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2,600 | a8d5ad9c-6ddd-11ea-b667-ccda262736ce | https://socratic.org/questions/if-28-0-g-of-methane-gas-ch-4-are-introduced-into-an-evacuated-2-00-l-gas-cylind | 22.06 atm | start physical_unit 27 28 pressure atm qc_end physical_unit 14 15 12 13 volume qc_end physical_unit 14 15 20 21 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] the cylinder [IN] atm"}] | [{"type":"physical unit","value":"22.06 atm"}] | [{"type":"physical unit","value":"Mass [OF] CH4 gas [=] \\pu{28.0 g}"},{"type":"physical unit","value":"Volume [OF] gas cylinder [=] \\pu{2.00 L}"},{"type":"physical unit","value":"Temperature [OF] gas cylinder [=] \\pu{35 ℃}"}] | <h1 class="questionTitle" itemprop="name"> If 28.0 g of methane gas (#CH_4#) are introduced into an evacuated 2.00-L gas cylinder at a temperature of 35°C, what is the pressure inside the cylinder? </h1> | null | 22.06 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> with the formula <mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p>Determine <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> of <mathjax>#"CH"_4"#</mathjax> by dividing the given mass by its molar mass, <mathjax>#"16.04246 g/mol"#</mathjax> <a href="https://pubchem.ncbi.nlm.nih.gov/compound/297" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/297</a></p>
<p><mathjax>#28.0cancel"g CH"_4xx(1"mol CH"_4)/(16.04246cancel"g CH"_4)="1.7454 mol CH"_4"#</mathjax></p>
<p><strong>Ideal Gas Law</strong></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V="2.00 L"#</mathjax><br/>
<mathjax>#n="1.7454 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)#</mathjax><br/>
<mathjax>#T="35"^@"C"+273.15="308 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure, <mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#P#</mathjax>. Substitute the given values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((1.7454"mol") xx (0.082057338"L atm K"^(-1) "mol"^(-1)) xx (308"K"))/(2.00"L")="22.2 atm"#</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">
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<div>
<div class="markdown"><p>The pressure of the methane gas will be 22.2 atm.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> with the formula <mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p>Determine <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> of <mathjax>#"CH"_4"#</mathjax> by dividing the given mass by its molar mass, <mathjax>#"16.04246 g/mol"#</mathjax> <a href="https://pubchem.ncbi.nlm.nih.gov/compound/297" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/297</a></p>
<p><mathjax>#28.0cancel"g CH"_4xx(1"mol CH"_4)/(16.04246cancel"g CH"_4)="1.7454 mol CH"_4"#</mathjax></p>
<p><strong>Ideal Gas Law</strong></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V="2.00 L"#</mathjax><br/>
<mathjax>#n="1.7454 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)#</mathjax><br/>
<mathjax>#T="35"^@"C"+273.15="308 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure, <mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#P#</mathjax>. Substitute the given values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((1.7454"mol") xx (0.082057338"L atm K"^(-1) "mol"^(-1)) xx (308"K"))/(2.00"L")="22.2 atm"#</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"> If 28.0 g of methane gas (#CH_4#) are introduced into an evacuated 2.00-L gas cylinder at a temperature of 35°C, what is the pressure inside the cylinder? </h1>
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<div class="markdown"><p>The pressure of the methane gas will be 22.2 atm.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> with the formula <mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is the gas constant, and <mathjax>#T#</mathjax> is temperature in Kelvins.</p>
<p>Determine <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> of <mathjax>#"CH"_4"#</mathjax> by dividing the given mass by its molar mass, <mathjax>#"16.04246 g/mol"#</mathjax> <a href="https://pubchem.ncbi.nlm.nih.gov/compound/297" rel="nofollow" target="_blank">https://pubchem.ncbi.nlm.nih.gov/compound/297</a></p>
<p><mathjax>#28.0cancel"g CH"_4xx(1"mol CH"_4)/(16.04246cancel"g CH"_4)="1.7454 mol CH"_4"#</mathjax></p>
<p><strong>Ideal Gas Law</strong></p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V="2.00 L"#</mathjax><br/>
<mathjax>#n="1.7454 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)#</mathjax><br/>
<mathjax>#T="35"^@"C"+273.15="308 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure, <mathjax>#P#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the formula to isolate <mathjax>#P#</mathjax>. Substitute the given values into the formula and solve.</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p><mathjax>#P=((1.7454"mol") xx (0.082057338"L atm K"^(-1) "mol"^(-1)) xx (308"K"))/(2.00"L")="22.2 atm"#</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> | If 28.0 g of methane gas (#CH_4#) are introduced into an evacuated 2.00-L gas cylinder at a temperature of 35°C, what is the pressure inside the cylinder? | null |
2,601 | a9539d88-6ddd-11ea-aab6-ccda262736ce | https://socratic.org/questions/how-many-valence-electrons-are-in-an-atom-of-phosphorus-1 | 5 | start physical_unit 2 3 number none qc_end end | [{"type":"physical unit","value":"Number [OF] valence electrons"}] | [{"type":"physical unit","value":"5"}] | [{"type":"physical unit","value":"Number [OF] phosphorus atom [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How many valence electrons are in an atom of phosphorus?</h1> | null | 5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>phosphorus has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a> <mathjax>#15#</mathjax>, and is in group <mathjax>#5#</mathjax> and period <mathjax>#3#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>.</p>
<p>it is in period <mathjax>#3#</mathjax>, because it has orbits at <mathjax>#3#</mathjax> energy levels.</p>
<p>there are <mathjax>#2#</mathjax> electrons at the first energy level, and <mathjax>#8#</mathjax> at the second energy level.</p>
<p>there are <mathjax>#15#</mathjax> electrons in total, there must be <mathjax>#5#</mathjax> electrons at the third energy level.</p>
<p>there are <mathjax>#3#</mathjax> energy levels, so the electrons at the third are outer, or valence, electrons.</p>
<p>this means that there are <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>.</p>
<p>all neutral atoms of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in group <mathjax>#5#</mathjax> have <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>.</p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>phosphorus has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a> <mathjax>#15#</mathjax>, and is in group <mathjax>#5#</mathjax> and period <mathjax>#3#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>.</p>
<p>it is in period <mathjax>#3#</mathjax>, because it has orbits at <mathjax>#3#</mathjax> energy levels.</p>
<p>there are <mathjax>#2#</mathjax> electrons at the first energy level, and <mathjax>#8#</mathjax> at the second energy level.</p>
<p>there are <mathjax>#15#</mathjax> electrons in total, there must be <mathjax>#5#</mathjax> electrons at the third energy level.</p>
<p>there are <mathjax>#3#</mathjax> energy levels, so the electrons at the third are outer, or valence, electrons.</p>
<p>this means that there are <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>.</p>
<p>all neutral atoms of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in group <mathjax>#5#</mathjax> have <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many valence electrons are in an atom of phosphorus?</h1>
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<div class="markdown"><p><mathjax>#5#</mathjax></p></div>
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<div class="markdown"><p>phosphorus has <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a> <mathjax>#15#</mathjax>, and is in group <mathjax>#5#</mathjax> and period <mathjax>#3#</mathjax> of <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a>.</p>
<p>it is in period <mathjax>#3#</mathjax>, because it has orbits at <mathjax>#3#</mathjax> energy levels.</p>
<p>there are <mathjax>#2#</mathjax> electrons at the first energy level, and <mathjax>#8#</mathjax> at the second energy level.</p>
<p>there are <mathjax>#15#</mathjax> electrons in total, there must be <mathjax>#5#</mathjax> electrons at the third energy level.</p>
<p>there are <mathjax>#3#</mathjax> energy levels, so the electrons at the third are outer, or valence, electrons.</p>
<p>this means that there are <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a>.</p>
<p>all neutral atoms of the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in group <mathjax>#5#</mathjax> have <mathjax>#5#</mathjax> <a href="https://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>.</p></div>
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</article> | How many valence electrons are in an atom of phosphorus? | null |
2,602 | a977ff3e-6ddd-11ea-b04d-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-a-gas-if-3-50-grams-of-the-gas-occupy-1-35-liters-of-s | 58.92 g/mol | start physical_unit 12 13 molar_mass g/mol qc_end physical_unit 12 13 9 10 mass qc_end physical_unit 12 13 15 16 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"58.92 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the gas [=] \\pu{3.50 grams}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{1.35 liters}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP? </h1> | null | 58.92 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You will need to use the equation for the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is gas constant, and <mathjax>#T#</mathjax> is temperature.</p>
<p><mathjax>#"STP"#</mathjax> is <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> (required for gas laws), and <mathjax>#10^5#</mathjax> <mathjax>#"Pa"#</mathjax> or <mathjax>#"100 kPa"#</mathjax>.</p>
<p>Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.</p>
<p><strong>Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#V="1.35 L"#</mathjax></p>
<p><mathjax>#R="8.31446 L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#n#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the ideal gas law equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol"#</mathjax> (rounded to three significant figures)</p>
<p>To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.</p>
<p><mathjax>#"molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol"#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>The molar mass of the gas is <mathjax>#"58.9 g/mol"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You will need to use the equation for the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is gas constant, and <mathjax>#T#</mathjax> is temperature.</p>
<p><mathjax>#"STP"#</mathjax> is <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> (required for gas laws), and <mathjax>#10^5#</mathjax> <mathjax>#"Pa"#</mathjax> or <mathjax>#"100 kPa"#</mathjax>.</p>
<p>Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.</p>
<p><strong>Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#V="1.35 L"#</mathjax></p>
<p><mathjax>#R="8.31446 L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#n#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the ideal gas law equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol"#</mathjax> (rounded to three significant figures)</p>
<p>To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.</p>
<p><mathjax>#"molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol"#</mathjax></p></div>
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<div class="markdown"><p>The molar mass of the gas is <mathjax>#"58.9 g/mol"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You will need to use the equation for the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>:</p>
<p><mathjax>#PV=nRT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#P#</mathjax> is pressure, <mathjax>#V#</mathjax> is volume, <mathjax>#n#</mathjax> is moles, <mathjax>#R#</mathjax> is gas constant, and <mathjax>#T#</mathjax> is temperature.</p>
<p><mathjax>#"STP"#</mathjax> is <mathjax>#0^@"C"#</mathjax> or <mathjax>#"273.15 K"#</mathjax> (required for gas laws), and <mathjax>#10^5#</mathjax> <mathjax>#"Pa"#</mathjax> or <mathjax>#"100 kPa"#</mathjax>.</p>
<p>Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.</p>
<p><strong>Known</strong></p>
<p><mathjax>#P="100 kPa"#</mathjax></p>
<p><mathjax>#V="1.35 L"#</mathjax></p>
<p><mathjax>#R="8.31446 L kPa K"^(-1) "mol"^(-1)"#</mathjax></p>
<p><mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#n#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the ideal gas law equation to isolate <mathjax>#n#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol"#</mathjax> (rounded to three significant figures)</p>
<p>To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.</p>
<p><mathjax>#"molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol"#</mathjax></p></div>
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</article> | What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP? | null |
2,603 | ab1ec914-6ddd-11ea-a334-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-the-sulfite-ion | SO3^2- | start chemical_formula qc_end substance 7 8 qc_end end | [{"type":"other","value":"Chemical Formula [OF] sulfite ion [IN] default"}] | [{"type":"chemical equation","value":"SO3^2-"}] | [{"type":"substance name","value":"Sulfite ion"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for the sulfite ion? </h1> | null | SO3^2- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Sulfur has a formal oxidation state of <mathjax>#+IV#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#SO_3^(2-)#</mathjax></p></div>
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<div class="markdown"><p>Sulfur has a formal oxidation state of <mathjax>#+IV#</mathjax>.</p></div>
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anor277
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<div class="markdown"><p><mathjax>#SO_3^(2-)#</mathjax></p></div>
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<div class="markdown"><p>Sulfur has a formal oxidation state of <mathjax>#+IV#</mathjax>.</p></div>
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</article> | What is the chemical formula for the sulfite ion? | null |
2,604 | ac612319-6ddd-11ea-80f1-ccda262736ce | https://socratic.org/questions/how-much-heat-is-required-to-warm-1-30-kg-of-sand-from-30-c-to-100-c | 75.99 kJ | start physical_unit 10 10 heat_energy kj qc_end physical_unit 10 10 7 8 mass qc_end physical_unit 10 10 12 13 temperature qc_end physical_unit 10 10 15 16 temperature qc_end end | [{"type":"physical unit","value":"Required heat [OF] sand [IN] kJ"}] | [{"type":"physical unit","value":"75.99 kJ"}] | [{"type":"physical unit","value":"Mass [OF] sand [=] \\pu{1.30 kg}"},{"type":"physical unit","value":"Temperature1 [OF] sand [=] \\pu{30 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] sand [=] \\pu{100 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much heat is required to warm 1.30 kg of sand from 30°C to 100°C?</h1> | null | 75.99 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat required to warm a substance is determined using heat capacity. The equation below allows us to solve for the heat released or absorbed when warming a substance.</p>
<p><mathjax>#q = mxxC_sxxDeltaT#</mathjax></p>
<p>For the equation above:</p>
<p><mathjax>#q#</mathjax> = heat, <mathjax>#m#</mathjax> = mass (in grams!), <mathjax>#C_s#</mathjax> = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax>= change in temperature. </p>
<p>Specific heat capacity (<mathjax>#C_s#</mathjax>) tells us how hard or easy it is to heat up a substance. You can look up the <mathjax>#C_s#</mathjax> for different substances in tables in your textbook or on the internet. </p>
<p>For sand, <mathjax>#C_s = 0.835 J/(gxx°C)#</mathjax></p>
<p>In your specific problem:<br/>
<mathjax>#DeltaT=T_f-T_i= 100 - 30 = 70 °C#</mathjax><br/>
<mathjax>#m = 1.30 kg = 1300 g#</mathjax> <br/>
(you must convert to grams when using specific heat capacities since they are in the units of grams)</p>
<p>Next we plug our values into the top equation listed:</p>
<p><mathjax>#q = mxxC_sxxDeltaT = 1300xx0.835xx70 =#</mathjax> 79,985 J</p>
<p>Since your temperature has only one significant figure we round to one sig fig:</p>
<p>q = 80,000 J.</p>
<p>For more examples see this video:</p>
<p>
<iframe src="https://www.youtube.com/embed/5FKFQsFDVl0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p>Heating 1.30 Kg of sand 70 degrees Celsius takes 80,000 Joules of heat. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat required to warm a substance is determined using heat capacity. The equation below allows us to solve for the heat released or absorbed when warming a substance.</p>
<p><mathjax>#q = mxxC_sxxDeltaT#</mathjax></p>
<p>For the equation above:</p>
<p><mathjax>#q#</mathjax> = heat, <mathjax>#m#</mathjax> = mass (in grams!), <mathjax>#C_s#</mathjax> = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax>= change in temperature. </p>
<p>Specific heat capacity (<mathjax>#C_s#</mathjax>) tells us how hard or easy it is to heat up a substance. You can look up the <mathjax>#C_s#</mathjax> for different substances in tables in your textbook or on the internet. </p>
<p>For sand, <mathjax>#C_s = 0.835 J/(gxx°C)#</mathjax></p>
<p>In your specific problem:<br/>
<mathjax>#DeltaT=T_f-T_i= 100 - 30 = 70 °C#</mathjax><br/>
<mathjax>#m = 1.30 kg = 1300 g#</mathjax> <br/>
(you must convert to grams when using specific heat capacities since they are in the units of grams)</p>
<p>Next we plug our values into the top equation listed:</p>
<p><mathjax>#q = mxxC_sxxDeltaT = 1300xx0.835xx70 =#</mathjax> 79,985 J</p>
<p>Since your temperature has only one significant figure we round to one sig fig:</p>
<p>q = 80,000 J.</p>
<p>For more examples see this video:</p>
<p>
<iframe src="https://www.youtube.com/embed/5FKFQsFDVl0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">How much heat is required to warm 1.30 kg of sand from 30°C to 100°C?</h1>
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<div class="markdown"><p>Heating 1.30 Kg of sand 70 degrees Celsius takes 80,000 Joules of heat. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The amount of heat required to warm a substance is determined using heat capacity. The equation below allows us to solve for the heat released or absorbed when warming a substance.</p>
<p><mathjax>#q = mxxC_sxxDeltaT#</mathjax></p>
<p>For the equation above:</p>
<p><mathjax>#q#</mathjax> = heat, <mathjax>#m#</mathjax> = mass (in grams!), <mathjax>#C_s#</mathjax> = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax>= change in temperature. </p>
<p>Specific heat capacity (<mathjax>#C_s#</mathjax>) tells us how hard or easy it is to heat up a substance. You can look up the <mathjax>#C_s#</mathjax> for different substances in tables in your textbook or on the internet. </p>
<p>For sand, <mathjax>#C_s = 0.835 J/(gxx°C)#</mathjax></p>
<p>In your specific problem:<br/>
<mathjax>#DeltaT=T_f-T_i= 100 - 30 = 70 °C#</mathjax><br/>
<mathjax>#m = 1.30 kg = 1300 g#</mathjax> <br/>
(you must convert to grams when using specific heat capacities since they are in the units of grams)</p>
<p>Next we plug our values into the top equation listed:</p>
<p><mathjax>#q = mxxC_sxxDeltaT = 1300xx0.835xx70 =#</mathjax> 79,985 J</p>
<p>Since your temperature has only one significant figure we round to one sig fig:</p>
<p>q = 80,000 J.</p>
<p>For more examples see this video:</p>
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<iframe src="https://www.youtube.com/embed/5FKFQsFDVl0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How much heat is required to warm 1.30 kg of sand from 30°C to 100°C? | null |
2,605 | a83b9f46-6ddd-11ea-aa3f-ccda262736ce | https://socratic.org/questions/2-30-l-of-air-at-4-53-atm-is-expanded-to-2219-mmhg-what-is-the-final-volume-in-m | 3570 mL | start physical_unit 3 3 volume ml qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 3 3 5 6 pressure qc_end physical_unit 3 3 10 11 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] air [IN] mL"}] | [{"type":"physical unit","value":"3570 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] air [=] \\pu{2.30 L}"},{"type":"physical unit","value":"Pressure1 [OF] air [=] \\pu{4.53 atm}"},{"type":"physical unit","value":"Pressure2 [OF] air [=] \\pu{2219 mmHg}"}] | <h1 class="questionTitle" itemprop="name"> 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?</h1> | null | 3570 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> is an absurd unit. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
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<h1 class="questionTitle" itemprop="name"> 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?</h1>
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<div class="markdown"><p><mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> is an absurd unit. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
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<div class="markdown"><p>The final volume will be <mathjax>#"3570 mL"#</mathjax>, rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First the units of pressure and volume need to be the same.</p>
<p><strong>Pressure</strong></p>
<p><mathjax>#"1 atm=760.0 mmHg"#</mathjax></p>
<p><strong>Convert mmHg to atm.</strong></p>
<p><mathjax>#2219color(red)(cancelcolor(black)("mmHg"))xx(1"atm")/(760.0color(red)(cancelcolor(black)("mmHg")))="2.920 atm"#</mathjax></p>
<p><strong>Volume</strong></p>
<p><mathjax>#"1 L=1000 mL"#</mathjax></p>
<p><strong>Convert <mathjax>#"L"#</mathjax> to <mathjax>#"mL"#</mathjax>.</strong></p>
<p><mathjax>#2.30color(red)(cancel(color(black)("L")))xx(1000"mL")/(1color(red)(cancel(color(black)("L"))))="2300 mL"=2.30xx10^3"mL"#</mathjax> </p>
<p>The number of mL above is given to three significant figures using <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. I will use <mathjax>#"2300 mL"#</mathjax> for convenience only.</p>
<p>This question is concerns <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> , which states that the volume <mathjax>#(V)#</mathjax> of a gas held at constant amount and temperature, is inversely proportional to the pressure <mathjax>#(P)#</mathjax>. This means that if the pressure goes up, the volume goes down, and vice-versa. The equation to use is:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><strong>Write what you know:</strong></p>
<p><mathjax>#P_1="4.53 atm"#</mathjax><br/>
<mathjax>#V_1=2.30xx10^3"mL"#</mathjax><br/>
<mathjax>#P_2="2.920 atm"#</mathjax></p>
<p><strong>Write what you don't know:</strong> <mathjax>#V_2#</mathjax>.</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known quantities into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=(4.53color(red)(cancel(color(black)("atm")))xx2300"mL")/(2.920color(red)(cancel(color(black)("atm"))))="3570 mL"#</mathjax> rounded to three significant figures</p></div>
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</article> | 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL? | null |
2,606 | ab1d69a2-6ddd-11ea-a6fb-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-that-consists-of-72-2-magnesium-and | Mg3N2 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"Mg3N2"}] | [{"type":"physical unit","value":"Percent by mass [OF] magnesium in the compound [=] \\pu{72.2%}"},{"type":"physical unit","value":"Percent by mass [OF] nitrogen in the compound [=] \\pu{27.8%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that consists of 72.2% magnesium and 27.8% nitrogen by mass?</h1> | null | Mg3N2 | <div class="answerDescription">
<div>
<div class="markdown"><p>It wouldn't be a bad idea to start your guessing with <mathjax>#"Mg"_3"N"_2#</mathjax>, as that is the expected ionic compound when <mathjax>#"Mg"^(2+)#</mathjax> forms a compound with <mathjax>#"N"^(3-)#</mathjax>. Ironically, that is the answer.</p>
<hr/>
<p>The molar masses of each atom are:</p>
<blockquote>
<p><mathjax>#"M"_"Mg" ~~ "24.305 g/mol"#</mathjax></p>
<p><mathjax>#"M"_"N" ~~ "14.007 g/mol"#</mathjax></p>
</blockquote>
<p>Using the knowledge that <mathjax>#72.2%#</mathjax> magnesium <strong>by mass</strong>, we can do this:</p>
<blockquote>
<p><mathjax>#%"Mg" = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722 = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" + 0.722"M"_"Mg,total" = "M"_"Mg,total"#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" = 0.278"M"_"Mg,total"#</mathjax></p>
</blockquote>
<p>We should get:</p>
<blockquote>
<p><mathjax>#color(green)("M"_"Mg,total"/"M"_"N,total") = 0.722/0.278#</mathjax></p>
<p><mathjax>#~~ color(green)(2.5971)#</mathjax></p>
</blockquote>
<p>So we have the approximate <strong>mass ratio of magnesium to nitrogen in the compound</strong>, but not the <mathjax>#"mol"#</mathjax> ratio. Note that if we have a 1:1 ratio of <mathjax>#"mol"#</mathjax>s of magnesium to nitrogen, we have:</p>
<blockquote>
<p><mathjax>#color(green)(("M"_"Mg")/("M"_"N")) = 24.305/14.007 ~~ color(green)(1.735)#</mathjax></p>
</blockquote>
<p>But the ratio we <em>have</em> is:</p>
<blockquote>
<p><mathjax>#("M"_"Mg,total")/("M"_"N,total") ~~ 2.5971#</mathjax></p>
</blockquote>
<p>So, we have <strong>more magnesium</strong> (or less nitrogen) than we would have in a 1:1 ratio. As a result, if we <em>divide</em> these two numbers like so, we get the <mathjax>#\mathbf"mol"#</mathjax> <strong>ratio of magnesium to nitrogen</strong>:</p>
<blockquote>
<p><mathjax>#2.5971/1.735 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1.5 = 3/2#</mathjax>, we can <em>scale this up</em> to determine the <strong>empirical formula</strong>. It's evidently:</p>
<blockquote>
<p><mathjax>#color(blue)("Mg"_3"N"_2)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerDescription">
<div>
<div class="markdown"><p>It wouldn't be a bad idea to start your guessing with <mathjax>#"Mg"_3"N"_2#</mathjax>, as that is the expected ionic compound when <mathjax>#"Mg"^(2+)#</mathjax> forms a compound with <mathjax>#"N"^(3-)#</mathjax>. Ironically, that is the answer.</p>
<hr/>
<p>The molar masses of each atom are:</p>
<blockquote>
<p><mathjax>#"M"_"Mg" ~~ "24.305 g/mol"#</mathjax></p>
<p><mathjax>#"M"_"N" ~~ "14.007 g/mol"#</mathjax></p>
</blockquote>
<p>Using the knowledge that <mathjax>#72.2%#</mathjax> magnesium <strong>by mass</strong>, we can do this:</p>
<blockquote>
<p><mathjax>#%"Mg" = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722 = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" + 0.722"M"_"Mg,total" = "M"_"Mg,total"#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" = 0.278"M"_"Mg,total"#</mathjax></p>
</blockquote>
<p>We should get:</p>
<blockquote>
<p><mathjax>#color(green)("M"_"Mg,total"/"M"_"N,total") = 0.722/0.278#</mathjax></p>
<p><mathjax>#~~ color(green)(2.5971)#</mathjax></p>
</blockquote>
<p>So we have the approximate <strong>mass ratio of magnesium to nitrogen in the compound</strong>, but not the <mathjax>#"mol"#</mathjax> ratio. Note that if we have a 1:1 ratio of <mathjax>#"mol"#</mathjax>s of magnesium to nitrogen, we have:</p>
<blockquote>
<p><mathjax>#color(green)(("M"_"Mg")/("M"_"N")) = 24.305/14.007 ~~ color(green)(1.735)#</mathjax></p>
</blockquote>
<p>But the ratio we <em>have</em> is:</p>
<blockquote>
<p><mathjax>#("M"_"Mg,total")/("M"_"N,total") ~~ 2.5971#</mathjax></p>
</blockquote>
<p>So, we have <strong>more magnesium</strong> (or less nitrogen) than we would have in a 1:1 ratio. As a result, if we <em>divide</em> these two numbers like so, we get the <mathjax>#\mathbf"mol"#</mathjax> <strong>ratio of magnesium to nitrogen</strong>:</p>
<blockquote>
<p><mathjax>#2.5971/1.735 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1.5 = 3/2#</mathjax>, we can <em>scale this up</em> to determine the <strong>empirical formula</strong>. It's evidently:</p>
<blockquote>
<p><mathjax>#color(blue)("Mg"_3"N"_2)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that consists of 72.2% magnesium and 27.8% nitrogen by mass?</h1>
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Truong-Son N.
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<div class="markdown"><p>It wouldn't be a bad idea to start your guessing with <mathjax>#"Mg"_3"N"_2#</mathjax>, as that is the expected ionic compound when <mathjax>#"Mg"^(2+)#</mathjax> forms a compound with <mathjax>#"N"^(3-)#</mathjax>. Ironically, that is the answer.</p>
<hr/>
<p>The molar masses of each atom are:</p>
<blockquote>
<p><mathjax>#"M"_"Mg" ~~ "24.305 g/mol"#</mathjax></p>
<p><mathjax>#"M"_"N" ~~ "14.007 g/mol"#</mathjax></p>
</blockquote>
<p>Using the knowledge that <mathjax>#72.2%#</mathjax> magnesium <strong>by mass</strong>, we can do this:</p>
<blockquote>
<p><mathjax>#%"Mg" = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722 = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" + 0.722"M"_"Mg,total" = "M"_"Mg,total"#</mathjax></p>
<p><mathjax>#0.722"M"_"N,total" = 0.278"M"_"Mg,total"#</mathjax></p>
</blockquote>
<p>We should get:</p>
<blockquote>
<p><mathjax>#color(green)("M"_"Mg,total"/"M"_"N,total") = 0.722/0.278#</mathjax></p>
<p><mathjax>#~~ color(green)(2.5971)#</mathjax></p>
</blockquote>
<p>So we have the approximate <strong>mass ratio of magnesium to nitrogen in the compound</strong>, but not the <mathjax>#"mol"#</mathjax> ratio. Note that if we have a 1:1 ratio of <mathjax>#"mol"#</mathjax>s of magnesium to nitrogen, we have:</p>
<blockquote>
<p><mathjax>#color(green)(("M"_"Mg")/("M"_"N")) = 24.305/14.007 ~~ color(green)(1.735)#</mathjax></p>
</blockquote>
<p>But the ratio we <em>have</em> is:</p>
<blockquote>
<p><mathjax>#("M"_"Mg,total")/("M"_"N,total") ~~ 2.5971#</mathjax></p>
</blockquote>
<p>So, we have <strong>more magnesium</strong> (or less nitrogen) than we would have in a 1:1 ratio. As a result, if we <em>divide</em> these two numbers like so, we get the <mathjax>#\mathbf"mol"#</mathjax> <strong>ratio of magnesium to nitrogen</strong>:</p>
<blockquote>
<p><mathjax>#2.5971/1.735 ~~ 1.5#</mathjax></p>
</blockquote>
<p>Since <mathjax>#1.5 = 3/2#</mathjax>, we can <em>scale this up</em> to determine the <strong>empirical formula</strong>. It's evidently:</p>
<blockquote>
<p><mathjax>#color(blue)("Mg"_3"N"_2)#</mathjax></p>
</blockquote></div>
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anor277
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<span class="dateCreated" datetime="2015-12-28T11:12:36" itemprop="dateCreated">
Dec 28, 2015
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<div class="markdown"><p><mathjax>#Mg_3N_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In 100 g of compound there are 72.2 g magnesium, and 27.8 g nitrogen.</p>
<p>We divide thru by the molar masses of each element:</p>
<p>For the metal: <mathjax>#(72.2*cancelg)/(24.31*cancel(g)*mol^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax>;</p>
<p>And for nitrogen: <mathjax>#(27.8*cancelg)/(14.01*cancel(g)*mol^-1)#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#2#</mathjax> <mathjax>#mol#</mathjax>.</p>
<p>So the empirical formula is <mathjax>#Mg_3N_2#</mathjax>. Because this is clearly NOT a molecular species, the empirical formula is all that is needed to identify it. </p></div>
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</article> | What is the empirical formula for a compound that consists of 72.2% magnesium and 27.8% nitrogen by mass? | null |
2,607 | ab452c9d-6ddd-11ea-a33a-ccda262736ce | https://socratic.org/questions/what-is-the-molar-solubility-of-pbcl2-with-a-ksp-of-1-6-x-10-6 | 1.59 × 10^(-2) M | start physical_unit 6 6 molar_solubility mol/l qc_end physical_unit 6 6 11 13 equilibrium_constant_k qc_end end | [{"type":"physical unit","value":"Molar solubility [OF] PbCl2 [IN] M"}] | [{"type":"physical unit","value":"1.59 × 10^(-2) M"}] | [{"type":"physical unit","value":"Ksp [OF] PbCl2 [=] \\pu{1.6 × 10^(-5)}"}] | <h1 class="questionTitle" itemprop="name"> What is the molar solubility of #PbCl_2# with a Ksp of #1.6 xx 10^(-5)#?
</h1> | null | 1.59 × 10^(-2) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Lead(II) chloride</em>, <mathjax>#"PbCl"_2#</mathjax>, is an <em>insoluble</em> <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a>, which means that it <strong>does not</strong> dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution. </p>
<p>Instead of dissociating completely, an <a href="http://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium rection</a> governed by the <em><a href="http://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a></em>, <mathjax>#K_"sp"#</mathjax>, will be established between the solid lead(II) chloride and the dissolved ions. </p>
<blockquote>
<p><mathjax>#"PbCl"_text(2(s]) rightleftharpoons "Pb"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the <strong>molar solubility</strong> of the compound, <mathjax>#s#</mathjax>, represents the number of moles of lead(II) chloride that will <em>dissolve</em> in aqueous solution at a particular temperature. </p>
<p>Notice that <em>every mole</em> of lead(II) chloride will produce <mathjax>#1#</mathjax> <strong>mole</strong> of lead(II) cations and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of chloride anions. Use an <strong>ICE table</strong> to find the molar solubility of the solid</p>
<blockquote>
<p><mathjax>#" " "PbCl"_text(2(s]) " "rightleftharpoons" " "Pb"_text((aq])^(2+) " "+" " color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "-" " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(x)-" " " " " " " " "(+s)" " " " " " " " "(+color(red)(2)s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(x)-" " " " " " " " " " "s" " " " " " " " " " "color(red)(2)s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_"sp" = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#K_"sp" = s * (color(red)(2)s)^color(red)(2) = 4s^3#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of lead(II) chloride will be </p>
<blockquote>
<p><mathjax>#4s^3 = 1.6 * 10^(-5) implies s = root(3)(1.6/4 * 10^(-5)) = color(green)("0.0159 M")#</mathjax></p>
</blockquote>
<p>
<iframe src="https://www.youtube.com/embed/tUnbm379Wfk?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.0159 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Lead(II) chloride</em>, <mathjax>#"PbCl"_2#</mathjax>, is an <em>insoluble</em> <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a>, which means that it <strong>does not</strong> dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution. </p>
<p>Instead of dissociating completely, an <a href="http://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium rection</a> governed by the <em><a href="http://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a></em>, <mathjax>#K_"sp"#</mathjax>, will be established between the solid lead(II) chloride and the dissolved ions. </p>
<blockquote>
<p><mathjax>#"PbCl"_text(2(s]) rightleftharpoons "Pb"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the <strong>molar solubility</strong> of the compound, <mathjax>#s#</mathjax>, represents the number of moles of lead(II) chloride that will <em>dissolve</em> in aqueous solution at a particular temperature. </p>
<p>Notice that <em>every mole</em> of lead(II) chloride will produce <mathjax>#1#</mathjax> <strong>mole</strong> of lead(II) cations and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of chloride anions. Use an <strong>ICE table</strong> to find the molar solubility of the solid</p>
<blockquote>
<p><mathjax>#" " "PbCl"_text(2(s]) " "rightleftharpoons" " "Pb"_text((aq])^(2+) " "+" " color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "-" " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(x)-" " " " " " " " "(+s)" " " " " " " " "(+color(red)(2)s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(x)-" " " " " " " " " " "s" " " " " " " " " " "color(red)(2)s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_"sp" = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#K_"sp" = s * (color(red)(2)s)^color(red)(2) = 4s^3#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of lead(II) chloride will be </p>
<blockquote>
<p><mathjax>#4s^3 = 1.6 * 10^(-5) implies s = root(3)(1.6/4 * 10^(-5)) = color(green)("0.0159 M")#</mathjax></p>
</blockquote>
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<h1 class="questionTitle" itemprop="name"> What is the molar solubility of #PbCl_2# with a Ksp of #1.6 xx 10^(-5)#?
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<div class="markdown"><p><mathjax>#"0.0159 M"#</mathjax></p></div>
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<div class="markdown"><p><em>Lead(II) chloride</em>, <mathjax>#"PbCl"_2#</mathjax>, is an <em>insoluble</em> <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a>, which means that it <strong>does not</strong> dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution. </p>
<p>Instead of dissociating completely, an <a href="http://socratic.org/chemistry/chemical-equilibrium/dynamic-equilibrium">equilibrium rection</a> governed by the <em><a href="http://socratic.org/chemistry/chemical-equilibrium/ksp">solubility product constant</a></em>, <mathjax>#K_"sp"#</mathjax>, will be established between the solid lead(II) chloride and the dissolved ions. </p>
<blockquote>
<p><mathjax>#"PbCl"_text(2(s]) rightleftharpoons "Pb"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>Now, the <strong>molar solubility</strong> of the compound, <mathjax>#s#</mathjax>, represents the number of moles of lead(II) chloride that will <em>dissolve</em> in aqueous solution at a particular temperature. </p>
<p>Notice that <em>every mole</em> of lead(II) chloride will produce <mathjax>#1#</mathjax> <strong>mole</strong> of lead(II) cations and <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of chloride anions. Use an <strong>ICE table</strong> to find the molar solubility of the solid</p>
<blockquote>
<p><mathjax>#" " "PbCl"_text(2(s]) " "rightleftharpoons" " "Pb"_text((aq])^(2+) " "+" " color(red)(2)"Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " " " "-" " " " " " " " " " "0" " " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(x)-" " " " " " " " "(+s)" " " " " " " " "(+color(red)(2)s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(x)-" " " " " " " " " " "s" " " " " " " " " " "color(red)(2)s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_"sp" = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)#</mathjax></p>
<p><mathjax>#K_"sp" = s * (color(red)(2)s)^color(red)(2) = 4s^3#</mathjax></p>
</blockquote>
<p>This means that the molar solubility of lead(II) chloride will be </p>
<blockquote>
<p><mathjax>#4s^3 = 1.6 * 10^(-5) implies s = root(3)(1.6/4 * 10^(-5)) = color(green)("0.0159 M")#</mathjax></p>
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</article> | What is the molar solubility of #PbCl_2# with a Ksp of #1.6 xx 10^(-5)#?
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2,608 | ac1ee911-6ddd-11ea-a63f-ccda262736ce | https://socratic.org/questions/563a8930581e2a05676df6a1 | +5 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] phosphorus"}] | [{"type":"physical unit","value":"+5"}] | [{"type":"chemical equation","value":"PCl5"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in the compound #"PCl"_5"#?</h1> | null | +5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in the compound #"PCl"_5"#?</h1>
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<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p>The oxidation number of <mathjax>#"P"#</mathjax> in the compound <mathjax>#"PCl"_5"#</mathjax> is <mathjax>#"+5"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sum of oxidation numbers in a compound must be zero. The oxidation number of chlorine in <mathjax>#"PCl"_5"#</mathjax> is <mathjax>#"-1"#</mathjax>, so there is an overall oxidation number of <mathjax>#"Cl"#</mathjax> is <mathjax>#"-5"#</mathjax>. In order for the sum of the oxidation numbers to equal zero, the oxidation number of <mathjax>#"P"#</mathjax> is <mathjax>#"+5"#</mathjax>.</p>
<p><mathjax>#(1xx+5)+(5xx-1)=0#</mathjax></p>
<blockquote>
<p><mathjax>#stackrel(+5)("P")stackrel(-1)("Cl"_5")#</mathjax></p>
</blockquote></div>
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</article> | What is the oxidation number of phosphorus in the compound #"PCl"_5"#? | null |
2,609 | ace381e4-6ddd-11ea-875b-ccda262736ce | https://socratic.org/questions/what-is-the-temperature-in-k-of-2-20-moles-of-gas-occupying-3-5-l-at-3-30-atm | 63.98 K | start physical_unit 10 10 temperature k qc_end physical_unit 10 10 7 8 mole qc_end physical_unit 10 10 12 13 volume qc_end physical_unit 10 10 15 16 pressure qc_end end | [{"type":"physical unit","value":"Temperature [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"63.98 K"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{2.20 moles}"},{"type":"physical unit","value":"Volume [OF] the gas [=] \\pu{3.5 L}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{3.30 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the temperature, in K, of 2.20 moles of gas occupying 3.5 L at 3.30 atm?</h1> | null | 63.98 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To figure this out you need to start with the ideal gas equation: </p>
<p><mathjax>#P*V = n*R*T#</mathjax></p>
<p>You have P (3.30 atm), V (3.5 l), n (2.20 moles) and you can look up the gas constant, R (0.082057 (l<em>atm)/(mol</em>K)). </p>
<p>We simply rearrange the ideal gas equation to get T by itself:</p>
<p><mathjax>#T = (P*V)/(n*R)#</mathjax></p>
<p>Then simply plug in the values you were given, making sure that the units are in liters, atmospheres, and moles (which they are, in this case):</p>
<p><mathjax>#T = (3.30" atm"*3.5" l")/(2.20" moles"*0.082057" (l*atm)/(mol*K)") = 63.9799" K"#</mathjax></p>
<p>Since volume was only given to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we can only report 2 significant figures for our answer:</p>
<p>64 K.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>64 K</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To figure this out you need to start with the ideal gas equation: </p>
<p><mathjax>#P*V = n*R*T#</mathjax></p>
<p>You have P (3.30 atm), V (3.5 l), n (2.20 moles) and you can look up the gas constant, R (0.082057 (l<em>atm)/(mol</em>K)). </p>
<p>We simply rearrange the ideal gas equation to get T by itself:</p>
<p><mathjax>#T = (P*V)/(n*R)#</mathjax></p>
<p>Then simply plug in the values you were given, making sure that the units are in liters, atmospheres, and moles (which they are, in this case):</p>
<p><mathjax>#T = (3.30" atm"*3.5" l")/(2.20" moles"*0.082057" (l*atm)/(mol*K)") = 63.9799" K"#</mathjax></p>
<p>Since volume was only given to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we can only report 2 significant figures for our answer:</p>
<p>64 K.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the temperature, in K, of 2.20 moles of gas occupying 3.5 L at 3.30 atm?</h1>
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<div class="markdown"><p>64 K</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To figure this out you need to start with the ideal gas equation: </p>
<p><mathjax>#P*V = n*R*T#</mathjax></p>
<p>You have P (3.30 atm), V (3.5 l), n (2.20 moles) and you can look up the gas constant, R (0.082057 (l<em>atm)/(mol</em>K)). </p>
<p>We simply rearrange the ideal gas equation to get T by itself:</p>
<p><mathjax>#T = (P*V)/(n*R)#</mathjax></p>
<p>Then simply plug in the values you were given, making sure that the units are in liters, atmospheres, and moles (which they are, in this case):</p>
<p><mathjax>#T = (3.30" atm"*3.5" l")/(2.20" moles"*0.082057" (l*atm)/(mol*K)") = 63.9799" K"#</mathjax></p>
<p>Since volume was only given to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>, we can only report 2 significant figures for our answer:</p>
<p>64 K.</p></div>
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</article> | What is the temperature, in K, of 2.20 moles of gas occupying 3.5 L at 3.30 atm? | null |
2,610 | aab4c49f-6ddd-11ea-a583-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-of-c6h6-o2-co2-h2o | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O"}] | [{"type":"chemical equation","value":"C6H6 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1> | null | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
</div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1>
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Alan P.
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Feb 8, 2017
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<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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Nikka C.
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<span class="dateCreated" datetime="2017-02-09T02:34:22" itemprop="dateCreated">
Feb 9, 2017
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<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p></div>
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<a href="https://socratic.org/answers/375411" itemprop="url">Answer link</a>
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<div class="answer" id="375412" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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<a class="topContributorPic" href="/users/nikka-c"><img alt="" class="" src="https://profilepictures.socratic.org/AQ8VOCyTM1J7U0TvIWhw_nkahara.jpg" title=""/></a>
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Nikka C.
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<span class="dateCreated" datetime="2017-02-09T02:34:35" itemprop="dateCreated">
Feb 9, 2017
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<div>
<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p>
<p>If you don't like fractions, you can always multiply the whole chemical equation by the denominator.</p>
<p>[<mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O] x 2</p>
<p><mathjax>#2#</mathjax> <mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(12)#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(6)#</mathjax> <mathjax>#H_2#</mathjax>O (also acceptable)</p></div>
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</article> | What is the balanced equation of C6H6+O2 = CO2+H2O? | null |
2,611 | ac6fc7f6-6ddd-11ea-bd7b-ccda262736ce | https://socratic.org/questions/57e7c8ab11ef6b17a64ad016 | 6 grams | start physical_unit 19 19 mass g qc_end physical_unit 7 8 2 3 mass qc_end physical_unit 7 7 11 12 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] oxygen [IN] grams"}] | [{"type":"physical unit","value":"6 grams"}] | [{"type":"physical unit","value":"Mass [OF] iron oxide [=] \\pu{10 g}"},{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{4 g}"}] | <h1 class="questionTitle" itemprop="name">In a #10*g# mass of an iron oxide there are #4*g# of iron. How many grams of oxygen are there?</h1> | null | 6 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have formed <mathjax>#10*g#</mathjax> of an oxide of iron; this could <mathjax>#FeO#</mathjax>, or <mathjax>#Fe_2O_3#</mathjax> or a mixed valence oxide. The point is that the constituent atoms are the metal and oxygen. If the iron content is <mathjax>#4*g#</mathjax> then surely there must be a <mathjax>#6*g#</mathjax> mass of oxygen.</p></div>
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<div class="markdown"><p>If I have <mathjax>#£10-00#</mathjax> and I give you <mathjax>#£4-00#</mathjax>, then how many pounds do I have left?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You have formed <mathjax>#10*g#</mathjax> of an oxide of iron; this could <mathjax>#FeO#</mathjax>, or <mathjax>#Fe_2O_3#</mathjax> or a mixed valence oxide. The point is that the constituent atoms are the metal and oxygen. If the iron content is <mathjax>#4*g#</mathjax> then surely there must be a <mathjax>#6*g#</mathjax> mass of oxygen.</p></div>
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<div class="markdown"><p>If I have <mathjax>#£10-00#</mathjax> and I give you <mathjax>#£4-00#</mathjax>, then how many pounds do I have left?</p></div>
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<div class="markdown"><p>You have formed <mathjax>#10*g#</mathjax> of an oxide of iron; this could <mathjax>#FeO#</mathjax>, or <mathjax>#Fe_2O_3#</mathjax> or a mixed valence oxide. The point is that the constituent atoms are the metal and oxygen. If the iron content is <mathjax>#4*g#</mathjax> then surely there must be a <mathjax>#6*g#</mathjax> mass of oxygen.</p></div>
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</article> | In a #10*g# mass of an iron oxide there are #4*g# of iron. How many grams of oxygen are there? | null |
2,612 | aaa91493-6ddd-11ea-a01e-ccda262736ce | https://socratic.org/questions/how-do-you-balance-mg-hno3-mg-no3-2-h2 | Mg + 2 HNO3 -> Mg(NO3)2 + 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":"Mg + 2 HNO3 -> Mg(NO3)2 + H2"}] | [{"type":"chemical equation","value":"Mg + HNO3 -> Mg(NO3)2 + H2"}] | <h1 class="questionTitle" itemprop="name">How do you balance Mg+HNO3 -->Mg(NO3)2 + H2?
</h1> | null | Mg + 2 HNO3 -> Mg(NO3)2 + H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>On the left side the are:<br/>
Mg : 1 Magnizium in elemental form <br/>
H : 2 There are 2 molecules of nitric acid each acid has one hidrogen so two molecules of nitric acid is 2 hidrogens<br/>
N : 2 There are 2 molecules of nitric acid each acid has one nitrogen so two molecules of nitric acid is 2 nitrogens<br/>
O : 6 There are 2 molecules of nitric acid each acid has 3 oxygens so two molecules of nitric acid is 6 oxygens</p>
<p>On the right side there are:<br/>
Mg : 1 There is one molecules of Magnesium nitrate and has 1 magnizium<br/>
H : 2 Hidrogen gas which has two hidrogen molecules<br/>
N : 2 There is one molecules of Magnesium nitrate and has 2 nitrogen atoms<br/>
O : 6 There is one molecules of Magnesium nitrate and has 6 oxygen atoms</p></div>
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<div class="markdown"><p><mathjax>#Mg + 2HNO_3 -> Mg(NO_3)_2 + H2#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>On the left side the are:<br/>
Mg : 1 Magnizium in elemental form <br/>
H : 2 There are 2 molecules of nitric acid each acid has one hidrogen so two molecules of nitric acid is 2 hidrogens<br/>
N : 2 There are 2 molecules of nitric acid each acid has one nitrogen so two molecules of nitric acid is 2 nitrogens<br/>
O : 6 There are 2 molecules of nitric acid each acid has 3 oxygens so two molecules of nitric acid is 6 oxygens</p>
<p>On the right side there are:<br/>
Mg : 1 There is one molecules of Magnesium nitrate and has 1 magnizium<br/>
H : 2 Hidrogen gas which has two hidrogen molecules<br/>
N : 2 There is one molecules of Magnesium nitrate and has 2 nitrogen atoms<br/>
O : 6 There is one molecules of Magnesium nitrate and has 6 oxygen atoms</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance Mg+HNO3 -->Mg(NO3)2 + H2?
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<div class="markdown"><p><mathjax>#Mg + 2HNO_3 -> Mg(NO_3)_2 + H2#</mathjax> </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>On the left side the are:<br/>
Mg : 1 Magnizium in elemental form <br/>
H : 2 There are 2 molecules of nitric acid each acid has one hidrogen so two molecules of nitric acid is 2 hidrogens<br/>
N : 2 There are 2 molecules of nitric acid each acid has one nitrogen so two molecules of nitric acid is 2 nitrogens<br/>
O : 6 There are 2 molecules of nitric acid each acid has 3 oxygens so two molecules of nitric acid is 6 oxygens</p>
<p>On the right side there are:<br/>
Mg : 1 There is one molecules of Magnesium nitrate and has 1 magnizium<br/>
H : 2 Hidrogen gas which has two hidrogen molecules<br/>
N : 2 There is one molecules of Magnesium nitrate and has 2 nitrogen atoms<br/>
O : 6 There is one molecules of Magnesium nitrate and has 6 oxygen atoms</p></div>
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</article> | How do you balance Mg+HNO3 -->Mg(NO3)2 + H2?
| null |
2,613 | acab6ad4-6ddd-11ea-b2e9-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-one-mole-of-magnesium-nitrate-mg-no-3-2 | 148.26 grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] Mg(NO3)2 [IN] grams"}] | [{"type":"physical unit","value":"148.26 grams"}] | [{"type":"physical unit","value":"Mole [OF] Mg(NO3)2 [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of one mole of magnesium nitrate #Mg(NO_3)_2#?</h1> | null | 148.26 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And I know that 1 mole of nitrogen has a mass of 14.01 g, and that 1 mole of oxygen has a mass of 15.99 g. Do you think that by adding the masses together, appropriately weighted, i.e. <mathjax>#1xxMg + 2xxN + 6xxO#</mathjax>, you would arrive at the formula mass of <mathjax>#Mg(NO_3)_2#</mathjax>?</p>
<p>After you do that, you could see if your summation is <a href="https://en.wikipedia.org/wiki/Magnesium_nitrate" rel="nofollow">right.</a> Good luck.</p></div>
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<div class="markdown"><p>Got me! I know that 1 mole of magnesium has a mass of 24.3 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And I know that 1 mole of nitrogen has a mass of 14.01 g, and that 1 mole of oxygen has a mass of 15.99 g. Do you think that by adding the masses together, appropriately weighted, i.e. <mathjax>#1xxMg + 2xxN + 6xxO#</mathjax>, you would arrive at the formula mass of <mathjax>#Mg(NO_3)_2#</mathjax>?</p>
<p>After you do that, you could see if your summation is <a href="https://en.wikipedia.org/wiki/Magnesium_nitrate" rel="nofollow">right.</a> Good luck.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of one mole of magnesium nitrate #Mg(NO_3)_2#?</h1>
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<div class="markdown"><p>Got me! I know that 1 mole of magnesium has a mass of 24.3 g.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And I know that 1 mole of nitrogen has a mass of 14.01 g, and that 1 mole of oxygen has a mass of 15.99 g. Do you think that by adding the masses together, appropriately weighted, i.e. <mathjax>#1xxMg + 2xxN + 6xxO#</mathjax>, you would arrive at the formula mass of <mathjax>#Mg(NO_3)_2#</mathjax>?</p>
<p>After you do that, you could see if your summation is <a href="https://en.wikipedia.org/wiki/Magnesium_nitrate" rel="nofollow">right.</a> Good luck.</p></div>
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<div class="markdown"><p>It is equal to the molecular weight, 147 grams. Use <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> definition.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Mole is defined as:</p>
<p><mathjax>#n=m/(Mr)#</mathjax></p>
<p>where:</p>
<p><mathjax>#m#</mathjax> is the mass in grams.<br/>
<mathjax>#Mr#</mathjax> is the molecular weight of the substance.</p>
<p>The molecular weight of a substance is equal to the sum of the molecular weights of the atoms that constitute it. For these atoms:</p>
<p><mathjax>#Mg:24.3#</mathjax><br/>
<mathjax>#N:14#</mathjax><br/>
<mathjax>#O:16#</mathjax></p>
<p>So the molecular weight:</p>
<p><mathjax>#Mg(NO_3)_2=MgN_2O_6#</mathjax><br/>
<mathjax>#Mr=23+2*14+6*16#</mathjax><br/>
<mathjax>#Mr=147#</mathjax></p>
<p>Therefore:</p>
<p><mathjax>#n=m/(Mr)#</mathjax></p>
<p><mathjax>#m=n*Mr=1*147=147grams#</mathjax></p></div>
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</article> | What is the mass of one mole of magnesium nitrate #Mg(NO_3)_2#? | null |
2,614 | ac82175c-6ddd-11ea-becc-ccda262736ce | https://socratic.org/questions/an-engagement-ring-contains-3-97-10-2-moles-of-gold-how-many-individual-gold-ato | 2.39 × 10^22 | start physical_unit 12 14 number none qc_end physical_unit 9 9 4 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] individual gold atoms"}] | [{"type":"physical unit","value":"2.39 × 10^22"}] | [{"type":"physical unit","value":"Mole [OF] gold [=] \\pu{3.97 × 10^(-2) moles}"}] | <h1 class="questionTitle" itemprop="name">An engagement ring contains #3.97 * 10^-2# moles of gold. How many individual gold atoms are in this ring?</h1> | null | 2.39 × 10^22 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have quoted a quantity of gold atoms in MOLES. Now it is a fact that if you have a MOLE of something (gold atoms, for instance, or even eggs) you have <mathjax>#6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> individual items of that substance. And you don't have to remember this number, because it will always be given to you in any test of chemistry or physics.</p>
<p>So the number of gold atoms:</p>
<p><mathjax>#"Number of gold atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.97xx10^(-2)cancel(mol)xx6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Please note the use of dimensions in this answer. I required a number, and got a dimensionless number. Capisce?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>If you had 30 eggs would you be able to tell me how many dozen eggs you had? I think you would.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have quoted a quantity of gold atoms in MOLES. Now it is a fact that if you have a MOLE of something (gold atoms, for instance, or even eggs) you have <mathjax>#6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> individual items of that substance. And you don't have to remember this number, because it will always be given to you in any test of chemistry or physics.</p>
<p>So the number of gold atoms:</p>
<p><mathjax>#"Number of gold atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.97xx10^(-2)cancel(mol)xx6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Please note the use of dimensions in this answer. I required a number, and got a dimensionless number. Capisce?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">An engagement ring contains #3.97 * 10^-2# moles of gold. How many individual gold atoms are in this ring?</h1>
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<div class="markdown"><p>If you had 30 eggs would you be able to tell me how many dozen eggs you had? I think you would.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have quoted a quantity of gold atoms in MOLES. Now it is a fact that if you have a MOLE of something (gold atoms, for instance, or even eggs) you have <mathjax>#6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> individual items of that substance. And you don't have to remember this number, because it will always be given to you in any test of chemistry or physics.</p>
<p>So the number of gold atoms:</p>
<p><mathjax>#"Number of gold atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.97xx10^(-2)cancel(mol)xx6.02214#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>Please note the use of dimensions in this answer. I required a number, and got a dimensionless number. Capisce?</p></div>
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</article> | An engagement ring contains #3.97 * 10^-2# moles of gold. How many individual gold atoms are in this ring? | null |
2,615 | a92d4e8c-6ddd-11ea-97e8-ccda262736ce | https://socratic.org/questions/for-the-reaction-ch-4-2o-2-co-2-h-2o-how-many-moles-of-carbon-dioxide-are-produc-1 | 6.25 moles | start physical_unit 15 16 mole mol qc_end chemical_equation 3 10 qc_end physical_unit 26 26 23 24 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon dioxide [IN] moles"}] | [{"type":"physical unit","value":"6.25 moles"}] | [{"type":"chemical equation","value":"CH4 + 2 O2 -> CO2 + H2O"},{"type":"physical unit","value":"Mass [OF] methane [=] \\pu{100 g}"}] | <h1 class="questionTitle" itemprop="name">For the reaction #CH_4 + 2O_2 -> CO_2 + H_2O#, how many moles of carbon dioxide are produced from the combustion of 100. g of methane?</h1> | null | 6.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mol of methane is 16 grams. Therefore, if you have 100 grams of methane, it means that you have <mathjax>#100/16=6.25#</mathjax> moles of methane.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide.</p>
<p>Your reaction is wrong though. I want to correct it.</p>
<p><mathjax>#CH_4 + 2O_2 -> CO_2 + 2H_2O#</mathjax> </p>
<p>When you burn 6.25 moles of methane, you need 12.5 moles of oxygen gas.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide (275 grams of <mathjax>#CO_2#</mathjax>) and 12.5 moles of water vapour (225 grams of <mathjax>#H_2O#</mathjax>).</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>1 mol of methane is 16 grams that produces 6.25 moles of <mathjax>#CO_2#</mathjax> when it is burnt </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mol of methane is 16 grams. Therefore, if you have 100 grams of methane, it means that you have <mathjax>#100/16=6.25#</mathjax> moles of methane.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide.</p>
<p>Your reaction is wrong though. I want to correct it.</p>
<p><mathjax>#CH_4 + 2O_2 -> CO_2 + 2H_2O#</mathjax> </p>
<p>When you burn 6.25 moles of methane, you need 12.5 moles of oxygen gas.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide (275 grams of <mathjax>#CO_2#</mathjax>) and 12.5 moles of water vapour (225 grams of <mathjax>#H_2O#</mathjax>).</p></div>
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<h1 class="questionTitle" itemprop="name">For the reaction #CH_4 + 2O_2 -> CO_2 + H_2O#, how many moles of carbon dioxide are produced from the combustion of 100. g of methane?</h1>
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<div class="markdown"><p>1 mol of methane is 16 grams that produces 6.25 moles of <mathjax>#CO_2#</mathjax> when it is burnt </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mol of methane is 16 grams. Therefore, if you have 100 grams of methane, it means that you have <mathjax>#100/16=6.25#</mathjax> moles of methane.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide.</p>
<p>Your reaction is wrong though. I want to correct it.</p>
<p><mathjax>#CH_4 + 2O_2 -> CO_2 + 2H_2O#</mathjax> </p>
<p>When you burn 6.25 moles of methane, you need 12.5 moles of oxygen gas.</p>
<p>When you burn 6.25 moles of methane, you produce 6.25 moles of carbondioxide (275 grams of <mathjax>#CO_2#</mathjax>) and 12.5 moles of water vapour (225 grams of <mathjax>#H_2O#</mathjax>).</p></div>
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</article> | For the reaction #CH_4 + 2O_2 -> CO_2 + H_2O#, how many moles of carbon dioxide are produced from the combustion of 100. g of methane? | null |
2,616 | ab30664d-6ddd-11ea-8a76-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-equation-h-2o-2-h-2o-o-2 | 2 H2O2 -> 2 H2O + O2 | start chemical_equation qc_end chemical_equation 6 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"2 H2O2 -> 2 H2O + O2"}] | [{"type":"chemical equation","value":"H2O2 -> H2O + O2"}] | <h1 class="questionTitle" itemprop="name">How do you balance this equation? #?H_2O_2 -> ?H_2O + ?O_2#</h1> | null | 2 H2O2 -> 2 H2O + O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#?H_2O_2 -> ?H_2O+?O_2#</mathjax></p>
</blockquote>
<p>Since we only have one molecule on the left, we can start using fractions:</p>
<blockquote>
<p><mathjax>#H_2O_2 -> H_2O + 1/2O_2#</mathjax></p>
</blockquote>
<p>Since the water molecule on the right of the equation contains the same number of hydrogens as the hydrogen peroxide on the left, its multiplier must be the same.</p>
<p>This leaves one oxygen atom - i.e. <mathjax>#1/2O_2#</mathjax> unaccounted for.</p>
<p>To give use integral multipliers, double all the quantities.</p>
<blockquote>
<p><mathjax>#2H_2O_2 -> 2H_2O + O_2#</mathjax></p>
</blockquote>
<p>Finally we can annotate with the probable phases: liquid for water and hydrogen peroxide, gas for oxygen.</p>
<blockquote>
<p><mathjax>#2H_2O_2 (l) -> 2H_2O (l) + O_2 (g)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2H_2O_2 (l) -> 2H_2O (l) + O_2 (g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#?H_2O_2 -> ?H_2O+?O_2#</mathjax></p>
</blockquote>
<p>Since we only have one molecule on the left, we can start using fractions:</p>
<blockquote>
<p><mathjax>#H_2O_2 -> H_2O + 1/2O_2#</mathjax></p>
</blockquote>
<p>Since the water molecule on the right of the equation contains the same number of hydrogens as the hydrogen peroxide on the left, its multiplier must be the same.</p>
<p>This leaves one oxygen atom - i.e. <mathjax>#1/2O_2#</mathjax> unaccounted for.</p>
<p>To give use integral multipliers, double all the quantities.</p>
<blockquote>
<p><mathjax>#2H_2O_2 -> 2H_2O + O_2#</mathjax></p>
</blockquote>
<p>Finally we can annotate with the probable phases: liquid for water and hydrogen peroxide, gas for oxygen.</p>
<blockquote>
<p><mathjax>#2H_2O_2 (l) -> 2H_2O (l) + O_2 (g)#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance this equation? #?H_2O_2 -> ?H_2O + ?O_2#</h1>
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George C.
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Mar 12, 2017
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<div class="markdown"><p><mathjax>#2H_2O_2 (l) -> 2H_2O (l) + O_2 (g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given:</p>
<blockquote>
<p><mathjax>#?H_2O_2 -> ?H_2O+?O_2#</mathjax></p>
</blockquote>
<p>Since we only have one molecule on the left, we can start using fractions:</p>
<blockquote>
<p><mathjax>#H_2O_2 -> H_2O + 1/2O_2#</mathjax></p>
</blockquote>
<p>Since the water molecule on the right of the equation contains the same number of hydrogens as the hydrogen peroxide on the left, its multiplier must be the same.</p>
<p>This leaves one oxygen atom - i.e. <mathjax>#1/2O_2#</mathjax> unaccounted for.</p>
<p>To give use integral multipliers, double all the quantities.</p>
<blockquote>
<p><mathjax>#2H_2O_2 -> 2H_2O + O_2#</mathjax></p>
</blockquote>
<p>Finally we can annotate with the probable phases: liquid for water and hydrogen peroxide, gas for oxygen.</p>
<blockquote>
<p><mathjax>#2H_2O_2 (l) -> 2H_2O (l) + O_2 (g)#</mathjax></p>
</blockquote></div>
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</article> | How do you balance this equation? #?H_2O_2 -> ?H_2O + ?O_2# | null |
2,617 | a9c4f940-6ddd-11ea-bc7d-ccda262736ce | https://socratic.org/questions/how-many-moles-of-al-are-needed-to-react-completely-with-1-2-mol-feo-in-the-reac | 0.80 moles | start physical_unit 4 4 mole mol qc_end physical_unit 13 13 11 12 mole qc_end chemical_equation 17 26 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] Al [IN] moles"}] | [{"type":"physical unit","value":"0.80 moles"}] | [{"type":"physical unit","value":"Mole [OF] FeO [=] \\pu{1.2 mol}"},{"type":"chemical equation","value":"2 Al + 3 FeO -> 3 Fe + Al2O3"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">How many moles of #Al# are needed to react completely with 1.2 mol #FeO# in the reaction #2Al + 3FeO -> 3Fe + Al_2O_3#?</h1> | null | 0.80 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>from the equation we can see,<br/>
<mathjax>#2#</mathjax> mole <mathjax>#Al#</mathjax> is needed to react completely with <mathjax>#3#</mathjax> mole <mathjax>#FeO#</mathjax></p>
<p>so,<br/>
<mathjax>#3#</mathjax> moles of <mathjax>#FeO#</mathjax> needs <mathjax>#2#</mathjax> moles <mathjax>#Al#</mathjax><br/>
so, <mathjax>#1#</mathjax> mole <mathjax>#FeO#</mathjax> needs <mathjax>#2/3#</mathjax> moles <mathjax>#Al#</mathjax></p>
<p>so, <mathjax>#1.2#</mathjax>mol <mathjax>#FeO#</mathjax> needs <mathjax>#(2xx1.2)/3#</mathjax>moles <mathjax>#Al#</mathjax></p>
<p><mathjax>#=0.8#</mathjax>moles <mathjax>#Al#</mathjax></p></div>
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<div class="markdown"><p>0.8 moles</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>from the equation we can see,<br/>
<mathjax>#2#</mathjax> mole <mathjax>#Al#</mathjax> is needed to react completely with <mathjax>#3#</mathjax> mole <mathjax>#FeO#</mathjax></p>
<p>so,<br/>
<mathjax>#3#</mathjax> moles of <mathjax>#FeO#</mathjax> needs <mathjax>#2#</mathjax> moles <mathjax>#Al#</mathjax><br/>
so, <mathjax>#1#</mathjax> mole <mathjax>#FeO#</mathjax> needs <mathjax>#2/3#</mathjax> moles <mathjax>#Al#</mathjax></p>
<p>so, <mathjax>#1.2#</mathjax>mol <mathjax>#FeO#</mathjax> needs <mathjax>#(2xx1.2)/3#</mathjax>moles <mathjax>#Al#</mathjax></p>
<p><mathjax>#=0.8#</mathjax>moles <mathjax>#Al#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #Al# are needed to react completely with 1.2 mol #FeO# in the reaction #2Al + 3FeO -> 3Fe + Al_2O_3#?</h1>
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Sihan Tawsik
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<div class="markdown"><p>0.8 moles</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>from the equation we can see,<br/>
<mathjax>#2#</mathjax> mole <mathjax>#Al#</mathjax> is needed to react completely with <mathjax>#3#</mathjax> mole <mathjax>#FeO#</mathjax></p>
<p>so,<br/>
<mathjax>#3#</mathjax> moles of <mathjax>#FeO#</mathjax> needs <mathjax>#2#</mathjax> moles <mathjax>#Al#</mathjax><br/>
so, <mathjax>#1#</mathjax> mole <mathjax>#FeO#</mathjax> needs <mathjax>#2/3#</mathjax> moles <mathjax>#Al#</mathjax></p>
<p>so, <mathjax>#1.2#</mathjax>mol <mathjax>#FeO#</mathjax> needs <mathjax>#(2xx1.2)/3#</mathjax>moles <mathjax>#Al#</mathjax></p>
<p><mathjax>#=0.8#</mathjax>moles <mathjax>#Al#</mathjax></p></div>
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</article> | How many moles of #Al# are needed to react completely with 1.2 mol #FeO# in the reaction #2Al + 3FeO -> 3Fe + Al_2O_3#? | null |
2,618 | aa203c19-6ddd-11ea-9a94-ccda262736ce | https://socratic.org/questions/how-do-you-balance-this-chemical-equation-p-4-no-3-h-2po-4-no | 3 P4 + 20 NO3- + 12 H2O -> 4 H2PO4- + 20 NO + 8 HPO4^2- + 8 OH- | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 P4 + 20 NO3- + 12 H2O -> 4 H2PO4- + 20 NO + 8 HPO4^2- + 8 OH-"}] | [{"type":"chemical equation","value":"P4 + NO3- -> H2PO4- + NO"}] | <h1 class="questionTitle" itemprop="name">How do you balance this chemical equation: #P_4 + NO_3 -> H_2PO_4^-##+ NO#?</h1> | null | 3 P4 + 20 NO3- + 12 H2O -> 4 H2PO4- + 20 NO + 8 HPO4^2- + 8 OH- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental phosphorus is oxidized to phosphoric acid <mathjax>#(P^0rarrP^V)#</mathjax>; nitrate is reduced to nitrous oxide, <mathjax>#(N^VrarrN^(II))#</mathjax></p>
<p>Oxidation half equation:</p>
<p><mathjax>#1/4P_4(s) +4H_2O rarr H_2PO_4^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Reduction half equation:</p>
<p><mathjax>#NO_3^(-) + 3e^(-) + 4H^+rarr NO(g) + 2H_2O #</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equation are balanced with respect to mass and charge. We cross multiply the redox equation to remove electrons:</p>
<p>i.e. <mathjax>#3xx(i)+5xx(ii)=#</mathjax></p>
<p><mathjax>#3/4P_4 +2H_2O+ 5NO_3^(-) +2H^(+) rarr3H_2PO_4^(-) + 5NO(g)#</mathjax></p>
<p>This is stoichiometrically balanced with respect to mass and charge. It is a fact, however, that these phosphorus digestions were conducted in alkaline media. So I simply have to add <mathjax>#2xx(OH)^-#</mathjax> to both sides:</p>
<p><mathjax>#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#</mathjax></p>
<p>Possibly, you have dragged this scheme from the old Russian literature (1960's-1980's); this chemistry was very hard to reproduce. As this scheme suggests, the nitrate anion can be a potent oxidant, but often its redox potential is masked in aqueous chemistry. </p></div>
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<div class="markdown"><p>I assume you mean <mathjax>#NO_3^-#</mathjax>, nitrate ion.</p>
<p><mathjax>#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental phosphorus is oxidized to phosphoric acid <mathjax>#(P^0rarrP^V)#</mathjax>; nitrate is reduced to nitrous oxide, <mathjax>#(N^VrarrN^(II))#</mathjax></p>
<p>Oxidation half equation:</p>
<p><mathjax>#1/4P_4(s) +4H_2O rarr H_2PO_4^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Reduction half equation:</p>
<p><mathjax>#NO_3^(-) + 3e^(-) + 4H^+rarr NO(g) + 2H_2O #</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equation are balanced with respect to mass and charge. We cross multiply the redox equation to remove electrons:</p>
<p>i.e. <mathjax>#3xx(i)+5xx(ii)=#</mathjax></p>
<p><mathjax>#3/4P_4 +2H_2O+ 5NO_3^(-) +2H^(+) rarr3H_2PO_4^(-) + 5NO(g)#</mathjax></p>
<p>This is stoichiometrically balanced with respect to mass and charge. It is a fact, however, that these phosphorus digestions were conducted in alkaline media. So I simply have to add <mathjax>#2xx(OH)^-#</mathjax> to both sides:</p>
<p><mathjax>#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#</mathjax></p>
<p>Possibly, you have dragged this scheme from the old Russian literature (1960's-1980's); this chemistry was very hard to reproduce. As this scheme suggests, the nitrate anion can be a potent oxidant, but often its redox potential is masked in aqueous chemistry. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance this chemical equation: #P_4 + NO_3 -> H_2PO_4^-##+ NO#?</h1>
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<div class="markdown"><p>I assume you mean <mathjax>#NO_3^-#</mathjax>, nitrate ion.</p>
<p><mathjax>#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental phosphorus is oxidized to phosphoric acid <mathjax>#(P^0rarrP^V)#</mathjax>; nitrate is reduced to nitrous oxide, <mathjax>#(N^VrarrN^(II))#</mathjax></p>
<p>Oxidation half equation:</p>
<p><mathjax>#1/4P_4(s) +4H_2O rarr H_2PO_4^(-) + 6H^(+) + 5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Reduction half equation:</p>
<p><mathjax>#NO_3^(-) + 3e^(-) + 4H^+rarr NO(g) + 2H_2O #</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equation are balanced with respect to mass and charge. We cross multiply the redox equation to remove electrons:</p>
<p>i.e. <mathjax>#3xx(i)+5xx(ii)=#</mathjax></p>
<p><mathjax>#3/4P_4 +2H_2O+ 5NO_3^(-) +2H^(+) rarr3H_2PO_4^(-) + 5NO(g)#</mathjax></p>
<p>This is stoichiometrically balanced with respect to mass and charge. It is a fact, however, that these phosphorus digestions were conducted in alkaline media. So I simply have to add <mathjax>#2xx(OH)^-#</mathjax> to both sides:</p>
<p><mathjax>#3/4P_4 +3H_2O+ 5NO_3^(-) rarrH_2PO_4^(-) +2HPO_4^(2-) + 5NO(g) +2HO^-#</mathjax></p>
<p>Possibly, you have dragged this scheme from the old Russian literature (1960's-1980's); this chemistry was very hard to reproduce. As this scheme suggests, the nitrate anion can be a potent oxidant, but often its redox potential is masked in aqueous chemistry. </p></div>
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</article> | How do you balance this chemical equation: #P_4 + NO_3 -> H_2PO_4^-##+ NO#? | null |
2,619 | a88c47d2-6ddd-11ea-9691-ccda262736ce | https://socratic.org/questions/at-35-c-argon-has-a-velocity-of-300-m-s-what-is-the-velocity-of-helium-at-the-sa | 947.8 m/s | start physical_unit 15 15 velocity m/s qc_end physical_unit 3 3 1 2 temperature qc_end physical_unit 3 3 8 9 velocity qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"velocity [OF] helium [IN] m/s"}] | [{"type":"physical unit","value":"947.8 m/s"}] | [{"type":"physical unit","value":"Temperature [OF] argon [=] \\pu{35 ℃}"},{"type":"physical unit","value":"Velocity [OF] argon [=] \\pu{300 m/s}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">At 35°C, argon has a velocity of 300 m/s. What is the velocity of helium at the same temperature?</h1> | null | 947.8 m/s | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Velocity of gas molecules at constant temperature is inversely proportional to square root of Molar mass of gas</p>
<blockquote>
<p><mathjax>#"v" ∝ 1/sqrt"M"#</mathjax></p>
</blockquote>
<p>From above</p>
<p><mathjax>#"v"_"He"/"v"_"Ar" = sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#"v"_"He" = "v"_"Ar" × sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = "300. m/s" × sqrt("39.948 g/mol"/"4.0026 g/mol")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = 947._8 "m/s"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"948 m/s"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Velocity of gas molecules at constant temperature is inversely proportional to square root of Molar mass of gas</p>
<blockquote>
<p><mathjax>#"v" ∝ 1/sqrt"M"#</mathjax></p>
</blockquote>
<p>From above</p>
<p><mathjax>#"v"_"He"/"v"_"Ar" = sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#"v"_"He" = "v"_"Ar" × sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = "300. m/s" × sqrt("39.948 g/mol"/"4.0026 g/mol")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = 947._8 "m/s"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At 35°C, argon has a velocity of 300 m/s. What is the velocity of helium at the same temperature?</h1>
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Truong-Son N.
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<span class="dateCreated" datetime="2018-06-11T20:28:55" itemprop="dateCreated">
Jun 11, 2018
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<div class="markdown"><p><mathjax>#"948 m/s"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Velocity of gas molecules at constant temperature is inversely proportional to square root of Molar mass of gas</p>
<blockquote>
<p><mathjax>#"v" ∝ 1/sqrt"M"#</mathjax></p>
</blockquote>
<p>From above</p>
<p><mathjax>#"v"_"He"/"v"_"Ar" = sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#"v"_"He" = "v"_"Ar" × sqrt("M"_"Ar"/"M"_"He")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = "300. m/s" × sqrt("39.948 g/mol"/"4.0026 g/mol")#</mathjax></p>
<p><mathjax>#color(white)("v"_"He") = 947._8 "m/s"#</mathjax></p></div>
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</article> | At 35°C, argon has a velocity of 300 m/s. What is the velocity of helium at the same temperature? | null |
2,620 | abe170d0-6ddd-11ea-a20a-ccda262736ce | https://socratic.org/questions/59117edd7c014926fb2517cb | 838.73 ℃ | start physical_unit 1 1 temperature °c qc_end physical_unit 1 1 3 4 volume qc_end physical_unit 1 1 9 10 pressure qc_end physical_unit 1 1 6 7 temperature qc_end physical_unit 1 1 20 21 volume qc_end physical_unit 1 1 23 24 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] ℃"}] | [{"type":"physical unit","value":"838.73 ℃"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1676 mL}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{225 mmHg}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{-85.1 ℃}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{838 mL}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{3.5 atm}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies 1676 mL at -85.1 °C and 225 mmHg. What will its temperature be when its volume is 838 mL at 3.5 atm?</h1> | null | 838.73 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To solve this problem, we can use the <strong>Combined <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">Gas Laws</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can solve this equation for <mathjax>#T_2#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#T_2 = T_1 × p_2/p_1 × V_2/V_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = 225 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.2960 atm"; p_2 = "3.5 atm"#</mathjax></p>
<p><mathjax>#V_1 = "1676 mL";color(white)(mmmmmmmmmmmmmmmll) V_2 = "838 mL"#</mathjax></p>
<p><mathjax>#T_1 = "(-85.1 + 273.15) K = 188.05 K";color(white)(mmmmm) T_2 = ?#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#T_2 = "188.05 K" × (3.5 color(red)(cancel(color(black)("atm"))))/(0.2960 color(red)(cancel(color(black)("atm")))) × (838 color(red)(cancel(color(black)("mL"))))/(1676 color(red)(cancel(color(black)("mL")))) = "1112 K" = "839 °C"#</mathjax></p>
<p>The new temperature is 839 °C.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new temperature would be 839 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To solve this problem, we can use the <strong>Combined <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">Gas Laws</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can solve this equation for <mathjax>#T_2#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#T_2 = T_1 × p_2/p_1 × V_2/V_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = 225 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.2960 atm"; p_2 = "3.5 atm"#</mathjax></p>
<p><mathjax>#V_1 = "1676 mL";color(white)(mmmmmmmmmmmmmmmll) V_2 = "838 mL"#</mathjax></p>
<p><mathjax>#T_1 = "(-85.1 + 273.15) K = 188.05 K";color(white)(mmmmm) T_2 = ?#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#T_2 = "188.05 K" × (3.5 color(red)(cancel(color(black)("atm"))))/(0.2960 color(red)(cancel(color(black)("atm")))) × (838 color(red)(cancel(color(black)("mL"))))/(1676 color(red)(cancel(color(black)("mL")))) = "1112 K" = "839 °C"#</mathjax></p>
<p>The new temperature is 839 °C.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas occupies 1676 mL at -85.1 °C and 225 mmHg. What will its temperature be when its volume is 838 mL at 3.5 atm?</h1>
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<div class="markdown"><p>The new temperature would be 839 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>To solve this problem, we can use the <strong>Combined <a href="https://socratic.org/chemistry/the-behavior-of-gases/gas-laws">Gas Laws</a></strong>:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can solve this equation for <mathjax>#T_2#</mathjax>.</p>
<blockquote>
<blockquote>
<p><mathjax>#T_2 = T_1 × p_2/p_1 × V_2/V_1#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#p_1 = 225 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.2960 atm"; p_2 = "3.5 atm"#</mathjax></p>
<p><mathjax>#V_1 = "1676 mL";color(white)(mmmmmmmmmmmmmmmll) V_2 = "838 mL"#</mathjax></p>
<p><mathjax>#T_1 = "(-85.1 + 273.15) K = 188.05 K";color(white)(mmmmm) T_2 = ?#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#T_2 = "188.05 K" × (3.5 color(red)(cancel(color(black)("atm"))))/(0.2960 color(red)(cancel(color(black)("atm")))) × (838 color(red)(cancel(color(black)("mL"))))/(1676 color(red)(cancel(color(black)("mL")))) = "1112 K" = "839 °C"#</mathjax></p>
<p>The new temperature is 839 °C.</p></div>
</div>
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</article> | A gas occupies 1676 mL at -85.1 °C and 225 mmHg. What will its temperature be when its volume is 838 mL at 3.5 atm? | null |
2,621 | ad0668e8-6ddd-11ea-b462-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-lithium-phosphate | Li3PO4 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] Lithium phosphate [IN] default"}] | [{"type":"chemical equation","value":"Li3PO4"}] | [{"type":"substance name","value":"Lithium phosphate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for Lithium phosphate?</h1> | null | Li3PO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, this would give <mathjax>#2xxLi^+#</mathjax>, and <mathjax>#HPO_4^(2-)#</mathjax>, and <mathjax>#HO^-#</mathjax>. If you have an old bottle it is probably <mathjax>#Li_2HPO_4#</mathjax>....</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Li_3PO_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, this would give <mathjax>#2xxLi^+#</mathjax>, and <mathjax>#HPO_4^(2-)#</mathjax>, and <mathjax>#HO^-#</mathjax>. If you have an old bottle it is probably <mathjax>#Li_2HPO_4#</mathjax>....</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for Lithium phosphate?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Li_3PO_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution, this would give <mathjax>#2xxLi^+#</mathjax>, and <mathjax>#HPO_4^(2-)#</mathjax>, and <mathjax>#HO^-#</mathjax>. If you have an old bottle it is probably <mathjax>#Li_2HPO_4#</mathjax>....</p></div>
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</article> | What is the formula for Lithium phosphate? | null |
2,622 | ac46dd50-6ddd-11ea-85a1-ccda262736ce | https://socratic.org/questions/how-do-you-balance-na-mgf-2-naf-mg | 2 Na + MgF2 -> 2 NaF + Mg | 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 + MgF2 -> 2 NaF + Mg"}] | [{"type":"chemical equation","value":"Na + MgF2 -> NaF + Mg"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Na + MgF_2 -> NaF + Mg#?</h1> | null | 2 Na + MgF2 -> 2 NaF + Mg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#2#</mathjax> F's on the left side (<mathjax>#MgF_color(red)(2)#</mathjax>) imply the that right side must have <mathjax>#2#</mathjax> F's (or a multiple of <mathjax>#2#</mathjax>).</p>
<p>The only F on the right side is in NaF<br/>
so to have <mathjax>#2#</mathjax> F's on the right side we will need <mathjax>#2#</mathjax> NaF's.<br/>
<mathjax>#color(white)("XXX")color(red)(2NaF)+Mg#</mathjax></p>
<p>This gives us <mathjax>#2#</mathjax> Na's on the right side (<mathjax>#color(blue)(2Na)F#</mathjax>)<br/>
which implies we will need <mathjax>#2#</mathjax> Na's on the left side.<br/>
<mathjax>#color(white)("XXX")color(blue)(2)Na + MgF_2rarr color(blue)(2Na)F + Mg#</mathjax></p>
<p>At this point the transformation becomes balanced:<br/>
On both sides we have:</p>
<blockquote>
<blockquote>
<p><mathjax>#2 Na#</mathjax>'s<br/>
<mathjax>#1 Mg#</mathjax>, and<br/>
<mathjax>#2 F#</mathjax>'s</p>
</blockquote>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2Na+MgF_2 rarr 2NaF + Mg#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#2#</mathjax> F's on the left side (<mathjax>#MgF_color(red)(2)#</mathjax>) imply the that right side must have <mathjax>#2#</mathjax> F's (or a multiple of <mathjax>#2#</mathjax>).</p>
<p>The only F on the right side is in NaF<br/>
so to have <mathjax>#2#</mathjax> F's on the right side we will need <mathjax>#2#</mathjax> NaF's.<br/>
<mathjax>#color(white)("XXX")color(red)(2NaF)+Mg#</mathjax></p>
<p>This gives us <mathjax>#2#</mathjax> Na's on the right side (<mathjax>#color(blue)(2Na)F#</mathjax>)<br/>
which implies we will need <mathjax>#2#</mathjax> Na's on the left side.<br/>
<mathjax>#color(white)("XXX")color(blue)(2)Na + MgF_2rarr color(blue)(2Na)F + Mg#</mathjax></p>
<p>At this point the transformation becomes balanced:<br/>
On both sides we have:</p>
<blockquote>
<blockquote>
<p><mathjax>#2 Na#</mathjax>'s<br/>
<mathjax>#1 Mg#</mathjax>, and<br/>
<mathjax>#2 F#</mathjax>'s</p>
</blockquote>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Na + MgF_2 -> NaF + Mg#?</h1>
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<div class="markdown"><p><mathjax>#2Na+MgF_2 rarr 2NaF + Mg#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <mathjax>#2#</mathjax> F's on the left side (<mathjax>#MgF_color(red)(2)#</mathjax>) imply the that right side must have <mathjax>#2#</mathjax> F's (or a multiple of <mathjax>#2#</mathjax>).</p>
<p>The only F on the right side is in NaF<br/>
so to have <mathjax>#2#</mathjax> F's on the right side we will need <mathjax>#2#</mathjax> NaF's.<br/>
<mathjax>#color(white)("XXX")color(red)(2NaF)+Mg#</mathjax></p>
<p>This gives us <mathjax>#2#</mathjax> Na's on the right side (<mathjax>#color(blue)(2Na)F#</mathjax>)<br/>
which implies we will need <mathjax>#2#</mathjax> Na's on the left side.<br/>
<mathjax>#color(white)("XXX")color(blue)(2)Na + MgF_2rarr color(blue)(2Na)F + Mg#</mathjax></p>
<p>At this point the transformation becomes balanced:<br/>
On both sides we have:</p>
<blockquote>
<blockquote>
<p><mathjax>#2 Na#</mathjax>'s<br/>
<mathjax>#1 Mg#</mathjax>, and<br/>
<mathjax>#2 F#</mathjax>'s</p>
</blockquote>
</blockquote></div>
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</article> | How do you balance #Na + MgF_2 -> NaF + Mg#? | null |
2,623 | aab2ef90-6ddd-11ea-8abb-ccda262736ce | https://socratic.org/questions/how-many-grams-of-solid-sodium-hydroxide-are-required-to-make-175-00-ml-of-a-0-5 | 3.57 grams | start physical_unit 4 6 mass g qc_end physical_unit 17 18 11 12 volume qc_end physical_unit 17 18 15 16 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] solid sodium hydroxide [IN] grams"}] | [{"type":"physical unit","value":"3.57 grams"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{175.00 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.5100 M }"}] | <h1 class="questionTitle" itemprop="name">How many grams of solid sodium hydroxide are required to make 175.00 mL of a 0.5100 M #NaOH# solution? </h1> | null | 3.57 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of</strong> <mathjax>#"NaOH"#</mathjax>.</p>
<p>You know there are 0.5100 mol of <mathjax>#"NaOH"#</mathjax> in 1 L of solution.</p>
<p>In 175.00 mL of solution there are</p>
<p><mathjax>#"0.175 00" cancel("L soln") × "0.5100 mol KCl"/(1 cancel("L soln")) = "0.089 25 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong>.</p>
<p>The mass of 1 mol of <mathjax>#"NaOH"#</mathjax> is <mathjax>#"(22.99 + 16.00+ 1.008) g = 40.00 g"#</mathjax>.</p>
<p>So the mass of <mathjax>#"0.089 25 mol of NaOH"#</mathjax> is</p>
<p><mathjax>#"0.089 25" cancel("mol NaOH") × "40.00 g NaOH"/(1 cancel("mol NaOH")) = "3.570 g NaOH"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You will require 3.570 g of <mathjax>#"NaOH"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of</strong> <mathjax>#"NaOH"#</mathjax>.</p>
<p>You know there are 0.5100 mol of <mathjax>#"NaOH"#</mathjax> in 1 L of solution.</p>
<p>In 175.00 mL of solution there are</p>
<p><mathjax>#"0.175 00" cancel("L soln") × "0.5100 mol KCl"/(1 cancel("L soln")) = "0.089 25 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong>.</p>
<p>The mass of 1 mol of <mathjax>#"NaOH"#</mathjax> is <mathjax>#"(22.99 + 16.00+ 1.008) g = 40.00 g"#</mathjax>.</p>
<p>So the mass of <mathjax>#"0.089 25 mol of NaOH"#</mathjax> is</p>
<p><mathjax>#"0.089 25" cancel("mol NaOH") × "40.00 g NaOH"/(1 cancel("mol NaOH")) = "3.570 g NaOH"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of solid sodium hydroxide are required to make 175.00 mL of a 0.5100 M #NaOH# solution? </h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-03-10T18:26:23" itemprop="dateCreated">
Mar 10, 2016
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<div class="markdown"><p>You will require 3.570 g of <mathjax>#"NaOH"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the moles of</strong> <mathjax>#"NaOH"#</mathjax>.</p>
<p>You know there are 0.5100 mol of <mathjax>#"NaOH"#</mathjax> in 1 L of solution.</p>
<p>In 175.00 mL of solution there are</p>
<p><mathjax>#"0.175 00" cancel("L soln") × "0.5100 mol KCl"/(1 cancel("L soln")) = "0.089 25 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Calculate the mass of <mathjax>#"NaOH"#</mathjax></strong>.</p>
<p>The mass of 1 mol of <mathjax>#"NaOH"#</mathjax> is <mathjax>#"(22.99 + 16.00+ 1.008) g = 40.00 g"#</mathjax>.</p>
<p>So the mass of <mathjax>#"0.089 25 mol of NaOH"#</mathjax> is</p>
<p><mathjax>#"0.089 25" cancel("mol NaOH") × "40.00 g NaOH"/(1 cancel("mol NaOH")) = "3.570 g NaOH"#</mathjax></p></div>
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</article> | How many grams of solid sodium hydroxide are required to make 175.00 mL of a 0.5100 M #NaOH# solution? | null |
2,624 | ac11a163-6ddd-11ea-b6f3-ccda262736ce | https://socratic.org/questions/590e3ee211ef6b54f0a456c6 | Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l) | start chemical_equation qc_end substance 6 7 qc_end substance 9 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l)"}] | [{"type":"substance name","value":"Magnesium hydroxide"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">Can you represent the reaction of magnesium hydroxide with hydrochloric acid?</h1> | null | Mg(OH)2(s) + 2 HCl(aq) -> MgCl2(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Is the given equation balanced with respect to mass and charge?</p></div>
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<div class="markdown"><p><mathjax>#Mg(OH)_2(s) + 2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is the given equation balanced with respect to mass and charge?</p></div>
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<div class="markdown"><p><mathjax>#Mg(OH)_2(s) + 2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>Is the given equation balanced with respect to mass and charge?</p></div>
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</article> | Can you represent the reaction of magnesium hydroxide with hydrochloric acid? | null |
2,625 | a8bf8d78-6ddd-11ea-8acf-ccda262736ce | https://socratic.org/questions/how-do-you-balance-h-2so-4-nh-4oh-h-2o-nh-4-2so-4 | H2SO4 + 2 NH4OH -> 2 H2O + (NH4)2SO4 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"H2SO4 + 2 NH4OH -> 2 H2O + (NH4)2SO4"}] | [{"type":"chemical equation","value":"H2SO4 + NH4OH -> H2O + (NH4)2SO4"}] | <h1 class="questionTitle" itemprop="name">How do you balance #H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#?</h1> | null | H2SO4 + 2 NH4OH -> 2 H2O + (NH4)2SO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> equation of an acid <mathjax>#H_2SO_4#</mathjax>, sulfuric acid with a base <mathjax>#NH_4OH#</mathjax> ammonium hydroxide to form water <mathjax>#H_2O#</mathjax> and a salt <mathjax>#(NH_4)_2SO_4#</mathjax> ammonium sulfate. </p>
<p>When balancing a neutralization, balance the water <mathjax>#H_2O#</mathjax> is <mathjax>#HOH#</mathjax> and the rest of the equation will balance. </p>
<p><mathjax>#H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>The 2 <mathjax>#H^+#</mathjax> ions in the acid need 2 <mathjax>#OH^-#</mathjax> ions to form 2 water molecules <mathjax>#HOH#</mathjax> or <mathjax>#H_2O#</mathjax></p>
<p>Therefore the equation balances at</p>
<p><mathjax>#H_2SO_4 + 2 NH_4OH -> 2 H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/AOz2rSLjI8I?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The balanced equation is <br/>
<mathjax>#H_2SO_4 + 2 NH_4OH -> 2 H_2O + (NH_4)_2SO_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> equation of an acid <mathjax>#H_2SO_4#</mathjax>, sulfuric acid with a base <mathjax>#NH_4OH#</mathjax> ammonium hydroxide to form water <mathjax>#H_2O#</mathjax> and a salt <mathjax>#(NH_4)_2SO_4#</mathjax> ammonium sulfate. </p>
<p>When balancing a neutralization, balance the water <mathjax>#H_2O#</mathjax> is <mathjax>#HOH#</mathjax> and the rest of the equation will balance. </p>
<p><mathjax>#H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>The 2 <mathjax>#H^+#</mathjax> ions in the acid need 2 <mathjax>#OH^-#</mathjax> ions to form 2 water molecules <mathjax>#HOH#</mathjax> or <mathjax>#H_2O#</mathjax></p>
<p>Therefore the equation balances at</p>
<p><mathjax>#H_2SO_4 + 2 NH_4OH -> 2 H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/AOz2rSLjI8I?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#?</h1>
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<div class="markdown"><p>The balanced equation is <br/>
<mathjax>#H_2SO_4 + 2 NH_4OH -> 2 H_2O + (NH_4)_2SO_4#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> equation of an acid <mathjax>#H_2SO_4#</mathjax>, sulfuric acid with a base <mathjax>#NH_4OH#</mathjax> ammonium hydroxide to form water <mathjax>#H_2O#</mathjax> and a salt <mathjax>#(NH_4)_2SO_4#</mathjax> ammonium sulfate. </p>
<p>When balancing a neutralization, balance the water <mathjax>#H_2O#</mathjax> is <mathjax>#HOH#</mathjax> and the rest of the equation will balance. </p>
<p><mathjax>#H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>The 2 <mathjax>#H^+#</mathjax> ions in the acid need 2 <mathjax>#OH^-#</mathjax> ions to form 2 water molecules <mathjax>#HOH#</mathjax> or <mathjax>#H_2O#</mathjax></p>
<p>Therefore the equation balances at</p>
<p><mathjax>#H_2SO_4 + 2 NH_4OH -> 2 H_2O + (NH_4)_2SO_4#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/AOz2rSLjI8I?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How do you balance #H_2SO_4 + NH_4OH -> H_2O + (NH_4)_2SO_4#? | null |
2,626 | ac108866-6ddd-11ea-b22c-ccda262736ce | https://socratic.org/questions/how-do-you-balance-h-2so-4-b-oh-3-b-2-so-4-3-h-2o | 3 H2SO4 + 2 B(OH)3 -> B2(SO4)3 + 6 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 H2SO4 + 2 B(OH)3 -> B2(SO4)3 + 6 H2O"}] | [{"type":"chemical equation","value":"H2SO4 + B(OH)3 -> B2(SO4)3 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #H_2SO_4 + B(OH)_3 -> B_2(SO_4)_3 + H_2O#?</h1> | null | 3 H2SO4 + 2 B(OH)3 -> B2(SO4)3 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + H_2O#</mathjax></p>
<p>We have:<br/>
LHS<br/>
H= 2 + 3= 5<br/>
S= 1<br/>
B= 1<br/>
O= 4 + 3= 7</p>
<p>RHS<br/>
H= 2<br/>
S= 3<br/>
B=2<br/>
O= 7 + 1= 8</p>
<p>We balance the equation by making the number of each element the same on both sides.</p>
<p>Starting with the Hydrogen:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
H= 6 (from <mathjax>#H_2SO_4#</mathjax>) + 6 (from <mathjax>#(OH)_3#</mathjax>)= 12</p>
<p>RHS<br/>
H= 12 (from <mathjax>#H_2O#</mathjax>)</p>
<p>Then the Sulphur:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
S= 3 (from adding the 3 in front of the <mathjax>#H_2SO_4#</mathjax>)</p>
<p>RHS<br/>
S= 3</p>
<p>Then the Boron:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>Add a 2 in front of <mathjax>#B(OH)_3#</mathjax> so that </p>
<p>LHS: B= 2 and RHS: B=2</p>
<p>Finally, the Oxygen:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>All we have to do here is make sure everything is adding up to the same number on both sides.</p>
<p>LHS<br/>
O= 7 (from <mathjax>#3H_2SO_4#</mathjax>) + 6 (from <mathjax>#2B(OH)_3#</mathjax>) = 13</p>
<p>RHS<br/>
O= 7 (from <mathjax>#B_2(SO_4)_3#</mathjax>) + 6 (from <mathjax>#6H_2O#</mathjax>) = 13</p>
<p>THE END whew!</p></div>
</div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + H_2O#</mathjax></p>
<p>We have:<br/>
LHS<br/>
H= 2 + 3= 5<br/>
S= 1<br/>
B= 1<br/>
O= 4 + 3= 7</p>
<p>RHS<br/>
H= 2<br/>
S= 3<br/>
B=2<br/>
O= 7 + 1= 8</p>
<p>We balance the equation by making the number of each element the same on both sides.</p>
<p>Starting with the Hydrogen:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
H= 6 (from <mathjax>#H_2SO_4#</mathjax>) + 6 (from <mathjax>#(OH)_3#</mathjax>)= 12</p>
<p>RHS<br/>
H= 12 (from <mathjax>#H_2O#</mathjax>)</p>
<p>Then the Sulphur:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
S= 3 (from adding the 3 in front of the <mathjax>#H_2SO_4#</mathjax>)</p>
<p>RHS<br/>
S= 3</p>
<p>Then the Boron:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>Add a 2 in front of <mathjax>#B(OH)_3#</mathjax> so that </p>
<p>LHS: B= 2 and RHS: B=2</p>
<p>Finally, the Oxygen:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>All we have to do here is make sure everything is adding up to the same number on both sides.</p>
<p>LHS<br/>
O= 7 (from <mathjax>#3H_2SO_4#</mathjax>) + 6 (from <mathjax>#2B(OH)_3#</mathjax>) = 13</p>
<p>RHS<br/>
O= 7 (from <mathjax>#B_2(SO_4)_3#</mathjax>) + 6 (from <mathjax>#6H_2O#</mathjax>) = 13</p>
<p>THE END whew!</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #H_2SO_4 + B(OH)_3 -> B_2(SO_4)_3 + H_2O#?</h1>
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<div class="markdown"><p><mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + H_2O#</mathjax></p>
<p>We have:<br/>
LHS<br/>
H= 2 + 3= 5<br/>
S= 1<br/>
B= 1<br/>
O= 4 + 3= 7</p>
<p>RHS<br/>
H= 2<br/>
S= 3<br/>
B=2<br/>
O= 7 + 1= 8</p>
<p>We balance the equation by making the number of each element the same on both sides.</p>
<p>Starting with the Hydrogen:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
H= 6 (from <mathjax>#H_2SO_4#</mathjax>) + 6 (from <mathjax>#(OH)_3#</mathjax>)= 12</p>
<p>RHS<br/>
H= 12 (from <mathjax>#H_2O#</mathjax>)</p>
<p>Then the Sulphur:<br/>
<mathjax>#3H_2SO_4 + B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>LHS<br/>
S= 3 (from adding the 3 in front of the <mathjax>#H_2SO_4#</mathjax>)</p>
<p>RHS<br/>
S= 3</p>
<p>Then the Boron:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>Add a 2 in front of <mathjax>#B(OH)_3#</mathjax> so that </p>
<p>LHS: B= 2 and RHS: B=2</p>
<p>Finally, the Oxygen:<br/>
<mathjax>#3H_2SO_4 + 2B(OH)_3 → B_2(SO_4)_3 + 6H_2O#</mathjax></p>
<p>All we have to do here is make sure everything is adding up to the same number on both sides.</p>
<p>LHS<br/>
O= 7 (from <mathjax>#3H_2SO_4#</mathjax>) + 6 (from <mathjax>#2B(OH)_3#</mathjax>) = 13</p>
<p>RHS<br/>
O= 7 (from <mathjax>#B_2(SO_4)_3#</mathjax>) + 6 (from <mathjax>#6H_2O#</mathjax>) = 13</p>
<p>THE END whew!</p></div>
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</article> | How do you balance #H_2SO_4 + B(OH)_3 -> B_2(SO_4)_3 + H_2O#? | null |
2,627 | a8cde58b-6ddd-11ea-b348-ccda262736ce | https://socratic.org/questions/how-much-heat-is-required-to-warm-1-60-l-of-water-from-26-0-c-to-100-0-c | 495.39 kJ | start physical_unit 10 10 heat_energy kj qc_end physical_unit 10 10 7 8 volume qc_end physical_unit 10 10 12 13 temperature qc_end physical_unit 10 10 15 16 temperature qc_end end | [{"type":"physical unit","value":"Required heat [OF] water [IN] kJ"}] | [{"type":"physical unit","value":"495.39 kJ"}] | [{"type":"physical unit","value":"Volume [OF] water [=] \\pu{1.60 L}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{26.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{100.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much heat is required to warm 1.60 L of water from 26.0°C to 100.0°C? </h1> | null | 495.39 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we need to know/look up this formula :</p>
<blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p>Where:</p>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:<br/>
V = 1.60L<br/>
<mathjax>#DeltaT#</mathjax> = (100-26 = 74)<br/>
Knowing that the liquid is water, we can use <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to get the mass.<br/>
<mathjax>#D#</mathjax> of water = 1.0g/mL</p>
<p>1.60L x 1.0g/mL x 1000mL/1L = <strong>1600g</strong> (We used 1000mL/1L conversion to get rid of any Volume Units)</p>
<p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> capacity is either given or you have to look this up also. Anyway, in this case we are dealing with water in a liquid phase. Therefore, <mathjax>#c = "4.184 J/g"cdot""^@"C"#</mathjax></p>
<p>Now all you have to do is plug everything in the formula above.</p>
<p><mathjax>#q#</mathjax> = 1600g x (100-26C) x (<mathjax>#"4.184 J/g"cdot""^@"C"#</mathjax>) = <strong>495385.6 <mathjax>#J#</mathjax></strong> or <strong>495.4 <mathjax>#kJ#</mathjax></strong></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>495385.6 <mathjax>#J#</mathjax></strong> or <strong>495.4 <mathjax>#kJ#</mathjax></strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we need to know/look up this formula :</p>
<blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p>Where:</p>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:<br/>
V = 1.60L<br/>
<mathjax>#DeltaT#</mathjax> = (100-26 = 74)<br/>
Knowing that the liquid is water, we can use <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to get the mass.<br/>
<mathjax>#D#</mathjax> of water = 1.0g/mL</p>
<p>1.60L x 1.0g/mL x 1000mL/1L = <strong>1600g</strong> (We used 1000mL/1L conversion to get rid of any Volume Units)</p>
<p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> capacity is either given or you have to look this up also. Anyway, in this case we are dealing with water in a liquid phase. Therefore, <mathjax>#c = "4.184 J/g"cdot""^@"C"#</mathjax></p>
<p>Now all you have to do is plug everything in the formula above.</p>
<p><mathjax>#q#</mathjax> = 1600g x (100-26C) x (<mathjax>#"4.184 J/g"cdot""^@"C"#</mathjax>) = <strong>495385.6 <mathjax>#J#</mathjax></strong> or <strong>495.4 <mathjax>#kJ#</mathjax></strong></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much heat is required to warm 1.60 L of water from 26.0°C to 100.0°C? </h1>
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David Tran
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<span class="dateCreated" datetime="2016-05-23T20:01:46" itemprop="dateCreated">
May 23, 2016
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<div class="markdown"><p><strong>495385.6 <mathjax>#J#</mathjax></strong> or <strong>495.4 <mathjax>#kJ#</mathjax></strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this problem, we need to know/look up this formula :</p>
<blockquote>
<p><mathjax>#q = m xx DeltaT xx c#</mathjax></p>
</blockquote>
<p>Where:</p>
<p><mathjax>#q#</mathjax> - amount of heat<br/>
<mathjax>#m#</mathjax> - mass<br/>
<mathjax>#c#</mathjax> - <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
<mathjax>#DeltaT#</mathjax> - Temperature difference</p>
<p>Given:<br/>
V = 1.60L<br/>
<mathjax>#DeltaT#</mathjax> = (100-26 = 74)<br/>
Knowing that the liquid is water, we can use <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> to get the mass.<br/>
<mathjax>#D#</mathjax> of water = 1.0g/mL</p>
<p>1.60L x 1.0g/mL x 1000mL/1L = <strong>1600g</strong> (We used 1000mL/1L conversion to get rid of any Volume Units)</p>
<p><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">Specific heat</a> capacity is either given or you have to look this up also. Anyway, in this case we are dealing with water in a liquid phase. Therefore, <mathjax>#c = "4.184 J/g"cdot""^@"C"#</mathjax></p>
<p>Now all you have to do is plug everything in the formula above.</p>
<p><mathjax>#q#</mathjax> = 1600g x (100-26C) x (<mathjax>#"4.184 J/g"cdot""^@"C"#</mathjax>) = <strong>495385.6 <mathjax>#J#</mathjax></strong> or <strong>495.4 <mathjax>#kJ#</mathjax></strong></p></div>
</div>
</div>
</div>
</div>
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</article> | How much heat is required to warm 1.60 L of water from 26.0°C to 100.0°C? | null |
2,628 | ac21f6d0-6ddd-11ea-a7c1-ccda262736ce | https://socratic.org/questions/a-standard-nitric-acid-solution-is-prepared-using-0-425-g-of-sodium-carbonate-na | 0.24 M | start physical_unit 1 4 molarity mol/l qc_end physical_unit 13 13 8 9 mass qc_end physical_unit 1 4 22 23 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] standard nitric acid solution [IN] M"}] | [{"type":"physical unit","value":"0.24 M"}] | [{"type":"physical unit","value":"Mass [OF] Na2CO3 [=] \\pu{0.425 g}"},{"type":"physical unit","value":"Volume [OF] standard nitric acid solution [=] \\pu{33.25 mL}"},{"type":"other","value":"Reach a permanent endpoint."}] | <h1 class="questionTitle" itemprop="name">A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, #Na_2CO_3#. What is the molarity of the acid if 33.25 mL are required to reach a permanent endpoint?</h1> | null | 0.24 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(Na_2CO_(3(s))+2HNO_(3(aq))rarr2NaNO_(3(aq))+CO_(2(g))+H_2O_((l)))#</mathjax></p>
<p>Find the no. moles of <mathjax>#sf(Na_2CO_3)#</mathjax>:</p>
<p><mathjax>#sf(n_(Na_2CO_3)=m/M_r=0.425/105.99=0.004)#</mathjax></p>
<p>From the equation you can see that the no. moles of <mathjax>#sf(HNO_3)#</mathjax> must be <strong>twice</strong> this amount.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n_(HNO_3)=0.004xx2=0.008)#</mathjax></p>
<p><mathjax>#sf(c=n/v=0.008/(33.75/1000)=0.241color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#sf(0.241color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(Na_2CO_(3(s))+2HNO_(3(aq))rarr2NaNO_(3(aq))+CO_(2(g))+H_2O_((l)))#</mathjax></p>
<p>Find the no. moles of <mathjax>#sf(Na_2CO_3)#</mathjax>:</p>
<p><mathjax>#sf(n_(Na_2CO_3)=m/M_r=0.425/105.99=0.004)#</mathjax></p>
<p>From the equation you can see that the no. moles of <mathjax>#sf(HNO_3)#</mathjax> must be <strong>twice</strong> this amount.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n_(HNO_3)=0.004xx2=0.008)#</mathjax></p>
<p><mathjax>#sf(c=n/v=0.008/(33.75/1000)=0.241color(white)(x)"mol/l")#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, #Na_2CO_3#. What is the molarity of the acid if 33.25 mL are required to reach a permanent endpoint?</h1>
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Michael
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<div class="markdown"><p><mathjax>#sf(0.241color(white)(x)"mol/l")#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(Na_2CO_(3(s))+2HNO_(3(aq))rarr2NaNO_(3(aq))+CO_(2(g))+H_2O_((l)))#</mathjax></p>
<p>Find the no. moles of <mathjax>#sf(Na_2CO_3)#</mathjax>:</p>
<p><mathjax>#sf(n_(Na_2CO_3)=m/M_r=0.425/105.99=0.004)#</mathjax></p>
<p>From the equation you can see that the no. moles of <mathjax>#sf(HNO_3)#</mathjax> must be <strong>twice</strong> this amount.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n_(HNO_3)=0.004xx2=0.008)#</mathjax></p>
<p><mathjax>#sf(c=n/v=0.008/(33.75/1000)=0.241color(white)(x)"mol/l")#</mathjax></p></div>
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</article> | A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, #Na_2CO_3#. What is the molarity of the acid if 33.25 mL are required to reach a permanent endpoint? | null |
2,629 | a86d6e26-6ddd-11ea-8531-ccda262736ce | https://socratic.org/questions/how-many-grams-of-chromium-are-needed-to-react-with-an-excess-of-cuso-4-to-produ-2 | 8.78 grams | start physical_unit 4 4 mass g qc_end physical_unit 18 18 16 17 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] chromium [IN] grams"}] | [{"type":"physical unit","value":"8.78 grams"}] | [{"type":"physical unit","value":"Mass [OF] Cu [=] \\pu{16.2 g}"},{"type":"other","value":"Chromium react with an excess of CuSO4."}] | <h1 class="questionTitle" itemprop="name">How many grams of chromium are needed to react with an excess of #CuSO_4# to produce 16.2 g Cu?</h1> | null | 8.78 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Write down the equation for this reaction and balance it </p>
<p><mathjax>#3CuSO_4 + 2Cr = 3Cu +Cr_2(SO_4)_3#</mathjax> <br/>
Calculate the amount of moles of copper. We have <mathjax>#16g#</mathjax>, and its molar mass is <mathjax>#64g#</mathjax>:</p>
<p><mathjax>#16/64 = 0.25#</mathjax> </p>
<p>And thus the mole for chromium will be <mathjax>#0.17#</mathjax> </p>
<p>As the ratio suggests <mathjax>#2:3#</mathjax> </p>
<p>Thus <mathjax>#(0.25*2)/ 3 = 0.17#</mathjax> </p>
<p>Use this mole and substitute in the equation <mathjax>#"moles"="mass"/"molar mass"#</mathjax> to calculate the mass which is <mathjax>#0.17 * 52= 8.775g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>its <mathjax>#8.775g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Write down the equation for this reaction and balance it </p>
<p><mathjax>#3CuSO_4 + 2Cr = 3Cu +Cr_2(SO_4)_3#</mathjax> <br/>
Calculate the amount of moles of copper. We have <mathjax>#16g#</mathjax>, and its molar mass is <mathjax>#64g#</mathjax>:</p>
<p><mathjax>#16/64 = 0.25#</mathjax> </p>
<p>And thus the mole for chromium will be <mathjax>#0.17#</mathjax> </p>
<p>As the ratio suggests <mathjax>#2:3#</mathjax> </p>
<p>Thus <mathjax>#(0.25*2)/ 3 = 0.17#</mathjax> </p>
<p>Use this mole and substitute in the equation <mathjax>#"moles"="mass"/"molar mass"#</mathjax> to calculate the mass which is <mathjax>#0.17 * 52= 8.775g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of chromium are needed to react with an excess of #CuSO_4# to produce 16.2 g Cu?</h1>
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Ayesha M.
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Surya K.
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<span class="dateCreated" datetime="2018-03-15T14:13:34" itemprop="dateCreated">
Mar 15, 2018
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<div class="markdown"><p>its <mathjax>#8.775g#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Write down the equation for this reaction and balance it </p>
<p><mathjax>#3CuSO_4 + 2Cr = 3Cu +Cr_2(SO_4)_3#</mathjax> <br/>
Calculate the amount of moles of copper. We have <mathjax>#16g#</mathjax>, and its molar mass is <mathjax>#64g#</mathjax>:</p>
<p><mathjax>#16/64 = 0.25#</mathjax> </p>
<p>And thus the mole for chromium will be <mathjax>#0.17#</mathjax> </p>
<p>As the ratio suggests <mathjax>#2:3#</mathjax> </p>
<p>Thus <mathjax>#(0.25*2)/ 3 = 0.17#</mathjax> </p>
<p>Use this mole and substitute in the equation <mathjax>#"moles"="mass"/"molar mass"#</mathjax> to calculate the mass which is <mathjax>#0.17 * 52= 8.775g#</mathjax></p></div>
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</article> | How many grams of chromium are needed to react with an excess of #CuSO_4# to produce 16.2 g Cu? | null |
2,630 | abf45168-6ddd-11ea-b5f8-ccda262736ce | https://socratic.org/questions/a-sample-of-chlorine-gas-occupies-a-volume-of-946-ml-at-a-pressure-of-726-mmhg-w | 4459.71 mmHg | start physical_unit 22 23 pressure mmhg qc_end physical_unit 1 4 9 10 volume qc_end physical_unit 1 4 15 16 pressure qc_end physical_unit 1 4 35 36 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] mmHg"}] | [{"type":"physical unit","value":"4459.71 mmHg"}] | [{"type":"physical unit","value":"Volume1 [OF] chlorine gas sample [=] \\pu{946 mL}"},{"type":"physical unit","value":"Pressure1 [OF] chlorine gas sample [=] \\pu{726 mmHg}"},{"type":"physical unit","value":"Volume2 [OF] chlorine gas sample [=] \\pu{154 mL}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?</h1> | null | 4459.71 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax> at <mathjax>#"constant temperature"#</mathjax>.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(946*cancel(mL)xx(726*cancel(mm*Hg))/(760*cancel(mm*Hg)*atm^-1))/(154*cancel(mL))=??atm#</mathjax></p>
<p>Note that you do NOT measure a pressure that is over <mathjax>#1*atm#</mathjax> in <mathjax>#mm*Hg#</mathjax>. You do this and you will get mercury all over the laboratory. It's a big deal and a big, messy clean up job if you spill it. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_2~=6*atm.#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax> at <mathjax>#"constant temperature"#</mathjax>.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(946*cancel(mL)xx(726*cancel(mm*Hg))/(760*cancel(mm*Hg)*atm^-1))/(154*cancel(mL))=??atm#</mathjax></p>
<p>Note that you do NOT measure a pressure that is over <mathjax>#1*atm#</mathjax> in <mathjax>#mm*Hg#</mathjax>. You do this and you will get mercury all over the laboratory. It's a big deal and a big, messy clean up job if you spill it. </p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?</h1>
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<div class="markdown"><p><mathjax>#P_2~=6*atm.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax> at <mathjax>#"constant temperature"#</mathjax>.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(946*cancel(mL)xx(726*cancel(mm*Hg))/(760*cancel(mm*Hg)*atm^-1))/(154*cancel(mL))=??atm#</mathjax></p>
<p>Note that you do NOT measure a pressure that is over <mathjax>#1*atm#</mathjax> in <mathjax>#mm*Hg#</mathjax>. You do this and you will get mercury all over the laboratory. It's a big deal and a big, messy clean up job if you spill it. </p></div>
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</article> | A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? | null |
2,631 | ac9c8315-6ddd-11ea-b89a-ccda262736ce | https://socratic.org/questions/what-is-the-oh-in-a-solution-that-has-a-h-3o-1-0-10-6-m | 1.0 × 10^(-8) M | start physical_unit 6 6 [oh-] mol/l qc_end physical_unit 6 6 12 15 [h3o+] qc_end end | [{"type":"physical unit","value":"[OH-] [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"1.0 × 10^(-8) M"}] | [{"type":"physical unit","value":"[H3O+] [OF] the solution [=] \\pu{1.0 × 10^(-6) M}"}] | <h1 class="questionTitle" itemprop="name">What is the #[OH^-]# in a solution that has a #[H_3O]=1.0*10^-6# #M#?</h1> | null | 1.0 × 10^(-8) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer can be obtained one of two ways:</p>
<ol>
<li><mathjax>#color(red)("Use the auto-ionization constant of water equation, Kw:"#</mathjax><br/>
<img alt="study.com" src="https://useruploads.socratic.org/vddVxwb9S2uSFVGMkcV6_kw-formula.jpg"/> <br/>
All you would have to do is rearrange the equation to solve for <mathjax>#[OH^(-)]#</mathjax> by dividing both sides of the equation by <mathjax>#[H_3O^(+)]:#</mathjax></li>
</ol>
<p><mathjax>#[OH^(-)] =(1.0xx10^(-14))/[[H_3O^(+)]]#</mathjax></p>
<p>Plug in the known concentration of <mathjax>#H_3O^(+)#</mathjax> ions:<br/>
<mathjax>#[OH^(-)] =(1.0xx10^(-14))/(1.0xx10^(-6))#</mathjax></p>
<p><mathjax>#[OH^(-)] = 1.0xx10^(-8)M#</mathjax></p>
<p>2.<br/>
Determine the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution by taking the negative logarithm <br/>
(-log) of the concentration of <mathjax>#H_3O^(+)#</mathjax> ions. </p>
<p><mathjax>#pH = -log(1.0xx10^(-6)) = 6#</mathjax></p>
<p>Then obtain the pOH using the equation: <mathjax>#pH + pOH = 14#</mathjax>. Rearrange to solve for pOH:</p>
<p><mathjax>#pOH = 14 - 6 = 8#</mathjax></p>
<p>Finally, you take the anti log (inverse of the natural logarithm), <mathjax>#10^("negative number")#</mathjax> of the pOH to obtain the concentration of <mathjax>#OH^(-)#</mathjax> ions.</p>
<p><mathjax>#10^(-pOH)#</mathjax><br/>
<mathjax>#10^(-8) = 1.0xx10^(-8)M#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The solution has a <mathjax>#[OH^(-)]#</mathjax> of <mathjax>#1.0xx10^(-8)M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer can be obtained one of two ways:</p>
<ol>
<li><mathjax>#color(red)("Use the auto-ionization constant of water equation, Kw:"#</mathjax><br/>
<img alt="study.com" src="https://useruploads.socratic.org/vddVxwb9S2uSFVGMkcV6_kw-formula.jpg"/> <br/>
All you would have to do is rearrange the equation to solve for <mathjax>#[OH^(-)]#</mathjax> by dividing both sides of the equation by <mathjax>#[H_3O^(+)]:#</mathjax></li>
</ol>
<p><mathjax>#[OH^(-)] =(1.0xx10^(-14))/[[H_3O^(+)]]#</mathjax></p>
<p>Plug in the known concentration of <mathjax>#H_3O^(+)#</mathjax> ions:<br/>
<mathjax>#[OH^(-)] =(1.0xx10^(-14))/(1.0xx10^(-6))#</mathjax></p>
<p><mathjax>#[OH^(-)] = 1.0xx10^(-8)M#</mathjax></p>
<p>2.<br/>
Determine the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution by taking the negative logarithm <br/>
(-log) of the concentration of <mathjax>#H_3O^(+)#</mathjax> ions. </p>
<p><mathjax>#pH = -log(1.0xx10^(-6)) = 6#</mathjax></p>
<p>Then obtain the pOH using the equation: <mathjax>#pH + pOH = 14#</mathjax>. Rearrange to solve for pOH:</p>
<p><mathjax>#pOH = 14 - 6 = 8#</mathjax></p>
<p>Finally, you take the anti log (inverse of the natural logarithm), <mathjax>#10^("negative number")#</mathjax> of the pOH to obtain the concentration of <mathjax>#OH^(-)#</mathjax> ions.</p>
<p><mathjax>#10^(-pOH)#</mathjax><br/>
<mathjax>#10^(-8) = 1.0xx10^(-8)M#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the #[OH^-]# in a solution that has a #[H_3O]=1.0*10^-6# #M#?</h1>
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<div class="markdown"><p>The solution has a <mathjax>#[OH^(-)]#</mathjax> of <mathjax>#1.0xx10^(-8)M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer can be obtained one of two ways:</p>
<ol>
<li><mathjax>#color(red)("Use the auto-ionization constant of water equation, Kw:"#</mathjax><br/>
<img alt="study.com" src="https://useruploads.socratic.org/vddVxwb9S2uSFVGMkcV6_kw-formula.jpg"/> <br/>
All you would have to do is rearrange the equation to solve for <mathjax>#[OH^(-)]#</mathjax> by dividing both sides of the equation by <mathjax>#[H_3O^(+)]:#</mathjax></li>
</ol>
<p><mathjax>#[OH^(-)] =(1.0xx10^(-14))/[[H_3O^(+)]]#</mathjax></p>
<p>Plug in the known concentration of <mathjax>#H_3O^(+)#</mathjax> ions:<br/>
<mathjax>#[OH^(-)] =(1.0xx10^(-14))/(1.0xx10^(-6))#</mathjax></p>
<p><mathjax>#[OH^(-)] = 1.0xx10^(-8)M#</mathjax></p>
<p>2.<br/>
Determine the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution by taking the negative logarithm <br/>
(-log) of the concentration of <mathjax>#H_3O^(+)#</mathjax> ions. </p>
<p><mathjax>#pH = -log(1.0xx10^(-6)) = 6#</mathjax></p>
<p>Then obtain the pOH using the equation: <mathjax>#pH + pOH = 14#</mathjax>. Rearrange to solve for pOH:</p>
<p><mathjax>#pOH = 14 - 6 = 8#</mathjax></p>
<p>Finally, you take the anti log (inverse of the natural logarithm), <mathjax>#10^("negative number")#</mathjax> of the pOH to obtain the concentration of <mathjax>#OH^(-)#</mathjax> ions.</p>
<p><mathjax>#10^(-pOH)#</mathjax><br/>
<mathjax>#10^(-8) = 1.0xx10^(-8)M#</mathjax></p></div>
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</article> | What is the #[OH^-]# in a solution that has a #[H_3O]=1.0*10^-6# #M#? | null |
2,632 | a8e81886-6ddd-11ea-bd0b-ccda262736ce | https://socratic.org/questions/what-is-the-volume-in-liters-of-2-5-moles-of-oxygen-gas-measured-at-250c-and-a-p | 100 liters | start physical_unit 10 11 volume l qc_end physical_unit 10 11 7 8 mole qc_end physical_unit 10 11 14 15 temperature qc_end physical_unit 10 11 20 21 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen gas [IN] liters"}] | [{"type":"physical unit","value":"100 liters"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{2.5 moles}"},{"type":"physical unit","value":"Temperature [OF] oxygen gas [=] \\pu{250 ℃ }"},{"type":"physical unit","value":"Pressure [OF] oxygen gas [=] \\pu{104.5 kpa}"}] | <h1 class="questionTitle" itemprop="name">What is the volume (in liters) of 2.5 moles of oxygen gas measured at 250C and a pressure of 104.5 kpa?</h1> | null | 100 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Of course you must choose the appropriate <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow">Gas constant.</a> Here it is <mathjax>#8.314*L*kPa*K^-1*mol^-1#</mathjax>.</p>
<p><mathjax>#"Volume"#</mathjax> <mathjax>#=(2.5*cancel(mol)xx8.314*L*cancel(kPa)*cancel(K^-1)*cancel(mol^-1)xx523*cancelK)/(104.5*cancel(kPa))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V=(nRT)/P#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#100*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Of course you must choose the appropriate <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow">Gas constant.</a> Here it is <mathjax>#8.314*L*kPa*K^-1*mol^-1#</mathjax>.</p>
<p><mathjax>#"Volume"#</mathjax> <mathjax>#=(2.5*cancel(mol)xx8.314*L*cancel(kPa)*cancel(K^-1)*cancel(mol^-1)xx523*cancelK)/(104.5*cancel(kPa))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume (in liters) of 2.5 moles of oxygen gas measured at 250C and a pressure of 104.5 kpa?</h1>
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<div class="markdown"><p><mathjax>#V=(nRT)/P#</mathjax> <mathjax>#~=#</mathjax> <mathjax>#100*L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Of course you must choose the appropriate <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow">Gas constant.</a> Here it is <mathjax>#8.314*L*kPa*K^-1*mol^-1#</mathjax>.</p>
<p><mathjax>#"Volume"#</mathjax> <mathjax>#=(2.5*cancel(mol)xx8.314*L*cancel(kPa)*cancel(K^-1)*cancel(mol^-1)xx523*cancelK)/(104.5*cancel(kPa))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</article> | What is the volume (in liters) of 2.5 moles of oxygen gas measured at 250C and a pressure of 104.5 kpa? | null |
2,633 | aab40179-6ddd-11ea-8e75-ccda262736ce | https://socratic.org/questions/consider-a-2-m-solution-of-magnesium-chloride-what-is-the-concentration-of-total | 6.00 M | start physical_unit 13 14 concentration mol/l qc_end physical_unit 6 7 2 3 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] total ions [IN] M"}] | [{"type":"physical unit","value":"6.00 M"}] | [{"type":"physical unit","value":"Concentration [OF] Magnesium Chloride solution [=] \\pu{2 M}"}] | <h1 class="questionTitle" itemprop="name">Consider a 2 M solution of Magnesium Chloride. What is the concentration of total ions (+ and -) in the solution?</h1> | null | 6.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If the solution is <mathjax>#2*mol*L^-1#</mathjax> with respect to <mathjax>#"magnesium chloride"#</mathjax>, the concentration of <mathjax>#"ions"#</mathjax> is <mathjax>#6*mol*L^-1#</mathjax>, i.e. <mathjax>#3#</mathjax> equiv ions per equiv of salt. </p></div>
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<div class="markdown"><p><mathjax>#MgCl_2 rightleftharpoons Mg^(2+) + 2Cl^(-)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If the solution is <mathjax>#2*mol*L^-1#</mathjax> with respect to <mathjax>#"magnesium chloride"#</mathjax>, the concentration of <mathjax>#"ions"#</mathjax> is <mathjax>#6*mol*L^-1#</mathjax>, i.e. <mathjax>#3#</mathjax> equiv ions per equiv of salt. </p></div>
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<h1 class="questionTitle" itemprop="name">Consider a 2 M solution of Magnesium Chloride. What is the concentration of total ions (+ and -) in the solution?</h1>
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<div class="markdown"><p><mathjax>#MgCl_2 rightleftharpoons Mg^(2+) + 2Cl^(-)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>If the solution is <mathjax>#2*mol*L^-1#</mathjax> with respect to <mathjax>#"magnesium chloride"#</mathjax>, the concentration of <mathjax>#"ions"#</mathjax> is <mathjax>#6*mol*L^-1#</mathjax>, i.e. <mathjax>#3#</mathjax> equiv ions per equiv of salt. </p></div>
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</article> | Consider a 2 M solution of Magnesium Chloride. What is the concentration of total ions (+ and -) in the solution? | null |
2,634 | a9b3487a-6ddd-11ea-b92f-ccda262736ce | https://socratic.org/questions/571f87b27c01492086f9ff38 | (NH4)2SO4(aq) + 2 NaOH -> Na2SO4(aq) + 2 NH3 + 2 H2O(l) | start chemical_equation qc_end substance 7 8 qc_end substance 10 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"(NH4)2SO4(aq) + 2 NaOH -> Na2SO4(aq) + 2 NH3 + 2 H2O(l)"}] | [{"type":"substance name","value":"Ammonium sulfate"},{"type":"substance name","value":"Sodium hydroxide"}] | <h1 class="questionTitle" itemprop="name">How do we represent that reaction between ammonium sulfate, and sodium hydroxide?</h1> | null | (NH4)2SO4(aq) + 2 NaOH -> Na2SO4(aq) + 2 NH3 + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NH_4^+ + HO^(-) rarr NH_3(aq) + H_2O#</mathjax>, this is the net ionic equation, sodium sulfate is simply along for the ride. </p></div>
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<div class="markdown"><p><mathjax>#(NH_4)_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2NH_3(aq) + 2H_2O(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#NH_4^+ + HO^(-) rarr NH_3(aq) + H_2O#</mathjax>, this is the net ionic equation, sodium sulfate is simply along for the ride. </p></div>
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<h1 class="questionTitle" itemprop="name">How do we represent that reaction between ammonium sulfate, and sodium hydroxide?</h1>
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<div class="markdown"><p><mathjax>#(NH_4)_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2NH_3(aq) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#NH_4^+ + HO^(-) rarr NH_3(aq) + H_2O#</mathjax>, this is the net ionic equation, sodium sulfate is simply along for the ride. </p></div>
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</article> | How do we represent that reaction between ammonium sulfate, and sodium hydroxide? | null |
2,635 | a95e1614-6ddd-11ea-ba21-ccda262736ce | https://socratic.org/questions/589aa059b72cff0695418f83 | AlF3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] an aluminum fluoride [IN] empirical"}] | [{"type":"chemical equation","value":"AlF3"}] | [{"type":"physical unit","value":"Percent by mass [OF] the metal in the aluminum fluoride [=] \\pu{32%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of an aluminum fluoride that is #32%# by mass with respect to the metal? </h1> | null | AlF3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of <mathjax>#"aluminum fluoride"#</mathjax>. And we work out the molar quantities of each constituent, i.e.</p>
<p><mathjax>#"Moles of metal"=(32.0*g)/(27.0*g*mol^-1)=1.19*mol.#</mathjax></p>
<p><mathjax>#"Moles of fluorine"=(68.0*g)/(19.0*g*mol^-1)=3.58*mol.#</mathjax></p>
<p>We divide thru each molar quantity thru by the smaller molar quantity, that of aluminum to give:</p>
<p><mathjax>#Al:(1.19*mol)/(1.19*mol)=1#</mathjax>; <mathjax>#F:(3.58*mol)/(1.19*mol)=3#</mathjax>, and thus we get an empirical formula of <mathjax>#AlF_3#</mathjax>. </p></div>
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<div class="markdown"><p>We find an empirical formula of <mathjax>#AlF_3#</mathjax>..........</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of <mathjax>#"aluminum fluoride"#</mathjax>. And we work out the molar quantities of each constituent, i.e.</p>
<p><mathjax>#"Moles of metal"=(32.0*g)/(27.0*g*mol^-1)=1.19*mol.#</mathjax></p>
<p><mathjax>#"Moles of fluorine"=(68.0*g)/(19.0*g*mol^-1)=3.58*mol.#</mathjax></p>
<p>We divide thru each molar quantity thru by the smaller molar quantity, that of aluminum to give:</p>
<p><mathjax>#Al:(1.19*mol)/(1.19*mol)=1#</mathjax>; <mathjax>#F:(3.58*mol)/(1.19*mol)=3#</mathjax>, and thus we get an empirical formula of <mathjax>#AlF_3#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of an aluminum fluoride that is #32%# by mass with respect to the metal? </h1>
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<div class="markdown"><p>We find an empirical formula of <mathjax>#AlF_3#</mathjax>..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We assume <mathjax>#100*g#</mathjax> of <mathjax>#"aluminum fluoride"#</mathjax>. And we work out the molar quantities of each constituent, i.e.</p>
<p><mathjax>#"Moles of metal"=(32.0*g)/(27.0*g*mol^-1)=1.19*mol.#</mathjax></p>
<p><mathjax>#"Moles of fluorine"=(68.0*g)/(19.0*g*mol^-1)=3.58*mol.#</mathjax></p>
<p>We divide thru each molar quantity thru by the smaller molar quantity, that of aluminum to give:</p>
<p><mathjax>#Al:(1.19*mol)/(1.19*mol)=1#</mathjax>; <mathjax>#F:(3.58*mol)/(1.19*mol)=3#</mathjax>, and thus we get an empirical formula of <mathjax>#AlF_3#</mathjax>. </p></div>
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</article> | What is the empirical formula of an aluminum fluoride that is #32%# by mass with respect to the metal? | null |
2,636 | ac04ebc1-6ddd-11ea-91f5-ccda262736ce | https://socratic.org/questions/2-30-l-of-air-at-4-53-atm-is-expanded-to-2219-mmhg-what-is-the-final-volume-in-m | 3570 mL | start physical_unit 3 3 volume ml qc_end physical_unit 3 3 0 1 volume qc_end physical_unit 3 3 5 6 pressure qc_end physical_unit 3 3 10 11 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] air [IN] mL"}] | [{"type":"physical unit","value":"3570 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] air [=] \\pu{2.30 L}"},{"type":"physical unit","value":"Pressure1 [OF] air [=] \\pu{4.53 atm}"},{"type":"physical unit","value":"Pressure2 [OF] air [=] \\pu{2219 mmHg}"}] | <h1 class="questionTitle" itemprop="name"> 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?</h1> | null | 3570 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> is an absurd unit. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
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<h1 class="questionTitle" itemprop="name"> 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?</h1>
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<div class="markdown"><p><mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> is an absurd unit. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1*atm#</mathjax> will support a column of mercury <mathjax>#760*mm#</mathjax> high. Often, we quote the pressure as <mathjax>#760*mm#</mathjax> <mathjax>#Hg#</mathjax> (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories. </p>
<p>It is tempting to say that <mathjax>#2219#</mathjax> <mathjax>#mm*Hg#</mathjax> </p>
<p><mathjax>#-=#</mathjax> <mathjax>#(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#</mathjax></p>
<p>But I would resist this temptation. </p></div>
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<div class="markdown"><p>The final volume will be <mathjax>#"3570 mL"#</mathjax>, rounded to three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First the units of pressure and volume need to be the same.</p>
<p><strong>Pressure</strong></p>
<p><mathjax>#"1 atm=760.0 mmHg"#</mathjax></p>
<p><strong>Convert mmHg to atm.</strong></p>
<p><mathjax>#2219color(red)(cancelcolor(black)("mmHg"))xx(1"atm")/(760.0color(red)(cancelcolor(black)("mmHg")))="2.920 atm"#</mathjax></p>
<p><strong>Volume</strong></p>
<p><mathjax>#"1 L=1000 mL"#</mathjax></p>
<p><strong>Convert <mathjax>#"L"#</mathjax> to <mathjax>#"mL"#</mathjax>.</strong></p>
<p><mathjax>#2.30color(red)(cancel(color(black)("L")))xx(1000"mL")/(1color(red)(cancel(color(black)("L"))))="2300 mL"=2.30xx10^3"mL"#</mathjax> </p>
<p>The number of mL above is given to three significant figures using <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. I will use <mathjax>#"2300 mL"#</mathjax> for convenience only.</p>
<p>This question is concerns <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> , which states that the volume <mathjax>#(V)#</mathjax> of a gas held at constant amount and temperature, is inversely proportional to the pressure <mathjax>#(P)#</mathjax>. This means that if the pressure goes up, the volume goes down, and vice-versa. The equation to use is:</p>
<p><mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><strong>Write what you know:</strong></p>
<p><mathjax>#P_1="4.53 atm"#</mathjax><br/>
<mathjax>#V_1=2.30xx10^3"mL"#</mathjax><br/>
<mathjax>#P_2="2.920 atm"#</mathjax></p>
<p><strong>Write what you don't know:</strong> <mathjax>#V_2#</mathjax>.</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known quantities into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=(4.53color(red)(cancel(color(black)("atm")))xx2300"mL")/(2.920color(red)(cancel(color(black)("atm"))))="3570 mL"#</mathjax> rounded to three significant figures</p></div>
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</article> | 2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL? | null |
2,637 | a9f67d82-6ddd-11ea-9d73-ccda262736ce | https://socratic.org/questions/the-h-h-concentration-in-a-solution-is-1-x-10-8-m-what-is-the-ph-of-the-solution | 8.00 | start physical_unit 17 18 ph none qc_end physical_unit 1 1 8 11 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"8.00"}] | [{"type":"physical unit","value":"Concentration [OF] H+ [=] \\pu{1 × 10^(-8) M}"}] | <h1 class="questionTitle" itemprop="name">The [#H^+#] (#H^+# concentration) in a solution is #1 x 10^-8# M. What is the pH of the solution?</h1> | null | 8.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydrogen ions that are in the solution:</p>
<p><mathjax>#pH = -log (1xx10^(-8)M) = 8.#</mathjax> </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is 8.</p></div>
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<div class="markdown"><p><mathjax>#pH = 8#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydrogen ions that are in the solution:</p>
<p><mathjax>#pH = -log (1xx10^(-8)M) = 8.#</mathjax> </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is 8.</p></div>
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<h1 class="questionTitle" itemprop="name">The [#H^+#] (#H^+# concentration) in a solution is #1 x 10^-8# M. What is the pH of the solution?</h1>
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<div class="markdown"><p><mathjax>#pH = 8#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> can be obtained directly from the concentration using the formula below:<br/>
<img alt="lpoli.50webs.com" src="https://useruploads.socratic.org/YkQhupNYRredCpWPbYR1_ph.jpg"/> </p>
<p>Take the -logarithm of the concentration of hydrogen ions that are in the solution:</p>
<p><mathjax>#pH = -log (1xx10^(-8)M) = 8.#</mathjax> </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is 8.</p></div>
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</article> | The [#H^+#] (#H^+# concentration) in a solution is #1 x 10^-8# M. What is the pH of the solution? | null |
2,638 | a991624b-6ddd-11ea-ad17-ccda262736ce | https://socratic.org/questions/what-is-the-molar-ratio-of-li-to-n-2-in-the-reaction-6li-n-2-2li-3n | 6:1 | start physical_unit 6 8 molar_ratio none qc_end chemical_equation 12 18 qc_end end | [{"type":"physical unit","value":"Molar ratio [OF] Li to N2"}] | [{"type":"physical unit","value":"6:1"}] | [{"type":"chemical equation","value":"6 Li + N2 -> 2 Li3N"}] | <h1 class="questionTitle" itemprop="name">What is the molar ratio of #Li# to #N_2# in the reaction #6Li + N_2 -> 2Li_3N#?</h1> | null | 6:1 | <div class="answerDescription">
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<div class="markdown"><p>When determining the molar ratio of two or more species, you just have to look at the coefficients that are given in the balanced chemical equation. In our case, the coefficient in front of lithium is 6 and the coefficient in front of nitrogen gas is 1. Therefore, the molar ratio is six-to-one.</p></div>
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<div class="markdown"><p>It's a 6:1 ratio. </p></div>
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<div class="markdown"><p>When determining the molar ratio of two or more species, you just have to look at the coefficients that are given in the balanced chemical equation. In our case, the coefficient in front of lithium is 6 and the coefficient in front of nitrogen gas is 1. Therefore, the molar ratio is six-to-one.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar ratio of #Li# to #N_2# in the reaction #6Li + N_2 -> 2Li_3N#?</h1>
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<div class="markdown"><p>It's a 6:1 ratio. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>When determining the molar ratio of two or more species, you just have to look at the coefficients that are given in the balanced chemical equation. In our case, the coefficient in front of lithium is 6 and the coefficient in front of nitrogen gas is 1. Therefore, the molar ratio is six-to-one.</p></div>
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</article> | What is the molar ratio of #Li# to #N_2# in the reaction #6Li + N_2 -> 2Li_3N#? | null |
2,639 | ac28d119-6ddd-11ea-9e37-ccda262736ce | https://socratic.org/questions/calculate-the-number-of-joules-given-off-when-32-0-grams-of-steam-cools-from-110 | 99600 joules | start physical_unit 11 11 heat_energy j qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 11 11 14 15 temperature qc_end physical_unit 17 17 19 20 temperature qc_end end | [{"type":"physical unit","value":"Given off energy [OF] steam [IN] joules"}] | [{"type":"physical unit","value":"99600 joules"}] | [{"type":"physical unit","value":"Mass [OF] steam [=] \\pu{32.0 grams}"},{"type":"physical unit","value":"Temperature1 [OF] steam [=] \\pu{110.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] ice [=] \\pu{-40.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C.</h1> | null | 99600 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are five heats to consider:</p>
<p><img alt="www.ausetute.com.au" src="http://www.ausetute.com.au/images/coolcurv.gif"/></p>
<blockquote></blockquote>
<p><mathjax>#q_1#</mathjax> = heat lost on <strong>cooling steam</strong> from 110.0 °C to 100 °C.</p>
<p><mathjax>#q_2#</mathjax> = heat lost on <strong>condensing steam to water</strong> at 100 °C.</p>
<p><mathjax>#q_3#</mathjax> = heat lost on <strong>cooling water</strong> from 100 °C to 0°C.</p>
<p><mathjax>#q_4#</mathjax> = heat lost on <strong>freezing water to ice</strong> at 0 °C.</p>
<p><mathjax>#q_5#</mathjax> = heat lost on <strong>cooling ice</strong> from 0 °C to -40.0 °C.</p>
<blockquote></blockquote>
<p>The total heat evolved is</p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax></p>
<blockquote></blockquote>
<p><strong>1. Cooling the Steam</strong> </p>
<p><mathjax># m = "32.0 g H"_2"O"#</mathjax></p>
<p>For steam, the specific heat capacity, <mathjax>#c = "2.010 J·g"^"-1""°C"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT#</mathjax> = <mathjax>#T_2 – T_1 = "(100.0 - 110.0) °C" = "-10.0 °C"#</mathjax></p>
<p><mathjax>#q_1 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "2.010 J·"color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-10.0" color(red)(cancel(color(black)("°C")))) = "-643 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Condensing the Steam</strong></p>
<p><mathjax>#"Heat of condensation = -Heat of vaporization"#</mathjax></p>
<p><mathjax>#Δ H_"cond" = ""-ΔH_"vap" = "-2260 J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = m Δ H_"cond" = 32.0 color(red)(cancel(color(black)("g")))×("-2260 J·"color(red)(cancel(color(black)("g"^"-1")))) = "-72 320 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Cooling the Water</strong></p>
<p>For liquid water, the specific heat capacity, <mathjax>#c = "4.184 J·°C"^"-1""g"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1")))× ("-100"color(red)(cancel(color(black)("°C")))) = "-13 389 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Freezing the Water</strong></p>
<p><mathjax>#"Heat of freezing = -Heat of fusion"#</mathjax></p>
<p><mathjax>#"-"ΔH_"fus" = "334 J·g"^"-1"#</mathjax></p>
<p><mathjax>#ΔH_"freeze" = "-"ΔH_"fus" = "-334 J·g"^-1"#</mathjax></p>
<p><mathjax>#q_4 = "-"m Δ H_"fus" = 32.0 color(red)(cancel(color(black)("g"))) × "-334 J·"color(red)(cancel(color(black)("g"^"-1"))) = "-10 689 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>5. Cooling the Ice</strong></p>
<p>The specific heat capacity of ice, <mathjax>#c = "2.03 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_2 – T_1 = "(-40.0 - 0) °C" = "-40.0 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × 2.03 "J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × (color(red)(cancel(color(black)("-40.0 °C")))) = "-2598 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Adding them all up</strong></p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-643 – 72 320 – 13 389 – 10 689 - 2598) J" = "-99 600 J"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The amount of energy given off is 99 600 J.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are five heats to consider:</p>
<p><img alt="www.ausetute.com.au" src="http://www.ausetute.com.au/images/coolcurv.gif"/></p>
<blockquote></blockquote>
<p><mathjax>#q_1#</mathjax> = heat lost on <strong>cooling steam</strong> from 110.0 °C to 100 °C.</p>
<p><mathjax>#q_2#</mathjax> = heat lost on <strong>condensing steam to water</strong> at 100 °C.</p>
<p><mathjax>#q_3#</mathjax> = heat lost on <strong>cooling water</strong> from 100 °C to 0°C.</p>
<p><mathjax>#q_4#</mathjax> = heat lost on <strong>freezing water to ice</strong> at 0 °C.</p>
<p><mathjax>#q_5#</mathjax> = heat lost on <strong>cooling ice</strong> from 0 °C to -40.0 °C.</p>
<blockquote></blockquote>
<p>The total heat evolved is</p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax></p>
<blockquote></blockquote>
<p><strong>1. Cooling the Steam</strong> </p>
<p><mathjax># m = "32.0 g H"_2"O"#</mathjax></p>
<p>For steam, the specific heat capacity, <mathjax>#c = "2.010 J·g"^"-1""°C"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT#</mathjax> = <mathjax>#T_2 – T_1 = "(100.0 - 110.0) °C" = "-10.0 °C"#</mathjax></p>
<p><mathjax>#q_1 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "2.010 J·"color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-10.0" color(red)(cancel(color(black)("°C")))) = "-643 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Condensing the Steam</strong></p>
<p><mathjax>#"Heat of condensation = -Heat of vaporization"#</mathjax></p>
<p><mathjax>#Δ H_"cond" = ""-ΔH_"vap" = "-2260 J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = m Δ H_"cond" = 32.0 color(red)(cancel(color(black)("g")))×("-2260 J·"color(red)(cancel(color(black)("g"^"-1")))) = "-72 320 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Cooling the Water</strong></p>
<p>For liquid water, the specific heat capacity, <mathjax>#c = "4.184 J·°C"^"-1""g"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1")))× ("-100"color(red)(cancel(color(black)("°C")))) = "-13 389 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Freezing the Water</strong></p>
<p><mathjax>#"Heat of freezing = -Heat of fusion"#</mathjax></p>
<p><mathjax>#"-"ΔH_"fus" = "334 J·g"^"-1"#</mathjax></p>
<p><mathjax>#ΔH_"freeze" = "-"ΔH_"fus" = "-334 J·g"^-1"#</mathjax></p>
<p><mathjax>#q_4 = "-"m Δ H_"fus" = 32.0 color(red)(cancel(color(black)("g"))) × "-334 J·"color(red)(cancel(color(black)("g"^"-1"))) = "-10 689 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>5. Cooling the Ice</strong></p>
<p>The specific heat capacity of ice, <mathjax>#c = "2.03 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_2 – T_1 = "(-40.0 - 0) °C" = "-40.0 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × 2.03 "J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × (color(red)(cancel(color(black)("-40.0 °C")))) = "-2598 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Adding them all up</strong></p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-643 – 72 320 – 13 389 – 10 689 - 2598) J" = "-99 600 J"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C.</h1>
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<div class="markdown"><p>The amount of energy given off is 99 600 J.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>There are five heats to consider:</p>
<p><img alt="www.ausetute.com.au" src="http://www.ausetute.com.au/images/coolcurv.gif"/></p>
<blockquote></blockquote>
<p><mathjax>#q_1#</mathjax> = heat lost on <strong>cooling steam</strong> from 110.0 °C to 100 °C.</p>
<p><mathjax>#q_2#</mathjax> = heat lost on <strong>condensing steam to water</strong> at 100 °C.</p>
<p><mathjax>#q_3#</mathjax> = heat lost on <strong>cooling water</strong> from 100 °C to 0°C.</p>
<p><mathjax>#q_4#</mathjax> = heat lost on <strong>freezing water to ice</strong> at 0 °C.</p>
<p><mathjax>#q_5#</mathjax> = heat lost on <strong>cooling ice</strong> from 0 °C to -40.0 °C.</p>
<blockquote></blockquote>
<p>The total heat evolved is</p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax></p>
<blockquote></blockquote>
<p><strong>1. Cooling the Steam</strong> </p>
<p><mathjax># m = "32.0 g H"_2"O"#</mathjax></p>
<p>For steam, the specific heat capacity, <mathjax>#c = "2.010 J·g"^"-1""°C"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT#</mathjax> = <mathjax>#T_2 – T_1 = "(100.0 - 110.0) °C" = "-10.0 °C"#</mathjax></p>
<p><mathjax>#q_1 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "2.010 J·"color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-10.0" color(red)(cancel(color(black)("°C")))) = "-643 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Condensing the Steam</strong></p>
<p><mathjax>#"Heat of condensation = -Heat of vaporization"#</mathjax></p>
<p><mathjax>#Δ H_"cond" = ""-ΔH_"vap" = "-2260 J·g"^"-1"#</mathjax></p>
<p><mathjax>#q_2 = m Δ H_"cond" = 32.0 color(red)(cancel(color(black)("g")))×("-2260 J·"color(red)(cancel(color(black)("g"^"-1")))) = "-72 320 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Cooling the Water</strong></p>
<p>For liquid water, the specific heat capacity, <mathjax>#c = "4.184 J·°C"^"-1""g"^"-1"#</mathjax>.</p>
<p><mathjax>#ΔT = T_2 – T_1 = "(0 - 100) °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × "4.184 J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1")))× ("-100"color(red)(cancel(color(black)("°C")))) = "-13 389 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Freezing the Water</strong></p>
<p><mathjax>#"Heat of freezing = -Heat of fusion"#</mathjax></p>
<p><mathjax>#"-"ΔH_"fus" = "334 J·g"^"-1"#</mathjax></p>
<p><mathjax>#ΔH_"freeze" = "-"ΔH_"fus" = "-334 J·g"^-1"#</mathjax></p>
<p><mathjax>#q_4 = "-"m Δ H_"fus" = 32.0 color(red)(cancel(color(black)("g"))) × "-334 J·"color(red)(cancel(color(black)("g"^"-1"))) = "-10 689 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>5. Cooling the Ice</strong></p>
<p>The specific heat capacity of ice, <mathjax>#c = "2.03 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#ΔT = T_2 – T_1 = "(-40.0 - 0) °C" = "-40.0 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 32.0 color(red)(cancel(color(black)("g"))) × 2.03 "J·"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × (color(red)(cancel(color(black)("-40.0 °C")))) = "-2598 J"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Adding them all up</strong></p>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5 = "(-643 – 72 320 – 13 389 – 10 689 - 2598) J" = "-99 600 J"#</mathjax></p></div>
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</article> | Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at -40.0 °C. | null |
2,640 | acd52db1-6ddd-11ea-92db-ccda262736ce | https://socratic.org/questions/how-would-you-balance-na2co3-s-hcl-aq-nacl-aq-h2o-l-co2-g | Na2CO3(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l) + CO2(g) | start chemical_equation qc_end chemical_equation 4 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Na2CO3(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l) + CO2(g)"}] | [{"type":"chemical equation","value":"Na2CO3(s) + HCl(aq) -> NaCl(aq) + H2O(l) + CO2(g)"}] | <h1 class="questionTitle" itemprop="name">How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?</h1> | null | Na2CO3(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l) + CO2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#</mathjax></p>
<p>The reasoning that you should follow in order to balance this reaction easily is:</p>
<p>1- Look at the <mathjax>#Na#</mathjax>. You have <mathjax>#2#</mathjax> in <mathjax>#Na_2CO_3#</mathjax> and <mathjax>#1#</mathjax> in <mathjax>#NaCl#</mathjax> then you should multiply <mathjax>#NaCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>2- When multiplying <mathjax>#NaCl#</mathjax> by 2, we will have to <mathjax>#Cl#</mathjax> in the products side, therefore, we should multiply <mathjax>#HCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>3- Look at the carbon atom <mathjax>#C#</mathjax> there is one atom at each side. No need for any action.</p>
<p>4- Look at the hydrogen atom <mathjax>#H#</mathjax> there is 2 atoms at each side. No need for any action.</p>
<p>5- Look at the oxygen atom <mathjax>#O#</mathjax> there is 3 atoms in <mathjax>#Na_2CO_3#</mathjax> and there is 1 atom in <mathjax>#H_2O#</mathjax> and 2 atoms in <mathjax>#CO_2#</mathjax> which makes it a total of 3 atoms. No need for any action.</p></div>
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<div class="markdown"><p><mathjax>#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#</mathjax></p>
<p>The reasoning that you should follow in order to balance this reaction easily is:</p>
<p>1- Look at the <mathjax>#Na#</mathjax>. You have <mathjax>#2#</mathjax> in <mathjax>#Na_2CO_3#</mathjax> and <mathjax>#1#</mathjax> in <mathjax>#NaCl#</mathjax> then you should multiply <mathjax>#NaCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>2- When multiplying <mathjax>#NaCl#</mathjax> by 2, we will have to <mathjax>#Cl#</mathjax> in the products side, therefore, we should multiply <mathjax>#HCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>3- Look at the carbon atom <mathjax>#C#</mathjax> there is one atom at each side. No need for any action.</p>
<p>4- Look at the hydrogen atom <mathjax>#H#</mathjax> there is 2 atoms at each side. No need for any action.</p>
<p>5- Look at the oxygen atom <mathjax>#O#</mathjax> there is 3 atoms in <mathjax>#Na_2CO_3#</mathjax> and there is 1 atom in <mathjax>#H_2O#</mathjax> and 2 atoms in <mathjax>#CO_2#</mathjax> which makes it a total of 3 atoms. No need for any action.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?</h1>
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<div class="markdown"><p><mathjax>#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)#</mathjax></p>
<p>The reasoning that you should follow in order to balance this reaction easily is:</p>
<p>1- Look at the <mathjax>#Na#</mathjax>. You have <mathjax>#2#</mathjax> in <mathjax>#Na_2CO_3#</mathjax> and <mathjax>#1#</mathjax> in <mathjax>#NaCl#</mathjax> then you should multiply <mathjax>#NaCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>2- When multiplying <mathjax>#NaCl#</mathjax> by 2, we will have to <mathjax>#Cl#</mathjax> in the products side, therefore, we should multiply <mathjax>#HCl#</mathjax> by <mathjax>#color(red)(2)#</mathjax>.</p>
<p>3- Look at the carbon atom <mathjax>#C#</mathjax> there is one atom at each side. No need for any action.</p>
<p>4- Look at the hydrogen atom <mathjax>#H#</mathjax> there is 2 atoms at each side. No need for any action.</p>
<p>5- Look at the oxygen atom <mathjax>#O#</mathjax> there is 3 atoms in <mathjax>#Na_2CO_3#</mathjax> and there is 1 atom in <mathjax>#H_2O#</mathjax> and 2 atoms in <mathjax>#CO_2#</mathjax> which makes it a total of 3 atoms. No need for any action.</p></div>
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</article> | How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)? | null |
2,641 | acf53a08-6ddd-11ea-8531-ccda262736ce | https://socratic.org/questions/how-many-moles-of-acetic-acid-are-present-in-100-0-ml-of-1-0-m-acetic-acid | 0.10 moles | start physical_unit 4 5 mole mol qc_end physical_unit 4 5 12 13 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] acetic acid [IN] moles"}] | [{"type":"physical unit","value":"0.10 moles"}] | [{"type":"physical unit","value":"Volume [OF] acetic acid solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Molarity [OF] acetic acid solution [=] \\pu{1.0 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of acetic acid are present in 100.0 mL of 1.0 M acetic acid? </h1> | null | 0.10 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax>. So <mathjax>#n = C xx V#</mathjax>. Thus, <mathjax>#0.1L xx 1.0 *mol L^(-1) = 0.1•mol#</mathjax>.</p></div>
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<div class="markdown"><p>Concentration = <mathjax>#1.0 *mol *L^-1#</mathjax>. Volume = <mathjax>#100 *mL#</mathjax>.</p>
<p>Moles of HOAc = 0.10 mol <mathjax>#L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax>. So <mathjax>#n = C xx V#</mathjax>. Thus, <mathjax>#0.1L xx 1.0 *mol L^(-1) = 0.1•mol#</mathjax>.</p></div>
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<div class="markdown"><p>Concentration = <mathjax>#1.0 *mol *L^-1#</mathjax>. Volume = <mathjax>#100 *mL#</mathjax>.</p>
<p>Moles of HOAc = 0.10 mol <mathjax>#L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Concentration <mathjax>#=#</mathjax> <mathjax>#n/V#</mathjax>. So <mathjax>#n = C xx V#</mathjax>. Thus, <mathjax>#0.1L xx 1.0 *mol L^(-1) = 0.1•mol#</mathjax>.</p></div>
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</article> | How many moles of acetic acid are present in 100.0 mL of 1.0 M acetic acid? | null |
2,642 | a8d3b1de-6ddd-11ea-b6fc-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-moles-of-o-2-that-will-react-with-10-0g-of-h-2-to-form-wat | 2.48 moles | start physical_unit 7 7 mole mol qc_end physical_unit 15 15 12 13 mass qc_end chemical_equation 22 28 qc_end end | [{"type":"physical unit","value":"Mole [OF] O2 [IN] moles"}] | [{"type":"physical unit","value":"2.48 moles"}] | [{"type":"physical unit","value":"Mass [OF] H2 [=] \\pu{10.0 g}"},{"type":"chemical equation","value":"2 H2 + O2 -> 2 H2O"}] | <h1 class="questionTitle" itemprop="name">What is the number of moles of #O_2# that will react with 10.0g of #H_2# to form water in the equation #2H_2 + O_2 -> 2H_2O#?</h1> | null | 2.48 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, we should implement the process of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. First, convert the grams of <mathjax>#H_2#</mathjax> into moles:</p>
<p><mathjax>#10.0g H_2xx(1molH_2)/(2.01588 g H_2)#</mathjax></p>
<p>Then multiply by your mole to mole ratio:</p>
<p><mathjax>#10.0 g H_2xx(1 mol H_2)/(2.01588 g H_2)xx(1 mol O_2)/(2 mol H_2)#</mathjax></p>
<p>Cancel out common terms and perform the math to get your answer:</p>
<p><mathjax>#10.0 cancel(g H_2)xx(1 cancel(mol H_2))/(2.01588 cancel(g H_2))xx(1 mol O_2)/(2 cancel(mol H_2))= 2.48 mol O_2#</mathjax></p>
<p>The answer has three sig figs because of the given 10.0g.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>2.48 moles of <mathjax>#O_2#</mathjax> will react with 10.0g of <mathjax>#H_2#</mathjax> to form water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, we should implement the process of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. First, convert the grams of <mathjax>#H_2#</mathjax> into moles:</p>
<p><mathjax>#10.0g H_2xx(1molH_2)/(2.01588 g H_2)#</mathjax></p>
<p>Then multiply by your mole to mole ratio:</p>
<p><mathjax>#10.0 g H_2xx(1 mol H_2)/(2.01588 g H_2)xx(1 mol O_2)/(2 mol H_2)#</mathjax></p>
<p>Cancel out common terms and perform the math to get your answer:</p>
<p><mathjax>#10.0 cancel(g H_2)xx(1 cancel(mol H_2))/(2.01588 cancel(g H_2))xx(1 mol O_2)/(2 cancel(mol H_2))= 2.48 mol O_2#</mathjax></p>
<p>The answer has three sig figs because of the given 10.0g.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of moles of #O_2# that will react with 10.0g of #H_2# to form water in the equation #2H_2 + O_2 -> 2H_2O#?</h1>
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<div class="markdown"><p>2.48 moles of <mathjax>#O_2#</mathjax> will react with 10.0g of <mathjax>#H_2#</mathjax> to form water.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, we should implement the process of <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>. First, convert the grams of <mathjax>#H_2#</mathjax> into moles:</p>
<p><mathjax>#10.0g H_2xx(1molH_2)/(2.01588 g H_2)#</mathjax></p>
<p>Then multiply by your mole to mole ratio:</p>
<p><mathjax>#10.0 g H_2xx(1 mol H_2)/(2.01588 g H_2)xx(1 mol O_2)/(2 mol H_2)#</mathjax></p>
<p>Cancel out common terms and perform the math to get your answer:</p>
<p><mathjax>#10.0 cancel(g H_2)xx(1 cancel(mol H_2))/(2.01588 cancel(g H_2))xx(1 mol O_2)/(2 cancel(mol H_2))= 2.48 mol O_2#</mathjax></p>
<p>The answer has three sig figs because of the given 10.0g.</p></div>
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</article> | What is the number of moles of #O_2# that will react with 10.0g of #H_2# to form water in the equation #2H_2 + O_2 -> 2H_2O#? | null |
2,643 | aa692913-6ddd-11ea-b53c-ccda262736ce | https://socratic.org/questions/the-mass-ratio-of-sodium-to-fluorine-in-sodium-fluoride-is-1-21-1-a-sample-of-so | 28.5 grams | start physical_unit 6 6 mass g qc_end physical_unit 4 6 11 11 mass_percent qc_end physical_unit 4 4 18 19 mass qc_end substance 8 9 qc_end end | [{"type":"physical unit","value":"Mass [OF] fluorine [IN] grams"}] | [{"type":"physical unit","value":"28.5 grams"}] | [{"type":"physical unit","value":"Mass ratio [OF] sodium to fluorine [=] \\pu{1.21:1}"},{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{34.5 g}"},{"type":"substance name","value":"Sodium fluoride"}] | <h1 class="questionTitle" itemprop="name">The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. A sample of sodium fluoride produces 34.5 g of sodium upon decomposition. How much fluorine (in grams) forms?</h1> | null | 28.5 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the given <em>mass ratio</em> to determine how much fluorine will be produced if the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition reaction</a> also produces <mathjax>#"34.5 g"#</mathjax> of sodium. </p>
<p>A <mathjax>#1.21:1#</mathjax> mass ratio between sodium and fluorine tells you that can expect to have a mass of sodium that is <mathjax>#1.21#</mathjax> times <strong>bigger</strong> than the mass of fluorine, <em>regardless</em> of the mass of the sodium fluoride. </p>
<p>So based on this information, you know that the mass of fluorine <em>must be</em> smaller than that of sodium. </p>
<p>More specifically, the mass of fluorine will come out to be </p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g Na"))) * "1 g F"/(1.21color(red)(cancel(color(black)("g Na")))) = color(green)("28.5 g F")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, we can double-check the result by finding the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between sodium and fluorine in a sample of sodium fluoride. </p>
<p>To do that, use the molar amsses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Let's say that you have a sample of sodium fluoride that contains <mathjax>#x#</mathjax> grams of sodium and <mathjax>#y#</mathjax> grams of fluoride. </p>
<p>You know that</p>
<blockquote>
<p><mathjax>#x/y = 1.21/1 = 1.21#</mathjax></p>
</blockquote>
<p>You can say that you have</p>
<blockquote>
<p><mathjax>#(x color(white)(x) color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = x/23.0"moles of Na"#</mathjax></p>
<p><mathjax>#(y color(white)(x) color(red)(cancel(color(black)("g"))))/(19.0color(red)(cancel(color(black)("g")))/"mol") = y/19.0"moles of F"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio will thus be </p>
<blockquote>
<p><mathjax>#x/23.0 * 19.0/y = overbrace(x/y)^(color(blue)(=1.21)) * 19/23 = 1.21 * 19/23 = 0.99956 ~~ 1#</mathjax> </p>
</blockquote>
<p>This tells you that you get <em>one mole</em> of fluorine <strong>for every</strong> <em>one mole</em> of sodium. </p>
<p>The decomposition reaction produced</p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23/0color(red)(cancel(color(black)("g")))) = "1.5 moles Na"#</mathjax></p>
</blockquote>
<p>Automatically, you know that it also produced <mathjax>#1.5#</mathjax> moles of fluorine. The mass of fluorine will thus be </p>
<blockquote>
<p><mathjax>#1.5color(red)(cancel(color(black)("moles"))) * "19.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("28.5 g F")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"28.5 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the given <em>mass ratio</em> to determine how much fluorine will be produced if the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition reaction</a> also produces <mathjax>#"34.5 g"#</mathjax> of sodium. </p>
<p>A <mathjax>#1.21:1#</mathjax> mass ratio between sodium and fluorine tells you that can expect to have a mass of sodium that is <mathjax>#1.21#</mathjax> times <strong>bigger</strong> than the mass of fluorine, <em>regardless</em> of the mass of the sodium fluoride. </p>
<p>So based on this information, you know that the mass of fluorine <em>must be</em> smaller than that of sodium. </p>
<p>More specifically, the mass of fluorine will come out to be </p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g Na"))) * "1 g F"/(1.21color(red)(cancel(color(black)("g Na")))) = color(green)("28.5 g F")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, we can double-check the result by finding the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between sodium and fluorine in a sample of sodium fluoride. </p>
<p>To do that, use the molar amsses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Let's say that you have a sample of sodium fluoride that contains <mathjax>#x#</mathjax> grams of sodium and <mathjax>#y#</mathjax> grams of fluoride. </p>
<p>You know that</p>
<blockquote>
<p><mathjax>#x/y = 1.21/1 = 1.21#</mathjax></p>
</blockquote>
<p>You can say that you have</p>
<blockquote>
<p><mathjax>#(x color(white)(x) color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = x/23.0"moles of Na"#</mathjax></p>
<p><mathjax>#(y color(white)(x) color(red)(cancel(color(black)("g"))))/(19.0color(red)(cancel(color(black)("g")))/"mol") = y/19.0"moles of F"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio will thus be </p>
<blockquote>
<p><mathjax>#x/23.0 * 19.0/y = overbrace(x/y)^(color(blue)(=1.21)) * 19/23 = 1.21 * 19/23 = 0.99956 ~~ 1#</mathjax> </p>
</blockquote>
<p>This tells you that you get <em>one mole</em> of fluorine <strong>for every</strong> <em>one mole</em> of sodium. </p>
<p>The decomposition reaction produced</p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23/0color(red)(cancel(color(black)("g")))) = "1.5 moles Na"#</mathjax></p>
</blockquote>
<p>Automatically, you know that it also produced <mathjax>#1.5#</mathjax> moles of fluorine. The mass of fluorine will thus be </p>
<blockquote>
<p><mathjax>#1.5color(red)(cancel(color(black)("moles"))) * "19.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("28.5 g F")#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. A sample of sodium fluoride produces 34.5 g of sodium upon decomposition. How much fluorine (in grams) forms?</h1>
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<div class="markdown"><p><mathjax>#"28.5 g"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the given <em>mass ratio</em> to determine how much fluorine will be produced if the <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition reaction</a> also produces <mathjax>#"34.5 g"#</mathjax> of sodium. </p>
<p>A <mathjax>#1.21:1#</mathjax> mass ratio between sodium and fluorine tells you that can expect to have a mass of sodium that is <mathjax>#1.21#</mathjax> times <strong>bigger</strong> than the mass of fluorine, <em>regardless</em> of the mass of the sodium fluoride. </p>
<p>So based on this information, you know that the mass of fluorine <em>must be</em> smaller than that of sodium. </p>
<p>More specifically, the mass of fluorine will come out to be </p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g Na"))) * "1 g F"/(1.21color(red)(cancel(color(black)("g Na")))) = color(green)("28.5 g F")#</mathjax></p>
</blockquote>
<p><strong>Alternatively</strong>, we can double-check the result by finding the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> that exists between sodium and fluorine in a sample of sodium fluoride. </p>
<p>To do that, use the molar amsses of the two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. Let's say that you have a sample of sodium fluoride that contains <mathjax>#x#</mathjax> grams of sodium and <mathjax>#y#</mathjax> grams of fluoride. </p>
<p>You know that</p>
<blockquote>
<p><mathjax>#x/y = 1.21/1 = 1.21#</mathjax></p>
</blockquote>
<p>You can say that you have</p>
<blockquote>
<p><mathjax>#(x color(white)(x) color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = x/23.0"moles of Na"#</mathjax></p>
<p><mathjax>#(y color(white)(x) color(red)(cancel(color(black)("g"))))/(19.0color(red)(cancel(color(black)("g")))/"mol") = y/19.0"moles of F"#</mathjax></p>
</blockquote>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio will thus be </p>
<blockquote>
<p><mathjax>#x/23.0 * 19.0/y = overbrace(x/y)^(color(blue)(=1.21)) * 19/23 = 1.21 * 19/23 = 0.99956 ~~ 1#</mathjax> </p>
</blockquote>
<p>This tells you that you get <em>one mole</em> of fluorine <strong>for every</strong> <em>one mole</em> of sodium. </p>
<p>The decomposition reaction produced</p>
<blockquote>
<p><mathjax>#34.5color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23/0color(red)(cancel(color(black)("g")))) = "1.5 moles Na"#</mathjax></p>
</blockquote>
<p>Automatically, you know that it also produced <mathjax>#1.5#</mathjax> moles of fluorine. The mass of fluorine will thus be </p>
<blockquote>
<p><mathjax>#1.5color(red)(cancel(color(black)("moles"))) * "19.0 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("28.5 g F")#</mathjax></p>
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</article> | The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. A sample of sodium fluoride produces 34.5 g of sodium upon decomposition. How much fluorine (in grams) forms? | null |
2,644 | ac4edfa4-6ddd-11ea-918c-ccda262736ce | https://socratic.org/questions/a-compressed-gas-has-a-pressure-of-4-882-atm-at-a-temperature-of-8-c-the-next-da | 270 K | start physical_unit 1 2 temperature k qc_end physical_unit 1 2 7 8 pressure qc_end physical_unit 1 2 13 14 temperature qc_end c_other OTHER qc_end physical_unit 1 2 27 28 pressure qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the compressed gas [IN] K"}] | [{"type":"physical unit","value":"270 K"}] | [{"type":"physical unit","value":"Pressure1 [OF] the compressed gas [=] \\pu{4.882 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the compressed gas [=] \\pu{8 ℃}"},{"type":"other","value":"The same container of gas."},{"type":"physical unit","value":"Pressure2 [OF] the compressed gas [=] \\pu{4.690 atm}"}] | <h1 class="questionTitle" itemprop="name">A compressed gas has a pressure of 4.882 atm at a temperature of 8° C. The next day, the same container of gas has a pressure of 4.690 atm. What is the new temperature?</h1> | null | 270 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Gay-Lussac's law, which states that when volume is kept constant, the pressure is directly proportional to the temperature in Kelvins.</p>
<p>The equation is <mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
pressure 1: <mathjax>#P_1="4.882 atm"#</mathjax><br/>
temperature 1: <mathjax>#T_1=8^"o""C"+273.15="281 K"#</mathjax><br/>
pressure 2: <mathjax>#P_2="4.690 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
temperature 2: <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(T_1P_2)/P_1#</mathjax></p>
<p><mathjax>#T_2=((281"K"xx4.690"atm"))/(4.882"atm")="270. K"#</mathjax> (rounded to three significant figures)</p></div>
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<div class="answerSummary">
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<div class="markdown"><p>The new temperature will be <mathjax>#"270. K"#</mathjax>, with three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Gay-Lussac's law, which states that when volume is kept constant, the pressure is directly proportional to the temperature in Kelvins.</p>
<p>The equation is <mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
pressure 1: <mathjax>#P_1="4.882 atm"#</mathjax><br/>
temperature 1: <mathjax>#T_1=8^"o""C"+273.15="281 K"#</mathjax><br/>
pressure 2: <mathjax>#P_2="4.690 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
temperature 2: <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(T_1P_2)/P_1#</mathjax></p>
<p><mathjax>#T_2=((281"K"xx4.690"atm"))/(4.882"atm")="270. K"#</mathjax> (rounded to three significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">A compressed gas has a pressure of 4.882 atm at a temperature of 8° C. The next day, the same container of gas has a pressure of 4.690 atm. What is the new temperature?</h1>
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<div class="markdown"><p>The new temperature will be <mathjax>#"270. K"#</mathjax>, with three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of Gay-Lussac's law, which states that when volume is kept constant, the pressure is directly proportional to the temperature in Kelvins.</p>
<p>The equation is <mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p><strong>Given/Known</strong><br/>
pressure 1: <mathjax>#P_1="4.882 atm"#</mathjax><br/>
temperature 1: <mathjax>#T_1=8^"o""C"+273.15="281 K"#</mathjax><br/>
pressure 2: <mathjax>#P_2="4.690 atm"#</mathjax></p>
<p><strong>Unknown</strong><br/>
temperature 2: <mathjax>#T_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#T_2=(T_1P_2)/P_1#</mathjax></p>
<p><mathjax>#T_2=((281"K"xx4.690"atm"))/(4.882"atm")="270. K"#</mathjax> (rounded to three significant figures)</p></div>
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</article> | A compressed gas has a pressure of 4.882 atm at a temperature of 8° C. The next day, the same container of gas has a pressure of 4.690 atm. What is the new temperature? | null |
2,645 | a919751f-6ddd-11ea-b7b4-ccda262736ce | https://socratic.org/questions/what-mass-of-solution-containing-5-00-potassium-iodide-ki-by-mass-contains-258-m | 5160 mg | start physical_unit 3 3 mass mg qc_end physical_unit 8 8 12 13 mass qc_end physical_unit 8 8 5 5 mass_percent qc_end end | [{"type":"physical unit","value":"Mass [OF] solution [IN] mg"}] | [{"type":"physical unit","value":"5160 mg"}] | [{"type":"physical unit","value":"Mass [OF] KI [=] \\pu{258 mg}"},{"type":"physical unit","value":"Percent by mass [OF] KI in solution [=] \\pu{5.00%}"}] | <h1 class="questionTitle" itemprop="name">What mass of solution containing 5.00% potassium iodide, #KI#, by mass contains 258 mg #KI#?</h1> | null | 5160 mg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out the mass of your <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution that contains <mathjax>#"258 mg"#</mathjax> of potassium iodide, which is your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>As you know, a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, is calculated by taking the number of <em>parts of solute</em> you get <strong>per</strong> <mathjax>#100#</mathjax> <strong>parts of solution</strong>. </p>
<p>In this case, you can say that a <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution contains <mathjax>#"5.00 mg"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mg"#</mathjax> <strong>of solution</strong>. </p>
<p>You can thus use the solution's percent concentration as a <strong>conversion factor</strong> to figure out how many milligrams of solution would contain <mathjax>#"258 mg"#</mathjax> of solute</p>
<blockquote>
<p><mathjax>#258 color(red)(cancel(color(black)("mg KI"))) * "100 mg solution"/(5.00color(red)(cancel(color(black)("mg KI")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("5160 mg solution")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>#"5160 mg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out the mass of your <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution that contains <mathjax>#"258 mg"#</mathjax> of potassium iodide, which is your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>As you know, a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, is calculated by taking the number of <em>parts of solute</em> you get <strong>per</strong> <mathjax>#100#</mathjax> <strong>parts of solution</strong>. </p>
<p>In this case, you can say that a <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution contains <mathjax>#"5.00 mg"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mg"#</mathjax> <strong>of solution</strong>. </p>
<p>You can thus use the solution's percent concentration as a <strong>conversion factor</strong> to figure out how many milligrams of solution would contain <mathjax>#"258 mg"#</mathjax> of solute</p>
<blockquote>
<p><mathjax>#258 color(red)(cancel(color(black)("mg KI"))) * "100 mg solution"/(5.00color(red)(cancel(color(black)("mg KI")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("5160 mg solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What mass of solution containing 5.00% potassium iodide, #KI#, by mass contains 258 mg #KI#?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-18T00:21:17" itemprop="dateCreated">
Jul 18, 2016
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<div class="markdown"><p><mathjax>#"5160 mg"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out the mass of your <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution that contains <mathjax>#"258 mg"#</mathjax> of potassium iodide, which is your <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. </p>
<p>As you know, a solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>, is calculated by taking the number of <em>parts of solute</em> you get <strong>per</strong> <mathjax>#100#</mathjax> <strong>parts of solution</strong>. </p>
<p>In this case, you can say that a <mathjax>#"5.00% m/m"#</mathjax> potassium iodide solution contains <mathjax>#"5.00 mg"#</mathjax> of potassium iodide <strong>for every</strong> <mathjax>#"100 mg"#</mathjax> <strong>of solution</strong>. </p>
<p>You can thus use the solution's percent concentration as a <strong>conversion factor</strong> to figure out how many milligrams of solution would contain <mathjax>#"258 mg"#</mathjax> of solute</p>
<blockquote>
<p><mathjax>#258 color(red)(cancel(color(black)("mg KI"))) * "100 mg solution"/(5.00color(red)(cancel(color(black)("mg KI")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("5160 mg solution")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> | What mass of solution containing 5.00% potassium iodide, #KI#, by mass contains 258 mg #KI#? | null |
2,646 | ac414c46-6ddd-11ea-af6b-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-hydrosulfuric-acid | H2S | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] hydrosulfuric acid [IN] default"}] | [{"type":"chemical equation","value":"H2S"}] | [{"type":"substance name","value":"Hydrosulfuric acid"}] | <h1 class="questionTitle" itemprop="name">What is the formula for hydrosulfuric acid? </h1> | null | H2S | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrosulphuric acid is nothing but it is Hydrogen Sulphide.<br/>
That is H2S.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>H2S</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrosulphuric acid is nothing but it is Hydrogen Sulphide.<br/>
That is H2S.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the formula for hydrosulfuric acid? </h1>
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Suresh Rawal
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<div class="markdown"><p>H2S</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrosulphuric acid is nothing but it is Hydrogen Sulphide.<br/>
That is H2S.</p></div>
</div>
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bubulaa
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<span class="dateCreated" datetime="2016-12-17T13:10:01" itemprop="dateCreated">
Dec 17, 2016
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<div class="markdown"><p><mathjax>#H_2S#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When you say that something is in acid, this means that there is Hydrogen ions. <br/>
Sulfur has a oxidation charge -2 Hydrogen has a charge of +1<br/>
You require two Hydrogen when you have one Sulfur<br/>
Therefore the equation is <mathjax>#H_2S#</mathjax> </p></div>
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</article> | What is the formula for hydrosulfuric acid? | null |
2,647 | ac9529e5-6ddd-11ea-ae96-ccda262736ce | https://socratic.org/questions/how-much-na2co3-is-needed-for-a-full-reaction-with-1-0g-cacl2-2h2o | 0.95 grams | start physical_unit 2 2 mass g qc_end physical_unit 12 12 10 11 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Na2CO3 [IN] grams"}] | [{"type":"physical unit","value":"0.95 grams"}] | [{"type":"physical unit","value":"Mass [OF] CaCl2.2H2O [=] \\pu{1.0 g}"}] | <h1 class="questionTitle" itemprop="name">How much NA2CO3 is needed for a full reaction with 1.0g CaCl2.2H2O?
</h1> | null | 0.95 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the chemical equation, now you have to determine the molar quantities of each reagent. Assume (reasonably) that calcium carbonate is insoluble, and will precipitate from solution.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + CaCl_2*2H_2O(aq) rarr 2NaCl(aq) + CaCO_3(s) darr +2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the chemical equation, now you have to determine the molar quantities of each reagent. Assume (reasonably) that calcium carbonate is insoluble, and will precipitate from solution.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How much NA2CO3 is needed for a full reaction with 1.0g CaCl2.2H2O?
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anor277
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Oct 22, 2015
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<div class="markdown"><p><mathjax>#Na_2CO_3(aq) + CaCl_2*2H_2O(aq) rarr 2NaCl(aq) + CaCO_3(s) darr +2H_2O(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the chemical equation, now you have to determine the molar quantities of each reagent. Assume (reasonably) that calcium carbonate is insoluble, and will precipitate from solution.</p></div>
</div>
</div>
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</article> | How much NA2CO3 is needed for a full reaction with 1.0g CaCl2.2H2O?
| null |
2,648 | aad1b82c-6ddd-11ea-998a-ccda262736ce | https://socratic.org/questions/how-do-you-complete-and-balance-be-o-2 | 2 Be + O2 -> 2 BeO | start chemical_equation qc_end chemical_equation 6 6 qc_end chemical_equation 8 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Be + O2 -> 2 BeO"}] | [{"type":"chemical equation","value":"Be"},{"type":"chemical equation","value":"O2"}] | <h1 class="questionTitle" itemprop="name">How do you complete and balance #Be + O_2 ->#?</h1> | null | 2 Be + O2 -> 2 BeO | <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 the metal beryllium reacts with oxygen gas to produce the solid beryllium oxide.</p></div>
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<div class="markdown"><p><mathjax>#"2Be(s) + O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2BeO(s)"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a synthesis reaction, also called a combination reaction, in which the metal beryllium reacts with oxygen gas to produce the solid beryllium oxide.</p></div>
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<div class="markdown"><p><mathjax>#"2Be(s) + O"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2BeO(s)"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a synthesis reaction, also called a combination reaction, in which the metal beryllium reacts with oxygen gas to produce the solid beryllium oxide.</p></div>
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</article> | How do you complete and balance #Be + O_2 ->#? | null |
2,649 | ac221a5e-6ddd-11ea-ba5d-ccda262736ce | https://socratic.org/questions/the-oh-of-an-aqueous-solution-is-4-62-10-4-m-what-is-the-ph-of-this-solution | 10.66 | start physical_unit 16 17 ph none qc_end physical_unit 4 5 7 10 [oh-] qc_end end | [{"type":"physical unit","value":"pH [OF] this solution"}] | [{"type":"physical unit","value":"10.66"}] | [{"type":"physical unit","value":"[OH-] [OF] aqueous solution [=] \\pu{4.62 × 10^(-4) M}"}] | <h1 class="questionTitle" itemprop="name">The [OH-] of an aqueous solution is #4.62 * 10^-4# M. What is the pH of this solution?</h1> | null | 10.66 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For pure water at <mathjax>#25^@"C"#</mathjax>, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, is <strong>equal</strong> to the concentration of <em>hydroxide ions</em>, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>More specifically, water undergoes a <strong>self-ionization reaction</strong> that results in the formation of equal concentrations of hydronium and hydroxide anions.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>At room temperature, the <em>self-ionization constant</em> of water is equal to </p>
<blockquote>
<p><mathjax>#K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)#</mathjax></p>
</blockquote>
<p>This means that neutral water at this temperature will have </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> and pOH are defined as</p>
<blockquote>
<p><mathjax>#"pH" = - log( ["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>and have the following relationship</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p>In your case, the concentration of hydroxide ions is <strong>bigger</strong> than <mathjax>#10^(-7)"M"#</mathjax>, which tells you that you're dealing with a <em>basic solution</em> and that you can expect the pH of the water to be <strong>higher</strong> than <mathjax>#7#</mathjax>.</p>
<p>A pH equal to <mathjax>#7#</mathjax> is characteristic of a neutral aqueous solution at room temperature. </p>
<p>So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first</p>
<blockquote>
<p><mathjax>#"pOH" = - log( 4.62 * 10^(-4)) = 3.34#</mathjax></p>
</blockquote>
<p>This means that the solution's pH will be </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 3.34 = color(green)(10.66)#</mathjax></p>
</blockquote>
<p>Indeed, the pH is <strong>higher</strong> than <mathjax>#7#</mathjax>, which confirms that you're dealing with a basic solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"pH" = 10.66#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For pure water at <mathjax>#25^@"C"#</mathjax>, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, is <strong>equal</strong> to the concentration of <em>hydroxide ions</em>, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>More specifically, water undergoes a <strong>self-ionization reaction</strong> that results in the formation of equal concentrations of hydronium and hydroxide anions.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>At room temperature, the <em>self-ionization constant</em> of water is equal to </p>
<blockquote>
<p><mathjax>#K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)#</mathjax></p>
</blockquote>
<p>This means that neutral water at this temperature will have </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> and pOH are defined as</p>
<blockquote>
<p><mathjax>#"pH" = - log( ["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>and have the following relationship</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p>In your case, the concentration of hydroxide ions is <strong>bigger</strong> than <mathjax>#10^(-7)"M"#</mathjax>, which tells you that you're dealing with a <em>basic solution</em> and that you can expect the pH of the water to be <strong>higher</strong> than <mathjax>#7#</mathjax>.</p>
<p>A pH equal to <mathjax>#7#</mathjax> is characteristic of a neutral aqueous solution at room temperature. </p>
<p>So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first</p>
<blockquote>
<p><mathjax>#"pOH" = - log( 4.62 * 10^(-4)) = 3.34#</mathjax></p>
</blockquote>
<p>This means that the solution's pH will be </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 3.34 = color(green)(10.66)#</mathjax></p>
</blockquote>
<p>Indeed, the pH is <strong>higher</strong> than <mathjax>#7#</mathjax>, which confirms that you're dealing with a basic solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The [OH-] of an aqueous solution is #4.62 * 10^-4# M. What is the pH of this solution?</h1>
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Stefan V.
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Dec 19, 2015
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<div class="markdown"><p><mathjax>#"pH" = 10.66#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For pure water at <mathjax>#25^@"C"#</mathjax>, the concentration of <em>hydronium ions</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, is <strong>equal</strong> to the concentration of <em>hydroxide ions</em>, <mathjax>#"OH"^(-)#</mathjax>. </p>
<p>More specifically, water undergoes a <strong>self-ionization reaction</strong> that results in the formation of equal concentrations of hydronium and hydroxide anions.</p>
<blockquote>
<p><mathjax>#2"H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>At room temperature, the <em>self-ionization constant</em> of water is equal to </p>
<blockquote>
<p><mathjax>#K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)#</mathjax></p>
</blockquote>
<p>This means that neutral water at this temperature will have </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"#</mathjax></p>
</blockquote>
<p>As you know, <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> and pOH are defined as</p>
<blockquote>
<p><mathjax>#"pH" = - log( ["H"_3"O"^(+)])#</mathjax></p>
<p><mathjax>#"pOH" = - log(["OH"^(-)])#</mathjax></p>
</blockquote>
<p>and have the following relationship</p>
<blockquote>
<p><mathjax>#"pH" + "pOH" = 14#</mathjax></p>
</blockquote>
<p>In your case, the concentration of hydroxide ions is <strong>bigger</strong> than <mathjax>#10^(-7)"M"#</mathjax>, which tells you that you're dealing with a <em>basic solution</em> and that you can expect the pH of the water to be <strong>higher</strong> than <mathjax>#7#</mathjax>.</p>
<p>A pH equal to <mathjax>#7#</mathjax> is characteristic of a neutral aqueous solution at room temperature. </p>
<p>So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first</p>
<blockquote>
<p><mathjax>#"pOH" = - log( 4.62 * 10^(-4)) = 3.34#</mathjax></p>
</blockquote>
<p>This means that the solution's pH will be </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 3.34 = color(green)(10.66)#</mathjax></p>
</blockquote>
<p>Indeed, the pH is <strong>higher</strong> than <mathjax>#7#</mathjax>, which confirms that you're dealing with a basic solution. </p></div>
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</article> | The [OH-] of an aqueous solution is #4.62 * 10^-4# M. What is the pH of this solution? | null |
2,650 | abe456be-6ddd-11ea-8e7b-ccda262736ce | https://socratic.org/questions/how-many-grams-of-water-could-be-made-from-5-0-mol-h-2-and-3-0-mol-o-2 | 90.05 grams | start physical_unit 4 4 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] grams"}] | [{"type":"physical unit","value":"90.05 grams"}] | [{"type":"physical unit","value":"Mole [OF] H2 [=] \\pu{5 mol}"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{3 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams of water could be made from 5.0 mol #H_2# and 3.0 mol #O_2#?</h1> | null | 90.05 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, dihydrogen is in deficiency, and ONLY <mathjax>#2.5*mol#</mathjax> dioxygen will react to give <mathjax>#5 *mol#</mathjax> water. </p>
<p>And this represents a mass of <mathjax>#5*cancel(mol)xx18.01*g*cancel(mol^-1)=90*g#</mathjax>.</p>
<p>What mas of dioxygen will remain after reaction?</p></div>
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<div class="markdown"><p>You need a stoichiometric equation to represent the formation of water:</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, dihydrogen is in deficiency, and ONLY <mathjax>#2.5*mol#</mathjax> dioxygen will react to give <mathjax>#5 *mol#</mathjax> water. </p>
<p>And this represents a mass of <mathjax>#5*cancel(mol)xx18.01*g*cancel(mol^-1)=90*g#</mathjax>.</p>
<p>What mas of dioxygen will remain after reaction?</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of water could be made from 5.0 mol #H_2# and 3.0 mol #O_2#?</h1>
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anor277
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<div class="markdown"><p>You need a stoichiometric equation to represent the formation of water:</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#H_2(g) + 1/2O_2(g) rarr H_2O(l)#</mathjax></p>
<p>Given the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>, dihydrogen is in deficiency, and ONLY <mathjax>#2.5*mol#</mathjax> dioxygen will react to give <mathjax>#5 *mol#</mathjax> water. </p>
<p>And this represents a mass of <mathjax>#5*cancel(mol)xx18.01*g*cancel(mol^-1)=90*g#</mathjax>.</p>
<p>What mas of dioxygen will remain after reaction?</p></div>
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<div class="markdown"><p>90 g</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The balanced equation is: <mathjax>#2H_2(g) + O_2(g) -> 2H_2O(l)#</mathjax></p>
<p>From this you can see that 5 moles of hydrogen gas would react with 2.5 of the available 3 moles of oxygen gas, to form 5 moles of water.</p>
<p>The molar mass of water is 18 g/mol, so you would end up with 5 x 18 = 90 g of water.</p></div>
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</article> | How many grams of water could be made from 5.0 mol #H_2# and 3.0 mol #O_2#? | null |
2,651 | aa105a63-6ddd-11ea-84ee-ccda262736ce | https://socratic.org/questions/what-is-the-propane-gas-chemical-formula | C3H8 | start chemical_formula qc_end substance 3 4 qc_end end | [{"type":"other","value":"Chemical Formula [OF] propane gas [IN] default"}] | [{"type":"chemical equation","value":"C3H8"}] | [{"type":"substance name","value":"Propane gas"}] | <h1 class="questionTitle" itemprop="name">What is the propane gas chemical formula?</h1> | null | C3H8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>OR, if we draw out the hydrocarbyl chain, we get <mathjax>#H_3C-CH_2-CH_3#</mathjax>.</p>
<p>And of course propane is an alkane, i.e. <mathjax>#C_nH_(2n+2)#</mathjax>. Would you expect propane to be more volatile than methane, <mathjax>#CH_4#</mathjax>; what about butane, <mathjax>#C_4H_10#</mathjax> Why or why not?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Propane is"#</mathjax> <mathjax>#C_3H_8#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>OR, if we draw out the hydrocarbyl chain, we get <mathjax>#H_3C-CH_2-CH_3#</mathjax>.</p>
<p>And of course propane is an alkane, i.e. <mathjax>#C_nH_(2n+2)#</mathjax>. Would you expect propane to be more volatile than methane, <mathjax>#CH_4#</mathjax>; what about butane, <mathjax>#C_4H_10#</mathjax> Why or why not?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the propane gas chemical formula?</h1>
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<div class="markdown"><p><mathjax>#"Propane is"#</mathjax> <mathjax>#C_3H_8#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>OR, if we draw out the hydrocarbyl chain, we get <mathjax>#H_3C-CH_2-CH_3#</mathjax>.</p>
<p>And of course propane is an alkane, i.e. <mathjax>#C_nH_(2n+2)#</mathjax>. Would you expect propane to be more volatile than methane, <mathjax>#CH_4#</mathjax>; what about butane, <mathjax>#C_4H_10#</mathjax> Why or why not?</p></div>
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</article> | What is the propane gas chemical formula? | null |
2,652 | a8e88b44-6ddd-11ea-923d-ccda262736ce | https://socratic.org/questions/if-a-buffer-solution-is-0-290-m-in-a-weak-base-kb-6-4-10-5-and-0-520-m-in-its-co | 9.56 | start physical_unit 2 3 ph none qc_end physical_unit 8 10 5 6 molarity qc_end physical_unit 20 22 17 18 molarity qc_end physical_unit 8 10 13 15 kb qc_end end | [{"type":"physical unit","value":"pH [OF] buffer solution"}] | [{"type":"physical unit","value":"9.56"}] | [{"type":"physical unit","value":"Molarity [OF] a weak base [=] \\pu{0.290 M}"},{"type":"physical unit","value":"Molarity [OF] its conjugate acid [=] \\pu{0.520 M}"},{"type":"physical unit","value":"Kb [OF] a weak base [=] \\pu{6.4 × 10^(−5)}"}] | <h1 class="questionTitle" itemprop="name">If a buffer solution is 0.290 M in a weak base (#K_b = 6.4 xx 10^(-5)#) and 0.520 M in its conjugate acid, what is the pH?</h1> | null | 9.56 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the <em>base dissociation constant</em>, <mathjax>#K_b#</mathjax>, to determine the <mathjax>#pK_b#</mathjax> of the weak base.</p>
<p>You should know that </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the given <mathjax>#K_b#</mathjax> to get</p>
<blockquote>
<p><mathjax>#pK_b = -log(6.4 * 10^(-5)) = 4.19#</mathjax></p>
</blockquote>
<p>So, your buffer solution contains a weak base and its conjugate acid in <em>comparable amounts</em>. As you know, the pOH of a buffer solution that contains a weak abase and its conjugate acid can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that when you have <strong>equal amounts</strong> of weak base and conjugate acid, the pOH of the solution is equal to <mathjax>#pK_b#</mathjax>. </p>
<p>In your case, you have <strong>more conjugate acid</strong> than weak base, so right from the start you should expect the pOH of the solution to be <strong>higher</strong> than <mathjax>#pK_b#</mathjax>. </p>
<p>The presence of more conjugate acid implies that the solution is <strong>more acidic</strong>, i.e. the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is <strong>lower</strong> than what corresponds to <mathjax>#"pOH" = pK_b#</mathjax>. </p>
<p>So, plug in your values and calculate the pOH of the solution </p>
<blockquote>
<p><mathjax>#"pOH" = pK_b + log( (0.520 color(red)(cancel(color(black)("M"))))/(0.290color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.19 + log(0.520/0.290) = 4.44#</mathjax></p>
</blockquote>
<p>Since an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>you can say that the pH of the buffer is </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 4.44 = color(green)(|bar(ul(color(white)(a/a)9.56color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 9.56#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The first thing to do here is use the <em>base dissociation constant</em>, <mathjax>#K_b#</mathjax>, to determine the <mathjax>#pK_b#</mathjax> of the weak base.</p>
<p>You should know that </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the given <mathjax>#K_b#</mathjax> to get</p>
<blockquote>
<p><mathjax>#pK_b = -log(6.4 * 10^(-5)) = 4.19#</mathjax></p>
</blockquote>
<p>So, your buffer solution contains a weak base and its conjugate acid in <em>comparable amounts</em>. As you know, the pOH of a buffer solution that contains a weak abase and its conjugate acid can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that when you have <strong>equal amounts</strong> of weak base and conjugate acid, the pOH of the solution is equal to <mathjax>#pK_b#</mathjax>. </p>
<p>In your case, you have <strong>more conjugate acid</strong> than weak base, so right from the start you should expect the pOH of the solution to be <strong>higher</strong> than <mathjax>#pK_b#</mathjax>. </p>
<p>The presence of more conjugate acid implies that the solution is <strong>more acidic</strong>, i.e. the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is <strong>lower</strong> than what corresponds to <mathjax>#"pOH" = pK_b#</mathjax>. </p>
<p>So, plug in your values and calculate the pOH of the solution </p>
<blockquote>
<p><mathjax>#"pOH" = pK_b + log( (0.520 color(red)(cancel(color(black)("M"))))/(0.290color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.19 + log(0.520/0.290) = 4.44#</mathjax></p>
</blockquote>
<p>Since an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>you can say that the pH of the buffer is </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 4.44 = color(green)(|bar(ul(color(white)(a/a)9.56color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If a buffer solution is 0.290 M in a weak base (#K_b = 6.4 xx 10^(-5)#) and 0.520 M in its conjugate acid, what is the pH?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-22T12:47:35" itemprop="dateCreated">
Mar 22, 2016
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<div class="markdown"><p><mathjax>#"pH" = 9.56#</mathjax></p></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 use the <em>base dissociation constant</em>, <mathjax>#K_b#</mathjax>, to determine the <mathjax>#pK_b#</mathjax> of the weak base.</p>
<p>You should know that </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_b = - log(K_b))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in the given <mathjax>#K_b#</mathjax> to get</p>
<blockquote>
<p><mathjax>#pK_b = -log(6.4 * 10^(-5)) = 4.19#</mathjax></p>
</blockquote>
<p>So, your buffer solution contains a weak base and its conjugate acid in <em>comparable amounts</em>. As you know, the pOH of a buffer solution that contains a weak abase and its conjugate acid can be calculated using the <strong>Henderson - Hasselbalch equation</strong></p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log( (["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Notice that when you have <strong>equal amounts</strong> of weak base and conjugate acid, the pOH of the solution is equal to <mathjax>#pK_b#</mathjax>. </p>
<p>In your case, you have <strong>more conjugate acid</strong> than weak base, so right from the start you should expect the pOH of the solution to be <strong>higher</strong> than <mathjax>#pK_b#</mathjax>. </p>
<p>The presence of more conjugate acid implies that the solution is <strong>more acidic</strong>, i.e. the <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> is <strong>lower</strong> than what corresponds to <mathjax>#"pOH" = pK_b#</mathjax>. </p>
<p>So, plug in your values and calculate the pOH of the solution </p>
<blockquote>
<p><mathjax>#"pOH" = pK_b + log( (0.520 color(red)(cancel(color(black)("M"))))/(0.290color(red)(cancel(color(black)("M")))))#</mathjax></p>
<p><mathjax>#"pOH" = 4.19 + log(0.520/0.290) = 4.44#</mathjax></p>
</blockquote>
<p>Since an aqueous solution at room temperature has </p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>you can say that the pH of the buffer is </p>
<blockquote>
<p><mathjax>#"pH" = 14 - "pOH"#</mathjax></p>
<p><mathjax>#"pH" = 14 - 4.44 = color(green)(|bar(ul(color(white)(a/a)9.56color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | If a buffer solution is 0.290 M in a weak base (#K_b = 6.4 xx 10^(-5)#) and 0.520 M in its conjugate acid, what is the pH? | null |
2,653 | ad291e5a-6ddd-11ea-b3fc-ccda262736ce | https://socratic.org/questions/how-many-gram-of-k-2cro-4-are-formed-per-gram-of-k-2co-3-used | 2.81 grams | start physical_unit 4 4 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] K2CrO4 [IN] grams"}] | [{"type":"physical unit","value":"2.81 grams"}] | [{"type":"physical unit","value":"Mass [OF] K2CO3 [=] \\pu{1 gram}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #K_2CrO_4# are formed per gram of #K_2CO_3# used?</h1> | null | 2.81 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"K"_2"Cr"_2"O"_7 + "K"_2"CO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2K"_2"CrO"_4 + "CO"_2"#</mathjax></p>
<p>Determine mol <mathjax>#"K"_2"CO"_3"#</mathjax> are in <mathjax>#"1.00 g"#</mathjax>. (<mathjax>#"1.00 g"#</mathjax> is arbitrary, since the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> were not specified. You can redo the calculations with the number of significant figures you wish.) Divide the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by its molar mass <mathjax>#("138.205 g/mol")#</mathjax> by multiplying the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by the inverse of its molar mass <mathjax>#("1 mol"/"138.205 g")#</mathjax>.</p>
<p><mathjax>#1.00color(red)cancel(color(black)("g K"_2"CO"_3))xx(1"mol K"_2"CO"_3)/(138.205color(red)cancel(color(black)("g K"_2"CO"_3)))="0.007236 mol K"_2"CO"_3"#</mathjax></p>
<p>I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.</p>
<p>Determine mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CO"_3"#</mathjax> by the mol ratio between the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in the balanced equation, with mol <mathjax>#"K"_2"CrO"_4"#</mathjax> in the numerator.</p>
<p><mathjax>#0.007236color(red)cancel(color(black)("mol K"_2"CO"_3))xx(2"mol K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CO"_3)))="0.01447 mol K"_2"CrO"_4"#</mathjax></p>
<p>Determine mass <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by its molar mass <mathjax>#("194.189 g/mol")#</mathjax>.</p>
<p><mathjax>#0.01447color(red)cancel(color(black)("mol K"_2"CrO"_4))xx(194.189"g K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CrO"_4)))="2.81 g K"_2"CrO"_4"#</mathjax> (rounded to three significant figures)</p>
<p>For every gram of <mathjax>#"K"_2"CO"_3"#</mathjax> used, <mathjax>#"2.81 g K"_2"CrO"_4"#</mathjax> is formed.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>For every gram of <mathjax>#"K"_2"CO"_3"#</mathjax> used, <mathjax>#"2.82 g K"_2"CrO"_4"#</mathjax> is formed.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"K"_2"Cr"_2"O"_7 + "K"_2"CO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2K"_2"CrO"_4 + "CO"_2"#</mathjax></p>
<p>Determine mol <mathjax>#"K"_2"CO"_3"#</mathjax> are in <mathjax>#"1.00 g"#</mathjax>. (<mathjax>#"1.00 g"#</mathjax> is arbitrary, since the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> were not specified. You can redo the calculations with the number of significant figures you wish.) Divide the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by its molar mass <mathjax>#("138.205 g/mol")#</mathjax> by multiplying the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by the inverse of its molar mass <mathjax>#("1 mol"/"138.205 g")#</mathjax>.</p>
<p><mathjax>#1.00color(red)cancel(color(black)("g K"_2"CO"_3))xx(1"mol K"_2"CO"_3)/(138.205color(red)cancel(color(black)("g K"_2"CO"_3)))="0.007236 mol K"_2"CO"_3"#</mathjax></p>
<p>I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.</p>
<p>Determine mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CO"_3"#</mathjax> by the mol ratio between the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in the balanced equation, with mol <mathjax>#"K"_2"CrO"_4"#</mathjax> in the numerator.</p>
<p><mathjax>#0.007236color(red)cancel(color(black)("mol K"_2"CO"_3))xx(2"mol K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CO"_3)))="0.01447 mol K"_2"CrO"_4"#</mathjax></p>
<p>Determine mass <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by its molar mass <mathjax>#("194.189 g/mol")#</mathjax>.</p>
<p><mathjax>#0.01447color(red)cancel(color(black)("mol K"_2"CrO"_4))xx(194.189"g K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CrO"_4)))="2.81 g K"_2"CrO"_4"#</mathjax> (rounded to three significant figures)</p>
<p>For every gram of <mathjax>#"K"_2"CO"_3"#</mathjax> used, <mathjax>#"2.81 g K"_2"CrO"_4"#</mathjax> is formed.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of #K_2CrO_4# are formed per gram of #K_2CO_3# used?</h1>
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<div class="markdown"><p>For every gram of <mathjax>#"K"_2"CO"_3"#</mathjax> used, <mathjax>#"2.82 g K"_2"CrO"_4"#</mathjax> is formed.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced equation</strong></p>
<p><mathjax>#"K"_2"Cr"_2"O"_7 + "K"_2"CO"_3"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2K"_2"CrO"_4 + "CO"_2"#</mathjax></p>
<p>Determine mol <mathjax>#"K"_2"CO"_3"#</mathjax> are in <mathjax>#"1.00 g"#</mathjax>. (<mathjax>#"1.00 g"#</mathjax> is arbitrary, since the number of <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a> were not specified. You can redo the calculations with the number of significant figures you wish.) Divide the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by its molar mass <mathjax>#("138.205 g/mol")#</mathjax> by multiplying the mass of <mathjax>#"K"_2"CO"_3"#</mathjax> by the inverse of its molar mass <mathjax>#("1 mol"/"138.205 g")#</mathjax>.</p>
<p><mathjax>#1.00color(red)cancel(color(black)("g K"_2"CO"_3))xx(1"mol K"_2"CO"_3)/(138.205color(red)cancel(color(black)("g K"_2"CO"_3)))="0.007236 mol K"_2"CO"_3"#</mathjax></p>
<p>I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.</p>
<p>Determine mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CO"_3"#</mathjax> by the mol ratio between the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in the balanced equation, with mol <mathjax>#"K"_2"CrO"_4"#</mathjax> in the numerator.</p>
<p><mathjax>#0.007236color(red)cancel(color(black)("mol K"_2"CO"_3))xx(2"mol K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CO"_3)))="0.01447 mol K"_2"CrO"_4"#</mathjax></p>
<p>Determine mass <mathjax>#"K"_2"CrO"_4"#</mathjax> by multiplying mol <mathjax>#"K"_2"CrO"_4"#</mathjax> by its molar mass <mathjax>#("194.189 g/mol")#</mathjax>.</p>
<p><mathjax>#0.01447color(red)cancel(color(black)("mol K"_2"CrO"_4))xx(194.189"g K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CrO"_4)))="2.81 g K"_2"CrO"_4"#</mathjax> (rounded to three significant figures)</p>
<p>For every gram of <mathjax>#"K"_2"CO"_3"#</mathjax> used, <mathjax>#"2.81 g K"_2"CrO"_4"#</mathjax> is formed.</p></div>
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</article> | How many grams of #K_2CrO_4# are formed per gram of #K_2CO_3# used? | null |
2,654 | a9a89b5a-6ddd-11ea-b403-ccda262736ce | https://socratic.org/questions/how-many-grams-in-3-75-mol-of-nabr | 386.88 grams | start physical_unit 7 7 mass g qc_end physical_unit 7 7 4 5 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] NaBr [IN] grams"}] | [{"type":"physical unit","value":"386.88 grams"}] | [{"type":"physical unit","value":"Mole [OF] NaBr [=] \\pu{3.75 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams in 3.75 mol of #NaBr#?</h1> | null | 386.88 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can find how many grams would be equivalent to <mathjax>#3.75#</mathjax> of <em>sodium bromide</em>, <mathjax>#"NaBr"#</mathjax>, by using the mass of <strong>one mole</strong> of this <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a></strong>.</p>
<p>As you know, the mass of <strong>one mole</strong> of a compound is expressed by that compound's <strong>molar mass</strong>. Sodium bromide has a molar mass <mathjax>#"102.9 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of sodium bromide has a mass of <mathjax>#"102.9 g"#</mathjax>. </p>
<p>You can thus use the molar mass of a compound to <strong>convert</strong> between its <em>mass</em> and the <em>number of moles</em> equivalent to that mass. In your case, you will have</p>
<blockquote>
<p><mathjax>#3.75 color(red)(cancel(color(black)("moles NaBr"))) * overbrace("102.9 g"/(1color(red)(cancel(color(black)("mole NaBr")))))^(color(purple)("molar mass")) = "385.875 g"#</mathjax></p>
</blockquote>
<p>You need to round this off 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 sodium bromide</p>
<blockquote>
<p><mathjax>#m_(NaBr) = color(green)("386 g")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"386 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can find how many grams would be equivalent to <mathjax>#3.75#</mathjax> of <em>sodium bromide</em>, <mathjax>#"NaBr"#</mathjax>, by using the mass of <strong>one mole</strong> of this <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a></strong>.</p>
<p>As you know, the mass of <strong>one mole</strong> of a compound is expressed by that compound's <strong>molar mass</strong>. Sodium bromide has a molar mass <mathjax>#"102.9 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of sodium bromide has a mass of <mathjax>#"102.9 g"#</mathjax>. </p>
<p>You can thus use the molar mass of a compound to <strong>convert</strong> between its <em>mass</em> and the <em>number of moles</em> equivalent to that mass. In your case, you will have</p>
<blockquote>
<p><mathjax>#3.75 color(red)(cancel(color(black)("moles NaBr"))) * overbrace("102.9 g"/(1color(red)(cancel(color(black)("mole NaBr")))))^(color(purple)("molar mass")) = "385.875 g"#</mathjax></p>
</blockquote>
<p>You need to round this off 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 sodium bromide</p>
<blockquote>
<p><mathjax>#m_(NaBr) = color(green)("386 g")#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">How many grams in 3.75 mol of #NaBr#?</h1>
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Stefan V.
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Feb 20, 2016
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<div class="markdown"><p><mathjax>#"386 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can find how many grams would be equivalent to <mathjax>#3.75#</mathjax> of <em>sodium bromide</em>, <mathjax>#"NaBr"#</mathjax>, by using the mass of <strong>one mole</strong> of this <strong><a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a></strong>.</p>
<p>As you know, the mass of <strong>one mole</strong> of a compound is expressed by that compound's <strong>molar mass</strong>. Sodium bromide has a molar mass <mathjax>#"102.9 g mol"^(-1)#</mathjax>, which means that <strong>one mole</strong> of sodium bromide has a mass of <mathjax>#"102.9 g"#</mathjax>. </p>
<p>You can thus use the molar mass of a compound to <strong>convert</strong> between its <em>mass</em> and the <em>number of moles</em> equivalent to that mass. In your case, you will have</p>
<blockquote>
<p><mathjax>#3.75 color(red)(cancel(color(black)("moles NaBr"))) * overbrace("102.9 g"/(1color(red)(cancel(color(black)("mole NaBr")))))^(color(purple)("molar mass")) = "385.875 g"#</mathjax></p>
</blockquote>
<p>You need to round this off 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 sodium bromide</p>
<blockquote>
<p><mathjax>#m_(NaBr) = color(green)("386 g")#</mathjax></p>
</blockquote></div>
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</article> | How many grams in 3.75 mol of #NaBr#? | null |
2,655 | a9709000-6ddd-11ea-9f6e-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-cesium-iodine-cesium-iodide | 2 Cs(s) + I2(g) -> 2 CsI(s) | start chemical_equation qc_end chemical_equation 6 11 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Cs(s) + I2(g) -> 2 CsI(s)"}] | [{"type":"chemical equation","value":"cesium + iodine -> cesium iodide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for cesium + iodine ---> cesium iodide?</h1> | null | 2 Cs(s) + I2(g) -> 2 CsI(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>2Cs(s) + I2(g) → CsI(s)<br/>
Notice in the equation that iodine is in it's gaseous state in order for the reaction to proceed. But you were 100% correct, the reaction between cesium and iodine is simply, cesium iodide. Very Good!!</p></div>
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<div class="markdown"><p>2Cs(s) + I2(g) → CsI(s) </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>2Cs(s) + I2(g) → CsI(s)<br/>
Notice in the equation that iodine is in it's gaseous state in order for the reaction to proceed. But you were 100% correct, the reaction between cesium and iodine is simply, cesium iodide. Very Good!!</p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical equation for cesium + iodine ---> cesium iodide?</h1>
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<div class="markdown"><p>2Cs(s) + I2(g) → CsI(s) </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>2Cs(s) + I2(g) → CsI(s)<br/>
Notice in the equation that iodine is in it's gaseous state in order for the reaction to proceed. But you were 100% correct, the reaction between cesium and iodine is simply, cesium iodide. Very Good!!</p></div>
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</article> | What is the chemical equation for cesium + iodine ---> cesium iodide? | null |
2,656 | acb89d15-6ddd-11ea-8318-ccda262736ce | https://socratic.org/questions/5905bcb3b72cff212c690bdf | NaClO4 | start chemical_formula qc_end physical_unit 11 11 8 9 mass qc_end physical_unit 15 15 12 13 mass qc_end physical_unit 20 20 17 18 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] this salt [IN] empirical"}] | [{"type":"chemical equation","value":"NaClO4"}] | [{"type":"physical unit","value":"Mass [OF] sodium [=] \\pu{0.92 g}"},{"type":"physical unit","value":"Mass [OF] chlorine [=] \\pu{1.42 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{2.56 g}"}] | <h1 class="questionTitle" itemprop="name">A salt is analyzed and found to contain #0.92*g# of sodium, #1.42*g# of chlorine, and #2.56*g# of oxygen. What is the empirical formula of this salt? </h1> | null | NaClO4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.92*g)/(22.99*g*mol^-1)=0.0400*mol#</mathjax></p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.42*g)/(35.45*g*mol^-1)=0.0400*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.56*g)/(15.99*g*mol^-1)=0.160*mol#</mathjax>.</p>
<p>Note that I DIVIDED THRU BY the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC MASS</a> of each constituent. And now, we divide thru by the SMALLEST molar quantity, that of sodium, to get an empirical formula of <mathjax>#NaClO_4#</mathjax>; and this is <mathjax>#"sodium perchlorate"#</mathjax>,</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well, we find the molar quantities of each constituent.........to get an empirical formula of <mathjax>#NaClO_4#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.92*g)/(22.99*g*mol^-1)=0.0400*mol#</mathjax></p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.42*g)/(35.45*g*mol^-1)=0.0400*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.56*g)/(15.99*g*mol^-1)=0.160*mol#</mathjax>.</p>
<p>Note that I DIVIDED THRU BY the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC MASS</a> of each constituent. And now, we divide thru by the SMALLEST molar quantity, that of sodium, to get an empirical formula of <mathjax>#NaClO_4#</mathjax>; and this is <mathjax>#"sodium perchlorate"#</mathjax>,</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A salt is analyzed and found to contain #0.92*g# of sodium, #1.42*g# of chlorine, and #2.56*g# of oxygen. What is the empirical formula of this salt? </h1>
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<div class="markdown"><p>Well, we find the molar quantities of each constituent.........to get an empirical formula of <mathjax>#NaClO_4#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of sodium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.92*g)/(22.99*g*mol^-1)=0.0400*mol#</mathjax></p>
<p><mathjax>#"Moles of chlorine"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.42*g)/(35.45*g*mol^-1)=0.0400*mol#</mathjax>.</p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.56*g)/(15.99*g*mol^-1)=0.160*mol#</mathjax>.</p>
<p>Note that I DIVIDED THRU BY the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">ATOMIC MASS</a> of each constituent. And now, we divide thru by the SMALLEST molar quantity, that of sodium, to get an empirical formula of <mathjax>#NaClO_4#</mathjax>; and this is <mathjax>#"sodium perchlorate"#</mathjax>,</p></div>
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</article> | A salt is analyzed and found to contain #0.92*g# of sodium, #1.42*g# of chlorine, and #2.56*g# of oxygen. What is the empirical formula of this salt? | null |
2,657 | acf757f4-6ddd-11ea-859c-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-percentage-by-mass-of-carbon-in-carbon-monoxide | 42.88% | start physical_unit 9 12 mass_percent none qc_end substance 11 12 qc_end substance 9 9 qc_end end | [{"type":"physical unit","value":"Percentage by mass [OF] carbon in carbon monoxide"}] | [{"type":"physical unit","value":"42.88%"}] | [{"type":"substance name","value":"Carbon monoxide"},{"type":"substance name","value":"Carbon"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the percentage by mass of carbon in carbon monoxide?</h1> | null | 42.88% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"So %C by mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.011*g*mol^-1)/((12.011+15.999)*g*mol^-1)xx100%#</mathjax></p>
<p>So <mathjax>#C-=O#</mathjax>, percentage carbon by mass is a bit under <mathjax>#50%#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"%C by mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of carbon"/"Mass of carbon + oxygen"xx100%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"So %C by mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.011*g*mol^-1)/((12.011+15.999)*g*mol^-1)xx100%#</mathjax></p>
<p>So <mathjax>#C-=O#</mathjax>, percentage carbon by mass is a bit under <mathjax>#50%#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate the percentage by mass of carbon in carbon monoxide?</h1>
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<div class="markdown"><p><mathjax>#"%C by mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of carbon"/"Mass of carbon + oxygen"xx100%#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"So %C by mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(12.011*g*mol^-1)/((12.011+15.999)*g*mol^-1)xx100%#</mathjax></p>
<p>So <mathjax>#C-=O#</mathjax>, percentage carbon by mass is a bit under <mathjax>#50%#</mathjax>.</p></div>
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</article> | How do you calculate the percentage by mass of carbon in carbon monoxide? | null |
2,658 | a9784b7b-6ddd-11ea-9b60-ccda262736ce | https://socratic.org/questions/what-mass-of-solute-is-present-in-250-ml-of-a-0-100-mol-l-solution-of-naoh-molar | 0.10 g | start physical_unit 3 3 mass g qc_end physical_unit 13 13 7 8 volume qc_end physical_unit 15 15 11 12 molarity qc_end physical_unit 15 15 21 22 molar_mass qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] g"}] | [{"type":"physical unit","value":"0.10 g"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{250 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.100 mol/L}"},{"type":"physical unit","value":"Molar mass [OF] NaOH [=] \\pu{40.000 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What mass of solute is present in 250. mL of a 0.100 mol/L solution of NaOH (molar mass of NaOH = 40.000 g/mol)?</h1> | null | 0.10 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we use the relationship.....</p>
<p><mathjax>#"Concentration"-="Amount of solute"/"Volume of solution"#</mathjax></p>
<p>And typically we report the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, where.....</p>
<p><mathjax>#"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"#</mathjax></p>
<p>And thus the units of <mathjax>#"molarity"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>.</p>
<p>And so....<mathjax>#"moles of solute"="Molarity"xx"Volume"#</mathjax></p>
<p><mathjax>#=0.100*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancel(L*mL^-1)#</mathjax>...</p>
<p><mathjax>#=2.50xx10^-3*mol#</mathjax></p>
<p>And this represents a mass of................</p>
<p><mathjax>#2.50xx10^-3*molxx40.00*g*mol^-1=??*g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#100*mg#</mathjax>.........</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we use the relationship.....</p>
<p><mathjax>#"Concentration"-="Amount of solute"/"Volume of solution"#</mathjax></p>
<p>And typically we report the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, where.....</p>
<p><mathjax>#"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"#</mathjax></p>
<p>And thus the units of <mathjax>#"molarity"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>.</p>
<p>And so....<mathjax>#"moles of solute"="Molarity"xx"Volume"#</mathjax></p>
<p><mathjax>#=0.100*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancel(L*mL^-1)#</mathjax>...</p>
<p><mathjax>#=2.50xx10^-3*mol#</mathjax></p>
<p>And this represents a mass of................</p>
<p><mathjax>#2.50xx10^-3*molxx40.00*g*mol^-1=??*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of solute is present in 250. mL of a 0.100 mol/L solution of NaOH (molar mass of NaOH = 40.000 g/mol)?</h1>
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<div class="markdown"><p><mathjax>#100*mg#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>As with all these problems, we use the relationship.....</p>
<p><mathjax>#"Concentration"-="Amount of solute"/"Volume of solution"#</mathjax></p>
<p>And typically we report the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>, where.....</p>
<p><mathjax>#"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"#</mathjax></p>
<p>And thus the units of <mathjax>#"molarity"#</mathjax> are <mathjax>#mol*L^-1#</mathjax>.</p>
<p>And so....<mathjax>#"moles of solute"="Molarity"xx"Volume"#</mathjax></p>
<p><mathjax>#=0.100*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancel(L*mL^-1)#</mathjax>...</p>
<p><mathjax>#=2.50xx10^-3*mol#</mathjax></p>
<p>And this represents a mass of................</p>
<p><mathjax>#2.50xx10^-3*molxx40.00*g*mol^-1=??*g#</mathjax></p></div>
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</article> | What mass of solute is present in 250. mL of a 0.100 mol/L solution of NaOH (molar mass of NaOH = 40.000 g/mol)? | null |
2,659 | ad081663-6ddd-11ea-8035-ccda262736ce | https://socratic.org/questions/58be1d1e11ef6b3c3dfe433b | 6.76 g | start physical_unit 3 3 mass g qc_end physical_unit 12 12 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] phosphorus [IN] g"}] | [{"type":"physical unit","value":"6.76 g"}] | [{"type":"physical unit","value":"Mass [OF] P2O5 [=] \\pu{15.5 g}"}] | <h1 class="questionTitle" itemprop="name">What mass of phosphorus is contained in a #15.5*g# mass of #P_2O_5#?</h1> | null | 6.76 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) to assess the molar quantity of <mathjax>#P_4O_10#</mathjax> (which of course is the same as <mathjax>#P_2O_5#</mathjax>, and which formula we can use without loss of generality).</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#P_4O_10#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15.5*g)/(283.89*g*mol^-1)=0.0546*mol#</mathjax>.</p>
<p>Now in one mole of <mathjax>#P_4O_10#</mathjax>, CLEARLY there are 4 moles of phosphorus. So (ii) we need to multiply the given molar quantity by <mathjax>#4xx30.9737*g*mol^-1#</mathjax>, where <mathjax>#30.9737*g*mol^-1#</mathjax> is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of phosphorus.</p>
<p><mathjax>#"Mass of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0546*molxx4xx30.9737*g*mol^-1#</mathjax></p>
<p><mathjax>#~=7*g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"Mass of phosphorus"~=7*g#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) to assess the molar quantity of <mathjax>#P_4O_10#</mathjax> (which of course is the same as <mathjax>#P_2O_5#</mathjax>, and which formula we can use without loss of generality).</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#P_4O_10#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15.5*g)/(283.89*g*mol^-1)=0.0546*mol#</mathjax>.</p>
<p>Now in one mole of <mathjax>#P_4O_10#</mathjax>, CLEARLY there are 4 moles of phosphorus. So (ii) we need to multiply the given molar quantity by <mathjax>#4xx30.9737*g*mol^-1#</mathjax>, where <mathjax>#30.9737*g*mol^-1#</mathjax> is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of phosphorus.</p>
<p><mathjax>#"Mass of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0546*molxx4xx30.9737*g*mol^-1#</mathjax></p>
<p><mathjax>#~=7*g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of phosphorus is contained in a #15.5*g# mass of #P_2O_5#?</h1>
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<div class="markdown"><p><mathjax>#"Mass of phosphorus"~=7*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) to assess the molar quantity of <mathjax>#P_4O_10#</mathjax> (which of course is the same as <mathjax>#P_2O_5#</mathjax>, and which formula we can use without loss of generality).</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#P_4O_10#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(15.5*g)/(283.89*g*mol^-1)=0.0546*mol#</mathjax>.</p>
<p>Now in one mole of <mathjax>#P_4O_10#</mathjax>, CLEARLY there are 4 moles of phosphorus. So (ii) we need to multiply the given molar quantity by <mathjax>#4xx30.9737*g*mol^-1#</mathjax>, where <mathjax>#30.9737*g*mol^-1#</mathjax> is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of phosphorus.</p>
<p><mathjax>#"Mass of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0546*molxx4xx30.9737*g*mol^-1#</mathjax></p>
<p><mathjax>#~=7*g#</mathjax>.</p></div>
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</article> | What mass of phosphorus is contained in a #15.5*g# mass of #P_2O_5#? | null |
2,660 | acfb7790-6ddd-11ea-b4e7-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-half-reaction-for-mg-s-zncl-2-ag-mgcl-2-ag-zn-s | Mg(s) -> Mg^2+ + 2e- | start chemical_equation qc_end chemical_equation 7 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation half reaction"}] | [{"type":"chemical equation","value":"Mg(s) -> Mg^2+ + 2e-"}] | [{"type":"chemical equation","value":"Mg(s) + ZnCl2(ag) -> MgCl2(ag) + Zn(s)"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation half reaction for #Mg(s) + ZnCl_2(ag) -> MgCl_2(ag) + Zn(s)#?</h1> | null | Mg(s) -> Mg^2+ + 2e- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you write what is known as the net ionic equation, it is a simpler matter to identify the oxidation (and the reduction form that matter).</p>
<p>First, write the two salts in aqueous ion form:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) +2Cl^-1 (aq) rarr Zn(s)+Mg^(2+)(aq) +2Cl^-1 (aq) #</mathjax></p>
<p>Since no change occurs in the chloride ion, these are "spectators" and can be omitted:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) rarr Zn(s)+Mg^(2+)(aq) #</mathjax></p>
<p>Now, we have a more clear description of what has taken place.</p>
<p><mathjax>#Zn^(2+)#</mathjax> has been reduced to <mathjax>#Zn(s)#</mathjax></p>
<p><mathjax>#Mg(s)#</mathjax> has been oxidized to <mathjax>#Mg^(2+)#</mathjax></p></div>
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<div class="markdown"><p>The oxidation half-reaction is</p>
<p><mathjax>#Mg(s) rarr Mg^(2+) + 2 e^-#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you write what is known as the net ionic equation, it is a simpler matter to identify the oxidation (and the reduction form that matter).</p>
<p>First, write the two salts in aqueous ion form:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) +2Cl^-1 (aq) rarr Zn(s)+Mg^(2+)(aq) +2Cl^-1 (aq) #</mathjax></p>
<p>Since no change occurs in the chloride ion, these are "spectators" and can be omitted:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) rarr Zn(s)+Mg^(2+)(aq) #</mathjax></p>
<p>Now, we have a more clear description of what has taken place.</p>
<p><mathjax>#Zn^(2+)#</mathjax> has been reduced to <mathjax>#Zn(s)#</mathjax></p>
<p><mathjax>#Mg(s)#</mathjax> has been oxidized to <mathjax>#Mg^(2+)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation half reaction for #Mg(s) + ZnCl_2(ag) -> MgCl_2(ag) + Zn(s)#?</h1>
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<div class="markdown"><p>The oxidation half-reaction is</p>
<p><mathjax>#Mg(s) rarr Mg^(2+) + 2 e^-#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you write what is known as the net ionic equation, it is a simpler matter to identify the oxidation (and the reduction form that matter).</p>
<p>First, write the two salts in aqueous ion form:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) +2Cl^-1 (aq) rarr Zn(s)+Mg^(2+)(aq) +2Cl^-1 (aq) #</mathjax></p>
<p>Since no change occurs in the chloride ion, these are "spectators" and can be omitted:</p>
<p><mathjax>#Mg(s)+Zn^(2+)(aq) rarr Zn(s)+Mg^(2+)(aq) #</mathjax></p>
<p>Now, we have a more clear description of what has taken place.</p>
<p><mathjax>#Zn^(2+)#</mathjax> has been reduced to <mathjax>#Zn(s)#</mathjax></p>
<p><mathjax>#Mg(s)#</mathjax> has been oxidized to <mathjax>#Mg^(2+)#</mathjax></p></div>
</div>
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<div class="markdown"><p>Anode: Mg → <mathjax>#Mg^(2+)#</mathjax> + 2e- 2.38V</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the equation we see that Mg metal (oxidation state 0) changes to the ion <mathjax>#Mg^(2+)#</mathjax>. It is thus “oxidized” and is the reducing agent. It is thus the anode of the cell. </p>
<p>Zn is changed from it’s ionic state, <mathjax>#Zn^(2+)#</mathjax> to its metallic state of 0. It is thus “reduced” and is the oxidizing agent. It is thus the cathode of the cell. </p>
<p>By convention, all half-cell emf's are compared to the emf of the standard hydrogen electrode. The emf of a half-cell, with respect to the standard hydrogen electrode, is called the reduction potential.<br/>
In an electrochemical cell, the general equation is:</p>
<p><mathjax>#E = E^o(red)#</mathjax> (cathode) - <mathjax>#E^o(red)#</mathjax> (anode) </p>
<p>The emf for the Mg-Zn cell described would be:</p>
<p><mathjax>#E = E^o(red)#</mathjax>(Zn) - <mathjax>#E^o(red)#</mathjax>(Mg) = -0.76 - (-2.38) = 1.62V if the solutions are 1.0 M.</p>
<p>The two “standard cell” cathode (reduction) half reactions are:<br/>
<mathjax>#Mg^(2+)#</mathjax> + 2e- → Mg - 2.38V<br/>
<mathjax>#Zn^(2+)#</mathjax> + 2e- → Zn - 0.76V<br/>
<a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html" rel="nofollow" target="_blank">http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html</a></p>
<p>Put in the general equation for the actual cell, the full reaction is:<br/>
Mg + <mathjax>#Zn^(2+)#</mathjax>→<mathjax>#Mg^(2+)#</mathjax> + Zn <br/>
-0.76 - (-2.38) = 1.62V</p></div>
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</article> | What is the oxidation half reaction for #Mg(s) + ZnCl_2(ag) -> MgCl_2(ag) + Zn(s)#? | null |
2,661 | aa8c1c46-6ddd-11ea-8a77-ccda262736ce | https://socratic.org/questions/57b79b5f7c014973f9cbe9fb | 10.57 | start physical_unit 11 12 poh none qc_end physical_unit 1 1 3 6 molarity qc_end end | [{"type":"physical unit","value":"pOH [OF] this solution"}] | [{"type":"physical unit","value":"10.57"}] | [{"type":"physical unit","value":"Molarity [OF] HO- [=] \\pu{2.7 × 10^(-11) mol/L}"}] | <h1 class="questionTitle" itemprop="name">If #HO^-=2.7xx10^-11*mol*L^-1#, what is #pOH# of this solution?</h1> | null | 10.57 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(2.7xx10^-11)#</mathjax>.</p>
<p>Is this solution acidic or basic? How do you know?</p></div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10.57#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(2.7xx10^-11)#</mathjax>.</p>
<p>Is this solution acidic or basic? How do you know?</p></div>
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<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10.57#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10[HO^-]#</mathjax>.</p>
<p>Thus <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(2.7xx10^-11)#</mathjax>.</p>
<p>Is this solution acidic or basic? How do you know?</p></div>
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</article> | If #HO^-=2.7xx10^-11*mol*L^-1#, what is #pOH# of this solution? | null |
2,662 | ab968b54-6ddd-11ea-89cd-ccda262736ce | https://socratic.org/questions/how-many-hydrogen-atoms-are-present-in-one-molecule-of-ammonium-acetate-nh-4c-2h-1 | 7 | start physical_unit 2 3 number none qc_end end | [{"type":"physical unit","value":"Number [OF] hydrogen atoms"}] | [{"type":"physical unit","value":"7"}] | [{"type":"physical unit","value":"Number [OF] NH4C2H3O2 particle [=] \\pu{1}"}] | <h1 class="questionTitle" itemprop="name">How many hydrogen atoms are present in one particle of ammonium acetate, #NH_4C_2H_3O_2#?</h1> | null | 7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The subscripts below the Hydrogen symbol are 4 and 3 <br/>
Add the subscripts <mathjax># 3 +4 = 7#</mathjax></p></div>
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<div class="markdown"><p>7</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The subscripts below the Hydrogen symbol are 4 and 3 <br/>
Add the subscripts <mathjax># 3 +4 = 7#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many hydrogen atoms are present in one particle of ammonium acetate, #NH_4C_2H_3O_2#?</h1>
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<div class="markdown"><p>7</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The subscripts below the Hydrogen symbol are 4 and 3 <br/>
Add the subscripts <mathjax># 3 +4 = 7#</mathjax></p></div>
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</article> | How many hydrogen atoms are present in one particle of ammonium acetate, #NH_4C_2H_3O_2#? | null |
2,663 | a91dd41c-6ddd-11ea-90e6-ccda262736ce | https://socratic.org/questions/a-45-0-ml-sample-of-water-is-heated-from-15-0-c-to-35-0-c-how-many-joules-of-ene | 4000 joules | start physical_unit 5 5 heat_energy j qc_end physical_unit 3 5 1 2 volume qc_end physical_unit 3 5 9 10 temperature qc_end physical_unit 3 5 12 13 temperature qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] water [IN] joules"}] | [{"type":"physical unit","value":"4000 joules"}] | [{"type":"physical unit","value":"Volume [OF] water sample [=] \\pu{45.0 mL}"},{"type":"physical unit","value":"Temperature1 [OF] water sample [=] \\pu{15.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water sample [=] \\pu{35.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 45.0 mL sample of water is heated from 15.0°C to 35.0°C. How many joules of energy have been absorbed by the water? </h1> | null | 4000 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now <mathjax>#C_p#</mathjax>, <mathjax>#"specific heat of water"#</mathjax>, is <mathjax>#4.18*J*g^-1*""^@C^-1#</mathjax>. We have the mass of water, and we have <mathjax>#DeltaT=20.0*""^@C#</mathjax>.</p>
<p>And so we solve the product.......</p>
<p><mathjax>#"Mass of water"xxC_pxxDeltaT#</mathjax></p>
<p><mathjax>#45.0*mL*xx1*g*mL^-1xx4.18*J*g^-1*""^@C^-1xx20.0*""^@C#</mathjax></p>
<p><mathjax>#~=4000*J#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We need the <a href="https://www.e-education.psu.edu/earth501/content/p5_p9.html" rel="nofollow">specific heat of water</a> in <mathjax>#J*g^-1*""^@C^-1#</mathjax>. These data should have been supplied with the question. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now <mathjax>#C_p#</mathjax>, <mathjax>#"specific heat of water"#</mathjax>, is <mathjax>#4.18*J*g^-1*""^@C^-1#</mathjax>. We have the mass of water, and we have <mathjax>#DeltaT=20.0*""^@C#</mathjax>.</p>
<p>And so we solve the product.......</p>
<p><mathjax>#"Mass of water"xxC_pxxDeltaT#</mathjax></p>
<p><mathjax>#45.0*mL*xx1*g*mL^-1xx4.18*J*g^-1*""^@C^-1xx20.0*""^@C#</mathjax></p>
<p><mathjax>#~=4000*J#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 45.0 mL sample of water is heated from 15.0°C to 35.0°C. How many joules of energy have been absorbed by the water? </h1>
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<div class="markdown"><p>We need the <a href="https://www.e-education.psu.edu/earth501/content/p5_p9.html" rel="nofollow">specific heat of water</a> in <mathjax>#J*g^-1*""^@C^-1#</mathjax>. These data should have been supplied with the question. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now <mathjax>#C_p#</mathjax>, <mathjax>#"specific heat of water"#</mathjax>, is <mathjax>#4.18*J*g^-1*""^@C^-1#</mathjax>. We have the mass of water, and we have <mathjax>#DeltaT=20.0*""^@C#</mathjax>.</p>
<p>And so we solve the product.......</p>
<p><mathjax>#"Mass of water"xxC_pxxDeltaT#</mathjax></p>
<p><mathjax>#45.0*mL*xx1*g*mL^-1xx4.18*J*g^-1*""^@C^-1xx20.0*""^@C#</mathjax></p>
<p><mathjax>#~=4000*J#</mathjax></p></div>
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</article> | A 45.0 mL sample of water is heated from 15.0°C to 35.0°C. How many joules of energy have been absorbed by the water? | null |
2,664 | a9cd3590-6ddd-11ea-b4ff-ccda262736ce | https://socratic.org/questions/what-is-the-final-temperature-of-the-copper-and-water-given-that-the-specific-he | 29.3 ℃ | start physical_unit 6 9 temperature °c qc_end physical_unit 7 7 18 21 specific_heat qc_end physical_unit 7 7 35 36 temperature qc_end physical_unit 7 7 26 27 mass qc_end physical_unit 43 43 40 41 volume qc_end physical_unit 43 43 46 47 temperature qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the copper and water [IN] ℃"}] | [{"type":"physical unit","value":"29.3 ℃"}] | [{"type":"physical unit","value":"Specific heat [OF] copper [=] \\pu{0.385 J/(g * ℃)}"},{"type":"physical unit","value":"Temperature1 [OF] copper [=] \\pu{76.8 ℃}"},{"type":"physical unit","value":"Mass [OF] copper [=] \\pu{48.7 g}"},{"type":"physical unit","value":"Volume [OF] H2O [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Temperature1 [OF] H2O [=] \\pu{25.0 ℃}"},{"type":"other","value":"Allow to reach thermal equilibrium."},{"type":"other","value":"Assume no heat is lost to surroundings."}] | <h1 class="questionTitle" itemprop="name">What is the final temperature of the copper and water given that the specific heat of copper is 0.385 #J/(g * "^oC)#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>A hot lump of 48.7 g of copper at an initial temperature of 76.8 C is placed in 50.0 mL of <mathjax>#H_2O#</mathjax> initially at 25.0°C and allowed to reach thermal equilibrium. Assume no heat is lost to surroundings.</p></div>
</h2>
</div>
</div> | 29.3 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, we're setting the heat values derived given the equation:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax> </p>
<p>equal to each other, as such,</p>
<p><mathjax>#q_(water)+q(Cu) = 0#</mathjax><br/>
<mathjax>#therefore q_(water) = -q(Cu)#</mathjax></p>
<p>Before we start, I will assume the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water is <mathjax>#"1.00 g/mL"#</mathjax>.</p>
<p><mathjax>#50.0 g*(4.184J)/(g*°C)*(T_f-25.0°C) = -[48.7g*(0.385J)/(g*°C)*(T_f-76.8°C)]#</mathjax></p>
<p><mathjax>#209.2T_f - 5230 °C = -18.75T_f + 1440 °C#</mathjax></p>
<p><mathjax>#227.95T_f = 6670 °C#</mathjax></p>
<p><mathjax>#therefore T_f = 29.3°C#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#T_f = 29.3 °C#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, we're setting the heat values derived given the equation:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax> </p>
<p>equal to each other, as such,</p>
<p><mathjax>#q_(water)+q(Cu) = 0#</mathjax><br/>
<mathjax>#therefore q_(water) = -q(Cu)#</mathjax></p>
<p>Before we start, I will assume the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water is <mathjax>#"1.00 g/mL"#</mathjax>.</p>
<p><mathjax>#50.0 g*(4.184J)/(g*°C)*(T_f-25.0°C) = -[48.7g*(0.385J)/(g*°C)*(T_f-76.8°C)]#</mathjax></p>
<p><mathjax>#209.2T_f - 5230 °C = -18.75T_f + 1440 °C#</mathjax></p>
<p><mathjax>#227.95T_f = 6670 °C#</mathjax></p>
<p><mathjax>#therefore T_f = 29.3°C#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the final temperature of the copper and water given that the specific heat of copper is 0.385 #J/(g * "^oC)#?</h1>
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<div class="markdown"><p>A hot lump of 48.7 g of copper at an initial temperature of 76.8 C is placed in 50.0 mL of <mathjax>#H_2O#</mathjax> initially at 25.0°C and allowed to reach thermal equilibrium. Assume no heat is lost to surroundings.</p></div>
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<div class="markdown"><p><mathjax>#T_f = 29.3 °C#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here, we're setting the heat values derived given the equation:</p>
<p><mathjax>#q = mC_sDeltaT#</mathjax> </p>
<p>equal to each other, as such,</p>
<p><mathjax>#q_(water)+q(Cu) = 0#</mathjax><br/>
<mathjax>#therefore q_(water) = -q(Cu)#</mathjax></p>
<p>Before we start, I will assume the <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water is <mathjax>#"1.00 g/mL"#</mathjax>.</p>
<p><mathjax>#50.0 g*(4.184J)/(g*°C)*(T_f-25.0°C) = -[48.7g*(0.385J)/(g*°C)*(T_f-76.8°C)]#</mathjax></p>
<p><mathjax>#209.2T_f - 5230 °C = -18.75T_f + 1440 °C#</mathjax></p>
<p><mathjax>#227.95T_f = 6670 °C#</mathjax></p>
<p><mathjax>#therefore T_f = 29.3°C#</mathjax></p></div>
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</article> | What is the final temperature of the copper and water given that the specific heat of copper is 0.385 #J/(g * "^oC)#? |
A hot lump of 48.7 g of copper at an initial temperature of 76.8 C is placed in 50.0 mL of #H_2O# initially at 25.0°C and allowed to reach thermal equilibrium. Assume no heat is lost to surroundings.
|
2,665 | a8c44876-6ddd-11ea-937c-ccda262736ce | https://socratic.org/questions/586d4558b72cff60dc4b5b13 | 232 L | start physical_unit 1 1 volume l qc_end physical_unit 1 1 6 7 pressure qc_end physical_unit 1 1 17 18 pressure qc_end physical_unit 1 1 9 10 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] gas [IN] L"}] | [{"type":"physical unit","value":"232 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{203 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] gas [=] \\pu{35.0 kPa}"},{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{40.0 L}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">A gas at a pressure of #"203 kPa"# occupies #"40.0 L"#. If the pressure is decreased to #"35.0 kPa"#, what is the new volume? Assume the temperature is held constant.</h1> | null | 232 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa</p>
<p>The equation to use is <mathjax>#"P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P#</mathjax> represents pressure, and <mathjax>#V#</mathjax> represents volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="203 kPa"#</mathjax><br/>
<mathjax>#V_1="40.0 L"#</mathjax><br/>
<mathjax>#P_2="35.0 kPa"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"#</mathjax></p>
<p>The final volume of helium gas will be <mathjax>#"232 L"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The final volume of helium gas will be <mathjax>#"232 L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa</p>
<p>The equation to use is <mathjax>#"P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P#</mathjax> represents pressure, and <mathjax>#V#</mathjax> represents volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="203 kPa"#</mathjax><br/>
<mathjax>#V_1="40.0 L"#</mathjax><br/>
<mathjax>#P_2="35.0 kPa"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"#</mathjax></p>
<p>The final volume of helium gas will be <mathjax>#"232 L"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas at a pressure of #"203 kPa"# occupies #"40.0 L"#. If the pressure is decreased to #"35.0 kPa"#, what is the new volume? Assume the temperature is held constant.</h1>
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<div class="markdown"><p>The final volume of helium gas will be <mathjax>#"232 L"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa</p>
<p>The equation to use is <mathjax>#"P_1V_1=P_2V_2#</mathjax>, where <mathjax>#P#</mathjax> represents pressure, and <mathjax>#V#</mathjax> represents volume.</p>
<p><strong>Known</strong><br/>
<mathjax>#P_1="203 kPa"#</mathjax><br/>
<mathjax>#V_1="40.0 L"#</mathjax><br/>
<mathjax>#P_2="35.0 kPa"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#V_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V_2#</mathjax>. Substitute the known values into the equation and solve.</p>
<p><mathjax>#V_2=(P_1V_1)/(P_2)#</mathjax></p>
<p><mathjax>#V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"#</mathjax></p>
<p>The final volume of helium gas will be <mathjax>#"232 L"#</mathjax>.</p></div>
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</article> | A gas at a pressure of #"203 kPa"# occupies #"40.0 L"#. If the pressure is decreased to #"35.0 kPa"#, what is the new volume? Assume the temperature is held constant. | null |
2,666 | ac332698-6ddd-11ea-9a17-ccda262736ce | https://socratic.org/questions/if-the-k-a-of-a-monoprotic-weak-acid-is-3-4-x10-6-what-is-the-ph-of-a-0-14-m-sol | 3.15 | start physical_unit 20 20 ph none qc_end physical_unit 5 7 9 11 ka qc_end physical_unit 7 7 18 19 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] acid solution"}] | [{"type":"physical unit","value":"3.15"}] | [{"type":"physical unit","value":"Ka [OF] monoprotic weak acid [=] \\pu{3.4 × 10^(-6)}"},{"type":"physical unit","value":"Molarity [OF] acid solution [=] \\pu{0.14 M}"}] | <h1 class="questionTitle" itemprop="name">If the #K_a# of a monoprotic weak acid is #3.4 x10^-6#, what is the pH of a 0.14 M solution of this acid?</h1> | null | 3.15 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let <mathjax>#sf(HX)#</mathjax> be the weak acid:</p>
<p><mathjax>#sf(HXrightleftharpoonsH^++X^-)#</mathjax></p>
<p>For which:</p>
<p><mathjax>#sf(K_a=([H^+][X^-])/([HX]))#</mathjax></p>
<p>If <mathjax>#sf(K_a)#</mathjax> lies in the range <mathjax>#sf(10^(-4))#</mathjax> - <mathjax>#sf(10^(-10))#</mathjax> we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.</p>
<p>This applies here.</p>
<p>Rearranging and taking negative logs of both sides gives:</p>
<p><mathjax>#sf(pH=1/2[pK_a-loga])#</mathjax></p>
<p><mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#sf(pK_a=-logK_a=-log(3.4xx10^(-6))=5.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[5.46-(-0.854)]=3.15)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#sf(pH=3.15)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let <mathjax>#sf(HX)#</mathjax> be the weak acid:</p>
<p><mathjax>#sf(HXrightleftharpoonsH^++X^-)#</mathjax></p>
<p>For which:</p>
<p><mathjax>#sf(K_a=([H^+][X^-])/([HX]))#</mathjax></p>
<p>If <mathjax>#sf(K_a)#</mathjax> lies in the range <mathjax>#sf(10^(-4))#</mathjax> - <mathjax>#sf(10^(-10))#</mathjax> we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.</p>
<p>This applies here.</p>
<p>Rearranging and taking negative logs of both sides gives:</p>
<p><mathjax>#sf(pH=1/2[pK_a-loga])#</mathjax></p>
<p><mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#sf(pK_a=-logK_a=-log(3.4xx10^(-6))=5.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[5.46-(-0.854)]=3.15)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If the #K_a# of a monoprotic weak acid is #3.4 x10^-6#, what is the pH of a 0.14 M solution of this acid?</h1>
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<div class="markdown"><p><mathjax>#sf(pH=3.15)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let <mathjax>#sf(HX)#</mathjax> be the weak acid:</p>
<p><mathjax>#sf(HXrightleftharpoonsH^++X^-)#</mathjax></p>
<p>For which:</p>
<p><mathjax>#sf(K_a=([H^+][X^-])/([HX]))#</mathjax></p>
<p>If <mathjax>#sf(K_a)#</mathjax> lies in the range <mathjax>#sf(10^(-4))#</mathjax> - <mathjax>#sf(10^(-10))#</mathjax> we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.</p>
<p>This applies here.</p>
<p>Rearranging and taking negative logs of both sides gives:</p>
<p><mathjax>#sf(pH=1/2[pK_a-loga])#</mathjax></p>
<p><mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#sf(pK_a=-logK_a=-log(3.4xx10^(-6))=5.46)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[5.46-(-0.854)]=3.15)#</mathjax></p></div>
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</article> | If the #K_a# of a monoprotic weak acid is #3.4 x10^-6#, what is the pH of a 0.14 M solution of this acid? | null |
2,667 | a9fbe0b6-6ddd-11ea-bc05-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-6-2-10-3-m-solution-of-hf | 2.77 | start physical_unit 10 10 ph none qc_end physical_unit 12 12 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] HF solution"}] | [{"type":"physical unit","value":"2.77"}] | [{"type":"physical unit","value":"Molarity [OF] HF solution [=] \\pu{6.2 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a #6.2 * 10^-3# M solution of #HF#?</h1> | null | 2.77 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The acid dissociation reaction for <mathjax>#HF#</mathjax> is the following:</p>
<p><mathjax>#HF->H^(+)+F^-#</mathjax></p>
<p>The <mathjax>#K_a#</mathjax> expression is: <mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of this solution could be found by: <mathjax>#pH=-log[H^(+)]#</mathjax>.</p>
<p>We will need to find the <mathjax>#[H^(+)]#</mathjax> using <mathjax>#ICE#</mathjax> table:</p>
<p><mathjax>#" " " " " " " " "HF" " " " " " "->" " " " H^(+) +" " " "F^-#</mathjax><br/>
<mathjax>#"Initial" " "6.2xx10^(-3)M" " " " " " " "0M" " " " " "0M#</mathjax><br/>
<mathjax>#"Change" " "-xM" " " " " " " " " "+xM" " " " "+xM#</mathjax><br/>
<mathjax>#"Equilibrium" ""(6.2xx10^(-3)-x)M" " "xM" " " " " " "xM#</mathjax></p>
<p>now, we can replace the concentrations by their values in the expression of <mathjax>#K_a#</mathjax>:</p>
<p><mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)=(x xx x)/(6.2xx10^(-3)-x)#</mathjax></p>
<p>Solve for <mathjax>#x=1.7xx10^(-3)M#</mathjax></p>
<p><mathjax>#=>pH=-log[H^(+)]=-log(1.7xx10^(-3))=2.77#</mathjax></p>
<p><strong>Acids & Bases | pH of a Weak Acid.</strong><br/>
<iframe src="https://www.youtube.com/embed/H6hZPq15a-E?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p><mathjax>#pH=2.77#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The acid dissociation reaction for <mathjax>#HF#</mathjax> is the following:</p>
<p><mathjax>#HF->H^(+)+F^-#</mathjax></p>
<p>The <mathjax>#K_a#</mathjax> expression is: <mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of this solution could be found by: <mathjax>#pH=-log[H^(+)]#</mathjax>.</p>
<p>We will need to find the <mathjax>#[H^(+)]#</mathjax> using <mathjax>#ICE#</mathjax> table:</p>
<p><mathjax>#" " " " " " " " "HF" " " " " " "->" " " " H^(+) +" " " "F^-#</mathjax><br/>
<mathjax>#"Initial" " "6.2xx10^(-3)M" " " " " " " "0M" " " " " "0M#</mathjax><br/>
<mathjax>#"Change" " "-xM" " " " " " " " " "+xM" " " " "+xM#</mathjax><br/>
<mathjax>#"Equilibrium" ""(6.2xx10^(-3)-x)M" " "xM" " " " " " "xM#</mathjax></p>
<p>now, we can replace the concentrations by their values in the expression of <mathjax>#K_a#</mathjax>:</p>
<p><mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)=(x xx x)/(6.2xx10^(-3)-x)#</mathjax></p>
<p>Solve for <mathjax>#x=1.7xx10^(-3)M#</mathjax></p>
<p><mathjax>#=>pH=-log[H^(+)]=-log(1.7xx10^(-3))=2.77#</mathjax></p>
<p><strong>Acids & Bases | pH of a Weak Acid.</strong><br/>
<iframe src="https://www.youtube.com/embed/H6hZPq15a-E?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">What is the pH of a #6.2 * 10^-3# M solution of #HF#?</h1>
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<div class="markdown"><p><mathjax>#pH=2.77#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The acid dissociation reaction for <mathjax>#HF#</mathjax> is the following:</p>
<p><mathjax>#HF->H^(+)+F^-#</mathjax></p>
<p>The <mathjax>#K_a#</mathjax> expression is: <mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)#</mathjax></p>
<p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of this solution could be found by: <mathjax>#pH=-log[H^(+)]#</mathjax>.</p>
<p>We will need to find the <mathjax>#[H^(+)]#</mathjax> using <mathjax>#ICE#</mathjax> table:</p>
<p><mathjax>#" " " " " " " " "HF" " " " " " "->" " " " H^(+) +" " " "F^-#</mathjax><br/>
<mathjax>#"Initial" " "6.2xx10^(-3)M" " " " " " " "0M" " " " " "0M#</mathjax><br/>
<mathjax>#"Change" " "-xM" " " " " " " " " "+xM" " " " "+xM#</mathjax><br/>
<mathjax>#"Equilibrium" ""(6.2xx10^(-3)-x)M" " "xM" " " " " " "xM#</mathjax></p>
<p>now, we can replace the concentrations by their values in the expression of <mathjax>#K_a#</mathjax>:</p>
<p><mathjax>#K_a=([H^(+)][F^-])/([HF])=6.6xx10^(-4)=(x xx x)/(6.2xx10^(-3)-x)#</mathjax></p>
<p>Solve for <mathjax>#x=1.7xx10^(-3)M#</mathjax></p>
<p><mathjax>#=>pH=-log[H^(+)]=-log(1.7xx10^(-3))=2.77#</mathjax></p>
<p><strong>Acids & Bases | pH of a Weak Acid.</strong><br/>
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</article> | What is the pH of a #6.2 * 10^-3# M solution of #HF#? | null |
2,668 | aa3a56cc-6ddd-11ea-bede-ccda262736ce | https://socratic.org/questions/when-cu-oh-2-is-heated-copper-ii-oxide-and-water-are-formed-how-would-you-write- | Cu(OH)2(s) ->[\delta] CuO(s) + H2O(g) | start chemical_equation qc_end chemical_equation 1 1 qc_end substance 4 5 qc_end substance 7 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Cu(OH)2(s) ->[\\delta] CuO(s) + H2O(g)"}] | [{"type":"chemical equation","value":"Cu(OH)2"},{"type":"substance name","value":"Copper(II) oxide"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">When #Cu(OH)_2#is heated, copper(II) oxide and water are formed. How would you write a balanced equation for the reaction?</h1> | null | Cu(OH)2(s) ->[\delta] CuO(s) + H2O(g) | <div class="answerDescription">
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<div class="markdown"><p><mathjax>#Cu(OH)_2(s) + Delta rarr CuO(s) + H_2O(g)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Cu(OH)_2(s) + Delta rarr CuO(s) + H_2O(g)#</mathjax></p></div>
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</article> | When #Cu(OH)_2#is heated, copper(II) oxide and water are formed. How would you write a balanced equation for the reaction? | null |
2,669 | aac54536-6ddd-11ea-af09-ccda262736ce | https://socratic.org/questions/a-chloride-of-rhenium-contains-63-6-rhenium-what-is-the-name-of-this-compound | ReCl3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"ReCl3"}] | [{"type":"physical unit","value":"Percent [OF] crhenium in chloride of rhenium [=] \\pu{63.6%}"}] | <h1 class="questionTitle" itemprop="name">A chloride of rhenium contains 63.6% rhenium. What is the name of this compound?</h1> | null | ReCl3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> of the metal species, there are <mathjax>#(63.6*g)/(186.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Re#</mathjax>, and <mathjax>#(36.4*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.03#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Cl#</mathjax>.</p>
<p>If we divide through by the lowest molar quantity (<mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax>), we get an empirical formula of <mathjax>#ReCl_3#</mathjax>, rhenium trichloride or rhenium (III) chloride. Do I win £5-00?</p></div>
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<div class="markdown"><p>Empirical formula is <mathjax>#ReCl_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> of the metal species, there are <mathjax>#(63.6*g)/(186.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Re#</mathjax>, and <mathjax>#(36.4*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.03#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Cl#</mathjax>.</p>
<p>If we divide through by the lowest molar quantity (<mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax>), we get an empirical formula of <mathjax>#ReCl_3#</mathjax>, rhenium trichloride or rhenium (III) chloride. Do I win £5-00?</p></div>
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<h1 class="questionTitle" itemprop="name">A chloride of rhenium contains 63.6% rhenium. What is the name of this compound?</h1>
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<div class="markdown"><p>Empirical formula is <mathjax>#ReCl_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> of the metal species, there are <mathjax>#(63.6*g)/(186.21*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Re#</mathjax>, and <mathjax>#(36.4*g)/(35.45*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.03#</mathjax> <mathjax>#mol#</mathjax> <mathjax>#Cl#</mathjax>.</p>
<p>If we divide through by the lowest molar quantity (<mathjax>#0.342#</mathjax> <mathjax>#mol#</mathjax>), we get an empirical formula of <mathjax>#ReCl_3#</mathjax>, rhenium trichloride or rhenium (III) chloride. Do I win £5-00?</p></div>
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<div class="markdown"><p>This compound is rhenium(III) chloride.</p>
<p><a href="http://www.sigmaaldrich.com/catalog/product/aldrich/309184?lang=en&region=US" rel="nofollow" target="_blank">http://www.sigmaaldrich.com/catalog/product/aldrich/309184?lang=en&region=US</a></p></div>
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<div class="markdown"><p>We will be determining the empirical formula for this compound containing the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> rhenium (Re) and chlorine (Cl).</p>
<p>Since the only other element is chlorine, we can subtract the percentage of rhenium from 100% to calculate the percent of chlorine in the compound.</p>
<p><mathjax>#100%.0-63.6%=36.4% "Cl"#</mathjax></p>
<p>Since the percentages add to 100, we can assume a 100-g sample of the compound, and the percentages become mass in grams.</p>
<p><strong>Moles of Each Element</strong><br/>
Determine the moles of each element by dividing the mass by its molar mass (atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol).</p>
<p><mathjax>#63.6cancel"g Re"xx(1"mol Re")/(186.207cancel"g Re")="0.342 mol Re"#</mathjax></p>
<p><mathjax>#36.4cancel"g Cl"xx(1"mol Cl")/(35.453cancel"g Cl")="1.027 mol Cl"#</mathjax></p>
<p><strong><a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">Mole Ratios</a> and Empirical Formula</strong><br/>
Determine <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio of each element by dividing the moles of each element by the least number of moles.</p>
<p><mathjax>#"Re":#</mathjax><mathjax>#(0.342cancel"mol")/(0.342cancel"mol")="1.00"#</mathjax></p>
<p><mathjax>#"Cl":#</mathjax><mathjax>#(1.027cancel"mol")/(0.342cancel"mol")="3.00"#</mathjax></p>
<p>The empirical formula is <mathjax>#"ReCl"_3"#</mathjax>. </p>
<p>This compound has an empirical formula mass of 292.6 g/mol. According to Sigma-Aldrich, a chemical supplier, this compound is rhenium(III) chloride.</p>
<p><a href="http://www.sigmaaldrich.com/catalog/product/aldrich/309184?lang=en&region=US" rel="nofollow" target="_blank">http://www.sigmaaldrich.com/catalog/product/aldrich/309184?lang=en&region=US</a></p></div>
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</article> | A chloride of rhenium contains 63.6% rhenium. What is the name of this compound? | null |
2,670 | ac59824c-6ddd-11ea-bf54-ccda262736ce | https://socratic.org/questions/what-is-the-formula-of-tin-1-chromate | Sn(CrO4)2 | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] tin (1) chromate [IN] default"}] | [{"type":"chemical equation","value":"Sn(CrO4)2"}] | [{"type":"substance name","value":"Tin (1) chromate"}] | <h1 class="questionTitle" itemprop="name">What is the formula of tin (1) chromate?</h1> | null | Sn(CrO4)2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula of tin chromate is</p>
<p><mathjax>#Sn(CrO4)2#</mathjax> </p>
<p>or</p>
<p><mathjax>#SnCr2O8#</mathjax></p>
<p>Molecular weight is 351.7 g/mol</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Sn(CrO4)2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula of tin chromate is</p>
<p><mathjax>#Sn(CrO4)2#</mathjax> </p>
<p>or</p>
<p><mathjax>#SnCr2O8#</mathjax></p>
<p>Molecular weight is 351.7 g/mol</p></div>
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<div class="markdown"><p><mathjax>#Sn(CrO4)2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The formula of tin chromate is</p>
<p><mathjax>#Sn(CrO4)2#</mathjax> </p>
<p>or</p>
<p><mathjax>#SnCr2O8#</mathjax></p>
<p>Molecular weight is 351.7 g/mol</p></div>
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</article> | What is the formula of tin (1) chromate? | null |
2,671 | a8ee2c58-6ddd-11ea-8d75-ccda262736ce | https://socratic.org/questions/l-have-56-l-of-h-2-gas-at-stp-haw-many-grams-do-i-have-of-h-2-gas | 5.0 grams | start physical_unit 5 6 mass g qc_end physical_unit 5 6 2 3 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mass [OF] H2 gas [IN] grams"}] | [{"type":"physical unit","value":"5.0 grams"}] | [{"type":"physical unit","value":"Volume [OF] H2 gas [=] \\pu{56 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">L have 56 L of #H_2# gas at STP. Haw many grams do I have of #H_2# gas?</h1> | null | 5.0 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine moles of <mathjax>#"H"_2"#</mathjax>. Then multiply the moles times the molar mass. <mathjax>#"STP=273.15 K and 100 kPa"#</mathjax>.</p>
<p><strong>Part 1: Ideal Gas Law</strong><br/>
<mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> represents moles and <mathjax>#R#</mathjax> is the gas constant.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#V="56 L"#</mathjax><br/>
<mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100cancel"kPa"xx56cancel"L")/(8.3144598cancel"L" cancel"kPa" cancel"K"^(-1) "mol"^(-1)xx273.15cancel"K")="2.5 mol H"_2"#</mathjax> (rounded to two significant figures)</p>
<p><strong>Part 2: Mass of <mathjax>#"H"_2"#</mathjax></strong><br/>
Multiply the moles of hydrogen gas times its molar mass, <mathjax>#"2.01588 g/mol"#</mathjax>.</p>
<p><mathjax>#2.5cancel"mol H"_2xx(2.01588"g H"_2)/(1cancel"mol H"_2)="5.0 g H"_2"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The mass of <mathjax>#"56 L H"_2"#</mathjax> at <mathjax>#"STP"#</mathjax> is <mathjax>#"5.0 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine moles of <mathjax>#"H"_2"#</mathjax>. Then multiply the moles times the molar mass. <mathjax>#"STP=273.15 K and 100 kPa"#</mathjax>.</p>
<p><strong>Part 1: Ideal Gas Law</strong><br/>
<mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> represents moles and <mathjax>#R#</mathjax> is the gas constant.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#V="56 L"#</mathjax><br/>
<mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100cancel"kPa"xx56cancel"L")/(8.3144598cancel"L" cancel"kPa" cancel"K"^(-1) "mol"^(-1)xx273.15cancel"K")="2.5 mol H"_2"#</mathjax> (rounded to two significant figures)</p>
<p><strong>Part 2: Mass of <mathjax>#"H"_2"#</mathjax></strong><br/>
Multiply the moles of hydrogen gas times its molar mass, <mathjax>#"2.01588 g/mol"#</mathjax>.</p>
<p><mathjax>#2.5cancel"mol H"_2xx(2.01588"g H"_2)/(1cancel"mol H"_2)="5.0 g H"_2"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">L have 56 L of #H_2# gas at STP. Haw many grams do I have of #H_2# gas?</h1>
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<div class="markdown"><p>The mass of <mathjax>#"56 L H"_2"#</mathjax> at <mathjax>#"STP"#</mathjax> is <mathjax>#"5.0 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine moles of <mathjax>#"H"_2"#</mathjax>. Then multiply the moles times the molar mass. <mathjax>#"STP=273.15 K and 100 kPa"#</mathjax>.</p>
<p><strong>Part 1: Ideal Gas Law</strong><br/>
<mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> represents moles and <mathjax>#R#</mathjax> is the gas constant.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="100 kPa"#</mathjax><br/>
<mathjax>#V="56 L"#</mathjax><br/>
<mathjax>#R="8.3144598 L kPa K"^(-1) "mol"^(-1)"#</mathjax><br/>
<mathjax>#T="273.15 K"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#n#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#n#</mathjax> and solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=(100cancel"kPa"xx56cancel"L")/(8.3144598cancel"L" cancel"kPa" cancel"K"^(-1) "mol"^(-1)xx273.15cancel"K")="2.5 mol H"_2"#</mathjax> (rounded to two significant figures)</p>
<p><strong>Part 2: Mass of <mathjax>#"H"_2"#</mathjax></strong><br/>
Multiply the moles of hydrogen gas times its molar mass, <mathjax>#"2.01588 g/mol"#</mathjax>.</p>
<p><mathjax>#2.5cancel"mol H"_2xx(2.01588"g H"_2)/(1cancel"mol H"_2)="5.0 g H"_2"#</mathjax></p></div>
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</article> | L have 56 L of #H_2# gas at STP. Haw many grams do I have of #H_2# gas? | null |
2,672 | aa0e1692-6ddd-11ea-a6a6-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-copper-i-oxide | Cu2O | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] copper(I) oxide [IN] default"}] | [{"type":"chemical equation","value":"Cu2O"}] | [{"type":"substance name","value":"Copper(I) oxide"}] | <h1 class="questionTitle" itemprop="name">What is the formula for copper(I) oxide?</h1> | null | Cu2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Cuprous oxide is a red powder, as opposed to cupric oxide, <mathjax>#CuO#</mathjax>, a black powder. Do I win £5-00?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#Cu_2O#</mathjax>, or cuprous oxide.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Cuprous oxide is a red powder, as opposed to cupric oxide, <mathjax>#CuO#</mathjax>, a black powder. Do I win £5-00?</p></div>
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<div class="markdown"><p><mathjax>#Cu_2O#</mathjax>, or cuprous oxide.</p></div>
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<div class="markdown"><p>Cuprous oxide is a red powder, as opposed to cupric oxide, <mathjax>#CuO#</mathjax>, a black powder. Do I win £5-00?</p></div>
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</article> | What is the formula for copper(I) oxide? | null |
2,673 | aa206454-6ddd-11ea-8654-ccda262736ce | https://socratic.org/questions/563a8930581e2a05676df6a1 | +5 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] phosphorus"}] | [{"type":"physical unit","value":"+5"}] | [{"type":"chemical equation","value":"PCl5"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in the compound #"PCl"_5"#?</h1> | null | +5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of phosphorus in the compound #"PCl"_5"#?</h1>
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<div class="markdown"><p><mathjax>#+5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">Oxidation numbers</a> are all about <a href="http://socratic.org/chemistry/the-periodic-table/periodic-trends-in-electronegativity">electronegativity</a> difference between covalently-bonded atoms. </p>
<p>You can assign <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> to the atoms that are a part of a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/covalent-compounds">covalent compound</a> by assuming that the <strong>more electronegative</strong> atom will <strong>take</strong> both <a href="http://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons.</p>
<p>In the case of <em>phosphorus pentachloride</em>, <mathjax>#"PCl"_5#</mathjax>, you have <strong>one</strong> phosphorus atom that forms covalent bonds with <strong>five</strong> chlorine atoms. </p>
<p>Since chlorine is <em>more electronegative</em> than phosphorus, you can assign its oxidation number by assuming that it takes <strong>all the bonding electrons</strong> used in the molecule. </p>
<p><img alt="http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonding" src="https://useruploads.socratic.org/oGQblqeNTMCjwWCvEkEu_pcl5.GIF"/> </p>
<p>Since each single bond contains <mathjax>#2#</mathjax> bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, <strong>and</strong> the electron phosphorus contributed to the bond, shown in red.</p>
<p>So, for <strong>each</strong> bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state <strong>of each</strong> chlorine atom will be <mathjax>#color(blue)(-1)#</mathjax>.</p>
<p>The phosphorus atom loses a total of <mathjax>#5#</mathjax> electrons, one to each chlorine atom, so its oxidation state will be <mathjax>#color(blue)(+5)#</mathjax>. </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p>The oxidation number of <mathjax>#"P"#</mathjax> in the compound <mathjax>#"PCl"_5"#</mathjax> is <mathjax>#"+5"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sum of oxidation numbers in a compound must be zero. The oxidation number of chlorine in <mathjax>#"PCl"_5"#</mathjax> is <mathjax>#"-1"#</mathjax>, so there is an overall oxidation number of <mathjax>#"Cl"#</mathjax> is <mathjax>#"-5"#</mathjax>. In order for the sum of the oxidation numbers to equal zero, the oxidation number of <mathjax>#"P"#</mathjax> is <mathjax>#"+5"#</mathjax>.</p>
<p><mathjax>#(1xx+5)+(5xx-1)=0#</mathjax></p>
<blockquote>
<p><mathjax>#stackrel(+5)("P")stackrel(-1)("Cl"_5")#</mathjax></p>
</blockquote></div>
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</article> | What is the oxidation number of phosphorus in the compound #"PCl"_5"#? | null |
2,674 | ab0c1ed3-6ddd-11ea-8cb5-ccda262736ce | https://socratic.org/questions/how-many-grams-of-ticl-4-are-needed-to-produce-44-9-g-of-titanium | 177.92 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 12 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] TiCl4 [IN] grams"}] | [{"type":"physical unit","value":"177.92 grams"}] | [{"type":"physical unit","value":"Mass [OF] titanium [=] \\pu{44.9 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #TiCl_4# are needed to produce 44.9 g of titanium? </h1> | null | 177.92 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of titanium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(44.9*g)/(47.87*g*mol^-1)=0.938*mol.#</mathjax></p>
<p>And thus we need AT LEAST an equimolar quantity of <mathjax>#TiCl_4#</mathjax>.</p>
<p>And this represents of equivalent mass of:</p>
<p><mathjax>#0.938*molxx189.68*g*mol^-1=??*g.#</mathjax></p>
<p>I think the industrial process utilizes either alkali metals or alkaline earth metals to reduce the <mathjax>#TiCl_4#</mathjax>. </p></div>
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<div class="markdown"><p>At least <mathjax>#178*g#</mathjax> of <mathjax>#"titanic chloride"#</mathjax> are required...........</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of titanium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(44.9*g)/(47.87*g*mol^-1)=0.938*mol.#</mathjax></p>
<p>And thus we need AT LEAST an equimolar quantity of <mathjax>#TiCl_4#</mathjax>.</p>
<p>And this represents of equivalent mass of:</p>
<p><mathjax>#0.938*molxx189.68*g*mol^-1=??*g.#</mathjax></p>
<p>I think the industrial process utilizes either alkali metals or alkaline earth metals to reduce the <mathjax>#TiCl_4#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #TiCl_4# are needed to produce 44.9 g of titanium? </h1>
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<div class="markdown"><p>At least <mathjax>#178*g#</mathjax> of <mathjax>#"titanic chloride"#</mathjax> are required...........</p></div>
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<div class="markdown"><p><mathjax>#"Moles of titanium"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(44.9*g)/(47.87*g*mol^-1)=0.938*mol.#</mathjax></p>
<p>And thus we need AT LEAST an equimolar quantity of <mathjax>#TiCl_4#</mathjax>.</p>
<p>And this represents of equivalent mass of:</p>
<p><mathjax>#0.938*molxx189.68*g*mol^-1=??*g.#</mathjax></p>
<p>I think the industrial process utilizes either alkali metals or alkaline earth metals to reduce the <mathjax>#TiCl_4#</mathjax>. </p></div>
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</article> | How many grams of #TiCl_4# are needed to produce 44.9 g of titanium? | null |
2,675 | ac2f2e8b-6ddd-11ea-bd65-ccda262736ce | https://socratic.org/questions/how-do-you-balance-c-7h-14-o-2-co-2-h-2o | C7H14(l) + 21/2 O2(g) -> 7 CO2(g) + 7 H2O(l) | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C7H14(l) + 21/2 O2(g) -> 7 CO2(g) + 7 H2O(l)"}] | [{"type":"chemical equation","value":"C7H14 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #C_7H_14 + O_2 -> CO_2 + H_2O#?</h1> | null | C7H14(l) + 21/2 O2(g) -> 7 CO2(g) + 7 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The products of complete combustion of ANY hydrocarbon are carbon dioxide and water. Above I balanced the carbons, and then the hydrogens, and then the oxygens. Is it balanced?</p>
<p>Note that you use this same principle of balancing, of stroichiometry, whenever you make a cash or electronic transaction. Garbage in must equal garbage out. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#C_7H_14(l) + 21/2O_2(g) rarr 7CO_2(g) + 7H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The products of complete combustion of ANY hydrocarbon are carbon dioxide and water. Above I balanced the carbons, and then the hydrogens, and then the oxygens. Is it balanced?</p>
<p>Note that you use this same principle of balancing, of stroichiometry, whenever you make a cash or electronic transaction. Garbage in must equal garbage out. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #C_7H_14 + O_2 -> CO_2 + H_2O#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#C_7H_14(l) + 21/2O_2(g) rarr 7CO_2(g) + 7H_2O(l)#</mathjax></p></div>
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<div class="markdown"><p>The products of complete combustion of ANY hydrocarbon are carbon dioxide and water. Above I balanced the carbons, and then the hydrogens, and then the oxygens. Is it balanced?</p>
<p>Note that you use this same principle of balancing, of stroichiometry, whenever you make a cash or electronic transaction. Garbage in must equal garbage out. </p></div>
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</article> | How do you balance #C_7H_14 + O_2 -> CO_2 + H_2O#? | null |
2,676 | ac8f161c-6ddd-11ea-946d-ccda262736ce | https://socratic.org/questions/how-much-of-8-00-mol-l-sulphuric-acid-must-you-add-to-an-empty-750-ml-bottle-to- | 140.6 mL | start physical_unit 5 6 volume ml qc_end physical_unit 5 6 3 4 concentration qc_end physical_unit 5 6 13 14 volume qc_end physical_unit 5 6 21 22 concentration qc_end end | [{"type":"physical unit","value":"Volume1 [OF] sulphuric acid [IN] mL"}] | [{"type":"physical unit","value":"140.6 mL"}] | [{"type":"physical unit","value":"Concentration1 [OF] sulphuric acid [=] \\pu{8.00 mol/L}"},{"type":"physical unit","value":"Volume2 [OF] sulphuric acid [=] \\pu{750 mL}"},{"type":"physical unit","value":"Concentration2 [OF] sulphuric acid [=] \\pu{1.50 mol/L}"}] | <h1 class="questionTitle" itemprop="name">How much of 8.00 mol/L sulphuric acid must you add to an empty 750. mL bottle to obtain a concentration of 1.50 mol/L?</h1> | null | 140.6 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that the number of moles of <mathjax>#H_2SO_4#</mathjax> that you will have in the 750 mL solution will be equal to the number of moles you must remove from the 8.00 M solution.</p>
<p>Since we know the desired concentration and volume (I assume we are to make a full 750 mL of the 1.50 M solution), we can use this to find the number of moles</p>
<p><mathjax>#"moles" H_2SO_4 = 1.50 "mol"/L xx 0.750 L = 1.125#</mathjax> mol</p>
<p>To find the volume of the concentrated solution needed, we start with the definition of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><mathjax>#"molar concentration" = ("number of moles")/ ("volume in litres")#</mathjax></p>
<p>and rewrite it as </p>
<p><mathjax>#"volume in litres" = ("number of moles")/("molar concentration")#</mathjax></p>
<p><mathjax>#"volume" = 1.125/8.00 = 0.1406 L#</mathjax> or 140.6 mL</p>
<p>Place this in the bottle and top it up to 750 mL with water.</p>
<p>(Practical note: Put about 500 mL of water into the bottle first, then add the acid before topping it up. It is safer that way! )</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>If you wish to produce 750 mL of the 1.50 M solution, you will need 140.6 mL of the 8.00 M solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that the number of moles of <mathjax>#H_2SO_4#</mathjax> that you will have in the 750 mL solution will be equal to the number of moles you must remove from the 8.00 M solution.</p>
<p>Since we know the desired concentration and volume (I assume we are to make a full 750 mL of the 1.50 M solution), we can use this to find the number of moles</p>
<p><mathjax>#"moles" H_2SO_4 = 1.50 "mol"/L xx 0.750 L = 1.125#</mathjax> mol</p>
<p>To find the volume of the concentrated solution needed, we start with the definition of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><mathjax>#"molar concentration" = ("number of moles")/ ("volume in litres")#</mathjax></p>
<p>and rewrite it as </p>
<p><mathjax>#"volume in litres" = ("number of moles")/("molar concentration")#</mathjax></p>
<p><mathjax>#"volume" = 1.125/8.00 = 0.1406 L#</mathjax> or 140.6 mL</p>
<p>Place this in the bottle and top it up to 750 mL with water.</p>
<p>(Practical note: Put about 500 mL of water into the bottle first, then add the acid before topping it up. It is safer that way! )</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How much of 8.00 mol/L sulphuric acid must you add to an empty 750. mL bottle to obtain a concentration of 1.50 mol/L?</h1>
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<div class="markdown"><p>If you wish to produce 750 mL of the 1.50 M solution, you will need 140.6 mL of the 8.00 M solution.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The key to solving this problem is to realize that the number of moles of <mathjax>#H_2SO_4#</mathjax> that you will have in the 750 mL solution will be equal to the number of moles you must remove from the 8.00 M solution.</p>
<p>Since we know the desired concentration and volume (I assume we are to make a full 750 mL of the 1.50 M solution), we can use this to find the number of moles</p>
<p><mathjax>#"moles" H_2SO_4 = 1.50 "mol"/L xx 0.750 L = 1.125#</mathjax> mol</p>
<p>To find the volume of the concentrated solution needed, we start with the definition of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>:</p>
<p><mathjax>#"molar concentration" = ("number of moles")/ ("volume in litres")#</mathjax></p>
<p>and rewrite it as </p>
<p><mathjax>#"volume in litres" = ("number of moles")/("molar concentration")#</mathjax></p>
<p><mathjax>#"volume" = 1.125/8.00 = 0.1406 L#</mathjax> or 140.6 mL</p>
<p>Place this in the bottle and top it up to 750 mL with water.</p>
<p>(Practical note: Put about 500 mL of water into the bottle first, then add the acid before topping it up. It is safer that way! )</p></div>
</div>
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</article> | How much of 8.00 mol/L sulphuric acid must you add to an empty 750. mL bottle to obtain a concentration of 1.50 mol/L? | null |
2,677 | a8de9f9c-6ddd-11ea-a9f6-ccda262736ce | https://socratic.org/questions/ethyne-is-completely-combusted-as-per-the-following-equation-c2h4-g-3-o2-g-2-co2 | 0.160 moles | start physical_unit 21 22 mole mol qc_end chemical_equation 34 43 qc_end physical_unit 15 15 10 13 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon dioxide [IN] moles"}] | [{"type":"physical unit","value":"0.160 moles"}] | [{"type":"chemical equation","value":"C2H4(g) + 3 O2(g) -> 2 CO2(g) + 2 H2O(g)"},{"type":"physical unit","value":"Mole [OF] oxygen [=] \\pu{2.40 × 10^(−1) moles}"}] | <h1 class="questionTitle" itemprop="name">Ethyne is completely combusted as per the following equation. If #2.40 xx 10^(-1) # moles of oxygen react, how many moles of carbon dioxide form? (Answer in mol to 3 s.d. in proper scientific notation.)</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I was trying to figure out how to start and if i needed to convert anything but my teacher wasnt very helpful and im struggling to understand what to do.</p>
<p><mathjax>#"C"_2"H"_4(g) + 3"O"_2(g) -> 2"CO"_2(g) + 2"H"_2"O(g)"#</mathjax></p></div>
</h2>
</div>
</div> | 0.160 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (4(g)) + color(blue)(3)"O"_ (2(g)) -> color(purple)(2)"CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that when the reaction consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#color(purple)(2)#</mathjax> <strong>moles</strong> of carbon dioxide are being produced. </p>
<p>This tells you that <strong>regardless</strong> of how many moles of oxygen gas react, the reaction will always produce carbon dioxide in a <mathjax>#color(blue)(3):color(purple)(2)#</mathjax> <strong>mole ratio</strong>. </p>
<p>In your case, you know that the reaction consumes</p>
<blockquote>
<p><mathjax>#n_( "O"_ 2) = 2.40 * 10^(-1)#</mathjax> <mathjax>#"moles O"_2#</mathjax></p>
</blockquote>
<p>You can now use the aforementioned mole ratio to determine how many moles of carbon dioxide will be produced</p>
<blockquote>
<p><mathjax>#2.40 * 10^(-1) color(red)(cancel(color(black)("moles O"_2))) * (color(purple)(2)color(white)(.)"moles CO"_2)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(darkgeen)(ul(color(black)(1.60 * 10^(-1)color(white)(.)"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> and in proper <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p>
<p>So remember, the stoichiometric coefficients that each compound has in the <strong>balanced</strong> chemical equation give you the <strong>mole ratio</strong> that exists when said compound are consumed or produced by the reaction. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.60 * 10^(-1)"moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (4(g)) + color(blue)(3)"O"_ (2(g)) -> color(purple)(2)"CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that when the reaction consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#color(purple)(2)#</mathjax> <strong>moles</strong> of carbon dioxide are being produced. </p>
<p>This tells you that <strong>regardless</strong> of how many moles of oxygen gas react, the reaction will always produce carbon dioxide in a <mathjax>#color(blue)(3):color(purple)(2)#</mathjax> <strong>mole ratio</strong>. </p>
<p>In your case, you know that the reaction consumes</p>
<blockquote>
<p><mathjax>#n_( "O"_ 2) = 2.40 * 10^(-1)#</mathjax> <mathjax>#"moles O"_2#</mathjax></p>
</blockquote>
<p>You can now use the aforementioned mole ratio to determine how many moles of carbon dioxide will be produced</p>
<blockquote>
<p><mathjax>#2.40 * 10^(-1) color(red)(cancel(color(black)("moles O"_2))) * (color(purple)(2)color(white)(.)"moles CO"_2)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(darkgeen)(ul(color(black)(1.60 * 10^(-1)color(white)(.)"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> and in proper <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p>
<p>So remember, the stoichiometric coefficients that each compound has in the <strong>balanced</strong> chemical equation give you the <strong>mole ratio</strong> that exists when said compound are consumed or produced by the reaction. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Ethyne is completely combusted as per the following equation. If #2.40 xx 10^(-1) # moles of oxygen react, how many moles of carbon dioxide form? (Answer in mol to 3 s.d. in proper scientific notation.)</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>I was trying to figure out how to start and if i needed to convert anything but my teacher wasnt very helpful and im struggling to understand what to do.</p>
<p><mathjax>#"C"_2"H"_4(g) + 3"O"_2(g) -> 2"CO"_2(g) + 2"H"_2"O(g)"#</mathjax></p></div>
</h2>
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Stefan V.
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<div class="markdown"><p><mathjax>#1.60 * 10^(-1)"moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look at the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#"C"_ 2"H"_ (4(g)) + color(blue)(3)"O"_ (2(g)) -> color(purple)(2)"CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that when the reaction consumes <mathjax>#color(blue)(3)#</mathjax> <strong>moles</strong> of oxygen gas, <mathjax>#color(purple)(2)#</mathjax> <strong>moles</strong> of carbon dioxide are being produced. </p>
<p>This tells you that <strong>regardless</strong> of how many moles of oxygen gas react, the reaction will always produce carbon dioxide in a <mathjax>#color(blue)(3):color(purple)(2)#</mathjax> <strong>mole ratio</strong>. </p>
<p>In your case, you know that the reaction consumes</p>
<blockquote>
<p><mathjax>#n_( "O"_ 2) = 2.40 * 10^(-1)#</mathjax> <mathjax>#"moles O"_2#</mathjax></p>
</blockquote>
<p>You can now use the aforementioned mole ratio to determine how many moles of carbon dioxide will be produced</p>
<blockquote>
<p><mathjax>#2.40 * 10^(-1) color(red)(cancel(color(black)("moles O"_2))) * (color(purple)(2)color(white)(.)"moles CO"_2)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(darkgeen)(ul(color(black)(1.60 * 10^(-1)color(white)(.)"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> and in proper <a href="https://socratic.org/chemistry/measurement-in-chemistry/scientific-notation">scientific notation</a>. </p>
<p>So remember, the stoichiometric coefficients that each compound has in the <strong>balanced</strong> chemical equation give you the <strong>mole ratio</strong> that exists when said compound are consumed or produced by the reaction. </p></div>
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</article> | Ethyne is completely combusted as per the following equation. If #2.40 xx 10^(-1) # moles of oxygen react, how many moles of carbon dioxide form? (Answer in mol to 3 s.d. in proper scientific notation.) |
I was trying to figure out how to start and if i needed to convert anything but my teacher wasnt very helpful and im struggling to understand what to do.
#"C"_2"H"_4(g) + 3"O"_2(g) -> 2"CO"_2(g) + 2"H"_2"O(g)"#
|
2,678 | a8df62ba-6ddd-11ea-b4ac-ccda262736ce | https://socratic.org/questions/if-an-engineer-needs-to-design-a-helium-tank-with-a-capacity-of-725-moles-at-stp | 16240.00 dm^(-3) | start physical_unit 7 7 volume dm^(-3) qc_end c_other STP qc_end physical_unit 7 7 13 14 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] helium [IN] dm^(-3)"}] | [{"type":"physical unit","value":"16240.00 dm^(-3)"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Mole [OF] helium [=] \\pu{725 moles}"}] | <h1 class="questionTitle" itemprop="name">If an engineer needs to design a helium tank with a capacity of 725 moles at STP, how big (volume) should it be? </h1> | null | 16240.00 dm^(-3) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So simply mulitply the molar quantity by the molar volume:</p>
<p><mathjax>#22.4*dm^3*mol^-1xx725*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??m^3#</mathjax></p>
<p>Remember that <mathjax>#1L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1xx10^-1*m)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/10^3)m^3#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>At <mathjax>#STP#</mathjax> <mathjax>#1#</mathjax> mole of Ideal gas occupies a volume of <mathjax>#22.4*dm^-3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So simply mulitply the molar quantity by the molar volume:</p>
<p><mathjax>#22.4*dm^3*mol^-1xx725*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??m^3#</mathjax></p>
<p>Remember that <mathjax>#1L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1xx10^-1*m)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/10^3)m^3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If an engineer needs to design a helium tank with a capacity of 725 moles at STP, how big (volume) should it be? </h1>
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anor277
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<div class="markdown"><p>At <mathjax>#STP#</mathjax> <mathjax>#1#</mathjax> mole of Ideal gas occupies a volume of <mathjax>#22.4*dm^-3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So simply mulitply the molar quantity by the molar volume:</p>
<p><mathjax>#22.4*dm^3*mol^-1xx725*mol#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??m^3#</mathjax></p>
<p>Remember that <mathjax>#1L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax> <mathjax>#dm^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1xx10^-1*m)^3#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/10^3)m^3#</mathjax></p></div>
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</article> | If an engineer needs to design a helium tank with a capacity of 725 moles at STP, how big (volume) should it be? | null |
2,679 | aafd078a-6ddd-11ea-90ff-ccda262736ce | https://socratic.org/questions/how-many-moles-of-hno-3-are-needed-to-prepare-5-0-liters-of-a-2-0-m-solution-of- | 10.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 4 4 13 14 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] HNO3 [IN] moles"}] | [{"type":"physical unit","value":"10.00 moles"}] | [{"type":"physical unit","value":"Volume [OF] HNO3 solution [=] \\pu{5.0 liters}"},{"type":"physical unit","value":"Molarity [OF] HNO3 solution [=] \\pu{2.0 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #HNO_3# are needed to prepare 5.0 liters of a 2.0 M solution of #HNO_3#?</h1> | null | 10.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution as a <strong>conversion factor</strong> to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that must be dissolved in <mathjax>#"5.0 L"#</mathjax> of solution in order to have a concentration of <mathjax>#"2.0 M"#</mathjax>.</p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of solute present in <mathjax>#"1 L"#</mathjax> of solution. In your case, a <mathjax>#"2.0 M"#</mathjax> solution will contain <mathjax>#2.0#</mathjax> <strong>moles</strong> of solute for every <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>Now, because <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are <em>homogeneous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">mixtures</a></em>, i.e. they have the same composition throughout, you can use the molarity of the solution as a conversion factor to get the moles of solute needed for your solution</p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("L solution"))) * overbrace("2.0 moles HNO"_3/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 2.0 M")) = color(darkgreen)(ul(color(black)("10. moles HNO"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>This tells you that dissolving <mathjax>#10.#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"5.0 L"#</mathjax> will give you <strong>the same molarity</strong> as dissolving <mathjax>#2.0#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"1.0 L"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"10. moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution as a <strong>conversion factor</strong> to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that must be dissolved in <mathjax>#"5.0 L"#</mathjax> of solution in order to have a concentration of <mathjax>#"2.0 M"#</mathjax>.</p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of solute present in <mathjax>#"1 L"#</mathjax> of solution. In your case, a <mathjax>#"2.0 M"#</mathjax> solution will contain <mathjax>#2.0#</mathjax> <strong>moles</strong> of solute for every <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>Now, because <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are <em>homogeneous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">mixtures</a></em>, i.e. they have the same composition throughout, you can use the molarity of the solution as a conversion factor to get the moles of solute needed for your solution</p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("L solution"))) * overbrace("2.0 moles HNO"_3/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 2.0 M")) = color(darkgreen)(ul(color(black)("10. moles HNO"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>This tells you that dissolving <mathjax>#10.#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"5.0 L"#</mathjax> will give you <strong>the same molarity</strong> as dissolving <mathjax>#2.0#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"1.0 L"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #HNO_3# are needed to prepare 5.0 liters of a 2.0 M solution of #HNO_3#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"10. moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is to use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution as a <strong>conversion factor</strong> to determine the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> that must be dissolved in <mathjax>#"5.0 L"#</mathjax> of solution in order to have a concentration of <mathjax>#"2.0 M"#</mathjax>.</p>
<p>As you know, <strong>molarity</strong> tells you the number of moles of solute present in <mathjax>#"1 L"#</mathjax> of solution. In your case, a <mathjax>#"2.0 M"#</mathjax> solution will contain <mathjax>#2.0#</mathjax> <strong>moles</strong> of solute for every <mathjax>#"1 L"#</mathjax> of solution. </p>
<p>Now, because <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are <em>homogeneous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mixtures">mixtures</a></em>, i.e. they have the same composition throughout, you can use the molarity of the solution as a conversion factor to get the moles of solute needed for your solution</p>
<blockquote>
<p><mathjax>#5.0 color(red)(cancel(color(black)("L solution"))) * overbrace("2.0 moles HNO"_3/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 2.0 M")) = color(darkgreen)(ul(color(black)("10. moles HNO"_3)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>This tells you that dissolving <mathjax>#10.#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"5.0 L"#</mathjax> will give you <strong>the same molarity</strong> as dissolving <mathjax>#2.0#</mathjax> <strong>moles</strong> of nitric acid in enough water to get the total volume of the solution to <mathjax>#"1.0 L"#</mathjax>.</p></div>
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</article> | How many moles of #HNO_3# are needed to prepare 5.0 liters of a 2.0 M solution of #HNO_3#? | null |
2,680 | ab234076-6ddd-11ea-b72c-ccda262736ce | https://socratic.org/questions/how-many-liters-of-oxygen-are-required-to-react-completely-with-2-4-liters-of-hy-1 | 1.20 liters | start physical_unit 4 4 volume l qc_end physical_unit 14 14 11 12 volume qc_end chemical_equation 18 24 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] liters"}] | [{"type":"physical unit","value":"1.20 liters"}] | [{"type":"physical unit","value":"Volume [OF] hydrogen [=] \\pu{2.4 liters}"},{"type":"chemical equation","value":"2 H2(g) + O2(g) -> 2 H2O(g)"},{"type":"other","value":"React completely."}] | <h1 class="questionTitle" itemprop="name">How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#?</h1> | null | 1.20 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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<div class="markdown"><p>Why <mathjax>#1.2*L#</mathjax> precisely..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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<h1 class="questionTitle" itemprop="name">How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#?</h1>
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anor277
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<div class="markdown"><p>Why <mathjax>#1.2*L#</mathjax> precisely..........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Recall old Avogadro's hypothesis...........</p>
<p><mathjax>#"Equal volumes of gases contain AN EQUAL NUMBER of PARTICLES"#</mathjax></p>
<p>And thus if there are <mathjax>#2.4*L#</mathjax> of dihydrogen, stoichiometric equivalence requires <mathjax>#1.2*L#</mathjax> of dioxygen. How many litres of water gas result?</p></div>
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Tartar C.
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Apr 30, 2017
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<div class="markdown"><p><mathjax>#1.2 L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Know that one mole of any gas at STP takes up 22.4 liters. So 2.4 L of hydrogen is <mathjax>#0.107142857 mols H_2#</mathjax>.</p>
<p>Since <mathjax>#H_2#</mathjax> reacts with <mathjax>#O_2#</mathjax> in a 2:1 ratio, to completely react the two you will need half as much oxygen as hydrogen.</p>
<p><mathjax>#"mols " O_2= (0.107142857 mols H_2)/2#</mathjax></p>
<p><mathjax>#"mols " O_2 = 0.053571429 mols#</mathjax></p>
<p>To find liters of oxygen, just multiply the moles above by 22.4 liters.</p>
<p><mathjax>#"liters "O_2 = 0.053571429 mols * 22.4 L/(mols)#</mathjax></p>
<p><mathjax>#"liters " O_2 = 1.2L#</mathjax></p></div>
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</article> | How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? #2H_2(g) + O_2(g) -> 2H_2O(g)#? | null |
2,681 | ac34fafa-6ddd-11ea-a8ed-ccda262736ce | https://socratic.org/questions/sylvanite-is-a-mineral-that-contains-28-0-gold-by-mass-how-much-sylvanite-would- | 236 g | start physical_unit 0 0 mass g qc_end physical_unit 7 7 21 22 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] sylvanite [IN] g"}] | [{"type":"physical unit","value":"236 g"}] | [{"type":"physical unit","value":"Percent by mass [OF] gold in sylvanite [=] \\pu{28.0%}"},{"type":"physical unit","value":"Mass [OF] gold [=] \\pu{66.0 g}"}] | <h1 class="questionTitle" itemprop="name">Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 66.0 g of gold?</h1> | null | 236 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And of course, <mathjax>#236*gxx28.0%=66*g#</mathjax>...</p>
<p>Of course, if you got such a rich gold ore....I would not stop at digging up just <mathjax>#236*g#</mathjax>. I think I could move <mathjax>#10*m^3#</mathjax> a day (well I bet I could when I was younger!)...and I would keep working....</p></div>
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<div class="markdown"><p>You need to dig up <mathjax>#236*g#</mathjax> of mineral...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And of course, <mathjax>#236*gxx28.0%=66*g#</mathjax>...</p>
<p>Of course, if you got such a rich gold ore....I would not stop at digging up just <mathjax>#236*g#</mathjax>. I think I could move <mathjax>#10*m^3#</mathjax> a day (well I bet I could when I was younger!)...and I would keep working....</p></div>
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<h1 class="questionTitle" itemprop="name">Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 66.0 g of gold?</h1>
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<div class="markdown"><p>You need to dig up <mathjax>#236*g#</mathjax> of mineral...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And of course, <mathjax>#236*gxx28.0%=66*g#</mathjax>...</p>
<p>Of course, if you got such a rich gold ore....I would not stop at digging up just <mathjax>#236*g#</mathjax>. I think I could move <mathjax>#10*m^3#</mathjax> a day (well I bet I could when I was younger!)...and I would keep working....</p></div>
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</article> | Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 66.0 g of gold? | null |
2,682 | a8dfbc38-6ddd-11ea-a23b-ccda262736ce | https://socratic.org/questions/59db9fe211ef6b5556960df3 | 98 μL | start physical_unit 17 17 volume μl qc_end physical_unit 5 5 1 2 volume qc_end physical_unit 5 5 13 13 dilution_factor qc_end end | [{"type":"physical unit","value":"Volume [OF] solvent [IN] μL"}] | [{"type":"physical unit","value":"98 μL"}] | [{"type":"physical unit","value":"Volume [OF] sample [=] \\pu{2.0 μL}"},{"type":"physical unit","value":"Dilute factor [OF] sample [=] \\pu{50}"}] | <h1 class="questionTitle" itemprop="name">If #2.0# #mu"L"# of a sample must be diluted by a factor of #50#, what volume of solvent should be added to accomplish this?</h1> | null | 98 μL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You can use the dilution formula:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)V_1c_1 = V_2c_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<p><mathjax>#V_1 = V_2 × c_2/c_1#</mathjax></p>
<blockquote>
<p><mathjax>#V_1 = ?; color(white)(mmll)" "" "c_1 = c_1#</mathjax></p>
<p><mathjax>#V_2 = "100 µL"; " "" "c_2 = c_1/50#</mathjax></p>
</blockquote>
<p><mathjax>#V_1 = "100 µL" × (stackrelcolor(blue)(1)color(red)(cancel(color(black)(c_1)))//50)/color(red)(cancel(color(black)(c_1))) = "100 µL" × 1/50 = "2 µL"#</mathjax></p>
</blockquote>
<p>∴ Add a bit less than <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> of solvent to <mathjax>#2#</mathjax> <mathjax>#µ"L"#</mathjax> of the sample, then fill up to the mark. </p>
<p>We don't quite add <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> right away because we want to account for any solution volume changes (up/down) due to mixing.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Mix <mathjax>#2.0#</mathjax> <mathjax>#µ"L"#</mathjax> of sample with a bit less than <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, then add solvent until you reach the <mathjax>#100#</mathjax> <mathjax>#µ"L"#</mathjax> mark.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You can use the dilution formula:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)V_1c_1 = V_2c_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<p><mathjax>#V_1 = V_2 × c_2/c_1#</mathjax></p>
<blockquote>
<p><mathjax>#V_1 = ?; color(white)(mmll)" "" "c_1 = c_1#</mathjax></p>
<p><mathjax>#V_2 = "100 µL"; " "" "c_2 = c_1/50#</mathjax></p>
</blockquote>
<p><mathjax>#V_1 = "100 µL" × (stackrelcolor(blue)(1)color(red)(cancel(color(black)(c_1)))//50)/color(red)(cancel(color(black)(c_1))) = "100 µL" × 1/50 = "2 µL"#</mathjax></p>
</blockquote>
<p>∴ Add a bit less than <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> of solvent to <mathjax>#2#</mathjax> <mathjax>#µ"L"#</mathjax> of the sample, then fill up to the mark. </p>
<p>We don't quite add <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> right away because we want to account for any solution volume changes (up/down) due to mixing.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">If #2.0# #mu"L"# of a sample must be diluted by a factor of #50#, what volume of solvent should be added to accomplish this?</h1>
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Ernest Z.
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Truong-Son N.
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Dec 29, 2017
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<div class="markdown"><p>Mix <mathjax>#2.0#</mathjax> <mathjax>#µ"L"#</mathjax> of sample with a bit less than <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, then add solvent until you reach the <mathjax>#100#</mathjax> <mathjax>#µ"L"#</mathjax> mark.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>You can use the dilution formula:</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)V_1c_1 = V_2c_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<p><mathjax>#V_1 = V_2 × c_2/c_1#</mathjax></p>
<blockquote>
<p><mathjax>#V_1 = ?; color(white)(mmll)" "" "c_1 = c_1#</mathjax></p>
<p><mathjax>#V_2 = "100 µL"; " "" "c_2 = c_1/50#</mathjax></p>
</blockquote>
<p><mathjax>#V_1 = "100 µL" × (stackrelcolor(blue)(1)color(red)(cancel(color(black)(c_1)))//50)/color(red)(cancel(color(black)(c_1))) = "100 µL" × 1/50 = "2 µL"#</mathjax></p>
</blockquote>
<p>∴ Add a bit less than <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> of solvent to <mathjax>#2#</mathjax> <mathjax>#µ"L"#</mathjax> of the sample, then fill up to the mark. </p>
<p>We don't quite add <mathjax>#98#</mathjax> <mathjax>#µ"L"#</mathjax> right away because we want to account for any solution volume changes (up/down) due to mixing.</p></div>
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</article> | If #2.0# #mu"L"# of a sample must be diluted by a factor of #50#, what volume of solvent should be added to accomplish this? | null |
2,683 | ad294067-6ddd-11ea-89fe-ccda262736ce | https://socratic.org/questions/what-is-the-standard-enthalpy-of-formation-of-methane-given-that-the-average-c-h | -67.2 kJ/mol | start physical_unit 8 8 standard_enthalpy_formation kj/mol qc_end physical_unit 13 14 17 18 enthalpy qc_end chemical_equation 22 28 qc_end chemical_equation 29 37 qc_end end | [{"type":"physical unit","value":"Standard enthalpy of formation [OF] methane [IN] kJ/mol"}] | [{"type":"physical unit","value":"-67.2 kJ/mol"}] | [{"type":"physical unit","value":"Average enthalpy [OF] C-H bond [=] \\pu{414 kJ/mol}"},{"type":"chemical equation","value":"C(s) -> C(g) H = 716 kJ/mol"},{"type":"chemical equation","value":"2 H2(g) -> 4 H(g) H = 872.8 kJ/mol"}] | <h1 class="questionTitle" itemprop="name">What is the standard enthalpy of formation of methane, given that the average C-H bond enthalpy is 414 kj/mol and the reactions:
C(s) -->C(g) H=716 kj/mol
2H2(g)--> 4H(g) H= 872.8 kj/mol?</h1> | null | -67.2 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of formation</strong>, <mathjax>#DeltaH_"f"^@#</mathjax>, for a given compound is defined as the enthalpy change of reaction when <strong>one mole</strong> of said compound is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable form. </p>
<p>In your case, the standard enthalpy of formation for methane, <mathjax>#"CH"_4#</mathjax>, is calculated for the reaction </p>
<blockquote>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])#</mathjax></p>
</blockquote>
<p>Your goal here will be to find a way to get the enthalpy change of reaction for the above reaction by using <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a>. </p>
<p>So, you know that the average <mathjax>#"C"-"H"#</mathjax> <strong>bond enthalpy</strong> is equal to <mathjax>#"414 kJ/mol"#</mathjax>. Notice that this value carries a <em>positive sign</em>. </p>
<p>As you know, <strong>bond breaking</strong> is an <a href="http://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic process</a> because it <strong>requires</strong> energy. This means that the value given to you corresponds to how much energy is <strong>needed</strong>, on average, in order to break a <mathjax>#"C"-"H"#</mathjax> bond. </p>
<p>However, you're interested <strong>forming</strong> methane, so you can say that the enthalpy change of reaction for </p>
<blockquote>
<p><mathjax>#"C"_text((g]) + 4"H"_text((g]) -> "CH"_text(4(g])" " " "color(purple)((1))#</mathjax></p>
</blockquote>
<p>will carry a <strong>negative sign</strong>, since this time you're <em>making bonds</em>, which is an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic process</a>.</p>
<p>Since you're making <strong>four</strong> <mathjax>#"C" - "H"#</mathjax> bonds, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn 1"^@ = 4 xx (-"414 kJ/mol") = -"1656 kJ/mol"#</mathjax></p>
</blockquote>
<p>The other two reactions given to you are</p>
<blockquote>
<p><mathjax>#"C"_text((s]) -> "C"_text((g]), " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"" " " "color(purple)((2))#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#2"H"_text(2(g]) -> 4"H"_text((g])," " DeltaH_text(rxn 3)^@ = "872.8 kJ/mol" " " " "color(purple)((3))#</mathjax></p>
</blockquote>
<p>Notice that equation <mathjax>#color(purple)((3))#</mathjax> represents the <strong>breaking</strong> of two <mathjax>#"H"-"H"#</mathjax> bonds, which is why it carries a positive sign.</p>
<p>Since <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> tells you that the overall enthalpy change for a reaction is <strong>independent</strong> of the pathway or the number of steps taken, you can add equations <mathjax>#color(purple)((1))#</mathjax>, <mathjax>#color(purple)((2))#</mathjax>, and <mathjax>#color(purple)((3))#</mathjax> to get </p>
<blockquote>
<p><mathjax>#color(white)(xxxxxxx)"C"_text((s]) -> color(red)(cancel(color(black)("C"_text((g])))), " " " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"#</mathjax><br/>
<mathjax>#color(red)(cancel(color(black)("C"_text((g])))) + color(blue)(cancel(color(black)(4"H"_text((g])))) -> "CH"_text(4(g]), " "color(white)(x)DeltaH_text(rxn 1)^@ = -"1656 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxx)2"H"_text(2(g]) -> color(blue)(cancel(color(black)(4"H"_text((g]))))," " " "DeltaH_text(rxn 3)^@ = "872.8 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx)/color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx)#</mathjax></p>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])," " " "DeltaH_text(rxn)^@ = DeltaH_text(f)^@#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#DeltaH_"f"^@ = DeltaH_"rxn 1"^@ + DeltaH_"rxn 2"^@ + DeltaH_"rxn 3"^@#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = "716 kJ/mol" + (-"1656 kJ/mol") + "872.8 kJ/mol"#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = color(green)(-"67.2 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#DeltaH_"f"^@ = -"67.2 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of formation</strong>, <mathjax>#DeltaH_"f"^@#</mathjax>, for a given compound is defined as the enthalpy change of reaction when <strong>one mole</strong> of said compound is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable form. </p>
<p>In your case, the standard enthalpy of formation for methane, <mathjax>#"CH"_4#</mathjax>, is calculated for the reaction </p>
<blockquote>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])#</mathjax></p>
</blockquote>
<p>Your goal here will be to find a way to get the enthalpy change of reaction for the above reaction by using <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a>. </p>
<p>So, you know that the average <mathjax>#"C"-"H"#</mathjax> <strong>bond enthalpy</strong> is equal to <mathjax>#"414 kJ/mol"#</mathjax>. Notice that this value carries a <em>positive sign</em>. </p>
<p>As you know, <strong>bond breaking</strong> is an <a href="http://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic process</a> because it <strong>requires</strong> energy. This means that the value given to you corresponds to how much energy is <strong>needed</strong>, on average, in order to break a <mathjax>#"C"-"H"#</mathjax> bond. </p>
<p>However, you're interested <strong>forming</strong> methane, so you can say that the enthalpy change of reaction for </p>
<blockquote>
<p><mathjax>#"C"_text((g]) + 4"H"_text((g]) -> "CH"_text(4(g])" " " "color(purple)((1))#</mathjax></p>
</blockquote>
<p>will carry a <strong>negative sign</strong>, since this time you're <em>making bonds</em>, which is an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic process</a>.</p>
<p>Since you're making <strong>four</strong> <mathjax>#"C" - "H"#</mathjax> bonds, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn 1"^@ = 4 xx (-"414 kJ/mol") = -"1656 kJ/mol"#</mathjax></p>
</blockquote>
<p>The other two reactions given to you are</p>
<blockquote>
<p><mathjax>#"C"_text((s]) -> "C"_text((g]), " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"" " " "color(purple)((2))#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#2"H"_text(2(g]) -> 4"H"_text((g])," " DeltaH_text(rxn 3)^@ = "872.8 kJ/mol" " " " "color(purple)((3))#</mathjax></p>
</blockquote>
<p>Notice that equation <mathjax>#color(purple)((3))#</mathjax> represents the <strong>breaking</strong> of two <mathjax>#"H"-"H"#</mathjax> bonds, which is why it carries a positive sign.</p>
<p>Since <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> tells you that the overall enthalpy change for a reaction is <strong>independent</strong> of the pathway or the number of steps taken, you can add equations <mathjax>#color(purple)((1))#</mathjax>, <mathjax>#color(purple)((2))#</mathjax>, and <mathjax>#color(purple)((3))#</mathjax> to get </p>
<blockquote>
<p><mathjax>#color(white)(xxxxxxx)"C"_text((s]) -> color(red)(cancel(color(black)("C"_text((g])))), " " " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"#</mathjax><br/>
<mathjax>#color(red)(cancel(color(black)("C"_text((g])))) + color(blue)(cancel(color(black)(4"H"_text((g])))) -> "CH"_text(4(g]), " "color(white)(x)DeltaH_text(rxn 1)^@ = -"1656 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxx)2"H"_text(2(g]) -> color(blue)(cancel(color(black)(4"H"_text((g]))))," " " "DeltaH_text(rxn 3)^@ = "872.8 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx)/color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx)#</mathjax></p>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])," " " "DeltaH_text(rxn)^@ = DeltaH_text(f)^@#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#DeltaH_"f"^@ = DeltaH_"rxn 1"^@ + DeltaH_"rxn 2"^@ + DeltaH_"rxn 3"^@#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = "716 kJ/mol" + (-"1656 kJ/mol") + "872.8 kJ/mol"#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = color(green)(-"67.2 kJ/mol")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the standard enthalpy of formation of methane, given that the average C-H bond enthalpy is 414 kj/mol and the reactions:
C(s) -->C(g) H=716 kj/mol
2H2(g)--> 4H(g) H= 872.8 kj/mol?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-01-07T01:35:11" itemprop="dateCreated">
Jan 7, 2016
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<div class="markdown"><p><mathjax>#DeltaH_"f"^@ = -"67.2 kJ/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>standard <a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of formation</strong>, <mathjax>#DeltaH_"f"^@#</mathjax>, for a given compound is defined as the enthalpy change of reaction when <strong>one mole</strong> of said compound is formed from its constituent <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in their most stable form. </p>
<p>In your case, the standard enthalpy of formation for methane, <mathjax>#"CH"_4#</mathjax>, is calculated for the reaction </p>
<blockquote>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])#</mathjax></p>
</blockquote>
<p>Your goal here will be to find a way to get the enthalpy change of reaction for the above reaction by using <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a>. </p>
<p>So, you know that the average <mathjax>#"C"-"H"#</mathjax> <strong>bond enthalpy</strong> is equal to <mathjax>#"414 kJ/mol"#</mathjax>. Notice that this value carries a <em>positive sign</em>. </p>
<p>As you know, <strong>bond breaking</strong> is an <a href="http://socratic.org/chemistry/thermochemistry/endothermic-processes">endothermic process</a> because it <strong>requires</strong> energy. This means that the value given to you corresponds to how much energy is <strong>needed</strong>, on average, in order to break a <mathjax>#"C"-"H"#</mathjax> bond. </p>
<p>However, you're interested <strong>forming</strong> methane, so you can say that the enthalpy change of reaction for </p>
<blockquote>
<p><mathjax>#"C"_text((g]) + 4"H"_text((g]) -> "CH"_text(4(g])" " " "color(purple)((1))#</mathjax></p>
</blockquote>
<p>will carry a <strong>negative sign</strong>, since this time you're <em>making bonds</em>, which is an <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic process</a>.</p>
<p>Since you're making <strong>four</strong> <mathjax>#"C" - "H"#</mathjax> bonds, you will have</p>
<blockquote>
<p><mathjax>#DeltaH_"rxn 1"^@ = 4 xx (-"414 kJ/mol") = -"1656 kJ/mol"#</mathjax></p>
</blockquote>
<p>The other two reactions given to you are</p>
<blockquote>
<p><mathjax>#"C"_text((s]) -> "C"_text((g]), " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"" " " "color(purple)((2))#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#2"H"_text(2(g]) -> 4"H"_text((g])," " DeltaH_text(rxn 3)^@ = "872.8 kJ/mol" " " " "color(purple)((3))#</mathjax></p>
</blockquote>
<p>Notice that equation <mathjax>#color(purple)((3))#</mathjax> represents the <strong>breaking</strong> of two <mathjax>#"H"-"H"#</mathjax> bonds, which is why it carries a positive sign.</p>
<p>Since <a href="http://socratic.org/chemistry/thermochemistry/hess-law">Hess' Law</a> tells you that the overall enthalpy change for a reaction is <strong>independent</strong> of the pathway or the number of steps taken, you can add equations <mathjax>#color(purple)((1))#</mathjax>, <mathjax>#color(purple)((2))#</mathjax>, and <mathjax>#color(purple)((3))#</mathjax> to get </p>
<blockquote>
<p><mathjax>#color(white)(xxxxxxx)"C"_text((s]) -> color(red)(cancel(color(black)("C"_text((g])))), " " " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"#</mathjax><br/>
<mathjax>#color(red)(cancel(color(black)("C"_text((g])))) + color(blue)(cancel(color(black)(4"H"_text((g])))) -> "CH"_text(4(g]), " "color(white)(x)DeltaH_text(rxn 1)^@ = -"1656 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxx)2"H"_text(2(g]) -> color(blue)(cancel(color(black)(4"H"_text((g]))))," " " "DeltaH_text(rxn 3)^@ = "872.8 kJ/mol"#</mathjax><br/>
<mathjax>#color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx)/color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxx)#</mathjax></p>
<p><mathjax>#"C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])," " " "DeltaH_text(rxn)^@ = DeltaH_text(f)^@#</mathjax></p>
</blockquote>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#DeltaH_"f"^@ = DeltaH_"rxn 1"^@ + DeltaH_"rxn 2"^@ + DeltaH_"rxn 3"^@#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = "716 kJ/mol" + (-"1656 kJ/mol") + "872.8 kJ/mol"#</mathjax></p>
<p><mathjax>#DeltaH_"f"^@ = color(green)(-"67.2 kJ/mol")#</mathjax></p>
</blockquote></div>
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</article> | What is the standard enthalpy of formation of methane, given that the average C-H bond enthalpy is 414 kj/mol and the reactions:
C(s) -->C(g) H=716 kj/mol
2H2(g)--> 4H(g) H= 872.8 kj/mol? | null |
2,684 | ab20c591-6ddd-11ea-80bc-ccda262736ce | https://socratic.org/questions/hydrogen-peroxide-h-2o-2-decomposes-to-form-water-and-oxygen-how-do-you-write-th | 2 H2O2(l) <=> 2 H2O(l) + O2(g) | start chemical_equation qc_end chemical_equation 2 2 qc_end substance 6 6 qc_end substance 8 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"2 H2O2(l) <=> 2 H2O(l) + O2(g)"}] | [{"type":"chemical equation","value":"H2O2"},{"type":"substance name","value":"Water"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">Hydrogen peroxide (#H_2O_2#) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?</h1> | null | 2 H2O2(l) <=> 2 H2O(l) + O2(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is in fact a disproportionation reaction in that oxygen in peroxide (<mathjax>#-I#</mathjax> oxidation state) has given water (<mathjax>#-II#</mathjax> oxidation state for oxygen) and zerovalent oxygen gas (<mathjax>#0#</mathjax> oxidation state for oxygen).</p>
<p>Often a <mathjax>#Mn^(2+)#</mathjax> salt is added to catalyze this reaction. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2H_2O_2(l) rightleftharpoons 2H_2O(l) + O_2(g)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is in fact a disproportionation reaction in that oxygen in peroxide (<mathjax>#-I#</mathjax> oxidation state) has given water (<mathjax>#-II#</mathjax> oxidation state for oxygen) and zerovalent oxygen gas (<mathjax>#0#</mathjax> oxidation state for oxygen).</p>
<p>Often a <mathjax>#Mn^(2+)#</mathjax> salt is added to catalyze this reaction. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Hydrogen peroxide (#H_2O_2#) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?</h1>
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anor277
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<div class="markdown"><p><mathjax>#2H_2O_2(l) rightleftharpoons 2H_2O(l) + O_2(g)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is in fact a disproportionation reaction in that oxygen in peroxide (<mathjax>#-I#</mathjax> oxidation state) has given water (<mathjax>#-II#</mathjax> oxidation state for oxygen) and zerovalent oxygen gas (<mathjax>#0#</mathjax> oxidation state for oxygen).</p>
<p>Often a <mathjax>#Mn^(2+)#</mathjax> salt is added to catalyze this reaction. </p></div>
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</article> | Hydrogen peroxide (#H_2O_2#) decomposes to form water and oxygen. How do you write the balanced equation for this reaction? | null |
2,685 | a8efdd63-6ddd-11ea-b446-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-the-covalent-compound-carbon-tetrachloride | CCl4 | start chemical_formula qc_end substance 8 9 qc_end end | [{"type":"other","value":"Chemical Formula [OF] the covalent compound [IN] default"}] | [{"type":"chemical equation","value":"CCl4"}] | [{"type":"substance name","value":"Carbon tetrachloride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for the covalent compound carbon tetrachloride?</h1> | null | CCl4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You will no longer find this <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> in undegraduate laboratories due to safety concerns. What is the structure of <mathjax>#"CCl"_4#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#"CCl"_4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You will no longer find this <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> in undegraduate laboratories due to safety concerns. What is the structure of <mathjax>#"CCl"_4#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#"CCl"_4#</mathjax></p></div>
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<div class="markdown"><p>You will no longer find this <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> in undegraduate laboratories due to safety concerns. What is the structure of <mathjax>#"CCl"_4#</mathjax>?</p></div>
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</article> | What is the formula for the covalent compound carbon tetrachloride? | null |
2,686 | ac896a3a-6ddd-11ea-b568-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-a-solution-which-contains-1-mole-of-cacl-2-dissolve | 0.50 M | start physical_unit 12 12 concentration mol/l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 6 6 15 16 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] CaCl2 solution [IN] M"}] | [{"type":"physical unit","value":"0.50 M"}] | [{"type":"physical unit","value":"Mole [OF] CaCl2 [=] \\pu{1 mole}"},{"type":"physical unit","value":"Volume [OF] CaCl2 solution [=] \\pu{2,000 milliliters}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of a solution which contains 1 mole of #CaCl_2# dissolved in 2,000 milliliters of solution?</h1> | null | 0.50 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(1*mol)/(2.000*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.5*mol*L^-1#</mathjax>.</p>
<p>In <mathjax>#"1 L"#</mathjax> of solution, what is the mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.5*mol*L^-1#</mathjax> with respect to <mathjax>#"CaCl"_2#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(1*mol)/(2.000*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.5*mol*L^-1#</mathjax>.</p>
<p>In <mathjax>#"1 L"#</mathjax> of solution, what is the mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of a solution which contains 1 mole of #CaCl_2# dissolved in 2,000 milliliters of solution?</h1>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.5*mol*L^-1#</mathjax> with respect to <mathjax>#"CaCl"_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=(1*mol)/(2.000*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.5*mol*L^-1#</mathjax>.</p>
<p>In <mathjax>#"1 L"#</mathjax> of solution, what is the mass of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>?</p></div>
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</article> | What is the concentration of a solution which contains 1 mole of #CaCl_2# dissolved in 2,000 milliliters of solution? | null |
2,687 | ac7f7f27-6ddd-11ea-85a4-ccda262736ce | https://socratic.org/questions/a-vinegar-bottle-has-a-hydroxide-concentration-of-5-0-10-12-m-what-is-the-hydron | 2.0 × 10^(-3) mol/L | start physical_unit 15 15 concentration mol/l qc_end physical_unit 5 5 8 11 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] hydronium [IN] mol/L"}] | [{"type":"physical unit","value":"2.0 × 10^(-3) mol/L"}] | [{"type":"physical unit","value":"Concentration [OF] hydroxide [=] \\pu{5.0 × 10^(-12) M}"}] | <h1 class="questionTitle" itemprop="name">A vinegar bottle has a hydroxide concentration of #5.0*10^-12# #M#. What is the hydronium concentration?</h1> | null | 2.0 × 10^(-3) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that water undergoes autoprotolysis, and that this reaction can be precisely quantified, i.e.</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p>This is an equilibrium reaction, and under standard conditions,</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax> (this is usually treated as a dimensionless quantity, and so.........</p>
<p>....when we take <mathjax>#log_10#</mathjax> of BOTH sides, we get......)</p>
<p><mathjax>#log_10{[H_3O^+][HO^-]}=log_10(10^-14)#</mathjax>....and</p>
<p><mathjax>#-14=log_10[H_3O^+]+log_10[HO^-]#</mathjax>.....OR</p>
<p><mathjax>#underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)=14#</mathjax></p>
<p>And thus we get the working relationship.....</p>
<p><mathjax>#pH+pOH=14#</mathjax> IN AQUEOUS SOLUTION UNDER STANDARD CONDITIONS. And this is something you will habitually use in acid-base equilibria. </p>
<p>So we gots an ACIDIC solution, where <mathjax>#[HO^-]=5.0xx10^-12*mol*L^-1#</mathjax>, and thus <mathjax>#pOH=-log_10(5.0xx10^-12)=-(11.30)=11.3#</mathjax>.</p>
<p>And <mathjax>#pH=14-pOH=14-11.3=2.7#</mathjax> And thus (FINALLY) <mathjax>#[H_3O^+]=10^(-pH)=....?#</mathjax></p>
<p>I acknowledge that I have included a lot of background in this question. But if you can follow it, and use logarithms effectively, these sorts of questions become quite trivial. If there is an issue for clarification, please report it. </p></div>
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<div class="markdown"><p><mathjax>#[H_3O^+]=10^(-2.70)*mol*L^-1=2.0xx10^-3*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We know that water undergoes autoprotolysis, and that this reaction can be precisely quantified, i.e.</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p>This is an equilibrium reaction, and under standard conditions,</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax> (this is usually treated as a dimensionless quantity, and so.........</p>
<p>....when we take <mathjax>#log_10#</mathjax> of BOTH sides, we get......)</p>
<p><mathjax>#log_10{[H_3O^+][HO^-]}=log_10(10^-14)#</mathjax>....and</p>
<p><mathjax>#-14=log_10[H_3O^+]+log_10[HO^-]#</mathjax>.....OR</p>
<p><mathjax>#underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)=14#</mathjax></p>
<p>And thus we get the working relationship.....</p>
<p><mathjax>#pH+pOH=14#</mathjax> IN AQUEOUS SOLUTION UNDER STANDARD CONDITIONS. And this is something you will habitually use in acid-base equilibria. </p>
<p>So we gots an ACIDIC solution, where <mathjax>#[HO^-]=5.0xx10^-12*mol*L^-1#</mathjax>, and thus <mathjax>#pOH=-log_10(5.0xx10^-12)=-(11.30)=11.3#</mathjax>.</p>
<p>And <mathjax>#pH=14-pOH=14-11.3=2.7#</mathjax> And thus (FINALLY) <mathjax>#[H_3O^+]=10^(-pH)=....?#</mathjax></p>
<p>I acknowledge that I have included a lot of background in this question. But if you can follow it, and use logarithms effectively, these sorts of questions become quite trivial. If there is an issue for clarification, please report it. </p></div>
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<h1 class="questionTitle" itemprop="name">A vinegar bottle has a hydroxide concentration of #5.0*10^-12# #M#. What is the hydronium concentration?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[H_3O^+]=10^(-2.70)*mol*L^-1=2.0xx10^-3*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We know that water undergoes autoprotolysis, and that this reaction can be precisely quantified, i.e.</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsH_3O^+ + HO^-#</mathjax></p>
<p>This is an equilibrium reaction, and under standard conditions,</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax> (this is usually treated as a dimensionless quantity, and so.........</p>
<p>....when we take <mathjax>#log_10#</mathjax> of BOTH sides, we get......)</p>
<p><mathjax>#log_10{[H_3O^+][HO^-]}=log_10(10^-14)#</mathjax>....and</p>
<p><mathjax>#-14=log_10[H_3O^+]+log_10[HO^-]#</mathjax>.....OR</p>
<p><mathjax>#underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)=14#</mathjax></p>
<p>And thus we get the working relationship.....</p>
<p><mathjax>#pH+pOH=14#</mathjax> IN AQUEOUS SOLUTION UNDER STANDARD CONDITIONS. And this is something you will habitually use in acid-base equilibria. </p>
<p>So we gots an ACIDIC solution, where <mathjax>#[HO^-]=5.0xx10^-12*mol*L^-1#</mathjax>, and thus <mathjax>#pOH=-log_10(5.0xx10^-12)=-(11.30)=11.3#</mathjax>.</p>
<p>And <mathjax>#pH=14-pOH=14-11.3=2.7#</mathjax> And thus (FINALLY) <mathjax>#[H_3O^+]=10^(-pH)=....?#</mathjax></p>
<p>I acknowledge that I have included a lot of background in this question. But if you can follow it, and use logarithms effectively, these sorts of questions become quite trivial. If there is an issue for clarification, please report it. </p></div>
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</article> | A vinegar bottle has a hydroxide concentration of #5.0*10^-12# #M#. What is the hydronium concentration? | null |
2,688 | aac37340-6ddd-11ea-9bf4-ccda262736ce | https://socratic.org/questions/what-is-the-percent-composition-by-mass-of-nitrogen-in-nh-4-2co-3-gram-formula-m | 29.19% | start physical_unit 8 10 mass_percent none qc_end physical_unit 10 10 13 14 molar_mass qc_end end | [{"type":"physical unit","value":"Percent composition by mass [OF] nitrogen in (NH4)2CO3"}] | [{"type":"physical unit","value":"29.19%"}] | [{"type":"physical unit","value":"Gram-formula mass [OF] (NH4)2CO3 [=] \\pu{96.0 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the percent composition by mass of nitrogen in #(NH_4)_2CO_3# (gram-formula mass 96.0 g/mo)?</h1> | null | 29.19% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the percentage by mass of <mathjax>#"N"#</mathjax> in ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, given that its molar mass is <mathjax>#96.0"g"/"mol"#</mathjax>.</p>
<p>To find the mass of <mathjax>#"N"#</mathjax> in the compound, we multiply its molar mas, <mathjax>#14.01"g"/"mol"#</mathjax>, by however many atoms of <mathjax>#"N"#</mathjax> are in the compound, which is <mathjax>#2#</mathjax>:</p>
<p><mathjax>#"mass of N" = (2)(14.01"g"/"mol") = 28.02"g"/"mol"#</mathjax></p>
<p>Thus, in one mole of <mathjax>#("NH"_4)_2"CO"_3#</mathjax> (which has a mass of <mathjax>#96.0#</mathjax> <mathjax>#"g"#</mathjax>), there are <mathjax>#28.02#</mathjax> <mathjax>#"g"#</mathjax> of nitrogen. The percentage by mass of <mathjax>#"N"#</mathjax> is the mass of <mathjax>#"N"#</mathjax> divided by the total mass, then multiplied by <mathjax>#100#</mathjax>:</p>
<p><mathjax>#% "N" = ("mass of N")/("mass of" ("NH"_4)_2"CO"_3) xx 100%#</mathjax></p>
<p><mathjax>#% "N" = (28.02"g N")/(96.0"g" ("NH"_4)_2"CO"_3) xx 100% = color(red)(29.2% "N"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#29.2% "N"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the percentage by mass of <mathjax>#"N"#</mathjax> in ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, given that its molar mass is <mathjax>#96.0"g"/"mol"#</mathjax>.</p>
<p>To find the mass of <mathjax>#"N"#</mathjax> in the compound, we multiply its molar mas, <mathjax>#14.01"g"/"mol"#</mathjax>, by however many atoms of <mathjax>#"N"#</mathjax> are in the compound, which is <mathjax>#2#</mathjax>:</p>
<p><mathjax>#"mass of N" = (2)(14.01"g"/"mol") = 28.02"g"/"mol"#</mathjax></p>
<p>Thus, in one mole of <mathjax>#("NH"_4)_2"CO"_3#</mathjax> (which has a mass of <mathjax>#96.0#</mathjax> <mathjax>#"g"#</mathjax>), there are <mathjax>#28.02#</mathjax> <mathjax>#"g"#</mathjax> of nitrogen. The percentage by mass of <mathjax>#"N"#</mathjax> is the mass of <mathjax>#"N"#</mathjax> divided by the total mass, then multiplied by <mathjax>#100#</mathjax>:</p>
<p><mathjax>#% "N" = ("mass of N")/("mass of" ("NH"_4)_2"CO"_3) xx 100%#</mathjax></p>
<p><mathjax>#% "N" = (28.02"g N")/(96.0"g" ("NH"_4)_2"CO"_3) xx 100% = color(red)(29.2% "N"#</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# (gram-formula mass 96.0 g/mo)?</h1>
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<div class="markdown"><p><mathjax>#29.2% "N"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We're asked to find the percentage by mass of <mathjax>#"N"#</mathjax> in ammonium carbonate, <mathjax>#("NH"_4)_2"CO"_3#</mathjax>, given that its molar mass is <mathjax>#96.0"g"/"mol"#</mathjax>.</p>
<p>To find the mass of <mathjax>#"N"#</mathjax> in the compound, we multiply its molar mas, <mathjax>#14.01"g"/"mol"#</mathjax>, by however many atoms of <mathjax>#"N"#</mathjax> are in the compound, which is <mathjax>#2#</mathjax>:</p>
<p><mathjax>#"mass of N" = (2)(14.01"g"/"mol") = 28.02"g"/"mol"#</mathjax></p>
<p>Thus, in one mole of <mathjax>#("NH"_4)_2"CO"_3#</mathjax> (which has a mass of <mathjax>#96.0#</mathjax> <mathjax>#"g"#</mathjax>), there are <mathjax>#28.02#</mathjax> <mathjax>#"g"#</mathjax> of nitrogen. The percentage by mass of <mathjax>#"N"#</mathjax> is the mass of <mathjax>#"N"#</mathjax> divided by the total mass, then multiplied by <mathjax>#100#</mathjax>:</p>
<p><mathjax>#% "N" = ("mass of N")/("mass of" ("NH"_4)_2"CO"_3) xx 100%#</mathjax></p>
<p><mathjax>#% "N" = (28.02"g N")/(96.0"g" ("NH"_4)_2"CO"_3) xx 100% = color(red)(29.2% "N"#</mathjax></p></div>
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</article> | What is the percent composition by mass of nitrogen in #(NH_4)_2CO_3# (gram-formula mass 96.0 g/mo)? | null |
2,689 | a9c18b24-6ddd-11ea-b46b-ccda262736ce | https://socratic.org/questions/58d5ddc5b72cff577db4fb0f | 2.27 × 10^(-5) mol/L | start physical_unit 1 2 concentration mol/l qc_end physical_unit 1 2 5 6 concentration qc_end substance 9 9 qc_end end | [{"type":"physical unit","value":"Concentration2 [OF] carbon dioxide solution [IN] mol/L"}] | [{"type":"physical unit","value":"2.27 × 10^(-5) mol/L"}] | [{"type":"physical unit","value":"Concentration1 [OF] carbon dioxide solution [=] \\pu{1 ppm}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">If carbon dioxide expresses a #1*"ppm"# concentration in water, what is its concentration in #mol*L^-1#?</h1> | null | 2.27 × 10^(-5) mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so you simply multiply the <mathjax>#"ppm"#</mathjax> concentration by <mathjax>#(1xx10^-3*g*L^-1)/(44.0*g*mol^-1)#</mathjax> to give a molar concentration in <mathjax>#mol*L^-1#</mathjax>. </p></div>
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<div class="markdown"><p>Well <mathjax>#"1 ppm "-=" 1 mg"*L^-1............#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And so you simply multiply the <mathjax>#"ppm"#</mathjax> concentration by <mathjax>#(1xx10^-3*g*L^-1)/(44.0*g*mol^-1)#</mathjax> to give a molar concentration in <mathjax>#mol*L^-1#</mathjax>. </p></div>
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<div class="markdown"><p>Well <mathjax>#"1 ppm "-=" 1 mg"*L^-1............#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And so you simply multiply the <mathjax>#"ppm"#</mathjax> concentration by <mathjax>#(1xx10^-3*g*L^-1)/(44.0*g*mol^-1)#</mathjax> to give a molar concentration in <mathjax>#mol*L^-1#</mathjax>. </p></div>
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</article> | If carbon dioxide expresses a #1*"ppm"# concentration in water, what is its concentration in #mol*L^-1#? | null |
2,690 | aba5618c-6ddd-11ea-b2b8-ccda262736ce | https://socratic.org/questions/given-the-balanced-equation-representing-a-reaction-2co-g-o-2-g-2co-2-g-what-is- | 1:1 | start physical_unit 20 22 mole_fraction none qc_end chemical_equation 7 13 qc_end end | [{"type":"physical unit","value":"Mole ratio [OF] CO(g) to CO2(g)"}] | [{"type":"physical unit","value":"1:1"}] | [{"type":"chemical equation","value":"2 CO(g) + O2(g) -> 2 CO2(g)"}] | <h1 class="questionTitle" itemprop="name">Given the balanced equation representing a reaction: #2CO(g) + O_2(g) -> 2CO_2(g)#. What is the mole ratio of #CO(g)# to #CO_2(g)# in this reaction?</h1> | null | 1:1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We could write...<mathjax>#CO(g) +1/2O_2 rarr CO_2(g)#</mathjax>, instead of ...</p>
<p><mathjax>#2CO(g) +O_2 rarr 2CO_2(g)#</mathjax></p>
<p>In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is why teachers go to such great efforts to teach <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>....and we could put in numbers to stress the mass balance of the equation....</p>
<p><mathjax>#underbrace(CO(g) +1/2O_2)_"28 g + 16 g = 44 g" rarr underbrace(CO_2(g))_"44 g"#</mathjax></p></div>
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<div class="markdown"><p>Is it not <mathjax>#1:1#</mathjax>...?</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We could write...<mathjax>#CO(g) +1/2O_2 rarr CO_2(g)#</mathjax>, instead of ...</p>
<p><mathjax>#2CO(g) +O_2 rarr 2CO_2(g)#</mathjax></p>
<p>In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is why teachers go to such great efforts to teach <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>....and we could put in numbers to stress the mass balance of the equation....</p>
<p><mathjax>#underbrace(CO(g) +1/2O_2)_"28 g + 16 g = 44 g" rarr underbrace(CO_2(g))_"44 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Given the balanced equation representing a reaction: #2CO(g) + O_2(g) -> 2CO_2(g)#. What is the mole ratio of #CO(g)# to #CO_2(g)# in this reaction?</h1>
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anor277
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<div class="markdown"><p>Is it not <mathjax>#1:1#</mathjax>...?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We could write...<mathjax>#CO(g) +1/2O_2 rarr CO_2(g)#</mathjax>, instead of ...</p>
<p><mathjax>#2CO(g) +O_2 rarr 2CO_2(g)#</mathjax></p>
<p>In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is why teachers go to such great efforts to teach <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>....and we could put in numbers to stress the mass balance of the equation....</p>
<p><mathjax>#underbrace(CO(g) +1/2O_2)_"28 g + 16 g = 44 g" rarr underbrace(CO_2(g))_"44 g"#</mathjax></p></div>
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</article> | Given the balanced equation representing a reaction: #2CO(g) + O_2(g) -> 2CO_2(g)#. What is the mole ratio of #CO(g)# to #CO_2(g)# in this reaction? | null |
2,691 | ac5cb667-6ddd-11ea-9730-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-formula-for-zinc-carbonate | ZnCO3 | start chemical_formula qc_end substance 6 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] zinc carbonate [IN] default"}] | [{"type":"chemical equation","value":"ZnCO3"}] | [{"type":"substance name","value":"Zinc carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the chemical formula for zinc carbonate?</h1> | null | ZnCO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The zinc ion has a charge of 2+<br/>
The carbonate ion has a charge of 2-</p>
<p>So if we take one of each the charges are in balance:</p>
<p><mathjax>#Zn^(2+)CO_3^(2-)or ZnCO_3#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#ZnCO_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The zinc ion has a charge of 2+<br/>
The carbonate ion has a charge of 2-</p>
<p>So if we take one of each the charges are in balance:</p>
<p><mathjax>#Zn^(2+)CO_3^(2-)or ZnCO_3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the chemical formula for zinc carbonate?</h1>
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<div class="markdown"><p><mathjax>#ZnCO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The zinc ion has a charge of 2+<br/>
The carbonate ion has a charge of 2-</p>
<p>So if we take one of each the charges are in balance:</p>
<p><mathjax>#Zn^(2+)CO_3^(2-)or ZnCO_3#</mathjax></p></div>
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</article> | What is the chemical formula for zinc carbonate? | null |
2,692 | a8e97702-6ddd-11ea-9625-ccda262736ce | https://socratic.org/questions/how-do-you-balance-the-following-equation-nh-3-o-2-no-h-2o | 4 NH3 + 5 O2 -> 4 NO + 6 H2O | start chemical_equation qc_end chemical_equation 7 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the following equation"}] | [{"type":"chemical equation","value":"4 NH3 + 5 O2 -> 4 NO + 6 H2O"}] | [{"type":"chemical equation","value":"? NH3 + ? O2 -> ? NO + ? H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance the following equation: #?NH_3 + ?O_2 -> ?NO + ?H_2O#?</h1> | null | 4 NH3 + 5 O2 -> 4 NO + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>These are standard <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>, nitrogen is oxidized, and oxygen is reduced:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#NH_3+H_2Orarr NO + 5H^+ + 5e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#O_2+ 4H^+ +4e^(-)rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We cross multiply and take <mathjax>#4xx(i)+5xx(ii)#</mathjax></p>
<p><mathjax>#4NH_3 +5O_2+4H_2O rarr 4NO +10H_2O#</mathjax> </p>
<p>Nitrogens, oxygens, and hydrogens all balance. Charges balance, so this is a valid representation. Of course we can substract 4 equiv of water from EACH side of the equation. </p>
<p>Ammonia might be fully oxidized up to <mathjax>#NO_2#</mathjax>. If you represent this reaction for practice, would you post the balanced equation in this thread?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>Overall,</p>
<p><mathjax>#4NH_3 +5O_2 rarr 4NO +6H_2O#</mathjax> </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>These are standard <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>, nitrogen is oxidized, and oxygen is reduced:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#NH_3+H_2Orarr NO + 5H^+ + 5e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#O_2+ 4H^+ +4e^(-)rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We cross multiply and take <mathjax>#4xx(i)+5xx(ii)#</mathjax></p>
<p><mathjax>#4NH_3 +5O_2+4H_2O rarr 4NO +10H_2O#</mathjax> </p>
<p>Nitrogens, oxygens, and hydrogens all balance. Charges balance, so this is a valid representation. Of course we can substract 4 equiv of water from EACH side of the equation. </p>
<p>Ammonia might be fully oxidized up to <mathjax>#NO_2#</mathjax>. If you represent this reaction for practice, would you post the balanced equation in this thread?</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance the following equation: #?NH_3 + ?O_2 -> ?NO + ?H_2O#?</h1>
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anor277
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<span class="dateCreated" datetime="2017-01-21T08:26:25" itemprop="dateCreated">
Jan 21, 2017
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<div class="markdown"><p>Overall,</p>
<p><mathjax>#4NH_3 +5O_2 rarr 4NO +6H_2O#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>These are standard <a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reactions</a>, nitrogen is oxidized, and oxygen is reduced:</p>
<p><mathjax>#"Oxidation:"#</mathjax></p>
<p><mathjax>#NH_3+H_2Orarr NO + 5H^+ + 5e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction:"#</mathjax></p>
<p><mathjax>#O_2+ 4H^+ +4e^(-)rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We cross multiply and take <mathjax>#4xx(i)+5xx(ii)#</mathjax></p>
<p><mathjax>#4NH_3 +5O_2+4H_2O rarr 4NO +10H_2O#</mathjax> </p>
<p>Nitrogens, oxygens, and hydrogens all balance. Charges balance, so this is a valid representation. Of course we can substract 4 equiv of water from EACH side of the equation. </p>
<p>Ammonia might be fully oxidized up to <mathjax>#NO_2#</mathjax>. If you represent this reaction for practice, would you post the balanced equation in this thread?</p></div>
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</article> | How do you balance the following equation: #?NH_3 + ?O_2 -> ?NO + ?H_2O#? | null |
2,693 | a90c07a5-6ddd-11ea-96fc-ccda262736ce | https://socratic.org/questions/a-4-0-l-sample-of-hydrogen-gas-at-700-mmhg-would-occupy-what-volume-at-250-mmhg | 11 L | start physical_unit 3 6 volume l qc_end physical_unit 3 6 1 2 volume qc_end physical_unit 3 6 8 9 pressure qc_end physical_unit 3 6 15 16 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] hydrogen gas sample [IN] L"}] | [{"type":"physical unit","value":"11 L"}] | [{"type":"physical unit","value":"Volume1 [OF] hydrogen gas sample [=] \\pu{4.0 L}"},{"type":"physical unit","value":"Pressure1 [OF] hydrogen gas sample [=] \\pu{700 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] hydrogen gas sample [=] \\pu{250 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A 4.0 L sample of hydrogen gas at 700 mmHg would occupy what volume at 250 mmHg? </h1> | null | 11 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can calculate the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is:<br/>
<img alt="study.com" src="https://useruploads.socratic.org/t4tHhzLTlm0WaJYf6dcR_boyles-law-equation.jpg"/> </p>
<p><mathjax>#color(hotpink)("Let's write down what we know and what we don't know based")#</mathjax><br/>
<mathjax>#color(hotpink) "on the information given to us:"#</mathjax></p>
<p>The first volume has a value of <mathjax>#4.0#</mathjax> L, the first <a href="https://socratic.org/physics/fluid-mechanics/pressure">pressure</a> is <mathjax>#700mmHg,#</mathjax> and the second pressure is <mathjax>#250 mmHg#</mathjax>. Our only unknown is the second volume.</p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#V_2#</mathjax></p>
<p>We do this by dividing both sides by <mathjax>#P_2#</mathjax> to get <mathjax>#V_2#</mathjax> by itself:<br/>
<mathjax>#V_2=(P_1xxV_1)/P_2#</mathjax></p>
<p>Now all we have to do is plug in the given values:<br/>
<mathjax>#V_2=(700\cancel"mmHg" xx 4.0\"L")/(250cancel"mmHg")#</mathjax> = <mathjax>#11 L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#color(indigo)("That sample of hydrogen would occupy a new volume of 11L")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can calculate the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is:<br/>
<img alt="study.com" src="https://useruploads.socratic.org/t4tHhzLTlm0WaJYf6dcR_boyles-law-equation.jpg"/> </p>
<p><mathjax>#color(hotpink)("Let's write down what we know and what we don't know based")#</mathjax><br/>
<mathjax>#color(hotpink) "on the information given to us:"#</mathjax></p>
<p>The first volume has a value of <mathjax>#4.0#</mathjax> L, the first <a href="https://socratic.org/physics/fluid-mechanics/pressure">pressure</a> is <mathjax>#700mmHg,#</mathjax> and the second pressure is <mathjax>#250 mmHg#</mathjax>. Our only unknown is the second volume.</p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#V_2#</mathjax></p>
<p>We do this by dividing both sides by <mathjax>#P_2#</mathjax> to get <mathjax>#V_2#</mathjax> by itself:<br/>
<mathjax>#V_2=(P_1xxV_1)/P_2#</mathjax></p>
<p>Now all we have to do is plug in the given values:<br/>
<mathjax>#V_2=(700\cancel"mmHg" xx 4.0\"L")/(250cancel"mmHg")#</mathjax> = <mathjax>#11 L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A 4.0 L sample of hydrogen gas at 700 mmHg would occupy what volume at 250 mmHg? </h1>
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<div class="markdown"><p><mathjax>#color(indigo)("That sample of hydrogen would occupy a new volume of 11L")#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can calculate the answer using <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.</p>
<p>The equation we use is:<br/>
<img alt="study.com" src="https://useruploads.socratic.org/t4tHhzLTlm0WaJYf6dcR_boyles-law-equation.jpg"/> </p>
<p><mathjax>#color(hotpink)("Let's write down what we know and what we don't know based")#</mathjax><br/>
<mathjax>#color(hotpink) "on the information given to us:"#</mathjax></p>
<p>The first volume has a value of <mathjax>#4.0#</mathjax> L, the first <a href="https://socratic.org/physics/fluid-mechanics/pressure">pressure</a> is <mathjax>#700mmHg,#</mathjax> and the second pressure is <mathjax>#250 mmHg#</mathjax>. Our only unknown is the second volume.</p>
<p>All we have to do is rearrange the equation to solve for <mathjax>#V_2#</mathjax></p>
<p>We do this by dividing both sides by <mathjax>#P_2#</mathjax> to get <mathjax>#V_2#</mathjax> by itself:<br/>
<mathjax>#V_2=(P_1xxV_1)/P_2#</mathjax></p>
<p>Now all we have to do is plug in the given values:<br/>
<mathjax>#V_2=(700\cancel"mmHg" xx 4.0\"L")/(250cancel"mmHg")#</mathjax> = <mathjax>#11 L#</mathjax></p></div>
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</article> | A 4.0 L sample of hydrogen gas at 700 mmHg would occupy what volume at 250 mmHg? | null |
2,694 | a97e17b0-6ddd-11ea-a4a2-ccda262736ce | https://socratic.org/questions/an-oven-cleaning-solution-is-40-0-by-mass-naoh-if-one-jar-of-this-product-contai | 182 g | start physical_unit 7 7 mass g qc_end physical_unit 2 2 15 16 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] NaOH [IN] g"}] | [{"type":"physical unit","value":"182 g"}] | [{"type":"physical unit","value":"Percent by mass [OF] NaOH in oven-cleaning solution [=] \\pu{40.0%}"},{"type":"physical unit","value":"Number [OF] jar [=] \\pu{1}"},{"type":"physical unit","value":"Mass [OF] solution [=] \\pu{454 g}"}] | <h1 class="questionTitle" itemprop="name">An oven-cleaning solution is 40.0% (by mass) #NaOH#. If one jar of this product contains 454 g of solution, how much #NaOH# does it contain?</h1> | null | 182 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find how much <mathjax>#"NaOH"#</mathjax> is in a <mathjax>#454"g"#</mathjax> solution containing <mathjax>#40% "NaOH"#</mathjax> by mass.</p>
<p>#40% in decimal form is </p>
<p><mathjax>#(40.0cancel(%))/(100cancel(%)) = 0.400#</mathjax></p>
<p>(you probably already know this, but it keeps calculations a little simpler:)</p>
<p>To find how much <mathjax>#"NaOH"#</mathjax> is in the solution, we take this fraction <mathjax>#(0.400)#</mathjax>, and multiply it by the total mass of the solution:</p>
<p>Mass <mathjax>#"NaOH" = (0.400)(454"g") = color(red)(182"g NaOH"#</mathjax></p>
<p>rounded to three significant figures, the amount given in the question.</p>
<p>You could also report this in <em>moles</em> of sodium hydroxide, using its molar mass to calculate the number of moles:</p>
<p><mathjax>#182 cancel("g NaOH")((1 "mol NaOH")/(40.00 cancel("g NaOH"))) = color(blue)(4.55 "mol NaOH"#</mathjax></p>
<p>However, since the amount of a substance is most often expressed in grams (because you can measure mass directly, but can't necessarily measure moles directly), it's more conventional to report your answer in grams, unless it says otherwise.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#182 "g NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find how much <mathjax>#"NaOH"#</mathjax> is in a <mathjax>#454"g"#</mathjax> solution containing <mathjax>#40% "NaOH"#</mathjax> by mass.</p>
<p>#40% in decimal form is </p>
<p><mathjax>#(40.0cancel(%))/(100cancel(%)) = 0.400#</mathjax></p>
<p>(you probably already know this, but it keeps calculations a little simpler:)</p>
<p>To find how much <mathjax>#"NaOH"#</mathjax> is in the solution, we take this fraction <mathjax>#(0.400)#</mathjax>, and multiply it by the total mass of the solution:</p>
<p>Mass <mathjax>#"NaOH" = (0.400)(454"g") = color(red)(182"g NaOH"#</mathjax></p>
<p>rounded to three significant figures, the amount given in the question.</p>
<p>You could also report this in <em>moles</em> of sodium hydroxide, using its molar mass to calculate the number of moles:</p>
<p><mathjax>#182 cancel("g NaOH")((1 "mol NaOH")/(40.00 cancel("g NaOH"))) = color(blue)(4.55 "mol NaOH"#</mathjax></p>
<p>However, since the amount of a substance is most often expressed in grams (because you can measure mass directly, but can't necessarily measure moles directly), it's more conventional to report your answer in grams, unless it says otherwise.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">An oven-cleaning solution is 40.0% (by mass) #NaOH#. If one jar of this product contains 454 g of solution, how much #NaOH# does it contain?</h1>
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Nathan L.
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May 27, 2017
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<div class="markdown"><p><mathjax>#182 "g NaOH"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We're asked to find how much <mathjax>#"NaOH"#</mathjax> is in a <mathjax>#454"g"#</mathjax> solution containing <mathjax>#40% "NaOH"#</mathjax> by mass.</p>
<p>#40% in decimal form is </p>
<p><mathjax>#(40.0cancel(%))/(100cancel(%)) = 0.400#</mathjax></p>
<p>(you probably already know this, but it keeps calculations a little simpler:)</p>
<p>To find how much <mathjax>#"NaOH"#</mathjax> is in the solution, we take this fraction <mathjax>#(0.400)#</mathjax>, and multiply it by the total mass of the solution:</p>
<p>Mass <mathjax>#"NaOH" = (0.400)(454"g") = color(red)(182"g NaOH"#</mathjax></p>
<p>rounded to three significant figures, the amount given in the question.</p>
<p>You could also report this in <em>moles</em> of sodium hydroxide, using its molar mass to calculate the number of moles:</p>
<p><mathjax>#182 cancel("g NaOH")((1 "mol NaOH")/(40.00 cancel("g NaOH"))) = color(blue)(4.55 "mol NaOH"#</mathjax></p>
<p>However, since the amount of a substance is most often expressed in grams (because you can measure mass directly, but can't necessarily measure moles directly), it's more conventional to report your answer in grams, unless it says otherwise.</p></div>
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</article> | An oven-cleaning solution is 40.0% (by mass) #NaOH#. If one jar of this product contains 454 g of solution, how much #NaOH# does it contain? | null |
2,695 | aa616918-6ddd-11ea-a172-ccda262736ce | https://socratic.org/questions/coal-gasification-is-a-process-that-converts-coal-into-methane-gas-if-this-react | 709.45 grams | start physical_unit 9 10 mass g qc_end physical_unit 12 13 19 19 percent_yield qc_end physical_unit 30 30 27 28 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] methane gas [IN] grams"}] | [{"type":"physical unit","value":"709.45 grams"}] | [{"type":"physical unit","value":"Percent yield [OF] this reaction [=] \\pu{85.0%}"},{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{1250 g}"}] | <h1 class="questionTitle" itemprop="name"> Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon?</h1> | null | 709.45 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"709 g CH"_4#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name"> Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"709 g CH"_4#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this reaction</p>
<blockquote>
<p><mathjax>#color(blue)(2)"C"_ ((s)) + 2"H"_ 2"O"_ ((g)) -> "CO"_ (2(g)) + "CH"_ (4(g))#</mathjax></p>
</blockquote>
<p>Notice that the reaction produces <mathjax>#1#</mathjax> <strong>mole</strong> of methane <strong>for every</strong> <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon that take part in the reaction. </p>
<p>This represents the reaction's <strong>theoretical yield</strong>, i.e. what you get if the reaction has a <mathjax>#100%#</mathjax> <strong>yield</strong>. </p>
<p>You can convert this mole ratio to a <strong>gram ratio</strong> by using the <strong>molar masses</strong> of carbon and methane</p>
<blockquote>
<p><mathjax>#2 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "24.022 g"#</mathjax></p>
<p><mathjax>#1 color(red)(cancel(color(black)("mole CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "16.04 g"#</mathjax></p>
</blockquote>
<p>You can thus say that the <strong>theoretical yield</strong> of the reaction will have it produce <mathjax>#"16.04 g"#</mathjax> of methane <strong>for every</strong> <mathjax>#"24.022 g"#</mathjax> of carbon that react. </p>
<p>This means that <mathjax>#"1250 g"#</mathjax> of carbon will <em>theoretically</em> produce</p>
<blockquote>
<p><mathjax>#1250 color(red)(cancel(color(black)("g C"))) * "16.04 g CH"_4/(24.022color(red)(cancel(color(black)("g C")))) = "834.65 g CH"_4#</mathjax></p>
</blockquote>
<p>Now, the reaction is said to have an <mathjax>#85.0%#</mathjax> yield, which basically means that <strong>for every</strong> <mathjax>#"100 g"#</mathjax> of methane that <em>could</em> be produced, you only get <mathjax>#"85.0 g"#</mathjax>. </p>
<p>As a result, the <strong>actual yield</strong> of the reaction will be</p>
<blockquote>
<p><mathjax>#834.65 color(red)(cancel(color(black)("g CH"_4))) * "85.0 g produced"/(100color(red)(cancel(color(black)("g CH"_4)))) = color(darkgreen)(ul(color(black)("709 g CH"_4)))#</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> | Coal gasification is a process that converts coal into methane gas. If this reaction has a percent yield of 85.0%, how much methane can be obtained from 1250 g of carbon? | null |
2,696 | abfa1852-6ddd-11ea-a598-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-mass-of-a-substance-if-22-5-g-dissolved-in-250-g-of-water- | 180 u | start physical_unit 7 7 molecular_weight amu qc_end physical_unit 7 7 9 10 mass qc_end physical_unit 16 16 13 14 mass qc_end physical_unit 19 19 24 25 freezing_point_temperature qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] the substance [IN] u"}] | [{"type":"physical unit","value":"180 u"}] | [{"type":"physical unit","value":"Mass [OF] the substance [=] \\pu{22.5 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{250 g}"},{"type":"physical unit","value":"Freezing point [OF] the solution [=] \\pu{-0.930 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular mass of a substance if 22.5 g dissolved in 250 g of water produces a solution whose freezing point is -0.930°C? </h1> | null | 180 u | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_"f"mcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> = the decrease in the freezing point<br/>
<mathjax>#K_"f"#</mathjax> = the molal freezing point depression constant of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#m#</mathjax> = the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<blockquote></blockquote>
<p>We can rearrange the formula to get</p>
<p><mathjax>#m = (ΔT_"f")/K_"f"#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#ΔT_"f" = "0.930 °C"#</mathjax><br/>
<mathjax>#K_"f" = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#m = (0.930 color(red)(cancel(color(black)("°C"))))/(1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.500 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>Now, <mathjax>#m = "0.500 mol"/(1 color(red)(cancel(color(black)("kg")))) = "22.5 g"/(0.250 color(red)(cancel(color(black)("kg"))))#</mathjax></p>
<p>∴ <mathjax>#"1 mol" = "22.5 g"/0.250 × 1/0.500 = "180 g"#</mathjax></p>
<p>The molar mass is 180 g/mol, so the molecular mass is 180 u.</p></div>
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<div class="markdown"><p>The molecular mass is 180 u.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_"f"mcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> = the decrease in the freezing point<br/>
<mathjax>#K_"f"#</mathjax> = the molal freezing point depression constant of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#m#</mathjax> = the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<blockquote></blockquote>
<p>We can rearrange the formula to get</p>
<p><mathjax>#m = (ΔT_"f")/K_"f"#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#ΔT_"f" = "0.930 °C"#</mathjax><br/>
<mathjax>#K_"f" = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#m = (0.930 color(red)(cancel(color(black)("°C"))))/(1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.500 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>Now, <mathjax>#m = "0.500 mol"/(1 color(red)(cancel(color(black)("kg")))) = "22.5 g"/(0.250 color(red)(cancel(color(black)("kg"))))#</mathjax></p>
<p>∴ <mathjax>#"1 mol" = "22.5 g"/0.250 × 1/0.500 = "180 g"#</mathjax></p>
<p>The molar mass is 180 g/mol, so the molecular mass is 180 u.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molecular mass of a substance if 22.5 g dissolved in 250 g of water produces a solution whose freezing point is -0.930°C? </h1>
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<div class="markdown"><p>The molecular mass is 180 u.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for freezing point depression is</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_"f"mcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
<p>where</p>
<p><mathjax>#ΔT_"f"#</mathjax> = the decrease in the freezing point<br/>
<mathjax>#K_"f"#</mathjax> = the molal freezing point depression constant of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a><br/>
<mathjax>#m#</mathjax> = the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a> of the solution</p>
<blockquote></blockquote>
<p>We can rearrange the formula to get</p>
<p><mathjax>#m = (ΔT_"f")/K_"f"#</mathjax></p>
<blockquote></blockquote>
<p>In your problem,</p>
<p><mathjax>#ΔT_"f" = "0.930 °C"#</mathjax><br/>
<mathjax>#K_"f" = "1.86 °C·kg·mol"^"-1"#</mathjax></p>
<p>∴ <mathjax>#m = (0.930 color(red)(cancel(color(black)("°C"))))/(1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.500 mol·kg"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p>Now, <mathjax>#m = "0.500 mol"/(1 color(red)(cancel(color(black)("kg")))) = "22.5 g"/(0.250 color(red)(cancel(color(black)("kg"))))#</mathjax></p>
<p>∴ <mathjax>#"1 mol" = "22.5 g"/0.250 × 1/0.500 = "180 g"#</mathjax></p>
<p>The molar mass is 180 g/mol, so the molecular mass is 180 u.</p></div>
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</article> | What is the molecular mass of a substance if 22.5 g dissolved in 250 g of water produces a solution whose freezing point is -0.930°C? | null |
2,697 | a8c6b97e-6ddd-11ea-b2bd-ccda262736ce | https://socratic.org/questions/how-many-resonance-structures-does-the-nitrate-ion-no-3-1-have | 3 | start physical_unit 2 3 number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Number [OF] resonance structures"}] | [{"type":"physical unit","value":"3"}] | [{"type":"chemical equation","value":"NO3-"}] | <h1 class="questionTitle" itemprop="name">How many resonance structures does the nitrate ion, #NO_3^-1# have?</h1> | null | 3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Nitrate ion is a good example to practise on, because a single resonance resonance isomer depicts 3 of the four atoms in the negative ion with formal charges:</p>
<p>i.e. <mathjax>#""^+N(=O)(-O^-)_2#</mathjax>. The nitrogen centre is quaternized with a formal positive charge, and the formally singly bound oxygen each bear a formal negative charge. How are the 24 <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> distributed? </p></div>
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<div class="markdown"><p><mathjax>#3#</mathjax> <a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/resonance">resonance</a> isomers are possible, but this is a formalism.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Nitrate ion is a good example to practise on, because a single resonance resonance isomer depicts 3 of the four atoms in the negative ion with formal charges:</p>
<p>i.e. <mathjax>#""^+N(=O)(-O^-)_2#</mathjax>. The nitrogen centre is quaternized with a formal positive charge, and the formally singly bound oxygen each bear a formal negative charge. How are the 24 <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> distributed? </p></div>
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<div class="markdown"><p><mathjax>#3#</mathjax> <a href="https://socratic.org/chemistry/covalent-bonds-and-formulas/resonance">resonance</a> isomers are possible, but this is a formalism.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Nitrate ion is a good example to practise on, because a single resonance resonance isomer depicts 3 of the four atoms in the negative ion with formal charges:</p>
<p>i.e. <mathjax>#""^+N(=O)(-O^-)_2#</mathjax>. The nitrogen centre is quaternized with a formal positive charge, and the formally singly bound oxygen each bear a formal negative charge. How are the 24 <a href="https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> distributed? </p></div>
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</article> | How many resonance structures does the nitrate ion, #NO_3^-1# have? | null |
2,698 | ab5682d8-6ddd-11ea-a092-ccda262736ce | https://socratic.org/questions/in-the-equation-2kclo-3-2kcl-3o-2-if-5-0-g-of-kclo-3-is-decomposed-what-volume-o | 1.37 L | start physical_unit 10 10 volume l qc_end chemical_equation 3 10 qc_end physical_unit 4 4 12 13 mass qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] O2 [IN] L"}] | [{"type":"physical unit","value":"1.37 L"}] | [{"type":"chemical equation","value":"2 KClO3 -> 2 KCl + 3 O2"},{"type":"physical unit","value":"Mass [OF] KClO3 [=] \\pu{5.0 g}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">In the equation #2KClO_3 -> 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?</h1> | null | 1.37 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Find the relative mass of the molecule <mathjax>#KClO_3#</mathjax> by adding together the masses of each atom in it. Find the masses from the <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a>.</p>
<p><mathjax>#39.01gmol^-1 + 35.45gmol^-1 + 3 * 16.00gmol^-1#</mathjax><br/>
<mathjax># = 122.46gmol^-1#</mathjax></p>
<p>Divide the total mass of the substance by its relative mass to find the number of moles of the substance. </p>
<p><mathjax>#(5.00g)/(122.46gmol^-1) = 0.041mol#</mathjax></p>
<p>There are two moles of <mathjax>#KClO_3#</mathjax> for every 3 moles of <mathjax>#O_2#</mathjax>, which we can see from the equation. Using the ratio <mathjax>#2:3#</mathjax> and inserting <mathjax>#0.0408mol#</mathjax>, we find that there must be <mathjax>#0.0612mol#</mathjax> of <mathjax>#O_2#</mathjax>.</p>
<p>One mole of an ideal gas at STP (standard temperature and pressure) occupies 22.4L. </p>
<p>Multiplying </p>
<p><mathjax>#22.4Lmol^-1 * 0.0612mol = 1.371L#</mathjax></p>
<p>which should be your final answer, depending on how you've rounded. </p></div>
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<div class="markdown"><p><mathjax>#1.371L#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Find the relative mass of the molecule <mathjax>#KClO_3#</mathjax> by adding together the masses of each atom in it. Find the masses from the <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a>.</p>
<p><mathjax>#39.01gmol^-1 + 35.45gmol^-1 + 3 * 16.00gmol^-1#</mathjax><br/>
<mathjax># = 122.46gmol^-1#</mathjax></p>
<p>Divide the total mass of the substance by its relative mass to find the number of moles of the substance. </p>
<p><mathjax>#(5.00g)/(122.46gmol^-1) = 0.041mol#</mathjax></p>
<p>There are two moles of <mathjax>#KClO_3#</mathjax> for every 3 moles of <mathjax>#O_2#</mathjax>, which we can see from the equation. Using the ratio <mathjax>#2:3#</mathjax> and inserting <mathjax>#0.0408mol#</mathjax>, we find that there must be <mathjax>#0.0612mol#</mathjax> of <mathjax>#O_2#</mathjax>.</p>
<p>One mole of an ideal gas at STP (standard temperature and pressure) occupies 22.4L. </p>
<p>Multiplying </p>
<p><mathjax>#22.4Lmol^-1 * 0.0612mol = 1.371L#</mathjax></p>
<p>which should be your final answer, depending on how you've rounded. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">In the equation #2KClO_3 -> 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP?</h1>
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reudhreghs
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<span class="dateCreated" datetime="2016-04-03T16:31:01" itemprop="dateCreated">
Apr 3, 2016
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<div class="markdown"><p><mathjax>#1.371L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Find the relative mass of the molecule <mathjax>#KClO_3#</mathjax> by adding together the masses of each atom in it. Find the masses from the <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">periodic table</a>.</p>
<p><mathjax>#39.01gmol^-1 + 35.45gmol^-1 + 3 * 16.00gmol^-1#</mathjax><br/>
<mathjax># = 122.46gmol^-1#</mathjax></p>
<p>Divide the total mass of the substance by its relative mass to find the number of moles of the substance. </p>
<p><mathjax>#(5.00g)/(122.46gmol^-1) = 0.041mol#</mathjax></p>
<p>There are two moles of <mathjax>#KClO_3#</mathjax> for every 3 moles of <mathjax>#O_2#</mathjax>, which we can see from the equation. Using the ratio <mathjax>#2:3#</mathjax> and inserting <mathjax>#0.0408mol#</mathjax>, we find that there must be <mathjax>#0.0612mol#</mathjax> of <mathjax>#O_2#</mathjax>.</p>
<p>One mole of an ideal gas at STP (standard temperature and pressure) occupies 22.4L. </p>
<p>Multiplying </p>
<p><mathjax>#22.4Lmol^-1 * 0.0612mol = 1.371L#</mathjax></p>
<p>which should be your final answer, depending on how you've rounded. </p></div>
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</article> | In the equation #2KClO_3 -> 2KCl + 3O_2#, if 5.0 g of #KClO_3# is decomposed, what volume of #O_2# is produced at STP? | null |
2,699 | a9ed13c9-6ddd-11ea-b51b-ccda262736ce | https://socratic.org/questions/how-many-grams-of-pso-3h-3-is-0-250-mol | 28.52 grams | start physical_unit 4 4 mass g qc_end physical_unit 4 4 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] PSO3H3 [IN] grams"}] | [{"type":"physical unit","value":"28.52 grams"}] | [{"type":"physical unit","value":"Mole [OF] PSO3H3 [=] \\pu{0.250 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #PSO_3H_3# is 0.250 mol?</h1> | null | 28.52 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given that <mathjax>#n_"noomber of moles"="mass"/"molar mass"#</mathjax>..</p>
<p>..then.... <mathjax>#"mass"=underbrace(0.250*cancel(mol))_"moles of stuff"xxunderbrace(114.06*g*cancel(mol^-1))_"molar mass of thiophosphoric acid"-=28.52*g#</mathjax></p></div>
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<div class="markdown"><p>So you got <mathjax>#"thiophosphoric acid...."#</mathjax>, <mathjax>#S=P(OH)_3#</mathjax>...</p></div>
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<div class="markdown"><p>Given that <mathjax>#n_"noomber of moles"="mass"/"molar mass"#</mathjax>..</p>
<p>..then.... <mathjax>#"mass"=underbrace(0.250*cancel(mol))_"moles of stuff"xxunderbrace(114.06*g*cancel(mol^-1))_"molar mass of thiophosphoric acid"-=28.52*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #PSO_3H_3# is 0.250 mol?</h1>
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<div class="markdown"><p>So you got <mathjax>#"thiophosphoric acid...."#</mathjax>, <mathjax>#S=P(OH)_3#</mathjax>...</p></div>
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<div class="markdown"><p>Given that <mathjax>#n_"noomber of moles"="mass"/"molar mass"#</mathjax>..</p>
<p>..then.... <mathjax>#"mass"=underbrace(0.250*cancel(mol))_"moles of stuff"xxunderbrace(114.06*g*cancel(mol^-1))_"molar mass of thiophosphoric acid"-=28.52*g#</mathjax></p></div>
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</article> | How many grams of #PSO_3H_3# is 0.250 mol? | null |
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