</p>
<p>A <strong>thermochemical equation</strong> is simply a <em>balanced chemical equation</em> that includes the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, for that given chemical reaction. </p>
<p>In this case, you have </p>
<blockquote>
<p><mathjax>#"Si"_text((s]) + color(red)(2)"Cl"_text(2(g]) -> "SiCl"_text(4(g])", "DeltaH_text(rxn) = -"657 kJ"#</mathjax></p>
</blockquote>
<p>This means that when <strong>one mole</strong> of silicon tetrachloride, <mathjax>#"SiCl"_4#</mathjax>, is formed, the reaction <strong>gives off</strong> <mathjax>#"657 kJ"#</mathjax> of heat. Since heat is being given off, hence the <em>negative sign</em> for the enthalpy change oi reaction, you know that you're dealing with an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic reaction</a>.</p>
<p>Your goal now will be to determine how many <strong>moles</strong> of silicon tetrachloride will be produced by that <mathjax>#"111-g"#</mathjax> sample of chlorine gas. </p>
<p>To do that, start by figuring out how many moles of chlorine gas are found in that many grams </p>
<blockquote>
<p><mathjax>#111 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("moles Cl"_2)))) = "1.565 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between chlorine gas and silicon tetrachloride. This means that the reaction will produce </p>
<blockquote>
<p><mathjax>#1.565 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole SiCl"_4/(color(red)(2)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7825 moles SiCl"_4#</mathjax></p>
</blockquote>
<p>So, if the reaction gives off <mathjax>#"657 kJ"#</mathjax> of heat when <strong>one mole</strong> of product is formed, it follows that producing <mathjax>#0.7825#</mathjax> <strong>moles</strong> of silicon tetrachloride will give off</p>
<blockquote>
<p><mathjax>#0.7825 color(red)(cancel(color(black)("moles SiCl"_4))) * "657 kJ"/(1color(red)(cancel(color(black)("mole SiCl"_4)))) = color(green)("514 kJ")#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>. </p>
<p>So, you can say that</p>
<blockquote>
<p><em>The reaction <strong>gives off</strong></em> <mathjax>#"514 kJ"#</mathjax> <em>of heat when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote>
<p><strong>or</strong></p>
<blockquote>
<p><em>The reactions has</em> <mathjax>#DeltaH_"rxn" = -"514 kJ"#</mathjax> <em>when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"514 kJ"#</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 the <strong>thermochemical equation</strong> for this <a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a>.</p>
<p>A <strong>thermochemical equation</strong> is simply a <em>balanced chemical equation</em> that includes the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, for that given chemical reaction. </p>
<p>In this case, you have </p>
<blockquote>
<p><mathjax>#"Si"_text((s]) + color(red)(2)"Cl"_text(2(g]) -> "SiCl"_text(4(g])", "DeltaH_text(rxn) = -"657 kJ"#</mathjax></p>
</blockquote>
<p>This means that when <strong>one mole</strong> of silicon tetrachloride, <mathjax>#"SiCl"_4#</mathjax>, is formed, the reaction <strong>gives off</strong> <mathjax>#"657 kJ"#</mathjax> of heat. Since heat is being given off, hence the <em>negative sign</em> for the enthalpy change oi reaction, you know that you're dealing with an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic reaction</a>.</p>
<p>Your goal now will be to determine how many <strong>moles</strong> of silicon tetrachloride will be produced by that <mathjax>#"111-g"#</mathjax> sample of chlorine gas. </p>
<p>To do that, start by figuring out how many moles of chlorine gas are found in that many grams </p>
<blockquote>
<p><mathjax>#111 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("moles Cl"_2)))) = "1.565 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between chlorine gas and silicon tetrachloride. This means that the reaction will produce </p>
<blockquote>
<p><mathjax>#1.565 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole SiCl"_4/(color(red)(2)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7825 moles SiCl"_4#</mathjax></p>
</blockquote>
<p>So, if the reaction gives off <mathjax>#"657 kJ"#</mathjax> of heat when <strong>one mole</strong> of product is formed, it follows that producing <mathjax>#0.7825#</mathjax> <strong>moles</strong> of silicon tetrachloride will give off</p>
<blockquote>
<p><mathjax>#0.7825 color(red)(cancel(color(black)("moles SiCl"_4))) * "657 kJ"/(1color(red)(cancel(color(black)("mole SiCl"_4)))) = color(green)("514 kJ")#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>. </p>
<p>So, you can say that</p>
<blockquote>
<p><em>The reaction <strong>gives off</strong></em> <mathjax>#"514 kJ"#</mathjax> <em>of heat when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote>
<p><strong>or</strong></p>
<blockquote>
<p><em>The reactions has</em> <mathjax>#DeltaH_"rxn" = -"514 kJ"#</mathjax> <em>when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many kilojoules are released when 111 g of #Cl_2# reacts with silicon in the reaction #Si(s)+2Cl_2(g) ->SiCl_4(g)# #ΔH = - 657kJ?#</h1>
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Stefan V.
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Jan 5, 2016
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<div class="markdown"><p><mathjax>#"514 kJ"#</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 the <strong>thermochemical equation</strong> for this <a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a>.</p>
<p>A <strong>thermochemical equation</strong> is simply a <em>balanced chemical equation</em> that includes the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>, for that given chemical reaction. </p>
<p>In this case, you have </p>
<blockquote>
<p><mathjax>#"Si"_text((s]) + color(red)(2)"Cl"_text(2(g]) -> "SiCl"_text(4(g])", "DeltaH_text(rxn) = -"657 kJ"#</mathjax></p>
</blockquote>
<p>This means that when <strong>one mole</strong> of silicon tetrachloride, <mathjax>#"SiCl"_4#</mathjax>, is formed, the reaction <strong>gives off</strong> <mathjax>#"657 kJ"#</mathjax> of heat. Since heat is being given off, hence the <em>negative sign</em> for the enthalpy change oi reaction, you know that you're dealing with an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic reaction</a>.</p>
<p>Your goal now will be to determine how many <strong>moles</strong> of silicon tetrachloride will be produced by that <mathjax>#"111-g"#</mathjax> sample of chlorine gas. </p>
<p>To do that, start by figuring out how many moles of chlorine gas are found in that many grams </p>
<blockquote>
<p><mathjax>#111 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("moles Cl"_2)))) = "1.565 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between chlorine gas and silicon tetrachloride. This means that the reaction will produce </p>
<blockquote>
<p><mathjax>#1.565 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole SiCl"_4/(color(red)(2)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7825 moles SiCl"_4#</mathjax></p>
</blockquote>
<p>So, if the reaction gives off <mathjax>#"657 kJ"#</mathjax> of heat when <strong>one mole</strong> of product is formed, it follows that producing <mathjax>#0.7825#</mathjax> <strong>moles</strong> of silicon tetrachloride will give off</p>
<blockquote>
<p><mathjax>#0.7825 color(red)(cancel(color(black)("moles SiCl"_4))) * "657 kJ"/(1color(red)(cancel(color(black)("mole SiCl"_4)))) = color(green)("514 kJ")#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>. </p>
<p>So, you can say that</p>
<blockquote>
<p><em>The reaction <strong>gives off</strong></em> <mathjax>#"514 kJ"#</mathjax> <em>of heat when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote>
<p><strong>or</strong></p>
<blockquote>
<p><em>The reactions has</em> <mathjax>#DeltaH_"rxn" = -"514 kJ"#</mathjax> <em>when</em> <mathjax>#0.7825#</mathjax> <em>moles of silicon tetrachloride are produced</em></p>
</blockquote></div>
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</article> | How many kilojoules are released when 111 g of #Cl_2# reacts with silicon in the reaction #Si(s)+2Cl_2(g) ->SiCl_4(g)# #ΔH = - 657kJ?# | null |
630 | a94347d4-6ddd-11ea-bf2d-ccda262736ce | https://socratic.org/questions/in-the-equation-mg-s-2hcl-aq-mgcl-2-aq-h-2-g-what-mass-of-hydrogen-will-be-obtai | 0.20 g | start physical_unit 14 14 mass g qc_end chemical_equation 3 10 qc_end physical_unit 6 6 22 23 molarity qc_end physical_unit 31 31 28 29 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrogen [IN] g"}] | [{"type":"physical unit","value":"0.20 g"}] | [{"type":"chemical equation","value":"Mg(s) + 2 HCl(aq) -> MgCl2(aq) + H2(g)"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{100 cm^3}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{2.00 mol/dm^3}"},{"type":"physical unit","value":"Mass [OF] magnesium [=] \\pu{4.86 g}"}] | <h1 class="questionTitle" itemprop="name">In the equation #Mg(s) + 2HCl(aq) -> MgCL_2(aq) + H_2(g)#, what mass of hydrogen will be obtained if #100# #cm^3# of 2.00 mol #dm^-3# #HCl# are added to 4.86 g of magnesium?</h1> | null | 0.20 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of metal, <mathjax>#=#</mathjax> <mathjax>#(4.86*g)/(24.305*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>Moles of <mathjax>#HCl#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100 *cm^-3xx2.00*mol*dm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Clearly, the acid is in <em>deficiency</em> ; i.e. it is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal. </p>
<p>So if <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax> acid react, then (by the stoichiometry), 1/2 this quantity, i.e. <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> of dihydrogen will evolve.</p>
<p>So, <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> dihydrogen are evolved; this has a mass of <mathjax>#0.100*molxx2.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax>.</p>
<p>If 1 mol dihydrogen gas occupies <mathjax>#24.5#</mathjax> <mathjax>#dm^3#</mathjax> at room temperature and pressure, what will be the VOLUME of gas evolved?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of metal, <mathjax>#=#</mathjax> <mathjax>#(4.86*g)/(24.305*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>Moles of <mathjax>#HCl#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100 *cm^-3xx2.00*mol*dm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Clearly, the acid is in <em>deficiency</em> ; i.e. it is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal. </p>
<p>So if <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax> acid react, then (by the stoichiometry), 1/2 this quantity, i.e. <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> of dihydrogen will evolve.</p>
<p>So, <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> dihydrogen are evolved; this has a mass of <mathjax>#0.100*molxx2.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax>.</p>
<p>If 1 mol dihydrogen gas occupies <mathjax>#24.5#</mathjax> <mathjax>#dm^3#</mathjax> at room temperature and pressure, what will be the VOLUME of gas evolved?</p></div>
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<h1 class="questionTitle" itemprop="name">In the equation #Mg(s) + 2HCl(aq) -> MgCL_2(aq) + H_2(g)#, what mass of hydrogen will be obtained if #100# #cm^3# of 2.00 mol #dm^-3# #HCl# are added to 4.86 g of magnesium?</h1>
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<div class="markdown"><p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of metal, <mathjax>#=#</mathjax> <mathjax>#(4.86*g)/(24.305*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>Moles of <mathjax>#HCl#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100 *cm^-3xx2.00*mol*dm^-3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax></p>
<p>Clearly, the acid is in <em>deficiency</em> ; i.e. it is the <a href="http://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a>, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal. </p>
<p>So if <mathjax>#0.200#</mathjax> <mathjax>#mol#</mathjax> acid react, then (by the stoichiometry), 1/2 this quantity, i.e. <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> of dihydrogen will evolve.</p>
<p>So, <mathjax>#0.100#</mathjax> <mathjax>#mol#</mathjax> dihydrogen are evolved; this has a mass of <mathjax>#0.100*molxx2.00*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax>.</p>
<p>If 1 mol dihydrogen gas occupies <mathjax>#24.5#</mathjax> <mathjax>#dm^3#</mathjax> at room temperature and pressure, what will be the VOLUME of gas evolved?</p></div>
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<div class="markdown"><p>The limiting reactant is <mathjax>#"HCl"#</mathjax>, which will produce <mathjax>#"0.202 g H"_2"#</mathjax> under the stated conditions.</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.</p>
<p><mathjax>#"Mg(s)" + "2HCl(aq)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"MgCl"_2("aq")"+ H"_2("g")"#</mathjax></p>
<p><strong>Determine Moles of Magnesium</strong><br/>
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).</p>
<p><mathjax>#4.86cancel"g Mg"xx(1"mol Mg")/(24.3050cancel"g Mg")="0.200 mol Mg"#</mathjax></p>
<p><strong>Determine Moles of 2M Hydrochloric Acid</strong><br/>
Convert <mathjax>#"100 cm"^3"#</mathjax> to <mathjax>#"100 mL"#</mathjax> and then to <mathjax>#"0.1 L"#</mathjax>. <br/>
<mathjax>#"1 dm"^3#</mathjax><mathjax>#=#</mathjax><mathjax>#"1 L"#</mathjax><br/>
Convert <mathjax>#"2.00 mol/dm"^3#</mathjax> to <mathjax>#"2.00 mol/L"#</mathjax><br/>
Multiply <mathjax>#0.1"L"#</mathjax> times <mathjax>#"2.00 mol/L"#</mathjax>.</p>
<p><mathjax>#100cancel"cm"^3xx(1cancel"mL")/(1cancel"cm"^3)xx(1"L")/(1000cancel"mL")="0.1 L HCl"#</mathjax></p>
<p><mathjax>#"2.00 mol/dm"^3"#</mathjax><mathjax>#=#</mathjax><mathjax>#"2.00 mol/L"#</mathjax></p>
<p><mathjax>#0.1cancel"L"xx(2.00"mol")/(1cancel"L")="0.200 mol HCl"#</mathjax></p>
<p>Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas, <mathjax>#"2.01588 g/mol"#</mathjax></p>
<p><mathjax>#0.200"mol Mg"xx(1"mol H"_2)/(1"mol Mg")xx(2.01588"g H"_2)/(1"mol H"_2)="0.403 g H"_2"#</mathjax></p>
<p><mathjax>#0.200"mol HCl"xx(1"mol H"_2)/(2"mol HCl")xx(2.01588"g H"_2)/(1"mol H"_2)="0.202 g H"_2"#</mathjax></p>
<p>The limiting reactant is <mathjax>#"HCl"#</mathjax>, which will produce <mathjax>#"0.202 g H"_2"#</mathjax> under the stated conditions.</p></div>
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</article> | In the equation #Mg(s) + 2HCl(aq) -> MgCL_2(aq) + H_2(g)#, what mass of hydrogen will be obtained if #100# #cm^3# of 2.00 mol #dm^-3# #HCl# are added to 4.86 g of magnesium? | null |
631 | aad42712-6ddd-11ea-a12b-ccda262736ce | https://socratic.org/questions/what-is-the-percent-composition-by-mass-of-nitrogen-in-nh-4-2co-3 | 29.16% | start physical_unit 8 10 mass_percent none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"percent composition by mass [OF] nitrogen in (NH4)2CO3"}] | [{"type":"physical unit","value":"29.16%"}] | [{"type":"chemical equation","value":"(NH4)2CO3"}] | <h1 class="questionTitle" itemprop="name">What is the percent composition by mass of nitrogen in #(NH_4)_2CO_3#?</h1> | null | 29.16% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus, <mathjax>#%N=(2xx14.01*g*mol^-1)/(96.09*g*mol^-1)~=27%#</mathjax> <mathjax>#"by mass."#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#%N="Mass of nitrogen"/"Molar mass of ammonium carbonate"xx100%#</mathjax></p></div>
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<div class="markdown"><p>And thus, <mathjax>#%N=(2xx14.01*g*mol^-1)/(96.09*g*mol^-1)~=27%#</mathjax> <mathjax>#"by mass."#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the percent composition by mass of nitrogen in #(NH_4)_2CO_3#?</h1>
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<div class="markdown"><p><mathjax>#%N="Mass of nitrogen"/"Molar mass of ammonium carbonate"xx100%#</mathjax></p></div>
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<div class="markdown"><p>And thus, <mathjax>#%N=(2xx14.01*g*mol^-1)/(96.09*g*mol^-1)~=27%#</mathjax> <mathjax>#"by mass."#</mathjax></p></div>
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</article> | What is the percent composition by mass of nitrogen in #(NH_4)_2CO_3#? | null |
632 | ace8b1ff-6ddd-11ea-843c-ccda262736ce | https://socratic.org/questions/how-do-you-balance-s8-o2-so3 | S8 + 12 O2 -> 8 SO3 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"S8 + 12 O2 -> 8 SO3"}] | [{"type":"chemical equation","value":"S8 + O2 -> SO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance S8 + O2 -->SO3? </h1> | null | S8 + 12 O2 -> 8 SO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing a chemical equation is actually an exercise where the balancing atoms is basically an adherence to the law of conservation of mass in Physics.</p>
<p>On the left hand side of the equation, there are 8 (moles of) <mathjax>#S#</mathjax> atoms. Therefore on the right hand side, there would need to be <mathjax>#8#</mathjax> (moles of) <mathjax>#SO_3#</mathjax> molecules.</p>
<p>If so, on the right hand side now, there would be a total of <mathjax>#24#</mathjax> (moles of) <mathjax>#O#</mathjax> atoms. Therefore we need to balance the O atoms on the left hand side giving us <mathjax>#12#</mathjax> (moles of) <mathjax>#O_2#</mathjax> gas.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#S_8#</mathjax> + <mathjax>#12O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8SO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing a chemical equation is actually an exercise where the balancing atoms is basically an adherence to the law of conservation of mass in Physics.</p>
<p>On the left hand side of the equation, there are 8 (moles of) <mathjax>#S#</mathjax> atoms. Therefore on the right hand side, there would need to be <mathjax>#8#</mathjax> (moles of) <mathjax>#SO_3#</mathjax> molecules.</p>
<p>If so, on the right hand side now, there would be a total of <mathjax>#24#</mathjax> (moles of) <mathjax>#O#</mathjax> atoms. Therefore we need to balance the O atoms on the left hand side giving us <mathjax>#12#</mathjax> (moles of) <mathjax>#O_2#</mathjax> gas.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance S8 + O2 -->SO3? </h1>
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Suryin =)
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Stefan V.
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<span class="dateCreated" datetime="2015-10-18T11:03:43" itemprop="dateCreated">
Oct 18, 2015
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<div class="markdown"><p><mathjax>#S_8#</mathjax> + <mathjax>#12O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#8SO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing a chemical equation is actually an exercise where the balancing atoms is basically an adherence to the law of conservation of mass in Physics.</p>
<p>On the left hand side of the equation, there are 8 (moles of) <mathjax>#S#</mathjax> atoms. Therefore on the right hand side, there would need to be <mathjax>#8#</mathjax> (moles of) <mathjax>#SO_3#</mathjax> molecules.</p>
<p>If so, on the right hand side now, there would be a total of <mathjax>#24#</mathjax> (moles of) <mathjax>#O#</mathjax> atoms. Therefore we need to balance the O atoms on the left hand side giving us <mathjax>#12#</mathjax> (moles of) <mathjax>#O_2#</mathjax> gas.</p></div>
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</article> | How do you balance S8 + O2 -->SO3? | null |
633 | aa92ab9c-6ddd-11ea-b7ab-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-titration-of-hcl-nh-3 | NH3(aq) + HCl(aq) -> NH4Cl(aq) | start chemical_equation qc_end chemical_equation 8 8 qc_end chemical_equation 10 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] titration"}] | [{"type":"chemical equation","value":"NH3(aq) + HCl(aq) -> NH4Cl(aq)"}] | [{"type":"chemical equation","value":"HCl"},{"type":"chemical equation","value":"NH3"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for titration of #HCl + NH_3#?</h1> | null | NH3(aq) + HCl(aq) -> NH4Cl(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrochloric acid</em>, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, will react with <em>ammonia</em>, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, according to the following chemical equation </p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>In this particular reaction, the chloride anions, <mathjax>#"Cl"^(-)#</mathjax>, act as <em>spectator ions</em>, which means that you can eliminate them from the balanced chemical equation to get the 8<em>net ionic equation</em>*</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote>
<p>It's worth mentioning that because you're titrating a <strong>strong acid</strong> with a <strong>weak base</strong>, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the resulting solution will be <strong>lower than</strong> <mathjax>#7#</mathjax> <em>at equivalence point</em>.</p>
<p>That is the case because this <strong><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a></strong> produces the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, which acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>Assuming that you're running ammonia into hydrochloric acid, the titration curve will look like this </p>
<blockquote>
<blockquote>
<p><img alt="http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html" src="https://useruploads.socratic.org/xAhPFUn1TieN5LC0D9cP_sawb2.gif"/> </p>
</blockquote>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrochloric acid</em>, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, will react with <em>ammonia</em>, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, according to the following chemical equation </p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>In this particular reaction, the chloride anions, <mathjax>#"Cl"^(-)#</mathjax>, act as <em>spectator ions</em>, which means that you can eliminate them from the balanced chemical equation to get the 8<em>net ionic equation</em>*</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote>
<p>It's worth mentioning that because you're titrating a <strong>strong acid</strong> with a <strong>weak base</strong>, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the resulting solution will be <strong>lower than</strong> <mathjax>#7#</mathjax> <em>at equivalence point</em>.</p>
<p>That is the case because this <strong><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a></strong> produces the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, which acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>Assuming that you're running ammonia into hydrochloric acid, the titration curve will look like this </p>
<blockquote>
<blockquote>
<p><img alt="http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html" src="https://useruploads.socratic.org/xAhPFUn1TieN5LC0D9cP_sawb2.gif"/> </p>
</blockquote>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for titration of #HCl + NH_3#?</h1>
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<div class="markdown"><p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrochloric acid</em>, <mathjax>#"HCl"#</mathjax>, a <strong>strong acid</strong>, will react with <em>ammonia</em>, <mathjax>#"NH"_3#</mathjax>, a <strong>weak base</strong>, to form aqueous ammonium chloride, <mathjax>#"NH"_4"Cl"#</mathjax>, according to the following chemical equation </p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "HCl"_ ((aq)) -> "NH"_ 4"Cl"_ ((aq))#</mathjax></p>
</blockquote>
<p>In this particular reaction, the chloride anions, <mathjax>#"Cl"^(-)#</mathjax>, act as <em>spectator ions</em>, which means that you can eliminate them from the balanced chemical equation to get the 8<em>net ionic equation</em>*</p>
<blockquote>
<p><mathjax>#"NH"_ (3(aq)) + "H"_ ((aq))^(+) -> "NH"_ (4(aq))^(+)#</mathjax></p>
</blockquote>
<p>It's worth mentioning that because you're titrating a <strong>strong acid</strong> with a <strong>weak base</strong>, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the resulting solution will be <strong>lower than</strong> <mathjax>#7#</mathjax> <em>at equivalence point</em>.</p>
<p>That is the case because this <strong><a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reaction</a></strong> produces the <em>ammonium cation</em>, <mathjax>#"NH"_4^(+)#</mathjax>, which acts as a <strong>weak acid</strong> in aqueous solution. </p>
<p>Assuming that you're running ammonia into hydrochloric acid, the titration curve will look like this </p>
<blockquote>
<blockquote>
<p><img alt="http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html" src="https://useruploads.socratic.org/xAhPFUn1TieN5LC0D9cP_sawb2.gif"/> </p>
</blockquote>
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</article> | What is the chemical equation for titration of #HCl + NH_3#? | null |
634 | abda3f70-6ddd-11ea-bd60-ccda262736ce | https://socratic.org/questions/24-2538-grams-of-an-unknown-compound-was-found-to-contain-50-15-sulfur-and-49-85 | SO2 | start chemical_formula qc_end physical_unit 4 5 0 1 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the unknown compound [IN] empirical"}] | [{"type":"chemical equation","value":"SO2"}] | [{"type":"physical unit","value":"Mass [OF] the unknown compound [=] \\pu{24.2538 grams}"},{"type":"physical unit","value":"Percent [OF] sulfur in the unknown compound [=] \\pu{50.15%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the unknown compound [=] \\pu{49.85%}"}] | <h1 class="questionTitle" itemprop="name">24.2538 grams of an unknown compound was found to contain 50.15% sulfur and 49.85% oxygen. What is the empirical formula?</h1> | null | SO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's a great example of an empirical formula problem that can be solved without doing any actual calculation. All you really need here is a <strong><a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a></strong>. </p>
<p>Notice that <em>atomic oxygen</em> has a molar mass of <em>a little under</em> <mathjax>#"16.0 g mol"^(-1)#</mathjax> and the <em>sulfur</em> has a molar mass of <em>a little over</em> <mathjax>#"32.0 g mol"^(-1)#</mathjax>. </p>
<p>This means that in order to have a <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of <strong>approximately</strong> <mathjax>#50%#</mathjax> for both <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, a given compound (that only contains sulfur and oxygen, of course) needs to have <strong>twice as many</strong> moles of oxygen than moles of sulfur. </p>
<p>Why <em>twice as many</em>? Because a mole of sulfur atoms is <strong>twice as heavy</strong> than a mole of oxygen atoms. </p>
<p>This means that the empirical formula <strong>for any</strong> compound that has a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#~~50%#</mathjax> for both oxygen and sulfur <strong>must be</strong></p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p>Now, let's do some calculations to prove that this is the case. Your sample of the unknown compound will contain </p>
<blockquote>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("50.15 g S"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("50.15% S")) = "12.2506 g S"#</mathjax></p>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("49.85 g O"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("49.85% O")) = "12.0905 g O"#</mathjax></p>
</blockquote>
<p>Use each element's <strong>molar mass</strong> to find how many moles you have present in the sample</p>
<blockquote>
<p><mathjax>#"For S: " 12.2506 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.38205 moles S"#</mathjax></p>
<p><mathjax>#"For O: " 12.0905color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.75568 moles O"#</mathjax></p>
</blockquote>
<p>To get the <em>mole ratio</em> that exists between the two elements in the sample, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For S: " (0.38205color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (0.75568color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1.9780 ~~ 2#</mathjax></p>
</blockquote>
<p>Once again, the <strong>empirical formula</strong> of the unknown compound comes out to be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"SO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's a great example of an empirical formula problem that can be solved without doing any actual calculation. All you really need here is a <strong><a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a></strong>. </p>
<p>Notice that <em>atomic oxygen</em> has a molar mass of <em>a little under</em> <mathjax>#"16.0 g mol"^(-1)#</mathjax> and the <em>sulfur</em> has a molar mass of <em>a little over</em> <mathjax>#"32.0 g mol"^(-1)#</mathjax>. </p>
<p>This means that in order to have a <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of <strong>approximately</strong> <mathjax>#50%#</mathjax> for both <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, a given compound (that only contains sulfur and oxygen, of course) needs to have <strong>twice as many</strong> moles of oxygen than moles of sulfur. </p>
<p>Why <em>twice as many</em>? Because a mole of sulfur atoms is <strong>twice as heavy</strong> than a mole of oxygen atoms. </p>
<p>This means that the empirical formula <strong>for any</strong> compound that has a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#~~50%#</mathjax> for both oxygen and sulfur <strong>must be</strong></p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p>Now, let's do some calculations to prove that this is the case. Your sample of the unknown compound will contain </p>
<blockquote>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("50.15 g S"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("50.15% S")) = "12.2506 g S"#</mathjax></p>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("49.85 g O"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("49.85% O")) = "12.0905 g O"#</mathjax></p>
</blockquote>
<p>Use each element's <strong>molar mass</strong> to find how many moles you have present in the sample</p>
<blockquote>
<p><mathjax>#"For S: " 12.2506 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.38205 moles S"#</mathjax></p>
<p><mathjax>#"For O: " 12.0905color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.75568 moles O"#</mathjax></p>
</blockquote>
<p>To get the <em>mole ratio</em> that exists between the two elements in the sample, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For S: " (0.38205color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (0.75568color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1.9780 ~~ 2#</mathjax></p>
</blockquote>
<p>Once again, the <strong>empirical formula</strong> of the unknown compound comes out to be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">24.2538 grams of an unknown compound was found to contain 50.15% sulfur and 49.85% oxygen. What is the empirical formula?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-06T00:36:26" itemprop="dateCreated">
Mar 6, 2016
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<div class="markdown"><p><mathjax>#"SO"_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's a great example of an empirical formula problem that can be solved without doing any actual calculation. All you really need here is a <strong><a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a></strong>. </p>
<p>Notice that <em>atomic oxygen</em> has a molar mass of <em>a little under</em> <mathjax>#"16.0 g mol"^(-1)#</mathjax> and the <em>sulfur</em> has a molar mass of <em>a little over</em> <mathjax>#"32.0 g mol"^(-1)#</mathjax>. </p>
<p>This means that in order to have a <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of <strong>approximately</strong> <mathjax>#50%#</mathjax> for both <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>, a given compound (that only contains sulfur and oxygen, of course) needs to have <strong>twice as many</strong> moles of oxygen than moles of sulfur. </p>
<p>Why <em>twice as many</em>? Because a mole of sulfur atoms is <strong>twice as heavy</strong> than a mole of oxygen atoms. </p>
<p>This means that the empirical formula <strong>for any</strong> compound that has a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <mathjax>#~~50%#</mathjax> for both oxygen and sulfur <strong>must be</strong></p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote>
<p><mathjax>#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p>Now, let's do some calculations to prove that this is the case. Your sample of the unknown compound will contain </p>
<blockquote>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("50.15 g S"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("50.15% S")) = "12.2506 g S"#</mathjax></p>
<p><mathjax>#24.2538color(red)(cancel(color(black)("g compound"))) * overbrace("49.85 g O"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("49.85% O")) = "12.0905 g O"#</mathjax></p>
</blockquote>
<p>Use each element's <strong>molar mass</strong> to find how many moles you have present in the sample</p>
<blockquote>
<p><mathjax>#"For S: " 12.2506 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.38205 moles S"#</mathjax></p>
<p><mathjax>#"For O: " 12.0905color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.75568 moles O"#</mathjax></p>
</blockquote>
<p>To get the <em>mole ratio</em> that exists between the two elements in the sample, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For S: " (0.38205color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For O: " (0.75568color(red)(cancel(color(black)("moles"))))/(0.38205color(red)(cancel(color(black)("moles")))) = 1.9780 ~~ 2#</mathjax></p>
</blockquote>
<p>Once again, the <strong>empirical formula</strong> of the unknown compound comes out to be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_2 implies "SO"_2color(white)(a/a)))|)#</mathjax></p>
</blockquote></div>
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</article> | 24.2538 grams of an unknown compound was found to contain 50.15% sulfur and 49.85% oxygen. What is the empirical formula? | null |
635 | ab6abb9c-6ddd-11ea-8bcc-ccda262736ce | https://socratic.org/questions/how-many-moles-of-naoh-are-present-in-15-0-ml-of-0-110-m-naoh | 1.65 × 10^(-3) moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] NaOH [IN] moles"}] | [{"type":"physical unit","value":"1.65 × 10^(-3) moles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{15.0 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.110 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #NaOH# are present in 15.0 mL of 0.110 M #NaOH#? </h1> | null | 1.65 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:</p>
<p><img alt="http://formulas.tutorvista.com/chemistry/molarity-formula.html" src="https://useruploads.socratic.org/E4haW5hRS0av5ZxOhN2X_molarity-formula.JPG"/> </p>
<p>In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 15.0 mL into liters by using the conversion factor 1000mL = 1L. When 15.0mL is divided by 1000 mL, we obtain a volume of 0.015 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of solute = (0.15 L) (0.110M) = 0.00165 moles</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>0.00165 moles of NaOH</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:</p>
<p><img alt="http://formulas.tutorvista.com/chemistry/molarity-formula.html" src="https://useruploads.socratic.org/E4haW5hRS0av5ZxOhN2X_molarity-formula.JPG"/> </p>
<p>In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 15.0 mL into liters by using the conversion factor 1000mL = 1L. When 15.0mL is divided by 1000 mL, we obtain a volume of 0.015 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of solute = (0.15 L) (0.110M) = 0.00165 moles</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of #NaOH# are present in 15.0 mL of 0.110 M #NaOH#? </h1>
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Kayla
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Jun 5, 2016
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<div class="markdown"><p>0.00165 moles of NaOH</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is represented by the following equation:</p>
<p><img alt="http://formulas.tutorvista.com/chemistry/molarity-formula.html" src="https://useruploads.socratic.org/E4haW5hRS0av5ZxOhN2X_molarity-formula.JPG"/> </p>
<p>In our case, we already have the molarity and the volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 15.0 mL into liters by using the conversion factor 1000mL = 1L. When 15.0mL is divided by 1000 mL, we obtain a volume of 0.015 L.</p>
<p>Now we can rearrange the equation to solve for the number of moles. We can do this by multiplying by L solution on both sides of the equation. The L solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:</p>
<p>Moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> = (L solution) * (Molarity)</p>
<p>Now we just plug the known values in!</p>
<p>Moles of solute = (0.15 L) (0.110M) = 0.00165 moles</p></div>
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</article> | How many moles of #NaOH# are present in 15.0 mL of 0.110 M #NaOH#? | null |
636 | ac108865-6ddd-11ea-8961-ccda262736ce | https://socratic.org/questions/how-would-you-balance-na-h2o-naoh-h2 | 2 Na + 2 H2O -> 2 NaOH + H2 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Na + 2 H2O -> 2 NaOH + H2"}] | [{"type":"chemical equation","value":"Na + H2O -> NaOH + H2"}] | <h1 class="questionTitle" itemprop="name">How would you balance: Na + H2O --> NaOH + H2?
</h1> | null | 2 Na + 2 H2O -> 2 NaOH + H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have to count the number of atoms in both sides of the reaction to be equal, to make it a little easier you can make a list. So in the equation <mathjax>#Na+H_2O->NaOH+H_2#</mathjax>, there are<br/>
<mathjax>#1-Na-1#</mathjax><br/>
<mathjax>#2-H-3#</mathjax><br/>
<mathjax>#1-O-1#</mathjax><br/>
So you multiply NaOH and H2O by 2, so you have 4 hydrogens in both sides, and then multiply Na by 2 to be equal to the NaOH you previously multiplied.</p></div>
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<div class="markdown"><p><mathjax>#2Na + 2H_2O -> 2NaOH +H_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have to count the number of atoms in both sides of the reaction to be equal, to make it a little easier you can make a list. So in the equation <mathjax>#Na+H_2O->NaOH+H_2#</mathjax>, there are<br/>
<mathjax>#1-Na-1#</mathjax><br/>
<mathjax>#2-H-3#</mathjax><br/>
<mathjax>#1-O-1#</mathjax><br/>
So you multiply NaOH and H2O by 2, so you have 4 hydrogens in both sides, and then multiply Na by 2 to be equal to the NaOH you previously multiplied.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance: Na + H2O --> NaOH + H2?
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<div class="markdown"><p><mathjax>#2Na + 2H_2O -> 2NaOH +H_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You have to count the number of atoms in both sides of the reaction to be equal, to make it a little easier you can make a list. So in the equation <mathjax>#Na+H_2O->NaOH+H_2#</mathjax>, there are<br/>
<mathjax>#1-Na-1#</mathjax><br/>
<mathjax>#2-H-3#</mathjax><br/>
<mathjax>#1-O-1#</mathjax><br/>
So you multiply NaOH and H2O by 2, so you have 4 hydrogens in both sides, and then multiply Na by 2 to be equal to the NaOH you previously multiplied.</p></div>
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<div class="markdown"><p><mathjax>#2"Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g))#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I'm assuming that you're supposed to balance this chemical equation <em>by inspection</em>. You could also balance it by using <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a>, but I don't think that you're supposed to go that route here. </p>
<p>So, your unbalanced chemical equation looks like this</p>
<blockquote>
<p><mathjax>#"Na"_ ((s)) + "H"_ 2"O"_ ((l)) -> "NaOH"_ ((aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Notice that you have <mathjax>#3#</mathjax> atoms of hydrogen on the products' side, but only <mathjax>#2#</mathjax> on the reactants' side. </p>
<p>Here is where a little experience can come in handy. You can multiply the sodium hydroxide by <mathjax>#2#</mathjax> to get a total of <mathjax>#4#</mathjax> atoms of hydrogen on the products' side. </p>
<p>This will allow you to simply double the number of water molecules to get <mathjax>#4#</mathjax> atoms of hydrogen on the reactants' side.</p>
<p>So you now have</p>
<blockquote>
<p><mathjax>#"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>Balance the atoms of sodium by multiplying sodium metal by <mathjax>#2#</mathjax></p>
<blockquote>
<p><mathjax>#2"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> color(blue)(2)"NaOH"_ ((aq)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>And now check to see if the atoms of oxygen are balanced. You have <mathjax>#2#</mathjax> atoms of hydrogen on the reactants' side and <mathjax>#2#</mathjax> on the products' side, which means that the equation is <strong>balanced</strong>.</p></div>
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</article> | How would you balance: Na + H2O --> NaOH + H2?
| null |
637 | a892e374-6ddd-11ea-b825-ccda262736ce | https://socratic.org/questions/598e9a0ab72cff426c677d8b | +2 | start physical_unit 6 8 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] the ipso carbon"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"HC(=O)O−"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of the ipso carbon in #HC(=O)O^(-)#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots <mathjax>#HC(=O)O^(-)Na^(+)#</mathjax>. The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> on the formate anion equals <mathjax>#-1#</mathjax>. The oxidation number of oxygen is usually <mathjax>#-II#</mathjax>, and it is here, and the oxidation number of hydrogen is usually <mathjax>#+I#</mathjax>, and it is here.....</p>
<p>And so <mathjax>#-1=C_"oxidation number"+1-2xx2#</mathjax></p>
<p><mathjax>#C_"oxidation number"=+II#</mathjax></p>
<p>Note that for acetate ion or acetic acid, the corresponding oxidation number is <mathjax>#+III#</mathjax> for the ipso carbon......</p>
<p>For general rules on the assignment of oxidation numbers see <a href="https://socratic.org/questions/how-do-you-find-oxidation-number-rules">here.</a> </p></div>
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<div class="markdown"><p>I make it <mathjax>#C(+II)#</mathjax>......</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots <mathjax>#HC(=O)O^(-)Na^(+)#</mathjax>. The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> on the formate anion equals <mathjax>#-1#</mathjax>. The oxidation number of oxygen is usually <mathjax>#-II#</mathjax>, and it is here, and the oxidation number of hydrogen is usually <mathjax>#+I#</mathjax>, and it is here.....</p>
<p>And so <mathjax>#-1=C_"oxidation number"+1-2xx2#</mathjax></p>
<p><mathjax>#C_"oxidation number"=+II#</mathjax></p>
<p>Note that for acetate ion or acetic acid, the corresponding oxidation number is <mathjax>#+III#</mathjax> for the ipso carbon......</p>
<p>For general rules on the assignment of oxidation numbers see <a href="https://socratic.org/questions/how-do-you-find-oxidation-number-rules">here.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of the ipso carbon in #HC(=O)O^(-)#?</h1>
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<div class="markdown"><p>I make it <mathjax>#C(+II)#</mathjax>......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots <mathjax>#HC(=O)O^(-)Na^(+)#</mathjax>. The sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> on the formate anion equals <mathjax>#-1#</mathjax>. The oxidation number of oxygen is usually <mathjax>#-II#</mathjax>, and it is here, and the oxidation number of hydrogen is usually <mathjax>#+I#</mathjax>, and it is here.....</p>
<p>And so <mathjax>#-1=C_"oxidation number"+1-2xx2#</mathjax></p>
<p><mathjax>#C_"oxidation number"=+II#</mathjax></p>
<p>Note that for acetate ion or acetic acid, the corresponding oxidation number is <mathjax>#+III#</mathjax> for the ipso carbon......</p>
<p>For general rules on the assignment of oxidation numbers see <a href="https://socratic.org/questions/how-do-you-find-oxidation-number-rules">here.</a> </p></div>
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<div class="markdown"><p>+2</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The Lewis Dot Structure analysis may be the best visual.</p>
<p>When a Lewis dot structure of a molecule is available, the oxidation states may be assigned by computing the difference between the number of valence electrons that a neutral atom of that element would have and the number of electrons that "belong" to it in the Lewis structure. </p>
<p>For purposes of computing oxidation states, electrons in a bond between atoms of different elements belong to the more electronegative atom; electrons in a bond between atoms of the same element are split equally, and electrons in alone pair belong only to the atom with the lone pair. </p>
<p><img alt="enter image source here" src="https://useruploads.socratic.org/WEOJr3mtR06YF5YueDCQ_Acetic_acid_oxidation_state_analysis%20-%20Lewis%20Structure.png"/> <br/>
From <a href="https://en.wikipedia.org/wiki/Oxidation_state" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Oxidation_state</a></p>
<p>The methyl group carbon atom has six valence electrons from its bonds to the hydrogen atoms because carbon is more electronegative than hydrogen. Also, one electron is gained from its bond with the other carbon atom because the electron pair in the C−C bond is split equally. </p>
<p>It therefore has a total of seven electrons, whereas a neutral carbon atom would have four valence electrons because carbon is in group 14 of the periodic table. The difference, 4 − 7 = −3, is the oxidation state of that carbon atom. That is, if it is assumed that all the bonds were 100% ionic (which in fact they are not), the carbon would be described as C3−<br/>
.</p></div>
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</article> | What is the oxidation number of the ipso carbon in #HC(=O)O^(-)#? | null |
638 | ad06239c-6ddd-11ea-9e45-ccda262736ce | https://socratic.org/questions/58eaee8f11ef6b7c0b766d68 | 250 mL | start physical_unit 2 2 volume ml qc_end physical_unit 9 10 8 8 percent qc_end physical_unit 9 10 12 12 percent qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] mL"}] | [{"type":"physical unit","value":"250 mL"}] | [{"type":"physical unit","value":"Percent1 [OF] acetic acid in solution [=] \\pu{2%}"},{"type":"physical unit","value":"Percent2 [OF] acetic acid in solution [=] \\pu{5%}"}] | <h1 class="questionTitle" itemprop="name">How much water should we add to prepare #2%# acetic acid from #5%# acetic acid?</h1> | null | 250 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#5%#</mathjax> acetic acid contains</p>
<p><mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>for <mathjax>#2%#</mathjax> of acetic acid we need</p>
<p><mathjax>#2#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>or <mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100/2xx5=250#</mathjax> units of water.</p>
<p>i.e. in each <mathjax>#100ml#</mathjax> of <mathjax>#5%#</mathjax> acetic acid we should add <mathjax>#150ml#</mathjax> of water i.e. one and a half times water</p>
<p>Hence, to prepare 2% acetic acid from <mathjax>#5%#</mathjax> acetic acid, we must add one and a half times water.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>To prepare 2% acetic acid from <mathjax>#5%#</mathjax> acetic acid, we must add one and a half times water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#5%#</mathjax> acetic acid contains</p>
<p><mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>for <mathjax>#2%#</mathjax> of acetic acid we need</p>
<p><mathjax>#2#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>or <mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100/2xx5=250#</mathjax> units of water.</p>
<p>i.e. in each <mathjax>#100ml#</mathjax> of <mathjax>#5%#</mathjax> acetic acid we should add <mathjax>#150ml#</mathjax> of water i.e. one and a half times water</p>
<p>Hence, to prepare 2% acetic acid from <mathjax>#5%#</mathjax> acetic acid, we must add one and a half times water.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much water should we add to prepare #2%# acetic acid from #5%# acetic acid?</h1>
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<div class="markdown"><p>To prepare 2% acetic acid from <mathjax>#5%#</mathjax> acetic acid, we must add one and a half times water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#5%#</mathjax> acetic acid contains</p>
<p><mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>for <mathjax>#2%#</mathjax> of acetic acid we need</p>
<p><mathjax>#2#</mathjax> units of acetic acid in <mathjax>#100#</mathjax> units of water</p>
<p>or <mathjax>#5#</mathjax> units of acetic acid in <mathjax>#100/2xx5=250#</mathjax> units of water.</p>
<p>i.e. in each <mathjax>#100ml#</mathjax> of <mathjax>#5%#</mathjax> acetic acid we should add <mathjax>#150ml#</mathjax> of water i.e. one and a half times water</p>
<p>Hence, to prepare 2% acetic acid from <mathjax>#5%#</mathjax> acetic acid, we must add one and a half times water.</p></div>
</div>
</div>
</div>
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</article> | How much water should we add to prepare #2%# acetic acid from #5%# acetic acid? | null |
639 | a942aa28-6ddd-11ea-962b-ccda262736ce | https://socratic.org/questions/what-is-the-net-ionic-equation-for-formation-of-an-aqueous-solution-of-nii-2-acc | CO3^2- + 2 H3O+ -> CO2(g) + 3 H2O(l) | start chemical_equation qc_end chemical_equation 13 13 qc_end chemical_equation 18 18 qc_end substance 21 21 qc_end chemical_equation 25 25 qc_end substance 27 29 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"CO3^2- + 2 H3O+ -> CO2(g) + 3 H2O(l)"}] | [{"type":"chemical equation","value":"NiI2"},{"type":"chemical equation","value":"CO2"},{"type":"substance name","value":"Water"},{"type":"chemical equation","value":"NiCO3"},{"type":"substance name","value":"Aqueous hydriodic acid"}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for formation of an aqueous solution of #NiI_2# accompanied by evolution of #CO_2# gas and water via mixing solid #NiCO_3# and aqueous hydriodic acid?</h1> | null | CO3^2- + 2 H3O+ -> CO2(g) + 3 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or</p>
<p><mathjax>#CO_3^(2-) + 2H^+ rarrCO_2(g)uarr + H_2O(l)#</mathjax></p>
<p>The iodide and nickel ions are just along for the ride.........</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#CO_3^(2-) + 2H_3O^+ rarr CO_2(g)uarr + 3H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or</p>
<p><mathjax>#CO_3^(2-) + 2H^+ rarrCO_2(g)uarr + H_2O(l)#</mathjax></p>
<p>The iodide and nickel ions are just along for the ride.........</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the net ionic equation for formation of an aqueous solution of #NiI_2# accompanied by evolution of #CO_2# gas and water via mixing solid #NiCO_3# and aqueous hydriodic acid?</h1>
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<div class="markdown"><p><mathjax>#CO_3^(2-) + 2H_3O^+ rarr CO_2(g)uarr + 3H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Or</p>
<p><mathjax>#CO_3^(2-) + 2H^+ rarrCO_2(g)uarr + H_2O(l)#</mathjax></p>
<p>The iodide and nickel ions are just along for the ride.........</p></div>
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</article> | What is the net ionic equation for formation of an aqueous solution of #NiI_2# accompanied by evolution of #CO_2# gas and water via mixing solid #NiCO_3# and aqueous hydriodic acid? | null |
640 | ac0ba229-6ddd-11ea-8fcf-ccda262736ce | https://socratic.org/questions/acetylene-is-used-in-blow-torches-and-burns-according-to-the-following-equation- | -2511.04 kJ | start physical_unit 33 33 heat_energy kj qc_end chemical_equation 13 23 qc_end physical_unit 23 23 36 37 deltahf^0 qc_end physical_unit 20 20 40 41 deltahf^0 qc_end physical_unit 14 14 44 45 deltahf^0 qc_end end | [{"type":"physical unit","value":"Heat [OF] reaction [IN] kJ"}] | [{"type":"physical unit","value":"-2511.04 kJ"}] | [{"type":"chemical equation","value":"2 C2H2(g) + 5 O2(g) -> 4 CO2(g) + 2 H2O(g)"},{"type":"physical unit","value":"DeltaHf^0 [OF] H2O(g) [=] \\pu{-241.82 kJ/mol}"},{"type":"physical unit","value":"DeltaHf^0 [OF] CO2(g) [=] \\pu{-393.5 kJ/mol}"},{"type":"physical unit","value":"DeltaHf^0 [OF] C2H2(g) [=] \\pu{226.77 kJ/mol}"}] | <h1 class="questionTitle" itemprop="name">Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Acetylene is used in blow torches, and burns according to the following<br/>
equation:<br/>
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)<br/>
Use the following information to calculate the heat of reaction:</p>
<p>Hfo<br/>
(H2O(g))= -241.82 kJ/mol</p>
<p>Hfo<br/>
(CO2(g))= -393.5 kJ/mol</p>
<p>Hfo<br/>
(C2H2(g))=226.77 kJ/mol</p></div>
</h2>
</div>
</div> | -2511.04 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> states that the overall <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of a reaction is independent of the route taken.</p>
<p>Thermodynamics is concerned with initial and final states and the law is a consequence of the conservation of energy.</p>
<p>You can solve this problem by constructing a Hess Cycle. </p>
<p>Write down the reaction you are interested in. Below this write down the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from which the reactants and products are made. </p>
<p>Then complete the cycle as shown:</p>
<p><img alt="MFCocs" src="https://useruploads.socratic.org/io7MArPjQgO6Q2DMA6Rx_hess3.jpg"/> </p>
<p>Notice I have multiplied the <mathjax>#Delta"H"_"f"#</mathjax> values by the relevant stoichiometric numbers.</p>
<p>In energy terms the <mathjax>#color(blue)"BLUE"#</mathjax> route must equal the <mathjax>#color(red)"RED"#</mathjax> route since the arrows start and finish in the same place.</p>
<p>So we can write:</p>
<p><mathjax>#(2xx226.77)+Delta"H"=(4xx-393.5)+(2xx-241.82)#</mathjax></p>
<p><mathjax>#:.453.4+Delta"H"=-1574-483.64#</mathjax></p>
<p><mathjax>#Delta"H"=-2511.04" ""kJ"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Delta"H"_(rxn)=-2511.04" ""kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> states that the overall <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of a reaction is independent of the route taken.</p>
<p>Thermodynamics is concerned with initial and final states and the law is a consequence of the conservation of energy.</p>
<p>You can solve this problem by constructing a Hess Cycle. </p>
<p>Write down the reaction you are interested in. Below this write down the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from which the reactants and products are made. </p>
<p>Then complete the cycle as shown:</p>
<p><img alt="MFCocs" src="https://useruploads.socratic.org/io7MArPjQgO6Q2DMA6Rx_hess3.jpg"/> </p>
<p>Notice I have multiplied the <mathjax>#Delta"H"_"f"#</mathjax> values by the relevant stoichiometric numbers.</p>
<p>In energy terms the <mathjax>#color(blue)"BLUE"#</mathjax> route must equal the <mathjax>#color(red)"RED"#</mathjax> route since the arrows start and finish in the same place.</p>
<p>So we can write:</p>
<p><mathjax>#(2xx226.77)+Delta"H"=(4xx-393.5)+(2xx-241.82)#</mathjax></p>
<p><mathjax>#:.453.4+Delta"H"=-1574-483.64#</mathjax></p>
<p><mathjax>#Delta"H"=-2511.04" ""kJ"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Acetylene is used in blow torches, and burns according to the following<br/>
equation:<br/>
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)<br/>
Use the following information to calculate the heat of reaction:</p>
<p>Hfo<br/>
(H2O(g))= -241.82 kJ/mol</p>
<p>Hfo<br/>
(CO2(g))= -393.5 kJ/mol</p>
<p>Hfo<br/>
(C2H2(g))=226.77 kJ/mol</p></div>
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Michael
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May 5, 2016
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<div class="markdown"><p><mathjax>#Delta"H"_(rxn)=-2511.04" ""kJ"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> states that the overall <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of a reaction is independent of the route taken.</p>
<p>Thermodynamics is concerned with initial and final states and the law is a consequence of the conservation of energy.</p>
<p>You can solve this problem by constructing a Hess Cycle. </p>
<p>Write down the reaction you are interested in. Below this write down the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> from which the reactants and products are made. </p>
<p>Then complete the cycle as shown:</p>
<p><img alt="MFCocs" src="https://useruploads.socratic.org/io7MArPjQgO6Q2DMA6Rx_hess3.jpg"/> </p>
<p>Notice I have multiplied the <mathjax>#Delta"H"_"f"#</mathjax> values by the relevant stoichiometric numbers.</p>
<p>In energy terms the <mathjax>#color(blue)"BLUE"#</mathjax> route must equal the <mathjax>#color(red)"RED"#</mathjax> route since the arrows start and finish in the same place.</p>
<p>So we can write:</p>
<p><mathjax>#(2xx226.77)+Delta"H"=(4xx-393.5)+(2xx-241.82)#</mathjax></p>
<p><mathjax>#:.453.4+Delta"H"=-1574-483.64#</mathjax></p>
<p><mathjax>#Delta"H"=-2511.04" ""kJ"#</mathjax></p></div>
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</article> | Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:? |
Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:
Hfo
(H2O(g))= -241.82 kJ/mol
Hfo
(CO2(g))= -393.5 kJ/mol
Hfo
(C2H2(g))=226.77 kJ/mol
|
641 | ab549fae-6ddd-11ea-ad0e-ccda262736ce | https://socratic.org/questions/if-you-started-with-7-461-g-of-magnesium-metal-how-much-oxygen-would-be-consumed | 4.91 g | start physical_unit 11 11 mass g qc_end physical_unit 7 8 4 5 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] g"}] | [{"type":"physical unit","value":"4.91 g"}] | [{"type":"physical unit","value":"Mass [OF] magnesium metal [=] \\pu{7.461 g}"},{"type":"other","value":"combustion"}] | <h1 class="questionTitle" itemprop="name">If you started with 7.461 g of magnesium metal, how much oxygen would be consumed during combustion?</h1> | null | 4.91 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s)+1/2O_2(g) rarr MgO(g)#</mathjax></p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.461*g)/(24.305*g*mol^-1)=0.3070*mol#</mathjax></p>
<p>And thus <mathjax>#0.1535*mol#</mathjax> dioxygen gas would be consumed. What is the mass of this quantity of gas? Under standard laboratory conditions, what would be its volume?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>A shade under <mathjax>#5*g#</mathjax> of <mathjax>#O_2#</mathjax> gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s)+1/2O_2(g) rarr MgO(g)#</mathjax></p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.461*g)/(24.305*g*mol^-1)=0.3070*mol#</mathjax></p>
<p>And thus <mathjax>#0.1535*mol#</mathjax> dioxygen gas would be consumed. What is the mass of this quantity of gas? Under standard laboratory conditions, what would be its volume?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If you started with 7.461 g of magnesium metal, how much oxygen would be consumed during combustion?</h1>
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anor277
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<span class="dateCreated" datetime="2016-09-08T04:09:39" itemprop="dateCreated">
Sep 8, 2016
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<div class="markdown"><p>A shade under <mathjax>#5*g#</mathjax> of <mathjax>#O_2#</mathjax> gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Mg(s)+1/2O_2(g) rarr MgO(g)#</mathjax></p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.461*g)/(24.305*g*mol^-1)=0.3070*mol#</mathjax></p>
<p>And thus <mathjax>#0.1535*mol#</mathjax> dioxygen gas would be consumed. What is the mass of this quantity of gas? Under standard laboratory conditions, what would be its volume?</p></div>
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<div class="markdown"><p>The amount of oxygen gas consumed is 4.911 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with a balanced equation.</p>
<p><mathjax>#"2Mg(s)" + "O"_2("g)"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2MgO(s)"#</mathjax></p>
<p>The coefficient in front of each formula is its number of moles. No coefficient is understood to be 1. <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio between magnesium and oxygen gas is <mathjax>#"2 mol Mg"/"1 mol O"_2"#</mathjax> or <mathjax>#"1 mol O"_2/"2 mol Mg"#</mathjax>.</p>
<p>You will need the molar masses of magnesium and oxygen gas from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>.</p>
<p>Magnesium: <mathjax>#"24.305 g/mol"#</mathjax><br/>
Oxygen gas (dioxygen): <mathjax>#(2xx15.999 "g/mol")="31.998 g/mol"#</mathjax></p>
<p>First determine the moles of Mg in 7.461 g by dividing the 7.461 g by its molar mass.</p>
<p><mathjax>#(7.461 cancel"g Mg")/(24.30cancel("g")/"mol Mg")="0.3069736 mol Mg"#</mathjax></p>
<p>Determine the moles <mathjax>#"O"_2"#</mathjax> by multiplying the moles Mg by the mole ratio with oxygen on top. </p>
<p><mathjax>#0.3069736 cancel"mol Mg"xx"1 mol O"_2/cancel"2mol Mg"="0.1534868 mol O"_2"#</mathjax></p>
<p>Determine the mass of <mathjax>#"O"_2"#</mathjax> by multiplying moles <mathjax>#"O"_2"#</mathjax> by its molar mass. </p>
<p><mathjax>#0.1534868 cancel"mol O"_2xx"31.998 g O"_2/cancel"1 mol O"_2"="4.911 "g O"_2"#</mathjax> (rounded to four <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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</article> | If you started with 7.461 g of magnesium metal, how much oxygen would be consumed during combustion? | null |
642 | a9eacddc-6ddd-11ea-9735-ccda262736ce | https://socratic.org/questions/when-aqueous-solutions-of-calcium-nitrate-and-ammonium-carbonate-are-combined-so | Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s) | start chemical_equation qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)"}] | [{"type":"substance name","value":"Calcium nitrate aqueous solution"},{"type":"substance name","value":"Ammonium carbonate aqueous solution"},{"type":"substance name","value":"Calcium carbonate solid"},{"type":"substance name","value":"Ammonium nitrate solution"}] | <h1 class="questionTitle" itemprop="name">When aqueous solutions of calcium nitrate and ammonium carbonate are combined, solid calcium carbonate and a solution of ammonium nitrate are formed?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>. The net ionic equation for this reaction is:</p></div>
</h2>
</div>
</div> | Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The fact that this <strong>double replacement reaction</strong> produces <em>solid</em> calcium carbonate tells you which two ions will combine to form the solid and which two ions will be <strong>spectator ions</strong>. </p>
<p>As its name suggests, calcium carbonate contains calcium cations, <mathjax>#"Ca"^(2+)#</mathjax>, and carbonate anions, <mathjax>#"CO"_3^(2-)#</mathjax>.</p>
<p>The calcium cations come from the <strong>soluble</strong> calcium nitrate, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, which implies that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, will act as spectator ions.</p>
<p>Similarly, the carbonate anions come from the <strong>soluble</strong> ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, which implies that the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, will act as spectator ions. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#"Ca"("NO"_3)_2 => {("Ca"^(2+) -> "part of the precipitate"), (2"NO"_3^(-) -> "spectator ions") :}#</mathjax></p>
<p><mathjax>#("NH"_4)_2"CO"_3 => {( 2"NH"_4^(+) -> "spectator ions"), ("CO"_3^(2-) -> "part of the precipitate") :}#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which does not include the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CaCO"_ (3(s)) darr#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong>, which includes the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-) -> "CacO"_ (3(s)) darr + 2"NH"_ 4"CO"_ (3(aq))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CaCO"_ (3(s)) darr#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The fact that this <strong>double replacement reaction</strong> produces <em>solid</em> calcium carbonate tells you which two ions will combine to form the solid and which two ions will be <strong>spectator ions</strong>. </p>
<p>As its name suggests, calcium carbonate contains calcium cations, <mathjax>#"Ca"^(2+)#</mathjax>, and carbonate anions, <mathjax>#"CO"_3^(2-)#</mathjax>.</p>
<p>The calcium cations come from the <strong>soluble</strong> calcium nitrate, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, which implies that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, will act as spectator ions.</p>
<p>Similarly, the carbonate anions come from the <strong>soluble</strong> ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, which implies that the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, will act as spectator ions. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#"Ca"("NO"_3)_2 => {("Ca"^(2+) -> "part of the precipitate"), (2"NO"_3^(-) -> "spectator ions") :}#</mathjax></p>
<p><mathjax>#("NH"_4)_2"CO"_3 => {( 2"NH"_4^(+) -> "spectator ions"), ("CO"_3^(2-) -> "part of the precipitate") :}#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which does not include the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CaCO"_ (3(s)) darr#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong>, which includes the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-) -> "CacO"_ (3(s)) darr + 2"NH"_ 4"CO"_ (3(aq))#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">When aqueous solutions of calcium nitrate and ammonium carbonate are combined, solid calcium carbonate and a solution of ammonium nitrate are formed?</h1>
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<div class="markdown"><p>. The net ionic equation for this reaction is:</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2017-10-22T22:55:54" itemprop="dateCreated">
Oct 22, 2017
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<div class="markdown"><p><mathjax>#"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CaCO"_ (3(s)) darr#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The fact that this <strong>double replacement reaction</strong> produces <em>solid</em> calcium carbonate tells you which two ions will combine to form the solid and which two ions will be <strong>spectator ions</strong>. </p>
<p>As its name suggests, calcium carbonate contains calcium cations, <mathjax>#"Ca"^(2+)#</mathjax>, and carbonate anions, <mathjax>#"CO"_3^(2-)#</mathjax>.</p>
<p>The calcium cations come from the <strong>soluble</strong> calcium nitrate, <mathjax>#"Ca"("NO"_3)_2#</mathjax>, which implies that the nitrate anions, <mathjax>#"NO"_3^(-)#</mathjax>, will act as spectator ions.</p>
<p>Similarly, the carbonate anions come from the <strong>soluble</strong> ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, which implies that the ammonium cations, <mathjax>#"NH"_4^(+)#</mathjax>, will act as spectator ions. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#"Ca"("NO"_3)_2 => {("Ca"^(2+) -> "part of the precipitate"), (2"NO"_3^(-) -> "spectator ions") :}#</mathjax></p>
<p><mathjax>#("NH"_4)_2"CO"_3 => {( 2"NH"_4^(+) -> "spectator ions"), ("CO"_3^(2-) -> "part of the precipitate") :}#</mathjax></p>
</blockquote>
<p>The <strong>net ionic equation</strong>, which does not include the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "CaCO"_ (3(s)) darr#</mathjax></p>
</blockquote>
<p>The <strong>complete ionic equation</strong>, which includes the spectator ions, will look like this</p>
<blockquote>
<p><mathjax>#"Ca"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"NH"_ (4(aq))^(+) + "CO"_ (3(aq))^(2-) -> "CacO"_ (3(s)) darr + 2"NH"_ 4"CO"_ (3(aq))#</mathjax></p>
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</article> | When aqueous solutions of calcium nitrate and ammonium carbonate are combined, solid calcium carbonate and a solution of ammonium nitrate are formed? |
. The net ionic equation for this reaction is:
|
643 | abd2e68a-6ddd-11ea-8943-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-7-83-10-23-atoms-cl | 40.09 g | start physical_unit 9 9 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] Cl [IN] g"}] | [{"type":"physical unit","value":"40.09 g"}] | [{"type":"physical unit","value":"Number [OF] Cl atoms [=] \\pu{7.83 × 10^23}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #7.83 * 10^23# atoms Cl?</h1> | null | 40.09 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I have Avogadro's number of chlorine atoms, I necessarily have 35.45 g chlorine.</p>
<p>Here we have a number that is in excess of Avogadro's number, <mathjax>#6.022xx10^(23)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Since I know that if I have <mathjax>#6.022xx10^(23)#</mathjax> atoms of chlorine I have a mass of <mathjax>#35.45#</mathjax> <mathjax>#g#</mathjax> of chlorine, then all I have to do is give the product, <mathjax>#35.45*g#</mathjax> <mathjax>#xx(7.83xx10^(23))/(6.022xx10^(23))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<div class="markdown"><p>Slightly over 35.5 grams.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I have Avogadro's number of chlorine atoms, I necessarily have 35.45 g chlorine.</p>
<p>Here we have a number that is in excess of Avogadro's number, <mathjax>#6.022xx10^(23)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Since I know that if I have <mathjax>#6.022xx10^(23)#</mathjax> atoms of chlorine I have a mass of <mathjax>#35.45#</mathjax> <mathjax>#g#</mathjax> of chlorine, then all I have to do is give the product, <mathjax>#35.45*g#</mathjax> <mathjax>#xx(7.83xx10^(23))/(6.022xx10^(23))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of #7.83 * 10^23# atoms Cl?</h1>
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anor277
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<div class="markdown"><p>Slightly over 35.5 grams.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If I have Avogadro's number of chlorine atoms, I necessarily have 35.45 g chlorine.</p>
<p>Here we have a number that is in excess of Avogadro's number, <mathjax>#6.022xx10^(23)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Since I know that if I have <mathjax>#6.022xx10^(23)#</mathjax> atoms of chlorine I have a mass of <mathjax>#35.45#</mathjax> <mathjax>#g#</mathjax> of chlorine, then all I have to do is give the product, <mathjax>#35.45*g#</mathjax> <mathjax>#xx(7.83xx10^(23))/(6.022xx10^(23))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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</article> | What is the mass of #7.83 * 10^23# atoms Cl? | null |
644 | ab72f1de-6ddd-11ea-8e08-ccda262736ce | https://socratic.org/questions/a-mixture-of-hydrazine-n2h4-hydrogen-peroxide-h2o2-is-used-as-a-fuel-for-rocket- | 7.26 grams | start physical_unit 3 3 mass g qc_end chemical_equation 33 41 qc_end physical_unit 5 6 25 26 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] hydrazine [IN] grams"}] | [{"type":"physical unit","value":"7.26 grams"}] | [{"type":"chemical equation","value":"N2H4(l) + 2 H2O2(l) -> N2(g) + 4 H2O(g)"},{"type":"physical unit","value":"Mole [OF] hydrogen peroxide [=] \\pu{0.453 moles}"}] | <h1 class="questionTitle" itemprop="name">A mixture of hydrazine and hydrogen peroxide is used as a fuel for rocket engines. How many grams of hydrazine are needed to react with #0.453# moles of hydrogen peroxide?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The reaction is </p>
<p><mathjax>#"N"_2"H"_4(l)+2"H"_2"O"_2(l) -> "N"_2(g)+4"H"_2"O"(g)#</mathjax></p></div>
</h2>
</div>
</div> | 7.26 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#"N"_ 2"H"_ (4(l)) + 2"H"_ 2"O"_ (2(l)) -> "N"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of hydrazine that takes part in the reaction consumes <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen peroxide. In other words, the two reactants take part in the reaction in a <mathjax>#1:2#</mathjax> <strong>mole ratio</strong>. </p>
<p>You already know that the reaction consumed <mathjax>#0.453#</mathjax> <strong>moles</strong> of hydrogen peroxide, so use this mole ratio to figure out how many <em>moles</em> of hydrazine reacted</p>
<blockquote>
<p><mathjax>#0.453 color(red)(cancel(color(black)("moles H"_2"O"_2))) * ("1 mole N"_2"H"_4)/(2color(red)(cancel(color(black)("moles H"_2"O"_2)))) = "0.2265 moles N"_2"H"_4#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrazine</p>
<blockquote>
<p><mathjax>#0.2265 color(red)(cancel(color(black)("moles N"_2"H"_4))) * "32.045 g"/(1color(red)(cancel(color(black)("mole N"_2"H"_4)))) = color(darkgreen)(ul(color(black)("7.26 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of moles of hydrogen peroxide. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"7.26 g N"_2"H"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#"N"_ 2"H"_ (4(l)) + 2"H"_ 2"O"_ (2(l)) -> "N"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of hydrazine that takes part in the reaction consumes <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen peroxide. In other words, the two reactants take part in the reaction in a <mathjax>#1:2#</mathjax> <strong>mole ratio</strong>. </p>
<p>You already know that the reaction consumed <mathjax>#0.453#</mathjax> <strong>moles</strong> of hydrogen peroxide, so use this mole ratio to figure out how many <em>moles</em> of hydrazine reacted</p>
<blockquote>
<p><mathjax>#0.453 color(red)(cancel(color(black)("moles H"_2"O"_2))) * ("1 mole N"_2"H"_4)/(2color(red)(cancel(color(black)("moles H"_2"O"_2)))) = "0.2265 moles N"_2"H"_4#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrazine</p>
<blockquote>
<p><mathjax>#0.2265 color(red)(cancel(color(black)("moles N"_2"H"_4))) * "32.045 g"/(1color(red)(cancel(color(black)("mole N"_2"H"_4)))) = color(darkgreen)(ul(color(black)("7.26 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of moles of hydrogen peroxide. </p></div>
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<h1 class="questionTitle" itemprop="name">A mixture of hydrazine and hydrogen peroxide is used as a fuel for rocket engines. How many grams of hydrazine are needed to react with #0.453# moles of hydrogen peroxide?</h1>
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<div class="markdown"><p>The reaction is </p>
<p><mathjax>#"N"_2"H"_4(l)+2"H"_2"O"_2(l) -> "N"_2(g)+4"H"_2"O"(g)#</mathjax></p></div>
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Stefan V.
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Jul 29, 2017
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<div class="markdown"><p><mathjax>#"7.26 g N"_2"H"_4#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, you know by looking at the balanced chemical equation </p>
<blockquote>
<p><mathjax>#"N"_ 2"H"_ (4(l)) + 2"H"_ 2"O"_ (2(l)) -> "N"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>that <strong>every mole</strong> of hydrazine that takes part in the reaction consumes <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen peroxide. In other words, the two reactants take part in the reaction in a <mathjax>#1:2#</mathjax> <strong>mole ratio</strong>. </p>
<p>You already know that the reaction consumed <mathjax>#0.453#</mathjax> <strong>moles</strong> of hydrogen peroxide, so use this mole ratio to figure out how many <em>moles</em> of hydrazine reacted</p>
<blockquote>
<p><mathjax>#0.453 color(red)(cancel(color(black)("moles H"_2"O"_2))) * ("1 mole N"_2"H"_4)/(2color(red)(cancel(color(black)("moles H"_2"O"_2)))) = "0.2265 moles N"_2"H"_4#</mathjax></p>
</blockquote>
<p>To convert this to <em>grams</em>, use the <strong>molar mass</strong> of hydrazine</p>
<blockquote>
<p><mathjax>#0.2265 color(red)(cancel(color(black)("moles N"_2"H"_4))) * "32.045 g"/(1color(red)(cancel(color(black)("mole N"_2"H"_4)))) = color(darkgreen)(ul(color(black)("7.26 g")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the number of moles of hydrogen peroxide. </p></div>
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</article> | A mixture of hydrazine and hydrogen peroxide is used as a fuel for rocket engines. How many grams of hydrazine are needed to react with #0.453# moles of hydrogen peroxide? |
The reaction is
#"N"_2"H"_4(l)+2"H"_2"O"_2(l) -> "N"_2(g)+4"H"_2"O"(g)#
|
645 | ab11e3e2-6ddd-11ea-8c24-ccda262736ce | https://socratic.org/questions/1-2-10-5-j-of-thermal-energy-are-added-to-a-sample-of-water-and-its-temperature- | 2.1 kg | start physical_unit 29 30 mass kg qc_end physical_unit 29 30 0 3 heat_energy qc_end physical_unit 29 30 19 20 temperature qc_end physical_unit 29 30 22 23 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] the water [IN] kg"}] | [{"type":"physical unit","value":"2.1 kg"}] | [{"type":"physical unit","value":"Added thermal energy [OF] the water [=] \\pu{1.2 × 10^5 J}"},{"type":"physical unit","value":"Temperature1 [OF] the water [=] \\pu{229 K}"},{"type":"physical unit","value":"Temperature2 [OF] the water [=] \\pu{243 K}"}] | <h1 class="questionTitle" itemprop="name">#1.2 xx 10^5# #J# of thermal energy are added to a sample of water and its temperature changes from #229# #K# to #243# #K#. What is the mass of the water?</h1> | null | 2.1 kg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the equation <mathjax>#Delta H = mCDeltaT#</mathjax> where <mathjax>#m#</mathjax> is the mass <mathjax>#(kg)#</mathjax>, <mathjax>#T#</mathjax> is the temperature <mathjax>#(K)#</mathjax>, <mathjax>#H#</mathjax> is the thermal energy <mathjax>#(J)#</mathjax> and <mathjax>#C#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#(Jkg^-1K^-1)#</mathjax>. <mathjax>#Delta#</mathjax> just means 'the change in'.</p>
<p>We know that for water, <mathjax>#C=4181#</mathjax> <mathjax>#Jkg^-1K^-1#</mathjax>. (Sometimes quoted as <mathjax>#4.2#</mathjax> <mathjax>#Jg^-1K^-1#</mathjax>, but it's better to work in SI units.)</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Delta H = mCDeltaT#</mathjax></p>
<p>Rearranging:</p>
<p><mathjax>#m = (Delta H)/(CDeltaT)=(1.2xx10^5)/(4181xx(243-229))=2.1#</mathjax> <mathjax>#kg#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the equation <mathjax>#Delta H = mCDeltaT#</mathjax> where <mathjax>#m#</mathjax> is the mass <mathjax>#(kg)#</mathjax>, <mathjax>#T#</mathjax> is the temperature <mathjax>#(K)#</mathjax>, <mathjax>#H#</mathjax> is the thermal energy <mathjax>#(J)#</mathjax> and <mathjax>#C#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#(Jkg^-1K^-1)#</mathjax>. <mathjax>#Delta#</mathjax> just means 'the change in'.</p>
<p>We know that for water, <mathjax>#C=4181#</mathjax> <mathjax>#Jkg^-1K^-1#</mathjax>. (Sometimes quoted as <mathjax>#4.2#</mathjax> <mathjax>#Jg^-1K^-1#</mathjax>, but it's better to work in SI units.)</p></div>
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<h1 class="questionTitle" itemprop="name">#1.2 xx 10^5# #J# of thermal energy are added to a sample of water and its temperature changes from #229# #K# to #243# #K#. What is the mass of the water?</h1>
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David G.
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<div class="markdown"><p><mathjax>#Delta H = mCDeltaT#</mathjax></p>
<p>Rearranging:</p>
<p><mathjax>#m = (Delta H)/(CDeltaT)=(1.2xx10^5)/(4181xx(243-229))=2.1#</mathjax> <mathjax>#kg#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the equation <mathjax>#Delta H = mCDeltaT#</mathjax> where <mathjax>#m#</mathjax> is the mass <mathjax>#(kg)#</mathjax>, <mathjax>#T#</mathjax> is the temperature <mathjax>#(K)#</mathjax>, <mathjax>#H#</mathjax> is the thermal energy <mathjax>#(J)#</mathjax> and <mathjax>#C#</mathjax> is the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> <mathjax>#(Jkg^-1K^-1)#</mathjax>. <mathjax>#Delta#</mathjax> just means 'the change in'.</p>
<p>We know that for water, <mathjax>#C=4181#</mathjax> <mathjax>#Jkg^-1K^-1#</mathjax>. (Sometimes quoted as <mathjax>#4.2#</mathjax> <mathjax>#Jg^-1K^-1#</mathjax>, but it's better to work in SI units.)</p></div>
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</article> | #1.2 xx 10^5# #J# of thermal energy are added to a sample of water and its temperature changes from #229# #K# to #243# #K#. What is the mass of the water? | null |
646 | acb874dc-6ddd-11ea-9721-ccda262736ce | https://socratic.org/questions/582a5578b72cff4084e1db32 | 784.33 J/(kg * ℃) | start physical_unit 10 10 specific_heat_capacity j/(°c_·_g) qc_end physical_unit 10 10 0 1 heat_energy qc_end physical_unit 10 10 15 16 mass qc_end physical_unit 10 10 18 19 temperature qc_end physical_unit 10 10 21 22 temperature qc_end end | [{"type":"physical unit","value":"Specific heat capacity [OF] the substance [IN] J/(kg * ℃)"}] | [{"type":"physical unit","value":"784.33 J/(kg * ℃)"}] | [{"type":"physical unit","value":"Required heat energy [OF] the substance [=] \\pu{2659 J}"},{"type":"physical unit","value":"Mass [OF] the substance [=] \\pu{24.25 g}"},{"type":"physical unit","value":"Temperature1 [OF] the substance [=] \\pu{23.7 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the substance [=] \\pu{163.5 ℃}"}] | <h1 class="questionTitle" itemprop="name">#"2659 J"# of heat energy are required to heat a substance with a mass of #"24.25 g"# from #"23.7"^@"C"# to #"163.5"^@"C"#. What is its specific heat capacity?</h1> | null | 784.33 J/(kg * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, the equation we are using is <mathjax>#Q = mcDeltaT#</mathjax>.</p>
<p>=> Where <mathjax>#Q#</mathjax> is the amount of thermal energy in joules, <mathjax>#J#</mathjax>. <br/>
=> Where <mathjax>#m#</mathjax> is the mass in kilograms, <mathjax>#kg#</mathjax>.<br/>
=> Where <mathjax>#c#</mathjax> is the specific heat capacity. <br/>
=> Where <mathjax>#DeltaT#</mathjax> is the change in temperature in degrees Celsius. </p>
<p><mathjax>#24.25 g = 0.02425 kg#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#Q = mcDeltaT#</mathjax></p>
<p><mathjax>#Q = mc (T_2-T_1)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"(163.5 - 23.7) °C"#</mathjax></p>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"139.8 °C"#</mathjax></p>
</blockquote>
<p><mathjax>#"19.02 J/°C"= "0.02425 kg"xxc#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c = "784.3 J/(°C·kg)"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The specific heat capacity is <mathjax>#784.3 J/(°C·kg)#</mathjax>.</p>
<p>Hope this helps :)</p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity is <mathjax>#784.3 J/(C·kg)#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, the equation we are using is <mathjax>#Q = mcDeltaT#</mathjax>.</p>
<p>=> Where <mathjax>#Q#</mathjax> is the amount of thermal energy in joules, <mathjax>#J#</mathjax>. <br/>
=> Where <mathjax>#m#</mathjax> is the mass in kilograms, <mathjax>#kg#</mathjax>.<br/>
=> Where <mathjax>#c#</mathjax> is the specific heat capacity. <br/>
=> Where <mathjax>#DeltaT#</mathjax> is the change in temperature in degrees Celsius. </p>
<p><mathjax>#24.25 g = 0.02425 kg#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#Q = mcDeltaT#</mathjax></p>
<p><mathjax>#Q = mc (T_2-T_1)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"(163.5 - 23.7) °C"#</mathjax></p>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"139.8 °C"#</mathjax></p>
</blockquote>
<p><mathjax>#"19.02 J/°C"= "0.02425 kg"xxc#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c = "784.3 J/(°C·kg)"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The specific heat capacity is <mathjax>#784.3 J/(°C·kg)#</mathjax>.</p>
<p>Hope this helps :)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">#"2659 J"# of heat energy are required to heat a substance with a mass of #"24.25 g"# from #"23.7"^@"C"# to #"163.5"^@"C"#. What is its specific heat capacity?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity is <mathjax>#784.3 J/(C·kg)#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So, the equation we are using is <mathjax>#Q = mcDeltaT#</mathjax>.</p>
<p>=> Where <mathjax>#Q#</mathjax> is the amount of thermal energy in joules, <mathjax>#J#</mathjax>. <br/>
=> Where <mathjax>#m#</mathjax> is the mass in kilograms, <mathjax>#kg#</mathjax>.<br/>
=> Where <mathjax>#c#</mathjax> is the specific heat capacity. <br/>
=> Where <mathjax>#DeltaT#</mathjax> is the change in temperature in degrees Celsius. </p>
<p><mathjax>#24.25 g = 0.02425 kg#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#Q = mcDeltaT#</mathjax></p>
<p><mathjax>#Q = mc (T_2-T_1)#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"(163.5 - 23.7) °C"#</mathjax></p>
<p><mathjax>#"2659 J" = "0.02425 kg"xxcxx"139.8 °C"#</mathjax></p>
</blockquote>
<p><mathjax>#"19.02 J/°C"= "0.02425 kg"xxc#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#c = "784.3 J/(°C·kg)"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The specific heat capacity is <mathjax>#784.3 J/(°C·kg)#</mathjax>.</p>
<p>Hope this helps :)</p></div>
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Nov 28, 2016
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> is <mathjax>#0.7843"J"/("g"*^@"C")#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Apply the equation <mathjax>#Q=mcDeltaT#</mathjax>, where <mathjax>#Q#</mathjax> is the energy in Joules, <mathjax>#m#</mathjax> is the mass in grams, <mathjax>#c#</mathjax> is the specific heat, and <mathjax>#DeltaT#</mathjax> is change in temperature. <mathjax>#DeltaT=T_"final"-T_"initial"#</mathjax></p>
<p><strong>Known</strong><br/>
<mathjax>#Q="2659 J"#</mathjax><br/>
<mathjax>#m="24.25 g"#</mathjax><br/>
<mathjax>#T_"initial"="23.7"^@"C"#</mathjax><br/>
<mathjax>#T_"final"="163.5"^@"C"#</mathjax><br/>
<mathjax>#DeltaT="163.5"^@"C"-"23.7"^@"C"="139.8"^@"C"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#c#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation <mathjax>#Q=mcDeltaT#</mathjax> to isolate <mathjax>#c#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#c=(Q)/(m*DeltaT)#</mathjax></p>
<p><mathjax>#c=(2659"J")/(24.25"g"*139.8^@"C")#</mathjax></p>
<p><mathjax>#c=0.7843"J"/("g"*^@"C")#</mathjax></p></div>
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</article> | #"2659 J"# of heat energy are required to heat a substance with a mass of #"24.25 g"# from #"23.7"^@"C"# to #"163.5"^@"C"#. What is its specific heat capacity? | null |
647 | aa5d64dd-6ddd-11ea-8a33-ccda262736ce | https://socratic.org/questions/what-is-the-correct-formula-for-barium-chlorate | Ba(ClO3)2 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] barium chlorate [IN] default"}] | [{"type":"chemical equation","value":"Ba(ClO3)2"}] | [{"type":"substance name","value":"Barium chlorate"}] | <h1 class="questionTitle" itemprop="name">What is the correct formula for barium chlorate?</h1> | null | Ba(ClO3)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The barium cation has the formula <mathjax>#"Ba"^(2+)"#</mathjax>. The chlorate polyatomic anion has the formula <mathjax>#"ClO"_3"^(-)#</mathjax>.</p>
<p>An ionic compound is neutral, which means that its overall charge is <mathjax>#0#</mathjax>. Since the barium ion has a charge of <mathjax>#2^(+)#</mathjax>, the negative charge will have to be <mathjax>#2^-#</mathjax>. We can't change the formula or charge of the chlorate ion, but we can change the number of chlorate ions. Two chlorate ions will have a charge of <mathjax>#2^(-)#</mathjax>. So the formula for barium chlorate is:</p>
<p><mathjax>#"Ba(ClO"_3)_2"#</mathjax></p>
<p>Now the formula represents a neutral compound. Notice that because there are two chlorate ions, which are polyatomic, we put the formula for the chlorate ion in parentheses with a subscript of <mathjax>#2#</mathjax>.</p>
<p>There is a method for writing formulas for <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> which is called the crisscross rule. The following video explains how it works.</p>
<p>
<iframe src="https://www.youtube.com/embed/edLBpAhndpw?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>If you're not sure how to determine the charge of a monatomic ion from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, the following video should be helpful.</p>
<p>
<iframe src="https://www.youtube.com/embed/URc75hoKGLY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<div class="markdown"><p><mathjax>#"Ba(ClO"_3)_2"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The barium cation has the formula <mathjax>#"Ba"^(2+)"#</mathjax>. The chlorate polyatomic anion has the formula <mathjax>#"ClO"_3"^(-)#</mathjax>.</p>
<p>An ionic compound is neutral, which means that its overall charge is <mathjax>#0#</mathjax>. Since the barium ion has a charge of <mathjax>#2^(+)#</mathjax>, the negative charge will have to be <mathjax>#2^-#</mathjax>. We can't change the formula or charge of the chlorate ion, but we can change the number of chlorate ions. Two chlorate ions will have a charge of <mathjax>#2^(-)#</mathjax>. So the formula for barium chlorate is:</p>
<p><mathjax>#"Ba(ClO"_3)_2"#</mathjax></p>
<p>Now the formula represents a neutral compound. Notice that because there are two chlorate ions, which are polyatomic, we put the formula for the chlorate ion in parentheses with a subscript of <mathjax>#2#</mathjax>.</p>
<p>There is a method for writing formulas for <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> which is called the crisscross rule. The following video explains how it works.</p>
<p>
<iframe src="https://www.youtube.com/embed/edLBpAhndpw?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>If you're not sure how to determine the charge of a monatomic ion from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, the following video should be helpful.</p>
<p>
<iframe src="https://www.youtube.com/embed/URc75hoKGLY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">What is the correct formula for barium chlorate?</h1>
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<div class="markdown"><p><mathjax>#"Ba(ClO"_3)_2"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The barium cation has the formula <mathjax>#"Ba"^(2+)"#</mathjax>. The chlorate polyatomic anion has the formula <mathjax>#"ClO"_3"^(-)#</mathjax>.</p>
<p>An ionic compound is neutral, which means that its overall charge is <mathjax>#0#</mathjax>. Since the barium ion has a charge of <mathjax>#2^(+)#</mathjax>, the negative charge will have to be <mathjax>#2^-#</mathjax>. We can't change the formula or charge of the chlorate ion, but we can change the number of chlorate ions. Two chlorate ions will have a charge of <mathjax>#2^(-)#</mathjax>. So the formula for barium chlorate is:</p>
<p><mathjax>#"Ba(ClO"_3)_2"#</mathjax></p>
<p>Now the formula represents a neutral compound. Notice that because there are two chlorate ions, which are polyatomic, we put the formula for the chlorate ion in parentheses with a subscript of <mathjax>#2#</mathjax>.</p>
<p>There is a method for writing formulas for <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a> which is called the crisscross rule. The following video explains how it works.</p>
<p>
<iframe src="https://www.youtube.com/embed/edLBpAhndpw?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p>
<p>If you're not sure how to determine the charge of a monatomic ion from <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, the following video should be helpful.</p>
<p>
<iframe src="https://www.youtube.com/embed/URc75hoKGLY?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | What is the correct formula for barium chlorate? | null |
648 | aa8c6ac1-6ddd-11ea-af6f-ccda262736ce | https://socratic.org/questions/59aeeffa11ef6b4c5d4a27d0 | H2C=CHCH2CH3(g) + 6 O2(g) -> 4 CO2(g) + 4 H2O(l) | start chemical_equation qc_end substance 6 6 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"H2C=CHCH2CH3(g) + 6 O2(g) -> 4 CO2(g) + 4 H2O(l)"}] | [{"type":"substance name","value":"1-butylene"}] | <h1 class="questionTitle" itemprop="name">Can you represent the oxidation of #"1-butylene"#?</h1> | null | H2C=CHCH2CH3(g) + 6 O2(g) -> 4 CO2(g) + 4 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The usual rigmarole is to balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens.........you can get very good at this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, and remember that you do this already every time you get change from a large bill....</p>
<p>For an alkanes, say butane, we follow the same rigmarole....</p>
<p><mathjax>#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#</mathjax></p>
<p>And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....</p>
<p><mathjax>#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#</mathjax></p>
<p>I think the arithmetic is easier in the former case.... Anyway try it out for <mathjax>#C_2H_6#</mathjax>, <mathjax>#H_2C=CHCH_3#</mathjax>, and <mathjax>#H_3C-CH_2OH#</mathjax>; these are probably all reasonable questions at your current level.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#H_2C=CHCH_2CH_3(g) + 6O_2(g)rarr4CO_2(g) + 4H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The usual rigmarole is to balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens.........you can get very good at this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, and remember that you do this already every time you get change from a large bill....</p>
<p>For an alkanes, say butane, we follow the same rigmarole....</p>
<p><mathjax>#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#</mathjax></p>
<p>And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....</p>
<p><mathjax>#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#</mathjax></p>
<p>I think the arithmetic is easier in the former case.... Anyway try it out for <mathjax>#C_2H_6#</mathjax>, <mathjax>#H_2C=CHCH_3#</mathjax>, and <mathjax>#H_3C-CH_2OH#</mathjax>; these are probably all reasonable questions at your current level.</p></div>
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<h1 class="questionTitle" itemprop="name">Can you represent the oxidation of #"1-butylene"#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#H_2C=CHCH_2CH_3(g) + 6O_2(g)rarr4CO_2(g) + 4H_2O(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The usual rigmarole is to balance the carbons as carbon dioxide, then balance the hydrogens as water, and then balance the oxygens.........you can get very good at this <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, and remember that you do this already every time you get change from a large bill....</p>
<p>For an alkanes, say butane, we follow the same rigmarole....</p>
<p><mathjax>#C_4H_10(g) + 13/2O_2(g)rarr4CO_2(g) + 5H_2O(l)#</mathjax></p>
<p>And thus here I took 4 equiv carbon dioxide, and then five equiv of water, and then balanced up the required dioxygens. You could double the entire equation to give....</p>
<p><mathjax>#2C_4H_10(g) + 13O_2(g)rarr8CO_2(g) + 10H_2O(l)#</mathjax></p>
<p>I think the arithmetic is easier in the former case.... Anyway try it out for <mathjax>#C_2H_6#</mathjax>, <mathjax>#H_2C=CHCH_3#</mathjax>, and <mathjax>#H_3C-CH_2OH#</mathjax>; these are probably all reasonable questions at your current level.</p></div>
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<div class="markdown"><p>The balanced equation is:</p>
<p><mathjax>#C_4H_8 + 6O_2 to 4CO_2+ 4H_2O#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula for <a href="https://en.wikipedia.org/wiki/Butene" rel="nofollow">butene</a> is</p>
<p><mathjax>#C_4H_8#</mathjax></p>
<p>You are going to bun it with plenty of oxygen:</p>
<p><mathjax>#C_4H_8 + (n_1)O_2#</mathjax></p>
<p>The products are carbon dioxide and water:</p>
<p><mathjax>#C_4H_8 + (n_1)O_2 to (n_2)CO_2+ (n_3)H_2O#</mathjax></p>
<p>Solved for the 3 variables:</p>
<p><mathjax>#C_4 = (n_2)C#</mathjax></p>
<p><mathjax>#n_2 = 4#</mathjax></p>
<p><mathjax>#H_8 = (n_3)H_2#</mathjax></p>
<p><mathjax>#n_3=4#</mathjax></p>
<p><mathjax>#(n_1)O_2 = (4)O_2+(4)O#</mathjax></p>
<p><mathjax>#(n_1)O_2 = (4)O_2+(2)O_2#</mathjax></p>
<p><mathjax>#n_1 = 6#</mathjax></p>
<p>The balanced equation is:</p>
<p><mathjax>#C_4H_8 + 6O_2 to 4CO_2+ 4H_2O#</mathjax></p></div>
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<div class="markdown"><p>The balanced equation is:</p>
<p><mathjax>#2C_4H_6 +11O_2 -> 8CO_2+6H_2O#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Unbalanced equation is:</p>
<p><mathjax>#C_4H_6+O_2 -> CO_2+ H_2O#</mathjax></p>
<p>First we can put <mathjax>#4#</mathjax> in front of <mathjax>#CO_2#</mathjax> to balance the number of carbon atoms:</p>
<p><mathjax>#C_4H_6+O_2 -> 4CO_2+ H_2O#</mathjax></p>
<p>Now we can put <mathjax>#3#</mathjax> next to water to balance hydrogene:</p>
<p><mathjax>#C_4H_6+O_2 -> 4CO_2+ 3H_2O#</mathjax></p>
<p>Hydrogen and carbon are now balanced, but not oxygen. On the right side there are <mathjax>#11#</mathjax> atoms, on the left side <mathjax>#2#</mathjax>. To balance it we have to put <mathjax>#5.5#</mathjax> on the left side:</p>
<p><mathjax>#C_4H_6+5.5O_2 -> 4CO_2+ 3H_2O#</mathjax> </p>
<p>The equation is now balanced, but it is not correct to put fractions as the coefficients in a chemical reactions. To keep the equation ballanced and make all coefficients integer we have to multiply all coefficients by <mathjax>#2#</mathjax></p>
<p><mathjax>#2C_4H_6+11O_2 -> 8CO_2+ 6H_2O#</mathjax></p></div>
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</article> | Can you represent the oxidation of #"1-butylene"#? | null |
649 | ab27da80-6ddd-11ea-87db-ccda262736ce | https://socratic.org/questions/a-20-0-ml-sample-of-a-gas-is-at-546-k-and-has-a-pressure-of-6-0-atm-if-the-tempe | 30.00 mL | start physical_unit 6 6 volume ml qc_end physical_unit 6 6 1 2 volume qc_end physical_unit 6 6 9 10 temperature qc_end physical_unit 6 6 16 17 pressure qc_end physical_unit 6 6 24 25 temperature qc_end physical_unit 6 6 30 31 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas sample [IN] mL"}] | [{"type":"physical unit","value":"30.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{20.0 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{546 K}"},{"type":"physical unit","value":"Pressure1 [OF] the gas sample [=] \\pu{6.0 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the gas sample [=] \\pu{273 K}"},{"type":"physical unit","value":"Pressure2 [OF] the gas sample [=] \\pu{2.0 atm}"}] | <h1 class="questionTitle" itemprop="name">A 20.0 mL sample of a gas is at 546 K and has a pressure of 6.0 atm. If the temperature is changed to 273 K and the pressure to 2.0 atm, the new volume of the gas is what?</h1> | null | 30.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#</mathjax></p>
<p>We immediately see that we will get an answer in <mathjax>#"terms of volume"#</mathjax>, so we're batting on a firm wicket.</p>
<p><mathjax>#V_2=(6.0*atmxx20.0*mLxx273*K)/(2.0*atmxx546*K)=??*mL#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, given constant <mathjax>#n#</mathjax>, so............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#</mathjax></p>
<p>We immediately see that we will get an answer in <mathjax>#"terms of volume"#</mathjax>, so we're batting on a firm wicket.</p>
<p><mathjax>#V_2=(6.0*atmxx20.0*mLxx273*K)/(2.0*atmxx546*K)=??*mL#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A 20.0 mL sample of a gas is at 546 K and has a pressure of 6.0 atm. If the temperature is changed to 273 K and the pressure to 2.0 atm, the new volume of the gas is what?</h1>
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<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, given constant <mathjax>#n#</mathjax>, so............</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2=(P_1xxV_1xxT_2)/(P_2xxT_1)#</mathjax></p>
<p>We immediately see that we will get an answer in <mathjax>#"terms of volume"#</mathjax>, so we're batting on a firm wicket.</p>
<p><mathjax>#V_2=(6.0*atmxx20.0*mLxx273*K)/(2.0*atmxx546*K)=??*mL#</mathjax></p></div>
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</article> | A 20.0 mL sample of a gas is at 546 K and has a pressure of 6.0 atm. If the temperature is changed to 273 K and the pressure to 2.0 atm, the new volume of the gas is what? | null |
650 | ac494e38-6ddd-11ea-8f97-ccda262736ce | https://socratic.org/questions/how-many-grams-of-oxygen-are-in-2-moles-of-potassium-dichromate-k-2cr-2o-7 | 223.99 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 12 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] grams"}] | [{"type":"physical unit","value":"223.99 grams"}] | [{"type":"physical unit","value":"Mole [OF] K2Cr2O7 [=] \\pu{2 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams of oxygen are in 2 moles of potassium dichromate, #K_2Cr_2O_7#?</h1> | null | 223.99 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are two simple steps you can take to answer your question. The number of grams of a substance can be determined by summing the atomic weights of the individual components and multiplying that result by the number of moles you have. </p>
<p>However, since you are only interested in the amount of grams of oxygen in 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you can focus only on the oxygen element. </p>
<p>There are 7 atoms of oxygen denoted by the subscript 7 in the compound. The <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of oxygen is 15.9994 grams per mole. </p>
<p>So in the <mathjax>#K_2Cr_2O_7#</mathjax> compound the atomic mass of the 7 oxygen atoms is:</p>
<blockquote>
<p><mathjax>#15.9994#</mathjax> <mathjax>#"grams"/"mole" * (7)#</mathjax></p>
</blockquote>
<p>which equals</p>
<blockquote>
<p><mathjax>#111.9958#</mathjax> <mathjax>#"grams"/"mole"#</mathjax> </p>
</blockquote>
<p>Since you have 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you multiple your atomic mass of the oxygen by 2 and with dimensional analysis you can see the moles cancel out and you are left with grams.</p>
<p>The answer being: <mathjax>#223.9916#</mathjax> grams</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>There are <mathjax>#223.9916#</mathjax> grams of oxygen in 2 moles of potassium dichromate. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are two simple steps you can take to answer your question. The number of grams of a substance can be determined by summing the atomic weights of the individual components and multiplying that result by the number of moles you have. </p>
<p>However, since you are only interested in the amount of grams of oxygen in 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you can focus only on the oxygen element. </p>
<p>There are 7 atoms of oxygen denoted by the subscript 7 in the compound. The <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of oxygen is 15.9994 grams per mole. </p>
<p>So in the <mathjax>#K_2Cr_2O_7#</mathjax> compound the atomic mass of the 7 oxygen atoms is:</p>
<blockquote>
<p><mathjax>#15.9994#</mathjax> <mathjax>#"grams"/"mole" * (7)#</mathjax></p>
</blockquote>
<p>which equals</p>
<blockquote>
<p><mathjax>#111.9958#</mathjax> <mathjax>#"grams"/"mole"#</mathjax> </p>
</blockquote>
<p>Since you have 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you multiple your atomic mass of the oxygen by 2 and with dimensional analysis you can see the moles cancel out and you are left with grams.</p>
<p>The answer being: <mathjax>#223.9916#</mathjax> grams</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of oxygen are in 2 moles of potassium dichromate, #K_2Cr_2O_7#?</h1>
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<div class="markdown"><p>There are <mathjax>#223.9916#</mathjax> grams of oxygen in 2 moles of potassium dichromate. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are two simple steps you can take to answer your question. The number of grams of a substance can be determined by summing the atomic weights of the individual components and multiplying that result by the number of moles you have. </p>
<p>However, since you are only interested in the amount of grams of oxygen in 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you can focus only on the oxygen element. </p>
<p>There are 7 atoms of oxygen denoted by the subscript 7 in the compound. The <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of oxygen is 15.9994 grams per mole. </p>
<p>So in the <mathjax>#K_2Cr_2O_7#</mathjax> compound the atomic mass of the 7 oxygen atoms is:</p>
<blockquote>
<p><mathjax>#15.9994#</mathjax> <mathjax>#"grams"/"mole" * (7)#</mathjax></p>
</blockquote>
<p>which equals</p>
<blockquote>
<p><mathjax>#111.9958#</mathjax> <mathjax>#"grams"/"mole"#</mathjax> </p>
</blockquote>
<p>Since you have 2 moles of <mathjax>#K_2Cr_2O_7#</mathjax>, you multiple your atomic mass of the oxygen by 2 and with dimensional analysis you can see the moles cancel out and you are left with grams.</p>
<p>The answer being: <mathjax>#223.9916#</mathjax> grams</p></div>
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</article> | How many grams of oxygen are in 2 moles of potassium dichromate, #K_2Cr_2O_7#? | null |
651 | ab815a3e-6ddd-11ea-8a15-ccda262736ce | https://socratic.org/questions/how-many-moles-are-there-in-100-g-of-copper-oxide-cuo | 1.25 moles | start physical_unit 11 11 mole mol qc_end physical_unit 11 11 6 7 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] CuO [IN] moles"}] | [{"type":"physical unit","value":"1.25 moles"}] | [{"type":"physical unit","value":"Mass [OF] CuO [=] \\pu{100 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are there in 100 g of copper oxide, #CuO#?</h1> | null | 1.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#(100*cancelg)/(79.95 *cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#100*g " cupric oxide"#</mathjax> constitutes a bit over <mathjax>#1*mol#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#(100*cancelg)/(79.95 *cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are there in 100 g of copper oxide, #CuO#?</h1>
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<div class="markdown"><p><mathjax>#100*g " cupric oxide"#</mathjax> constitutes a bit over <mathjax>#1*mol#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax> <mathjax>#=#</mathjax></p>
<p><mathjax>#(100*cancelg)/(79.95 *cancelg*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol#</mathjax>.</p></div>
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</article> | How many moles are there in 100 g of copper oxide, #CuO#? | null |
652 | aa3a7e10-6ddd-11ea-ba8e-ccda262736ce | https://socratic.org/questions/58cb03c97c014912542570f1 | 5.20 | start physical_unit 6 7 ph none qc_end physical_unit 7 7 10 11 volume qc_end physical_unit 23 23 18 19 molarity qc_end physical_unit 7 7 25 26 volume qc_end physical_unit 38 38 33 34 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the buffer solution"}] | [{"type":"physical unit","value":"5.20"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{50 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.220 mol/L}"},{"type":"physical unit","value":"Volume [OF] HOAc solution [=] \\pu{100 mL}"},{"type":"physical unit","value":"Molarity [OF] HOAc solution [=] \\pu{0.150 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the #pH# of a buffer solution prepared from #50*mL# of sodium hydroxide solution that is #0.220*mol*L^-1# with respect to #NaOH#, and #100*mL# of acetic acid solution that is #0.150*mol*L^-1# with respect to #HOAc#?
</h1> | null | 5.20 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the buffer <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">equation</a> such that,</p>
<p><mathjax>#pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}#</mathjax></p>
<p>But of course, we have to calculate <mathjax>#[""^(-)OAc]#</mathjax> and <mathjax>#[HOAc]#</mathjax>. The volume of the solution is clearly <mathjax>#150*mL#</mathjax>.</p>
<p><mathjax>#"Moles of NaOH"-=(50.0xx10^-3Lxx0.220*mol*L^-1)=0.011*mol.#</mathjax></p>
<p><mathjax>#"Moles of HOAc"-=(100.0xx10^-3Lxx0.150*mol*L^-1)=0.015*mol.#</mathjax></p>
<p>And since <mathjax>#NaOH#</mathjax> reacts quantitatively with <mathjax>#HOAc#</mathjax>, we have a solution that is nominally, </p>
<p><mathjax>#[HOAc]=((0.015-0.011)*mol)/(150xx10^-3L)=0.0267*mol*L^-1#</mathjax></p>
<p>And <mathjax>#[""^(-)OAc]=(0.011*mol)/(150xx10^-3L)=0.0733*mol*L^-1#</mathjax></p>
<p>And so we fill in the dots; the <mathjax>#pH#</mathjax> should be elevated from the <mathjax>#pK_a#</mathjax> (which I happen to know is <mathjax>#4.76#</mathjax>; these data really should have been supplied with the question!). </p>
<p><mathjax>#pH=4.76+log_10{(0.0733*mol*L^-1)/(0.0267*mol*L^-1)}#</mathjax></p>
<p><mathjax>#=4.76+0.439=5.20#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pH=5.20#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the buffer <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">equation</a> such that,</p>
<p><mathjax>#pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}#</mathjax></p>
<p>But of course, we have to calculate <mathjax>#[""^(-)OAc]#</mathjax> and <mathjax>#[HOAc]#</mathjax>. The volume of the solution is clearly <mathjax>#150*mL#</mathjax>.</p>
<p><mathjax>#"Moles of NaOH"-=(50.0xx10^-3Lxx0.220*mol*L^-1)=0.011*mol.#</mathjax></p>
<p><mathjax>#"Moles of HOAc"-=(100.0xx10^-3Lxx0.150*mol*L^-1)=0.015*mol.#</mathjax></p>
<p>And since <mathjax>#NaOH#</mathjax> reacts quantitatively with <mathjax>#HOAc#</mathjax>, we have a solution that is nominally, </p>
<p><mathjax>#[HOAc]=((0.015-0.011)*mol)/(150xx10^-3L)=0.0267*mol*L^-1#</mathjax></p>
<p>And <mathjax>#[""^(-)OAc]=(0.011*mol)/(150xx10^-3L)=0.0733*mol*L^-1#</mathjax></p>
<p>And so we fill in the dots; the <mathjax>#pH#</mathjax> should be elevated from the <mathjax>#pK_a#</mathjax> (which I happen to know is <mathjax>#4.76#</mathjax>; these data really should have been supplied with the question!). </p>
<p><mathjax>#pH=4.76+log_10{(0.0733*mol*L^-1)/(0.0267*mol*L^-1)}#</mathjax></p>
<p><mathjax>#=4.76+0.439=5.20#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the #pH# of a buffer solution prepared from #50*mL# of sodium hydroxide solution that is #0.220*mol*L^-1# with respect to #NaOH#, and #100*mL# of acetic acid solution that is #0.150*mol*L^-1# with respect to #HOAc#?
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anor277
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<div class="markdown"><p><mathjax>#pH=5.20#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We use the buffer <a href="https://socratic.org/questions/how-do-buffers-maintain-ph">equation</a> such that,</p>
<p><mathjax>#pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}#</mathjax></p>
<p>But of course, we have to calculate <mathjax>#[""^(-)OAc]#</mathjax> and <mathjax>#[HOAc]#</mathjax>. The volume of the solution is clearly <mathjax>#150*mL#</mathjax>.</p>
<p><mathjax>#"Moles of NaOH"-=(50.0xx10^-3Lxx0.220*mol*L^-1)=0.011*mol.#</mathjax></p>
<p><mathjax>#"Moles of HOAc"-=(100.0xx10^-3Lxx0.150*mol*L^-1)=0.015*mol.#</mathjax></p>
<p>And since <mathjax>#NaOH#</mathjax> reacts quantitatively with <mathjax>#HOAc#</mathjax>, we have a solution that is nominally, </p>
<p><mathjax>#[HOAc]=((0.015-0.011)*mol)/(150xx10^-3L)=0.0267*mol*L^-1#</mathjax></p>
<p>And <mathjax>#[""^(-)OAc]=(0.011*mol)/(150xx10^-3L)=0.0733*mol*L^-1#</mathjax></p>
<p>And so we fill in the dots; the <mathjax>#pH#</mathjax> should be elevated from the <mathjax>#pK_a#</mathjax> (which I happen to know is <mathjax>#4.76#</mathjax>; these data really should have been supplied with the question!). </p>
<p><mathjax>#pH=4.76+log_10{(0.0733*mol*L^-1)/(0.0267*mol*L^-1)}#</mathjax></p>
<p><mathjax>#=4.76+0.439=5.20#</mathjax>.</p></div>
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</article> | What is the #pH# of a buffer solution prepared from #50*mL# of sodium hydroxide solution that is #0.220*mol*L^-1# with respect to #NaOH#, and #100*mL# of acetic acid solution that is #0.150*mol*L^-1# with respect to #HOAc#?
| null |
653 | ac5ed91e-6ddd-11ea-93e2-ccda262736ce | https://socratic.org/questions/an-aqueous-solution-is-40-0-by-mass-hydrochloric-acid-hcl-and-has-a-density-of-1 | 13.17 M | start physical_unit 7 8 molarity mol/l qc_end end | [{"type":"physical unit","value":"Molarity [OF] hydrochloric acid [IN] M"}] | [{"type":"physical unit","value":"13.17 M"}] | [{"type":"physical unit","value":"Percent by mass [OF] HCl in aqueous solution [=] \\pu{40.0%}"},{"type":"physical unit","value":"Density [OF] HCl aqueous solution [=] \\pu{1.20 g/mL}"}] | <h1 class="questionTitle" itemprop="name">An aqueous solution is 40.0% by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. What is the molarity of hydrochloric acid in the solution?</h1> | null | 13.17 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is pick a sample of this solution. </p>
<p>Since you're going to find its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, you can make the calculations easier by picking a <mathjax>#"1.00-L"#</mathjax> sample. Keep in mind ,the answer will be the same <em>regardless</em> of the volume you pick. </p>
<p>So, use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution to determine how many grams you get in that respective volume</p>
<blockquote>
<p><mathjax>#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.20 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"#</mathjax></p>
</blockquote>
<p>Now, you that the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> for this solution is <mathjax>#"40.0% HCl"#</mathjax>. This means that you get <mathjax>#"40.0 g"#</mathjax> of hydrochloric acid <strong>for every</strong> <mathjax>#"100.0 g"#</mathjax> of solution. </p>
<p>The mass of hydrochloric acid you get in that sample will thus be </p>
<blockquote>
<p><mathjax>#1200color(red)(cancel(color(black)("g solution"))) * "40.0 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "480 g HCl"#</mathjax></p>
</blockquote>
<p>Use hydrochloric acid's molar mass to determine how many moles you have in that many grams </p>
<blockquote>
<p><mathjax>#480color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "13.17 moles HCl"#</mathjax> </p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will thus be - remember that we started with a <mathjax>#"1.00-L"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#c = n/V#</mathjax></p>
<p><mathjax>#c = "13.17 moles"/"1.00 L" = color(green)("13.2 M") ->#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>This is very close to the maximum possible concentration for hydrochloric acid in aqueous solution. At this concentration, the solution is actually fuming at a significant evaporation rate.</em></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"13.2 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is pick a sample of this solution. </p>
<p>Since you're going to find its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, you can make the calculations easier by picking a <mathjax>#"1.00-L"#</mathjax> sample. Keep in mind ,the answer will be the same <em>regardless</em> of the volume you pick. </p>
<p>So, use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution to determine how many grams you get in that respective volume</p>
<blockquote>
<p><mathjax>#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.20 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"#</mathjax></p>
</blockquote>
<p>Now, you that the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> for this solution is <mathjax>#"40.0% HCl"#</mathjax>. This means that you get <mathjax>#"40.0 g"#</mathjax> of hydrochloric acid <strong>for every</strong> <mathjax>#"100.0 g"#</mathjax> of solution. </p>
<p>The mass of hydrochloric acid you get in that sample will thus be </p>
<blockquote>
<p><mathjax>#1200color(red)(cancel(color(black)("g solution"))) * "40.0 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "480 g HCl"#</mathjax></p>
</blockquote>
<p>Use hydrochloric acid's molar mass to determine how many moles you have in that many grams </p>
<blockquote>
<p><mathjax>#480color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "13.17 moles HCl"#</mathjax> </p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will thus be - remember that we started with a <mathjax>#"1.00-L"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#c = n/V#</mathjax></p>
<p><mathjax>#c = "13.17 moles"/"1.00 L" = color(green)("13.2 M") ->#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>This is very close to the maximum possible concentration for hydrochloric acid in aqueous solution. At this concentration, the solution is actually fuming at a significant evaporation rate.</em></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">An aqueous solution is 40.0% by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. What is the molarity of hydrochloric acid in the solution?</h1>
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Stefan V.
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Nov 13, 2015
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<div class="markdown"><p><mathjax>#"13.2 M"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is pick a sample of this solution. </p>
<p>Since you're going to find its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, you can make the calculations easier by picking a <mathjax>#"1.00-L"#</mathjax> sample. Keep in mind ,the answer will be the same <em>regardless</em> of the volume you pick. </p>
<p>So, use the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of the solution to determine how many grams you get in that respective volume</p>
<blockquote>
<p><mathjax>#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.20 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"#</mathjax></p>
</blockquote>
<p>Now, you that the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> for this solution is <mathjax>#"40.0% HCl"#</mathjax>. This means that you get <mathjax>#"40.0 g"#</mathjax> of hydrochloric acid <strong>for every</strong> <mathjax>#"100.0 g"#</mathjax> of solution. </p>
<p>The mass of hydrochloric acid you get in that sample will thus be </p>
<blockquote>
<p><mathjax>#1200color(red)(cancel(color(black)("g solution"))) * "40.0 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "480 g HCl"#</mathjax></p>
</blockquote>
<p>Use hydrochloric acid's molar mass to determine how many moles you have in that many grams </p>
<blockquote>
<p><mathjax>#480color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "13.17 moles HCl"#</mathjax> </p>
</blockquote>
<p>The <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution will thus be - remember that we started with a <mathjax>#"1.00-L"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#c = n/V#</mathjax></p>
<p><mathjax>#c = "13.17 moles"/"1.00 L" = color(green)("13.2 M") ->#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>This is very close to the maximum possible concentration for hydrochloric acid in aqueous solution. At this concentration, the solution is actually fuming at a significant evaporation rate.</em></p></div>
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</article> | An aqueous solution is 40.0% by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. What is the molarity of hydrochloric acid in the solution? | null |
654 | a8504888-6ddd-11ea-a66c-ccda262736ce | https://socratic.org/questions/in-photosynthesis-glucose-c6h12o6-and-o2-are-produced-from-co2-and-h2o-6co-g-6h2 | 14.32 grams | start physical_unit 3 3 mass g qc_end physical_unit 9 9 33 34 mass qc_end chemical_equation 12 24 qc_end end | [{"type":"physical unit","value":"Mass [OF] C6H12O6 [IN] grams"}] | [{"type":"physical unit","value":"14.32 grams"}] | [{"type":"physical unit","value":"mass [OF] CO2 [=] \\pu{21.0 g}"},{"type":"chemical equation","value":"6 CO2(g) + 6 H2O(l) + 680 kcal -> C6H12O6(aq) + 6 O2(g)"}] | <h1 class="questionTitle" itemprop="name">In photosynthesis glucose c6h12o6 and o2 are produced from co2 and h2o.
6CO(g)+6H2O(l)+680kcal=C6H12O6(aq)+6O2(g)
How many grams of glucose are produced from 21.0g of CO2?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>How to set problem, I believe I'm going from grams to grams</p>
<p><img alt="" src="https://s3.amazonaws.com/socratic-ocr-uploads/prod/27365bd6-2833-44de-ae45-320fc77052e0/processed.jpeg"/></p></div>
</h2>
</div>
</div> | 14.32 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#6CO_2(g) + 6H_2O(l) + Delta rarr C_6H_12O_6(aq) + 6O_2(g)#</mathjax></p>
<p><mathjax>#"Moles of "CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.0*g)/(44.0*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.477#</mathjax> <mathjax>#mol#</mathjax></p>
<p><mathjax>#180.16*g*mol^-1xx1/6xx0.477*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#6CO_2(g) + 6H_2O(l) + Delta rarr C_6H_12O_6(aq) + 6O_2(g)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#6CO_2(g) + 6H_2O(l) + Delta rarr C_6H_12O_6(aq) + 6O_2(g)#</mathjax></p>
<p><mathjax>#"Moles of "CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.0*g)/(44.0*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.477#</mathjax> <mathjax>#mol#</mathjax></p>
<p><mathjax>#180.16*g*mol^-1xx1/6xx0.477*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">In photosynthesis glucose c6h12o6 and o2 are produced from co2 and h2o.
6CO(g)+6H2O(l)+680kcal=C6H12O6(aq)+6O2(g)
How many grams of glucose are produced from 21.0g of CO2?</h1>
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<div class="markdown"><p>How to set problem, I believe I'm going from grams to grams</p>
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<div class="markdown"><p><mathjax>#6CO_2(g) + 6H_2O(l) + Delta rarr C_6H_12O_6(aq) + 6O_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#6CO_2(g) + 6H_2O(l) + Delta rarr C_6H_12O_6(aq) + 6O_2(g)#</mathjax></p>
<p><mathjax>#"Moles of "CO_2(g)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(21.0*g)/(44.0*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.477#</mathjax> <mathjax>#mol#</mathjax></p>
<p><mathjax>#180.16*g*mol^-1xx1/6xx0.477*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
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</article> | In photosynthesis glucose c6h12o6 and o2 are produced from co2 and h2o.
6CO(g)+6H2O(l)+680kcal=C6H12O6(aq)+6O2(g)
How many grams of glucose are produced from 21.0g of CO2? |
How to set problem, I believe I'm going from grams to grams
|
655 | ac490086-6ddd-11ea-9364-ccda262736ce | https://socratic.org/questions/a-gas-occupies-a-volume-of-0-140-l-at-35-0-c-and-97-0-kpa-what-is-the-volume-of- | 0.12 L | start physical_unit 19 20 volume l qc_end physical_unit 19 20 6 7 volume qc_end physical_unit 19 20 9 10 temperature qc_end physical_unit 19 20 12 13 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.12 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.140 L}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{35.0 ℃}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{97.0 kPa}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A gas occupies a volume of 0.140 L at 35.0°C and 97.0 kPa. What is the volume of the gas at STP?</h1> | null | 0.12 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>STP currently has the values of 273.15 K and 100 kPa.</p>
<p>This is a question involving the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> as shown below.</p>
<p><img alt="http://me-mechanicalengineering.com/gas-laws/" src="https://useruploads.socratic.org/OkU45BVcQN671ga8WkLI_unnamed.png"/> </p>
<p><mathjax>#V_1="0.140 L"#</mathjax><br/>
<mathjax>#P_1="97.0 kPa"#</mathjax><br/>
<mathjax>#T_1="35.0"^@"C"+273.15=308.15 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p>Rearrange the combined gas law to isolate <mathjax>#V_2#</mathjax>, plug in the known values, then solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=((97.0cancel"kPa")xx(0.140"L")xx(273.15cancel"K"))/((308.15cancel"K")xx(100cancel"kPa"))="0.120 L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The volume of the gas at STP of 273.15 K and 100 kPa is 0.120 L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>STP currently has the values of 273.15 K and 100 kPa.</p>
<p>This is a question involving the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> as shown below.</p>
<p><img alt="http://me-mechanicalengineering.com/gas-laws/" src="https://useruploads.socratic.org/OkU45BVcQN671ga8WkLI_unnamed.png"/> </p>
<p><mathjax>#V_1="0.140 L"#</mathjax><br/>
<mathjax>#P_1="97.0 kPa"#</mathjax><br/>
<mathjax>#T_1="35.0"^@"C"+273.15=308.15 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p>Rearrange the combined gas law to isolate <mathjax>#V_2#</mathjax>, plug in the known values, then solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=((97.0cancel"kPa")xx(0.140"L")xx(273.15cancel"K"))/((308.15cancel"K")xx(100cancel"kPa"))="0.120 L"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A gas occupies a volume of 0.140 L at 35.0°C and 97.0 kPa. What is the volume of the gas at STP?</h1>
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<div class="markdown"><p>The volume of the gas at STP of 273.15 K and 100 kPa is 0.120 L.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>STP currently has the values of 273.15 K and 100 kPa.</p>
<p>This is a question involving the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> as shown below.</p>
<p><img alt="http://me-mechanicalengineering.com/gas-laws/" src="https://useruploads.socratic.org/OkU45BVcQN671ga8WkLI_unnamed.png"/> </p>
<p><mathjax>#V_1="0.140 L"#</mathjax><br/>
<mathjax>#P_1="97.0 kPa"#</mathjax><br/>
<mathjax>#T_1="35.0"^@"C"+273.15=308.15 K"#</mathjax><br/>
<mathjax>#P_2="100 kPa"#</mathjax><br/>
<mathjax>#T_2="273.15 K"#</mathjax><br/>
<mathjax>#V_2="???"#</mathjax></p>
<p>Rearrange the combined gas law to isolate <mathjax>#V_2#</mathjax>, plug in the known values, then solve.</p>
<p><mathjax>#V_2=(P_1V_1T_2)/(T_1P_2)#</mathjax></p>
<p><mathjax>#V_2=((97.0cancel"kPa")xx(0.140"L")xx(273.15cancel"K"))/((308.15cancel"K")xx(100cancel"kPa"))="0.120 L"#</mathjax></p></div>
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</article> | A gas occupies a volume of 0.140 L at 35.0°C and 97.0 kPa. What is the volume of the gas at STP? | null |
656 | abd4bb1e-6ddd-11ea-b0b0-ccda262736ce | https://socratic.org/questions/how-can-i-balance-this-equation-n2-h2-nh3 | N2 + 3 H2 -> 2 NH3 | start chemical_equation qc_end chemical_equation 6 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"}] | [{"type":"chemical equation","value":"N2 + H2 -> NH3"}] | <h1 class="questionTitle" itemprop="name">How can I balance this equation? ____ N2 + ____ H2 ---> ____ NH3</h1> | null | N2 + 3 H2 -> 2 NH3 | <div class="answerDescription">
<div>
<div class="markdown"><p>This reaction is the synthesis of Ammonia using Nitrogen and Hydrogen gas.</p>
<p><mathjax>#N + H -> NH_3#</mathjax></p>
<p>We must remember that Nitrogen and Hydrogen are both diatomic molecules in their standard gas form. This adjusts the equation to</p>
<p><mathjax>#N_2 + H_2 -> NH_3#</mathjax></p>
<p>Now we need to adjust coefficients in order to balance the atoms on each side of the equation. Currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side. </p>
<p>We can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen. </p>
<p><mathjax>#N_2 + 3H_2 -> 2NH_3#</mathjax></p>
<p>This gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side. </p>
<p>I hope this was helpful.<br/>
SMARTERTEACHER</p>
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<div class="markdown"><p>This reaction is the synthesis of Ammonia using Nitrogen and Hydrogen gas.</p>
<p><mathjax>#N + H -> NH_3#</mathjax></p>
<p>We must remember that Nitrogen and Hydrogen are both diatomic molecules in their standard gas form. This adjusts the equation to</p>
<p><mathjax>#N_2 + H_2 -> NH_3#</mathjax></p>
<p>Now we need to adjust coefficients in order to balance the atoms on each side of the equation. Currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side. </p>
<p>We can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen. </p>
<p><mathjax>#N_2 + 3H_2 -> 2NH_3#</mathjax></p>
<p>This gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side. </p>
<p>I hope this was helpful.<br/>
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<h1 class="questionTitle" itemprop="name">How can I balance this equation? ____ N2 + ____ H2 ---> ____ NH3</h1>
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<div class="markdown"><p>This reaction is the synthesis of Ammonia using Nitrogen and Hydrogen gas.</p>
<p><mathjax>#N + H -> NH_3#</mathjax></p>
<p>We must remember that Nitrogen and Hydrogen are both diatomic molecules in their standard gas form. This adjusts the equation to</p>
<p><mathjax>#N_2 + H_2 -> NH_3#</mathjax></p>
<p>Now we need to adjust coefficients in order to balance the atoms on each side of the equation. Currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side. </p>
<p>We can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen. </p>
<p><mathjax>#N_2 + 3H_2 -> 2NH_3#</mathjax></p>
<p>This gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side. </p>
<p>I hope this was helpful.<br/>
SMARTERTEACHER</p>
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<div class="markdown"><p><mathjax>#N_2#</mathjax> + 3<mathjax>#H_2#</mathjax> -----> 2N<mathjax>#H_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#N_2#</mathjax> + <mathjax>#H_2#</mathjax> -----> N<mathjax>#H_3#</mathjax></p>
<p>Let us balance this equation by counting the number of atoms on both sides of the arrow.</p>
<p><mathjax>#N_2#</mathjax> + <mathjax>#H_2#</mathjax> -----> N<mathjax>#H_3#</mathjax></p>
<p>N=2 , H=2 N=1, H=3</p>
<p>To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N<mathjax>#H_3#</mathjax> on RHS </p>
<p><mathjax>#N_2#</mathjax> + <mathjax>#H_2#</mathjax> -----> 2N<mathjax>#H_3#</mathjax></p>
<p>N=2 , H=2 N=2 , H= 6 </p>
<p>To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of <mathjax>#H_2#</mathjax> on LHS </p>
<p><mathjax>#N_2#</mathjax> + 3<mathjax>#H_2#</mathjax> -----> 2N<mathjax>#H_3#</mathjax></p>
<p>N=2 , H=6 N=2 , H= 6 </p></div>
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</article> | How can I balance this equation? ____ N2 + ____ H2 ---> ____ NH3 | null |
657 | ad0eccc8-6ddd-11ea-8456-ccda262736ce | https://socratic.org/questions/nitrogen-n-and-hydrogen-h-react-wth-each-other-to-produce-ammonia-nh-3-this-reac | 0.33 | start physical_unit 33 37 ratio none qc_end chemical_equation 21 27 qc_end end | [{"type":"physical unit","value":"Ratio [OF] nitrogen atoms to hydrogen atoms"}] | [{"type":"physical unit","value":"0.33"}] | [{"type":"chemical equation","value":"N2 + 3 H2 -> 2 NH3"}] | <h1 class="questionTitle" itemprop="name">Nitrogen (N) and hydrogen (H) react wth each other to produce ammonia (#NH_3#). This reactor is shown in an equation as #N_2 + 3H_2 -> 2NH_3#. What is the ratio of nitrogen atoms to hydrogen atoms in ammonia?</h1> | null | 0.33 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical formula for ammonia is <br/>
<mathjax>#2NH_3#</mathjax> <br/>
meaning that we have two nitrogen atoms and six hydrogen atoms.<br/>
To find the ratio of nitrogen atoms to hydrogen atoms we say:</p>
<p><mathjax>#(2/6) = 0.33#</mathjax></p></div>
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<div class="markdown"><p>0.33 </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical formula for ammonia is <br/>
<mathjax>#2NH_3#</mathjax> <br/>
meaning that we have two nitrogen atoms and six hydrogen atoms.<br/>
To find the ratio of nitrogen atoms to hydrogen atoms we say:</p>
<p><mathjax>#(2/6) = 0.33#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Nitrogen (N) and hydrogen (H) react wth each other to produce ammonia (#NH_3#). This reactor is shown in an equation as #N_2 + 3H_2 -> 2NH_3#. What is the ratio of nitrogen atoms to hydrogen atoms in ammonia?</h1>
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<div class="markdown"><p>0.33 </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The chemical formula for ammonia is <br/>
<mathjax>#2NH_3#</mathjax> <br/>
meaning that we have two nitrogen atoms and six hydrogen atoms.<br/>
To find the ratio of nitrogen atoms to hydrogen atoms we say:</p>
<p><mathjax>#(2/6) = 0.33#</mathjax></p></div>
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</article> | Nitrogen (N) and hydrogen (H) react wth each other to produce ammonia (#NH_3#). This reactor is shown in an equation as #N_2 + 3H_2 -> 2NH_3#. What is the ratio of nitrogen atoms to hydrogen atoms in ammonia? | null |
658 | ab6e65e6-6ddd-11ea-a7fd-ccda262736ce | https://socratic.org/questions/if-2-35-moles-of-an-ideal-gas-has-a-pressure-of-3-63-atm-and-a-volume-of-68-93-l | 1023.85 degrees Celsius | start physical_unit 24 25 temperature °c qc_end physical_unit 5 6 1 2 mole qc_end physical_unit 5 6 11 12 pressure qc_end physical_unit 5 6 17 18 volume qc_end end | [{"type":"physical unit","value":"Temperature [OF] the sample [IN] degrees Celsius"}] | [{"type":"physical unit","value":"1023.85 degrees Celsius"}] | [{"type":"physical unit","value":"Mole [OF] ideal gas [=] \\pu{2.35 moles}"},{"type":"physical unit","value":"Pressure [OF] ideal gas [=] \\pu{3.63 atm}"},{"type":"physical unit","value":"Volume [OF] ideal gas [=] \\pu{68.93 L}"}] | <h1 class="questionTitle" itemprop="name">If 2.35 moles of an ideal gas has a pressure of 3.63 atm, and a volume of 68.93 L, what is the temperature of the sample in degrees Celsius?</h1> | null | 1023.85 degrees Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong>, which establishes a relationship between the <em>volume</em>, <em>pressure</em>, <em>number of moles</em>, and <em>temperature</em> of a gas. </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>
<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>
<p>Notice that the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation uses <strong>absolute temperature</strong>, which is another term used for temperature expressed in <em>Kelvin</em>. </p>
<p>In your case, the problem wants you to find the temperature in <em>degrees Celsius</em>, so right from the start you know that you must <strong>convert</strong> the temperature from Kelvin to degrees Celsius. </p>
<p>So, rearrange the ideal gas law equation and solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>The units given to you for pressure and volume <strong>match</strong> those used in the expression of the universal gas constant, so plug them into the equation to find</p>
<blockquote>
<p><mathjax>#T = (3.63 color(red)(cancel(color(black)("atm"))) * 68.93color(red)(cancel(color(black)("L"))))/(2.35color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#T = "1297 K"#</mathjax></p>
</blockquote>
<p>Now you're ready to convert this to degrees Celsius by using </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = T["K"] - 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#t = "1297 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)color(black)(1020^@"C")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> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1020^@"C"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong>, which establishes a relationship between the <em>volume</em>, <em>pressure</em>, <em>number of moles</em>, and <em>temperature</em> of a gas. </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>
<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>
<p>Notice that the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation uses <strong>absolute temperature</strong>, which is another term used for temperature expressed in <em>Kelvin</em>. </p>
<p>In your case, the problem wants you to find the temperature in <em>degrees Celsius</em>, so right from the start you know that you must <strong>convert</strong> the temperature from Kelvin to degrees Celsius. </p>
<p>So, rearrange the ideal gas law equation and solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>The units given to you for pressure and volume <strong>match</strong> those used in the expression of the universal gas constant, so plug them into the equation to find</p>
<blockquote>
<p><mathjax>#T = (3.63 color(red)(cancel(color(black)("atm"))) * 68.93color(red)(cancel(color(black)("L"))))/(2.35color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#T = "1297 K"#</mathjax></p>
</blockquote>
<p>Now you're ready to convert this to degrees Celsius by using </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = T["K"] - 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#t = "1297 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)color(black)(1020^@"C")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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<h1 class="questionTitle" itemprop="name">If 2.35 moles of an ideal gas has a pressure of 3.63 atm, and a volume of 68.93 L, what is the temperature of the sample in degrees Celsius?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-06-08T01:12:14" itemprop="dateCreated">
Jun 8, 2016
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<div class="markdown"><p><mathjax>#1020^@"C"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This looks like a job for the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong>, which establishes a relationship between the <em>volume</em>, <em>pressure</em>, <em>number of moles</em>, and <em>temperature</em> of a gas. </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>
<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>
<p>Notice that the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation uses <strong>absolute temperature</strong>, which is another term used for temperature expressed in <em>Kelvin</em>. </p>
<p>In your case, the problem wants you to find the temperature in <em>degrees Celsius</em>, so right from the start you know that you must <strong>convert</strong> the temperature from Kelvin to degrees Celsius. </p>
<p>So, rearrange the ideal gas law equation and solve for <mathjax>#T#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies T = (PV)/(nR)#</mathjax></p>
</blockquote>
<p>The units given to you for pressure and volume <strong>match</strong> those used in the expression of the universal gas constant, so plug them into the equation to find</p>
<blockquote>
<p><mathjax>#T = (3.63 color(red)(cancel(color(black)("atm"))) * 68.93color(red)(cancel(color(black)("L"))))/(2.35color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))#</mathjax></p>
<p><mathjax>#T = "1297 K"#</mathjax></p>
</blockquote>
<p>Now you're ready to convert this to degrees Celsius by using </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = T["K"] - 273.15)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#t = "1297 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)color(black)(1020^@"C")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> | If 2.35 moles of an ideal gas has a pressure of 3.63 atm, and a volume of 68.93 L, what is the temperature of the sample in degrees Celsius? | null |
659 | ab6871a2-6ddd-11ea-9927-ccda262736ce | https://socratic.org/questions/5816f6dab72cff0c9182352d | 9.05 × 10^(-5) | start physical_unit 38 39 equilibrium_constant_k none qc_end physical_unit 4 6 1 2 mass qc_end end | [{"type":"physical unit","value":"Kc [OF] this reaction"}] | [{"type":"physical unit","value":"9.05 × 10^(-5)"}] | [{"type":"physical unit","value":"Mass [OF] ammonium chloride solid [=] \\pu{2.83 g}"},{"type":"physical unit","value":"Percent [OF] ammonium chloride solid decomposed [=] \\pu{40%}"}] | <h1 class="questionTitle" itemprop="name">Suppose #"2.83 g"# of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is #K_c# for this reaction?</h1> | null | 9.05 × 10^(-5) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step in any equilibrium problem like this is to write the complete balanced equation:</p>
<p><mathjax>#NH_4Cl(s) harr NH_3(g) + HCl(g)#</mathjax></p>
<p>The form of <mathjax>#K_c#</mathjax> for this reaction (the subscript c means concentration, and the standard concentration is in units of <mathjax>#"mol/L"#</mathjax>) is:</p>
<p><mathjax>#K_c = [NH_3][HCl]#</mathjax></p>
<p>The <mathjax>#NH_4Cl#</mathjax> does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.</p>
<p>The starting amount of reactant is <mathjax>#(2.83g)/(59.49 g/(mol))=0.0476 mol#</mathjax></p>
<p>Because 40% of the reactant decomposed, the equilibrium concentrations are:<br/>
<mathjax>#[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M#</mathjax></p>
<p>The value of <mathjax>#K_c#</mathjax> is therefore <mathjax>#K_c=(0.00951)^2=9.05 times 10^-5#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The equilibrium constant, <mathjax>#K_c=9.05 times 10^-5#</mathjax> is calculated from the equilibrium concentrations of gases. If more <mathjax>#NH_4Cl(s)#</mathjax> is added, nothing happens because the gases are already at their equilibrium concentrations.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step in any equilibrium problem like this is to write the complete balanced equation:</p>
<p><mathjax>#NH_4Cl(s) harr NH_3(g) + HCl(g)#</mathjax></p>
<p>The form of <mathjax>#K_c#</mathjax> for this reaction (the subscript c means concentration, and the standard concentration is in units of <mathjax>#"mol/L"#</mathjax>) is:</p>
<p><mathjax>#K_c = [NH_3][HCl]#</mathjax></p>
<p>The <mathjax>#NH_4Cl#</mathjax> does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.</p>
<p>The starting amount of reactant is <mathjax>#(2.83g)/(59.49 g/(mol))=0.0476 mol#</mathjax></p>
<p>Because 40% of the reactant decomposed, the equilibrium concentrations are:<br/>
<mathjax>#[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M#</mathjax></p>
<p>The value of <mathjax>#K_c#</mathjax> is therefore <mathjax>#K_c=(0.00951)^2=9.05 times 10^-5#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Suppose #"2.83 g"# of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is #K_c# for this reaction?</h1>
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Chuck W.
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Truong-Son N.
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<span class="dateCreated" datetime="2016-11-04T23:38:41" itemprop="dateCreated">
Nov 4, 2016
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<div class="markdown"><p>The equilibrium constant, <mathjax>#K_c=9.05 times 10^-5#</mathjax> is calculated from the equilibrium concentrations of gases. If more <mathjax>#NH_4Cl(s)#</mathjax> is added, nothing happens because the gases are already at their equilibrium concentrations.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first step in any equilibrium problem like this is to write the complete balanced equation:</p>
<p><mathjax>#NH_4Cl(s) harr NH_3(g) + HCl(g)#</mathjax></p>
<p>The form of <mathjax>#K_c#</mathjax> for this reaction (the subscript c means concentration, and the standard concentration is in units of <mathjax>#"mol/L"#</mathjax>) is:</p>
<p><mathjax>#K_c = [NH_3][HCl]#</mathjax></p>
<p>The <mathjax>#NH_4Cl#</mathjax> does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.</p>
<p>The starting amount of reactant is <mathjax>#(2.83g)/(59.49 g/(mol))=0.0476 mol#</mathjax></p>
<p>Because 40% of the reactant decomposed, the equilibrium concentrations are:<br/>
<mathjax>#[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M#</mathjax></p>
<p>The value of <mathjax>#K_c#</mathjax> is therefore <mathjax>#K_c=(0.00951)^2=9.05 times 10^-5#</mathjax></p></div>
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</article> | Suppose #"2.83 g"# of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is #K_c# for this reaction? | null |
660 | acbce1ec-6ddd-11ea-bc62-ccda262736ce | https://socratic.org/questions/how-many-moles-of-n-2-are-produced-by-the-decomposition-of-1-25-mol-of-sodium-az | 1.88 moles | start physical_unit 4 4 mole mol qc_end physical_unit 14 15 11 12 mole qc_end chemical_equation 19 26 qc_end end | [{"type":"physical unit","value":"Mole [OF] N2 [IN] moles"}] | [{"type":"physical unit","value":"1.88 moles"}] | [{"type":"physical unit","value":"Mole [OF] sodium azide [=] \\pu{1.25 mol}"},{"type":"chemical equation","value":"2 NaN3(s) -> 2 Na(s) + 3 N2(g)"}] | <h1 class="questionTitle" itemprop="name">How many moles of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#?</h1> | null | 1.88 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"2NaN"_3("s")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2Na("s")+3N"_2("g")"#</mathjax></p>
<p>Multiply the given moles of sodium azide by <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"NaN"_3"#</mathjax> and <mathjax>#"N"_2"#</mathjax> from the balanced equation, with <mathjax>#"N"_2"#</mathjax> in the numerator. <mathjax>#"2 mol N"_2:#</mathjax><mathjax>#"3 mol N"_2"#</mathjax></p>
<p><mathjax>#1.25cancel"mol NaN"_3xx(3"mol N"_2)/(2cancel"mol NaN"_3)="1.88 mol N"_2"#</mathjax> rounded to three <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>#"1.25 mol NaN"_3"#</mathjax> will produce <mathjax>#"1.88 mol N"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"2NaN"_3("s")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2Na("s")+3N"_2("g")"#</mathjax></p>
<p>Multiply the given moles of sodium azide by <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"NaN"_3"#</mathjax> and <mathjax>#"N"_2"#</mathjax> from the balanced equation, with <mathjax>#"N"_2"#</mathjax> in the numerator. <mathjax>#"2 mol N"_2:#</mathjax><mathjax>#"3 mol N"_2"#</mathjax></p>
<p><mathjax>#1.25cancel"mol NaN"_3xx(3"mol N"_2)/(2cancel"mol NaN"_3)="1.88 mol N"_2"#</mathjax> rounded to three <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 of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#?</h1>
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<div class="markdown"><p><mathjax>#"1.25 mol NaN"_3"#</mathjax> will produce <mathjax>#"1.88 mol N"_2"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"2NaN"_3("s")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2Na("s")+3N"_2("g")"#</mathjax></p>
<p>Multiply the given moles of sodium azide by <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"NaN"_3"#</mathjax> and <mathjax>#"N"_2"#</mathjax> from the balanced equation, with <mathjax>#"N"_2"#</mathjax> in the numerator. <mathjax>#"2 mol N"_2:#</mathjax><mathjax>#"3 mol N"_2"#</mathjax></p>
<p><mathjax>#1.25cancel"mol NaN"_3xx(3"mol N"_2)/(2cancel"mol NaN"_3)="1.88 mol N"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | How many moles of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#? | null |
661 | aa90fe42-6ddd-11ea-8996-ccda262736ce | https://socratic.org/questions/580ff69a11ef6b31b6c2953d | 316.55 K | start physical_unit 12 13 temperature k qc_end physical_unit 12 13 9 10 activation_barrier qc_end physical_unit 12 13 16 17 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the reaction [IN] K"}] | [{"type":"physical unit","value":"316.55 K"}] | [{"type":"physical unit","value":"Activation energy [OF] the reaction [=] \\pu{125 kJ/mol}"},{"type":"physical unit","value":"Temperature1 [OF] the reaction [=] \\pu{312.00 K}"},{"type":"other","value":"Its rate constant is twice as large."}] | <h1 class="questionTitle" itemprop="name">If the activation energy of a certain reaction is #"125 kJ/mol"#, and the reaction is at #"312.00 K"#, at what new temperature would its rate constant be twice as large?</h1> | null | 316.55 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The <strong>Arrhenius equation</strong> gives the relation between temperature and reaction rates:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#k#</mathjax> = the rate constant<br/>
<mathjax>#A#</mathjax> = the pre-exponential factor<br/>
<mathjax>#E_"a"#</mathjax> = the activation energy<br/>
<mathjax>#R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature</p>
<blockquote></blockquote>
<p>If we take the logarithms of both sides, we get</p>
<blockquote>
<blockquote>
<p><mathjax>#lnk = lnA - E_"a"/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p>Finally, if we have the rates at two different temperatures, we can derive the expression</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem, </p>
<p><mathjax>#k_2 = 2k_1#</mathjax><br/>
<mathjax>#k_1 = "0.200 s"^"-1"#</mathjax><br/>
<mathjax>#E_"a" = "125 kJ/mol" = "125 000 J/mol"#</mathjax><br/>
<mathjax>#R = "8.314 J·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T_1 = "312 K"#</mathjax></p>
<blockquote></blockquote>
<p>Now, let's insert the numbers.</p>
<p><mathjax>#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#</mathjax></p>
<p><mathjax>#ln((2color(red)(cancel(color(black)(k_1))))/color(red)(cancel(color(black)(k_1)))) = ("125 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) (1/("312 K") - 1/(T_2 ))#</mathjax></p>
<p><mathjax>#ln2 = ("15 035" color(red)(cancel(color(black)("K"))))/(312 color(red)(cancel(color(black)("K")))) - "15 035 K"/T_2 = 48.19 - "15 035 K"/T_2#</mathjax></p>
<p><mathjax>#"15 035 K"/T_2 = 48.19 - ln2 = 47.50#</mathjax></p>
<p><mathjax>#T_2 = "15 035 K"/47.50 = "317 K"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The rate constant will double at 317 K.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The <strong>Arrhenius equation</strong> gives the relation between temperature and reaction rates:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#k#</mathjax> = the rate constant<br/>
<mathjax>#A#</mathjax> = the pre-exponential factor<br/>
<mathjax>#E_"a"#</mathjax> = the activation energy<br/>
<mathjax>#R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature</p>
<blockquote></blockquote>
<p>If we take the logarithms of both sides, we get</p>
<blockquote>
<blockquote>
<p><mathjax>#lnk = lnA - E_"a"/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p>Finally, if we have the rates at two different temperatures, we can derive the expression</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem, </p>
<p><mathjax>#k_2 = 2k_1#</mathjax><br/>
<mathjax>#k_1 = "0.200 s"^"-1"#</mathjax><br/>
<mathjax>#E_"a" = "125 kJ/mol" = "125 000 J/mol"#</mathjax><br/>
<mathjax>#R = "8.314 J·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T_1 = "312 K"#</mathjax></p>
<blockquote></blockquote>
<p>Now, let's insert the numbers.</p>
<p><mathjax>#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#</mathjax></p>
<p><mathjax>#ln((2color(red)(cancel(color(black)(k_1))))/color(red)(cancel(color(black)(k_1)))) = ("125 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) (1/("312 K") - 1/(T_2 ))#</mathjax></p>
<p><mathjax>#ln2 = ("15 035" color(red)(cancel(color(black)("K"))))/(312 color(red)(cancel(color(black)("K")))) - "15 035 K"/T_2 = 48.19 - "15 035 K"/T_2#</mathjax></p>
<p><mathjax>#"15 035 K"/T_2 = 48.19 - ln2 = 47.50#</mathjax></p>
<p><mathjax>#T_2 = "15 035 K"/47.50 = "317 K"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If the activation energy of a certain reaction is #"125 kJ/mol"#, and the reaction is at #"312.00 K"#, at what new temperature would its rate constant be twice as large?</h1>
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Ernest Z.
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Oct 30, 2016
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<div class="markdown"><p>The rate constant will double at 317 K.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The <strong>Arrhenius equation</strong> gives the relation between temperature and reaction rates:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#k#</mathjax> = the rate constant<br/>
<mathjax>#A#</mathjax> = the pre-exponential factor<br/>
<mathjax>#E_"a"#</mathjax> = the activation energy<br/>
<mathjax>#R#</mathjax> = the Universal Gas Constant<br/>
<mathjax>#T#</mathjax> = the temperature</p>
<blockquote></blockquote>
<p>If we take the logarithms of both sides, we get</p>
<blockquote>
<blockquote>
<p><mathjax>#lnk = lnA - E_"a"/(RT)#</mathjax></p>
</blockquote>
</blockquote>
<p>Finally, if we have the rates at two different temperatures, we can derive the expression</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>In your problem, </p>
<p><mathjax>#k_2 = 2k_1#</mathjax><br/>
<mathjax>#k_1 = "0.200 s"^"-1"#</mathjax><br/>
<mathjax>#E_"a" = "125 kJ/mol" = "125 000 J/mol"#</mathjax><br/>
<mathjax>#R = "8.314 J·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T_1 = "312 K"#</mathjax></p>
<blockquote></blockquote>
<p>Now, let's insert the numbers.</p>
<p><mathjax>#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#</mathjax></p>
<p><mathjax>#ln((2color(red)(cancel(color(black)(k_1))))/color(red)(cancel(color(black)(k_1)))) = ("125 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) (1/("312 K") - 1/(T_2 ))#</mathjax></p>
<p><mathjax>#ln2 = ("15 035" color(red)(cancel(color(black)("K"))))/(312 color(red)(cancel(color(black)("K")))) - "15 035 K"/T_2 = 48.19 - "15 035 K"/T_2#</mathjax></p>
<p><mathjax>#"15 035 K"/T_2 = 48.19 - ln2 = 47.50#</mathjax></p>
<p><mathjax>#T_2 = "15 035 K"/47.50 = "317 K"#</mathjax></p></div>
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Truong-Son N.
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<span class="dateCreated" datetime="2016-10-30T23:02:21" itemprop="dateCreated">
Oct 30, 2016
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<div class="markdown"><p>I got <mathjax>#"316.55 K"#</mathjax>.</p>
<hr/>
<p>This is a fairly straightforward problem; the challenge is to interpret what the variables we are given mean, which equation to use, and how to use it.</p>
<blockquote>
<p><mathjax>#E_a = "125 kJ/mol"#</mathjax><br/>
<mathjax>#k_1 = "0.200 s"^(-1)#</mathjax><br/>
<mathjax>#k_2 = 2k_1#</mathjax><br/>
<mathjax>#T_1 = "312.00 K"#</mathjax><br/>
<mathjax>#T_2 = ?#</mathjax></p>
</blockquote>
<p>Each of these variables can be found in the <strong>Arrhenius equation</strong></p>
<blockquote>
<p><mathjax>#k = Ae^(-E_a"/"RT)#</mathjax>,</p>
</blockquote>
<p>which can be written in the form of an initial <mathjax>#(1)#</mathjax> and final state <mathjax>#(2)#</mathjax>:</p>
<blockquote>
<p><mathjax>#k_1 = Ae^(-E_a"/"RT_1)#</mathjax>,<br/>
<mathjax>#k_2 = Ae^(-E_a"/"RT_2)#</mathjax>,</p>
</blockquote>
<p>noting that for the <strong>same</strong> reaction at two <strong>different</strong> temperatures <mathjax>#T_1#</mathjax> and <mathjax>#T_2#</mathjax>, the activation energy <mathjax>#E_a#</mathjax>, frequency factor <mathjax>#A#</mathjax>, and universal gas constant <mathjax>#R#</mathjax> are <strong>identical</strong>. </p>
<p>Therefore, we can solve for <mathjax>#T_2#</mathjax> by dividing these equations and taking their natural log:</p>
<blockquote>
<p><mathjax>#k_1/k_2 = e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2)#</mathjax></p>
<p><mathjax>#ln(k_1/k_2) = ln(e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2))#</mathjax></p>
<p><mathjax>#= ln(e^(-E_a"/"RT_1)) - ln(e^(-E_a"/"RT_2))#</mathjax></p>
<p><mathjax>#= -(E_a)/(RT_1) + (E_a)/(RT_2)#</mathjax></p>
<p><mathjax>#= -(E_a)/(R)[1/(T_1) - 1/(T_2)]#</mathjax></p>
</blockquote>
<p>which we can recognize as the Arrhenius equation for <em>two different temperatures</em> <mathjax>#T_1#</mathjax> and <mathjax>#T_2#</mathjax>:</p>
<blockquote>
<p><mathjax>#bb(ln(k_1/k_2) = -(E_a)/(R)[1/(T_1) - 1/(T_2)])#</mathjax></p>
</blockquote>
<p>Now, setting <mathjax>#k_2 = 2k_1#</mathjax> as our desired result, we get:</p>
<blockquote>
<p><mathjax>#ln(cancel(k_1)/(2cancel(k_1))) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#</mathjax></p>
<p><mathjax>#-ln(2) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#</mathjax></p>
</blockquote>
<p>Moving <mathjax>#E_a/R#</mathjax> to the other side:</p>
<blockquote>
<p><mathjax>#(Rln(2))/(E_a) = 1/(T_1) - 1/(T_2)#</mathjax></p>
</blockquote>
<p>Subtracting <mathjax>#1/T_1#</mathjax> to the other side and cross-multiplying:</p>
<blockquote>
<p><mathjax>#(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1) = -1/(T_2)#</mathjax></p>
</blockquote>
<p>Reciprocating both sides of the equation:</p>
<blockquote>
<p><mathjax>#=> T_2 = -1/[(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1)]#</mathjax></p>
<p><mathjax>#= -1/[(Rln(2)T_1 - E_a)/(E_aT_1)] = 1/[(E_a - Rln(2)T_1)/(E_aT_1)]#</mathjax></p>
</blockquote>
<p>Therefore:</p>
<blockquote>
<p><mathjax>#color(blue)(T_2) = [E_aT_1]/(E_a - Rln(2)T_1)#</mathjax></p>
<p><mathjax>#= [("125 kJ/mol")("312.00 K")]/(("125 kJ/mol") - ("0.008314472 kJ/mol"cdot"K")ln(2)("312.00 K"))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#color(blue)("316.55 K")#</mathjax></p>
</blockquote>
<p>(Note that we needed five sig figs to get the necessary precision. If we had written <mathjax>#T = "317 K"#</mathjax>, as we would for <mathjax>#3#</mathjax> sig figs, we would have gotten that <mathjax>#k_2#</mathjax> is about <mathjax>#2.14#</mathjax> times <mathjax>#k_1#</mathjax> instead, when we wanted <mathjax>#2#</mathjax> times <mathjax>#k_1#</mathjax>.)</p></div>
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</article> | If the activation energy of a certain reaction is #"125 kJ/mol"#, and the reaction is at #"312.00 K"#, at what new temperature would its rate constant be twice as large? | null |
662 | ac0c45d9-6ddd-11ea-99bc-ccda262736ce | https://socratic.org/questions/how-many-grams-are-there-in-5-0-moles-of-h-2o | 90.08 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] H2O [IN] grams"}] | [{"type":"physical unit","value":"90.08 grams"}] | [{"type":"physical unit","value":"Mole [OF] H2O [=] \\pu{5.0 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams are there in 5.0 moles of #H_2O#?</h1> | null | 90.08 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We first determine the molar mass of <mathjax>#H_2O#</mathjax> bij adding the atomic masses:<br/>
<mathjax>#2xx1.008+1xx16.00=18.016g//mol#</mathjax></p>
<p>Multiply by the number of mols: </p>
<p><mathjax>#5.0cancel(mol)xx18.016g/cancel(mol)=90.080g~~90g#</mathjax></p>
<p>(we have to round to two significant digits).</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#90g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We first determine the molar mass of <mathjax>#H_2O#</mathjax> bij adding the atomic masses:<br/>
<mathjax>#2xx1.008+1xx16.00=18.016g//mol#</mathjax></p>
<p>Multiply by the number of mols: </p>
<p><mathjax>#5.0cancel(mol)xx18.016g/cancel(mol)=90.080g~~90g#</mathjax></p>
<p>(we have to round to two significant digits).</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams are there in 5.0 moles of #H_2O#?</h1>
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<div class="markdown"><p><mathjax>#90g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We first determine the molar mass of <mathjax>#H_2O#</mathjax> bij adding the atomic masses:<br/>
<mathjax>#2xx1.008+1xx16.00=18.016g//mol#</mathjax></p>
<p>Multiply by the number of mols: </p>
<p><mathjax>#5.0cancel(mol)xx18.016g/cancel(mol)=90.080g~~90g#</mathjax></p>
<p>(we have to round to two significant digits).</p></div>
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<div class="markdown"><p>The formula for calculating moles = <mathjax>#"mass"/"molar mass"#</mathjax> <br/>
Therefore = <mathjax>#" molar mass * moles" = mass #</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>molar mass of water = 18g/mol<br/>
moles of water = 5<br/>
<mathjax># 5 "*" 18 = 90#</mathjax><br/>
Therefore there are 90 grams of water </p></div>
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</article> | How many grams are there in 5.0 moles of #H_2O#? | null |
663 | ab804a68-6ddd-11ea-af4e-ccda262736ce | https://socratic.org/questions/how-many-grams-of-nacl-are-required-to-prepare-985-ml-of-0-77-m-nacl-solution | 44.32 grams | start physical_unit 4 4 mass g qc_end physical_unit 14 15 12 13 molarity qc_end physical_unit 14 15 9 10 volume qc_end end | [{"type":"physical unit","value":"Mass [OF] NaCl [IN] grams"}] | [{"type":"physical unit","value":"44.32 grams"}] | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [=] \\pu{0.77 M}"},{"type":"physical unit","value":"Volume [OF] NaCl solution [=] \\pu{985 mL}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #NaCl# are required to prepare 985 mL of 0.77 M #NaCl# solution?</h1> | null | 44.32 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the target solution, so your first step here will be to use this information to figure out how many <strong>moles</strong> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, it must contain. </p>
<p>Once you know that, use the compound's <strong>molar mass</strong> to convert the number of moles to <em>grams</em>. </p>
<p>So, <strong>molarity</strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>per liter of solution</strong>. In your case, a <mathjax>#"0.77 M"#</mathjax> solution will contain <mathjax>#0.77#</mathjax> <strong>moles</strong> of sodium chloride, your solute, in <mathjax>#"1.0 L"#</mathjax> of solution. </p>
<p>Your target solution has a volume of </p>
<blockquote>
<p><mathjax>#985 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.985 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.985 color(red)(cancel(color(black)("L solution"))) * overbrace("0.77 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.77 M")) = "0.75845 moles NaCl"#</mathjax></p>
</blockquote>
<p>Sodium chloride has a <strong>molar mass</strong> of <mathjax>#"58.44 g mol"^(-1)#</mathjax>, which basically means that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>In your case, <mathjax>#0.75845#</mathjax> <strong>moles</strong> of sodium chloride will have a mass of </p>
<blockquote>
<p><mathjax>#0.75845 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"44 g NaCl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the target solution, so your first step here will be to use this information to figure out how many <strong>moles</strong> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, it must contain. </p>
<p>Once you know that, use the compound's <strong>molar mass</strong> to convert the number of moles to <em>grams</em>. </p>
<p>So, <strong>molarity</strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>per liter of solution</strong>. In your case, a <mathjax>#"0.77 M"#</mathjax> solution will contain <mathjax>#0.77#</mathjax> <strong>moles</strong> of sodium chloride, your solute, in <mathjax>#"1.0 L"#</mathjax> of solution. </p>
<p>Your target solution has a volume of </p>
<blockquote>
<p><mathjax>#985 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.985 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.985 color(red)(cancel(color(black)("L solution"))) * overbrace("0.77 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.77 M")) = "0.75845 moles NaCl"#</mathjax></p>
</blockquote>
<p>Sodium chloride has a <strong>molar mass</strong> of <mathjax>#"58.44 g mol"^(-1)#</mathjax>, which basically means that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>In your case, <mathjax>#0.75845#</mathjax> <strong>moles</strong> of sodium chloride will have a mass of </p>
<blockquote>
<p><mathjax>#0.75845 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of #NaCl# are required to prepare 985 mL of 0.77 M #NaCl# solution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"44 g NaCl"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the target solution, so your first step here will be to use this information to figure out how many <strong>moles</strong> of sodium chloride, <mathjax>#"NaCl"#</mathjax>, it must contain. </p>
<p>Once you know that, use the compound's <strong>molar mass</strong> to convert the number of moles to <em>grams</em>. </p>
<p>So, <strong>molarity</strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> <strong>per liter of solution</strong>. In your case, a <mathjax>#"0.77 M"#</mathjax> solution will contain <mathjax>#0.77#</mathjax> <strong>moles</strong> of sodium chloride, your solute, in <mathjax>#"1.0 L"#</mathjax> of solution. </p>
<p>Your target solution has a volume of </p>
<blockquote>
<p><mathjax>#985 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.985 L"#</mathjax></p>
</blockquote>
<p>which means that it will contain </p>
<blockquote>
<p><mathjax>#0.985 color(red)(cancel(color(black)("L solution"))) * overbrace("0.77 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.77 M")) = "0.75845 moles NaCl"#</mathjax></p>
</blockquote>
<p>Sodium chloride has a <strong>molar mass</strong> of <mathjax>#"58.44 g mol"^(-1)#</mathjax>, which basically means that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>In your case, <mathjax>#0.75845#</mathjax> <strong>moles</strong> of sodium chloride will have a mass of </p>
<blockquote>
<p><mathjax>#0.75845 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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</article> | How many grams of #NaCl# are required to prepare 985 mL of 0.77 M #NaCl# solution? | null |
664 | a94c234d-6ddd-11ea-b5d1-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-100-ml-solution-containing-0-0040-g-of-hcl | 3.0 | start physical_unit 8 8 ph none qc_end physical_unit 8 8 6 7 volume qc_end physical_unit 13 13 10 11 mass qc_end end | [{"type":"physical unit","value":"ph [OF] solution"}] | [{"type":"physical unit","value":"3.0"}] | [{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{100 ml}"},{"type":"physical unit","value":"Mass [OF] HCl [=] \\pu{0.0040 g}"}] | <h1 class="questionTitle" itemprop="name">What is the ph of a 100 ml solution containing 0.0040 g of HCl?</h1> | null | 3.0 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to calculate the <em>number of moles</em> of hydrochloric acid present in that sample. To do that, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.0040 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46 color(red)(cancel(color(black)("g")))) = "0.0001097 moles HCl"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates 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>This basically means that every mole of hydrochloric acid dissolved in water will produce <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<blockquote>
<p><mathjax>#"no. of moles of H"_ 3"O"^(+) = "0.0001097 moles"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydronium cations must be calculated for <mathjax>#"1 L"#</mathjax> of solution, so do</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * ("0.0001097 moles H"_3"O"^(+))/(100color(red)(cancel(color(black)("mL solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.001097 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Since that is how many moles of hydronium cations you have in <mathjax>#"1 L"#</mathjax> of solution, you can say that the concentration of the hydronium cations will be</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.001097 M"#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is calculated by taking the negative log of the concentration of hydronium cations</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.001097) = color(darkgreen)(ul(color(black)(3.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <em>sig figs</em> you have for the volume of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 3.0#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to calculate the <em>number of moles</em> of hydrochloric acid present in that sample. To do that, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.0040 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46 color(red)(cancel(color(black)("g")))) = "0.0001097 moles HCl"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates 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>This basically means that every mole of hydrochloric acid dissolved in water will produce <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<blockquote>
<p><mathjax>#"no. of moles of H"_ 3"O"^(+) = "0.0001097 moles"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydronium cations must be calculated for <mathjax>#"1 L"#</mathjax> of solution, so do</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * ("0.0001097 moles H"_3"O"^(+))/(100color(red)(cancel(color(black)("mL solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.001097 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Since that is how many moles of hydronium cations you have in <mathjax>#"1 L"#</mathjax> of solution, you can say that the concentration of the hydronium cations will be</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.001097 M"#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is calculated by taking the negative log of the concentration of hydronium cations</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.001097) = color(darkgreen)(ul(color(black)(3.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <em>sig figs</em> you have for the volume of the solution. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the ph of a 100 ml solution containing 0.0040 g of HCl?</h1>
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<div class="markdown"><p><mathjax>#"pH" = 3.0#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is to calculate the <em>number of moles</em> of hydrochloric acid present in that sample. To do that, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.0040 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46 color(red)(cancel(color(black)("g")))) = "0.0001097 moles HCl"#</mathjax></p>
</blockquote>
<p>Now, hydrochloric acid is a <strong>strong acid</strong>, which means that it dissociates 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>This basically means that every mole of hydrochloric acid dissolved in water will produce <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. </p>
<blockquote>
<p><mathjax>#"no. of moles of H"_ 3"O"^(+) = "0.0001097 moles"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the hydronium cations must be calculated for <mathjax>#"1 L"#</mathjax> of solution, so do</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * ("0.0001097 moles H"_3"O"^(+))/(100color(red)(cancel(color(black)("mL solution"))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= "0.001097 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Since that is how many moles of hydronium cations you have in <mathjax>#"1 L"#</mathjax> of solution, you can say that the concentration of the hydronium cations will be</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.001097 M"#</mathjax></p>
</blockquote>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is calculated by taking the negative log of the concentration of hydronium cations</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find</p>
<blockquote>
<p><mathjax>#"pH" = - log(0.001097) = color(darkgreen)(ul(color(black)(3.0)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <em>decimal place</em>, the number of <em>sig figs</em> you have for the volume of the solution. </p></div>
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</article> | What is the ph of a 100 ml solution containing 0.0040 g of HCl? | null |
665 | ace556d9-6ddd-11ea-8730-ccda262736ce | https://socratic.org/questions/what-is-the-equation-for-solid-potassium-chloride-and-oxygen-gas-combine-to-form | 2 KCl(s) + 3 O2(g) -> 2 KClO3(s) | start chemical_equation qc_end substance 5 7 qc_end substance 9 10 qc_end substance 14 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 KCl(s) + 3 O2(g) -> 2 KClO3(s)"}] | [{"type":"substance name","value":"Solid potassium chloride"},{"type":"substance name","value":"Oxygen gas"},{"type":"substance name","value":"Solid potassium chlorate"}] | <h1 class="questionTitle" itemprop="name">What is the equation for solid potassium chloride and oxygen gas combine to form solid potassium chlorate?</h1> | null | 2 KCl(s) + 3 O2(g) -> 2 KClO3(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a synthesis reaction, also called a combination reaction in which two or more smaller reactants combine to form a single, more complex compound.</p></div>
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<div class="markdown"><p><mathjax>#"2KCl(s) + 3O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2KClO"_3("s")"#</mathjax></p></div>
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<div>
<div class="markdown"><p>This is a synthesis reaction, also called a combination reaction in which two or more smaller reactants combine to form a single, more complex compound.</p></div>
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<div class="markdown"><p><mathjax>#"2KCl(s) + 3O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2KClO"_3("s")"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a synthesis reaction, also called a combination reaction in which two or more smaller reactants combine to form a single, more complex compound.</p></div>
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</article> | What is the equation for solid potassium chloride and oxygen gas combine to form solid potassium chlorate? | null |
666 | a9eecab6-6ddd-11ea-bc1c-ccda262736ce | https://socratic.org/questions/an-organic-compound-has-the-following-percentage-composition-c-12-36-h-2-13-br-8 | C2H4Br2 | start chemical_formula qc_end physical_unit 1 2 21 21 vapour_density qc_end end | [{"type":"other","value":"Chemical Formula [OF] the organic compound [IN] molecular"}] | [{"type":"chemical equation","value":"C2H4Br2"}] | [{"type":"physical unit","value":"Percentage composition [OF] C in the organic compound [=] \\pu{12.36%}"},{"type":"physical unit","value":"Percentage composition [OF] H in the organic compound [=] \\pu{2.13%}"},{"type":"physical unit","value":"Percentage composition [OF] Br in the organic compound [=] \\pu{85%}"},{"type":"physical unit","value":"Vapour density [OF] the organic compound [=] \\pu{94}"}] | <h1 class="questionTitle" itemprop="name">An organic compound has the following percentage composition C = 12.36%, H = 2.13%, Br = 85%. Its vapour density is 94. what is its molecular formula? </h1> | null | C2H4Br2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always, we assume a <mathjax>#100*g#</mathjax> mass of compound, and interrogate its elemental composition in terms of moles.....</p>
<p><mathjax>#"Moles of C"=(12.36*g)/(12.01*g*mol^-1)=1.029*mol#</mathjax></p>
<p><mathjax>#"Moles of Br"=(85*g)/(79.90*g*mol^-1)=1.064*mol#</mathjax></p>
<p><mathjax>#"Moles of H"=(2.13*g)/(1.00794*g*mol^-1)=2.11*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity, to give an empirical formula of <mathjax>#CH_2Br#</mathjax></p>
<p>Now <mathjax>#"vapour density"#</mathjax> is an uncommon measurement, and reports the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of a vapour in relation to that of dihydrogen gas. And thus the molecular mass of the gas is <mathjax>#2xx94*g*mol^-1=188*g*mol^-1#</mathjax>.</p>
<p>As always, <mathjax>#"molecular formula"#</mathjax> is a MULTIPLE of the <mathjax>#"empirical formula"#</mathjax>; so <mathjax>#{12.011+2xx1.00794+79.9}*g*mol^-1xxn=188*g*mol^-1#</mathjax></p>
<p>So, clearly, <mathjax>#n=2#</mathjax>, and our MOLECULAR formula is <mathjax>#C_2H_4Br_2#</mathjax>, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. <mathjax>#Br_2HC-CH_3#</mathjax> or isomeric <mathjax>#BrH_2C-CH_2Br#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>We get a molecular formula of <mathjax>#C_2H_4Br_2#</mathjax>........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always, we assume a <mathjax>#100*g#</mathjax> mass of compound, and interrogate its elemental composition in terms of moles.....</p>
<p><mathjax>#"Moles of C"=(12.36*g)/(12.01*g*mol^-1)=1.029*mol#</mathjax></p>
<p><mathjax>#"Moles of Br"=(85*g)/(79.90*g*mol^-1)=1.064*mol#</mathjax></p>
<p><mathjax>#"Moles of H"=(2.13*g)/(1.00794*g*mol^-1)=2.11*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity, to give an empirical formula of <mathjax>#CH_2Br#</mathjax></p>
<p>Now <mathjax>#"vapour density"#</mathjax> is an uncommon measurement, and reports the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of a vapour in relation to that of dihydrogen gas. And thus the molecular mass of the gas is <mathjax>#2xx94*g*mol^-1=188*g*mol^-1#</mathjax>.</p>
<p>As always, <mathjax>#"molecular formula"#</mathjax> is a MULTIPLE of the <mathjax>#"empirical formula"#</mathjax>; so <mathjax>#{12.011+2xx1.00794+79.9}*g*mol^-1xxn=188*g*mol^-1#</mathjax></p>
<p>So, clearly, <mathjax>#n=2#</mathjax>, and our MOLECULAR formula is <mathjax>#C_2H_4Br_2#</mathjax>, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. <mathjax>#Br_2HC-CH_3#</mathjax> or isomeric <mathjax>#BrH_2C-CH_2Br#</mathjax>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">An organic compound has the following percentage composition C = 12.36%, H = 2.13%, Br = 85%. Its vapour density is 94. what is its molecular formula? </h1>
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anor277
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<div class="markdown"><p>We get a molecular formula of <mathjax>#C_2H_4Br_2#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As always, we assume a <mathjax>#100*g#</mathjax> mass of compound, and interrogate its elemental composition in terms of moles.....</p>
<p><mathjax>#"Moles of C"=(12.36*g)/(12.01*g*mol^-1)=1.029*mol#</mathjax></p>
<p><mathjax>#"Moles of Br"=(85*g)/(79.90*g*mol^-1)=1.064*mol#</mathjax></p>
<p><mathjax>#"Moles of H"=(2.13*g)/(1.00794*g*mol^-1)=2.11*mol#</mathjax></p>
<p>We divide thru by the smallest molar quantity, to give an empirical formula of <mathjax>#CH_2Br#</mathjax></p>
<p>Now <mathjax>#"vapour density"#</mathjax> is an uncommon measurement, and reports the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of a vapour in relation to that of dihydrogen gas. And thus the molecular mass of the gas is <mathjax>#2xx94*g*mol^-1=188*g*mol^-1#</mathjax>.</p>
<p>As always, <mathjax>#"molecular formula"#</mathjax> is a MULTIPLE of the <mathjax>#"empirical formula"#</mathjax>; so <mathjax>#{12.011+2xx1.00794+79.9}*g*mol^-1xxn=188*g*mol^-1#</mathjax></p>
<p>So, clearly, <mathjax>#n=2#</mathjax>, and our MOLECULAR formula is <mathjax>#C_2H_4Br_2#</mathjax>, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. <mathjax>#Br_2HC-CH_3#</mathjax> or isomeric <mathjax>#BrH_2C-CH_2Br#</mathjax>.</p></div>
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</article> | An organic compound has the following percentage composition C = 12.36%, H = 2.13%, Br = 85%. Its vapour density is 94. what is its molecular formula? | null |
667 | ac3d28ca-6ddd-11ea-9252-ccda262736ce | https://socratic.org/questions/a-gas-at-52-c-occupies-6-8-l-at-a-pressure-of-1-5-atm-what-will-the-pressure-be- | 3.28 atm | start physical_unit 1 1 pressure atm qc_end physical_unit 1 1 3 4 temperature 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 temperature qc_end physical_unit 1 1 28 29 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"3.28 atm"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{52 ℃}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{6.8 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.5 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{-43 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{2.2 L}"}] | <h1 class="questionTitle" itemprop="name">A gas at 52°C occupies 6.8 L at a pressure of 1.5 atm. What will the pressure be at -43°C if the volume is reduced to 2.2 L?</h1> | null | 3.28 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_2=(P_1V_1T_2)/(T_1V_2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.5*atmxx6.8*Lxx230*K)/(325*Kxx2.2*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*atm#</mathjax></p>
<p>Note that we had to convert all temperature readings to the absolute scale, as we typically do for gas problems.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_2V_2)/T_2#</mathjax>. Thus <mathjax>#P_2~=3*atm#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_2=(P_1V_1T_2)/(T_1V_2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.5*atmxx6.8*Lxx230*K)/(325*Kxx2.2*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*atm#</mathjax></p>
<p>Note that we had to convert all temperature readings to the absolute scale, as we typically do for gas problems.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas at 52°C occupies 6.8 L at a pressure of 1.5 atm. What will the pressure be at -43°C if the volume is reduced to 2.2 L?</h1>
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anor277
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> holds that <mathjax>#(P_1V_1)/T_1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(P_2V_2)/T_2#</mathjax>. Thus <mathjax>#P_2~=3*atm#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#P_2=(P_1V_1T_2)/(T_1V_2)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.5*atmxx6.8*Lxx230*K)/(325*Kxx2.2*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*atm#</mathjax></p>
<p>Note that we had to convert all temperature readings to the absolute scale, as we typically do for gas problems.</p></div>
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</article> | A gas at 52°C occupies 6.8 L at a pressure of 1.5 atm. What will the pressure be at -43°C if the volume is reduced to 2.2 L? | null |
668 | aa15b1a6-6ddd-11ea-b637-ccda262736ce | https://socratic.org/questions/what-mass-of-aluminum-is-required-if-40-0-grams-of-iron-iii-oxide-is-to-be-compl | 13.50 grams | start physical_unit 3 3 mass g qc_end physical_unit 10 12 7 8 mass qc_end chemical_equation 21 29 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] aluminum [IN] grams"}] | [{"type":"physical unit","value":"13.50 grams"}] | [{"type":"physical unit","value":"Mass [OF] iron (III) oxide [=] \\pu{40.0 grams}"},{"type":"chemical equation","value":"Fe2O3(s) + 2 Al(s) -> 2 Fe(l) + Al2O3(s)"},{"type":"other","value":"Completely consume."}] | <h1 class="questionTitle" itemprop="name">What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?</h1> | null | 13.50 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n_(Fe_2O_3)=m/(m_r)=40/(2xx55.8+3xx16)=0.25mol#</mathjax>.</p>
<p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Hence 1 mole iron(III)oxide requires 2 moles aluminium.<br/>
Therefore by ratio and proportion, <mathjax>#0.25 mol#</mathjax> iron(III)oxide requires <mathjax>#0.50#</mathjax> moles aluminium.<br/>
This corresponds to a mass of :</p>
<p><mathjax>#m_(Al)=nxxM_r=0.50xx27=13.5g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#13.5g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n_(Fe_2O_3)=m/(m_r)=40/(2xx55.8+3xx16)=0.25mol#</mathjax>.</p>
<p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Hence 1 mole iron(III)oxide requires 2 moles aluminium.<br/>
Therefore by ratio and proportion, <mathjax>#0.25 mol#</mathjax> iron(III)oxide requires <mathjax>#0.50#</mathjax> moles aluminium.<br/>
This corresponds to a mass of :</p>
<p><mathjax>#m_(Al)=nxxM_r=0.50xx27=13.5g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?</h1>
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<div class="markdown"><p><mathjax>#13.5g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n_(Fe_2O_3)=m/(m_r)=40/(2xx55.8+3xx16)=0.25mol#</mathjax>.</p>
<p>The balanced chemical equation represents <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio in which chemicals combine.</p>
<p>Hence 1 mole iron(III)oxide requires 2 moles aluminium.<br/>
Therefore by ratio and proportion, <mathjax>#0.25 mol#</mathjax> iron(III)oxide requires <mathjax>#0.50#</mathjax> moles aluminium.<br/>
This corresponds to a mass of :</p>
<p><mathjax>#m_(Al)=nxxM_r=0.50xx27=13.5g#</mathjax>.</p></div>
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</article> | What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#? | null |
669 | aab64ba7-6ddd-11ea-a1bc-ccda262736ce | https://socratic.org/questions/a-gas-has-a-volume-of-39-liters-at-stp-what-will-its-volume-be-at-4-atm-and-25-c | 10.64 liters | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 19 20 temperature qc_end physical_unit 1 1 16 17 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"10.64 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{39 liters}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{4 atm}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 39 liters at STP. What will its volume be at 4 atm and 25°c?</h1> | null | 10.64 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new volume of a gas after it is subdued to changes in pressure and temperature.</p>
<p>To solve this problem, we can use the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></em>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>This equation allows us to find how one of these variables changes if the other two both change by known amounts.</p>
<p>The first thing we must do is convert the temperature from degrees Celsius to Kelvin:</p>
<p><mathjax>#"K" = ""^"o""C" + 273 = 25^"o""C" + 273 = color(red)(298"K"#</mathjax></p>
<p>Now, we have all our variables needed to solve this problem [remember that for <em>standard temperature and pressure</em>, <mathjax>#P = 1#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#T= 273#</mathjax> <mathjax>#"K"(0^"o""C")#</mathjax>]. Let's rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>, and plug in our known variables:</p>
<p><mathjax>#V_2 = (P_1V_1T_2)/(P_2T_1) = ((1cancel("atm"))(39"L")(color(red)(298cancel("K"))))/((4cancel("atm"))(273cancel("K"))) = color(blue)(11"L")#</mathjax></p>
<p>The new volume of the gas after the changes in temperature and pressure will thus be <mathjax>#11"L"#</mathjax>, rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (the number given in the problem).</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#11"L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new volume of a gas after it is subdued to changes in pressure and temperature.</p>
<p>To solve this problem, we can use the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></em>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>This equation allows us to find how one of these variables changes if the other two both change by known amounts.</p>
<p>The first thing we must do is convert the temperature from degrees Celsius to Kelvin:</p>
<p><mathjax>#"K" = ""^"o""C" + 273 = 25^"o""C" + 273 = color(red)(298"K"#</mathjax></p>
<p>Now, we have all our variables needed to solve this problem [remember that for <em>standard temperature and pressure</em>, <mathjax>#P = 1#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#T= 273#</mathjax> <mathjax>#"K"(0^"o""C")#</mathjax>]. Let's rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>, and plug in our known variables:</p>
<p><mathjax>#V_2 = (P_1V_1T_2)/(P_2T_1) = ((1cancel("atm"))(39"L")(color(red)(298cancel("K"))))/((4cancel("atm"))(273cancel("K"))) = color(blue)(11"L")#</mathjax></p>
<p>The new volume of the gas after the changes in temperature and pressure will thus be <mathjax>#11"L"#</mathjax>, rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (the number given in the problem).</p></div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 39 liters at STP. What will its volume be at 4 atm and 25°c?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#11"L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the new volume of a gas after it is subdued to changes in pressure and temperature.</p>
<p>To solve this problem, we can use the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a></em>:</p>
<p><mathjax>#(P_1V_1)/(T_1) = (P_2V_2)/(T_2)#</mathjax></p>
<p>This equation allows us to find how one of these variables changes if the other two both change by known amounts.</p>
<p>The first thing we must do is convert the temperature from degrees Celsius to Kelvin:</p>
<p><mathjax>#"K" = ""^"o""C" + 273 = 25^"o""C" + 273 = color(red)(298"K"#</mathjax></p>
<p>Now, we have all our variables needed to solve this problem [remember that for <em>standard temperature and pressure</em>, <mathjax>#P = 1#</mathjax> <mathjax>#"atm"#</mathjax> and <mathjax>#T= 273#</mathjax> <mathjax>#"K"(0^"o""C")#</mathjax>]. Let's rearrange this equation to solve for the final volume, <mathjax>#V_2#</mathjax>, and plug in our known variables:</p>
<p><mathjax>#V_2 = (P_1V_1T_2)/(P_2T_1) = ((1cancel("atm"))(39"L")(color(red)(298cancel("K"))))/((4cancel("atm"))(273cancel("K"))) = color(blue)(11"L")#</mathjax></p>
<p>The new volume of the gas after the changes in temperature and pressure will thus be <mathjax>#11"L"#</mathjax>, rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> (the number given in the problem).</p></div>
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</article> | A gas has a volume of 39 liters at STP. What will its volume be at 4 atm and 25°c? | null |
670 | ad1ffa71-6ddd-11ea-bffc-ccda262736ce | https://socratic.org/questions/a-flask-contains-85-5-g-c-12h-22o-11-sucrose-in-1-00-liters-of-solution-what-is- | 0.25 M | start physical_unit 5 5 molarity mol/l qc_end physical_unit 5 5 3 4 mass qc_end physical_unit 11 11 8 9 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] C12H22O11 solution [IN] M"}] | [{"type":"physical unit","value":"0.25 M"}] | [{"type":"physical unit","value":"Mass [OF] C12H22O11 [=] \\pu{85.5 g}"},{"type":"physical unit","value":"Volume [OF] C12H22O11 solution [=] \\pu{1.00 liter}"}] | <h1 class="questionTitle" itemprop="name">A flask contains 85.5 g #C_12H_22O_11# (sucrose) in 1.00 liters of solution. What is the molarity?</h1> | null | 0.25 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar concentration</strong> (<strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>) of a sucrose solution, given the volume of solution and mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>The equation for molarity is</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p>We're given the volume is <mathjax>#1.00#</mathjax> <mathjax>#"L"#</mathjax>, so we just need to find the <em>moles</em> of solute (we're given grams) via the <em>molar mass</em> of sucrose (<mathjax>#342.30#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#85.5cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.250#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>Therefore, we have</p>
<p><mathjax>#"molarity" = (color(red)(0.250color(white)(l)"mol sucrose"))/(1.00color(white)(l)"L soln") = color(blue)(0.250M#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"molarity" = 0.250M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar concentration</strong> (<strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>) of a sucrose solution, given the volume of solution and mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>The equation for molarity is</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p>We're given the volume is <mathjax>#1.00#</mathjax> <mathjax>#"L"#</mathjax>, so we just need to find the <em>moles</em> of solute (we're given grams) via the <em>molar mass</em> of sucrose (<mathjax>#342.30#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#85.5cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.250#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>Therefore, we have</p>
<p><mathjax>#"molarity" = (color(red)(0.250color(white)(l)"mol sucrose"))/(1.00color(white)(l)"L soln") = color(blue)(0.250M#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A flask contains 85.5 g #C_12H_22O_11# (sucrose) in 1.00 liters of solution. What is the molarity?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#"molarity" = 0.250M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar concentration</strong> (<strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong>) of a sucrose solution, given the volume of solution and mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p>
<p>The equation for molarity is</p>
<p><mathjax>#"molarity" = "mol solute"/"L soln"#</mathjax></p>
<p>We're given the volume is <mathjax>#1.00#</mathjax> <mathjax>#"L"#</mathjax>, so we just need to find the <em>moles</em> of solute (we're given grams) via the <em>molar mass</em> of sucrose (<mathjax>#342.30#</mathjax> <mathjax>#"g/mol"#</mathjax>):</p>
<p><mathjax>#85.5cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.250#</mathjax> <mathjax>#color(red)("mol sucrose"#</mathjax></p>
<p>Therefore, we have</p>
<p><mathjax>#"molarity" = (color(red)(0.250color(white)(l)"mol sucrose"))/(1.00color(white)(l)"L soln") = color(blue)(0.250M#</mathjax></p></div>
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</article> | A flask contains 85.5 g #C_12H_22O_11# (sucrose) in 1.00 liters of solution. What is the molarity? | null |
671 | a9f3a80a-6ddd-11ea-a0b1-ccda262736ce | https://socratic.org/questions/if-22-15ml-of-0-100m-sulfuric-acid-is-required-to-neutralize-10-0ml-of-lithium-h | 0.44 M | start physical_unit 24 25 molarity mol/l qc_end physical_unit 17 17 1 2 volume qc_end physical_unit 15 17 12 13 volume qc_end chemical_equation 32 41 qc_end end | [{"type":"physical unit","value":"Molar concentration [OF] the base [IN] M"}] | [{"type":"physical unit","value":"0.44 M"}] | [{"type":"physical unit","value":"Molar concentration [OF] sulfuric acid solution [=] \\pu{0.100 M}"},{"type":"physical unit","value":"Volume [OF] sulfuric acid solution [=] \\pu{22.15 mL}"},{"type":"physical unit","value":"Volume [OF] lithium hydroxide solution [=] \\pu{10.0 mL}"},{"type":"chemical equation","value":"H2SO4 (aq) + 2 LiOH(aq) -> Li2SO4 + 2 H2O"}] | <h1 class="questionTitle" itemprop="name">If 22.15mL of 0.100M sulfuric acid is required to neutralize 10.0mL of lithium hydroxide solution, what is the molar concentration of the base?
How would I approach this problem?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#H_2SO_4#</mathjax> (aq) + 2LiOH(aq) <mathjax>#rarr#</mathjax> <mathjax>#Li_2SO_4#</mathjax> + <mathjax>#2H_2O#</mathjax> </p></div>
</h2>
</div>
</div> | 0.44 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy when it comes to <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reactions</a> is to use the <em>balanced chemical equation</em> to help you determine how many moles of each reactant must be present in solution in order for <strong>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> to take place. </p>
<p>In your case, the balanced chemical equation looks like this </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_text(4(aq]) + color(red)(2)"LiOH"_text((aq]) -> "Li"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sulfuric acid and lithium hydroxide. This tells you that the reaction will <strong>always</strong> consume <mathjax>#color(red)(2)#</mathjax> <strong>times more moles</strong> of lithium hydroxide than moles of sulfuric acid. </p>
<p>Now, the problem provides you with the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the sulfuric acid solution, which is one way of providing with the <strong>number of moles</strong> of sulfuric acid, since </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>If you determine the number of moles of sulfuric acid present in the initial solution, you can sue the aforementioned mole ratio to figure out how many moles of lithium hydroxide would be needed in order to make sure that <strong>all the moles</strong> of acid are neutralized. </p>
<p>So, the number of moles of sulfuric acid will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.100 M" * 22.15 * 10^(-3)"L" = "0.002215 moles H"_2"SO"_4#</mathjax> </p>
</blockquote>
<p>This means that you'd need </p>
<blockquote>
<p><mathjax>#0.002215 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)" moles LiOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00443 moles LiOH"#</mathjax></p>
</blockquote>
<p>Now all you need to do is use the volume of the solution to figure out the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> </p>
<blockquote>
<p><mathjax>#c = "0.00443 moles"/(10.0 * 10^(-3)"L") = color(green)("0.443 M")#</mathjax></p>
</blockquote>
<p>So remember, chemical reactions are all about <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a>. As far as <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are concerned, start by focusing on moles, the use molarity and volume to help you find what you need. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.443 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy when it comes to <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reactions</a> is to use the <em>balanced chemical equation</em> to help you determine how many moles of each reactant must be present in solution in order for <strong>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> to take place. </p>
<p>In your case, the balanced chemical equation looks like this </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_text(4(aq]) + color(red)(2)"LiOH"_text((aq]) -> "Li"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sulfuric acid and lithium hydroxide. This tells you that the reaction will <strong>always</strong> consume <mathjax>#color(red)(2)#</mathjax> <strong>times more moles</strong> of lithium hydroxide than moles of sulfuric acid. </p>
<p>Now, the problem provides you with the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the sulfuric acid solution, which is one way of providing with the <strong>number of moles</strong> of sulfuric acid, since </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>If you determine the number of moles of sulfuric acid present in the initial solution, you can sue the aforementioned mole ratio to figure out how many moles of lithium hydroxide would be needed in order to make sure that <strong>all the moles</strong> of acid are neutralized. </p>
<p>So, the number of moles of sulfuric acid will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.100 M" * 22.15 * 10^(-3)"L" = "0.002215 moles H"_2"SO"_4#</mathjax> </p>
</blockquote>
<p>This means that you'd need </p>
<blockquote>
<p><mathjax>#0.002215 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)" moles LiOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00443 moles LiOH"#</mathjax></p>
</blockquote>
<p>Now all you need to do is use the volume of the solution to figure out the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> </p>
<blockquote>
<p><mathjax>#c = "0.00443 moles"/(10.0 * 10^(-3)"L") = color(green)("0.443 M")#</mathjax></p>
</blockquote>
<p>So remember, chemical reactions are all about <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a>. As far as <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are concerned, start by focusing on moles, the use molarity and volume to help you find what you need. </p></div>
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<h1 class="questionTitle" itemprop="name">If 22.15mL of 0.100M sulfuric acid is required to neutralize 10.0mL of lithium hydroxide solution, what is the molar concentration of the base?
How would I approach this problem?</h1>
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<div class="markdown"><p><mathjax>#H_2SO_4#</mathjax> (aq) + 2LiOH(aq) <mathjax>#rarr#</mathjax> <mathjax>#Li_2SO_4#</mathjax> + <mathjax>#2H_2O#</mathjax> </p></div>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.443 M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy when it comes to <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization reactions</a> is to use the <em>balanced chemical equation</em> to help you determine how many moles of each reactant must be present in solution in order for <strong>complete <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> to take place. </p>
<p>In your case, the balanced chemical equation looks like this </p>
<blockquote>
<p><mathjax>#"H"_2"SO"_text(4(aq]) + color(red)(2)"LiOH"_text((aq]) -> "Li"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#</mathjax></p>
</blockquote>
<p>Notice that you have a <mathjax>#1:color(red)(2)#</mathjax> <a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sulfuric acid and lithium hydroxide. This tells you that the reaction will <strong>always</strong> consume <mathjax>#color(red)(2)#</mathjax> <strong>times more moles</strong> of lithium hydroxide than moles of sulfuric acid. </p>
<p>Now, the problem provides you with the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the sulfuric acid solution, which is one way of providing with the <strong>number of moles</strong> of sulfuric acid, since </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>If you determine the number of moles of sulfuric acid present in the initial solution, you can sue the aforementioned mole ratio to figure out how many moles of lithium hydroxide would be needed in order to make sure that <strong>all the moles</strong> of acid are neutralized. </p>
<p>So, the number of moles of sulfuric acid will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = "0.100 M" * 22.15 * 10^(-3)"L" = "0.002215 moles H"_2"SO"_4#</mathjax> </p>
</blockquote>
<p>This means that you'd need </p>
<blockquote>
<p><mathjax>#0.002215 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(red)(2)" moles LiOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.00443 moles LiOH"#</mathjax></p>
</blockquote>
<p>Now all you need to do is use the volume of the solution to figure out the solution's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> </p>
<blockquote>
<p><mathjax>#c = "0.00443 moles"/(10.0 * 10^(-3)"L") = color(green)("0.443 M")#</mathjax></p>
</blockquote>
<p>So remember, chemical reactions are all about <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a>. As far as <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are concerned, start by focusing on moles, the use molarity and volume to help you find what you need. </p></div>
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</article> | If 22.15mL of 0.100M sulfuric acid is required to neutralize 10.0mL of lithium hydroxide solution, what is the molar concentration of the base?
How would I approach this problem? |
#H_2SO_4# (aq) + 2LiOH(aq) #rarr# #Li_2SO_4# + #2H_2O#
|
672 | aa33a0cb-6ddd-11ea-9107-ccda262736ce | https://socratic.org/questions/a-patient-s-urine-sample-gave-a-ph-reading-of-8-2-what-is-the-value-of-h-3o-in-t | 6.31 × 10^(-9) M | start physical_unit 2 3 [h3o+] mol/l qc_end physical_unit 2 3 9 9 ph qc_end end | [{"type":"physical unit","value":"Value of [H3O+] [OF] urine sample [IN] M"}] | [{"type":"physical unit","value":"6.31 × 10^(-9) M"}] | [{"type":"physical unit","value":"pH [OF] urine sample [=] \\pu{8.2}"}] | <h1 class="questionTitle" itemprop="name">A patient's urine sample gave a pH reading of 8.2. What is the value of #[H_3O^+]# in this?</h1> | null | 6.31 × 10^(-9) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward problem that can be solved by using the equation </p>
<blockquote>
<p><mathjax>#color(blue)(["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-8.2) = 6.31 * 10^(-9)"M"#</mathjax></p>
</blockquote>
<p>Interestingly enough, the normal <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> range for urine lies between <mathjax>#4.8#</mathjax> and <mathjax>#8.0#</mathjax>, so your patient's urine sample could be an indicator of a metabolic problem. </p>
<p>By comparison, the normal pH range for blood is <strong>much</strong> less susceptible to variation. For a healthy individual, blood pH ranges from <mathjax>#7.35#</mathjax> to <mathjax>#7.45#</mathjax>.</p>
<p>This pH range is controlled by the <strong>carbonic acid/bicarbonate buffer</strong>, one of the body's main <a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer systems</a>.</p>
<p><img alt="https://fmss12uchemd.wordpress.com/2013/05/06/carbonic-acid-bicarbonate-buffer-system/" src="https://useruploads.socratic.org/g1FqwwDyQryrxV5s4QEp_screen-shot-2013-05-07-at-5-04-55-pm.png"/> </p></div>
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<div>
<div class="markdown"><p><mathjax>#6.31 * 10^(-9)"M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward problem that can be solved by using the equation </p>
<blockquote>
<p><mathjax>#color(blue)(["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-8.2) = 6.31 * 10^(-9)"M"#</mathjax></p>
</blockquote>
<p>Interestingly enough, the normal <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> range for urine lies between <mathjax>#4.8#</mathjax> and <mathjax>#8.0#</mathjax>, so your patient's urine sample could be an indicator of a metabolic problem. </p>
<p>By comparison, the normal pH range for blood is <strong>much</strong> less susceptible to variation. For a healthy individual, blood pH ranges from <mathjax>#7.35#</mathjax> to <mathjax>#7.45#</mathjax>.</p>
<p>This pH range is controlled by the <strong>carbonic acid/bicarbonate buffer</strong>, one of the body's main <a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer systems</a>.</p>
<p><img alt="https://fmss12uchemd.wordpress.com/2013/05/06/carbonic-acid-bicarbonate-buffer-system/" src="https://useruploads.socratic.org/g1FqwwDyQryrxV5s4QEp_screen-shot-2013-05-07-at-5-04-55-pm.png"/> </p></div>
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<h1 class="questionTitle" itemprop="name">A patient's urine sample gave a pH reading of 8.2. What is the value of #[H_3O^+]# in this?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#6.31 * 10^(-9)"M"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a pretty straightforward problem that can be solved by using the equation </p>
<blockquote>
<p><mathjax>#color(blue)(["H"_3"O"^(+)] = 10^(-"pH"))#</mathjax></p>
</blockquote>
<p>In your case, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, will be equal to </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-8.2) = 6.31 * 10^(-9)"M"#</mathjax></p>
</blockquote>
<p>Interestingly enough, the normal <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> range for urine lies between <mathjax>#4.8#</mathjax> and <mathjax>#8.0#</mathjax>, so your patient's urine sample could be an indicator of a metabolic problem. </p>
<p>By comparison, the normal pH range for blood is <strong>much</strong> less susceptible to variation. For a healthy individual, blood pH ranges from <mathjax>#7.35#</mathjax> to <mathjax>#7.45#</mathjax>.</p>
<p>This pH range is controlled by the <strong>carbonic acid/bicarbonate buffer</strong>, one of the body's main <a href="https://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer systems</a>.</p>
<p><img alt="https://fmss12uchemd.wordpress.com/2013/05/06/carbonic-acid-bicarbonate-buffer-system/" src="https://useruploads.socratic.org/g1FqwwDyQryrxV5s4QEp_screen-shot-2013-05-07-at-5-04-55-pm.png"/> </p></div>
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</article> | A patient's urine sample gave a pH reading of 8.2. What is the value of #[H_3O^+]# in this? | null |
673 | abe652a1-6ddd-11ea-b525-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-h-in-a-0-121-m-hcl-solution | 0.12 M | start physical_unit 5 5 concentration mol/l qc_end physical_unit 10 11 8 9 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] H+ [IN] M"}] | [{"type":"physical unit","value":"0.12 M"}] | [{"type":"physical unit","value":"Concentration [OF] HCl solution [=] \\pu{0.121 M}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of H+ in a 0.121 M HCl solution?</h1> | null | 0.12 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As <mathjax>#"HCl"#</mathjax> is a strong acid and hence a strong electrolyte, it will dissociate as </p>
<p><mathjax>#"HCl" rarr "H"^+ + "Cl"^-#</mathjax></p>
<p>So, the concentration of <mathjax>#"H"^+#</mathjax> will be <mathjax>#"0.121 M"#</mathjax> (the same as <mathjax>#"HCl"#</mathjax>). </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The concentration of <mathjax>#"H"^+#</mathjax> will be <mathjax>#"0.121 M"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As <mathjax>#"HCl"#</mathjax> is a strong acid and hence a strong electrolyte, it will dissociate as </p>
<p><mathjax>#"HCl" rarr "H"^+ + "Cl"^-#</mathjax></p>
<p>So, the concentration of <mathjax>#"H"^+#</mathjax> will be <mathjax>#"0.121 M"#</mathjax> (the same as <mathjax>#"HCl"#</mathjax>). </p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of H+ in a 0.121 M HCl solution?</h1>
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Ratnesh Bhosale
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Stefan V.
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<div class="markdown"><p>The concentration of <mathjax>#"H"^+#</mathjax> will be <mathjax>#"0.121 M"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As <mathjax>#"HCl"#</mathjax> is a strong acid and hence a strong electrolyte, it will dissociate as </p>
<p><mathjax>#"HCl" rarr "H"^+ + "Cl"^-#</mathjax></p>
<p>So, the concentration of <mathjax>#"H"^+#</mathjax> will be <mathjax>#"0.121 M"#</mathjax> (the same as <mathjax>#"HCl"#</mathjax>). </p></div>
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</article> | What is the concentration of H+ in a 0.121 M HCl solution? | null |
674 | a920b47e-6ddd-11ea-b993-ccda262736ce | https://socratic.org/questions/5977517911ef6b7fd22ecab3 | 7920 mL | start physical_unit 7 7 volume ml qc_end physical_unit 7 7 11 12 volume qc_end physical_unit 7 7 14 15 temperature qc_end physical_unit 7 7 17 18 pressure qc_end physical_unit 7 7 24 25 temperature qc_end physical_unit 7 7 31 32 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"7920 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1000 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{396 K}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{2 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{200 K}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{0.5 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the volume occupied by a gas that has a #1000*mL# at #396*K# under #2*atm# pressure, that is cooled to #200*K#, and the pressure decreased to #0.5*atm#?</h1> | null | 7920 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, for a given quantity of gas according to the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>.....</p>
<p>We solve for <mathjax>#V_2=(P_1V_1T_2)/(P_2T_1)=??*mL#</mathjax></p>
<p><mathjax>#(2*atm*1000*mL*396*K)/(0.5*atm*200*K)=??*mL#</mathjax>....</p>
<p>Clearly we get an answer with units of volume as required. Do you agree? And remember <mathjax>#1*L-=1000*mL#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#V_2~=8000*mL#</mathjax> as required........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>, for a given quantity of gas according to the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>.....</p>
<p>We solve for <mathjax>#V_2=(P_1V_1T_2)/(P_2T_1)=??*mL#</mathjax></p>
<p><mathjax>#(2*atm*1000*mL*396*K)/(0.5*atm*200*K)=??*mL#</mathjax>....</p>
<p>Clearly we get an answer with units of volume as required. Do you agree? And remember <mathjax>#1*L-=1000*mL#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume occupied by a gas that has a #1000*mL# at #396*K# under #2*atm# pressure, that is cooled to #200*K#, and the pressure decreased to #0.5*atm#?</h1>
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<div class="markdown"><p><mathjax>#V_2~=8000*mL#</mathjax> as required........</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>, for a given quantity of gas according to the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a>.....</p>
<p>We solve for <mathjax>#V_2=(P_1V_1T_2)/(P_2T_1)=??*mL#</mathjax></p>
<p><mathjax>#(2*atm*1000*mL*396*K)/(0.5*atm*200*K)=??*mL#</mathjax>....</p>
<p>Clearly we get an answer with units of volume as required. Do you agree? And remember <mathjax>#1*L-=1000*mL#</mathjax>.</p></div>
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</article> | What is the volume occupied by a gas that has a #1000*mL# at #396*K# under #2*atm# pressure, that is cooled to #200*K#, and the pressure decreased to #0.5*atm#? | null |
675 | a978739e-6ddd-11ea-943b-ccda262736ce | https://socratic.org/questions/how-many-moles-of-gas-occupy-a-3-45-l-container-at-a-pressure-of-150-kpa-and-a-t | 0.03 moles | start physical_unit 4 4 mole mol qc_end physical_unit 9 9 7 8 volume qc_end physical_unit 4 4 14 15 pressure qc_end physical_unit 4 4 20 21 temperature qc_end end | [{"type":"physical unit","value":"Mole [OF] gas [IN] moles"}] | [{"type":"physical unit","value":"0.03 moles"}] | [{"type":"physical unit","value":"Volume [OF] container [=] \\pu{3.45 L}"},{"type":"physical unit","value":"Pressure [OF] gas [=] \\pu{150 kPa}"},{"type":"physical unit","value":"Temperature [OF] gas [=] \\pu{45.6 ℃}"}] | <h1 class="questionTitle" itemprop="name">How many moles of gas occupy a 3.45-L container at a pressure of 150 kPa and a temperature of 45.6°C?</h1> | null | 0.03 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> equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Now, it's important to realize that the units you have for the volume, pressure, and temperature of the gas <strong>must match</strong> the unit used in the expression of the universal gas constant.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaacolor(black)("kilopascals " ["kPa"])aaaaaaaaacolor(black)("atmospheres " ["atm"])aaa)color(red)(xx)#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>This means that in order to use the ideal gas law equation with the given value for the universal gas constant, you must convert the pressure and the temperature of the gas by using the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 atm = 760 kPa")))" "#</mathjax> and <mathjax>#" " color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law 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 -- <strong>do not</strong> forget the conversion factors!</p>
<blockquote>
<p><mathjax>#n = ( 150/760 color(red)(cancel(color(black)("atm"))) * 3.45 color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (45.6 + 273.15)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("0.026 moles")))#</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 pressure of the gas. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.026 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> equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Now, it's important to realize that the units you have for the volume, pressure, and temperature of the gas <strong>must match</strong> the unit used in the expression of the universal gas constant.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaacolor(black)("kilopascals " ["kPa"])aaaaaaaaacolor(black)("atmospheres " ["atm"])aaa)color(red)(xx)#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>This means that in order to use the ideal gas law equation with the given value for the universal gas constant, you must convert the pressure and the temperature of the gas by using the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 atm = 760 kPa")))" "#</mathjax> and <mathjax>#" " color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law 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 -- <strong>do not</strong> forget the conversion factors!</p>
<blockquote>
<p><mathjax>#n = ( 150/760 color(red)(cancel(color(black)("atm"))) * 3.45 color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (45.6 + 273.15)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("0.026 moles")))#</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 pressure of the gas. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of gas occupy a 3.45-L container at a pressure of 150 kPa and a temperature of 45.6°C?</h1>
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Stefan V.
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Feb 16, 2017
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<div class="markdown"><p><mathjax>#"0.026 moles"#</mathjax></p></div>
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<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> equation</strong>, which looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>Now, it's important to realize that the units you have for the volume, pressure, and temperature of the gas <strong>must match</strong> the unit used in the expression of the universal gas constant.</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaacolor(black)("kilopascals " ["kPa"])aaaaaaaaacolor(black)("atmospheres " ["atm"])aaa)color(red)(xx)#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>This means that in order to use the ideal gas law equation with the given value for the universal gas constant, you must convert the pressure and the temperature of the gas by using the conversion factors</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("1 atm = 760 kPa")))" "#</mathjax> and <mathjax>#" " color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Rearrange the ideal gas law 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 -- <strong>do not</strong> forget the conversion factors!</p>
<blockquote>
<p><mathjax>#n = ( 150/760 color(red)(cancel(color(black)("atm"))) * 3.45 color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (45.6 + 273.15)color(red)(cancel(color(black)("K"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("0.026 moles")))#</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 pressure of the gas. </p></div>
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</article> | How many moles of gas occupy a 3.45-L container at a pressure of 150 kPa and a temperature of 45.6°C? | null |
676 | aab5eb54-6ddd-11ea-a644-ccda262736ce | https://socratic.org/questions/how-much-energy-is-generated-from-freezing-2-5-g-of-water | 200 cal | start physical_unit 10 10 heat_energy cal qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Generated energy [OF] water [IN] cal"}] | [{"type":"physical unit","value":"200 cal"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{2.5 g}"}] | <h1 class="questionTitle" itemprop="name">How much energy is generated from freezing 2.5 g of water?</h1> | null | 200 cal | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>(latent) heat of fusion</strong> of a material, is either one of:</p>
<p>1) the heat required to melt the material without temperature change or</p>
<p>2) the heat removed from the material to freeze it without temperature change.</p>
<p>For water this latent heat is <mathjax>#80" cal/g"#</mathjax>. Multiply this by <mathjax>#2.5" g"#</mathjax> to get <mathjax>#200" cal"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#200#</mathjax> calories.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>(latent) heat of fusion</strong> of a material, is either one of:</p>
<p>1) the heat required to melt the material without temperature change or</p>
<p>2) the heat removed from the material to freeze it without temperature change.</p>
<p>For water this latent heat is <mathjax>#80" cal/g"#</mathjax>. Multiply this by <mathjax>#2.5" g"#</mathjax> to get <mathjax>#200" cal"#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#200#</mathjax> calories.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <strong>(latent) heat of fusion</strong> of a material, is either one of:</p>
<p>1) the heat required to melt the material without temperature change or</p>
<p>2) the heat removed from the material to freeze it without temperature change.</p>
<p>For water this latent heat is <mathjax>#80" cal/g"#</mathjax>. Multiply this by <mathjax>#2.5" g"#</mathjax> to get <mathjax>#200" cal"#</mathjax>.</p></div>
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</article> | How much energy is generated from freezing 2.5 g of water? | null |
677 | ab004752-6ddd-11ea-83dd-ccda262736ce | https://socratic.org/questions/what-pressure-is-exerted-by-0-450-mol-of-a-gas-at-25-degrees-celsius-if-the-gas- | 22.02 atm | start physical_unit 15 16 pressure atm qc_end physical_unit 15 16 5 6 mole qc_end physical_unit 15 16 11 13 temperature qc_end physical_unit 22 22 20 21 volume qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"22.02 atm"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{0.450 mol}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{25 degrees Celsius}"},{"type":"physical unit","value":"Volume [OF] the container [=] \\pu{0.50 L}"}] | <h1 class="questionTitle" itemprop="name">What pressure is exerted by 0.450 mol of a gas at 25 degrees Celsius if the gas is in a 0.50 L container? </h1> | null | 22.02 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming that the gas is behaving ideally, then we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>.</p>
<p>Therefore, <mathjax>#P=(nRT)/V=(0.450cancel(mol)xx0.0821(cancel(L)*atm)/(cancel(mol)*cancel(K))xx298cancel(K))/(0.50cancel(L))=22atm#</mathjax></p></div>
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<div>
<div class="markdown"><p><mathjax>#P=22atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming that the gas is behaving ideally, then we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>.</p>
<p>Therefore, <mathjax>#P=(nRT)/V=(0.450cancel(mol)xx0.0821(cancel(L)*atm)/(cancel(mol)*cancel(K))xx298cancel(K))/(0.50cancel(L))=22atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What pressure is exerted by 0.450 mol of a gas at 25 degrees Celsius if the gas is in a 0.50 L container? </h1>
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Dr. Hayek
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<div>
<div class="markdown"><p><mathjax>#P=22atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming that the gas is behaving ideally, then we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: </p>
<p><mathjax>#PV=nRT#</mathjax>.</p>
<p>Therefore, <mathjax>#P=(nRT)/V=(0.450cancel(mol)xx0.0821(cancel(L)*atm)/(cancel(mol)*cancel(K))xx298cancel(K))/(0.50cancel(L))=22atm#</mathjax></p></div>
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</article> | What pressure is exerted by 0.450 mol of a gas at 25 degrees Celsius if the gas is in a 0.50 L container? | null |
678 | abb02dfe-6ddd-11ea-bf98-ccda262736ce | https://socratic.org/questions/a-gas-occupies-100-0-ml-at-a-pressure-of-780-mm-hg-what-is-the-volume-of-the-gas | 87.6 mL | start physical_unit 16 17 volume ml qc_end physical_unit 16 17 3 4 volume qc_end physical_unit 16 17 9 10 pressure qc_end physical_unit 16 17 24 25 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"87.6 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{780 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{880 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies 100.0 mL at a pressure of 780 mm Hg. What is the volume of the gas when its pressure is increased to 880 mm Hg?</h1> | null | 87.6 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A manometer is thus a highly visual measure of the strength of a vacuum, or as a measure of <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>.</p>
<p>If you put the column under a pressure greater than <mathjax>#1*atm#</mathjax> you are likely to get mercury all over the laboratory, where it will inhabit every crack and every cranny. This is a MAJOR cleanup job, which contract cleaners would not touch.</p>
<p>And so <mathjax>#P_1="780 mm Hg"/("760 mm Hg"*atm^-1)=1.026*atm.#</mathjax></p>
<p>And so <mathjax>#P_2="880 mm Hg"/("760 mm Hg"*atm^-1)=1.158*atm.#</mathjax></p>
<p>And now we use old <mathjax>#"Boyle's Law"#</mathjax>, <mathjax>#P_1V_1=P_2V_2.....#</mathjax>, and solve for </p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1.026*atmxx100.0*mL)/(1.158*atm)#</mathjax></p>
<p><mathjax>#=87.6*mL#</mathjax></p>
<p>The volume is slightly decreased, as we would expect, why?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This question has been set by someone who has never used a mercury manometer............we gets <mathjax>#V_2=87.6*mL#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A manometer is thus a highly visual measure of the strength of a vacuum, or as a measure of <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>.</p>
<p>If you put the column under a pressure greater than <mathjax>#1*atm#</mathjax> you are likely to get mercury all over the laboratory, where it will inhabit every crack and every cranny. This is a MAJOR cleanup job, which contract cleaners would not touch.</p>
<p>And so <mathjax>#P_1="780 mm Hg"/("760 mm Hg"*atm^-1)=1.026*atm.#</mathjax></p>
<p>And so <mathjax>#P_2="880 mm Hg"/("760 mm Hg"*atm^-1)=1.158*atm.#</mathjax></p>
<p>And now we use old <mathjax>#"Boyle's Law"#</mathjax>, <mathjax>#P_1V_1=P_2V_2.....#</mathjax>, and solve for </p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1.026*atmxx100.0*mL)/(1.158*atm)#</mathjax></p>
<p><mathjax>#=87.6*mL#</mathjax></p>
<p>The volume is slightly decreased, as we would expect, why?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas occupies 100.0 mL at a pressure of 780 mm Hg. What is the volume of the gas when its pressure is increased to 880 mm Hg?</h1>
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anor277
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<div class="markdown"><p>This question has been set by someone who has never used a mercury manometer............we gets <mathjax>#V_2=87.6*mL#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One atmosphere of pressure will support a column of mercury that is <mathjax>#760*mm#</mathjax> high. A manometer is thus a highly visual measure of the strength of a vacuum, or as a measure of <a href="https://socratic.org/chemistry/the-behavior-of-gases/atmospheric-pressure">atmospheric pressure</a>.</p>
<p>If you put the column under a pressure greater than <mathjax>#1*atm#</mathjax> you are likely to get mercury all over the laboratory, where it will inhabit every crack and every cranny. This is a MAJOR cleanup job, which contract cleaners would not touch.</p>
<p>And so <mathjax>#P_1="780 mm Hg"/("760 mm Hg"*atm^-1)=1.026*atm.#</mathjax></p>
<p>And so <mathjax>#P_2="880 mm Hg"/("760 mm Hg"*atm^-1)=1.158*atm.#</mathjax></p>
<p>And now we use old <mathjax>#"Boyle's Law"#</mathjax>, <mathjax>#P_1V_1=P_2V_2.....#</mathjax>, and solve for </p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(1.026*atmxx100.0*mL)/(1.158*atm)#</mathjax></p>
<p><mathjax>#=87.6*mL#</mathjax></p>
<p>The volume is slightly decreased, as we would expect, why?</p></div>
</div>
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<a href="https://socratic.org/answers/411450" itemprop="url">Answer link</a>
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</article> | A gas occupies 100.0 mL at a pressure of 780 mm Hg. What is the volume of the gas when its pressure is increased to 880 mm Hg? | null |
679 | ac330198-6ddd-11ea-b042-ccda262736ce | https://socratic.org/questions/how-many-moles-of-coo-s-will-be-formed-for-every-mole-of-cocl-2-6h-2o-s | 1 moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] CoO(s) [IN] moles"}] | [{"type":"physical unit","value":"1 moles"}] | [{"type":"physical unit","value":"Mole [OF] CoCl2.6H2O(s) [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #CoO(s)# will be formed for every mole of #CoCl_2 * 6H_2O(s)#?</h1> | null | 1 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming you meant how many moles of <mathjax>#CoO#</mathjax> will be formed from the reaction of <mathjax>#CoCl_2#</mathjax> with a strong base such as <mathjax>#NaOH#</mathjax>, we can start working this out, well, at first we have</p>
<p><mathjax>#CoCl_2 + NaOH rarr Co(OH)_(2(s)) + NaCl#</mathjax><br/>
(The water was ignored because it's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, so, even if we do add it it's not gonna participate in the reaction)</p>
<p>Balancing that equation we see we need to have 2 sodiums in the LHS and 2 chlorides in the RHS, which we can solve by putting a 2 in front of <mathjax>#NaOH#</mathjax> and <mathjax>#NaCl#</mathjax></p>
<p><mathjax>#CoCl_2 + 2NaOH rarr Co(OH)_(2(s)) + 2NaCl#</mathjax></p>
<p>So we know that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) hydroxide, next we need to see the dehydration reaction.</p>
<p><mathjax>#Co(OH)_2 rarr CoO + H_2O#</mathjax></p>
<p>Which, if we check is already balanced, so we can say that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) oxide.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The ratio is <mathjax>#1:1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming you meant how many moles of <mathjax>#CoO#</mathjax> will be formed from the reaction of <mathjax>#CoCl_2#</mathjax> with a strong base such as <mathjax>#NaOH#</mathjax>, we can start working this out, well, at first we have</p>
<p><mathjax>#CoCl_2 + NaOH rarr Co(OH)_(2(s)) + NaCl#</mathjax><br/>
(The water was ignored because it's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, so, even if we do add it it's not gonna participate in the reaction)</p>
<p>Balancing that equation we see we need to have 2 sodiums in the LHS and 2 chlorides in the RHS, which we can solve by putting a 2 in front of <mathjax>#NaOH#</mathjax> and <mathjax>#NaCl#</mathjax></p>
<p><mathjax>#CoCl_2 + 2NaOH rarr Co(OH)_(2(s)) + 2NaCl#</mathjax></p>
<p>So we know that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) hydroxide, next we need to see the dehydration reaction.</p>
<p><mathjax>#Co(OH)_2 rarr CoO + H_2O#</mathjax></p>
<p>Which, if we check is already balanced, so we can say that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) oxide.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of #CoO(s)# will be formed for every mole of #CoCl_2 * 6H_2O(s)#?</h1>
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<div class="markdown"><p>The ratio is <mathjax>#1:1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Assuming you meant how many moles of <mathjax>#CoO#</mathjax> will be formed from the reaction of <mathjax>#CoCl_2#</mathjax> with a strong base such as <mathjax>#NaOH#</mathjax>, we can start working this out, well, at first we have</p>
<p><mathjax>#CoCl_2 + NaOH rarr Co(OH)_(2(s)) + NaCl#</mathjax><br/>
(The water was ignored because it's <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, so, even if we do add it it's not gonna participate in the reaction)</p>
<p>Balancing that equation we see we need to have 2 sodiums in the LHS and 2 chlorides in the RHS, which we can solve by putting a 2 in front of <mathjax>#NaOH#</mathjax> and <mathjax>#NaCl#</mathjax></p>
<p><mathjax>#CoCl_2 + 2NaOH rarr Co(OH)_(2(s)) + 2NaCl#</mathjax></p>
<p>So we know that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) hydroxide, next we need to see the dehydration reaction.</p>
<p><mathjax>#Co(OH)_2 rarr CoO + H_2O#</mathjax></p>
<p>Which, if we check is already balanced, so we can say that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) oxide.</p></div>
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</article> | How many moles of #CoO(s)# will be formed for every mole of #CoCl_2 * 6H_2O(s)#? | null |
680 | a93bf35c-6ddd-11ea-a769-ccda262736ce | https://socratic.org/questions/how-much-energy-is-required-to-melt-1-5-kg-of-lead | 34.53 kJ | start physical_unit 10 10 energy kj qc_end physical_unit 10 10 7 8 mass qc_end c_other melt qc_end end | [{"type":"physical unit","value":"Required energy [OF] lead [IN] kJ"}] | [{"type":"physical unit","value":"34.53 kJ"}] | [{"type":"physical unit","value":"Mass [OF] lead [=] \\pu{1.5 kg}"},{"type":"other","value":"melt"}] | <h1 class="questionTitle" itemprop="name"> How much energy is required to melt 1.5 kg of lead?</h1> | null | 34.53 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The enthalpy of fusion (energy need to change state from solid to liquid) of lead is 4.77 kJ/mol (so it takes 4.77 kJ to melt one mole of lead).<br/>
We need to find the number of moles (<mathjax>#n#</mathjax>) of lead in 1.5kg (<mathjax>#m#</mathjax>).</p>
<p><mathjax>#n=m/M#</mathjax><br/>
From periodic table, M=207.2g/mol<br/>
Therefore, <mathjax>#n=(1500g)/(207.2g.mol^-1)#</mathjax><br/>
<mathjax>#n=7.24#</mathjax> moles<br/>
Energy <mathjax>#=7.24#</mathjax>moles<mathjax>#*4.77kJ.mol^-1#</mathjax><br/>
Energy <mathjax>#34.53#</mathjax>kJ</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#34.53#</mathjax>kJ</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The enthalpy of fusion (energy need to change state from solid to liquid) of lead is 4.77 kJ/mol (so it takes 4.77 kJ to melt one mole of lead).<br/>
We need to find the number of moles (<mathjax>#n#</mathjax>) of lead in 1.5kg (<mathjax>#m#</mathjax>).</p>
<p><mathjax>#n=m/M#</mathjax><br/>
From periodic table, M=207.2g/mol<br/>
Therefore, <mathjax>#n=(1500g)/(207.2g.mol^-1)#</mathjax><br/>
<mathjax>#n=7.24#</mathjax> moles<br/>
Energy <mathjax>#=7.24#</mathjax>moles<mathjax>#*4.77kJ.mol^-1#</mathjax><br/>
Energy <mathjax>#34.53#</mathjax>kJ</p></div>
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<h1 class="questionTitle" itemprop="name"> How much energy is required to melt 1.5 kg of lead?</h1>
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Isini M.
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<span class="dateCreated" datetime="2017-06-06T00:25:55" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#34.53#</mathjax>kJ</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The enthalpy of fusion (energy need to change state from solid to liquid) of lead is 4.77 kJ/mol (so it takes 4.77 kJ to melt one mole of lead).<br/>
We need to find the number of moles (<mathjax>#n#</mathjax>) of lead in 1.5kg (<mathjax>#m#</mathjax>).</p>
<p><mathjax>#n=m/M#</mathjax><br/>
From periodic table, M=207.2g/mol<br/>
Therefore, <mathjax>#n=(1500g)/(207.2g.mol^-1)#</mathjax><br/>
<mathjax>#n=7.24#</mathjax> moles<br/>
Energy <mathjax>#=7.24#</mathjax>moles<mathjax>#*4.77kJ.mol^-1#</mathjax><br/>
Energy <mathjax>#34.53#</mathjax>kJ</p></div>
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</article> | How much energy is required to melt 1.5 kg of lead? | null |
681 | aa90d83f-6ddd-11ea-8930-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-percentage-composition-of-oxygen-in-zn-no-3-2 | 51.67% | start physical_unit 8 10 percent_composition none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Percentage composition [OF] oxygen in Zn(NO3)2"}] | [{"type":"physical unit","value":"51.67%"}] | [{"type":"chemical equation","value":"Zn(NO3)2"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the percentage composition of Oxygen in #Zn(NO_3)_2#?</h1> | null | 51.67% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#15.999*g*mol^-1#</mathjax> is of course the molar mass of the oxygen <strong>atom</strong>. <mathjax>#189.36*g*mol^-1#</mathjax> is the molar mass of zinc nitrate. So it is a bit under 50% by mass. </p>
<p><strong>Molar mass of zinc nitrate:</strong><br/>
You can calculate this by adding all the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in this compound:<br/>
<mathjax>#1xxZn+2xxN+6xxO#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#(6xx15.999*g*mol^-1)/(189.36*g*mol^-1)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#15.999*g*mol^-1#</mathjax> is of course the molar mass of the oxygen <strong>atom</strong>. <mathjax>#189.36*g*mol^-1#</mathjax> is the molar mass of zinc nitrate. So it is a bit under 50% by mass. </p>
<p><strong>Molar mass of zinc nitrate:</strong><br/>
You can calculate this by adding all the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in this compound:<br/>
<mathjax>#1xxZn+2xxN+6xxO#</mathjax> </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the percentage composition of Oxygen in #Zn(NO_3)_2#?</h1>
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<div class="markdown"><p><mathjax>#(6xx15.999*g*mol^-1)/(189.36*g*mol^-1)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#100%#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#15.999*g*mol^-1#</mathjax> is of course the molar mass of the oxygen <strong>atom</strong>. <mathjax>#189.36*g*mol^-1#</mathjax> is the molar mass of zinc nitrate. So it is a bit under 50% by mass. </p>
<p><strong>Molar mass of zinc nitrate:</strong><br/>
You can calculate this by adding all the atomic masses of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in this compound:<br/>
<mathjax>#1xxZn+2xxN+6xxO#</mathjax> </p></div>
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</article> | How do you calculate the percentage composition of Oxygen in #Zn(NO_3)_2#? | null |
682 | ac486596-6ddd-11ea-a005-ccda262736ce | https://socratic.org/questions/what-volume-of-oxygen-can-be-obtained-by-the-addition-of-excess-water-to-1-0g-of | 0.15 L | start physical_unit 3 3 volume l qc_end physical_unit 17 18 14 15 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] L"}] | [{"type":"physical unit","value":"0.15 L"}] | [{"type":"physical unit","value":"Mass [OF] sodium peroxide [=] \\pu{1.0 g}"},{"type":"other","value":"Excess water."}] | <h1 class="questionTitle" itemprop="name">What volume of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide?</h1> | null | 0.15 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, you can't calculate the volume of oxygen produced by the reaction if you don't know the conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>SInce no mention of this was made, I'll assume that the reaction takes place at <strong>STP</strong> - Standard Temperature and Pressure, which implies a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</p>
<p>Now focus on writing the balanced chemical equation for this reaction. </p>
<p><em>Sodium peroxide</em>, <mathjax>#"Na"_2"O"_2#</mathjax>, will react with water to form <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, and <em>hydrogen peroxide</em>, <mathjax>#"H"_2"O"_2#</mathjax>. </p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_2"O"_text(2(aq])#</mathjax></p>
</blockquote>
<p>The oxygen gas will actually come from the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> of the hydrogen peroxide.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>This equation can be rewritten as </p>
<blockquote>
<p><mathjax>#"H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Combine these two equations to get the overall reaction</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>which leads to</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + "H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1/2#</mathjax> moles of oxygen gas <em>for every</em> mole of sodium peroxide. </p>
<p>To determine how many moles of sodium peroxide you have, use the compound's molar mass</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"O"_2)/(77.978color(red)(cancel(color(black)("g")))) = "0.0128 moles Na"_2"O"_2#</mathjax></p>
</blockquote>
<p>The reaction will produce</p>
<blockquote>
<p><mathjax>#0.0128color(red)(cancel(color(black)("moles Na"_2"O"_2))) * ("1/2 moles O"_2)/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = "0.00640 moles O"_2#</mathjax></p>
</blockquote>
<p>At STP, <em>one mole</em> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> - this is known as the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. </p>
<p>In your case, this many moles of oxygen gas would occupy</p>
<blockquote>
<p><mathjax>#0.00640color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.15 L")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"0.15 L " ->#</mathjax> at STP</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, you can't calculate the volume of oxygen produced by the reaction if you don't know the conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>SInce no mention of this was made, I'll assume that the reaction takes place at <strong>STP</strong> - Standard Temperature and Pressure, which implies a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</p>
<p>Now focus on writing the balanced chemical equation for this reaction. </p>
<p><em>Sodium peroxide</em>, <mathjax>#"Na"_2"O"_2#</mathjax>, will react with water to form <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, and <em>hydrogen peroxide</em>, <mathjax>#"H"_2"O"_2#</mathjax>. </p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_2"O"_text(2(aq])#</mathjax></p>
</blockquote>
<p>The oxygen gas will actually come from the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> of the hydrogen peroxide.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>This equation can be rewritten as </p>
<blockquote>
<p><mathjax>#"H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Combine these two equations to get the overall reaction</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>which leads to</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + "H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1/2#</mathjax> moles of oxygen gas <em>for every</em> mole of sodium peroxide. </p>
<p>To determine how many moles of sodium peroxide you have, use the compound's molar mass</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"O"_2)/(77.978color(red)(cancel(color(black)("g")))) = "0.0128 moles Na"_2"O"_2#</mathjax></p>
</blockquote>
<p>The reaction will produce</p>
<blockquote>
<p><mathjax>#0.0128color(red)(cancel(color(black)("moles Na"_2"O"_2))) * ("1/2 moles O"_2)/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = "0.00640 moles O"_2#</mathjax></p>
</blockquote>
<p>At STP, <em>one mole</em> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> - this is known as the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. </p>
<p>In your case, this many moles of oxygen gas would occupy</p>
<blockquote>
<p><mathjax>#0.00640color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.15 L")#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What volume of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide?</h1>
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Stefan V.
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Oct 23, 2015
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<div class="markdown"><p><mathjax>#"0.15 L " ->#</mathjax> at STP</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, you can't calculate the volume of oxygen produced by the reaction if you don't know the conditions for <em>pressure</em> and <em>temperature</em>. </p>
<p>SInce no mention of this was made, I'll assume that the reaction takes place at <strong>STP</strong> - Standard Temperature and Pressure, which implies a pressure of <mathjax>#"100 kPa"#</mathjax> and a temperature of <mathjax>#0^@"C"#</mathjax>.</p>
<p>Now focus on writing the balanced chemical equation for this reaction. </p>
<p><em>Sodium peroxide</em>, <mathjax>#"Na"_2"O"_2#</mathjax>, will react with water to form <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, and <em>hydrogen peroxide</em>, <mathjax>#"H"_2"O"_2#</mathjax>. </p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_2"O"_text(2(aq])#</mathjax></p>
</blockquote>
<p>The oxygen gas will actually come from the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> of the hydrogen peroxide.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((l]) + "O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>This equation can be rewritten as </p>
<blockquote>
<p><mathjax>#"H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Combine these two equations to get the overall reaction</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>which leads to</p>
<blockquote>
<p><mathjax>#"Na"_2"O"_text(2(s]) + "H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + 1/2"O"_text(2(g]) uarr#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1/2#</mathjax> moles of oxygen gas <em>for every</em> mole of sodium peroxide. </p>
<p>To determine how many moles of sodium peroxide you have, use the compound's molar mass</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"O"_2)/(77.978color(red)(cancel(color(black)("g")))) = "0.0128 moles Na"_2"O"_2#</mathjax></p>
</blockquote>
<p>The reaction will produce</p>
<blockquote>
<p><mathjax>#0.0128color(red)(cancel(color(black)("moles Na"_2"O"_2))) * ("1/2 moles O"_2)/(1color(red)(cancel(color(black)("mole Na"_2"O"_2)))) = "0.00640 moles O"_2#</mathjax></p>
</blockquote>
<p>At STP, <em>one mole</em> of any ideal gas occupies exactly <mathjax>#"22.7 L"#</mathjax> - this is known as the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. </p>
<p>In your case, this many moles of oxygen gas would occupy</p>
<blockquote>
<p><mathjax>#0.00640color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.15 L")#</mathjax></p>
</blockquote></div>
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</article> | What volume of oxygen can be obtained by the addition of excess water to 1.0g of sodium peroxide? | null |
683 | a9eb181c-6ddd-11ea-874d-ccda262736ce | https://socratic.org/questions/how-many-moles-of-co-2-are-produced-from-0-110-mole-o-2 | 0.11 moles | start physical_unit 4 4 mole mol qc_end physical_unit 10 10 8 9 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] CO2 [IN] moles"}] | [{"type":"physical unit","value":"0.11 moles"}] | [{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{0.110 moles}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #CO_2# are produced from 0.110 mole #O_2#?</h1> | null | 0.11 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since no chemical equation was provided, I will assume that you're dealing with this one</p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>Here carbon undergoes combustion to form carbon dioxide. Notice that it takes <mathjax>#1#</mathjax> <strong>mole</strong> of carbon to react with <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>This basically means that the number of moles of carbon dioxide produced by the reaction will be <strong>equal</strong> to the number of moles of oxygen gas that react, provided, of course, that you have <em>enough</em> moles of carbon. </p>
<p>Since no mention of the number of moles of carbon was made, you can safely assume that carbon is <em>in excess</em>. </p>
<p>Therefore, you have</p>
<blockquote>
<p><mathjax>#0.110 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("0.110 moles CO"_2)))#</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>#"0.110 moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since no chemical equation was provided, I will assume that you're dealing with this one</p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>Here carbon undergoes combustion to form carbon dioxide. Notice that it takes <mathjax>#1#</mathjax> <strong>mole</strong> of carbon to react with <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>This basically means that the number of moles of carbon dioxide produced by the reaction will be <strong>equal</strong> to the number of moles of oxygen gas that react, provided, of course, that you have <em>enough</em> moles of carbon. </p>
<p>Since no mention of the number of moles of carbon was made, you can safely assume that carbon is <em>in excess</em>. </p>
<p>Therefore, you have</p>
<blockquote>
<p><mathjax>#0.110 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("0.110 moles CO"_2)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #CO_2# are produced from 0.110 mole #O_2#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.110 moles CO"_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since no chemical equation was provided, I will assume that you're dealing with this one</p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>Here carbon undergoes combustion to form carbon dioxide. Notice that it takes <mathjax>#1#</mathjax> <strong>mole</strong> of carbon to react with <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas in order to produce <mathjax>#1#</mathjax> <strong>mole</strong> of carbon dioxide. </p>
<p>This basically means that the number of moles of carbon dioxide produced by the reaction will be <strong>equal</strong> to the number of moles of oxygen gas that react, provided, of course, that you have <em>enough</em> moles of carbon. </p>
<p>Since no mention of the number of moles of carbon was made, you can safely assume that carbon is <em>in excess</em>. </p>
<p>Therefore, you have</p>
<blockquote>
<p><mathjax>#0.110 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("0.110 moles CO"_2)))#</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 #CO_2# are produced from 0.110 mole #O_2#? | null |
684 | ab9816c0-6ddd-11ea-9011-ccda262736ce | https://socratic.org/questions/5925f54711ef6b72af01fa47 | 16.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 13 8 9 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] hydrogen [IN] moles"}] | [{"type":"physical unit","value":"16.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] ammonium hydroxide [=] \\pu{3.2 mol}"}] | <h1 class="questionTitle" itemprop="name">How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#?</h1> | null | 16.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
</div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3.2xxN_Axx5=??#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#?</h1>
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<div class="markdown"><p><mathjax>#3.2xxN_Axx5=??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Well, <mathjax>#N_A=6.022xx10^23*mol^-1#</mathjax>. In one mole of <mathjax>#"ammonium hydroxide"#</mathjax>, <mathjax>#NH_3*H_2O#</mathjax>, there are clearly 5 moles of hydrogens..........</p>
<p>And so we take the product................</p>
<p><mathjax>#3.2*molxx6.022xx10^23*mol^-1xx5#</mathjax> <mathjax>#"hydrogen atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Note that ammonium hydroxide is likely better represented as <mathjax>#NH_3*H_2O#</mathjax> than <mathjax>#NH_4^+HO^-#</mathjax>...</p></div>
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Nathan L.
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<div class="markdown"><p><mathjax>#9.6xx10^24#</mathjax> <mathjax>#"atoms H"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to calculate the number of atoms of <mathjax>#"H"#</mathjax> in <mathjax>#3.2#</mathjax> <mathjax>#"mol NH"_4"OH"#</mathjax>.</p>
<p>We can do this two different ways. The first is to calculate the number of molecules of <mathjax>#"NH"_4"OH"#</mathjax>, and then atoms of <mathjax>#"H"#</mathjax>, and the second method is to first calculate the number of moles of <mathjax>#"H"#</mathjax>, and then atoms of <mathjax>#"H"#</mathjax>.</p>
<p>Method 1:</p>
<p>Using <em>Avogadro's number</em> (<mathjax>#6.022xx10^23"particles"/"mol"#</mathjax>), let's calculate the number of <em>molecules</em> of <mathjax>#"NH"_4"OH"#</mathjax>:</p>
<p><mathjax>#3.2cancel("mol NH"_4"OH")((6.022xx10^23"molecules NH"_4"OH")/(1 cancel("mol NH"_4"OH")))#</mathjax></p>
<p><mathjax>#= 1.927xx10^24#</mathjax> <mathjax>#"molecules NH"_4"OH"#</mathjax></p>
<p>Now, using the fact that there are <mathjax>#5#</mathjax> atoms of <mathjax>#"H"#</mathjax> per molecule of <mathjax>#"NH"_4"OH"#</mathjax>, we have</p>
<p><mathjax>#"atoms H" =#</mathjax><br/>
<mathjax>#1.927xx10^24cancel("molecules NH"_4"OH")((5"atoms H")/(1 cancel("molecule NH"_4"OH")))#</mathjax></p>
<p><mathjax>#= color(red)(9.6xx10^24#</mathjax> <mathjax>#color(red)("atoms H"#</mathjax></p>
<p>rounded to <mathjax>#2#</mathjax> <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, the amount given in the problem.</p>
<blockquote></blockquote>
<p>Method 2:</p>
<p>For the second method, we can calculate the number of moles of <mathjax>#"H"#</mathjax> per mole of <mathjax>#"NH"_4"OH"#</mathjax>, and then atoms <mathjax>#"H"#</mathjax> from moles:</p>
<p><mathjax>#3.2#</mathjax> <mathjax>#cancel("mol NH"_4"OH")((5cancel("mol H"))/(1cancel("mol NH"_4"OH")))((6.022xx10^23"atoms H")/(1cancel("mol H")))#</mathjax></p>
<p><mathjax>#= color(red)(9.6xx10^24#</mathjax> <mathjax>#color(red)("atoms H"#</mathjax></p>
<p>rounded to <mathjax>#2#</mathjax> significant figures, the amount given in the problem.</p>
<p>As you can see, both methods yield the same result, so either one can be used (I would probably prefer the second method; I find it to be a little more exact).</p></div>
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</article> | How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#? | null |
685 | a8f1144d-6ddd-11ea-9c58-ccda262736ce | https://socratic.org/questions/what-mass-of-silver-chloride-can-be-produced-from-1-09-l-of-a-0-118-m-solution-o | 18.49 g | start physical_unit 3 4 mass g qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 17 18 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] silver chloride [IN] g"}] | [{"type":"physical unit","value":"18.49 g"}] | [{"type":"physical unit","value":"Volume [OF] silver nitrate solution [=] \\pu{1.09 L}"},{"type":"physical unit","value":"Molarity [OF] silver nitrate solution [=] \\pu{0.118 M}"}] | <h1 class="questionTitle" itemprop="name">What mass of silver chloride can be produced from 1.09 L of a 0.118 M solution of silver nitrate?</h1> | null | 18.49 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want to perform a precipitation reaction........</p>
<p><mathjax>#Ag^(+) + Cl^(-)rarrAgCl(s)darr#</mathjax></p>
<p>And thus moles of <mathjax>#Ag^(+)-=Ag#</mathjax></p>
<p>We gots a molar quantity of .................</p>
<p><mathjax>#1.09*Lxx0.118*mol*L^-1=0.129*mol#</mathjax> with respect to <mathjax>#Ag^+#</mathjax>.</p>
<p>And thus an equimolar quantity of <mathjax>#AgCl#</mathjax> may be precipitated, which is as soluble as a brick (but much more difficult to handle!).</p>
<p>And thus mass of <mathjax>#"silver chloride"#</mathjax> possible is....</p>
<p><mathjax>#143.32*g·mol^-1xx0.129*mol-=??g#</mathjax></p></div>
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<div class="markdown"><p>Under <mathjax>#20*g#</mathjax> of the halide may be precipitated.....</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want to perform a precipitation reaction........</p>
<p><mathjax>#Ag^(+) + Cl^(-)rarrAgCl(s)darr#</mathjax></p>
<p>And thus moles of <mathjax>#Ag^(+)-=Ag#</mathjax></p>
<p>We gots a molar quantity of .................</p>
<p><mathjax>#1.09*Lxx0.118*mol*L^-1=0.129*mol#</mathjax> with respect to <mathjax>#Ag^+#</mathjax>.</p>
<p>And thus an equimolar quantity of <mathjax>#AgCl#</mathjax> may be precipitated, which is as soluble as a brick (but much more difficult to handle!).</p>
<p>And thus mass of <mathjax>#"silver chloride"#</mathjax> possible is....</p>
<p><mathjax>#143.32*g·mol^-1xx0.129*mol-=??g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of silver chloride can be produced from 1.09 L of a 0.118 M solution of silver nitrate?</h1>
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<div class="markdown"><p>Under <mathjax>#20*g#</mathjax> of the halide may be precipitated.....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We want to perform a precipitation reaction........</p>
<p><mathjax>#Ag^(+) + Cl^(-)rarrAgCl(s)darr#</mathjax></p>
<p>And thus moles of <mathjax>#Ag^(+)-=Ag#</mathjax></p>
<p>We gots a molar quantity of .................</p>
<p><mathjax>#1.09*Lxx0.118*mol*L^-1=0.129*mol#</mathjax> with respect to <mathjax>#Ag^+#</mathjax>.</p>
<p>And thus an equimolar quantity of <mathjax>#AgCl#</mathjax> may be precipitated, which is as soluble as a brick (but much more difficult to handle!).</p>
<p>And thus mass of <mathjax>#"silver chloride"#</mathjax> possible is....</p>
<p><mathjax>#143.32*g·mol^-1xx0.129*mol-=??g#</mathjax></p></div>
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</article> | What mass of silver chloride can be produced from 1.09 L of a 0.118 M solution of silver nitrate? | null |
686 | a922fe9a-6ddd-11ea-b7b9-ccda262736ce | https://socratic.org/questions/under-a-pressure-of-200-kpa-a-confined-gas-has-a-volume-of-2-500-cubic-meters-th | 1000 cubic meters | start physical_unit 20 21 volume m^3 qc_end physical_unit 20 21 13 15 volume qc_end physical_unit 20 21 4 5 pressure qc_end physical_unit 20 21 25 26 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] cubic meters"}] | [{"type":"physical unit","value":"1000 cubic meters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{2,500 cubic meters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{200 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{500 kPa}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">Under a pressure of 200 kPa, a confined gas has a volume of 2,500 cubic meters. The pressure acting on the gas is increased to 500 kPa. What is the new volume of the gas if the temperature remains the same? </h1> | null | 1000 cubic meters | <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/boyle-s-law">Boyle's Law</a>: <mathjax>#P_1 V_1 = P_2 V_2#</mathjax> . substituting and making the units consistent: <mathjax>#(200xx10^3)(2500)=(500xx10^3)(V_2)#</mathjax>. solve for <mathjax>#V_2#</mathjax> . </p></div>
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<div class="markdown"><p><mathjax>#V_2 = 1000 m^3#</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/boyle-s-law">Boyle's Law</a>: <mathjax>#P_1 V_1 = P_2 V_2#</mathjax> . substituting and making the units consistent: <mathjax>#(200xx10^3)(2500)=(500xx10^3)(V_2)#</mathjax>. solve for <mathjax>#V_2#</mathjax> . </p></div>
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<h1 class="questionTitle" itemprop="name">Under a pressure of 200 kPa, a confined gas has a volume of 2,500 cubic meters. The pressure acting on the gas is increased to 500 kPa. What is the new volume of the gas if the temperature remains the same? </h1>
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<div class="markdown"><p><mathjax>#V_2 = 1000 m^3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>: <mathjax>#P_1 V_1 = P_2 V_2#</mathjax> . substituting and making the units consistent: <mathjax>#(200xx10^3)(2500)=(500xx10^3)(V_2)#</mathjax>. solve for <mathjax>#V_2#</mathjax> . </p></div>
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</article> | Under a pressure of 200 kPa, a confined gas has a volume of 2,500 cubic meters. The pressure acting on the gas is increased to 500 kPa. What is the new volume of the gas if the temperature remains the same? | null |
687 | a93cb718-6ddd-11ea-9bd3-ccda262736ce | https://socratic.org/questions/what-is-the-h-3o-ion-concentration-of-a-solution-whose-oh-ion-concentration-is-1 | 10^(-11) M | start physical_unit 3 4 concentration mol/l qc_end physical_unit 10 11 14 17 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] H3O+ ion [IN] M"}] | [{"type":"physical unit","value":"10^(-11) M"}] | [{"type":"physical unit","value":"Concentration [OF] OH- ion [=] \\pu{1 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the #H_3O^+# ion concentration of a solution whose #OH^- # ion concentration is # 1 xx 10^-3 M#?</h1> | null | 10^(-11) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the following equilibrium..........</p>
<p><mathjax>#H_2OrightleftharpoonsH^+ +HO^(-)#</mathjax>, and further that we know the equilibrium constant for this expression is......</p>
<p><mathjax>#[H^+]xx[HO^-]=10^-14#</mathjax> under standard conditions.</p>
<p>Alternatively, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>. We can take <mathjax>#log_10#</mathjax> of both sides and get.........</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)#</mathjax>, </p>
<p>or......<mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#</mathjax>,</p>
<p>or......<mathjax>#-log_10(10^-14)=pH+pOH=14#</mathjax></p>
<p>But the expression on the left hand side is simply <mathjax>#14#</mathjax>, and this goes back to definition of the log function, the <mathjax>#log#</mathjax> of something is the exponent to which you raise the <mathjax>#"base"#</mathjax> (normally <mathjax>#10#</mathjax> or <mathjax>#e#</mathjax>) to get the bracketed term. </p>
<p>or......<mathjax>#-log_10(10^-14)=-(-14)=14#</mathjax></p>
<p>And so finally <mathjax>#pH+pOH=14#</mathjax>, you must know this for A-level....</p>
<p>Here <mathjax>#[HO^-]=10^-3*mol*L^-1#</mathjax>, so <mathjax>#pOH=-log_10(10^-3)=-(-3)=3#</mathjax> </p>
<p><mathjax>#pH=11#</mathjax>, and <mathjax>#pOH=3#</mathjax>, i.e. taking antilogs.......</p>
<p><mathjax>#[H_3O^+]=10^-11*mol*L^-1.#</mathjax></p></div>
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<div class="markdown"><p>We know that <mathjax>#pH+pOH=14#</mathjax>..........so <mathjax>#pH=11#</mathjax>, and <mathjax>#[H_3O^+]=10^-11*mol*L^-1#</mathjax>.</p></div>
</div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the following equilibrium..........</p>
<p><mathjax>#H_2OrightleftharpoonsH^+ +HO^(-)#</mathjax>, and further that we know the equilibrium constant for this expression is......</p>
<p><mathjax>#[H^+]xx[HO^-]=10^-14#</mathjax> under standard conditions.</p>
<p>Alternatively, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>. We can take <mathjax>#log_10#</mathjax> of both sides and get.........</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)#</mathjax>, </p>
<p>or......<mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#</mathjax>,</p>
<p>or......<mathjax>#-log_10(10^-14)=pH+pOH=14#</mathjax></p>
<p>But the expression on the left hand side is simply <mathjax>#14#</mathjax>, and this goes back to definition of the log function, the <mathjax>#log#</mathjax> of something is the exponent to which you raise the <mathjax>#"base"#</mathjax> (normally <mathjax>#10#</mathjax> or <mathjax>#e#</mathjax>) to get the bracketed term. </p>
<p>or......<mathjax>#-log_10(10^-14)=-(-14)=14#</mathjax></p>
<p>And so finally <mathjax>#pH+pOH=14#</mathjax>, you must know this for A-level....</p>
<p>Here <mathjax>#[HO^-]=10^-3*mol*L^-1#</mathjax>, so <mathjax>#pOH=-log_10(10^-3)=-(-3)=3#</mathjax> </p>
<p><mathjax>#pH=11#</mathjax>, and <mathjax>#pOH=3#</mathjax>, i.e. taking antilogs.......</p>
<p><mathjax>#[H_3O^+]=10^-11*mol*L^-1.#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #H_3O^+# ion concentration of a solution whose #OH^- # ion concentration is # 1 xx 10^-3 M#?</h1>
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<div class="markdown"><p>We know that <mathjax>#pH+pOH=14#</mathjax>..........so <mathjax>#pH=11#</mathjax>, and <mathjax>#[H_3O^+]=10^-11*mol*L^-1#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the following equilibrium..........</p>
<p><mathjax>#H_2OrightleftharpoonsH^+ +HO^(-)#</mathjax>, and further that we know the equilibrium constant for this expression is......</p>
<p><mathjax>#[H^+]xx[HO^-]=10^-14#</mathjax> under standard conditions.</p>
<p>Alternatively, <mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax>. We can take <mathjax>#log_10#</mathjax> of both sides and get.........</p>
<p><mathjax>#log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)#</mathjax>, </p>
<p>or......<mathjax>#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#</mathjax>,</p>
<p>or......<mathjax>#-log_10(10^-14)=pH+pOH=14#</mathjax></p>
<p>But the expression on the left hand side is simply <mathjax>#14#</mathjax>, and this goes back to definition of the log function, the <mathjax>#log#</mathjax> of something is the exponent to which you raise the <mathjax>#"base"#</mathjax> (normally <mathjax>#10#</mathjax> or <mathjax>#e#</mathjax>) to get the bracketed term. </p>
<p>or......<mathjax>#-log_10(10^-14)=-(-14)=14#</mathjax></p>
<p>And so finally <mathjax>#pH+pOH=14#</mathjax>, you must know this for A-level....</p>
<p>Here <mathjax>#[HO^-]=10^-3*mol*L^-1#</mathjax>, so <mathjax>#pOH=-log_10(10^-3)=-(-3)=3#</mathjax> </p>
<p><mathjax>#pH=11#</mathjax>, and <mathjax>#pOH=3#</mathjax>, i.e. taking antilogs.......</p>
<p><mathjax>#[H_3O^+]=10^-11*mol*L^-1.#</mathjax></p></div>
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</article> | What is the #H_3O^+# ion concentration of a solution whose #OH^- # ion concentration is # 1 xx 10^-3 M#? | null |
688 | aa5f89e8-6ddd-11ea-81d2-ccda262736ce | https://socratic.org/questions/how-many-ml-of-0-655-m-phosphoric-acid-solution-are-required-to-neutralize-15-ml | 9.80 mL | start physical_unit 6 8 volume ml qc_end physical_unit 18 19 13 14 volume qc_end physical_unit 6 8 4 5 molarity qc_end physical_unit 18 19 16 17 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] phosphoric acid solution [IN] mL"}] | [{"type":"physical unit","value":"9.80 mL"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{15 mL}"},{"type":"physical unit","value":"Molarity [OF] phosphoric acid solution [=] \\pu{0.655 M}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{1.284 M}"}] | <h1 class="questionTitle" itemprop="name">How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M #NaOH# solution? </h1> | null | 9.80 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we write the stoichiometric equation as...</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>And so .................</p>
<p><mathjax>#n_(NaOH)=15*mLxx10^-3*L*mL^-1xx1.284*mol*L^-1=0.01926*mol#</mathjax>...and so we need HALF of this molar quantity, i.e. <mathjax>#9.63xx10^-3*mol#</mathjax>, with respect to <mathjax>#H_3PO_4#</mathjax>.</p>
<p>And so we take the quotient....<mathjax>#(9.63xx10^-3*mol)/(0.655*mol*L^-1)=14.7xx10^-3*L=14.7*mL#</mathjax> to get the volume required for equivalence. </p></div>
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<div class="markdown"><p>Well, you need to know that phosphoric acid is a DIACID in aqueous solution... The third proton is rarely removed. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we write the stoichiometric equation as...</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>And so .................</p>
<p><mathjax>#n_(NaOH)=15*mLxx10^-3*L*mL^-1xx1.284*mol*L^-1=0.01926*mol#</mathjax>...and so we need HALF of this molar quantity, i.e. <mathjax>#9.63xx10^-3*mol#</mathjax>, with respect to <mathjax>#H_3PO_4#</mathjax>.</p>
<p>And so we take the quotient....<mathjax>#(9.63xx10^-3*mol)/(0.655*mol*L^-1)=14.7xx10^-3*L=14.7*mL#</mathjax> to get the volume required for equivalence. </p></div>
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<h1 class="questionTitle" itemprop="name">How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M #NaOH# solution? </h1>
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anor277
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<div class="markdown"><p>Well, you need to know that phosphoric acid is a DIACID in aqueous solution... The third proton is rarely removed. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And we write the stoichiometric equation as...</p>
<p><mathjax>#H_3PO_4(aq) + 2NaOH(aq) rarr Na_2HPO_4(aq) + 2H_2O(l)#</mathjax></p>
<p>And so .................</p>
<p><mathjax>#n_(NaOH)=15*mLxx10^-3*L*mL^-1xx1.284*mol*L^-1=0.01926*mol#</mathjax>...and so we need HALF of this molar quantity, i.e. <mathjax>#9.63xx10^-3*mol#</mathjax>, with respect to <mathjax>#H_3PO_4#</mathjax>.</p>
<p>And so we take the quotient....<mathjax>#(9.63xx10^-3*mol)/(0.655*mol*L^-1)=14.7xx10^-3*L=14.7*mL#</mathjax> to get the volume required for equivalence. </p></div>
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Ratnesh Bhosale
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<div class="markdown"><p><mathjax>#V=9.80 mL#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of <mathjax>#NaOH#</mathjax> <mathjax>#= 1.284/1000 xx 15#</mathjax></p>
<p>Moles of <mathjax>#NaOH#</mathjax> <mathjax>#= 0.01926#</mathjax></p>
<p>Reaction will be -<br/>
<mathjax>#H_3PO_4 + 3NaOH rarr Na_3PO_4 +3H_2O #</mathjax> </p>
<p>So, <mathjax>#0.01926 /3#</mathjax> moles are required to neutralize <mathjax>#H_3PO_4#</mathjax></p>
<p>Moles required to neutralize <mathjax>#H_3PO_4 = 0.00642#</mathjax></p>
<p>Let the volume required to neutralize <mathjax>#H_3PO_4#</mathjax> be <mathjax>#V#</mathjax> mL.</p>
<p>Moles of <mathjax>#H_3PO_4#</mathjax> <mathjax>#= 0.655/1000xxV#</mathjax></p>
<p><mathjax>#0.00642 = 0.655/1000xxV#</mathjax></p>
<p><mathjax>#V= 0.00642xx 1000/0.655#</mathjax></p>
<p><mathjax>#V= 6.42/0.655#</mathjax></p>
<p><mathjax>#V=9.80 mL#</mathjax></p></div>
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</article> | How many mL of 0.655 M phosphoric acid solution are required to neutralize 15 mL of 1.284 M #NaOH# solution? | null |
689 | ac6df8b4-6ddd-11ea-b926-ccda262736ce | https://socratic.org/questions/when-the-ph-of-a-solution-is-12-83-what-is-h | 1.48 × 10^(-13) mol/L | start physical_unit 5 5 [h+] mol/l qc_end physical_unit 5 5 7 7 ph qc_end end | [{"type":"physical unit","value":"[H+] [OF] the solution [IN] mol/L"}] | [{"type":"physical unit","value":"1.48 × 10^(-13) mol/L"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{12.83}"}] | <h1 class="questionTitle" itemprop="name">When the pH of a solution is 12.83, what is [H+]? </h1> | null | 1.48 × 10^(-13) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is a logarthmic scale, we can use the formula: <mathjax>#[H^+] = 10^("-pH")#</mathjax>. Where...<br/>
=> <mathjax>#[H^+]#</mathjax> is the concentration of hydrogen ions in the solution.<br/>
=> <mathjax>#pH#</mathjax> is the pH of the solution. <br/>
=> Where <mathjax>#10#</mathjax> is the base of the power - it's a logarithm formula.</p>
<p>We can now just sub in the values and solve for [<mathjax>#H^+]#</mathjax>. </p>
<blockquote>
<p><mathjax>#[H^+] = 10^("-pH")#</mathjax> </p>
<blockquote>
<blockquote>
<p><mathjax># = 10^(-(12.83))#</mathjax> </p>
<p><mathjax># = 1.479108388 xx 10^-13#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
<p>We can round (if required) to <mathjax>#1.479 xx 10^-13#</mathjax>. </p>
<p>Thus, there are <mathjax>#1.479 xx 10^-13#</mathjax> concentration of hydrogen ions <mathjax>#(mol)/L#</mathjax>. </p>
<p>Hope this helps :)</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>There are <mathjax>#1.479 xx 10^-13#</mathjax> concentration of hydrogen ions <mathjax>#mol/L#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is a logarthmic scale, we can use the formula: <mathjax>#[H^+] = 10^("-pH")#</mathjax>. Where...<br/>
=> <mathjax>#[H^+]#</mathjax> is the concentration of hydrogen ions in the solution.<br/>
=> <mathjax>#pH#</mathjax> is the pH of the solution. <br/>
=> Where <mathjax>#10#</mathjax> is the base of the power - it's a logarithm formula.</p>
<p>We can now just sub in the values and solve for [<mathjax>#H^+]#</mathjax>. </p>
<blockquote>
<p><mathjax>#[H^+] = 10^("-pH")#</mathjax> </p>
<blockquote>
<blockquote>
<p><mathjax># = 10^(-(12.83))#</mathjax> </p>
<p><mathjax># = 1.479108388 xx 10^-13#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
<p>We can round (if required) to <mathjax>#1.479 xx 10^-13#</mathjax>. </p>
<p>Thus, there are <mathjax>#1.479 xx 10^-13#</mathjax> concentration of hydrogen ions <mathjax>#(mol)/L#</mathjax>. </p>
<p>Hope this helps :)</p></div>
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<h1 class="questionTitle" itemprop="name">When the pH of a solution is 12.83, what is [H+]? </h1>
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<div class="markdown"><p>There are <mathjax>#1.479 xx 10^-13#</mathjax> concentration of hydrogen ions <mathjax>#mol/L#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is a logarthmic scale, we can use the formula: <mathjax>#[H^+] = 10^("-pH")#</mathjax>. Where...<br/>
=> <mathjax>#[H^+]#</mathjax> is the concentration of hydrogen ions in the solution.<br/>
=> <mathjax>#pH#</mathjax> is the pH of the solution. <br/>
=> Where <mathjax>#10#</mathjax> is the base of the power - it's a logarithm formula.</p>
<p>We can now just sub in the values and solve for [<mathjax>#H^+]#</mathjax>. </p>
<blockquote>
<p><mathjax>#[H^+] = 10^("-pH")#</mathjax> </p>
<blockquote>
<blockquote>
<p><mathjax># = 10^(-(12.83))#</mathjax> </p>
<p><mathjax># = 1.479108388 xx 10^-13#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
<p>We can round (if required) to <mathjax>#1.479 xx 10^-13#</mathjax>. </p>
<p>Thus, there are <mathjax>#1.479 xx 10^-13#</mathjax> concentration of hydrogen ions <mathjax>#(mol)/L#</mathjax>. </p>
<p>Hope this helps :)</p></div>
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</article> | When the pH of a solution is 12.83, what is [H+]? | null |
690 | a9e74776-6ddd-11ea-a282-ccda262736ce | https://socratic.org/questions/zinc-oxide-reacts-with-hydrochloric-acid-to-produce-zinc-chloride-and-dihydrogen | 490.7 grams | start physical_unit 8 9 mass g qc_end physical_unit 0 1 14 15 mole qc_end c_other OTHER qc_end substance 11 12 qc_end end | [{"type":"physical unit","value":"Mass [OF] zinc chloride [IN] grams"}] | [{"type":"physical unit","value":"490.7 grams"}] | [{"type":"physical unit","value":"Mole [OF] zinc oxide [=] \\pu{3.6 moles}"},{"type":"other","value":"Excess hydrochloric acid."},{"type":"substance name","value":"Dihydrogen monoxide"}] | <h1 class="questionTitle" itemprop="name">Zinc oxide reacts with hydrochloric acid to produce zinc chloride and dihydrogen monoxide. If 3.6 moles of zinc oxide reacts with an excess of hydrochloric acid, then how many grams of zinc chloride were produced?</h1> | null | 490.7 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction happening in solution is the following:</p>
<p><mathjax>#ZnO(aq)+2HCl(aq)->ZnCl_2(aq)+H_2O(l)#</mathjax></p>
<p>Since the hydrochloric acid (<mathjax>#HCl#</mathjax>) is in excess, therefore, the number of moles of Zinc oxide (<mathjax>#ZnO#</mathjax>) will determine the mass of Zinc chloride (<mathjax>#ZnCl#</mathjax>).</p>
<p>Using dimensional analysis, we find:</p>
<p><mathjax>#?g ZnCl_2=3.6cancel("mol " ZnO)xx(1 cancel ("mol "ZnCl_2))/(1cancel("mol " ZnO))xx(136.3"g "ZnCl_2)/(1 cancel ("mol "ZnCl_2))=490.7"g "ZnCl_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#490.7"g "ZnCl_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction happening in solution is the following:</p>
<p><mathjax>#ZnO(aq)+2HCl(aq)->ZnCl_2(aq)+H_2O(l)#</mathjax></p>
<p>Since the hydrochloric acid (<mathjax>#HCl#</mathjax>) is in excess, therefore, the number of moles of Zinc oxide (<mathjax>#ZnO#</mathjax>) will determine the mass of Zinc chloride (<mathjax>#ZnCl#</mathjax>).</p>
<p>Using dimensional analysis, we find:</p>
<p><mathjax>#?g ZnCl_2=3.6cancel("mol " ZnO)xx(1 cancel ("mol "ZnCl_2))/(1cancel("mol " ZnO))xx(136.3"g "ZnCl_2)/(1 cancel ("mol "ZnCl_2))=490.7"g "ZnCl_2#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Zinc oxide reacts with hydrochloric acid to produce zinc chloride and dihydrogen monoxide. If 3.6 moles of zinc oxide reacts with an excess of hydrochloric acid, then how many grams of zinc chloride were produced?</h1>
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Dr. Hayek
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<span class="dateCreated" datetime="2016-01-10T20:36:38" itemprop="dateCreated">
Jan 10, 2016
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<div class="markdown"><p><mathjax>#490.7"g "ZnCl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction happening in solution is the following:</p>
<p><mathjax>#ZnO(aq)+2HCl(aq)->ZnCl_2(aq)+H_2O(l)#</mathjax></p>
<p>Since the hydrochloric acid (<mathjax>#HCl#</mathjax>) is in excess, therefore, the number of moles of Zinc oxide (<mathjax>#ZnO#</mathjax>) will determine the mass of Zinc chloride (<mathjax>#ZnCl#</mathjax>).</p>
<p>Using dimensional analysis, we find:</p>
<p><mathjax>#?g ZnCl_2=3.6cancel("mol " ZnO)xx(1 cancel ("mol "ZnCl_2))/(1cancel("mol " ZnO))xx(136.3"g "ZnCl_2)/(1 cancel ("mol "ZnCl_2))=490.7"g "ZnCl_2#</mathjax></p></div>
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</article> | Zinc oxide reacts with hydrochloric acid to produce zinc chloride and dihydrogen monoxide. If 3.6 moles of zinc oxide reacts with an excess of hydrochloric acid, then how many grams of zinc chloride were produced? | null |
691 | a9b25e26-6ddd-11ea-9e17-ccda262736ce | https://socratic.org/questions/how-many-moles-of-h-2so-4-are-present-in-0-500-l-of-a-0-150-m-h-2so-4-solution | 0.08 moles | start physical_unit 4 4 mole mol qc_end physical_unit 14 15 8 9 volume qc_end physical_unit 14 15 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] H2SO4 [IN] moles"}] | [{"type":"physical unit","value":"0.08 moles"}] | [{"type":"physical unit","value":"Volume [OF] H2SO4 solution [=] \\pu{0.500 L}"},{"type":"physical unit","value":"Molarity [OF] H2SO4 solution [=] \\pu{0.150 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #H_2SO_4# are present in 0.500 L of a 0.150 M #H_2SO_4# solution? </h1> | null | 0.08 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Numer of moles of <mathjax>#H_2SO_4#</mathjax> are present in 0.500 L of a 0.150 M <mathjax>#H_2SO_4#</mathjax> solution is <br/>
<mathjax>#="concentration in molarity"xx "Volume in liter" #</mathjax><br/>
<mathjax>#=0.15 (mol)/Lxx0.500L=0.075mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#0.075mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Numer of moles of <mathjax>#H_2SO_4#</mathjax> are present in 0.500 L of a 0.150 M <mathjax>#H_2SO_4#</mathjax> solution is <br/>
<mathjax>#="concentration in molarity"xx "Volume in liter" #</mathjax><br/>
<mathjax>#=0.15 (mol)/Lxx0.500L=0.075mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#0.075mol#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Numer of moles of <mathjax>#H_2SO_4#</mathjax> are present in 0.500 L of a 0.150 M <mathjax>#H_2SO_4#</mathjax> solution is <br/>
<mathjax>#="concentration in molarity"xx "Volume in liter" #</mathjax><br/>
<mathjax>#=0.15 (mol)/Lxx0.500L=0.075mol#</mathjax></p></div>
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</article> | How many moles of #H_2SO_4# are present in 0.500 L of a 0.150 M #H_2SO_4# solution? | null |
692 | a919ea54-6ddd-11ea-9b6c-ccda262736ce | https://socratic.org/questions/how-many-moles-of-oxygen-are-present-in-33-6-l-of-the-gas-at-1-atm-and-0-c | 1.5 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 12 8 9 volume qc_end physical_unit 11 12 17 18 temperature qc_end physical_unit 11 12 14 15 pressure qc_end end | [{"type":"physical unit","value":"Mole [OF] oxygen [IN] moles"}] | [{"type":"physical unit","value":"1.5 moles"}] | [{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{33.6 L}"},{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{1 atm}"}] | <h1 class="questionTitle" itemprop="name">How many moles of oxygen are present in 33.6 L of the gas at 1 atm and 0°C?</h1> | null | 1.5 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that dioxygen gas behaves ideally. Under the given conditions, an ideal gas has a volume of <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p>So we simply divide the volume by the molar volume under standard conditions, <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p><mathjax>#(33.6*cancelL)/(22.4*cancelL*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax></p>
<p>How did I know that the molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> was <mathjax>#22.4*L#</mathjax>? You would be given this information as supplementary material in any examination.</p></div>
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<div class="markdown"><p><mathjax>#1.5#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#O_2#</mathjax> gas. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that dioxygen gas behaves ideally. Under the given conditions, an ideal gas has a volume of <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p>So we simply divide the volume by the molar volume under standard conditions, <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p><mathjax>#(33.6*cancelL)/(22.4*cancelL*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax></p>
<p>How did I know that the molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> was <mathjax>#22.4*L#</mathjax>? You would be given this information as supplementary material in any examination.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of oxygen are present in 33.6 L of the gas at 1 atm and 0°C?</h1>
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<div class="markdown"><p><mathjax>#1.5#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#O_2#</mathjax> gas. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that dioxygen gas behaves ideally. Under the given conditions, an ideal gas has a volume of <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p>So we simply divide the volume by the molar volume under standard conditions, <mathjax>#22.4*L*mol^-1#</mathjax>.</p>
<p><mathjax>#(33.6*cancelL)/(22.4*cancelL*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax></p>
<p>How did I know that the molar volume of an ideal gas at <mathjax>#"STP"#</mathjax> was <mathjax>#22.4*L#</mathjax>? You would be given this information as supplementary material in any examination.</p></div>
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</article> | How many moles of oxygen are present in 33.6 L of the gas at 1 atm and 0°C? | null |
693 | ac86fda4-6ddd-11ea-b88e-ccda262736ce | https://socratic.org/questions/a-volume-of-32-45-ml-of-0-122-m-naoh-was-required-to-titration-4-89-g-of-vinegar | 3.96 × 10^(-3) moles | start physical_unit 8 8 mole mol qc_end physical_unit 8 8 6 7 molarity qc_end physical_unit 16 16 13 14 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] NaOH [IN] moles"}] | [{"type":"physical unit","value":"3.96 × 10^(-3) moles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{32.45 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.122 M}"},{"type":"physical unit","value":"Mass [OF] vinegar [=] \\pu{4.89 g}"},{"type":"other","value":"Titration to phenolphthalein end point."}] | <h1 class="questionTitle" itemprop="name">A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added? </h1> | null | 3.96 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Vxxc#</mathjax> <mathjax>#=#</mathjax> <mathjax>#32.45#</mathjax> <mathjax>#xx10^(-3)cancelL#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#0.122#</mathjax> <mathjax>#mol*cancelL^-1#</mathjax></p>
<p><mathjax>#= 3.96xx10^(-3) mol#</mathjax> <mathjax>#NaOH#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Concentration (<mathjax>#c#</mathjax>) is always moles (or sometimes mass) per unit volume. If <mathjax>#c#</mathjax> <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax>, then <mathjax>#n#</mathjax> (molar quantity) <mathjax>#=#</mathjax> volume <mathjax>#(V)#</mathjax> <mathjax>#xx#</mathjax> concentration <mathjax>#n/V#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Vxxc#</mathjax> <mathjax>#=#</mathjax> <mathjax>#32.45#</mathjax> <mathjax>#xx10^(-3)cancelL#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#0.122#</mathjax> <mathjax>#mol*cancelL^-1#</mathjax></p>
<p><mathjax>#= 3.96xx10^(-3) mol#</mathjax> <mathjax>#NaOH#</mathjax> </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added? </h1>
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anor277
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<div class="markdown"><p>Concentration (<mathjax>#c#</mathjax>) is always moles (or sometimes mass) per unit volume. If <mathjax>#c#</mathjax> <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax>, then <mathjax>#n#</mathjax> (molar quantity) <mathjax>#=#</mathjax> volume <mathjax>#(V)#</mathjax> <mathjax>#xx#</mathjax> concentration <mathjax>#n/V#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#Vxxc#</mathjax> <mathjax>#=#</mathjax> <mathjax>#32.45#</mathjax> <mathjax>#xx10^(-3)cancelL#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#0.122#</mathjax> <mathjax>#mol*cancelL^-1#</mathjax></p>
<p><mathjax>#= 3.96xx10^(-3) mol#</mathjax> <mathjax>#NaOH#</mathjax> </p></div>
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</article> | A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added? | null |
694 | ab45c87a-6ddd-11ea-9df6-ccda262736ce | https://socratic.org/questions/what-is-the-ka-for-1-0moll-1-nh4-with-a-ph-4-64 | 5.3 × 10^(-10) | start physical_unit 8 9 ka none qc_end physical_unit 8 9 6 7 molarity qc_end physical_unit 8 9 14 14 ph qc_end end | [{"type":"physical unit","value":"Ka [OF] NH4+ solution"}] | [{"type":"physical unit","value":"5.3 × 10^(-10)"}] | [{"type":"physical unit","value":"Molarity [OF] NH4+ solution [=] \\pu{1.0 mol/L}"},{"type":"physical unit","value":"pH [OF] NH4+ solution [=] \\pu{4.64}"}] | <h1 class="questionTitle" itemprop="name">What is the #K_a# for a #"1.0-mol L"^(-1)# #"NH"_4^(+)# solution with a #"pH"# of #4.64# ?</h1> | null | 5.3 × 10^(-10) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The ammonium cation will act as a <em>weak acid</em> in aqueous solution, as you can undoubtedly tell from the fact that the <mathjax>#"pH"#</mathjax> of the solution is <mathjax>#<7#</mathjax>. </p>
<p>The balanced chemical equation that describes the partial ionization of the ammonium cation looks like this</p>
<blockquote>
<p><mathjax>#"NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, will take the form</p>
<blockquote>
<p><mathjax>#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of the hydronium cations, since</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of ammonium cations <strong>that ionizes</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of ammonia and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. This tells you that, at equilibrium, the concentration of ammonia will be <strong>equal</strong> to the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#["NH"_3] = ["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>The equilibrium concentration of the ammonium cations will be equal to</p>
<blockquote>
<p><mathjax>#["NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
<blockquote>
<p>This is the case because in order for the reaction to <strong>produce</strong> <mathjax>#["H"_3"O"^(+)] =10^(-"pH") quad "M"#</mathjax>, the initial concentration of the ammoniumc cations, <mathjax>#["NH"_4^(+)]_0#</mathjax>, must <strong>decrease</strong> by <mathjax>#10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
</blockquote>
<p>The acid dissociation constant will thus be equal to--I'll do the calculations <em>without added units</em>! </p>
<blockquote>
<p><mathjax>#K_ a = (10^(-"pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH"))#</mathjax></p>
<p><mathjax>#K_a = 10^(-2"pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(- 2 * 4.64)/(1.0 - 10^(-4.64)) = color(darkgreen)(ul(color(black)(5.3 * 10^(-10))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_a = 5.3 * 10^(-10)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The ammonium cation will act as a <em>weak acid</em> in aqueous solution, as you can undoubtedly tell from the fact that the <mathjax>#"pH"#</mathjax> of the solution is <mathjax>#<7#</mathjax>. </p>
<p>The balanced chemical equation that describes the partial ionization of the ammonium cation looks like this</p>
<blockquote>
<p><mathjax>#"NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, will take the form</p>
<blockquote>
<p><mathjax>#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of the hydronium cations, since</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of ammonium cations <strong>that ionizes</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of ammonia and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. This tells you that, at equilibrium, the concentration of ammonia will be <strong>equal</strong> to the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#["NH"_3] = ["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>The equilibrium concentration of the ammonium cations will be equal to</p>
<blockquote>
<p><mathjax>#["NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
<blockquote>
<p>This is the case because in order for the reaction to <strong>produce</strong> <mathjax>#["H"_3"O"^(+)] =10^(-"pH") quad "M"#</mathjax>, the initial concentration of the ammoniumc cations, <mathjax>#["NH"_4^(+)]_0#</mathjax>, must <strong>decrease</strong> by <mathjax>#10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
</blockquote>
<p>The acid dissociation constant will thus be equal to--I'll do the calculations <em>without added units</em>! </p>
<blockquote>
<p><mathjax>#K_ a = (10^(-"pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH"))#</mathjax></p>
<p><mathjax>#K_a = 10^(-2"pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(- 2 * 4.64)/(1.0 - 10^(-4.64)) = color(darkgreen)(ul(color(black)(5.3 * 10^(-10))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the #K_a# for a #"1.0-mol L"^(-1)# #"NH"_4^(+)# solution with a #"pH"# of #4.64# ?</h1>
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Stefan V.
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Feb 15, 2018
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<div class="markdown"><p><mathjax>#K_a = 5.3 * 10^(-10)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The ammonium cation will act as a <em>weak acid</em> in aqueous solution, as you can undoubtedly tell from the fact that the <mathjax>#"pH"#</mathjax> of the solution is <mathjax>#<7#</mathjax>. </p>
<p>The balanced chemical equation that describes the partial ionization of the ammonium cation looks like this</p>
<blockquote>
<p><mathjax>#"NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote>
<p>By definition, the <strong>acid dissociation constant</strong>, <mathjax>#K_a#</mathjax>, will take the form</p>
<blockquote>
<p><mathjax>#K_a = (["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)])#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution will give you the <strong>equilibrium concentration</strong> of the hydronium cations, since</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>Notice that <strong>every mole</strong> of ammonium cations <strong>that ionizes</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of ammonia and <mathjax>#1#</mathjax> <strong>mole</strong> of hydronium cations. This tells you that, at equilibrium, the concentration of ammonia will be <strong>equal</strong> to the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#["NH"_3] = ["H"_3"O"^(+)] = 10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
<p>The equilibrium concentration of the ammonium cations will be equal to</p>
<blockquote>
<p><mathjax>#["NH"_ 4^(+)] = ["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
<blockquote>
<p>This is the case because in order for the reaction to <strong>produce</strong> <mathjax>#["H"_3"O"^(+)] =10^(-"pH") quad "M"#</mathjax>, the initial concentration of the ammoniumc cations, <mathjax>#["NH"_4^(+)]_0#</mathjax>, must <strong>decrease</strong> by <mathjax>#10^(-"pH") quad "M"#</mathjax></p>
</blockquote>
</blockquote>
<p>The acid dissociation constant will thus be equal to--I'll do the calculations <em>without added units</em>! </p>
<blockquote>
<p><mathjax>#K_ a = (10^(-"pH") * 10^(-"pH"))/(["NH"_ 4^(+)]_ 0 - 10^(-"pH"))#</mathjax></p>
<p><mathjax>#K_a = 10^(-2"pH")/(["NH"_ 4^(+)]_ 0 - 10^(-"pH")#</mathjax></p>
</blockquote>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#K_a = 10^(- 2 * 4.64)/(1.0 - 10^(-4.64)) = color(darkgreen)(ul(color(black)(5.3 * 10^(-10))))#</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> | What is the #K_a# for a #"1.0-mol L"^(-1)# #"NH"_4^(+)# solution with a #"pH"# of #4.64# ? | null |
695 | a9863d02-6ddd-11ea-a614-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-3-moles-of-c2h2 | 78.12 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] C2H2 [IN] g"}] | [{"type":"physical unit","value":"78.12 g"}] | [{"type":"physical unit","value":"Mole [OF] C2H2 [=] \\pu{3 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 3 moles of #"C"_2"H"_2# ? </h1> | null | 78.12 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every mole of <mathjax>#"C"_2"H"_2#</mathjax> (acetylene) has a mass of <mathjax>#"26.04 g"#</mathjax>, i.e. acetylene has a molar mass of <mathjax>#"26.04 g/mol"#</mathjax>.</p>
<p>Thus, </p>
<p><mathjax>#"26.04 g/mol " * " 3 mol"#</mathjax> </p>
<p>gives you the mass of <mathjax>#3#</mathjax> moles, <mathjax>#"78 g"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"78 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every mole of <mathjax>#"C"_2"H"_2#</mathjax> (acetylene) has a mass of <mathjax>#"26.04 g"#</mathjax>, i.e. acetylene has a molar mass of <mathjax>#"26.04 g/mol"#</mathjax>.</p>
<p>Thus, </p>
<p><mathjax>#"26.04 g/mol " * " 3 mol"#</mathjax> </p>
<p>gives you the mass of <mathjax>#3#</mathjax> moles, <mathjax>#"78 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 3 moles of #"C"_2"H"_2# ? </h1>
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<div class="markdown"><p><mathjax>#"78 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Every mole of <mathjax>#"C"_2"H"_2#</mathjax> (acetylene) has a mass of <mathjax>#"26.04 g"#</mathjax>, i.e. acetylene has a molar mass of <mathjax>#"26.04 g/mol"#</mathjax>.</p>
<p>Thus, </p>
<p><mathjax>#"26.04 g/mol " * " 3 mol"#</mathjax> </p>
<p>gives you the mass of <mathjax>#3#</mathjax> moles, <mathjax>#"78 g"#</mathjax></p></div>
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Sahar Mulla ❤
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<div class="markdown"><p>Molar mass of <mathjax>#C_2H_2 => 26g#</mathjax>, that's the mass of 1 mole.</p>
<p>Therefore, mass of 3 moles of <mathjax>#C_2H_2#</mathjax> would be <mathjax>#26 xx 3 = color(red)(78g#</mathjax></p></div>
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Nam D.
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<span class="dateCreated" datetime="2018-02-27T00:25:53" itemprop="dateCreated">
Feb 27, 2018
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<div class="markdown"><p><mathjax>#78 \ "g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Acetylene <mathjax>#(C_2H_2)#</mathjax> has a molar mass of <mathjax>#26.04 \ "g/mol"#</mathjax>.</p>
<p>To find out the mass of moles, we have to multiply the number of moles by the molar mass of the substance, i.e. we can form an equation like this:</p>
<p><mathjax>#"mass"="moles"*"molar mass"#</mathjax></p>
<p>So, the mass of three moles of acetylene is</p>
<p><mathjax>#"mass"_(C_2H_2)=3color(red)cancelcolor(black)"mol"*(26.04 \ "g")/(color(red)cancelcolor(black)"mol")=78.12 \ "g"#</mathjax></p>
<p>Now, we can round off to the nearest whole number to make the answer more reasonable, so it'll be</p>
<p><mathjax>#78.12 \ "g"~~78 \ "g"#</mathjax></p></div>
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</article> | What is the mass of 3 moles of #"C"_2"H"_2# ? | null |
696 | aa5f89e6-6ddd-11ea-9a3a-ccda262736ce | https://socratic.org/questions/how-do-you-write-a-balanced-chemical-equation-for-the-reaction-in-which-dinitrog | N2O5(g) + H2O(g) <=> 2 HNO3(g) | start chemical_equation qc_end substance 13 15 qc_end substance 18 18 qc_end substance 21 22 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"N2O5(g) + H2O(g) <=> 2 HNO3(g)"}] | [{"type":"substance name","value":"Dinitrogen pentoxide gas"},{"type":"substance name","value":"Water"},{"type":"substance name","value":"Nitric acid"}] | <h1 class="questionTitle" itemprop="name">How do you write a balanced chemical equation for the reaction in which dinitrogen pentoxide gas reacts with water to produce nitric acid?</h1> | null | N2O5(g) + H2O(g) <=> 2 HNO3(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Dinitrogen pentoxide"#</mathjax> is the <mathjax>#"acid anhydride"#</mathjax> of nitric acid. </p></div>
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<div class="markdown"><p><mathjax>#N_2O_5(g) + H_2O(g) rightleftharpoons2HNO_3(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Dinitrogen pentoxide"#</mathjax> is the <mathjax>#"acid anhydride"#</mathjax> of nitric acid. </p></div>
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<div class="markdown"><p><mathjax>#N_2O_5(g) + H_2O(g) rightleftharpoons2HNO_3(g)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Dinitrogen pentoxide"#</mathjax> is the <mathjax>#"acid anhydride"#</mathjax> of nitric acid. </p></div>
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</article> | How do you write a balanced chemical equation for the reaction in which dinitrogen pentoxide gas reacts with water to produce nitric acid? | null |
697 | ab028cd8-6ddd-11ea-8afe-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-cupric-cyanide | Cu(CN)2 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] cupric cyanide [IN] default"}] | [{"type":"chemical equation","value":"Cu(CN)2"}] | [{"type":"substance name","value":"Cupric cyanide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for cupric cyanide? </h1> | null | Cu(CN)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The cupric ion (from the latin "cuprum", corresponding to <mathjax>#"Cu"#</mathjax>) is the <strong>copper(II)</strong> ion, with a charge of <mathjax>#2^+#</mathjax>, because the "-ic" ending implies the higher of two oxidation states (<mathjax>#+1,+2#</mathjax>), as opposed to "cuprous". Its symbol is therefore <mathjax>#"Cu"^(2+)#</mathjax>. </p>
<p>The cyanide ion has a charge of <mathjax>#1^-#</mathjax>, and its formula is <mathjax>#"CN"^(-)#</mathjax>.</p>
<p>All <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>' charges must <strong>add up to</strong> the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>, so the number of positive ions and negative ions must be such that the total positive charge plus the total negative charge <strong>equals</strong> the net charge.</p>
<p>Since the cupric ion has a charge of <mathjax>#2^+#</mathjax>, and the cyanide ion has a charge of <mathjax>#1^(-)#</mathjax>, there must be <strong>two</strong> cyanide ions <strong>for every</strong> cupric ion. </p>
<p>Therefore, the formula for <strong>cupric cyanide</strong> contains <em>one</em> cupric ion and <em>two</em> cyanide ions, and the formula is <mathjax>#"Cu(CN)"_2"#</mathjax>. This compound is also called <em>copper(II) cyanide</em>.</p>
<blockquote>
<p><mathjax>#"Cu"^(2+) + "2CN"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Cu(CN)"_2#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The formula for cupric cyanide is <mathjax>#"Cu(CN)"_2"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The cupric ion (from the latin "cuprum", corresponding to <mathjax>#"Cu"#</mathjax>) is the <strong>copper(II)</strong> ion, with a charge of <mathjax>#2^+#</mathjax>, because the "-ic" ending implies the higher of two oxidation states (<mathjax>#+1,+2#</mathjax>), as opposed to "cuprous". Its symbol is therefore <mathjax>#"Cu"^(2+)#</mathjax>. </p>
<p>The cyanide ion has a charge of <mathjax>#1^-#</mathjax>, and its formula is <mathjax>#"CN"^(-)#</mathjax>.</p>
<p>All <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>' charges must <strong>add up to</strong> the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>, so the number of positive ions and negative ions must be such that the total positive charge plus the total negative charge <strong>equals</strong> the net charge.</p>
<p>Since the cupric ion has a charge of <mathjax>#2^+#</mathjax>, and the cyanide ion has a charge of <mathjax>#1^(-)#</mathjax>, there must be <strong>two</strong> cyanide ions <strong>for every</strong> cupric ion. </p>
<p>Therefore, the formula for <strong>cupric cyanide</strong> contains <em>one</em> cupric ion and <em>two</em> cyanide ions, and the formula is <mathjax>#"Cu(CN)"_2"#</mathjax>. This compound is also called <em>copper(II) cyanide</em>.</p>
<blockquote>
<p><mathjax>#"Cu"^(2+) + "2CN"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Cu(CN)"_2#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for cupric cyanide? </h1>
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<div class="markdown"><p>The formula for cupric cyanide is <mathjax>#"Cu(CN)"_2"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The cupric ion (from the latin "cuprum", corresponding to <mathjax>#"Cu"#</mathjax>) is the <strong>copper(II)</strong> ion, with a charge of <mathjax>#2^+#</mathjax>, because the "-ic" ending implies the higher of two oxidation states (<mathjax>#+1,+2#</mathjax>), as opposed to "cuprous". Its symbol is therefore <mathjax>#"Cu"^(2+)#</mathjax>. </p>
<p>The cyanide ion has a charge of <mathjax>#1^-#</mathjax>, and its formula is <mathjax>#"CN"^(-)#</mathjax>.</p>
<p>All <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>' charges must <strong>add up to</strong> the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a>, so the number of positive ions and negative ions must be such that the total positive charge plus the total negative charge <strong>equals</strong> the net charge.</p>
<p>Since the cupric ion has a charge of <mathjax>#2^+#</mathjax>, and the cyanide ion has a charge of <mathjax>#1^(-)#</mathjax>, there must be <strong>two</strong> cyanide ions <strong>for every</strong> cupric ion. </p>
<p>Therefore, the formula for <strong>cupric cyanide</strong> contains <em>one</em> cupric ion and <em>two</em> cyanide ions, and the formula is <mathjax>#"Cu(CN)"_2"#</mathjax>. This compound is also called <em>copper(II) cyanide</em>.</p>
<blockquote>
<p><mathjax>#"Cu"^(2+) + "2CN"^(-)#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Cu(CN)"_2#</mathjax></p>
</blockquote></div>
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</article> | What is the formula for cupric cyanide? | null |
698 | aade2fc6-6ddd-11ea-8def-ccda262736ce | https://socratic.org/questions/given-the-empirical-formula-nh2cl-and-a-molar-mass-of-51-5-g-mol-what-is-the-mol | NH2Cl | start chemical_formula qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"NH2Cl"}] | [{"type":"other","value":"The compound has an empirical formula NH2Cl."},{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{51.5 g/mol}"}] | <h1 class="questionTitle" itemprop="name">Given the empirical formula NH2Cl and a molar mass of 51.5 g/mol, what is the molecular formula?</h1> | null | NH2Cl | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <em>empirical formula</em> tells you what the <strong>smallest whole number ratio</strong> that exists between the atoms that make up a compound is. </p>
<p>In your case, you know that the empirical formula is <mathjax>#"NH"_2"Cl"#</mathjax>, which means that the <strong>regardles</strong> of how many atoms of each element you get in the actual compound, the ratio that exists between them will <strong>always</strong> be <mathjax>#1 : 2 : 1#</mathjax>.</p>
<p>What you actually need to determine is <em>how many empirical formulas</em> are needed to get to the molecular formula. </p>
<p>Notice that the problem provides you with the <em>molar mass</em> of the compound. This means that you can use the molar mass of the <strong>empirical formula</strong> to determine exactly how many atoms you need to form the compound's molecule. </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass empirical formula" xx n = "molar mass compound")#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, use the molar masses of its <em>constituent atoms</em></p>
<blockquote>
<p><mathjax>#"14.0067 g/mol" + 2 xx "1.00794 g/mol" + "35.453 g/mol" = "51.48 g/mol" ~~ "51.5 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#51.5 color(red)(cancel(color(black)("g/mol"))) xx n = 51.5color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>As you can see, you have <mathjax>#n=1#</mathjax>. </p>
<p>This means that the empirical formula and the molecular formula are equivalent, <mathjax>#"NH"_2"Cl"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"NH"_2"Cl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <em>empirical formula</em> tells you what the <strong>smallest whole number ratio</strong> that exists between the atoms that make up a compound is. </p>
<p>In your case, you know that the empirical formula is <mathjax>#"NH"_2"Cl"#</mathjax>, which means that the <strong>regardles</strong> of how many atoms of each element you get in the actual compound, the ratio that exists between them will <strong>always</strong> be <mathjax>#1 : 2 : 1#</mathjax>.</p>
<p>What you actually need to determine is <em>how many empirical formulas</em> are needed to get to the molecular formula. </p>
<p>Notice that the problem provides you with the <em>molar mass</em> of the compound. This means that you can use the molar mass of the <strong>empirical formula</strong> to determine exactly how many atoms you need to form the compound's molecule. </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass empirical formula" xx n = "molar mass compound")#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, use the molar masses of its <em>constituent atoms</em></p>
<blockquote>
<p><mathjax>#"14.0067 g/mol" + 2 xx "1.00794 g/mol" + "35.453 g/mol" = "51.48 g/mol" ~~ "51.5 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#51.5 color(red)(cancel(color(black)("g/mol"))) xx n = 51.5color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>As you can see, you have <mathjax>#n=1#</mathjax>. </p>
<p>This means that the empirical formula and the molecular formula are equivalent, <mathjax>#"NH"_2"Cl"#</mathjax>. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">Given the empirical formula NH2Cl and a molar mass of 51.5 g/mol, what is the molecular formula?</h1>
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Stefan V.
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Nov 6, 2015
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<div class="markdown"><p><mathjax>#"NH"_2"Cl"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, the <em>empirical formula</em> tells you what the <strong>smallest whole number ratio</strong> that exists between the atoms that make up a compound is. </p>
<p>In your case, you know that the empirical formula is <mathjax>#"NH"_2"Cl"#</mathjax>, which means that the <strong>regardles</strong> of how many atoms of each element you get in the actual compound, the ratio that exists between them will <strong>always</strong> be <mathjax>#1 : 2 : 1#</mathjax>.</p>
<p>What you actually need to determine is <em>how many empirical formulas</em> are needed to get to the molecular formula. </p>
<p>Notice that the problem provides you with the <em>molar mass</em> of the compound. This means that you can use the molar mass of the <strong>empirical formula</strong> to determine exactly how many atoms you need to form the compound's molecule. </p>
<blockquote>
<p><mathjax>#color(blue)("molar mass empirical formula" xx n = "molar mass compound")#</mathjax></p>
</blockquote>
<p>To get the molar mass of the empirical formula, use the molar masses of its <em>constituent atoms</em></p>
<blockquote>
<p><mathjax>#"14.0067 g/mol" + 2 xx "1.00794 g/mol" + "35.453 g/mol" = "51.48 g/mol" ~~ "51.5 g/mol"#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#51.5 color(red)(cancel(color(black)("g/mol"))) xx n = 51.5color(red)(cancel(color(black)("g/mol")))#</mathjax></p>
</blockquote>
<p>As you can see, you have <mathjax>#n=1#</mathjax>. </p>
<p>This means that the empirical formula and the molecular formula are equivalent, <mathjax>#"NH"_2"Cl"#</mathjax>. </p></div>
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</article> | Given the empirical formula NH2Cl and a molar mass of 51.5 g/mol, what is the molecular formula? | null |
699 | a97712c8-6ddd-11ea-af25-ccda262736ce | https://socratic.org/questions/a-compound-is-found-to-be-40-0-carbon-6-7-hydrogen-and-53-5-oxygen-its-molecular | C2H4O2 | start chemical_formula qc_end physical_unit 0 1 17 18 molecular_weight qc_end end | [{"type":"other","value":"Chemical Formula [OF] a compound [IN] molecular"}] | [{"type":"chemical equation","value":"C2H4O2"}] | [{"type":"physical unit","value":"Percent [OF] carbon in compound [=] \\pu{40.0%}"},{"type":"physical unit","value":"Percent [OF] hydrogen in compound [=] \\pu{6.7%}"},{"type":"physical unit","value":"Percent [OF] oxygen in compound [=] \\pu{53.5%}"},{"type":"physical unit","value":"Molecular mass [OF] a compound [=] \\pu{60.0 g/mol}"}] | <h1 class="questionTitle" itemprop="name">A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula?</h1> | null | C2H4O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of the unknown, there are <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#C#</mathjax>; <mathjax>#(6.7*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#H#</mathjax>; and <mathjax>#(53.5*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax>.</p>
<p>We divide thru to get, <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33:6.65:3.34#</mathjax>. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of <mathjax>#CH_2O#</mathjax>, i.e. near enough to WHOLE numbers. </p>
<p>Now the molecular formula is always a multiple of the empirical formula; i.e. <mathjax>#("EF")_n = "MF"#</mathjax>.</p>
<p>So <mathjax>#60.0*g*mol^-1 = nxx(12.011+2xx1.00794+16.00)g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>, and the molecular formula is <mathjax>#2xx(CH_2O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_xH_yO_z#</mathjax>.</p></div>
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<div class="markdown"><p>The empirical formula is <mathjax>#CH_2O#</mathjax>, and the molecular formula is some multiple of this.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of the unknown, there are <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#C#</mathjax>; <mathjax>#(6.7*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#H#</mathjax>; and <mathjax>#(53.5*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax>.</p>
<p>We divide thru to get, <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33:6.65:3.34#</mathjax>. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of <mathjax>#CH_2O#</mathjax>, i.e. near enough to WHOLE numbers. </p>
<p>Now the molecular formula is always a multiple of the empirical formula; i.e. <mathjax>#("EF")_n = "MF"#</mathjax>.</p>
<p>So <mathjax>#60.0*g*mol^-1 = nxx(12.011+2xx1.00794+16.00)g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>, and the molecular formula is <mathjax>#2xx(CH_2O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_xH_yO_z#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula?</h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#CH_2O#</mathjax>, and the molecular formula is some multiple of this.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In 100 g of the unknown, there are <mathjax>#(40.0*g)/(12.011*g*mol^-1)#</mathjax> <mathjax>#C#</mathjax>; <mathjax>#(6.7*g)/(1.00794*g*mol^-1)#</mathjax> <mathjax>#H#</mathjax>; and <mathjax>#(53.5*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#O#</mathjax>.</p>
<p>We divide thru to get, <mathjax>#C:H:O#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.33:6.65:3.34#</mathjax>. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of <mathjax>#CH_2O#</mathjax>, i.e. near enough to WHOLE numbers. </p>
<p>Now the molecular formula is always a multiple of the empirical formula; i.e. <mathjax>#("EF")_n = "MF"#</mathjax>.</p>
<p>So <mathjax>#60.0*g*mol^-1 = nxx(12.011+2xx1.00794+16.00)g*mol^-1#</mathjax>.</p>
<p>Clearly <mathjax>#n=2#</mathjax>, and the molecular formula is <mathjax>#2xx(CH_2O)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_xH_yO_z#</mathjax>.</p></div>
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</article> | A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula? | null |
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